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# Microindentation Hardness Testing
## Andr Lus de Brito Baptsta
UFF/EEIMVR - Brazil
Microindentation hardness testing is a very valuable tool for the materials engineer, but it must be
used with care and full understanding of potential problems.
The purpose of microindentation hardness testing is to study fine scale changes in hardness,
either intentional or accidental. The technique is also commonly known as microhardness testing,
but this term is misleading because it implies that the hardness is extremely low, which is not the
case. The applied load and the resulting indent size are small relative to bulk tests, but the same
hardness number is derived. Consequently, ASTM Committee E-4 on Metallography recommends
the term "microindentation hardness testing," which could be given the acronym MHT.
This article describes the two most common microindentation tests -- the Vickers and the
Knoop tests, which are currently being updated as ASTM Standard E 384.
## The Vickers test
In 1925, Smith and Sandland of the United Kingdom
developed a new indentation test for metals that were too
hard to evaluate by the Brinell test, whose hardened steel ball
was limited to steels with hardnesses below ~ 450 HBS (~ 48
HRC). In designing the new indenter, a square-based
diamond pyramid (Fig. 1), they chose a geometry that would
produce hardness numbers nearly identical to Brinell
numbers within the range of both tests. This was a very wise
decision, because it made the Vickers hardness test very easy
## Fig. 1- Schematic diagram
showing the shape of the
Vickers indenter and
impression. The Vickers
hardness number is
calculated based on the
surface area of the indent.
The ideal d / D ratio (d = impression diameter, D = ball diameter) for a spherical indenter is
0.375.
If tangents are drawn to the ball at the impression edges for d / D = 0.375, they meet below
the center of the impression at an angle of 136 degrees, the angle chosen for the Vickers indenter.
Diamond allows the Vickers test to evaluate any material and, furthermore, has the very
important advantage of placing the hardness of all materials on one continuous scale. This is a
major disadvantage of Rockwell type tests, for which 15 standard and 15 superficial scales were
developed. Not one of these scales can cover the full hardness range. The HRA scale covers the
broadest hardness range, but this scale is not commonly used.
In the Vickers test, the load is applied smoothly, without impact, and held in place for 10 or
15 seconds. The physical quality of the indenter and the accuracy of the applied load (defined in E
384) must be controlled to get the correct results. After the load is removed, the two impression
diagonals are measured, usually with a filar micrometer, to the nearest 0.1 m, and then averaged.
The Vickers hardness (HV) is calculated by: HV = 1854.4L / d2 where the load L is in grams-force
and the average diagonal d is in m (although the hardness number units are expressed in units of
kgf / mm2 rather than the equivalent gf / m2).
The original Vickers testers were developed for test loads of 1 to 120 kgf, which produce
rather large indents. Recognizing the need for lower test loads, the National Physical Laboratory
(U.K.) experimented with lower test loads in 1932. The first low-load Vickers tester was described
by Lips and Sack in 1936.
Because the shape of the Vickers indentation is geometrically similar at all test loads, the
HV value is constant, within statistical precision, over a very wide test load range, as long as the
test specimen is reasonably homogeneous.
However, studies of microindentation hardness test results conducted over the past several
years on a wide range of loads have shown that results are not constant at very low loads. This
problem, called the "indentation size effect," or ISE, has been attributed to fundamental
characteristics of the material. In fact, the same effect is observed at the low load test range of bulk
Vickers testers. Furthermore, an ASTM inter-laboratory "round robin" of indents made at one
laboratory but measured by twelve different people, reported all three possible ISE responses for the
same set of indents!
Since the 1960s, the standard symbol for Vickers hardness per ASTM E 92 and E 384, has
been HV. This nomenclature is preferred to the older, obsolete symbols DPN or VPN. The hardness
is expressed in a standard format. For example, if a 300 gf load reveals a hardness of 375 HV, the
hardness is expressed as 375 HV300. Note that ASTM recommends a "soft" metric approach in this
case, because rigorous application of the SI system would result in hardness units expressed not in
the standard, understandable kgf / mm2 values, but in GPa units, which are entirely meaningless to
engineers and technicians.
## The Knoop test
The Knoop test is conducted in the same manner, and with the same tester, as the Vickers
test. However, only the long diagonal is measured, except for the projected area hardness (PAH)
test recommended by Blau. This, of course, saves some time.
The Knoop hardness is calculated from HK = 14229L / d2 where the load L is in gf and the
long diagonal d is in m. Again, the symbol HK was adopt in the early 1960s while other terms,
such as HKN or KHN, are obsolete. The Knoop hardnness is expressed in the same manner as the
Vickers hardness: 375 HK300 means that a 300 gf load produced a Knoop hardness of 375 kgf /
mm2.
## In the Vickers test, it is assumed that
recovery is not elastic after the load is
removed. However, in reality recovery is
elastic, and sometimes its influence is quite
pronounced. Generally, the impression (Fig.
2) appears to be square, and the two
diagonals have similar lengths.
Fig. 2 - Example of Properly formed indents
with excellent image contrast (400X)
## Fig. 3 - Example of a distorted Vickers
indent in an austenitic stainless steel
specimen (400X)
## As with the Brinell test, the Vickers hardness
number is calculated based on the surface
area of the indent rather than the projected
area. However, if the impression shape is
distorted by elastic recovery, a very common
result in anisotropic materials (Fig. 3), should
the hardness be based on the average of the
two diagonals? It is possible to calculate the
Vickers hardness based on the projected area
of the impression, which can be measured by
image analysis. Although rigorous studies of
this problem are seldom found in the
literature, it appears that the diagonal
measurement is the preferred approach even
for distorted indents.
## The Knoop test
As an alternative to the Vickers test, particularly for very
thin layers, Fredrick Knoop and his associates at the former
National Bureau of Standards (now NIST) developed a lowload test with a rhombohedral-shaped diamond indenter,
Fig. 4. The long diagonal is seven times (7.114 actually) as
long as the short diagonal. With this indenter shape, elastic
recovery can be held to a minimum. Some investigators
claim no elastic recovery with the Knoop indent, but this
cannot be true, because measurements of the ratio of longto-short diagonal often reveal results substantially different
than the ideal 7.114 value.
## Fig. 4 - Schematic showing the
shape of the Knoop indenter
and impression.
The Knoop test is conducted in the same manner, and with the same tester, as the Vickers
test. However, only the long diagonal is measured, except for the projected area hardness (PAH)
test recommended by Blau. This, of course, saves some time. The Knoop hardness is calculated
from HK = 14229L / d2 where the load L is in gf and the long diagonal d is in m. Again, the
symbol HK was adopt in the early 1960s while other terms, such as HKN or KHN, are obsolete.
The Knoop hardnness is expressed in the same manner as the Vickers hardness: 375 HK300 means
that a 300 gf load produced a Knoop hardness of 375 kgf / mm2.
Aside from a minor savings of time, one chief merit of the Knoop test is the ability to test
thin layers more easily. For surfaces with varying hardness, such as case hardened parts, Knoop
indents can be spaced closer together than Vickers indents. Thus, a single Knoop traverse can
define a hardness gradient more simply than a series of two or three parallel Vickers traverses in
which each indent is made at different depths. Furthermore, if the hardness varies strongly with the
depth, the Vickers indent is distorted by this change; that is, the diagonal parallel to the hardness
change is affected by the hardness gradient, while the diagonal perpendicular to the hardness
gradient remains unaffected (both halves of this diagonal are of the same approximate length).
The shortcoming of the Knoop indent is that the three-dimensional indent shape changes
with test load and, consequently, HK varies with load. In fact, HK values may be reliably converted
to other test scales only for HK values produced at the standard load, generally 500 gf, that was
used to develop the correlations. However, at high loads the variation is not substantial. Note that
all hardness scale conversions are based on empirical data; consequently, conversions are not
precise but are estimates.
## Accuracy, precision, and bias
Many factors (see Table) can influence the quality of microindentation test results.
Table: Factors affecting precision and bias in microindentation hardness testing
Instrument Factors
Measurement Factors
Material Factors
Calibration of the
Angle of indentation.
measurement system. Heterogeneity in composition or
Lateral movement of the indenter Resolving power of the
microstructure.
or specimen.
objective.
Crystallographic texture.
Indentation time.
Magnification.
Quality of the specimen
Indenter shape deviations.
Operator bias in sizing.
preparation.
Damage to the indenter.
Inadequate image quality. Low reflectivity or transparency.
Insufficient spacing between
Nonuniform illumination.
indents or from edges.
In the early days of low-load (<100 gf) hardness testing, it was quickly recognized that
improper specimen preparation can influence hardness test results. Most texts state that improper
preparation yields higher test results because the surface contains excessive preparation-induced
deformation. While this is certainly true, improper preparation may also create excessive heat,
which reduces the hardness and strength of many metals and alloys. Either problem may be
encountered due to faulty preparation.
For many years, it was considered necessary to electrolytically polish specimens so that the
preparation-induced damage could be removed, thus permitting bias-free low-load testing.
However, the science behind mechanical specimen preparation, chiefly due to the work of Len
Samuels, has led to development of excellent mechanical specimen preparation procedures, and
electropolishing is no longer required.
In addition, several operational factors must be controlled for optimum test results. First, it is
good practice to inspect the indenter periodically for damage; for example, cracking or chipping of
the diamond. If you have metrology equipment, you can measure the face angles and the sharpness
of the tip. Specifications for Vickers and Knoop indenter geometries are given in E 384.
A prime source of error is the alignment of specimen surface relative to the indenter. The
indenter itself must be properly aligned perpendicular (1) to the stage plate. Next, the surface
must be perpendicular to the indenter. Most testers provide holders that align the polished face
perpendicular to the indenter (parallel to the stage).
If a specimen is simply placed on the stage surface, its back surface must be parallel to its
polished surface. Tilting the surface more than one degree from perpendicular results in
nonsymmetrical impressions, and can produce lateral movement between specimen and indenter.
However, in most cases, indenting procedures are not the major source of error. For
example, the writer has encountered units that were not applying the correct load, as shown in Fig.
5. Tester A produced nearly constant results over the full load range, while tester B produced the
correct results only at 1000 gf. As the applied load decreased, the hardness decreased to less than
25% of the correct value! Apparently, the load being applied, for loads under 1000gf, must have
been substantially greater than specified. After such an evaluation, it is easy to decide which tester
to purchase!
As this experience shows, it is important to regularly check the performance of your tester
with a certified test block. The safest choice is a test block manufactured for microindentation
testing and certified for the test (Vickers or Knoop) as well as the specified load. Strictly speaking,
a block certified for Vickers testing at 300 or 500 gf (commonly chosen loads) should yield
essentially the same hardness with loads from about 50 to 1000 gf. That is, if you take the average
(knowing the standard deviation of the test results), statistical tests can tell you (at any desired
confidence level) if the difference between the mean values of the tests at the two loads is
statistically significant or not.
The greatest source of error is measuring the indent, as documented in an ASTM interlaboratory test. Place the indent in the center of the measuring field, because lens image quality is
best in the center. The light source should provide adequate, even illumination to provide maximum
contrast and resolution. The accuracy of the filar micrometer, or other measuring device, should be
verified by a stage micrometer.
Specimen preparation quality becomes more important as the load decreases, and it must be
at an acceptable level. Specimen thickness must be at least 2.5 times the Vickers diagonal length.
Because the Knoop indent is shallower than the Vickers at the same load, somewhat thinner
specimens can be tested.
Spacing of indents is important because indenting produces plastic deformation and a strain
field around the indent. If the spacing is too small, the new indent will be affected by the strain field
around the last indent. ASTM recommends a minimum spacing (center to edge of adjacent indent)
of 2.5 times the Vickers diagonal.
For the Knoop test, in which the long diagonals are parallel, the spacing is 2.5 times the
short diagonal. The minimum recommended spacing between the edge of the specimen and the
center of the indent should be 2.5 times. Again, Knoop indents can be placed closer to the surface
than Vickers indents.
## When considering a new tester, it is prudent to perform a series
shown in Fig. 5). Then, plot the mean and 95% confidence
limits (not shown in Fig. 5) of each test as a function of load.
Because of the method of defining HV and HK, which involves
dividing by d2, measurement errors become more critical as d
gets smaller; that is, as L decreases and the material's hardness
increases.
Therefore, departure from a constant hardness for the Vickers or
Knoop tests as a function of load becomes a greater problem as
the hardness increases. For the Knoop test, HK increases as L
decreases because the indent geometry changes with indent
depth and width. But the change in HK varies with the test load
-- at a higher hardness, the change is greater as L decreases.
## Fig. 5 - Load vs. Vickers
hardness test results for two
testers using a quenched and
tempered 440C martensitic
stainless steel specimen. Tester
A is red, tester B is blue.
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3.2k views
Calculate the order of leaf ($P_{leaf}$) and non leaf (P) nodes of a $B^{+}$ tree based on the information given below.
Search key field = $12$ field
Record pointer = $10$ bytes
Block pointer = $8$ bytes
Block size = $1$KB
1. $P_{leaf}$ = 51 & p = 46
2. $P_{leaf}$ = 47 & p = 52
3. $P_{leaf}$ = 46 & p = 51
4. $P_{leaf}$ = 52 & p = 47
| 3.2k views
option c is the ans
by Boss (11.1k points)
selected
0
correct
0
what is the meaning of the word Order .. ?
Is it Number of key values in a node .. ?
0
Isn't it supposed to be (n-1)(searchKey + RecordPointer) + BlkPtr <= BlockSize instead of n(searchKey + RecordPointer) + BlkPtr <= BlockSize? Here n is order of leaf node.
Pleaf
1*BP + (Pleaf )*(key+Recrd Pointer) <= Block size
Pleaf<=46.18
Pnonleaf
P*Block Pointer +(P-1)*(Key) <= Block Size
Pnonleaf <=51.8
by Veteran (119k points)
+1
Ans should be c)
You made mistake in order of leaf node of b+ ...
For leaf node order is pleaf(key+record pointer)+ block pointer (next)<= block size
0
tnks .ans c) rt?
+1
o yes , i swiped :P
0
Could you let me know why we take block pointer into consideration in leaf node??
Won't the pointer that points to the leaf node be in int's parent what is the need to include it.
Is the block pointer that is included the one pointing to the sibling node to the right of the current node??
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# Variance in Online Slots
The second you start to understand how variance/volatility works – these terms are used interchangeably in the industry – the better your grasp on which slot games you should target.
The majority of people have heard of these terms, but they don’t really know how they are applied to an online casino. A basic explanation of variance would be how often a result occurs that is different to the average or mean.
One of the easiest ways to get to grips with it is by taking a simple coin toss. The likelihood of landing on heads or tails is 50/50. Although no casino games offer a true 50/50 shot, it’s still relevant, so bear with us.
A coin toss would mean that the game is low variance, in that you’ve just as good a chance of getting one result as the other. The string of results in reality, however, isn’t necessarily going to be heads one time, then tails the next, and so on.
The fact of the matter is that you could easily throw 10 heads in a row or 10 tails in a row. Given that this is against what the statistical numbers for each result would be (50/50), we would then put this down to variance.
## Can you Reduce Variance?
You can’t physically change how variance is applied to each game as this is pre-programmed, but you can alter how affects you. The simple way of doing this is by increasing the frequency with which you play a particular game.
By doing this, you are going to be drastically decreasing the effect that variance has on your results.
Let’s go back to the coin toss quickly. If you flipped the coin 10 times, it’s plausible that you could flip 10 of either heads or tails in a row. If you flipped the coin a million times, then you would be more likely to see a more even split between the two as it worked itself out over time.
The sample size gives the variance a chance to catch up. The more you play, the more chances you give the game to equalize and the less variance will come into play.
The fact of the matter is that most casino players aren’t going to stick around for long enough to cancel variance out. We are likely talking millions of spins of the same game to get a truly level playing field.
This is one of the reasons that slots in particular are so successful. The games are only slightly tilted in the casino’s favour, but it’s this small edge that cumulates over millions and millions of spins by many players which means variance isn’t really a factor for them, only for us, so they will be able to take the house edge for that particular game or bet.
So, the player is seeing a huge amount of variance given their relatively low number of games played, and the casino is seeing next to no variance given the cumulative number of hands played throughout their casino from thousands of different players.
## What Levels of Variance are There?
It’s widely accepted that there are three different levels of variance within slots:
• Low variance
• Medium variance
• High variance
Let’s break these down and look at each one in turn in a little more detail. We will include some examples so that you can take a look for yourself.
### Low Variance
With low variance games, you are looking at smaller win amounts monetarily, but these wins will come more often. These slots have been designed to allow people to play for long periods of time without losing too much too soon, but also not allowing them to win particularly big payouts.
These types of games are perfect for people who are trying to clear bonuses and just bet for the sake of betting and make their bankroll last. They know that if they play long enough the chances are that they won’t be down an awful lot, and if they are the bonus that they are clearing will cover that loss. Any winnings on top will be a welcome extra.
A good way to check for low variance games is to check the pay table that comes with all online slots. You need to be looking for games that don’t have huge payouts in terms of multipliers from one level to another. So, if jumps from 4 to 5 symbol wins are below 5x, then it’s likely you will have a low variance game on your hands.
Another way to check for low variance games is the number of paylines. You will find that games with more paylines are generally lower variance due to the fact that you can land so many more winning combinations.
Thunderstruck II is one of the most popular low variance games on the market right now. The game, developed by Microgaming, has been around since 2010 and comes with 243 paylines across 5 reels. The maximum win is 6,000 coins with a maximum bet toping just 15 coins (note that ‘coins’ will literally just transfer across to the currency that you use). This is not to be confused with Thunderstruck I, which was a notoriously high-variance game.
### Medium Variance
The medium variance games are often the most popular in the casino. They offer a better balance of winning a reasonable number of times per session, and also offering the chance to win wat the average plyer would consider large amounts.
You’re going to find that these types of games are where you start to see more elaborate bonus rounds and are more feature-heavy. They draw players in by making the games a little bit more exciting, whilst still offering some longevity in that you won’t blow your whole bankroll within just a few minutes – providing you are following proper bankroll management, of course.
In terms of picking these games, you want to be looking at multipliers between 4 and 5 symbol payouts to be between 5x and 10x, any larger and you will start to venture into high variance territory.
Starburst is probably one of the most well-known medium variance slots on the market right now. The game comes with just 10 paylines, but offers a 5,000x jackpot along with a number of bonus features and free spins rounds.
Starburst is also a super common inclusion for many casinos when offering free spins as a welcome bonus when players first join. All those ‘sign up today and get 50 free spins’ type promotions that you’ve likely seen are often linked to the Starburst game because of its mid-variance playing field; people can win large sums which entices them to join the casino and those that lose will pass it off as free spins anyway.
### High Variance
High variance games are where you are playing for large jackpots and the ability to win big with one spin. You can burn through money very quickly on these, but the rewards are like no other online slots.
A good indication of these types of games are jackpot slots, and in particular, progressive jackpots. Games like Mega Moolah and Major Millions offer up progressive jackpots worth millions of pounds. The jackpots are really high but Mega Moolah, for example, will pay out on average just 5 times per year. When you consider how many spins are made on one of the world’s most popular online slots, this is a staggeringly small amount.
Another way to spot high variance games are the multipliers that are on offer. Generally, the larger the multiplier the bigger the variance. Bonanza Megaways by Microgaming has one of the biggest multipliers that we’ve ever seen with 120,000x with a max bet of 500 coins and a massive 117,649 paylines over 6 reels. You would need quite a bankroll and a lot of luck to trigger that.
## Is RTP the Same as Variance?
No, RTP is different from variance.
For those that aren’t aware, the RTP percentage is a number that is set on all online slots and it’s basically the amount the slot will pay out per 100. The numbers can vary, with anywhere from 90-98% being fairly common.
A simplistic way of looking at is that the percentage represents the money that gets paid back. So, if you had a slot that had an RTP of 95%, then that slot would pay out 95p for every £1 staked. This is a theoretical number based on a huge number of spins, however, and not the exact return that you can expect, otherwise there would be no point playing. In short bursts of play the RTP doesn’t really apply, people just hope they get a lucky streak.
Variance is more about the frequency with which you will win, rather than the amount. If you had two slots that both paid out 95% RTP, then the only difference will be how you experience those wins or losses, rather than the amount that gets paid out.
For example, if you won £1 from a 50p spin on your first spin, then this would be a 200% RTP. We know that the actual RTP is 95%, meaning that the fact we are winning is down to variance.
If we had 1 million spins on two different games with 95% RTP, it’s likely that by the end the RTP we experienced would be very close to the 95% stated on both, as the length of play has given variance a chance to catch up. If we looked at the numbers after only 10 spins on each, they could be wildy different.
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# Methods of Setting Stop Loss Indices Trading Orders In Indices Trading
Traders using a indices trading system must have mathematical calculations that reveal where the order must be placed.
A Indices trader can also place a stop loss indices trading order according to the technical indicators used to set these orders. Certain technical indicators use mathematical equations to calculate where the stop loss stock indices order should be set so as to provide an optimal exit point. These indicators can be used as the basis for setting these orders.
Other indices traders also place these orders according to a predetermined risk to reward ratio. This method of setting is dependent upon certain mathematical equations. For example a ratio of 50 pips stop loss can be used by a Indices trader if the trade has the potential to make 100 pips in profit; this is a risk reward ratio of 2:1
Others just use a predetermined percentage of their total trading account balance.
To set a stop loss order it is best to use one of the following methods:
## 1. Percentage of indices trading account balance
This is based on the percent of account balance that the indices trader is willing to risk.
If a indices trader is willing to risk 2% of account balance then the indices trader determines how far he will set the order level based on the position size that he has bought or sold.
Example:
If a indices trader has a \$100,000 account and is willing to risk 2% then the position size of the trade that they will open for Indices will be determined by this 2% stop loss level.
## 2. Setting Stop Loss Indices Trading Order using Support and Resistance Levels
Another way of setting stop loss stock indices orders is to use supports and resistance levels, on the trading charts.
Given that stop loss indices trading orders tend to congregate at key points, when one of these levels is touched by the indices price, others are set off, like dominos. Stop loss orders tend to accumulate just above or below the resistance or support levels, respectively.
A resistance or a support level should act like a barrier for stock indexes price movement, this is why they are used to set stop losses, if this barrier is broken the stock indexes price movement can go towards the opposite direction of the original indices trade, but if this barriers (support and resistance levels) are not broken the stock indexes price will continue moving in the intended direction.
### Stop Loss Indices Trading Order level using a resistance level
Setting order above the resistance
### Stop Loss Indices Trading Order level using a support Level
Setting order below the Support Line
## 3. Indices Trend Lines
A indices trend line can be used to set stop losses where the order is set just below the indices trend line. As long as the indices trend line holds the indices trader will be able to continue making profits while at the same time set this order that will lock his profit once the indices trend line is broken.
Setting order below the indices trend line
Example of where to set this order using indices trend lines.
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# How to select only one solution from infinite list
Using NSolveValues on certain equations returns lists of solutions of the following kind:
For each of these rules I want to select only one value, namely the one absolutely closest to $$2\pi/10$$, but without selecting this manually. How would I find this?
Edit:
I have many of these equations which depend on a lot of different parameters, so I won't put all of that here. But the point is that I want to solve a large number of equations iteratively, where each equation depends on the solution of the previous one. However each equation results in a list ofinfinite solutions as above, but I only want to use the solution that is absolutely nearest to $$2\pi/10$$.
f[x_]:=ArcCos[0.5 (-0.00756467 +
0.749491 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) +
0.503279 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) +
0.538571 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) +
0.430085 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) -
0.0857465 (0.227498 Cos[45. x] + 0.457042 Sin[45. x]) -
0.838206 (-0.423397 Cos[45. x] +
0.841604 Sin[45. x]))] ((0.566265 (0. +
0.503279 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) -
0.749491 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.0857465 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) -
0.538571 (0.227498 Cos[45. x] +
0.457042 Sin[45. x])))/(\[Sqrt](Abs[
0. + 0.503279 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) -
0.749491 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.0857465 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) -
0.538571 (0.227498 Cos[45. x] + 0.457042 Sin[45. x])]^2 +
Abs[0.112051 +
0.430085 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.503279 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) -
0.838206 (0.227498 Cos[45. x] + 0.457042 Sin[45. x]) +
0.0857465 (-0.423397 Cos[45. x] + 0.841604 Sin[45. x])]^2 +
Abs[0.050167 -
0.430085 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) +
0.838206 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) +
0.749491 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) +
0.538571 (-0.423397 Cos[45. x] +
0.841604 Sin[45. x])]^2)) + (0.336803 (0.112051 +
0.430085 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.503279 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) -
0.838206 (0.227498 Cos[45. x] + 0.457042 Sin[45. x]) +
0.0857465 (-0.423397 Cos[45. x] +
0.841604 Sin[45. x])))/(\[Sqrt](Abs[
0. + 0.503279 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) -
0.749491 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.0857465 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) -
0.538571 (0.227498 Cos[45. x] + 0.457042 Sin[45. x])]^2 +
Abs[
0.112051 +
0.430085 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.503279 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) -
0.838206 (0.227498 Cos[45. x] + 0.457042 Sin[45. x]) +
0.0857465 (-0.423397 Cos[45. x] + 0.841604 Sin[45. x])]^2 +
Abs[0.050167 -
0.430085 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) +
0.838206 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) +
0.749491 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) +
0.538571 (-0.423397 Cos[45. x] +
0.841604 Sin[45. x])]^2)) - (0.752268 (0.050167 -
0.430085 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) +
0.838206 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) +
0.749491 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) +
0.538571 (-0.423397 Cos[45. x] +
0.841604 Sin[45. x])))/(\[Sqrt](Abs[
0. + 0.503279 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) -
0.749491 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.0857465 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) -
0.538571 (0.227498 Cos[45. x] + 0.457042 Sin[45. x])]^2 +
Abs[0.112051 +
0.430085 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.503279 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) -
0.838206 (0.227498 Cos[45. x] + 0.457042 Sin[45. x]) +
0.0857465 (-0.423397 Cos[45. x] + 0.841604 Sin[45. x])]^2 +
Abs[0.050167 -
0.430085 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) +
0.838206 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) +
0.749491 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) +
0.538571 (-0.423397 Cos[45. x] + 0.841604 Sin[45. x])]^2)))
NSolveValues[f[x] == 2 \[Pi]/10, x, Reals]
• Can you put the code that produced this result? Dec 9, 2021 at 9:57
• @polfosol Thank you for your comment, please see my edit. Dec 9, 2021 at 10:04
• Also I was thinking about using something like NearestTo, but that doesn't work for the kind of object that I have from NSolveValues. Dec 9, 2021 at 10:08
• What is the meaning of argument 45. x ? Dec 9, 2021 at 10:28
• You could try to give an additional constrain like: NSolveValues[{f[x] == 2 \[Pi]/10, (x - 2 Pi/10)^2 < .005}, x] Dec 9, 2021 at 10:49
If I understand you right, you are looking for a solution f[x]==2Pi/10 with constraint (x-2Pi/10)^2 minimal?
Try NMinimize
mini=NMinimize[{(x - 2 Pi/10)^2, f[x] == 2 Pi/10}, x]
(*{0.0327761, {x -> 0.80936}}*)
• Thank you for your answer, I tried it however it as you say it give 0.80936, but there is another solution of f[x]==2Pi/10, which is 0.669734. Shouldn't it return that instead? Dec 9, 2021 at 10:33
• Interessting, don't know why! If you restrict the solution range mini = NMinimize[{(x - 2 Pi/10)^2, f[x] == 2 Pi/10, 0.6 < x < 1 }, x ] the correct result follows Dec 9, 2021 at 10:53
• Or FindRoot[f[x] == 2 Pi/10, {x, 2 Pi/10}] seems to be more reliable. Dec 9, 2021 at 15:48
Since the desired root is near 2Pi/10 use FindRoot with an initial estimate of 2Pi/10
Clear["Global*"]
f[x_] = ArcCos[
0.5 (-0.00756467 +
0.749491 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) +
0.503279 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) +
0.538571 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) +
0.430085 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) -
0.0857465 (0.227498 Cos[45. x] + 0.457042 Sin[45. x]) -
0.838206 (-0.423397 Cos[45. x] +
0.841604 Sin[45. x]))] ((0.566265 (0. +
0.503279 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) -
0.749491 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.0857465 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) -
0.538571 (0.227498 Cos[45. x] +
0.457042 Sin[45. x])))/(\[Sqrt](Abs[
0. + 0.503279 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) -
0.749491 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.0857465 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) -
0.538571 (0.227498 Cos[45. x] + 0.457042 Sin[45. x])]^2 +
Abs[
0.112051 +
0.430085 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.503279 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) -
0.838206 (0.227498 Cos[45. x] + 0.457042 Sin[45. x]) +
0.0857465 (-0.423397 Cos[45. x] + 0.841604 Sin[45. x])]^2 +
Abs[
0.050167 -
0.430085 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) +
0.838206 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) +
0.749491 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) +
0.538571 (-0.423397 Cos[45. x] +
0.841604 Sin[45. x])]^2)) + (0.336803 (0.112051 +
0.430085 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.503279 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) -
0.838206 (0.227498 Cos[45. x] + 0.457042 Sin[45. x]) +
0.0857465 (-0.423397 Cos[45. x] +
0.841604 Sin[45. x])))/(\[Sqrt](Abs[
0. + 0.503279 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) -
0.749491 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.0857465 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) -
0.538571 (0.227498 Cos[45. x] + 0.457042 Sin[45. x])]^2 +
Abs[
0.112051 +
0.430085 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.503279 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) -
0.838206 (0.227498 Cos[45. x] + 0.457042 Sin[45. x]) +
0.0857465 (-0.423397 Cos[45. x] + 0.841604 Sin[45. x])]^2 +
Abs[
0.050167 -
0.430085 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) +
0.838206 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) +
0.749491 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) +
0.538571 (-0.423397 Cos[45. x] +
0.841604 Sin[45. x])]^2)) - (0.752268 (0.050167 -
0.430085 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) +
0.838206 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) +
0.749491 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) +
0.538571 (-0.423397 Cos[45. x] +
0.841604 Sin[45. x])))/(\[Sqrt](Abs[
0. + 0.503279 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) -
0.749491 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.0857465 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) -
0.538571 (0.227498 Cos[45. x] + 0.457042 Sin[45. x])]^2 +
Abs[
0.112051 +
0.430085 (0.457042 Cos[45. x] - 0.227498 Sin[45. x]) -
0.503279 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) -
0.838206 (0.227498 Cos[45. x] + 0.457042 Sin[45. x]) +
0.0857465 (-0.423397 Cos[45. x] + 0.841604 Sin[45. x])]^2 +
Abs[
0.050167 -
0.430085 (0.287778 Cos[45. x] - 0.876915 Sin[45. x]) +
0.838206 (0.876915 Cos[45. x] + 0.287778 Sin[45. x]) +
0.749491 (0.841604 Cos[45. x] + 0.423397 Sin[45. x]) +
0.538571 (-0.423397 Cos[45. x] + 0.841604 Sin[45. x])]^2))) //
Rationalize[#, 0] & // Simplify
(* -((ArcCos[(1/
2000000000000)(-7564670000 + 1615335655046 Cos[45 x] -
1179276556501 Sin[45 x])] (-84197000000 +
1139633204749573157 Cos[45 x] +
1561033413550243039 Sin[45 x]))/(500000 \[Sqrt](Abs[
224102000000 - 907979420789 Cos[45 x] - 1243722090718 Sin[45 x]]^2 +
Abs[790864917931 Cos[45 x] + 1083301448052 Sin[45 x]]^2 +
4 Abs[
50167000000 + 1014011691237 Cos[45 x] +
1388960982854 Sin[45 x]]^2))) *)
Solving,
sol = FindRoot[f[x] == 2 Pi/10, {x, 2 Pi/10}]
(* {x -> 0.669734} *)
Graphically,
Plot[f[x] - 2 Pi/10, {x, 0.4, 0.9},
PlotPoints -> 200,
MaxRecursion -> 5,
GridLines -> {{2 Pi/10}, None},
GridLinesStyle -> Directive[Gray, Thick, Dashed],
Epilog -> {Red, AbsolutePointSize[4], Point[{x /. sol, 0}]}]
`
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https://www.numbersaplenty.com/24802882267
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|
Search a number
24802882267 = 106323332909
BaseRepresentation
bin10111000110010111…
…011111001011011011
32101000112222222010011
4113012113133023123
5401244014213032
615221055502351
71535432010052
oct270627371333
971015888104
1024802882267
11a578642609
1249825279b7
13245374b0c7
1412b4115399
159a2737247
hex5c65df2db
24802882267 has 4 divisors (see below), whose sum is σ = 24826216240. Its totient is φ = 24779548296.
The previous prime is 24802882261. The next prime is 24802882277. The reversal of 24802882267 is 76228820842.
It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 76228820842 = 238114410421.
It is a cyclic number.
It is not a de Polignac number, because 24802882267 - 23 = 24802882259 is a prime.
It is a Duffinian number.
It is a self number, because there is not a number n which added to its sum of digits gives 24802882267.
It is not an unprimeable number, because it can be changed into a prime (24802882261) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 11665392 + ... + 11667517.
It is an arithmetic number, because the mean of its divisors is an integer number (6206554060).
Almost surely, 224802882267 is an apocalyptic number.
24802882267 is a deficient number, since it is larger than the sum of its proper divisors (23333973).
24802882267 is a wasteful number, since it uses less digits than its factorization.
24802882267 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 23333972.
The product of its (nonzero) digits is 1376256, while the sum is 49.
The spelling of 24802882267 in words is "twenty-four billion, eight hundred two million, eight hundred eighty-two thousand, two hundred sixty-seven".
Divisors: 1 1063 23332909 24802882267
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| 11,294
|
## Acceleration in Physics at a Glance
From the above mentioned equations, you may also imagine what are velocity units. In physics there are a couple of intriguing experiments to measure acceleration. Acceleration is the speed at which velocity changes.
## New Questions About Acceleration in Physics
The student’s size, however, ought to be considered just to the extent that competitive sport may be looked at as essential in later decades. Socially and emotionally, he should be free professional lab report of any serious adjustment problems. He should be aware that if the trial period is not a success, they will return to the original grade placement.
When you’re solving a problem in which you want to find one of these variables but you’re lacking another one, you should combine both formulae to get rid of the unknown variable. When you begin to address the equations, you will discover that you have too many unknowns and you may use this understanding to reduce them at that point. Give explanation to strengthen your diagram.
The very first alternative is to add more productive force. Physical quantities that are completely specified by just giving out there magnitude are called scalars. It is https://samedayessays.org/lab-report-help/ possible to calculate velocity by employing a very simple formula which uses rate, distance, and time.
## A Secret Weapon for Acceleration in Physics
Then establish a table to keep an eye on whatever you know. Next time you’re out walking, imagine you’re still and it’s the world that moves under your feet. If you wish to understand how fast or slow something is moving, you want to learn its speed.
## Vital Pieces of Acceleration in Physics
Remember you may always easily switch between them all in our tool! Within this software you click the job of the object you are attempting to analyze during each frame. When an object isn’t moving then it is known as stationary.
## Facts, Fiction and Acceleration in Physics
Matrices may be used to carry out many operations on objects like translation, rotation, scaling and sheer. Utilizing these equations provides you with the ability to discover information concerning the motion whilst lacking a variable. Velocity refer to the speed at which an object changes position in a sure direction.
Accordingly, to be able http://people.fas.harvard.edu/~lsci1a/lab3.pdf to keep the applied force constant, you might need to change the hanging mass slightly. If a individual in motion wants to maximize their velocity, then that person must make every attempt to make the most of the quantity that they’re displaced from their original position. Sign and Direction Perhaps the most significant point to note about such examples is the symptoms of the answers.
For the rest of the kinds of motion, both effects add to the acceleration. In many instances only a single answer will be relevant, and you are going to have to determine which. There are various sorts of F-type cars.
## Acceleration in Physics Ideas
Walking A range of principles of physics take part in simple act of walking. As a consequence of athletes’ strong urge to compete and return to play, there’s an inclination within the sports community to reduce the seriousness of injuries. Air drag is a sort of frictional force, and you may even care for the interaction of a good body moving on or through water as a frictional interaction.
## The Fundamentals of Acceleration in Physics Revealed
Within this case we’ll examine the region below the curve. The computer calculates the error at each point, and attempts to minimize the square of that error. Velocity is like speed.
This usually means that you want to specify the direction of motion too. Galileo investigated the dilemma of relative velocities. The gravities of different planets are not the same as the planet’s gravity since they have various masses.
Therefore a net external force is necessary to cause a centripetal acceleration. Negative velocity together with positive acceleration is one particular form of deceleration, even though a positive velocity with negative acceleration is another type. So you’re accelerating to boost your speed.
## The Lost Secret of Acceleration in Physics
The typical velocity formula describes the association between the duration of your route and the time that it can take to travel. Distance traveled is the complete length of the path traveled between the 2 positions. Constant or uniform acceleration is as soon as the velocity changes the identical amount in every equal time frame.
## The Appeal of Acceleration in Physics
It would be far more advisable to analyze the issue with regard to velocity (not speed) plus vector acceleration. The utmost height attained is dependent upon the velocity of the object. The definition of instantaneous velocity does not mean that time is made up of instants.
## The War Against Acceleration in Physics
When motion or speed rises, the body is supposed to accelerate. Constant velocity on the opposite hand usually means that the speed isn’t changing! Exactly the way the bat swing speed is connected to bat weight for any given player is a bit more difficult to determine.
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http://www.chegg.com/homework-help/applied-statics-strength-of-materials-and-building-structure-design-1st-edition-chapter-28-solutions-9780136746317
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# Applied Statics Strength of Materials and Building Structure Design (1st Edition)Solutions for Chapter 28
• 301 step-by-step solutions
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Chapter: Problem:
A 10 in × 10 in tied concrete column with four No. 14 longitudinal bars and No. 3 ties is under consideration. Is the total cross-sectional area of the longitudinal bars acceptable under the procedures outlined?
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 1
Calculate the using the equation,
Substitute for , and for
As per the ACI standards must not be less than 0.01 or greater than the 0.08
Here, 0.084 is greater than the 0.08.
Therefore, the total cross section area of longitudinal bars is not .
Corresponding Textbook
Applied Statics Strength of Materials and Building Structure Design | 1st Edition
9780136746317ISBN-13: 0136746314ISBN: Joseph B WujekAuthors:
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# Investigate convergence of the series Σ e^n/n^2 from n=1 to ∞ | test the convergence of series Σ e^n/n^2 from n=1 to ∞
\lim_{n\rightarrow \infty}\frac{e^n}{n^2}
## Solution
Calculate the limit \lim_{n\rightarrow \infty}\frac{e^n}{n^2} Using L’Hopital’s rule, we find
\lim_{x\rightarrow \infty}\frac{e^x}{x^2}=\lim_{x\rightarrow \infty}\frac{e^x}{2x}=\lim_{x\rightarrow \infty}\frac{e^x}{2}=\infty
Hence, the original series diverges by the nth term test
## infinite series problems and solutions
Share
Related Show that the infinite decimal 0.9999 • • • is equal to 1. How Can 0.999... = 1? | The Decimal 0.999... is Equivalent to 1
Top 5 Most Expensive Domains Ever Sold 4 Must-Try ChatGPT Alternatives: Perplexity AI, BardAI, Pi, and More! Types of Trading Techniques in the Stock Market. ChatGPT app now available in India this AI chatbot can help you make your life more productive. What is wrong with following function code?
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# Language Logic and Proof Q. 6.26
Using the natural deduction rules, give a formal proof of
A ∨ D
from the premises
1. A ∨ (B ∧ C)
2. (¬ B ∨ ¬ C) ∨ D
Now my approach is to prove that when ¬A then (B ∧ C). Then I want to prove that ¬(B ∧ C) = (¬ B ∨ ¬ C) and then through disjunction introduction A ∨ (B ∧ C) ∨ D. Which then through disjunction elimination get to A ∨ D
But I am getting nowhere with this approach. Can somebody point me in the right direction here
¬(B ∧ C) = (¬ B ∨ ¬ C) is DeMorgan's Law.
``````1. A ∨ (B ∧ C) Premise
2. (¬ B ∨ ¬ C) ∨ D Premise
3. | ¬A Assumption
4. | (B ∧ C) DS 3,1
5. | ¬¬(B ∧ C) DN 4
6. | ¬(¬ B ∨ ¬ C) DeM 5
7. | D DS 2,6
8. ¬A -> D Proof 3-6
9. ¬¬A v D Mat. Imp 8
10. A v D DN 9
``````
Without DeMorgan's,
``````1. A ∨ (B ∧ C) Premise
2. (¬ B ∨ ¬ C) ∨ D Premise
3. | ¬A Assumption
4. | (B ∧ C) DS 3,1
5. | | ¬ B ∨ ¬ C Assumption
6. | | B CE 4
7. | | ¬¬B DN 6
8. | | ¬ C DS 5,7
9. | | C CE 4
10.| ¬(¬ B ∨ ¬ C) Proof by contradiction 5-9
11.| D DS 2,10
12. ¬A -> D Proof 3-11
13. ¬¬A v D Mat. Imp 12
14. A v D DN 13
``````
I'm not super-familiar with Fitch, which apparently requires two proofs by assumption and a v to do vE. but DS and vE are not identical. But in any case, we can do without both:
``````1. A ∨ (B ∧ C) Premise
2. (¬ B ∨ ¬ C) ∨ D Premise
3. | ¬A Assumption
4. | | (B ∧ C) Assumption
5. | | (B ∧ C) R
----
6. | | A Assumption
7. | | A ∧ ¬A ∧I 3,6
8. | | [Contra. Intr.] 7
9. | | (B ∧ C) Contra Elim 8
10.| (B ∧ C) vE 1,9,5
11.| | ¬ B ∨ ¬ C Assumption
12.| | | ¬ B Assumption
13.| | | B CE 10
14.| | | B ∧ ¬ B ∧ Introduction
15.| | | D Contradiction Elimination 14
------------------
15.| | | ¬ C Assumption
16.| | | C CE 10
17.| | | C ∧ ¬ C ∧ Introduction
18.| | | D Contradiction Elimination 17
19.| | D vE 11,15,18
--------------------
20.| | D Assumption
21.| | D Repetition
22.| D vE 2,19,21
23. ¬ A -> D Proof 3-22
24. ¬¬ A v D Material Implication 23
25. A v D DN 24
``````
And that's closer but looking at Fitch, you may not have access to material implication.
• you beat me to it. – shane Sep 3 '14 at 11:59
• My problem is that Fitch for some reason does not allow step 4. I used disjunction elim 3, 1 – Leon Sep 3 '14 at 12:17
• @Leon I've never used Fitch so I can't say, but vE / DS is valid in normal sentential logic and one of the most basic forms. I guess if it doesn't have that, then change it to an implication first, through material implication. – virmaior Sep 3 '14 at 13:22
• @Leon did it without DS. Still using material implication. – virmaior Sep 3 '14 at 15:19
• Disjunctive Syll : ¬B ∨ C, B ⊢ C is easily derived in Natural Deduction with ∨-E : C ⊢ C and ¬B, B ⊢ ⊥ ⊢ C. Thus, from C ⊢ C and ¬B, B ⊢ C, we conclude by ∨-E with : ¬B ∨ C, B ⊢ C. – Mauro ALLEGRANZA Sep 3 '14 at 15:40
This will be very similar to the above poster, but I will share it too. Corrections are welcome:
1. A v (B ^ C)
2. (~B v ~C) v D / A v D
3. ~(B ^ C) v D 2, De M.
4. (B ^ C) > D 3, Impl.
5. ~A > (B ^ C) 1, Impl.
6. ~A > D 4, 5, H.S.
7. A v D 6, Impl.
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# 15.64 kg to lbs - 15.64 kilograms into pounds
Before we move on to the more practical part - it means 15.64 kg how much lbs conversion - we are going to tell you some theoretical information about these two units - kilograms and pounds. So let’s move on.
How to convert 15.64 kg to lbs? 15.64 kilograms it is equal 34.4802977768 pounds, so 15.64 kg is equal 34.4802977768 lbs.
## 15.64 kgs in pounds
We are going to begin with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, known also as International System of Units (in short form SI).
From time to time the kilogram can be written as kilogramme. The symbol of the kilogram is kg.
The kilogram was defined first time in 1795. The kilogram was described as the mass of one liter of water. First definition was simply but hard to use.
Later, in 1889 the kilogram was described using the International Prototype of the Kilogram (in abbreviated form IPK). The IPK was made of 90% platinum and 10 % iridium. The IPK was used until 2019, when it was replaced by a new definition.
Nowadays the definition of the kilogram is build on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It could be also divided into 100 decagrams and 1000 grams.
## 15.64 kilogram to pounds
You know some facts about kilogram, so now we can go to the pound. The pound is also a unit of mass. We want to point out that there are more than one kind of pound. What are we talking about? For example, there are also pound-force. In this article we want to concentrate only on pound-mass.
The pound is used in the Imperial and United States customary systems of measurements. To be honest, this unit is in use also in another systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is exactly 0.45359237 kilograms. One avoirdupois pound can be divided to 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was placed in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 15.64 kg?
15.64 kilogram is equal to 34.4802977768 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 15.64 kg in lbs
Theoretical section is already behind us. In this section we are going to tell you how much is 15.64 kg to lbs. Now you know that 15.64 kg = x lbs. So it is time to know the answer. Let’s see:
15.64 kilogram = 34.4802977768 pounds.
That is an exact result of how much 15.64 kg to pound. It is possible to also round off this result. After it your result is as following: 15.64 kg = 34.408 lbs.
You know 15.64 kg is how many lbs, so have a look how many kg 15.64 lbs: 15.64 pound = 0.45359237 kilograms.
Naturally, in this case you may also round off this result. After rounding off your outcome will be exactly: 15.64 lb = 0.45 kgs.
We are also going to show you 15.64 kg to how many pounds and 15.64 pound how many kg outcomes in tables. Have a look:
We want to begin with a table for how much is 15.64 kg equal to pound.
### 15.64 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
15.64 34.4802977768 34.4080
Now look at a table for how many kilograms 15.64 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
15.64 0.45359237 0.45
Now you know how many 15.64 kg to lbs and how many kilograms 15.64 pound, so we can move on to the 15.64 kg to lbs formula.
### 15.64 kg to pounds
To convert 15.64 kg to us lbs you need a formula. We will show you two versions of a formula. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 34.4802977768 result in pounds
The first formula will give you the most accurate outcome. In some situations even the smallest difference can be significant. So if you need an accurate result - this formula will be the best solution to calculate how many pounds are equivalent to 15.64 kilogram.
So move on to the shorer version of a formula, which also enables calculations to know how much 15.64 kilogram in pounds.
The second version of a formula is as following, have a look:
Amount of kilograms * 2.2 = the result in pounds
As you can see, this formula is simpler. It can be the best choice if you want to make a conversion of 15.64 kilogram to pounds in easy way, for instance, during shopping. Just remember that final outcome will be not so correct.
Now we are going to show you these two versions of a formula in practice. But before we are going to make a conversion of 15.64 kg to lbs we want to show you another way to know 15.64 kg to how many lbs without any effort.
### 15.64 kg to lbs converter
An easier way to know what is 15.64 kilogram equal to in pounds is to use 15.64 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Converter is based on longer formula which we showed you above. Due to 15.64 kg pound calculator you can effortless convert 15.64 kg to lbs. Just enter number of kilograms which you need to convert and click ‘convert’ button. You will get the result in a flash.
So try to convert 15.64 kg into lbs with use of 15.64 kg vs pound calculator. We entered 15.64 as an amount of kilograms. This is the result: 15.64 kilogram = 34.4802977768 pounds.
As you can see, this 15.64 kg vs lbs converter is user friendly.
Now we are going to our chief issue - how to convert 15.64 kilograms to pounds on your own.
#### 15.64 kg to lbs conversion
We are going to begin 15.64 kilogram equals to how many pounds conversion with the first version of a formula to get the most correct result. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 34.4802977768 the outcome in pounds
So what need you do to know how many pounds equal to 15.64 kilogram? Just multiply amount of kilograms, in this case 15.64, by 2.20462262. It is exactly 34.4802977768. So 15.64 kilogram is exactly 34.4802977768.
You can also round it off, for instance, to two decimal places. It is equal 2.20. So 15.64 kilogram = 34.4080 pounds.
It is high time for an example from everyday life. Let’s calculate 15.64 kg gold in pounds. So 15.64 kg equal to how many lbs? And again - multiply 15.64 by 2.20462262. It is exactly 34.4802977768. So equivalent of 15.64 kilograms to pounds, when it comes to gold, is 34.4802977768.
In this case it is also possible to round off the result. It is the result after rounding off, this time to one decimal place - 15.64 kilogram 34.408 pounds.
Now we can move on to examples calculated with a short version of a formula.
#### How many 15.64 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 34.408 the outcome in pounds
So 15.64 kg equal to how much lbs? As in the previous example you need to multiply amount of kilogram, in this case 15.64, by 2.2. Have a look: 15.64 * 2.2 = 34.408. So 15.64 kilogram is exactly 2.2 pounds.
Let’s do another conversion using shorer formula. Now convert something from everyday life, for example, 15.64 kg to lbs weight of strawberries.
So let’s convert - 15.64 kilogram of strawberries * 2.2 = 34.408 pounds of strawberries. So 15.64 kg to pound mass is 34.408.
If you know how much is 15.64 kilogram weight in pounds and can convert it using two different formulas, let’s move on. Now we are going to show you all results in tables.
#### Convert 15.64 kilogram to pounds
We know that results shown in tables are so much clearer for most of you. It is totally understandable, so we gathered all these results in tables for your convenience. Thanks to this you can easily compare 15.64 kg equivalent to lbs results.
Begin with a 15.64 kg equals lbs chart for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
15.64 34.4802977768 34.4080
And now have a look at 15.64 kg equal pound chart for the second version of a formula:
Kilograms Pounds
15.64 34.408
As you see, after rounding off, when it comes to how much 15.64 kilogram equals pounds, the results are the same. The bigger number the more significant difference. Remember it when you need to make bigger number than 15.64 kilograms pounds conversion.
#### How many kilograms 15.64 pound
Now you learned how to convert 15.64 kilograms how much pounds but we are going to show you something more. Do you want to know what it is? What about 15.64 kilogram to pounds and ounces conversion?
We want to show you how you can calculate it step by step. Let’s start. How much is 15.64 kg in lbs and oz?
First thing you need to do is multiply number of kilograms, this time 15.64, by 2.20462262. So 15.64 * 2.20462262 = 34.4802977768. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To know how much 15.64 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is equal 327396192 ounces.
So final result is exactly 2 pounds and 327396192 ounces. It is also possible to round off ounces, for instance, to two places. Then final result will be equal 2 pounds and 33 ounces.
As you can see, conversion 15.64 kilogram in pounds and ounces quite easy.
The last conversion which we will show you is conversion of 15.64 foot pounds to kilograms meters. Both of them are units of work.
To convert it you need another formula. Before we show you this formula, look:
• 15.64 kilograms meters = 7.23301385 foot pounds,
• 15.64 foot pounds = 0.13825495 kilograms meters.
Now have a look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to calculate 15.64 foot pounds to kilograms meters you need to multiply 15.64 by 0.13825495. It is exactly 0.13825495. So 15.64 foot pounds is 0.13825495 kilogram meters.
You can also round off this result, for example, to two decimal places. Then 15.64 foot pounds is 0.14 kilogram meters.
We hope that this conversion was as easy as 15.64 kilogram into pounds calculations.
We showed you not only how to do a conversion 15.64 kilogram to metric pounds but also two another conversions - to know how many 15.64 kg in pounds and ounces and how many 15.64 foot pounds to kilograms meters.
We showed you also other solution to make 15.64 kilogram how many pounds conversions, this is using 15.64 kg en pound converter. This will be the best choice for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs conversions on your own.
We hope that now all of you can make 15.64 kilogram equal to how many pounds calculation - on your own or using our 15.64 kgs to pounds calculator.
So what are you waiting for? Let’s calculate 15.64 kilogram mass to pounds in the way you like.
Do you need to make other than 15.64 kilogram as pounds calculation? For example, for 10 kilograms? Check our other articles! We guarantee that conversions for other amounts of kilograms are so easy as for 15.64 kilogram equal many pounds.
### How much is 15.64 kg in pounds
To quickly sum up this topic, that is how much is 15.64 kg in pounds , we prepared for you an additional section. Here you can find all you need to know about how much is 15.64 kg equal to lbs and how to convert 15.64 kg to lbs . You can see it down below.
How does the kilogram to pound conversion look? To make the kg to lb conversion it is needed to multiply 2 numbers. Let’s see 15.64 kg to pound conversion formula . See it down below:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 15.64 kilogram to pounds? The exact answer is 34.4802977768 pounds.
There is also another way to calculate how much 15.64 kilogram is equal to pounds with another, shortened type of the equation. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So this time, 15.64 kg equal to how much lbs ? The answer is 34.4802977768 pounds.
How to convert 15.64 kg to lbs in an easier way? You can also use the 15.64 kg to lbs converter , which will make the rest for you and give you an exact result .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
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user1234440
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# 44 Actions
May 6 revised Summation to Matrix added 4 characters in body May 6 asked Summation to Matrix Dec 11 comment Big Omega Proof Switch i just editted it Dec 11 revised Big Omega Proof Switch edited body Dec 11 comment Big Omega Proof Switch oops, sorry it should be omega Dec 11 asked Big Omega Proof Switch Dec 10 accepted Antisymmetric Relations Dec 10 comment Antisymmetric Relations So it is similar to the idea of implications, if we don't know if the premise is true for sure, the conclusion is always true? Dec 10 asked Antisymmetric Relations Dec 9 asked Combinatoric Selection of Passwords Dec 5 awarded Supporter Dec 5 accepted Probability that z precedes both a and b in the permutation? Dec 1 asked Probability that z precedes both a and b in the permutation? Nov 13 accepted Simplifying Boolean Function with Karnaugh Maps Nov 13 comment Simplifying Boolean Function with Karnaugh Maps Thanks for the answer. The problem I am having is know which to include inside the circles. How come the blue circle not include the first column too? Wouldn't such grouping making it a bigger block? Nov 13 awarded Student Nov 13 asked Simplifying Boolean Function with Karnaugh Maps Oct 17 comment Proving that floor(n/2)=n/2 if n is an even integer and floor(n/2)=(n-1)/2 if n is an odd integer. for the odd case, its n-1/2 in terms of m Oct 17 revised Proving that floor(n/2)=n/2 if n is an even integer and floor(n/2)=(n-1)/2 if n is an odd integer. deleted 8 characters in body; edited title Oct 17 awarded Scholar
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# Attic numerals
Detail of stela showing tributes paid by allies of Athens in the League of Delos. The amounts are in Attic numerals, using the drachma sign "𐅂" instead of the geenric unit sign "Ι". Some amounts are "𐅄" = 50, "ΗΗΗ" = 300, "𐅅ΗΗΗ" = 800, "ΔΔΔ𐅂𐅂𐅂" = 33, "Χ" = 1000, and "Χ𐅅𐅄Δ𐅂𐅂"? = 1562?.
The Attic numerals are a symbolic number notation used by the ancient Greeks. They were also known as Herodianic numerals because they were first described in a 2nd-century manuscript by Herodian; or as acrophonic numerals (from acrophony) because the basic symbols derive from the first letters of the (ancient) Greek words that the symbols represented.
The Attic numerals were a decimal (base 10) system, like the older Egyptian and the later Etruscan, Roman, and Hindu-Arabic systems. Namely, the number to be represented was broken down into simple multiples (1 to 9) of powers of ten — units, tens, hundred, thousands, etc.. Then these parts were written down in sequence, in order of decreasing value. As in the basic Roman system, each part was written down using a combination of two symbols, representing one and five times that power of ten.
Attic numerals were adopted possibly starting in the 7th century BCE, and were eventually replaced by the classic Greek numerals around the 3rd century BCE. They are believed to have served as model for the Etruscan number system, although the two were nearly contemporary and the symbols are not obviously related.[citation needed]
## The system
### Symbols
The Attic numerals used the following main symbols, with the given values:
Value Symbol Talents Staters Notes Etruscan Roman
1 Ι Tally mark? 𐌠 I
5 Π 𐅈 𐅏 Old Greek: ΠΕΝΤΕ [pɛntɛ] Modern: πέντε 𐌡 V
10 Δ 𐅉 𐅐 Old Greek: ΔΕΚΑ [deka] Modern: δέκα 𐌢 X
50 𐅄 𐅊 𐅑 "Δ" in "Π": 10 × 5 = 50 𐌣 L
100 Η 𐅋 𐅒 Old Greek: ΗΕΚΑΤΟΝ [hɛkaton] Modern: ἑκατόν 𐌟 C
500 𐅅 𐅌 𐅓 "Η" in "Π": 100 × 5 = 500 ? D
1000 Χ 𐅍 𐅔 Old Greek: ΧΙΛΙΟΙ [kʰilioi] Modern: χίλιοι ? M
5000 𐅆 𐅎 "Χ" in "Π": 1000 × 5 = 5000 ? V
10000 Μ 𐅕 Old Greek: ΜΥΡΙΟΝ [myrion] Modern: μύριον ? X
50000 𐅇 𐅖 "Μ" in "Π": 10000 × 5 = 50000 ? X
The symbols representing 50, 500, 5000, and 50000 were composites of an old form of the capital letter pi (with a short right leg) and a tiny version of the applicable power of ten. For example, 𐅆 was five times one thousand.
#### Special simbols
The fractions "one half" and "one quarter" were written "𐅁" and "𐅀", respectively.
The symbols were slightly modified when used to encode amounts in talents (with a small capital tau, "Τ") or in staters (with a small capital sigma, "Σ"). Specific numeral symbols were used to represent one drachma ("𐅂") and ten minas "𐅗".
#### The symbol for 100
The use of "Η" (capital eta) for 100 reflects the early date of this numbering system. In the Greek language of the time, the word for a hundred would be pronounced [hɛkaton] (with a "rough aspirated" sound /h/) and written "ΗΕΚΑΤΟΝ", because "Η" represented the sound /h/ in the Attic alphabet. In later, "classical" Greek, with the adoption of the Ionic alphabet throughout the majority of Greece, the letter eta had come to represent the long e sound while the rough aspiration was no longer marked.[1][2] It was not until Aristophanes of Byzantium introduced the various accent markings during the Hellenistic period that the spiritus asper began to represent /h/, resulting in the modern Greek spelling ἑκατόν. In modern Greek the /h/ phoneme has disappeared altogether, but the accent on the is retained in the standard spelling.
### Simple multiples of powers of ten
Multiples 1 to 9 of each power of ten were written by combining the two corresponding "1" and "5" digits, namely:
Units Ι II III IIII Π ΠI ΠII ΠIII ΠIIII 1 2 3 4 5 6 7 8 9 Tens Δ ΔΔ ΔΔΔ ΔΔΔΔ 𐅄 𐅄Δ 𐅄ΔΔ 𐅄ΔΔΔ 𐅄ΔΔΔΔ 10 20 30 40 50 60 70 80 90 Hundreds Η ΗΗ ΗΗΗ ΗΗΗΗ 𐅅 𐅅Η 𐅅ΗΗ 𐅅ΗΗΗ 𐅅ΗΗΗΗ 100 200 300 400 500 600 700 800 900 Thousands Χ ΧΧ ΧΧΧ ΧΧΧΧ 𐅆 𐅆Χ 𐅆ΧΧ 𐅆ΧΧΧ 𐅆ΧΧΧΧ 1000 2000 3000 4000 5000 6000 7000 8000 9000 Tens of thousands Μ ΜΜ ΜΜΜ ΜΜΜΜ 𐅇 𐅇Μ 𐅇ΜΜ 𐅇ΜΜΜ 𐅇ΜΜΜΜ 10000 20000 30000 40000 50000 60000 70000 80000 90000
Unlike the more familiar Roman numeral system, the Attic system used only the so-called "additive" notation. Thus, the numbers 4 and 9 were written ΙΙΙΙ and ΠΙΙΙΙ, not ΙΠ and ΙΔ.
### General numbers
In general, the number to be represented was broken down into simple multiples (1 to 9) of powers of ten — units, tens, hundred, thousands, etc.. Then these parts would be written down in sequence, from largest to smallest value. For example:
• 49 = 40 + 9 = ΔΔΔΔ + ΠΙΙΙΙ = ΔΔΔΔΠΙΙΙΙ
• 2001 = 2000 + 1 = ΧΧ + I = ΧΧΙ
• 1982 = 1000 + 900 + 80 + 2 = Χ + 𐅆ΗΗΗΗ + 𐅄ΔΔΔ + ΙΙ = Χ𐅆ΗΗΗΗ𐅄ΔΔΔΙΙ
• 62708 = 60000 + 2000 + 700 + 8 = 𐅇Μ + ΧΧ + 𐅅ΗΗ + ΠIII = 𐅇ΜΧΧ𐅅ΗΗΠIII.
Attic numerals are available in Unicode in the Ancient Greek Numbers block (U+10140 to U+1018F).
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## Tuesday, March 13, 2018
### Walk the Line
I keep a close watch on this number line
I keep my eyes wide open for the sign
Making sense, connections all the time
For number sense, I walk the line
Apologies to Johnny Cash.
At Math In Action two weeks ago, I presented on number talks in the middle grades. (Here's the handout/resources, including a link to the slides.) I was quoting Pam Harris, then the keynote speaker was quoting Pam, so I made a comment on Twitter, badda bing, she turned out to be coming to my neighborhood the next week. Muskegon Regional Math Science Center let me crash, thanks Kristin Frang, so I got to crash for 1.5/2 days.
Pam was presenting on secondary number strings (thread 1 and thread 2 for my notes). Mini-review: I wouldn't hesitate to bring Pam in to work with teachers. Great energy, hilarious, solid ideas presented in a way that invites teacher access, and great modeling of instruction. All building on a great central message about giving learners an opportunity to mathematize.
I'm teaching Introduction to Mathematics this semester, a gen ed math class with a lot of freedom. I'm using Anna Weltman's This Is Not a Math Book as a text, mostly as a resource for learners and introduction activities. In my head it's about redeeming mathematics for this successful students who have (mostly) learned to dislike mathematics. While we're mostly doing math and art, emphasizing problem posing and solving. But we take some time to redeem arithmetic, and we need to do algebra before we do patterning.
So, Pam's problem strings (modified because I can't help), followed by some clothesline math... rare day when I didn't have a way to take pictures. Sorry! The theme of the lesson is what else do we know?, which is one of my main understandings of math.
I drew a line, put on a hashmark, and labeled above it x, and below it 3. What else do we know? Some discussion about putting on a scale, vs knowing left or right. Finally someone shares -x is -3, and says it would go somewhere to the left. The someone said that means we know zero, which I encouraged as a good mathematician question.
First I did Pam's coordinates problem string.
(-2,5) show in graph, table, function notation
(2,-3)
(3,-9)
(1, __) which first brought up approximating with the line, then the idea of slope.
( __, 0) which brought a lot of people to a halt as they tried to remember an algorithm, then a remembered algorithm for solving equations. I made a bit of a joke how I was not interested in a memorized method, sorry, but only making sense.
Next, x is -2. Where is zero, left or right? What is x? Some discussion but pretty quick.
x-4 is 6. Is x left or right? What is x? How do you know? Quick discussion.
x+4 is -6. Heated discussion. -10 and -2 mentioned frequently. More and less start getting used more than left an d right. Interesting symmetry comparison between the last and this.
Then I introduced the clothesline, strung across the front of the room, with just an x in the middle. What else do we know? I had cards to ask them about. 0, -x, x+3, 5. We know where zero is, then dissuaded. What if it's negative? We know where -x is, then dissuaded. Then x+3, that has to be to the right. We don't know how far, but definitely right. What else? x-3 we would know. Excellent! Finally I put 0 up (left of x-3 by less than the distance to x) and ask what x could be?
Then I gave them cards. Make some cards, figure out the order you're going to reveal them and the questions you'll ask. Two really interesting situations came up.
One group had a couple of variable expressions, and then x+200. Really nicely subverting the sense of scale, and a great numeracy discussion of what x could be then. They wanted to just toss off a big number, but other learners argued for more precision.
Another group introduced a new variable, g, and then blew our minds with g^2. Has to be to the right, because squares are bigger. But what if it's negative? It's still to the right? Always? Then a lot of discussion if placement of it was setting the scale. Not until 0 is placed. It was amazing.
So that's my story. Thanks to Pam Harris, Chris Shore and these great learners.
## Saturday, March 10, 2018
### Let's Discuss Professional Development
One of my favorite math ed profs is Sam Otten at Missouri (and the Lois Knowles Faculty Fellow). His research is interesting and situated, he holds teachers in high regard and listens to their ideas, and he illuminates research through the Mathed podcast. He has definitely enriched my practice. In addition, he's just a lovely and creative guy, as well as a world class expert on the DC Comics film universe. Beyond that, he's a GVSU grad, so I knew him when.
He is a part of the team that produced some new professional development materials, and I had a few questions for him about it. Mathematics Discourse in Secondary Classrooms, MDISC, is based on research and developed with teachers in the field. I'm a big believer in the importance of discourse in learning, and know that secondary mathematics has been one the places where traditional teaching has included the least discourse. I also think people need support to make changes, so something like this project is needed.
What inspired these materials? Was it an idea you wanted to develop or a response to situations you saw in the classroom?
The MDISC materials came from a group of math education scholars at Michigan State University and the University of Delaware, led by Beth Herbel-Eisenmann, who were passionate about the role of discourse in math classrooms. We all believed that there was profound value in students discussing mathematical ideas and building meaning together as a community. So at its core, MDISC is a set of professional development materials that are intended to help teachers increase the quantity and quality of discourse in their classrooms.
As we set out to create these materials, we tried to draw on other work that already existed in the math ed literature. Some of that work was Beth's own research with Michelle Cirillo. They had worked for years with a group of secondary teachers, examining discourse patterns and power dynamics. We also drew on the work of Chapin, O'Connor, and Anderson, who wrote a great book called Classroom Discussions that focused on mathematical discourse at the elementary level. They had some really amazing results with respect to student achievement scores that stemmed from a new emphasis on discourse. With MDISC, we tried to take some of those ideas from the elementary level and reinterpret them in ways that made sense at the secondary level -- focusing on middle school and high school classes.
Overall, the MDISC PD materials equip secondary math teachers to think about discourse in productive ways and it also provides them with specific tools for changing the discourse in their classrooms so that it really empowers students. It helps move us beyond teaching-as-telling.
What are some of the different ways these materials might be used?
The MDISC materials include a physical facilitator's guide and then digital versions of all the participant materials as well as sample videos. It could be used by a teacher leader, facilitating sessions with secondary math teachers, or by a PLC of teachers who want to work through it on their own. It could also serve as a textbook for a graduate-level course, so a teacher educator going through the activities with practicing teachers, for example in a Master's course or an Ed Specialist course. The materials are designed to be a year-long study, with connections to everyday classroom practice, but it's flexible -- so with some adjustments, it could also be used in one semester. Or people could select which components they want to focus on.
There's also an optional follow-up where teachers can be guided through some action research, if they want to continue making purposeful efforts toward shaping their classroom discourse. There are several different options, and the MDISC team is very open to communicating with people if they have questions about enactment. We've also enacted the materials many times in many different settings, so we have a lot of experiences to share.
As you piloted these materials, what were some of the changes you saw in classroom discourse?
We have piloted the materials and had others pilot them in both Michigan and Delaware, with several different groups of teachers. They have been very well received thus far, with some teachers willingly joining in for a second and third year because once they start, they don't want to stop thinking about their classroom discourse. Some of the teachers have called it the most important learning experience in their teaching career, and this even came from a 30-year veteran.
The most visible changes have been the number of students talking in class. They open up more and share their ideas, and the great thing is that they're sharing mathematical ideas. I think this comes from MDISC's dual approach of not only providing insight into the nature of discourse but also providing specific moves for teachers to use. For example, MDISC develops six teacher discourse moves that include inviting student participation and also probing a student's thinking. These are concrete ways to get the discussion going and keep it directed toward important mathematics.
Another big change that is noticeable is that more wrong answers come to the surface -- it's not that MDISC leads to student confusion (just to be clear), it's that an increase in discourse helps more student ideas to come to the surface. And of course some of those ideas are incorrect or imprecise, and that can lead to good discussions and good learning opportunities for the group.
What’s one feature of these materials or an example experience that might help teachers understand how they will support their teaching?
One feature of the MDISC materials is that they are practice-based and case-based. So teachers will get to make constant connections to their own instructional practices and their own students. Those connections are built right into the materials. And there is also the chance to see and discuss detailed cases of other teachers. Rather than lots of little isolated examples, MDISC instead is built around larger cases of real teaching. So for example, when you're learning about the transition from small-group work to whole-class discussions, you can actually see a middle school teacher as she circulates among her students and selects certain ideas to be shared later, telling the students that she'd really like them to bring it up in front of the whole class. Then you can follow the case to see how it played out in the discussion.
Another important feature is that the MDISC materials integrate an emphasis on equity. Powerful discourse means that everyone has an opportunity to be heard and to learn from the conversations. So there is a lot of attention paid to how teachers can use a discourse-based approach to reach more students, including those with traditionally marginalized backgrounds.
What movie would you like to see DC make next?
Great question! When I'm not working in math ed or spending time with family, I love watching and analyzing DC superhero movies. I really loved Man of Steel and then I thought Batman v Superman took it up another notch, with great themes about immigrant experiences and the danger of overt masculinity having to face feelings of powerlessness. So although I'm excited about Aquaman and the Wonder Woman sequel, I would really like to see another Superman solo film make it onto the slate. And it would also be great if the Cyborg standalone would get the green light because I thought he was a really intriguing character in Justice League and I think his story could be used not only as a commentary about race in modern society but also about our increasing dependence on technology.
(Back to me) There's so much promising here. Use of real classroom discussions with connections to your own. The focus on equity. And the idea that in running it with teachers there's a measurable change in the number of kids participating in discussion, as well as the frequency and quality of discussions - that's a dream. I'd love a chance to work through this with teachers.
Find out more:
This has been a long time coming. Funville Adventures by A.O. (Sasha) Fradkin and A.B. Bishop is full of fun adventures.
Sasha is a Twitter acquaintance, an elementary math enrichment teacher with an amazing personal math journey, and I probably heard about the Kickstarter from there. I love to encourage these passion projects in general, but this book is especially delightful. (Sasha on Twitter & her blog.)
As a story, it may remind you in flavor of The Phantom Tollboth or Dragon Tales. Emmy and Leo are kids transported to an allegorical land, Funville. Kids in Funville each have a special ability. The story makes sense and is enjoyable without even knowing the math in a formal way, because the math is the idea behind the people they meet, but not how it's discussed. These quirky characters are brought to life in quick vignettes and charming illustrations.
Part of the charm is that, since there are mathematical ideas behind the kids of Funville, the way they work and interact is surprising but logical. Readers can predict what's going to happen or wonder what would happen.
“Yeah,” said Harvey as if it was the most natural thing in the world to have a power. “My power is to halve things in size.”
“Halve?” inquired Emmy. “But he is no more than a tenth of what he was!”
“That’s because ...
What do you think happened? Small mysteries like that. Big mysteries, like how will Leo get back to full size? Harvey has a brother Doug - will that relate? And the biggest: how will they get home?
Want to know more? There's a whole Funville Blog Tour, with lots of perspectives. As with a couple of those, you might find yourself wanting to make your own Funville characters. Even now, whom do you think Emmy and Leo might meet?
I have one here... Fan Funville Fiction!
Dylan's Dangling
Emmy and Leo worked hard all week to get done with school work and chores, so they would have an afternoon free to visit their friends in Funville. They had developed the habit of tucking things into an old rucksack that would be interesting to see just how their friends' powers would work on them. This time the rucksack held a tiny Ant Man action figure, an elephant toy, a stretchable rubber snake, and an assortment of snacks.
Emmy was particularly interested in Fay's and Randy's powers and how they interacted with other powers. So she was always glad to see them at the other end of the slide down the Thief. But she and Leo were both surprised to see someone new in the playground.
He was sitting on one half of a see-saw, but was up in the air instead of down on the ground. Maybe Heather had been here? He had on a shirt that was way too long, but otherwise seemed to fit him well. Adding to his stretched out appearance was a very tall cylinder of curly black hair.
Leo ran over to him immediately. "I'm Leo!" he announced, and the boy answered him. "Oh, I know. I came down here to meet you two because I was so interested in the stories that everyone told about you. People without powers, but you're fun anyway? And Pencilvania? Wherever that is!"
"Pennsylvania," Emmy corrected, "but that's a good synonym! You probably know I'm Emmy, but who are you?"
"Dylan," Dylan answered, "and I'm stuck up here. I was playing with- "
"Heather?" Emmy interrupted.
"Exactly!" said Dylan. "But her mother was calling, and she hopped off, and didn't notice that she made that end so heavy, and ... long story short, here I am."
"How can we help?" asked Leo.
"Do you have anything small enough to stand up under the other end?" wondered Dylan.
"Sure!" said Leo, and rummaged in the rucksack for a perfect skipping stone he was hoping Cory would be willing to use his power on. He slid it under the down side of the see saw. "Like this?"
"Mmmmmm hmmmmm," said Dylan, who was already concentrating. Slowly the see saw seat lifted up, pushed by the stone, which was growing. But not getting like a bigger stone - more like a tree. Once the see saw got level, Dylan hopped off.
"So your power is growing things?" asked Emmy.
"Not exactly..."
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# Thread: sums and differnces of different angles
1. ## sums and differnces of different angles
the question is:
find the exact value of
sin(x-y) when sinx=8/17 and cos=5/13
need help
2. Use the formula $\displaystyle \sin(x-y)=\sin x\cos y-\sin y\cos x$.
3. Originally Posted by red_dog
Use the formula $\displaystyle \sin(x-y)=\sin x\cos y-\sin y\cos x$.
but you only have sinx and cosy you dont have siny and cosx
4. $\displaystyle \cos x=\pm\sqrt{1-\sin^2x}, \ \sin y=\pm\sqrt{1-\cos^2y}$
To choose the sign before the radical, you have to know in which quadrant are x and y.
5. i wasnt taught the radical yet, is there another way like guess and check?
6. Originally Posted by pogiphilip
i wasnt taught the radical yet, is there another way like guess and check?
Hi
Do you know the domain for x and y ? ie 0<x<90 ??
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#### hupfer
##### New member
Hello,
Could someone help me with a simple problem?
I would like to know how to solve a fraction within a fraction.
Example:
10/4/x
The issue I'm having is with a formula from a physics problem. I have this formula and I need to rearrange like this . I don't quite understand how to get there.
#### pka
##### Elite Member
The issue I'm having is with a formula from a physics problem. I have this formula View attachment 12181
I need to rearrange like this View attachment 12182. I don't quite understand how to get there.
\displaystyle \begin{align*}\Delta x &= \left( {\frac{{v + {v_0}}}{2}} \right)t\\{\left( {\frac{{v + {v_0}}}{2}} \right)^{ - 1}}\Delta x &= t\\\left( {\frac{2}{{v + {v_0}}}} \right)\Delta x &= t \end{align*}
#### hupfer
##### New member
Would you care to elaborate a little bit more?
I'm having trouble understanding how you got the negative exponent
#### Riman643
##### New member
Hello,
Could someone help me with a simple problem?
I would like to know how to solve a fraction within a fraction.
Example:
10/4/x
In this example the answer would be $$\displaystyle \frac{10x}{4}$$ which can be simplified to $$\displaystyle \frac{5x}{2}$$. When dividing a fraction by a number, you keep the denominator the same (in this case 4) and multiply the numerator (10) by the number on the bottom (x).
The issue I'm having is with a formula from a physics problem. I have this formula View attachment 12181 and I need to rearrange like this View attachment 12182. I don't quite understand how to get there.
For this physics equation, you want to isolate t, so you need to divide both sides by $$\displaystyle \frac{v + v_{o}}{2}$$. This will give you $$\displaystyle t = \frac{Δx}{\frac{v + v_{o}}{2}}$$. Using the same technique as above, we keep the denominator the same (v + vo) and multiply the numerator (Δx) by the number on the bottom (2). This will give you $$\displaystyle t = \frac{2 Δx}{v + v_{o}}$$ or as your equation shows $$\displaystyle t = Δx(\frac{2}{v + v_{o}})$$. I hope this clears up some of the confusion.
#### Subhotosh Khan
##### Super Moderator
Staff member
Hello,
Could someone help me with a simple problem?
I would like to know how to solve a fraction within a fraction.
Example:
10/4/x
The issue I'm having is with a formula from a physics problem. I have this formula View attachment 12181 and I need to rearrange like this View attachment 12182. I don't quite understand how to get there.
$$\displaystyle Δx=\left(\dfrac{v+v_0}{2}\right)t$$
Multiply both sides by 2 - to get,,
$$\displaystyle 2 * Δx = (v+v_0) * t$$
can you continue?
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# discrete function examples
Note that since the domain is discrete, the range is also discrete. There are more properties of mgf's that allow us to find moments for functions of random variables. The Dirac comb of period 2 π although not strictly a function, is a limiting form of many directional distributions. The syntax for creating discrete-time models is similar to that for continuous-time models, except that you must also provide a sample time (sampling interval in seconds). From Wikibooks, open books for an open world < Discrete Mathematics. In mathematics, we can create recursive functions, which depend on its previous values to create new ones. It is essentially a wrapped Dirac delta function. Example: A clock stops at any random time during the day. They are uniquely characterized by a cumulative distribution function that can be used to calculate the probability for each subset of the support. DTFT is a frequency analysis tool for aperiodic discrete-time signals The DTFT of , , has been derived in (5.4): (6.1) The derivation is based on taking the Fourier transform of of (5.2) As in Fourier transform, is also called spectrum and is a continuous function of the frequency parameter Is DTFT complex? We often call these recurrence relations . PDF for the above example. You can see in the two examples above that there are functions which are surjective but not injective, injective but not surjective, both, or neither. Transfer functions are a frequency-domain representation of linear time-invariant systems. Understanding Discrete Distributions. Have a look at the previously shown output of the RStudio console. It supports almost all common properties from MATLAB that are supported by a continuous plotting function plot(). define function and give examples of functions; find the domain, codomain and range of a function; define the different types of functions such as injective function (one-to-one function), surjective function (onto function), bijective function, give examples of each kind of function… Translations of the phrase DISCRETE FUNCTIONS from english to french and examples of the use of "DISCRETE FUNCTIONS" in a sentence with their translations: A llows for 3 discrete functions only( no shared functions). The SAS INTCK Function: Examples. discrete creates a discrete vector which is distinct from a continuous vector, or a factor/ordered vector. Using the moment generating function, we can now show, at least in the case of a discrete random variable with finite range, that its distribution function is completely determined by its moments. The variable x contains numeric values and the variable y is a factor consisting of four different categories. S-functions that use the variable-step sample time can be used only with variable-step solvers. For instance, consider a continuous-time SISO dynamic system represented by the transfer function sys(s) = N(s)/D(s), where s = jw and N(s) and D(s) are called the numerator and denominator polynomials, respectively. The vsfunc.c example is a discrete S-function that delays its first input by an amount of time determined by the second input. By taking the contrapositive of the implication in this definition, a function is injective if … Note that although we sayX is 3.5 on the average, we must keep in mind that our X never actually equals 3.5 (in fact, it is impossible forX to equal 3.5). In this paper we start with brie°y surveying two related topics: harmonic functions on graphs and discrete analytic functions on grids. The other function are tools for manipulating descrete vectors. Discrete Distribution. Discrete Mathematics/Functions and relations. However, if the arguments aren’t … A discrete distribution, as mentioned earlier, is a distribution of values that are countable whole numbers. And the density curve is given by. Solution: We observe that the graph corresponds to a continuous set of input values, from \(- 2\) to 3. The PDF for X is. Discretized function representation¶ Shows how to make a discretized representation of a function. A function f from A to B is said to be one-to-one, or injective, if and only if f(a) = f(b) implies that a = b for all a and b in the domain A. For example, we can have the function : f ( x )=2 f ( x -1), with f (1)=1 If we calculate some of f 's values, we get In the case when a function is both one-to-one and onto (an injection and surjection), we say the function is a bijection , or that the function is a bijective function. It shows that our example data has two columns. Worked examples on identifying valid discrete probability distributions. Example 2: The plot of a function f is shown below: Find the domain and range of the function. In this section, we give examples of the most common uses of the SAS INTCK function. Specifying Discrete-Time Models. Discrete functions may be represented by a discrete Fourier transform, which also we shall not look at in this book. These components consist of a fundamental frequency component, multiples of the fundamental frequency, called the harmonics and a bias term, which represents the average off-set from zero. Note that the mgf of a random variable is a function of \(t\). discrete example sentences. Together, we will learn how to create a joint probability mass function and find probability, marginal probability, conditional probability, and mean and variance. Discrete values are countable, finite, non-negative integers, such as 1, 10, 15, etc. 5.1. Let X be the time (Hours plus fractions of hours ) at which the clock stops.
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Next: 3. The General Case Up: Solving One-Sided Equations in Previous: 1. Introduction
# 2. The Special Case of One Equation
Let and be the equation we want to solve. In order to describe the generating set of solutions we have to find one solution of the inhomogeneous equation
(1)
and if possible a finite set of generators for the solutions of the homogeneous equation
(2)
The set of solutions of equation 2 forms a right -module which is called the right module of syzygies of according to the term used for ordinary Gröbner bases in the literature (see e.g. [2]). To find a generating set for this right module we proceed as suggested in [1]:
1. Let be a finite reduced prefix Gröbner basis of the right ideal generated by in , and , corresponding vectors. There are two linear mappings given by matrices1 , such that and .
2. Equation 1 is solvable if and only if . This is equivalent to and the reduction sequence gives rise to a representation where . Then, as , we get and is such a solution of equation 1.
3. Let be a generating set for the solutions of the homogeneous equation
(3)
and let be the identity matrix. Further let be the columns of the matrix . Since these are solutions of the homogeneous equation 2 as well. We can even show that the set generates all solutions of 2:
Let be an arbitrary solution of equation 2, i.e. . Then is a solution of equation 3 as . Hence there are such that . Further we find
and hence is a right linear combination of elements in .
Now the important part is to find a generating set for the solutions of the homogeneous equation 3. Let be a finite reduced prefix Gröbner basis of the right ideal generated by and let , for . In [1] the following situations define a set of generating zeros:
For every such that is a prefix (as a word) of , i.e. for some , by Lemma 8 (see Section 5) we know for some . We determine vectors as follows:
where the polynomials are due to the reduction sequence .
Then , where
, is a solution of 3 as .
If no such polynomials exist in , Baader concluded that the homogeneous equation 3 in the free monoid ring had no solution. This is no longer true for arbitrary monoid rings.
Example 1 Let be a monoid ring where is presented by the string rewriting system , . Then for the homogeneous equation
we find that the set is a reduced prefix Gröbner basis of the right ideal it generates. Moreover neither of the head terms of the polynomials in this basis is prefix of the other. Still the equation can be solved: is a solution since .
This phenomenon is due to the fact that in most monoid rings s-polynomials are not sufficient for a Gröbner basis test. In [6] such a test for monoid rings as described here is presented. Besides testing s-polynomials the following test set has to be considered2:
For let for some . Additionally, we define vectors for and as follows: Let . For every we know as is a prefix Gröbner basis. Then , where
, is a solution of equation 3 as .
Lemma 2 The finitely many vectors , , form a right generating set for all solutions of equation 3.
Proof : 1.1
Let be an arbitrary (non-trivial) solution of 3. Let be the maximal term when concatenating the head terms of the multiples in the sum and the number of multiples with . A solution is called smaller than if either or and . We will prove our claim by induction on and 3. Since we assume to be a non-trivial solution, we know . Then we can distinguish two cases:
1. If there is such that , then , , and for some , . Then with
, is again a solution of 3. It remains to show that it is a smaller one. To see this we have to examine the multiples for all . Remember that since we get as the arise from the reduction sequence .
1. For we get and as and the resulting monomials add up to zero we get .
2. For we get and either if or else .
Hence either or is decreased.
2. Let us now assume there are such that . Without loss of generality we can assume that for some . Then by Lemma 8 we know that for some . For the corresponding s-polynomial we have a vector and we can define where with
. It remains to show that this solution indeed is smaller. To do this we examine the multiples for all .
1. For we get and by Lemma 8 . Hence implying either if or else .
2. For we get . Since we find .
3. For we get . Hence either if or .
Hence we find that either or in case , . Therefore, in all cases the solution derived from is smaller and we can show our claim by induction.
Next: 3. The General Case Up: Solving One-Sided Equations in Previous: 1. Introduction
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These Thurston maps are NET maps for every choice of translation term. They are primitive and have degree 35. PURE MODULAR GROUP HURWITZ EQUIVALENCE CLASSES FOR TRANSLATIONS {0} {lambda1} {lambda2} {lambda1+lambda2} These pure modular group Hurwitz classes each contain infinitely many Thurston equivalence classes. The number of pure modular group Hurwitz classes in this modular group Hurwitz class is 8. ALL THURSTON MULTIPLIERS c/d IN UNREDUCED FORM 0/35, 1/35, 1/7, 1/5, 2/7, 2/5, 3/7, 4/7, 3/5, 4/5, 1/1, 5/5, 6/5, 2/1, 3/1 4/1, 6/1, 8/1, 9/1, 11/1, 12/1, 13/1, 16/1, 17/1, 18/1, 19/1, 22/1, 23/1 24/1, 26/1, 27/1, 29/1, 31/1, 32/1, 33/1, 34/1 EXCLUDED INTERVALS FOR THE HALF-SPACE COMPUTATION (-117.713406,-0.017473 ) ( 0.017473,117.713406) 1/0 is the slope of a Thurston obstruction with c = 2 and d = 1. These NET maps are not rational. SLOPE FUNCTION INFORMATION NUMBER OF FIXED POINTS: 2 EQUATOR? FIXED POINT c d 0 lambda1 lambda2 lambda1+lambda2 1/0 2 1 No No No No 0/1 1 35 Yes Yes No No NUMBER OF EQUATORS: 1 1 0 0 There are no more slope function fixed points. Number of excluded intervals computed by the fixed point finder: 1897 No nontrivial cycles were found. Here is the action of the slope function on an invariant set S of slopes. N/1 -> 2N/1 The set S contains infinitely many infinite slope function trajectory tails. Here is the action of the slope function on an invariant set S of slopes. (3N+1)/3 -> (6N+2)/3, (3N+2)/3 -> (6N+4)/3 The set S contains infinitely many infinite slope function trajectory tails. Here is the action of the slope function on an invariant set S of slopes. (11N+1)/11 -> (22N+2)/11, (11N+2)/11 -> (22N+4)/11, (11N+3)/11 -> (22N+6)/11, (11N+4)/11 -> (22N+8)/11, (11N+5)/11 -> (22N+10)/11, (11N+6)/11 -> (22N+12)/11, (11N+7)/11 -> (22N+14)/11, (11N+8)/11 -> (22N+16)/11, (11N+9)/11 -> (22N+18)/11, (11N+10)/11 -> (22N+20)/11 The set S contains infinitely many infinite slope function trajectory tails. Here is the action of the slope function on an invariant set S of slopes. (33N+1)/33 -> (66N+2)/33, (33N+2)/33 -> (66N+4)/33, (33N+4)/33 -> (66N+8)/33, (33N+8)/33 -> (66N+16)/33, (33N+16)/33 -> (66N+32)/33, (33N+17)/33 -> (66N+34)/33, (33N+25)/33 -> (66N+50)/33, (33N+29)/33 -> (66N+58)/33, (33N+31)/33 -> (66N+62)/33, (33N+32)/33 -> (66N+64)/33 The set S contains infinitely many infinite slope function trajectory tails. Here is the action of the slope function on an invariant set S of slopes. (33N+5)/33 -> (66N+10)/33, (33N+7)/33 -> (66N+14)/33, (33N+10)/33 -> (66N+20)/33, (33N+13)/33 -> (66N+26)/33, (33N+14)/33 -> (66N+28)/33, (33N+19)/33 -> (66N+38)/33, (33N+20)/33 -> (66N+40)/33, (33N+23)/33 -> (66N+46)/33, (33N+26)/33 -> (66N+52)/33, (33N+28)/33 -> (66N+56)/33 The set S contains infinitely many infinite slope function trajectory tails. The slope function maps some slope to the nonslope. If the slope function maps slope s to a slope s' and if the intersection pairing of s with 1/0 is n, then the intersection pairing of s' with 1/0 is at most n. The slope function orbit of every slope whose intersection pairing pairing with 1/0 is at most 50 either ends in the nonslope or ends in one of the slopes described above or it has an infinite tail in one of the infinite sets described above. FUNDAMENTAL GROUP WREATH RECURSIONS When the translation term of the affine map is 0: NewSphereMachine( "a=(2,35)(3,34)(4,33)(5,32)(6,31)(7,30)(8,29)(9,28)(10,27)(11,26)(12,25)(13,24)(14,23)(15,22)(16,21)(17,20)(18,19)", "b=(1,35)(2,34)(3,33)(4,32)(5,31)(6,30)(7,29)(8,28)(9,27)(10,26)(11,25)(12,24)(13,23)(14,22)(15,21)(16,20)(17,19)", "c=(1,35)(2,34)(3,33)(4,32)(5,31)(6,30)(7,29)(8,28)(9,27)(10,26)(11,25)(12,24)(13,23)(14,22)(15,21)(16,20)(17,19)", "d=(2,35)(3,34)(4,33)(5,32)(6,31)(7,30)(8,29)(9,28)(10,27)(11,26)(12,25)(13,24)(14,23)(15,22)(16,21)(17,20)(18,19)", "a*b*c*d"); When the translation term of the affine map is lambda1: NewSphereMachine( "a=(1,2)(3,35)(4,34)(5,33)(6,32)(7,31)(8,30)(9,29)(10,28)(11,27)(12,26)(13,25)(14,24)(15,23)(16,22)(17,21)(18,20)", "b=(2,35)(3,34)(4,33)(5,32)(6,31)(7,30)(8,29)(9,28)(10,27)(11,26)(12,25)(13,24)(14,23)(15,22)(16,21)(17,20)(18,19)", "c=(2,35)(3,34)(4,33)(5,32)(6,31)(7,30)(8,29)(9,28)(10,27)(11,26)(12,25)(13,24)(14,23)(15,22)(16,21)(17,20)(18,19)", "d=(1,2)(3,35)(4,34)(5,33)(6,32)(7,31)(8,30)(9,29)(10,28)(11,27)(12,26)(13,25)(14,24)(15,23)(16,22)(17,21)(18,20)", "a*b*c*d"); When the translation term of the affine map is lambda2: NewSphereMachine( "a=<1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,c,c*d*c^-1>(1,34)(2,33)(3,32)(4,31)(5,30)(6,29)(7,28)(8,27)(9,26)(10,25)(11,24)(12,23)(13,22)(14,21)(15,20)(16,19)(17,18)", "b=(1,35)(2,34)(3,33)(4,32)(5,31)(6,30)(7,29)(8,28)(9,27)(10,26)(11,25)(12,24)(13,23)(14,22)(15,21)(16,20)(17,19)", "c=(1,35)(2,34)(3,33)(4,32)(5,31)(6,30)(7,29)(8,28)(9,27)(10,26)(11,25)(12,24)(13,23)(14,22)(15,21)(16,20)(17,19)", "d=(1,34)(2,33)(3,32)(4,31)(5,30)(6,29)(7,28)(8,27)(9,26)(10,25)(11,24)(12,23)(13,22)(14,21)(15,20)(16,19)(17,18)", "a*b*c*d"); When the translation term of the affine map is lambda1+lambda2: NewSphereMachine( "a=(1,35)(2,34)(3,33)(4,32)(5,31)(6,30)(7,29)(8,28)(9,27)(10,26)(11,25)(12,24)(13,23)(14,22)(15,21)(16,20)(17,19)", "b=(2,35)(3,34)(4,33)(5,32)(6,31)(7,30)(8,29)(9,28)(10,27)(11,26)(12,25)(13,24)(14,23)(15,22)(16,21)(17,20)(18,19)", "c=(2,35)(3,34)(4,33)(5,32)(6,31)(7,30)(8,29)(9,28)(10,27)(11,26)(12,25)(13,24)(14,23)(15,22)(16,21)(17,20)(18,19)", "d=(1,35)(2,34)(3,33)(4,32)(5,31)(6,30)(7,29)(8,28)(9,27)(10,26)(11,25)(12,24)(13,23)(14,22)(15,21)(16,20)(17,19)", "a*b*c*d"); ****************************INTEGER OVERFLOW REPORT***************************** Imminent integer overflow caused the modular group computation to abort.
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# Direction cosines
We can find the angle between two nonzero vectors using the dot product:
$\displaystyle \vec{u} \cdot \vec{v} = \left \lvert \vec{u} \right \rvert \left \lvert \vec{v} \right \rvert \cos{\theta} \qquad \Rightarrow \qquad \cos{\theta} = \frac{\vec{u} \cdot \vec{v}}{\left ( \left \lvert \vec{u} \mid \left \lvert \vec{v} \right \rvert \right ) \right.}$.
Often, we want to determine the angle that a certain vector makes with the coordinate axes. To do this, we just put $\displaystyle \hat{i}$, $\displaystyle \hat{j}$, and $\displaystyle \hat{k}$ in the place of $\displaystyle \vec{v}$ one at a time and solve for the angle. Since this operation is so common, it is worthwhile to work out specific equations.
Let’s start with the x-axis. If we have a vector $\displaystyle \vec{u}$ in component form, then
$\displaystyle \cos{\theta} = \frac{\vec{u} \cdot \hat{i}}{\left \lvert \vec{u} \right \rvert \left \lvert \hat{i} \right \rvert} = \frac{\left [ u_{1} , u_{2} , u_{3} \right ] \cdot \left [ 1 , 0 , 0 \right ]}{\left \lvert \vec{u} \right \rvert \left ( 1 \right )} = \frac{u_{1}}{\left \lvert \vec{u} \right \rvert}$.
When we dot $\displaystyle \vec{u}$ with one of the standard basis vectors, we are effectively choosing one of its components. Simplifying the equation for the other two axes is just as easy. To avoid confusion, we use three different Greek letters to represent the three angles:
Axis Basis Cosine Value
x $\displaystyle \hat{i}$ $\displaystyle \cos{\alpha}$ $\displaystyle u_{1} / \left \lvert \vec{u} \right \rvert$
y $\displaystyle \hat{j}$ $\displaystyle \cos{\beta}$ $\displaystyle u_{2} / \left \lvert \vec{u} \right \rvert$
z $\displaystyle \hat{k}$ $\displaystyle \cos{\gamma}$ $\displaystyle u_{3} / \left \lvert \vec{u} \right \rvert$
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# How many four-letter words w using an n-letter alphabet satisfy...
This question has two parts. The first part was easy.
a) How many four-letter words w using an n-letter alphabet satisfy $w_i \neq w_{i+1}$ for $i=1,2,3$
Simple. Choose the first letter in $n$ ways. Because no consecutive letter can have the same letter as the previous, we have $n-1$ ways for the last three for a total:
$$n \times (n-1) \times (n-1) \times (n-1) = n(n-1)^3$$
b) How many of the words in (a) also satisfy $w_4 \neq w_1$?
I'm struggling here. I thought of the possibilites, but I'm having trouble counting them.
Consider when $w_3 = w_1$ and when $w_3 \neq w_1$. Obviously, if $w_3 = w_1$, $w_4$ can be chosen in $n-1$ ways. Why? Take for example the standard alphabet. If we have the incomplete word:
A E A _
$w_4$ can be every letter except "A" to satisfy $w_4 \neq w_1$. Hence, $w_4$ has $n-1$ possibilities.
If $w_3 \neq w_1$ there are $n-2$ ways to choose $w_4$ by the same reasoning. Certainly, if we have the incomplete word:
A E C _
$w_4$ can be every letter except "A" and "C." $n-2$ ways.
But how can I count this? I thought it would be:
$$n \times (n-1)^2 \times (n-2)$$
But that doesn't take into account the special cases I considered above, does it?
You have also correctly identified the cases in the second question.
Case 1: The third letter is the same as the first letter.
We have $n$ ways to select the first letter. Since the second letter must be different from the first, we can select it in $n - 1$ ways. We have only one choice for the third letter since it must be the same as the first letter. Since the last letter is different from the third letter, it must also be different from the first letter. Thus, there are $n - 1$ choices for the fourth letter. Hence, there are $$n \cdot (n - 1) \cdot 1 \cdot (n - 1) = n(n - 1)^2$$ such words.
Case 2: The third letter is different from the first letter.
We have $n$ ways to select the first letter. Since the second letter must be different from the first, we can select it in $n - 1$ ways. Since the third letter must be different from both the first letter and the second letter, the third letter can be selected in $n - 2$ ways. Since the fourth letter must be different from both the third letter and the first letter, we can select the fourth letter in $n - 2$ ways. Hence, there are $$n(n - 1)(n - 2)(n - 2) = n(n - 1)(n - 2)^2$$ such words.
Total: Since the cases are mutually exclusive, there are $$n(n - 1)^2 + n(n - 1)(n - 2)^2$$
four letter words in which each letters differs from the preceding letter and the last letter is different from the first letter.
• Ah thank you. This clears everything up! Commented Mar 14, 2017 at 23:50
The answer is $n(n-1)^2 + n(n-1)(n-2)^2$. Maybe a tree chart makes this clear but a don't know how to use tikz here :(
However I describe what I mean. You already figured out that there a two different cases: Either $w_3 = w_1$ or $w_3 \neq w_1$.
Case 1: In the first case that $w_3 = w_1$ you already know that $w_3 = w_1$ so you don't have a choice for the third latter which gives $$n\cdot (n-1) \cdot 1 \cdot (n-1) \quad \text{Opportunities}$$
Case 2: In the other case you know $w_3 \neq w_1$ and $w_3 \neq w_2$ so you have $(n-2)$ opportunities for $w_3$ and you also have $w_4 \neq w_3$ and $w_4 \neq w_1$ which gives altogether (in this case) $$n\cdot (n-1) \cdot (n-2) \cdot (n-2) \quad \text{Opportunities}$$
When you sum all opportunities together you receive what I already mentioned in the beginning.
• You can insert diagrams by selecting the button that looks like the image of a mountain. In case 2, note that the fourth letter must be different from both the first letter and the third letter. Commented Mar 14, 2017 at 23:50
• Your answer was basically the same, however I can only choose one. Thank you! Commented Mar 14, 2017 at 23:50
• @N.F.Taussig lol I had that on my paper and then I forgot about the question and corrected that to $n-1$. facepalm... I will edit it Commented Mar 14, 2017 at 23:54
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# Chapter 11
## Chapter 11
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
2. Chapter 11: Counting Methods 11.1 Counting by Systematic Listing 11.2 Using the Fundamental Counting Principle 11.3 Using Permutations and Combinations 11.4 Using Pascal’s Triangle 11.5 Counting Problems Involving “Not” and “Or” © 2008 Pearson Addison-Wesley. All rights reserved
5. Uniformity Criterion for Multiple-Part Tasks A multiple-part task is said to satisfy the uniformity criterion if the number of choices for any particular part is the same no matter which choices were selected for the previous parts. © 2008 Pearson Addison-Wesley. All rights reserved
6. Fundamental Counting Principle When a task consists of k separate parts and satisfies the uniformity criterion, if the first part can be done in n1 ways, the second part can be done in n2 ways, and so on through the kth part, which can be done in nk ways, then the total number of ways to complete the task is given by the product © 2008 Pearson Addison-Wesley. All rights reserved
8. Example: Two-Digit Numbers with Restrictions How many two-digit numbers that do not contain repeated digits can be made from the set {0, 1, 2, 3, 4, 5} ? Solution There are 5(5) = 25 two-digit numbers. © 2008 Pearson Addison-Wesley. All rights reserved
9. Example: Two-Digit Numbers with Restrictions How many ways can you select two letters followed by three digits for an ID? Solution There are 26(26)(10)(10)(10) = 676,000 IDs possible. © 2008 Pearson Addison-Wesley. All rights reserved
10. Factorials For any counting number n, the product of all counting numbers from n down through 1 is called n factorial, and is denoted n!. © 2008 Pearson Addison-Wesley. All rights reserved
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# How can I convert to seamless pattern in Illustrator?
After following many tutorials about pattern making using pattern maker and brush pattern in Illustrator and then offset in Photoshop, I couldn't make the below image into a usable seamless pattern :(
It is the complexity of the "wave" shape that makes it difficult to achieve; most other tutorials and questions here in Stackexchange talk about simple/square patterns.. this one is different.
Any suggestions?
• Create the wave at a 90° vertical or horizontal.... that will be much easier. Then you can rotate the pattern when it's applied to an object (Photoshop can't do that). Dec 17, 2013 at 23:58
• Unless you can make a pattern then rotate it as Scott suggests, making tiles with rotated/diagonal images is much more complicated than if they are straight. See How to create a pattern or tiles from rotated elements for some detailed explanations on why Dec 20, 2013 at 16:30
• Thanks guys for the comments. The thing is I have this image as a base, and I want to convert it to a seamless pattern without having to create the waves from scratch I already tried duplicating a curve line in AI, but it failed to create a seamless pattern :( Dec 23, 2013 at 12:22
• Did you try doing as Scott suggests? Dec 23, 2013 at 15:47
I'm prepared to eat my words here, but unless I'm greatly mistaken that is NOT a repeating pattern - despite appearances. It may have been one once, but not in its current form.
I'll try to explain;
`1.` (if you want to repeat my test) Rotate the image so that one wavy line is exactly horizontal. You can do this by finding the high point of one line and aligning it horizontally against its next high point.
`2.` If you draw a line through these high points, and then do the same with two OTHER wavy lines, you will see that they arc away from one another - like a cone, or a pint glass on its side.
The image below demonstrates this - the top horizontal line travels at approximately 3.3 degrees, the middle line at 0 degrees (obviously, because its been lined up) and the lower line at approximately -3.8 degrees.
`3.` To make things even more complex, the lines aren't traveling in a straight line, either. Looking at the middle line more closely, we can see it actually travels in an arc.
In other words, the image appears to be warped - perhaps by design. There are also some forms of DRM that protect images by warping them subtly so that smaller pieces of a larger whole image can't be pieced back together again (I don't particularly think that's happening here, but I'm just throwing that in there as an aside).
To my mind, your only real option here is to create the pattern yourself, so I'll explain how you can achieve a very similar effect.
`1)` Open a new Illustrator document at 900px by 900px in size (I only used this size so that I could use your source image as a reference - and it happens to be that size).
`2)` Drag in your PNG pattern or go to FILE > PLACE, then select the pattern file (either achieves the same thing). Keep the image on its own layer to keep it out of the way.
`3)` Create some new color swatches so that you can exactly reference the colors of your source pattern (if you want things to look the same). One way to do this is by selecting the eyedropper tool and clicking on each of the two predominant colors of your image - each time dragging the color to the swatches panel once you're happy with the color (you can drag the color directly from the fill or stroke boxes).
A few pointers;
• If the eyedropper tool isn't picking up the color you want, double-click on the tool and make sure that everything is unticked in 'Eyedropper Picks Up' except Appearance > Focal Fill > Color. You can always put things back again afterwards.
• If the wavy line color isn't coming out right, make sure you zoom right in to eyedrop the brightest, most vivid part of the line. Because it's a rasterized image, there are lots of different shades at the edge of the line, which may not be too helpful.
• finally, if that seems too much like hard work, you can just copy the value I picked up. Background Color (RGB): (R) 35, (G) 106, (B) 85, and Line Color: (R) 57, (G) 121, (B) 102.
`4)` Now for the effect. Use the ellipse tool to create a symmetrical circle (hold down shift when dragging). The size doesn't matter too much at this stage. For ease of working, remove any fill and give it 1pt of stroke so you can see it easily.
`5)` With the circle selected, use the scissors tool (might be hidden under the eraser tool) and snip both the top and bottom anchor points of the circle. Effectively, this will chop it into two.
`6)` With the Selection Tool (black arrow), pull the halves apart. Then drag the top-most anchor point of one half directly on top of the lower-most anchor point of the other. If you have Smart Guides enabled (VIEW > SMART GUIDES), the two points should snap on top of each other very easily.
`6)` With the Direct Selection Tool (white arrow), marquee over the now-overlapping anchor points and join them together (OBJECT > PATH > JOIN). You should now have a single open path (either an 'S' shape or a reverse 'S', depending on which points you joined.
`7)` With the Direct Selection Tool (white arrow), select just the top anchor point and delete it. Then do the same with the lowest anchor point. You should end up with something like this (again, yours may or may not be in reverse). This is the basis for our wavy line;
`8)` In order to get a similar effect to the existing pattern, we need to reshape this line a little. First, with the line selected, double-click on the rotate tool in order to bring up the dialog box. Set the rotation to 45 degrees (or -45 degrees if you're going the other way) so that the line is running exactly horizontally. This is important so that we don't warp the line in an irregular way in the next step.
`9)` Find the Transform controls in Illustrator's toolbar (you will either need to click on the word 'Transform' if you're working on a smaller screen - or the controls will already be visible). Set the width to 364px and the height to 26.5px. These values are simply those that I came up with when comparing the shape of the original pattern against my custom wavy line.
`10)` With your wavy line selected, change the stroke to the lighter shade of green we defined in step 3 and set the stroke weight to 1.5pt. Remove any fill if one is present.
`11)` Go to the appearance panel (WINDOW > APPEARANCE, if you can't already see it), make sure your stroke is highlighted, and add a second stroke by clicking on the hollow square in the bottom left-hand corner. If your existing stroke was highlighted, this will create a duplicate.
`12)` Select the lower of the two strokes and set its weight to 9pt and change the color to the darker shade of green. You should see something like this;
`13)` In the appearance panel, select 'Path' (you DON'T want either of the strokes selected - this needs to be applied to the whole thing), and then go to 'EFFECT > DISTORT & TRANSFORM > TRANSFORM'. Use the following settings;
• Leave the scale values alone (they should both be 100%)
• Set MOVE > VERTICAL to 7px (horizontal should remain at 0px)
• Rotate angle should be left alone (0 degrees)
• Under OPTIONS, make sure 'Transform Objects' is ticked (untick everything else, if in doubt)
• Set 'Copies' to 50
Enable 'Preview' if you like - but your results should look like this;
`14)` With your path selected, go to OBJECT > PATTERN > MAKE. Use the following settings;
• Tile Type: Grid
• Width: 364px
• Height: 357px
Then click 'Done' at the top of the screen. Your new pattern will appear in your swatches window.
`15)` Test your pattern! Create a shape of some sort (a rectangle will do), apply the new pattern to the fill and there you go. If you want, use the rotate tool to put your pattern at an angle (make sure that 'Transform Patterns' is ticked and 'Transform Objects' is UNticked).
Additional things you might want to try;
• the wavy line I've created with this example isn't perfect (though close enough for the purposes of explanation) - use the control handles to get an even more perfect curve.
• If you find that your repeating pattern displays an ever-so slight seam at the edge, go back to your path (the complete item immediately before using 'Make Pattern' - at the end of step 13) and DUPLICATE the path. Place the two paths directly alongside one another so that the farthest anchor points overlap and join them together (joining is important to remove any seam). The shape will be twice as wide, but otherwise identical. Now create the pattern again (step 14) - you can use exactly the same settings - but the difference is that your wider pattern will now overlap with half of itself, and in doing so hide any seam beneath it.
I know that's a lot of steps, but it really doesn't take too long once you're familiar with the concepts being used. The advantages are that because this is entirely vector based, you can scale, tweak, rotate and recolor the pattern to your heart's content without any loss of quality, as well.
Hope that's of some help.
• Waw! if everyone else in stackexchange.com writes answers like the one you posted above.. Then we can forget about Wikipedia! Thanks @Hobbes Jun 30, 2014 at 12:50
• haha ^_^ If everyone wrote answers like the one posted above we'd probably see a decline in the world's human population on account of everyone being chronically unsociable. Glad to be of use, nume. -_^ Jun 30, 2014 at 16:11
• Excellent, encyclopedic answer. +1 for stuffed tigers. Jul 1, 2014 at 8:24
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# Multimeter reading double voltage [duplicate]
Using a multimeter and set it on 200V DC - that's the lowest on this meter.
Tested a 1.5V and it reads 2.8
Tested a 3V and it reads 6.
Just replaced the battery.
The meter is ETEKCITY.
Any suggestions?
• Buy a different meter? Is 200VDC really the smallest range? Can you post a picture? Commented Sep 14, 2020 at 15:04
• Well I'm measuring a 3V CR2032 battery - it says 3V on it. Yes 200V is the smallest - 200 and 500. I can see some inaccuracies but 2x - most probably something wrong with the meter. Time to buy another cheap one. Commented Sep 14, 2020 at 15:19
• And if you measured 0 volts it might read 1.3 volts - that's an absolute error of infinity or, relative to full-scale, an error of 0.65%. It's a numbers game. Commented Sep 14, 2020 at 15:24
• Replaced the battery. With a new one? Cheap multimeters read high when the battery is too low. Commented Sep 14, 2020 at 15:50
• Now we know you were measuring a DC voltage, using the AC voltage range of the DMM, then this existing question provides a good answer to explain that behavior: "Measuring AC voltage from DC battery". Voting to close this question as a duplicate of that one. Commented Sep 14, 2020 at 20:52
Figure 1. A comparison with an analogue meter. 1.5 volt will barely move the pointer and any reading will have little precision.
A 200 V meter is not a suitable instrument for measuring a 1.5 V cell. The resolution isn't available at the accuracy you require. Digital meters have specifications for error typically as a percentage of full scale ± a couple of digits of the display. That's what you're seeing.
I've never seen a "multimeter" that had a minimum V DC range of 200 V. For low voltage measurement you'll need a more versatile meter.
I had the multimeter on AC voltage - duh!
I posted this also for other newbies running into the same issue.
I've added the correct symbols to look for in the image below
Now why in AC it reads double of what it should be in DC - there should be an interesting reason.
Thanks guys.
• Just gave this a try and can confirm a somewhat similar result. Likely it's a side effect of the method used to estimate RMS voltage when measuring an AC source, in the unexpected case where there's no sinusoidal variation. Also interesting is that I get no reading with the leads reversed! If someone wanted to have some fun, they could try pulsing a DC source at various rates, or researching how common meters actually do the measurement and math. Commented Sep 14, 2020 at 19:25
• dashman - Hi, Regarding: "Now why in AC it reads double of what it should be in DC" You can't ask questions in an answer. However I have linked some previous questions above, where some answers to those (especially this one) explain why you can see approximately 2x the DC voltage, on some DMMs, when measuring a DC voltage using the AC voltage ranges. Commented Sep 14, 2020 at 20:51
• @SamGibson - if that "i can't answer within a question" was serious. My "answer" was really I was a boob and had it in the wrong setting. Everyone can stand down. Re the question within the answer - I was just curious and thanks for the link. Commented Sep 14, 2020 at 22:19
• I don't see this as asking a question so much as explaining what happened and admitting that the precise reason why this mistake gives this reading is unknown. This is a valid answer as it resolves the asker's issue, but yes, closing the question as a duplicate of another where it has been previously observed and where there may be more explanation is appropriate. Commented Sep 14, 2020 at 23:07
if the minimum voltage range is 200V, the precision is a bit lower in your test range. the catalog should mention the resolution in every range. and there are a few calibration pots on the multimeter board if you open it up, and if the manufacturer deisignates them to find out which one controls which parameter.
I searched up etekcity and it seems they make a few applicanes and ordinary things, but not precision things.
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# Distance between Shigatse (RKZ) and Arxan (YIE)
Flight distance from Shigatse to Arxan (Shigatse Peace Airport – Arxan Yi'ershi Airport) is 2051 miles / 3301 kilometers / 1782 nautical miles. Estimated flight time is 4 hours 23 minutes.
Driving distance from Shigatse (RKZ) to Arxan (YIE) is 2886 miles / 4644 kilometers and travel time by car is about 52 hours 26 minutes.
## Map of flight path and driving directions from Shigatse to Arxan.
Shortest flight path between Shigatse Peace Airport (RKZ) and Arxan Yi'ershi Airport (YIE).
## How far is Arxan from Shigatse?
There are several ways to calculate distances between Shigatse and Arxan. Here are two common methods:
Vincenty's formula (applied above)
• 2051.129 miles
• 3300.972 kilometers
• 1782.382 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 2049.301 miles
• 3298.031 kilometers
• 1780.794 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Shigatse Peace Airport
City: Shigatse
Country: China
IATA Code: RKZ
ICAO Code: ZURK
Coordinates: 29°21′6″N, 89°18′41″E
B Arxan Yi'ershi Airport
City: Arxan
Country: China
IATA Code: YIE
ICAO Code: ZBES
Coordinates: 47°18′38″N, 119°54′42″E
## Time difference and current local times
There is no time difference between Shigatse and Arxan.
CST
CST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 223 kg (492 pounds).
## Frequent Flyer Miles Calculator
Shigatse (RKZ) → Arxan (YIE).
Distance:
2051
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
2051
Round trip?
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### Author Topic: Ejs Open Source Gravitational Field & Potential of Earth and Moon Java Applet (Read 22552 times)
#### lookang
• Moderator
• Hero Member
• Posts: 1774
• http://weelookang.blogspot.com
##### Ejs Open Source Gravitational Field & Potential of Earth and Moon Java Applet
« on: August 10, 2010, 05:49:01 pm »
Ejs Open Source Gravitational Field & Potential of Earth and Moon Java Applet.
based on Real Data!
customized by lookang based on an applet by Professor Andrew Duffy, remixed by lookang
Full screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
• Please feel free to post your ideas about how to use the simulation for better teaching and learning.
• Post questions to be asked to help students to think, to explore.
• Upload worksheets as attached files to share with more users.
Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!
« Last Edit: June 28, 2012, 10:09:58 am by lookang »
#### lookang
• Moderator
• Hero Member
• Posts: 1774
• http://weelookang.blogspot.com
##### Re: Ejs Open Source Gravitational Field & Potential of Earth and Moon Java Applet
« Reply #1 on: August 10, 2010, 05:55:54 pm »
1 key in real data of Earth Moon using new format 0.00E00 like in tracker
2 redrawn and scale the world view to fit both g field graph and potential graph.
3 insert earth and moon images
4 design menu for "Moon Surface;Net Force Zero;Earth Surface;Random"
5 design features for exploring escape velocity concept select position from drop down menu and key in velocity
v_escape_moon = 2280 m s-1
v_escape_earth = 11300 m s-1
6 dt is in 60 sec interval
7 can explore the gravitational constant on Earth surface = 9.81 m/s^2 and Moon = 1.6 m/s^2 approximately
« Last Edit: August 11, 2010, 12:25:19 am by lookang »
#### lookang
• Moderator
• Hero Member
• Posts: 1774
• http://weelookang.blogspot.com
##### Re: Ejs Open Source Gravitational Field & Potential of Earth and Moon Java Applet
« Reply #2 on: August 11, 2010, 11:47:05 am »
exercises by lookang:
activity A
unselect the M2=5.97E24 kg checkbox
drag the red test mass m, on the world side view closer to the Moon.
notice the value of g1 varies according to the distance away from the center of the Moon
since M2 is unselected, gnet = g1 + g2 and g2 = 0
therefore gnet = g1
note down the value of gnet when m =1kg
note down the value of Fnet when m=1 kg
now change the value of m and record down the values of g1 , gnet and Fnet. do this for a few readings.
suggest a relationship between Fnet and gnet.
by manipulating the relationship variables above, write down the form that best describe the concept of a gravitational field as an example of field of force.
hence, derive the meaning that gravitational field strength as force per unit mass
Activity B
reset the simulation if need
unselect the test mass, m and select M2.
notice the green vector drawn on the center of the Moon and Earth.
the readings are as shown as F1= 1.98E20N and F2=-1.98E20N
using the real life data that you can get from textbook, lecture notes or/and the internet, verify the equation
F = G M1M2/r^2
suggest what does F1 represent?
hint: force on ________ exerted by ___________
suggest what does F2 represent?
hint: force on ________ exerted by ___________
drag on the Moon and Earth to move along the horizontal line, observe what happens to the magnitude and direction of the forces F1 and F2.
hint: magnitude, direction and different bodies?
What is the name of this physics idea?
What is the meaning of the negative sign on the force that points in the direction opposite to x-axis direction?
Activity C
Given that Newton's law of gravitation in the form F = G M1M2/r^2 and derive the equation for gravitational field strength, g.
hint: select the g field checkbox to reveal the graph of g vs r for a system of M1 alone.
select the M2 checkbox and deduce the relationship when the system is 2 mass, M1 and M2
you may use the data from the applet to verify your equation.
Activity D
apply the equation for gravitational field strength, g = G M/r^2 to the situation of the applet.
write the meaning of g1
write the meaning of g2
hence, suggest what is the net gravitational field strength for the case of a Earth and Moon system.
gnet =
select the gravity g field checkbox
vary the left slider to the bottom to change the scale of the y axis to -1.2 to 1.2 N/kg
notice the shape of the graph of g vs r. sketch it on your worksheet or lecture.
select and deselect the M2 to test your understanding.
Activity E nil
(e) show an appreciation that on the surface of the Earth g is approximately constant and equal
to the acceleration of free fall. another applet perhaps?
Activity F
let the infinity point be i
let the final position of the point be f
write down the energies of a mass m an infinity,
hint: KEi + PEi = 0 + (-G M / infinity) = 0
write down the the energies of a mass m an a point r away from source of gravity field say M.
hint: KEf + PEf = 0 + (-G M / r)
use conservation of energy or otherwise, WDpropulsion + KEi + PEi = KEf + PEf
derive WDpropulsion in terms of G, M and r
define M
define r
hence or otherwise, verify whether you can define potential ? at a point as work done in bringing unit mass from infinity to the point.
write down the equation that shows this clearly.
select the gravity ? potential checkbox
vary the left slider to the bottom to change the scale of the y axis to high value J/kg
sketch the shape of the ? potential vs r.
select and deselect the M2 to test your understanding.
Activity G
solve problems using the equation ? = - G M/r for the potential in the field of a point mass.
for example,
Certain meteorites (tektites) found on the Earth have a composition identical with that of lunar granite. It is thought that they may be debris from volcanic eruption on the Moon. The applet shows how the gravitational potential between the surface of the Moon and the surface of the Earth varies along the line joining their centres. At the point P, the gravitational potential is a maximum.
By considering the separate contributions of earth and Moon to the gravitational potential, explain why the graph has a maximum and why the curve is asymmetrical
State how the resultant gravitational force on the tektite at any point between the Moon and the Earth could be deduced
When a tektite is at P ( drop menu select "Net Force Zero) , the gravitational forces on it due to Moon and Earth are F_M and F_E respectively. State the relation which applies between F_M and F_E.
F_Moon is which color force ?
F_Earth is which color force ?
given that the distance between Earth and Moon used in the applet is 384 403 000 m
determine the distance between test mass m and M1 (moon)
determine the distance between test mass m and M2 (earth)
verify whether the applet is accurate, which the uncertainty error between the 2 values?
If the tektite is to reach Earth, it must be projected from the volcano on the Moon with a minimum speed v0. Making use of appropriate values from the applet, find this speed. Explain your reasoning.
suggest why you cannot use the value derived theoretically, but it should be a value greater or lesser? explain.
Run the simulation with an escape velocity from Moon as v =2500 m/s, Predict and discuss very briefly whether a tektite will reach the Earth’s surface with a speed less than, equal to or greater than the speed of projection v =2500 m/s.
vary the simulation to test out the v =2500 m/s.
what is the value of velocity of test mass impacting earth?
change the values of test mass, m and rerun the sim, what is the velocity of impact on Earth?
by using equation of conservation of energy or otherwise, calculate the velocity of impact on Earth of test mass m.
(h) recognise the analogy between certain qualitative and quantitative aspects of gravitational
and electric fields.
Ejs Open Source Electric Field & Potential of 2 Charged Particles Java Applet
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1918msg6989;topicseen#msg6989
Design an experiment to verify that following table of data
Gravitational Fields Electric Fields
Due to mass interaction, m Due to charge interaction, +q and -q
Only attractive Either attractive or repulsive
Newton’s Law of gravitation Coulomb’s Law
Gravitational Field Strength Electric Field Strength
Gravitational Potential Electric Potential
« Last Edit: August 11, 2010, 02:31:11 pm by lookang »
#### lookang
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##### Re: Ejs Open Source Gravitational Field & Potential of Earth and Moon Java Applet
« Reply #3 on: April 27, 2012, 01:35:53 pm »
discussion with teacher
Can you make the moon and Earth more obvious? Moon is too small and not too visible.
Thanks.
it is currently showing real data of distance and radius of earth and moon.
Can explain how to make bigger while still showing real distances ?
I not sure how and what to change
http://weelookang.blogspot.com
#### lookang
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##### Re: Ejs Open Source Gravitational Field & Potential of Earth and Moon Java Applet
« Reply #4 on: May 07, 2012, 01:25:16 pm »
changes
made consistent to original simulation http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1921.0
#### lookang
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##### Re: Ejs Open Source Gravitational Field & Potential of Earth and Moon Java Applet
« Reply #5 on: June 08, 2012, 02:36:53 pm »
discussion
I can't reliably get the mass to be launched at the speed i want.
When i key v = -1.11E4 and click play, it doesnt even go halfway. Nothing happens if i continue to change the velocity value.
But i found i was able to make it work again if i always "reset" it by selecting "random spot" then return back to "Earth surface". However, the value of -1.11E4 never works. In fact, -1.15E5 will also cause it to return back to Earth surface. Only value of -1.16E4 onwards will it reliably reach the moon.
made the menu bar remember but the first time need to set manually
theory says 11200 m/s, to reach infinity as speed 0
but in practice it could be larger like 11500 m/s, say to reach a very far place with a speed of 100 to 1000 m/s
i have change the launch position to be slightly above sea level surface of Earth .
i think it works for 11400 m/s now.
http://en.wikipedia.org/wiki/Escape_velocity also say it is roughly 11200 m/s, so it should be larger than 11200 m/s according to the computer model about 11500 m/s should escape
So:
1. I wonder if there's anything wrong in the applet calculation values (rounding off issues) that causes the value to be different from that calculated in the worksheet.
i checked, the values i used are pretty accurate, it is the theory that is problematic.
error analysis
(11500-11200)/11200 = 2.7 % error is it acceptable?
2. Can it be done such that i need not do the "manual reset" each time i want to test a new velocity?
done, the menu remember past last values, but need to set it for first time.
« Last Edit: June 08, 2012, 05:45:43 pm by lookang »
#### lookang
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##### Re: Ejs Open Source Gravitational Field & Potential of Earth and Moon Java Applet
« Reply #6 on: June 12, 2012, 12:23:47 pm »
thanks to prof hwang's expert and masterful tips, here http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=2461.msg9275#msg9275
attached is the latest model that has the corrected escape velocity from Earth as -11200 m/s.
enjoy!
the new refinement is thanks to teacher feedback
"As for the simulation speed itself . . . i think this one is a bit slower? I dont recall having to wait for 3-4 minutes for the entire launch from Earth to moon, but apparently that's what is happening to the critical speeds (1.12 m/s to 1.15 m/s) Can this be addressed?"
done thanks to Prof Hwang's tips http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=2461.msg9275#msg9275
also added a dt for time step
« Last Edit: June 12, 2012, 12:37:10 pm by lookang »
#### lookang
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##### Re: Ejs Open Source Gravitational Field & Potential of Earth and Moon Java Applet
« Reply #7 on: June 26, 2012, 05:40:27 pm »
fixed a bug with the values of g
and the accln function with scale1 and scle2 included to control the checkbox of the M1 and M2
« Last Edit: June 26, 2012, 05:46:25 pm by lookang »
#### lookang
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##### Re: Ejs Open Source Gravitational Field & Potential of Earth and Moon Java Applet
« Reply #8 on: June 28, 2012, 10:09:29 am »
changes
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# Convert hour angle to octant
Learn how to convert 1 hour angle to octant step by step.
## Calculation Breakdown
Set up the equation
$$1.0\left(hour \text{ } angle\right)={\color{rgb(20,165,174)} x}\left(octant\right)$$
Define the base values of the selected units in relation to the SI unit $$\left(radian\right)$$
$$\text{Left side: 1.0 } \left(hour \text{ } angle\right) = {\color{rgb(89,182,91)} \dfrac{π}{12.0}\left(radian\right)} = {\color{rgb(89,182,91)} \dfrac{π}{12.0}\left(rad\right)}$$
$$\text{Right side: 1.0 } \left(octant\right) = {\color{rgb(125,164,120)} \dfrac{π}{4.0}\left(radian\right)} = {\color{rgb(125,164,120)} \dfrac{π}{4.0}\left(rad\right)}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(hour \text{ } angle\right)={\color{rgb(20,165,174)} x}\left(octant\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} \dfrac{π}{12.0}} \times {\color{rgb(89,182,91)} \left(radian\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} \dfrac{π}{4.0}}} \times {\color{rgb(125,164,120)} \left(radian\right)}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} \dfrac{π}{12.0}} \cdot {\color{rgb(89,182,91)} \left(rad\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} \dfrac{π}{4.0}} \cdot {\color{rgb(125,164,120)} \left(rad\right)}$$
$$\text{Cancel SI units}$$
$$1.0 \times {\color{rgb(89,182,91)} \dfrac{π}{12.0}} \cdot {\color{rgb(89,182,91)} \cancel{\left(rad\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} \dfrac{π}{4.0}} \times {\color{rgb(125,164,120)} \cancel{\left(rad\right)}}$$
$$\text{Conversion Equation}$$
$$\dfrac{π}{12.0} = {\color{rgb(20,165,174)} x} \times \dfrac{π}{4.0}$$
Cancel factors on both sides
$$\text{Cancel factors}$$
$$\dfrac{{\color{rgb(255,204,153)} \cancel{π}}}{12.0} = {\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{π}}}{4.0}$$
Switch sides
$${\color{rgb(20,165,174)} x} \times \dfrac{1.0}{4.0} = \dfrac{1.0}{12.0}$$
Isolate $${\color{rgb(20,165,174)} x}$$
Multiply both sides by $$\left(\dfrac{4.0}{1.0}\right)$$
$${\color{rgb(20,165,174)} x} \times \dfrac{1.0}{4.0} \times \dfrac{4.0}{1.0} = \dfrac{1.0}{12.0} \times \dfrac{4.0}{1.0}$$
$$\text{Cancel}$$
$${\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{1.0}} \times {\color{rgb(99,194,222)} \cancel{4.0}}}{{\color{rgb(99,194,222)} \cancel{4.0}} \times {\color{rgb(255,204,153)} \cancel{1.0}}} = \dfrac{{\color{rgb(255,204,153)} \cancel{1.0}} \times 4.0}{12.0 \times {\color{rgb(255,204,153)} \cancel{1.0}}}$$
$$\text{Simplify}$$
$${\color{rgb(20,165,174)} x} = \dfrac{4.0}{12.0}$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x}\approx0.3333333333\approx3.3333 \times 10^{-1}$$
$$\text{Conversion Equation}$$
$$1.0\left(hour \text{ } angle\right)\approx{\color{rgb(20,165,174)} 3.3333 \times 10^{-1}}\left(octant\right)$$
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# DoITPoMS
TLP Library
Teaching and learning packages (TLPs) are self-contained, interactive resources, each focusing on one area of Materials Science.
Introduction To Anisotropy
It is common in basic analysis to treat bulk materials as isotropic - their properties are independent of the direction in which they are measured. However the atomic scale structure can result in properties that vary with direction. This teaching and learning package (TLP) looks into typical examples of such anisotropy and gives a brief mathematical look into modelling the behaviour.
Atomic Scale Structure of Materials
This teaching and learning package provides an introduction to crystalline, polycrystalline and amorphous solids, and how the atomic-level structure has radical consequences for some of the properties of the material. It introduces the use of polarised light to examine the optical properties of materials, and shows how a variety of simple models can be used to visualise important features of the microstructure of materials.
Avoidance of Crystallization in Biological Systems
This teaching and learning package discusses the two main environmental threats leading to crystallization in plants and animals, and the ways in which organisms have adapted to avoid this crystallization. As part of this discussion, there is coverage of some of the theory of nucleation and crystallization.
Crystallinity in Polymers
An understanding of polymer crystallinity is important because the mechanical properties of crystalline polymers are different from those of amorphous polymers. Polymer crystals are much stiffer and stronger than amorphous regions of polymer.
Crystallography
Crystalline materials are characterised by a regular atomic structure that repeats itself in all three dimensions. In other words the structure displays translational symmetry.
Diffusion
An introduction to the mechanisms and driving forces of diffusion, and some of the processes in which it is observed.
Elasticity in Biological Materials
This teaching and learning package (TLP) discusses the elasticity of biological materials. Whilst some show Hookean elasticity, the vast majority do not. Non-linear elasticity is considered, in particular J-shaped and S-shaped curves. Viscoelasticity is also discussed, using hair and spiders' silk as examples.
Ferromagnetic Materials
How many ferromagnets do you think you own? Maybe many more than you realise. Ferromagnetic materials lie at the heart not just of the humble compass, but also of many loudspeakers and of computer memory. This teaching and learning package outlines the microscopic basis of magnetism and some of the conquences of ferromagnetic order in real materials.
Liquid Crystals
This Teaching and Learning Package provides an introduction to liquid crystals, their physical properties and their modern-day applications.
Mechanics of Fibre-reinforced Composites
This teaching and learning package (TLP) gives an introduction to the nature of fibre-reinforced composite materials and their basic mechanical characteristics.
Optimisation of Materials Properties in Living Systems
This teaching and learning package discusses the uses of merit indices in conjunction with materials-selection maps focusing on biomaterials. The derivation of merit indices is discussed and biological examples are shown.
Introduction To Photoelasticity
This tutorial is based on lab work within the Department of Materials Science and Metallurgy at the University of Cambridge. The tutorial provides an introduction to the topic of photoelasticity and preparation for lab work. Photographs illustrate many features of birefringence in polymers under polarised light.
Polymer Basics
This teaching and learning package is an introduction to the basic concepts of polymer science. It includes molecular structure, synthesis and tests for identification.
The Stereographic Projection
This TLP covers the use of the Stereographic projection and Wulff nets.
The Structure and Mechanical Behaviour of Wood
This teaching and learning package discusses the structure of wood, focusing on the structure of the tree trunk and the differences between hardwoods and softwoods. The stiffness and strength of different types of wood are discussed, and the different behaviour of wood when wet is investigated.
Structure of Bone and Implant Materials
This teaching and learning package (TLP) describes the structure of bone from the macro-scale to the micro-scale and considers its description as a biological composite. The structure of hip replacements is described and common implant materials are discussed in relation to the mechanical properties of bone.
Superconductivity
Electrons in pairs? Levitating trains? Superconductivity - the combination of lossless electrical conduction and the ability of a material to expel a magnetic field - is a property that excites interest in fundamental science whilst offering tantalising prospects for a range of applications. In this teaching and learning package (TLP), we trace the history of superconductivity, outline some fundamental properties of superconductors, and describe current and potential applications of materials with this unusual property.
Tensors
This TLP offers an introduction to the mathematics of tensors rather than the intricacies of their applications. Its aims are to familiarise the learner with tensor notation, how they can be constructed and how they can be manipulated to give numerical answers to problems.
Thermal Expansion and the Bi-material Strip
This teaching and learning package (TLP) is based on lab work in the Department of Materials Science and Metallurgy at the University of Cambridge. The TLP provides an introduction to the topic of thermal expansion, and its application, together with the different stiffness of materials, in the bi-material strip. The TLP leads you through experiments to measure Young's Modulus from the deflection of a cantilever beam, and to estimate the boiling temperature of nitrogen and the expansivity of a polycarbonate material from the curvature of a bi-material strip immersed in liquid nitrogen.
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# Capacitance
## 8. Effect of Dielectric
• Dielectric are non conducting materials for ex- Glass,mica,wood etc.
• What happened when space between the two plates of the capacitor is filled by a dielectric was first discovered by faraday.
• Faraday discovered that if the space between conductors of the capacitor is occupied by the dielectric,the capacitance of capacitor is increased.
• If the dielectric completely fills the space between the conductors of the capacitor ,the capacitance is increased by an factor K which is characterstics of the dielectric and This factor is known as the dielectric constan.
• Dielectric constant of vaccum is unity.
• Consider a capacitor of capacitance C0 is being charged by the connecting it to a battery.
• If Q0 is the amount of charged on the capacitor at the end of the charging and V0 is potental diffrence across the plates of the capacitor then
C0=Q0 /V0 ----(17)
Thus charge being placed on the capacitor is
Q0=C0V0
• If the battery is diconnected and space between the capacitor is filled by a dielectric the P.D decrease to a new value
V=V0/K.
• Since the original charge is still on the capacitor,the new capacitance will be
C=Q0/V=KQ0/V0=KC0----(19)
• From equation 19 it follows that C is greater then C0.
• Again if the dielectric is inserted while the battery is still connected then battery would have to supply some amount of charge to maintain the P.D between the plates and then total charge on the plates would be Q=KQ0.
• In either of the cases ,capacitance of the capacitor is increase by the amount K.
• For a parallel plate capacitor with dielectric of dielectric constant K between its plates its capacitance becomes
C=εA/D ----(20)
where ε=Kε0
• When a sufficiently strong electric field is applied to any dielectric material it becomes a conductor and this phenomenon is known as dielectric breakdown.
• The maximun electric field a material can withstand without the occurence of breakdown is called dielectric strength of that material.
• Thus field across the capacitor should never exceed breakdown limits in order to store charge on capacitor without leaking.
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# RL's Digit Master Pro Big Game software download.
Topic closed. 2092 replies. Last post 4 years ago by sandnan.
Page 114 of 140
Krakow
Poland
Member #86302
February 2, 2010
892 Posts
Offline
Posted: September 6, 2012, 10:00 am - IP Logged
dr san,
No offence, but as I've already written I have problems with your posts. I'd better read them in Spanish, French, Russian or even Portuguese as I think I would be in a much better position to get what you mean. It' so ambiguous and abstruse that it helps me in no way.
san diego
United States
Member #68237
December 16, 2008
97 Posts
Offline
Posted: September 6, 2012, 11:41 am - IP Logged
Thanks for the tips of just using W-wide open and Blocking of groups. Double check my TG is set correctly. I was able to reduce RFW Temp file from 65 to 18 still retaining the 5 number win.
FA
Pennsylvania
United States
Member #2218
September 1, 2003
5396 Posts
Online
Posted: September 6, 2012, 11:57 am - IP Logged
This was a post RL made a few pages back.
RL, If anything is incorrect I will change it.
===========================================================================================================
You don't have to set these filters they are the ones you need to consider when choosing the Groups and by analyzing both you can make better choices.
(BN) - Base Number to Non-Base Number and (NB) Non-Base Number to Base Number both have a Max value of 2 (0, 1, or 2).
There are only 9 ways to set both the (BN) - Base Number to Non-Base Number and (NB) Non-Base Number to Base Number: 00, 10, 01, 11, 12, 21, 02, 20 , 22
Since all 4 of these filters must total 4 then setting a couple of the easy ones will give you a good idea what the others will be.
If (NN) Non-Base to Non-Base = 4 then this would be amazing information to the digits setup and would also mean that (DB) Double Base would be set to (0).
If (NN) Non-Base to Non-Base = 4 then BB BN NB (Base Number to Base Number Base Number to Non-Base Number Non-Base Number to Base NUmber) will also all be (0).
If (NN) Non-Base to Non-Base = 4 then all the 2nd digits will be greater than 3 (4,6,6,7,8,9,0)
If (NN) Non-Base to Non-Base = 4 then (DB) Double Base will always = 0
If (NN) Non-Base to Non-Base = 4 in the next draw then that means that all the 2nd Digits are going to be above 3 (4,5,6,7,8,9,0) so no way can you have any (DB) Double Base numbers.
If (NN) Non-Base to Non-Base = 3 then either 1 of the following BB BN NB (Base Number to Base Number or Base Number to Non-Base Number or Non-Base Number to Base Number) would = 1 meaning that 2 of them have
to hit 0 in the next draw.
If (NN) Non-Base to Non-Base = 3 then we know that 1 of the 2nd digits must be a (BD) Base Digit.
If (NN) Non-Base to Non-Base = 2 then we know that 1 or 2 (BD) Base Digits 1,2,3 will hit as 2nd digits.
If (NN) Non-Base to Non-Base = 2 then we know that we need 3 or 4 numbers 2nd digit is (NB) Non-Base.
If (BN) Base Number to Non-Base Number = 1 then it will help us place our Base and Non-base numbers.
In reverse if (DB) Double Base was thought to be a (0) for the next draw then NN = 4 (Non-Base to Non-Base) so if (DB) Double Base = 0 then BB, BN, and NB all = 0.
Using both filters together to find a common setting just makes sense.
If DB is expected to show greater than > zero then we don't need to consider 4 as a possibility for NN.
If DB (Double Base) = 2 in the next draw then this can help with all 4 of the Base & Non Base filters.
Connections, connections, connections.
Everything in Digit Master is there for a reason of helping pick a value for something else and vice versa.
Two heads are better than one and three are better than two.
When one filters data does not give enough data to make a choice then look at the other filters that are related to it for some help.
==================================================================================================
Pennsylvania
United States
Member #2218
September 1, 2003
5396 Posts
Online
Posted: September 6, 2012, 12:00 pm - IP Logged
This was a post RL made a few pages back.
RL, If anything is incorrect I will change it.
===========================================================================================================
You don't have to set these filters they are the ones you need to consider when choosing the Groups and by analyzing both you can make better choices.
(BN) - Base Number to Non-Base Number and (NB) Non-Base Number to Base Number both have a Max value of 2 (0, 1, or 2).
There are only 9 ways to set both the (BN) - Base Number to Non-Base Number and (NB) Non-Base Number to Base Number: 00, 10, 01, 11, 12, 21, 02, 20 , 22
Since all 4 of these filters must total 4 then setting a couple of the easy ones will give you a good idea what the others will be.
If (NN) Non-Base to Non-Base = 4 then this would be amazing information to the digits setup and would also mean that (DB) Double Base would be set to (0).
If (NN) Non-Base to Non-Base = 4 then BB BN NB (Base Number to Base Number Base Number to Non-Base Number Non-Base Number to Base NUmber) will also all be (0).
If (NN) Non-Base to Non-Base = 4 then all the 2nd digits will be greater than 3 (4,6,6,7,8,9,0)
If (NN) Non-Base to Non-Base = 4 then (DB) Double Base will always = 0
If (NN) Non-Base to Non-Base = 4 in the next draw then that means that all the 2nd Digits are going to be above 3 (4,5,6,7,8,9,0) so no way can you have any (DB) Double Base numbers.
If (NN) Non-Base to Non-Base = 3 then either 1 of the following BB BN NB (Base Number to Base Number or Base Number to Non-Base Number or Non-Base Number to Base Number) would = 1 meaning that 2 of them have
to hit 0 in the next draw.
If (NN) Non-Base to Non-Base = 3 then we know that 1 of the 2nd digits must be a (BD) Base Digit.
If (NN) Non-Base to Non-Base = 2 then we know that 1 or 2 (BD) Base Digits 1,2,3 will hit as 2nd digits.
If (NN) Non-Base to Non-Base = 2 then we know that we need 3 or 4 numbers 2nd digit is (NB) Non-Base.
If (BN) Base Number to Non-Base Number = 1 then it will help us place our Base and Non-base numbers.
In reverse if (DB) Double Base was thought to be a (0) for the next draw then NN = 4 (Non-Base to Non-Base) so if (DB) Double Base = 0 then BB, BN, and NB all = 0.
Using both filters together to find a common setting just makes sense.
If DB is expected to show greater than > zero then we don't need to consider 4 as a possibility for NN.
If DB (Double Base) = 2 in the next draw then this can help with all 4 of the Base & Non Base filters.
Connections, connections, connections.
Everything in Digit Master is there for a reason of helping pick a value for something else and vice versa.
Two heads are better than one and three are better than two.
When one filters data does not give enough data to make a choice then look at the other filters that are related to it for some help.
==================================================================================================
Eugene Oregan
United States
Member #128629
May 29, 2012
419 Posts
Offline
Posted: September 6, 2012, 9:36 pm - IP Logged
Winsum
Are we still sending you our choices as usual for the mm draw fri? dld
cleveland ohio
United States
Member #65897
October 9, 2008
275 Posts
Offline
Posted: September 6, 2012, 10:23 pm - IP Logged
This was a post RL made a few pages back.
RL, If anything is incorrect I will change it.
===========================================================================================================
You don't have to set these filters they are the ones you need to consider when choosing the Groups and by analyzing both you can make better choices.
(BN) - Base Number to Non-Base Number and (NB) Non-Base Number to Base Number both have a Max value of 2 (0, 1, or 2).
There are only 9 ways to set both the (BN) - Base Number to Non-Base Number and (NB) Non-Base Number to Base Number: 00, 10, 01, 11, 12, 21, 02, 20 , 22
Since all 4 of these filters must total 4 then setting a couple of the easy ones will give you a good idea what the others will be.
If (NN) Non-Base to Non-Base = 4 then this would be amazing information to the digits setup and would also mean that (DB) Double Base would be set to (0).
If (NN) Non-Base to Non-Base = 4 then BB BN NB (Base Number to Base Number Base Number to Non-Base Number Non-Base Number to Base NUmber) will also all be (0).
If (NN) Non-Base to Non-Base = 4 then all the 2nd digits will be greater than 3 (4,6,6,7,8,9,0)
If (NN) Non-Base to Non-Base = 4 then (DB) Double Base will always = 0
If (NN) Non-Base to Non-Base = 4 in the next draw then that means that all the 2nd Digits are going to be above 3 (4,5,6,7,8,9,0) so no way can you have any (DB) Double Base numbers.
If (NN) Non-Base to Non-Base = 3 then either 1 of the following BB BN NB (Base Number to Base Number or Base Number to Non-Base Number or Non-Base Number to Base Number) would = 1 meaning that 2 of them have
to hit 0 in the next draw.
If (NN) Non-Base to Non-Base = 3 then we know that 1 of the 2nd digits must be a (BD) Base Digit.
If (NN) Non-Base to Non-Base = 2 then we know that 1 or 2 (BD) Base Digits 1,2,3 will hit as 2nd digits.
If (NN) Non-Base to Non-Base = 2 then we know that we need 3 or 4 numbers 2nd digit is (NB) Non-Base.
If (BN) Base Number to Non-Base Number = 1 then it will help us place our Base and Non-base numbers.
In reverse if (DB) Double Base was thought to be a (0) for the next draw then NN = 4 (Non-Base to Non-Base) so if (DB) Double Base = 0 then BB, BN, and NB all = 0.
Using both filters together to find a common setting just makes sense.
If DB is expected to show greater than > zero then we don't need to consider 4 as a possibility for NN.
If DB (Double Base) = 2 in the next draw then this can help with all 4 of the Base & Non Base filters.
Connections, connections, connections.
Everything in Digit Master is there for a reason of helping pick a value for something else and vice versa.
Two heads are better than one and three are better than two.
When one filters data does not give enough data to make a choice then look at the other filters that are related to it for some help.
==================================================================================================
Winsome,
This is what I was looking at on my last set up the first one agreed but he last 2 couldnt have happend. Not
with those filters but with others I started to make me think about how Im putting my choices together.
Now im my case I just sent them in but we all need to look at the filters in a manner that as discribed
will be a playable play. We will get this no doubt.
United States
Member #59354
March 13, 2008
4092 Posts
Offline
Posted: September 7, 2012, 1:58 am - IP Logged
Winsum
Are we still sending you our choices as usual for the mm draw fri? dld
did
We are going to do something a little different for friday. If you have your choices then go ahead
and send them but they won't be needed for the next draw. Steve and I are working together on
the setup and I have a couple new tools that I will be using for the groups and digits setups. I am
still testing the programs right now but hope to be finished by early morning. I will then take a break
for several hours before I do the final setup. Every thing is looking good in back test so maybe we will
do well.
RL
Working on my Ph.D. "University of hard Knocks"
I will consider the opinion that my winnings are a product of chance if you are willing to consider
they are not. Many great discoveries come while searching for something else
Trump / 2016 & 2020
Eugene Oregan
United States
Member #128629
May 29, 2012
419 Posts
Offline
Posted: September 7, 2012, 7:43 am - IP Logged
did
We are going to do something a little different for friday. If you have your choices then go ahead
and send them but they won't be needed for the next draw. Steve and I are working together on
the setup and I have a couple new tools that I will be using for the groups and digits setups. I am
still testing the programs right now but hope to be finished by early morning. I will then take a break
for several hours before I do the final setup. Every thing is looking good in back test so maybe we will
do well.
RL
RL
Did send last night, thanks for the heads up. Do appreciate all of the hard work you and winsum are doing. dld
United States
Member #59354
March 13, 2008
4092 Posts
Offline
Posted: September 7, 2012, 7:53 am - IP Logged
RL
Did send last night, thanks for the heads up. Do appreciate all of the hard work you and winsum are doing. dld
did
I got the programs finished and ran many back-test but got mixed results. I have been at this for
around 18 hours now and I need to get some rest before I do any more. If everything goes well
then we will play tonight but if not we may sit the game out. If enough people sent in their choices
I will compare everything to see what I can come up with.
Craig
Working on my Ph.D. "University of hard Knocks"
I will consider the opinion that my winnings are a product of chance if you are willing to consider
they are not. Many great discoveries come while searching for something else
Trump / 2016 & 2020
Pennsylvania
United States
Member #2218
September 1, 2003
5396 Posts
Online
Posted: September 7, 2012, 10:08 am - IP Logged
Sorry for the late notice.
I did receive filter selections from a handful of pool members. If those who have not sent in their filter selections can send them in by 2PM today (New York) time.
Again, sorry for the laste notce.
From Denver, Rocky Mountain Empire,
United States
Member #49750
February 13, 2007
449 Posts
Offline
Posted: September 7, 2012, 11:02 am - IP Logged
did
We are going to do something a little different for friday. If you have your choices then go ahead
and send them but they won't be needed for the next draw. Steve and I are working together on
the setup and I have a couple new tools that I will be using for the groups and digits setups. I am
still testing the programs right now but hope to be finished by early morning. I will then take a break
for several hours before I do the final setup. Every thing is looking good in back test so maybe we will
do well.
RL
RL, I think only you and winsum should work on the setup for the game for the next three draws or so, while the rest of the pool members hone their skills. While I am getting better each day with the groups and digits part, my filter selections are for the most part based on what the "suggestion" button recommends! While it is a good thing to ask pool members to participate in play selection, it doesn't have to be EVERYTIME! If the desired results are not there, then it is ok to try different strategies. Change is good!
In this case, what we have here is "too many cooks spoiling the broth". All this confusing selections sent by pool members is interfering with your thought process and hindering your judgement. I believe, you and winsum should be given at least a chance to work on this thing thoroughly, without interference from other pool members and see if we can get better results. Thanx!
Georgia
United States
Member #129908
July 1, 2012
208 Posts
Offline
Posted: September 7, 2012, 11:22 am - IP Logged
RL, I think only you and winsum should work on the setup for the game for the next three draws or so, while the rest of the pool members hone their skills. While I am getting better each day with the groups and digits part, my filter selections are for the most part based on what the "suggestion" button recommends! While it is a good thing to ask pool members to participate in play selection, it doesn't have to be EVERYTIME! If the desired results are not there, then it is ok to try different strategies. Change is good!
In this case, what we have here is "too many cooks spoiling the broth". All this confusing selections sent by pool members is interfering with your thought process and hindering your judgement. I believe, you and winsum should be given at least a chance to work on this thing thoroughly, without interference from other pool members and see if we can get better results. Thanx!
RL and Winsum
I'm in 100% agreement on what adulane62 said. Guys take your time and don't worry about us. It wouldn't hurt my feelings if you didn't ask us to participate in selections for a few months.
Hey, just do it.
Just do it......
Braselton Ga
United States
Member #121665
January 14, 2012
35 Posts
Offline
Posted: September 7, 2012, 11:57 am - IP Logged
RL & Winsum
Do what you do. Let's just win i am in for the long Haul
MA
United States
Member #89094
March 30, 2010
245 Posts
Offline
Posted: September 7, 2012, 1:57 pm - IP Logged
RL, I think only you and winsum should work on the setup for the game for the next three draws or so, while the rest of the pool members hone their skills. While I am getting better each day with the groups and digits part, my filter selections are for the most part based on what the "suggestion" button recommends! While it is a good thing to ask pool members to participate in play selection, it doesn't have to be EVERYTIME! If the desired results are not there, then it is ok to try different strategies. Change is good!
In this case, what we have here is "too many cooks spoiling the broth". All this confusing selections sent by pool members is interfering with your thought process and hindering your judgement. I believe, you and winsum should be given at least a chance to work on this thing thoroughly, without interference from other pool members and see if we can get better results. Thanx!
Yes, by all means let's give them a try!
No feelings hurt by any means.
Let's give it the old college try!
M
san diego
United States
Member #68237
December 16, 2008
97 Posts
Offline
Posted: September 7, 2012, 2:57 pm - IP Logged
By all means do what you have to do. If the setup and data says "GO" then play. If data is mixed and a "NO-GO" then we just sit and we just do a dry run.
FA
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# Encoder
Discussion in 'Homework Help' started by digitalize, Apr 16, 2015.
1. ### digitalize Thread Starter New Member
Apr 15, 2015
9
0
I'm currently working on an assignment where a 3-input encoder that assigns a 2-bit code to each of the three input combinations. Here I'm trying to figure out how many implementations I could have. So, for example I have 2^3, which is 8 possibilities.
Here is my current three input combinations:
|X|Y|Z | A|B|
|0|0|1 | 0|1|
|0|1|0 | 1|0|
|1|0|0 | 1|1|
My question is, if there are a total of 8 total possibilities and 5 of them are don't care, is it possible to have more than three implementations? I feel as though I'm confusing myself with implementation and possibilities.
2. ### ericgibbs Moderator
Jan 29, 2010
8,534
1,713
hi,
Are you sure that your table includes all the combinations.?
E
3. ### digitalize Thread Starter New Member
Apr 15, 2015
9
0
These are the only cases i'm currently aware of where either a or b equals 1
4. ### WBahn Moderator
Mar 31, 2012
24,555
7,691
Not sure what you are asking. Where does the "three implementations" come from to begin with?
Let's simply things and use two inputs and say that I'm only interested in having the output, Y, follow the following two states for the inputs, A and B (with all the others being don't cares):
|A|B|Y|
|1|0|0|
|1|1|1|
I could implement this a number of ways, including just setting Y=B. But I could also use an AND gate or an XNOR gate. I could also implement it using a circuit that has ten thousand gates, which gets into the distinction between a logic function and an implementation of that function.
In this case, I have two Boolean inputs so I have four possible combinations. Since two of those combinations are don't cares, those two lines could take on any of four possible combination of output values. Hence there are four Boolean functions on those two inputs that will give me what I want. But each of those functions can be implemented in an infinite number of ways.
As an example, consider the function Y = A+B. I can implement that direction using an OR gate. But I can also implement it as Y=(A'B')'. It's a different implementation of the exact same function.
5. ### digitalize Thread Starter New Member
Apr 15, 2015
9
0
I actually meant 4 implementations. Here are the four:
a=z'
b=y'
a=x'yz'+xy'z'
b=x'y'z +xy'z'
I was wondering if more were were possible?
6. ### WBahn Moderator
Mar 31, 2012
24,555
7,691
Sure more are possible.
Take your first function for A, namely A=Z'. Enumerate the truth table for it. Now flip the state of any on of the "don't care" rows and you have a different function that, if implemented, gives you the correct output for the three rows you happen to care about.
How many different ways can I change the function outputs for five rows? That's how many different functions there are that will produce the correct output for the three rows you care about -- and that's just for the A output. You have the same for the B output. Now consider that you get to mix and match A and B outputs.
7. ### digitalize Thread Starter New Member
Apr 15, 2015
9
0
Ok, I think I get it, so if I have 5 don't care states, then couldn't I just take 2^5=32, which give me possible implementations for A only? Then I would have to do the same for B.
8. ### WBahn Moderator
Mar 31, 2012
24,555
7,691
That will give you the number of unique functions for each output. But there are literally in infinite number of implementations.
Consider the function Y=A+B. I could implement that as
Y = A+B
Y = A'B + AB' + AB (canonical SOP)
Y = (A'B')'
and the list could keep going.
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PDF Lesson
The odds of predicting the type of natural disaster that will occur are rare, but predicting the severity can be even harder. If you eliminate determining the severity, will your likelihood of correctly predicting the type and location of a natural disasters increase? Spin the slots to test your luck!
Doing the science
1. Start the Trouble Incorporated Simulation by clicking on the "Sim" tab.
4. Choose a natural disaster (bridge collapse, earthquake, flood, hurricane, or tornado) to predict the disaster by clicking on the yellow icon.
6. Choose a city (Orlando, New York, Los Angeles, Dallas, or Chicago) to predict the location of the disaster by clicking on the yellow icon.
7. Click on the golden "Spin" button to start the machine.
8. In Table 1 below, record the predictions in each category that you got correct with a check mark "c", incorrect with an "x."
9. Repeat steps 3 – 8 nine more times.
Table 1.
Spin 1 2 3 4 5 6 7 8 9 10 Event Where
Do You Understand?
1. Discuss your ability to correctly predict the outcome of each spin. Suggest a reason for your ability to correctly/incorrectly predict the spin outcomes.
2. Was there a category (Event or Where) that you predicted the spin outcome correctly more often than the other categories? If so, state a reason for your improved prediction ability in this category.
3. What is the probability of getting both your Event and Where predictions correct on each spin?
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# Commits
committed 94c4466
More stuff in slides.
• Participants
• Parent commits d19d10a
# File doc/clin-2012/slides.tex
\DeclareSymbolFont{AMSb}{U}{msb}{m}{n}
\DeclareMathOperator*{\argmax}{argmax}
-\newcommand{\tick}{\color{blue}\ding{51}}
-\newcommand{\cross}{\color{red}\ding{55}}
+\newcommand{\tick}{{\color{blue}\ding{51}}}
+\newcommand{\cross}{{\color{red}\ding{55}}}
\mode<presentation>{
Soft classes & \tick & \tick \\\hline
Bayesian & \cross & \tick \\
\bf Online & \tick & \cross \\
- Unbounded & \tick & \cross \\
\bf Parameters & \cross & \tick \\
+ Adaptive K & \tick & \cross \\
Fast & \cross & \tick \\
\end{tabular}
\end{center}
\item Incrementally optimizes a joint entropy criterion:
\begin{small}
\begin{equation*}
- H(X,Y) = {\color{red} H(Y)} + {\color{blue} H(X|Y)}
+ H(X,Y) = {\color{blue} H(X|Y)} + {\color{red} H(Y)}
\end{equation*}
\end{small}
\begin{itemize}
\end{center}
\end{frame}
+
+\begin{frame}
+ \frametitle{Word class LDA}
+ \begin{itemize}
+ \item Number of classes K is specified as a parameter
+ \item $\alpha$ and $\beta$ control sparsity of priors
+ \item Inference using Gibbs sampler (batch)
+ \end{itemize}
+\end{frame}
+
+
+\begin{frame}
+ \frametitle{Model evaluation}
+ \begin{center}
+ \begin{block}{}
+ Evaluate \vskip 0.5cm
+ \begin{itemize}
+ \item {\bf Parameterized} $\Delta$H
+ \item {\bf Online} Gibbs sampler for word class LDA
+ \end{itemize}\vskip 0.5cm
+ on the {\bf same task} and the {\bf same dataset}.
+ \end{block}
+ \end{center}
+\end{frame}
+
+\begin{frame}
+ \frametitle{Dataset}
+ \begin{itemize}
+ \item Manchester portion of CHILDES (mothers)
+ \item Discard one-word sentences and punctuation
+ \end{itemize}
+ \begin{center}
+ \begin{tabular}[!t]{l r r r }
+ \hline
+ {\bf Data Set} & {\bf Sessions} & {\bf \#Sent} & {\bf \#Words} \\
+ \hline
+ Training & 26--28 & 22,491 & 125,339 \\
+ Development & 29--30 & 15,193 & 85,361 \\
+ \hline
+ \end{tabular}
+ \end{center}
+\end{frame}
+
+
+\begin{frame}
+ \begin{itemize}
+ \item Relevant for cognitive modeling
+ \item Used in NLP -- language model evaluation
+ \end{itemize}
+\end{frame}
+
+\begin{frame}
+ \frametitle{Word prediction}
+ \begin{itemize}\small
+ \item (Soft)-assign classes from context
+ \item Rank words based on predicted class
+ \end{itemize}
+\begin{block}{Reciprocal rank}
+ \small
+ \begin{tabular}{cc|c|cc}
+ want & to & \color{gray} put & them & on \\
+ \pause
+ & & $y_{123}$ & & \\
+ \end{tabular}
+ \begin{tabular}{l|l|r}
+ $y_{123}$
+ & make & \\
+ & take & \\
+ & \color{red}put& $\textit{rank}^{-1}=\frac{1}{3}$\\
+ & get & \\
+ & sit & \\
+ & eat & \\
+ & let & \\
+ \end{tabular}
+\end{block}
+\end{frame}
+
+\begin{frame}
+ \frametitle{Parametrizing $\Delta$H}
+ \begin{itemize}
+ \item No free parameters in $\Delta$H
+ \begin{itemize}
+ \item[\tick] No need to optimize them separately
+ \item[\cross] Lack of flexibility
+ \end{itemize}
+ \item If we force parameterization
+ \begin{itemize}
+ \item Is the algorithm well-behaved?
+ \item Can we smoothly control the tradeoff?
+ \end{itemize}
+ \end{itemize}
+\end{frame}
+
+\begin{frame}
+ \begin{block}{Parametrized $\Delta$H}
+ \begin{small}
+ \begin{equation*}
+ H_{\alpha}(X,Y) = {\color{blue} \alpha H(X|Y)} + {\color{red} (1-\alpha) H(Y)}
+ \end{equation*}
+ \end{small}
+ \end{block}
+ \end{frame}
+
+
+
+
+
+\begin{frame}
+ \begin{center}
+ \large Thank you
+ \end{center}
+\end{frame}
+
+\begin{frame}
+ \frametitle{Word prediction: variants}
+ \begin{itemize}
+ \item $\Delta H_{\max}$
+ $+ P(w|h) = P(w|\argmax_i R(y_i|h)^{-1}) +$
+ \item $\Delta H_\Sigma$
+ $+ P(w | h) = \sum_{i=1}^N P(w | y_i) \frac{\mathrm{R}(y_i|h)^{-1}}{\sum_{i=1}^N \mathrm{R}(y_i | h)^{-1}} +$
+ \end{itemize}
+ \end{frame}
\end{document}
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What Is a Real Number?
What is a number? Well that depends. There are a variety of different kinds of numbers, each with their own particular properties. One sort of number, upon which statistics, probability, and much of mathematics is based upon, is called a real number.
To learn what a real number is, we will first take a brief tour of other kinds of numbers.
Types of Numbers
We first learn about numbers in order to count.
We began with matching the numbers 1, 2, and 3 with our fingers. Then we and kept going as high as we could, which probably wasn't that high. These counting numbers or natural numbers were the only numbers that we knew about.
Later, when dealing with subtraction, negative whole numbers were introduced. The set of positive and negative whole numbers is called the set of integers. Shortly after this, rational numbers, also called fractions were considered. Since every integer can be written as a fraction with 1 in the denominator, we say that the integers form a subset of the rational numbers.
The ancient Greeks realized that not all numbers can be formed as a fraction. For example, the square root of 2 cannot be expressed as a fraction. These kinds of numbers are called irrational numbers. Irrational numbers abound, and somewhat surprisingly in a certain sense there are more irrational numbers than rational numbers.
Other irrational numbers include pi and e.
Decimal Expansions
Every real number can be written as a decimal. Different kinds of real numbers have different kinds of decimal expansions. The decimal expansion of a rational number is terminating, such as 2, 3.25, or 1.2342, or repeating, such as .33333.
. . Or .123123123. . . In contrast to this, the decimal expansion of an irrational number is nonterminating and nonrepeating. We can see this in the decimal expansion of pi. There is a never ending string of digits for pi, and what's more, there is no string of digits that indefinitely repeats itself.
Visualization of Real Numbers
The real numbers can be visualized by associating each one of them to one of the infinite number of points along a straight line. The real numbers have an order, meaning that for any two distinct real numbers we can say that one is greater than the other. By convention, moving to the left along on the real number line corresponds to lesser and lesser numbers. Moving to the right along the real number line corresponds to greater and greater numbers.
Basic Properties of the Real Numbers
The real numbers behave like other numbers that we are used to dealing with. We can add, subtract, multiply and divide them (as long as we don't divide by zero). The order of addition and multiplication is unimportant, as there is a commutative property. A distributive property tells us how multiplication and addition interact with one another.
As mentioned before, the real numbers possess an order.
Given any two real numbers x and y, we know that one and only one of the following is true:
x = y, x < y or x > y.
Another Property - Completeness
The property that sets the real numbers apart from other sets of numbers, like the rationals, is a property known as completeness. Completeness is a bit technical to explain, but the intuitive notion is that the set of rational numbers has gaps in it. The set of real numbers does not have any gaps, because it is complete.
As an illustration, we will look at the sequence of rational numbers 3, 3.1, 3.14, 3.141, 3.1415, . . . Each term of this sequence is an approximation to pi, obtained by truncating the decimal expansion for pi. The terms of this sequence get closer and closer to pi. However, as we have mentioned, pi is not a rational number. We need to use irrational numbers to plug in the holes of the number line that occur by only considering the rational numbers.
How Many Real Numbers?
It should be no surprise that there are an infinite number of real numbers. This can be seen fairly easily when we consider that whole numbers form a subset of the real numbers. We could also see this by realizing that the number line has an infinite number of points.
What is surprising is that the infinity used to count the real numbers is of a different kind than the infinity used to count the whole numbers. Whole numbers, integers and rationals are countably infinite. The set of real numbers is uncountably infinite.
Why Call Them Real?
Real numbers get their name to set them apart from an even further generalization to the concept of number. The imaginary number i is defined to be the square root of negative one. Any real number multiplied by i is also known as an imaginary number. Imaginary numbers definitely stretch our conception of number, as they are not at all what we thought about when we first learned to count.
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## Patents
Publication number US7184992 B1 Publication type Grant Application number US 10/285,461 Publication date Feb 27, 2007 Filing date Nov 1, 2002 Priority date Nov 1, 2001 Fee status Lapsed Publication number 10285461, 285461, US 7184992 B1, US 7184992B1, US-B1-7184992, US7184992 B1, US7184992B1 Inventors Original Assignee George Mason Intellectual Properties, Inc. Export Citation Patent Citations (8), Referenced by (46), Classifications (11), Legal Events (5) External Links:
Constrained optimization tool
US 7184992 B1
Abstract
A system for finding a solution to a constrained optimization problem is disclosed. The system uses a mathematical formulation describing the constrained optimization problem. A transformer builds a specific function using the mathematical formulation, variables; an objective function; at least one constraint; a class of transformation functions with a predefined set of properties; and Lagrange multipliers. The specific function may include a nonlinear resealing part; and an augmented Lagrangian part. An iterative solver uses the specific function, and Lagrange multipliers to generate a solution.
Images(14)
Claims(30)
1. A tangible computer readable medium containing a computer program comprising:
a) a mathematical formulation describing a constrained optimization problem for a physical system; said mathematical formulation including:
i) variables x=(x1, . . . , xn);
ii) an objective function f(x); and
iii) at least one constraint selected from the group consisting of:
(1) an inequality constraint ci(x)≧0, i=1, . . . , p; and
(2) an equality constraint ej(x)=0, j=1, . . . , q;
b) a class of transformation functions with a predefined set of properties;
c) Lagrange multipliers λ=(λ1, . . . , λp), v=(v1, . . . , vq)
d) a scaling parameter k;
e) a transformer capable of building a specific function L(x,λ,v,k) from said mathematical formulation, said Lagrange multipliers and said scaling parameter using said class of transformation functions, said specific function including
$f ( x ) - k - 1 ∑ i = 1 p λ i ψ ( kc i ( x ) ) - ∑ j = 1 q v j e j ( x ) + 0.5 ∑ j = 1 q ke i 2 ( x ) ;$
f) a Lagrange multipliers updater capable of calculating updated Lagrange multipliers {circumflex over (λ)}=({circumflex over (λ)}1, . . . , {circumflex over (λ)}m), {circumflex over (v)}=({circumflex over (v)}1, . . . , {circumflex over (v)}q);
g) a scaling parameter updater capable of calculating an updated scaling parameter {circumflex over (k)};
h) a merit function calculator capable of calculating a merit function μ(x,λ,v);
i) a general stopping criteria verifier;
j) an accuracy of solution parameter (ε); and
k) an iterative solver capable using said specific function, said Lagrange multipliers updater and said scaling parameter updater to generate a solution; and
wherein said solution is applied to said physical system.
2. A medium according to claim 1, wherein said predefined set of properties includes:
i) ψ(t): 1 1 defined on −∞≦a<t<b≦∞;
ii) ψ(0)=0;
iii) ψ′(0)=1;
iv) ψ′(t)>0;
v) ψ″(t)<0; and
vi)
$lim t -> b ψ ' ( t ) = 0.$
3. A medium according to claim 2, wherein said predefined set of properties further includes:
i) ψ″(t)<0; and
ii)
$lim t -> b ψ ' ( t ) = 0.$
4. A medium according to claim 2, wherein said inequality constraint is ci(x)≦0; i=1, . . . , p and said predefined set of properties further includes:
i) ψ″(t)>0; and
ii)
$lim t -> a ψ ' ( t ) = 0.$
5. A medium according to claim 1, wherein said mathematical formulation includes said objective function being minimized, subject to said inequality constraints and said equality constraints.
6. A medium according to claim 1, wherein said Lagrange multipliers updater includes Lagrange multipliers update formulas.
7. A medium according to claim 6, wherein said Lagrange multipliers update formulas include {circumflex over (λ)}iiψ′(kci({circumflex over (x)})), i=1, . . . , p.
8. A medium according to claim 6, wherein said Lagrange multipliers update formulas include {circumflex over (v)}j=vj−kej({circumflex over (x)}), j=1, . . . , q.
9. A medium according to claim 1, wherein said scaling parameter updater includes scaling parameter update formula
{circumflex over (k)}=max {kγ,μ({circumflex over (x)},{circumflex over (λ)},{circumflex over (v)})−δ}, δ>0, γ≧1.
10. A medium according to claim 1, wherein said iterative solver includes a general stopping criteria verifier for determining when said iterative solver should terminate.
11. A medium according to claim 1, wherein said iterative solver includes a general stopping criteria verifier uses a general stopping criteria.
12. A medium according to claim 10, wherein said general stopping criteria verifier uses said accuracy of solution parameter.
13. A medium according to claim 12, wherein said general stopping criteria verifier considers when said merit function is equal or less than said accuracy of solution parameter.
14. A medium according to claim 1, wherein said iterative solver is a MIN_UPDATE solver.
15. A medium according to claim 1, wherein said iterative solver is a PRIMAL_DUAL solver.
16. A medium according to claim 14, wherein said MIN_UPDATE solver includes:
a) a specific function minimizer for performing an unconstrained minimization in variables of said specific function with determined said Lagrange multipliers and said scaling parameter; said specific function minimizer capable of producing updated variables;
b) a minimizer stopping criteria checker for determining when said specific function minimizer should terminate;
c) said Lagrange multipliers updater for calculating said updated Lagrange multipliers using said updated variables, said Lagrange multipliers and said Lagrange multipliers update formula;
d) said scaling parameter updater for calculating said updated scaling parameter using said updated variables, said updated Lagrange multipliers, said scaling parameter and said scaling parameter update formula; and
e) said general stopping criteria verifier.
17. A medium according to claim 16, wherein said specific function minimizer uses a Newton's method.
18. A medium according to claim 16, wherein said MIN_UPDATE solver further includes an accuracy of minimizer parameter (εm) and a minimizer stopping criteria uses said accuracy of minimizer parameter.
19. A medium according to claim 16, wherein said updated variables ({circumflex over (x)}) must satisfy a minimizer stopping criteria.
20. A medium according to claim 16, wherein said scaling parameter updater uses said merit function.
21. A medium according to claim 16, wherein a minimizer stopping criteria includes at least one of the following selected from the group consisting of:
a) said updated variables;
b) said Lagrange multipliers;
c) said scaling parameter;
d) said updated Lagrange multipliers; and
e) a stopping inequality verifier.
22. A medium according to claim 21, wherein said stopping inequality verifier is select from the group consisting of:
$a ) x ^ : ∇ x L ( x ^ , λ , v , k ) ≤ ɛ m ; and b ) x ^ : ∇ x L ( x ^ , λ , v , k ) ≤ α ( k 1 + θ ) - 1 max { λ ^ - λ , v ^ - v } , α > 0 , 0 < θ < 0.1 .$
23. A medium according to claim 15, wherein said PRIMAL_DUAL solver includes:
a) a primal-dual system of equations, including:
i) a gradient of classical Lagrangian
$∇ f ( x ) - ∑ i = 1 p λ i ∇ c i ( x ) - ∑ j = 1 q v j ∇ e j ( x ) ;$
and
ii) said Lagrange multipliers update formulas;
b) a linear system for Newton directions;
c) a predictor-corrector algorithm for solving said linear system for Newton directions; and
d) said scaling parameter update formula.
24. A medium according to claim 23, wherein said gradient of classical Lagrangian equals zero.
25. A medium according to claim 23, wherein said predictor-corrector algorithm includes:
a) inequality predictors ( λ) for said Lagrange multipliers associated with inequality constraints;
b) equality predictors ( v) for said Lagrange multipliers associated with equality constraints;
c) a primal corrector (Δx);
d) a primal linear system for finding said primal corrector;
e) inequality correctors (Δλ) for said Lagrange multipliers associated with inequality constraints;
f) equality correctors (Δv) for said Lagrange multipliers associated with equality constraints; and
g) a primal-dual updater.
26. A tangible computer readable medium containing a computer program, the computer program implementing a method for finding a solution to a constrained optimization problem for a physical system, the method comprising the steps of:
a) defining a mathematical formulation describing said problem; said mathematical formulation including:
i) variables x=(x1, . . . , xn);
ii) an objective function f(x);
iii) inequality constraints ci(x)≧0, i=1, . . . , p; and
iv) equality constraints ej(x)=0, j=1, . . . , q ;
b) selecting initial Lagrange multipliers λ=(λ1, . . . , λp), v=(v1, . . . , vq);
c) selecting an initial scaling parameter k;
d) selecting initial variables x=(x1, . . . , xn);
e) building a specific function L(x,λ,v,k) from said mathematical formulation, said initial Lagrange multipliers and said initial scaling parameter using an element of a class of transformation functions having a predefined set of properties, said specific function further includes
$f ( x ) - k - 1 ∑ i = 1 p λ i ψ ( kc i ( x ) ) - ∑ j = 1 q v j e j ( x ) + 0.5 ∑ j = 1 q ke i 2 ( x ) ;$
f) performing a nonlinear resealing augmented Lagrangian iterative algorithm; and
g) applying said solution to said physical system.
27. The medium according to claim 26, wherein said step of performing a nonlinear resealing augmented Lagrangian iterative algorithm includes iteratively:
a) performing a minimization of said specific function;
b) calculating updated Lagrange multipliers {circumflex over (λ)}=({circumflex over (λ)}1, . . . , {circumflex over (λ)}m), {circumflex over (v)}=({circumflex over (v)}1, . . . , {circumflex over (v)}q
c) calculating an updated scaling parameter {circumflex over (k)};
d) calculating a merit function μ(x,λ,v); and
e) stopping iterations when a general stopping criteria using an accuracy of solution parameter (ε) is satisfied.
28. The medium according to claim 26, wherein said step of performing a nonlinear resealing augmented Lagrangian iterative algorithm (PRIMAL_DUAL) includes iteratively:
a) solving a primal-dual system of equations, said primal-dual system of equations including:
i) a gradient of classical Lagrangian
$∇ f ( x ) - ∑ i = 1 p λ i ∇ c i ( x ) - ∑ j = 1 q v j ∇ e j ( x ) ;$
and
ii) said Lagrange multipliers update formulas;
b) solving a linear system for Newton directions;
c) performing a predictor-corrector algorithm for solving said linear system for Newton directions;
d) updating primal dual variables; and
e) updating said initial scaling parameter.
29. A medium according to claim 1, wherein said physical system is an electrical transmission network.
30. A medium according to claim 26, wherein said physical system is an electrical transmission network.
Description
CROSS-REFERENCE TO RELATED APPLICATIONS
The present application claims the benefit of provisional patent applications:
• a) Ser. No. 60/330,864 to POLYAK et al., filed on Nov. 1, 2001; and
• b) Ser. No. 60/340,861 to POLYAK et al., filed on Dec. 19, 2001, entitled “Primal-Dual Modified Barrier Augmented Lagrangian system and method for solving constrained optimization problems”,
which are hereby incorporated by reference.
BACKGROUND OF THE INVENTION
The present invention relates to the field of constraint optimization. More specifically, the present invention provides for finding optimized solutions of constrained problems using a Nonlinear resealing-Augmented Lagrangian solver.
BRIEF SUMMARY OF THE INVENTION
One advantage of the invention is it may be applied to many different types of constrained problems.
Another advantage of the invention is that it may be implemented in software.
Another advantage of the invention is that it may handle problems with both inequality constraints and equations without transforming one set of constraints into another one.
Another advantage of the invention is that it does not require the unbounded increase of the barrier-penalty parameter to guarantee convergence as it happened in the Interior Point and Penalty type methods.
Yet another advantage of the invention is that it allows to avoid the ill conditioning, i.e. the condition number of the MBAL Hessian is stable when the primal-dual approximation approaches the primal-dual solution. In turn, it keeps stable the area where Newton's method is well defined.
Yet another advantage of the invention is that under the standard second order optimality conditions it converges with Q-linear rate when the barrier-penalty parameter is fixed. The ratio may be made as small as one wants by choosing a fixed but large enough parameter.
Yet another advantage of the invention is that it eliminates the combinatorial nature of constrained optimization with inequality constraints.
Yet another advantage of the invention is that it does not require an interior starting point for constrained optimization with inequality constraints.
To achieve the foregoing and other advantages, in accordance with all of the invention as embodied and broadly described herein, a system for finding a solution to a constrained optimization problem comprising: a mathematical formulation describing the constrained optimization problem; the mathematical formulation including: variables; an objective function; at least one constraint selected from the group consisting of: an inequality constraint; and an equality constraint; a class of transformation functions with a predefined set of properties; Lagrange multipliers; a scaling parameter; a transformer capable of building a specific function from the mathematical formulation, the Lagrange multipliers and the scaling parameter using the class of transformation functions, the specific function further including: a nonlinear resealing part; and a augmented Lagrangian part; a Lagrange multipliers updater capable of calculating updated Lagrange multipliers; a scaling parameter updater capable of calculating an updated scaling parameter; a merit function calculator capable of calculating a merit function; a general stopping criteria verifier; an accuracy of solution parameter; and an iterative solver capable using the specific function, the Lagrange multipliers updater and the scaling parameter updater to generate the solution.
In yet a further aspect of the invention, a method for finding a solution to a constrained optimization problem comprising the steps of: defining a mathematical formulation describing said problem; said mathematical formulation including: variables; an objective function; inequality constraints; and equality constraints; selecting initial Lagrange multipliers; select an initial scaling parameter; selecting initial variables; building a specific function from said mathematical formulation, said Lagrange multipliers and said scaling parameter using an element of a class of transformation functions having a predefined set of properties, said specific function further including: a nonlinear resealing part; and an augmented Lagrangian part; perform a nonlinear resealing augmented Lagrangian iterative algorithm.
Additional objects, advantages and novel features of the invention will be set forth in part in the description which follows, and in part will become apparent to those skilled in the art upon examination of the following or may be learned by practice of the invention. The objects and advantages of the invention may be realized and attained by means of the instrumentalities and combinations particularly pointed out in the appended claims.
BRIEF DESCRIPTION OF THE SEVERAL VIEWS OF THE DRAWINGS
The accompanying drawings, which are incorporated in and form a part of the specification, illustrate an embodiment of the present invention and, together with the description, serve to explain the principles of the invention.
FIG. 1 is a block diagram of a constrained optimizer as per an aspect of an embodiment the present invention.
FIG. 2 is a block diagram of a mathematical formulation of a constrained optimization problem as per an aspect of an embodiment of the present invention.
FIG. 3 is a block diagram of transformation function properties as per an aspect of an embodiment of the present invention.
FIG. 4A is a block diagram of additional transformation function properties as per an aspect of an embodiment of the present invention.
FIG. 4B is a block diagram of additional transformation function properties as per an aspect of an embodiment of the present invention.
FIG. 5A is a block diagram of Lagrange multipliers as per an aspect of an embodiment of the present invention.
FIG. 5B is a block diagram of a Lagrange multipliers updater as per an aspect of an embodiment of the present invention.
FIG. 5C is a Lagrange multipliers formula that responds to inequality constraints as per an aspect of an embodiment of the present invention.
FIG. 5D is a Lagrange multipliers formula that responds to equality constraints as per an aspect of an embodiment of the present invention.
FIG. 6A is a block diagram showing a scaling parameter as per an aspect of an embodiment of the present invention.
FIG. 6B is a block diagram of a scaling parameter updater as per an aspect of an embodiment of the present invention.
FIG. 6C is a scaling parameter formula as per an aspect of an embodiment of the present invention.
FIG. 7A is a block diagram showing a specific function as per an aspect of an embodiment of the present invention.
FIG. 7B is a block diagram showing parts of a specific function as per an aspect of an embodiment of the present invention.
FIG. 8A is a block diagram showing a merit function calculator as per an aspect of an embodiment of the present invention.
FIG. 8B is a block diagram showing an accuracy of solution parameter as per an aspect of an embodiment of the present invention.
FIG. 9 is a flow diagram of a of a constrained optimizer as per an aspect of an embodiment of the present invention.
FIG. 10 is a block diagram of a MIN_UPDATE iterative solver as per an aspect of the present invention.
FIG. 11 is a flow diagram of a MIN_UPDATE iterative solver as per an aspect of an embodiment of the present invention.
FIG. 12 is a block diagram of a PRIMAL DUAL iterative solver as per an aspect of the present invention.
FIG. 13 is a flow diagram of a PRIMAL DUAL iterative solver as per an aspect of an embodiment of the present invention.
DETAILED DESCRIPTION OF THE INVENTION
The present invention is a system for finding a solution 100 to a constrained optimization problem 110. Since most problems have constraints of one type or another, one may realize that many problems may be described as a constrained optimization problem 110. Examples may include power flow problems, image reconstruction problems, pattern recognition problems, data processing problems, network flow problems, optimal control problems, optimal design problems, etc. Further, one skilled in the art will recognize that may other types of problems may be expressed as a constrained optimization problem for which the present invention may be used.
Solution(s) 100 to the constrained optimization problem 110 may be a vector of decision variables that minimize given criteria and satisfy given constraints. The given criteria may be provided as part of the problem description and may describe a preferred outcome. This description is preferably provided as a mathematical formulation 120. For the purposes of the present invention, alternate descriptions may be translated into a mathematical formulation 120.
The system includes but is not limited to: a mathematical formulation 120 describing the constrained optimization problem 110; a class of transformation functions 170; Lagrange multipliers 190; a scaling parameter 200; a transformer 220; a Lagrange multipliers updater 260; a scaling parameter updater 280; a merit function calculator 300; a general stopping criteria verifier 800; an accuracy of solution parameter 320; and an iterative solver 330. One skilled in the art will recognize that aspects of this system may be reproduced using equivalent components or methods. This invention is intended to cover these equivalent embodiments. Further, it may be possible to practice the invention without all of the listed elements. For example, the accuracy of solution parameter 320 is useful in providing input into the system in deciding when a solution is close enough. One skilled in the art will recognize that this function may be built into the system without necessarily needing a parameter.
As illustrated in FIG. 2, the system may include a mathematical formulation 120. The mathematical formulation 120 should include an objective function 140 f(x); and at least one constraint 145. The objective function 140 may be a function of many variables. Both the constraint(s) 145 and objective function 140 may depend upon variables 130 x=(x1, . . . , xn). The constraint(s) 150 may further include at least one inequality constraint 150 ci(x)≧0, i=1, . . . , p; and/or at least one equality constraint 160 ej(x)=0, j=1, . . . , q. It is possible that problem will only have one constraint 145, or only one type of constraint. For example, for p=0, the problem may have no inequality constraints 150. Similarly, q=0 may result in no equality constraints 160. Further, the mathematical formulation 120 may include the objective function 140 being minimized, subject to the inequality constraints 150 and the equality constraints 160.
One skilled in the art will recognize that much of this disclosure uses notation that should not be construed literally. For example, variables have just been noted as variables x. However, variables could just as easily be noted as variables y, or k, or z.
Further, the system may include a class of transformation functions 170 having certain properties 180. Referring to FIG. 3, we may see an example of a transformation function 170 exhibiting properties 180 as per an embodiment of the present invention. Preferably, at least some of the properties will be a predefined set of properties 180. A first predefined property 400 may be that the transformation function is a function of one variable between a and b ψ(t):
1 1 defined on −∞≦a<t<b≦∞ 400. Point ‘a’ may be a negative point or minus infinity and point ‘b’ may be a positive point or plus infinity. A second property 410 may be that the function passes through zero coordinate ψ(0)=0. A third property 420 may be that the slope of the function as it passes through zero be 45 degrees ψ′(0)=1. A fourth property 430 may be that the function is increasing ψ″(t)>0 430.
FIG. 4A shows a fifth property 440 and sixth property 450. These properties constrain the function to be concave ψ″(t)<0; and have a slope of zero at point ‘b’. Alternately, the properties are shown in FIG. 4B to respond to an alternative mathematical formulation with the following inequality constraints for 150 ci(x)≦0; i=1, . . . , p of the same problem. The fifth property 440 a and sixth property 450 a constrain the function to be convex ψ″(t)>0; and have with a slope of zero at point ‘a’
$lim t -> a ψ ' ( t ) = 0.$
The transformer 220 preferably builds a specific function 230 L(x,λ,v,k) from the mathematical formulation 120, the Lagrange multipliers 190 and the scaling parameter 200 using the class of transformation functions 170. The specific function 230 may then depend upon variables 130, Lagrange multipliers 190 and the scaling parameter 200. The specific function 230 preferably has a nonlinear resealing part 240 and a augmented Lagrangian part 250. The specific function 230 may include
$f ( x ) - k - 1 ∑ i = 1 p λ i ψ ( k c i ( x ) ) - ∑ j = 1 q v j e j ( x ) + 0.5 ∑ j = 1 q k e i 2 ( x )$
The scaling parameter 200 k maybe a singular 1 dimensional scaling variable. Although the present invention is described using a single scaling parameter, in fact, a system as per this invention may use many scaling parameters or no scaling parameters. In the case of using many scaling parameters, it is possible that a separate scaling parameter could be used for each and every inequality and/or equality constraint. The use of a scaling factor may be totally dependent upon the specific implementation, and/or the power of the solving system.
FIG. 5A shows Lagrange multipliers 190 λ=(λ1, . . . , λp) v=(v1, . . . , vq) which are multi-dimensional parameters. Multipliers λ=(λ1, . . . , λp) may respond to inequality constraints. Similarly, v=(v1, . . . , vq) may respond to equality constraints. As shown in FIG. 5B, the Lagrange multipliers updater 260 capable of calculating updated Lagrange multipliers 270 {circumflex over (λ)}=({circumflex over (λ)}1, . . . , {circumflex over (λ)}m), {circumflex over (v)}=({circumflex over (v)}1, . . . , {circumflex over (v)}q) from Lagrange multiplier 190. In some embodiments, it is possible for updated Lagrange multiplier 270 to be used as Lagrange multipliers 190, especially when the Lagrange multiplier updater 260 is used in an iterative solver 330. The Lagrange multipliers updater 260 may include Lagrange multipliers update formulas 500. FIG. 5C is an example from a best embodiment of an update formula for the Lagrange multipliers 190 which respond to inequality constraints. Similarly, FIG. 5D shows an example from a best embodiment of an update formula for the Lagrange multipliers 190 which respond to equality constraints.
FIG. 6A is a block diagram showing a scaling parameter as per an aspect of an embodiment of the present invention. FIG. 6B is a block diagram of a scaling parameter updater as per an aspect of an embodiment of the present invention. The scaling parameter updater 280 which may generate an updated scaling parameter 290 from a scaling parameter 200. In some embodiments, it is possible for an updated scaling parameter 290 to be used as a scaling parameter 200, especially when the scaling parameter updater 280 is used in an iterative solver 330. The scaling parameter updater 280 may include a scaling parameter update formula. FIG. 6C is an example from a best embodiment of a formula for updating a scaling parameter 200.
FIG. 7A is a block diagram showing a specific function as per an aspect of an embodiment of the present invention. The specific function 230 may depend upon variables 130, Lagrange multipliers 190 and scaling parameter 200. FIG. 7B is a block diagram showing parts of a specific function as per an aspect of an embodiment of the present invention. The specific function 230 preferably includes the objective function 140, and has a nonlinear resealing part 240 and a augmented Lagrangian part 250. The nonlinear resealing part 240 may includes
$k - 1 ∑ i = 1 p λ i ψ ( kc i ( x ) )$
and the augmented Lagrangian part 250 may include
$- ∑ j = 1 q v j e j ( x ) + 0.5 ∑ j = 1 q ke i 2 ( x ) .$
FIG. 8A is a block diagram showing a merit function calculator as per an aspect of an embodiment of the present invention. The merit function calculator 300 preferably uses a merit function 305 μ(x,λ,v) dependent upon variables 130, and Lagrange multipliers 190 to calculate a merit value. FIG. 8B is a block diagram showing a general stopping criteria verifier 800 as per an aspect of an embodiment of the present invention. The general stopping criteria 800 may be used in determining when the iterative solver 330 should terminate and may use an accuracy of solution parameter. The general stopping criteria verifier 800 may compare the merit function 305 against the accuracy of solution parameter to decide when solution is accurate enough. The general stopping criteria 310 may consider whether the merit function 305 is equal or less than the accuracy of solution parameter. The scaling parameter updater 280 may also use the merit function 305.
A constrained optimization problem 110 may be solved using the present invention by the method shown in FIG. 9. At step S10, a mathematical formulation 120 describing the constrained optimization problem 110 may be generated. The mathematical formulation 120 may include variables 130, an objective function 140, inequality constraints 150, and equality constraints 160. Next at step S12, initial variables 130 and Lagrange multipliers 190 may be selected. An initial scaling parameter 200 k may be selected at step S14, and transformation functions 170 may be selected at step S16.
A specific function 230 may them be built at step S18 from the mathematical formulation 120, the Lagrange multipliers 190, and the scaling parameter 200 using an element of a class of transformation functions 170 having a predefined set of properties 180. The specific function 230 may further include a nonlinear resealing part 240; and a augmented Lagrangian part 250. At step S20, a nonlinear resealing augmented Lagrangian iterative algorithm may be performed.
The iterative solver 330 may be a MIN_UPDATE solver 600. FIG. 10 is a block diagram of a MIN_UPDATE iterative solver 600 as per an aspect of the present invention. FIG. 11 is a flow diagram of a MIN_UPDATE iterative solver as per an aspect of an embodiment of the present invention. This MIN_UPDATE iterative solver 600 is a best mode iterative solver 330. At step S30, a specific function 230 may be inputted into a specific function minimizer in variables device 610. At step S32, initial variables 130 may be inputted into the specific function minimizer in variables device 610. At step S34, initial Lagrange multipliers 190 may be inputted into the specific function minimizer in variables device 610. At step S36, a scaling parameter 280 may be inputted into the specific function minimizer in variables device 610.
Next, at step S38, the specific function minimizer in variables device 610 preferably performs an unconstrained minimization of the specific function 230. This minimization may keep occurring until a minimizer stopping criteria checker 220 decides that the minimization has satisfied a minimizer stopping criteria 630. The minimizer stopping criteria 630 may uses an accuracy of minimizer parameter. The updated variables ({circumflex over (x)}) may need to satisfy the minimizer stopping criteria 630. Parameters may then be updated at step S40 by a Lagrange multipliers parameter updater 260 and a scaling parameter updater 280. The minimizer stopping criteria 630 may use updated variables 620, Lagrange multipliers 190, scaling parameter 200, updated Lagrange multipliers 270, or a stopping inequality verifier 650 when making its' determination.
At step S42, a general stopping criteria verifier 800 may make a determination as to whether the parameters and variables satisfy a general stop criteria. The general stopping criteria verifier 800 may use updated variables 620, updated Lagrange multipliers 270, or a stopping criteria 310 when making its' determination. If the determination is negative, then the updated Lagrange multipliers 270 and the updated scaling factor 290 may be input to the specific function minimizer in variables device 610 for another iteration. If the determination is positive, then the optimization may end and an optimized solution 230 outputted.
The iterative solver 330 may be a PRIMAL DUAL iterative solver 700. FIG. 12 is a block diagram of a PRIMAL DUAL iterative solver 700 as per an aspect of the present invention. FIG. 13 is a flow diagram of an PRIMAL DUAL iterative solver 700 as per an aspect of an embodiment of the present invention. This PRIMAL DUAL iterative solver 700 may include a primal-dual system of equations which may also include a gradient of classical Lagrangian and Lagrange multipliers update formulas. Variable together with Langrange multipliers may be alternatively be called primal dual variables. Further, the gradient of classical Lagrangian may equal zero. At step S52, initial variables 130 may be input to a linear system to find Newton directions 710. The linear system to find Newton directions 610 may be implemented in hardware or software. Next at step S54, largrange multipliers 190 may be input to the linear system for Newton directions device 710. A scaling parameter 280 may be input to the linear system for Newton directions device at step S56.
The linear system to find Newton directions 710 may then solve the linear system at step S58. A predictor corrector algorithm 740 may be used to help solve the linear system for Newton directions. The predictor corrector algorithm 740 may use inequality predictors ( λ), equality predictors ( v), a primal corrector (Δx), a primal linear system, inequality correctors (Δλ), equality correctors (Δv), and/or a primal-dual updater.
Parameters may then be updated at step S60 by a primal dual updater 790 and a scaling parameter updater 280. Then, at step S62, a general stopping criteria verifier 800 may make a determination as to whether the parameters and variables satisfy a general stopping criteria. If the determination is negative, then the updated Lagrange multipliers 270, updated variables 130, and the updated scaling factor 290 may be input to the linear system for Newton directions device 710 for another iteration. If the determination is positive, then the optimization may end and an optimized solution 230 outputted.
A mathematical explanation of a nonlinear resealing-augmented Lagrangian solver (MRALS) as per the present invention will now be disclosed as. Let f(x), ci(x), i=1, . . . , p and ej(x), j=1, . . . , q are smooth enough functions. Constrained optimization problem with both inequality and equality constraints consists in finding a solution of the following primal problem (mathematical formulation)
f(x*)={f(x)|c i(x)≧0, i=1, . . . , p; e j(x)=0, j=1, . . . , q} (1)
The classical Lagrangian associated with problem (1) may be given by the formula:
$L ( x , λ , v ) = f ( x ) - ∑ i = 1 p λ i c i ( x ) - ∑ j = 1 q v j e j ( x ) ( 2 )$
where λ=(λ1, . . . , λp) is a vector of the Lagrange multipliers associated with inequality constraints and v=(v1, . . . , vq) is a vector of the Lagrange multipliers associated with equality constraints.
Along with the primal problem (1), the present invention also considers the dual problem for y=(λ,v)
d(y*)={d(y)≡d(x,λ)|λi≧0, i=1, . . . , p} (3)
where
$d ( y ) = d ( λ , v ) = inf x L ( x , y )$
is the dual function.
The present solver may solve both the primal (1) and the dual (3) problems. Let −∞≦a<0<b≦∞. we consider a class class Ψ of twice continuously differential transformation functions, which satisfy the following properties:
i) ψ(t):
1 1 defined on −∞≦a<t<b≦∞
ii) ψ(0)=0
iii) ψ′(0)=1
iv) ψ′(t)>0
v) ψ″(t)<0 and
vi)
$lim t → b ψ ′ ( t ) = 0.$
The functions ψεΨ we use to transform inequality constraints of a given constrained optimization problem into equivalent set of constraints.
Let us consider a few examples ψεΨ.
Logarithmic Modified Barrier Function (MBF): ψ1(t)=ln(t+1),
Hyperbolic MBF:
$ψ 1 ( t ) = t t + 1 .$
Each of above transformations may be modified in the following way. For a given −1<τ<0 a quadratic extrapolation of the transformations 1–2 may be defined by formulas.
A third example may be described by:
$ψ qi ( t ) = { ψ i ( t ) , t ≥ τ q i ( t ) = a i t 2 + b i t + c i t ≤ τ ,$
where ai, bi and ci may be found from the following equations: ψi(τ)=qi(τ), ψ′1(τ)=q′1(τ) and ψ″1(τ)=q″1(τ). Equations ai=0.5ψ″1(τ), b1=ψ′(τ)−τψ″i(τ), and ciψ(τ)−τψ′(τ)+0.5τ2ψ″i(τ), may be obtained so that ψq i (t)εC2.
The specific function
(x,λ,v,k) is preferably defined by the formula:
$( x , λ , v , k ) = f ( x ) - k - 1 ∑ i = 1 p λ i ψ ( k c i ( x ) ) - ∑ j = 1 q v j e j ( x ) + 0.5 k ∑ j = 1 q e j 2 ( x )$
This formula has three parts:
• 1. An objective function f(x);
• 2. a nonlinear resealing part:
$k - 1 ∑ t = 1 p λ i ψ ( kc i ( x ) ) ,$
• which corresponds to the inequality constraints; and
• 3. an augmented Lagrangian part:
$∑ j = 1 q v j e j ( x ) + 0.5 k ∑ j = 1 q e j 2 ( x ) ,$
• which corresponds to equality constraints.
A transformer preferably uses the primal variables x=(x1, . . . , xn), dual variables (Lagrange multipliers) y=(λ,v), where: λ=(λ1, . . . , λp) is a vector of nonnegative components corresponding to inequality constraints; v=(v1, . . . , vq) is a vector with unrestricted components corresponding to equality constraints; and k is a scaling parameter.
Disclosed herein are two iterative solvers: MIN_UPDATE and PRIMAL_DUAL. MIN_UPDATE generates a primal-dual sequence {xs,ys,ks} by the following formulas.
$x s + 1 : ( x s + 1 , λ s , v s , k s ) = min n ( x , λ s , v s , k s ) , λ i s + 1 = ψ ' ( kc i ( x s + 1 ) ) λ i s , i = 1 , … , p , v j s + 1 = v j s - ke j ( x s + 1 ) , j = 1 , … , q , k s + 1 := γ k s , γ ≥ 1.$
A specific function minimizer for finding xs+1 may be based on Newton method for minimizing
(x,λs,vs,ks) in x. Let x:=xs. By solving the following linear system
xx 2 (x,λ s ,v s ,k sx=−∇ x (x,λ s ,v s ,k s)
for Δx, the Newton direction may be found. The new approximation x:=x+tΔx where the step length t>0 may be found by the following procedure.
t:=1, y=(λs,vs), k=ks, 0<η<0.5.
while
(x+tΔx,y,k)−(x,y,k)≧ηt(∇x (x,y,k),Δx) do
begin
$t := t l ,$
l>1 end
The minimizer stopping criteria may be defined by the following inequality
$x ^ : ∇ x ( x , λ , v , k ) ≤ α ( k 1 + θ ) - 1 max { λ ^ - λ , v ^ - v }$
where a>0, 0<θ<0.1, {circumflex over (λ)}i={circumflex over (λ)}iψ′(kci({circumflex over (x)})), i=1, . . . , p; {circumflex over (v)}j=vj−kej({circumflex over (x)}), j=1, . . . , q;
The iterative solvers MIN_UPDATE and PRIMAL_DUAL produce the primal-dual sequence {xs, ys} up to the point when the general stopping criteria is satisfied. To describe the stopping criteria we may first introduce a merit function such as:
$μ ( x , λ , v ) = max { ∇ x L ( x , λ , v ) , - min 1 ≤ i ≤ p c i ( x ) , max 1 ≤ j ≤ q e j ( x ) , max 1 ≤ i ≤ p λ i c i ( x ) } .$
For any (x,λ,y) the merit function μ(x,λ,y)≧0 and μ(x,λ,v)=0
x=x*, λ=λ*, v=v*.
If ε>0 is the accuracy of solution, then the general stopping criteria may be for the pair (x,y) to satisfy the inequality: μ(x,λ,y)≦ε.
The following pseudo-code describes an embodiment of a Newton Nonlinear Rescaling method as per the present invention.
Input
An accuracy parameter ε>0
Primal x0ε
n, Dual λ0=(1, . . . , 1)Tε m
Scaling parameter k>0, τ>0, 0<η<0.5
Begin
x:=x0, λ:=λ0, v:=v0, μ:=μ(x,λ,v,k)
while μ>ε do
• begin
• while
∥∇xL( x,λ,ν,k)∥>τk−1min{∥diagψ′(kci( x))λ—λ∥,k∥e( x)∥}
• do
• begin
• find Δx:∇xx (x,λ,v,k)Δx=−∇x (x,λ,v,k)
• t:=1
• while (x+tΔx,λ,v,k)−(x,λ,v,k)>ηt(∇x (x,λ,v,k),Δx) do
• t:=t/2
• x:=x+tΔx
• end
• λi:=ψ′(kci(x))λi, i=1, . . . , p; vj=vj−kej(x), j=1, . . . , q
• μ=μ(x,λ,v,k)
• end
end
Output x,λ,v,μ
The iterative solver MIN_UPDATE may be based on Newton's method for minimization of the specific function L(x,y,k) in x followed by the Lagrange multipliers update.
Another iterative solver PRIMAL_DUAL may be based on solving a primal-dual system. Unconstrained minimization of the specific function and updating of the Lagrange multipliers may be equivalent to solving the following primal-dual system of equations for {circumflex over (x)} and ŷ=({circumflex over (λ)},{circumflex over (v)})
$∇ f ( x ^ ) - ∑ i = 1 p λ ^ i ∇ c i ( x ^ ) - ∑ j = 1 q v ^ j ∇ e j ( x ^ ) = 0$
{circumflex over (λ)}i=ψ′(kc({circumflex over (x)}))λi , i=1, . . . , p
{circumflex over (v)} j =v j −ke j({circumflex over (x)}), j=1, . . . , q
Newton's method may be performed as part of solving the primal-dual system for ({circumflex over (x)},ŷ). After linearizing the system for ({circumflex over (x)},ŷ)=({circumflex over (x)},{circumflex over (λ)},{circumflex over (v)}),the following linear system of equations for finding the primal-dual Newton direction (Δx,Δy)=(Δx, Δλ,Δv) may be obtained.
xx 2 L(x,yx−∇c(x)T Δλ−∇e(x)T Δv=−∇ x L(x,y) (4)
kΛΨ″(kc(x))∇c(xx+Δλ= λ−λ (5)
k∇e(xx+Δv={circumflex over (v)}−v (6)
where y=( λ,{circumflex over (v)}), λ i=ψ′(kci(x))λ, i=1, . . . , p, v l=vj−kej( x), j=1, . . . , q Λ=diag(λi)i=1 p, Ψ′(kc(x))=diag(ψ′(kci(x)))i=1 p, Ψ″(kci(x))=diag(ψ″(kci(x)))i=1 p, ∇c(x) is Jacobian of vector function c(x)=(c1(x), . . . , cp(x))T and ∇e(x) is Jacobian of vector function e(x)=(e1(x), . . . , eq(x))T System (4)–(6) may be rewritten as follows:
$[ ∇ xx 2 L ( x , y ) - ∇ c ( x ) T - ∇ e ( x ) T - ∇ c ( x ) ( k ⩓ Ψ ″ ( kc ( x ) ) ) - 1 0 - ∇ e ( x ) 0 - k - 1 I q ] [ Δ x Δ λ Δ v ] = [ - ∇ x L ( x , y ) ( k ⩓ Ψ ″ ( kc ( x ) ) ) - 1 ( λ _ - λ ) - k - 1 ( v _ - v ) ] ( 7 )$
The matrix on the left hand side is symmetric and usually sparse. It may allow solving the system efficiently using modern linear algebra technique. Also, system (4) and (5) may be used to find Δλ and Δv. After substituting Δλand Δv into (4), the following system for finding the primal Newton direction Δx may be obtained.
(∇xx 2 L(x,y)+zI−k∇c(x)TΛΨ″(kc(x))∇c(x)+k∇e(x)T ∇e(x))Δx=−∇ x L(x, y ) (8)
where parameter z≧0. Matrix
M(x,y,k)=∇xx 2 L(x,y)+zI−k∇c(x)TΛΨ″(kc(x))∇c(x)+k∇e(x)T ∇e(x)
is symmetric, positive semi-definite for z≧0 and positive definite for z>0. The Newton direction Δλ for the Lagrange multipliers associated with the inequality constraints may be found by formula
Δλ=kΛΨ″(kc(x))∇c(xx−λ+ λ (9)
The Newton direction Δv for the Lagrange multipliers associated with the equality constraints may be found by formula
Δv=−k∇e(xx+ v−v (10)
So one step of the PRIMAL_DUAL iterative algorithm may include:
• 1. Finding the dual predictor for the Lagrange multipliers associated with the inequality constraints by formulas
λ i=ψ′(kc i(x))λ, i=1, . . . , p
• 2. Finding the dual predictor for the Lagrange multipliers associated with the equality constraints by formulas
v j =v j −ke j( x ), j=1, . . . , q
• 3. Finding the primal corrector Δx by solving the system (8).
• 4. Finding the dual corrector Δλ associated with the inequality constraints by formula (9).
• 5. Finding the dual corrector Δv associated with the equality constraints by formula (10).
• 6. Finding the new primal-dual variables by formulas (primal-dual updater):
• a) primal: x:=x+Δx.
• b) new Lagrange multipliers associated with the inequality constraints:
λ:=λ+Δλ= λ+kΛΨ″(kc(x))∇c(x)Δx
• c) new Lagrange multipliers associated with the equality constraints:
v:=v+Δv= v−k∇e(xx
• 7. Updating the scaling parameter by formula
k:=max {kγ, μ(x,λ,v)−δ}, δ>0 γ≧1
We will now discuss an embodiment of the present invention which solves Power Flow Optimization Problems and general constrained optimization problems using a Newton Modified Barrier-Augmented Lagrangian algorithm. This example is a particular embodiment of the present invention in which the Nonlinear Rescaling Lagrangian is the Newton Modified Barrier-Augmented Lagrangian. This example is presented purely for illustrative purposes of this particular embodiment. One skilled in the art will recognize that this embodiment is only illustrative of how the present invention may be used and not inclusive of all embodiments. It is not intended that the present invention be limited to or any specific embodiment.
The electrical utilities worldwide seek cost effective means to increase the operational efficiency of their transmission networks. Two research areas are of great interests for power system analysis: the study of voltage collapse and Optimal Power Flow (OPF) solutions. One of the OPF criteria is loadability limit of the power system. The purpose is to find such OPF, which maximize the loadability. The other possible criteria is the cost of power restoration in case of voltage collapse. The third area of great interest is to find such transformers tap positions that the system may withstand light and peak load.
All these problems may be solved using OPF, a static Nonlinear Optimization (NLP) problem with both inequality constraints and equations. These problems may be defined by a set of physical and operational constraints imposed by equipment limitation and security requirement. Among variables are bus voltages, bus angles, real and reactive generation tap settings of transformers etc. The constraints may include Kirghoff equations, upper and lower bounds for particular variables, as well as limitations imposed by security requirements.
Electric network is operating heavily loaded systems, therefore planning and operation tools preferably address strong nonlinearities in the system's behavior.
Over the last three decades OPF methods used different mathematical programming technique: sequential linear programming, and Newton's based NLP methods. In the last decade, it became very popular to use a different version of the Interior Point Methods (IPM).
The present embodiment of the present invention uses a new approach for constrained optimization problems with both inequality constraints and equations, which is based on Modified Barrier and Augmented Lagrangian (MBAL) theory. This MBAL theory [1] provides the basic framework for the algorithms, which we used to solve the OPF problems, which arise in the first phase of our research.
Let f, ci and dj are smooth enough functions. We consider the following problem
x*εX*=Arg min {f(x)|c i(x)≧0, i=1, . . . , p; d j(x)=0, j=1, . . . , q}. (1)
Since the mid 80s the Interior Point Methods (IPMs) became the most popular tool for solving constrained optimization problems with inequality constraints. The IPMs not only have become the mainstream in the Optimization Community, but also have been widely used in applications and in particular for solving OPF problems [4].
The IPMs for Linear Programming (LP) calculations is a great success story (see [8]).
The situation in Nonlinear Optimization (NLP) is not as bright as in LP. Although the Primal-Dual IPM [7] produced reasonable results for a number of NLP, the ill conditioning, which is typical for the Barrier and Penalty methods, remains a serious issue. The unbounded increase of the barrier parameter, which is the only tool to control the computational process in the IPM calculations, leads to the unbounded increase of the condition number of the correspondent Hessian. So the area, where Newton's method is well-defined shrinks to zero. Therefore to stay in the Newton area one has to increase the barrier parameter very careful. It leads to slow convergence. More drastic increase of the barrier parameter leads to numerical instability.
In LP calculations the ill-conditioning phenomenon was possible to avoid due to the special structure of the correspondent Hessian as well as substantial advances in the numerical linear algebra [5].
In the NLP the structure of the Classical Barrier Function Hessian is fundamentally different than in LP. Therefore there is a substantial gap between efficiency of the IPM for LP and NLP calculations.
The problems, which the electric power industry faces, are nonlinear and large scale by nature.
The OPF problems along with inequality constraints have equations. Therefore it is hard to expect that the IPM may be used efficiently in particular in case of nonlinear equality constraints. Also the ill conditioning effect is much more troubling in NLP calculations than in LP.
Hence there is a need for new mathematical approaches, which may deal with large scale OPF, successfully handle both equations and inequality constraints and avoid the ill-conditioning effect. But, most of all the methods have to be fast and numerically stable.
In the following we describe such an approach, which is based on the Modified Barrier Function (MBF) [2] and Augmented Lagrangian (AL) [6] theories.
The MBF approach may eliminate the basic problems related to the Classical Log-Barrier functions for inequality constraints, whereas the AL eliminates the basic problems associated with the penalty type functions for equality constraints.
By applying the MBF methodology [2] for inequality constraints and by treating the equations with Augmented Lagrangian term [6] we may obtain the MBAL, which is our main tool for solving problem (1). The MBAL function
: n× + p× q× +is given by formula
$( x , λ , v , k ) = f ( x ) - k - 1 ∑ i = 1 p λ i ln ( kc i ( x ) + 1 ) - ∑ i = 1 q v i d i ( x ) + 0.5 k ∑ i = 1 q d i 2 ( x )$
The function was introduced in [1]. The first two terms represent the Lagrangian for the equivalent problems in the absence of the equality constraints, because for any fixed k>0 the system ln(kci(x)+1)≧0, i=1, . . . , p is equivalent to ci(x)≧0, i=1, . . . , p.
The last two terms represent the Augmented Lagrangian for equality constraints [6]. Along with the Classical Lagrangian term
$- ∑ i = 1 q v i d i ( x )$
there is a penalty part
$0.5 k ∑ i = 1 q d i 2 ( x ) ,$
which is design to penalize the violation of the equality constraints. Keeping in mind that the MBF function
$F ( x , λ , k ) = f ( x ) - k - 1 ∑ i = 1 p λ i ln ( kc i ( x ) + 1 )$
has all the characteristics of the Interior Augmented Lagrangian (see [2]), the MBAL
(x,λ,v,k) may be viewed as Interior-Exterior Augmented Lagrangian.
Before we will describe the MBAL multipliers method we would like to emphasize a few important characteristics of the MBAL at the primal-dual solution.
In contrast to the Classical Barrier Function, the MBF exists at the solution together with its derivatives of any order. Moreover for any k>0 MBAL possesses the following important properties at the primal-dual solution.
10.
(x*,λ*,v*,k)=f(x*).
20. ∇x
(x*,λ*,v*,k)=∇xL(x*,λ*,v*)=0 ,
where
$L ( x , λ , v ) = f ( x ) - ∑ i = 1 p λ i c i ( x ) - ∑ i = 1 q v i d i ( x )$
is the Classical Lagrangian for the original problem (1).
30.
xx 2
(x*,λ* ,v*,k)=∇xx 2L(x*,λ*,v*)+k∇cT(x*)Λ*∇c(x*)+k∇dT(x*)∇d(x*),
where ∇c(x)=J(c(x)) and ∇d(x)=J(d(x)) are Jacobians of c(x) and d(x).
So the MBAL not only exists at the primal-dual solution together with its derivatives of any order but for λ=λ* and v=v* the MBAL is the exact smooth approximation of a nonsmooth function, minimum of which coincides with x*.
Also under the standard second order optimality conditions for (1) the MBAL Hessian ∇xx 2
(x*,λ*,v*,k) is positive definite independently from convexity of f, ci, dj for any k≧k0>0 if k0 is large enough.
Therefore for any pair (λ,v) close enough to (λ*,v*) the MBAL
(x,λ,v,k) is strongly convex in x no matter whether the objective function and the constraints are convex or not. Keeping in mind the smoothness of (x,λv,k) in xε n we may expect that Newton's method for primal minimization MBAL will be efficient.
The MBAL method generates the primal-dual sequence {xss,vs} while k>0 may be fixed or one may change k>0 from step to step. We will describe the MBAL method under the fixed barrier-penalty parameter, k>0.
For any x0ε
n one may find k>0 such that kci(x0)+1>0 Hence, the numerical realization of the MBAL method in contrast to IPM does not require finding initial interior point. The initial dual approximation is not an issue either, so we may take λ0=eε p and v0ε q as an initial approximation for dual vectors. We assume ln t=−∞, t≦0.
Let's assume that (xss,vs) has been already found. The next approximation we find by formulas
x s+1 =arg min {
(x,λ s ,v s ,k)| n} (2)
i.e.
x s+1:∇x (x s+1s ,v s ,k)=0 (3)
or
$∇ f ( x s + 1 ) - ∑ i = 1 p λ i s kc i ( x s + 1 ) + 1 ∇ c i ( x s + 1 ) - ∑ j = 1 q ( v j s - kd j ( x s + 1 ) ) ∇ d j ( x s + 1 ) = ∇ f ( x s + 1 ) - ∑ i = 1 p λ i s + 1 ∇ c i ( x s + 1 ) - ∑ j = 1 q λ j s + 1 ∇ d j ( x s + 1 ) = ∇ x L ( x s + 1 , λ s + 1 , v s + 1 ) ( 4 )$
The new Lagrange multipliers may be found by formulas
λi s+1i s(kc i(x s+1)+1)−1 , i=1, . . . , p (5)
v j s+1 =v j s −kd j(x s+1), j=1, . . . , q (6)
In other words the Lagrange multipliers for inequality constraints we update as in MBF method [2] while the Lagrange multipliers, which correspond the equality constraints, we update as in Augmented Lagrangian methods [6].
The convergence of the MBAL method is just due to the Lagrange multipliers update while k>0 may be fixed, so the condition number of the MBAL Hessian remains stable and the area where Newton's method for the primal minimization converges does not shrink to a point. So we may expect that the area where Newton's method is “well defined” will remain large enough up to the end of the computational process.
It makes the computation process robust and eventually produces very accurate results.
So MBAL is an exterior point method in primal, because xs usually does not satisfy neither the primal inequality nor the equations.
It was proven in [1] that both the primal sequence and the dual sequence {ys}={λs,vs} converges to the primal-dual solution under the standard second order optimality conditions. Moreover the rate of convergence are Q-linear, i.e. for primal {xs} and dual {ys}={λs, vs} sequences, which were generated by formulas (3)-(6) the following estimation takes place
$ x s + 1 - x * ≤ c k y s - y * , y s + 1 - y * ≤ c k y s - y * ,$
where c>0 is independent of k≧k0>0 and k0>0 is large enough.
Each step has two basic components: unconstrained minimization of the MBAL
(x,λs,vs,k) in x under fixed vector of Lagrange multipliers ys=(λs,vs) and fixed scaling parameter k>0 and update by (5)–(6) of the Lagrange multipliers, using the unconstrained minimizer xs+1.
In the next section we describe the Newton MBAL algorithm, which uses Newton's method with the step length for unconstrained minimization and introduce a stopping criteria for the unconstrained minimization, which makes the algorithm practical.
We will now provide a formal description of the Newton MBAL algorithm. To make the MBAL method practical we have to replace the infinite procedure of finding the primal minimizer by a finite procedure. It may be done by introducing well grounded stopping criteria for the primal minimization.
To satisfy the stopping criteria we may use Newton's method with step length for finding an approximation for
{circumflex over (x)}=arg min {
(x,λ,v,k)| n}
or we may use Newton's method for solving the primal-dual system.
The first direction leads to Newton's MBAL, the second leads to the Primal-Dual MBAL method.
The stopping criteria is based on comparison the norm of the gradient ∇x
(x,λ,v,k) with the norm of the difference between the new and old Lagrange multipliers. Let τ>0 be a chosen parameter. Let us assume that for xε n the following inequality is satisfied
$ ∇ x ( x _ , λ , v , k ) ≤ τ k - 1 min { diag ( kc i ( x _ ) + 1 ) - 1 λ - λ , k d ( x _ ) } ( 7 )$
then for the pair of vectors ( x, y), where y=( λ, v), λ=(kC( x)+I)−1λ, v=v−kd( x) the following bound
$max { x _ - x * , y _ - y * } ≤ c ( 1 + τ ) k y - y * $
holds true. The formula (7) is the foundation for the numerical realization of the MBAL method. We start with x0ΔΩk={x:kci(x)+1≧0, i=1, . . . , p}, λ0=eε
++ p and v0ε q. Let us assume that pair ( x s, y s) has been found already. The next approximation ( x s+1, y s+1) we find by the following operations:
1. Find
$x _ s + 1 : ∇ y ( x _ s + 1 , λ _ s v _ s , k ) ≤ τ k - 1 min { diag ( kc i ( x _ s + 1 ) + 1 ) - 1 λ _ s - λ _ s , k d ( x _ s - 1 ) } ( 8 )$
2. Find new Lagrange multipliers
λ i s+1=(kc i( x s+1)+1)−1 λ i s , i=1, . . . , p
v l s+1 = v j s −kd j( x s+1), j=1, . . . , q
To find the approximation for the primal minimizer x s+1 we used Newton's method with step-length for minimizing
(x, λ s, v s, k) in x. In other words the primal Newton's direction Δx we find from the following system
xx (x, λ s , v s ,kx=−∇ x (x, λ s , v s , k) (9)
The Hessian ∇xx
(x, λ s, v s,k) we compute by formula
xx (x, λ s , v s ,k)=∇xx L(x, λ s , v s)+k∇c(x)T Λ s(kC(x)+I)−1 ∇c(x)+k∇d T(x)∇d(x)
where λ s+1=(kC( x s+1)+I)−1 λ s, v s+1= v s−kd( x s+1), (kC(x)+I)=diag(kci(x)+1)t=1 p.
(x, λ s, v s,k)=∇xL(x, λ s+1, v s+1).
The numerical realization of the MBAL leads to Newton's MBAL method.
The Newton's MBAL method consists of using Newton's method with steplength for minimization of
(x, λ s, v s,k) in xε n following by the Lagrange multipliers update using approximation x s+1, which satisfies the criteria (8). The chart below describes Newton NR method.
We will first introduce a merit function, which may measure the violation of the optimality condition.
Let us define the merit function μ:
n× + p× q× ++ 1 1 by the following formula
$μ ( · ) = μ ( x , λ , v , k ) = max { - min 1 < _ i < _ p c 1 ( x ) , max 1 < _ j < _ q d j ( x ) , ∇ ( x , λ , v , k ) , ∑ i = 1 p λ i c i ( x ) , ∑ i = 1 q v i d i ( x ) , }$
It is easy to see, that for any k>0 the function μ(.) is nonnegative and
μ(x,λ,v,k)=0 iff x=x*,λ=λ*,v=v*
Input
An accuracy parameter ε>0
Primal x0ε
n, Dual λ0=(1, . . . , 1)Tε m
Scaling parameter k>0, τ>0, 0<η<0.5
Begin
x:=x0, λ:=λ0:v:=v0, μ:=μ(x,λ,v,k)
while μ>ε do
• begin
• while
$ ∇ x ( x , λ , v , k ) > τ k - 1 min { diag ( kc i ( x _ ) + 1 ) - 1 λ - λ , k d ( x _ ) }$
• do
• begin
• find Δx:∇xx (x,λ,v,k)Δx=−∇x (x,λ,v,k)
• t:=1
• while (x+tΔx,λ,v,k)−(x,λ,v,k)>ηt(∇x (x,λ,v,k),Δx) do
• t:=t/2
• x:=x+tΔx
• end
• λi:=(1+kci(x))−1, i=1, . . . , p; vj=vj−kdj(x), j=1, . . . , q
• μ=μ(x,λ,v,k)
• end
end
Output x,λ,v,μ
As we may see from the MBAL algorithm the most costly operation is finding Newton's direction. It requires solving system (9). The dimension of the primal space defines the size of the system. Therefore the dimension of the primal space is much more important than the number of constraints both inequalities and equations.
The most time consuming part is the first few MBAL primal minimizations. After few Lagrange multipliers update the Primal optimization does not require much effort, because the Lagrange multipliers become close to their optimal value.
The main task of the proposed research is to improve the first phase of the Newton MBAL. The second task is to find an optimal strategy for the penalty barrier parameter k>0 update, which allows to reach the “hot start” as soon as possible.
From “hot start” on, it requires very few and often even one Newton's step for the Lagrange multipliers update, which reduces the distance between the primal-dual approximation to the primal-dual solution by a factor γ=ck−1, where c>0 is independent on k≧k0. It happens because Newton's area, i.e. the area where Newton's method is “well defined” [2] does not shrink to a point, it remains stable when the primal dual approximation approaches the primal-dual solution. It distinguishes fundamentally the MBAL approach from the IPM or penalty type methods.
The following references are included in this specification by reference to provide support for the state of art relied upon in describing the present invention.
• [1] D. Goldfarb, K. Mints, R. Polyak, I. Yuzefovich, Modified Barrier-Augmented Lagrangian Method for Constrained Minimization, Computational Optimization and Applications, 14, p 55-74, 1999.
• [2] R. Polyak, Modified Barrier Functions, Mathematical Programming 54 (1992) 177-222 North-Holland.
• [3] S. Nash and J. Nacedal, A numerical study of the limited memory BFGS method and the truncated-Newton method for large Scale Optimization, SIAM Journal on Optimization 1 (1991) pp. 358-372.
• [4] I. Nejdawi, K. Clements, P. Davis, An Efficient Interior Point Method for Sequential Quadratic Programming Based on Optimal Power Flow, IEEE Transactions on Power Systems, 15(4), 1179-1183, 2000.
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• [8] S. Wright, Primal-Dual Interior Points methods, SIAM, 1997.
The foregoing descriptions of the preferred embodiments of the present invention have been presented for purposes of illustration and description. They are not intended to be exhaustive or to limit the invention to the precise forms disclosed, and obviously many modifications and variations are possible in light of the above teaching. The illustrated embodiments were chosen and described in order to best explain the principles of the invention and its practical application to thereby enable others skilled in the art to best utilize the invention in various embodiments and with various modifications as are suited to the particular use contemplated.
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International ClassificationG06F17/00, G06N5/00, G06N5/02
Cooperative ClassificationG05B13/021, G06Q10/04
European ClassificationG06Q10/04, G05B13/02A1
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# Math, Mindset, and Attribution Retraining
If you dig the growth mindset and Jo Boaler’s course, this post is for you. Here are a few activities to get students thinking about their mindset. The ideas are derived from Carol Dweck’s work which is referenced extensively in Jo Boaler’s course How to Learn Math. With more than 20,000 enrolled chances are you are taking the course with me, but I thought it would be helpful to share a few activities on mindset and attribution retraining—a fancy phrase for how to move students from a fixed to growth mindset.
Since 2011, the district I work in has offered a graduate level course called The Skillful Teacher. I’ve taken the class and I’m finding strong similarities between the two courses with respect to mindset and feedback. The Skillful Teacher required extensive “homework” so that’s given me an opportunity to share a few activities that you can use, abuse, or refuse.
By the way, Algebra’s Friend has written a fine overview of Session 1 on Boaler’s course. I would appreciate everyone who’s taking the course to chime in there and here so we can learn even more.
As promised:
## Activity #1 Mindset Quiz
Here’s a Mindset Quiz I retyped from this document. It’s a self assessment that students score themselves.
## Activity #2 Fixed vs. growth mindset card sort
The card sort activity introduces students to growth and fixed mindsets. Cut the statements into strips. Mix them up for students to sort into two categories.
“The basic principle of attribution theory as it applies to motivation is that a person’s own perceptions or attributions for success or failure determine the amount of effort the person will expend on that activity in the future.”– via
Using a T-chart, students will brainstorm what makes a successful and unsuccessful student. From the list the teacher will frame the rest of the period doing 4 corners—asking who thinks success is due to effort; who thinks success is due to luck, who thinks success is due to ability, who thinks success is due to how easy or hard the task was.
Assign a Think-Pair-Share activity to create situations where only effort was needed to complete the task, only luck, only ability, only the difficulty of the task at hand. Students share scenarios and agree or disagree using a human continuum.
This handout is a related activity. Note: it doesn’t get into stable or unstable causes of success or failure.
## Activity #4 You Can Grow Your Own Intelligence
This brief article from Health and Science News You Can Use can be used for a class discussion on how the brain learns. Here are some comprehension questions as well.
## Activity #5 Math Attitude Scale
To be honest I haven’t used this. It is something I stumbled upon a couple of years ago. The intent was to give it to students and score it using Mastery Manager.
If you have resources to share I would love to hear about them.
## 11 thoughts on “Math, Mindset, and Attribution Retraining”
1. Mary!!! I just worked through session 2 … and then took my Mindset book outside and skimmed a few chapters. As I came back in the house, I was thinking I needed to do some activities related to mindset on the very first day of school. And bless you … there they are! I’ll be using some of your ideas for sure.
2. Thank you for posting this! I’m also taking the course (like you said, plenty of people are). I could definitely see myself using these.
1. Hi Rachel!
Some activities could be used “right off the shelf”, others may need to be adapted. We’re in this together 🙂 so I’m happy to help. Thanks for taking the time to check out the post. We’re all incredibly busy and I appreciate it.
3. Kathryn Laster (@kklaster) says:
Thanks for all of these great resources! I’m just finishing part one of the course and after reading your post, I am really looking forward to the next session. Our campus studied Mindset for the past two years, and our kids (and faculty) also responded well to the “Famous Failures” types of commercials. Nike has a bunch of awesome ones, and after the Olympics, we found a ton of great examples, but here are a few links, in case you want to add these to your list of resources. (Sorry if these are already mentioned in #howtolearnmath.)
1. I appreciate the links! For the reading crowd, literature teachers may want to carve time from their curriculum to share excerpts from the book Right Words at the Right Time. It features brief, celebrity essays in which the celebs describe words that lingered and inspired them. Given my 7th graders, there are some that I wouldn’t use in the classroom, such as a teacher telling Mohammed Ali he wouldn’t amount to anything or Carlos Santana’s family upbringing, but there are several worth sharing in class then having students write their own Right Words at the Right Time.
Your mention of the #howtolearnmath twitter community got me thinking that I should share this post. Perhaps I will. Thanks for stopping by!
4. Thanks for the post, I am in the course with you and know that these resources will help me implement the class into my curriculum. I currently have my kids write math autobiographies. This work will help them understand who they are as learners (and the type of learner they can become)
1. Math autobiographies are a great idea! It would be interesting for them to write about the math messages they’ve heard over the course of their lives. Identify some turning point either in a positive or negative direction, etc.
Thanks for sharing!
5. Michelle says:
Mary, thanks so much for sharing! This is awesome! Do you have the article on growing your own intelligence? The link takes me to your other document. Thanks!
1. Oops! I fixed the link, but here it is. Thanks for proofreading!!! I try to check all the links before publishing, but this one got past me. I appreciate you stopping by.
6. Jill Brooks says:
Mary,
We were looking for examples of attribution retraining during an Administrator’s meeting and I found this post. Thanks so much for a great example!
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Learn advanced calculus problems and solutions FAQ, engg math FAQ, competency based interview questions with MCQs based online test prep. These frequently asked questions has multiple choice questions (MCQ), engg math quiz questions and answers: ode which contain initial condition must have, answer key with options final values, mid values, initial values and arbitrary constant for competitive exam preparation. Free FAQ, situational interview questions are to learn advanced calculus problems and solutions quiz online with MCQs to practice test questions with answers.
MCQ: ODE which contain initial condition must have
1. final values
2. mid values
3. initial values
4. arbitrary constant
C
MCQ: Euler Method is a
1. second order method
2. first order method
3. third order method
4. fourth order method
B
MCQ: A first-order numerical procedure for solving ordinary differential equations with a given initial value is
1. Newton's Method
2. Euler's Method
3. Miller's Method
4. Separable Method
B
MCQ: Which of following solution contains arbitrary constant?
1. general solution
2. logical solution
3. particular solution
4. binary solution
A
MCQ: Mathematical equation with particular initial condition is termed as
1. finial value problem
2. initial value problem
3. dynamic value problem
4. static value problem
B
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# `int sqrt(25-x^2) dx` Find the indefinite integral
given to solve ,
`int sqrt(25-x^2) dx`
using the Trig Substitutions we can solve these type of integrals easily and the solution is as follows
for `sqrt(a-bx^2) ` we can take `x= sqrt(a/b) sin(u)`
so ,
`int sqrt(25-x^2) dx`
the `x = sqrt(25/1) sin(u) = 5sin(u)`
=> `dx =...
## Unlock This Answer Now
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given to solve ,
`int sqrt(25-x^2) dx`
using the Trig Substitutions we can solve these type of integrals easily and the solution is as follows
for `sqrt(a-bx^2) ` we can take `x= sqrt(a/b) sin(u)`
so ,
`int sqrt(25-x^2) dx`
the `x = sqrt(25/1) sin(u) = 5sin(u)`
=> `dx = 5 cos(u) du`
so ,
`int sqrt(25-x^2) dx`
= `int sqrt(25-(5sin(u))^2) (5 cos(u) du)`
= `int sqrt(25-25(sin(u))^2) (5 cos(u) du)`
= `int 5 sqrt(1-sin^2 u )(5 cos(u) du)`
= `int 25 (cos(u))(cos(u)) du`
= `25 int cos^2(u) du`
= `25 int(1+cos(2u))/2 du`
= `(25/2) int (1+cos(2u)) du`
= `(25/2) [u+(1/2)(sin(2u))]+c`
but `x=5sin(u)`
=>` x/5 = sin(u)`
=> `u= arcsin(x/5)`
so,
`(25/2) [u+(1/2)(sin(2u))]+c`
=`(25/2) [(arcsin(x/5))+(1/2)(sin(2(arcsin(x/5))))]+c`
Approved by eNotes Editorial Team
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https://documen.tv/question/someone-help-pls-a-bag-contains-1-crayon-of-each-color-red-orange-yellow-green-blue-pink-maroon-19828761-46/
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## someone help pls A bag contains 1 crayon of each color: red, orange, yellow, green, blue, pink, maroon, and purple. Event
Question
someone help pls
A bag contains 1 crayon of each color: red, orange, yellow, green, blue, pink, maroon, and purple.
Event B; A person chooses a crayon at random out of the bag and walks off to use it. A second person comes to get a crayon chosen at random out of the bag.
What is the probability the second person gets the yellow crayon?
in progress 0
2 months 2021-07-25T16:27:58+00:00 1 Answers 4 views 0
The answer is: 1/4 chance witch is 25% also
Step-by-step explanation:
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LAB1 - Objective The objective of this lab is to learn how to use the Quartus Schematic ModelSim and the Intel DE10-Lite Board We also expect to design
# LAB1 - Objective The objective of this lab is to learn how...
• Lab Report
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Objective: The objective of this lab is to learn how to use the Quartus Schematic, ModelSim and the Intel DE10-Lite Board. We also expect to design a circuit that counts from 0 to 9. Design and Test Procedure: (a) The first step of this lab was to setup Quartus correctly for our board and the correct simulation too (ModelSim-Altera). After that we can start building the first circuit (Figure 1). Figure 1: The Initial Schematic (b) After the circuit is built we can use the System Builder program to generate a System Builder that can be assigned to our circuit. After that we can compile and program it to our DE10-Lite Board. We notice that it worked as expected - it count from 0 to 9 and then roll over to 0 again. (c) After that we change our circuit to the Figure 2 one and we programmed the board for the new circuit. We noticed that since the frequency is too high, the board only show the number 8, but we can see that turning the clock off the others numbers can be seen. Figure 2: Using a 50 Mhz Clock Signal
#### You've reached the end of your free preview.
Unformatted text preview: (d) Since this testing is not effective we used ModelSim for a more effective testing. We 1 first setup ModelSim and open the Wave View where we can see the waves of GPIO, SW (force mode), HEX0 and MAX10_CLK1_50 (clock mode). Result and Answers to Questions: The result from the simulation can be seen on Figure 3: Figure 3: ModelSim Wave View As can be seen the circuit behaves as expected (match the figure from the lab manual) and the GPIO output is counting from 0 to 9, changing the number at the clock frequency of 50MHz. Each waveform does match the behavior of the corresponding signal when the circuit operates on the DE10-LITEboard since we see the expected output on the board as expected based on the simulations. Conclusions: In this lab we learned how to use the the Quartus Schematic Capture to design circuits, ModelSim Simulation to simulate the design and the Intel DE10-Lite Board to test and verify the design. 2...
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# If Carla, Nora, and Wanda have a total of \$48, how much money does Car
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If Carla, Nora, and Wanda have a total of \$48, how much money does Car [#permalink]
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Updated on: 14 Oct 2018, 08:31
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If Carla, Nora, and Wanda have a total of \$48, how much money does Carla have?
(1) The positive difference between the amounts of money that Carla and Nora have is \$12.
(2) Nora and Wanda have the same amount of money.
Difficulty Level: 700
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Last edited by SajjadAhmad on 14 Oct 2018, 08:31, edited 1 time in total.
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Re: If Carla, Nora, and Wanda have a total of \$48, how much money does Car [#permalink]
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26 May 2017, 10:28
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If Carla, Nora, and Wanda have a total of \$48, how much money does Carla have?
(1) The positive difference between the amounts of money that Carla and Nora have is \$12.
(2) Nora and Wanda have the same amount of money.
Target question: How much money does Carla have?
Given: Carla, Nora, and Wanda have a total of \$48
Statement 1: The positive difference between the amounts of money that Carla and Nora have is \$12
So, EITHER Carla has \$12 more than Nora OR Carla has \$12 less than Nora.
This certainly doesn't seem very helpful.
ALSO, there's no information about Wanda.
Statement 1 is clearly NOT SUFFICIENT
Statement 2: Nora and Wanda have the same amount of money.
Statement 2 is clearly NOT SUFFICIENT
Statements 1 and 2 combined
Let's examine the 2 possible cases from statement 1 in conjunction with statement 2.
Case a: Carla has \$12 more than Nora
Let x = the \$ Nora has
This means that x+12 = the \$ Carla has
And x = the \$ Wanda has (since Nora and Wanda have the same amount of money)
Since the TOTAL = \$48, we can write: x + (x+12) + x = 48
Simplify: 3x + 12 = 48
Solve: x = 12
Since x+12 = the \$ Carla has, we can conclude that 12+12 = Carla's money.
In other words, Carla has \$24
Case b: Carla has \$12 less than Nora
Let x = the \$ Carla has
This means that x+12 = the \$ Nora has
And x+12 = the \$ Wanda has (since Nora and Wanda have the same amount of money)
Since the TOTAL = \$48, we can write: x + (x+12) + (x+12) = 48
Simplify: 3x + 24 = 48
Solve: x = 8
In other words, Carla has \$8
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT
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Re: If Carla, Nora, and Wanda have a total of \$48, how much money does Car [#permalink]
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14 Oct 2018, 07:06
Twisted language in Statement 1.
Actually it is saying that absolute value of C-N= 12
So here are two cases.
C-N= 12
or
N-C =12
Nice learning from this question.
Re: If Carla, Nora, and Wanda have a total of \$48, how much money does Car [#permalink] 14 Oct 2018, 07:06
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Calculators » Home & Lifestyle » Electricity Consumption
## Energy Consumption Calculator
Our online tools will provide quick answers to your calculation and conversion needs. On this page, you can calculate energy consumption by electric appliances. You can then, determine the energy cost to use an electrical appliance by multiplying Units consumed with charges applicable per unit.
Appliance power rating (in watts):
Usage (minutes per day):
1 unit of electricity is equal to consumption of 1 kilowatt (kW) of power in 1 hour, where 1kW = 1000W. One kilowatt-hour is thus equal to the power used by ten 100-watt bulbs for one hour. View typical wattages of common home appliances.
Android: Use this electricity calculator offline with our all-in-one calculator app.
#### Related Calculators
Power consumption formula:
U = W*h/1000
Where,
U = Units consumed
W = Appliance power in watts
h = Usage time in hours
#### Typical wattages of various household appliances
• Coffee maker = 900-1200
• Clothes washer = 350-500
• Clothes dryer = 1800-5000
• Dishwasher = 1200-2400 (using the drying feature greatly increases energy consumption)
• Dehumidifier = 785
• Electric blanket (Single/Double) = 60 / 100
• Fans
Ceiling = 65-175
Window = 55-250
Furnace = 750
Whole house = 240-750
• Hair dryer = 1200-1875
• Heater (portable) = 750-1500
• Clothes iron = 1000-1800
• Microwave oven = 750-1100
• Personal computer
CPU - awake / asleep = 120 / 30 or less
Monitor - awake / asleep = 150 / 30 or less
Laptop = 50
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https://brilliant.org/problems/blood-relation-2/
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# Blood Relation
Logic Level 2
If $P + Q$ means $P$ is the brother of $Q$; $P \times Q$ means $P$ is the father of $Q$ and $P - Q$ means $P$ is the sister of $Q$, which of the following relation shows that $I$ is the niece of $K$?
Details
• The operations $+,-,\times$ are binary relations. I.e. $P+Q\times R$ means that $P$ is the brother of $Q$, and $Q$ is the father of $R$.
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Forces
Any object which is free of external forces, or on which the sum of external forces is zero, will either be at rest or will be moving with a constant velocity.
A bicycle seems to disobey this principle because when we stop pushing the pedals we will slow down and ultimately come to rest. This simply means that there are a number of external forces that we need to overcome in order to maintain a constant velocity. These forces are caused by the resistance of the air, by the resistance to rolling, and to the change in gravitational energy on an inclined road.
The force on the pedals is transmitted to the contact between the tire and the ground by means of the chain ring, the chain and the cog.
The reaction of the ground generates the driving power. To say it the way of Newton, the rider pushes the ground backwards, the ground pushes the bike forwards!
In normal conditions the length of the cranks is much smaller than the radius of the wheel, and the chain ring is bigger than the cogs, thus the driving force is much smaller than the force applied to the pedals.
In physics a force is expressed in units Newton (N). Sometimes the kilogram-force ( kgf) is used as a practical indication of force. This is the force that is caused by gravity on a mass of 1 kilogram. The kgf equates to 9.8 Newton.
The force of the foot on the pedal is decomposed into a radial force which is parallel to the crank and is directed towards the centre of the bottom bracket, and a tangential force which is perpendicular to the crank. The radial force is completely useless and a good rider will rotate the pedals elegantly which means that this radial force is minimized. When the rider has a bad technique or when he is tired he will have an ugly and less efficient pedalling cycle.
The relation between the useful tangential force on the pedals Ft and the driving force F is given by;
Here R is the radius of the wheel, L is the length of the crank, nv is the number of teeth on the chain wheel and na is number of teeth on the chosen cog.
When using the big gear e.g. 54 x 11 the driving force may be 10 times smaller than the pedalling force, yet we are riding at high speed!
However when climbing and using a mini gear of 30 x 25 the force ratio is only 2.43
Luckily we do not need a high driving force in order to go fast on a flat road. A standard person (not too fat nor too slender) may need only 17 N to be riding at 32 km/h. With a gear ratio 52 x 16 this corresponds to a mean pedal force of 107 N which is approximately only 16 % of the weight of an adult person..
Work and Energy
Work is done only when the force causes a displacement i.e. Work = Force x Displacement
The displacement has to be measured in the direction of the force which means that when a displacement is perpendicular to the force, no work is done!
Sometimes even a very strong force does not deliver any work. No matter how hard you push against a wall, as long as the wall does not move you are not "working". For this reason bicycles should be as stiff as possible, the forces applied to the pedals are to be used to move the bicycles forward, not to deform the frame.
Energy is simply another word for work, work is transformed into energy and energy is transformed into work. We can distinguish 3 main forms of energy, Kinetic energy, potential energy and radiative energy such as heat and light. Let us illustrate the work-energy duality with help of a cyclist riding at constant speed on a flat road.
The work he does is used to displace the air, so the air gains some kinetic energy. Also the tires are deformed through the contact with the road. This deformation causes internal and external friction that is transformed into heat and causes the rolling resistance.
The physical unit of work and energy is the Joule (J) but for practical reasons we frequently use the secondary unit calorie (cal)
One calorie equates the amount of energy needed to increase the temperature of 1 gram of water by one degree centigrade.
One calorie is equal to 4.182 Joule. One kilocalorie (kcal or Cal) is equal to 1000 calorie. Please not that in the U.S.A the kilo cal is noted Cal with capital C
An adult person will need a food intake of approximately 1600 kilocalorie per day in order to stay alive without any extra effort or activity.
Power
Power is the amount of work done during one second.
Power = Force x displacement per second = Force x velocity
The unit of power is Watt (W)
Suppose a person of 70 kg walks up the stairs of 2.70 meter. In order to do this he produces an amount of work70 x 9.81 x 2.70 = 1854 Joule. In order to walk this steps in 5 seconds he needed a power of 1854/5 = 370 Watts during the 5 seconds.
Professional cyclists typically produce 200 to 300 Watts on easy endurance training, and between 350 to 450 Watts in long time trials, depending mainly on the weight and talent of the cyclist.
It is not my purpose to go deep into the many physiological aspects of power generation such as the the rate of intake of oxygen (VO2), the blood concentration of lactate, the hart rate. These are used as indicators of the "how good" the human machine is running. We are more interested in the amount of work the human body can produce in a steady state i.e. for an almost indefinitely long time. We can qualify the best possible performance with the aid of 2 measurable parameters, the Critical Power and the total anaerobic work capacity..
Het kritisch vermogen (KV) is het vermogen dat de fietser theoretisch oneindig lang kan volhouden. Dit is dus het maximaal aėroob vermogen, waarvoor de type 1 spieren verantwoordelijk zijn. Wij aanzien kritisch vermogen PKV en maximaal aėroob vermogen Pae als synoniemen.
Het specifiek vermogen is het maximaal aėroob vermogen gedeeld door de lichaamsmassa. Dit specifiek vermogen is allesbepalend voor klimmers.
Het oppervlak-specifiek vermogen is het maximaal aėroob vermogen gedeeld door het frontaal oppervlak.
De anaėrobe capaciteit (AEWC = AnaErobic Work Capacity) wordt uitgedrukt in kJ of kcal en is de totale hoeveelheid werk dat de renner op anaėrobe manier kan leveren. Wanneer deze capaciteit wordt overschreden verzuren de spieren, gaat de renner door de muur, en valt stil.
Het oppervlak-specifiek vermogen en de juiste dosering van kritisch vermogen en AEWC is zeer belangrijk voor tijdrijders
Sommigen durven wel eens de begrippen Kracht en Vermogen (Power) door elkaar halen. Zo kunnen we horen dat Vinokourov met groot verzet en veel "power" klimt, daar waar in werkelijkheid een renner met een "klein molentje" sneller gaat en dus meer vermogen met minder kracht ontwikkelt.
Het totale vermogen (P) dat een fietser levert wordt gebruikt om de 3 weerstanden te overwinnen, m.a.w. de rolweerstand, de luchtweerstand en de klim- of zwaartekrachtsweerstand.
P = Prol + Plucht + Pklim
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# Calculator search results
Formula
Expand the expression
Factorize the expression
$35x ^{ 2 } y+56x ^{ 3 } y ^{ 2 } +63x ^{ 4 } y ^{ 5 }$
$63 x ^ { 4 } y ^ { 5 } + 56 x ^ { 3 } y ^ { 2 } + 35 x ^ { 2 } y$
Organize polynomials
$\color{#FF6800}{ 35 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 56 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 63 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 5 } }$
Sort the polynomial expressions in descending order
$\color{#FF6800}{ 63 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 5 } } \color{#FF6800}{ + } \color{#FF6800}{ 56 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 35 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ y }$
$7 x ^ { 2 } y \left ( 9 x ^ { 2 } y ^ { 4 } + 8 x y + 5 \right )$
Arrange the expression in the form of factorization..
$\color{#FF6800}{ 35 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 56 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 63 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 5 } }$
Expand the expression
$\color{#FF6800}{ 63 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 5 } } \color{#FF6800}{ + } \color{#FF6800}{ 56 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 35 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ y }$
$\color{#FF6800}{ 63 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 5 } } \color{#FF6800}{ + } \color{#FF6800}{ 56 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 35 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ y }$
Tie a common factor
$\color{#FF6800}{ 7 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ y } \left ( \color{#FF6800}{ 9 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right )$
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# Week 1 – Problem Set >> Cryptography I
## Week 1 – Problem Set >> Cryptography I
Week 1 – Problem Set LATEST SUBMISSION GRADE 80% 1. Question 1 Data compression is often used in data storage and transmission. Suppose you want to use data compression in conjunction with encryption. Does it make more sense to: 0 / 1 point The order does not matter — either one is fine. Encrypt then compress. The order does not matter — neither one will compress the data. Compress then encrypt. Incorrect Ciphertexts tend to look like random strings and therefore compressing after encryption will not compress the data. 2. Question 2 Let G:\{0,1\}^s \to \{0,1\}^nG:{0,1} s →{0,1} n be a secure PRG. Which of the following is a secure PRG (there is more than one correct answer): 1 / 1 point G'(k) = G(k) \bigoplus 1^nG ′ (k)=G(k)⨁1 n Correct a distinguisher for G’G ′ gives a distinguisher for GG. G'(k) = G(k \oplus 1^s)G ′ (k)=G(k⊕1 s ) Correct a distinguisher for G’G ′ gives a distinguisher for GG. G'(k) = G(k)[0,\ldots,n-2]G ′ (k)=G(k)[0,…,n−2] (i.e., G'(k)G ′ (k) drops the last bit of G(k)G(k)) Correct a distinguisher for G’G ′ gives a distinguisher for GG. G'(k) = G(k) \;\big\|\; 0G ′ (k)=G(k) ∥ ∥ ∥ 0 (here \big\| ∥ ∥ ∥ denotes concatenation) G'(k) = G(0)G ′ (k)=G(0) G'(k) = G(k) \;\big\|\; G(k)G ′ (k)=G(k) ∥ ∥ ∥ G(k) (here \big\| ∥ ∥ ∥ denotes concatenation) 3. Question 3 Let G:K \to \{0,1\}^nG:K→{0,1} n be a secure PRG. Define G'(k_1,k_2) = G(k_1) \;\bigwedge\; G(k_2)G ′ (k 1 ,k 2 )=G(k 1 )⋀G(k 2 ) where \bigwedge⋀ is the bit-wise AND function. Consider the following statistical test AA on \{0,1\}^n{0,1} n : A(x)A(x) outputs \text{LSB}(x)LSB(x), the least significant bit of xx. What is Adv_{\text{PRG}}[A,G’]Adv PRG [A,G ′ ] ? You may assume that \text{LSB}(G(k))LSB(G(k)) is 0 for exactly half the seeds kk in KK. Note: Please enter the advantage as a decimal between 0 and 1 with a leading 0. If the advantage is 3/4, you should enter it as 0.75 1 / 1 point 0.25 Correct for a random string x we have Pr[A(x)=1]=1/2Pr[A(x)=1]=1/2 but for a pseudorandom string G'(k_1,k_2)G ′ (k 1 ,k 2 ) we have Pr_{k_1,k_2}[A(G'(k_1,k_2))=1]=1/4Pr k 1 ,k 2 [A(G ′ (k 1 ,k 2 ))=1]=1/4. 4. Question 4 Let (E,D)(E,D) be a (one-time) semantically secure cipher with key space K = \{0,1\}^\ellK={0,1} ℓ . A bank wishes to split a decryption key k \in \{0,1\}^\ellk∈{0,1} ℓ into two pieces p_1p 1 and p_2p 2 so that both are needed for decryption. The piece p_1p 1 can be given to one executive and p_2p 2 to another so that both must contribute their pieces for decryption to proceed. The bank generates random k_1k 1 in \{0,1\}^\ell{0,1} ℓ and sets k_1′ \gets k \oplus k_1k 1 ′ ←k⊕k 1 . Note that k_1 \oplus k_1′ = kk 1 ⊕k 1 ′ =k. The bank can give k_1k 1 to one executive and k_1’k 1 ′ to another. Both must be present for decryption to proceed since, by itself, each piece contains no information about the secret key kk (note that each piece is a one-time pad encryption of kk). Now, suppose the bank wants to split kk into three pieces p_1,p_2,p_3p 1 ,p 2 ,p 3 so that any two of the pieces enable decryption using kk. This ensures that even if one executive is out sick, decryption can still succeed. To do so the bank generates two random pairs (k_1,k_1′)(k 1 ,k 1 ′ ) and (k_2,k_2′)(k 2 ,k 2 ′ ) as in the previous paragraph so that k_1 \oplus k_1′ = k_2 \oplus k_2′ = kk 1 ⊕k 1 ′ =k 2 ⊕k 2 ′ =k. How should the bank assign pieces so that any two pieces enable decryption using kk, but no single piece can decrypt? 1 / 1 point p_1 = (k_1,k_2),\quad p_2 = (k_1′,k_2), \quad p_3 = (k_2′)p 1 =(k 1 ,k 2 ),p 2 =(k 1 ′ ,k 2 ),p 3 =(k 2 ′ ) p_1 = (k_1,k_2),\quad p_2 = (k_1′,k_2′), \quad p_3 = (k_2′)p 1 =(k 1 ,k 2 ),p 2 =(k 1 ′ ,k 2 ′ ),p 3 =(k 2 ′ ) p_1 = (k_1,k_2),\quad p_2 = (k_1′), \quad p_3 = (k_2′)p 1 =(k 1 ,k 2 ),p 2 =(k 1 ′ ),p 3 =(k 2 ′ ) p_1 = (k_1,k_2),\quad p_2 = (k_1,k_2), \quad p_3 = (k_2′)p 1 =(k 1 ,k 2 ),p 2 =(k 1 ,k 2 ),p 3 =(k 2 ′ ) p_1 = (k_1,k_2),\quad p_2 = (k_2,k_2′), \quad p_3 = (k_2′)p 1 =(k 1 ,k 2 ),p 2 =(k 2 ,k 2 ′ ),p 3 =(k 2 ′ ) Correct executives 1 and 2 can decrypt using k_1,k_1’k 1 ,k 1 ′ , executives 1 and 3 can decrypt using k_2,k_2’k 2 ,k 2 ′ , and executives 2 and 3 can decrypt using k_2,k_2’k 2 ,k 2 ′ . Moreover, a single executive has no information about $k$. 5. Question 5 Let M=C=K=\{0,1,2,\ldots,255\}M=C=K={0,1,2,…,255} and consider the following cipher defined over (K,M,C)(K,M,C): E(k,m) = m+k \pmod{256} \qquad;\qquad D(k,c) = c-k \pmod{256} \ .E(k,m)=m+k(mod256);D(k,c)=c−k(mod256) . Does this cipher have perfect secrecy? 1 / 1 point Yes. No, only the One Time Pad has perfect secrecy. No, there is a simple attack on this cipher. Correct as with the one-time pad, there is exactly one key mapping a given message m to a given ciphertext c. 6. Question 6 Let (E,D)(E,D) be a (one-time) semantically secure cipher where the message and ciphertext space is \{0,1\}^n{0,1} n . Which of the following encryption schemes are (one-time) semantically secure? 1 / 1 point E'(\ (k,k’),\ m) = E(k,m) \;\big\|\; E(k’,m) E ′ ( (k,k ′ ), m)=E(k,m) ∥ ∥ ∥ E(k ′ ,m) Correct an attack on E’E ′ gives an attack on EE. E'(k,m) = \text{reverse}(E(k,m))E ′ (k,m)=reverse(E(k,m)) Correct an attack on E’E ′ gives an attack on EE. E'(k,m) = E(k,m) \;\big\|\; k E ′ (k,m)=E(k,m) ∥ ∥ ∥ k E'(k,m) = E(k,m) \;\big\|\; \text{LSB}(m) E ′ (k,m)=E(k,m) ∥ ∥ ∥ LSB(m) E'(k,m) = 0 \;\big\|\; E(k,m)E ′ (k,m)=0 ∥ ∥ ∥ E(k,m) (i.e. prepend 0 to the ciphertext) Correct an attack on E’E ′ gives an attack on EE. E'(k,m) = E(0^n,m) E ′ (k,m)=E(0 n ,m) 7. Question 7 Suppose you are told that the one time pad encryption of the message “attack at dawn” is 6c73d5240a948c86981bc294814d (the plaintext letters are encoded as 8-bit ASCII and the given ciphertext is written in hex). What would be the one time pad encryption of the message “attack at dusk” under the same OTP key? 1 / 1 point 6c73d5240a948c86981bc2808548 Correct 8. Question 8 The movie industry wants to protect digital content distributed on DVD’s. We develop a variant of a method used to protect Blu-ray disks called AACS. Suppose there are at most a total of nn DVD players in the world (e.g. n = 2^{32}n=2 32 ). We view these nn players as the leaves of a binary tree of height \log_2{n}log 2 n. Each node in this binary tree contains an AES key k_ik i . These keys are kept secret from consumers and are fixed for all time. At manufacturing time each DVD player is assigned a serial number i \in [0, n − 1]i∈[0,n−1]. Consider the set of nodes S_iS i along the path from the root to leaf number ii in the binary tree. The manufacturer of the DVD player embeds in player number ii the keys associated with the nodes in the set S_iS i . A DVD movie mm is encrypted as E(k_{\text{root}},k) \big\| E(k,m)E(k root ,k) ∥ ∥ ∥ E(k,m) where kk is a random AES key called a content-key and k_{\text{root}}k root is the key associated with the root of the tree. Since all DVD players have the key k_{\text{root}}k root all players can decrypt the movie mm. We refer to E(k_{\text{root}},k)E(k root ,k) as the header and E(k,m)E(k,m) as the body. In what follows the DVD header may contain multiple ciphertexts where each ciphertext is the encryption of the content-key kk under some key k_ik i in the binary tree. Suppose the keys embedded in DVD player number rr are exposed by hackers and published on the Internet. In this problem we show that when the movie industry distributes a new DVD movie, they can encrypt the contents of the DVD using a slightly larger header (containing about \log_2 nlog 2 n keys) so that all DVD players, except for player number rr, can decrypt the movie. In effect, the movie industry disables player number rr without affecting other players. As shown below, consider a tree with n=16n=16 leaves. Suppose the leaf node labeled 25 corresponds to an exposed DVD player key. Check the set of keys below under which to encrypt the key kk so that every player other than player 25 can decrypt the DVD. Only four keys are needed. 1 / 1 point 11 Correct You cannot encrypt kk under key 5, but 11’s children must be able to decrypt kk. 29 1 Correct You cannot encrypt kk under the root, but 1’s children must be able to decrypt kk. 5 7 6 Correct You cannot encrypt kk under 2, but 6’s children must be able to decrypt kk. 26 Correct You cannot encrypt kk under any key on the path from the root to node 25. Therefore 26 can only decrypt if you encrypt kk under key k_{26}k 26 . 10 9. Question 9 Continuing with the previous question, if there are nn DVD players, what is the number of keys under which the content key kk must be encrypted if exactly one DVD player’s key needs to be revoked? 1 / 1 point \log_2{n}log 2 n n-1n−1 22 n/2n/2 \sqrt{n} n Correct That’s right. The key will need to be encrypted under one key for each node on the path from the root to the revoked leaf. There are \log_2{n}log 2 n nodes on the path. 10. Question 10 Continuing with question 8, suppose the leaf nodes labeled 16, 18, and 25 correspond to exposed DVD player keys. Check the smallest set of keys under which to encrypt the key k so that every player other than players 16,18,25 can decrypt the DVD. Only six keys are needed. 0 / 1 point 4 Correct Yes, this will let players 19-22 decrypt. 6 11 Correct Yes, this will let players 23,24 decrypt. 15 Correct Yes, this will let players 15 decrypt. 17 Correct Yes, this will let players 17 decrypt. 26 Correct Yes, this will let players 26 decrypt. 8 13 14 20 You didn’t select all the correct answers
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# Maths Coursework
The issues we are studying at our maths classes turn out to be helpful in our everyday life. So, we should be attentive and try to make our maths courseworks perfect if we want to obtain desirable results.
When we talk about quantity, or space, or structure we talk about maths principles and riles. The world of Mathematics is so complicated but if you want you can comprehend it. You get the chance to investigate a little part of this huge system as Mathematics when you get the task to write a maths coursework.
You are welcome to the great variety of maths coursework topics. Maybe some of you think that maths is a concrete science full of certain established facts and rules. But if you can add a bit of imagination to your maths coursework you can turn maths coursework writing process into the catching and thrilling process. The results will be noticed.
The number of the topics for your maths coursework is great. The issues from trigonometry and statistics, geometry and space can be developed certainly by you.
What you need is to create a title and a thesis statement and try to disclose them in your maths coursework.
Do not forget about time you need for your perfect maths coursework. You are to set up the timetable with your maths coursework tutor and follow it.
In the maths courseworks you should not only retell already known facts. Try to find out some catching not very popular ideas and develop them. Also, it is possible to use the established theories in order to find the answers to the up-do-date questions.
In your maths coursework you can even describe the lives and the creations made by the famous scientists. But do not be carried away by the trite facts. Do not be boring.
It goes without saying that it is very difficult to turn a maths coursework full of numbers and theories into a perfect academic writing. You should follow the rules and be as creative as possible. We wish you good luck in such a hard task!
This post originally appeared on http://writing-services.org/blog/2007/11/08/maths-coursework/
Maths Coursework 9.9 of 10 on the basis of 1647 Review.
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#### Details of this Paper
##### Pantanal, Inc., manufactures car seats in a local...
Description
Solution
Question
Pantanal, Inc., manufactures car seats in a local factory. For costing purposes, it uses a first-in, first-out (FIFO) process costing system. The factory has three departments: Molding, Assembling, and Finishing. Following is information on the beginning work-in-process inventory in the Assembling Department on August 1: Costs Degree of Completion Work-in-process beginning inventory (10,000 units) Transferred-in from Molding \$ 99,000 100 % Direct materials costs 170,800 70 Conversion costs 57,000 40 -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- Work-in-process balance (August 1) \$ 326,800 -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- During August, 105,000 units were transferred in from the Molding Department at a cost of \$2,163,000 and started in Assembling. The Assembling Department incurred other costs of \$1,073,270 in August as follows: August Costs Direct materials costs \$ 933,680 Conversion costs 139,590 -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- Total August costs \$ 1,073,270 -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- At the end of August, 19,000 units remained in inventory that were 90 percent complete with respect to direct materials and 60 percent complete with respect to conversion. Required: Compute the cost of goods transferred out in August and the cost of work-in-process ending inventory. (Round the unit costs for prior department costs, materials, and conversion to 2 decimals and the final answers to the nearest dollar amount. Omit the "\$" sign in your response.) Cost of goods transferred out \$ Cost of WIP ending inventory \$
Paper#9607 | Written in 18-Jul-2015
Price : \$25
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# praharsh
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# First Optimization
Usually, Optuna is used to optimize hyper-parameters, but as an example, let us directly optimize a quadratic function in an IPython shell.
`In [1]: import optuna`
The objective function is what will be optimized.
```In [2]: def objective(trial):
...: x = trial.suggest_uniform('x', -10, 10)
...: return (x - 2) ** 2
...:```
This function returns the value of (x - 2)^2. Our goal is to find the value of `x` that minimizes the output of the `objective` function. This is the "optimization." During the optimization, Optuna repeatedly calls and evaluates the objective function with different values of `x`.
A :class:`~optuna.trial.Trial` object corresponds to a single execution of the objective function and is internally instantiated upon each invocation of the function.
The suggest APIs (for example, :func:`~optuna.trial.Trial.suggest_uniform`) are called inside the objective function to obtain parameters for a trial. :func:`~optuna.trial.Trial.suggest_uniform` selects parameters uniformly within the range provided. In our example, from -10 to 10.
To start the optimization, we create a study object and pass the objective function to method :func:`~optuna.study.Study.optimize` as follows.
```In [3]: study = optuna.create_study()
In [4]: study.optimize(objective, n_trials=100)
[I 2018-05-09 10:03:22,469] Finished trial#0 resulted in value: 52.9345515866657. Current best value is 52.9345515866657 with parameters: {'x': -5.275613485244093}.
[I 2018-05-09 10:03:22,474] Finished trial#1 resulted in value: 32.82718929591965. Current best value is 32.82718929591965 with parameters: {'x': -3.7295016620924066}.
[I 2018-05-09 10:03:22,475] Finished trial#2 resulted in value: 46.89428737068025. Current best value is 32.82718929591965 with parameters: {'x': -3.7295016620924066}.
[I 2018-05-09 10:03:22,476] Finished trial#3 resulted in value: 100.99613064563654. Current best value is 32.82718929591965 with parameters: {'x': -3.7295016620924066}.
[I 2018-05-09 10:03:22,477] Finished trial#4 resulted in value: 110.56391159932272. Current best value is 32.82718929591965 with parameters: {'x': -3.7295016620924066}.
[I 2018-05-09 10:03:22,478] Finished trial#5 resulted in value: 42.486606942847395. Current best value is 32.82718929591965 with parameters: {'x': -3.7295016620924066}.
[I 2018-05-09 10:03:22,479] Finished trial#6 resulted in value: 1.130813338091735. Current best value is 1.130813338091735 with parameters: {'x': 3.063397074517198}.
...
[I 2018-05-09 10:03:23,431] Finished trial#99 resulted in value: 8.760381111220335. Current best value is 0.0026232243068543526 with parameters: {'x': 1.9487825780924659}.
In [5]: study.best_params
Out[5]: {'x': 1.9487825780924659}```
We can see that Optuna found the best `x` value `1.9487825780924659`, which is close to the optimal value of `2`.
Note
When used to search for hyper-parameters in machine learning, usually the objective function would return the loss or accuracy of the model.
## Study Object
Let us clarify the terminology in Optuna as follows:
• Trial: A single call of the objective function
• Study: An optimization session, which is a set of trials
• Parameter: A variable whose value is to be optimized, such as `x` in the above example
In Optuna, we use the study object to manage optimization. Method :func:`~optuna.study.create_study` returns a study object. A study object has useful properties for analyzing the optimization outcome.
```In [5]: study.best_params
Out[5]: {'x': 1.9926578647650126}
In [6]: study.best_value
Out[6]: 5.390694980884334e-05
In [7]: study.best_trial
Out[7]: FrozenTrial(number=26, state=<TrialState.COMPLETE: 1>, params={'x': 1.9926578647650126}, user_attrs={}, system_attrs={'_number': 26}, value=5.390694980884334e-05, intermediate_values={}, datetime_start=datetime.datetime(2018, 5, 9, 10, 23, 0, 87060), datetime_complete=datetime.datetime(2018, 5, 9, 10, 23, 0, 91010), trial_id=26)
In [8]: study.trials # all trials
Out[8]:
[FrozenTrial(number=0, state=<TrialState.COMPLETE: 1>, params={'x': -4.219801301030433}, user_attrs={}, system_attrs={'_number': 0}, value=38.685928224299865, intermediate_values={}, datetime_start=datetime.datetime(2018, 5, 9, 10, 22, 59, 983824), datetime_complete=datetime.datetime(2018, 5, 9, 10, 22, 59, 984053), trial_id=0),
...
user_attrs={}, system_attrs={'_number': 99}, value=8.2881000286123179, intermediate_values={}, datetime_start=datetime.datetime(2018, 5, 9, 10, 23, 0, 886434), datetime_complete=datetime.datetime(2018, 5, 9, 10, 23, 0, 891347), trial_id=99)]
In [9]: len(study.trials)
Out[9]: 100```
By executing :func:`~optuna.study.Study.optimize` again, we can continue the optimization.
```In [10]: study.optimize(objective, n_trials=100)
...
In [11]: len(study.trials)
Out[11]: 200```
You can’t perform that action at this time.
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2024-07-18 14:28:08 +0200 received badge ● Popular Question (source) 2024-07-07 00:35:18 +0200 received badge ● Notable Question (source) 2024-07-02 15:02:44 +0200 commented answer How to calculate rotation and spin coefficients of tetrads in SageMath? This is exactly what I was looking for. Thanks a lot Prof. Eric, I'll try it out. 2024-07-02 14:59:36 +0200 marked best answer How to calculate rotation and spin coefficients of tetrads in SageMath? I'm trying to reproduce the results of this paper where the author calculates the Ricci rotation and spin coefficients with Mathematica, when provided a metric and a choice of tetrads. I looked through the literature and references of SageMath and I couldn't find any mention of tetrad methods. I came across this post in this forum from 3 years ago where the answer mentions that general tetrads are implemented but spin coefficient methods are not. It links to the documentation page with a lot of info and since I'm an amateur in SageMath, I'm finding it very difficult to understand what needs to be done. Can anyone help me with how to do this or maybe refer me to a notebook example where this is done? Please let me know. Thanks. 2024-07-01 16:19:15 +0200 edited question How to calculate rotation and spin coefficients of tetrads in SageMath? How to calculate rotation and spin coefficients of tetrads in SageMath? I'm trying to reproduce the results of this pape 2024-07-01 16:16:21 +0200 asked a question How to calculate rotation and spin coefficients of tetrads in SageMath? How to calculate rotation and spin coefficients of tetrads in SageMath? I'm trying to reproduce the results of this pape 2024-05-03 12:39:06 +0200 received badge ● Popular Question (source) 2024-05-03 12:32:38 +0200 received badge ● Organizer (source) 2024-05-03 12:31:42 +0200 asked a question Defining NP formalism and checking consistency Defining NP formalism and checking consistency I tried to define a Newman-Penrose tetrad and check the dot products and 2024-03-01 13:30:19 +0200 marked best answer Confused about covariant derivatives and tensors in SageMath I'm trying to calculate covariant derivatives of tensor fields in SageMath, I established the metric connection using g.connection(). But I cannot find any specific method to calculate the covariant derivatives unlike Lie derivative which is an inbuilt method. I looked up documentations and I found that I have to apply the connection (which is just the christoffel symbol) to the tensor field to calculate its covariant derivative. Is that the correct way or is there an inbuilt method that I'm missing? I tried diff() to calculate partial derivative of a tensor defined on a chart on a manifold but it is not working. So ultimately I want to know the correct method to calculate partial and covariant derivatives of vector fields and tensors on a manifold. On a related note, I want to know how to save a tensor quantity which I transformed from one coord to another using Tensor.display_comp(chart=differentOne). This method seems to take the original and just display the terms in a different coord system, how do I save the transformed one as a different tensor defined on the differentOnechart? Please let me know. Thanks in advance for responding! 2024-03-01 13:30:19 +0200 received badge ● Scholar (source) 2024-02-29 09:41:15 +0200 commented answer Confused about covariant derivatives and tensors in SageMath Thanks a lot! This is very helpful. It would be great if you can also explain how to save a coord transformed tensor as 2024-02-27 17:43:07 +0200 edited question Confused about covariant derivatives and tensors in SageMath Confused about covariant derivatives and tensors in SageMath I'm trying to calculate covariant derivatives of tensor fie 2024-02-27 17:39:30 +0200 received badge ● Editor (source) 2024-02-27 17:39:30 +0200 edited question Confused about covariant derivatives and tensors in SageMath Confused about covariant derivatives and tensors in SageMath I'm trying to calculate covariant derivatives of tensor fie 2024-02-27 17:37:56 +0200 asked a question Confused about covariant derivatives and tensors in SageMath Confused about covariant derivatives and tensors in SageMath I'm trying to calculate covariant derivatives of tensor fie 2024-02-20 07:20:17 +0200 commented question Comparison between SageMath ODE solvers and Scipy ODE solvers Okay so I will implement Scipy solvers as I need numerical solutions for my work. Thanks for pointing out the symbolic s 2024-02-19 16:46:11 +0200 asked a question Comparison between SageMath ODE solvers and Scipy ODE solvers Comparison between SageMath ODE solvers and Scipy ODE solvers I'm pretty new to SageMath and I'm currently trying to wor 2024-02-17 06:35:05 +0200 received badge ● Popular Question (source) 2024-02-15 15:22:47 +0200 received badge ● Supporter (source) 2024-02-13 19:42:30 +0200 received badge ● Nice Question (source) 2024-02-13 09:38:25 +0200 received badge ● Student (source) 2024-02-13 09:36:52 +0200 asked a question How to make ubuntu launch jupyter(sage) in windows browser? How to make ubuntu launch jupyter(sage) in windows browser? I installed sagemath in WSL from source. I followed the ste
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Calculus AB Section I Part A A CALCULATOR MAY NOT BE USED ON THIS PART OF THE EXAMINATION
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1 lculus Section I Prt LULTOR MY NOT US ON THIS PRT OF TH XMINTION In this test: Unless otherwise specified, the domin of function f is ssumed to e the set of ll rel numers for which f () is rel numer.. Wht is the coordinte of the point of inflection on the grph of y ? If f is continuous for < < nd differentile for < <, which of the following could e flse? f ' c f f for some c such tht < < f ' c 0 for some c such tht < < f hs minimum vlue on < < f hs mimum vlue on < < 3. If y 0, then when =, dy d
2 4. Let f nd g e differentile functions with the following properties: i) g () > 0 for ll ii) f (0) = If h () = f () g () nd h' f g', then f f ' g () e 0 5. Wht is the instntneous rte of chnge t = of the function f given y f? 6 6 f ln for 0 < = ln for < 4 6. If ( ), then lim f ( ) = ln ln 8 ln 6 4 noneistent
3 7. The grph of the function f shown in the figure elow hs verticl tngent t the point (, 0) nd horizontl tngents t the points (, ) nd (3, ). y For wht vlues of, < < 4, is f not differentile? 0 only 0 nd only nd 3 only 0,, nd 3 only 0,,, nd 3 8. prticle moves long the -is so tht its position t time t is given y t t 6t 5. For wht vlue of t is the velocity of the prticle zero? 3 4 5, then f ' 9. If f sine cose cose cose e e e cose e cose
4 0. The grph of twice-differentile function f is shown in the figure elow. Which of the following is true? y f f ' f '' f f '' f ' f ' f f '' f '' f f ' f '' f ' f. n eqution of the line tngent to the grph of y cos t the point (0, ) is y = + y = + y = y = y = 0. If f '', then the grph of f hs inflection points when = only only nd 0 only nd only, 0, nd only
5 4 3. The function f is given y f. On which of the following intervls is f incresing?,, 0,,0, 4. The grph of f is shown elow. Which of the following could e the grph of the derivtive of f?
6 5. The mimum ccelertion ttined on the intervl 0 < t < 3 y the prticle whose velocity is given y 3 vt t3t t 4 is The function f is continuous on the closed intervl [0, ] nd hs vlues tht re given in the tle elow. 0 f () k The eqution f ( ) = must hve t lest two solutions in the intervl [0, ] if k = If f tn, then f '
7 lculus Section I Prt LULTOR IS RQUIR FOR SOM QUSTIONS ON THIS PRT OF TH XMINTION In this test: ) The ect numericl vlue of the correct nswer does not lwys pper mong the choices given. When this hppens, select from mong the choices the numer tht est pproimtes the ect numericl vlue. ) Unless otherwise specified, the domin of function f is ssumed to e the set of ll rel numers for which f () is rel numer. 76. The grph of function f is shown elow. y Which of the following sttements out f is flse? f is continuous t =. f hs reltive mimum t =. = is in the domin of f. lim f is equl to lim f lim f eists 77. Let f e the function given y f 3e 3 nd let g e the function given y g 6. t wht vlue of do the grph of f nd g hve prllel tngent lines?
8 78. The grphs of the derivtives of the functions f, g, nd h re shown elow. y y y f ' y g' y y h' Which of the functions f, g, or h hve reltive mimum on the open intervl < <? f only g only h only f nd g only f, g, nd h cos 79. The first derivtive of the function f is given y f '. How mny criticl vlues does f hve on the 5 open intervl (0, 0)? One Three Four Five Seven 80. Let f e the function given y f. Which of the following sttements out f re true? I. f is continuous t = 0. II. f is differentile t = 0. III. f hs n solute minimum t = 0. I only II only III only I nd III only II nd III only
9 8. If 0, then lim 4 4 = 6 0 noneistent 4 8. Which of the following is n eqution of the line tngent to the grph of f t the point where f '? y = 8 5 y = + 7 y = y = 0. y = If g is differentile function such tht g () < 0 for ll rel numers nd if f ' 4g following is true? f hs reltive mimum t = nd reltive minimum t =. f hs reltive minimum t = nd reltive mimum t =. f hs reltive minim t = nd t =. f hs reltive mim t = nd t =., which of the It cnnot e determined if f hs ny reltive etrem.
10 84. Let f e function tht is differentile on the open intervl (, 0). If f 5, f 5 5, nd f 9 5, which of the following must e true? I. f hs t lest zeros. II. The grph of f hs t lest one horizontl tngent. III. For some c, < c < 5, f (c) = 3. None I only I nd II only I nd III only I, II, nd III Solutions: Prt : Prt :
11 lculus Section II PRT GRPHING LULTOR IS RQUIR FOR SOM PROLMS OR PRTS OF PROLMS. n isosceles tringle, whose se is the intervl from (0, 0) to (c, 0), hs its verte on the grph of f. For wht vlue of c does the tringle hve mimum re? y y (0, 0) (c, 0) Justify your nswer:
12 . Given the following tle of vlues t = nd =, find the indicted derivtives in prts l. f f ' g g' d f g d ) 3 d g) g f d d ) f g d d h) g g d c) d f d g d i) f g 46 d d) d g d f d j) g 3 d d e) f g d d k) f d d f) f g d l) d f d
13 for 3. Let f e function defined y f. k p for ) For wht vlues of k nd p will f e continuous nd differentile t =? ) For the vlues of k nd p found in prt, on wht intervl or intervls is f incresing? c) Using the vlues of k nd p found in prt, find ll points of inflection of the grph of f. Support your conclusion.
14 4. onsider the curve defined y y y 7. ) Write n epression for the slope of the curve t ny point (, y). ) etermine whether the lines tngent to the curve t the intercepts of the curve re prllel. Show the nlysis tht leds to your conclusion. c) Find the points on the curve where the lines tngent to the curve re verticl.
15 5. The figure elow shows the grph of f ', the derivtive of function f. The domin of the function is the set of ll such tht 3 3. y y f ' Note: This is the grph of the derivtive of f, not the grph of f. ) For wht vlues of, 3 < < 3, does f hve reltive minimum? reltive mimum? Justify your nswer. ) For wht vlues of is the grph of f concve up? Justify your nswer. c) Use the informtion found in prts nd nd the fct tht f ( 3) = 0 to sketch possile grph of f on the es provided elow.
16 3 6. Find the vlue of c tht stisfies the Men Vlue Theorem for f 5 over [ 4, ] n oject moves long the -is with velocity vt t t 5t where t 0,. ) When is the oject stopped? Justify your nswer. ) When is the oject moving right? Justify your nswer. c) Wht is the ccelertion t time t =.3? Show ll your work. d) When is the oject speeding up? Justify your nswer.
17 8. Find the derivtive of the given function. ) g( ) = log ( ) 3 4 ) f tn = d) ( ) = ( 5 6) c) y cot ( ln(5 ) ) 3 cos h e 9. vlute ech limit. If limit N, eplin why. Show ll work ) lim 4 sin(3 ) 0 ) lim e + c) lim 4 + d) lim For the limits in question 9, prts c nd d, drw conclusion out the ehvior of the function s pproches the given vlue. Justify, with clculus, your conclusions.
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MATH SS124 Sec 39 Concepts summary with examples
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BTS and the Wiseman Group found each other through our mutual customers. Our studies of the leaders identified as multipliers revealed some of the unique ways they build collective, viral intelligence in teams. However, most books tend to be quite technical and move quickly from an introduction to more complex applications. What Multipliers Do Differently. Videos, worksheets, 5-a-day and much more The BTS-Wiseman Group Partnership. Exercise $$\PageIndex{1}$$ Use the method of Lagrange multipliers to find the maximum value of $f(x,y)=9x^2+36xy−4y^2−18x−8y \nonumber$ subject to the constraint $$3x+4y=32.$$ Hint. 1. Her work, including her latest books Multipliers How the Best Leaders Make Everyone Smarter and The Multiplier Effect: Tapping the Genius Inside Our School. These include team leaders, project managers, and other heads or managers of business or groups. Use the problem-solving strategy for the method of Lagrange multipliers. However, it will also be valuable for any professional or individual in leadership position. However, most books tend to be quite technical and move quickly from an introduction to more complex applications. The book also provides tips and worksheets for leaders on how to develop these five practices; effective multipliers excel in at least three of these practices. In gearing up for ASTD 2013 and her keynote, I got really excited about the concept of leadership multipliers—what kinds of … Answer. Liz Wiseman is a bestselling author, speaker and executive advisor. The 30-Day Multiplier Challenge. By contrast, this guidebook uses a series of hands-on exercises to gradually introduce SAMs and multiplier analysis. Don’t try to change everything at once. Another reason this book is great is that the authors set out a challenge for their readers, suggesting you focus on a single discipline for thirty days. The Corbettmaths Textbook Exercise on Increasing/Decreasing by a Percentage. Multipliers is a book for organization leaders such as CEOs and other C-suite executives.
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# Solve a problem with a condition. The economy was sent to vegetable warehouses by 3 cars with cabbage in the first day and 8 same loaded cars in the second day. In the second day is sent to 15 t of cabbage more. How many tons of cabbage are sent in the second day?
391
It is necessary to learn how many tons of cabbage are located in one car. Let's accept this size for x. In the first day took away 3 tons. In the second day took away the 8th tons, and it is 15 more, than in the first day. Let's work out the equation: the 8th - 15 = 3 5th = 15 x = 3 (cabbage tons in one car) In the second day were taken away by 8 cars, and these are 8 * 3 = 24 tons. Answer: 24
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Cody
# Problem 411. Back to basics 21 - Matrix replicating
Solution 1164636
Submitted on 17 Apr 2017 by Abhishek Sahoo
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
x = [1]; y_correct = [1 1;1 1]; assert(isequal(matrix_replication(x),y_correct))
2 Pass
x = [1 2;3 4]; y_correct = [1 2 1 2; 3 4 3 4; 1 2 1 2; 3 4 3 4]; assert(isequal(matrix_replication(x),y_correct))
3 Pass
x = [1 2]; y_correct = [1 2 1 2; 1 2 1 2]; assert(isequal(matrix_replication(x),y_correct))
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# Logical Reasoning :: Letter Series Problems And Answers
17. Q1F, S2E, U6D, W21C, ?
A. Z88B B. Y66B C. Y88B D. Y44B
Explanation:
The first letter moves two steps forward. The last letter moves one step backward. The number series runs as below:
1 x 1 + 1 = 2
2 x 2 + 2 = 6
6 x 3 + 3 = 21
21 x 4 + 4 = 88.
Who are all can get the advantages from this Letter Series Question and Answers section?
Those are all planning for any competitive examinations can use this segment to enhance their abilities.
• L.I.C/ G. I.C Competitive Exams
• Career Aptitude Test (IT Companies) and etc.
• Defence Competitive Exams
• Common Aptitude Test (CAT)
• UPSC Competitive Exams
• Bank Competitive Exam
• SSC Competitive Exams
• Railway Competitive Exam
• University Grants Commission (UGC)
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# Math Cheat Sheet Free Homeschool Resource For Grades Measuring Worksheet Freebie Metric Conversion
Maryl Flavie October 21, 2020 Math Worksheet
I believe in the importance of mathematics in our daily lives and it is critical that we nurture our kids with a proper math education. Mathematics involves pattern and structure; it’s all about logic and calculation. Understanding of these math concepts are also needed in understanding science and technology. Learning math is quite difficult for most kids. As a matter of fact, it causes stress and anxiety to parents. How much stress our kids go through? Parents and teachers are aware of the importance of math as well as all of the benefits. Taken in the account how important math is, parents will do whatever it takes to help their struggling children to effectively manage math anxiety. By using worksheets, it can play a major role in helping your kids cope with these stressful. This is a good way to show our children that practicing their math skills will help them improve. Here are some of the advantages using math and worksheets.
First, the Basics! The x axis of a graph refers to the horizontal line while the y axis refers to the vertical line. Together these lines form a cross and the point where they both meet is called the origin. The value of the origin is always 0. So if you move your pencil from the origin to the right, you are drawing a line across the positive values of the x axis, i.e., 1, 2, 3 and so on. From the origin to the left, you’re moving across the negative values of the x axis, i.e., -1, -2, -3 and so on. If you go up from the origin, you are covering the positive values of the y axis. Going down from the origin, will take you to the negative values of the y axis.
A lot of the websites charge high dollar amount for these math worksheets. Considering your willingness to spend the high dollar, you run in to a new problem – the uniqueness of the material, a lot of websites offer sheets packed with identical problems leading the child to almost rote memorization of the problems and solution. This leads to deprivation of student and his/her ability to solve the problems logically. And some websites offer sheets in portable document format. And to open such sheets you are forced to download special software.
Once downloaded, you can customize the math worksheet to suit your kid. The level of the child in school will determine the look and content of the worksheet. Use the school textbook that your child uses at school as a reference guide to help you in the creation of the math worksheet. This will ensure that the worksheet is totally relevant to the kid and will help the child improve his or her grades in school. The math worksheet is not only for the young children in kindergarten and early primary school; they are also used for tutoring high school and university students to keep the students’ math skills sharp. The sites that offer these worksheets have helped a lot and this resource is now a common thing to use for all kinds and levels of educators. The formats for the worksheets differ according to the level and content of the worksheets. For the young kids it is preferable to have the worksheet in large print, while the older students commonly use the small print ones that are simple and uncluttered.
### Math Course 1 Printable Worksheets
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I remember that with my Mom everything was somehow connected to math. She made me count the buttons in my shirt as she dressed me up, asked questions that demanded answers that are related to sums, like how many pair of shoes do you have? How many buttons are there on your Daddy’s shirt? Count all the furniture in the living room and several math games. All my toys were one way or the other math related. I had puzzles, and tons of things Mom had me do as games on daily basis at home to get me ready for kindergarten! In fact, she continued guiding me towards being math friendly throughout kindergarten and first grade during which time 1st grade math worksheets was my constant companion.
The use of math worksheets can help solve numerous arithmetic problems. ”Practice makes an individual perfect,” is the best motto to be kept in mind while studying math. The motto will help a person to reinforce his desire to better himself in the subject. Without the help of these online resources, one will not be able to achieve the mastery of math. Since education is one of those areas which receive little or no funding from the government, it is essential for parents to look for various options that can help give their child a better education. Some sites do offer math online quiz that is sure to bring about an inclination towards math among children. It is a common practice for parents around the world to send their children to special math training centers. Invariably, every parent is unaware of the actual quality of training provided by these centers. To help parents combat this problem, there are a lot of online resources available that offer math assignment help exclusively for children.
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# Find the smallest number among the following choices.
Welcome to Warren Institute, your go-to source for all things Mathematics education! In this article, we will delve into the intriguing question: "Which of the following numbers is the smallest?" Join us as we explore the concept of comparing numbers and discover clever techniques to identify the smallest number in a set. Whether you're a student looking to enhance your problem-solving skills or an educator seeking innovative teaching strategies, this article will provide valuable insights. So, put on your thinking caps and get ready to dive into the fascinating world of number comparison!
## Comparing Numbers: Determining the Smallest Number
When it comes to comparing numbers in mathematics education, it is essential to understand how to determine the smallest number among a given set. This involves analyzing the values of each number and identifying the one with the least value. To make this process easier, students can use comparison symbols such as "<" (less than) or ">" (greater than) to indicate the relationship between numbers. By carefully evaluating the numerical values, students can confidently identify the smallest number.
## Strategies for Identifying the Smallest Number
There are several strategies that students can employ to identify the smallest number among a given set. One effective approach is to arrange the numbers in ascending order from least to greatest. This allows students to visually compare the values and quickly determine the smallest number. Alternatively, students can utilize number lines, which provide a visual representation of the numerical sequence. By plotting the numbers on a number line, students can easily identify the smallest number based on its position towards the left end of the line.
## Understanding Place Value: a Key Factor in Determining the Smallest Number
Understanding place value is crucial in determining the smallest number among a set of numbers. Place value refers to the value of a digit within a number based on its position. By recognizing the significance of each digit's position, students can accurately compare numbers and identify the one with the smallest value. For instance, in the number 325, the digit "3" holds more value than the digit "2" because it is in the hundreds place. This knowledge helps students differentiate between the numbers and determine which one is the smallest.
## Real-World Applications: Importance of Identifying the Smallest Number
The ability to identify the smallest number is not only essential in mathematics education but also in real-world applications. In various scenarios, such as budgeting, comparison shopping, or analyzing data, individuals need to determine the smallest value among a set of numbers. By applying the skills learned in mathematics education, individuals can make informed decisions and choose the most cost-effective option or identify the least significant data point. Understanding how to find the smallest number is a valuable skill that can be applied in various practical situations.
## frequently asked questions
### Which of the following numbers is the smallest: 0, 5, 10, or 15?
The smallest number among 0, 5, 10, and 15 is 0.
### In the set {2, 4, 6, 8}, which of the following numbers is the smallest?
The smallest number in the set {2, 4, 6, 8} is 2.
### Among the numbers -3, -2, -1, and 0, which one is the smallest?
-3 is the smallest number among -3, -2, -1, and 0.
### Consider the fractions 1/4, 1/3, 1/2, and 2/
In Mathematics education, the fractions 1/4, 1/3, 1/2, and 2/3 are commonly taught as foundational concepts. They represent different parts of a whole or a group and are important for understanding fractions, decimals, and percentages. These fractions are often used in various problem-solving activities and real-life contexts to develop students' mathematical reasoning and critical thinking skills.
### Which of these numbers is the smallest?
The smallest number among these options would be the one with the lowest value.
### Given the decimal numbers
Given the decimal numbers, it is important to understand their place value and how to perform operations such as addition, subtraction, multiplication, and division.
### 1,
1 Mathematics education refers to the teaching and learning of mathematics in schools and educational institutions.
### 01,
01 In Mathematics education, the term "01" refers to the representation of the number one in binary code.
### 001, and
001, in the context of Mathematics education, refers to a specific topic or concept being discussed or taught.
### 0001, which one is the smallest?
0001 is the smallest number in the context of Mathematics education.
In conclusion, understanding the concept of comparing numbers is crucial in Mathematics education. By identifying which of the following numbers is the smallest, students can develop their critical thinking skills and enhance their mathematical reasoning abilities. This fundamental skill lays the foundation for more complex mathematical concepts and problem-solving strategies. Moreover, teachers play a vital role in guiding students towards mastering this skill by incorporating interactive and engaging activities into their lesson plans. By providing ample practice opportunities and utilizing visual aids, such as number lines or manipulatives, educators can effectively support students' understanding of numerical comparisons. Ultimately, by emphasizing the importance of comparing numbers, we can help students build a strong mathematical foundation and foster a lifelong love for learning.
See also Master fractions with our free PDF worksheet on adding and subtracting fractions with unlike denominators!
If you want to know other articles similar to Find the smallest number among the following choices. you can visit the category General Education.
Michaell Miller
Michael Miller is a passionate blog writer and advanced mathematics teacher with a deep understanding of mathematical physics. With years of teaching experience, Michael combines his love of mathematics with an exceptional ability to communicate complex concepts in an accessible way. His blog posts offer a unique and enriching perspective on mathematical and physical topics, making learning fascinating and understandable for all.
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# Please help me to resolve that sevent questions.Many people are interested in losing weight through
exercise. An important fact to consider is that a person
needs to burn off 3,500 calories more than he or she takes
in to lose 1 pound, according to the American Dietetic
Association.
The following table shows the number of calories burned per hour (cal/h) for a variety
of activities, where the figures are based on a 150-pound person.
Chart:
Activity Cal/h
Bicycling 6 mi/h 240
Bicycling 12 mi/h 410
Cross-country skiing 700
Jogging 5 1/2 mi/h 740
Jogging 7 mi/h 920
Jumping rope 750
Running in place 650
Running 10 mi/h 1280
Swimming 25 yd/min 275
Swimming 50 yd/min 500
Tennis (single) 400
Walking 2 mi/h 240
Walking 3 mi/h 320
Walking 4 1/2 mi/h 440
Work with your group members to solve the following problems. You may find that
setting up proportions is helpful.
For problems 1 through 4, assume a 150-pound person.
1. If a person jogs at a rate of 5 1/2 mi/h for 31/2 h in a week, how many calories do they
burn?
2. If a person runs in place for 15 minutes, how many calories will be burned?
3. Iersoncross-countryskis for35minutes,howmanycalories willbeburned?
4. How many hours would a person have to jump rope in order to lose 1 pound? (Assume
calorie consumption is just enough to maintain weight, with no activity.)
Heavier people burn more calories (for the same activity), and lighter people burn
fewer. In fact, you can calculate similar figures for burning calories by setting up the
appropriate proportion.
5. At what rate would a 120-pound person burn calories while bicycling at 12 mi/h?
6. At what rate would a 180-pound person burn calories while bicycling at 12 mi/h?
7. How many hours of jogging at 5 mi/h would be needed for a 200-pound person to
lose 5 pounds? (Again, assume calorie consumption is just enough to maintain weight,
with no activity.)
1
2
1
2
1
2
Activity Cal/h Activity Cal/h
Bicycling 6 mi/h 240 Running 10 mi/h 1,280
Bicycling 12 mi/h 410 Swimming 25 yd/min 275
Cross-country skiing 700 Swimming 50 yd/min 500
Jogging 5 mi/h 740 Tennis (singles) 400
Jogging 7 mi/h 920 Walking 2 mi/h 240
Jumping rope 750 Walking 3 mi/h 320
Running in place 650 Walking 4 mi/h 440
1
2
1
2
Question 7 is 5 1/2 not 5
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# Sudoku puzzle solver using awk
awk is a fun and powerful language available in the command-line of Linux, Mac OS X, and even Windows (with a little help from Cygwin). In fact, our own Dale Dougherty co-authored one of the classic books on awk (and sed, another great Unix power tool) back in 1990 (the second edition was released in 1997), sed & awk.
At The Geek Stuff, Bill Duncan has posted a fun awk program that can solve Sudoku puzzles:
The application I chose to use as an example is “yet another sudoku puzzle solver”. I must confess at the outset that I have never sat down to solve one of these puzzles myself, but sketched this out over a few days while commuting on a train and watching other people work on them. It was far more fun I think than actually doing any of the puzzles..
[…]
This program uses a very simple depth-first recursive backtracking algorithm with up-front and ongoing elimination of invalid entries. Awk may not have the expressive power for representing complex data that perl or other languages have, but with care, many moderate sized problems and data sets can be used. This algorithm may not be the best one around, but it is certainly fast enough for most problems and is easy to implement.
When you strip out blank lines and comments it’s only 67 lines! Keeping in the awk spirit, that would be:
`awk '!/^[ t]*#/ && !/^\$/' solve.awk | wc -l`
Yet Another Sudoku Puzzle Solver Using AWK
If you need to generate some puzzles to throw at it, try this Sudoku Generator written in Python.
## 14 thoughts on “Sudoku puzzle solver using awk”
1. charlie says:
you don’t need to clutter up windows by adding cygwin, there are a few native ones, gnuish, unxutils, even a native port of the one true awk, as well as lookalikes like gawk.
Features vary.
2. Brian Jepson says:
Thanks, Charlie; I usually install cygwin on many of my machines, but true, it’s not the simplest way to get your hands on awk.
3. darkonn says:
Oh, i like sudoku! I posted a small review of the sudoku on my blog a week or 2 ago so. You can read it by visiting: http://ambienbuy.net
4. David Wong says:
5. Ramesh Dahal says:
You may use harder puzzle visit: http://www.ambienbuy.org/
6. larrywall says:
hehe yandex
7. larrywall says:
hehe yandex
Discuss this article with the rest of the community on our Discord server!
Tagged
### Brian Jepson
I'm a tinkerer and finally reached the point where I fix more things than I break. When I'm not tinkering, I'm probably editing a book for Maker Media.
View more articles by Brian Jepson
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# Why is the display resolution not exactly 16:9 or 4:3?
Most displays get advertised with either 16:9 or 4:3 display ratio. However, if you compare the resolution with the display ratio, it's most often neither of both.
For example, the resolution of my notebook display is 1366x768.
But 1366/768 = 683/384 != 688/387 = 16/9
Another common resolution is 1920/1200 = 8/5
But for some resolutions it's correct:
• 800/600 = 4/3
Is there a technical reason / user experience reason for this? Why do displays have other ratios than what they get advertised?
(I assume that every pixel is a perfect square. Is this assumption wrong?)
• <irony>In love and in advertising lying is not only allowed, but expected...</irony>;
– lexu
Commented Jan 6, 2013 at 10:41
• I would say 1366 x 768 is close enough 16:9. To be 16:9 exactly, it would have to be 1365 1/3 x 768 or 1366 x 768 3/8. Commented Jan 6, 2013 at 20:39
• @moose Which is a horrible choice because the height is not an even number. This breaks a lot of applications, in particular graphics rendering, where many techniques, both software and hardware, implicitly require height in pixels to be a multiple of two. Commented Jan 10, 2013 at 4:04
• @Thomas I didn't know that. This is the kind of answer I've expected to get with this question. Can you tell me an example for such a technique? Commented Jan 10, 2013 at 8:50
• Are you sure that the pixels are square? You might actually have the exact ratio but not realise it... Commented Apr 9, 2019 at 16:52
Not every display resolution has to be 16:9 or 4:3.
My laptop and my TV have the well known 16:9 ratio.
My regular display has 16:10, at least they are marketed as 16:10, however the image below has them as 8:5. The broken screen that still sits on top of the locker behind me has a resolution of 5:4.
The image below shows most of the standard resolutions that are available.
I actually like 16:10 more than 16:9 and would pay a fair amount more money to get one of these instead. This however is personal opinion but should exemplary show you why there are not only two but a lot more standards to choose from.
Why do I like it so much? Not all movies are 16:9, there are a lot of 4:3 shows out.
When playing games I like it more to have a bit more vertical space to place menus, HUDs etc.
This of course comes down to personal preference. Personal preference between individuals is different and so are displays.
Why are displays marketed as 16:9 if they are not?
If this is done knowingly, I'd call that a scam.
• Great & comprehensive answer. Re: your comment "marketed as 16:10, however the image below has them as 8:5", those are, obviously, the same ratio. I reckon calling it 16:10 is market-speak to make it sound similar to (and comparable to) the common 16:9 aspect ratio. Commented Jan 6, 2013 at 17:11
• -1: Using the picture of CRT phosphor dots to say "pixels aren't squares" is misleading. Pixels refers to the rectangular grid of picture elements in the computer video memory. This doesn't always line up with the color dots on the screen. A CRT monitor doesn't line up pixels to phosphor dots in any particular way, so on a CRT, a pixel is usually not the same as a trio of phosphor dots. An LCD screen is able to line up a pixel with a trio of color elements exactly (when you use the LCD's native resolution). Subpixel methods only work well on LCD screens because of this. ... Commented Jan 7, 2013 at 0:07
• ... I drew some white pixels on a black background in Paint, then photographed them on a CRT at its best resolution, then an LCD at its best resolution: image. I don't know how to take macro pictures well, so the LCD color elements are blured together, but you should be able to get the idea. ... Commented Jan 7, 2013 at 0:08
• I rolled back the revision that suggested the picture of the phosphor dots and LCD segments illustrated non-square pixels, and removed my downvote. Commented Jan 10, 2013 at 3:05
• this doesn't explain why the resolution is not exactly 16:9 or 4:3 Commented Aug 9, 2014 at 19:32
The exact ratio can only be obtained if the denominator is divisible by the denominator of the aspect ratio you want. 768 isn't divisible by 9, so there won't be any 16:9 integer resolution with that height. So why wasn't 1360:765 chosen?
Because dimensions of display resolutions tend to be a power of 2 (or a multiple of a power of 2 that is as large as possible), possibly because powers of 2 work better for a binary computer
• 2D image formats as well as video codecs process the images in blocks instead of pixel-by-pixel individually or line line-by-line. The block sizes are always powers of 2 like 8×8, 16×16 or less frequently 4×8, 8×16, 4×16 because they're easier to arrange in memory, and also more suitable for the CPU's SIMD unit... That's why you'll see blocky artifacts when viewing a low quality image or video file.
• 3D graphics renderers often use a technique called mipmapping that involves using images with sizes that are powers of 2 of each other, to increase rendering speed and reduce aliasing artifacts. If you're interested, check out How does Mipmapping improve performance?
So regardless of the graphics type, using powers of 2 eases the job of the encoder/decoder and/or GPU/CPU. Images with a non-power-of-2 side length will always have the corresponding side rounded up to a power of 2 (which you'll see later on the example of 1920×1080) and you'll end up wasting some memory at the edges for storing those dummy pixels. Transforming those odd-sized images like that also introduces artifacts (which are sometimes unavoidable) due to the dummy values. For example rotating odd-sized JPEGs will introduce noise to the result
Rotations where the image is not a multiple of 8 or 16, which value depends upon the chroma subsampling, are not lossless. Rotating such an image causes the blocks to be recomputed which results in loss of quality.
https://en.wikipedia.org/wiki/JPEG#Lossless_editing
See
Now obviously 1360:765 is precisely 16:9 as you said, but 765 isn't divisible by any power of 2, while 768 can be divisible by 256 (28), so 768 for the height is a better choice. Moreover using 768 as the height has the advantage of being able to display the old 1024×768 natively without scaling
`768/(16/9) = 1365.333...`, so if you round it down, you'll get a value that's closest to 16:9. However it's an odd value, so people round it up to 1366×768, which is quite close to 16:9. But again, 1366 is only divisible by 2 so some screen manufacturers use 1360×768 instead since 1360 is divisible by 16 which is much better. 1360/768 = 1.7708333... which approximates 16/9 to about 2 decimal places, and that's enough. 1360×768 also has the bonus that it fits nicely inside 1MB of RAM (whereas 1366×768 doesn't). 1344×768, another less commonly used resolution, is also divisible by 16.
WXGA can also refer to a 1360×768 resolution (and some others that are less common), which was made to reduce costs in integrated circuits. 1366×768 8-bit pixels would take just above 1-MiB to be stored (1024.5KiB), so that would not fit into an 8-Mbit memory chip and you would have to have a 16-Mbit memory chip just to store a few pixels. That is why something a bit lower that 1366 was chosen. Why 1360? Because you can divide it by 8 (or even 16) which is far simpler to handle when processing graphics (and could bring to optimized algorithms).
Why Does the 1366×768 Screen Resolution Exist?
Many 12MP cameras have effective resolution of 4000×3000, and when shooting in 16:9, instead of using the resolution 4000×2250 which is exactly 16:9, they use 4000×2248 because 2248 is divisible by 8 (which is the common block size in many video codecs), and 2250 is divisible by 2.
Some Kodak cameras use 4000×2256 too, since 2256 is divisible by 16, and 4000/2256 still approximates 16/9 to about 2 decimal places. If shooting in 3:2 they'll use 4000×2664, not 4000×2667 or 4000×2666 which are closer to 3:2, for the same reason.
Another example is 848×480 (divisible by 16) or 854×480 (divisible by 4) instead of 853×480
The 480 denotes a vertical resolution of 480 pixels, usually with a horizontal resolution of 640 pixels and 4:3 aspect ratio (480 × 4⁄3 = 640) or a horizontal resolution of 854 or less (848 should be used for mod16 compatibility) pixels for an approximate 16:9 aspect ratio (480 × 16⁄9 = 853.3).
https://en.wikipedia.org/wiki/480p
And this is true for other resolutions too. You'll almost never find any image resolutions that are odd. Most will be at least divisible by 4 - or better, 8. The full HD resolution, 1920×1080, has a height not divisible by 16, so many codecs will round it up to 1920×1088 instead, with 8 dummy lines of pixels, then crop it down when displaying or after processing. But sometimes it's not cropped so you can see there are many 1920×1088 videos on the net. Some files are reported as 1080 but actually 1088 inside.
You may also find the option to crop 1088 to 1080 in various video decoder's settings.
1080-line video is actually encoded with 1920×1088 pixel frames, but the last eight lines are discarded prior to display. This is due to a restriction of the MPEG-2 video format, which requires the height of the picture in luma samples (i.e. pixels) to be divisible by 16.
https://en.wikipedia.org/wiki/ATSC_standards#MPEG-2
Back to your example 1920/1200 = 8/5, it's not strange at all because it's the common 16:10 aspect ratio that is close to the golden ratio. You can find it in 1280×800, 640×400, 2560×1600, 1440×900, 1680×1050... No one would advertised it as 16:9 because they're clearly 16:10
I assume that every pixel is a perfect square. Is this assumption wrong?
That's wrong. In the past pixels are often not a square but a rectangular shape. Other pixel arrangements like hexagon do exist although not very common. See Why are pixels square?
Yeah, it's to do with manufacturing.
We already made loads of 1024x768 panels, so why not just make them wider so they are 1366x768.
I'm not sure about the other one, I haven't come across panels with that resolution.
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"Kbar" redirects here. For the knife, see KA-BAR.
An aluminium cylinder
(5 mm or 0.197 in thickness) after 700 bar (10,153 psi) pressure.
The bar is a metric (but not SI) unit of pressure, defined by the IUPAC as exactly equal to 100,000 Pa.[1] It is about equal to the atmospheric pressure on Earth at sea level, and since 1982 the IUPAC has recommended that the standard for atmospheric pressure should be harmonized to 100,000 Pa = 1 bar ≈ 750.0616827 Torr.[2] The same definition is used in the compressor and the pneumatic tool industries (ISO 2787).
The bar and the millibar were introduced by the British meteorologist William Napier Shaw in 1909, while he was the director of the Meteorological Office in London.[3]
Units derived from the bar are the megabar (symbol: Mbar), kilobar (symbol: kbar), decibar (symbol: dbar), centibar (symbol: cbar), and millibar (symbol: mbar or mb). These are not SI or cgs units, but they are accepted by the BIPM for use with the SI.[4] The bar is legally recognized in countries of the European Union.[5]
The bar unit is considered deprecated by some entities. While the BIPM includes it under the class "Non-SI units accepted for use with the SI",[4] the NIST includes it in the list of units to avoid and recommends the use of kilopascals (kPa) and megapascals (MPa) instead.[6] The IAU also lists it under "Non-SI units and symbols whose continued use is deprecated."[7]
Bar(g) is a unit of gauge pressure, i.e., pressure in bars above ambient or atmospheric pressure; see absolute pressure and gauge pressure below.
## Definition and conversion
The bar is defined using the SI unit pascal, namely: 1 bar100,000 Pa. 1 bar is therefore equal to:
• 100 kPa (in SI units)
• 1×105 N/m2 (alternative representation in SI units)
• 1,000,000 dyn/cm2 (barye) (in cgs units)
• 0.987 atm
• 14.5038 psi absolute
• 29.53 inHg
• 750.06 mmHg
• 750.06 torr
• 1,019.72 cmH2O
## Origin
The word bar has its origin in the Greek word βάρος (baros), meaning weight. The unit's official symbol is bar; the earlier symbol b is now deprecated, and conflicts with the use of b as a unit symbol to denote the barn, but it is still encountered, especially as mb (rather than the proper mbar) to denote the millibar.
## Usage
Atmospheric air pressure is often given in millibars where standard sea level pressure is defined as 1000 mbar, 100 (kPa), or 1 bar. This should be distinguished from the now deprecated unit of pressure, known as the "atmosphere" (atm), which is equal to 1.01325 bar. Despite the millibar not being an SI unit, meteorologists and weather reporters worldwide have long measured air pressure in millibars as the values are convenient. After the advent of SI units, some meteorologists began using hectopascals (symbol hPa) which are numerically equivalent to millibars; for the same reason, the hectopascal is now the standard unit used to express barometric pressures in aviation in most countries. For example, the weather office of Environment Canada uses kilopascals and hectopascals on their weather maps.[8][9] In contrast, Americans are familiar with the use of the millibar in US reports of hurricanes and other cyclonic storms.[citation needed]
In fresh water, there is an approximate numerical equivalence between the change in pressure in decibars and the change in depth from the water surface in metres. Specifically, an increase of 1 decibar occurs for every 1.019716 m increase in depth. In sea water with respect to the gravity variation, the latitude and the geopotential anomaly the pressure can be converted into meters depth according to an empirical formula (UNESCO Tech. Paper 44, p. 25).[10] As a result, decibars are commonly used in oceanography.
Many engineers worldwide use the bar as a unit of pressure because, in much of their work, using pascals would involve using very large numbers.
In the automotive field, turbocharger boost is often described in bars in the metric part of the world (i.e. outside the USA).
Unicode has characters for "mb" (, U+33D4) and "bar" (, U+3374), but they exist only for compatibility with legacy Asian encodings and are not intended to be used in new documents.
The kilobar, equivalent to 100 MPa, is commonly used in geological systems, particularly in experimental petrology.
## Absolute pressure and gauge pressure
Bourdon tube pressure gauges, vehicle tire gauges, and many other types of pressure gauges are zero-referenced to atmospheric pressure, which means that they measure the pressure above atmospheric pressure (which is around 1 bar); this is gauge pressure and is often referred to in writing as barg or bar(g), spoken as "bar gauge". In contrast, absolute pressures are zero-referenced to a complete vacuum and when expressed in bars are often referred to as bara or bar(a). Thus, the absolute pressure of any system is the gauge pressure of the system plus atmospheric pressure. The usage of bara and barg is now deprecated, with qualification of the physical property being preferred, e.g., "The gauge pressure is 2.3 bar; the absolute pressure is 3.3 bar".[5]
In the United States, where pressures are still often expressed in pounds per square inch (symbol psi), gauge pressures are referred to as psig and absolute pressures are referred to as psia. Gauge pressure is also sometimes spelled as gage pressure.
Sometimes, the context in which the word pressure is used helps to identify it as meaning either the absolute or gauge pressure. However, for best practice, whenever a pressure is expressed in any units (bar, Pa, psi, atm, etc.), it should be denoted in some manner as being either absolute or gauge pressure to avoid any possible misunderstanding. One recommended way of doing so is to spell out what is meant, for example as bar gauge or kPa absolute.[11]
## References
1. ^ IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "IUPAC Gold Book - bar".
2. ^ IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "IUPAC Gold Book - standard pressure".
3. ^ Sir William Napier Shaw
4. ^ a b International Bureau of Weights and Measures (2006), The International System of Units (SI) (8th ed.), p. 127, ISBN 92-822-2213-6.
5. ^ a b British Standard BS 350:2004 Conversion Factors for Units
6. ^ NIST Special Publication 1038, Sec. 4.3.2 NIST Special Publication 811, 2008 edition, Sec. 5.2
7. ^ International Astronomical Union Style Manual. Comm. 5 in IAU Transactions XXB, 1989, Table 6
8. ^ Environment Canada Weather Map
9. ^ Weather - Environment Canada
10. ^ http://unesdoc.unesco.org/images/0005/000598/059832eb.pdf
11. ^ FAQ (from the website of the National Physics Laboratory, United Kingdom)
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7 news items
Chattanooga Times Free Press NRC chairwoman sees 'no show stoppers' at Watts Bar nuclear plant Chattanooga Times Free Press Tue, 24 Jun 2014 21:02:21 -0700 TVA's critical path toward starting the Watts Bar Unit 2 still requires regulators to also adopt a court-accepted plan for storing or disposing of highly radioactive wastes generated at nuclear plants. Macfarlane said the NRC will present such a plan ... Environment & Energy Publishing In 'Fukushima Building,' TVA tests a new age of nuclear construction Environment & Energy Publishing Tue, 22 Jul 2014 07:44:46 -0700 "All of this can be used to prevent a Fukushima event," said Bob Williams, the Fukushima project manager for Watts Bar Unit 2. The FLEX buildings are the nuclear industry's answer to the Nuclear Regulatory Commission's safety directives that followed ... Chattanooga Times Free Press WATCH: 185000 gallons pour into TVA nuclear reactor in this time-lapse video Chattanooga Times Free Press Thu, 10 Jul 2014 05:37:49 -0700 TVA filled the new reactor vessel in the Watts Bar Nuclear Plant reactor being built near Spring City, Tenn., this week with 185,000 gallons of water in just over 23 minutes. The flow filled 18 feet of water into a space 60 feet long and 22 feet wide ... New York Times On Britain's South Coast, 'Playground' for the Rich New York Times Thu, 10 Jul 2014 03:52:30 -0700 There also is an open-plan living area, with a large kitchen and Deco-style bar unit made of maple wood standing in the center. “It's top of the range because we thought we would be here for good,” said Jimmy James, who, with his wife, Theresa, is ... Chattanooga Times Free Press Carbon limits could boost nuclear power, former EPA head says Chattanooga Times Free Press Thu, 26 Jun 2014 20:56:15 -0700 At the Watts Bar Unit 2 reactor under construction, for instance, 3,200 construction workers are now on site and the average pay for those in the nuclear industry is 36 percent above the U.S. median, Whitman said. While nuclear power will add jobs, a ... MMA Corner ONE FC's Rob Lisita: No Doubts This Will Be First of Many Main Events MMA Corner Mon, 07 Jul 2014 15:02:08 -0700 Rob would like to thank Haleo, Muscle Bar, Unit 27, Phuket Top Team, Phuket Pro Nutrition, Vicious Circle and Pokerstars. He would also like to thank God and his family, who believe in him and support him so much. He adds, “I love them with all my heart. Båtliv Italienska Rivas första och största yacht helt byggd i aluminium är en riktig ... Båtliv Mon, 23 Jun 2014 02:08:46 -0700 The first 122' Mythos yacht features a large Jacuzzi, sided by a comfortable aft sun pad and a bar unit to the bow. In front of the bar another spacious C-shaped aft-facing sofa. Steering from the forward flybridge offers perfect viewing of yacht ...
Limit to books that you can completely read online Include partial books (book previews) .gsc-branding { display:block; }
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Home >> AIMS >> Class11 >> Physics
# The pressure exerted by an object on a given surface is 2.4 Pa. The equivalent amount of pressure in CGS unit will be
$\begin {array} {1 1} (A)\;0.024 g \: cm^{−1} s^{−2} & \quad (B)\;0.24 g \: cm^{−1} s^{−2} \\ (C)\;24 g \: cm^{−1} s^{−2} & \quad (D)\;240 g \: cm^{−1} s^{−2} \end {array}$
$24 g \: cm^{−1} s^{−2}$
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Precalculus (6th Edition)
RECALL: The Commutative Property of multiplication states that for any real numbers $a$ and $b$, $a(b) = b(a)$. With $a=8$ and $b=(5+9)$, the given expression illustrates the Commutative Property of Multiplication.
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display | more...
Numbers divisible by the product of their prime factors.
Salacious numbers are rare and fascinating things indeed. Although there exist many superficially similar classes of numbers, such as 'perfect' numbers (equal to the sum of their prime factors), 'friendly' numbers (pairs of numbers equal to the sum of each other's prime factors), 'greedy' numbers (equal to the sum of all prime factors except their own), 'comely' numbers (equal to the product of all of their prime factors that look good in base 23), etc., these have no real mathematical significance and are really just numerological fancies.
Salacious numbers however, are different. They have relevance to several areas of modern research, and have many tantalising properties that make them rewarding, if elusive, beasts to study. For example, the following things are known about them:
• There are infinitely many salacious numbers.
• For any n, there are infinitely many salacious numbers that are multiples of n.
• The proportion of salacious numbers less than n remains roughly constant as n grows (this is Stobbs's Salacious Distribution Theorem).
But lest you think that these are well-understood little integers, the following hypotheses should snap you out of your complacent day-dreaming. None of these are proven, but all have extensive evidence in their favour and few (if any) counter-examples:
• There is at least one salacious number between any two non-consecutive primes.
• Every even number can be written as the sum of two salacious numbers (Gubbon's Conjecture)
• For all n there is a solution to xn + yn = zn where x,y,z are salacious numbers.
An intruiging topic then, but one containing many traps for the casual dummlichter!.
Log in or register to write something here or to contact authors.
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## Everyday Mathematics 4th Grade Answer Key Unit 3 Fractions and Decimals
Sharing Equally
Use drawings to help you solve the problems. Solve each problem in more than one way. Show your work.
Question 1.
Four friends shared 5 pizzas equally. How much pizza did each friend get?
____ pizzas
One way:
Another way:
Explanation :
First Way :
Number of Pizzas = 5
Number of people = 4
Each Person take a pizza, number of pizzas taken = 4 pizzas
Number of pizzas left = 1
Fraction of pizza each person get = $$\frac{1}{4}$$.
Fraction of pizza each person get = 1 + $$\frac{1}{4}$$ = $$\frac{5}{4}$$ .
Second Way :
Number of Pizzas = 5
Number of people = 4
Each pizza is divided into 4 parts .
Each person takes 1 pizza slice .
Fraction of each pizza slice = $$\frac{1}{4}$$
Total Fraction of 5 pizza slices = 5 × $$\frac{1}{4}$$ = $$\frac{5}{4}$$
Therefore , Each person gets $$\frac{5}{4}$$ fraction of pizza that means each person gets 1 slice from 5 pizzas .
it is represented in above figure with different colors .
Question 2.
Five kittens are sharing 6 cups of milk equally. How much milk does each kitten get?
____ cups of milk
One way:
Another way:
One Way :
Number of Kittens = 5
Number of cups of milk = 6
Each kittens gets 1 cup that means 5 cups of milk is given to 5 kittens .
1 cup of milk is left which is shared by 5 kittens .
Fraction of milk given to each kitten in 1 cup = number of cups  ÷ number of kittens = 1 ÷ 5 =
Total cups of milk given each kitten = 1 + $$\frac{1}{5}$$ = $$\frac{5}{5}$$ + $$\frac{1}{5}$$ = $$\frac{6}{5}$$ cups .
Another Way :
Number of Kittens = 5
Number of cups of milk = 6
Each cup of milk is divided into 5 parts .
Each kitten is given $$\frac{1}{5}$$Â cup of milk
As there are 6 cups Each kitten gets 6 × $$\frac{1}{5}$$ cup of milk = $$\frac{6}{5}$$ cups of milk .
Therefore each kitten gets $$\frac{6}{5}$$ cups of milk .
Practice
Question 3.
Name the next 4 multiples of 7. 7, ___, ___, ___, ___
The next 4 multiples of 7 are :
7, 14, 21, 28 .
Question 4.
List all the factors of 18. ____
All factors of 18 are : 1, 2, 3, 6, 9 , 18 .
Question 5.
List all the factors of 18 that are prime.
All factors of 18 are : 1, 2, 3, 6, 9 , 18 .
All factors of 18 that are prime : 1, 2, 3 .
Explanation :
A number that is divisible only by itself and 1 (e.g. 2, 3, 5, 7, 11).
Question 6.
List all the factor pairs of 40.
___ and ___; ___ and ; ___
___ and ___; ___ and ; ___
The pair factors of 40 are (1, 40), (2, 20), (4, 10), and (5, 8).
Fraction Circles
Question 1.
Divide into 4 equal parts. Shade $$\frac{1}{4}$$.
Explanation :
Circle is divided into 4 parts . Each part is a fraction of $$\frac{1}{4}$$ and and 1 part is shaded .
Question 2.
Divide into 8 equal parts. Shade $$\frac{2}{8}$$.
Explanation :
Circle is divided into 8 parts . Each part is a fraction of $$\frac{1}{8}$$ and and 2 parts are shaded .
Question 3.
Divide into 12 equal parts. Shade $$\frac{3}{12}$$.
Explanation :
Circle is divided into 12 parts . Each part is a fraction of $$\frac{1}{12}$$ and and 3 parts are shaded .
Question 4.
Explanation :
Circle is divided into 2 parts . Each part is a fraction of $$\frac{1}{2}$$ and 1 part is shaded which is a fraction of $$\frac{1}{2}$$Â .
Question 5.
What patterns do you notice in Problems 1 through 3?
The pattern we notice in problem 1 through 3 is equivalent fraction .
$$\frac{1}{4}$$ = $$\frac{2}{8}$$ = $$\frac{3}{12}$$ .
Practice
Question 6.
List the next 4 multiples of 5. 20, __, ___, ___, ___
4 multiples of 5 are from 20 , 25, 30, 35, 40 .
Question 7.
List all the factors of 48. __________
All Factors of 48 are : 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
Question 8.
List the factors of 48 that are composite. __________
All Factors of 48 are : 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
composite numbers are :Â 4, 6, 8, 12, 16, 24 and 48.
Explanation :
AÂ composite number is a natural number or a positive integer which has more than two factors. For example, 15 has factors 1, 3, 5 .
Finding Equivalent Fractions
1 . Fill in the blank with = or ≠.
a. $$\frac{2}{3}$$ __ $$\frac{1}{3}$$
$$\frac{2}{3}$$ > $$\frac{1}{3}$$
Explanation :
1. A fraction is larger if it’s farther from 0 on the number line.
2. A fraction is smaller if it’s closer to 0 on the number line.
$$\frac{2}{3}$$ is farther from 0 so, it is greater number .
Therefore, $$\frac{2}{3}$$ > $$\frac{1}{3}$$ .
b. $$\frac{2}{6}$$ ____ $$\frac{1}{3}$$
$$\frac{2}{6}$$ =Â $$\frac{1}{3}$$
Explanation :
$$\frac{2}{6}$$ = $$\frac{1}{3}$$ as both locate at the same point as shown in above figure .
c. $$\frac{2}{6}$$ __ $$\frac{2}{5}$$
$$\frac{2}{5}$$ > $$\frac{2}{6}$$
Explanation :
1. A fraction is larger if it’s farther from 0 on the number line.
2. A fraction is smaller if it’s closer to 0 on the number line.
$$\frac{2}{5}$$ is farther from 0 than $$\frac{2}{6}$$ so, it is a greater number
Therefore, $$\frac{2}{5}$$ > $$\frac{2}{6}$$ .
d. $$\frac{1}{5}$$ ___ $$\frac{2}{10}$$
Explanation :
$$\frac{2}{6}$$ = $$\frac{1}{3}$$ as both locate at the same point as shown in above figure .
Question 2.
Fill in the missing numbers.
a.
$$\frac{1}{5}$$ = $$\frac{2}{10}$$
Explanation :
The fraction $$\frac{1}{5}$$ can be renamed as an infinite number of equivalent fractions.
When you multiply the numerator 1 by 2, the result is 2. When you multiply the denominator 5 by 2, the result is 10.
$$\frac{1 \times 2}{5 \times 2}$$ = $$\frac{2}{10}$$
This results in the number sentence $$\frac{1}{5}$$ = $$\frac{2}{10}$$.
b.
$$\frac{4}{12}$$ = $$\frac{1}{3}$$
Explanation :
The fraction $$\frac{4}{12}$$ can be renamed as an infinite number of equivalent fractions.
When you divide the numerator 4 by 4, the result is 1. When you divide the denominator 12 by 4, the result is 3.
$$\frac{4  ÷ 4 }{12 ÷ 4 }$$= $$\frac{1}{3}$$
This results in the number sentence $$\frac{4}{12}$$ = $$\frac{1}{3}$$.
c.
$$\frac{5}{10}$$ = $$\frac{1}{2}$$
Explanation :
The fraction $$\frac{5}{10}$$ can be renamed as an infinite number of equivalent fractions.
When you divide the numerator 5 by 5, the result is 1. When you divide the denominator 10 by 5, the result is 2.
$$\frac{5 ÷ 5 }{10 ÷ 5 }$$= $$\frac{1}{2}$$
This results in the number sentence $$\frac{5}{10}$$ = $$\frac{1}{2}$$.
d.
$$\frac{3}{6}$$ = $$\frac{6}{12}$$
Explanation :
The fraction $$\frac{3}{6}$$ can be renamed as an infinite number of equivalent fractions.
When you multiply the numerator 3 by 2, the result is 6. When you multiply the denominator 6 by 2, the result is 12.
$$\frac{3 \times 2}{6 \times 2}$$ = $$\frac{6}{12}$$
This results in the number sentence $$\frac{3}{6}$$ = $$\frac{6}{12}$$.
e.
$$\frac{4}{6}$$ = $$\frac{2}{3}$$
Explanation :
The fraction $$\frac{4}{6}$$ can be renamed as an infinite number of equivalent fractions.
When you divide the numerator 4 by 2, the result is 2. When you divide the denominator 6 by 2, the result is 3.
$$\frac{4  ÷ 2 }{6 ÷ 2 }$$= $$\frac{2}{3}$$
This results in the number sentence $$\frac{4}{6}$$ = $$\frac{2}{3}$$.
Question 3.
Circle the number sentences that are NOT true.
a. $$\frac{3}{12}$$ = $$\frac{1}{4}$$
Yes, it is true
$$\frac{3}{12}$$ = $$\frac{1}{4}$$
Explanation :
The fraction $$\frac{3}{12}$$ can be renamed as an infinite number of equivalent fractions.
When you divide the numerator 3 by 3, the result is 1. When you divide the denominator 12 by 3, the result is 4.
$$\frac{3 ÷ 3 }{12 ÷ 3 }$$= $$\frac{1}{4}$$
This results in the number sentence $$\frac{3}{12}$$ = $$\frac{1}{4}$$.
b. $$\frac{1}{2}$$ = $$\frac{5}{10}$$
Yes, it is true
$$\frac{1}{2}$$ = $$\frac{5}{10}$$
Explanation :
The fraction $$\frac{1}{2}$$ can be renamed as an infinite number of equivalent fractions.
When you multiply the numerator 1 by 5, the result is 5. When you multiply the denominator 2 by 5, the result is 10.
$$\frac{1 \times 5}{2 \times 5}$$ = $$\frac{5}{10}$$
This results in the number sentence $$\frac{1}{2}$$ = $$\frac{5}{10}$$.
c. $$\frac{2}{6}$$ = $$\frac{2}{5}$$
No it is not true $$\frac{2}{6}$$ can’t be equal to $$\frac{2}{5}$$
d. $$\frac{7}{10}$$ = $$\frac{4}{6}$$
No it is not true $$\frac{7}{10}$$ can’t be equal to $$\frac{4}{6}$$
e. $$\frac{9}{10}$$ = $$\frac{11}{12}$$
No it is not true $$\frac{9}{10}$$ can’t be equal to $$\frac{11}{12}$$
Practice
Question 4.
____ = 989 + 657
1646 = 989 + 657
Explanation :
Step 1
Add the 1s: 9 + 7 = 16.
16 ones = 1 ten and 6 ones
Write 1 in the 6s place below the line.
Write 1 above the digits in the 10s place.
Step 2
Add the 10s: 8 + 5 + 1 = 14
14 tens = 1 hundred + 4 tens
Write 4 in the 10s place below the line.
Write 1 above the digits in the 100s place.
Step 3
Add the 100s: 9 + 6 + 1 = 16
16 hundreds = 1 thousand + 6 hundred
Write 6 in the 100s place below the line.
Write 1 above the digits in the 1000s place.
No values to add in the thousand’s place so write 1 below the line in 1000s place .
Question 5.
3,314 + 4,719 = ____
3,314 + 4,719 = 8033
Explanation :
Step 1
Add the 1s: 4 + 9 = 13.
13 ones = 1 ten and 3 ones
Write 3 in the 1s place below the line.
Write 1 above the digits in the 10s place.
Step 2
Add the 10s: 1 + 1 + 1 = 3
Write 3 in the 10s place below the line.
Step 3
Add the 100s: 3 + 7Â = 10
10 hundreds = 1 thousand + 0 hundred
Write 0 in the 100s place below the line.
Write 1 above the digits in the 1000s place.
Step 4
Add the 1000s: 3 + 4 + 1 = 8
Write 8 in the 1000s place below the line.
Question 6.
5,887 – 3,598 = ___
5,887 – 3,598 = 2289
Question 7.
____ = 2,004 – 1,716
288= 2,004 – 1,716
Finding Equivalent Fractions
Family Note Today students learned about an Equivalent Fractions Rule, which can be used to rename any fraction as an equivalent fraction. The rule for multiplication states that if the numerator and denominator are multiplied by the same nonzero number, the result is a fraction that is equivalent to the original fraction.
For example, the fraction $$\frac{1}{2}$$ can be renamed as an infinite number of equivalent fractions.
When you multiply the numerator 1 by 5, the result is 5. When you multiply the denominator 2 by 5, the result is 10.
$$\frac{1 \times 5}{2 \times 5}$$ = $$\frac{5}{10}$$
This results in the number sentence $$\frac{1}{2}$$ = $$\frac{5}{10}$$. If you multiplied both the numerator and denominator in $$\frac{1}{2}$$ by 3, the result would be $$\frac{3}{6}$$, which is also equal to $$\frac{1}{2}$$.
Fill in the boxes to complete the equivalent fractions.
Example:
Question 1.
$$\frac{1}{2}$$ = $$\frac{6}{12}$$
Explanation :
The fraction $$\frac{1}{2}$$ can be renamed as an infinite number of equivalent fractions.
When you multiply the numerator 1 by 6, the result is 6. When you multiply the denominator 2 by 6, the result is 12.
$$\frac{1 \times 6}{2 \times 6}$$ = $$\frac{6}{12}$$
This results in the number sentence $$\frac{1}{2}$$ = $$\frac{6}{12}$$.
Question 2.
$$\frac{1}{4}$$ = $$\frac{3}{12}$$
Explanation :
The fraction $$\frac{1}{4}$$ can be renamed as an infinite number of equivalent fractions.
When you multiply the numerator 1 by 3, the result is 3. When you multiply the denominator 4 by 3, the result is 12.
$$\frac{1 \times 3}{4 \times 3}$$ = $$\frac{3}{12}$$
This results in the number sentence $$\frac{1}{4}$$ = $$\frac{3}{12}$$.
Question 3.
$$\frac{1}{3}$$ = $$\frac{2}{6}$$
Explanation :
The fraction $$\frac{1}{3}$$ can be renamed as an infinite number of equivalent fractions.
When you multiply the numerator 1 by 2, the result is 2. When you multiply the denominator 3 by 2, the result is 6.
$$\frac{1 \times 2}{3 \times 2}$$ = $$\frac{2}{6}$$
This results in the number sentence $$\frac{1}{3}$$ = $$\frac{2}{6}$$.
Question 4.
$$\frac{2}{3}$$ = $$\frac{8}{12}$$
Explanation :
The fraction $$\frac{2}{3}$$ can be renamed as an infinite number of equivalent fractions.
When you multiply the numerator 2 by 4, the result is 8. When you multiply the denominator 3 by 4, the result is 12.
$$\frac{2 \times 4}{3 \times 4}$$ = $$\frac{8}{12}$$
This results in the number sentence $$\frac{2}{3}$$ = $$\frac{8}{12}$$.
Question 5.
$$\frac{1}{5}$$ = $$\frac{2}{10}$$
Explanation :
The fraction $$\frac{1}{5}$$ can be renamed as an infinite number of equivalent fractions.
When you multiply the numerator 1 by 2, the result is 2. When you multiply the denominator 5 by 2, the result is 10.
$$\frac{1 \times 2}{5 \times 2}$$ = $$\frac{2}{10}$$
This results in the number sentence $$\frac{1}{5}$$ = $$\frac{2}{10}$$.
Question 6.
$$\frac{2}{5}$$ = $$\frac{4}{10}$$
Explanation :
The fraction $$\frac{2}{5}$$ can be renamed as an infinite number of equivalent fractions.
When you multiply the numerator 2 by 2, the result is 4. When you multiply the denominator 5 by 2, the result is 10.
$$\frac{2 \times 2}{5 \times 2}$$ = $$\frac{4}{10}$$
This results in the number sentence $$\frac{2}{5}$$ = $$\frac{4}{10}$$.
Question 7.
$$\frac{3}{4}$$ = $$\frac{9}{12}$$
Explanation :
The fraction $$\frac{3}{4}$$ can be renamed as an infinite number of equivalent fractions.
When you multiply the numerator 3 by 3, the result is 9. When you multiply the denominator 4 by 3, the result is 12.
$$\frac{3 \times 3}{4 \times 3}$$ = $$\frac{9}{12}$$
This results in the number sentence $$\frac{3}{4}$$ = $$\frac{9}{12}$$.
Question 8.
$$\frac{5}{6}$$ = $$\frac{10}{12}$$
Explanation :
The fraction $$\frac{5}{6}$$ can be renamed as an infinite number of equivalent fractions.
When you multiply the numerator 5 by 2, the result is 10. When you multiply the denominator 6 by 2, the result is 12.
$$\frac{5 \times 2}{6 \times 2}$$ = $$\frac{10}{12}$$
This results in the number sentence $$\frac{5}{6}$$ = $$\frac{10}{12}$$.
Question 9.
$$\frac{6}{9}$$ = $$\frac{2}{3}$$
Explanation :
The fraction $$\frac{6}{9}$$ can be renamed as an infinite number of equivalent fractions.
When you divide the numerator 6 by 3, the result is 2. When you divide the denominator 9 by 3, the result is 3.
$$\frac{6 ÷ 3 }{9 ÷ 3 }$$= $$\frac{2}{3}$$
This results in the number sentence $$\frac{2}{3}$$ = $$\frac{6}{9}$$.
Question 10.
$$\frac{4}{6}$$ = $$\frac{8}{12}$$
Explanation :
The fraction $$\frac{8}{12}$$ can be renamed as an infinite number of equivalent fractions.
When you divide the numerator 8 by 2, the result is 4. When you divide the denominator 12 by 2, the result is 6.
$$\frac{8 ÷ 2 }{12 ÷ 2 }$$= $$\frac{4}{6}$$
This results in the number sentence $$\frac{4}{6}$$ = $$\frac{8}{12}$$.
Question 11.
Name 3 equivalent fractions for $$\frac{1}{2}$$. ___
The fraction $$\frac{8}{12}$$ can be renamed as an infinite number of equivalent fractions.
The 3 equivalent fractions for $$\frac{1}{2}$$ = $$\frac{2}{4}$$ = $$\frac{3}{6}$$ = $$\frac{4}{8}$$ .
Practice
Question 12.
List all the factors of 56.
Factors of 56: 1, 2, 4, 7, 8, 14, 28 and 56
Question 13.
Write the factor pairs for 30.
___ and ___, ____ and ___, ___ and, ___.
___ and ____
The factor pairs for 30 are 2 and 15, 3 and 10, 5 and 6 , 1 and 30 .
Question 14.
Is 30 prime or composite? ____
30 is a Composite .
Explanation :
A composite number is a number that can be divided evenly by more numbers than 1 and itself. It is the opposite of a prime number. The number 30 can be evenly divided by 1 2 3 5 6 10 15 and 30, with no remainder. Since 30 cannot be divided by just 1 and 30, it is a composite number.
Sharing Veggie Pizza
Question 1.
Karen and her 3 friends want to share 3 small veggie pizzas equally. Karen tried to figure out how much pizza each of the 4 children would get. She drew this picture and wrote two answers.
a. Which of Karen’s answers is correct? ___
b. Draw on Karen’s diagram to make it clear how the pizza should be distributed among the 4 children.
a. $$\frac{3}{4}$$ is the correct answer .
b.
Number of children = 4
Number of pizzas = 3
Each Pizza is divided into 4 parts. Each children get $$\frac{1}{4}$$ part of pizza .
Fraction of pizza received by each children from 3 pizzas= 3 × $$\frac{1}{4}$$ part of pizza . = $$\frac{3}{4}$$
Each children get respective number written on pizzas as shown in below diagram .
Question 2.
Erin and her 7 friends want to share 6 small veggie pizzas equally. How much pizza will each of the 8 children get ? ___
Erin and here 7 friends are sharing pizza
⇒8 people are sharing pizza.
6 pizzas divided among 8 people:
Fraction of pizza each children get = $$\frac{6 pizzas }{8 people}$$ = $$\frac{6}{8}$$ = $$\frac{3}{4}$$ .
That means 4 children share 3 pizzas .
Each Pizza is divided into 4 parts. Each children get $$\frac{1}{4}$$ part of pizza .
Fraction of pizza received by each children from 3 pizzas= 3 × $$\frac{1}{4}$$ part of pizza . = $$\frac{3}{4}$$
Each children get respective number written on pizzas as shown in below diagram .
Question 3.
Who will get more pizza, Karen or Erin? ____ Explain or show how you know.
Both Karen and Erin gets same amount of Pizza that is $$\frac{3}{4}$$ of the pizza .
Practice
Question 4.
List all the factors of 50. __________
Factors of 50 : 1, 2, 5, 10 and 50 .
Question 5.
Is 50 prime or composite? __________
50 is a composite number .
Explanation :
A composite number is a number that can be divided evenly by more numbers than 1 and itself. … The number 50 can be evenly divided by 1, 2, 5, 10, 25 and 50, with no remainder. Since 50 cannot be divided by just 1 and 50, it is a composite number
Question 6.
Write the factor pairs for 75.
___ and ___
___ and ___
___ and ___
The factor pairs for 75 are
3 and 25.
5 and 15.
1 and 75 .
Solving Fraction Comparison Number Stories
Solve the problems below.
Question 1.
Tenisha and Christa were each reading the same book. Tenisha said she was $$\frac{3}{4}$$ of the way done with it, and Christa said she was $$\frac{6}{8}$$ of the way finished. Who has read more, or have they read the same amount? Ho w do you know?
Fraction of book read by Tenisha = $$\frac{3}{4}$$
Fraction of book read by christa = $$\frac{6}{8}$$ = $$\frac{3}{4}$$ .
$$\frac{3}{4}$$ = $$\frac{6}{8}$$
Explanation :
The fraction $$\frac{6}{9}$$ can be renamed as an infinite number of equivalent fractions.
When you divide the numerator 6 by 2, the result is 3. When you divide the denominator 8 by 2, the result is 4.
$$\frac{6 ÷ 2 }{8 ÷ 2 }$$= $$\frac{3}{4}$$
This results in the number sentence $$\frac{3}{4}$$ = $$\frac{6}{8}$$.
Question 2.
Heather and Jerry each bought an ice cream bar. Although the bars were the same size, they were different flavors. Heather ate $$\frac{5}{8}$$ of her ice cream bar, and Jerry ate $$\frac{5}{10}$$ of his.
Who ate more, or did they eat the same amount? ___
Write a number sentence to show this. ___
Explanation :
Fraction of ice cream ate by Heather = $$\frac{5}{8}$$
Fraction of ice cream ate by Jerry = $$\frac{5}{10}$$
Both the ice creams bar are represented in the above figure .
from the above figure we notice that $$\frac{5}{8}$$ > $$\frac{5}{10}$$ that means Jerry ate more ice cream bar than Heather .
Question 3.
Howard’s baseball team won $$\frac{7}{10}$$ of its games. Jermaine’s team won $$\frac{2}{5}$$ of its games. They both played the same number of games.
Whose team won more games, or did they win the same amount? ___
How do you know? __________
Fraction of games won by Howard’s = $$\frac{7}{10}$$
Fraction of games won by Jermaine’s = $$\frac{2}{5}$$
$$\frac{7}{10}$$Â >$$\frac{2}{5}$$Â that means Howard won more games than Jermaine as it shown in the above figure .
Question 4.
Write your own fraction number story. Ask someone at home to solve it.
Cristine had 4/5 m of cloth. She used 3/4 of it to make handkerchiefs. How much cloth had she left?
Explanation :
Fraction of total cloth with Cristine = $$\frac{4}{5}$$Â m
Fraction of cloth used = $$\frac{3}{4}$$ m
Fraction of cloth left = $$\frac{4}{5}$$Â – $$\frac{3}{4}$$ m =Â $$\frac{16}{20}$$Â – $$\frac{15}{20}$$ = $$\frac{1}{20}$$ m .
Therefore, Fraction of cloth left = $$\frac{1}{20}$$ m .
Practice
Write T for true or F for false.
Question 5.
1,286 + 2,286 = 3,752 ___
1,286 + 2,286 = 3,752 is not true as
Explanation :
1,286 + 2,286 = 3,572
Question 6.
9,907 – 9,709 = 200 ___
9,907 – 9,709 = 200 is not true as 9,907 – 9,709 = 198 .
Explanation :
Question 7.
2,641 + 4,359 = 2,359 + 4,641 ___
2,641 + 4,359 = 2,359 + 4,641 yes, it is true .
Explanation :
Question 8.
2,345 – 198 = 2,969 – 822 ____
2,345 – 198 = 2,969 – 822 Yes, it is true .
Explanation :
Comparing and Ordering Fractions
Write the fractions from smallest to largest, and then justify your conclusions by placing the numbers in the correct places on the number lines.
Question 1.
Explanation :
1. A fraction is larger if it’s farther from 0 on the number line.
2. A fraction is smaller if it’s closer to 0 on the number line.
$$\frac{5}{6}$$ is farther from 0 so, it is greater number .
$$\frac{2}{6}$$ is closer to 0 so, it is smaller number .
Therefore, $$\frac{2}{6}$$ < $$\frac{4}{6}$$Â < $$\frac{5}{6}$$ .
Question 2.
Explanation :
1. A fraction is larger if it’s farther from 0 on the number line.
2. A fraction is smaller if it’s closer to 0 on the number line.
$$\frac{9}{10}$$ is farther from 0 so, it is greater number .
$$\frac{1}{4}$$ is closer to 0 so, it is smaller number .
Therefore, $$\frac{1}{4}$$ < $$\frac{5}{12}$$Â < $$\frac{3}{5}$$ < $$\frac{9}{10}$$ .
Question 3.
Explanation :
1. A fraction is larger if it’s farther from 0 on the number line.
2. A fraction is smaller if it’s closer to 0 on the number line.
$$\frac{2}{3}$$ is farther from 0 so, it is greater number .
$$\frac{1}{6}$$ is closer to 0 so, it is smaller number .
Therefore, $$\frac{1}{6}$$ < $$\frac{4}{10}$$Â < $$\frac{1}{2}$$ < $$\frac{7}{12}$$ < $$\frac{2}{3}$$ .
Practice
Question 4.
___ = 5,494 + 3,769
9263 = 5,494 + 3,769
Explanation :
Question 5.
5,853 + 4,268 = ___
5,853 + 4,268 = 10,121
Explanation :
Question 6.
__ = 8,210 – 6,654
1556 = 8,210 – 6,654
Explanation :
Question 7.
7,235 – 5,906 = ___
7,235 – 5,906 = 1329
Explanation :
Names for Fractions and Decimals
Question 1.
Fill in the blanks in the table below.
Question 2.
Name two ways you might see decimals used outside of school.
The Two ways we see decimals used outside of school are Money and to represent Weight .
Question 3.
What decimal is represented by the tick mark labeled M?
The M and P are marked as per given table which in the fractions of 10 .
The fraction of tick Mark labeled M =Â $$\frac{5}{10}$$Â = 0.5
Question 4.
What fraction is represented by the tick mark labeled M?
The fraction of tick Mark labeled M =Â $$\frac{5}{10}$$
Question 5.
What decimal is represented by the tick mark labeled P?
The M and P are marked as per given table which in the fractions of 10 .
The fraction of tick Mark labeled P =Â $$\frac{9}{10}$$Â = 0.9
Question 6.
What fraction is represented by the tick mark labeled P?
The fraction of tick Mark labeled P =Â $$\frac{9}{10}$$
Practice
Question 7.
List all the factors of 100.
The Factors of 100 : 1, 2, 4, 5, 10, 20, 25, 50, and 100
Question 8.
List the factors of 100 that are prime.
The Factors of 100 : 1, 2, 4, 5, 10, 20, 25, 50, and 100
where 2 and 5 are the prime numbers.
Explanation :
Prime numbers are special numbers, greater than 1, that have exactly two factors, themselves and 1.
Question 9.
Write the factor pairs for 42.
___ and ___ ___and ___
___ and ___ ___and ___
The factor pairs for 42 are
2 and 21.
3 and 14 .
6 and 7 .
Representing Fractions and Decimals
If the grid is the whole, then what part of each grid is shaded?
Write a fraction and a decimal below each grid.
Question 1.
fraction: __________
decimal: __________
Total Number of boxes = 100
Number of boxes shaded = 30 .
Fraction = Number of boxes shaded by Total Number of boxes in a grid .
fraction: $$\frac{30}{100}$$
decimal: 0.30
Explanation :
Division by 10, 100 ,1000, etc. Shift the decimal point to the left by as many digits as there are zeroes in the divisor.
Question 2.
fraction: __________
decimal: __________
Total Number of boxes = 100
Number of boxes shaded = 9 .
Fraction = Number of boxes shaded by Total Number of boxes in a grid .
fraction: $$\frac{9}{100}$$
decimal: 0.09
Explanation :
Division by 10, 100 ,1000, etc. Shift the decimal point to the left by as many digits as there are zeroes in the divisor.
Question 3.
fraction: __________
decimal: __________
Total Number of boxes = 100
Number of boxes shaded = 65 .
Fraction = Number of boxes shaded by Total Number of boxes in a grid .
fraction: $$\frac{65}{100}$$
decimal: 0.65
Explanation :
Division by 10, 100 ,1000, etc. Shift the decimal point to the left by as many digits as there are zeroes in the divisor.
Question 4.
Color 0.8 of the grid.
Explanation :
decimal: 0.8
Multiplying by 100 to numerator and denominator we get fraction as $$\frac{80}{100}$$
Total Number of boxes = 100
Number of boxes shaded = 80 .
Fraction = Number of boxes shaded by Total Number of boxes in a grid .
fraction: $$\frac{80}{100}$$
Question 5.
Color 0.04 of the grid.
Explanation :
0.04
Multiplying by 100 to numerator and denominator we get fraction as $$\frac{80}{100}$$
Total Number of boxes = 100
Number of boxes shaded = 4 .
Fraction = Number of boxes shaded by Total Number of boxes in a grid .
fraction: $$\frac{4}{100}$$
Question 6.
Color 0.53 of the grid.
Explanation :
decimal: 0.53
Multiplying by 100 to numerator and denominator we get fraction as $$\frac{80}{100}$$
Total Number of boxes = 100
Number of boxes shaded = 53 .
Fraction = Number of boxes shaded by Total Number of boxes in a grid .
fraction: $$\frac{53}{100}$$
Practice
Question 7.
The numbers 81, 27, and 45 are all multiples of 1, ___, and ___.
The numbers 81, 27, and 45 are all multiples of 1, 3, and 9 .
Question 8.
List the first ten multiples of 6.
___, ___, ___, ___, ___, ___, ___, ___, ___, ___.
The first ten multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60 , 66 .
Tenths and Hundredths
Family Note Your child continues to work with decimals. Encourage him or her to think about ways to write money amounts. This is called dollars-and-cents notation. For example, $0.07 (7 cents),$0.09 (9 cents), and so on.
Write the decimal numbers that represent the shaded part in each diagram .
Question 1.
__ hundredths
__ tenths __ hundredths
Total Number of boxes = 100
Number of boxes shaded = 57 .
Fraction = Number of boxes shaded by Total Number of boxes in a grid .
fraction: $$\frac{57}{100}$$
decimal = 0.53 .
5 tenths 7 hundredths.
57 hundredths
Explanation :
To the right of the decimal point are the tenths and hundredths, where you put digits that represent numbers that are fractional parts of one, numbers that are more than zero and less than one
Question 2.
__ hundredths
__ tenths __ hundredths
Total Number of boxes = 100
Number of boxes shaded = 70 .
Fraction = Number of boxes shaded by Total Number of boxes in a grid .
fraction: $$\frac{70}{100}$$
decimal = 0.70 .
7 tenths 0 hundredths.
0 hundredths
Explanation :
To the right of the decimal point are the tenths and hundredths, where you put digits that represent numbers that are fractional parts of one, numbers that are more than zero and less than one
Question 3.
__ hundredths
__ tenths __ hundredths
Total Number of boxes = 100
Number of boxes shaded = 4 .
Fraction = Number of boxes shaded by Total Number of boxes in a grid .
fraction: $$\frac{4}{100}$$
decimal = 0.04 .
0 tenths 4 hundredths.
4 hundredths .
Explanation :
To the right of the decimal point are the tenths and hundredths, where you put digits that represent numbers that are fractional parts of one, numbers that are more than zero and less than one
Write the words as decimal numbers.
Question 4.
twenty-three hundredths ___
twenty-three hundredths = 0.23
Explanation :
23 hundredths would be written as . 23 as a decimal. 23 hundredths is the same as the fraction 23/100, or 23 out of 100 parts.
Question 5.
eight and four-tenths
______
eight and four-tenths = 8.4
Question 6.
thirty and twenty-hundredths
____
thirty and twenty-hundredths = 30.20
Question 7.
five-hundredths
______
five-hundredths = 0.05
Continue each pattern.
Question 8.
0.1, 0.2, 0.3, ____, ___, ___, ___, ____, ___
0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9
Question 9.
0.01, 0.02, 0.03, ____, ___, ___, ___, ____, ___
0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09
Practice
Question 10.
Round 7,604 to the nearest thousand. ______
7,604 to the nearest thousand is 8000 .
Question 11.
Round 46,099 to the nearest thousand. ____
46,099 to the nearest thousand is 46,000
Question 12.
Round 8,500,976 three ways: nearest thousand, hundred-thousand, and million.
_____ ____ ____
8,500,976 three ways: nearest
thousand = 8,501,000
hundred-thousand = 8,500,000
million = 9,000,000
Practice with Decimals
Fill in the missing numbers.
Question 1.
Question 2.
Follow these directions on the ruler below.
Question 3.
Make a dot at 7 cm and label it with the letter A.
Question 4.
Make a dot at 90 mm and label it with the letter B.
Explanation :
1cm = 10mm
9 cm = 90 mm
Question 5.
Make a dot at 0.13 m and label it with the letter C.
Explanation :
1m = 100 cm
0.13 m = 13 cm
Question 6.
Make a dot at 0.06 m and label it with the letter D.
Explanation :
1m = 100 cm
0.06 m = 6 cm
Question 7.
Write <, >, or =.
a. 1.2 ___ 0.12
1.2 > 0.12
Explanation :
when decimals are compared start with tenths place and then hundredths place, etc. If one decimal has a higher number in the tenths place then it is larger than a decimal with fewer tenths. If the tenths are equal compare the hundredths, then the thousandths etc. until one decimal is larger or there are no more places to compare. If each decimal place value is the same then the decimals are equal.
b. 0.3 __ 0.38
0.3 < 0.38
Explanation :
when decimals are compared start with tenths place and then hundredths place, etc. If one decimal has a higher number in the tenths place then it is larger than a decimal with fewer tenths. If the tenths are equal compare the hundredths, then the thousandths etc. until one decimal is larger or there are no more places to compare. If each decimal place value is the same then the decimals are equal.
c. 0.80 __ 0.08
0.80 >Â 0.08
Explanation :
when decimals are compared start with tenths place and then hundredths place, etc. If one decimal has a higher number in the tenths place then it is larger than a decimal with fewer tenths. If the tenths are equal compare the hundredths, then the thousandths etc. until one decimal is larger or there are no more places to compare. If each decimal place value is the same then the decimals are equal.
Question 8.
Complete.
Practice
Question 9.
6,366 + 7,565 = ____
6,366 + 7,565 = 13,931
Question 10.
3,238 + 29,784 = ___
3,238 + 29,784 = 33022
Question 11.
9,325 – 7,756 = ___
9,325 – 7,756 = 1569
Explanation :
Question 12.
14,805 – 2,927 = ___
14,805 – 2,927 = 11878
Measuring Centimeters and Millimeters
Question 1.
Find 6 objects in your home to measure. Use the ruler from the bottom of the page to measure them, first in centimeters and then in millimeters. Record your objects and their measurements.
Fill in the tables.
Question 2.
Explanation :
1 cm is 10 mm
multiply the given cm value with 10 to get the value in mm .
Respective given values are converted into mm and written in the above tabular column .
Question 3.
Explanation :
1 cm is 10 mm
multiply the given cm value with 10 to get the value in mm .
divide the given mm value with 10 to get the value in cm .
Respective given values are converted into mm and cm , written in the above tabular column .
Practice
Question 4.
List the factors for 63. ________
The factors of 63 are 1, 3, 7, 9, 21 and 63
Question 5.
Write the factor pairs for 60.
The factor pairs for 60 are
1 and 60
2 and 30
3 and 20
4 and 15
5 and 12
6 and 10
Comparing Decimals
2. Say and for the decimal point.
3. Read the digits after the decimal point as though they form their own number.
4. Say tenths or hundredths, depending on the placement of the right-hand digit. Encourage your child to exaggerate the -ths sound. For example, 2.37 is read as “two and thirty-seven hundredths.”
Write >, <, or =.
Question 1.
2.35 ___ 2.57
2.35 < 2.57
Explanation :
when decimals are compared start with tenths place and then hundredths place, etc. If one decimal has a higher number in the tenths place then it is larger than a decimal with fewer tenths. If the tenths are equal compare the hundredths, then the thousandths etc. until one decimal is larger or there are no more places to compare. If each decimal place value is the same then the decimals are equal.
Question 2.
1.08 __ 1.8
1.08 < 1.8
Explanation :
when decimals are compared start with tenths place and then hundredths place, etc. If one decimal has a higher number in the tenths place then it is larger than a decimal with fewer tenths. If the tenths are equal compare the hundredths, then the thousandths etc. until one decimal is larger or there are no more places to compare. If each decimal place value is the same then the decimals are equal.
Question 3.
0.64 __ 0.46
0.64 > 0.46
Explanation :
when decimals are compared start with tenths place and then hundredths place, etc. If one decimal has a higher number in the tenths place then it is larger than a decimal with fewer tenths. If the tenths are equal compare the hundredths, then the thousandths etc. until one decimal is larger or there are no more places to compare. If each decimal place value is the same then the decimals are equal.
Question 4.
0.90 ___ 0.9
0.90 = 0.9
Explanation :
when decimals are compared start with tenths place and then hundredths place, etc. If one decimal has a higher number in the tenths place then it is larger than a decimal with fewer tenths. If the tenths are equal compare the hundredths, then the thousandths etc. until one decimal is larger or there are no more places to compare. If each decimal place value is the same then the decimals are equal.
Question 5.
42.1 ___ 42.09
42.1 > 42.09
Explanation :
when decimals are compared start with tenths place and then hundredths place, etc. If one decimal has a higher number in the tenths place then it is larger than a decimal with fewer tenths. If the tenths are equal compare the hundredths, then the thousandths etc. until one decimal is larger or there are no more places to compare. If each decimal place value is the same then the decimals are equal.
Question 6.
7.09 __ 7.54
7.09 < 7.54
Explanation :
when decimals are compared start with tenths place and then hundredths place, etc. If one decimal has a higher number in the tenths place then it is larger than a decimal with fewer tenths. If the tenths are equal compare the hundredths, then the thousandths etc. until one decimal is larger or there are no more places to compare. If each decimal place value is the same then the decimals are equal.
Question 7.
0.4 __ 0.40
0.4 = 0.40
Explanation :
when decimals are compared start with tenths place and then hundredths place, etc. If one decimal has a higher number in the tenths place then it is larger than a decimal with fewer tenths. If the tenths are equal compare the hundredths, then the thousandths etc. until one decimal is larger or there are no more places to compare. If each decimal place value is the same then the decimals are equal.
Question 8.
0.26 __ 0.21
0.26 > 0.21
Explanation :
when decimals are compared start with tenths place and then hundredths place, etc. If one decimal has a higher number in the tenths place then it is larger than a decimal with fewer tenths. If the tenths are equal compare the hundredths, then the thousandths etc. until one decimal is larger or there are no more places to compare. If each decimal place value is the same then the decimals are equal.
Question 9.
The 9 in 4.59 stands for 9 ___ or ____.
The 9 in 4.59 stands for 9 Hundredths or 0.09 .
Question 10.
The 3 in 3.62 stands for 3 ___ or ___.
The 3 in 3.62 stands for 3Â Ones or 3 .
Continue each number pattern.
Question 11.
6.56, 6.57, 6.58, ___, ___, ___
6.56, 6.57, 6.58, 6.59, 6.60, 6.61
Explanation :
0.01 more
Question 12.
0.73, 0.83, 0.93, __, ___, ___
0.73, 0.83, 0.93, 1.03, 1.13, 1.23
Explanation :
0.10 more
Write the number that is 0.1 more.
Question 13.
4.3 ___
4.3 + 0.1 = 4.4
Question 14.
4.07 ___
4.07 + 0.1 = 4.17
Write the number that is 0.1 less.
Question 15.
8.2 ___
8.2 – 0.1 = 8.1
Question 16.
5.63 ___
5.63 – 0.1 =5.53
Practice
Question 17.
43,589 + 12,641 = ____
43,589 + 12,641 = 56230
Explanation :
Question 18.
63,274 + 97,047 = ____
63,274 + 97,047 = 1,60,321
Explanation :
Question 19.
41,805 – 26,426 = ____
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+0
0
276
1
3
over
1 - sqrt of 2
Plz work out the steps
Don't do the sqrt of 2 on a calculator to where u get a decimal!!
Guest Apr 10, 2017
#1
+7340
+3
$$\frac{3}{1-\sqrt2}$$
Multiply the numerator and denominator by the conjugate of the denominator.
$$=\frac{3}{1-\sqrt2}\cdot\frac{1+\sqrt2}{1+\sqrt2} \\~\\ =\frac{3+3\sqrt2}{1+\sqrt2-\sqrt2-2} \\~\\ =\frac{3+3\sqrt2}{-1} \\~\\ =-3-3\sqrt2$$
And that is approximately -7.243
hectictar Apr 10, 2017
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# Converting inches to feet
This was a homework assignment that I'm now done with - I submitted it as is. However the fact that I needed to use the same code twice bugged me... The double code is:
printf("Enter a distance in inches (0 to quit): ");
scanf("%f",&input);
Is there a better way to do the same thing in my loop instead of the double scanf/printf? It does need to quit the program immediately if a 0 is entered.
#include <stdio.h>
int main()
{
float distance, floatFeet, input;
int feet;
printf("Enter a distance in inches (0 to quit): ");
scanf("%f", &input);
while (input != 0)
{
feet = input/12;
distance = (input-feet*12);
floatFeet = input/12;
printf("%d feet and %f inches or %f feet \n\n", feet, distance, floatFeet);
printf("Enter a distance in inches (0 to quit): ");
scanf("%f", &input);
}
}
• there is a codereview.SO where questions like are better suited
– ratchet freak
Commented Feb 21, 2013 at 14:29
• And off we go...
– Michael Myers
Commented Feb 22, 2013 at 15:41
• I would flush the output before reading for the input.. Commented Feb 26, 2013 at 9:10
In general, you would need to insert an explicit conditional somewhere in order to distinguish between the first iteration (when no results should be displayed before the prompt) and any subsequent ones. That's because the "do not display output" condition is different from the "exit program" condition, so you can't just shove both into the loop condition.
Since you are going to insert an if no matter what, you might as well make it break out of the loop and make the loop infinite:
while (1)
{
printf("Enter a distance in inches (0 to quit): ");
scanf("%f",&input);
if (input == 0) break;
feet = input/12;
inches = (input-feet*12);
floatFeet = input/12;
printf("%d feet and %f inches or %f feet \n\n",feet,inches,floatFeet);
}
• In C, you'd probably want while(1), not while(true). Commented Feb 21, 2013 at 13:52
• @Antimony: So true, thanks (no pun intended). I 've blown my cover! ;-)
– Jon
Commented Feb 21, 2013 at 13:53
• while (true) throws an 'undeclared' error at me, should I just use while (1)?
– Symon
Commented Feb 21, 2013 at 13:57
• ah okay thanks antimony, comments didn't refresh right away :P
– Symon
Commented Feb 21, 2013 at 13:57
• In C, the idiom for "infinite loop" is for(;;) { .... }
– vonbrand
Commented Feb 22, 2013 at 1:56
You could move the duplicated lines into a function, something like this should work:
#include <stdio.h>
float get_input(void)
{
float input;
printf("Enter a distance in inches (0 to quit): ");
scanf("%f", &input);
return input;
}
int main()
{
float inches, floatFeet, input;
int feet;
while (input = get_input())
{
feet = input/12;
inches = (input-feet*12);
floatFeet = input/12;
printf("%d feet and %f inches or %f feet \n\n",feet,inches,floatFeet);
}
printf("Goodbye!\n");
}
• @mskfisher, thanks! Maybe out of scope though, if the OP has not learned about functions and such yet. Commented Feb 21, 2013 at 14:37
• @Marton while(input = get_input()) checks whether input is 0. Commented Feb 21, 2013 at 14:43
• @Marton It's not while(0), it's while(input) - except that input also gets assigned when the loop condition is evaluated. Commented Feb 21, 2013 at 14:56
• @marco-fiset This is possible in C99 but not in classic C. C99 doesn't really seem to have set off yet Commented Feb 21, 2013 at 15:13
• Better would be while ((input = get_input()) != 0) which won't give a compiler warning and doesn't confuse the assignment with a boolean. Commented Feb 22, 2013 at 16:31
You can use a do while loop, which will always execute once and check the condition at the end of the loop.
If you need the program to exit immediately without printing when a 0 is entered, you could just put the calculation and printing code in an if statement within the loop:
do {
printf("Enter a distance in inches (0 to quit): ");
scanf("%f",&input);
if( input != 0 ) {
feet = input/12;
inches = (input-feet*12);
floatFeet = input/12;
printf("%d feet and %f inches or %f feet \n\n",feet,inches,floatFeet);
}
} while (input != 0)
Or, to avoid repeating anything:
while(true) {
printf("Enter a distance in inches (0 to quit): ");
scanf("%f",&input);
if( input == 0 ) {
break;
}
feet = input/12;
inches = (input-feet*12);
floatFeet = input/12;
printf("%d feet and %f inches or %f feet \n\n",feet,inches,floatFeet);
}
You can replace this loop with a "forever" loop that exits from the middle of its body, like this:
for (;;) {
printf("Enter a distance in inches (0 to quit): ");
scanf("%f",&input);
if (input == 0) break;
feet = input/12;
inches = (input-feet*12);
floatFeet = input/12;
printf("%d feet and %f inches or %f feet \n\n",feet,inches,floatFeet);
}
There are multiple ways to do a "forever" loop; this one is borrowed from K&R's book.
• Any reason to use a for loop instead of a while loop when making a 'forever' loop?
– Symon
Commented Feb 21, 2013 at 13:55
• @Symon I use for(;;) because it is a very common idiom popularized by "the" book on the C language. Other than that, there is no reason to go one way or the other: the compiler will generate identical binary code for both loops. Commented Feb 21, 2013 at 13:57
• Some people even define: #define ever (;;)
– mouviciel
Commented Feb 21, 2013 at 13:58
• @mouviciel: People who have just not learned to leave the preprocessor alone if at all possible.
– Jon
Commented Feb 21, 2013 at 13:59
• @Symon In a compiler which does not optimize at all, while (1) might generate a comparison to a constant which is omitted in for (;;). However, all normal compilers would optimize that comparison away. Commented Feb 21, 2013 at 15:06
Use do-while instead while.
#include <stdio.h>
int main()
{
float inches, floatFeet, input;
int feet;
do
{
printf("Enter a distance in inches (0 to quit): ");
scanf("%f",&input);
if (input == 0) break;
feet = input/12;
inches = (input-feet*12);
floatFeet = input/12;
printf("%d feet and %f inches or %f feet \n\n",feet,inches,floatFeet);
}
while (input != 0);
}
• "It does need to quit the program immediately if a 0 is entered." This would print "0 feet and 0 inches or 0 feet".
– lc.
Commented Feb 21, 2013 at 13:49
• That depends on what you mean by immediately. Since the code executed before quitting has no side effects, it shouldn't matter. Commented Feb 21, 2013 at 13:50
• @Antimony The side effect is printing a line. I don't think that "shouldn't matter"...the code as edited will work though (although there is an extra comparison every time the loop loops).
– lc.
Commented Feb 21, 2013 at 13:50
• Also, if you're breaking out explicitly, there's no point in a do loop. Just make it while(1). Commented Feb 21, 2013 at 13:51
• @Antimony The two methods may be functionally equivalent; but the do-while construct makes it explicit that it's not intended to be an infinite loop. That said, I'd probably replace the if (input == 0) break; construct with if(input !=0) { ...}and put the rest of the loop inside the {}'s to eliminate the second exit point from the loop as well. Commented Feb 21, 2013 at 14:05
This is exactly where you use the:
do {
//stuff
} while (condition)
... where you need to do something once, at least, but conditionally repeat the same task...
Symon, here is another alternative. Notice the following things:
1. I have separated getting input into a function, get_input. The function takes a pointer to a variable to accept the input value and checks whether scanf actually reads a value from the user input. It returns an integer value that is non-zero (true) if scanf really did read a value. The exit condition has changed to ctrl-d which is the standard UNIX way of indicating end-of-input. Equally, typing a non-numeric will also cause scanf to return 0.
2. The while loop now exits on failure of the input function.
3. The variables are defined at the point of use. This is better practice as it reduces the scope of the variable (ie. the amount of the code where it is valid).
4. distance is renamed inches, which seems more appropriate.
5. The division is done only once and there is a cast (int) to avoid a conversion compiler warning (turn many warnings ON in your compiler).
6. A new-line is printed on exit to be nice to the user.
#include <stdio.h>
static inline int get_input(float *input)
{
printf("\nEnter a distance in inches (ctrl-d to quit): ");
return scanf("%f", input) == 1;
}
int main(void)
{
float input;
while (get_input(&input) != 0) {
float float_feet = input / 12;
int feet = (int) float_feet;
float inches = input - (feet * 12);
printf("%d feet, %f inches or %f feet \n", feet, inches, float_feet);
}
putchar('\n');
return 0;
}
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# Binary Math
## Binary numbers and mathematics.
Let's first take a look at decimal addition.
As an example we have 26 plus 36,
26
+36
To add these two numbers, we first consider the "ones" column and calculate 6 plus 6, which results in 12. Since 12 is greater than 9 (remembering that base 10 operates with digits 0-9), we "carry" the 1 from the "ones" column to the "tens column" and leave the 2 in the "ones" column.
Considering the "tens" column, we calculate 1 + (2 + 3), which results in 6. Since 6 is less than 9, there is nothing to "carry" and we leave 6 in the "tens" column.
26
+36
62
works in the same way, except that only 0's and 1's can be used, instead of the whole spectrum of 0-9. This actually makes binary addition much simpler than decimal addition, as we only need to remember the following:
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 10
As an example of binary addition we have,
101
+101
a) To add these two numbers, we first consider the "ones" column and calculate 1 + 1, which (in binary) results in 10. We "carry" the 1 to the "tens" column, and the leave the 0 in the "ones" column.
b) Moving on to the "tens" column, we calculate 1 + (0 + 0), which gives 1. Nothing "carries" to the "hundreds" column, and we leave the 1 in the "tens" column.
c) Moving on to the "hundreds" column, we calculate 1 + 1, which gives 10. We "carry" the 1 to the "thousands" column, leaving the 0 in the "hundreds" column.
101
+101
1010
1011
+1011
10110
Note that in the "tens" column, we have 1 + (1 + 1), where the first 1 is "carried" from the "ones" column. Recall that in binary,
1 + 1 + 1 = 10 + 1
= 11
## Binary subtraction
is simplified as well, as long as we remember how subtraction and the base 2 number system. Let's first look at an easy example.
111
- 10
101
Note that the difference is the same if this was decimal subtraction. Also similar to decimal subtraction is the concept of "borrowing." Watch as "borrowing" occurs when a larger digit, say 8, is subtracted from a smaller digit, say 5, as shown below in decimal subtraction.
35
- 8
27
For 10 minus 1, 1 is borrowed from the "tens" column for use in the "ones" column, leaving the "tens" column with only 2. The following examples show "borrowing" in binary subtraction.
10 100 1010
- 1 - 10 - 110
1 10 100
To practice binary addition and subtraction, visit the Practice Exercises page.
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| 1,561,394,533,000,000,000
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| 789,206,427
| 5,518
|
Scientific Method
Joe Student rolled a toy car down a ramp and measured how far it traveled along the floor.
He recorded the distance along with the original mass of the toy car. Then, he taped a
penny to the roof of the same toy car, measured its new mass, and rolled it down the same
ramp from the same starting position. Again, he measured the distance the toy car traveled
and recorded the data. He taped a 2
nd
penny to the toy car’s roof, measured its new mass,
and rolled it down the same ramp from the same starting position. And again, he measured
the distance the toy car traveled and recorded the data. Based on this data, he concluded
that the greater mass caused the car to roll a greater distance.
What was the error in the experiment?
1. 👍 0
2. 👎 0
3. 👁 92
1. How did he roll it? Did he give it a push or just let it roll by itself? If he just let it roll by itself, I don't detect any error, except that he might want to repeat the rolls many times, to make sure that the differences were not merely chance occurrences.
1. 👍 0
2. 👎 0
posted by PsyDAG
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# How do you find the vertex and intercepts for y=2x^2-16x+27?
May 7, 2016
$y = 2 {\left(x - 4\right)}^{2} - 5 \text{ }$ Vertex is at $\left(4 , - 5\right)$
$y -$intercept is at 27
#### Explanation:
$y = 2 {x}^{2} - 16 x + 27$
$y = 2 \left({x}^{2} - 8 x + \frac{27}{2}\right) \text{ }$now complete the square
$y = 2 \left[\textcolor{b l u e}{{x}^{2} - 8 x + 16} - 16 + \frac{27}{2}\right] \text{ } \Rightarrow$ +16-16 = 0
$y = 2 \left[\textcolor{b l u e}{{\left(x - 4\right)}^{2}} - 16 + \frac{27}{2}\right] \text{ }$ $- 16 + \frac{27}{2} = - \frac{5}{2}$
$y = 2 {\left(x - 4\right)}^{2} - 2 \times \frac{5}{2}$
$y = 2 {\left(x - 4\right)}^{2} - 5$
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| 222,975,508
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# Tag Info
## Hot answers tagged vectors
21
Sorry folks for posting such a trivial issue! The issue is solved. I was using the wrong function. Here goes the correct one: glm::vec2 testVec(6,-4); float len = glm::length(testVec); The member function of the same name returns the number of components instead (i.e. vec2::length will always yield 2, vec3::length will always yield 3, etc.).
9
Here's the simple answer. In 4D, to be able to multiply them by a 4x4 matrix, vectors are represented as (x,y,z,0) and points are represented as (x,y,z,1). Since the 4th row of a 4x4 matrix represents the translation of the matrix, the above representations make it so points are affected by translation, but vectors are not. Both vectors and points are ...
6
I (believe) I've solved this (even if it has taken 2 days). My problem was essentially I wanted to take the dot product of the face normal, and line-of-sight vector like below And determine the angle to see if the face was looking towards or away from the view point. My erroneous step was that I was doing this AFTER transforming from world-space to view-...
5
Let's say you have your two points that define the line segment: $A$ and $B$, and a point $P$ that you are testing to see if it is on the line segment. Firstly, you get a normalized vector from $A$ to $B$, as well as the distance from $A$ to $B$. direction = normalize(B - A); length = length(B-A); Next, you use dot product to project the point $P$ onto ...
5
I did some research and found the answer I was looking for. The three most common ways to interpolate vectors are: Slerp - short for "spherical interpolation", this is the most correct way, but is also the costliest. In practice you likely do not need the precision. lerp - short for "linear interpolation", you just do a regular linear interpolation ...
5
In a traditional camera, the photons from the scene travel through the lens of the camera, then hit the sensor at the focal length. A consequence of the lens is that the image is upside down and backwards. In ray tracing, we can optimize this by imagining the focal plane is in front of the camera. You can think that the photons from the scene travel towards ...
4
I suggest computing the normal vector for each polygon and compute the angle between the normal vector of one triangle and the normal vector of a plane perpendicular to the other surface that passes through the common edge. See the following illustration, where the red arrows are the normal vectors of the red polygons. The blue vector is the plane ...
4
What you are looking for is a way to get access to individual pixels in an image, in a way that you can modify those pixels with CPU code. Once you have that you'll want to find software rendering / software rasterization methods. A great place to start would be reading up in the various bresenham algorithms. There is a lot of info out there on that. Here ...
4
I've done some small changes to how I usually construct my view matrix, here is what I've modified: // put it after N = ... U = cross(N, Vspec); U = U / sqrt(dot(U,U)); % normalize V = cross(U, N); Then I also set N vector as -N to R matrix, like this: R= [U(1),U(2),U(3),0; % U is direction of camera space X axis V(1),V(2),V(3),0; % V is direction of ...
4
Short answer: yes. Longer answer: yes, because the vectors you’re using are meant to represent directions, not directions-and-distances. Think of it in terms of light: it doesn’t matter how far a photon’s traveled, whether it’s from the sun or from a lamp on your desk—once it arrives at a surface, it’s going to get reflected in exactly the same way. From a ...
4
Well there's no way to know for sure unless you look at the source code but my guess is that they do zooming by lowering the FOV (field of view) of the camera. This is easy to implement and actually sort of represents how a real scope works (in the way that light rays are bent through an optical lens). A zoom effect where you change the actual position of ...
3
I think there might be a misprinting in the book. I am getting this $((A-C) + tB)\cdot((A-C) + tB) = R\cdot R$ Let A-C = Y $(Y + tB)\cdot(Y + tB) = R\cdot R$ $Y\cdot Y + Y\cdot tB + tB\cdot Y + t^2B\cdot B = R\cdot R$ Substituting back $(A-C)\cdot(A-C) + 2t * B\cdot(A-C) + t^2B\cdot B = R\cdot R$
3
It looks like the way you're using smoothstep isn't quite right. With lerp, the first two parameters are the endpoints of the output range, and the third is a 0–1 value specifying how far to interpolate. It looks like you're trying to use smoothstep the same way, but it doesn't work like that: the first two parameters to smoothstep are endpoints of the ...
3
Good answer above. (I like that the book adds in p then subtracts it.) This is a hard-coded grey world as a "hello world" kind of ray tracer. The 0.5 is the hard-coded reflectance. Keep working through book and the object reflectance and max-recursion will be addressed. Ray branching I avoid like the plague. In cost benefit I am always a skeptic ...
3
Instead of a screen plane in front of the eye, it describes a film plane, where the image is projected, to explicitly model camera optics. You don't need to compute the focal point in doing ray tracing - it's just a way to find the plane of focus for depth of field effects. For depth of field effects I use a standard perspective projection but jitter the ...
2
Your transform looks correct. To transform from world to eye coordinates, I I always use a "lookat" transform, defined by 3 vectors: $\bf{e}$, $\bf{a}$ and $\bf{u}$; in english, the eye position, the point it's looking at, and an up vector, which must not be in the same direction as $\bf{a} - \bf{e}$ (more specifically, not a multiple of it). The space is ...
2
vec3 target = rec.p + rec.normal + random_in_unit_sphere(); return 0.5*color(ray(rec.p, target-rec.p), world); this is the exact same as vec3 targetDir = rec.normal + random_in_unit_sphere(); return 0.5*color(ray(rec.p, targetDir), world); In other words the sphere is a way to pick a random direction which is biased towards the normal direction without ...
2
If you would look up the definition of a vector and a point, then a vector is: A quantity, such as velocity, completely specified by a magnitude and a direction. http://www.thefreedictionary.com/vector And a point is: A dimensionless geometric object having no properties except location. http://www.thefreedictionary.com/point So you could say ...
2
Trying to explain a problem seems to help the thought process. This is what eventually worked for me: I realized that I can easily find the point (in 3D space) where a line between two of the corners brakes through one of the walls in my pyramid field of view, given the coordinates of the out-of-view corner and one of the visible ones. It is only a matter ...
2
Since the question was somewhat clarified I will formalize both the question and the answer for future readers. Having a differentiable scalar field $f : \mathbb{R}^4 \rightarrow \mathbb{R}$ we want to find the gradient of the field with respect to $\theta, \phi$ on the 2-manifold defined parametrically by: $$(x(\theta,\phi), y(\theta,\phi) z(\theta,\phi), ... 2 See equation (16) in Microfacet Models for Refraction through Rough Surfaces : -(\eta_i w_i + \eta_o w_o) which you'll probably want to normalize. \eta are the two indices of refraction and w the two vectors (they use i and o in the paper). Also look at the left part of figure 7 on the same page to see where all the vectors are pointing. 2 There is very little geometrical meaning to the multiplication of two vectors and there are no well defined identities describing the result. But that doesn't make it illegal, and is an important part of many well defined functions such as the dot product. Multiplication of vectors also have several important identities that allow terms to be moved around in ... 1 The most common use is almost certainly computing normal vectors for a surface, using a cross product. Although strictly speaking, this does not quite qualify as the two initial vectors are usually not orthogonal. This is used for shading calculations and various other algorithms. I know of at least one where the initial surface is explicitly a rectangle so ... 1 Since we have:$$\pmb{v}_{\parallel} \parallel \pmb{n},$$then:$$\pmb{v}_{\parallel} = k\pmb{n}$$We want that the length of \pmb{v}_{\parallel} is \|\pmb{v}_{\parallel}\|, then:$$\|\pmb{v}_{\parallel}\| =|k|\|\pmb{n}\| \implies |k| = \frac{\|\pmb{v}_{\parallel}\|}{\|\pmb{n}\|} Then the sign of $k$ is simply chosen to match the direction of $\... 1 Since v∥ is parallel to n, it can be expressed as some multiple of n. The multiple is the ratio of the length of v∥ to the length of n. The n vector doesn’t have to be normalized, but if it is, then |n| is 1, so the equation becomes v∥ = n * |v∥|. 1 It would seem that, in the second example, the morphNormals are targets while in in the first they are deltas. This explains the difference in math used between the two pieces of code. 1 I'm not going to dive into much details about affine transformations and such, you are better off reading up on these concepts from good graphics books. I think The matrix you need is this.$\begin{bmatrix} 0.5 & 0 & 0 & 4\\ 0 & 0.5 & 0 & 6\\ 0 & 0 & 0.5 & 2\\ 0 & 0 & 0 & 1\\ \end{bmatrix}\$ The reason why ...
1
Rotation matrix | wikipedia.org in section that says "in three dimensions". It's the first search result when searching about rotating a vector about an axis. Let me know if you have any questions. For your specific problem: bring the point P to the origin, then rotate the vector about the Z axis, then bring the point P back to its original location.
1
The simplest solution is to use a lookat matrix. This is a matrix calculated from a "eye" point, a "target" point and an up vector. The resulting matrix will then make the Z axis point from the eye to the target and the y axis in the general direction of up.
Only top voted, non community-wiki answers of a minimum length are eligible
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Anomali Interview Questions, Process, and Tips
Ques:- Who is the information and technology minister of tamilnadu
Ques:- A reduction of 25% in the price of oil enables a house wife to obtain 5kgs more for Rs.800, what is the reduced price for kg?
Explanation:
800*(25/100) = 200 —- 5
? —- 1 => Rs.40
Ques:- Find out the wrong number in the series:
40960, 10240, 2560, 640, 200, 40, 10
A. 40
B. 200
C. 640
D. 2560
200
Ques:- Describe a situation where you influenced others to achieve a common goal or objective?
Ques:- Two ladies and two men are playing cards and are seated at North, East, South and West of a table. No lady is facing East. Persons sitting opposite to each other are not of the same sex. One man is facing South. Which directions are the ladies facing?
M
M. L
L
Ques:- What is the largest prime number stored in 8 bit memory?
251
Ques:- Given a credit card number with a certain number of fixed digits and a certain number of digit placeholders, find the number of solutions such that the entire number yields a remainder of 3 when divided by 13. Use properties of modulus
Ques:- Why are you interested working here?
Ques:- Ayesha’s father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents?
6
Ques:- What is the planning of your future?
Ques:- Tom has three boxes with fruits in his barn: one box with apples, one box with pears, and one box with both apples and pears. The boxes have labels that describe the contents, but none of these labels is on the right box. How can Tom, by taking only one piece of fruit from one box, determine what each of the boxes contains?
Ques:- What do you mean by GSM, CDMA?
Ques:- Imagine you are standing in front of a mirror, facing it. Raise your left hand. Raise your right hand. Look at your reflection. When you raise your left hand your reflection raises what appears to be his right hand. But when you tilt your head up, your reflection does too, and does not appear to tilt his/her head down. Why is it that the mirror appears to reverse left and right, but not up and down
The definition of left and right depends on the observer and
is reversed when facing the opposite direction. The
definition of up and down does not depend on the orientation
of the observer.
Ques:- Healing of wound is accelerated by?
Ques:- A room is 15m long, 4m broad and 3m height. Find the cost of white washing its four walls at 50p per m2Â ?
2*3(15+4)=114
114*1/2=57 rs
Ques:- The number of sequences in which 7 players can throw a ball, so that the youngest player may not be the last is -.
Ques:- A train moves fast a telegraph post and a bridge 264 m long in 8 sec and 20 sec respectively. What is the speed of the train?
237.6 Kmh
Ques:- “your director is in another part of the region when you receive an email from another dept requiring the return of a draft framework for a 2 yr strategic plan by 5pm the same day. You realize the originals are with your director.what would you do?
i will drop a mail saying that can you pleaae meal me the the originals of the stregic plan
Ques:- A fast typist can type some matter in 2 hours and a slow typist can type the same in 3 hours. If both type combinely, in how much time will they finish?
Ques:- Whats ur career objective?
Ques:- A jogger running at 9 km/hr along side a railway track is 240 m ahead of the engine of a 120 m long train running at 45 km/hr in the same direction. In how much time will the train pass the jogger?
– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.
– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m
To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec
In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m
at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m
If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)
Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.
For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.
This will take an additional (9.6 + 2) seconds.
Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds
Ques:- In a class , except 18 all are above 50 years. 15 are below 50 years of age. how many people are there
None of these
Ques:- a student is ranked 13th from right and 8th from left.how many are there?
twenty members
Ques:- The current of a stream at 1 kmph. A motor boat goes 35 km upstream and back to the starting point in 12 hours. The speed of the motor boat in still water is?
Ques:- How long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 170 m in length?
16.8 sec
Ques:- How does this position compare with others you’re applying for?
Ques:- What would u looking in your job ?
Ques:- Eight persons – P, Q, R, S, T, U, V and W are sitting around a circular table. S is to the immediate right of W. V is not next to either R or T. W is to the immediate right of T, who is sitting opposite to R. U and W are sitting opposite to each other. Who is sitting two places to the right of W?
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https://cheapnursingtutors.com/a-company-has-just-negotiated-a-contract-to-produce-a-part-for-another-firm/
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# A company has just negotiated a contract to produce a part for another firm.
July 30, 2019
###### Advanced Time Value Of Money
July 30, 2019
A company has just negotiated a contract to produce a part for another firm. In the process of manufacturing the part, the inside diameter of successive parts becomes smaller and smaller as the cutting tool wears. However, the specs are so wide relative to machine capabilities that it is possible to set the diameter initially at a large value and let the process run for a while before replacing the cutting tool. The inside diameter decreases at an average rate of .001 cm per part, and the process has a standard deviation of .05 cm. The variability is approximately normal. Assuming a three-sigma buffer at each end, how frequently must the tool be replaced if the process specs are 3 cm and 3.5 cm. Use (Number of shafts)n = 1.
Determine how many pieces can be produced before the LCL just crosses the lower tolerance of 3 cm. (Do not round your intermediate calculations.)
After [removed] pieces the cutting tool should be replaced.
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https://numberworld.info/4341104
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# Number 4341104
### Properties of number 4341104
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 181 * 1499
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
423d70
Base 32:
44fbg
sin(4341104)
0.84458369569608
cos(4341104)
0.53542355286666
tan(4341104)
1.5774122956941
ln(4341104)
15.283639251608
lg(4341104)
6.6376001903734
sqrt(4341104)
2083.5316172307
Square(4341104)
### Number Look Up
Look Up
4341104 which is pronounced (four million three hundred forty-one thousand one hundred four) is a impressive number. The cross sum of 4341104 is 17. If you factorisate the figure 4341104 you will get these result 2 * 2 * 2 * 2 * 181 * 1499. 4341104 has 20 divisors ( 1, 2, 4, 8, 16, 181, 362, 724, 1448, 1499, 2896, 2998, 5996, 11992, 23984, 271319, 542638, 1085276, 2170552, 4341104 ) whith a sum of 8463000. 4341104 is not a prime number. The number 4341104 is not a fibonacci number. The figure 4341104 is not a Bell Number. The number 4341104 is not a Catalan Number. The convertion of 4341104 to base 2 (Binary) is 10000100011110101110000. The convertion of 4341104 to base 3 (Ternary) is 22011112212122. The convertion of 4341104 to base 4 (Quaternary) is 100203311300. The convertion of 4341104 to base 5 (Quintal) is 2102403404. The convertion of 4341104 to base 8 (Octal) is 20436560. The convertion of 4341104 to base 16 (Hexadecimal) is 423d70. The convertion of 4341104 to base 32 is 44fbg. The sine of the figure 4341104 is 0.84458369569608. The cosine of the figure 4341104 is 0.53542355286666. The tangent of the figure 4341104 is 1.5774122956941. The root of 4341104 is 2083.5316172307.
If you square 4341104 you will get the following result 18845183938816. The natural logarithm of 4341104 is 15.283639251608 and the decimal logarithm is 6.6376001903734. You should now know that 4341104 is very amazing number!
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# Computer organisation
1. Using a "word" of 5 bits, list all of the possible signed binary numbers and their decimal equivalents that are represent able in: [3 Marks]
a. Signed magnitude
b. One's complement
c. Two's complement
2. A Computer uses IEEE-754 format to represent floating points. What value (in decimal) the computer represents if the floating point is represented using the following binary digits: [3 Marks]
1 10000011 11000000000000000000000
3. Using Boolean algebra, prove that: [2 marks]
X’Y + XYZ’ + Y’ + XZ (Y+Y’) = 1
( 4个评论 ) melbourne, Australia
\$24AUD 在1天里
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2.3
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https://www.teacherspayteachers.com/Browse/Search:counting%20money%20activities
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showing 1-52 of 9,362 results
This counting money pack has 2 activities. Activity one: counting bills and coins -great for scoot, walk around, math center, or small group instruction -16 piggy bank cards with bills and coins for counting -some coins equal an additional dollar -features heads only and front of bills only -answer
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This activity is intended to be used as a scoot, walkaround, or math center. Only quarters, dimes, nickels, and pennies are used on the cards. There are no amounts above \$1.00 on this task. Includes: 20 turkey coin cards 1 student answer sheet 1 answer key Thanks for stopping by 2nd grade Snicke
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This activity is designed to help engage students during the holiday season as well as support their learning of counting money. The value of money does not exceed \$60.00 in this activity. Students will be creating their 'shopping list' by picking from the 30 different holiday item cards. Each ca
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This is a worksheet to be completed by students after reading Amy Axelrod's 'Pigs Will Be Pigs: Fun with Math and Money.' Student will perform tasks based on information from the book. First, students must add up how much money the Pig family found. Next, Student will determine how much money the
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One of our single preschool printable activities from our full curriculum CD-R. A single PDF file that prints 1 page. Here is what you get! 1 Money Count and Clip Card activity that prints 11 clip it cards and 1 instructions card. Prints 3 pages. Use an ink dauber or colored pencil to fill in the
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This is a great little set for math stations or to do with a partner. Students will build the number in the piggy bank 4 different ways. This is a good activity for students to practice making combinations of coins. Color and blackline versions are included! Included in this set is building: •1
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Use these fun and engaging activities to have your students practice their money skills. All of the activities involve students counting combinations of U.S. dollars and coins or solving word problems with dollars, quarters, dimes, nickels, and/or pennies. Included are task cards, no prep worksheets
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This packet includes three separate activities to help students -count money -understand the value of money -determine how much change they should receive when purchasing various St. Patrick’s Day related items. Materials required: Play or real money (totaling \$5.00 for each student) Note: Play
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This packet includes three separate activities to help students -count money -understand the value of money -determine how much change they should receive when purchasing various Easter related items. Materials required: Play or real money (totaling \$5.00 for each student) Note: Play dollar bill
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This product includes 32 different cards with varying money amounts from 14 cents to \$1.48. Students can count out change to match the money amounts on each card. These can be used in a math center or small group.
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Three game boards with copies of real coins. One game board encompasses counting pennies and nickels, another, pennies, nickels and dimes. The last one has pennies, nickels, dimes, and quarters to count. Game directions are at the top of the game boards. The winner is the one with the most money at
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Students will learn to count money by using the money ladder. The ladder increases in increments of 5's or nickels. Students quickly learn that a dime is worth 2 nickels or 2 jumps on the ladder and a quarter is worth 5 nickels or 5 jumps on the ladder. Through repeated practice my students were
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This packet includes three separate activities to help students -count money -understand the value of money -determine how much change they should receive when purchasing various spring related items. Materials required: An assortment of play or real money (totaling \$5.00 for each student) Note
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This resource has two pages of counting coins and one page of counting out coins to "buy" items.
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The Grocery Game is a highly engaging math activity that provides students with a hands-on way to practice counting money! To play, players take turns picking food cards to fill up their grocery bags. Every card has a price, so students must pay for each item by counting out the correct amount of
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Get this as part of a bundle: 2nd Grade Math Bundle - Games, Centers, Activities for the Whole Year
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The Magic Of Math Unit 5 for SECOND GRADE focuses on: Week 1: Counting Coins (to 99 cents) Week 2: Activities are Differentiated Option 1- Counting Coins, Option 2- Using Bills and Coins Week 3: Saving and Spending with word problems Week 4: Personal Financial Literacy/Economics *Pleas
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This money unit for grades 2-3 is filled with challenging yet fun centers, games and activities to supplement a study of money for students in grades 2-3. It just got even better because I gave it a facelift with new fonts and clip art and added some new activities, so be sure and re-download if y
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This packet includes US money worksheets and activities for your students to learn about money using coins (Q, N, D, and P) to 100 cents. There are 55 worksheets and 4 colored as well as black and white posters for penny, nickel, dime and quarter. Please check the preview! You may also like: Dog-t
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This money unit has many lessons to enrich your lessons. Here's what is included: 11 pages of counting money with answer sheets 10 pages of word problems with answer sheets 4 money centers with recording sheets and answer sheets 4 posters Common Core Math Standard: 2.MD.8 Click on preview for FR
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These 3 money activities reinforce skills that accompany the money unit from Smart Board that I have uploaded. These activities follow the first grade standards of learning in Virginia. In this unit you get - - tree map sort (sort between 4 coins) - count how much money is in each box (mix of pen
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Counting Money Activities: You will receive no prep printable worksheets for students to practice counting money. NOTE: This product was updated April 2018 to add several more practice pages. You will now receive: 6 cut and paste worksheets 6 draw a line to match the money to the number 6 circle
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Get this as part of a bundle: Counting Money Task Cards & Worksheets, 2nd Grade Math Review Bundle
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This packet has 31 practice sheets for your students to practice counting US coins - pennies, nickels, dimes, and quarters. THE MAXIMUM AMOUNT TO BE COUNTED IS \$3.For money task cards, please click here. Here is a brief description: 3 pages to count the pennies. 3 pages to count the nickels. 3 pa
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Scoot is faced paced, whole group activity. It can be used as a fun review, practice, assessment of prior knowledge or test. Play it once and it will quickly become a classroom favorite! Topics covered include: Counting coins, one and five dollar bills. Both font and back of American coins are pict
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| Money | Counting Bills | Coins | Life Skills | Math | Real World | Year after year, my students struggle with concepts of money. The struggles range from bill and coin identification all the way up to adding mixed bills/coin totals up successfully. I have heard from other teachers as well that th
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If you are a second grade teacher, you will love my Second Day In Second Grade Unit: A Back-To-School Packet! This September packet is "2" good "2" pass up! It is over 60 pages of games, activities and printables to help celebrate the second day of second grade and beyond. Welcome to the wonderf
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The “Canadian Money” activity packet is designed to meet the curriculum expectations for the Ontario Grade 4 Mathematics curriculum. This packet has students: -Read and write money amounts to \$100; -Add and subtract money amounts by making simulated purchases and providing change for amounts up to \$
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Canadian Money. Start up kit for teaching younger students about Canadian coins. Included are the following: Money posters (Penny, nickel, dime, quarter, loonie, toonie) Anchor chart (large coins and values to make a great chart for your class) Counting by 5s, 10s, 25s chart Strategies for counti
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A money unit on introducing children to identifying and counting coins! This resource contains lessons, activities, centers, games, posters, and everything you need for a successful unit. Skills: Coin recognition and counting by pennies! Coins: Penny, Nickel, Dime, and Quarter Click here to get a
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These FUN activities can be used to accompany any money unit or to serve as a review of money! All activities/centers include a task card, materials, and recording sheet. * Coin Match-Up - Match the coin cards to the piggy bank amounts. Recording sheet included. * Shopping Time! - Use the coins t
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Money Money Counting Coins Counting Money Worksheets 12 worksheets: *** counting coins, writing total amount, 1 page each, total: 10 pages - pennies - nickels - dimes - quarters - quarters and dimes - quarters and pennies - dimes and nickels - dimes and pennies - nickels and pennies *** coloring c
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Help teach your students how to count, add and calculate change with these activities! 35 differentiated worksheets, including some word problem activities. Have fun learning about money with catalogues from supermarkets or other stores by cutting out items and learning to work out change, add item
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** Please note that since the Ontario mathematics curriculum has not been updated, this activity pack still uses the penny to match the expectations. ** This activity pack is aligned with the Ontario Grade 3 Mathematics curriculum expectations for money. Included in this activity pack are a variety
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Counting Change Task Box Clip-It Cards This bundle of counting change clip-it task cards is a great way to differentiate math centers or task box times. Prep once and use over and over again! This is a bundle of all 6 levels 54 Clip-It Task Cards for all levels of students learning to count chan
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Money Activities In this money pack, you will find different supplemental activities to use when teaching how to identify coins, identify coin worth, counting money, and solve money word problems! Money Activities Included: Money Posters Money Mini Books Money Printables Money Centers Money Inte
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Canadian Money Activities – Counting Canadian Coins To \$1.00 These activities will be a great addition to your money unit! Coins included – penny, nickel, dime and quarter Included: * Mini Money Book – copy and paste the coins needed to buy the school supplies. Each school item can be purchased
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**SEE THE PREVIEW AND FIND OUT FOR YOURSELF WHY WE ARE THE ORIGINAL AND THE #1 ESCAPE ROOM MATERIAL USED BY THOUSANDS OF TEACHERS AND SCHOOLS** INCLUDES: 1) Escape Room Activity 2) Handout Material 3) Lesson Plan On Subject 4) Power Point Presentation 5) Promotional Videos 6) How-To Video 7) Cou
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*** Before you purchase this mat, please go check out my Math mats bundle. You'll get one of these mats plus more than 20 others for a GREAT deal! http://www.teacherspayteachers.com/Product/Math-Mats-Bundle-K-2-over-20-Math-mats These money mats help organize money for counting. The touch points
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Money Activities and Assessments Included: Page 3...Large coins instructions Pages 4-11...Large coins Page 12...Student Coin Chart Page 13--14...Penny Value Cards Page 15-28...Coin Books (three different levels) Page 29...Coin rubbings page Page 30...Coin Recognition Concentration game cards Page 3
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********Updated 2/19/2014********* This 42 page money activity unit is a supplement to the Smart Board money unit I have created. The Smart Board unit can be seen here. In this unit: - counting pennies worksheet - 2 pages of counting pennies picture cards - counting nickels worksheet - 2 pages of
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Have fun shopping with this money game! Students count their money and record how much they have on their shopping list. Then, they go shopping to see what they can buy. Students record their choices and add up the prices. Then, they have to answer the question - Do you have enough money? This mon
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Money Madness-Pennies, Nickels, Dimes Kids are mad about money! Money Madness is a fun, engaging introduction to counting money. It focuses on pennies, nickels, and dimes. Your students will become proficient with these 3 coins before quarters are introduced! I have included One QUARTER poster in t
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Money Activity contains printable money word posters in which each letter is assigned a monetary value. Posters are available with money symbols: dollars, pounds and euros. There are also posters with coin illustrations for: US, Canada, UK, Australia, and Euro coins. Checkout this resource in the
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Counting Money: Extensions for Your Money Unit! - Riddles for Counting Money ★★★ Leveled Tasks for Easy Differentiation! ★★★ Extend and supplement your money unit with these fun money riddles! Kids love them and use thinking and reasoning skills as they count coins and problem solve, "Which coin
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Find Someone Who...can count the coins and bills This is part of my Find Someone Who Math Bundle, which includes 8 activities for \$5. This is a cooperative learning activity that gets the whole class up and moving while practicing their money-counting skills. It can be used at the end of the unit
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Your students will be begging to do these math centers! The candy theme math activities will get them excited to add money amounts, learn coin values (compared to a dollar), count back change, and even have their own candy store. Teacher directions included for these center activities. Give your s
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This fun adapted book is perfect for teaching the extended Math standards for grades 3-5: MD.35.4c Solve addition 1-step, real-world word problems involving mass, volume or money (e.g., following a recipe, paying for groceries). The activity pack includes 3 books: What Can I Buy? - Grocery Stor
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120 U.S. Currency Coins Money Task Cards: 36 cards of Unmixed Coins 20 cards of Nickels and Pennies 20 cards of Dimes and Pennies 12 cards of Quarters and Pennies 12 cards of Nickels, Dimes, and Quarters 20 cards of Mixed Coins All cards use cents symbol. No card answer exceeds 100 cents. Look
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Scoot is a fun whole-group activity that can be used in many different ways. It can be used as a review game, skill practice or even a test! You could even use these cards as task cards for a center or for early finishers. This version of Scoot tests students' knowledge of counting bills and coins
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Get this as part of a bundle: Math SCOOT! Games Bundle- 20 GAMES!
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This download includes games and worksheets to help your 2nd grade students practice counting money and solving word problems using various money amounts. Activities and games are aligned with the 2nd grade common core math standard 2.MD.8 (solve word problems involving dollar bills, quarters, dime
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Looking for a fun interactive teaching idea for representing decimal numbers? Well look no further as Decimals Game Puzzles, for CCSS 4.NF.6, will serve as an exciting lesson plan for 4th grade elementary school classrooms. This is a great resource to incorporate into your unit as a guided math cent
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** Please note that since the Ontario mathematics curriculum has not been updated, this activity pack still uses the penny to match the expectations. ** This activity pack is aligned with the Grade 2 Ontario Mathematics curriculum guide. This activity pack explores money up to 100¢. The concepts ad
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56 U.S. Currency Money Task Cards with dollar bills and coins; pennies, nickels, dimes, and quarters. Lowest amount is \$1.01 and the highest amount with one dollar bills is \$6.00. Also included: 8 cards with five dollar bills and various coins, for differentiation. *** Answer Key and Recording Shee
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```Elements of Computing Systems, Nisan & Schocken, MIT Press
www.idc.ac.il/tecs
Virtual Machine
Part I: Stack Arithmetic
This presentation contains lecture materials that accompany the textbook “The Elements of
Computing Systems” by Noam Nisan & Shimon Schocken, MIT Press, 2005.
The book web site, www.idc.ac.il/tecs , features 13 such presentations, one for each book
chapter. Each presentation is designed to support about 3 hours of classroom or self-study
instruction.
You are welcome to use or edit this presentation as you see fit for instructional and noncommercial purposes.
If you use our materials, we will appreciate it if you will include in them a reference to the book’s
web site.
If you have any questions or comments, you can reach us at tecs.ta@gmail.com
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 1
Where we are at:
Human
Thought
Abstract design
Software
hierarchy
abstract interface
Chapters 9, 12
H.L. Language
&
Operating Sys.
Compiler
abstract interface
Chapters 10 - 11
Virtual
Machine
VM Translator
abstract interface
Chapters 7 - 8
Assembly
Language
Assembler
Chapter 6
abstract interface
Machine
Language
Computer
Architecture
abstract interface
Chapters 4 - 5
Hardware
Platform
Hardware
hierarchy
Gate Logic
abstract interface
Chapters 1 - 3
Chips &
Logic Gates
Electrical
Engineering
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
Physics
slide 2
Motivation
class Main {
static int x;
function void main() {
// Input and multiply 2 numbers
var int a, b, x;
let a = Keyboard.readInt(“Enter a number”);
let b = Keyboard.readInt(“Enter a number”);
let x = mult(a,b);
return;
}
}
// Multiplies two numbers.
function int mult(int x, int y) {
var int result, j;
let result = 0; let j = y;
while not(j = 0) {
let result = result + x;
let j = j – 1;
}
return result;
}
Ultimate goal:
Translate highlevel programs
into executable
code.
Compiler
...
@a
M=D
@b
M=0
(LOOP)
@a
D=M
@b
D=D-A
@END
D;JGT
@j
D=M
@temp
M=D+M
@j
M=M+1
@LOOP
0;JMP
(END)
@END
0;JMP
...
}
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 3
Compilation models
direct compilation:
language 1
language 2
...
2-tier compilation:
language n
language 1
language 2
...
language n
intermediate language
hardware
platform 1
hardware
platform 2
.
...
requires n m translators
hardware
platform m
hardware
platform 1
hardware
platform 2
...
hardware
platform m
requires n + m translators
Two-tier compilation:
First compilation stage depends only on the details of the source language
Second compilation stage depends only on the details of the target platform.
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 4
The big picture
Some
language
...
Some
compiler
...
Some Other
language
Jack
language
Jack
compiler
Some Other
compiler
CISC
machine
language
VM imp.
over RISC
platforms
RISC
machine
language
CISC
machine
RISC
machine
VM imp.
over the Hack
platform
VM
emulator
written in
a high-level
language
...
...
The interface between
the 2 compilation stages
Must be sufficiently
general to support many
<high-level language,
machine language> pairs
Can be modeled as the
language of an abstract
virtual machine (VM)
Intermediate code
VM
implementation
over CISC
platforms
The intermediate code:
Can be implemented in
many different ways.
Hack
machine
language
...
other digital platforms, each equipped
with its own VM implementation
Any
computer
Hack
computer
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 5
The big picture
Some
language
...
Some
compiler
...
Some Other
language
Jack
language
Chapters
9-13
Jack
compiler
Some Other
compiler
VM language
VM
implementation
over CISC
platforms
CISC
machine
language
VM imp.
over RISC
platforms
RISC
machine
language
RISC
machine
written in
a high-level
language
...
...
CISC
machine
VM imp.
over the Hack
platform
VM
emulator
Chapters
7-8
Hack
machine
language
Chapters
1-6
...
other digital platforms, each equipped
with its VM implementation
Any
computer
Hack
computer
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 6
Lecture plan
Goal: Specify and implement a VM model and language
Arithmetic / Boolean commands
sub
neg
eq
gt
lt
Program flow commands
label
(declaration)
goto
(label)
if-goto
(label)
Next lecture
This lecture
Function calling commands
and
or
function
(declaration)
call
(a function)
return
(from a function)
not
Memory access commands
pop x (variable)
push y
(variable or constant)
Method: (a) specify the abstraction (model’s constructs and commands)
(b) propose how to implement it over the Hack platform.
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 7
The VM language
Important:
From here till the end of this and the next lecture we describe the VM
model used in the Hack-Jack platform
Other VM models (like JVM/JRE and IL/CLR) are similar in spirit and
different in scope and details.
Our VM features a single 16-bit data type that can be used as:
Integer
Boolean
Pointer.
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 8
Stack arithmetic
17
4
17
9
5
SP
SP
Typical operation:
Pops topmost values x,y from the stack
Computes the value of some function f(x,y)
Pushes the result onto the stack
(Unary operations are similar, using x and f(x) instead)
Impact: the operands are replaced with the operation’s result
In general: all arithmetic and Boolean operations are implemented similarly.
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 9
Memory access (first approximation)
Stack
Memory
Stack
...
121
a
5
b
SP
6
push b
6
...
17
108
b
108
108
...
SP
(before)
(after)
Stack
Memory
Stack
...
121
a
b
Memory
...
121
6
...
17
SP
a
5
...
5
...
121
...
17
Memory
5
pop a
a
...
SP
108
...
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
17
b
108
...
slide 10
Memory access (first approximation)
Stack
Memory
Stack
...
121
a
5
...
b
SP
push b
6
...
17
108
b
108
108
...
SP
(before)
(after)
Stack
Memory
Stack
...
121
a
b
Memory
...
121
6
...
17
SP
a
5
...
5
...
121
6
17
Memory
5
pop a
a
...
SP
108
...
17
b
108
...
Classical data structure
Elegant and powerful
Many implementation options.
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 11
Evaluation of arithmetic expressions
// d=(2-x)*(y+5)
push 2
push x
sub
push y
push 5
mult
pop d
Memory
...
5
y
9
2
push 2
SP
x
...
Stack
sub
2
push x
5
SP
SP
-3
-3
push y
SP
9
-3
push 5
9
SP
5
SP
Memory
Stack
-42
-3
14
mult
SP
pop d
...
SP
SP
5
x
9
y
...
d
-42
...
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 12
Evaluation of Boolean expressions
// if (x<7) or (y=8)
push x
push 7
lt
push y
push 8
eq
or
Memory
...
x
y
Stack
SP
...
12
push x
12
12
push 7
lt
7
SP
8
SP
false
false
push y
SP
8
false
push 8
8
SP
eq
8
SP
true
false
true
or
SP
SP
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 13
Arithmetic and Boolean commands (wrap-up)
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 14
Memory access (motivation)
Modern programming languages normally feature the following variable kinds:
Class level
Static variables
Private variables (AKA “object variabls” / “fields” / “properties”)
Method level:
Local variables
Argument variables
A VM abstraction must support (at least) all these variable kinds.
The memory of our VM model consists of 8 memory segments:
static, argument, local, this, that, constant, pointer, and temp.
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 15
Memory access commands
Command format:
pop segment i
push segment i
(Rather than pop x and push y,
as was shown in previous slides,
which was a conceptual simplification)
Where i is a non-negative integer and segment is one of the following:
static:
holds values of global variables, shared by all functions in the same class
argument: holds values of the argument variables of the current function
local:
holds values of the local variables of the current function
this:
holds values of the private (“object”) variables of the current object
that:
holds array values
constant: holds all the constants in the range 0…32767 (pseudo memory segment)
pointer:
used to align this and that with different areas in the heap
temp:
fixed 8-entry segment that holds temporary variables for general
use; Shared by all VM functions in the program.
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 16
VM programming
VM programs are normally written by compilers, not by humans
In order to write compilers, it helps to understand the spirit of VM
programming. So we will now see how some common programming tasks can be
implemented in the VM abstraction:
Disclaimer:
These programming examples don’t belong here; They belong to the compiler
chapter, since expressing programming tasks in the VM language is the business
of the compiler (e.g., translating Java programs to Bytecode programs)
We discuss them here to give some flavor of programming at the VM level.
(One can safely skip from here to slide 21)
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 17
Arithmetic example
High-level code
VM code (first approx.)
function mult(x,y) {
int result, j;
result=0;
j=y;
while ~(j=0) {
result=result+x;
j=j-1;
}
return result;
}
Just after mult(7,3) is entered:
Stack
argument
SP
7
x
0
0
sum
1
3
y
1
0
j
...
Just after mult(7,3) returns:
Stack
21
SP
local
0
...
function mult(x,y)
push 0
pop result
push y
pop j
label loop
push j
push 0
eq
if-goto end
push result
push x
pop result
push j
push 1
sub
pop j
goto loop
label end
push result
return
VM code
function mult 2
push
constant
pop
local 0
push
argument
pop
local 1
label
loop
push
local 1
push
constant
eq
if-goto end
push
local 0
push
argument
pop
local 0
push
local 1
push
constant
sub
pop
local 1
goto
loop
label
end
push
local 0
return
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
0
1
0
0
1
slide 18
Object handling example
High level program view
b
object
x:
y:
color:
0
120
following
compilation
80
50
412
...
3012
...
3
(Actual RAM locations of program variables are
run-time dependent, and thus the addresses shown
here are arbitrary examples.)
3012
120
3013
80
3014
50
3015
3
argument
pointer
0
3012
0
1
17
1
this
0
...
b
b
object
push argument 0
// Point the this seg. to b:
pop pointer 0
// Get r's value
push argument 1
// Set b's third field to r:
pop this 2
...
Virtual memory segments just before
...
/* Assume that b and r were
passed to the function as
its first two arguments.
The following code
implements the operation
RAM view
Virtual memory segments just after
argument
pointer
0
3012
0
1
17
1
...
3012
this
0
120
1
80
2
17
3
3
(this 0
is now
alligned with
RAM[3012])
...
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 19
Array handling example
High-level program view
bar
array
0
1
2
3
RAM view
0
7
53
following
compilation
121
398
8
...
9
19
(Actual RAM locations of program variables are
run-time dependent, and thus the addresses shown
here are arbitrary examples.)
...
4315 bar
...
4315
7
4316
53
4317
4318
121
8
local
4315
1
...
pointer
0
1
4324
...
bar
array
19
Virtual memory segments
Just after the bar[2]=19 operation:
local
that
0
0
1
push local 0
push constant 2
// Set that’s base to (bar+2):
pop pointer 1
push constant 19
// *(bar+2)=19:
pop that 0
...
Virtual memory segments
Just before the bar[2]=19 operation:
0
/* Assume that bar is the
first local variable declared
in the high-level program. The
code below implements
bar[2]=19, or *(bar+2)=19. */
...
1
4315
...
pointer
0
1
4317
that
0
1
19
...
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
(that 0
is now
alligned with
RAM[4317])
slide 20
Lecture plan
Goal: Specify and implement a VM model and language
Arithmetic / Boolean commands
sub
neg
eq
gt
lt
Program flow commands
label
(declaration)
goto
(label)
if-goto
(label)
Next lecture
This lecture
and
or
Function calling commands
function
(declaration)
call
(a function)
return
(from a function)
not
Memory access commands
pop segment i
push segment i
Method: (a) specify the abstraction (model’s constructs and commands)
(b) propose how to implement it over the Hack platform.
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 21
Implementation
VM implementation options:
Software-based (emulation)
Translator-based (e.g., to the Hack language)
Hardware-based (CPU-level)
Well-known translator-based implementations:
JVM (runs bytecode programs in the Java platform)
CLR (runs IL programs in the .NET platform).
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 22
Our VM emulator (part of the course software suite)
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 23
VM implementation on the Hack platform
SP
0
LCL
1
ARG
2
THIS
3
THAT
4
5
TEMP
...
The challenge: (i) map the VM constructs on the
host RAM, and (ii) given this mapping, figure out
how to implement each VM command using
assembly commands that operate on the RAM
Host
RAM
12
13
General
purpose
static: static variable number j in a VM file f is
implemented by the assembly language symbol
f.j (and recall that the assembler maps such
symbols to the RAM starting from address 16)
14
15
16
...
Statics
255
256
...
Stack
2047
2048
...
local,argument,this,that: mapped on the
heap. The base addresses of these segments are
entry of a segment is implemented by accessing
the segment’s (base + i) word in the RAM
Heap
constant i is implemented by supplying the
constant i
pointer,temp: see the book
Exercise: given the above game rules, write the
Hack commands that implement, say,
push constant 5 and pop local 2.
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 24
Parser module (proposed design)
Parser: Handles the parsing of a single .vm file, and encapsulates access to the input code. It reads VM commands, parses them, and
Routine
Arguments
Returns
Function
Input file / stream
--
Opens the input file/stream and gets ready to parse it.
--
boolean
Are there more commands in the input?
--
--
Reads the next command from the input and makes it
the current command. Should be called only if
hasMoreCommands() is true. Initially there is no
current command.
--
C_ARITHMETIC,
C_PUSH, C_POP,
C_LABEL, C_GOTO,
C_IF, C_FUNCTION,
C_RETURN, C_CALL
Returns the type of the current VM command.
C_ARITHMETIC is returned for all the arithmetic
commands.
Constructor
hasMoreCommands
commandType
arg1
--
string
Returns the first argument of the current command. In
the case of C_ARITHMETIC, the command itself
(add, sub, etc.) is returned. Should not be called if the
current command is C_RETURN.
arg2
--
int
Returns the second argument of the current command.
Should be called only if the current command is
C_PUSH, C_POP, C_FUNCTION, or C_CALL.
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 25
CodeWriter module (proposed design)
CodeWriter: Translates VM commands into Hack assembly code.
Routine
Arguments
Returns
Function
Constructor
Output file / stream
--
Opens the output file/stream and gets ready to write
into it.
setFileName
fileName (string)
--
Informs the code writer that the translation of a new
VM file is started.
writeArithmetic
command (string)
--
Writes the assembly code that is the translation of the
given arithmetic command.
WritePushPop
Command (C_PUSH or
C_POP),
--
Writes the assembly code that is the translation of the
given command, where command is either C_PUSH
or C_POP.
--
Closes the output file.
segment (string),
index (int)
Close
--
Comment: More routines will be added to this module in chapter 8.
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 26
Some
language
Perspective
...
Some
compiler
...
Some Other
language
Jack
compiler
Some Other
compiler
VM language
In this lecture we began the process of
building a compiler
VM
implementation
over CISC
platforms
CISC
machine
language
VM imp.
over RISC
platforms
RISC
machine
language
Modern compiler architecture:
VM
emulator
...
...
Front end (translates from high level language to a VM language)
Back end (implements the VM language on a target platform)
Translator
written in
a high-level
language
Hack
...
Brief history of virtual machines:
1970’s: p-Code
1990’s: Java’s JVM
2000’s: Microsoft .NET
A full blown VM implementation typically includes a common software library
(can be viewed as a mini, portable OS).
We will build such a mini OS later in the course.
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 27
Conceptually
similar to:
And to:
Complete the VM
specification and
implementation
(chapters 7,8)
JVM
CLR
Introduce Jack, a
high-level programming
language (chapter 9)
Java
C#
Build a compiler for it
(chapters 10,11)
Java compiler
C# compiler
Finally, build a mini-OS,
i.e. a run-time library
(chapter 12).
JRE
.NET base class library
Elements of Computing Systems, Nisan & Schocken, MIT Press, www.idc.ac.il/tecs , Chapter 7: Virutal Machine, Part I
slide 28
```
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https://www.quesba.com/questions/calculator-exercise-8-22-presented-below-is-information-related-to-tamaris--4514
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# CALCULATOR Exercise 8-22 Presented below is information related to Tamaris Company FULL SCREEN TER V
CALCULATOR Exercise 8-22 Presented below is information related to Tamaris Company FULL SCREEN TER VERSION Ending Inventory Date (End-of-Year Price Prices) Index December 31, 2017 \$ 87,400 100 December 31, 2018 212.625 189 December 31, 2019 209,304 216 December 31, 2020 235,170 234 December 31, 2021 281,736 252 December 31, 2022 339,039 261 Compute the ending inventory for Tamaris Company for 2017 throu2022 using the dollar-value LIFO method Ending Inventory 2017 2018 2019 2020 2022 Click if you would like to show Work for this question: Open Show Work Policy | 2000-2019 John Wiley Sons, Inc. All Rights Reserved. A Division of the Wix Sons, Inc
Feb 21 2020| 12:50 PM |
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# Race building
Hi, I am new here and Newish to ACKs. I am having a bit of an issue with creating new race values. I have seen a few posts by Alex, but I am still unclear on the value xp.
Now I am using a post in this thread as a guideline http://www.autarch.co/forum/gnomish-infravision-0 post number 8
Now My race has the following at Value zero
• Claws and kick( claws and fang) at a value of 2
• Powerful leaper( custom power I valued at 2)
• Low light vision ( Custom Power i valued at 1)
So going by the steps
1: The number of powers is 3
2:-1 giving us 2
3: add 3 for 3 powers above 1+ humans do not have. Total of 5
or do I Add 1 point per point of power humans do not have in that cases its 7 not 5
4: 5 or 7 x 40 giving 200 or maybe 280
5: round to nearest 25( up only?) which is 200 or 275 or 300?
Ok six and seven is where it really breaks down…How do I find values for race 1-4? Do I simply add it up as steps 1-5?
If anyone can help, I would be very grateful.
That is what I linked in my post. It does not answer my questions. It helped but it does not tell me
1:On step 3 I do it the 1st or 2nd way I listed
2:On Step 5 do I round up only?
3:On step 6 and 7…how do I found the Values of Race 1 though 4?
1: Step 3 “add 1 if”, not “add 1 for each”, so you just add 1 if the condition is met (powers not available to humans).
2: Step 5, “round nearest”, nothing about up or down. 280 is closer to 275 than 300, so it becomes 275.
You seem to be missing racial values 1-4 entirely. You can’t figure out Racial Value 4 XP cost unless you create “content” for those values.
Thanks,for the reply. That helps, so my total would
*3 powers-1
*3 powers over 1
*total 5x40= 200
Correct?
I have the values created, I just posted the zero value as an example. I just did not know if I added up values 1-4 as steps 1 though 4. I am now Guessing I just follow steps 1-4 to find the values correct?
Yeah, 200 XP looks correct.
And you basically just repeat the same thing independently for each Racial Value: (# of powers) - 1, +1 if there are powers not available to humans, x40, round to nearest interval of 25…
A lot of races also just straight up give you the equivalent of one of the other Values, in which case you’d add that cost straight to your racial value XP cost: granting Fighting +1 adds 500 XP, Fighting +2 adds 1000 XP…
Many thanks. I stuck +1 fighting on value 4 as honestly I had no clue what else to add that fit. Seemed like what all the other races did.
It all depends on what sort of classes you want to build. I think when building races past Racial Value 0, you need to think less about simulation and more about what classes you want to use it for.
Elves are supposed to make “multi-class” mages - they combine X with “Arcane Magician” - so they get Arcane Value.
Thrassians are supposed to make “Monster” warriors with good HD and I suppose warrior-priests and warrior-mages - they combine X with “Warrior” - so they get Fighting Value.
Dwarves are supposed to be skillful, so their Racial Value just adds proficiencies.
I’ve created Half-Giants, who get free HD Value with their Racial Value, because I wanted to make all Half-Giants very tough. (The only class so far is the Half-Giant Brute, with d12 HD and Hero fighting ability.)
I’ve also made Mantis Men, who get Fighting Value +1 at Racial Value 2 and higher; most of them are Explorer types, so this leaves me more class points to give them more skills or better HD.
Good points. These guys are feathered Dino people, they are the settings 'Lizard men" and well, elves. Many have a spiritual side, so maybe cleric works best at rank 4. Ranks 1-3 just up the claws, adds attuned to nature, alertness and scent tracking to reflect their predatory nature.
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### 2.3.3.5 Six Moments Transport Model - Closure at
Taking the first six moments, eqns. (2.99) to (2.104), into account give three balance and three flux equations
(2.168) (2.169) (2.170)
(2.171) (2.172) (2.173)
By using just a MAXWELL distribution function to close the system one would not obtain any additional information as compared to the energy transport model. A shifted MAXWELL distribution function has only three independent parameters, namely its amplitude, the displacement, and the standard deviation, which correspond to the carrier concentration , the carrier velocity , and the carrier temperature , respectively. By simply increasing the number of considered moments of the distribution function no additional independent variables can be found.
In analogy to statistical mathematics a quantity called kurtosis has been introduced, which is in this work defined as the deviation of the fourth moment of the non-MAXWELL distribution function from the fourth moment of a MAXWELL distribution function with the same standard deviation
(2.174)
The system is now closed at . Eqn. (2.126) is one possible closure relation obtained from a MAXWELL distribution function. Other empirical closures are also possible (eqn. (2.176)). By introducing an additional temperature 2.11
(2.175)
the third power of the temperature in eqn. (2.126) is substituted by empirically combining different powers of and
(2.176)
Simulations have shown, that the combination with fits best to Monte Carlo data [39, G5]. This is depicted in Fig. 2.4 where the different closure relations are compared with the sixth moment obtained from a Monte Carlo simulation of a one-dimensional -- test structure. As can be seen, the closure for the case gives the smallest error within the channel. The convergence behavior of the resulting discretized equation system also appeared most stable when using . Especially for , which corresponds to closing the system with a MAXWELLian distribution function eqn. (2.126) [40] the NEWTON procedure failed to converge in most cases.
Using the closure relation becomes
(2.177)
and the full six moments transport model reads
(2.178) (2.179)
(2.180) (2.181)
(2.182) (2.183)
In the following the equations for the six moments transport model are rewritten by introducing the charge sign for electrons and the coefficients to . The balance equations become
(2.184) (2.185) (2.186)
with
(2.187)
and the following flux equations:
(2.188) (2.189) (2.190)
The equations for holes are obtained by replacing by and taking into account that :
(2.191) (2.192)
(2.193) (2.194)
(2.195) (2.196)
M. Gritsch: Numerical Modeling of Silicon-on-Insulator MOSFETs PDF
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# Ticket #2103: 2103-rebased.patch
File 2103-rebased.patch, 47.2 KB (added by davidloeffler, 5 years ago)
rebased to 3.1.4 (also works in 3.2.alpha1)
• ## sage/modular/abvar/abvar.py
# HG changeset patch
# User David Loeffler <D.Loeffler@dpmms.cam.ac.uk>
# Date 1225191768 0
# Node ID dc06bf9ff94210b7ba42e52999af0b6d27d2ffab
# Parent 3859ace869681b4efc4ded59cf84a24e26383e9f
#2103: equivalence classes of cusps for congruence subgroups
diff -r 3859ace86968 -r dc06bf9ff942 sage/modular/abvar/abvar.py
a class that matters: sage: D = (J0(11)*J1(11)).decomposition(); D [ Simple abelian subvariety 11a(1,11) of dimension 1 of J0(11) x J1(11), Simple abelian subvariety 11aG1(1,11) of dimension 1 of J0(11) x J1(11) Simple abelian subvariety 11aG1(1,11) of dimension 1 of J0(11) x J1(11), Simple abelian subvariety 11a(1,11) of dimension 1 of J0(11) x J1(11) ] sage: D[0]._complement_shares_no_factors_with_same_label() True sage: D[0].newform_label() '11a' '11aG1' sage: D[1].newform_label() '11aG1' '11a' """ try: return self.__complement_shares
• ## sage/modular/congroup.py
diff -r 3859ace86968 -r dc06bf9ff942 sage/modular/congroup.py
a import sage.rings.arith as arith from sage.rings.integer_mod_ring import IntegerModRing from sage.rings.all import QQ, ZZ, divisors from sage.matrix.matrix_space import MatrixSpace from congroup_element import CongruenceSubgroupElement import cusps from sage.sets.set import Set import sage.modular.modsym.p1list return x raise NotImplementedError def reduce_cusp(self, c): r""" Given a cusp $c \in \mathbb{P}^1(\mathbb{Q})$, return the unique reduced cusp equivalent to c under the action of self, where a reduced cusp is an element r/s with r,s coprime integers, s as small as possible, and r as small as possible for that s. NOTE: This function should be overridden by all subclasses. EXAMPLES: sage: sage.modular.congroup.CongruenceSubgroup(5).reduce_cusp(1/4) Traceback (most recent call last): ... NotImplementedError """ raise NotImplementedError def cusps(self, bdmap=False): r""" Return a set of inequivalent cusps for self, i.e. a set of representatives for the orbits of self on $\mathbb{P}^1(\mathbb{Q})$. These should be returned in a reduced form. INPUTS: (bool) -- whether to directly calculate the cusps, or to calculate them as a by-product of computing the boundary map on the associated space of modular symbols. The latter is slower, but is older and presumably better-tested code. EXAMPLES: sage: Gamma0(36).cusps() {1/6, 1/4, 1/3, 1/2, 1/9, 0, 1/18, 5/12, Infinity, 1/12, 2/3, 5/6} sage: Gamma0(36).cusps(True) == Gamma0(36).cusps(False) True sage: GammaH(36, [19,29]).cusps() == Gamma0(36).cusps() True sage: Gamma0(1).cusps() {Infinity} """ # special case for SL2Z as modular symbols space is zero if bdmap == True: # special case for SL2Z as modular symbols space is zero if self == SL2Z: return Set([cusps.Cusp(1,0)]) return Set([self.reduce_cusp(c) for c in self.modular_symbols().cusps()]) else: return self._find_cusps() def _find_cusps(self): r""" Calculate a list of inequivalent cusps. EXAMPLES: sage: sage.modular.congroup.CongruenceSubgroup(5)._find_cusps() Traceback (most recent call last): ... NotImplementedError NOTE: This function should be overridden by all subclasses. """ raise NotImplementedError def lift_to_sl2z(c, d, N): """ Given a vector (c, d) in $(Z/NZ)^2$, this function computes and return [z2, -z1, c, d] def is_Gamma0(x): """ Return True if x is a congruence subgroup of type Gamma0. EXAMPLES: sage: from sage.modular.congroup import is_Gamma0 sage: is_Gamma0(SL2Z) True sage: is_Gamma0(Gamma0(13)) True sage: is_Gamma0(Gamma1(6)) False """ return isinstance(x, Gamma0_class) _gamma0_cache = {} def Gamma0(N): """ Return the congruence subgroup Gamma0(N). EXAMPLES: sage: G = Gamma0(51) ; G Congruence Subgroup Gamma0(51) sage: G == Gamma0(51) True sage: G is Gamma0(51) True """ try: return _gamma0_cache[N] except KeyError: _gamma0_cache[N] = Gamma0_class(N) return _gamma0_cache[N] class Gamma0_class(CongruenceSubgroup): def __init__(self, level): r""" The congruence subgroup $\Gamma_0(N)$. EXAMPLES: sage: G = Gamma0(11); G Congruence Subgroup Gamma0(11) sage: loads(G.dumps()) == G True """ CongruenceSubgroup.__init__(self, level) def _repr_(self): """ Return the string representation of self. EXAMPLES: sage: Gamma0(98)._repr_() 'Congruence Subgroup Gamma0(98)' """ return "Congruence Subgroup Gamma0(%s)"%self.level() def __reduce__(self): """ Used for pickling self. EXAMPLES: sage: Gamma0(22).__reduce__() (, (22,)) """ return Gamma0, (self.level(),) def _latex_(self): r""" Return the \LaTeX representation of self. EXAMPLES: sage: Gamma0(20)._latex_() '\\Gamma_0(20)' sage: latex(Gamma0(20)) \Gamma_0(20) """ return "\\Gamma_0(%s)"%self.level() def _generators_for_H(self): """ Return generators for the subgroup H of the units mod self.level() that defines self. EXAMPLES: sage: Gamma0(15)._generators_for_H() [11, 7] """ try: return self.__generators_for_H except AttributeError: self.__generators_for_H = [int(x) for x in IntegerModRing(self.level()).unit_gens()] return self.__generators_for_H def _list_of_elements_in_H(self): """ Returns a sorted list of Python ints that are representatives between 0 and N-1 of the elements of H. EXAMPLES: sage: G = Gamma0(11) sage: G._list_of_elements_in_H() [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] sage: G = Gamma0(6) sage: G._list_of_elements_in_H() [1, 5] sage: G = Gamma0(1) sage: G._list_of_elements_in_H() [1] """ N = self.level() if N != 1: gcd = arith.gcd return [ x for x in range(1, N) if gcd(x, N) == 1 ] else: return [1] def __cmp__(self, right): """ Compare self to right. EXAMPLES: sage: Gamma0(21).__cmp__(Gamma0(21)) 0 sage: Gamma0(21) < Gamma0(32) True """ if not is_Gamma0(right): if is_CongruenceSubgroup(right): c = cmp(self.level(), right.level()) if c: return c return cmp(type(self), type(right)) return cmp(self.level(), right.level()) def is_even(self): """ Return True precisely if this subgroup contains the matrix -1. Since Gamma0(N) always, contains the matrix -1, this always returns True. EXAMPLES: sage: Gamma0(12).is_even() True sage: SL2Z.is_even() True """ return True def is_subgroup(self, right): """ Return True if self is a subgroup of right. EXAMPLES: sage: G = Gamma0(20) sage: G.is_subgroup(SL2Z) True sage: G.is_subgroup(Gamma0(4)) True sage: G.is_subgroup(Gamma0(20)) True sage: G.is_subgroup(Gamma0(7)) False sage: Gamma0(2).is_subgroup(Gamma1(2)) True """ if right.level() == 1: return True if is_Gamma0(right): return self.level() % right.level() == 0 if is_Gamma1(right): if right.level() >= 3: return False elif right.level() == 2: return self.level() == 2 # case level 1 dealt with above raise NotImplementedError def coset_reps(self): r""" Return representatives for the right cosets of this congruence subgroup in ${\rm SL}_2(\Z)$ as a generator object. Use \code{list(self.coset_reps())} to obtain coset reps as a list. EXAMPLES: sage: list(Gamma0(5).coset_reps()) [[1, 0, 0, 1], [0, -1, 1, 0], [1, 0, 1, 1], [1, 1, 1, 2], [1, 2, 1, 3], [1, 3, 1, 4]] sage: list(Gamma0(4).coset_reps()) [[1, 0, 0, 1], [0, -1, 1, 0], [1, 0, 1, 1], [1, 1, 1, 2], [1, 2, 1, 3], [-1, -1, 2, 1]] sage: list(Gamma0(1).coset_reps()) [[1, 0, 0, 1]] """ N = self.level() for z in sage.modular.modsym.p1list.P1List(N): yield lift_to_sl2z(z[0], z[1], N) def generators(self): r""" Return generators for this congruence subgroup. The result is cached. EXAMPLE: sage: for g in Gamma0(3).generators(): ... print g ... print '---' [1 1] [0 1] --- [-1 0] [ 0 -1] --- ... --- [-2 1] [-3 1] --- """ try: return self.__gens except AttributeError: from sage.modular.modsym.p1list import P1List from congroup_pyx import generators_helper level = self.level() gen_list = generators_helper(P1List(level), level, Mat2Z) self.__gens = [self(g, check=False) for g in gen_list] return self.__gens def gamma_h_subgroups(self): r""" Return the subgroups of the form $\Gamma_H(N)$ contained in self, where $N$ is the level of self. EXAMPLES: sage: G = Gamma0(11) sage: G.gamma_h_subgroups() [Congruence Subgroup Gamma_H(11) with H generated by [2], Congruence Subgroup Gamma_H(11) with H generated by [4], Congruence Subgroup Gamma_H(11) with H generated by [10], Congruence Subgroup Gamma_H(11) with H generated by []] sage: G = Gamma0(12) sage: G.gamma_h_subgroups() [Congruence Subgroup Gamma_H(12) with H generated by [5, 7], Congruence Subgroup Gamma_H(12) with H generated by [7], Congruence Subgroup Gamma_H(12) with H generated by [5], Congruence Subgroup Gamma_H(12) with H generated by []] """ N = self.level() R = IntegerModRing(N) return [GammaH(N, H) for H in R.multiplicative_subgroups()] def __call__(self, x, check=True): r""" Create an element of this congruence subgroup from x. If the optional flag check is True (default), check whether x actually gives an element of self. EXAMPLES: sage: G = Gamma0(12) sage: G([1, 0, 24, 1]) [ 1 0] [24 1] sage: G(matrix(ZZ, 2, [1, 1, -12, -11])) [ 1 1] [-12 -11] sage: G([1, 0, 23, 1]) Traceback (most recent call last): ... TypeError: matrix must have lower left entry (=23) divisible by 12 """ if isinstance(x, CongruenceSubgroupElement) and x.parent() == self: return x x = CongruenceSubgroupElement(self, x, check=check) if not check: return x c = x.c() N = self.level() if c%N == 0: return x else: raise TypeError, "matrix must have lower left entry (=%s) divisible by %s" %(c, N) def is_SL2Z(x): """ Return True if x is the modular group ${\rm SL}_2(\Z)$. EXAMPLES: sage: from sage.modular.congroup import is_SL2Z sage: is_SL2Z(SL2Z) True sage: is_SL2Z(Gamma0(6)) False """ return isinstance(x, SL2Z_class) class SL2Z_class(Gamma0_class): def __init__(self): r""" The modular group ${\rm SL}_2(\Z)$. EXAMPLES: sage: G = SL2Z; G Modular Group SL(2,Z) sage: G.gens() ([ 0 -1] [ 1 0], [1 1] [0 1]) sage: G.0 [ 0 -1] [ 1 0] sage: G.1 [1 1] [0 1] sage: latex(G) \mbox{\rm SL}_2(\mathbf{Z}) sage: G([1,-1,0,1]) [ 1 -1] [ 0 1] sage: loads(G.dumps()) == G True sage: SL2Z.0 * SL2Z.1 [ 0 -1] [ 1 1] sage: SL2Z == loads(dumps(SL2Z)) True sage: SL2Z is loads(dumps(SL2Z)) True """ Gamma0_class.__init__(self, 1) def __reduce__(self): """ Used for pickling self. EXAMPLES: sage: SL2Z.__reduce__() (, ()) """ return _SL2Z_ref, () def _repr_(self): """ Return the string representation of self. EXAMPLES: sage: SL2Z._repr_() 'Modular Group SL(2,Z)' """ return "Modular Group SL(2,Z)" def _latex_(self): r""" Return the \LaTeX representation of self. EXAMPLES: sage: SL2Z._latex_() '\\mbox{\\rm SL}_2(\\mathbf{Z})' sage: latex(SL2Z) \mbox{\rm SL}_2(\mathbf{Z}) """ return "\\mbox{\\rm SL}_2(%s)"%(ZZ._latex_()) def is_subgroup(self, right): """ Return True if self is a subgroup of right. EXAMPLES: sage: SL2Z.is_subgroup(SL2Z) True sage: SL2Z.is_subgroup(Gamma1(1)) True sage: SL2Z.is_subgroup(Gamma0(6)) False """ return right.level() == 1 SL2Z = SL2Z_class() def _SL2Z_ref(): """ Return SL2Z. (Used for pickling SL2Z.) EXAMPLES: sage: sage.modular.congroup._SL2Z_ref() Modular Group SL(2,Z) sage: sage.modular.congroup._SL2Z_ref() is SL2Z True """ return SL2Z def is_Gamma1(x): """ Return True if x is a congruence subgroup of type Gamma1. EXAMPLES: sage: from sage.modular.congroup import is_Gamma1 sage: is_Gamma1(SL2Z) True sage: is_Gamma1(Gamma1(13)) True sage: is_Gamma1(Gamma0(6)) False """ return (isinstance(x, Gamma1_class) or is_SL2Z(x)) _gamma1_cache = {} def Gamma1(N): r""" Return the congruence subgroup $\Gamma_1(N)$. EXAMPLES: sage: Gamma1(5) Congruence Subgroup Gamma1(5) sage: G = Gamma1(23) sage: G is Gamma1(23) True sage: G == loads(dumps(G)) True sage: G is loads(dumps(G)) True """ try: return _gamma1_cache[N] except KeyError: _gamma1_cache[N] = Gamma1_class(N) return _gamma1_cache[N] class Gamma1_class(CongruenceSubgroup): def __init__(self, level): r""" The congruence subgroup $\Gamma_1(N)$. EXAMPLES: sage: G = Gamma1(11); G Congruence Subgroup Gamma1(11) sage: loads(G.dumps()) == G True """ CongruenceSubgroup.__init__(self, level) def _repr_(self): """ Return the string representation of self. EXAMPLES: sage: Gamma1(133)._repr_() 'Congruence Subgroup Gamma1(133)' """ return "Congruence Subgroup Gamma1(%s)"%self.level() def __reduce__(self): """ Used for pickling self. EXAMPLES: sage: Gamma1(82).__reduce__() (, (82,)) """ return Gamma1, (self.level(),) def _latex_(self): r""" Return the \LaTeX representation of self. EXAMPLES: sage: Gamma1(3)._latex_() '\\Gamma_1(3)' sage: latex(Gamma1(3)) \Gamma_1(3) """ return "\\Gamma_1(%s)"%self.level() def __cmp__(self, right): """ Compare self to right. EXAMPLES: sage: G = Gamma1(111) sage: G.__cmp__(Gamma1(111)) 0 sage: G.__cmp__(135) is not 0 True """ if not is_Gamma1(right): if is_CongruenceSubgroup(right): c = cmp(self.level(), right.level()) if c: return c return cmp(type(self), type(right)) return cmp(self.level(), right.level()) def is_even(self): """ Return True precisely if this subgroup contains the matrix -1. EXAMPLES: sage: Gamma1(1).is_even() True sage: Gamma1(2).is_even() True sage: Gamma1(15).is_even() False """ return self.level() in [1,2] def is_subgroup(self, right): """ Return True if self is a subgroup of right. EXAMPLES: sage: Gamma1(3).is_subgroup(SL2Z) True sage: Gamma1(3).is_subgroup(Gamma1(5)) False sage: Gamma1(3).is_subgroup(Gamma1(6)) False sage: Gamma1(6).is_subgroup(Gamma1(3)) True sage: Gamma1(6).is_subgroup(Gamma0(2)) True """ if right.level() == 1: return True if is_Gamma0(right) or is_Gamma1(right): return self.level() % right.level() == 0 raise NotImplementedError def generators(self): r""" Return generators for this congruence subgroup. The result is cached. EXAMPLE: sage: for g in Gamma1(3).generators(): ... print g ... print '---' [1 1] [0 1] --- [ 31 -14] [ 51 -23] --- [-5 4] [-9 7] --- ... --- [4 3] [9 7] --- [ -5 -2] [-12 -5] --- """ try: return self.__gens except AttributeError: from sage.modular.modsym.g1list import G1list from congroup_pyx import generators_helper level = self.level() gen_list = generators_helper(G1list(level), level, Mat2Z) self.__gens = [self(g, check=False) for g in gen_list] return self.__gens def __call__(self, x, check=True): r""" Create an element of this congruence subgroup from x. If the optional flag check is True (default), check whether x actually gives an element of self. EXAMPLES: sage: G = Gamma1(5) sage: G([1, 0, -10, 1]) [ 1 0] [-10 1] sage: G(matrix(ZZ, 2, [6, 1, 5, 1])) [6 1] [5 1] sage: G([1, 1, 6, 7]) Traceback (most recent call last): ... TypeError: matrix must have diagonal entries (=1, 7) congruent to 1 modulo 5, and lower left entry (=6) divisible by 5 """ if isinstance(x, CongruenceSubgroupElement) and x.parent() == self: return x x = CongruenceSubgroupElement(self, x, check=check) if not check: return x a = x.a() c = x.c() d = x.d() N = self.level() if (a%N == 1) and (c%N == 0) and (d%N == 1): return x else: raise TypeError, "matrix must have diagonal entries (=%s, %s) congruent to 1 modulo %s, and lower left entry (=%s) divisible by %s" %(a, d, N, c, N) _gammaH_cache = {} def GammaH(level, H): r""" Return the congruence subgroup $\Gamma_H(N)$. Return the congruence subgroup $\Gamma_H(N)$, which is the subgroup of $SL_2(\Z)$ consisting of matrices of the form $\begin{pmatrix} a & b \\ c & d \end{pmatrix$ with $N | c$ and $a, b \in H$, for $H$ a specified subgroup of $(\Z/N\Z)^\times$. INPUT: level -- an integer Return True if x is a congruence subgroup of type GammaH. EXAMPLES: sage: from sage.modular.congroup import is_GammaH sage: from sage.modular.congroup import is_GammaH sage: is_GammaH(GammaH(13, [2])) True sage: is_GammaH(Gamma0(6)) True sage: is_GammaH(sage.modular.congroup.CongruenceSubgroup(5)) False """ return isinstance(x, GammaH_class) def __cmp__(self, other): """ Compare self to right. Compare self to right. The ordering on congruence subgroups is first by level, then by the set of elements of H. Note that Gamma1 and Gamma0 groups are treated as instances of GammaH for the purposes of comparison. EXAMPLES: sage: G = GammaH(86, [9]) 0 sage: G.__cmp__(GammaH(86, [11])) is not 0 True sage: Gamma1(11) < Gamma0(11) True sage: Gamma1(11) == GammaH(11, []) True sage: Gamma0(11) == GammaH(11, [2]) True """ if not is_CongruenceSubgroup(other): return cmp(type(self), type(other)) c = cmp(self.level(), other.level()) if c: return c if is_GammaH(other): c = cmp(self.level(), other.level()) if c: return c return cmp(self._list_of_elements_in_H(), other._list_of_elements_in_H()) return cmp(type(self), type(other)) return (new_u, new_v) def reduce_cusp(self, c): r""" Compute a minimal representative for the given cusp c. Returns a cusp c' which is equivalent to the given cusp, and is in lowest terms with minimal positive denominator, and minimal positive numerator for that denominator. Two cusps $u1/v1$ and $u2/v2$ are equivalent modulo $\Gamma_H(N)$ if and only if $v1 = h*v2 (mod N)$ and $u1 = h^(-1)*u2 (mod gcd(v1,N))$ or $v1 = -h*v2 (mod N)$ and $u1 = -h^(-1)*u2 (mod gcd(v1,N))$ for some $h \in H$. EXAMPLES: sage: GammaH(6,[5]).reduce_cusp(Cusp(5,3)) 1/3 sage: GammaH(12,[5]).reduce_cusp(Cusp(8,9)) 1/3 sage: GammaH(12,[5]).reduce_cusp(Cusp(5,12)) Infinity sage: GammaH(12,[]).reduce_cusp(Cusp(5,12)) 5/12 sage: GammaH(21,[5]).reduce_cusp(Cusp(-9/14)) 1/7 """ return self._reduce_cusp(c)[0] def _reduce_cusp(self, c): r""" Compute a minimal representative for the given cusp c. Returns a pair (c', t), where c' is the minimal representative for the given cusp, and t is either 1 or -1, as explained below. Compute a minimal representative for the given cusp c. Returns a pair (c', t), where c' is the minimal representative for the given cusp, and t is either 1 or -1, as explained below. Largely for internal use. The minimal representative for a cusp is the element in $P^1(Q)$ in lowest terms with minimal positive denominator, and minimal positive numerator for that denominator. sage: GammaH(21,[5])._reduce_cusp(Cusp(-9/14)) (1/7, 1) """ N = int(self.level()) Cusps = c.parent() v = int(c.denominator() % N) sign = s return Cusps((u_min, val_min)), sign def _find_cusps(self): r""" Return a set of inequivalent cusps for self, i.e. a set of representatives for the orbits of self on $\mathbb{P}^1(\mathbb{Q})$. These are returned in a reduced form; see self.reduce_cusp for the definition of reduced. ALGORITHM: Lemma 3.2 in Cremona's 1997 book shows that for the action of Gamma1(N) on "signed projective space" $\Q^2 / (\Q_{\ge 0}^+)$, we have $u_1/v_1 \sim u_2 / v_2$ if and only if $v_1 = v_2 \bmod N$ and $u_1 = u_2 \bmod gcd(v_1, N)$. It follows that every orbit has a representative $u/v$ with $v \le N$ and $0 \le u \le gcd(v, N)$. We iterate through all pairs $(u,v)$ satisfying this. Having found a set containing at least one of every equivalence class modulo Gamma1(N), we can be sure of picking up every class modulo GammaH(N) since this contains Gamma1(N); and the reduce_cusp call does the checking to make sure we don't get too many duplicates. EXAMPLES: sage: Gamma1(5)._find_cusps() {0, Infinity, 1/2, 2/5} sage: Gamma1(35)._find_cusps() {3/35, 9/10, 9/14, 11/35, 3/14, 9/35, 3/10, 11/14, 8/35, 4/7, 8/15, Infinity, 13/14, 16/35, 13/35, 0, 4/15, 1/13, 2/35, 1/11, 1/10, 1/17, 1/16, 1/15, 1/14, 1/7, 1/6, 1/5, 1/4, 1/3, 1/2, 6/35, 1/9, 1/8, 6/7, 3/5, 3/7, 7/10, 4/35, 2/5, 17/35, 2/7, 5/7, 1/12, 4/5, 5/14, 12/35, 2/15} sage: Gamma1(24)._find_cusps() == Gamma1(24).cusps(bdmap=True) True sage: GammaH(24, [13,17])._find_cusps() == GammaH(24,[13,17]).cusps(bdmap=True) True """ s = [] N=self.level() for d in xrange(1, 1+N): w = arith.gcd(d, N) for a in xrange(1,(w==1 and 2) or w): if arith.gcd([a, d, w]) != 1: continue while arith.gcd(a, d) != 1: a += w s.append(self.reduce_cusp(cusps.Cusp(a,d))) return Set(s) def __call__(self, x, check=True): r""" Create an element of this congruence subgroup from x. raise TypeError, "matrix must have lower right entry (=%s) congruent modulo %s to some element of H" %(d, N) def is_Gamma0(x): """ Return True if x is a congruence subgroup of type Gamma0. EXAMPLES: sage: from sage.modular.congroup import is_Gamma0 sage: is_Gamma0(SL2Z) True sage: is_Gamma0(Gamma0(13)) True sage: is_Gamma0(Gamma1(6)) False """ return isinstance(x, Gamma0_class) _gamma0_cache = {} def Gamma0(N): """ Return the congruence subgroup Gamma0(N). EXAMPLES: sage: G = Gamma0(51) ; G Congruence Subgroup Gamma0(51) sage: G == Gamma0(51) True sage: G is Gamma0(51) True """ try: return _gamma0_cache[N] except KeyError: _gamma0_cache[N] = Gamma0_class(N) return _gamma0_cache[N] class Gamma0_class(GammaH_class): def __init__(self, level): r""" The congruence subgroup $\Gamma_0(N)$. EXAMPLES: sage: G = Gamma0(11); G Congruence Subgroup Gamma0(11) sage: loads(G.dumps()) == G True """ GammaH_class.__init__(self, level, [int(x) for x in IntegerModRing(level).unit_gens()]) def _repr_(self): """ Return the string representation of self. EXAMPLES: sage: Gamma0(98)._repr_() 'Congruence Subgroup Gamma0(98)' """ return "Congruence Subgroup Gamma0(%s)"%self.level() def __reduce__(self): """ Used for pickling self. EXAMPLES: sage: Gamma0(22).__reduce__() (, (22,)) """ return Gamma0, (self.level(),) def _latex_(self): r""" Return the \LaTeX representation of self. EXAMPLES: sage: Gamma0(20)._latex_() '\\Gamma_0(20)' sage: latex(Gamma0(20)) \Gamma_0(20) """ return "\\Gamma_0(%s)"%self.level() def _generators_for_H(self): """ Return generators for the subgroup H of the units mod self.level() that defines self. EXAMPLES: sage: Gamma0(15)._generators_for_H() [11, 7] """ try: return self.__generators_for_H except AttributeError: self.__generators_for_H = [int(x) for x in IntegerModRing(self.level()).unit_gens()] return self.__generators_for_H def _list_of_elements_in_H(self): """ Returns a sorted list of Python ints that are representatives between 0 and N-1 of the elements of H. EXAMPLES: sage: G = Gamma0(11) sage: G._list_of_elements_in_H() [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] sage: G = Gamma0(6) sage: G._list_of_elements_in_H() [1, 5] sage: G = Gamma0(1) sage: G._list_of_elements_in_H() [1] """ N = self.level() if N != 1: gcd = arith.gcd return [ x for x in range(1, N) if gcd(x, N) == 1 ] else: return [1] def is_even(self): """ Return True precisely if this subgroup contains the matrix -1. Since Gamma0(N) always contains the matrix -1, this always returns True. EXAMPLES: sage: Gamma0(12).is_even() True sage: SL2Z.is_even() True """ return True def is_subgroup(self, right): """ Return True if self is a subgroup of right. EXAMPLES: sage: G = Gamma0(20) sage: G.is_subgroup(SL2Z) True sage: G.is_subgroup(Gamma0(4)) True sage: G.is_subgroup(Gamma0(20)) True sage: G.is_subgroup(Gamma0(7)) False sage: Gamma0(2).is_subgroup(Gamma1(2)) True """ if right.level() == 1: return True if is_Gamma0(right): return self.level() % right.level() == 0 if is_Gamma1(right): if right.level() >= 3: return False elif right.level() == 2: return self.level() == 2 # case level 1 dealt with above raise NotImplementedError def coset_reps(self): r""" Return representatives for the right cosets of this congruence subgroup in ${\rm SL}_2(\Z)$ as a generator object. Use \code{list(self.coset_reps())} to obtain coset reps as a list. EXAMPLES: sage: list(Gamma0(5).coset_reps()) [[1, 0, 0, 1], [0, -1, 1, 0], [1, 0, 1, 1], [1, 1, 1, 2], [1, 2, 1, 3], [1, 3, 1, 4]] sage: list(Gamma0(4).coset_reps()) [[1, 0, 0, 1], [0, -1, 1, 0], [1, 0, 1, 1], [1, 1, 1, 2], [1, 2, 1, 3], [-1, -1, 2, 1]] sage: list(Gamma0(1).coset_reps()) [[1, 0, 0, 1]] """ N = self.level() for z in sage.modular.modsym.p1list.P1List(N): yield lift_to_sl2z(z[0], z[1], N) def generators(self): r""" Return generators for this congruence subgroup. The result is cached. EXAMPLE: sage: for g in Gamma0(3).generators(): ... print g ... print '---' [1 1] [0 1] --- [-1 0] [ 0 -1] --- ... --- [-2 1] [-3 1] --- """ try: return self.__gens except AttributeError: from sage.modular.modsym.p1list import P1List from congroup_pyx import generators_helper level = self.level() gen_list = generators_helper(P1List(level), level, Mat2Z) self.__gens = [self(g, check=False) for g in gen_list] return self.__gens def gamma_h_subgroups(self): r""" Return the subgroups of the form $\Gamma_H(N)$ contained in self, where $N$ is the level of self. EXAMPLES: sage: G = Gamma0(11) sage: G.gamma_h_subgroups() [Congruence Subgroup Gamma_H(11) with H generated by [2], Congruence Subgroup Gamma_H(11) with H generated by [4], Congruence Subgroup Gamma_H(11) with H generated by [10], Congruence Subgroup Gamma_H(11) with H generated by []] sage: G = Gamma0(12) sage: G.gamma_h_subgroups() [Congruence Subgroup Gamma_H(12) with H generated by [5, 7], Congruence Subgroup Gamma_H(12) with H generated by [7], Congruence Subgroup Gamma_H(12) with H generated by [5], Congruence Subgroup Gamma_H(12) with H generated by []] """ N = self.level() R = IntegerModRing(N) return [GammaH(N, H) for H in R.multiplicative_subgroups()] def __call__(self, x, check=True): r""" Create an element of this congruence subgroup from x. If the optional flag check is True (default), check whether x actually gives an element of self. EXAMPLES: sage: G = Gamma0(12) sage: G([1, 0, 24, 1]) [ 1 0] [24 1] sage: G(matrix(ZZ, 2, [1, 1, -12, -11])) [ 1 1] [-12 -11] sage: G([1, 0, 23, 1]) Traceback (most recent call last): ... TypeError: matrix must have lower left entry (=23) divisible by 12 """ if isinstance(x, CongruenceSubgroupElement) and x.parent() == self: return x x = CongruenceSubgroupElement(self, x, check=check) if not check: return x c = x.c() N = self.level() if c%N == 0: return x else: raise TypeError, "matrix must have lower left entry (=%s) divisible by %s" %(c, N) def _find_cusps(self): r""" Return a set of inequivalent cusps for self, i.e. a set of representatives for the orbits of self on $\mathbb{P}^1(\mathbb{Q})$. These are returned in a reduced form; see self.reduce_cusp for the definition of reduced. ALGORITHM: Uses explicit formulae specific to $\Gamma_0(N)$: a reduced cusp on $\Gamma_0(N)$ is always of the form $a/d$ where $d | N$, and $a_1/d \sim a_2/d$ if and only if $a_1 \cong a_2 \bmod {\rm gcd}(d, N/d)$. EXAMPLES: sage: Gamma0(90)._find_cusps() {1/6, 1/5, 1/3, 1/2, 11/30, 1/9, 2/3, 1/30, Infinity, 5/6, 1/45, 0, 1/18, 1/10, 1/15, 2/15} sage: Gamma0(1).cusps() {Infinity} sage: Gamma0(180).cusps() == Gamma0(180).cusps(bdmap=True) True """ N = self.level() s = [] for d in divisors(N): w = arith.gcd(d, N/d) if w == 1: if d == 1: s.append(cusps.Cusp(1,0)) elif d == N: s.append(cusps.Cusp(0,1)) else: s.append(cusps.Cusp(1,d)) else: for a in xrange(1, w): if arith.gcd(a, w) == 1: while arith.gcd(a, d/w) != 1: a += w s.append(cusps.Cusp(a,d)) return Set(s) def is_SL2Z(x): """ Return True if x is the modular group ${\rm SL}_2(\Z)$. EXAMPLES: sage: from sage.modular.congroup import is_SL2Z sage: is_SL2Z(SL2Z) True sage: is_SL2Z(Gamma0(6)) False """ return isinstance(x, SL2Z_class) class SL2Z_class(Gamma0_class): def __init__(self): r""" The modular group ${\rm SL}_2(\Z)$. EXAMPLES: sage: G = SL2Z; G Modular Group SL(2,Z) sage: G.gens() ([ 0 -1] [ 1 0], [1 1] [0 1]) sage: G.0 [ 0 -1] [ 1 0] sage: G.1 [1 1] [0 1] sage: latex(G) \mbox{\rm SL}_2(\mathbf{Z}) sage: G([1,-1,0,1]) [ 1 -1] [ 0 1] sage: loads(G.dumps()) == G True sage: SL2Z.0 * SL2Z.1 [ 0 -1] [ 1 1] sage: SL2Z == loads(dumps(SL2Z)) True sage: SL2Z is loads(dumps(SL2Z)) True """ Gamma0_class.__init__(self, 1) def __reduce__(self): """ Used for pickling self. EXAMPLES: sage: SL2Z.__reduce__() (, ()) """ return _SL2Z_ref, () def _repr_(self): """ Return the string representation of self. EXAMPLES: sage: SL2Z._repr_() 'Modular Group SL(2,Z)' """ return "Modular Group SL(2,Z)" def _latex_(self): r""" Return the \LaTeX representation of self. EXAMPLES: sage: SL2Z._latex_() '\\mbox{\\rm SL}_2(\\mathbf{Z})' sage: latex(SL2Z) \mbox{\rm SL}_2(\mathbf{Z}) """ return "\\mbox{\\rm SL}_2(%s)"%(ZZ._latex_()) def is_subgroup(self, right): """ Return True if self is a subgroup of right. EXAMPLES: sage: SL2Z.is_subgroup(SL2Z) True sage: SL2Z.is_subgroup(Gamma1(1)) True sage: SL2Z.is_subgroup(Gamma0(6)) False """ return right.level() == 1 def reduce_cusp(self, c): r""" Return the unique reduced cusp equivalent to c under the action of self. Always returns Infinity, since there is only one equivalence class of cusps for $SL_2(Z)$. EXAMPLES: sage: SL2Z.reduce_cusp(Cusps(-1/4)) Infinity """ return cusps.Cusp(1,0) SL2Z = SL2Z_class() def _SL2Z_ref(): """ Return SL2Z. (Used for pickling SL2Z.) EXAMPLES: sage: sage.modular.congroup._SL2Z_ref() Modular Group SL(2,Z) sage: sage.modular.congroup._SL2Z_ref() is SL2Z True """ return SL2Z def is_Gamma1(x): """ Return True if x is a congruence subgroup of type Gamma1. EXAMPLES: sage: from sage.modular.congroup import is_Gamma1 sage: is_Gamma1(SL2Z) True sage: is_Gamma1(Gamma1(13)) True sage: is_Gamma1(Gamma0(6)) False """ return (isinstance(x, Gamma1_class) or is_SL2Z(x)) _gamma1_cache = {} def Gamma1(N): r""" Return the congruence subgroup $\Gamma_1(N)$. EXAMPLES: sage: Gamma1(5) Congruence Subgroup Gamma1(5) sage: G = Gamma1(23) sage: G is Gamma1(23) True sage: G == loads(dumps(G)) True sage: G is loads(dumps(G)) True """ try: return _gamma1_cache[N] except KeyError: _gamma1_cache[N] = Gamma1_class(N) return _gamma1_cache[N] class Gamma1_class(GammaH_class): def __init__(self, level): r""" The congruence subgroup $\Gamma_1(N)$. EXAMPLES: sage: G = Gamma1(11); G Congruence Subgroup Gamma1(11) sage: loads(G.dumps()) == G True """ GammaH_class.__init__(self, level, []) def _repr_(self): """ Return the string representation of self. EXAMPLES: sage: Gamma1(133)._repr_() 'Congruence Subgroup Gamma1(133)' """ return "Congruence Subgroup Gamma1(%s)"%self.level() def __reduce__(self): """ Used for pickling self. EXAMPLES: sage: Gamma1(82).__reduce__() (, (82,)) """ return Gamma1, (self.level(),) def _latex_(self): r""" Return the \LaTeX representation of self. EXAMPLES: sage: Gamma1(3)._latex_() '\\Gamma_1(3)' sage: latex(Gamma1(3)) \Gamma_1(3) """ return "\\Gamma_1(%s)"%self.level() def is_even(self): """ Return True precisely if this subgroup contains the matrix -1. EXAMPLES: sage: Gamma1(1).is_even() True sage: Gamma1(2).is_even() True sage: Gamma1(15).is_even() False """ return self.level() in [1,2] def is_subgroup(self, right): """ Return True if self is a subgroup of right. EXAMPLES: sage: Gamma1(3).is_subgroup(SL2Z) True sage: Gamma1(3).is_subgroup(Gamma1(5)) False sage: Gamma1(3).is_subgroup(Gamma1(6)) False sage: Gamma1(6).is_subgroup(Gamma1(3)) True sage: Gamma1(6).is_subgroup(Gamma0(2)) True """ if right.level() == 1: return True if is_Gamma0(right) or is_Gamma1(right): return self.level() % right.level() == 0 raise NotImplementedError def generators(self): r""" Return generators for this congruence subgroup. The result is cached. EXAMPLE: sage: for g in Gamma1(3).generators(): ... print g ... print '---' [1 1] [0 1] --- [ 31 -14] [ 51 -23] --- [-5 4] [-9 7] --- ... --- [4 3] [9 7] --- [ -5 -2] [-12 -5] --- """ try: return self.__gens except AttributeError: from sage.modular.modsym.g1list import G1list from congroup_pyx import generators_helper level = self.level() gen_list = generators_helper(G1list(level), level, Mat2Z) self.__gens = [self(g, check=False) for g in gen_list] return self.__gens def __call__(self, x, check=True): r""" Create an element of this congruence subgroup from x. If the optional flag check is True (default), check whether x actually gives an element of self. EXAMPLES: sage: G = Gamma1(5) sage: G([1, 0, -10, 1]) [ 1 0] [-10 1] sage: G(matrix(ZZ, 2, [6, 1, 5, 1])) [6 1] [5 1] sage: G([1, 1, 6, 7]) Traceback (most recent call last): ... TypeError: matrix must have diagonal entries (=1, 7) congruent to 1 modulo 5, and lower left entry (=6) divisible by 5 """ if isinstance(x, CongruenceSubgroupElement) and x.parent() == self: return x x = CongruenceSubgroupElement(self, x, check=check) if not check: return x a = x.a() c = x.c() d = x.d() N = self.level() if (a%N == 1) and (c%N == 0) and (d%N == 1): return x else: raise TypeError, "matrix must have diagonal entries (=%s, %s) congruent to 1 modulo %s, and lower left entry (=%s) divisible by %s" %(a, d, N, c, N) import congroup_pyx degeneracy_coset_representatives_gamma0 = congroup_pyx.degeneracy_coset_representatives_gamma0 degeneracy_coset_representatives_gamma1 = congroup_pyx.degeneracy_coset_representatives_gamma1
• ## sage/modular/dims.py
diff -r 3859ace86968 -r dc06bf9ff942 sage/modular/dims.py
a else: raise TypeError def dimension_modular_forms_H(X, k=2): return dimension_cusp_forms_H(X, k) + dimension_eis_H(X, k) def dimension_modular_forms(X, k=2): r""" 1 sage: dimension_modular_forms(Gamma1(13),2) 13 sage: dimension_modular_forms(GammaH(11, [10]), 2) 10 sage: e = DirichletGroup(20).1 sage: dimension_modular_forms(e,3) 9 raise TypeError, "Argument 1 must be a congruence subgroup or Dirichlet character." if k == 0: return 1 if congroup.is_GammaH(X): if congroup.is_Gamma0(X) or congroup.is_Gamma1(X) or isinstance(X, dirichlet.DirichletCharacter): return dimension_cusp_forms(X, k) + dimension_eis(X, k) elif congroup.is_GammaH(X): return dimension_modular_forms_H(X, k) return dimension_cusp_forms(X, k) + dimension_eis(X, k) else: raise NotImplementedError, "Computation of dimensions for congruence subgroups other than Gamma0, Gamma1 or GammaH not implemented" def dimension_modular_forms_H(X, k=2): return dimension_cusp_forms_H(X, k) + dimension_eis_H(X, k) def sturm_bound(level, weight=2): r"""
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CC-MAIN-2013-20
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https://nl.mathworks.com/matlabcentral/cody/problems/41-cell-joiner/solutions/1856383
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Cody
# Problem 41. Cell joiner
Solution 1856383
Submitted on 24 Jun 2019 by Monika Phadnis
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
x = {'hello', 'basic', 'test', 'case'}; y_correct = 'hello basic test case'; assert(isequal(cellstr_joiner(x, ' '),y_correct))
2 Pass
x = {'this', 'one', '', 'has', ' ', 'some tricky', 'stuff'}; y_correct = 'this one has some tricky stuff'; assert(isequal(cellstr_joiner(x, ' '),y_correct))
3 Pass
x = {'delimiters', 'are', 'not', 'always', 'spaces'}; y_correct = 'delimiters?are?not?always?spaces'; assert(isequal(cellstr_joiner(x, '?'),y_correct))
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CC-MAIN-2019-51
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latest
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https://programmer.ink/think/artificial-mentally-retarded-three-piece-chess.html
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Artificial mentally retarded three piece chess
Posted by Zyxist on Tue, 22 Feb 2022 06:01:09 +0100
After learning arrays and functions, we can do some interesting things, such as the Gobang game mentioned in this article.
First of all, before we start to write code, we need to know what the game needs and divide the modules.
2. How do players choose game mode
3. game board
4. How players play chess
5. Computer chess method
6. How to judge winning or losing
Next, let's improve the content of the module
The game menu of Sanzi chess is very simple. It only takes 1 Start the game and 2 Just end the game
```void menu()
{
printf(" 1.Start the game ");
printf(" 0.End the game ");
}```
2. How do players choose game mode
We can use the switch statement to perform different game modes according to the value selected by the player
```void test()
{
int input = 0;
do
{
scanf("%d", &input);
switch (input)
{
case 1:
game();
break;
case 0:
printf("End the game\n");
break;
default:
printf("Wrong choice\n");
break;
}
} while (input);
}```
3. Game board
Use the two-dimensional array to store the chess pieces. According to the Sanzi game, you can print a 3 by output × 3 chessboard, and its contents need to be saved. Therefore, a 3 can be used × 3 to save.
1) Initialization of chessboard
```void InitBoard(char Board[3][3], int row, int col)
{
for (int i = 0; i < row; i++)
{
for (int j = 0; j < col; j++)
{
Board[i][j] = ' ';
}
}
}```
2) Presentation of chessboard: the chessboard of Sanzi is a tic tac toe, so it needs to reasonably arrange | symbols and - -- Symbols in order to print correctly
```void DiplayBoard(char Board[ROW][COL], int row, int col)
{
for (int i = 0; i < row; i++)
{
for (int j = 0; j < col; j++)
{
printf(" %c ", Board[i][j]);
if (j < col - 1)
{
printf("|");
}
}
printf("\n");
if (i < row - 1)
{
for (int j = 0; j < col; j++)
{
printf("---");
if (j != col - 1)
{
printf("|");
}
}
printf("\n");
}
}
}```
4. How players play chess
Factors to consider before settling
1) The player is not a programmer and doesn't know that the array starts from 0, so it needs to be optimized slightly
2) Before falling, you need to check whether the coordinates entered by the player are legal
3) Is there a piece in the player's position before falling
The way to play chess is to enter the elements of the array to drop the seeds
```Playmove(char Board[ROW][COL], int row, int col)
{
int x = 0, y = 0;
while (1)
{
scanf("%d %d", &x, &y);
if (x >= 1 && x <= row && y>=1 && y <= col)
{
if (Board[x - 1][y - 1] == ' ')
{
Board[x - 1][y - 1] = '*';
break;
}
else
{
}
}
else
{
printf("Input error\n");
}
}
}```
5. Methods of computer playing chess
What we use here is a chess playing method of artificial mental retardation: that is, we can use the time stamp to deal with it to make the number within the range of the array.
```void Computermove(char Board[ROW][COL], int row, int col)
{
int x = 0;
int y = 0;
while (1)
{
x = rand() % row;
y = rand() % col;
if (Board[x - 1][y - 1] == ' ')
{
Board[x - 1][y - 1] = '#';
break;
}
}
}```
5. How to judge the winner
The situation of winning Sanzi chess
1) There are three pieces in a line
2) There are three pieces in a line
3) There are three pieces in the diagonal line
```char is_win(char Board[ROW][COL], int row, int col)
{
//Judgment line
for (int i = 0; i < row; i++)
{
if (Board[i][0] == Board[i][1] && Board[i][1] == Board[i][2] && Board[i][1] != ' ')
{
return Board[i][1];
}
}
//Judgment column
for (int j = 0; j < col; j++)
{
if (Board[0][j] == Board[1][j] && Board[1][j] == Board[2][j] && Board[1][j] != ' ')
{
return Board[1][j];
}
}
//Judge diagonal
if (Board[0][0] == Board[1][1] && Board[1][1] == Board[2][2] && Board[1][1] != ' ')
{
return Board[1][1];
}
if (Board[0][2] == Board[1][1] && Board[1][1] == Board[2][0] && Board[1][1] != ' ')
{
return Board[1][1];
}
//it ends in a draw
if (is_full(Board,ROW,COL) == 1)
{
return 'Q';
}
//continue
return 'C';
}```
In Sanzi chess, it is mainly the application of array, and then it is to clarify the logic of chessboard and the logic of winning or losing. After figuring out these, you can easily realize Sanzi chess
Topics: C C++
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https://www.convertunits.com/from/lot/to/ton+%5Bmetric%5D
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## ››Convert lot [Germany] to ton [metric]
lot ton [metric]
Did you mean to convert lot to ton [short, US] long ton ton [metric] ton [long, UK]
How many lot in 1 ton [metric]? The answer is 66666.666666667.
We assume you are converting between lot [Germany] and ton [metric].
You can view more details on each measurement unit:
lot or ton [metric]
The SI base unit for mass is the kilogram.
1 kilogram is equal to 66.666666666667 lot, or 0.001 ton [metric].
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between lot [Germany] and tons.
Type in your own numbers in the form to convert the units!
## ››Want other units?
You can do the reverse unit conversion from ton [metric] to lot, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Ton
A tonne (also called metric ton) is a non-SI unit of mass, accepted for use with SI, defined as: 1 tonne = 1000 kg (= 106 g).
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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https://isabelle.in.tum.de/repos/isabelle/rev/4aaeb9427c96
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author wenzelm Sun, 11 Sep 2016 00:14:37 +0200 changeset 63833 4aaeb9427c96 parent 63832 a400b127853c child 63834 6a757f36997e
misc tuning and modernization;
```--- a/src/HOL/Old_Number_Theory/Pocklington.thy Sun Sep 11 00:13:25 2016 +0200
+++ b/src/HOL/Old_Number_Theory/Pocklington.thy Sun Sep 11 00:14:37 2016 +0200
@@ -492,8 +492,8 @@
{fix m assume m: "0 < m" "m < n" "\<not> coprime m n"
hence mS': "m \<notin> ?S'" by auto
have "insert m ?S' \<le> ?S" using m by auto
- from m have "card (insert m ?S') \<le> card ?S"
- by - (rule card_mono[of ?S "insert m ?S'"], auto)
+ have "card (insert m ?S') \<le> card ?S"
+ by (rule card_mono[of ?S "insert m ?S'"]) (use m in auto)
hence False
unfolding card_insert_disjoint[of "?S'" m, OF fS' mS'] ceq
by simp }
@@ -767,7 +767,7 @@
hence "(n - 1) mod m = 0" by auto
then have mn: "m dvd n - 1" by presburger
then obtain r where r: "n - 1 = m*r" unfolding dvd_def by blast
- from n01 r m(2) have r01: "r\<noteq>0" "r\<noteq>1" by - (rule ccontr, simp)+
+ from n01 r m(2) have r01: "r\<noteq>0" "r\<noteq>1" by auto
from prime_factor[OF r01(2)] obtain p where p: "prime p" "p dvd r" by blast
hence th: "prime p \<and> p dvd n - 1" unfolding r by (auto intro: dvd_mult)
have "(a ^ ((n - 1) div p)) mod n = (a^(m*r div p)) mod n" using r
@@ -800,8 +800,8 @@
moreover
{assume "n\<noteq>0 \<and> n\<noteq>1" hence n2:"n \<ge> 2" by arith
from na have na': "coprime a n" by (simp add: coprime_commute)
- from phi_lowerbound_1[OF n2] fermat_little[OF na']
- have ex: "\<exists>m>0. ?P m" by - (rule exI[where x="\<phi> n"], auto) }
+ have ex: "\<exists>m>0. ?P m"
+ by (rule exI[where x="\<phi> n"]) (use phi_lowerbound_1[OF n2] fermat_little[OF na'] in auto) }
ultimately have ex: "\<exists>m>0. ?P m" by blast
from nat_exists_least_iff'[of ?P] ex na show ?thesis
unfolding o[symmetric] by auto
@@ -992,7 +992,7 @@
from prime_factor[OF d(3)]
obtain p where p: "prime p" "p dvd d" by blast
from n have np: "n > 0" by arith
- from d(1) n have "d \<noteq> 0" by - (rule ccontr, auto)
+ have "d \<noteq> 0" by (rule ccontr) (use d(1) n in auto)
hence dp: "d > 0" by arith
from mult_mono[OF dvd_imp_le[OF p(2) dp] dvd_imp_le[OF p(2) dp]] d(2)
have "p\<^sup>2 \<le> n" unfolding power2_eq_square by arith
@@ -1029,7 +1029,7 @@
from d dvd_mult[OF P(2), of "ord p (a^r)"] have Pq: "P dvd q" by simp
from aq P(1) Pq have caP:"coprime (a^ ((n - 1) div P) - 1) n" by blast
from Pq obtain s where s: "q = P*s" unfolding dvd_def by blast
- have P0: "P \<noteq> 0" using P(1) prime_0 by - (rule ccontr, simp)
+ have P0: "P \<noteq> 0" by (rule ccontr) (use P(1) prime_0 in simp)
from P(2) obtain t where t: "d = P*t" unfolding dvd_def by blast
from d s t P0 have s': "ord p (a^r) * t = s" by algebra
have "ord p (a^r) * t*r = r * ord p (a^r) * t" by algebra
@@ -1052,8 +1052,8 @@
with divides_rexp[OF d(2)[unfolded dp], of "n - 2"]
have th0: "p dvd a ^ (n - 1)" by simp
from n have n0: "n \<noteq> 0" by simp
- from d(2) an n12[symmetric] have a0: "a \<noteq> 0"
- by - (rule ccontr, simp add: modeq_def)
+ have a0: "a \<noteq> 0"
+ by (rule ccontr) (use d(2) an n12[symmetric] in \<open>simp add: modeq_def\<close>)
have th1: "a^ (n - 1) \<noteq> 0" using n d(2) dp a0 by auto
from coprime_minus1[OF th1, unfolded coprime]
dvd_trans[OF pn cong_1_divides[OF an]] th0 d(3) dp
@@ -1064,7 +1064,7 @@
from fermat_little[OF arp, simplified ord_divides] o phip
have "q dvd (p - 1)" by simp
then obtain d where d:"p - 1 = q * d" unfolding dvd_def by blast
- from prime_0 pp have p0:"p \<noteq> 0" by - (rule ccontr, auto)
+ have p0: "p \<noteq> 0" by (rule ccontr) (use prime_0 pp in auto)
from p0 d have "p + q * 0 = 1 + q * d" by simp
with nat_mod[of p 1 q, symmetric]
show ?thesis by blast```
```--- a/src/HOL/Old_Number_Theory/Primes.thy Sun Sep 11 00:13:25 2016 +0200
+++ b/src/HOL/Old_Number_Theory/Primes.thy Sun Sep 11 00:14:37 2016 +0200
@@ -406,37 +406,44 @@
lemma prime_Suc0[simp]: "~ prime (Suc 0)" by (simp add: prime_def)
lemma prime_ge_2: "prime p ==> p \<ge> 2" by (simp add: prime_def)
-lemma prime_factor: assumes n: "n \<noteq> 1" shows "\<exists> p. prime p \<and> p dvd n"
-using n
-proof(induct n rule: nat_less_induct)
+
+lemma prime_factor: "n \<noteq> 1 \<Longrightarrow> \<exists>p. prime p \<and> p dvd n"
+proof (induct n rule: nat_less_induct)
fix n
assume H: "\<forall>m<n. m \<noteq> 1 \<longrightarrow> (\<exists>p. prime p \<and> p dvd m)" "n \<noteq> 1"
- let ?ths = "\<exists>p. prime p \<and> p dvd n"
- {assume "n=0" hence ?ths using two_is_prime by auto}
- moreover
- {assume nz: "n\<noteq>0"
- {assume "prime n" hence ?ths by - (rule exI[where x="n"], simp)}
- moreover
- {assume n: "\<not> prime n"
- with nz H(2)
- obtain k where k:"k dvd n" "k \<noteq> 1" "k \<noteq> n" by (auto simp add: prime_def)
+ show "\<exists>p. prime p \<and> p dvd n"
+ proof (cases "n = 0")
+ case True
+ with two_is_prime show ?thesis by auto
+ next
+ case nz: False
+ show ?thesis
+ proof (cases "prime n")
+ case True
+ then have "prime n \<and> n dvd n" by simp
+ then show ?thesis ..
+ next
+ case n: False
+ with nz H(2) obtain k where k: "k dvd n" "k \<noteq> 1" "k \<noteq> n"
+ by (auto simp: prime_def)
from dvd_imp_le[OF k(1)] nz k(3) have kn: "k < n" by simp
from H(1)[rule_format, OF kn k(2)] obtain p where p: "prime p" "p dvd k" by blast
- from dvd_trans[OF p(2) k(1)] p(1) have ?ths by blast}
- ultimately have ?ths by blast}
- ultimately show ?ths by blast
+ from dvd_trans[OF p(2) k(1)] p(1) show ?thesis by blast
+ qed
+ qed
qed
-lemma prime_factor_lt: assumes p: "prime p" and n: "n \<noteq> 0" and npm:"n = p * m"
+lemma prime_factor_lt:
+ assumes p: "prime p" and n: "n \<noteq> 0" and npm:"n = p * m"
shows "m < n"
-proof-
- {assume "m=0" with n have ?thesis by simp}
- moreover
- {assume m: "m \<noteq> 0"
- from npm have mn: "m dvd n" unfolding dvd_def by auto
- from npm m have "n \<noteq> m" using p by auto
- with dvd_imp_le[OF mn] n have ?thesis by simp}
- ultimately show ?thesis by blast
+proof (cases "m = 0")
+ case True
+ with n show ?thesis by simp
+next
+ case m: False
+ from npm have mn: "m dvd n" unfolding dvd_def by auto
+ from npm m have "n \<noteq> m" using p by auto
+ with dvd_imp_le[OF mn] n show ?thesis by simp
qed
lemma euclid_bound: "\<exists>p. prime p \<and> n < p \<and> p <= Suc (fact n)"
@@ -491,7 +498,7 @@
lemma coprime_bezout_strong: assumes ab: "coprime a b" and b: "b \<noteq> 1"
shows "\<exists>x y. a * x = b * y + 1"
proof-
- from ab b have az: "a \<noteq> 0" by - (rule ccontr, auto)
+ have az: "a \<noteq> 0" by (rule ccontr) (use ab b in auto)
from bezout_gcd_strong[OF az, of b] ab[unfolded coprime_def]
show ?thesis by auto
qed
@@ -577,15 +584,15 @@
lemma distinct_prime_coprime: "prime p \<Longrightarrow> prime q \<Longrightarrow> p \<noteq> q \<Longrightarrow> coprime p q"
unfolding prime_def coprime_prime_eq by blast
-lemma prime_coprime_lt: assumes p: "prime p" and x: "0 < x" and xp: "x < p"
+lemma prime_coprime_lt:
+ assumes p: "prime p" and x: "0 < x" and xp: "x < p"
shows "coprime x p"
-proof-
- {assume c: "\<not> coprime x p"
- then obtain g where g: "g \<noteq> 1" "g dvd x" "g dvd p" unfolding coprime_def by blast
+proof (rule ccontr)
+ assume c: "\<not> ?thesis"
+ then obtain g where g: "g \<noteq> 1" "g dvd x" "g dvd p" unfolding coprime_def by blast
from dvd_imp_le[OF g(2)] x xp have gp: "g < p" by arith
- from g(2) x have "g \<noteq> 0" by - (rule ccontr, simp)
- with g gp p[unfolded prime_def] have False by blast}
-thus ?thesis by blast
+ have "g \<noteq> 0" by (rule ccontr) (use g(2) x in simp)
+ with g gp p[unfolded prime_def] show False by blast
qed
lemma prime_odd: "prime p \<Longrightarrow> p = 2 \<or> odd p" unfolding prime_def by auto
@@ -755,28 +762,30 @@
shows "\<exists>i j. x = p ^i \<and> y = p^ j"
using xy
proof(induct k arbitrary: x y)
- case 0 thus ?case apply simp by (rule exI[where x="0"], simp)
+ case 0
+ thus ?case apply simp by (rule exI[where x="0"], simp)
next
case (Suc k x y)
+ have p0: "p \<noteq> 0" by (rule ccontr) (use p in simp)
from Suc.prems have pxy: "p dvd x*y" by auto
- from prime_divprod[OF p pxy] have pxyc: "p dvd x \<or> p dvd y" .
- from p have p0: "p \<noteq> 0" by - (rule ccontr, simp)
- {assume px: "p dvd x"
+ from prime_divprod[OF p pxy] show ?case
+ proof
+ assume px: "p dvd x"
then obtain d where d: "x = p*d" unfolding dvd_def by blast
from Suc.prems d have "p*d*y = p^Suc k" by simp
hence th: "d*y = p^k" using p0 by simp
from Suc.hyps[OF th] obtain i j where ij: "d = p^i" "y = p^j" by blast
with d have "x = p^Suc i" by simp
- with ij(2) have ?case by blast}
- moreover
- {assume px: "p dvd y"
+ with ij(2) show ?thesis by blast
+ next
+ assume px: "p dvd y"
then obtain d where d: "y = p*d" unfolding dvd_def by blast
from Suc.prems d have "p*d*x = p^Suc k" by (simp add: mult.commute)
hence th: "d*x = p^k" using p0 by simp
from Suc.hyps[OF th] obtain i j where ij: "d = p^i" "x = p^j" by blast
with d have "y = p^Suc i" by simp
- with ij(2) have ?case by blast}
- ultimately show ?case using pxyc by blast
+ with ij(2) show ?thesis by blast
+ qed
qed
lemma prime_power_exp: assumes p: "prime p" and n:"n \<noteq> 0" ```
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# Lester Hill’s “The checking of the accuracy …”, part 3
Backstory: Lester Saunders Hill wrote unpublished notes, about 40 pages long, probably in the mid- to late 1920s. They were titled “The checking of the accuracy of transmittal of telegraphic communications by means of operations in finite algebraic fields”. In the 1960s this manuscript was given to David Kahn by Hill’s widow. The notes were typewritten, but mathematical symbols, tables, insertions, and some footnotes were often handwritten. The manuscript is being LaTeXed “as we speak”. I thank Chris Christensen, the National Cryptologic Museum, and NSA’s David Kahn Collection, for their help in obtaining these notes. Many thanks also to Rene Stein of the NSA Cryptologic Museum library and David Kahn for permission to publish this transcription. Comments by transcriber will look his this: [This is a comment. – wdj]. I used Sage (www.sagemath.org) to generate the tables in LaTeX.
Here is just the third section of his paper. I hope to post more later. (Part 2 is here.)
Section 3: Preliminary characterization of checking procedure
Our problem is to provide convenient and practical accuracy checks upon a sequence of $n$ elements $f_1$, $f_2$, $\dots$, $f_n$ in a finite algebraic field $F$.
To fix the ideas, let us assume that we are to employ $q$-element checks $c_1$, $c_2$, $\dots$, $c_q$ upon the sequence $f_1$, $f_2$, $\dots$, $f_n$. The checks are to be determined by means of a fixed reference matrix
$Q = \left( \begin{array}{cccc} k_{11} & k_{12} & \dots & k_{1n} \\ k_{21} & k_{22} & \dots & k_{2n} \\ \vdots & & & \vdots \\ k_{q1} & k_{q2} & \dots & k_{qn} \\ \end{array} \right)$
of elements of $F$, the matrix having been suitably constructed according to criteria which will be developed in the following pages. We send, in place of the simple sequence $f_1$, $f_2$, $\dots$, $f_n$, the amplified sequence
$f_1, f_2, \dots, f_n, c_1, c_2, \dots, c_q$
consisting of the “operand” sequence and the “checking” sequence. The checking sequence contains $q$ elements of $F$ as follows:
$c_j = \sum_{i=1}^n k_{ji}f_i,$
for $j = 1, 2, \dots, q$. Considerations of telegraphic economy dictate the assumption, made throughout the paper, that $q\leq n$.
Before laying down specifications for the reference matrix $Q$, we define a matrix of “index” $q$ as one in which no $q$-rowed determinant vanishes.
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×
Back to all chapters
# Solving Triangles
Trigonometric problem solving culminates in this chapter. Leave no side and no angle unmeasured!
# Solving Triangles: Level 4 Challenges
In a triangle $$ABC$$, the measures of the angles are $$\angle A=30^\circ,$$ $$\angle B=80^\circ.$$ Point $$M$$ lies inside the triangle, such that $$\angle MAC = 10^\circ,$$ $$\angle MCA = 30^\circ.$$ What is the measure (in degrees) of $$\angle BMC?$$
Triangle $$ABC$$ is isosceles with $$AC = BC$$ and $$\angle ACB = 106^\circ$$ Point $$M$$ is inside the triangle so that $$\angle MAC = 7^\circ$$ and $$\angle MCA = 23^\circ$$. Find the measure of $$\angle CMB$$ in degrees.
In a triangle $$ABC$$ $$\angle A =84^\circ,$$ $$\angle C=78^\circ.$$ Points $$D$$ and $$E$$ are taken on the sides $$AB$$ and $$BC,$$ so that $$\angle ACD =48^\circ,$$ $$\angle CAE =63^\circ.$$ What is the measure (in degrees) of $$\angle CDE$$ ?
In triangle $$ABC$$, $$\cos A : \cos B : \cos C = 2 : 9 : 12.$$
If $$\sin A : \sin B : \sin C = p : q : r$$, where $$p$$, $$q$$ and $$r$$ are coprime positive integers, find the value of $$pqr-(p+q+r).$$
Triangle $$ABC$$ has area equal to $$\dfrac {90 \sqrt{3}}{4}$$ and perimeter equal to $$30.$$ Also, one of its angles is equal to $$60^\circ.$$ What is the product of the sides of $$ABC?$$
×
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Number 36801875
Properties of number 36801875
Cross Sum:
Factorization:
5 * 5 * 5 * 5 * 11 * 53 * 101
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
2318d53
Base 32:
1333aj
sin(36801875)
0.90131895742892
cos(36801875)
-0.43315601921161
tan(36801875)
-2.0808182674442
ln(36801875)
17.421059352928
lg(36801875)
7.5658699458857
sqrt(36801875)
6066.4548955712
Square(36801875)
Number Look Up
Look Up
36801875 (thirty-six million eight hundred one thousand eight hundred seventy-five) is a unique number. The cross sum of 36801875 is 38. If you factorisate 36801875 you will get these result 5 * 5 * 5 * 5 * 11 * 53 * 101. 36801875 has 40 divisors ( 1, 5, 11, 25, 53, 55, 101, 125, 265, 275, 505, 583, 625, 1111, 1325, 1375, 2525, 2915, 5353, 5555, 6625, 6875, 12625, 14575, 26765, 27775, 33125, 58883, 63125, 72875, 133825, 138875, 294415, 364375, 669125, 694375, 1472075, 3345625, 7360375, 36801875 ) whith a sum of 51620976. The figure 36801875 is not a prime number. 36801875 is not a fibonacci number. The figure 36801875 is not a Bell Number. The figure 36801875 is not a Catalan Number. The convertion of 36801875 to base 2 (Binary) is 10001100011000110101010011. The convertion of 36801875 to base 3 (Ternary) is 2120020201200102. The convertion of 36801875 to base 4 (Quaternary) is 2030120311103. The convertion of 36801875 to base 5 (Quintal) is 33410130000. The convertion of 36801875 to base 8 (Octal) is 214306523. The convertion of 36801875 to base 16 (Hexadecimal) is 2318d53. The convertion of 36801875 to base 32 is 1333aj. The sine of the figure 36801875 is 0.90131895742892. The cosine of the figure 36801875 is -0.43315601921161. The tangent of 36801875 is -2.0808182674442. The root of 36801875 is 6066.4548955712.
If you square 36801875 you will get the following result 1354378003515625. The natural logarithm of 36801875 is 17.421059352928 and the decimal logarithm is 7.5658699458857. that 36801875 is very impressive figure!
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Medication Drug Calculations Self-Study Course
Despite all recent advances in medication administration, drug-related adverse events and error rates remain a concern. Many medication errors are related to administering incorrect dosages.
Often some type of calculation prior to medication administration is needed. Drug calculations can be simple or more complex. Regardless of the drug to be administered, careful and accurate calculations are necessary to help prevent medication errors. Practicing drug calculations will develop stronger and more confident math skills.
The purpose of this module is to provide Licensed Practical Nurses (LPNs) with an opportunity to practice medication drug calculations. This module will be of interest to any nurse – newly graduated or experienced – whose responsibilities include medication administration.
If you find that you are having difficulty with the practice quizzes and/or games, we recommend you engage in self-study or complete an educational course that will assist you in developing proficiency with medication drug calculation skills.
This module consists of several practice quizzes, learning games and resources. The quizzes and games focus on six calculation areas: basic mathematics, metric conversions, tablet dosage, fluid dosage, injection dosage and infusion rates. To access a learning activity, simply click on the Link. After you have completed an activity, return to this page to proceed to the next activity.
Introduction
Before you begin the practice quizzes and games, you should review these helpful tips about how to accurately calculate drug dosages:
Dosage Calculations Practice Quizzes *
Basic Mathematics – This quiz is designed to test and refresh your knowledge and skills of basic mathematics used in medication dosage calculations. (25 questions randomly selected from 100 items). (PDF/PRINT Versions: Quiz 1 | Quiz 2 | Quiz 3 | Quiz 4)
Metric Conversions – This quiz tests and reviews your knowledge of the metric system and how to accurately make conversions between the various units. (25 questions randomly selected from 60 items). (PDF/PRINT Versions: Quiz 1 | Quiz 2)
Tablet Dosage Calculations – Test and refresh your knowledge and skills about how to calculate the correct tablet medication dosages. (25 questions randomly selected from 50 items). (PDF/PRINT Versions: Quiz 1 | Quiz 2)
Fluid Dosage Calculations – This quiz is designed to assess and review your knowledge and skills in calculating the correct fluid dosages for medications. (25 questions randomly selected from 75 items). (PDF/PRINT Versions: Quiz 1 | Quiz 2 | Quiz 3)
Injection Dosage Calculations – Test and refresh your knowledge and skills in calculating the correct dosages for medication administration by injection. (25 questions randomly selected from 75 items). (PDF/PRINT Versions: Quiz 1 | Quiz 2 | Quiz 3)
Infusion Rate Calculations – This quiz assesses your knowledge and skills in calculating the correct intravenous (IV) dosages, flow rates and administration times. (25 questions randomly selected from 100 items). (PDF/PRINT Versions: Quiz 1 | Quiz 2 | Quiz 3 | Quiz 4)
Dosage Calculations Games **
Metric Measurements – This educational nursing game is designed to test and refresh your knowledge of metric volume and weight units used in nursing care. (15 questions randomly selected from 82 items).
Oral Dosages – A game about calculating the correct dosages for medications administered orally using solutions, suspensions, mixtures and syrups. (15 questions randomly selected from 35 items).
Tablet Dosages I – Review your skills about calculating the correct tablet dosages for medications and drugs. (15 questions randomly selected from 30 items).
Tablet Dosages II – A media-enhanced version of the game calculating correct tablet dosages for medications and drugs. (15 questions randomly selected from 30 items).
Final Examination *
Final Examination – This final exam consists of 60 randomly selected questions from a pool of 350 items. A printable CLPNA Certificate of Completion is available upon passing this exam with a score of 80% or better. If you are an Alberta LPN, please enter your Registration Number when requested. Everyone else, leave in the default number. If you want a copy of the Final Exam Results e-mailed to yourself (or anyone else), enter the appropriate e-mail address. Otherwise, leave it blank. (Results are ONLY e-mailed if you have passed the exam).
Module Evaluation
Medication Drug Calculation Module Evaluation – Please complete this online survey after finishing this Module. Your feedback and comments will greatly assist CLPNA in developing and delivering effective, quality nursing educational resources.
Resources
MacEwan University, NURS 0164 Medication Administration and Calculations
NorQuest College, XHLT1530 – Pharmacology Therapeutics and Medication Administration
DosageHelp.com
Apps
References
Lapham, R. (2015). Drug Calculations for Nurses: A Step-By-Step Approach, (4th ed.). Boca Raton, FL: CRC Press.
Mulholland, J. & Turner, S. (2015). The Nurse, The Math, The Meds: Drug Calculations using Dimensional Analysis, (3rd ed.). St. Louis, MS: Mosby.
Pickar, G.D., Pickar Abernethy, A., Swart, B., Graham, H., & Swedish, M. (2015). Dosage Calculations, (3rd ed.). Toronto, Canada: Nelson Education.
Technology Notes
* The Practice Quizzes and Final Exam can be completed on regular browsers, tablets and smart phones. For iPhones and iPads, download and install the FREE iSpring viewer from the iTunes Store. For Android devices, iSpring Play is available here. The quizzes can also be downloaded and saved on your mobile device and completed off-line at your convenience.
** The Calculations Games can be played on regular browsers, tablets and mobile phones. For best viewing on Apple's devices, download and install the FREE Articulate Player App from the iTunes Store. If you are using an Android tablet or phone, download and install the FREE Android Player App from Google Play. The games can also be downloaded to your mobile device for off-line use.
If you are using these quizzes / games / final exam for class assignments or continuing education credits, and want a copy of the Results for your records, enter your name when requested. Otherwise, use Guest and proceed with the quiz / game / exam.
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# Lesson 7
Juntemos figuras planas
## Warm-up: Observa y pregúntate: Perros (10 minutes)
### Narrative
The purpose of this warm-up is to elicit student knowledge about composing and decomposing composite shapes, which will be useful when students build their own composite shapes later in this lesson. While students may notice and wonder many things about this image, the idea that shapes, such as hexagons, can be composed and decomposed into other shapessuch as multiple triangles, rhombuses, trapezoids, or a combination of triangles, rhombuses, or trapezoidsis the important discussion point. Students may refer to the pattern blocks by their colors when describing them or sharing different combinations that may form the same shape. Use this as an opportunity to ask the students if they know the names of the shapes and review shape vocabulary.
### Launch
• Groups of 2
• Display the image.
• “¿Qué observan? ¿Qué se preguntan?” // “What do you notice? What do you wonder?”
• 1 minute: quiet think time
### Activity
• “Discutan con un compañero lo que pensaron” // “Discuss your thinking with your partner.”
• 1 minute: partner discussion
• Share and record responses.
### Student Facing
¿Qué observas?
¿Qué te preguntas?
### Activity Synthesis
• “Algunos de ustedes se dieron cuenta de que la imagen de la parte de abajo se parece a la de arriba, pero se usaron figuras diferentes para formar los hexágonos” // “Some of you noticed that the image at the bottom looks like the one at the top, but they used different shapes to make up the hexagons.”
• “¿De qué otras maneras podrían armar un hexágono con las fichas geométricas?” // “What other ways could you make a hexagon with the pattern blocks?” (2 trapezoids, 1 trapezoid and 3 triangles, 2 rhombuses and 2 triangles)
• “¿De qué otras maneras podrían armar un rombo?, ¿un trapecio?” // "What other ways could you make a rhombus? a trapezoid?”
## Activity 1: Construyamos con fichas geométricas (15 minutes)
### Narrative
The purpose of this activity is for students to compose two-dimensional shapes into larger shapes in different ways. As students work, consider asking them to identify the shapes they are using and their attributes. During this activity, the teacher may consider taking photos of the way students composed the large triangle (third problem) and the large rectangle (fourth problem) to share during the synthesis so that students do not need to move their pattern blocks to share with the class.
MLR8 Discussion Supports. Synthesis: When asking questions, invite students to take turns sharing their responses. Ask students to restate what they heard using precise mathematical language and their own words. Display the sentence frame: “Te escuché decir . . .” // “I heard you say . . .” Original speakers can agree or clarify for their partner.
### Required Materials
Materials to Gather
Materials to Copy
• Flat Shapes Puzzles, Spanish
### Required Preparation
• Consider making the Flat Shapes Puzzles blackline master into a packet for each student.
### Launch
• Give students pattern blocks and the flat shape puzzles.
### Activity
• “Usen las fichas geométricas para llenar la figura de diferentes maneras. Cada vez, usen imágenes, números o palabras para mostrar cómo llenaron la figura” // “Use the pattern blocks to fill the outline in different ways. Each time, record how you filled in the shape with pictures, numbers, or words.”
• 8 minutes: independent work time
• Monitor for different ways that students filled in the large triangle and large rectangle puzzles.
### Activity Synthesis
• Display work from two students for the large triangle.
• “¿Cómo armaron el triángulo de maneras diferentes? ¿Por qué las dos maneras funcionan?” // “How was the triangle created differently? Why do both ways work?” (One way used all triangles, the other used some triangles and a trapezoid. They both work because a trapezoid is the same shape as three triangles put together.)
• Display student work for the large rectangle.
• “¿Alguien encontró una manera diferente de construir esta figura? ¿Por qué no pudimos encontrar otras maneras?” // “Did anyone find a different way to build this shape? Why didn’t we find any other ways?” (The rectangle has square corners and the square is the only pattern block that also has square corners.)
## Activity 2: Imágenes de figuras (10 minutes)
### Narrative
The purpose of this activity is for students to compose two-dimensional shapes into composite shapes and find other smaller shapes that can make the same composite shape. Students use pattern blocks to compose a new shape, then recreate the new shape using smaller shapes.
Action and Expression: Internalize Executive Functions. Check for understanding by inviting students to rephrase directions in their own words.
Supports accessibility for: Memory, Organization
### Required Materials
Materials to Gather
### Launch
• Give students pattern blocks.
• “¿Qué tipos de figuras pueden armar con estas fichas geométricas?” // “What kinds of shapes can you make from pattern blocks?” (triangles, rectangles, flowers)
• 30 seconds: quiet think time
• Share responses.
### Activity
• “Usen las fichas geométricas para construir algo nuevo. Pueden construir lo que quieran. Después de construir el objeto nuevo, trácenlo y cuenten el número de fichas que usaron. Después, construyan el mismo objeto usando fichas diferentes y escriban cuántas fichas usaron” // “Use the pattern blocks to build something new. You can build whatever you want. After you build your new object, trace it and count the number of pattern blocks you used. Then build the same object using different pattern blocks, and record how many blocks you used.”
• 6 minutes: independent work time
• Monitor for 2–3 students to share their two shapes.
### Student Facing
1. Construye algo usando las fichas geométricas.
Traza lo que hiciste y escribe cuántas fichas geométricas usaste.
2. Arma el mismo objeto que hiciste usando fichas geométricas diferentes.
Escribe cuántas fichas geométricas usaste.
Si te queda tiempo: construye el mismo objeto otra vez usando la mayor cantidad de fichas geométricas que puedas.
Escribe cuántas fichas geométricas usaste.
### Activity Synthesis
• Invite previously identified students to share.
• “¿Qué observaron sobre las figuras que ellos construyeron?” // “What do you notice about the shapes they built?” (They made a rocket. They used six triangles to make a hexagon in their first shape but used hexagons in the second shape.)
• Repeat as time allows.
## Activity 3: Conozcamos “Libros de imágenes: Encuentra figuras” (15 minutes)
### Narrative
The purpose of this activity is for students to learn stage 3 of the Picture Books center introduced in kindergarten. Students look through picture books and notice and describe shapes they see in the pictures. Students record the shapes they see with drawings or words.
### Required Materials
Materials to Gather
Materials to Copy
• Picture Books Stage 3 Recording Sheet, Spanish
### Required Preparation
• Each group of 2 needs at least one picture book that shows a variety of shapes throughout the book.
### Launch
• Groups of 2
• Give each group two recording sheets and access to picture books.
• “Vamos a buscar figuras en libros de imágenes. Hablen con su pareja sobre las figuras que ven. Describan las figuras y nómbrenlas si pueden. Después, usen dibujos o palabras para registrar las figuras que ven” // “We are going to look for shapes in picture books. Talk to your partner about the shapes you see. Describe the shapes and name them if you can. Then record the shapes you see with a drawing or words.”
### Activity
• 10 minutes: partner work time
### Activity Synthesis
• Display a page from a picture book with lots of recognizable shapes.
• “¿Qué figuras ven en esta página?” // “What shapes do you see on this page?” (The sun looks like a circle. The blanket is a rectangle. The ball is a sphere.)
## Lesson Synthesis
### Lesson Synthesis
Display a chart titled “Lo que hemos aprendido sobre las figuras” // “What We've Learned About Shapes.”
“Reflexionemos sobre lo que hicimos con figuras en esta sección. Primero, van a pensar individualmente. Después, van a compartir sus ideas con un compañero” // “Let’s reflect on the work we did in this section with shapes. First you will think on your own. Then you will share your ideas with a partner.”
Share and record responses.
## Student Section Summary
### Student Facing
Aprendimos sobre figuras sólidas.
Aprendimos sobre figuras planas.
Describimos y nombramos figuras.
Esta figura es un triángulo porque tiene 3 lados rectos y 3 esquinas.
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# Simulating n-bodies and functional programming
### The n-bodies problem
There’s been some activity in the #haskell IRC channel recently towards coming up with a better solution to the n-bodies problem. In this problem, you’re given a set of bodies, and are asked to simulate Newtonian gravity on them. Most of the solutions I’ve seen go past have involved some sort of crazy state monad based stuff, reading from and writing to STUArrays and other imperative ideas. So, I thought I’d try and solve the problem using a pure functional style instead.
### Reactive
I’ve recently been playing with Reactive a lot, initially for my work, but also just because it’s so damned cool. Reactive is a library for Haskell that lets you describe time varying values in a purely functional way. That is, out goes the concept of state, and in comes the concept of describing exactly what is going on.
### Back to n-bodies
In order to simulate newtonian physics, we need the bodies’ mass,the positions of the bodies, and the velocities of the bodies:
```data Body = Body { mass :: Double
, position :: Point3 Double
, velocity :: Vector3 Double
}```
That was pretty straight forward, lets get down to the core of the problem – simulating gravity:
```accelBetween :: Body -> Body -> Vector3 Double
accelBetween p1 p2 =
(mass p2 *^ vecBetween) / ((mag vecBetween) ^ 3)
where
vecBetween = position p2 ^-^ position p1
computeAccels :: Body -> [Body] -> Vector3 Double
computeAccels p = sum . map (accelBetween p)```
Well, that was surprisingly easy! The `accelBetween` function describes the newtonian gravity equation just as the maths does, and summing up the acceleration due to all the other planets in the system is fairly straightforward.
What we’ve not seen yet though, is any of that lovely functional reactive programming I was talking about. We know now how to compute the acceleration affecting any one body at any one time, but what we want to know is the acceleration affecting a body at *any* time. To do this, we’re going to need to move from doing computations on `Body`s, to instead doing computation on `Behavior`s of `Body`s.
```bodyAccels :: [Behaviour Body] -> [Behaviour (Vector3 Double)]
bodyAccels ps = withOthers (liftA2 computeAccels) ps
withOthers :: (a -> [a] -> b) -> [a] -> [b]
withOthers f xs =
withOthers' f [] xs
where
withOthers' _ _ [] = []
withOthers' f os (x : xs) = f x (os ++ xs) : withOthers' f (x : os) xs
```
The `withOthers` function here is just like map, but it passes in all the other values in the list in a second argument to the function.
So then, `bodyAccels` computes a continuous acceleration function for all the bodies in the system. For each body, it runs the `computeAccels` function, giving it all other bodies in the system as its second argument. Crucially, `liftA2` allows us to do this in the `Behavior` Applicative, so we are no longer computing it on rigid, static bodies, but instead, on all the positions and velocities the bodies in the system will ever have (isn’t lazyness great!).
Finally, we can get from these accelerations down to the velocities, and then positions of the bodies using integration on the acceleration:
```bodyVel :: Body -> Behaviour (Vector3 Double) -> Behaviour (Vector3 Double)
bodyVel p acc = (velocity p ^+^) <\$> integral dt acc```
I’m sure you can imagine what the bodyPos function looks like. You may wonder what the `dt` in here is talking about. This is an unfortunate effect of not being able to mathematically integrate arbitrary functions. Instead, we must use euler integration, and that requires us to provide times at which to take samples. The `dt` argument is an event which ticks reasonably fast, and progresses our simulation:
```dt :: Event ()
dt = atTimes [0,0.01..]```
So, now we are able to combine all our efforts together, and solve the whole n-bodies problem:
```nbodies :: [Body] -> [Behaviour Body]
nbodies ps = pbs
where
pbs = Body <\$> (mass <\$> ps) <*> pps <*> pvs
pps = zipWith bodyPos ps pvs
pvs = zipWith bodyVel ps pas
pas = bodyAccels pbs```
### Conclusions
A lot of assumptions are made about how we must write programs. Often, even beautiful mathematical problems end up described as horrible state-full masses of code that obscure what it is we’re trying to compute. I’ve presented a solution to the n-bodies problem using the Reactive library to get a handle on time in a purely functional setting, it turned out that this was rather beautiful!
Full code for my solution can be found below.
```module NBodies (Body(..), nbodies) where
import Control.Applicative
import Data.VectorSpace
import Data.AffineSpace
import FRP.Reactive
import Graphics.FieldTrip
data Body = Body { mass :: Double
, position :: Point3 Double
, velocity :: Vector3 Double
}
nbodies :: [Body] -> [Behaviour Body]
nbodies ps = pbs
where
pbs = Body <\$> (mass <\$> ps) <*> pps <*> pvs
pps = zipWith bodyPos ps pvs
pvs = zipWith bodyVel ps pas
pas = bodyAccels pbs
accelBetween :: Body -> Body -> Vector3 Double
accelBetween p1 p2 =
(mass p2 *^ vecBetween) / ((mag vecBetween) ^ 3)
where
vecBetween = position p2 ^-^ position p1
computeAccels :: Body -> [Body] -> Vector3 Double
computeAccels p = sum . map (accelBetween p)
bodyAccels :: [Behaviour Body] -> [Behaviour (Vector3 Double)]
bodyAccels ps = withOthers (liftA2 computeAccels) ps
withOthers :: (a -> [a] -> b) -> [a] -> [b]
withOthers f xs =
withOthers' f [] xs
where
withOthers' _ _ [] = []
withOthers' f os (x : xs) = f x (os ++ xs) : withOthers' f (x : os) xs
bodyVel :: Body -> Behaviour (Vector3 Double) -> Behaviour (Vector3 Double)
bodyVel p acc = (velocity p ^+^) <\$> integral dt acc
bodyPos :: Body -> Behaviour (Vector3 Double) -> Behaviour (Point3 Double)
bodyPos p vel = (position p .+^) <\$> integral dt vel
dt :: Event ()
dt = atTimes [0,0.01..]
```
## 10 comments on “Simulating n-bodies and functional programming”
1. axman6 says:
I think it should be mentioned that the original problem being discussed was the shootout [1] nbodies problem, who’s soul purpose is to iterate n bidoes 50,000,000 times in the shortest possible time. But good work man, I’ll have to play with this sometime.
• beelsebob says:
The reason I didn’t mention that is that the n-bodies problem is a pretty general problem, it’s not constrained solely to the great language shootout. I intend to take this further, and use it within an app for simulating planetary physics.
2. Shane says:
It’s very neat to see the n-bodies problem done in reactive form. Thank you for the article. I’ve ported a simple physics engine to Haskell, and one thing I was wondering about is how one might optimize for anti-symmetric forces like gravity? For example, Suppose the force on body 2 wrt 1 is the negative of the force on body 1 wrt 2, F_{12} = -F_{21}. Is there a nice way to functionally avoid having two function evaluations?
• beelsebob says:
That’s a really good question, and was one of the first things I was thinking of doing with this when I started optimising it – I’ll get back to you.
3. Chris Dew says:
I think I may have just started to (needlessly) re-invent reactive programming, as I hadn’t heard of it before I read this post. Thanks for the enlightenment.
Is what I’m attempting (in the link above) reactive programming, or have I misunderstood?
• beelsebob says:
Yes, that looks very much like FRP in the making 🙂
4. wagstaff says:
This code appears to be a long way from working. So fas as I can see, almost no part of it typechecks.
• beelsebob says:
Yep, the code was written when I didn’t even have a working Reactive version, so it’s not bang on right. It only takes a few tweaks to get it there though. Having said that, if you do tweak it, unfortunately, Reactive is still a little too buggy to run it. Close, but no cigar.
5. wagstaff says:
“A few tweaks” – perhaps depends on your level of expertise!
I’ve got most of this compiling, but I don’t suppose you’d care to provide hints for corrected versions of bodyAccels and nbodies, would you?
6. agrigioli says:
What Word Press theme do you use?
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# Tableau – Operators
Tableau is the easy-to-use Business Intelligence tool used in data visualization. Its unique feature is, to allow data real-time collaboration and data blending, etc. Through Tableau, users can connect databases, files, and other big data sources and can create a shareable dashboard through them. Tableau is mainly used by researchers, professionals, and government organizations for data analysis and visualization.
In this article, we are going to discuss Tableau Operators and types of Tableau operators.
The symbols which tell the compiler to execute specific mathematical or logical manipulations are known as Operators. Tableau too has a good number of operators, which help in the formation of calculated fields and formulas.
Tableau has 4 basic types of Operators as mentioned below:
• General Tableau Operators
• Arithmetic Tableau Operators
• Relational Tableau Operators
• Logical Tableau Operators
GENERAL TABLEAU OPERATORS –
The operator which we generally use in mathematics computation comes under this category, like Addition and Subtraction.
• Subtraction: –
ARITHMETIC TABLEAU OPERATORS –
The operators which help to perform arithmetic calculations on different fields of uploaded data source are known as Arithmetic operators.
• Multiply: *
• Divide: /
• Modulation: %
• Power: ^
RELATIONAL TABLEAU OPERATORS –
A relational operator or comparison operator is a binary operator that takes two operands as input, compare their values and returns the output. A comparison operator is generally used in conditional statements, loops, where the comparison result will decide, whether execution should go ahead or not.
• Equality Operator: =
• Not Equality Operator: < >, !=, ^=
• Greater Than Operator: >
• Less Than Operator: <
• Greater Than Equal to Operator: >=
• Less Than Equal to Operator: <=
LOGICAL TABLEAU OPERATORS –
In Logical operator, there are three basic commands –
• Logical conjunction operator or “AND” operator
• Logical disjunction operator or “ OR” operator
• Logical negation operator or “NOT” operator
Logical AND operator: The operator will return True, if both conditions are true. If either condition is false, it will return false else return Null.
Logical OR operator: The operator will return True, if either of the condition is True, else returns False.
Logical NOT operator: The operator will return True if condition is False and vice versa.
My Personal Notes arrow_drop_up
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# 4.17: Acceleration Due to Gravity
Difficulty Level: At Grade Created by: CK-12
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It’s a good thing this mountain climber’s safety gear is working, because it’s a long way down to the ground! If he were to fall, he’d be moving really fast by the time he got there. The higher any object starts falling from above Earth’s surface, the faster it’s traveling by the time it reaches the ground. Do you know why? The reason is gravity.
### Speeding up While Falling Down
Gravity is a force that pulls objects down toward the ground. When objects fall to the ground, gravity causes them to accelerate. Acceleration is a change in velocity, and velocity, in turn, is a measure of the speed and direction of motion. Gravity causes an object to fall toward the ground at a faster and faster velocity the longer the object falls. In fact, its velocity increases by 9.8 m/s2, so by 1 second after an object starts falling, its velocity is 9.8 m/s. By 2 seconds after it starts falling, its velocity is 19.6 m/s (9.8 m/s + 9.8 m/s), and so on. The acceleration of a falling object due to gravity is illustrated in the Fiugre below. For a more detailed explanation of acceleration due to gravity, watch this video:
You can compare the acceleration due to gravity on Earth, the moon, and Mars with the interactive animation called “Freefall” at this URL:
Q: In this diagram, the boy drops the object at time t= 0 s. By t = 1 s, the object is falling at a velocity of 9.8 m/s. What is its velocity by t = 5 s? What will its velocity be at t = 6 s if it keeps falling?
A: Its velocity at t = 5 s is 49.0 m/s, and at t = 6 s, it will be 58.8 m/s (49.0 m/s + 9.8 m/s).
### Mass and Acceleration Due to Gravity
What if you were to drop a bowling ball and a soccer ball at the same time from the same distance above the ground? The bowling ball has greater mass than the basketball, so the pull of gravity on it is greater. Would it fall to the ground faster? No, the bowling ball and basketball would reach the ground at the same time. The reason? The more massive bowling ball is also harder to move because of its greater mass, so it ends up moving at the same acceleration as the soccer ball. This is true of all falling objects. They all accelerate at the same rate due to gravity, unless air resistance affects one object more than another. For example, a falling leaf is slowed down by air resistance more than a falling acorn because of the leaf’s greater surface area. You can simulate the effect of air resistance on acceleration due to gravity by doing the interactive animation at this URL: http://www.science-animations.com/support-files/freefall.swf
Q: If a leaf and an acorn were to fall to the ground in the absence of air (that is, in a vacuum), how would this affect their acceleration due to gravity?
A: They would both accelerate at the same rate and reach the ground at the same time.
### Summary
• When gravity pulls objects toward the ground, it always causes them to accelerate at a rate of 9.8 m/s2.
• Regardless of differences in mass, all objects accelerate at the same rate due to gravity unless air resistance affects one more than another.
### Vocabulary
• acceleration: Measure of the change in velocity of a moving object.
• gravity: As traditionally defined, force of attraction between things that have mass.
### Practice
Watch the video at the following URL. Based on the teens interviewed in the video, identify two common misconceptions about gravity, mass, and acceleration due to gravity. Then explain why each misconception is false.
### Review
1. Why do objects fall faster the longer they fall toward Earth?
2. What is the rate of acceleration due to gravity?
3. How does mass affect acceleration due to gravity?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
acceleration Measure of the change in velocity of a moving object.
gravity As traditionally defined, force of attraction between things that have mass.
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# Betting Tips, Free Betting Predictions
With American odds, the number represented is how much you have to Mma Betting Calculator bet to win \$100. If the number is preceded with a “+” sign, then you would risk less than \$100 to win a wager worth \$100. And, the opposite is true when the number is preceded with a “-” where the bettor would have to risk more than \$100 just to win \$100 of the wager. Parlays are essentially a combination of two or more bets into one, such as three teams winning their respective matches. These parlays are usually a long shot since so many teams have to win, but in the end, they are also some of the highest paying games.
## More From Betfair
A -1 to -3 rating is usually for replacement level bench players; anything below that is fit for guys at informative post the end of the bench. It is a simple calculation of a player’s contribution to the team’s overall score when he is on the floor. While +/- is an individual stat, meaning different players have different plus/minus ratings, it is actually a good indication of how well a lineup plays together.
## What Do Nba Vegas Odds Mean?
This way, it levels the playing field enough to give each team the appearance of an equal shot. Betting odds provide bettors with a clear idea of what the winnings will be if the bet wins. A favorite’s ML odds are based on the amount that a bettor would have to bet to win \$100. An underdog’s ML odds are based on the amount they’d win when betting \$100. A bet calculator is ideal for providing an accurate idea of the bettors’ potential winnings for all types of sports wagers, including standard single bets, using the three different odd types.
If you see a plus symbol before a set of odds, this means that the team is an underdog to win. You’re met with a wall of numbers, dots and dashes before you even get started. The plus indicates the potential payoff of a \$100 bet while the minus shows the amount needed to realize a \$100 payoff.
## Vi Consensus Nfl Line
With up to the minute presidential bet odds bookmark this page and check back often. If you want a way to root for both teams to score — or root against both teams scoring — betting the total is the way to go. Long after a game’s outcome is decided, even after the point spread has been covered, you’ll find bettors in the sportsbook sweating every meaningless basket or run scored.
Only the 1-2 match would have ended as a draw with Everton getting a 1-goal start, which means that a draw should be 5/1, yet it is only 3-1. For example, for Liverpool to win conceding two goals you might get 11/2, three goals 14/1, and four goals 33/1. It is useful to be able to do, if you don’t have a conversion table or tool to hand. Most jurisdictions that allow gambling require participants to be above a certain age.
## Calculating Payouts From Fractional Odds
In the example below, there’s no value to betting on the Baltimore Ravens with the moneyline. I wrote up a guide to golf betting that explains these unique features so you can feel comfortable wagering on PGA events too. Use the rotation number/numbers and the name of the team/teams you want to bet on. The rotation number appears to the left of each team and is used as a unique identifier so that there is no confusion about what you want.
Bettors also lose if they did not select “draw” for a game or fight that ends with such an outcome if provided as an option. All money line bets are expressed in units of \$100, however you should note that wagers of all sizes can be placed on money lines. As an example, if you wanted to place a \$5 wager on Buster Douglas in the above example, then you’d make a profit of \$200. (\$5 is 5% of \$100, so you’ll receive 5% of the return, – 5% of \$4000 is \$200).
Hence, Green Bay entering as favorites, but favorites expected to prevail by just 3 points. Thus, even on the road, the 49ers rate as -7.5 favorites at DK Sportsbook. In other words, they’re expected to win by more than a touchdown.
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# GATE 2022 Champion Study plan Compressible Flow Study notes
By Apoorbo Roy|Updated : September 9th, 2021
In this article, we are providing you with the Compressible flow study notes for GATE, BARC other Mechanical exams. Itis an important part of Thermodynamics & its Application has a good weightage of the questions to be asked in various mechanical engineering exams.
In this article, we are providing you with the Compressible flow study notes for GATE, BARC other Mechanical exams. It is an important part of Thermodynamics & its Applications Subject that has a good weightage of the questions to be asked in various mechanical engineering exams.
COMPRESSIBLE FLOW THROUGH NOZZLE
When the density of the fluid is function pressure it is called compressible fluid. It is Ma > 0.3.
Stagnation point
It is possessing the fluid flow where velocity is 0.
H/mass(h) = u + PV
For high-speed flows [like jet engines], it is current to combine enthalpy and K.E. into single term called stagnation or total enthalpy.
Advantages and Applications of Convergent-Divergent Type of Nozzles
Convergent-Divergent(C-D) type of nozzles have a lot of application as a propelling nozzle in automobiles and jets. Few examples of the application of convergent-divergent type of nozzles in engineering are:
Steam turbines: In power plants.
Rockets: for providing sufficient thrust to move upwards.
The supersonic gas turbine engine: for the air intake when the air requirement of the engine is high.
C-D nozzles can be seen in water supply pumps or in formula car intake systems or jet engines for providing sufficient thrust to propel at high speeds mostly in supersonic jets.
If you are preparing for ESE/ GATE or other PSU Exams (Mechanical Engineering), then avail Online Classroom Program for ESE and GATE ME:
## Comprehensive Preparation for GATE & ESE ME Exams(500+ Hours of Live Classes, 300+ Quizzes for Practice and 60+ Mock Tests)
You can avail Test Series specially designed for all Mechanical Engineering Exams:
All the Best!
#DreamStriveSucceed.
Update BYJU’S Exam Prep, Best GATE exam app for Preparation
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# When Is A Profit, A Profit? 2 (Success Series)
BarroMetrics Views: When Is A Profit, A Profit? 2 (Success Series)
In the last issue, I considered the question: when do we add profits to capital? When do we deduct losses from capital? I introduced Ryan Jones’ approach.
(An aside.
Speaking of Ryan Jones, he has a sale on for his course starting Tuesday, April 19 and running for two days: see Fixed Ratio on Tuesday.
His course is normally USD 797.00; on the sale, his price will be, USD 197.00. And ‘no’ to those who will ask, “I am not receiving a fee for referring the course’. In fact, Ryan does not even know I have mentioned the saving).
Turning back to my subject……
Ryan’s approach does not suit me. I prefer to use another method – not better, just one with which I am more comfortable. My take requires we know:
1. Number of historical consecutive losses
2. Avg consecutive losses
3. Loss Rate
4. Population Size
5. Avg pa return on capital (Avg ROI)
Let’s take my trading. Taking my trading results since 1990 (when I began the private limited partnership fund), we have the following results:
• Historical consecutive losses: 16
• Avg Consecutive losses: 4
• Loss Rate: .485
• Avg pa return on capital: 27.63%
Firstly, the data allows me to calculate the theoretical consecutive loss: 10.64 (say 11). So, I would not consider the consecutive loss sequence of 16 as an outlier. (If I did, I would need to reassess the data. For example does the data show a period of extraordinary losses?) With a run of 16 losses, if I risk 2% per trade, I’d risk a loss of 32%. Since my average profit is only 27.63%, I am not prepared to run that risk.
What if I risked 1%? My possible loss would be 16%. Still a little high – I am aiming for a risk of around 50% of average ROI (14%). So, I’d bring down the risk per trade to around 0.90% which I find an acceptable risk; I treat the 0.9% as my ‘normal’ size.
Secondly, when to add to capital? I’d add on any increase above 50% of average ROI = 14%. And finally, when to subtract losses from capital? I’d subtract losses whenever I’d lose 50% of the increased value (i.e. 50% of 14 = 7%).
So far we have determined what I would consider ‘normal size.’ I have one more step to determine my position size for the current trade: I have to decide whether my trading is currently in Ebb, Flow or Normal State.
More on this tomorrow.
## 10 thoughts on “When Is A Profit, A Profit? 2 (Success Series)”
1. ray says:
Apologies if you have posted comments.
We migrated to a new platform. In the move, we lost your comments Do please post again.
Thanks
ray
2. chrisj says:
re
“5.Avg pa return on capital”
I would like to confirm my understanding of your post.
By capital you mean the \$ amount that you are basing your return calculation on at any given time and the \$amount used for position sizing calc as opposed to the actual account balance.
For example, the actual account balance could be \$107,115 but the “capital” figure being used for position sizing and return on capital calculation might be \$102,150.
3. chrisj says:
Re
“Firstly, the data allows me to calculate the theoretical consecutive loss: 10.64 (say 11)”.
How was that calculated?
4. ray says:
Hi Chris
Best explain by way of example.
Let’s say I start with \$100,000.00 – that’s my capital. I make \$8,000.00. That’s less than 14% so, my capital is still \$100,000.00.
If I now lose 7% of \$100,000.00, I reduce my capital
to \$93,000.00. The \$8,000.00 is held in reserve as it were. I would now need to increase my \$93,000.00 by 14% (including the \$8,000.00) to add profits to the \$93,000.00
But, let’s then say instead of losing, I make another S7,000.00. I’d this profit to the \$8,0000.00. I have now made 15%. So, I add the \$15,000.00 to \$100,000.00.
I know have \$115,000.00 as my capital base.
5. chrisj says:
Thanks Ray.
Perfectly clear.
6. ray says:
chrij posted this comment. I approved it but the post has not published:
““chrisj said, in April 19th, 2016 at 7:33 am
Re
“Firstly, the data allows me to calculate the theoretical consecutive loss: 10.64 (say 11)”.
How was that calculated?”
Hi Chris
The formula I use:
(log of runs)/(-log(winRate)
7. chrisj says:
Can you provide a working example using the stats in your original post as inputs to the formula.
8. ray says:
Hi Chris
I’ll try – all I can do is give you some inputs, Excel does the rest for me.
Let’s say my population is 2210 and my loss rate is .485. The formula I use is:
(log of runs)/(-log(loss rate) = longest losing streak.
The quotient gives an occurrence of 11.
I have seen another formula: (log(1-DC)/log(1-PW)
Where DC = Confidence Level (usually .99). The result gives an occurrence of 7 where the LossRate is .485.
I prefer to use the first formula.
A Google search will throw up probability calculators. I usually don’t use them unless they disclose the formula they are using.
9. chrisj says:
For any one else who may have gotten confused this may help.
(log of runs)/(-log(winRate)
Later corrected to:
(log of runs)/(-log(loss rate)
i.e. loss rate.
I did a little spreadsheet and with the revised formula I get the same result as in Ray’s original post.
I will send my spreadsheet to Ray so he can upload it as an attachment if he is okay with doing so.
Chris
1. ray says:
Hi Chris
Thanks.
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It's not as hard as you think it is. The first step is to combine like terms....what are those? Like terms are constants and variables that share qualities. For example you have two terms her with the variable "x". to combine them, look at the sign that is in front of them and add them together....
Here are some etymology references on the orign of "hello".. http://www.etymonline.com/index.php?term=hello http://en.wikipedia.org/wiki/Hello Hope that helps. Please remember to click the thumbs up if this helped you! Thank you :)
Just to add to Nataliya's anwer. The set of numbers given identify a point on the coordinate grid, and the m represents the slope of a line on which the identified point lies. So if you were to graph this line, you would start at (2, -5) then move 2 up and 1 to the right (slope is always rise over...
I won't do your homework for you :) But here are a couple links to websites with this information. http://webspace.ship.edu/cgboer/siddhartha.html http://en.wikipedia.org/wiki/Gautama_Buddha http://buddhism.about.com/od/lifeofthebuddha/a/buddhalife.htm
This looks much more confusing than it really is. Let's look at the problem step-by-step and I'll explain it as I go along. First, you have x that stand for a number that is unknown for now, but look closly at the problem. You have more than 1 x, and they are all being added. So we can use the...
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# Four easy ways to set the baseline for your time series forecasts.
Tealfeed Guest Blog
3 years ago | 4 min read
A simple, common-sense approach will establish a baseline that you’ll have to beat in order to demonstrate the usefulness of more-advanced machine-learning models.
For data scientists to keep their sanity when building models, they set a baseline — a score that the model must outperform. Normally, the state-of-the-art is used as the baseline but for problems with no existing solutions yet, one should build their own baseline.
As Francois Chollet put it, a baseline is used to demonstrate the usefulness of more advanced forecasting techniques.
In this article, we will review four elementary baselines applied in time series forecasting problems.
To demonstrate each of the baseline method, let’s use the historical monthly prices of wheat from World Bank between 2001 and 2019. Prices in 2019 will be forecasted.
Wheat prices from 2001–2019 in \$/mt
To quantify the performance of the baseline methods, I’ll use mean squared error (MSE). As the name suggests, MSE, is an error function that measures the average squared difference between forecasted and true values.
Mean Squared Error
MSE can be computed using the equation above where N is the number of observations, Yi is the true value and Yi(hat) is the forecasted value. And since it is an error function, you’ll want it to be as small as possible.
# Average Method
This method simply takes the average (or “mean”) value of the entire historical data and use that to forecast future values. Very useful for data with small variance or whose value lies close to the mean.
Using the wheat price data, the forecasts for 2019 prices is equal to the average monthly price from year 2001 to 2018. MSE for this method is 19.7. To make the visualizations clear, only the last 60 observations including the forecasted values are shown. The blue flat line in the figure below represents the average value from year 2001 to 2018, while the orange one represents true values.
Average Method
# Drift Method
Drift is the amount of change observed from the data. In this method, drift is set to be the average change seen in the whole historical data and uses that to forecast values in the future. Basically, this just means drawing a straight line using the first and last values and extend that line into the future. This method works well on data that follows a general trend over time.
Drift method forecast is shown as the blue line in the figure below. Its slope follows the slope of the line drawn between the first price in the data and the last price from year 2018. MSE using this method on wheat price data is lower than the average method at 12.4.
Drift Method
# Moving Average
A variation of the average method is to use a series of averages from a fixed number of recent values, this is called moving average method. This method is often used in technical analysis of stock prices. This is useful if you are more concerned about long-term trends rather than short-term ones.
If we set the size of the moving average window to 24, then we will use the last 24 values to forecast the next time step. Repeat this step by moving the window one step forward, where the previous forecast will be included in the average. Keep repeating until the desired forecast length is achieved. Using a window of 24, the moving average forecast for year 2019 is shown in the figure below as the blue line. MSE for wheat price data is much better at 5.84.
Moving Average Method
# Naïve Method
This method uses the most recent value as the forecasted value for the next time step. The assumption followed by this method is that its value tomorrow is equal to its value today.
To do this using the wheat prices data, we simply shift the 2019 prices one time step forward and use that as the forecast prices. Notice that the blue line leads the orange line by a single time step in the figure below.
For many economic and financial time series, this method works outstandingly well. MSE for wheat data is even better at 4.
Naïve Method
# Conclusion
In most cases and from what we have seen in the demonstration above, Naïve method performs best among the four methods discussed and is commonly used as the baseline for forecasting tasks. These methods appear to be very simple, but such elementary baselines are sometimes hard to beat.
Though one have the choice to directly select any of the baseline methods discussed above based on the characteristics of the data being analyzed, it is a good practice to exhaust all known methods so long as the resources permit.
For example, if you use deep learning for forecasting tasks, an ARIMA model might be a good choice for a baseline as it can give even better results than any of the methods discussed in this article.
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