url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
values | snapshot_type stringclasses 2
values | language stringclasses 1
value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.intel.com/content/www/us/en/docs/programmable/683406/17-1/small-rom-architecture.html | 1,679,356,356,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943562.70/warc/CC-MAIN-20230320211022-20230321001022-00220.warc.gz | 944,288,201 | 27,120 | ID 683406
Date 11/06/2017
Public
## 3.1.2. Small ROM Architecture
.To reduce memory usage (but increase LE usage) and increase output frequency, use the small ROM architecture.
In a small ROM architecture, the device memory only stores 45 degrees of the sine and cosine waveforms. All other output values are derived from these values based on the position of the rotating phasor on the unit circle.
Table 6. Derivation of Output Values
Position in Unit Circle Range for Phase x sin(x) cos(x)
1 0 <= x < π/4 sin(x) cos(x)
2 π/4 <= x < π/2 cos(π/2-x) sin(π/2-x)
3 π/2 <= x < 3π/4 cos(x-π/2) -sin(x-π/2)
4 3π/4 <= x < π sin(π-x) -cos(π-x)
5 π <= x < 5π/4 -sin(x-π) -cos(x-π)
6 5π/4 <= x < 3π/2 -cos(3π/2-x) -sin(3π/2-x)
7 3π/2 <= x < 7π/4 -cos(x-3π/2) sin(x-3π/2)
8 7π/4 <= x < 2π -sin(2π-x) cos(2π-x)
A small ROM implementation is more likely to have periodic value repetition, so the resulting waveform's SFDR is lower than that of the large ROM architecture. However, you can often mitigate this reduction in SFDR by using phase dithering.
Figure 9. Derivation of output Values | 377 | 1,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-14 | longest | en | 0.836104 |
https://www.rapidtables.com/calc/math/average-calculator.html | 1,717,106,856,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971684053.99/warc/CC-MAIN-20240530210614-20240531000614-00488.warc.gz | 821,510,018 | 6,126 | # Average Calculator
* SD = standard deviation
Weighted average calculator ►
## Average calculation
The average (arithmetic mean) is equal to the sum of the n numbers divided by n:
Average =
sumcount
=
a1+a2+...+ann
#### Example
The average of 1,2,5 is:
Average =
1+2+53
= 2.667
## Weighted average calculator
Weight Data value
## Weighted average calculation
The weighted average (x) is equal to the sum of the product of the weight (wi) times the data number (xi) divided by the sum of the weights:
#### Example
Find the weighted average of class grades (with equal weight) 70,70,80,80,80,90:
Since the weight of all grades are equal, we can calculate these grades with simple average or we can count how many times each grade appears and use weighted average.
2×70,3×80,1×90
x =
2×70+3×80+1×902+3+1
=
4706
= 78.33333
Weighted average calculator ► | 241 | 868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-22 | latest | en | 0.856021 |
http://www.stuffintheair.com/maths-for-quantifying-things-in-real-life.html | 1,519,588,165,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816912.94/warc/CC-MAIN-20180225190023-20180225210023-00266.warc.gz | 531,593,505 | 8,445 | # Maths for Quantifying Things in Real Life
Some help
I guess without Mathematics we cannot quantify things that we look and feel in this world. In order to quantify various things like solid, liquid and gas we need to have different units to each object. The metric system helps us to use different units to quantify objects. In order to learn the metric system, one should practice all the units in real life when we look at a particular object.
For example, we travel daily from one place to another. We must first get familiar with the units involved in travelling alone like feet, speed, time, inches, metres or miles. Unless we practice the metric units in day to day life we will automatically forget and get confused with things.
Also, good knowledge in metric units make you a person with good common sense.
Barry's Response - Everyday math is something we use more than we realize. Thanks for shedding light on this idea.
If you do anything like shopping or photography, you will use latent math in your pursuits almost inadvertently. If you don't somebody else just might...and it might not be to your advantage.
You have a right to know if something is trying to take advantage of you, and you have a right to make decisions accordingly and do something about it if you see fit. Mathematics provides one (of many possible) tools to make these decisions intelligently. You don't need expertise in algebra, statistics or calculus to do so, but these disciplines can help in specialized situations of course.
You might want to understand risk, often tied to probability, in many life situations. You want to learn to recognize the limitations your predictive calculations may implicitly possess. A volcano, for example, may suddenly throw off anybody's long term weather forecast. There are lots of things to consider...more than listed here.
ADD TO OTHER SOCIAL BOOKMARKS: Del.icio.us DiggSpurl | 382 | 1,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-09 | latest | en | 0.932289 |
politicalmyths.wordpress.com | 1,498,695,143,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323808.56/warc/CC-MAIN-20170629000723-20170629020723-00356.warc.gz | 808,356,520 | 32,609 | ## Polling Results Within the “Margin Of Error” Are Not a “Statistical Tie.”
Political Myth: If the difference between two candidates’ polling numbers is less than the “margin of error” for the poll, then it’s a “statistical tie.”
Reality: Even if the results are “within the margin of error,” the candidate with higher polling numbers is still most likely to be in the lead.
Explanation
Suppose you have a bag with 100 million balls. Some of them are blue, and some of them are red. You want to know how many of your 100 million balls are red and how many are blue, but it would take too long to count them all. So, you thoroughly mix up the balls in the bag, close your eyes, and pick out 500 balls at random. By counting those 500 balls, assuming you truely did pick them randomly, you can come up with a pretty close approximation of how many of the 100 million balls in the bag are blue, and how many are red?
So the question is, how close an approximation can you get, and how reliable is it? If 250 of the balls you chose were blue and 250 were red, then it is theoretically possible that the total number of blue balls in the bag was 250, and you just happened to randomly select all of them. It is, similarly, theoretically possible that there were only 250 red balls in the bag. The chances against either being the case are astronomically small, but it’s possible.
If you choose 500 balls at random from a bag with 100 million balls, there is a 95% chance that the 500 balls you picked will accurately represent the colors of the balls in the bag +/- 4.4%. The “95%” figure is called the “Confidence Level.” The “4.4% figure is called the “margin of error.” A 4.4% margin of error at a 95% confidence level means 95% of the time, the result will be accurate within + or – 4.4%, but it also means that 5% of the time it will be more than 4.4% inaccurate.
So, suppose Candidate A is polling at 47% and Candidate B is polling at 50%, with a poll that has a margin of error of +/- 4.4% at a 95% confidence level. Since the two are only 3% apart, we can’t be 95% cure that Candidate B is beating Candidate A. But does that mean each candidate is equally likely to be winning? Not by a long shot. We’re 95% sure that the sample is accurate within +/- 4.4%, but we’re 90% sure that the sample is accurate within +/-3.7%, and 80% sure it’s accurate to within 2.9%. The “margin of error” and “confidence level” of a poll go hand-in-hand. One number is meaningless without the other. Unfortunately, most polls published in the press do not specify what their confidence level is, but the industry standard is 95%.
Of course, the closer the two candidates’ polling numbers are, the more likely it is that the candidate who appears to be behind is actually ahead, but just because we can’t be 95% sure that Candidate B is ahead doesn’t mean that we might not be 90% sure, or 80% sure, or even 75% sure. But, to stretch a point, even if a poll with a margin of error of +/- 10% at a 90% confidence leval has Candidate A beating Candidate B by 51%/49%, there is always more than a 50% chance that Candidate A is winning. Depending on the poll it may be a 60% chance, an 80% chance, or an 85% chance, but it’s always more likely than not that the candidate who appears to be ahead actually is ahead.
That said, “margin of error” is far from being the biggest issue in accurate polling. The only thing “margin of error” accounts for is random variation — i.e. the possibility that a random sample might not be completely representative of the population the sample is taken from. But if you’re confident that you can determine what an entire pot of soup will taste like by tasting only one spoonful, you can also be generally confident that you can determine how 100 million people will vote by asking a randomly selected 1,000 of them. | 985 | 3,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-26 | longest | en | 0.962286 |
http://nrich.maths.org/7301/note?nomenu=1 | 1,498,207,844,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320040.36/warc/CC-MAIN-20170623082050-20170623102050-00530.warc.gz | 312,393,830 | 3,632 | ## Shaping It
These pictures were made quite simply by starting with a square, finding the half-way point on each side and joining those points up. This creates a smaller shape (which also happens to be a square) inside the original. The half-way points of this new shape are then joined up to make a third shape. This way of making new shapes is continued until it gets too small to do properly.
Here's one where I've coloured each new halving line to help to see what has happened more clearly.
So, it's your turn to have a go.
It's probably good to start with a fairly large shape since it's going to get smaller and smaller each time.
Here are some challenges for you to pursue:
• Having made a design like one above, cut out the triangles and the smallest inner shape and rearrange the pieces to form a new shape/design.
• Talk about and record the things you notice as you have drawn more and more halving lines.
• What is happening to the enclosed area each time the sides are halved? (Try investigating a regular shape first.)
This problem is based on an idea suggested by Ian Short.
### Why do this problem?
This activity enables pupils with a wide range of attainment level to work on the same challenge to improve their concepts of shape and space. It also offers opportunities for further investigation and for pupils to create their own challenges.
### Possible approach
This might need to vary according to your learners' experiences. For those who have good pencil, ruler and measuring skills, some of the examples here (Word document or pdf) could be presented. You could ask children to describe what they see, with a partner first and then open it out to the whole group. Discussion could follow that would allow pupils to decide what their own explorations would be. Some may want to create their own whereas those who struggle with motor control may wish to explore the given ones.
It could be that you ask children to feed back about their discoveries orally, or you may wish them to create a poster of some kind. Encourage them to explain their observations. This activity could lend itself to being investigated over an extended period of time (a 'simmering activity') and it would be useful to dedicate a space on the wall for learners to contribute their work during that time.
### Key questions
What have you been exploring?
Tell me what you have found.
### Possible extension
Some pupils may be encouraged to use a spreadsheet to explore sizes of different lengths/areas. Others may be encouraged to compare results when the new lines are placed a quarter (or other fraction) along the previous lines instead of half way.
### Possible support
Some pupils may need to have help with the fine motor skills required whilst being interested in the overall effect of performing this kind of action on a shape. | 586 | 2,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-26 | latest | en | 0.96994 |
https://www.calculushowto.com/ampersand-curve/ | 1,643,051,138,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304600.9/warc/CC-MAIN-20220124185733-20220124215733-00234.warc.gz | 722,258,517 | 24,151 | # Ampersand Curve
Share on
The ampersand curve, which resembles an ampersand “&”, is a quartic curve with the algebraic equation:
6x4 + 4y4 – 21x3 + 6x2y2 + 19x2 – 11xy2 – 3y2 = 0.
Alternatively, the implicit equation is [1]:
(y2 – x2)(x – 1)(2x – 3) = 4(x2 + y2 – 2x)2.
The curve has three ordinary double points in the real plane at (x, y) = (0, 0), (1, 1) and (1, -1) [2].
## History of the Ampersand Curve
The named “ampersand curve” was popularized by Cundy and Rollett [3], but they weren’t the first to study it. According to John Derbyshire, author of Unknown Quantity: A Real and Imaginary History of Algebra [4], the curve can be found in a 19th century book by Percival Frost called An Elementary Treatise on Curve Tracing which was published in 1892 and showed the reader a myriad of methods for turning algebraic equations into drawn curves. The ampersand figure can be found in Frost’s book as Plate VII, Figure 27.
The ampersand curve is sometimes called the Plücker quartic, after J. Plücker who constructed a version of the ampersoid in his Theorie der algebraischen Curven [5].
## References
[1] Knill, O. Curves.
[2] Bobenko, A. (Ed.). (2011). Computational Approach to Riemann Surfaces. Springer.
[3] H.M. Cundy, A.P. Rollet, mathematical models, Oxford, Clarendon Press (1972), traduit en français : modèles mathématiques, Cédic (1978).
[4] Derbyshire, J. (2006). Unknown Quantity: A Real and Imaginary History of Algebra. Joseph Henry Press.
[5] Plücker, J. Theorie der algebraischen Curven: Gegründet auf eine neue Behandlungsweise der analytischen Geometrie. Berlin: Adolph Marcus, 1839.
CITE THIS AS:
Stephanie Glen. "Ampersand Curve" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/ampersand-curve/
---------------------------------------------------------------------------
Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free! | 584 | 2,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2022-05 | longest | en | 0.785804 |
https://wizbangblog.com/2004/06/14/scientists-take/ | 1,560,890,880,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998817.58/warc/CC-MAIN-20190618203528-20190618225528-00188.warc.gz | 653,732,038 | 12,832 | Scientists Take The Fun Out Of Funny.
The Guardian reports that scientists expressed the prefect joke mathematically.
The mathematical equation for the perfect joke was revealed by scientists in London yesterday. The calculation is c=(m nO)/p.
In the formula, c is the funniness of the joke; m is the “comic moment” (arrived at by multiplying the punchline’s funniness rating by the length of the joke’s buildup); nO is the number of times the subject undergoes a pratfall, multiplied by the “ouch factor” – the social and physical pain of the indignity involved. The total is divided by the number of puns, p.
The formula was worked out by Helen Pilcher and Timandra Harkness – both scientists and stand-up comedians – who make up the Comedy Research Project, which they run in collaboration with the Science Museum’s Dana Centre in London.
Puns are seen as dissipating the power of a joke because they tend to encourage groans rather than laughter.For some reason I am reminded of the story of the new prisoner in the penitentiary, in with a bunch of old timers. He notices an odd ritual going on among the inmates. During dinner and other times, some one will suddenly look around at the other guys and say, “. . #37”, or some other number, apparently at random. The other guys will all grin, or sometimes laugh. He asks somebody what is going on. “Well, we’ve all been here so long, that we’ve all heard everybody’s jokes many times. So we’ve just assigned numbers to each of them. Now we just say the number instead of bothering with the whole joke.” The new guy decides to give it a try. At dinner he looks around and says, “. . #54!” But no one smiles or laughs. He asks the fellow next to him, “What’s wrong? Did I pick a bad joke?” The other guy shakes his head and says, “Well, don’t take it wrong, but some folks just can’t tell a joke.” | 426 | 1,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-26 | latest | en | 0.95429 |
http://books.google.com/books?id=A-FEAAAAIAAJ&lr= | 1,394,376,971,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999679206/warc/CC-MAIN-20140305060759-00044-ip-10-183-142-35.ec2.internal.warc.gz | 25,538,912 | 22,815 | # Elements of Geometry and Trigonometry: With Applications in Mensuration (Google eBook)
A.S. Barnes, 1886 - Geometry - 324 pages
### What people are saying -Write a review
We haven't found any reviews in the usual places.
### Contents
BOOK 9 Axioms 16 BOOK II 38 Problems relating to the First and Second Books 53 53 BOOK III 69 BOOK IV 82 Measurement of Areas and Proportions of Figures 88108 88 Problems relating to the Fourth Book 109113 109
BOOK V 116 BOOK VI 126 Appendix 103164 163 Pure 165 APPLICATIONS OF GEOMETRY 211 Mensuration or Solids 289 240 Mensuration or the Round Bodies 248
### Popular passages
Page 12 - For this purpose it is divided into 360 equal parts called degrees, each degree into 60 equal parts called minutes, and each minute into 60 equal parts called seconds. The degrees, minutes, and seconds are marked thus ° ' " ; and 9° 18' 16", are read, 9 degrees 18 minutes and 16 seconds.
Page 60 - After remarking that the mathematician positively knows that the sum of the three angles of a triangle is equal to two right angles...
Page 84 - If two triangles have two sides and the included angle of the one, equal to two sides and the included angle of the other, each to each, the two triangles will be equal in all their parts." Axiom 1. "Things which are equal to the same thing, are equal to each other.
Page 130 - ... or cylinder be cut by a plane parallel to the base, the section is a figure parallel and similar to the base. The one point a...
Page 49 - The angle formed by a tangent and a chord is measured by half the intercepted arc.
Page 209 - Being on a horizontal plane, and wanting to ascertain the height of a tower, standing on the top of an inaccessible hill, there were measured, the angle of elevation of the top of the hill 40°, and of the top of the tower 51° ; then measuring in a direct line 180 feet farther from the hill, the angle of elevation of the top of the tower was 33° 45' ; required the height of the tower.
Page 30 - The perpendicular is the shortest straight line that can be drawn from a given point to a given straight line; and...
Page 70 - To express that the ratio of A to B is equal to the ratio of C to D, we write the quantities thus : A : B : : C : D; and read, A is to B as C to D.
Page 12 - For the purpose of measuring angles, the circumference is divided into 360 equal parts, called degrees ; each degree into 60 equal parts, called minutes; each minute into 60 equal parts, called seconds.
Page 59 - If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are congruent.
### References from web pages
Free Books > Computers & Internet > Web Development > HTML ...
Free Books > Computers & Internet > Web Development > HTML, Graphics, & Design > Web Graphics > Elements Of Geometry And Trigonometry
2020ok.com/ books/ 29/ elements-of-geometry-and-trigonometry-19129.htm
The Online Books Page: Elements of Geometry and Trigonometry, by ...
Link: frame- and javascript-dependent page images at Cornell · Stable link here: http://onlinebooks.library.upenn.edu/webbin/book/lookupid?key=olbp21963 ...
onlinebooks.library.upenn.edu/ webbin/ book/ lookupid?key=olbp21963
No. 472: Legendre's Math Text
About an old geometry text at the Battle of Charleston
www.uh.edu/ engines/ epi472.htm
JSTOR: American Contributions to Mathematical Symbolism
American Contributions to Mathematical Symbolism. Florian Cajori. The American Mathematical Monthly, Vol. 32, No. 8, 414-416. Oct., 1925. ...
Collection of Early American Mathematics Books
Collection of Early American Mathematics Books. The collection is in the Bancroft Library at the University of California in Berkeley. ...
www.math.gatech.edu/ ~hill/ publications/ books/ booklist.html
Elements of geometry and trigonometry de Davies Charles, Libro Usado
Compra el libro Elements of geometry and trigonometry de Davies Charles
www.buscalibros.cl/ libro.php?libro=220992
Amy Ackerberg-Hastings
Chapter Five. The Two Circles Will Touch Each Other Internally:. Charles Davies at the Art and Business of Teaching Geometry. Ò[W]ith scientific attainments ...
www.math.usma.edu/ people/ Rickey/ dms/ DeptHeads/ Davies%20by%20Amy%20Ackerberg-Hastings.rtf
Earliest Known Uses of Some of the Words of Mathematics (C)
CALCULUS. In Latin calculus means "pebble." It is the diminutive of calx, meaning a piece of limestone. The counters of a Roman abacus were originally made ...
members.aol.com/ jeff570/ c.html | 1,110 | 4,498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2014-10 | latest | en | 0.909696 |
https://mathoverflow.net/questions/321955/lower-bound-to-epsilon-expansion-of-a-subset-of-a-half-sphere | 1,553,406,577,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203326.34/warc/CC-MAIN-20190324043400-20190324065400-00518.warc.gz | 568,584,721 | 28,508 | # Lower bound to $\epsilon$-expansion of a subset of a half-sphere
Below are two known lemmas on a $$d$$-dimensional sphere (related to the isoperimetric inequality). I would like to know: does a similar statement like this holds for a $$d$$-dimensional dome also (i.e. half the sphere)
Lemma 1: Suppose $$s$$ is a subset of the $$d$$-sphere with normalised measure $$\mu(s)=1/2$$. The $$\epsilon$$-expansion of the set $$s$$ is then at least as large as the $$\epsilon$$-expansion of a half sphere, if we use the geodesic metric.
Lemma 2: An $$\epsilon$$-expansion of a half sphere in the geodesic metric has a normalised measure $$\geq 1-\sqrt{\pi/8} \exp(-d \epsilon^2/2)$$
Definition used: $$\epsilon$$-expansion of a set $$s$$ contains all points that are at most $$\epsilon$$ units away from $$s$$ according to a specified metric.
Suppose we don't have a sphere to begin with, but a $$d$$-dimensional dome or half a sphere instead. We can keep using the geodesic metric. Then we take a subset $$s'$$ of this dome with normalised measure $$\mu(s')=0.5$$. What is the normalised measure of the $$\epsilon$$-expansion of this set $$s'$$? Can it be lower-bounded by some function $$f(d, \epsilon)$$ just like lemma 2 for the sphere?
Any leads on this problem would be much appreciated! Thank you in advance! | 371 | 1,315 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-13 | latest | en | 0.916466 |
http://sciencedocbox.com/Physics/77082537-Chapter-13-small-signal-modeling-and-linear-amplification.html | 1,563,896,627,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529480.89/warc/CC-MAIN-20190723151547-20190723173547-00277.warc.gz | 139,205,622 | 29,224 | # Chapter 13 Small-Signal Modeling and Linear Amplification
Save this PDF as:
Size: px
Start display at page:
## Transcription
1 Chapter 13 Small-Signal Modeling and Linear Amplification Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock 1/4/12 Chap 13-1
2 Chapter Goals Understanding of concepts related to: Transistors as linear amplifiers dc and ac equivalent circuits Use of coupling and bypass capacitors and inductors to modify dc and ac equivalent circuits Concept of small-signal voltages and currents Small-signal models for diodes and transistors Identification of common-source and common-emitter amplifiers Amplifier characteristics such as voltage gain, input and output resistances, and linear signal range Rule-of-thumb estimates for the voltage gain of common-emitter and common-source amplifiers. 1/4/12 Chap 13-2
3 Introduction to Amplifiers Amplifiers usually use electronic devices operating in the Active Region A BJT is used as an amplifier when biased in the forward-active region The FET can be used as amplifier if operated in the pinch-off or saturation region In these regions, transistors can provide high voltage, current and power gains Bias is provided to stabilize the operating point (the Q-Point) in the desired region of operation Q-point also determines Small-signal parameters of transistor Voltage gain, input resistance, output resistance Maximum input and output signal amplitudes Power consumption 1/4/12 Chap 13-3
4 Transistor Amplifiers BJT Amplifier Concept The BJT is biased in the active region by dc voltage source V BE. Q-point is set at (I C, V CE ) = (1.5 ma, 5 V) with I B = 15 µa (β F = 100) Total base-emitter voltage is: v BE = V BE + v be Collector-emitter voltage is: v CE = 10 i C R C This is the load line equation. 1/4/12 Chap 13-4
5 Transistor Amplifiers BJT Amplifier (cont.) If changes in operating currents and voltages are small enough, then i C and v CE waveforms are undistorted replicas of the input signal. A small voltage change at the base causes a large voltage change at collector. Voltage gain is given by: 8 mv peak change in v BE gives 5 µa change in i B and 0.5 ma change in i C. 0.5 ma change in i C produces a 1.65 V change in v CE. A v = V ce V be = o o = o = 206 Minus sign indicates 180 o phase shift between th einput and output signals. 1/4/12 Chap 13-5
6 Transistor Amplifiers MOSFET Amplifier Concept A v = V ds V gs A v = 4 180o 1 0 o A v = 4.00 MOSFET is biased in active region by dc voltage source V GS. Q-point is set at (I D, V DS ) = (1.56 ma, 4.8 V) with V GS = 3.5 V Total gate-source voltage is: v GS = V GS + v gs 1 V p-p change in v GS yields 1.25 ma p-p change in i D and a 4 V p-p change in v DS 1/4/12 Chap 13-6
7 Transistor Amplifiers Coupling and Bypass Capacitors Capacitors are designed to provide negligible impedance at frequencies of interest and provide open circuits at dc. ac coupling through capacitors is used to inject ac input signal and extract output signal without disturbing Q-point C 1 and C 2 are low impedance coupling capacitors or dc blocking capacitors whose reactance at the signal frequency is designed to be negligible. C 3 is a bypass capacitor that provides a low impedance path for ac current from emitter to ground, thereby removing R E (required for good Q-point stability) from the circuit when ac signals are considered. 1/9/12 Chap 13-7
8 Transistor Amplifiers dc and ac Analysis Two Step Analysis dc analysis: Find dc equivalent circuit by replacing all capacitors by open circuits and inductors by short circuits. Find Q-point from dc equivalent circuit by using appropriate largesignal transistor model. ac analysis: Find ac equivalent circuit by replacing all capacitors by short circuits, inductors by open circuits, dc voltage sources by ground connections and dc current sources by open circuits. Replace transistor by its small-signal model Use small-signal ac equivalent to analyze ac characteristics of amplifier. Combine end results of dc and ac analysis to yield total voltages and currents in the network. 1/9/12 Chap 13-8
9 Transistor Amplifiers dc Equivalent Circuit for BJT Amplifier All capacitors in the original amplifier circuit are replaced by open circuits, disconnecting v I, R I, and R 3 from circuit. 1/9/12 Chap 13-9
10 Transistor Amplifiers ac Equivalent Circuit for BJT Amplifier Capacitors are replaced by short circuits 1/9/12 Chap 13-10
11 Transistor Amplifiers dc and ac Equivalents for a MOSFET Amplifier Full circuit dc equivalent ac equivalent Simplified ac equivalent 01/10/12 Chap 13-11
12 Small-Signal Operation Diode Small-Signal Model The slope of the diode characteristic at the Q-point is called the diode conductance and is given by: Diode resistance is given by: r d = 1 g d 01/10/12 Chap 13-12
13 Small-Signal Operation Diode Small-Signal Model (cont.) g d is small but non-zero for I D = 0 because slope of diode equation is nonzero at the origin. At the origin, the diode conductance and resistance are given by: g d = I S V T and r d = V T I S 01/10/12 Chap 13-13
14 Small-Signal Operation Definition of a Small-Signal Now let s determine the largest magnitude of v d that represents a small signal. V T i D = I S exp v D 1 V T I D +i d = I S exp V D 1 +I S exp v D V T I D +i d = I S exp V +v D d 1 v d V T Subtracting I D from both sides of the equation, i d =(I D + I S ) v d + V 1 T 2 2 v d V T v d V T V T v 2 d V T v 3 d V T For i d to be a linear function of signal voltage v d, v d << 2V T = 0.05 V or v d 5 mv. Thus v d 5 mv represents the requirement for small-signal (linear) operation of the diode i d =(I D + I S ) v d = g v i d d D = I D + g d v d V T 01/10/12 Chap 13-14
15 Small-Signal Operation BJT Small-Signal Model (The Hybrid-Pi Model) The bipolar transistor is assumed to be operating in the Forward-Active Region: i C I S exp v BE 1+ v CE and i B i C V T V A Using a two-port y-parameter network: i c = g m v be + g o v ce i b = g π v be + g r v ce The port variables can represent either time-varying part of total voltages and currents or small changes in them away from Q-point values. β F ( V CE V BE ) 01/10/12 Chap 13-15
16 Small-Signal Operation BJT Small-Signal Model (The Hybrid-Pi Model) i c = g m v be + g o v ce i b = g π v be + g r v ce i C I S exp v BE 1+ v CE i B i C β F V T V A β o is called the small-signal commonemitter current gain of the BJT. g m = i c = i C = I C v be v v ce=0 BE V Q-point T v be g o = i c v ce v be=0 g π = i b v be v ce=0 g r = i b v ce v be=0 v BE = i C v CE Q-point = i B v BE Q-point = i B v CE Q-point V T I C = V A +V T = I C β o V T = 0 01/10/12 Chap 13-16
17 BJT Small-Signal Operation Current Gain & Intrinsic Voltage Gain Intrinsic voltage gain is defined by: β o = g m r π = β F 1 β 1 I F C β F i C Q point β o > β F for i C < I M, and β o < β F for i C > I M. However, for simplicity β F and β o will be assumed to be equal µ f = g m r o = I C V T For V CE << V A µ f V A V T V A +V CE I C 40V A = V A +V CE V T µ f represents the maximum voltage gain an individual BJT can provide and does not change with operating point. 01/10/12 Chap 13-17
18 Small-Signal Operation BJT Hybrid-Pi Model - Summary Transconductance: g m = I C V T 40I C Input resistance: The hybrid-pi small-signal model is the intrinsic representation of the BJT. Small-signal parameters are controlled by the Q-point and are independent of geometry of the BJT r π = β o V T I C Output resistance: r o = V A +V CE I C = β o g m V A I C or β o = g m r π 01/10/12 Chap 13-18
19 BJT Small-Signal Operation Equivalent Forms of Small-Signal Model Voltage-controlled current source g mv be can be transformed into current-controlled current source, v be = i b r π g m v be = g m i b r π = β o i b i c = β o i b + v ce r o β o i b Basic relationship i c = βi b is useful in both dc and ac analysis when the BJT is in the forward-active region. 01/10/12 Chap 13-19
20 BJT Small-Signal Operation Small-Signal Definition i C = I S exp v BE = I S exp V BE + v be i C = I C + i c = I S exp V BE exp v be = I C exp v be V T I C + i c = I C 1+ v be + 1 V T 2 v be V T V T v i c = I be C v be + 1 v be V T 2 V T 6 V T 3 v be V T For linearity, i c should be proportional to v be with v be << 2V T or v be V. i C I C 1+ v be v = I C + I be C The change in i c that corresponds to small-signal operation is: V T V T i c I C = v be V T 0.005V 0.025V = V T V T = I C + g m v be V T 01/12/12 Chap 13-20
21 BJT Small-Signal Operation Small-Signal Model for pnp Transistor For the pnp transistor Signal current injected into base causes decrease in total collector current which is equivalent to increase in signal current entering collector. So the small-signal models for the npn and pnp devices are identical! 01/12/12 Chap 13-21
22 Common-Emitter Amplifiers Small-Signal Analysis - ac Equivalent Circuit ac equivalent circuit is constructed by assuming that all capacitances have zero impedance at signal frequency and dc voltage sources are ac ground. Assume that Q-point is already known. 01/12/12 Chap 13-22
23 Common-Emitter Amplifiers Small-Signal Equivalent Circuit Input voltage is applied to the base terminal Output signal appears at collector terminal Emitter is common to both input and output signals Thus circuit is termed a Common-Emitter (C-E) Amplifier. The terminal gain of the C-E amplifier is the gain from the base terminal to the collector terminal A vt CE = v c v b = g m R L R L = r o R C R 3 01/12/12 Chap 13-23
24 Common-Emitter Amplifiers Input Resistance and Signal Source Gain (β o +1)R E Define R ib as the input resistance looking into the base of the transistor: R ib = v b i b = r π The input resistance presented to v i is: R in = R I + R B R ib = R I + R B r π The signal source voltage gain is: A v CE = v o v i = v o v b v b v i = A vt CE R B r π R I + R B r π 01/12/12 Chap 13-24
25 Common-Emitter Amplifiers Rule of Thumb Design Estimate A CE CE R v = A B r π CE vt A vt R I + R B r π A CE vt = g m R L R L = r o R C R 3 Typically: r o >> R C and R 3 >> R C A v CE g m R C = 40I C R C I C R C represents the voltage dropped across collector resistor R C A typical design point is I C R C = V CC 3 A v CE 40 V CC 3 =13.3V CC To help account for all the approximations and have a number that is easy to remember, our "rule-of-thumb" estimate for the voltage gain becomes A v CE 10V CC 01/13/12 Chap 13-25
26 Common-Emitter Amplifiers Voltage Gain Example Problem: Calculate voltage gain, input resistance and maximum input signal level for a common-emitter amplifier with a specified Q-point Given data: β F = 100, V A = 75 V, Q-point is (0.245 ma, 3.39 V) Assumptions: Transistor is in active region, β O = β F. Signals are low enough to be considered small signals. Room temperature. Analysis: g m = 40I C = mA r o = V A +V CE I C = ( ) = 9.80 ms r π = β o = 100 =10.2 kω g m 9.8mS 75V V 0.245mA = 320 kω R B = R 1 R 2 =160kΩ 300kΩ =104 kω R L = r o R C R L = 320kΩ 22kΩ 100kΩ =17.1 kω R B r π =104kΩ 10.2kΩ = 9.29 kω 7/20/10 Chap 13-26
27 Common-Emitter Amplifiers Voltage Gain Example (cont.) Analysis (cont): R A v = g m R B r π L R I + R B R in = R I + R B r π r π =10.3 kω = 9.8mS 17.1kΩ ( ) 9.29kΩ = kΩ+ 9.29kΩ R v be = v B r π i R I + R B r v be 0.005V v i 5mV 10.3kΩ = 5.54 mv π 9.29kΩ ( ) = 151 Check the rule-of-thumb estimate: A v CE ( ) = 120 (ballpark estimate) ( ) = V What is the maximum amplitude of the output signal: v o 5.54mV 151 1/13/12 Chap 13-27
28 Common-Emitter Amplifiers Voltage Gain Example (cont.) Simulation Results: The graph below presents the output voltage for an input voltage that is a 5-mV, 10-kHz sine wave. Note that although the sine wave at first looks good, the positive and negative peak amplitudes are different indicating the presence of distortion. The input is near our small-signal limit for linear operation. 1/13/12 Chap 13-28
29 Common-Emitter Amplifiers Dual Supply Operation - Example Analysis: To find the Q-point, the dc equivalent circuit is constructed I B +V BE +(β F +1)I B ( )= 5 I B =3.71 µa I C =65I B =241 µa Problem: Find voltage gain, input and output resistances for the circuit above Given data: β F = 65, V A = 50 V Assumptions: Active-region operation, V BE = 0.7 V, small signal operating conditions. I E =66I B =245 µa I C V CE ( )I E ( 5)=0 V CE =3.67 V 1/28/12 Chap 13-29
30 Common-Emitter Amplifiers Dual Supply Operation - Example (cont.) Next we construct the ac equivalent and simplify it kω R in = R B r π =6.31 kω R out = R C r o = 9.57 kω ( ) A v CE = v o v i = g m R out R 3 R in = 84.0 R I + R in Gain Estimate: A CE v 10( V CC +V EE ) = 100 1/28/12 Chap 13-30
31 Small-Signal Operation MOSFET Small-Signal Model g π = i g v gs vds =0 g r = i g v ds v ds =0 = i G v GS Q-point = i G v DS Q-point Using a two-port y-parameter network, i d g m = vgs vds=0 i D = vgs Q-point i g = g π v gs + g r v ds i d = g m v gs + g o v ds The port variables can represent either time-varying part of total voltages and currents or small changes in them away from Q-point values. g o = i d v ds v ds =0 I G = 0 = i D v DS Q-point I D = K n ( 2 V V GS TN ) 2 ( 1+ λv DS ) 1/17/12 Chap 13-31
32 Small-Signal Operation MOSFET Small-Signal Model (cont.) I G = 0 I D = K n ( 2 V V GS TN ) 2 ( 1+ λv DS ) g π = i G v GS Q-point g r = i G v DS Q-point = 0 = 0 i g = g π v gs + g r v ds i d = g m v gs + g o v ds g m = i D v GS Q-point g o = i D v DS Q-point = K n ( 2 V V GS TN )( 1+ λv DS ) = = λ K n ( 2 V GS V TN ) 2 = λi D = 1+ λv DS 2I D V GS V TN I D 1 λ +V DS 1/17/12 Chap 13-32
33 Small-Signal Operation MOSFET Small-Signal Model - Summary Transconductance: g m = 2I D = 2K n I D V GS V TN Since gate is insulated from channel by gate-oxide input resistance of transistor is infinite. Small-signal parameters are controlled by the Q-point. For the same operating point, MOSFET has lower transconductance and an output resistance that is similar to the BJT. Output resistance: r o = 1 g o = 1+λV DS λi D 1 λi D Amplification factor for λv DS <<1: µ f =g m r o = 1+λV DS λi D 1 λ 2K n I D 1/17/12 Chap 13-33
34 MOSFET Small-Signal Operation Small-Signal Definition Assume λv DS <<1. Then i D K n ( 2 v V GS TN ) 2 for v DS v GS V TN For v GS = V GS + v gs, i D = I D +i d = K n 2 2 ( ) + v gs ( v GS V TN ) 2 + 2v gs V GS V TN 2 ( ) + v gs i d = K n 2 2v gs V GS V TN For linearity, i d should be proportional to v gs and we require v 2 gs << 2v gs V GS V TN or v gs << 2( V GS V TN ) v gs 0.2( V GS V TN ) ( ) Since the MOSFET can be biased with (V GS - V TN ) equal to several volts, it can handle much larger values of v gs than corresponding the values of v be for the BJT. The change in drain current that corresponds to small-signal operation is: i d = g m 2 v gs I D I D V GS V TN ( ) 0.2 V GS V TN i d 0.4 I D 1/17/12 Chap 13-34
35 MOSFET Small-Signal Operation Body Effect in Four-terminal MOSFETs Drain current depends on threshold voltage which in turn depends on v SB. Back-gate transconductance is: g mb = i D = i D v BS v Q-point SB Q-point i D g mb = VTN V TN v SB Q-point = ( g m η)=g m η 0 < η < 3 is called the back-gate transconductance parameter. The bulk terminal is a reverse-biased diode. Hence, no conductance from the bulk terminal to other terminals. 01/18/12 Chap 13-35
36 MOSFET Small-Signal Operation Small-Signal Model for PMOS Transistor For a PMOS transistor v SG = V GG v gg i D = I D i d Positive signal voltage v gg reduces sourcegate voltage of the PMOS transistor causing decrease in total current exiting the drain, equivalent to an increase in the signal current entering the drain. The NMOS and PMOS small-signal models are the same! 01/18/12 Chap 13-36
37 Common-Source Amplifiers Small-Signal Analysis - ac Equivalent Circuit ac equivalent circuit is constructed by assuming that all capacitances have zero impedance at signal frequency and dc voltage sources are ac ground. 01/19/12 Chap 13-37
38 Common-Source Amplifiers Small-Signal Equivalent Circuit Input voltage is applied to the gate terminal Output signal appears at the drain terminal Source is common to both input and output signals Thus circuit is termed a Common-Source (C-S) Amplifier. The terminal gain of the C-S amplifier is the gain from the gate terminal to the drain terminal A vt CE = v d v g = g m R L R L = r o R D R 3 01/19/12 Chap 13-38
39 Common-Source Amplifiers Input Resistance and Signal-Source Gain Define R ig as the input resistance looking into the base of the transistor. R vg ig = = R G ii R in is the resistance presented to v i. R in = R I + R G The signal source voltage gain is: A v CS = v o v i = v o v g v g v i CS = A vt A CS R v = g m R G L R I + R G R G R I + R G 01/12/12 Chap 13-39
40 Common-Source Amplifiers Rule of Thumb Design Estimate R A CS v = g m R G L A vt R I + R G CS A CS vt = g m R L R L = r o R D R 3 I Typically: r o >> R D and R 3 >> R D A CS v g m R D = D R D V GS V TN 2 I D R D represents the voltage dropped across drain resistor R D A typical design point is I D R D = V DD 2 with V GS V TN =1 V A v CS V DD Our rule-of-thumb estimate for the C-S amplifier: the voltage gain equals the power supply voltage. Note that this is 10 times smaller than that for the BJT! 01/19/12 Chap 13-40
41 Common-Source Amplifiers Voltage Gain Example Problem: Calculate voltage gain, input resistance and maximum input signal level for a common-source amplifier with a specified Q-point Given data: Κ n = 0.50 ma/v 2, V TN = 1 V, λ = V -1, Q-point is (0.241 ma, 3.81 V) Assumptions: Transistor is in the active region. Signals are low enough to be considered small signals. Analysis: g m = 2K n I D ( 1+ λv DS ) = ms r o = λ 1 +V DS I D = 328 kω R G = R 1 R 2 = 892 kω R L = r o R D R 3 =17.1 kω 1/25/12 Chap 13-41
42 Common-Source Amplifiers Voltage Gain Example (cont.) g m = ms r o = 328 kω R G = 892 kω R L =17.1 kω R A CS v = g m R G 892kΩ L = 0.503mS ( 17.1kΩ) = R I + R G 1kΩ+892kΩ R R in = R I + R G = 893 kω v gs = v G R i v G i 0.2 V GS V TN R I + R G R I + R G V GS V TN 2I D K n = V v i V ( ) 893kΩ ( ) = 8.59 ( ) = V 892kΩ Check the rule-of-thumb estimate: A v CS V DD = 12 V (ballpark estimate) 1/25/12 Chap 13-42
43 Common-Source Amplifiers Voltage Gain Example (cont.) Simulation Results: The graph below presents the output voltage for an input voltage that is a 0.15-V, 10-kHz sine wave. The expected output voltage amplitude is vo = 8.59(0.15) = 1.29 V. Note that although the sine wave at first looks good, the positive and negative peak amplitudes are different indicating the presence of distortion, and the amplitude is actually larger than expected. The input is near our small-signal limit for linear operation. 1/25/12 Chap 13-43
44 C-E and C-S Amplifiers Output Resistance 01/19/12 Chap 13-44
45 C-E and C-S Amplifiers Output Resistance (cont.) Apply test source v x and find i x (with v i = 0) v be = 0 g m v be = 0 R out = v x i x = R C r o R out R C for r o >> R C v gs = 0 g m v gs = 0 R out = v x i x = R D r o R out R D for r o >> R D For comparable bias points, output resistances of C-S and C-E amplifiers are similar. 01/19/12 Chap 13-45
46 JFET Small-Signal Operation Small-Signal Model g g = 1 r g = i G v GS Q-point = I G + I SG V T 0 for I G = 0 i D = I DSS 1 v GS V P for v DS v GS V P i G = I SG exp v GS 1 V T 2 ( 1+ λv DS ) g m = i D = v GS Q-point g o = 1 r o = i D v DS Q-point 2I D 2 I DSS V 2 GS V P V GS V P V P = I D 1 λ +V DS ( ) 01/18/12 Microelectronic Circuit Design Chap 13-46
47 JFET Small-Signal Operation Small-Signal Model (cont.) For small-signal operation, the input signal limit is: v gs <0.2( V GS V P ) Since JFET is normally operated with gate junction reverse-biased, I G = I SG r g = The amplification factor is given by: 1 µ f = g m r o = 2 λ +V DS 2 V GS V λv P P I DSS I D 01/18/12 Microelectronic Circuit Design Chap 13-47
48 Common-Source Amplifiers JFET Example Analysis: Dc equivalent circuit is constructed. Ι G = 0, Ι S = Ι D. V GS = 2000I D V GS = ( )( )1 V GS ( 1) Choose V GS less negative than V P. 2 Problem: Find voltage gain, input and output resistances. Given data: Ι DSS = 1 ma, V P = -1V, λ = 0.02 V -1 Assumptions: Pinch-off region of operation. V GS = 0.5 V I D = ma 12= 27,000I D +V DS +2000I S V DS = 4.75V 1/28/12 Microelectronic Circuit Design Chap 13-48
49 Common-Source Amplifier JFET Example (cont.) Next we construct the ac equivalent and simplify it. 1/28/12 Microelectronic Circuit Design Chap 13-49
50 BJT and FET Small-Signal Model Summary 01/18/12 Chap 13-50
51 Common-Emitter/Common-Source Amplifiers Summary 1/25/11 Chap 13-51
52 Amplifier Power Dissipation Static power dissipation in amplifiers is found from their dc equivalent circuits. (a) Total power dissipated in the C-B and E-B junctions is: P D = V CE I C + V BE I B Total power supplied is: P S = V CC I C + V EE I E where V CE = V CB + V BE (b) Total power dissipated in the transistor is: P D = V DS I D + V GS I G = V DS I D Total power supplied is: P S = V DD I D +V DD2 /(R 1 +R 2 ) 1/28/12 Chap 13-52
53 Amplifier Signal Range v CE = V CE V m sinωt where V m is the output signal. Active region operation requires v CE v BE So: V m V CE V BE Also: v Rc ( t) = I C R C V m sinωt 0 ( ) V m min I C R C, V CE V BE Similarly for MOSFETs and JFETs: V M min I D R D,(V DS (V GS V TN )) V M min I D R D,(V DS (V GS V P )) 1/28/12 Chap 13-53
54 End of Chapter 13 1/28/12 Chap 13-54
### Biasing the CE Amplifier
Biasing the CE Amplifier Graphical approach: plot I C as a function of the DC base-emitter voltage (note: normally plot vs. base current, so we must return to Ebers-Moll): I C I S e V BE V th I S e V th
### Chapter 4 Field-Effect Transistors
Chapter 4 Field-Effect Transistors Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock 5/5/11 Chap 4-1 Chapter Goals Describe operation of MOSFETs. Define FET characteristics in operation
### EE105 Fall 2014 Microelectronic Devices and Circuits
EE05 Fall 204 Microelectronic Devices and Circuits Prof. Ming C. Wu wu@eecs.berkeley.edu 5 Sutardja Dai Hall (SDH) Terminal Gain and I/O Resistances of BJT Amplifiers Emitter (CE) Collector (CC) Base (CB)
### Lecture 15: MOS Transistor models: Body effects, SPICE models. Context. In the last lecture, we discussed the modes of operation of a MOS FET:
Lecture 15: MOS Transistor models: Body effects, SPICE models Context In the last lecture, we discussed the modes of operation of a MOS FET: oltage controlled resistor model I- curve (Square-Law Model)
### CHAPTER 3: TRANSISTOR MOSFET DR. PHAM NGUYEN THANH LOAN. Hà Nội, 9/24/2012
1 CHAPTER 3: TRANSISTOR MOSFET DR. PHAM NGUYEN THANH LOAN Hà Nội, 9/24/2012 Chapter 3: MOSFET 2 Introduction Classifications JFET D-FET (Depletion MOS) MOSFET (Enhancement E-FET) DC biasing Small signal
### Chapter 5. BJT AC Analysis
Chapter 5. Outline: The r e transistor model CB, CE & CC AC analysis through r e model common-emitter fixed-bias voltage-divider bias emitter-bias & emitter-follower common-base configuration Transistor
### CHAPTER.4: Transistor at low frequencies
CHAPTER.4: Transistor at low frequencies Introduction Amplification in the AC domain BJT transistor modeling The re Transistor Model The Hybrid equivalent Model Introduction There are three models commonly
### ESE319 Introduction to Microelectronics. Output Stages
Output Stages Power amplifier classification Class A amplifier circuits Class A Power conversion efficiency Class B amplifier circuits Class B Power conversion efficiency Class AB amplifier circuits Class
### DEPARTMENT OF ECE UNIT VII BIASING & STABILIZATION AMPLIFIER:
UNIT VII IASING & STAILIZATION AMPLIFIE: - A circuit that increases the amplitude of given signal is an amplifier - Small ac signal applied to an amplifier is obtained as large a.c. signal of same frequency
### ECE 342 Electronic Circuits. 3. MOS Transistors
ECE 342 Electronic Circuits 3. MOS Transistors Jose E. Schutt-Aine Electrical & Computer Engineering University of Illinois jschutt@emlab.uiuc.edu 1 NMOS Transistor Typically L = 0.1 to 3 m, W = 0.2 to
### Mod. Sim. Dyn. Sys. Amplifiers page 1
AMPLIFIERS A circuit containing only capacitors, amplifiers (transistors) and resistors may resonate. A circuit containing only capacitors and resistors may not. Why does amplification permit resonance
### Microelectronic Circuit Design 4th Edition Errata - Updated 4/4/14
Chapter Text # Inside back cover: Triode region equation should not be squared! i D = K n v GS "V TN " v & DS % ( v DS \$ 2 ' Page 49, first exercise, second answer: -1.35 x 10 6 cm/s Page 58, last exercise,
### Forward-Active Terminal Currents
Forward-Active Terminal Currents Collector current: (electron diffusion current density) x (emitter area) diff J n AE qd n n po A E V E V th ------------------------------ e W (why minus sign? is by def.
### DC Biasing. Dr. U. Sezen & Dr. D. Gökçen (Hacettepe Uni.) ELE230 Electronics I 15-Mar / 59
Contents Three States of Operation BJT DC Analysis Fixed-Bias Circuit Emitter-Stabilized Bias Circuit Voltage Divider Bias Circuit DC Bias with Voltage Feedback Various Dierent Bias Circuits pnp Transistors
### At point G V = = = = = = RB B B. IN RB f
Common Emitter At point G CE RC 0. 4 12 0. 4 116. I C RC 116. R 1k C 116. ma I IC 116. ma β 100 F 116µ A I R ( 116µ A)( 20kΩ) 2. 3 R + 2. 3 + 0. 7 30. IN R f Gain in Constant Current Region I I I C F
### figure shows a pnp transistor biased to operate in the active mode
Lecture 10b EE-215 Electronic Devices and Circuits Asst Prof Muhammad Anis Chaudhary BJT: Device Structure and Physical Operation The pnp Transistor figure shows a pnp transistor biased to operate in the
### 6.012 MICROELECTRONIC DEVICES AND CIRCUITS
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.012 MICROELECTRONIC DEVICES AND CIRCUITS Answers to Exam 2 Spring 2008 Problem 1: Graded by Prof. Fonstad
### Chapter7. FET Biasing
Chapter7. J configurations Fixed biasing Self biasing & Common Gate Voltage divider MOS configurations Depletion-type Enhancement-type JFET: Fixed Biasing Example 7.1: As shown in the figure, it is the
### ECE-305: Fall 2017 MOS Capacitors and Transistors
ECE-305: Fall 2017 MOS Capacitors and Transistors Pierret, Semiconductor Device Fundamentals (SDF) Chapters 15+16 (pp. 525-530, 563-599) Professor Peter Bermel Electrical and Computer Engineering Purdue
### ECE-343 Test 1: Feb 10, :00-8:00pm, Closed Book. Name : SOLUTION
ECE-343 Test : Feb 0, 00 6:00-8:00pm, Closed Book Name : SOLUTION C Depl = C J0 + V R /V o ) m C Diff = τ F g m ω T = g m C µ + C π ω T = g m I / D C GD + C or V OV GS b = τ i τ i = R i C i ω H b Z = Z
### Lecture 23: Negative Resistance Osc, Differential Osc, and VCOs
EECS 142 Lecture 23: Negative Resistance Osc, Differential Osc, and VCOs Prof. Ali M. Niknejad University of California, Berkeley Copyright c 2005 by Ali M. Niknejad A. M. Niknejad University of California,
### EE105 - Fall 2006 Microelectronic Devices and Circuits
EE105 - Fall 2006 Microelectronic Devices and Circuits Prof. Jan M. Rabaey (jan@eecs) Lecture 7: MOS Transistor Some Administrative Issues Lab 2 this week Hw 2 due on We Hw 3 will be posted same day MIDTERM
### EE 321 Analog Electronics, Fall 2013 Homework #8 solution
EE 321 Analog Electronics, Fall 2013 Homework #8 solution 5.110. The following table summarizes some of the basic attributes of a number of BJTs of different types, operating as amplifiers under various
### Fig. 1 CMOS Transistor Circuits (a) Inverter Out = NOT In, (b) NOR-gate C = NOT (A or B)
1 Introduction to Transistor-Level Logic Circuits 1 By Prawat Nagvajara At the transistor level of logic circuits, transistors operate as switches with the logic variables controlling the open or closed
### Chapter 2. - DC Biasing - BJTs
Chapter 2. - DC Biasing - BJTs Objectives To Understand : Concept of Operating point and stability Analyzing Various biasing circuits and their comparison with respect to stability BJT A Review Invented
### ECE-342 Test 2 Solutions, Nov 4, :00-8:00pm, Closed Book (one page of notes allowed)
ECE-342 Test 2 Solutions, Nov 4, 2008 6:00-8:00pm, Closed Book (one page of notes allowed) Please use the following physical constants in your calculations: Boltzmann s Constant: Electron Charge: Free
### 1/13/12 V DS. I d V GS. C ox ( = f (V GS ,V DS ,V SB = I D. + i d + I ΔV + I ΔV BS V BS. 19 January 2012
/3/ 9 January 0 Study the linear model of MOS transistor around an operating point." MOS in saturation: V GS >V th and V S >V GS -V th " VGS vi - I d = I i d VS I d = µ n ( L V V γ Φ V Φ GS th0 F SB F
### The Common-Emitter Amplifier
c Copyright 2009. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering. The Common-Emitter Amplifier Basic Circuit Fig. shows the circuit diagram
### Lecture 29 - The Long Metal-Oxide-Semiconductor Field-Effect Transistor (cont.) April 20, 2007
6.720J/3.43J - Integrated Microelectronic Devices - Spring 2007 Lecture 29-1 Lecture 29 - The Long Metal-Oxide-Semiconductor Field-Effect Transistor (cont.) April 20, 2007 Contents: 1. Non-ideal and second-order
### Chapter 10 Instructor Notes
G. izzoni, Principles and Applications of lectrical ngineering Problem solutions, hapter 10 hapter 10 nstructor Notes hapter 10 introduces bipolar junction transistors. The material on transistors has
### Spring Semester 2012 Final Exam
Spring Semester 2012 Final Exam Note: Show your work, underline results, and always show units. Official exam time: 2.0 hours; an extension of at least 1.0 hour will be granted to anyone. Materials parameters
### Integrated Circuits & Systems
Federal University of Santa Catarina Center for Technology Computer Science & Electronics Engineering Integrated Circuits & Systems INE 5442 Lecture 10 MOSFET part 1 guntzel@inf.ufsc.br ual-well Trench-Isolated
### CMPEN 411 VLSI Digital Circuits. Lecture 03: MOS Transistor
CMPEN 411 VLSI Digital Circuits Lecture 03: MOS Transistor Kyusun Choi [Adapted from Rabaey s Digital Integrated Circuits, Second Edition, 2003 J. Rabaey, A. Chandrakasan, B. Nikolic] CMPEN 411 L03 S.1
### Lecture 4: CMOS Transistor Theory
Introduction to CMOS VLSI Design Lecture 4: CMOS Transistor Theory David Harris, Harvey Mudd College Kartik Mohanram and Steven Levitan University of Pittsburgh Outline q Introduction q MOS Capacitor q
### VI. Transistor amplifiers: Biasing and Small Signal Model
VI. Transistor amplifiers: iasing and Small Signal Model 6.1 Introduction Transistor amplifiers utilizing JT or FET are similar in design and analysis. Accordingly we will discuss JT amplifiers thoroughly.
### 4.4 The MOSFET as an Amp and Switch
10/31/004 section 4_4 The MSFET as an Amp and Switch blank 1/1 44 The MSFET as an Amp and Switch Reading Assignment: pp 70-80 Now we know how an enhancement MSFET works! Q: A: 1 H: The MSFET as an Amp
### Common Drain Stage (Source Follower) Claudio Talarico, Gonzaga University
Common Drain Stage (Source Follower) Claudio Talarico, Gonzaga University Common Drain Stage v gs v i - v o V DD v bs - v o R S Vv IN i v i G C gd C+C gd gb B&D v s vv OUT o + V S I B R L C L v gs - C
### MOS Transistor Theory
CHAPTER 3 MOS Transistor Theory Outline 2 1. Introduction 2. Ideal I-V Characteristics 3. Nonideal I-V Effects 4. C-V Characteristics 5. DC Transfer Characteristics 6. Switch-level RC Delay Models MOS
### à FIELD EFFECT TRANSISTORS
Prof.M.G.Guvench à FIELD EFFECT TRANSISTORS ü FET: CONTENTS Principles of Operation Models: DC, S.S.A.C. and SPICE Applications: AC coupled S.S. Amplifiers ü FET: NAMES JFET Junction Field Effect Transistor
### Transistor amplifiers: Biasing and Small Signal Model
Transistor amplifiers: iasing and Small Signal Model Transistor amplifiers utilizing JT or FT are similar in design and analysis. Accordingly we will discuss JT amplifiers thoroughly. Then, similar FT
### 7. DESIGN OF AC-COUPLED BJT AMPLIFIERS FOR MAXIMUM UNDISTORTED VOLTAGE SWING
à 7. DESIGN OF AC-COUPLED BJT AMPLIFIERS FOR MAXIMUM UNDISTORTED VOLTAGE SWING Figure. AC coupled common emitter amplifier circuit ü The DC Load Line V CC = I CQ + V CEQ + R E I EQ I EQ = I CQ + I BQ I
### EE 330 Lecture 16. MOSFET Modeling CMOS Process Flow
EE 330 Lecture 16 MOSFET Modeling CMOS Process Flow Model Extensions 300 Id 250 200 150 100 50 300 0 0 1 2 3 4 5 Vds Existing Model 250 200 Id 150 100 50 Slope is not 0 0 0 1 2 3 4 Actual Device Vds Model
### EE 560 MOS TRANSISTOR THEORY PART 2. Kenneth R. Laker, University of Pennsylvania
1 EE 560 MOS TRANSISTOR THEORY PART nmos TRANSISTOR IN LINEAR REGION V S = 0 V G > V T0 channel SiO V D = small 4 C GC C BC substrate depletion region or bulk B p nmos TRANSISTOR AT EDGE OF SATURATION
### MOSFET: Introduction
E&CE 437 Integrated VLSI Systems MOS Transistor 1 of 30 MOSFET: Introduction Metal oxide semiconductor field effect transistor (MOSFET) or MOS is widely used for implementing digital designs Its major
### ELEC 3908, Physical Electronics, Lecture 26. MOSFET Small Signal Modelling
ELEC 3908, Physical Electronics, Lecture 26 MOSFET Small Signal Modelling Lecture Outline MOSFET small signal behavior will be considered in the same way as for the diode and BJT Capacitances will be considered
### UNI-JUNCTION TRANSISTOR
UNI-JUNCTION TRANSISTOR The UJT as the name implies, is characterized by a single pn junction. It exhibits negative resistance characteristic that makes it useful in oscillator circuits. The symbol for
### EE5311- Digital IC Design
EE5311- Digital IC Design Module 1 - The Transistor Janakiraman V Assistant Professor Department of Electrical Engineering Indian Institute of Technology Madras Chennai October 28, 2017 Janakiraman, IITM
### EECS 141: FALL 05 MIDTERM 1
University of California College of Engineering Department of Electrical Engineering and Computer Sciences D. Markovic TuTh 11-1:3 Thursday, October 6, 6:3-8:pm EECS 141: FALL 5 MIDTERM 1 NAME Last SOLUTION
### Section 1: Common Emitter CE Amplifier Design
ECE 3274 BJT amplifier design CE, CE with Ref, and CC. Richard Cooper Section 1: CE amp Re completely bypassed (open Loop) Section 2: CE amp Re partially bypassed (gain controlled). Section 3: CC amp (open
### Semiconductor Physics fall 2012 problems
Semiconductor Physics fall 2012 problems 1. An n-type sample of silicon has a uniform density N D = 10 16 atoms cm -3 of arsenic, and a p-type silicon sample has N A = 10 15 atoms cm -3 of boron. For each
### Advanced Current Mirrors and Opamps
Advanced Current Mirrors and Opamps David Johns and Ken Martin (johns@eecg.toronto.edu) (martin@eecg.toronto.edu) slide 1 of 26 Wide-Swing Current Mirrors I bias I V I in out out = I in V W L bias ------------
### E2.2 Analogue Electronics
E2.2 Analogue Electronics Instructor : Christos Papavassiliou Office, email : EE 915, c.papavas@imperial.ac.uk Lectures : Monday 2pm, room 408 (weeks 2-11) Thursday 3pm, room 509 (weeks 4-11) Problem,
### Lecture 12 Digital Circuits (II) MOS INVERTER CIRCUITS
Lecture 12 Digital Circuits (II) MOS INVERTER CIRCUITS Outline NMOS inverter with resistor pull-up The inverter NMOS inverter with current-source pull-up Complementary MOS (CMOS) inverter Static analysis
### CHAPTER 5 MOS FIELD-EFFECT TRANSISTORS
CHAPTER 5 MOS FIELD-EFFECT TRANSISTORS 5.1 The MOS capacitor 5.2 The enhancement-type N-MOS transistor 5.3 I-V characteristics of enhancement mode MOSFETS 5.4 The PMOS transistor and CMOS technology 5.5
### Semiconductor Physics Problems 2015
Semiconductor Physics Problems 2015 Page and figure numbers refer to Semiconductor Devices Physics and Technology, 3rd edition, by SM Sze and M-K Lee 1. The purest semiconductor crystals it is possible
### FIELD-EFFECT TRANSISTORS
FIEL-EFFECT TRANSISTORS 1 Semiconductor review 2 The MOS capacitor 2 The enhancement-type N-MOS transistor 3 I-V characteristics of enhancement MOSFETS 4 The output characteristic of the MOSFET in saturation
### Basics of Network Theory (Part-I)
Basics of Network Theory (PartI). A square waveform as shown in figure is applied across mh ideal inductor. The current through the inductor is a. wave of peak amplitude. V 0 0.5 t (m sec) [Gate 987: Marks]
### MOSFET and CMOS Gate. Copy Right by Wentai Liu
MOSFET and CMOS Gate CMOS Inverter DC Analysis - Voltage Transfer Curve (VTC) Find (1) (2) (3) (4) (5) (6) V OH min, V V OL min, V V IH min, V V IL min, V OHmax OLmax IHmax ILmax NM L = V ILmax V OL max
### Institute of Solid State Physics. Technische Universität Graz. Exam. Feb 2, 10:00-11:00 P2
Technische Universität Graz nstitute of Solid State Physics Exam Feb 2, 10:00-11:00 P2 Exam Four questions, two from the online list. Calculator is ok. No notes. Explain some concept: (tunnel contact,
### Microelectronic Devices and Circuits Lecture 9 - MOS Capacitors I - Outline Announcements Problem set 5 -
6.012 - Microelectronic Devices and Circuits Lecture 9 - MOS Capacitors I - Outline Announcements Problem set 5 - Posted on Stellar. Due net Wednesday. Qualitative description - MOS in thermal equilibrium
### CMPEN 411 VLSI Digital Circuits. Lecture 04: CMOS Inverter (static view)
CMPEN 411 VLSI Digital Circuits Lecture 04: CMOS Inverter (static view) Kyusun Choi [Adapted from Rabaey s Digital Integrated Circuits, Second Edition, 2003 J. Rabaey, A. Chandrakasan, B. Nikolic] CMPEN
### IXBT24N170 IXBH24N170
High Voltage, High Gain BIMOSFET TM Monolithic Bipolar MOS Transistor IXBT24N17 IXBH24N17 S 11 = 1 = 24A (sat) 2. TO-26 (IXBT) Symbol Test Conditions Maximum Ratings S = 25 C to 15 C 17 V V CGR = 25 C
### Lecture 9 MOSFET(II) MOSFET I V CHARACTERISTICS(contd.)
Lecture 9 MOSFET(II) MOSFET I V CHARACTERISTICS(contd.) Outline 1. The saturation region 2. Backgate characteristics Reading Assignment: Howe and Sodini, Chapter 4, Section 4.4 6.012 Spring 2009 Lecture
### Lecture 6 Power Zhuo Feng. Z. Feng MTU EE4800 CMOS Digital IC Design & Analysis 2010
EE4800 CMOS Digital IC Design & Analysis Lecture 6 Power Zhuo Feng 6.1 Outline Power and Energy Dynamic Power Static Power 6.2 Power and Energy Power is drawn from a voltage source attached to the V DD
### CMOS INVERTER. Last Lecture. Metrics for qualifying digital circuits. »Cost» Reliability» Speed (delay)»performance
CMOS INVERTER Last Lecture Metrics for qualifying digital circuits»cost» Reliability» Speed (delay)»performance 1 Today s lecture The CMOS inverter at a glance An MOS transistor model for manual analysis
### Errata of CMOS Analog Circuit Design 2 nd Edition By Phillip E. Allen and Douglas R. Holberg
Errata 2 nd Ed. (5/22/2) Page Errata of CMOS Analog Circuit Design 2 nd Edition By Phillip E. Allen and Douglas R. Holberg Page Errata 82 Line 4 after figure 3.2-3, CISW CJSW 88 Line between Eqs. (3.3-2)
### DC and Transient Responses (i.e. delay) (some comments on power too!)
DC and Transient Responses (i.e. delay) (some comments on power too!) Michael Niemier (Some slides based on lecture notes by David Harris) 1 Lecture 02 - CMOS Transistor Theory & the Effects of Scaling
### EKV MOS Transistor Modelling & RF Application
HP-RF MOS Modelling Workshop, Munich, February 15-16, 1999 EKV MOS Transistor Modelling & RF Application Matthias Bucher, Wladek Grabinski Electronics Laboratory (LEG) Swiss Federal Institute of Technology,
### PURPOSE: See suggested breadboard configuration on following page!
ECE4902 Lab 1 C2011 PURPOSE: Determining Capacitance with Risetime Measurement Reverse Biased Diode Junction Capacitance MOSFET Gate Capacitance Simulation: SPICE Parameter Extraction, Transient Analysis
### ACADAMIC CHAPTER OF SWECHA September- 2010
Swecha Documents SF-SAC/ ECE / II-II/LM/2010 /ver. 1.0 LABMANAUALS DEPARTMENT : ECE ELECTRONIC CIRCUITS ANALYSIS LABORATORY MANUAL ACADAMIC CHAPTER OF SWECHA September- 2010 INDEX S.NO NAME OF THE EXPERIMENT
### High-to-Low Propagation Delay t PHL
High-to-Low Propagation Delay t PHL V IN switches instantly from low to high. Driver transistor (n-channel) immediately switches from cutoff to saturation; the p-channel pull-up switches from triode to
### ) t 0(q+ t ) dt n t( t) dt ( rre i dq t 0 u = = t l C t) t) a i( ( q tric c le E
EE70 eview Electrical Current i ( t ) dq ( t ) dt t q ( t ) i ( t ) dt + t 0 q ( t 0 ) Circuit Elements An electrical circuit consists o circuit elements such as voltage sources, resistances, inductances
### Monolithic N-Channel JFET Dual
N9 Monolithic N-Channel JFET Dual V GS(off) (V) V (BR)GSS Min (V) g fs Min (ms) I G Max (pa) V GS V GS Max (mv). to. Monolithic Design High Slew Rate Low Offset/Drift Voltage Low Gate Leakage: pa Low Noise:
### CHAPTER 14 SIGNAL GENERATORS AND WAVEFORM SHAPING CIRCUITS
CHAPTER 4 SIGNA GENERATORS AND WAEFORM SHAPING CIRCUITS Chapter Outline 4. Basic Principles of Sinusoidal Oscillators 4. Op Amp RC Oscillators 4.3 C and Crystal Oscillators 4.4 Bistable Multivibrators
### Review - Differential Amplifier Basics Difference- and common-mode signals: v ID
6.012 Microelectronic Devices and Circuits Lecture 20 DiffAmp Anal. I: Metrics, Max. Gain Outline Announcements Announcements D.P.: No Early effect in large signal analysis; just LECs. Lec. 21 foils useful;
### Lecture 13 - Digital Circuits (II) MOS Inverter Circuits. March 20, 2003
6.012 Microelectronic Devices and Circuits Spring 2003 Lecture 131 Lecture 13 Digital Circuits (II) MOS Inverter Circuits March 20, 2003 Contents: 1. NMOS inverter with resistor pullup (cont.) 2. NMOS
### Active Circuits: Life gets interesting
Actie Circuits: Life gets interesting Actie cct elements operational amplifiers (P AMPS) and transistors Deices which can inject power into the cct External power supply normally comes from connection
### Transistors - a primer
ransistors - a primer What is a transistor? Solid-state triode - three-terminal device, with voltage (or current) at third terminal used to control current between other two terminals. wo types: bipolar
### 4.10 The CMOS Digital Logic Inverter
11/11/2004 section 4_10 The CMOS Digital Inverter blank.doc 1/1 4.10 The CMOS Digital Logic Inverter Reading Assignment: pp. 336346 Complementary MOSFET (CMOS) is the predominant technology for constructing
### AON V Common-Drain Dual N-Channel MOSFET
2V CommonDrain Dual NChannel MOSFET General Description Low R DS(ON) With ESD Protection to improve battery performance and safety Common drain configuration for design simplicity RoHS and HalogenFree
### Lecture 310 Open-Loop Comparators (3/28/10) Page 310-1
Lecture 310 Open-Loop Comparators (3/28/10) Page 310-1 LECTURE 310 OPEN-LOOP COMPARATORS LECTURE ORGANIZATION Outline Characterization of comparators Dominant pole, open-loop comparators Two-pole, open-loop
### Lecture #23. Warning for HW Assignments and Exams: Make sure your writing is legible!! OUTLINE. Circuit models for the MOSFET
Lecture #23 arning for H Assignments and Exams: Make sure your writing is legible!! OUTLINE MOFET I s. V characteristic Circuit models for the MOFET resistie switch model small-signal model Reference Reading
### ECE 438: Digital Integrated Circuits Assignment #4 Solution The Inverter
ECE 438: Digital Integrated Circuits Assignment #4 The Inverter Text: Chapter 5, Digital Integrated Circuits 2 nd Ed, Rabaey 1) Consider the CMOS inverter circuit in Figure P1 with the following parameters.
### Switched-Capacitor Circuits David Johns and Ken Martin University of Toronto
Switched-Capacitor Circuits David Johns and Ken Martin University of Toronto (johns@eecg.toronto.edu) (martin@eecg.toronto.edu) University of Toronto 1 of 60 Basic Building Blocks Opamps Ideal opamps usually
### IXTK5N250 IXTX5N250 = 2500V = 5A < 8.8Ω. High Voltage Power MOSFET w/ Extended FBSOA. Advance Technical Information. R DS(on)
High Voltage Power MOSFET w/ Extended FBSOA N-Channel Enhancement Mode Avalanche Rated Guaranteed FBSOA Advance Technical Information IXTKN IXTXN S = V I D = A R DS(on)
### CCS050M12CM2 1.2kV, 50A Silicon Carbide Six-Pack (Three Phase) Module Z-FET TM MOSFET and Z-Rec TM Diode
CCS5M2CM2.2kV, 5A Silicon Carbide Six-Pack (Three Phase) Module Z-FET TM MOSFET and Z-Rec TM Diode Features Ultra Low Loss Zero Reverse Recovery Current Zero Turn-off Tail Current High-Frequency Operation
### Exact Analysis of a Common-Source MOSFET Amplifier
Exact Analysis of a Common-Source MOSFET Amplifier Consider the common-source MOSFET amplifier driven from signal source v s with Thévenin equivalent resistance R S and a load consisting of a parallel
### Chapter 13 Bipolar Junction Transistors
Chapter 3 ipolar Junction Transistors Goal. ipolar Junction Transistor Operation in amplifier circuits. 2. Load-line Analysis & Nonlinear Distortion. 3. Large-signal equialent circuits to analyze JT circuits.
### ECE 145A/218A Power Amplifier Design Lectures. Power Amplifier Design 1
Power Amplifiers; Part 1 Class A Device Limitations Large signal output match Define efficiency, power-added efficiency Class A operating conditions Thermal resistance We have studied the design of small-signal
### EE 435. Lecture 2: Basic Op Amp Design. - Single Stage Low Gain Op Amps
EE 435 ecture 2: Basic Op Amp Design - Single Stage ow Gain Op Amps 1 Review from last lecture: How does an amplifier differ from an operational amplifier?? Op Amp Amplifier Amplifier used in open-loop
### Lecture 5 Review Current Source Active Load Modified Large / Small Signal Models Channel Length Modulation
Lecture 5 Review Current Source Active Load Modified Large / Small Signal Models Channel Length Modulation Text sec 1.2 pp. 28-32; sec 3.2 pp. 128-129 Current source Ideal goal Small signal model: Open
### Lecture 140 Simple Op Amps (2/11/02) Page 140-1
Lecture 40 Simple Op Amps (2//02) Page 40 LECTURE 40 SIMPLE OP AMPS (READING: TextGHLM 425434, 453454, AH 249253) INTRODUCTION The objective of this presentation is:.) Illustrate the analysis of BJT and
### Digital Integrated Circuits
Chapter 6 The CMOS Inverter 1 Contents Introduction (MOST models) 0, 1 st, 2 nd order The CMOS inverter : The static behavior: o DC transfer characteristics, o Short-circuit current The CMOS inverter :
### AO V Dual N-Channel MOSFET
AO688 V Dual NChannel MOSFET General Description The AO688 uses advanced trench technology to provide excellent R DS(ON), low gate charge and operation with gate voltages as low as.5v. This device is suitable
### Chapter 10 Feedback. PART C: Stability and Compensation
1 Chapter 10 Feedback PART C: Stability and Compensation Example: Non-inverting Amplifier We are analyzing the two circuits (nmos diff pair or pmos diff pair) to realize this symbol: either of the circuits
### The Miller Approximation
The Miller Approximation The exact analysis is not particularly helpful for gaining insight into the frequency response... consider the effect of C µ on the input only I t C µ V t g m V t R'out = r o r
### PRACTICE PROBLEMS FOR CMOS ANALOG CIRCUIT DESIGN, 2 ND EDITION
Practice Problems (5/27/07) Page PRACTICE PROBLEMS FOR CMOS ANALOG CIRCUIT DESIGN, 2 ND EDITION TECHNOLOGY Problem (044430E3P5) The following questions pertain to a standard npn BJT process. a.) Give the
### V DS I D (at V GS =10V) R DS(ON) (at V GS = 4.5V) 100% UIS Tested 100% R g Tested. Top View
4V Dual NChannel MOSFET General Description The AON64 uses advanced trench technology to provide excellent R DS(ON) with low gate charge. This is an all purpose device that is suitable for use in a wide | 13,813 | 47,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2019-30 | longest | en | 0.820643 |
https://www.airmilescalculator.com/distance/bfi-to-mob/ | 1,603,492,641,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107865665.7/warc/CC-MAIN-20201023204939-20201023234939-00334.warc.gz | 619,132,999 | 79,960 | # Distance between Seattle, WA (BFI) and Mobile, AL (MOB)
Flight distance from Seattle to Mobile (Seattle Boeing Field – Mobile Regional Airport) is 2142 miles / 3447 kilometers / 1861 nautical miles. Estimated flight time is 4 hours 33 minutes.
Driving distance from Seattle (BFI) to Mobile (MOB) is 2690 miles / 4329 kilometers and travel time by car is about 45 hours 23 minutes.
## Map of flight path and driving directions from Seattle to Mobile.
Shortest flight path between Seattle Boeing Field (BFI) and Mobile Regional Airport (MOB).
## How far is Mobile from Seattle?
There are several ways to calculate distances between Seattle and Mobile. Here are two common methods:
Vincenty's formula (applied above)
• 2141.845 miles
• 3446.966 kilometers
• 1861.213 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
• 2139.167 miles
• 3442.656 kilometers
• 1858.886 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 Seattle Boeing Field
City: Seattle, WA
Country: United States
IATA Code: BFI
ICAO Code: KBFI
Coordinates: 47°31′47″N, 122°18′7″W
B Mobile Regional Airport
City: Mobile, AL
Country: United States
IATA Code: MOB
ICAO Code: KMOB
Coordinates: 30°41′28″N, 88°14′34″W
## Time difference and current local times
The time difference between Seattle and Mobile is 2 hours. Mobile is 2 hours ahead of Seattle.
PDT
CDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 234 kg (515 pounds).
## Frequent Flyer Miles Calculator
Seattle (BFI) → Mobile (MOB).
Distance:
2142
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
2142
Round trip? | 493 | 1,910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-45 | longest | en | 0.834766 |
https://techcommunity.microsoft.com/t5/excel/yearfrac-stopping-dates-in-future/td-p/3652881 | 1,713,229,369,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817036.4/warc/CC-MAIN-20240416000407-20240416030407-00057.warc.gz | 515,579,171 | 57,187 | SOLVED
# Yearfrac stopping dates in future
Copper Contributor
# Yearfrac stopping dates in future
Want to use yearfrac instead of datedif and it works but we want to have it give a value of 0 if one of the dates is in the future. I want to figure out how long someone has worked for us based on todays date and their start date.
4 Replies
best response confirmed by mathetes (Silver Contributor)
Solution
# Re: Yearfrac stopping dates in future
Let's say you have dates in D2 and E2.
=IF(OR(D2:E2>TODAY()),0,YEARFRAC(D2,E2))
This can be filled down.
If you want to compare D2 to today's date:
=IF(D2>TODAY(),0,YEARFRAC(D2,TODAY()))
# Re: Yearfrac stopping dates in future
Awesome. I use today in a seperate tab so will give this a whirl tomorrow.
# Re: Yearfrac stopping dates in future
@Hans Vogelaar seems my colleague is happy with your solution. Thank you
# Re: Yearfrac stopping dates in future
Good to hear that. Thanks for the feedback.
1 best response
Accepted Solutions
best response confirmed by mathetes (Silver Contributor)
Solution
# Re: Yearfrac stopping dates in future
Let's say you have dates in D2 and E2.
=IF(OR(D2:E2>TODAY()),0,YEARFRAC(D2,E2))
This can be filled down.
If you want to compare D2 to today's date:
=IF(D2>TODAY(),0,YEARFRAC(D2,TODAY())) | 358 | 1,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-18 | latest | en | 0.885223 |
https://www.mathworks.com/matlabcentral/answers/161027-multiple-ode-s-that-are-coupled?s_tid=prof_contriblnk | 1,632,808,563,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060201.9/warc/CC-MAIN-20210928032425-20210928062425-00668.warc.gz | 883,178,764 | 24,266 | # Multiple ODE's that are coupled
6 views (last 30 days)
Rick on 2 Nov 2014
Edited: Rick on 2 Nov 2014
Hello,
I attached a PDF of the coupled ODE's that I am trying to solve. I can make a function that runs, but I don't get the same output as the PDF for my graphs, what is going wrong?? The reaction rate plot is clearly wrong
function dz = semibatch(t,z)
k = 2.2;
V0 = 5;
v0 = 0.05;
V = @(t) V0 + v0*t;
CB0 = 0.025;
CA0 = 0.05;
dz = zeros(4,1);
dz(1) = -k*z(1).*z(2) - (v0.*z(1)./V(t));
dz(2) = -k*z(1).*z(2) + ((CB0-z(2)).*v0./V(t));
dz(3) = k*z(1).*z(2) - (v0*z(3)./V(t));
dz(4) = k*z(1).*z(2) - (v0*z(4)./V(t));
end
This is what is in my script file,
CA0 = 0.05;
CB0 = 0.025;
k = 2.2;
v0 = 0.05;
V0 = 5;
V = @(t) V0 + v0*t;
[T,Z] = ode45(@semibatch,[0 500],[CA0, CB0, 0, 0]);
plot(T,Z(:,1),T,Z(:,2),T,Z(:,3))
xlabel('time')
ylabel('C_{i}')
legend('C_{A}','C_{B}','C_{C}')
X = [];
rA = [];
for i = 1:size(Z,1)
X(i) = (CA0*V0 - Z(i,1)*V(T(i)))/(CA0*V0);
rA(i) = -k*Z(i,1)*Z(i,2);
end
figure
subplot(1,2,1)
plot(T,X)
xlabel('time')
ylabel('conversion')
subplot(1,2,2)
plot(T,rA)
xlabel('time')
ylabel('reaction rate') | 501 | 1,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-39 | latest | en | 0.643994 |
https://discuss.leetcode.com/topic/64620/java-6ms-solution-easy-understanding | 1,516,270,651,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887224.19/warc/CC-MAIN-20180118091548-20180118111548-00595.warc.gz | 661,295,676 | 8,749 | # Java 6ms solution, easy understanding
1. get the smallest m with the same number of digits as n, ps: n=356, m=100; n=35000, m=10000
2. from m to n, count x (in [m,n]) and when x has a '0' ending, insert y=x/10 before x
3. if y still ends with 0, continue inserting action
4. repeat [n/10+1,m-1] with step 2 and 3
``````public class Solution {
public int findKthNumber(int n, int k) {
int m = 1, tmp=n/10;
while (tmp>0) {
tmp /= 10;
m *= 10;
}
int firstPartNumber = count(m,n,m);
if (k<=firstPartNumber) return findKthNumber(m,n,m,k);
if (k<=n) return findKthNumber(n/10+1,m-1,m/10,k-firstPartNumber);
return 0;
}
public int count(int start, int end, int flag) {
// assume start and end has same amount of digits, flag represents the smallest number with the same length of digits, such as 10,100,1000,...
int result = 0;
while (flag>0) {
result += (end/flag-start/flag+((start%flag==0)?1:0));
flag /= 10;
}
return result;
}
public int findKthNumber(int start, int end, int flag, int k) {
int left = start, right = end;
// b-search
while (left<=right) {
int mid = (left+right)/2;
int x = count(start,mid,flag);
if (x==k) return mid;
if (x<k)
left=mid+1;
else
right=mid-1;
}
int t = right+1;
int zeroToBedeleted = count(start,t,flag)-k;
for (int i=0; i<zeroToBedeleted; i++)
t /= 10;
return t;
}
}
``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 472 | 1,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-05 | latest | en | 0.587696 |
http://hitsradio.us/congruent-quadrilaterals-worksheets/ | 1,591,016,401,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347417746.33/warc/CC-MAIN-20200601113849-20200601143849-00038.warc.gz | 58,203,206 | 30,109 | In Free Printable Worksheets216 views
4.07 / 5 ( 201votes )
Top Suggestions Congruent Quadrilaterals Worksheets :
Congruent Quadrilaterals Worksheets After learning about this geometry concept students will demonstrate what they know as they identify congruent 2d shapes in a multiple choice style format this math worksheet pairs well with the Learn about some shape basics congruent shapes it may sound complicated students will demonstrate what they know as they identify congruent 2d shapes in a multiple choice style format this math.
## Quadrilateral Worksheets Math Worksheets 4 Kids
Perimeter Of Quadrilaterals Worksheets Calculate The Perimeter Of Various Special Quadrilaterals Like Squares Rectangles Parallelograms Rhombuses Kites And Trapezoids With This Array Of Worksheets With Dimensions Presented As Integers And Decimals Comprehend The Congruent Property And Solve Algebraic Expressions To Find The Side Length
Bolster Practice With This Compilation Of Printable Perimeter Of Quadrilaterals Worksheets For Students Of Grade 3 Through Grade 8 Calculate The Perimeter Of Quadrilaterals With Dimensions Expressed As Integers And Decimals Comprehend The Congruent Properties Of Quadrilaterals And Solve Algebraic Expressions To Determine The Side Lengths Of The Quadrilaterals Presented As Geometrical Shapes
##### Angles In A Quadrilateral Worksheets Math Worksheets 4 Kids
Special Quadrilaterals Vertex Angles The Congruent Parts Are Marked And The Measures Of One Or Two Angles Of The Special Quadrilaterals Are Provided Apply Relevant Properties To Find The Measure Of The Indicated Vertex Angles Of Quadrilaterals Download The Set 3 Worksheets
###### Area Of Quadrilaterals Worksheets Tutoringhour
Area Of A Quadrilateral Worksheets Calculating The Area Of These Four Sided Polygons Is Really A Piece Of Cake If You Work Out The Sums In This Exclusive Set Of Worksheets On Area Of Quadrilaterals Squares Rectangles Rhombuses Etc Are Only A Few Examples Of Quadrilaterals That Have Four Sides And Four Vertices
Quadrilaterals Sheet 1 Math Worksheets 4 Kids
Congruent Shapes Worksheets Questions And Revision Mme
Congruence In Triangles There Is One Shape That Has Its Own Special Tests For Congruence And That Is The Triangle If We Have Two Triangles Which Pass One Of The 4 Tests Listed Below Then They Must Be Congruent Each Is A Unique Test If Two Triangles Have All 3 Things Listed On One Bullet Point In Common Then That Is Enough To Prove They Are Congruent
Congruent Shapes 6th Grade Worksheets Lesson Worksheets
Congruent Shapes 6th Grade Displaying All Worksheets Related To Congruent Shapes 6th Grade Worksheets Are Congruent Shapes Similar Congruent Shapes 6th Grade Practice Test Georgia Standards Of Excellence Curriculum Frameworks Grade 3 Geometry Work Identifying Quadrilaterals Working With Polygons Ms Work 132 153 Geometry 06 Polygons Quadrilaterals And Special Parallelograms
Describing Quadrilaterals Grade 3 Geometry Worksheet Answers Quadrilaterals A Quadrilateral Is A Shape That Has 4 Sides And 4 Angles Two Types Of Quadrilaterals Are Trapezoids And Parallelograms A Trapezoid Has One Pair Of Parallel Sides A Parallelogram Has Two Pairs Of Parallel Sides Types Of Parallelograms A Rectangle Is A Parallelogram With All Angles Equal And Opposite Sides Equal A
80 Printable Geometry Worksheets Helpingwithmath
Free Printable Geometry Worksheets The Worksheets Below Can Be Used As Part Of Extra Math Homework These Geometry Worksheets Are Free And Easily Printable They Cover Typical School Work From 4th Through 8th Grade They Include Questions On Polygons 3d Objects Angles And Calculations Of Area Volume Coordinate Geometry Etc List Of Geometry Worksheets The Printable Worksheets Are
Congruent Quadrilaterals Worksheets. The worksheet is an assortment of 4 intriguing pursuits that will enhance your kid's knowledge and abilities. The worksheets are offered in developmentally appropriate versions for kids of different ages. Adding and subtracting integers worksheets in many ranges including a number of choices for parentheses use.
You can begin with the uppercase cursives and after that move forward with the lowercase cursives. Handwriting for kids will also be rather simple to develop in such a fashion. If you're an adult and wish to increase your handwriting, it can be accomplished. As a result, in the event that you really wish to enhance handwriting of your kid, hurry to explore the advantages of an intelligent learning tool now!
Consider how you wish to compose your private faith statement. Sometimes letters have to be adjusted to fit in a particular space. When a letter does not have any verticals like a capital A or V, the very first diagonal stroke is regarded as the stem. The connected and slanted letters will be quite simple to form once the many shapes re learnt well. Even something as easy as guessing the beginning letter of long words can assist your child improve his phonics abilities. Congruent Quadrilaterals Worksheets.
There isn't anything like a superb story, and nothing like being the person who started a renowned urban legend. Deciding upon the ideal approach route Cursive writing is basically joined-up handwriting. Practice reading by yourself as often as possible.
Research urban legends to obtain a concept of what's out there prior to making a new one. You are still not sure the radicals have the proper idea. Naturally, you won't use the majority of your ideas. If you've got an idea for a tool please inform us. That means you can begin right where you are no matter how little you might feel you've got to give. You are also quite suspicious of any revolutionary shift. In earlier times you've stated that the move of independence may be too early.
Each lesson in handwriting should start on a fresh new page, so the little one becomes enough room to practice. Every handwriting lesson should begin with the alphabets. Handwriting learning is just one of the most important learning needs of a kid. Learning how to read isn't just challenging, but fun too.
The use of grids The use of grids is vital in earning your child learn to Improve handwriting. Also, bear in mind that maybe your very first try at brainstorming may not bring anything relevant, but don't stop trying. Once you are able to work, you might be surprised how much you get done. Take into consideration how you feel about yourself. Getting able to modify the tracking helps fit more letters in a little space or spread out letters if they're too tight. Perhaps you must enlist the aid of another man to encourage or help you keep focused.
Congruent Quadrilaterals Worksheets. Try to remember, you always have to care for your child with amazing care, compassion and affection to be able to help him learn. You may also ask your kid's teacher for extra worksheets. Your son or daughter is not going to just learn a different sort of font but in addition learn how to write elegantly because cursive writing is quite beautiful to check out. As a result, if a kid is already suffering from ADHD his handwriting will definitely be affected. Accordingly, to be able to accomplish this, if children are taught to form different shapes in a suitable fashion, it is going to enable them to compose the letters in a really smooth and easy method. Although it can be cute every time a youngster says he runned on the playground, students want to understand how to use past tense so as to speak and write correctly. Let say, you would like to boost your son's or daughter's handwriting, it is but obvious that you want to give your son or daughter plenty of practice, as they say, practice makes perfect.
Without phonics skills, it's almost impossible, especially for kids, to learn how to read new words. Techniques to Handle Attention Issues It is extremely essential that should you discover your kid is inattentive to his learning especially when it has to do with reading and writing issues you must begin working on various ways and to improve it. Use a student's name in every sentence so there's a single sentence for each kid. Because he or she learns at his own rate, there is some variability in the age when a child is ready to learn to read. Teaching your kid to form the alphabets is quite a complicated practice.
Author: Lilli Kauz
Have faith. But just because it's possible, doesn't mean it will be easy. Know that whatever life you want, the grades you want, the job you want, the reputation you want, friends you want, that it's possible.
Top | 1,761 | 8,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-24 | latest | en | 0.623804 |
https://www.numbersaplenty.com/518400 | 1,679,302,091,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943471.24/warc/CC-MAIN-20230320083513-20230320113513-00684.warc.gz | 977,369,265 | 4,071 | Search a number
518400 = 283452
BaseRepresentation
bin1111110100100000000
3222100010000
41332210000
5113042100
615040000
74256241
oct1764400
9870100
10518400
11324533
12210000
13151c5c
14d6cc8
15a3900
hex7e900
518400 has 135 divisors (see below), whose sum is σ = 1916761. Its totient is φ = 138240.
The previous prime is 518389. The next prime is 518411. The reversal of 518400 is 4815.
518400 = 253 + 263 + ... + 393.
The square root of 518400 is 720.
It is a perfect power (a square), and thus also a powerful number.
It is a Jordan-Polya number, since it can be written as (6!)2.
It is an interprime number because it is at equal distance from previous prime (518389) and next prime (518411).
It can be written as a sum of positive squares in only one way, i.e., 186624 + 331776 = 432^2 + 576^2 .
It is a tau number, because it is divible by the number of its divisors (135).
It is a Harshad number since it is a multiple of its sum of digits (18).
It is a Duffinian number.
It is a nialpdrome in base 12.
It is an unprimeable number.
518400 is an untouchable number, because it is not equal to the sum of proper divisors of any number.
It is a polite number, since it can be written in 14 ways as a sum of consecutive naturals, for example, 103678 + ... + 103682.
2518400 is an apocalyptic number.
518400 is a gapful number since it is divisible by the number (50) formed by its first and last digit.
518400 is the 720-th square number.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 518400
518400 is an abundant number, since it is smaller than the sum of its proper divisors (1398361).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
518400 is an equidigital number, since it uses as much as digits as its factorization.
518400 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 38 (or 10 counting only the distinct ones).
The product of its (nonzero) digits is 160, while the sum is 18.
The cubic root of 518400 is about 80.3319540099.
518400 divided by its product of nonzero digits (160) gives a triangular number (3240 = T80).
The spelling of 518400 in words is "five hundred eighteen thousand, four hundred". | 657 | 2,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-14 | latest | en | 0.915439 |
https://stats.stackexchange.com/questions/549110/why-is-a-2nd-order-derivative-optimization-better-for-no-hidden-layer-neural-net/549543 | 1,638,584,341,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362923.11/warc/CC-MAIN-20211204003045-20211204033045-00231.warc.gz | 607,003,527 | 34,574 | # Why is a 2nd order derivative optimization better for no hidden layer neural networks?
I was reading in this blog. That first order derivative SGD optimization methods are worse for neural networks without hidden layers and 2nd order is better, because that's what regression uses. Why is 2nd order derivative optimization methods better for NN without hidden layers?
• The observation "That first order derivative SGD optimization methods are worse for neural networks without hidden layers and 2nd order is better" may not be universally correct. Are there any authoritative publication showing this? Oct 20 at 22:02
• @MehmetSüzen not that I know of. The argument in the blog is that regressions are not trained using SGD and 2nd order derivative methods therefore since a simple regression is most similar to a non-hidden layered NN that it's probably better to use this 2nd order derivative method Oct 20 at 22:08
• Interesting, this could be an open research problem. Oct 20 at 22:37
• There is a good study on this topic by Tom Minka tminka.github.io/papers/logreg/minka-logreg.pdf (spoiler: conjugate gradient descent and quasi-Newton algorithms work well). I suspect simply second order methods (e.g. Iteratively Reweighted Least Squares) are only good for rather small models. Oct 24 at 14:30
• I suspect for large models with very large datasets, it may be better to start off with SGD to get the weights into the ballpark and then polish them with SCGD, because you don't need all of the data to find the ballpark. Just GD rather than SGD is likely to get you to a good solution fairly quickly, especially if you use something like RPROP. Oct 24 at 14:52
## 1 Answer
The optimization task in the blog post, a classification task with cross-entropy loss, is convex when there are no hidden layers, so you might expect both first and second order optimization methods to be able to converge arbitrarily well.
However, you would expect second order methods to converge in fewer iterations -- for example, a second order method can locate the min/max of a quadratic function with just one step. On the other hand, a first order method needs many iterations, but this is balanced by the fact that computing the first derivative is much cheaper than computing the hessian, especially when the dimensionality becomes big. The author didn't seem to take this into account, and ran both methods for the same number of iterations.
Another factor might be that L-BFGS implementations are often written with convergence as a goal, whereas no one really expects NN optimizers like Adam to "converge" -- when training a deep neural network, convergence to a local minimum isn't the goal, so the default parameters of Adam are probably not tuned accordingly -- for example, for gradient descent to converge, it's usually necessary to decay the step-sizes to 0, but this not the default behavior in most machine learning libraries.
• I suspect it is the "stochastic" in SGD that is preventing it from converging to the exact minima as the updates are noisy as they depend on the batch used in each iteration. I don't think it is a clean test of first-order versus second order optimisation. Oct 24 at 14:55
• ah yes, that is a good catch Oct 24 at 14:58 | 709 | 3,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-49 | latest | en | 0.931475 |
http://www.gmathacks.com/quantreview2ed/problem-solving-013-explanation.html | 1,501,043,666,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425766.58/warc/CC-MAIN-20170726042247-20170726062247-00552.warc.gz | 430,608,647 | 5,512 | Bookshelf
Total GMAT Math Jeff's complete Quant guide, on sale now!
Total GMAT Verbal Everything you need to ace GMAT Verbal!
New: GMAT 111 Improve every aspect of your GMAT prep!
Guides To the Official Guide
(Updated for new editions!)
1,800 Practice Quant Questions
GMAT Question of the Day
GMAT Official Guide
OG Math | OG Verbal
Beginner's Guide to the GMAT
GMAT Hacks Affiliate Program
Resources
MBA.com
GMAC Official Site
Free GMATPrep Practice Tests
Stacy Blackman Consulting
Book | Essay Guides
GRE HQ
Total GRE Math
Ultimate SAT Verbal
Official Guide Explanation:Problem Solving #13
Background
This is just one of hundreds of free explanations I've created to the quantitative questions in The Official Guide for GMAT Quantitative Review (2nd ed.). Click the links on the question number, difficulty level, and categories to find explanations for other problems.
These are the same explanations that are featured in my "Guides to the Official Guide" PDF booklets. However, because of the limitations of HTML and cross-browser compatibility, some mathematical concepts, such as fractions and roots, do not display as clearly online.
Question: 13
Page: 63
Difficulty: 4 (Moderately Easy)
Category 1: Algebra > Linear Equations-Two Unk >
Category 2: Word Problems > Other >
Explanation: Convert the two sentences into algebra:
g = 2h - 10
g + h = 425
Substitute the value for g given in the first equation into the second equation:
(2h - 10) + h = 425
3h - 10 = 425
3h = 435
h = (435/3) = 145
We're looking for g, so substitute the value of h back into the second equation (either one would work):
g + h = 425
g + 145 = 425
g = 425 - 145
g = 280, choice (E).
You should follow me on Twitter. While you're at it, take a moment to subscribe to GMAT Hacks via RSS or Email.
Total GMAT Math The comprehensive guide to the GMAT Quant section. It's "far and away the best study material available," including over 300 realistic practice questions and more than 500 exercises! Click to read more. | 511 | 2,032 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-30 | longest | en | 0.876605 |
https://www.weegy.com/?ConversationId=777A7DB4 | 1,702,126,304,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100909.82/warc/CC-MAIN-20231209103523-20231209133523-00031.warc.gz | 1,174,495,626 | 11,310 | What is roman numerals
Question
Asked 10/6/2008 10:28:11 AM
Updated 10/19/2022 3:42:09 PM
Flagged by SCM [11/30/2015 7:06:16 AM], Edited by jeifunk [11/30/2015 7:16:54 AM], Edited by jeifunk [11/30/2015 7:16:57 AM]
s
Original conversation
User: What is roman numerals
Question
Asked 10/6/2008 10:28:11 AM
Updated 10/19/2022 3:42:09 PM
Flagged by SCM [11/30/2015 7:06:16 AM], Edited by jeifunk [11/30/2015 7:16:54 AM], Edited by jeifunk [11/30/2015 7:16:57 AM]
Rating
8
Roman numerals, the numeric system used in ancient Rome, employs combinations of letters from the Latin alphabet to signify values.
Added 11/30/2015 7:06:17 AM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [11/30/2015 7:08:43 AM], Rated good by jeifunk
0
Roman numeral, any of the symbols used in a system of numerical notation based on the ancient Roman system. The symbols are I, V, X, L, C, D, and M, standing respectively for 1, 5, 10, 50, 100, 500, and 1,000 in the Hindu-Arabic numeral system.
Added 10/19/2022 3:42:09 PM
There are no comments.
Questions asked by the same visitor
what are roman numerals
Weegy: 'M' (More)
Question
Updated 3/8/2014 1:00:04 PM
Roman numerals, the numeric system used in ancient Rome, employs combinations of letters from the Latin alphabet to signify values.
Added 3/8/2014 1:00:03 PM
This answer has been confirmed as correct and helpful.
Confirmed by selymi [3/8/2014 1:01:22 PM]
38,688,143
Popular Conversations
programmed cell death to removed damaged cells so they can't pass on ...
Weegy: Programmed cell death to remove damaged cells so they can't pass on their mutations is also called Apoptosis. ...
11/29/2023 6:53:58 PM| 5 Answers
whats the purpose of the conclusion in a persuasive document?
Weegy: 2 + 2 = 4
12/1/2023 3:35:57 PM| 5 Answers
The _______ optimizes airflow, keeps cylinder pressure in a certain ...
Weegy: The valve timing optimizes airflow, keeps cylinder pressure in a certain range, and determines the engine's ...
11/29/2023 6:45:41 PM| 4 Answers
A writer might use a Venn Diagram to?
Weegy: A writer might use a Venn diagram to compare and contrast two or more ideas. User: A thesis statement should ...
12/1/2023 3:22:13 PM| 4 Answers
14. An emergency plan provides a methodical way to think through the ...
Weegy: Schools must plan for all emergencies for all the reasons below EXCEPT: Planning allows the local and state ...
12/6/2023 5:33:41 AM| 4 Answers
Progression through the stages of change is
Weegy: Progression through the stages of change is linear.
12/6/2023 5:31:30 PM| 4 Answers
30. When identifying stakeholders, you should: A. include only ...
Weegy: When identifying stakeholders, you should: consider internal and external individuals who may play a role in ...
12/6/2023 5:23:40 AM| 3 Answers
44. Which of the following is an example of a natural hazard? A. ...
Weegy: Earthquake - is an example of a natural hazard. User: 45. Which of the following is NOT a benefit of ...
12/6/2023 5:26:22 AM| 3 Answers
*
Get answers from Weegy and a team of really smart live experts.
S
L
P
Points 556 [Total 4540] Ratings 0 Comments 556 Invitations 0 Offline
S
L
P
R
L
P
L
P
P
Points 221 [Total 14228] Ratings 0 Comments 221 Invitations 0 Offline
S
L
Points 144 [Total 1157] Ratings 1 Comments 134 Invitations 0 Offline
S
L
R
R
L
Points 82 [Total 6012] Ratings 0 Comments 82 Invitations 0 Offline
S
L
Points 79 [Total 1236] Ratings 0 Comments 79 Invitations 0 Offline
S
Points 50 [Total 50] Ratings 0 Comments 50 Invitations 0 Offline
S
L
Points 22 [Total 822] Ratings 0 Comments 22 Invitations 0 Offline
S
L
L
Points 15 [Total 6393] Ratings 0 Comments 15 Invitations 0 Offline
S
L
1
1
Points 12 [Total 1962] Ratings 1 Comments 2 Invitations 0 Offline
S
L
R
P
L
P
P
P
Points 10 [Total 8661] Ratings 0 Comments 10 Invitations 0 Offline
* Excludes moderators and previous
winners (Include)
Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,271 | 3,934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-50 | latest | en | 0.88011 |
https://forums.aat.org.uk/Forum/discussion/443296/financial-performance | 1,708,905,947,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474649.44/warc/CC-MAIN-20240225234904-20240226024904-00449.warc.gz | 266,663,820 | 73,572 | # Financial performance
Registered Posts: 16 New contributor ?
Hi I am doing financial performance assignment 1 and I am stuck on task 1
The following cost have all been identified as semi variable
Analyse each item into the fixed cost element and variable cost per unit
Item 3000 units. 10000 units. Total fixed cost. Variable cost per unit
17750. 44000
9000. 24750
12900. 25500
How do I work this out please
• Registered, Moderator Posts: 2,034
Lindas1967
The method to use is called Hi-Lo
I put the figures into a table to help me
If you know for example that cost £17,750 to make 3,000 and £44,000 to make 10,000, you can calculate the extra cost of the extra units.
The extra 7,000 units [10,000 - 3,000] cost £26,250 to make. In other words we can say the average extra cost per unit as production goes up from 3,000 units is £3.75
The £17,750 we spend making 3,000 includes a fixed cost and a variable cost for each unit produced. The extra £3.75 we have identified is the variable cost per unit.
Knowing the variable cost of one unit we can multiply that figure by one of the production totals. I choose the 10,000 but you can do this sum using 3,000. In my case, the answer was £37,500.
This means that when we produce 10,000 units the total variable cost is £37,500.
We already know that the total cost [fixed + variable] for 10,000 units is £44,000
So the difference between these two amounts will be the fixed cost: £6,500
Do the same: take the 3,000 units x £3.75 and find the difference from the total cost of 3,000 units. I'm expecting £6,500 to be your answer too.
I hope this helps you.
Sandy
[email protected]
www.sandyhood.com
• Registered Posts: 16 New contributor ?
Thank you for that | 467 | 1,714 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-10 | latest | en | 0.9197 |
https://www.bdword.com/english-to-bengali-meaning-watt | 1,713,264,706,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817081.52/warc/CC-MAIN-20240416093441-20240416123441-00877.warc.gz | 609,059,815 | 16,636 | # English to Bangla Meaning of watt - ত্তঅট্
Watt :
ত্তঅট্
ত্তঅট্
ত্তঅট্ওয়াট
Definitions of watt in English
Noun(1) a unit of power equal to 1 joule per second; the power dissipated by a current of 1 ampere flowing across a resistance of 1 ohm(2) Scottish engineer and inventor whose improvements in the steam engine led to its wide use in industry (1736-1819
Examples of watt in English
(1) Changing to public ownership of the grid will not add a single watt of power.(2) When is it necessary to have a 350 watt power supply compared to just a 300 watt ?(3) The original half watt of power will diminish to less than a billionth of a watt at the point of arrival.(4) It generated hundreds of watts of power in usable form, actually running, and we were astonished to see this.(5) The chip will have a power consumption of under two watts .(6) The result is power densities are going from 40 watts per square fool to 68 watts per square foot and higher.(7) The smallest laser printers use at least 500 watts and the higher-end ones use at least a kilowatt.(8) Capable of producing microwatts to watts of power, it is designed for spectroscopy applications.(9) One pump requires 28,000 watts of electricity and the main pool has a capacity of 10,000 cubic meters of water.(10) The unit has a power output of 150 watts and is small enough to fit in your plane's glove compartment when it's not in use.(11) With 460 watts of power its sure to provide even the hungriest of systems with stable voltages.(12) Instead, a few tens of watts are enough to power a chip with transistors numbering in the tens of millions.(13) Just for a bit of perspective, a typical microwave easily consumes 1,200 watts .(14) But it is estimated that the net heating is at least one watt, perhaps closer to two watts per square yard.(15) Now it doesn't take a genius to figure out if I put in 100 watts and get out 350 watts I am getting a lot of free energy.(16) Since 1987, the government has subsidized the use of electricity up to 220 watts for each unit in the building.
Related Phrases of watt
(1) watt-hour meter ::
ওয়াট-ঘণ্টা মিটার
(2) watt-hour ::
ওয়াট-ঘণ্টা
Synonyms
M
1. watt ::
ত্তঅট্
Different Forms
watt, watts
English to Bangla Dictionary: watt
Meaning and definitions of watt, translation in Bangla language for watt with similar and opposite words. Also find spoken pronunciation of watt in Bangla and in English language.
Tags for the entry 'watt'
What watt means in Bangla, watt meaning in Bangla, watt definition, examples and pronunciation of watt in Bangla language. | 699 | 2,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-18 | latest | en | 0.937599 |
https://converterin.com/acceleration/kilometer-square-second-to-mile-hour-second.html | 1,566,122,139,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313747.38/warc/CC-MAIN-20190818083417-20190818105417-00392.warc.gz | 417,888,925 | 6,846 | # KILOMETER/SQUARE SECOND TO MILE/HOUR SECOND CONVERTER
FROM
TO
The result of your conversion between kilometer/square second and mile/hour second appears here
## KILOMETER/SQUARE SECOND TO MILE/HOUR SECOND (km/s² TO mph/s) FORMULA
To convert between Kilometer/square Second and Mile/hour Second you have to do the following:
First divide 1000 / 0.44704 = 2236.93629205
Then multiply the amount of Kilometer/square Second you want to convert to Mile/hour Second, use the chart below to guide you.
## KILOMETER/SQUARE SECOND TO MILE/HOUR SECOND (km/s² TO mph/s) CHART
• 1 kilometer/square second in mile/hour second = 2236.93629205 km/s²
• 10 kilometer/square second in mile/hour second = 22369.36292054 km/s²
• 50 kilometer/square second in mile/hour second = 111846.81460272 km/s²
• 100 kilometer/square second in mile/hour second = 223693.62920544 km/s²
• 250 kilometer/square second in mile/hour second = 559234.0730136 km/s²
• 500 kilometer/square second in mile/hour second = 1118468.1460272 km/s²
• 1,000 kilometer/square second in mile/hour second = 2236936.2920544 km/s²
• 10,000 kilometer/square second in mile/hour second = 22369362.92054402 km/s²
Symbol: km/s²
No description
Symbol: mph/s
No description | 385 | 1,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-35 | latest | en | 0.676462 |
https://randnyevg34.com/1-what-volume-of-water-is-displaced-by-a-submerged-2-0-kg-cylinder-made-of-solid-aluminum-aluminum-density-2-7-103-kg-m3-and-water-density-1-0-103kg-m3/ | 1,675,632,280,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00736.warc.gz | 497,542,844 | 17,777 | # 1. What volume of water is displaced by a submerged 2.0-kg cylinder made of solid aluminum? (aluminum density = 2.7 103 kg/m3 and water density = 1.0 103kg/m3)
Here is the answer for the question – 1. What volume of water is displaced by a submerged 2.0-kg cylinder made of solid aluminum? (aluminum density = 2.7 103 kg/m3 and water density = 1.0 103kg/m3). You’ll find the correct answer below
7.4 l0-4 m3
### Reason Explained
7.4 l0-4 m3 is correct for 1. What volume of water is displaced by a submerged 2.0-kg cylinder made of solid aluminum? (aluminum density = 2.7 103 kg/m3 and water density = 1.0 103kg/m3)
Thankyou for using answerout. We hope you get all your answers here. If you have any special questions, you can comment to ask us.
See also In a standard monohybrid cross between purple-flowered and white-flowered peas, in which purple flowers are dominant, what fraction of the purple-flowered peas in the F2 generation would you expect to be true-breeding? | 280 | 983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-06 | latest | en | 0.903249 |
https://gmatclub.com/forum/meet-yale-s-mba-class-of-206370.html | 1,503,147,488,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105451.99/warc/CC-MAIN-20170819124333-20170819144333-00288.warc.gz | 763,784,601 | 45,551 | It is currently 19 Aug 2017, 05:58
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Meet Yale's MBA class of 2017
Author Message
Joined: 23 Dec 2014
Posts: 285
Kudos [?]: 43 [0], given: 5
Concentration: Social Entrepreneurship, Healthcare
GMAT 1: 700 Q47 V38
GMAT 2: 700 Q44 V40
GMAT 3: 710 Q44 V42
GPA: 3.3
Meet Yale's MBA class of 2017 [#permalink]
### Show Tags
30 Sep 2015, 14:34
Interesting article from Poets and Quants' on the SOM's latest intake.
http://poetsandquants.com/2015/09/07/me ... of-2017/2/
Kudos [?]: 43 [0], given: 5
Current Student
Joined: 18 Mar 2013
Posts: 88
Kudos [?]: 23 [0], given: 6
Concentration: Finance
GMAT 1: 740 Q48 V42
GPA: 3.57
Re: Meet Yale's MBA class of 2017 [#permalink]
### Show Tags
07 Oct 2015, 15:03
good thing you posted this after the submission date. otherwise, i would've just thrown in the towel right then and there.
Kudos [?]: 23 [0], given: 6
Joined: 23 Dec 2014
Posts: 285
Kudos [?]: 43 [1] , given: 5
Concentration: Social Entrepreneurship, Healthcare
GMAT 1: 700 Q47 V38
GMAT 2: 700 Q44 V40
GMAT 3: 710 Q44 V42
GPA: 3.3
Re: Meet Yale's MBA class of 2017 [#permalink]
### Show Tags
08 Oct 2015, 08:04
1
KUDOS
dolongdao wrote:
good thing you posted this after the submission date. otherwise, i would've just thrown in the towel right then and there.
Kudos [?]: 43 [1] , given: 5
Re: Meet Yale's MBA class of 2017 [#permalink] 08 Oct 2015, 08:04
Similar topics Replies Last post
Similar
Topics:
2017-2018 Yale SOM MBA Deadlines 0 19 May 2017, 05:15
Calling all Yale Executive MBA Applicants: (2017 Intake) Class of 2019 10 26 Apr 2017, 10:17
4 Yale SOM Class of 2017 - Calling all Admitted Students 3 27 May 2015, 13:57
1 Calling all Yale SOM EMBA Applicants (2015 Intake) Class of 2017!! 3 31 May 2015, 15:39
345 Calling all Yale SOM Applicants -(2015 Intake) Class of 2017 1012 16 Jun 2015, 14:50
Display posts from previous: Sort by
# Meet Yale's MBA class of 2017
Moderators: OasisGC, aerien, BrushMyQuant | 822 | 2,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2017-34 | longest | en | 0.785107 |
https://www.physicsforums.com/threads/forces-of-3-charges-in-an-equilateral-triangle.922181/ | 1,723,094,078,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640719674.40/warc/CC-MAIN-20240808031539-20240808061539-00215.warc.gz | 735,338,845 | 18,190 | Forces of 3 charges in an equilateral triangle?
• Medeiros
In summary, the forces of 3 charges in an equilateral triangle can be calculated using Coulomb's Law. The magnitude and direction of the forces depend on the distance between the charges and the relative positions of the charges within the triangle. The net force on any one charge is zero due to the symmetry of the triangle, but the forces between the three charges can still be significant. The forces can also be affected by the presence of an external electric field.
Medeiros
1. The problem statement: in the first photo
all variables and given/known data,
The Attempt at a Solution
:
i wrote it all down in the second photo
[/B]
Attachments
• 20170808_152219.jpg
24.6 KB · Views: 427
• 20170808_152510.jpg
44.9 KB · Views: 419
Question 6 asks for the electric field not force?
1. How do you calculate the net force on a charge in an equilateral triangle with three charges?
The net force on a charge in an equilateral triangle with three charges can be calculated by using Coulomb's Law, which states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In an equilateral triangle, the distance between each charge is the same, so the net force can be found by adding the individual forces between each pair of charges.
2. Can the net force on a charge in an equilateral triangle ever be zero?
Yes, the net force on a charge in an equilateral triangle can be zero if the three charges are equal in magnitude and are arranged symmetrically in the triangle. In this case, the forces between each pair of charges will cancel out, resulting in a net force of zero on the third charge.
3. How does the distance between the charges affect the net force on a charge in an equilateral triangle?
The distance between the charges has a significant impact on the net force on a charge in an equilateral triangle. According to Coulomb's Law, the force between two charges decreases as the distance between them increases. Therefore, if the distance between the charges in the equilateral triangle is increased, the net force on the third charge will decrease.
4. Does the orientation of the charges in an equilateral triangle affect the net force on a charge?
Yes, the orientation of the charges in an equilateral triangle can affect the net force on a charge. If the charges are arranged in a straight line, the net force on the third charge will be greater compared to when they are arranged in a triangle. This is because the distance between the charges in a straight line is smaller, resulting in a stronger force.
5. Can the net force on a charge in an equilateral triangle ever be negative?
Yes, the net force on a charge in an equilateral triangle can be negative if the charges have opposite signs and are arranged in a way that the forces between them are in opposite directions. In this case, the net force will be in the direction of the charge with the larger magnitude, and the sign of the force will be negative.
• Precalculus Mathematics Homework Help
Replies
14
Views
863
• Introductory Physics Homework Help
Replies
27
Views
4K
• Introductory Physics Homework Help
Replies
2
Views
3K
• Introductory Physics Homework Help
Replies
1
Views
3K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Precalculus Mathematics Homework Help
Replies
5
Views
543
• Introductory Physics Homework Help
Replies
2
Views
1K
• General Math
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
15
Views
3K
• Introductory Physics Homework Help
Replies
13
Views
3K | 828 | 3,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-33 | latest | en | 0.922701 |
http://stackoverflow.com/questions/8406148/how-to-do-reverse-memcmp | 1,469,281,305,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257822598.11/warc/CC-MAIN-20160723071022-00160-ip-10-185-27-174.ec2.internal.warc.gz | 237,999,895 | 19,278 | Dismiss
Announcing Stack Overflow Documentation
We started with Q&A. Technical documentation is next, and we need your help.
Whether you're a beginner or an experienced developer, you can contribute.
# How to do reverse memcmp?
How can I do reverse memory comparison? As in, I give the ends of two sequences and I want the pointer to be decremented towards the beginning, not incremented towards the end.
-
Oops, sorry, wrong marking – user405725 Dec 6 '11 at 20:12
@VladLazarenko even though this is about comparison and not copying, I guess the answer is the same - hence this could be regarded as a dupe. – Till Dec 6 '11 at 20:12
I actually made a little mistake, I take my vote back, sorry guys – user405725 Dec 6 '11 at 20:12
There's no built-in function in the C standard library to do it. Here's a simple way to roll your own:
``````int memrcmp(const void *s1, const void *s2, size_t n)
{
if(n == 0)
return 0;
// Grab pointers to the end and walk backwards
const unsigned char *p1 = (const unsigned char*)s1 + n - 1;
const unsigned char *p2 = (const unsigned char*)s2 + n - 1;
while(n > 0)
{
// If the current characters differ, return an appropriately signed
// value; otherwise, keep searching backwards
if(*p1 != *p2)
return *p1 - *p2;
p1--;
p2--;
n--;
}
return 0;
}
``````
If you something high-performance, you should compare 4-byte words at a time instead of individual bytes, since memory latency will be the bottleneck; however, that solution is significantly more complex and not really worth it.
-
should be const unsigned char p1 = (const unsigned char)s1 + n - 1; const unsigned char p2 = (const unsigned char)s2 + n - 1; for example consider n=1. – OSH Dec 6 '11 at 20:55
@Oren: Oops thanks, good catch. Fixed. – Adam Rosenfield Dec 7 '11 at 2:23
Much like in a post (C memcpy in reverse) originally linked by Vlad Lazarenko, here is a solution based on that, that I haven't yet tested but should get you started.
``````int reverse_memcmp(const void *s1, const void *s2, size_t n)
{
unsigned char *a, *b;
a = s1;
b = s2;
size_t i = 0;
// subtracting i from last position and comparing
for (i = 0; i < n; i++) {
if (a[n-1-i] != b[n-1-i]) {
// return differences between different byte, strcmp()-style
return (a[n-1-i] - b[n-1-i]);
}
}
return 0;
}
``````
-
This doesn't compile -- `void*` pointers cannot be dereferenced. You should instead cast to `unsigned char*` before dereferencing. – Adam Rosenfield Dec 6 '11 at 20:31
Good catch. Fixed. – Dan Fego Dec 6 '11 at 20:34
All you need to do is specify your two ends and the size that you'd like to compare, as well as a step size. Please note especially that the step size may be the most important part for getting the expected results. It will greatly ease the implementation if you restrict the sizes. For the size of a char you could do something like:
``````int compare (void *one, void *two, size_t size)
{
char *one_char = (char *)one;
char *two_char = (char *)two;
size_t i;
for (i = 0; i < size; i++)
{
if (*(one_char - i) != *(two_char - i))
return(NOT_EQUAL);
}
return(EQUAL);
}
``````
-
shorter code (C code doesn't need force pointer type cast):
``````int reverse_memcmp(const void *end1, const void *end2, size_t n) {
const unsigned char *a = end1, *b = end2;
for (; n; --n)
if (*--a != *--b) return *a - *b;
return 0;
}
``````
- | 965 | 3,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2016-30 | latest | en | 0.861028 |
https://stats.stackexchange.com/questions/205414/handling-missing-continuous-attribute-values-in-id3 | 1,571,102,938,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655735.13/warc/CC-MAIN-20191015005905-20191015033405-00126.warc.gz | 656,677,023 | 31,428 | # Handling missing continuous attribute values in ID3
I'm implementing the ID3 algorithm. I have an attribute which happens to be continuous like 12.21, 3.01, etc. AND have missing values which are marked as "NA".
How I'm discretizing the data: I'm finding the optimal split which results in the max information gain. How I'm dealing with missing values: I will use the most probable attribute value to replace the "?".
Of course I can do either process in both ways, and this is where my confusion arises. Is there a correct way in handling this?
ID3 should be able to automatically handle missing values. So you don't need to manually impute the missing values. Just make sure this attribute is in numeric format (and not in a character/string format), and replace the 'NA' with null (blank) values before running ID3.
Update #1: Since you're implementing ID3 manually in Java, here are a couple of options I'd suggest:
1. The goal of each split is to minimize entropy (or alternatively, maximize information gain). For each splitting criterion, you can do an additional test by alternatively putting the missing instances into each child node. And then chose the one that produces lower entropy (for each splitting rule). Finally, you can choose the most optimal split across all such splitting criteria.
Update #2: More clarification on how to do this:
When calculating $gain$ as outlined below, $$Gain \ (S, A) = Entropy\ (S) \ - \sum_{v\in A}\frac{|S_v|}{|S|} \ Entropy(S_v)$$
Where $v$ is a value (cut-off point for a continuous variable) of $A$, $|Sv|$ is the subset of instances of $S$ where $A <= v$, and $|S|$ is the number of instances.
You can calculate two $gain$ values for each variable in two different ways: (1) include the missing instances in $S_v$, and (2) exclude the missing instances in $S_v$. You can then choose one of these two options that has the higher $gain$ value. (Consequently, the missing value handling is dictated by the approach that yields a higher value for $gain$.
1. A better option is to utilize the approach that's used by ID3's more popular extension: the C4.5 algorithm. You can distribute missing instances across all children node, and assign weights that are proportional to the number of instances in each child node.
• Sorry I think I didn't explain myself clearly, I am writing code in Java to implement the ID3 Algorithm. – Edqu3 Apr 4 '16 at 14:33
• "you can do an additional test by alternatively putting the missing instances into each child node". what do you mean by this? say I turn the classify the continuous attribute values to 2 subsets: >= Threshold and < Threshold. how can I put the missing instances into the child nodes? – Edqu3 Apr 4 '16 at 15:09
• I've updated my answer to clarify further. – Vishal Apr 5 '16 at 1:49 | 666 | 2,798 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-43 | latest | en | 0.881994 |
https://math.stackexchange.com/questions/2784782/about-the-solutions-of-x-varphiyzy-varphixz-z-varphixy-being | 1,643,177,582,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304915.53/warc/CC-MAIN-20220126041016-20220126071016-00718.warc.gz | 433,434,832 | 37,213 | # About the solutions of $x^{\varphi(yz)}+y^{\varphi(xz)}=z^{\varphi(xy)}$, being $\varphi(n)$ the Euler's totient
In this post we denote the Euler's totient function as $\varphi(n)$ for integers $n\geq 1$. I wondered about the solutions of the equation $$x^{\varphi(yz)}+y^{\varphi(xz)}=z^{\varphi(xy)}\tag{1}$$ for integers $x\geq 1$, $y\geq 1$ and $z\geq 1$.
Question. I think that $(1)$ has a finite number of such solutions $(x,y,z)$. Am I right? Many thanks.
I am waiting about a rigorous proof or a remarkable reasoning that can invoke heuristics or conjectures.
Computational fact. For integers $1\leq x,y,z\leq 100$ we can find (using a Pari-GP program; these can also be found by inspection) the solutions $(x,y,z)=(1,1,1),(1,3,2)$ and $(x,y,z)=(3,1,2)$.
• What led you to asking this question? May 17 '18 at 9:22
• Many thanks for your attention @Mastrem . I was playing with variants of an equation from Guy's book (Unsolved Problems in Number Theory). After that I ran the Pari-GP program I would like to know why previous equation seems to have a finite number of solutions. It is not clear to me if only by using Euler-Fermat theorem can it be solved.
– user243301
May 17 '18 at 9:25
• Let $d:=\gcd(x,y,z)$, so that there exist integers $a,b,c\geq1$ such that $$\varphi(yz)=a\varphi(d^2),\qquad \varphi(xz)=b\varphi(d^2),\qquad \varphi(xy)=c\varphi(d^2).$$ If $d\geq3$ we get a contradiction with Fermat's last theorem as then $$(x^a)^{\varphi(d^2)}+(y^b)^{\varphi(d^2)}=(z^c)^{\varphi(d^2)},$$ where $x^a,y^b,z^c\geq1$ and $\varphi(d^2)>2$. Hence $d\leq2$. May 17 '18 at 10:10
• I think that your comment has the merit to be written as an answer. Thus many thanks and feel free to edit it as an answer @Servaes
– user243301
May 17 '18 at 10:12
• I think it is helpful, but it does not answer the question. I will give the problem some more thought and hopefully expand this into an answer soon. May 17 '18 at 10:13
The only solutions are those you have found. If we are careful, we can avoid using Fermat's last theorem or Catalan's conjecture:
Claim 1: The only solutions with one of $x,y,z$ equal to $1$ are those you found.
Proof. Clearly $z>1$. Wlog $x=1$ and we are left with $1 + y^{\phi(z)} = z^{\phi(y)}$. We can observe that $\phi(n)$ is even for $n>2$, and that there are no nonzero consecutive squares. Hence either:
• $\phi(z)=1$ so $z=2$, and $1+y = 2^{\phi(y)}$. One shows that for $y$ odd the RHS is strictly larger unless $y=1$ or $y=3$, in which case there is equality. The details are at the bottom.
• $\phi(y)=1$ so either $y=1$ which gives $z=2$, or $y=2$ which gives $1+2^{\phi(z)}=z$. The LHS is now strictly larger, by the same inequalities as above.
We will suppose $x, y, z > 1$ from now on.
Claim 2: Every prime divisor $p$ of $z$ divides $x$ and $y$.
Proof. For $p>2$ this follows by Fermat's little theorem. Let $p=2$. Suppose $x,y \geq 3$ are odd. We have $xy, yz, xy > 2$ so all exponents are even. Odd squares are $1$ mod $4$, hence the LHS is $2$ mod $4$, while the RHS is $0$.
Claim 3: $z$ is a power of $2$.
Proof. With FLT: If $p>2$ is a prime divisor of $z$, then $p$ is a divisor of $\phi(xy),\phi(yz),\phi(xy)$, which contradicts Fermat's last theorem. Without FLT, we can proceed as follows: If $z$ has two prime divisors, then $4$ divides $\phi(xy),\phi(yz),\phi(xy)$ contradicting FLT for exponent $4$. Hence $z=p^k$ is a prime power. Also by FLT(4), $p=2$ or $p \equiv 3 \pmod 4$. But $p \equiv 3 \pmod 4$ is a contradiction with the fact that $p^{k\phi(xy)}$ is not a sum of two nonzero squares. Thus $p=2$.
Claim 4: There are no other solutions.
Proof. We have that $x,y,z$ are even. By FLT for exponent $4$, we cannote have $4 \mid z$. So $z=2$. By FLT for exponent $4$, not both $x$ and $y$ can have an odd prime divisor or be divisible by $4$. Wlog, $y=2$. But for $y=z$ the equation is $$x^{\phi(yz)} + y^{\phi(yx)} = y^{\phi(yx)}$$ which would imply $x=0$, contradiction.
On the comparison of $2^{\phi(y)}$ and $y$. We use the following inequalities:
Fact 1: $2^n \geq n+2$ for $n \geq 2$ with equality only for $n=2$.
Fact 2: $\prod b_i \geq \sum b_i$ for $b_i \geq 2$ with equality only for $b_1 \geq b_1$ and $2\cdot2 \geq 2+2$.
(You can show both by induction.)
Back to the problem. Let $y$ be odd and $y = \prod p_i^{a_i}$. By fact 2, we have $\phi(y) \geq \sum a_i(p_i-1)$ with equality only for $y=1$ or $y$ prime. By fact 1 applied to $n=p_i-1$ and mulitplying the inequalities: $$2^{\phi(y)} \geq 2^{\sum a_i(p_i-1)} > \prod (p_i+1)^{a_i} \geq y+1$$ with equality only if $y=1$ or if $y=p$ is prime and $p=3$.
• Many thanks I am going to study your answer and previous answers. You and the other users answering this question are true mathematicians.
– user243301
May 17 '18 at 11:23
• I am going to need some days to undestand your proof and the proof below, many thanks to you and all users that did contributions in answers and comments. Many thanks for your great answer and share the calculations in this site, you're generous.
– user243301
May 17 '18 at 13:57
Hint: Let $p$ be a prime factor of $z$ not dividing $x$ or $y$ and compute $x^{\varphi(yz)}+y^{\varphi(zx)}$ modulo $p$.
• ... which leaves the case where there is no such prime factor... May 17 '18 at 9:31
• From your hint I see it. $x^{\varphi(yz)}+y^{\varphi(zx)}\equiv 2\text{ mod }p$, that is zero if $p=2$.
– user243301
May 17 '18 at 9:32
• @barto yes, you're right. For a second, I thought that would mean that $z$ divided $x$ and $y$, which is obviously ffalse for anything divisible by a square. May 17 '18 at 9:32
• @user243301 no, I'm saying that any prime factor $p$ of $z$ must divide both $x$ and $y$ except if it's equal to $2$. Sadly, that doesn't solve the problem. As barto pointed out, my hint is a good start, but doesn't solve the problem. May 17 '18 at 9:36
• @servaes corrected. For about the 918,752nd time I would have preferred to be able to edit my own comments instead of having to copy, delete and paste. May 17 '18 at 11:31
Observation 1: The greatest common divisor of $x$, $y$ and $z$ divides $2$.
Let $d:=\gcd(x,y,z)$, so that there exist integers $a,b,c\geq1$ such that $$\varphi(yz)=a\varphi(d^2),\qquad \varphi(xz)=b\varphi(d^2),\qquad \varphi(xy)=c\varphi(d^2).$$ If $d\geq3$ we get a contradiction with Fermat's last theorem because then $$(x^a)^{\varphi(d^2)}+(y^b)^{\varphi(d^2)}=(z^c)^{\varphi(d^2)},$$ where $x^a,y^b,z^c\geq1$ and $\varphi(d^2)>2$. Hence $d\leq2$.
Observation 2: $z$ is a power of $2$.
Let $p>2$ be a prime dividing $z$. Then by the above $p$ does not divide $x$ or $y$, and there are integers $a,b\geq1$ such that $\varphi(xz)=a\varphi(p)$ and $\varphi(yz)=b\varphi(p)$, so by Fermat's little theorem we have $$z^{\varphi(xy)}=x^{\varphi(yz)}+y^{\varphi(xz)}=(x^b)^{\varphi(p)}+(y^a)^{\varphi(p)}\equiv2\pmod{p},$$ contradicting the fact that $p$ divides $z$ and $p>2$. Hence no such $p$ exists, so $z$ is a power of $2$.
Observation 3: $z=2$.
Clearly $z=1$ is impossible, and if $4\mid z$ then $2$ divides both $\varphi(xz)$ and $\varphi(yz)$, so the congruence $$x^{\varphi(yz)}+y^{\varphi(xz)}=z^{\varphi(xy)}\equiv0\pmod{4}$$ implies that $4$ divides both $x$ and $y$. But then $4$ divides both $\varphi(xz)$ and $\varphi(yz)$, and $2$ divides $\varphi(xy)$, yielding $u,v,w\geq1$ such that $$u^4+v^4=w^2,$$ a contradiction. Hence $z=2$. | 2,585 | 7,371 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-05 | latest | en | 0.908265 |
http://www.smokingmeatforums.com/index.php?threads/the-money-jar.78121/ | 1,513,040,106,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948514238.17/warc/CC-MAIN-20171212002021-20171212022021-00128.warc.gz | 474,007,308 | 12,552 | # The Money Jar!
Discussion in 'Jokes' started by jsbkrs, Jun 18, 2009.
1. ### jsbkrsNewbie
A fellow walks into a bar,
notices a very large jar on the counter,
and sees that it's filled to the brim with \$10 bills.
He guesses there must be
more than ten thousand dollars in it.
He approaches the bartender and asks,
'What's with the money in the jar?'
'Well... you pay \$10 and
if you pass three tests,
you get all the money and
the keys to a brand new Lexus.'
The man certainly isn't going to pass this up.
And so he asks, 'What are the three tests?'
'You must pay first...
Those are the rules,' says the bartender.
So, after thinking it over a while,
the man gives the bartender the \$10
and the bartender drops it into the jar.
'Okay,' the bartender says,
'Here's what you need to do:
First - You have to drink a whole quart of tequila,
in a minute or less, and
you can't make a face while doing it.
Second - There's a pit bull chained in the back with a bad tooth. You have to remove that tooth with your bare hands.
Third - There's a 90-year old lady upstairs
You have to take care of that problem!'
The man is stunned.
'I know I paid my \$10, but I'm not an idiot!
I won't do it!
You'd have to be nuts
to drink a quart of tequila, and
then do all those other things...'
'But, your money stays where it is.'
As time goes on, and
the man has a few more drinks,
he finally says,
'Where's the damn tequila?'
He grabs the bottle with both hands
and drinks it as fast as he can.
Tears stream down both cheeks...
but he doesn't make a face, and
he did it in fifty-eight seconds!
Next, he staggers out the back door,
where he sees the pit bull chained to a pole.
Soon the people inside the bar hear
growling , biting, and screaming sounds...
then nothing but silence!
Just when they think that
the man surely must be dead,
he staggers back into the bar,
with his shirt ripped open
and there are scratches and
he's bleeding all over his body.
He says,
'Now where's that old woman
The lesson in this story:
Listen carefully to the directions, | 526 | 2,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-51 | latest | en | 0.966008 |
https://mathematica.stackexchange.com/questions/258309/how-to-adjust-the-limits-of-this-numerical-integral | 1,716,266,122,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058383.61/warc/CC-MAIN-20240521025434-20240521055434-00335.warc.gz | 336,285,254 | 40,115 | # how to adjust the limits of this numerical integral?
I have a complicated function as follows:
f[rp_] := (E^(-2 rp^2 \[Gamma]) M^2 norm^2 \[Pi]^(
3/2) (E^((M \[Alpha] - 2 me rp \[Gamma])^2/(
2 (M^2 \[Beta] +
me^2 \[Gamma]))) ((-1 + E^((4 M me rp \[Alpha] \[Gamma])/(
M^2 \[Beta] + me^2 \[Gamma]))) M \[Alpha] +
2 (1 + E^((4 M me rp \[Alpha] \[Gamma])/(
M^2 \[Beta] + me^2 \[Gamma]))) me rp \[Gamma]) +
E^((M \[Alpha] - 2 me rp \[Gamma])^2/(
2 (M^2 \[Beta] + me^2 \[Gamma]))) (M \[Alpha] -
2 me rp \[Gamma]) Erf[(M \[Alpha] - 2 me rp \[Gamma])/(
Sqrt[2] M Sqrt[\[Beta] + (me^2 \[Gamma])/M^2])] -
E^((M \[Alpha] + 2 me rp \[Gamma])^2/(
2 (M^2 \[Beta] + me^2 \[Gamma]))) (M \[Alpha] +
2 me rp \[Gamma]) Erf[(M \[Alpha] + 2 me rp \[Gamma])/(
Sqrt[2] M Sqrt[\[Beta] + (me^2 \[Gamma])/M^2])]))/(8 Sqrt[2]
me rp \[Gamma] (M^2 \[Beta] + me^2 \[Gamma]) Sqrt[\[Beta] + (
me^2 \[Gamma])/M^2]);
I want to do some calculations on f function to obtain my desire final number. In this vein I need to integrate over rp as follows $$\int_{|re-r|}^{re+r} f(r_p) dr_p \tag{1}$$ As Mathematica can't solve (1) analytically, I have to use NIntegrate, but when I try (1) numerically
NIntegrate[f[rp], {rp, Abs[re - r], r + re}, {re, 0, \[Infinity]}, {r, 0, \[Infinity]}]
where the numerical values of constants are
me = 1;
mp = 1;
M = mp + me;
\[Omega] = 100;
\[Gamma] = SetPrecision[0.5*M*\[Omega], 50];
{\[Alpha], \[Beta]} = \
{0.40731020595362082969970174417539965361356735229492187550.,
24.58661827021306933716005005408078432083129882812550.};
norm = 189.2253326188998254888908479720045476056553654261503687655661\
49.454966552225144;
I get this error:
NIntegrate::nlim: rp = Abs[-1. r+re] is not a valid limit of integration.
So how can I solve my integral?! Any idea?
• You doesn't show the call to NIntegrate: Perhaps re,r isn't defined yet? You should also provide numerical values for all the parameters! Nov 15, 2021 at 14:42
• Please see the update Nov 15, 2021 at 14:44
• NIntegrate is a numerical function and needs numerical parameter values! Nov 15, 2021 at 14:48
• the situation is the same when I add limits of re and r. Please see the update again Nov 15, 2021 at 14:51
• Have you defined numerical values for $\alpha$, $\beta$, $\gamma$, M, norm, and me? You haven't provided them in the code above, and NIntegrate definitely won't work if you don't. Nov 15, 2021 at 14:54
Changing the order of the integration seems to help. I'm not sure why; it probably has something to do with how Mathematica does its sampling. The documentation for NIntegrate says that "The first variable given [for the integration region] corresponds to the outermost integral and is done last", so it is plausible that the limits of the first integration variable must be numeric.
NIntegrate[f[rp], {re, 0, \[Infinity]}, {r, 0, \[Infinity]}, {rp, Abs[re - r], r + re}]
(* 3.98439 *)
` | 1,021 | 2,875 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-22 | longest | en | 0.54974 |
https://gmatclub.com/forum/modern-navigation-systems-which-are-found-in-most-of-todays-7422.html?kudos=1 | 1,508,476,279,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823731.36/warc/CC-MAIN-20171020044747-20171020064747-00349.warc.gz | 688,070,080 | 47,042 | It is currently 19 Oct 2017, 22:11
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Modern navigation systems, which are found in most of todays
Author Message
Senior Manager
Joined: 01 May 2004
Posts: 335
Kudos [?]: 115 [0], given: 0
Location: USA
### Show Tags
25 Jun 2004, 11:42
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 1 sessions
### HideShow timer Statistics
2. Modern navigation systems, which are found in most of today’s commercial aircraft, are made with low-power circuitry, which is more susceptible to interference than the vacuum-tube circuitry found in older planes. <During landing, navigation systems receive radio signals from the airport to guide the plane to the runway>. Recently, one plane with low-power circuitry veered off course during landing, its dials dimming, when a passenger turned on a laptop computer. <Clearly, modern aircraft navigation systems are being put at risk by the electronic devices that passengers carry on board, such as cassette players and laptop computers>.
The two portions in boldface play which of the following roles?
(A) The first is a principle that the argument relies on and the second is a conclusion that can be drawn from the first.
(B) The first is a fact that argument relies on and the second is a conclusion that must be drawn from this argument.
(C) The first acknowledges a consideration that supports that main position; the second is that conclusion.
(D) The first is an evidence that supports the conclusion, the second is that conclusion.
(E) The first is a principle that is necessary for this argument, the second is a conclusion that could be drawn from this argument
Kudos [?]: 115 [0], given: 0
Manager
Joined: 08 Jun 2004
Posts: 245
Kudos [?]: 9 [0], given: 0
Location: INDIA
### Show Tags
25 Jun 2004, 12:24
ans E...
first BOLD face... -->navigation systems help in the landing of the plane on the run-way
second BOLD face...--> navigation systems are ineterfered in their functioning by electronic devices ''
first one= premise and principle told...
second one= cretainly the conlcusion...
A) parlty right as for the frst part but certainly we cannot get the conclusion from the FISRT bold face alone....
B)very close ... but it s too extreme when it says ''argument rlies on ''and 'must b drawn'' if I m wrong this cud be the right ans...
C ) incosistent and cofusing...
D) partly wrong... the frst does noway support the conclusion...
only E goes right when it says'' frst= a principle , seocnd= a conclusion that can be derived ''...
hope that helps!
Have fun
_________________
the whole worldmakes way for the man who knows wer he's going... good luck
Kudos [?]: 9 [0], given: 0
Manager
Joined: 02 Jun 2004
Posts: 184
Kudos [?]: 7 [0], given: 0
Location: Kiev, Ukraine
### Show Tags
25 Jun 2004, 12:58
boksana wrote:
2. Modern navigation systems, which are found in most of today’s commercial aircraft, are made with low-power circuitry, which is more susceptible to interference than the vacuum-tube circuitry found in older planes. <During landing, navigation systems receive radio signals from the airport to guide the plane to the runway>. Recently, one plane with low-power circuitry veered off course during landing, its dials dimming, when a passenger turned on a laptop computer. <Clearly, modern aircraft navigation systems are being put at risk by the electronic devices that passengers carry on board, such as cassette players and laptop computers>.
The two portions in boldface play which of the following roles?
(A) The first is a principle that the argument relies on and the second is a conclusion that can be drawn from the first.
(B) The first is a fact that argument relies on and the second is a conclusion that must be drawn from this argument.
(C) The first acknowledges a consideration that supports that main position; the second is that conclusion.
(D) The first is an evidence that supports the conclusion, the second is that conclusion.
(E) The first is a principle that is necessary for this argument, the second is a conclusion that could be drawn from this argument
Another difficult one.
It looks like B to me.
During landing, navigation systems receive radio signals from the airport to guide the plane to the runway - to me, that sounds like a fact. The author tells us what happens during landing, there's no doubt that this is the way it happens.
Clearly, modern aircraft navigation systems are being put at risk by the electronic devices that passengers carry on board, such as cassette players and laptop computers - this sounds like a conclusion to me, and it must be drawn from the argument, there's no alternative course for this argument to run.
Fact 2 - low-power circuitry is more susceptible to interference
Evidence 1 - plane steered off course during landing, because a laptop was turned on
Assumption - use of laptops & other electronic equipment interferes with the low-power circuitry
Conclusion - modern aircraft navigation systems are being put at risk by the electronic devices [that interfere with low-power circuitry while the plane receives radio signals during landing]that passengers carry on board
_________________
Kudos [?]: 7 [0], given: 0
Intern
Joined: 05 Jan 2004
Posts: 27
Kudos [?]: [0], given: 0
Location: Los Angeles
### Show Tags
25 Jun 2004, 13:08
To me E sounds better
No where in the passage is it stated that the laptop computers and the personal items belonging to passengers interfere with the system. They could be ASSUMED to do so and hence this is one of the conclusions that can be drawn.
It could be that when plane was landing there were other equipments that malfunctioned because of the laptop.
For this conclusion to be drawn absolutely certainly, there should be a fact that states that these devices produce signals that interefere with low power circuitry
Kudos [?]: [0], given: 0
Senior Manager
Joined: 25 May 2004
Posts: 281
Kudos [?]: 20 [0], given: 0
Location: Ukraine
### Show Tags
25 Jun 2004, 14:39
boksana wrote:
2. Modern navigation systems, which are found in most of today’s commercial aircraft, are made with low-power circuitry, which is more susceptible to interference than the vacuum-tube circuitry found in older planes. <During landing, navigation systems receive radio signals from the airport to guide the plane to the runway>. Recently, one plane with low-power circuitry veered off course during landing, its dials dimming, when a passenger turned on a laptop computer. <Clearly, modern aircraft navigation systems are being put at risk by the electronic devices that passengers carry on board, such as cassette players and laptop computers>.
The two portions in boldface play which of the following roles?
(A) The first is a principle that the argument relies on and the second is a conclusion that can be drawn from the first.
(B) The first is a fact that argument relies on and the second is a conclusion that must be drawn from this argument.
(C) The first acknowledges a consideration that supports that main position; the second is that conclusion.
(D) The first is an evidence that supports the conclusion, the second is that conclusion.
(E) The first is a principle that is necessary for this argument, the second is a conclusion that could be drawn from this argument
This one is not very difficult. I guess a course in Writing covers such questions.
During landing, navigation systems receive radio signals from the airport to guide the plane to the runway is a fact on which the argumet rests.
The first and the third sentences provide specific details/facts that build to a logical conclusion. Hence B.
Kudos [?]: 20 [0], given: 0
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4285
Kudos [?]: 527 [0], given: 0
### Show Tags
25 Jun 2004, 18:13
http://www.gmatclub.com/phpbb/viewtopic ... ght=#28240
_________________
Best Regards,
Paul
Kudos [?]: 527 [0], given: 0
Senior Manager
Joined: 01 May 2004
Posts: 335
Kudos [?]: 115 [0], given: 0
Location: USA
### Show Tags
29 Jun 2004, 13:10
I don't think it's B. Without assumption the conclusion falls apart
I believe the answer should be E
Kudos [?]: 115 [0], given: 0
Senior Manager
Joined: 07 Oct 2003
Posts: 349
Kudos [?]: 21 [0], given: 0
Location: Manhattan
### Show Tags
30 Jun 2004, 12:19
I also went with E on this one. What tipped me is that B states: "conclusion that must be drawn from this argument". However there could be other reasons why the plane's controls were off. I would've felt a lot more comfortable if B stated "conclusion that may be drawn from this argument".
I do acknowledge that the first statement is a fact, but what's wrong with calling it a principle?
Principle -- "A basic truth, law, or assumption"
Kudos [?]: 21 [0], given: 0
Director
Joined: 20 Apr 2005
Posts: 584
Kudos [?]: 280 [0], given: 0
### Show Tags
28 May 2005, 08:23
Kudos [?]: 280 [0], given: 0
28 May 2005, 08:23
Display posts from previous: Sort by
# Modern navigation systems, which are found in most of todays
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,354 | 9,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-43 | latest | en | 0.94648 |
https://robotics.stackexchange.com/questions/8372/denavit-hartenberg-parameters | 1,656,266,383,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103271763.15/warc/CC-MAIN-20220626161834-20220626191834-00132.warc.gz | 555,496,319 | 68,074 | # Denavit Hartenberg parameters
Can anybody help figure out HD parameters for the case where two links with a revolute joint are in the same plane, thus that the variable angle is 0, but the twist is not 0. This is a simple drawing. I think that x-axis that is perpendicular to both z-axis, points away and goes through the intercection of z-axis. The link length is 0, the twist is a and the offset is d. Whould it be correct? Thanks.
• Your question needs a lot of clarification. Each individual link in a manipulator has its own joint -- so two links have two joints. Do you mean the two links have revolute joints whose axes are parallel? If that is the case then the relevant $\alpha$ transformation should be 0 ("twist angle"?). Meanwhile, $\theta$ is the joint angle and can be any value within the limits of the joint ("variable angle"?). Your drawing has no x-axes or z-axes, so your reference to them is confusing. Also, $d$ and $a$ are themselves DH parameters already, so you shouldn't use them for new dimensions. Nov 2, 2015 at 9:57
• The links lie in the plane of the drawing I would call it xy-plane, but it would be confusing cause it would look like z-axis lies in the xy-plane. The second joint is revolute, but imagine that the first link is welded at a certain angle that differs from the right angle. Thus the axis are not parallel, they intersect at the point where i would put an vertical x-axis. The directions of the joints are represented by the short thik lines at the ends of the links. The third one would be parallel to the second one, but i am asking about the first and second ones. Nov 2, 2015 at 10:51
• If the first "link" is welded and fixed then it is not a link. Are you saying that $\alpha$ is not 0 for the first link? So the joint axes are not parallel? Please replace your diagram with something in 3D that is more clear. Nov 2, 2015 at 11:08
• A link is always welded to a joint, it is a joint that does all DOFs and a link is just attached to it, I said "welded". This configuration lies in the plane of the drawing. If it was the right angle this would be an ordinary revolute joint. Nov 2, 2015 at 11:20
• Don't confuse "joint" with the parts that rotate and "link" with the parts attached to that. In terms of a DH manipulator configuration, a link is a single degree-of-freedom that is attached to a previous link through a joint. The first link is attached to some kind of base, but we need not consider that in the model since it is usually fixed in the world. See this diagram. Nov 2, 2015 at 11:46
You can use dummy transformations when the DH parameterization cannot get you from one axis to the other. Think of it as two succesive DH transformations where the first one does not have a moving joint. It is just there to get to the other joint the the second DH transformation
EDIT:
to complete the answer with reference:
http://petercorke.com/doc/rtb_dh.pdf
If your previous Z axis is not intersecting the next X axis and/or they are not perpendicular, you will not be able to use one DH transformation matrix from one coordiante system to the other
• I dont quite understand what you mean. It is an ordinary revolute joint. If two consecutive joints had the same direction for z-axes, it would be no problem to find the parameters, cause they are parallel and you can get a perpendicular line to both of them. But if those z-axes intersect one another the perpendicular line (x-axis) can be put only like in the drawing. The twist - a rotation from one z-axis to another about x-axis can seem to be found easily - it is represented by a, r-parameter seem to be zero and I am not sure what I would call d in this case. Nov 3, 2015 at 21:12
• The point is this: DH parameters allow you to transform from one joint to another using a common set of rotations and translations (see this video for clarification). Your particular configuration may need a modified version of DH parameters -- namely a y-component for the link length (usually has only x- and z-components, $a$ and $d$, respectively). This is because the new rotation axis for a joint should rotate about the previous x-axis, but your configuration cannot be defined as such. Nov 4, 2015 at 1:12
• As Brian Lynch pointed out, the DH transformation matrix has only 4 parameters, so you can only do 4 "motions" from one coordinate system to the next one. In your case it seems that with one DH transformation you cannot get from where you are to where you want to be. In these cases, if you still want to use the DH convention, you can use two DH transformations Use the first one just to rotate the coordinate system in the first joint to an orientaion the allows you to use a second DH transofmration to transittion to the desired coordinate system at your joint in queston.
– 50k4
Nov 4, 2015 at 8:54
• To Brian Lynch: I saw this video. It deals with the convenient case I can deal with without the video. Is there a formal formulation for this y-parameter and does it mean that HD parameters can not deal with all cases because actually just slitest deviation from a right angle when z-axes are not parallel but lie in the same plane means I can not use the standard HD parameters? Nov 4, 2015 at 10:55
• To 50k4: >>if you still want to use the DH convention. What would be another set of parameters or another convention that I could use insted? Nov 4, 2015 at 10:58 | 1,292 | 5,392 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2022-27 | longest | en | 0.956777 |
https://www.bartleby.com/solution-answer/chapter-83-problem-17e-calculus-10th-edition/9781285057095/773ddc51-57d1-11e9-8385-02ee952b546e | 1,708,741,686,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474482.98/warc/CC-MAIN-20240224012912-20240224042912-00035.warc.gz | 670,387,534 | 101,351 | Calculus
10th Edition
ISBN: 9781285057095
Author: Ron Larson, Bruce H. Edwards
Publisher: Cengage Learning
#### Videos
Question
Chapter 8.3, Problem 17E
To determine
## To calculate: The value of integral ∫0π2sin6xdx using Walli’s formula.
Students have asked these similar questions
Integration techniques Use the methods introduced evaluate the following integrals. ∫cos2 4θ dθ
Using Cauchy's Theorem calculate the following integral and the singular points of the function, where C: z(t) = 3*cost(t) + i*(3+ 3*sin(t)) 0 < t < 2π
a) Use the Trapezoid Rule to estimate the following integral. Use n=5 integral of (e^(x^2)) dx from 0 to 1 b) Use the Error Bound Formula for the Trapezoid Rule to find a best estimate for the error involved in part a
• Copyright, Community Guidelines, DSA & other Legal Resources: Learneo Legal Center
• bartleby, a Learneo, Inc. business | 247 | 877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-10 | latest | en | 0.684313 |
http://math.stackexchange.com/questions/246289/transpose-of-block-matrix/246305 | 1,435,987,616,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375096504.12/warc/CC-MAIN-20150627031816-00004-ip-10-179-60-89.ec2.internal.warc.gz | 145,467,754 | 17,595 | # Transpose of block matrix
I'm attempting to prove that
$$\left[ \begin{array}{c c} A & B \\ C & D \\ \end{array} \right]^\top = \left[ \begin{array}{c c} A^\top & C^\top \\ B^\top & D^\top \\ \end{array} \right].$$
Intuitively, I can see that it's true. However, when I try to formally prove it, I quickly get lost in the indices. What tricks can I use to keep things straight?
Source: Exercise 2.6.16, P116, Intro to Linear Algebra, 4th Ed by Strang
-
Most people would just claim this is obvious and omit the proof, but if you don't want to do that then perhaps you could first prove that $$\begin{bmatrix} M & N \end{bmatrix}^T = \begin{bmatrix} M^T \\ N^T \end{bmatrix}$$ and $$\begin{bmatrix} M \\ N \end{bmatrix}^T = \begin{bmatrix} M^T & N^T \end{bmatrix}.$$ Then \begin{align} \begin{bmatrix} A & B \\ C & D \end{bmatrix}^T &= \begin{bmatrix} \begin{bmatrix} A \\ C \end{bmatrix}^T \\ \begin{bmatrix} B \\ D \end{bmatrix}^T \end{bmatrix} \\ &= \begin{bmatrix} A^T & C^T \\ B^T & D^T \end{bmatrix}. \end{align}
-
Definition of Transpose is $(A^T)_{ij} = A_{ji}$
Why $\begin{bmatrix} M & N \end{bmatrix}^T = \begin{bmatrix} M^T \\ N^T \end{bmatrix}$? Why NOT $\begin{bmatrix} M & N \end{bmatrix}^T = \begin{bmatrix} M \\ N\end{bmatrix}$?
After transpose, $M$ is in (1, 1) position and $N$ is in (2,1) position. Why still keep the $^T$?
Why $\begin{bmatrix} M \\ N \end{bmatrix}^T = \begin{bmatrix} M^T & N^T \end{bmatrix}$? Why NOT $\begin{bmatrix} M \\ N \end{bmatrix}^T = \begin{bmatrix} M & N \end{bmatrix}$ ?
After transpose - $M$ is in (1, 1) position and $N$ is in (1,2) position. Why still keep the $^T$?
Example $\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}^T = \begin{bmatrix} a_{11}^T & a_{12}^T \\ a_{21}^T & a_{22}^T \end{bmatrix} = \begin{bmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{bmatrix}$. No $^T$ at the end.
-
In your example at the end, the $a$s are scalar values, so $a_{11}^T = a_{11}$. But the question is about block matrices and therefore $M = M^T$ doesn’t need to be true, so you cannot omit the $^T$. – Eike Schulte Oct 1 '13 at 9:14 | 806 | 2,108 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 2, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2015-27 | longest | en | 0.670147 |
http://www.convertunits.com/from/grams+OSi3/to/moles | 1,386,380,037,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163052970/warc/CC-MAIN-20131204131732-00077-ip-10-33-133-15.ec2.internal.warc.gz | 284,973,480 | 5,123 | grams OSi3
moles
# ››More information from the unit converter
How many grams OSi3 in 1 moles? The answer is 100.2559.
We assume you are converting between grams OSi3 and mole.
You can view more details on each measurement unit:
molecular weight of OSi3 or moles
The SI base unit for amount of substance is the mole.
1 grams OSi3 is equal to 0.00997447531766 mole.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between grams OSi3 and mole.
Type in your own numbers in the form to convert the units!
# ››Similar chemical formulas
Note that all formulas are case-sensitive. Did you mean to convert one of these similar formulas?
grams OSI3 to moles
grams OsI3 to moles
grams OSi3 to moles
# ››Convert another chemical substance
Convert grams to moles
I'm feeling lucky, show me a random compound.
# ››Want other units?
You can do the reverse unit conversion from moles to grams OSi3, or enter any two units below:
Convert to
I'm feeling lucky, show me some random units.
# ››Details on molecular weight calculations
In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together.
Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. These relative weights computed from the chemical equation are sometimes called equation weights.
Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance.
If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100.
Finding molar mass starts with units of grams per mole (g/mol). When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula.
The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. We use the most common isotopes. This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass.
A common request on this site is to convert grams to moles. To complete this calculation, you have to know what substance you are trying to convert. The reason is that the molar mass of the substance affects the conversion. This site explains how to find molar mass.
# ››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!
This page was loaded in 0.0035 seconds. | 795 | 3,641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2013-48 | latest | en | 0.888659 |
http://www.jiskha.com/display.cgi?id=1302891131 | 1,498,554,490,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321306.65/warc/CC-MAIN-20170627083142-20170627103142-00603.warc.gz | 578,937,524 | 3,977 | # physics
posted by .
Suppose that a 1.00 cm3 box in the shape of a cube is perfectly evacuated, except for a single particle of mass 1.13 x 10^-3 g. The particle is initially moving perpendicular to one of the walls of the box at a speed of 525 m/s. Assume that the collisions of the particle with the walls are elastic.
(a) Find the mass density inside the box.
(b) Find the average pressure on the walls perpendicular to the particle's path.
(c) Find the average pressure on the other walls.
(d) Find the temperature inside the box.
• physics -
a) d=m/v
d= (1.13*10^-3)/1
d= 1.13e-3 g/cm^3
• physics -
c) 0 Pa | 172 | 622 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-26 | latest | en | 0.81374 |
https://www.bartleby.com/questions-and-answers/c3-nn/010992d5-f9a3-4494-867a-9b250978d4bd | 1,585,454,962,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493684.2/warc/CC-MAIN-20200329015008-20200329045008-00032.warc.gz | 849,177,658 | 27,961 | Question
1 views
In the figure below, a 16.0 V battery is connected across capacitors of capacitances with the following values.
C1 = C6 = 2.70 μF
C3 = C5 = 4.10 μF
C2 = C4 = 2.05 μF
(d) What is q1 of capacitor 1?
______________C
(e) What is V2 of capacitor 2?
______________V
(f) What is q2 of capacitor 2?
_______________C
check_circle
Step 1
Draw the equivalent circuit for the given circuit diagram:
Step 2
Step 3
...
### Want to see the full answer?
See Solution
#### Want to see this answer and more?
Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*
See Solution
*Response times may vary by subject and question.
Tagged in | 213 | 708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-16 | latest | en | 0.749384 |
https://electronics.stackexchange.com/questions/536091/issues-with-simple-nicd-battery-charger-circuit | 1,719,273,991,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865490.6/warc/CC-MAIN-20240624214047-20240625004047-00867.warc.gz | 196,212,037 | 43,463 | # Issues with simple NiCd battery charger circuit
Inspired by this circuit, I'm trying the following one for charging 2 NiCd batteries:
simulate this circuit – Schematic created using CircuitLab
There is a current limiter part, with a red LED (D3) indicating that the charging is in progress. The overvoltage protection should be provided by a 2.7 V Zener diode (D1).
The current limiter part is working as expected, delivering about a constant 120 mA current. The problem comes with the overvoltage protection, since the zener appears to conduct enough current to turn on Q2, with the batteries still at 2.7 V. More precisely, I measure a current of 0.35 mA on the zener, with a voltage drop of 1.9 V on it, and so D3 is always off.
My try to fix this was to replace R3 with a lower resistance, namely 20 Ω. With this new configuration Q1 is turned on, with the zener conducting 15 mA over a 3.3 V voltage drop on it (this Zener diode is pretty far from being ideal, isn't it?). Voltage drop on R3 is then 0.45 V, not enough to turn on Q2.
Based on these experiments, I suspect that the value of R3 can probably be arranged in order to turn off Q1 at an appropriate voltage of the batteries (e.g. 2.9 V). Yet, this seems to me too much dependant on the particular components I'm using, and so difficult to predict and calculate. Also, I wouldn't like to rely on such fine-tuning arrangements in order to prevent overvoltage on my batteries.
My question is: is there a way to properly fix this circuit, or is it simply wrong by design?
Thanks
• Looks like you've discovered that a lot of circuits on the internet cannot be trusted. Commented Dec 7, 2020 at 15:41
• Today, I have done a circuit very close to that one you sketched. These is 2 major differences from yours, as I use an AZ431/TL431 as precision Zener and it is desined to charge 5x NiCd: V_final = 6.87V and i_lim = 140mA. It is working flawlessly. If it is still worth for you, I can answer with it too.
– EJE
Commented Aug 4, 2022 at 21:47
• And about your questions, yes I believe it can be fixed as it is, where I suggest 2 changes: #1 (easier) try first R3= 330R, as mid point between Zz and Zzk in BZX55C datasheet. (vishay.com/docs/85604/bzx55.pdf). #2 replace 2V7 Zener by any TL431 as 2.495V precision reference (joint Ref+Cathode). For this, I guesstimate any 220R < R3 < 2K2 will work fine.
– EJE
Commented Aug 4, 2022 at 22:03
Q1: Is there a way to properly fix this circuit?
Q2: Or is it simply wrong by design?
A1 = Yes there is. See detailed answer, with 4 Checks/Changes and 2 Tips.
A2 = No, the circuit is not ‘wrong by design’, but needs to be calculated properly and are ways to improve its performance. Then you proto-board it too and you get something that works as intended
A1: Yes there are ways you can ‘fix’ it, if you accept a not-so-defined knee from “constant” current to “constant” voltage.
Between quotes, because is not really constant, but would work ok.
I already built some circuits like that, obviously changing values and components. But in terms of topology/configuration, they are quite similar.
I suggest a few changes:
Q1 = 2N2222 is borderline adequate, it will “fry” due to power dissipation. Assume as worst case, V_in = 8V and dead batteries (0V). In this case Power = 8V x 120mA = 960mW.
If you use a BD135 or MJE340 assuming HFE > 30, or in a better TO-220 as TIP31 or even a TIP110 Darlington with HFE = 2500, you have several design BJT options that will dissipate 1W without heatsink.
Please observe that using a Darlington, it may be necessary to raise V_in.
#2 Review & Change R1.
Base current resistor, we could think of using
R1 = ((8 - 0 - Vbe) / 0.120) * (HFE_min / 3).
R1_BD135 = ~620R (commercial). R1_TIP110 = ~47K.
Current in R1 is important because will drive CC LED: However, as we wish i_LED ~= 10mA, by inspection we could see R1 is far from the determined values FDTE to polarization requirements of Q1. Recalculating by LED current.
In this case, for V_in = 8V, delta_V_NPN+LED+Bat = 5.8V.
R1 = (V_in - delta_V)/i_LED = 220.
R1 = 220R.
#3 - Review R2 & D3.
Using a LED_RED or Green, with V_f ~=2.0V, V_R2 = (2.0 - Vbe) R2 = V_R2/i_bat = ((2.0-0.7)/0.12) = 10R8.
R2 = 10R.
D3 = RED(new) or GREEN.
#4 Check ZenerDiode D1 and R3.
Zener Diode BZX55C in Vz = 2.7V, has an equivalent resistance between 85R (@ 5mA) and 600R (@ 1mA).
R3 voltage at 0.7V will make Q3 conduct; assuming Zener current = 5mA, we expect Vz ~= 2.7V.
So say, i_D1 = 2mA and for that, R3 = 0.7/0.005.
R3 = 150R.
Tip #1: R3 and V_out fine-tuning.
Curiously, Vbe_Q2 and Vf_D2 are not much different, so it is expected that
V_out ~= Vzener_D1.
A trick to fine-tune the Output voltage would be to change R3 in such a way that current would swing from 2mA and 30mA, which would increase the Vz by tenths of volts around the nominal 2V7.
Using R3 = 22R (fixed) + 330R (trimpot) would be a first guess.
Tip #2: Improving Voltage stability.
If you wish to get a very sharp knee with really Flat plateau for the Constant Voltage part, Zener is not the best option, as it varies the impedance much more than an “integrated Zener” as the TL-431. Another point is related to Thermal drift, which a 2V7 Zener is not the most precise if it is left close to the power transistor getting hot.
Don’t get me wrong, the Zener circuit will work as a Voltage reference, and even can be tricked (as above) because it has a comparative large impedance at a given operating point; but you can use other references too.
So, if you replace the Zener by a TL431 and use Datasheet’s Figure 10-3 or 10-5 or 10-6, it works beautifully.
Just put an 4K7 in parallel with your D4 and replace Q1, D1 and R3 by the TL431 and its voltage divider. Oh, and leave the previous C.C. Circuit as it is.
A2: So, can work and is not ‘wrong by design’. It just need some recalculations, but the circuit works.
That is exactly what I have done, for a friend of mine that needed a 5 cell NiCd charger. I made one running at V_out = 6.88V & i_cc = 140mA.
See the assembly for that, using:
• Vref = TL431 + 12K & 6K8 (V_divider)
• NPN (MJE13003+2N3904)= Darlington parts from C.F.Lamp.
• Blue LED for D3 (compensate using Darlington).
• Any LED (white) as D4; too bright, ‘painted’ as RED: nail enamel used for that.
• My R1 = 820R, as I work with V_in = 12V.
• I decided to not use that output diode (your D2) as the back-voltage is not high to cause damages to the circuit and it is being used to charge a toy, so it is just connected when another recharging is needed. If you don’t feel this as adequate for you, maintain the “D2” and comped paste that in the TL’s voltage divider.
## CC-CV Charger - Update with Test Data
Using the charger from the Picture, I tested it using 5 AAA advertised as 800mAh but tested individually (La Crosse smart charger) and showed actual 450-500 mAh. Battery pack (5x AAA) was fully and deeply discharged, using initially a 5R6 and then a 10R power resistor. Recovery voltage (instantaneous and after 1 minute was tabulated too.
Then a charging was monitored measuring selected Input Voltages (Nominal V_in is 12V), V out, I_out, LED status, and Temperature @ heatsink.
The Excel spreadsheet (Made in Brazil, so using comma as decimal separator and DD/MM/YYYY) then discounts the stand-by current (last line) to estimate the actual charging current. Accumulated mAh was "integrated" by trapezoidal segments (Romberg's method) for each measured delta Time.
Then general curves showing some key values are plotted - consider that Nominal C = 400mAh and i_cc = 140 mA, charging rate is about C/3:
## Summary and conclusions for this CC-CV Charger at C/3
Using a charger that limits CC to C/3 (C = Capacity), and CV set at 1.38V/cell:
• 25% C occurs at CC-CV switch-over (seen in LEDs).
• 50% C is achieved at 2 Hours mark.
• 80% C is reached in 6 Hours.
• 90% C is reached in 9H.
• 95% C is reached in 16H.
• 100% C is reached in 20H, with "final" charging current of 1 mA.
Here is a simple answer. You could try using three or four silicon diodes in series for D1 (about 2.5V) although with a negative temperature coefficient. Or a white LED, perhaps in series with a diode, may have the voltage you need. You might also be able to replace the zener with a resistor of about 3k, which will forward bias Q2 at about 2.8 volts. You could replace D1 and R3 with a 5k trimpot and adjust output voltage.
Here's my simulation: | 2,425 | 8,432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.955278 |
https://rbjlabs.com/form/integration-formulas/ | 1,719,309,108,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865694.2/warc/CC-MAIN-20240625072502-20240625102502-00138.warc.gz | 437,492,815 | 44,738 | # Integration formulas
It is considered a and \text{C} as real or constant numbers, u is the function of x and u' is the derivative of u.
## Basic forms and properties of integrals
• \displaystyle \int \ u^{n} \cdot du = \cfrac{u^{(n+1)}}{n+1} + \text{C},\qquad n \neq -1
• \displaystyle \int \ du = u + \text{C}
• \displaystyle \int a \cdot (u^{n}) \cdot du = \displaystyle a \cdot \int u^{n} \cdot du = a \cdot \Bigg ( \cfrac{u^{(n + 1)}}{n + 1} \Bigg ) + \text{C}, \qquad[katex] n \neq -1[/katex]
• \displaystyle \int \cfrac{du}{u} = \ln |u| + \text{C}
• \displaystyle \int e^{u} \cdot du = e^{u} + \text{C}
• \displaystyle \int a^{u} \cdot du = \cfrac{a^{u}}{\ln a} + \text{C}
#### The integral of an addition / subtraction is equal to the addition / subtraction of the integrals
• \displaystyle \int \big (u^{n} \pm u^{n\pm 1} \pm u^{n\pm 2} \big ) \cdot du = \displaystyle \int u^{n} \cdot du \pm \int u^{n \pm 1} \cdot du \pm \int u^{n\pm 2} \cdot du
#### Integral of sine
• \displaystyle \int \sin u \cdot du = - \cos u + \text{C}
#### Integral of cosine
• \displaystyle \int \cos u \cdot du = \sin u + \text{C}
#### Integral of tangent
• \displaystyle \int \tan u \cdot du = \ln |\sec u| + \text{C} = -\ln |\cos u| + \text{C}
#### Integral of cotangent
• \displaystyle \int \cot u \cdot du = \ln |\sin u|+ \text{C}
#### Integral of secant
• \displaystyle \int \sec u \cdot du = \log\left(\sec u + \tan u\right)+ \text{C}
#### Integral of cosecant
• \displaystyle \int \csc u \cdot du = \ln |\csc u - \cot u|+ \text{C} = -\csc u \cdot \cot u + \text{C} = -\log\left( \cot u + \csc u\right)
#### Integral of secant by tangent
• \displaystyle \int \sec u \cdot \tan u \cdot du = \sec u + \text{C}
#### Integral of cosecant by cotangent
• \displaystyle \int \csc u \cdot \cot u \cdot du = - \csc u + \text{C}
## Several integrals
• \displaystyle \int \cfrac{du}{u^{2} + a^{2}} = \cfrac{1}{a} \cdot \tan^{-1} \cfrac{u}{a} + \text{C}
• \displaystyle \int \cfrac{du}{u^{2} - a^{2}} = \cfrac{1}{2 \cdot a} \cdot \ln \bigg | \cfrac{u - a}{u + a} \bigg | + \text{C}
• \displaystyle \int \cfrac{du}{a^{2} - u^{2}} = \cfrac{1}{2a} \cdot \ln \bigg |\cfrac{u + a}{u - a} \bigg | + \text{C}
• \displaystyle \int \cfrac{du}{\sqrt{a^{2} - u^{2}}} = \sin^{-1} \cfrac{u}{a} + \text{C}, \qquad a > 0
• \displaystyle \int \cfrac{du}{\sqrt{u^{2} - a^{2}}} = \ln \big ( u + \sqrt{u^{2} - a^{2}} \big ) + \text{C}
• \displaystyle \int \cfrac{du}{\sqrt{u^{2} + a^{2}}} = \ln \big ( u + \sqrt{u^{2} + a^{2}} \big ) + \text{C} = \sinh^{-1} \cfrac{u}{a} + \text{C}
• \displaystyle \int \cfrac{du}{u \cdot \sqrt{u^{2} - a^{2}}} = \cfrac{1}{a} \cdot \sec^{-1} \bigg | \cfrac{u}{a} \bigg | + \text{C}
• \displaystyle \int \cfrac{du}{u \cdot \sqrt{u^{2} + a^{2}}} = - \cfrac{1}{a} \cdot \ln \Bigg ( \cfrac{a + \sqrt{u^{2} + a^{2}}}{u} \Bigg ) + \text{C}
• \displaystyle \int \cfrac{du}{u \cdot \sqrt{a^{2} - u^{2}}} = - \cfrac{1}{a} \cdot \ln \Bigg ( \cfrac{a + \sqrt{a^{2} - u^{2}}}{u} \Bigg ) + \text{C}
## Integration formula by parts
• \displaystyle \int u \cdot dv = u \cdot v - \int v \cdot du
## Formulas of trigonometric integrals
#### Integral square sine
• \displaystyle \int \sin^{2}u \cdot du = \cfrac{1}{2} \cdot u - \cfrac{1}{4} \cdot \sin (2u) + \text{C} = \cfrac{1}{2} \cdot (u - \sin u \cdot \cos u) + \text{C}
#### Integral of square cosine
• \displaystyle \int \cos^{2}u \cdot du = \cfrac{1}{2} \cdot u + \cfrac{1}{4} \cdot \sin(2u) + \text{C} = \cfrac{1}{2} \cdot (u + \sin u \cdot \cos u)
#### Integral square tangent
• \displaystyle \int \tan^{2}u \cdot du = \tan u - u + \text{C}
#### Integral square cotangent
• \displaystyle \int \cot^{2}u \cdot du = - \cot u - u + \text{C}
#### Integral square secant
• \displaystyle \int \sec^{2}u \cdot du = \tan u + \text{C}
#### Integral square cosecant
• \displaystyle \int \csc^{2}u \cdot du = -\cot u + \text{C}
#### Integral cubic sinus
• \displaystyle \int \sin^{3}u \cdot du = - \cfrac{1}{3} \cdot (2 + \sin^{2} u) \cdot \cos u + \text{C}
#### Integral cubic cosine
• \displaystyle \int \cos^{3}u \cdot du = \cfrac{1}{3} \cdot (2 + \cos^{2} u) \cdot \sin u + \text{C}
#### Integral cubic tangent
• \displaystyle \int \tan^{3}u \cdot du = \cfrac{1}{2} \cdot \tan^{2}u + \ln |\cos u| + \text{C}
#### Integral of cubic cotangent
• \displaystyle \int \cot^{3}u \cdot du = - \cfrac{1}{2}\cdot \cot^{2}u - \ln |\sin u| + \text{C}
#### Integral cubic secant
• \displaystyle \int \sec^{3}u \cdot du = \cfrac{1}{2} \cdot \sec u \cdot \tan u + \cfrac{1}{2} \cdot \ln |\sec u + \tan u| + \text{C}
#### Integral cubic cosecant
• \displaystyle \int \csc^{3}u \cdot du = - \cfrac{1}{2} \cdot \csc u \cdot \cot u + \cfrac{1}{2} \cdot \ln |\csc u - \cot u| + \text{C}
## Several integrals
• \displaystyle \int \sin au \cdot \sin bu \cdot du = \cfrac{\sin(a - b)\cdot u}{2 \cdot (a - b)} - \cfrac{\sin(a + b) \cdot u}{2 \cdot (a+b)}+ \text{C}
• \displaystyle \int \cos au \cdot \cos bu \cdot du = \cfrac{\sin(a - b) \cdot u}{2 \cdot (a - b)} + \cfrac{\sin(a + b) \cdot u}{2 \cdot (a+b)} + \text{C}
• \displaystyle \int \sin au \cdot \cos bu \cdot du = - \cfrac{\cos(a - b) \cdot u}{2 \cdot (a - b)} - \cfrac{\cos(a + b) \cdot u}{2 \cdot (a + b)} + \text{C}
• \displaystyle \int u \cdot \sin u \cdot du = \sin u - u \cdot \cos u + \text{C}
• \displaystyle \int u \cdot \cos u \cdot du = \cos u + u \cdot \sin u + \text{C}
## Formulas of trigonometric reduction integrals
• \displaystyle \int \sin^{n}u \cdot du = - \cfrac{1}{n} \cdot \sin^{n - 1} u \cdot \cos u + \displaystyle \cfrac{n-1}{n} \cdot \int \sin^{n-2}u \cdot du + \text{C}
• \displaystyle \int \cos^{n}u \cdot du = \cfrac{1}{n} \cdot \cos^{n-1} u \cdot \sin u + \displaystyle \cfrac{n-1}{n} \cdot \int \cos^{n-2}u \cdot du + \text{C}
• \displaystyle \int \tan^{n}u \cdot du = \displaystyle \cfrac{1}{1-n} \cdot \tan^{n - 1}u - \int \tan^{n - 2} u \cdot du + \text{C}
• \displaystyle \int \cot^{n}u \cdot du = \displaystyle - \cfrac{1}{n - 1} \cdot \cos^{n - 1}u - \displaystyle \int \cot^{n-2}u \cdot du + \text{C}
• \displaystyle \int \sec^{n}u \cdot du = \displaystyle \cfrac{1}{n-1} \cdot \tan u \cdot \sec^{n - 2}u + \displaystyle \cfrac{n-2}{n-1} \cdot \int \sec^{n-2}u \cdot du + \text{C}
• \displaystyle \int \csc^{n}u \cdot du = \displaystyle - \cfrac{1}{n-1} \cdot \cot u \cdot \csc^{n - 2}u + \displaystyle \cfrac{n-2}{n-1} \cdot \int \csc^{n - 2}u \cdot du + \text{C}
• \displaystyle \int u^{n} \cdot \sin u \cdot du = \displaystyle - u^{n} \cdot \cos u + n \cdot \int u^{n - 1} \cdot \cos u \cdot du + \text{C}
• \displaystyle \int u^{n} \cdot \cos u \cdot du = \displaystyle u^{n} \cdot \sin u - n \cdot \int u^{n - 1} \cdot \sin u \cdot du + \text{C}
• \displaystyle \int \sin^{n}u \cdot \cos^{m}u \cdot du = \displaystyle - \cfrac{\sin^{n - 1}u \cdot \cos^{m + 1} u}{n + m} + \cfrac{n - 1}{n + m} \cdot \int \sin^{n-2}u \cdot \cos^{m}u \cdot du + \text{C} = \displaystyle \cfrac{\sin^{n + 1}u \cdot \cos^{m - 1} u}{n + m} + \cfrac{m-1}{n+m} \cdot \int \sin^{n}u \cdot \cos^{m-2}u \cdot du + \text{C}
## Formulas of inverse trigonometric integrals
• \displaystyle \int \sin^{-1}u \cdot du = u \cdot \sin^{-1}u + \sqrt{1 - u^{2}} + \text{C}
• \displaystyle \int \cos^{-1}u \cdot du = u \cdot \cos^{-1}u - \sqrt{1 - u^{2}} + \text{C}
• \displaystyle \int \tan^{-1}u \cdot du = u \cdot \tan^{-1}u - \cfrac{1}{2} \cdot \ln \big ( 1+u^{2} \big ) + \text{C}
• \displaystyle \int u \cdot \sin^{-1}u \cdot du = \cfrac{2 \cdot u^{2} - 1}{4} \cdot \sin^{-1} u + \cfrac{u \cdot \sqrt{1 - u^{2}}}{4} + \text{C}
• \displaystyle \int u \cdot \cos^{-1}u \cdot du = \cfrac{2 \cdot u^{2} - 1}{4} \cdot \cos^{-1} u - \cfrac{ u \cdot \sqrt{1-u^{2}}}{4} + \text{C}
• \displaystyle \int u \cdot \tan^{-1}u \cdot du = \cfrac{u^{2} + 1}{4} \cdot \tan^{-1}u - \cfrac{u}{2} + \text{C}
• \displaystyle \int u^{n} \cdot \sin^{-1} u \cdot du = \displaystyle \cfrac{1}{n + 1} \cdot \Bigg [ u^{n+1} \cdot \sin^{-1}u - \int \cfrac{u^{n + 1} \cdot du}{\sqrt{1 - u^{2}}} \Bigg ] + \text{C}, \quad n \neq -1
• \displaystyle \int \ u^{n} \cdot \cos^{-1}u \cdot du = \displaystyle \cfrac{1}{n + 1} \cdot \Bigg [ u^{n + 1} \cdot \cos^{-1}u + \int \cfrac{u^{n + 1} \cdot du}{ \sqrt{1 - u^{2}}} \Bigg ] + \text{C}, \quad n \neq -1
• \displaystyle \int u^{n} \cdot \tan^{-1}u \cdot du = \displaystyle \cfrac{1}{n + 1} \cdot \Bigg [ u^{n + 1} \cdot \tan^{-1}u - \int \cfrac{u^{n + 1} \cdot du}{1 + u^{2}} \Bigg ] + \text{C}, \quad n \neq -1
## Substantial substitutions of the integrals
• It has u = a \cdot x + b
\displaystyle \int F(a \cdot x + b) \cdot dx = \displaystyle \cfrac{1}{a} \cdot \int F(u) \cdot du + \text{C}
• It has u = \sqrt{a \cdot x + b}
\displaystyle \int F \big ( \sqrt{a \cdot x + b} \big ) \cdot dx = \displaystyle \cfrac{2}{a} \cdot \int u \cdot F(u) \cdot du + \text{C}
• It has u = (a \cdot x + b)^{1/n}
\displaystyle \int F \big ( (a \cdot x + b)^{1/n} \big ) \cdot dx = \displaystyle \cfrac{n}{a} \cdot \int u^{n-1} \cdot F(u) \cdot du + \text{C}
• It has u = a \cdot \sin u
\displaystyle \int F \big ( \sqrt{a^{2} - x^{2}} \big ) \cdot dx = \displaystyle a \cdot \int F(a \cdot \cos u) \cdot \cos u \cdot du + \text{C}
• It has u = a \cdot \tan u
\displaystyle \int F \big ( \sqrt{x^{2} + a^{2}} \big ) \cdot dx = \displaystyle a \cdot \int F(a \cdot \sec u) \cdot \sec^{2} u \cdot du + \text{C}
• It has u = a \cdot \sec u
\displaystyle \int \ F \big ( \sqrt{x^{2} - a^{2}} \big ) \cdot dx = \displaystyle a \cdot \int \ F(a \cdot \tan u) \cdot \sec u \cdot \tan u \cdot du + \text{C}
• It has u = e^{a \cdot x}
\displaystyle \int F(e^{a\cdot x}) \cdot dx = \displaystyle \cfrac{1}{a} \cdot \int \cfrac{F(u)}{u} \cdot du + \text{C}
• It has u = \ln x
\displaystyle \int F(\ln x) \cdot dx = \displaystyle \cfrac{1}{a} \cdot \int F(u) \cdot e^{u} \cdot du + \text{C}
• It has u = \sin^{-1} \cfrac{x}{a}
\displaystyle \int F \bigg ( \sin^{-1} \cfrac{x}{a} \bigg ) \cdot dx = \displaystyle a \cdot \int F(u) \cdot \cos u \cdot du + \text{C} | 4,229 | 9,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-26 | latest | en | 0.267328 |
https://cracku.in/52-john-takes-twice-as-much-time-as-jack-to-finish-a--x-cat-2020-slot-2?utm_source=blog&utm_medium=video&utm_campaign=video_solution | 1,708,658,796,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474360.86/warc/CC-MAIN-20240223021632-20240223051632-00023.warc.gz | 194,407,026 | 23,720 | Question 52
# John takes twice as much time as Jack to finish a job. Jack and Jim together take one-thirds of the time to finish the job than John takes working alone. Moreover, in order to finish the job, John takes three days more than that taken by three of them working together. In how many days will Jim finish the job working alone?
Solution
Let Jack take "t" days to complete the work, then John will take "2t" days to complete the work. So work done by Jack in one day is (1/t) and John is (1/2t) .
Now let Jim take "m" days to complete the work. According to question, $$\frac{1}{t}+\frac{1}{m}=\frac{3}{2t}\ or\ \frac{1}{m}=\frac{1}{2t\ }or\ m=2t$$ Hence Jim takes "2t" time to complete the work.
Now let the three of them complete the work in "p" days. Hence John takes "p+3" days to complete the work.
$$\frac{1}{2t}\left(m+3\right)=\left(\frac{4}{2t}\right)m$$
$$\frac{1}{2t}\left(m+3\right)=\left(\frac{4}{2t}\right)m$$
or m=1. Hence JIm will take (1+3)=4 days to complete the work. Similarly John will also take 4 days to complete the work | 331 | 1,063 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-10 | latest | en | 0.915888 |
https://muhaz.org/experiment-7-newtons-second-law-materials-v2.html | 1,652,748,532,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515466.5/warc/CC-MAIN-20220516235937-20220517025937-00763.warc.gz | 461,602,409 | 7,363 | # Experiment 7: Newton's Second Law Materials
Yüklə 92,57 Kb.
tarix 14.08.2018 ölçüsü 92,57 Kb. #70655
Experiment 7: Newton's Second Law
Experiment 7: Newton's Second Law
Materials
Dynamics Track with Feet and End Stop Super Pulley with Clamp PAScar Stopwatch Mass hanger and mass set Mass balance string (about 2 m)
Purpose
• In this experiment, you will verify Newton's Second Law, Fnet = ma.
• You will also investigate the direct relationship between force and acceleration and the indirect relationship between mass and acceleration.
Theory
According to Newton's Second Law, Fnet = ma, where Fnet is the net force acting on the object of mass m, and a is the resulting acceleration of the object.
For a cart of mass m1 on a horizontal track with a string attached over a pulley to a hanging mass m2 (see Figure 7.1), the net force Fnet on the entire system (cart and hanging mass) is the weight of hanging mass, Fg(masses) = m2g, (assuming that friction is negligible).
According to Newton's Second Law, this net force should be equal to ma, where m is the total mass that is being accelerated, which in this case is m1 + m2. You will check to see if m2g = (m1 + m2)a as predicted by theory.
To determine the acceleration, you will release the cart from rest and measure the time(t) for it to travel a certain distance (d). Since d = ½at2, the acceleration can be calculated using, a = 2d/t2.
Procedure I: The Effect of Mass on Acceleration
1. Install the feet on the track and level it.
1. Install the end stop on the track near one end with the magnets facing away from the track.
1. Measure the mass of the cart and record it in Table 7.1.
1. Attach the pulley and end stop to the track as shown in Figure 7.l. Place the cart on the track. Tie a string to the lower attachment point of the cart. Tie a mass hanger on the other end of the string. Run the string under the end stop and over the pulley. Adjust the pulley so that the string runs parallel to the track. The string must be just long enough so the mass hanger just strikes the floor before the cart reaches the end stop.
1. Pull the cart back until the mass hanger reaches the pulley. Record this initial release position in Table 7.1. This will be the release position for all the trials. Add 20 grams to the mass hanger and record the hanging mass in Table 7.1(do not forget the mass of the hanger).
1. Place the cart against the end stop on the pulley end of the track and record the final position of the cart in Table 7.1.
1. Pull the cart back to the initial release position. Release it and time how long it takes to reach the end stop. Record the time in Table 7.1.
1. Measure the time at least 5 times with the same mass and record these values in Table 7.1.
Table 7.1: Effect of Mass on Acceleration
Initial release position = cm Final position = cm Distance traveled (d) = cm Cart Mass (m1) (kg) Hanging Mass (m2) (kg) Time (s) Average Time (s) Trial 1 Trial 2 Trial 3 Trial 4 Trial 5
1. Add a 250 g mass to the cart and repeat the procedure. Continue this process until you have added a total of 1,000 g to the cart.
Procedure II: The Effect of Force on Acceleration
1. Start with 1,000g of mass in the cart, and then tape four 20g masses to the top. Record this value Table 7.2. Don’t forget the mass of the cart.
1. As before, pull the cart back to the initial release position. Release it and time how long it takes to reach the end stop. Record the time in Table 7.2.
1. Measure the time at least 5 times with the same mass and record these values in Table 7.2.
1. Remove a 20 g mass from the top of the cart and add it to the hanger and repeat the procedure. Continue this process until you have added a total of 80 g to the hanger. As before, do not forget to include the mass of the hanger.
Table 7.2: Effect of Force on Acceleration
Initial release position = cm Final position = cm Distance traveled (d) = cm Cart Mass (m1) (kg) Hanging Mass (m2) (kg) Time (s) Average Time Trial 1 Trial 2 Trial 3 Trial 4 Trial 5
Data Analysis
1. Calculate the average times and record them in Table 7.1 and Table 7.2.
2. Record the distance traveled (from initial to final position) in both tables.
3. Calculate the accelerations and record them Tables 7. 3 and 7.4.
4. For each case, calculate (m1 + m2)a and record in Tables 7.3 and 7.4.
5. For each case, calculate the net force, Fnet = m2g and record in Tables 7.3 and 7.4.
1. For each case, calculate the percent difference between Fnet and (m1 + m2)a and
record in Tables 7.3 and 7.4. Use Fnet as the accepted value.
* Note: For numbers 3 - 6, you must show a sample calculation, showing all work, including the formula and substitution with units for each step. You only need to do this for one data point.
Table 7.3: Effect of Mass on Acceleration
Hanger Mass = Cart Mass (kg) Acceleration (m/s2) (m1 + m2)a (N) Fnet = m2g (N) % Difference
Table 7.4: Effect of Force on Acceleration
Mass of Cart + Hanging Mass (m1 + m2) = Mass of Hanger (kg) Acceleration (m/s2) (m1 + m2)a (N) Fnet = m2g (N) % Difference
1. Plot the cart mass vs. acceleration on a graph using the data from Table 7.3.
• Based upon your graph, what relationship exists between cart mass and acceleration?
1. Plot Fnet vs acceleration on a graph using the data from Table 7.4.
• Determine the slope of your line and compare to (m1 + m2). Must show all work if doing by hand.
• Based upon your graph, what is the basic relationship that exists between force and acceleration?
1. Did the results of this experiment verify that F = ma?
1. Why must the mass in F = ma include the hanging mass as well as the mass of the cart?
Error Analysis & Conclusions
Yüklə 92,57 Kb.
Dostları ilə paylaş:
Verilənlər bazası müəlliflik hüququ ilə müdafiə olunur ©muhaz.org 2022
rəhbərliyinə müraciət | 1,571 | 5,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2022-21 | longest | en | 0.886332 |
https://www.physicsclassroom.com/mop/Refraction-and-Lenses/TIR-and-Critical-Angle/QG7help | 1,708,862,387,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474595.59/warc/CC-MAIN-20240225103506-20240225133506-00605.warc.gz | 938,384,986 | 26,885 | # Refraction and Lenses - Mission RL6 Detailed Help
Indices of refraction for various materials are shown in the table below. Use these values to rank the critical angles of the following boundaries in descending order. List the boundary with the greatest critical angle first; the smallest critical angle last.
Snell's Law of Refraction: The refraction of light towards or away from the normal follows a very predictable mathematical relationship known as Snell's law. n1 • sin Θ1 = n2 • sin Θ2 where n1 and n2 are the indices of refraction of the two individual media and Θ1 and Θ2 are the angles of incidence and refraction within those media. Knowing any three of the four quantities in the equation allows one to predict the fourth variable.
The most sure approach to this problem involves performing calculations of the critical angle for each of the four boundaries. Get organized and record the results in a legible manner. Then rank the four boundaries from the largest value to the smallest value. The only alternative involves doing the same as above, but simply finding the ratio of indices as opposed to the actual critical angle. The ranking in terms of critical angle will follow the same order as ranking the boundaries in terms of the n2/n1 ratio.
On most calculators, the sin-1 function can be used by pushing two buttons. If you've never used it, then look above the button labeled sin (for sine). You will likely see the sin-1label. Usually, the angle can be determined by pressing the 2nd-Sin button in consecutive fashion. On simpler, 1-line calculators, it is usually necessary that you first determine the ratio of n2/n1 and then click the sin-1 label. On more complex, multiline calculators, the 2nd-Sin buttons are first pressed and then the n2/n1 ratio is entered.
Tired of Ads?
Go ad-free for 1 year | 414 | 1,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-10 | latest | en | 0.842641 |
http://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/chapter-12-review-true-false-quiz-page-842/19 | 1,524,360,431,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945484.57/warc/CC-MAIN-20180422002521-20180422022521-00158.warc.gz | 432,191,795 | 12,565 | ## Calculus: Early Transcendentals 8th Edition
Consider $u=\lt1,1\gt$ and $v=\lt1, -1\gt$ Then $u.v=1(1)+1(-1)=0$ But neither $u$ nor $v$ are zero. Hence, the statement is false. | 72 | 179 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-17 | latest | en | 0.554404 |
http://www.dmnews.com/dataanalytics/the-failure-of-crm-mathematics/article/77353/ | 1,524,297,633,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945082.84/warc/CC-MAIN-20180421071203-20180421091203-00363.warc.gz | 412,166,314 | 24,326 | # The Failure of CRM Mathematics
In a previous article (DM News, March 18), I discussed why so many customer relationship management projects have failed. I listed the four assumptions underlying CRM and discussed three of them. This article covers the fourth: that CRM has a positive return on investment. Why is this seldom true?
Let's start with some basic principles. Customers do respond to communications, offers and promotions. For the communications to work the seller needs some sort of established business. He has a product or service to sell, and a working distribution channel. For CRM to be used, he has to have a revenue stream (otherwise, where would he get the money to build his warehouse?).
The reason for using CRM is to make incremental increases sales beyond what is being made with mass marketing, retail stores, catalogs and database marketing. Somehow, the marketer assumes, if he just knew more about the decision-making process in the customers' minds, he could increase sales (to customers and prospects) by as much as 5 percent or 10 percent. Let's say he is selling some product or service for \$150. It could be rental cars, small appliances or carpets. He contacts 20 million customers per year, with sales of \$900 million. His current marketing methods involve mass marketing and direct marketing to customer segments, including a loyalty program. Assume that of the \$150, his net profits a sale are 10 percent, after deducting all costs including marketing.
Assume a modest CRM warehouse containing data on 14 million prospects and past customers and 6 million current customers. Assume that you can build it for \$10 million including the software for access, with annual maintenance costs of \$1.5 million that includes the cost for the staff, NCOA and appended external data. Amortized over three years, the warehouse will cost \$5.5 million per year.
Compare this to the cost of database marketing in the same situation. It costs far less:
To these CRM or database costs, you must add the cost of communications with customers and prospects. After all, if you are going to make the right offer to the right customers at the right time, you have to communicate the offer to them personally, or what is the warehouse for? So assume that your firm has been collecting e-mails to keep the cost of communications down. The firm has e-mails on half of all its customers.
So with these costs in hand, what will be the result of using CRM or database marketing?
By using CRM or database marketing you have managed to increase the average sale per customer by 10 percent to \$165. You may have done this by getting some customers to place larger orders, by getting some to place more orders or by reducing the attrition rate, or a combination of these and other methods. Sales have increased by \$90 million with increased gross profits of \$9 million. But when you factor in the cost of CRM, net profits have gone down by \$580,000. That is what virtually all companies installing CRM have discovered. Note that these figures assume that CRM is working -- that it is increasing sales by 10 percent. But, in many cases, for reasons shown earlier, the CRM has no such valuable benefit. In the majority of companies, it makes only a very small improvement in the sales rate.
There is a fundamental disconnect between the goals of CRM and the results. If you have invested the millions and have a wonderful data warehouse and analytic software, then what? How do the data warehouse and software increase the retention rate? Somehow, you are supposed to take the software and change customer behavior. The CRM Forum (www.crm-forum.com) lists three reasons why CRM usually fails:
· Problems in organizational change and internal politics.
· Lack of the right skills and enterprisewide understanding of the initiatives.
· Poor initiative planning.
Unfortunately, many CRM analysts write a type of dense prose that is almost impossible for normal people to understand. Take this example from a Gartner report on the CRM Forum:
"Enterprises struggling with lack of coordination across departments should consider the needs of individual groups and work on a top-down/bottom-up strategy. Because department-level CRM deployments may be more cost-effectively supported by a Tier 2 system integrator with domain-specific skill sets and software relationships, the enterprise should identify a master list of approved smaller service providers to work on individual projects."
It really should not be so complicated. The goal of CRM is to select the right prospects and to get customers to be more loyal and buy more. A warehouse is of only marginal help in doing this. The goals can be achieved by coming up with products, services, tactics, strategies and communications that the customers will like and respond to. The CRM warehouse and software do none of these things. The goal can be realized by building a database, determining customer segments and designing a marketing program for each segment, using personal data from the database for each communication.
Many marketers are doing database marketing and calling it CRM. They do not have a multimillion-dollar warehouse. They are marketing to segments and making money doing it. That is great. But it is not CRM.
## Company of the Week
We recently were named B2B Magazine's Direct Marketing Agency of the Year, and with good reason: We make real, measureable, positive change happen for our clients. A full-service agency founded in 1974, Bader Rutter expertly helps you get the right message to the right audience at the right time through the right channels. As we engage our clients' audiences along their journey, direct marketing (email, direct mail, phone, SMS) and behavioral marketing (SEM, retargeting, contextual) channels deliver information relevant to the needs of each stage. We are experts at implementing and leveraging marketing technologies such as CRM and marketing automation in order to synchronize sales and marketing communications. Our team of architects and activators plan, execute, measure and adjust in real time to ensure the strategy is working as needed and change things if it's not.
Find out more here »
### Career Center
Check out hundreds of exciting professional opportunities available on DMN's Career Center.
Explore careers in digital marketing, sales, eCommerce, marketing communications, IT, data strategies, and much more. And don't forget to update your resume so employers can contact you privately about job opportunities. | 1,290 | 6,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-17 | latest | en | 0.960459 |
https://jp.mathworks.com/matlabcentral/answers/1748025-60k-lines-on-one-plot-too-slow | 1,660,470,772,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572021.17/warc/CC-MAIN-20220814083156-20220814113156-00570.warc.gz | 327,845,717 | 34,808 | # +60k Lines on one plot - Too slow
1 ビュー (過去 30 日間)
Ömer Yaman 2022 年 6 月 25 日
コメント済み: Ömer Yaman 2022 年 6 月 27 日
Dear all,
I need to plot more than 60k lines. I'm doing it in a for loop. It takes too much time, about 170 seconds with the rest of the code. Is there any quicker way?
And also saveas(fig ... command takes too much time as well. About 60 seconds. I would like to save my plot as a png. file.
What are your suggestions to reduce time cost.
Best Regards,
##### 4 件のコメント表示非表示 3 件の古いコメント
Ömer Yaman 2022 年 6 月 25 日
At that moment, I have my code on the computer that I have in my university. I'will share the plot asap from here.
Task is given to me is like that, I have about 60500 different points which has 6 different value for different time domain. I am supposed to visualize that the generated lines (for each point for 6 different increasing values, basicaly y=mx+n) has slopes that it should not be higher than a certain amount of mean slope value among all slopes and should not be lower too.
Calculations are correct but the time cost is too much. I'm seeking more optimized coding to save time.
サインインしてコメントする。
### 採用された回答
Adam Danz 2022 年 6 月 25 日
Another idea is to plot your 60k lines as a 2D density plot (initially suggested in this thread).
Pros:
• 1 graphics object, light weight plot
• simple to compute & simple to plot
• you can see the density patterns created by overlapping lines etc.
• If there is a strong median curve, you'll see it in the plot
Cons:
• Single curves won't appear in the plot (but you wouldn't see many single curves anyway with 60k lines)
• Outliers may become difficult or impossible to see
Example
Create 10k line series that follow a general pattern with varying amounts of noise. I'll use the same x-coordinates but this works just as well if the x-values vary too.
gaus = @(x,mu,sig,amp,vo)amp*exp(-(((x-mu).^2)/(2*sig.^2)))+vo;
x = 0:850;
n = 1e4;
y = zeros(n,numel(x));
rng('default') % to reproduce these results
for i = 1:n
mu = rand(1)*175+300;
sig = rand(1)*100+150;
amp = randn(1)*20+30;
vo = rand(1)*10;
y(i,:) = gaus(x,mu,sig,amp,vo);
end
Now, let's plot 50 random lines to see what they look like. Use randperm to do the random selection.
figure
m = 50;
selection = randperm(n,m);
plot(x, y(selection,:))
title(sprintf('%d random curves from %d',m,n))
axis tight
Plot the 2D density using histogram2()
% Replicate the X-vector (skip this if your x values vary for each curve)
xRep = repmat(x, n, 1);
% plot the bivariate historgram
figure
h = histogram2(xRep(:),y(:),'DisplayStyle','tile','ShowEmptyBins','on');
title('Bivariate density of curves')
cb = colorbar();
ylabel(cb,'Number of curves')
Alternatively, you could turn off the empty bins to see the shape of the entire dataset
figure
h = histogram2(xRep(:),y(:),'DisplayStyle','tile');
title('...ignoring empty bins')
cb = colorbar();
ylabel(cb,'Number of curves')
##### 2 件のコメント表示非表示 1 件の古いコメント
Ömer Yaman 2022 年 6 月 27 日
Thank you, I will go with this solution.
サインインしてコメントする。
### その他の回答 (2 件)
dpb 2022 年 6 月 25 日
This is way too much data to plot on single figure realistically, of course it's going to take time.
But, the use of a loop and plot is the slowest way possible -- you're computing/drawing every single line individually instead of using the power of MATLAB in being vectorized...
y=[slopes(:,1)*energy+slopes(i,2)].'; % generate y(i,j) by column
hL=plot(energy,y); % plot the result
Here on a not terribly upscale system,
>> tic,y=m*e;toc
Elapsed time is 0.001305 seconds.
>> tic,hL=plot(e,y);toc
Elapsed time is 15.350194 seconds.
>> delete(hL)
>> close
although it took the renderer quite some additional time to actually draw and display the figure -- and at least as long to delete 60K graphics handles.
Since there are only some 1-2K pixels on a monitor anyway, 60K points is some 60X the possible density to be able to display; I would strongly suggest decimating the y array by at least a factor of 10 and likely you'll be unable to see any difference in the result at factors approaching 100X.
##### 1 件のコメント表示非表示 なし
Ömer Yaman 2022 年 6 月 25 日
サインインしてコメントする。
Image Analyst 2022 年 6 月 25 日
Plot just a subsample of the curves. You'll never be able to see all 60k lines anyway, so just pick, say 1000 of them at random and plot those. You'll probably still get the general idea of what all the data would look like if you could have plotted them.
##### 1 件のコメント表示非表示 なし
Ömer Yaman 2022 年 6 月 27 日
Thank you for your help, I will keep the math but reduce the number of lines which is plotted.
サインインしてコメントする。
R2020a
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by | 1,350 | 4,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-33 | latest | en | 0.862664 |
http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=OCN | 1,524,284,082,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944982.32/warc/CC-MAIN-20180421032230-20180421052230-00582.warc.gz | 410,535,455 | 94,397 | Back to list of Stocks See Also: Seasonal Analysis of OCNGenetic Algorithms Stock Portfolio Generator, and Best Months to Buy/Sell Stocks
# Fourier Analysis of OCN (Ocwen Financial Corporation)
OCN (Ocwen Financial Corporation) appears to have interesting cyclic behaviour every 45 weeks (.9031*sine), 53 weeks (.7335*cosine), and 56 weeks (.6481*sine).
OCN (Ocwen Financial Corporation) has an average price of 12.22 (topmost row, frequency = 0).
Click on the checkboxes shown on the right to see how the various frequencies contribute to the graph. Look for large magnitude coefficients (sine or cosine), as these are associated with frequencies which contribute most to the associated stock plot. If you find a large magnitude coefficient which dramatically changes the graph, look at the associated "Period" in weeks, as you may have found a significant recurring cycle for the stock of interest.
## Fourier Analysis
Using data from 1/3/2000 to 4/16/2018 for OCN (Ocwen Financial Corporation), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
012.22427 0
1-2.91861 -8.19114 (1*2π)/955955 weeks
2-8.09069 .52614 (2*2π)/955478 weeks
31.25224 7.05695 (3*2π)/955318 weeks
44.66018 1.07791 (4*2π)/955239 weeks
5.20357 -3.74624 (5*2π)/955191 weeks
6-3.29055 1.01487 (6*2π)/955159 weeks
7-1.1229 .45641 (7*2π)/955136 weeks
82.13723 .10425 (8*2π)/955119 weeks
9-.01294 -.62887 (9*2π)/955106 weeks
10-.25013 -.03049 (10*2π)/95596 weeks
11-.2784 .32317 (11*2π)/95587 weeks
12.10013 .37601 (12*2π)/95580 weeks
13.5118 .35522 (13*2π)/95573 weeks
14.36076 -.01483 (14*2π)/95568 weeks
15-.14376 .16727 (15*2π)/95564 weeks
16.52657 .19452 (16*2π)/95560 weeks
17.32681 -.64808 (17*2π)/95556 weeks
18-.73352 .08257 (18*2π)/95553 weeks
19-.45965 .62823 (19*2π)/95550 weeks
20.33385 .46012 (20*2π)/95548 weeks
21.20954 -.90308 (21*2π)/95545 weeks
22-.51782 -.03611 (22*2π)/95543 weeks
23-.21356 .52513 (23*2π)/95542 weeks
24.24048 .14275 (24*2π)/95540 weeks
25.25216 -.02225 (25*2π)/95538 weeks
26-.06493 -.05184 (26*2π)/95537 weeks
27.0981 .03134 (27*2π)/95535 weeks
28-.16109 .07749 (28*2π)/95534 weeks
29.00854 .07683 (29*2π)/95533 weeks
30.03866 .05139 (30*2π)/95532 weeks
31-.11597 .07681 (31*2π)/95531 weeks
32.19069 .16396 (32*2π)/95530 weeks
33.22101 -.0835 (33*2π)/95529 weeks
34-.10865 -.1838 (34*2π)/95528 weeks
35-.27542 -.01249 (35*2π)/95527 weeks
36-.07675 .07561 (36*2π)/95527 weeks
37.23149 .10914 (37*2π)/95526 weeks
38.11853 .12973 (38*2π)/95525 weeks
39-.02186 -.1733 (39*2π)/95524 weeks
40-.23583 -.102 (40*2π)/95524 weeks
41-.1853 .13131 (41*2π)/95523 weeks
42.1218 .15873 (42*2π)/95523 weeks
43.08404 .00362 (43*2π)/95522 weeks
44-.17941 -.34758 (44*2π)/95522 weeks
45-.05757 .02787 (45*2π)/95521 weeks
46.01988 .04353 (46*2π)/95521 weeks
47.12469 .04421 (47*2π)/95520 weeks
48-.05112 .01631 (48*2π)/95520 weeks
49-.00868 -.10631 (49*2π)/95519 weeks
50.13658 .00547 (50*2π)/95519 weeks
51-.16211 .00206 (51*2π)/95519 weeks
52-.0203 .02541 (52*2π)/95518 weeks
53.02622 .0556 (53*2π)/95518 weeks
54.03137 -.20791 (54*2π)/95518 weeks
55.01053 .08994 (55*2π)/95517 weeks
56-.03588 -.02251 (56*2π)/95517 weeks
57-.03694 -.09175 (57*2π)/95517 weeks
58-.12551 .05907 (58*2π)/95516 weeks
59.21301 .04243 (59*2π)/95516 weeks
60.08282 -.14336 (60*2π)/95516 weeks
61-.23264 -.19485 (61*2π)/95516 weeks
62-.14918 .27691 (62*2π)/95515 weeks
63.25191 .07564 (63*2π)/95515 weeks
64.06954 -.01329 (64*2π)/95515 weeks
65-.13093 -.12503 (65*2π)/95515 weeks
66.09106 .20116 (66*2π)/95514 weeks
67.12179 -.07191 (67*2π)/95514 weeks
68-.00986 -.14728 (68*2π)/95514 weeks
69-.11933 -.06976 (69*2π)/95514 weeks
70.04331 .16118 (70*2π)/95514 weeks
71.1935 .01958 (71*2π)/95513 weeks
72.06253 -.09986 (72*2π)/95513 weeks
73.00476 -.14325 (73*2π)/95513 weeks
74-.14112 .06996 (74*2π)/95513 weeks
75-.00829 .19744 (75*2π)/95513 weeks
76.11439 .11546 (76*2π)/95513 weeks
77.08524 -.1351 (77*2π)/95512 weeks
78-.07554 -.19376 (78*2π)/95512 weeks
79-.20533 .15531 (79*2π)/95512 weeks
80-.00887 .10974 (80*2π)/95512 weeks
81.25858 .07641 (81*2π)/95512 weeks
82.02262 -.18694 (82*2π)/95512 weeks
83-.09296 -.0095 (83*2π)/95512 weeks
84-.16577 .11568 (84*2π)/95511 weeks
85.0571 .08607 (85*2π)/95511 weeks
86.16999 .03296 (86*2π)/95511 weeks
87-.00976 -.10928 (87*2π)/95511 weeks
88-.07218 -.06089 (88*2π)/95511 weeks
89-.02562 .02838 (89*2π)/95511 weeks
90.08343 -.04454 (90*2π)/95511 weeks
91-.05241 .07738 (91*2π)/95510 weeks
92.07082 .08345 (92*2π)/95510 weeks
93.07399 -.09054 (93*2π)/95510 weeks
94.00452 -.06532 (94*2π)/95510 weeks
95-.15563 .09568 (95*2π)/95510 weeks
96.00587 .11131 (96*2π)/95510 weeks
97.06558 .006 (97*2π)/95510 weeks
98-.07005 -.00035 (98*2π)/95510 weeks
99.02626 .00597 (99*2π)/95510 weeks
100.01133 .05203 (100*2π)/95510 weeks
101.0129 -.05991 (101*2π)/9559 weeks
102.03071 .01852 (102*2π)/9559 weeks
103.05664 .0603 (103*2π)/9559 weeks
104.01384 .00287 (104*2π)/9559 weeks
105.03122 -.0352 (105*2π)/9559 weeks
106-.03258 .0237 (106*2π)/9559 weeks
107.02292 .01808 (107*2π)/9559 weeks
108-.0179 .0238 (108*2π)/9559 weeks
109.05571 -.01991 (109*2π)/9559 weeks
110-.02626 -.00365 (110*2π)/9559 weeks
111.02687 .00657 (111*2π)/9559 weeks
112-.0023 .02048 (112*2π)/9559 weeks
113.03805 -.0464 (113*2π)/9558 weeks
114-.16133 .06619 (114*2π)/9558 weeks
115.01057 .1058 (115*2π)/9558 weeks
116.07202 .03715 (116*2π)/9558 weeks
117-.04847 -.03604 (117*2π)/9558 weeks
118-.05008 .08475 (118*2π)/9558 weeks
119.05371 -.01551 (119*2π)/9558 weeks
120.00737 -.03251 (120*2π)/9558 weeks
121-.03305 -.01631 (121*2π)/9558 weeks
122.03184 -.02173 (122*2π)/9558 weeks
123.00884 .01774 (123*2π)/9558 weeks
124-.03094 .01473 (124*2π)/9558 weeks
125.05271 .07511 (125*2π)/9558 weeks
126.03235 -.01425 (126*2π)/9558 weeks
127-.01917 -.02026 (127*2π)/9558 weeks
128-.06062 -.01011 (128*2π)/9557 weeks
129-.05349 .03483 (129*2π)/9557 weeks
130-.01633 -.01194 (130*2π)/9557 weeks
131-.04955 .04626 (131*2π)/9557 weeks
132.089 .03734 (132*2π)/9557 weeks
133.00138 -.03595 (133*2π)/9557 weeks
134-.06529 .03717 (134*2π)/9557 weeks
135-.01819 .06343 (135*2π)/9557 weeks
136.04161 -.01855 (136*2π)/9557 weeks
137-.0058 .00608 (137*2π)/9557 weeks
138-.01664 -.04041 (138*2π)/9557 weeks
139.0105 -.00662 (139*2π)/9557 weeks
140-.05948 -.01298 (140*2π)/9557 weeks
141.06838 .0103 (141*2π)/9557 weeks
142.0445 -.02539 (142*2π)/9557 weeks
143.01095 -.04738 (143*2π)/9557 weeks
144-.06456 .01052 (144*2π)/9557 weeks
145.02176 .01801 (145*2π)/9557 weeks
146.06832 .00694 (146*2π)/9557 weeks
147-.02918 -.02838 (147*2π)/9556 weeks
148-.11794 .02439 (148*2π)/9556 weeks
149-.04615 .0604 (149*2π)/9556 weeks
150.05843 -.00934 (150*2π)/9556 weeks
151.06928 -.04167 (151*2π)/9556 weeks
152-.0608 -.03143 (152*2π)/9556 weeks
153-.00667 -.02168 (153*2π)/9556 weeks
154.03203 .0448 (154*2π)/9556 weeks
155.12821 -.02225 (155*2π)/9556 weeks
156-.03614 -.06039 (156*2π)/9556 weeks
157-.04725 .00619 (157*2π)/9556 weeks
158-.02087 .06536 (158*2π)/9556 weeks
159.12361 .03943 (159*2π)/9556 weeks
160-.02904 -.05794 (160*2π)/9556 weeks
161-.03124 .00417 (161*2π)/9556 weeks
162-.02575 -.0561 (162*2π)/9556 weeks
163.01152 -.00986 (163*2π)/9556 weeks
164-.04433 -.02387 (164*2π)/9556 weeks
165.01524 .0753 (165*2π)/9556 weeks
166.08505 -.00312 (166*2π)/9556 weeks
167.03939 -.06917 (167*2π)/9556 weeks
168-.0233 -.02846 (168*2π)/9556 weeks
169-.04086 .07392 (169*2π)/9556 weeks
170.06612 .00914 (170*2π)/9556 weeks
171-.00956 -.0119 (171*2π)/9556 weeks
172-.00763 -.02315 (172*2π)/9556 weeks
173.02301 .07641 (173*2π)/9556 weeks
174.05676 -.00943 (174*2π)/9555 weeks
175.0304 -.06287 (175*2π)/9555 weeks
176-.01759 -.04812 (176*2π)/9555 weeks
177-.01602 .05441 (177*2π)/9555 weeks
178.06391 .07816 (178*2π)/9555 weeks
179.01658 -.04086 (179*2π)/9555 weeks
180-.01026 -.04572 (180*2π)/9555 weeks
181-.0564 .02856 (181*2π)/9555 weeks
182.02522 .03875 (182*2π)/9555 weeks
183.02971 -.00057 (183*2π)/9555 weeks
184-.00931 -.06886 (184*2π)/9555 weeks
185-.01174 .03567 (185*2π)/9555 weeks
186.0296 .05353 (186*2π)/9555 weeks
187.03909 -.02135 (187*2π)/9555 weeks
188-.02399 -.00416 (188*2π)/9555 weeks
189-.01802 .02765 (189*2π)/9555 weeks
190-.02705 .053 (190*2π)/9555 weeks
191.07254 .02956 (191*2π)/9555 weeks
192-.00569 -.04247 (192*2π)/9555 weeks
193-.04131 .0077 (193*2π)/9555 weeks
194-.00865 .07188 (194*2π)/9555 weeks
195.08654 -.00968 (195*2π)/9555 weeks
196.01667 -.06643 (196*2π)/9555 weeks
197-.09557 -.0461 (197*2π)/9555 weeks
198-.05157 .10266 (198*2π)/9555 weeks
199.07868 .00395 (199*2π)/9555 weeks
200.03706 -.06323 (200*2π)/9555 weeks
201-.05229 -.02756 (201*2π)/9555 weeks
202-.05404 .04829 (202*2π)/9555 weeks
203.06264 .05986 (203*2π)/9555 weeks
204.01019 -.04402 (204*2π)/9555 weeks
205-.01527 .01354 (205*2π)/9555 weeks
206-.00987 .03008 (206*2π)/9555 weeks
207-.00155 .01556 (207*2π)/9555 weeks
208.00398 -.02898 (208*2π)/9555 weeks
209-.03598 .04347 (209*2π)/9555 weeks
210.03 .05523 (210*2π)/9555 weeks
211.07025 -.03891 (211*2π)/9555 weeks
212-.01728 -.03991 (212*2π)/9555 weeks
213-.05089 -.00592 (213*2π)/9554 weeks
214.05722 .04417 (214*2π)/9554 weeks
215-.01686 -.07394 (215*2π)/9554 weeks
216-.03616 .01226 (216*2π)/9554 weeks
217-.00024 .07613 (217*2π)/9554 weeks
218.06704 -.05817 (218*2π)/9554 weeks
219-.02033 -.02164 (219*2π)/9554 weeks
220-.03056 .04464 (220*2π)/9554 weeks
221.02019 .0382 (221*2π)/9554 weeks
222-.00569 -.01765 (222*2π)/9554 weeks
223-.0324 .01457 (223*2π)/9554 weeks
224.04878 .04497 (224*2π)/9554 weeks
225-.0086 -.02419 (225*2π)/9554 weeks
226-.04483 -.02202 (226*2π)/9554 weeks
227-.01047 .05206 (227*2π)/9554 weeks
228.07727 .03105 (228*2π)/9554 weeks
229.03715 -.0593 (229*2π)/9554 weeks
230-.04865 -.0294 (230*2π)/9554 weeks
231-.04951 -.0063 (231*2π)/9554 weeks
232.01354 .03553 (232*2π)/9554 weeks
233.00301 -.00768 (233*2π)/9554 weeks
234.01412 .00706 (234*2π)/9554 weeks
235-.00735 -.01183 (235*2π)/9554 weeks
236-.00726 -.01015 (236*2π)/9554 weeks
237-.01269 -.01541 (237*2π)/9554 weeks
238-.03098 .05087 (238*2π)/9554 weeks
239.04136 .02814 (239*2π)/9554 weeks
240.00751 -.05144 (240*2π)/9554 weeks
241-.01622 -.02703 (241*2π)/9554 weeks
242-.05484 .05392 (242*2π)/9554 weeks
243.03694 .07286 (243*2π)/9554 weeks
244.02512 -.0418 (244*2π)/9554 weeks
245.01301 -.05596 (245*2π)/9554 weeks
246-.00595 .01968 (246*2π)/9554 weeks
247.01443 .01097 (247*2π)/9554 weeks
248-.00996 -.02377 (248*2π)/9554 weeks
249-.01599 -.0002 (249*2π)/9554 weeks
250-.0044 .03008 (250*2π)/9554 weeks
251.0387 .0676 (251*2π)/9554 weeks
252.02944 -.02692 (252*2π)/9554 weeks
253.00854 -.03634 (253*2π)/9554 weeks
254-.0433 -.03091 (254*2π)/9554 weeks
255-.02925 .068 (255*2π)/9554 weeks
256.0207 .03584 (256*2π)/9554 weeks
257.0251 -.03024 (257*2π)/9554 weeks
258-.05102 -.01101 (258*2π)/9554 weeks
259-.00253 .04781 (259*2π)/9554 weeks
260.06036 -.03109 (260*2π)/9554 weeks
261-.02201 -.02189 (261*2π)/9554 weeks
262-.02338 .02226 (262*2π)/9554 weeks
263.04947 .04794 (263*2π)/9554 weeks
264.05285 -.05099 (264*2π)/9554 weeks
265-.03147 .00774 (265*2π)/9554 weeks
266-.02718 .04416 (266*2π)/9554 weeks
267.0209 .02324 (267*2π)/9554 weeks
268.03035 -.04527 (268*2π)/9554 weeks
269-.02236 .00747 (269*2π)/9554 weeks
270-.0004 .03371 (270*2π)/9554 weeks
271-.00144 -.02244 (271*2π)/9554 weeks
272-.00646 .01071 (272*2π)/9554 weeks
273.02028 .00554 (273*2π)/9553 weeks
274.01993 -.02282 (274*2π)/9553 weeks
275-.06853 -.01201 (275*2π)/9553 weeks
276-.027 .06168 (276*2π)/9553 weeks
277.067 .04663 (277*2π)/9553 weeks
278.02039 -.04741 (278*2π)/9553 weeks
279-.01266 -.02295 (279*2π)/9553 weeks
280.0142 -.00903 (280*2π)/9553 weeks
281.02207 .00478 (281*2π)/9553 weeks
282.04123 -.00583 (282*2π)/9553 weeks
283.01365 .0182 (283*2π)/9553 weeks
284.02411 -.01478 (284*2π)/9553 weeks
285.01462 .00412 (285*2π)/9553 weeks
286-.03938 .03581 (286*2π)/9553 weeks
287.01292 .05118 (287*2π)/9553 weeks
288.0293 .00376 (288*2π)/9553 weeks
289.01321 -.02459 (289*2π)/9553 weeks
290-.01741 -.0325 (290*2π)/9553 weeks
291-.05104 .01555 (291*2π)/9553 weeks
292.02245 .03002 (292*2π)/9553 weeks
293.03746 .02241 (293*2π)/9553 weeks
294.01764 -.00707 (294*2π)/9553 weeks
295-.02228 .02017 (295*2π)/9553 weeks
296.04005 -.00097 (296*2π)/9553 weeks
297-.0154 -.01532 (297*2π)/9553 weeks
298-.0168 -.01093 (298*2π)/9553 weeks
299-.00486 .06026 (299*2π)/9553 weeks
300.04733 .01511 (300*2π)/9553 weeks
301-.00466 -.03289 (301*2π)/9553 weeks
302-.00881 -.00925 (302*2π)/9553 weeks
303-.03107 .02834 (303*2π)/9553 weeks
304.00924 .01786 (304*2π)/9553 weeks
305.01162 -.00328 (305*2π)/9553 weeks
306.04292 -.02439 (306*2π)/9553 weeks
307.00466 -.05268 (307*2π)/9553 weeks
308-.04849 .00446 (308*2π)/9553 weeks
309-.02613 .02002 (309*2π)/9553 weeks
310.04374 .04907 (310*2π)/9553 weeks
311.00669 -.00801 (311*2π)/9553 weeks
312.03174 -.01683 (312*2π)/9553 weeks
313-.0154 -.03755 (313*2π)/9553 weeks
314-.03881 -.00565 (314*2π)/9553 weeks
315-.01742 .03475 (315*2π)/9553 weeks
316.06568 .03078 (316*2π)/9553 weeks
317.02492 -.03214 (317*2π)/9553 weeks
318-.04128 .00504 (318*2π)/9553 weeks
319.00305 .03299 (319*2π)/9553 weeks
320.00869 .00862 (320*2π)/9553 weeks
321.06136 .00037 (321*2π)/9553 weeks
322-.01727 -.05913 (322*2π)/9553 weeks
323-.02488 -.03538 (323*2π)/9553 weeks
324-.0381 .0358 (324*2π)/9553 weeks
325.01032 .02249 (325*2π)/9553 weeks
326.05273 .01738 (326*2π)/9553 weeks
327-.00259 -.029 (327*2π)/9553 weeks
328-.03395 -.012 (328*2π)/9553 weeks
329.01176 .01763 (329*2π)/9553 weeks
330-.02286 -.01086 (330*2π)/9553 weeks
331-.00579 .04241 (331*2π)/9553 weeks
332.03734 .0034 (332*2π)/9553 weeks
333.02648 -.03824 (333*2π)/9553 weeks
334-.0478 -.01256 (334*2π)/9553 weeks
335-.04753 .05066 (335*2π)/9553 weeks
336.04052 .03267 (336*2π)/9553 weeks
337.03239 -.01043 (337*2π)/9553 weeks
338.01224 -.01288 (338*2π)/9553 weeks
339-.03775 -.02194 (339*2π)/9553 weeks
340-.01521 -.02256 (340*2π)/9553 weeks
341-.03414 .00838 (341*2π)/9553 weeks
342.03735 .04267 (342*2π)/9553 weeks
343.02146 -.02491 (343*2π)/9553 weeks
344-.0406 -.04516 (344*2π)/9553 weeks
345-.0112 .04918 (345*2π)/9553 weeks
346.02173 .01537 (346*2π)/9553 weeks
347.00566 -.02131 (347*2π)/9553 weeks
348-.03715 -.01052 (348*2π)/9553 weeks
349.03203 .00757 (349*2π)/9553 weeks
350.01605 -.00529 (350*2π)/9553 weeks
351-.02785 .00689 (351*2π)/9553 weeks
352.01531 .02947 (352*2π)/9553 weeks
353.01535 -.01003 (353*2π)/9553 weeks
354-.00393 .00404 (354*2π)/9553 weeks
355.01527 .01123 (355*2π)/9553 weeks
356.01961 -.02751 (356*2π)/9553 weeks
357-.00557 -.01893 (357*2π)/9553 weeks
358-.01047 .01289 (358*2π)/9553 weeks
359.03288 .00339 (359*2π)/9553 weeks
360-.00797 -.0204 (360*2π)/9553 weeks
361-.0365 .02982 (361*2π)/9553 weeks
362.03113 .02908 (362*2π)/9553 weeks
363.01786 .00189 (363*2π)/9553 weeks
364-.02228 -.00711 (364*2π)/9553 weeks
365.00075 .03747 (365*2π)/9553 weeks
366.00587 -.01556 (366*2π)/9553 weeks
367-.00758 -.00795 (367*2π)/9553 weeks
368-.0333 .00975 (368*2π)/9553 weeks
369.01956 .02246 (369*2π)/9553 weeks
370.01347 -.02837 (370*2π)/9553 weeks
371.00268 -.01867 (371*2π)/9553 weeks
372.0099 -.00294 (372*2π)/9553 weeks
373.02631 -.0024 (373*2π)/9553 weeks
374-.02177 -.00539 (374*2π)/9553 weeks
375.02487 .02089 (375*2π)/9553 weeks
376.00637 -.01005 (376*2π)/9553 weeks
377-.0336 -.00421 (377*2π)/9553 weeks
378-.01599 .03019 (378*2π)/9553 weeks
379.0262 .02404 (379*2π)/9553 weeks
380.01185 -.03999 (380*2π)/9553 weeks
381-.02717 .00989 (381*2π)/9553 weeks
382.01652 -.003 (382*2π)/9553 weeks
383.00494 .00137 (383*2π)/9552 weeks
384.00457 .01072 (384*2π)/9552 weeks
385-.00858 .01311 (385*2π)/9552 weeks
386.02578 -.00064 (386*2π)/9552 weeks
387-.01747 -.01923 (387*2π)/9552 weeks
388-.01402 .00848 (388*2π)/9552 weeks
389.02785 .01007 (389*2π)/9552 weeks
390.00892 -.02509 (390*2π)/9552 weeks
391-.01396 -.01688 (391*2π)/9552 weeks
392.02735 -.00121 (392*2π)/9552 weeks
393-.01537 -.00531 (393*2π)/9552 weeks
394-.00892 .05407 (394*2π)/9552 weeks
395.04247 .00971 (395*2π)/9552 weeks
396.00525 -.05418 (396*2π)/9552 weeks
397-.05218 .01431 (397*2π)/9552 weeks
398.00716 .07208 (398*2π)/9552 weeks
399.05668 .00132 (399*2π)/9552 weeks
400-.01325 -.06747 (400*2π)/9552 weeks
401-.0441 -.02547 (401*2π)/9552 weeks
402-.01049 .05816 (402*2π)/9552 weeks
403-.0053 .01089 (403*2π)/9552 weeks
404-.00863 .01186 (404*2π)/9552 weeks
405.01702 -.02225 (405*2π)/9552 weeks
406.02273 -.02283 (406*2π)/9552 weeks
407-.05114 .00439 (407*2π)/9552 weeks
408-.01426 .03115 (408*2π)/9552 weeks
409.03847 .01044 (409*2π)/9552 weeks
410.00574 -.04366 (410*2π)/9552 weeks
411.01618 -.00564 (411*2π)/9552 weeks
412-.01852 -.01517 (412*2π)/9552 weeks
413-.03464 .03199 (413*2π)/9552 weeks
414.04758 .03326 (414*2π)/9552 weeks
415.04737 -.02885 (415*2π)/9552 weeks
416-.03905 -.04759 (416*2π)/9552 weeks
417-.0517 .01375 (417*2π)/9552 weeks
418-.00648 .02614 (418*2π)/9552 weeks
419.01258 -.0147 (419*2π)/9552 weeks
420-.00833 .02813 (420*2π)/9552 weeks
421.02888 -.00458 (421*2π)/9552 weeks
422.02937 -.01535 (422*2π)/9552 weeks
423-.01749 -.02765 (423*2π)/9552 weeks
424-.01657 .0377 (424*2π)/9552 weeks
425.02062 .02948 (425*2π)/9552 weeks
426.04986 -.01267 (426*2π)/9552 weeks
427-.02679 -.04484 (427*2π)/9552 weeks
428-.02514 .02188 (428*2π)/9552 weeks
429-.00848 .00585 (429*2π)/9552 weeks
430.00711 .00999 (430*2π)/9552 weeks
431-.00139 .00978 (431*2π)/9552 weeks
432.03273 -.01789 (432*2π)/9552 weeks
433-.00465 -.0493 (433*2π)/9552 weeks
434-.04315 .01536 (434*2π)/9552 weeks
435-.00919 .01519 (435*2π)/9552 weeks
436.01236 .02723 (436*2π)/9552 weeks
437.0046 .00687 (437*2π)/9552 weeks
438.001 .00352 (438*2π)/9552 weeks
439-.01223 -.00132 (439*2π)/9552 weeks
440.01427 -.00707 (440*2π)/9552 weeks
441-.00041 .00276 (441*2π)/9552 weeks
442.01687 .01471 (442*2π)/9552 weeks
443.00965 -.01321 (443*2π)/9552 weeks
444.00269 -.01905 (444*2π)/9552 weeks
445-.03888 .02861 (445*2π)/9552 weeks
446.02038 .04599 (446*2π)/9552 weeks
447.04499 .01347 (447*2π)/9552 weeks
448-.00332 -.05664 (448*2π)/9552 weeks
449-.0606 -.01122 (449*2π)/9552 weeks
450-.02644 .05375 (450*2π)/9552 weeks
451.04919 .00776 (451*2π)/9552 weeks
452.00364 -.03987 (452*2π)/9552 weeks
453-.01392 -.0204 (453*2π)/9552 weeks
454-.00936 .01042 (454*2π)/9552 weeks
455.01051 .00619 (455*2π)/9552 weeks
456-.0028 .01645 (456*2π)/9552 weeks
457.00661 .03275 (457*2π)/9552 weeks
458.03105 -.00745 (458*2π)/9552 weeks
459-.02063 -.00511 (459*2π)/9552 weeks
460-.01513 .00911 (460*2π)/9552 weeks
461.0099 .02951 (461*2π)/9552 weeks
462.01051 -.03545 (462*2π)/9552 weeks
463.00278 -.00209 (463*2π)/9552 weeks
464.01408 .00791 (464*2π)/9552 weeks
465.02137 -.02006 (465*2π)/9552 weeks
466-.00553 -.01792 (466*2π)/9552 weeks
467-.00046 .00251 (467*2π)/9552 weeks
468.01153 -.00324 (468*2π)/9552 weeks
469-.00384 -.00591 (469*2π)/9552 weeks
470-.04703 -.00357 (470*2π)/9552 weeks
471-.02513 .0245 (471*2π)/9552 weeks
472.01499 .04507 (472*2π)/9552 weeks
473.02458 -.02704 (473*2π)/9552 weeks
474-.02022 -.02373 (474*2π)/9552 weeks
475-.01747 -.00353 (475*2π)/9552 weeks
476.00867 .03267 (476*2π)/9552 weeks
477.02147 -.00447 (477*2π)/9552 weeks
478.02147 .00447 (478*2π)/9552 weeks
479.00867 -.03267 (479*2π)/9552 weeks
480-.01747 .00353 (480*2π)/9552 weeks
481-.02022 .02373 (481*2π)/9552 weeks
482.02458 .02704 (482*2π)/9552 weeks
483.01499 -.04507 (483*2π)/9552 weeks
484-.02513 -.0245 (484*2π)/9552 weeks
485-.04703 .00357 (485*2π)/9552 weeks
486-.00384 .00591 (486*2π)/9552 weeks
487.01153 .00324 (487*2π)/9552 weeks
488-.00046 -.00251 (488*2π)/9552 weeks
489-.00553 .01792 (489*2π)/9552 weeks
490.02137 .02006 (490*2π)/9552 weeks
491.01408 -.00791 (491*2π)/9552 weeks
492.00278 .00209 (492*2π)/9552 weeks
493.01051 .03545 (493*2π)/9552 weeks
494.0099 -.02951 (494*2π)/9552 weeks
495-.01513 -.00911 (495*2π)/9552 weeks
496-.02063 .00511 (496*2π)/9552 weeks
497.03105 .00745 (497*2π)/9552 weeks
498.00661 -.03275 (498*2π)/9552 weeks
499-.0028 -.01645 (499*2π)/9552 weeks
500.01051 -.00619 (500*2π)/9552 weeks
501-.00936 -.01042 (501*2π)/9552 weeks
502-.01392 .0204 (502*2π)/9552 weeks
503.00364 .03987 (503*2π)/9552 weeks
504.04919 -.00776 (504*2π)/9552 weeks
505-.02644 -.05375 (505*2π)/9552 weeks
506-.0606 .01122 (506*2π)/9552 weeks
507-.00332 .05664 (507*2π)/9552 weeks
508.04499 -.01347 (508*2π)/9552 weeks
509.02038 -.04599 (509*2π)/9552 weeks
510-.03888 -.02861 (510*2π)/9552 weeks
511.00269 .01905 (511*2π)/9552 weeks
512.00965 .01321 (512*2π)/9552 weeks
513.01687 -.01471 (513*2π)/9552 weeks
514-.00041 -.00276 (514*2π)/9552 weeks
515.01427 .00707 (515*2π)/9552 weeks
516-.01223 .00132 (516*2π)/9552 weeks
517.001 -.00352 (517*2π)/9552 weeks
518.0046 -.00687 (518*2π)/9552 weeks
519.01236 -.02723 (519*2π)/9552 weeks
520-.00919 -.01519 (520*2π)/9552 weeks
521-.04315 -.01536 (521*2π)/9552 weeks
522-.00465 .0493 (522*2π)/9552 weeks
523.03273 .01789 (523*2π)/9552 weeks
524-.00139 -.00978 (524*2π)/9552 weeks
525.00711 -.00999 (525*2π)/9552 weeks
526-.00848 -.00585 (526*2π)/9552 weeks
527-.02514 -.02188 (527*2π)/9552 weeks
528-.02679 .04484 (528*2π)/9552 weeks
529.04986 .01267 (529*2π)/9552 weeks
530.02062 -.02948 (530*2π)/9552 weeks
531-.01657 -.0377 (531*2π)/9552 weeks
532-.01749 .02765 (532*2π)/9552 weeks
533.02937 .01535 (533*2π)/9552 weeks
534.02888 .00458 (534*2π)/9552 weeks
535-.00833 -.02813 (535*2π)/9552 weeks
536.01258 .0147 (536*2π)/9552 weeks
537-.00648 -.02614 (537*2π)/9552 weeks
538-.0517 -.01375 (538*2π)/9552 weeks
539-.03905 .04759 (539*2π)/9552 weeks
540.04737 .02885 (540*2π)/9552 weeks
541.04758 -.03326 (541*2π)/9552 weeks
542-.03464 -.03199 (542*2π)/9552 weeks
543-.01852 .01517 (543*2π)/9552 weeks
544.01618 .00564 (544*2π)/9552 weeks
545.00574 .04366 (545*2π)/9552 weeks
546.03847 -.01044 (546*2π)/9552 weeks
547-.01426 -.03115 (547*2π)/9552 weeks
548-.05114 -.00439 (548*2π)/9552 weeks
549.02273 .02283 (549*2π)/9552 weeks
550.01702 .02225 (550*2π)/9552 weeks
551-.00863 -.01186 (551*2π)/9552 weeks
552-.0053 -.01089 (552*2π)/9552 weeks
553-.01049 -.05816 (553*2π)/9552 weeks
554-.0441 .02547 (554*2π)/9552 weeks
555-.01325 .06747 (555*2π)/9552 weeks
556.05668 -.00132 (556*2π)/9552 weeks
557.00716 -.07208 (557*2π)/9552 weeks
558-.05218 -.01431 (558*2π)/9552 weeks
559.00525 .05418 (559*2π)/9552 weeks
560.04247 -.00971 (560*2π)/9552 weeks
561-.00892 -.05407 (561*2π)/9552 weeks
562-.01537 .00531 (562*2π)/9552 weeks
563.02735 .00121 (563*2π)/9552 weeks
564-.01396 .01688 (564*2π)/9552 weeks
565.00892 .02509 (565*2π)/9552 weeks
566.02785 -.01007 (566*2π)/9552 weeks
567-.01402 -.00848 (567*2π)/9552 weeks
568-.01747 .01923 (568*2π)/9552 weeks
569.02578 .00064 (569*2π)/9552 weeks
570-.00858 -.01311 (570*2π)/9552 weeks
571.00457 -.01072 (571*2π)/9552 weeks
572.00494 -.00137 (572*2π)/9552 weeks
573.01652 .003 (573*2π)/9552 weeks
574-.02717 -.00989 (574*2π)/9552 weeks
575.01185 .03999 (575*2π)/9552 weeks
576.0262 -.02404 (576*2π)/9552 weeks
577-.01599 -.03019 (577*2π)/9552 weeks
578-.0336 .00421 (578*2π)/9552 weeks
579.00637 .01005 (579*2π)/9552 weeks
580.02487 -.02089 (580*2π)/9552 weeks
581-.02177 .00539 (581*2π)/9552 weeks
582.02631 .0024 (582*2π)/9552 weeks
583.0099 .00294 (583*2π)/9552 weeks
584.00268 .01867 (584*2π)/9552 weeks
585.01347 .02837 (585*2π)/9552 weeks
586.01956 -.02246 (586*2π)/9552 weeks
587-.0333 -.00975 (587*2π)/9552 weeks
588-.00758 .00795 (588*2π)/9552 weeks
589.00587 .01556 (589*2π)/9552 weeks
590.00075 -.03747 (590*2π)/9552 weeks
591-.02228 .00711 (591*2π)/9552 weeks
592.01786 -.00189 (592*2π)/9552 weeks
593.03113 -.02908 (593*2π)/9552 weeks
594-.0365 -.02982 (594*2π)/9552 weeks
595-.00797 .0204 (595*2π)/9552 weeks
596.03288 -.00339 (596*2π)/9552 weeks
597-.01047 -.01289 (597*2π)/9552 weeks
598-.00557 .01893 (598*2π)/9552 weeks
599.01961 .02751 (599*2π)/9552 weeks
600.01527 -.01123 (600*2π)/9552 weeks
601-.00393 -.00404 (601*2π)/9552 weeks
602.01535 .01003 (602*2π)/9552 weeks
603.01531 -.02947 (603*2π)/9552 weeks
604-.02785 -.00689 (604*2π)/9552 weeks
605.01605 .00529 (605*2π)/9552 weeks
606.03203 -.00757 (606*2π)/9552 weeks
607-.03715 .01052 (607*2π)/9552 weeks
608.00566 .02131 (608*2π)/9552 weeks
609.02173 -.01537 (609*2π)/9552 weeks
610-.0112 -.04918 (610*2π)/9552 weeks
611-.0406 .04516 (611*2π)/9552 weeks
612.02146 .02491 (612*2π)/9552 weeks
613.03735 -.04267 (613*2π)/9552 weeks
614-.03414 -.00838 (614*2π)/9552 weeks
615-.01521 .02256 (615*2π)/9552 weeks
616-.03775 .02194 (616*2π)/9552 weeks
617.01224 .01288 (617*2π)/9552 weeks
618.03239 .01043 (618*2π)/9552 weeks
619.04052 -.03267 (619*2π)/9552 weeks
620-.04753 -.05066 (620*2π)/9552 weeks
621-.0478 .01256 (621*2π)/9552 weeks
622.02648 .03824 (622*2π)/9552 weeks
623.03734 -.0034 (623*2π)/9552 weeks
624-.00579 -.04241 (624*2π)/9552 weeks
625-.02286 .01086 (625*2π)/9552 weeks
626.01176 -.01763 (626*2π)/9552 weeks
627-.03395 .012 (627*2π)/9552 weeks
628-.00259 .029 (628*2π)/9552 weeks
629.05273 -.01738 (629*2π)/9552 weeks
630.01032 -.02249 (630*2π)/9552 weeks
631-.0381 -.0358 (631*2π)/9552 weeks
632-.02488 .03538 (632*2π)/9552 weeks
633-.01727 .05913 (633*2π)/9552 weeks
634.06136 -.00037 (634*2π)/9552 weeks
635.00869 -.00862 (635*2π)/9552 weeks
636.00305 -.03299 (636*2π)/9552 weeks
637-.04128 -.00504 (637*2π)/9551 weeks
638.02492 .03214 (638*2π)/9551 weeks
639.06568 -.03078 (639*2π)/9551 weeks
640-.01742 -.03475 (640*2π)/9551 weeks
641-.03881 .00565 (641*2π)/9551 weeks
642-.0154 .03755 (642*2π)/9551 weeks
643.03174 .01683 (643*2π)/9551 weeks
644.00669 .00801 (644*2π)/9551 weeks
645.04374 -.04907 (645*2π)/9551 weeks
646-.02613 -.02002 (646*2π)/9551 weeks
647-.04849 -.00446 (647*2π)/9551 weeks
648.00466 .05268 (648*2π)/9551 weeks
649.04292 .02439 (649*2π)/9551 weeks
650.01162 .00328 (650*2π)/9551 weeks
651.00924 -.01786 (651*2π)/9551 weeks
652-.03107 -.02834 (652*2π)/9551 weeks
653-.00881 .00925 (653*2π)/9551 weeks
654-.00466 .03289 (654*2π)/9551 weeks
655.04733 -.01511 (655*2π)/9551 weeks
656-.00486 -.06026 (656*2π)/9551 weeks
657-.0168 .01093 (657*2π)/9551 weeks
658-.0154 .01532 (658*2π)/9551 weeks
659.04005 .00097 (659*2π)/9551 weeks
660-.02228 -.02017 (660*2π)/9551 weeks
661.01764 .00707 (661*2π)/9551 weeks
662.03746 -.02241 (662*2π)/9551 weeks
663.02245 -.03002 (663*2π)/9551 weeks
664-.05104 -.01555 (664*2π)/9551 weeks
665-.01741 .0325 (665*2π)/9551 weeks
666.01321 .02459 (666*2π)/9551 weeks
667.0293 -.00376 (667*2π)/9551 weeks
668.01292 -.05118 (668*2π)/9551 weeks
669-.03938 -.03581 (669*2π)/9551 weeks
670.01462 -.00412 (670*2π)/9551 weeks
671.02411 .01478 (671*2π)/9551 weeks
672.01365 -.0182 (672*2π)/9551 weeks
673.04123 .00583 (673*2π)/9551 weeks
674.02207 -.00478 (674*2π)/9551 weeks
675.0142 .00903 (675*2π)/9551 weeks
676-.01266 .02295 (676*2π)/9551 weeks
677.02039 .04741 (677*2π)/9551 weeks
678.067 -.04663 (678*2π)/9551 weeks
679-.027 -.06168 (679*2π)/9551 weeks
680-.06853 .01201 (680*2π)/9551 weeks
681.01993 .02282 (681*2π)/9551 weeks
682.02028 -.00554 (682*2π)/9551 weeks
683-.00646 -.01071 (683*2π)/9551 weeks
684-.00144 .02244 (684*2π)/9551 weeks
685-.0004 -.03371 (685*2π)/9551 weeks
686-.02236 -.00747 (686*2π)/9551 weeks
687.03035 .04527 (687*2π)/9551 weeks
688.0209 -.02324 (688*2π)/9551 weeks
689-.02718 -.04416 (689*2π)/9551 weeks
690-.03147 -.00774 (690*2π)/9551 weeks
691.05285 .05099 (691*2π)/9551 weeks
692.04947 -.04794 (692*2π)/9551 weeks
693-.02338 -.02226 (693*2π)/9551 weeks
694-.02201 .02189 (694*2π)/9551 weeks
695.06036 .03109 (695*2π)/9551 weeks
696-.00253 -.04781 (696*2π)/9551 weeks
697-.05102 .01101 (697*2π)/9551 weeks
698.0251 .03024 (698*2π)/9551 weeks
699.0207 -.03584 (699*2π)/9551 weeks
700-.02925 -.068 (700*2π)/9551 weeks
701-.0433 .03091 (701*2π)/9551 weeks
702.00854 .03634 (702*2π)/9551 weeks
703.02944 .02692 (703*2π)/9551 weeks
704.0387 -.0676 (704*2π)/9551 weeks
705-.0044 -.03008 (705*2π)/9551 weeks
706-.01599 .0002 (706*2π)/9551 weeks
707-.00996 .02377 (707*2π)/9551 weeks
708.01443 -.01097 (708*2π)/9551 weeks
709-.00595 -.01968 (709*2π)/9551 weeks
710.01301 .05596 (710*2π)/9551 weeks
711.02512 .0418 (711*2π)/9551 weeks
712.03694 -.07286 (712*2π)/9551 weeks
713-.05484 -.05392 (713*2π)/9551 weeks
714-.01622 .02703 (714*2π)/9551 weeks
715.00751 .05144 (715*2π)/9551 weeks
716.04136 -.02814 (716*2π)/9551 weeks
717-.03098 -.05087 (717*2π)/9551 weeks
718-.01269 .01541 (718*2π)/9551 weeks
719-.00726 .01015 (719*2π)/9551 weeks
720-.00735 .01183 (720*2π)/9551 weeks
721.01412 -.00706 (721*2π)/9551 weeks
722.00301 .00768 (722*2π)/9551 weeks
723.01354 -.03553 (723*2π)/9551 weeks
724-.04951 .0063 (724*2π)/9551 weeks
725-.04865 .0294 (725*2π)/9551 weeks
726.03715 .0593 (726*2π)/9551 weeks
727.07727 -.03105 (727*2π)/9551 weeks
728-.01047 -.05206 (728*2π)/9551 weeks
729-.04483 .02202 (729*2π)/9551 weeks
730-.0086 .02419 (730*2π)/9551 weeks
731.04878 -.04497 (731*2π)/9551 weeks
732-.0324 -.01457 (732*2π)/9551 weeks
733-.00569 .01765 (733*2π)/9551 weeks
734.02019 -.0382 (734*2π)/9551 weeks
735-.03056 -.04464 (735*2π)/9551 weeks
736-.02033 .02164 (736*2π)/9551 weeks
737.06704 .05817 (737*2π)/9551 weeks
738-.00024 -.07613 (738*2π)/9551 weeks
739-.03616 -.01226 (739*2π)/9551 weeks
740-.01686 .07394 (740*2π)/9551 weeks
741.05722 -.04417 (741*2π)/9551 weeks
742-.05089 .00592 (742*2π)/9551 weeks
743-.01728 .03991 (743*2π)/9551 weeks
744.07025 .03891 (744*2π)/9551 weeks
745.03 -.05523 (745*2π)/9551 weeks
746-.03598 -.04347 (746*2π)/9551 weeks
747.00398 .02898 (747*2π)/9551 weeks
748-.00155 -.01556 (748*2π)/9551 weeks
749-.00987 -.03008 (749*2π)/9551 weeks
750-.01527 -.01354 (750*2π)/9551 weeks
751.01019 .04402 (751*2π)/9551 weeks
752.06264 -.05986 (752*2π)/9551 weeks
753-.05404 -.04829 (753*2π)/9551 weeks
754-.05229 .02756 (754*2π)/9551 weeks
755.03706 .06323 (755*2π)/9551 weeks
756.07868 -.00395 (756*2π)/9551 weeks
757-.05157 -.10266 (757*2π)/9551 weeks
758-.09557 .0461 (758*2π)/9551 weeks
759.01667 .06643 (759*2π)/9551 weeks
760.08654 .00968 (760*2π)/9551 weeks
761-.00865 -.07188 (761*2π)/9551 weeks
762-.04131 -.0077 (762*2π)/9551 weeks
763-.00569 .04247 (763*2π)/9551 weeks
764.07254 -.02956 (764*2π)/9551 weeks
765-.02705 -.053 (765*2π)/9551 weeks
766-.01802 -.02765 (766*2π)/9551 weeks
767-.02399 .00416 (767*2π)/9551 weeks
768.03909 .02135 (768*2π)/9551 weeks
769.0296 -.05353 (769*2π)/9551 weeks
770-.01174 -.03567 (770*2π)/9551 weeks
771-.00931 .06886 (771*2π)/9551 weeks
772.02971 .00057 (772*2π)/9551 weeks
773.02522 -.03875 (773*2π)/9551 weeks
774-.0564 -.02856 (774*2π)/9551 weeks
775-.01026 .04572 (775*2π)/9551 weeks
776.01658 .04086 (776*2π)/9551 weeks
777.06391 -.07816 (777*2π)/9551 weeks
778-.01602 -.05441 (778*2π)/9551 weeks
779-.01759 .04812 (779*2π)/9551 weeks
780.0304 .06287 (780*2π)/9551 weeks
781.05676 .00943 (781*2π)/9551 weeks
782.02301 -.07641 (782*2π)/9551 weeks
783-.00763 .02315 (783*2π)/9551 weeks
784-.00956 .0119 (784*2π)/9551 weeks
785.06612 -.00914 (785*2π)/9551 weeks
786-.04086 -.07392 (786*2π)/9551 weeks
787-.0233 .02846 (787*2π)/9551 weeks
788.03939 .06917 (788*2π)/9551 weeks
789.08505 .00312 (789*2π)/9551 weeks
790.01524 -.0753 (790*2π)/9551 weeks
791-.04433 .02387 (791*2π)/9551 weeks
792.01152 .00986 (792*2π)/9551 weeks
793-.02575 .0561 (793*2π)/9551 weeks
794-.03124 -.00417 (794*2π)/9551 weeks
795-.02904 .05794 (795*2π)/9551 weeks
796.12361 -.03943 (796*2π)/9551 weeks
797-.02087 -.06536 (797*2π)/9551 weeks
798-.04725 -.00619 (798*2π)/9551 weeks
799-.03614 .06039 (799*2π)/9551 weeks
800.12821 .02225 (800*2π)/9551 weeks
801.03203 -.0448 (801*2π)/9551 weeks
802-.00667 .02168 (802*2π)/9551 weeks
803-.0608 .03143 (803*2π)/9551 weeks
804.06928 .04167 (804*2π)/9551 weeks
805.05843 .00934 (805*2π)/9551 weeks
806-.04615 -.0604 (806*2π)/9551 weeks
807-.11794 -.02439 (807*2π)/9551 weeks
808-.02918 .02838 (808*2π)/9551 weeks
809.06832 -.00694 (809*2π)/9551 weeks
810.02176 -.01801 (810*2π)/9551 weeks
811-.06456 -.01052 (811*2π)/9551 weeks
812.01095 .04738 (812*2π)/9551 weeks
813.0445 .02539 (813*2π)/9551 weeks
814.06838 -.0103 (814*2π)/9551 weeks
815-.05948 .01298 (815*2π)/9551 weeks
816.0105 .00662 (816*2π)/9551 weeks
817-.01664 .04041 (817*2π)/9551 weeks
818-.0058 -.00608 (818*2π)/9551 weeks
819.04161 .01855 (819*2π)/9551 weeks
820-.01819 -.06343 (820*2π)/9551 weeks
821-.06529 -.03717 (821*2π)/9551 weeks
822.00138 .03595 (822*2π)/9551 weeks
823.089 -.03734 (823*2π)/9551 weeks
824-.04955 -.04626 (824*2π)/9551 weeks
825-.01633 .01194 (825*2π)/9551 weeks
826-.05349 -.03483 (826*2π)/9551 weeks
827-.06062 .01011 (827*2π)/9551 weeks
828-.01917 .02026 (828*2π)/9551 weeks
829.03235 .01425 (829*2π)/9551 weeks
830.05271 -.07511 (830*2π)/9551 weeks
831-.03094 -.01473 (831*2π)/9551 weeks
832.00884 -.01774 (832*2π)/9551 weeks
833.03184 .02173 (833*2π)/9551 weeks
834-.03305 .01631 (834*2π)/9551 weeks
835.00737 .03251 (835*2π)/9551 weeks
836.05371 .01551 (836*2π)/9551 weeks
837-.05008 -.08475 (837*2π)/9551 weeks
838-.04847 .03604 (838*2π)/9551 weeks
839.07202 -.03715 (839*2π)/9551 weeks
840.01057 -.1058 (840*2π)/9551 weeks
841-.16133 -.06619 (841*2π)/9551 weeks
842.03805 .0464 (842*2π)/9551 weeks
843-.0023 -.02048 (843*2π)/9551 weeks
844.02687 -.00657 (844*2π)/9551 weeks
845-.02626 .00365 (845*2π)/9551 weeks
846.05571 .01991 (846*2π)/9551 weeks
847-.0179 -.0238 (847*2π)/9551 weeks
848.02292 -.01808 (848*2π)/9551 weeks
849-.03258 -.0237 (849*2π)/9551 weeks
850.03122 .0352 (850*2π)/9551 weeks
851.01384 -.00287 (851*2π)/9551 weeks
852.05664 -.0603 (852*2π)/9551 weeks
853.03071 -.01852 (853*2π)/9551 weeks
854.0129 .05991 (854*2π)/9551 weeks
855.01133 -.05203 (855*2π)/9551 weeks
856.02626 -.00597 (856*2π)/9551 weeks
857-.07005 .00035 (857*2π)/9551 weeks
858.06558 -.006 (858*2π)/9551 weeks
859.00587 -.11131 (859*2π)/9551 weeks
860-.15563 -.09568 (860*2π)/9551 weeks
861.00452 .06532 (861*2π)/9551 weeks
862.07399 .09054 (862*2π)/9551 weeks
863.07082 -.08345 (863*2π)/9551 weeks
864-.05241 -.07738 (864*2π)/9551 weeks
865.08343 .04454 (865*2π)/9551 weeks
866-.02562 -.02838 (866*2π)/9551 weeks
867-.07218 .06089 (867*2π)/9551 weeks
868-.00976 .10928 (868*2π)/9551 weeks
869.16999 -.03296 (869*2π)/9551 weeks
870.0571 -.08607 (870*2π)/9551 weeks
871-.16577 -.11568 (871*2π)/9551 weeks
872-.09296 .0095 (872*2π)/9551 weeks
873.02262 .18694 (873*2π)/9551 weeks
874.25858 -.07641 (874*2π)/9551 weeks
875-.00887 -.10974 (875*2π)/9551 weeks
876-.20533 -.15531 (876*2π)/9551 weeks
877-.07554 .19376 (877*2π)/9551 weeks
878.08524 .1351 (878*2π)/9551 weeks
879.11439 -.11546 (879*2π)/9551 weeks
880-.00829 -.19744 (880*2π)/9551 weeks
881-.14112 -.06996 (881*2π)/9551 weeks
882.00476 .14325 (882*2π)/9551 weeks
883.06253 .09986 (883*2π)/9551 weeks
884.1935 -.01958 (884*2π)/9551 weeks
885.04331 -.16118 (885*2π)/9551 weeks
886-.11933 .06976 (886*2π)/9551 weeks
887-.00986 .14728 (887*2π)/9551 weeks
888.12179 .07191 (888*2π)/9551 weeks
889.09106 -.20116 (889*2π)/9551 weeks
890-.13093 .12503 (890*2π)/9551 weeks
891.06954 .01329 (891*2π)/9551 weeks
892.25191 -.07564 (892*2π)/9551 weeks
893-.14918 -.27691 (893*2π)/9551 weeks
894-.23264 .19485 (894*2π)/9551 weeks
895.08282 .14336 (895*2π)/9551 weeks
896.21301 -.04243 (896*2π)/9551 weeks
897-.12551 -.05907 (897*2π)/9551 weeks
898-.03694 .09175 (898*2π)/9551 weeks
899-.03588 .02251 (899*2π)/9551 weeks
900.01053 -.08994 (900*2π)/9551 weeks
901.03137 .20791 (901*2π)/9551 weeks
902.02622 -.0556 (902*2π)/9551 weeks
903-.0203 -.02541 (903*2π)/9551 weeks
904-.16211 -.00206 (904*2π)/9551 weeks
905.13658 -.00547 (905*2π)/9551 weeks
906-.00868 .10631 (906*2π)/9551 weeks
907-.05112 -.01631 (907*2π)/9551 weeks
908.12469 -.04421 (908*2π)/9551 weeks
909.01988 -.04353 (909*2π)/9551 weeks
910-.05757 -.02787 (910*2π)/9551 weeks
911-.17941 .34758 (911*2π)/9551 weeks
912.08404 -.00362 (912*2π)/9551 weeks
913.1218 -.15873 (913*2π)/9551 weeks
914-.1853 -.13131 (914*2π)/9551 weeks
915-.23583 .102 (915*2π)/9551 weeks
916-.02186 .1733 (916*2π)/9551 weeks
917.11853 -.12973 (917*2π)/9551 weeks
918.23149 -.10914 (918*2π)/9551 weeks
919-.07675 -.07561 (919*2π)/9551 weeks
920-.27542 .01249 (920*2π)/9551 weeks
921-.10865 .1838 (921*2π)/9551 weeks
922.22101 .0835 (922*2π)/9551 weeks
923.19069 -.16396 (923*2π)/9551 weeks
924-.11597 -.07681 (924*2π)/9551 weeks
925.03866 -.05139 (925*2π)/9551 weeks
926.00854 -.07683 (926*2π)/9551 weeks
927-.16109 -.07749 (927*2π)/9551 weeks
928.0981 -.03134 (928*2π)/9551 weeks
929-.06493 .05184 (929*2π)/9551 weeks
930.25216 .02225 (930*2π)/9551 weeks
931.24048 -.14275 (931*2π)/9551 weeks
932-.21356 -.52513 (932*2π)/9551 weeks
933-.51782 .03611 (933*2π)/9551 weeks
934.20954 .90308 (934*2π)/9551 weeks
935.33385 -.46012 (935*2π)/9551 weeks
936-.45965 -.62823 (936*2π)/9551 weeks
937-.73352 -.08257 (937*2π)/9551 weeks
938.32681 .64808 (938*2π)/9551 weeks
939.52657 -.19452 (939*2π)/9551 weeks
940-.14376 -.16727 (940*2π)/9551 weeks
941.36076 .01483 (941*2π)/9551 weeks
942.5118 -.35522 (942*2π)/9551 weeks
943.10013 -.37601 (943*2π)/9551 weeks
944-.2784 -.32317 (944*2π)/9551 weeks
945-.25013 .03049 (945*2π)/9551 weeks
946-.01294 .62887 (946*2π)/9551 weeks
9472.13723 -.10425 (947*2π)/9551 weeks
948-1.1229 -.45641 (948*2π)/9551 weeks
949-3.29055 -1.01487 (949*2π)/9551 weeks
950.20357 3.74624 (950*2π)/9551 weeks
9514.66018 -1.07791 (951*2π)/9551 weeks
9521.25224 -7.05695 (952*2π)/9551 weeks
953-8.09069 -.52614 (953*2π)/9551 weeks | 16,480 | 37,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-17 | latest | en | 0.784522 |
https://www.urionlinejudge.com.br/judge/en/profile/404203 | 1,623,810,441,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621699.22/warc/CC-MAIN-20210616001810-20210616031810-00558.warc.gz | 927,152,020 | 6,746 | # PROFILE
Check out all the problems this user has already solved.
Problem Problem Name Ranking Submission Language Runtime Submission Date
1134 Type of Fuel 12025º 18691332 C 0.000 6/27/20, 11:10:00 AM
1151 Easy Fibonacci 00086º 18684489 JavaScript 0.092 6/26/20, 6:20:24 PM
1165 Prime Number 00160º 18682646 JavaScript 0.132 6/26/20, 3:05:41 PM
1164 Perfect Number 00112º 18682509 JavaScript 0.080 6/26/20, 2:52:10 PM
1156 S Sequence II 00049º 18682399 JavaScript 0.072 6/26/20, 2:41:34 PM
1155 S Sequence 00063º 18682349 JavaScript 0.096 6/26/20, 2:37:27 PM
1116 Dividing X by Y 00072º 18602456 JavaScript 0.260 6/19/20, 8:10:30 AM
1157 Divisors I 00023º 18596459 JavaScript 0.060 6/18/20, 5:47:56 PM
1074 Even or Odd 00135º 18595354 JavaScript 0.088 6/18/20, 4:14:29 PM
1079 Weighted Averages 00141º 18595134 JavaScript 0.072 6/18/20, 4:08:03 PM
1153 Simple Factorial 00031º 18594149 JavaScript 0.052 6/18/20, 2:52:33 PM
1132 Multiples of 13 00029º 18594077 JavaScript 0.056 6/18/20, 2:47:36 PM
1133 Rest of a Division 00045º 18593995 JavaScript 0.060 6/18/20, 2:40:59 PM
1040 Average 3 00183º 18591839 JavaScript 0.072 6/18/20, 11:06:02 AM
1072 Interval 2 00075º 18605736 JavaScript 0.052 6/17/20, 7:20:53 PM
1143 Squared and Cubic 00028º 18582762 JavaScript 0.060 6/17/20, 5:20:24 PM
1098 Sequence IJ 4 00042º 18581726 JavaScript 0.060 6/17/20, 3:53:32 PM
1097 Sequence IJ 3 00103º 18581409 JavaScript 0.084 6/17/20, 3:43:45 PM
1096 Sequence IJ 2 00032º 18581344 JavaScript 0.052 6/17/20, 3:41:25 PM
1095 Sequence IJ 1 00059º 18581302 JavaScript 0.060 6/17/20, 3:36:29 PM
1078 Multiplication Table 00097º 18569463 JavaScript 0.052 6/16/20, 7:22:58 PM
1073 Even Square 00124º 18569336 JavaScript 0.080 6/16/20, 7:16:40 PM
1071 Sum of Consecutive Odd... 00101º 18569097 JavaScript 0.080 6/16/20, 7:00:08 PM
1070 Six Odd Numbers 00213º 18568966 JavaScript 0.084 6/16/20, 6:56:38 PM
1075 Remaining 2 00061º 18568649 JavaScript 0.060 6/16/20, 6:35:38 PM
1067 Odd Numbers 00123º 18568593 JavaScript 0.060 6/16/20, 6:32:47 PM
1066 Even, Odd, Positive and... 00202º 18691291 JavaScript 0.092 6/16/20, 4:12:09 PM
1065 Even Between five Numbers 00078º 18565814 JavaScript 0.060 6/16/20, 4:03:53 PM
1060 Positive Numbers 00198º 18565423 JavaScript 0.068 6/16/20, 3:36:37 PM
1061 Event Time 08720º 18565144 C++ 0.000 6/16/20, 3:17:43 PM
1 of 3 | 1,036 | 2,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-25 | latest | en | 0.304497 |
https://www.mersenneforum.org/showthread.php?s=60ba68c34917dd55eee2df1c89e7d471&t=26477 | 1,642,937,623,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304261.85/warc/CC-MAIN-20220123111431-20220123141431-00645.warc.gz | 950,819,913 | 12,243 | mersenneforum.org Searching Mersenne Primes with Mathematica
Register FAQ Search Today's Posts Mark Forums Read
2021-02-09, 12:55 #1 mattprim Feb 2021 Salt Lake City, UT 111012 Posts Searching Mersenne Primes with Mathematica I independently confirmed two Mersenne non-primes below 1GHz x 1 min https://www.mersenne.ca/exponent/172279571 and https://www.mersenne.ca/exponent/172279589 with the following Mathematica code: Do[a = RandomInteger[100000]; b = PrimeQ[a*2*172279571 + 1]; If[b == True, Print[b, " ", a, " ", IntegerQ[(2^(172279571) - 1)/(a*2*172279571 + 1)]]], {i, 1, 10000}];
2021-02-09, 13:37 #2
paulunderwood
Sep 2002
Database er0rr
3×1,327 Posts
Quote:
Originally Posted by mattprim I independently confirmed two Mersenne non-primes below 1GHz x 1 min https://www.mersenne.ca/exponent/172279571 and https://www.mersenne.ca/exponent/172279589 with the following Mathematica code: Do[a = RandomInteger[100000]; b = PrimeQ[a*2*172279571 + 1]; If[b == True, Print[b, " ", a, " ", IntegerQ[(2^(172279571) - 1)/(a*2*172279571 + 1)]]], {i, 1, 10000}];
Code:
p=17227971;for(a=1,100000,if(Mod(2,2*a*p+1)^p==1,print([p,a])))
Pari/GP runs this in 1/7 of a second and gives a=25 for a factor.
Last fiddled with by paulunderwood on 2021-02-09 at 13:40
2021-02-09, 15:19 #3
mattprim
Feb 2021
Salt Lake City, UT
358 Posts
Quote:
Originally Posted by paulunderwood Code: p=172279571;for(a=1,100000,if(Mod(2,2*a*p+1)^p==1,print([p,a]))) Pari/GP runs this in 1/7 of a second and gives a=25 for a factor.
Sure, https://www.mersenne.ca/exponent/172279571 and my Pari/GP run:
? p=172279571
%9 = 172279571
? for(a=1,100000000,if(Mod(2,2*a*p+1)^p==1,print([p,a])))
[172279571, 25]
[172279571, 1138120]
I mean I did not know and did not check in the databases if they were primes and found them independently simply guessing them primes and then finding that those factors are small recovering the old result by random toss of a in a narrow belt:
Last fiddled with by mattprim on 2021-02-09 at 16:18
2021-02-09, 15:45 #4
paulunderwood
Sep 2002
Database er0rr
3×1,327 Posts
Quote:
Originally Posted by mattprim Sure, I mean I did not know and did not check in the databases if they were primes and found them independently simply guessing them primes and then finding that those factors are small recovering the old result by random toss of a in a narrow belt.
Taking a 10000 samples in a 100000 space is silly!? You may as well iterate over the whole space.
The heavy lifting is done with Mod(2,2*a*p+1)^p==1. There is no need for lots of memory to hold the full Mp number.
I am not sure testing the primaiity of a potential factor is quicker.
Perhaps you could translate my Pari/GP code into Mathematica code as an exercise and let us how fast it runs.
Last fiddled with by paulunderwood on 2021-02-09 at 15:51
2021-02-09, 16:07 #5
mattprim
Feb 2021
Salt Lake City, UT
29 Posts
Quote:
Originally Posted by paulunderwood Taking a 10000 samples in a 100000 space is silly!? You may as well iterate over the whole space. The heavy lifting is done with Mod(2,2*a*p+1)^p==1. There is no need for lots of memory to hold the full Mp number. I am not sure testing the primaiity of a potential factor is quicker. Perhaps you could translate my Pari/GP code into Mathematica code as an exercise and let us how fast it runs.
This was one of my fist runs on my own guessed numbers to check if the line of code works. The test was showing me the estimated time on Prime95 https://www.mersenne.org/download/ clustering software as above 1 year for both.
Last fiddled with by mattprim on 2021-02-09 at 16:16
2021-02-09, 16:16 #6
paulunderwood
Sep 2002
Database er0rr
3·1,327 Posts
Quote:
Originally Posted by mattprim This was one of my fist runs on my own guessed numbers to check if the line of code works. The test was showing me the estimated time on Prim95 clustering software as above 1 year.
In order: TF (trial factoring), P-1 factoring and finally a lengthy PRP (probable prime) test is done (and an LL (Lucas Lehmer) test if successful).
2021-02-10, 17:06 #7 mattprim Feb 2021 Salt Lake City, UT 29 Posts Giant 2^58501120901-1 is divisible by 13384*2*58501120901+1 = 1565958004277969 (about 1Ghz x hour to test it knowing it from PariGP on the same Ghz) PrimeQ[58501120901] True IntegerQ[(2^(58501120901) - 1)/(13384*2*58501120901 + 1)] True Giant 2^13730539469-1 is divisible by 19*2*13730539469+1 = 521760499823 (1 Ghz x min Mathematica): Do[c = False; a = RandomInteger[100]; b = PrimeQ[a*2*13730539469 + 1]; If[b == True, c = IntegerQ[(2^(13730539469) - 1)/(a*2*13730539469 + 1)]]; If[b == True, Print[" ", a, " ", c]], {i, 1, 100}] 34 False 30 False 79 False 45 False 79 False 12 False 12 False 34 False 19 True 30 False 30 False
2021-02-10, 17:53 #8
Batalov
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
25DB16 Posts
Quote:
Originally Posted by mattprim Giant 2^58501120901-1 is divisible by 13384*2*58501120901+1 = 1565958004277969 (about 1Ghz x hour to test it knowing it from PariGP on the same Ghz)
This is immensely slow. (like others already told you)
You are trying to fix a wristwatch with a bread knife.
2021-02-11, 06:57 #9
LaurV
Romulan Interpreter
"name field"
Jun 2011
Thailand
71·139 Posts
Quote:
Originally Posted by paulunderwood I am not sure testing the primaiity [sic] of a potential factor is quicker.
Your feeling is right, it is not quicker. It is in fact way WAY slower. When you get to reasonable large factors, even a much faster 2-PRP test is slower than doing directly the modular exponentiation. That is because if you want to test if q=2kp+1 is (2-probable-)prime, you need to compute 2^(q-1)==1 (mod q), i.e. 2^(2kp)==1 (mod q) which is much slower than testing the divisibility, for which you do 2^p==1 (mod q). As the candidate gets larger, this gets slower/heavier, i.e. for larger k, needs additional iterations.
Even mfakt[c|o], they won't test if the potential factors are prime, they just make a list of factors, sieve them for a while (which is much faster than x-PRP testing) and then do exponentiation (divisibility test) for the surviving candidates. This way, few percents of the surviving candidates are not prime, but it becomes faster to just do the exponentiation for them, than continue sieving (percent of composite factors depending on how large the sieving prime was). I think there was once a composite factor found this way, but I don't remember the exponent.
Last fiddled with by LaurV on 2021-02-11 at 07:07
2021-02-11, 07:06 #10 paulunderwood Sep 2002 Database er0rr 1111100011012 Posts ta!
2021-02-11, 11:12 #11 mattprim Feb 2021 Salt Lake City, UT 1D16 Posts Mathematica has the native prime Number Theory routines: PrimeQ, NextPrime, PrimePi etc. Just was mainly curious how PrimeQ[2^n-1] giving the answer if 2^n-1 is prime. You can actually link stuff like PariGP to Mathematica using Mathlink so you can include own faster routines for Mersenne. Last fiddled with by mattprim on 2021-02-11 at 11:16
Similar Threads Thread Thread Starter Forum Replies Last Post joblack Lounge 20 2009-01-05 15:18 flosculus Information & Answers 6 2008-11-10 18:59 davieddy Math 7 2007-08-21 04:51 Erasmus Factoring 3 2004-05-14 09:26 daxm Miscellaneous Math 5 2003-07-20 19:32
All times are UTC. The time now is 11:33.
Sun Jan 23 11:33:43 UTC 2022 up 184 days, 6:02, 0 users, load averages: 1.51, 1.32, 1.20 | 2,338 | 7,422 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-05 | latest | en | 0.751722 |
doc.opendatadsl.com | 1,716,548,169,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00696.warc.gz | 182,798,764 | 28,060 | # TimeSeries Scaling
This guide explains how to rescale TimeSeries to a different calendar.
## Introduction
Rescaling TimeSeries is the process of changing the changing the calendar of the TimeSeries and fitting the data using that new calendar.
Rescaling from a high frequency calendar to a low frequency calendar, e.g. Daily to Monthly is called `aggregation`
Rescaling from a low frequency calendar to a high frequency calendar, e.g. Monthly to Daily is called `distribution`
## Aggregation
When you rescale data from a high frequency calendar to a low frequency calendar, we need to reduce or aggregate the data. The way the data is aggregated is determined by the `observed` setting.
### Observed setting
#### beginning
An observation made at the beginning of the period
#### end
An observation made at the end of the period.
This is typically used for end of day settlement prices where the settlement value is either the last traded price or a calculation using the last tradded prices.
#### averaged
An average of the values observed throughout the period
#### summed
A sum of the values observed throughout the period
This is typically used for volumes, e.g. the number of traded items in a day, so aggregating to a lower frequency will sum up these values to give you a total volume for the period.
#### high
The highest value observed throuhout the period
#### low
The lowest value observed throughout the period
#### delta
The change of value from the start to the end of the period.
This is typically used for metered data where the values are meter readings which always increase in value. Scaling using delta allows you to see the amount used per period.
#### none
A point-in-time observation
### Implicit observed
When you scale a TimeSeries without specifying the observed option, OpenDataDSL first checks the global observed setting which is set using:
``set observed value``
If this hasn't been set, it uses the observed attribute of the TimeSeries which defaults to `end`.
## Distribution
When you rescale data from a low frequency calendar to a high frequency calendar, we need to distribute the observed value across a range of values. The way the data is distributed is determined by the `distribution` setting.
### Distribution setting
#### constant
The value is constant across the whole period.
This can be seen like a bar chart where the tops of the bars are flat and the value jumps at each source period start.
#### linear
The values are interpolated using a linear spline between the source values.
This can be seen as a simple line chart where the values are points on the line
#### cubic
The values are interpolated using a cubic spline between the source values
This can be seen as a smoothed line chart.
#### none
A single value is used at the source index, the rest of the values are filled in with NaN or Missing
### Implicit distribution
When you scale a TimeSeries without specifying the distribution method, OpenDataDSL first checks the global distribution setting which is set using:
``set distribution value``
If this hasn't been set then `constant` distribution is used.
### Distribution value
The value used to calculate the distributed values is the observed value in the source TimeSeries except for the following cases:
#### Summed observed
If the TimeSeries is observed as `summed` then the distributed values are divided by the number of observations so that the sum of those observations matches the original observed value.
## Scaling
To rescale a TimeSeries, you using the `scale` function which has 3 signatures:
``ts = scale(input, calendar)ts = scale(input, calendar, observed)ts = scale(input, calendar, observed, distribution)``
### Examples
#### Monthly to Daily TimeSeries
``// Create a monthly TimeSeriesmts = TimeSeries("2021-01-01", MonthlyCalendar(), [1,3,7,5,4,5])// Scale to daily using a cubic splinects = scale(mts, DailyCalendar())print cts.values``
#### Monthly to Daily TimeSeries - cubic spline
``// Create a monthly TimeSeriesmts = TimeSeries("2021-01-01", MonthlyCalendar(), [1,3,7,5,4,5])// Scale to daily using a cubic splinects = scale(mts, DailyCalendar(), "beginning", "cubic")print cts.values`` | 903 | 4,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-22 | longest | en | 0.839125 |
http://www.slideshare.net/metalkid132/experimental-design-15771913 | 1,467,211,945,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397748.48/warc/CC-MAIN-20160624154957-00094-ip-10-164-35-72.ec2.internal.warc.gz | 857,208,601 | 33,361 | Upcoming SlideShare
×
# Experimental design
2,078 views
1,818 views
Published on
1 Like
Statistics
Notes
• Full Name
Comment goes here.
Are you sure you want to Yes No
• Be the first to comment
Views
Total views
2,078
On SlideShare
0
From Embeds
0
Number of Embeds
5
Actions
Shares
0
36
0
Likes
1
Embeds 0
No embeds
No notes for slide
### Experimental design
1. 1. Intact Group Design andTrue ExperimentaldesignPresenter : Minh Sang
2. 2. Intact Group Design This is the design that most classroom researchers use. Step 1 : Select 2 classes to make 2 groups. One is experimental group and the other is control group ( You can decide it by the flip of a coin ) Step 2 : You give the treatment ( experimental instruction ) to the experimental group, not the control group
3. 3. Intact Group Design Step 3 : Give the 2 groups a posttest. After the posttest, we can have the result for the research. In short, an intact group design is : G1 x T1 G2 T1G1 : Experimental GroupG2 : Control GroupT1 : Posttest
4. 4. Intact Group Design Example : You want to investigate the effect of grammar correction on the writing skills of ESL students Step 1 : Select two similar groups of ESL ss. Flip the coin to decide which group is the experimental and control group Step 2 : Give the Exp Group the treatment ( grammar correction ) and do nothing with the Control group.
5. 5. Intact Group Design Step 3 : Give a pottest to 2 groups. After the test, you will have a conclusion that grammar correction is effective or not ( If the Exp Group has the higher scores, it means that your treatment – grammar correction is effective )
6. 6. True Experimental Design This design is used for situations in real life, when we don’t have any particular groups or classes, teams for our research. It is similar to the intact group design. and you may have a pretest for the Pretest pottest control group design
7. 7. True Experimental Design Pottest only control group : this is nearly the same as the the intact group design. The difference is that we choose the members for the group randomly :So we have : G1 ( random ) X T1 G2 ( random ) T1
8. 8. True Experimental Design Pretest pottest control group design : We may have a pretest for this design : G1 ( random ) T1 X T2 G2 ( random ) T1 T2So why do we have the T2 ( prettest ) ?
9. 9. True Experimental Design The T2 ( prettest ) is given when the time you have between the prettest and pottest is not considerable ( not sufficient ) and it may affect the conclusion of your research. It is when you give a prettest to test the knowledge, ability…of the 2 groups that you have. After all, your conclusion should be much more defensible.
10. 10. QUASI-EXPERIMENTAL DESIGN Presenter: Minh Dang
11. 11. • Quasi-experimental design is practical compromises between true experimentation and which we wish to investigate.
12. 12. • Quasi-experimental design is susceptible (easily effected) to some of the questions of internal and external validity
13. 13. • By using Quasi-experimental design, we control as many variables as we can and also limit the kinds of interpretations we make about cause-effect relationships and hedge the power of our generalization statements
14. 14. Time-series design• Because of limitations sometimes it is impossible to have a control group• use time-series design to deal with the lack of control group
15. 15. • Time-series design use several pretest and several postest• No treatment during the pretests know the changes when there are no treatments• After some pretests treatment some posttests changes from the treatment• more accurate comparison, conclusion
16. 16. • Line 1: no effect• Line 2: negative effect• Line 3: positive effect treatment is effective
17. 17. Equivalent time sample design• The treatment is introduced and reintroduced between every other pretests and posttests• Test 1 treatment test 2 treatment test 3 treatment …
18. 18. In short• Quasi-experimental design: control many variables and reduce limitations• Time-series design: pretests treatment posttests• Equivalent time-sample design: test treatment test treatment …
19. 19. EX POST FACTO DESIGNS Presenter: Huu Loc
20. 20. EX POST FACTO DESIGNS When researchers control the threats to internal and external validity, they are trying to find a direct relationship between the independent and dependent variables. In other words, they select the population, sample, treatments, and variables in order to find a cause-and- effect relationship between the variables.
21. 21. exampleYou may have created a series of media lessonson how to say no to requests in English.Not randomly select Can not draw causalyour Ss, organize relationshipsyour control and between your mediatreatment groups, materials and Ssand control for improvement infactors aside from ability to turn downthe media lessons requests gracefully inwhich might English.influence the results
22. 22. When considering all the factors that you would need to control, you might think that designing a true experimental research project is almost impossible.But , it should not mean that we have to giveup approximating the ideal as much aspossible. X causes Y is an extremely difficult thingto do unless the research is carefullydesigned and as many extraneous factorsare controlled as possible.
23. 23. When there is no possibility of randomselection of Ss, instead of abandoning theresearch, we simply have to limit the domainof our claims.oWe have to avoid making cause and effectstatements.
24. 24. EX POST FACTO designs are often usedwhen the researcher does not havecontrol over the selection andmanipulation or the independentvariable. Researchers look at the type and/or degree of relationship between the two variables rather than at a cause-and- effect relationship.
25. 25. example We can study the relationship between scores on a school- leaving exam in ESL and teachers ratings for the Ss using an ex post facto design. We can see if there is a certain amount of agreement between the two sets of scores. Any relationship between the scores of the groups would not be related to any instructional program we had given them before the test. The designs are called ex post facto. The researcher has no control over what has already happened to the Ss. The treatment has been given prior to the research project.
26. 26. There’re 2 EX POST FACTO designs o Correlational designs o Criterion group design
27. 27. Correlational designs are the mostcommonly used subset, in which a group ofSs may give us data on two differentvariables.o For example, students planing to study in the US take the TOEFL. Many universities also have entrance exam to administer to students. We can then look at the relationship of Ss’ scores on one test to their scores on the other.o Or, foreign students may be asked to take both the Graduate Record Exam (GRE) and an English placement exam prior to admission to a university.o The score for each S on one test can be compared with the score on the other, allowing us to see whether whose students who score high on one lest also score high on the other.
28. 28. The schematic representation of thisdesign would be T1 T 2
29. 29. • It’s no causal relationship between the two variables --> the distinction between independent and dependent variables is not well defined.• It is arbitrary to call one or the other the independent variable.• But, it is usually the case that the investigator may be more concerned with one than the other and may therefore label the first the independent variable and the second the dependent variable and show this by the labels X and Y.
30. 30. In a CRITERION GROUP DESIGN, twogroups of Ss are compared on onemeasure. In this design, two groups of Ss are compared on one measure. With this design, you might, for example, measure the reading peed of Iranian and French students, assuming you want to see how related or different they might be.
31. 31. The design would look like this:G1 T1G2 T1
32. 32. You can change the design into a two-criterion design by considering level oflanguage proficiency as well as their nativelanguage.In this case the criterion group design formsa factorial design.
33. 33. FACTORIAL DESIGNS Presenter: Ngan Giang
34. 34. DEFINITION• Is simply the addition of more variables to the other designs• There will be more than one independent variable considered• The variables may have one or many levels
35. 35. 2 x 2 example Room Temperature Test Difficulty (Level) 50 degrees (Level) 90 degrees (Level) Hard Test Hard Test in 50 degrees Hard Test in 90 degrees (Level) Easy Test Easy Test in 50 degrees Easy Test in 90 degreesWe are interested in studying the effect of room temperatureon test taking. To do this, we compare test scores of studentswho take a test in a 90 degree room vs. those who take a testin a 50 degree room.
36. 36. • Factor 1: Treatment – psychothera py – behavior modification• Factor 2: Setting – inpatient – day treatment – outpatient
37. 37. • higher scores mean the patient is doing worse. • day treatment is never the best condition.• psychotherapy works best with inpatient care and behavior modification works best with outpatient care.
38. 38. THE PROS AND CONS• Factorial designs are extremely useful to psychologists and field scientists as a preliminary study, allowing them to judge whether there is a link between variables, whilst reducing the possibility of experimental error and confounding variables .• The factorial design, as well as simplifying the process and making research cheaper, allows many levels of analysis. As well as highlighting the relationships between variables, it also allows the effects of manipulating a single variable to be isolated and analyzed singly.• The main disadvantage is the difficulty of experimenting with more than two factors, or many levels. A factorial design has to be planned meticulously, as an error in one of the levels, will jeopardize a great amount of work.
39. 39. Summary1. What is experimental design?2. Types of experimental design: – Pre-experimental design – True experimental design – Quasi-experimental design – Ex post facto design – Factorial design
40. 40. Thanks for your attention!• Group 5: 1. Dinh Quoc Minh Dang 2. Vo Huu Loc 3. Nguyen Dinh Minh Sang 4. Nguyen Ngoc Cam 5. Tran Thi Ngan Giang | 2,467 | 10,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2016-26 | latest | en | 0.834606 |
http://www.jiskha.com/members/profile/posts.cgi?name=charles | 1,493,208,078,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121305.61/warc/CC-MAIN-20170423031201-00309-ip-10-145-167-34.ec2.internal.warc.gz | 580,398,355 | 10,254 | # Posts by charles
Total # Posts: 385
Chemistry
How do you prepare 17.2cm³ of 36% conc HCL per dm³ of solution?
math
The length by the width of Roxie's photo is 24 cm long and 14 cm wide. In order to make a photo that will fit into a wallet, the photo needs to be reduced to a length of 7 cm. Determine the width of the wallet size photo.
math
The mean of Alice's last 5 math test scores is 88. The first 4 of these 5 math test scores were 87, 75, 93, and 85. What score did Alice earn on her fifth math test?
Math
A parking garage charges \$2.00 to enter the garage and \$1.50 an hour to park. Use the coordinate plane to graph this situation.
chemistry
a gas occupies a volume of 50.0mL at degrees Celsius and 630mmHg.at what temperature in degrees Celsius would the pressure be 101.3kPa if the volume remains constant?
geometry
Determine the Cartesian form of the equation r^2 cos theta -cos3theta
Chemistry
What is the morality of a solution prepared by dissolving 10.60g of sodium trioxocarbonate (v) in enough distilled water to obtain 250.0 ml of a solution?
Physics
Steve, if you can, look at the answer given to me by Chegg. It is much more explanatory how 100 comes from than from what you wrote and just told me? 100 is not in the initial problem, so how am I to assume 100 is part of it? The only number given to me is 1%.
Physics
Where does 100 come from?
Physics
A certain number of identical resistors are connected in series and their total resistance is measured. When the resistors are disconnected and re-connected in parallel, their total resistance drops to one-percent of the value measured when they were connected in series. How ...
chemistry
math
two strips of fabric measure 14 2/3 inches and 20 7/8 inches in length. to form one length, the strips are sew together using 3/8 inch from each strip to make the seam. what is the final length of the fabric after the two strips are sewn together.
Math
How many times does a palindromic number appear on a 12 hour digital clock between 12 midnight and 12 noon?
Chemistry
The atomic mass of CU 63 is 62.9296 amu with a natural abudance of 0.692. The average mass of copper is 63.546amu.What is the mass of copper's other isotope?
Statistics
Bill,mary,and Joe belong to a club of nineteen people. A committee of twelve is to be selected at random from the membership. How many different arrangements are possible? How many of these committees will contain Bill,Mary, and Joe at the same time?
Statistics
Bill, Mary, and Joe belong to a club of nineteen people. A committee of twelve is to be selected at random from the membership. How many different arrangements are possible. How many of these committees will contain Bill, Mary and Joe at the same time?
Statistics
Two committees, one with six and one with seven people, are to be selected from a group of eighteen people, with no one person serving on both committees at the same time. In how many ways can this be done?
math
195%
Physics
W=f (s2- s1 -s3)
Science
A 0.5kg mass connected to a light spring of constant 20N/M occilates on a frictionless track. Calculate the total energy of the system
Math
3. The population of Columbus, Ohio, in 2014 was about 823,000. The population of the United States was about 319 million in 2014. (a) Write each population in scientific notation. (b) About how many times larger was the population of the United States than the population of ...
geography
Beauty spot
probability statistics
A jar contains 6 red, 7 white, and 7 blue marbles. If a marble chosen at random. Find the following probabilities? A)P(red) B)P(white) C)P(red or blue) D)P(red and white)
math
In isosceles right triangle ABC, point D is on hypotenuse BC such that AD is an altitude of triangle ABC and DC = 5. What is the area of triangle ABC?
Chemistry
Here is my work and I keep getting the hundredth value to be incorrect: 10^-4.5 = [1.8E-5 (1.8-x)]/x 10^-4.5 = [3.24E-5 - 1.8E-5]/x 10^-4.5x = 3.24E-5 - 1.8E-5x 4.96E-5x = 3.24E-5 x= 0.653
Chemistry
How do you get 0.657 mols? I solved for x and got 0.6529 mols
Chemistry
What volume of 1.0 M NaOH must be added to 1.5 L of 1.20 M CH3COOH (ka=1.8x10^-5) to make the solution a buffer for pH 4.50? I do not understand how to determine the mols of NaOH. I tried to use the equation where: [H3O+] = ka (nHA/nA-) but it is not giving me the correct ...
math
I need help making this question into a equation? A cable television company charges \$15 plus \$5 each for extra services and now write an equation to show how much a person would pay for television if they had x number of extra services?
chemistry
A 0.109g sample of a pure gaseous compound occupies 112ml at 100degrees and 750torr: what is the molecular weight of the compound. ANS=(30.2g,mol). Please I need the solution
Physics
A man walks 3km E and then 4km N. What is his resultant displacement?
Physics
just 20oN
Physics
An airplane is heading 20degree north flying at 200km/hr relative to the air. Simultaneously the wind carries the plane 10degree, east of south at the speed of 60km/hr. (a) find the velocity of the plane relative to the ground. (b) find the component (east and south) of the ...
Physics
Find the vector cross product of two vectors C=2i-j and D=i+2j-3k. What is the angle between the vectors
Physics
Find the magnitude of the dot product of the two vectors B=6i+j+2k, D=-1+3j. What is the angle between B and D
Physics
Consider two vectors one of magnitude 3 and the other 4 unit, which of the following is not correct? We can combine these vectors to yield a resultant of. A.7 B.5 C.1 D.0
Physics
If D=7i-3j+2k and E=4i+5j-3k. Deduce angle between D and E
Physics
If D=7i-3j+2k. Find the magnitude of D
Physics
If D=7i-3j+2k and E=4i+5j-3k. Find the scalar product of D.E
Physics
Find the sum of vectors A=5m<60degree and B=6m<130degree
Thermodynamic
One mole of CO2 is to be compressed adiabatically from 1 bar and 25 C. Because of irreversibilities and poor design of the compressor the compressor work required is found to be 25 %greater than that for a well-designed (reversible) compressor. Compute the outlet temperature ...
Physics
Am I to substitute it this way 2[5(sin2)]/9.8.? Or will I make use of tan instead, since tan=2
Physics
I don't really get you scott
Physics
A body is projected such that the horizontal range is twice the maximum height of projection. Find the angle of projection; and the time of flight if the velocity of projection is 5m/s
Physics
What is the total time of flight of an object projected vertically upwards with a speed of 30m/s? (g=10m/s^2
physics
the surface tension Y, of liquid depends on the density,P, acceleration of gravity,g, and the height, h. use the method of dimension to derive the actual relationship between these quantities
physics
a sphere of radius r moving with velocity v through a fluid experiences a retarding force f given by F=6r^yv^zn^x where n is the viscosity of the liquid. use the method of dimensions to find the values of x,y,and z.
Physics
A brass cube 2cm on each edge is held in the jaw of a wire with a force 1600N. By aw much is the cube compressed. (Y for brass is 9.1×10^10pa)
physics
The vertical and horizontal component of the initial velocity of a projectile are 36m/s and 64m/s. Find the initial velocity and angle inclination
maths
Stationery #84 000 ;sports equipments #262 000;computers #672 000; school lockers #157 500.
maths
A contractor supplied some items to a school. The contractor's bill is shown below. If each amount includes 5% VAT calculate how much the government receives as VAT. Electrical fittings = #52 500.
maths
God bless you
maths
A rectangle of length y m and breadth 4 m less than the length. What is the perimeter??
math-finance
suppose you are offered two identical jobs: one paying a starting salary of \$40,000 with yearly raises of \$2,000 and one paying a starting salary of \$36,000 with yearly raises of \$25,000. Which job will pay more over a 10 year period? I=PRT = 40,000 x 0.05 x 10 = 20,000. S = ...
math
How do find determine fractions from the greatest to least
physcis
a girl throws a ball vertically upward with a speed of 20 ft/sec from the roof of a building 60 ft high.(a)how long will it take the ball to reach the ground? (b)what will its speed be when it strikes the ground?
math
jane gets 1/2 of a gold star each time she helps her teacher clean up the classroom after class. How many times would Jane need to help her teacher to earn 3 whole gold stars? Draw fraction. Strips to help.
English
to watch
Programming Logic and Design
Design an application that accepts eight friends' first names and displays them in alphabetical order
derivrtives
Find the derivative of d^2Y/dx^2 of y=e^2t and x=cost^2 and leave your answer in terms of x and y
math
Air is pumped into a spherical balloon at the rate of 20cm^3/s^-1.find the rate of change of radius of the balloon when the radius is 6cm
probability theory
3 students; A, B and C, are in a swimming race. A and B have the same probability of winning and each is twice as likely to win as C. Find the probability that B or C wins the swimming race. 3) Given three boxes as follows: Box A contains 3 red, 2 white and 2 blue balls. Box B...
English
if you want to say he left the stage then your answer should be "He disappeared from stage"..... else both are correct with different meanings..
SPR,PHYSICS
ijhljh
chem
How would your experimental formula of magnesium chloride “MgClx” have been affected if your product was not dried completely before weighing it? Would “x” be too high or two low?
Biology
Some people experience “heartburn” when eating fatty, greasy foods (similar to the effect that is felt when eating acidic foods). Why would this occur with fats? I know fatty acids are made of carboxylic acid and if you have too much, would it make sense to say that ...
PHYSICS
An automobile accident causes both the driver and passenger front airbags to deploy. a. If the vehicle travelled at a speed of 88.6 km/h is now at rest, find the change in momentum for both the 68.4 kg adult driver and the 34.2 kg child passenger. b. The adult took 0.564 s and...
Combinations and Permutations
A committee of 3 members is selected from a club made up of 10 junior members and 20 senior members. What is the expected number of juniors in the committee?
Science
The solubility of a compound in water is 6.8g/100ml at 25°C.. Calculate the amount of water required to crystallize 10g of the compound. If the pure compound is collected at 25°C what is the maximum possible yield percent.
life science
investigation to determine whether expired air contains carbondioxide
Math
A toy company manufactures miniature cars. They sell miniature sports cars in a box. They sell 12 miniature trucks in a box. Mrs. Murphy wants to buy an equal number of sports cars and trucks for the Boys and Girls Club. What is the least number of miniature sports cars she ...
maths
A box contains 5 red bowls, 6 yellow bowls and 3 white bowls. If 4 bowls are randomly drawn from the box at the same time, (a)find the probability that exactly 2 red bowls are drawn; (b)find the probability that at least 2 red bowls are drawn.
math
A box contains 5 red bowls, 6 yellow bowls and 3 white bowls. If 4 bowls are randomly drawn from the box at the same time, (a)find the probability that exactly 2 red bowls are drawn; (b)find the probability that at least 2 red bowls are drawn. My work: (a) (5/14)^2(6/14)(3/14...
Statistics
6739.2
physics
consider a vertical piston system. the piston has a 50g mass and a given radius of 3mm. what amount of pressure does the piston exert? ( use g==9.8 m/s^2 )
math
what is the surface area of a cube with sides measuring 2 inches?
Managerial Economics
1. A principal-agent problems occur when managerial decisions are not consistent with the firm's shareholders' interests. T 2. A firm making more than a normal profit may still be experiencing an economic loss. T 3. An inferior good is a good whose demand decreases as ...
english
Choose the correct sentence pattern for the sentence below. I hereby declare Tuesday Shakespeare Day.
Chemistry
What volume of carbon tetrachloride (density = 2.71 g/mL) contains 2.46x10^25 chlorine atoms? I did: 2.46x10^25 Cl atoms x 1 atom/4 atoms x 1 mol/6.02x10^23 x 154 g/1 mol x 1 mL/2.71 g And the correct answer is apparently 572 mL when I got 581 mL.
Chemistry
but how do you get the right amount of mL? I did: 2.46x10^25 Cl atoms x 1 atom/4 atoms x 1 mol/6.02x10^23 x 154 g/1 mol x 1 mL/2.71 g And the correct answer is apparently 572 mL when I got 581 mL.
Chemistry
What volume of carbon tetrachloride (density = 2.71 g/mL) contains 2.46x10^25 chlorine atoms?
Chemistry
If 84.1 g of sodium hydroxide are reacted with 51.0 g of aluminum in the following equation: 2NaOH + 2Al + 2H2O --> 2NaAlO2 + 3H2 a) what mass of excess reactant remains? b) what mass of water would react?
Physics
A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 30.0o from the horizontal with a velocity of 40.0 m/s. How far will it travel in the air? Working out please
math
I don't know or get it can show me how yougot the answer?
Maths
e) A fair coin is tossed 6 times. What is the probability of getting exactly 2 heads? What is the mean of this experiment?
Math
I have 14 hundreds, 16 tens and 9 ones. What number am I?
Math
Math
Math
The winner of a raffle will receive 3/4 of the 530.40 raised from raffle ticket sales
Math
Delia has 493 stamps in her stamp collection. She can put 16 stamps on each page of an album. How many pages can she fill completely?
GRANTHAM
Find the present value of the ordinary annuity. Please round the answer to the nearest cent.\$2,000 per semiannual period for 7 yr at 12%/year compounded semiannually
Math
I can't. This is how my math teacher put in the question. Its confusing, right? Its just messed up. I was just hoping if someone can plug in the fraction. If I saw how it looks when its all plugged in I might get it.
Math
The denominator of a fraction is one more then the numerator. If 2 were added to both the numerator and denominator, the new fraction equals 4/5. Find the original fraction.
Cardiovascular Respirtory
In which condition is the patient most likely to hemorrhage? A. Thalassemia B. Erythroblastosis fetalis C. Thrombocytopenia D. Polycythemia
Chem
(V1/n1)=(V2/n2) (5.12÷7.51)=18.3÷n2 n2=26.84mol
physics
A 75-g piece of aluminum at 89 °C is dropped in 1.0 L of water at 1.3 101 °C, which is in an insulated beaker. Assuming that there is negligible heat loss to the surroundings, determine the equilibrium temperature of the system
math
Gina's art teacher mixes 9 pints of yellow paint with 6 pints of blue paint. Did Gina use the same ratio of yellow paint to blue paint instructed by her teacher.
math
taryn and alastair both mow lawns. Each charges a flat fee to mow a lawn. The table shows the number of lawns in the past week, the time spent mowing lawns and the money earned. Taryn mowed 9 lawns. She spent 7.5 hrs mowing lawns. Taryn earned \$112.50 Alastair mowed 7 lawns. ...
math
or is this right? 1/3 * 9/7 =7/21 * 27/21 =189/21 =9
math
1/3 / 7/9 =1/3 * 9/7 =9/21 is this right??
math
thanks! practice will make perfect. thanks for helping me.
1. Pages:
2. 1
3. 2
4. 3
5. 4
6. Next>>
Post a New Question | 4,208 | 15,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-17 | latest | en | 0.943699 |
http://www.maplesoft.com/support/help/Maple/view.aspx?path=Statistics/MillsRatio | 1,500,662,180,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423808.34/warc/CC-MAIN-20170721182450-20170721202450-00615.warc.gz | 501,268,900 | 26,932 | Statistics - Maple Programming Help
Home : Support : Online Help : Statistics and Data Analysis : Statistics Package : Quantities : Statistics/MillsRatio
Statistics
MillsRatio
compute the Mills ratio
Calling Sequence MillsRatio(X, t, options)
Parameters
X - algebraic; random variable or distribution t - algebraic; point options - (optional) equation of the form numeric=value; specifies options for computing the Mills ratio of a random variable
Description
• The MillsRatio ratio computes the Mills ratio of the specified random variable at the specified point.
• The first parameter can be a distribution (see Statistics[Distribution]), a random variable, or an algebraic expression involving random variables (see Statistics[RandomVariable]).
Computation
• By default, all computations involving random variables are performed symbolically (see option numeric below).
Options
The options argument can contain one or more of the options shown below. More information for some options is available in the Statistics[RandomVariables] help page.
• numeric=truefalse -- By default, the Mills ratio is computed using exact arithmetic. To compute the Mills ratio numerically, specify the numeric or numeric = true option.
Examples
> $\mathrm{with}\left(\mathrm{Statistics}\right):$
Compute the Mills ratio of the beta distribution with parameters p and q.
> $\mathrm{MillsRatio}\left('\mathrm{Β}'\left(p,q\right),t\right)$
$\frac{{1}{-}\left({{}\begin{array}{cc}{0}& {t}{<}{0}\\ \frac{{{t}}^{{p}}{}{\mathrm{hypergeom}}{}\left(\left[{p}{,}{1}{-}{q}\right]{,}\left[{1}{+}{p}\right]{,}{t}\right)}{{\mathrm{Β}}{}\left({p}{,}{q}\right){}{p}}& {t}{<}{1}\\ {1}& {\mathrm{otherwise}}\end{array}\right)}{{{}\begin{array}{cc}{0}& {t}{<}{0}\\ \frac{{{t}}^{{-}{1}{+}{p}}{}{\left({1}{-}{t}\right)}^{{q}{-}{1}}}{{\mathrm{Β}}{}\left({p}{,}{q}\right)}& {t}{<}{1}\\ {0}& {\mathrm{otherwise}}\end{array}}$ (1)
Use numeric parameters.
> $\mathrm{MillsRatio}\left('\mathrm{Β}'\left(3,5\right),\frac{1}{2}\right)$
$\frac{{64}}{{105}}{-}\frac{{8}}{{3}}{}{\mathrm{hypergeom}}{}\left(\left[{-}{4}{,}{3}\right]{,}\left[{4}\right]{,}\frac{{1}}{{2}}\right)$ (2)
> $\mathrm{MillsRatio}\left('\mathrm{Β}'\left(3,5\right),\frac{1}{2},\mathrm{numeric}\right)$
${0.138095238095238}$ (3)
Define new distribution.
> $T≔\mathrm{Distribution}\left(\mathrm{PDF}=\left(t→\frac{1}{\mathrm{π}\left({t}^{2}+1\right)}\right)\right):$
> $X≔\mathrm{RandomVariable}\left(T\right):$
> $\mathrm{CDF}\left(X,t\right)$
$\frac{{1}}{{2}}{}\frac{{\mathrm{π}}{+}{2}{}{\mathrm{arctan}}{}\left({t}\right)}{{\mathrm{π}}}$ (4)
> $\mathrm{MillsRatio}\left(X,t\right)$
$\left({1}{-}\frac{{1}}{{2}}{}\frac{{\mathrm{π}}{+}{2}{}{\mathrm{arctan}}{}\left({t}\right)}{{\mathrm{π}}}\right){}\left({{t}}^{{2}}{+}{1}\right){}{\mathrm{π}}$ (5)
Another distribution
> $U≔\mathrm{Distribution}\left(\mathrm{CDF}=\left(t→F\left(t\right)\right),\mathrm{PDF}=\left(t→f\left(t\right)\right)\right):$
> $Y≔\mathrm{RandomVariable}\left(U\right):$
> $\mathrm{CDF}\left(Y,t\right)$
${F}{}\left({t}\right)$ (6)
> $\mathrm{MillsRatio}\left(Y,t\right)$
$\frac{{1}{-}{F}{}\left({t}\right)}{{f}{}\left({t}\right)}$ (7)
References
Stuart, Alan, and Ord, Keith. Kendall's Advanced Theory of Statistics. 6th ed. London: Edward Arnold, 1998. Vol. 1: Distribution Theory. | 1,058 | 3,333 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 19, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-30 | latest | en | 0.419329 |
http://ashutoshmehra.net/blog/2008/12/ | 1,534,822,949,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221217951.76/warc/CC-MAIN-20180821034002-20180821054002-00417.warc.gz | 35,195,562 | 7,396 | # Fun with ZDDs: Notes from Knuth’s 14th Annual Christmas Tree Lecture
In this entry, I’ll attempt to record the important ideas Knuth presented in his 14th Annual Christmas Tree Lecture, part of his regular Computer Musings.
While I wasn’t fortunate enough to attend the lecture in person, I did go through the recording that, thankfully, was quite well done. Like all other “musings”, this one included some fascinating anecdotal bits (no pun intended) and Knuth’s good sense of humor was sprinkled throughout; but most noticeable was his infectious enthusiasm for the topics he spoke about. Continue reading
# Analytically evaluating a limiting probability (June 2007 Ponder This)
In this post, I’ll describe my solution to June 2007’s Ponder This. I had felt that my solution was kind of nifty, and different from the one that was published. It had actually taken me several days to work the whole thing out.
Here’s part (2), the tougher part, of the problem: Values for a random variable are generated independently and uniformly over $\left[0,1\right)$. By accumulating these values, your job is to reach a sum between $n+x$ and $n+1$, where $n$ is a positive integer, and $0\,\textless\,x\,\textless\,1$. Continue reading
# Certain autorickshaw phenomena
Some time back, I had read a hilarious piece by Shashank, a friend of mine from the days at BITS, Pilani. The auto-podal-tow phenomena that he describes has to be seen to be believed: an autorickshaw towing another with one of the drivers using his leg as a connection! A good soul captured the moment and licensed it under Creative Commons Attribution-Noncommercial 2.0 Generic. Continue reading
# In lieu of a foreword
In this blog, I plan to write about the fascinating/curious stuff that I come across. My plan, for the moment, is to stick to topics I have some familiarity with: CS/math, programming, tools. I’ve a deep interest in certain areas of mathematics and computer science — analysis, combinatorics, probability theory, design & analysis of algorithms, graph theory & network effects, computational problems. Many of my posts would be on these topics. Some would be about EMACS, the one true editor. The remainder would be on programming languages, performance hacks, bitwise tricks… Continue reading | 516 | 2,285 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-34 | longest | en | 0.950407 |
https://www.inquirymaths.org/home/geometry-prompts/pythagorean-triples-inquiry | 1,718,714,504,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861752.43/warc/CC-MAIN-20240618105506-20240618135506-00557.warc.gz | 737,940,307 | 24,143 | # The prompt
Mathematical inquiry processes: Identify and extend patterns; make connections; generalise and prove. Conceptual field of inquiry: Pythagoras' Theorem; quadratic and cubic sequences; substitution.
Zack Miodownik, a secondary school teacher of mathematics, devised the prompt for his year 10 class. The students had already learnt about Pythagoras' theorem earlier in their schooling. Zack hoped the prompt would encourage students to make mathematical connections between the theorem and algebraic expressions for quadratic sequences.
Zack invited the students to question, notice and wonder. This is how they responded to the prompt:
The class decided to work collaboratively to find the next cases in the sequence. The students used different approaches:
Students then debated what triangles come before those in the prompt and what orientation they might have .
Collaborative inquiry
Zack reports that students collaborated throughout the inquiry as they followed different lines of inquiry (see below): "The students and I really enjoyed exploring the prompt. The whole class came together to collaborate on the inquiry. The energy in the room contributed to a brilliant sense of community."
May 2024
# Lines of inquiry
In the second half of the lesson the class explored different lines of inquiry.
One group of students derived algebraic expressions from the sequences on each side of the triangle. They then used Pythagoras' theorem to prove that the expressions always give right-angled triangles.
Another group used the expression for the nth term of the 'base sequence' (2n + 1) to generate algebraic expressions for the other two sides.
They attempted to prove the expressions always give right-angled triangles by using Pythagoras' theorem - this time by finding the length of the hypotenuse (c).
A third group used Euclid's formula, which Zack introduced into the inquiry, to find Pythagorean triples that did not follow the rules of those in the prompt. They asked if the new triples follow their own rules.
As they explored systematically and recorded their results in a table, the students noticed that some triples were multiples of others. Zack told them about the concept of primitive triples to help them categorise the results.
# Misconception
Zack cautions teachers who use the prompt about sowing a misconception. Students might generalise from the prompt that for all right-angled triangles the square of the length of the base equals the sum of the lengths of the other two sides. The teacher must address this idea whether it arises explicitly or not.
The best way is through the presentation and discussion of counter-examples, such as those illustrated. (The fact that, in the second counter-example, the square of the length of the base is twice the sum of the lengths of the other two sides can lead to further exploration.) | 564 | 2,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-26 | latest | en | 0.937357 |
https://terrorsofjoy.com/doing-299 | 1,675,063,093,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00830.warc.gz | 584,037,893 | 5,049 | # Geometry tutors near me
One tool that can be used is Geometry tutors near me. We can help me with math work.
## The Best Geometry tutors near me
Geometry tutors near me can be found online or in math books. In order to solve inequality equations, you have to first make sure that every variable is listed. This will ensure that you are accounting for all of the relevant information. Once you have accounted for all variables, you can start to solve the equation. When solving inequality equations, keep in mind that multiplication and division are not commutative operations. For example, if you want to find the value of x in an inequality equation, you should not just divide both sides by x. Instead, you should multiply both sides by the reciprocal of x: To solve inequality equations, it is best to use graphing calculators because they can handle more complex mathematics than simple hand-held calculators can. Graphing calculators can also be used to graph inequalities and other functions such as t and ln(x).
The best domain solver is a tool that helps you find the most relevant domains for your business. It does this by using complex algorithms to analyze a vast array of data about different websites and their respective categories. As the owner of your own company, you need to make sure that you’re putting the right words in front of the right people. Doing so is critical to your online success. On top of this, it’s also important to keep an eye on changes in the market. This means staying up-to-date on industry trends, as well as keeping tabs on your competitors. By doing so, you can make sure that you’re always ahead of the game. Once you have a solid handle on this important aspect of your business, it’s time to find a domain solver that works for you.
The Mathpapa area can be tricky to navigate if you're not familiar with the layout of a square. Here's a quick guide to make sure you're getting everything right: You start at (0, 0), so you can't go off the grid. The scale bar is at the top-left corner. Each quarter of an inch represents one foot of length. The "squared area" value is found by multiplying the length by itself, then adding 1/4th of that value for each quarter inch you add to your length measurement. Round all measurements to whole numbers! The Mathpapa area can be tricky to navigate if you're not familiar with the layout of a square. Here's a quick guide to make sure you're getting everything right:
Over the last few years, there has been an increase in the number of people who are living with disabilities. The number has been reported to be as high as 12.7% in the United States and between 13-20% across all age groups. This is a significant increase from previous years and poses many challenges for those involved. Some of the most common issues that people with disabilities face include limited mobility, difficulty performing daily tasks, and struggles with social interactions. These challenges can have a significant impact on their quality of life and their ability to participate fully in society. There are a number of steps that can be taken to address these issues and help improve the quality of life for people with disabilities. In addition to supporting individuals with disabilities through government programs, it is important for families to take an active role in supporting them as well. There are a number of ways that they can do this, including setting up accessible environments, helping them navigate social settings, providing assistance when needed, and being open about their needs so that they feel supported and included in the community. By taking these steps, families can help ensure that they are able to support their loved ones and make sure that they have a fulfilling life.
This is perfect for me, I'm now in high school and math is my least favorite subject and this app helps. I purchased everything on this app so am using the app plus and it really helps
Floss Rivera
This is a great app, although it may seem lazy it gives you a full explanation on hot to get that answer. They also give you the option for different solutions. Highly recommend especially if you are confused!!!
Luciana Barnes
Solution College Values Math problems to do Hard equations to solve How to do pre algebra word problems Enter math problems and get answers Algebra solver with solutions | 882 | 4,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-06 | latest | en | 0.954866 |
https://scienceoxygen.com/how-do-you-find-surface-area-to-volume-ratio/ | 1,685,835,808,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00220.warc.gz | 539,563,719 | 24,030 | # How do you find surface area to volume ratio?
## What is the formula for surface area in biology?
Why are cells, the basic units of life, so small? The answer lies in the relationship between a cell’s surface area and its volume. Surface area is the amount of surface an object has. For a cube, the formula for area is (length of a side)2 x 6.
## What is the formula for the surface area to volume ratio of a rectangular prism?
Use A = 2 l w + 2 w h + 2 l h A=2lw+2wh+2lh A=2lw+2wh+2lh to find the surface area. Use V = l w h V=lwh V=lwh to find the volume.
## Why is the ratio of surface area to volume important?
The important point is that the surface area to the volume ratio gets smaller as the cell gets larger. Thus, if the cell grows beyond a certain limit, not enough material will be able to cross the membrane fast enough to accommodate the increased cellular volume.
## How do you find the surface area and volume of a triangular prism?
The two most basic equations are: volume = 0.5 * b * h * length , where b is the length of the base of the triangle, h is the height of the triangle and length is prism length. area = length * (a + b + c) + (2 * base_area) , where a, b, c are sides of the triangle and base_area is the triangular base area.
## What is the relationship between size and surface area to volume ratio?
The smaller-sized organisms have a higher surface area than their volume. When we compare two cubes one smaller one and one bigger one, we can come to a conclusion that when the size of the organism increases lower is its SA: V ratio.
## What is the formula for finding the surface area of a triangular pyramid?
Thus, the surface area of a triangular pyramid formula is 1⁄2(a × b) + 3⁄2(b × s) in squared units.
## Is surface area the same as volume?
The surface area of any given object is the area or region occupied by the surface of the object. Whereas volume is the amount of space available in an object. In geometry, there are different shapes and sizes such as sphere, cube, cuboid, cone, cylinder, etc. Each shape has its surface area as well as volume.
## What is prism formula?
The Prism Formula is as follows, The surface area of a prism = (2×BaseArea) +Lateral Surface Area. The volume of a prism =Base Area× Height.
## What is the formula for finding the area of a prism?
The formula for the surface area of a prism is obtained by taking the sum of (twice the base area) and (the lateral surface area of the prism). The surface area of a prism is given as S = (2 × Base Area) + (Base perimeter × height) where “S” is the surface area of the prism.
## What is the formula in finding the surface area of a sphere?
And the formula for the surface area of a sphere of radius R is 4*Pi*R2.
## What is the formula in finding the volume of pyramid?
A pyramid is a polyhedron formed by connecting a polygonal base and an apex. The basic formula for pyramid volume is the same as for a cone: volume = (1/3) * base_area * height , where height is the height from the base to the apex.
## How do you find the surface area of a square?
How to find the area of a square – formulas. The area of a square is the product of the length of its sides: area = a * a = a² , where a is a square side.
## How do you find the surface area of a pyramid with slant height?
To find the surface area using the slant height, we use the formula: SA = a2 + 2×a×l.
## What is the easiest way to learn surface area and volume formulas?
1. To find the surface area of a solid, add the areas of all the faces. You can remember the formula as sum of areas of the all the faces.
2. Volume of a prism is area of its base times height.
3. Volume of a pyramid is one third of area of its base times height.
## What is the formula for area and volume?
Whereas the basic formula for the area of a rectangular shape is length × width, the basic formula for volume is length × width × height. How you refer to the different dimensions does not change the calculation: you may, for example, use ‘depth’ instead of ‘height’.
## What is prism in biology?
Prism is a homogeneous solid transparent and refracting medium bounded by two plane surfaces inclined at an angle. The commonly used prism has two triangular faces that are parallel to each other and three rectangular surfaces. | 1,028 | 4,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-23 | latest | en | 0.917527 |
https://chandoo.org/forum/threads/how-to-filter-out-multiple-vacation-ranges-dates-in-networkdays-holiday-function-by-employee.41843/ | 1,569,141,262,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575402.81/warc/CC-MAIN-20190922073800-20190922095800-00193.warc.gz | 421,833,230 | 12,325 | # How to filter out multiple vacation ranges/dates in NETWORKDAYS holiday function by employee
#### orangehybrid
##### New Member
I have a workload spreadsheet where has one tab that shows the estimated hours per week by employee by project. I want to be able to discount the hours in a given week if a certain employee is planning to be away for a portion or all of that time.
I have another tab (let call it "Vacation1") that has a matrix of employees and vacation dates, with X's marked for the days during which each employee will be away. Vacation1 is organized with the first column containing the list of all dates in a given year (or beyond); the first row (starting in B1) showing the names of each employee; and then an x under each employee column for the dates in which that person would be away.
Alternatively, I also have another tab (let's call it "Vacation 2") that shows the same information, but in a much more compact form where the first column is for employee name, second column for the leave date, and third column for return date. Old vacation dates are not removed so that past estimates of hours are preserved, meaning that the same employee can show up more than once in the first column.
The Problem: The issue I'm having is with the NETWORKDAYS function that I'm using as part of a larger formula. I want to be able to reference a dynamic array for the "holidays" portion of the formula. This dynamic array would either look at the "x"'s in Vacation1 or the multiple ranges shown in Vacation2. Right now I have the following sub-formula that should give me the number of working days in a given week minus the number days an employee is on vacation during that work week:
=NETWORKDAYS(March 3, 2019,March 3, 2019 +5,ROW(INDIRECT(VLOOKUP("John Smith",Vacation2,2,FALSE)&":"&VLOOKUP("John Smith",Vacation2,3,FALSE))))
(Note: for the purposes of this example, please disregard regular holidays and weekends. I have accounted for those separately.)
The problem with this equation is that it only accounts for the first vacation range it finds in Vacation2 for "John Smith". But if John Smith adds any additional vacation information in Vacation2, those ranges get ignored by this function.
My failed solution: I have tried to rethink this problem by coming up with the Vacation1 matrix approach, but I am running into the issue of my own limited understanding of excel's abilities with INDEX, MATCH, INDIRECT, OFFSET, etc. I have tried the following array function (entering CSE), but keep getting #VALUE errors.:
={NETWORKDAYS(March 3, 2019,March 3, 2019+5,INDEX(Vacation1_Column1,if(JohnSmith="x",Row(JohnSmith)-Row(Index(JohnSmith,1,1))+1)))}
Where JohnSmith = the column range associated with John Smith in the Vacation 1 tab.
The idea behind this approach is to use the INDEX function to return the array of dates associated with John Smith for the dates he has marked "X" in Vacation1. This approach was roughly based on the example here: https://exceljet.net/formula/get-nth-match-with-index-match
Help me: I'm not sure if I'm applying the example from ExcelJet correctly. Please advise. Thank you!!
#### GraH - Guido
##### Well-Known Member
Hi @orangehybrid , welcome to the forum. Very exhaustive explanation of your issue, that's a nice change. If you could also upload your sample workbook, as explained in the forum's best practices, we can more easily assist you in finding the correct formulae. Not seeing the data often leads to ill guidance.
#### orangehybrid
##### New Member
See the uploaded example. Note that there's some conditional formatting included, but those are not related to the issues I'm having.
#### Attachments
• 75.7 KB Views: 9
#### p45cal
##### Well-Known Member
In the array-entered formula:
Code:
``=NETWORKDAYS(43402,43416,((INDEX(Vacation1!\$C\$4:\$F\$1214,,MATCH(\$A6,Vacation1!\$C\$3:\$F\$3,0))="x")*(Vacation1!\$B\$4:\$B\$1214)))``
the 43402 is the start date Oct 29 2018, the 43416 is end date Nov 12 2018, the rest will be a long list of dates, many of which will be 0 but won't matter since that date is some time around 1/1/1900 which is out of your range of interest. I added a few xs into Brenda's (the name at A6 of the Workload Planning Spreadsheet sheet) column (F) on the Vacation1 sheeet in the pertinent date range to test.
If you have more than 4 people, obviously you'll have to extend
Vacation1!\$C\$3:\$F\$3
in:
MATCH(\$A6,Vacation1!\$C\$3:\$F\$3,0)
and in:
=NETWORKDAYS(43402,43416,((INDEX(Vacation1!\$C\$4:\$F\$1214,,MAT
and all the names must be distinct.
Not using OFFSET should make this formula less resource-hungry.
Last edited:
#### orangehybrid
##### New Member
I will try this. Thank you so much!!
#### orangehybrid
##### New Member
This worked! THANKS!!! | 1,200 | 4,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-39 | latest | en | 0.941278 |
https://lydianrain.wordpress.com/2008/11/ | 1,516,409,305,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888341.28/warc/CC-MAIN-20180120004001-20180120024001-00564.warc.gz | 733,864,786 | 14,404 | ## The Group of Units (mod n)
November 24, 2008
The ring $\mathbb{Z}/n\mathbb{Z}$ is more interesting than it is often given credit for. This post will tell you the isomorphism type of its group of units.
Why care about such a simple thing? Well first, the rings $\mathbb{Z}/n\mathbb{Z}$ are universal, in the following sense. If $R$ is any ring with unity, there exists a unique $n\geq 0$ and injective homomorphism $f: \mathbb{Z}/n\mathbb{Z} \to R$. $n$ here is called the characteristic of $R$. It’s easy to find this homomorphism: there is a natural map $\mathbb{Z}\to R$, and we factor out its kernel (which is of the form $n\mathbb{Z}$, $\mathbb{Z}$ being a principal ideal domain). So we should expect $\mathbb{Z}/n\mathbb{Z}$ to come up anywhere that rings do (though $n=1$ only applies to a single ring, and $n=0$ is disproportionately common).
Theorem: If $n=2^l \prod_i p_i^{a_i}$, where the product is over odd primes, the group of units $(\mathbb{Z}/n\mathbb{Z})^\times$ can be written as a product of cyclic groups $\prod_i Z_{p^i - p^{i-1}}$ when $l\leq 1$ and $Z_2 \times Z_{2^{l-2}} \times \prod_i Z_{p^i - p^{i-1}}$ when $l\geq 2$. In particular, the group of units is cyclic exactly when $n$ is a power of an odd prime, twice a power of an odd prime, or $4$.
The rest of this post will prove this theorem. First, the Chinese remainder theorem tells us that if $n$ factors into primes as $n=\prod_i p_i^{a_i}$, then $\mathbb{Z}/n\mathbb{Z} \cong \prod_i \mathbb{Z}/p_i^{a_i} \mathbb{Z}$. So we only need to address the question when $n=p^k$ is a power of a prime.
What are the invertible elements in $\mathbb{Z}/p^k \mathbb{Z}$? If $p\bigm| a$, then $p^{k-1} a =0$, so $a$ cannot be invertible. Those who are quick on the draw with the number theory will be able to quickly show that if $p$ does not divide $a$, then $a$ is invertible in $\mathbb{Z}/p^k \mathbb{Z}$. (the standard technique here is to note that $a$ and $p^k$ are relatively prime, so we can choose integers $x$ and $y$ such that $ax + p^k y = 1$)
Since there are $p^{k-1}$ elements divisible by $p$, it follows that $\mathbb{Z}/p^k\mathbb{Z}$ has $p^k-p^{k-1}$ units. (notice that this, along with the Chinese remainder theorem, gives us a nice formula for Euler’s phi function)
Proposition 1: When $p$ is odd, the group of units of $\mathbb{Z}/p^k\mathbb{Z}$ is a cyclic group of order $p^k-p^{k-1}$.
Proof: The case $k=1$ is actually the most difficult. It is a special case of a theorem which says that a finite subgroup of the multiplicative group of any field is cyclic. For a proof of this case, see e.g. here. (I will certainly discuss this topic in detail in a future post)
Suppose that $\sigma$ is a generator for the group of units of $\mathbb{Z}/p \mathbb{Z}$. Then the order of $\sigma\ (\rm{mod}\ p^k)$ is divisible by $p-1$, so it is of the form $p^j (p-1)$. Then $\sigma^{p^j}$ has order $p-1$. We claim that $1+p$ has order $p^{k-1}$, from which it follows that $\tau = \sigma^{p^j} (1+p)$ has order $p^{k-1} (p-1)=p^k -p^{k-1}$.
What we really need to show here is that $(1+p)^{p^{k-2}} - 1$ is divisible by $p^{k-1}$ but not $p^k$. But this is straightforward induction: if $(1+p)^{p^{k-2}} = 1 + a p^{k-1}$ with $a$ not divisible by $p$, then taking $p$-th powers we see that $(1+p)^{p^{k-1}} = 1 + ap^k + (\ldots)p^{k+1}$.
Proposition 2: If $k\geq 3$, the group of units of $\mathbb{Z}/2^k\mathbb{Z}$ is the product of two cyclic groups, one of order $2$, and one of order $2^{k-2}$. If $k\leq 2$, the group of units of $\mathbb{Z}/2^k\mathbb{Z}$ is cyclic.
Proof In fact, we will show that $5$ generates a cyclic group of order $2^{k-2}$ that does not contain $- 1$, an element of order $2$. (there is nothing special about $5$ here—any choice that is $5\ (\rm{mod}\ 8)$ will work) We first show, by induction on $k$, that $5^{2^{k-3}} \equiv 1+2^{k-1}\ (\rm{mod}\ 2^k)$, the case $k=3$ being clear.
For $k\geq 3$, the inductive hypothesis now tells us that $5^{2^{k-3}} \equiv 1+2^{k-1}\ (\rm{mod}\ 2^k)$, so there is some $a\in\{0,1\}$ such that $5^{2^{k-3}} \equiv 1+2^{k-1} + a 2^k\ (\rm{mod}\ 2^{k+1})$. Squaring, we get $5^{2^{k-2}} \equiv 1+2^k\ (\rm{mod}\ 2^{k+1})$.
Since $1+2^{k-1}$ is an element of order $2$, this shows that $5$ generates a cyclic group of order $2^{k-2}$. So we just need to make sure that $-1\not\in \langle 5\rangle$. But a cyclic group can contain at most one element of order $2$, and we have already shown $1+2^{k-1}\in\langle 5\rangle$.
The second statement in the proposition is easily verified.
Exercise: Prove the second sentence of the theorem! (hint: for one direction, use the Chinese remainder theorem for cyclic groups, and for the other, count elements of order $2$)
## Fun With Group Actions
November 10, 2008
I had to take a bit of a break to move into my new apartment in the Castro, (which has, incidentally, been a pretty exciting place to live this past week) but now I’m all settled down and ready to tell you about group actions.
The basic idea of a group action is to visualize a group as a set of permutations of some set $X$, giving a homomorphism $\phi: G\to S_{|X|}$. A basic example would be the cyclic group $Z_n$ acting on an $n$-gon by rotation. But here are some more interesting and general group actions for a group $G$:
• $G$ acts on $G$ by left multiplication.
• $G$ acts on a normal subgroup of $G$ by conjugation.
• $G$ acts on the set of subgroups of $G$ of a fixed order, by conjugation.
• If $H\leq G$, $G$ acts on the set of left cosets of $H$ by left multiplication.
All of these can provide useful information through the existence of homomorphisms to some symmetric group $S_n$. For an example of this technique, let’s consider this classic problem:
Problem: Show that any simple group of order $60$ is isomorphic to $A_5$. (recall that a simple group is one with no nontrivial normal subgroups)
We make use of some basic Sylow theory. Let $G$ be a simple group of order $60$ and let $k$ be the number of subgroups of $G$ that have order $5$. Sylow’s theorems tell us that $k$ is a factor of $12$ and is congruent to $1\ (\rm{mod}\ 5)$, so $k\in \{1,6\}$. If $k=1$, then the group of order $5$ is normal, (do you see why?) so $k=6$.
Let $G$ act on the set of subgroups of $G$ of order $5$, by conjugation. This gives a homomorphism $\phi: G\to S_6$. Because none of these subgroups can be normal (or again appealing to Sylow theory) it is easy to see that $\phi\neq 1$, so $\ker{\phi}\neq G$. But $\ker{\phi}$ is a normal subgroup of $G$, so $\ker{\phi}=1$ and $\phi$ is injective.
So we can imagine $G$ as a subgroup of $S_6$. In fact, $\phi(G)\leq A_6$, as the following lemma will tell us:
Lemma: If $H\leq S_n$ is a simple group of order larger than $2$, $H\leq A_n$.
Proof: Let $\sigma: S_n\to Z_2$ be the sign homomorphism, so $A_n = \ker{\sigma}$. Then $\ker{\sigma_H}$ is a normal subgroup of $H$. But $\sigma_H$ cannot be injective, as $|H|>|Z_2|$, so $\ker{\sigma_H} = H$, in other words $H\leq \ker{\sigma}=A_n$.
Returning to our solution, we can now assume that $G\leq A_6$. Counting orders, we see that $G$ has index $360/60=6$. Let $A_6$ act on the set of left cosets of $G$ by left multiplication, giving a homomorphism $\psi: A_6 \to S_6$. Pausing for a moment to verify that $\psi\neq 1$, the fact that $A_6$ is simple tells us that $\psi$ is injective, and applying the lemma tells us easily that $\psi:A_6\to A_6$ is an isomorphism.
What is the image of $G$ under this isomorphism? Well, let’s think about the action again. We have some cosets $x_1 G, x_2 G, x_3 G, x_4 G, x_5 G, G$. Multiplication on the left by elements of $G$ is certainly not guaranteed to fix the first five, but it will always fix the last one. In other words, $\psi(G)\leq S_5$. So $\psi(G)\leq S_5 \cap A_6=A_5$.
Since $|G|=|A_5|$, it follows that $\psi$ is an isomorphism carrying $G$ to $A_5$, and we are done.
As we can see, group actions provide an excellent if somewhat magical way of reasoning about finite groups that are too big to be easily understood by hand. Make them part of your group theory toolbox. | 2,672 | 8,125 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 165, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-05 | longest | en | 0.831605 |
https://www.aqua-calc.com/one-to-one/density/kilogram-per-cubic-decimeter/long-ton-per-cubic-inch/1 | 1,571,590,147,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986717235.56/warc/CC-MAIN-20191020160500-20191020184000-00467.warc.gz | 803,892,135 | 8,600 | # 1 kilogram per cubic decimeter in long tons per cubic inch
## kg/dm³ to long tn/in³ unit converter of density
1 kilogram per cubic decimeter [kg/dm³] = 2 × 10-5 long ton per cubic inch [long tn/in³]
### kilograms per cubic decimeter to long tons per cubic inch density conversion cards
• 1
through
25
kilograms per cubic decimeter
• 1 kg/dm³ to long tn/in³ = 2 × 10-5 long tn/in³
• 2 kg/dm³ to long tn/in³ = 3 × 10-5 long tn/in³
• 3 kg/dm³ to long tn/in³ = 5 × 10-5 long tn/in³
• 4 kg/dm³ to long tn/in³ = 6 × 10-5 long tn/in³
• 5 kg/dm³ to long tn/in³ = 8 × 10-5 long tn/in³
• 6 kg/dm³ to long tn/in³ = 0.0001 long tn/in³
• 7 kg/dm³ to long tn/in³ = 0.00011 long tn/in³
• 8 kg/dm³ to long tn/in³ = 0.00013 long tn/in³
• 9 kg/dm³ to long tn/in³ = 0.00015 long tn/in³
• 10 kg/dm³ to long tn/in³ = 0.00016 long tn/in³
• 11 kg/dm³ to long tn/in³ = 0.00018 long tn/in³
• 12 kg/dm³ to long tn/in³ = 0.00019 long tn/in³
• 13 kg/dm³ to long tn/in³ = 0.00021 long tn/in³
• 14 kg/dm³ to long tn/in³ = 0.00023 long tn/in³
• 15 kg/dm³ to long tn/in³ = 0.00024 long tn/in³
• 16 kg/dm³ to long tn/in³ = 0.00026 long tn/in³
• 17 kg/dm³ to long tn/in³ = 0.00027 long tn/in³
• 18 kg/dm³ to long tn/in³ = 0.00029 long tn/in³
• 19 kg/dm³ to long tn/in³ = 0.00031 long tn/in³
• 20 kg/dm³ to long tn/in³ = 0.00032 long tn/in³
• 21 kg/dm³ to long tn/in³ = 0.00034 long tn/in³
• 22 kg/dm³ to long tn/in³ = 0.00035 long tn/in³
• 23 kg/dm³ to long tn/in³ = 0.00037 long tn/in³
• 24 kg/dm³ to long tn/in³ = 0.00039 long tn/in³
• 25 kg/dm³ to long tn/in³ = 0.0004 long tn/in³
• 26
through
50
kilograms per cubic decimeter
• 26 kg/dm³ to long tn/in³ = 0.00042 long tn/in³
• 27 kg/dm³ to long tn/in³ = 0.00044 long tn/in³
• 28 kg/dm³ to long tn/in³ = 0.00045 long tn/in³
• 29 kg/dm³ to long tn/in³ = 0.00047 long tn/in³
• 30 kg/dm³ to long tn/in³ = 0.00048 long tn/in³
• 31 kg/dm³ to long tn/in³ = 0.0005 long tn/in³
• 32 kg/dm³ to long tn/in³ = 0.00052 long tn/in³
• 33 kg/dm³ to long tn/in³ = 0.00053 long tn/in³
• 34 kg/dm³ to long tn/in³ = 0.00055 long tn/in³
• 35 kg/dm³ to long tn/in³ = 0.00056 long tn/in³
• 36 kg/dm³ to long tn/in³ = 0.00058 long tn/in³
• 37 kg/dm³ to long tn/in³ = 0.0006 long tn/in³
• 38 kg/dm³ to long tn/in³ = 0.00061 long tn/in³
• 39 kg/dm³ to long tn/in³ = 0.00063 long tn/in³
• 40 kg/dm³ to long tn/in³ = 0.00065 long tn/in³
• 41 kg/dm³ to long tn/in³ = 0.00066 long tn/in³
• 42 kg/dm³ to long tn/in³ = 0.00068 long tn/in³
• 43 kg/dm³ to long tn/in³ = 0.00069 long tn/in³
• 44 kg/dm³ to long tn/in³ = 0.00071 long tn/in³
• 45 kg/dm³ to long tn/in³ = 0.00073 long tn/in³
• 46 kg/dm³ to long tn/in³ = 0.00074 long tn/in³
• 47 kg/dm³ to long tn/in³ = 0.00076 long tn/in³
• 48 kg/dm³ to long tn/in³ = 0.00077 long tn/in³
• 49 kg/dm³ to long tn/in³ = 0.00079 long tn/in³
• 50 kg/dm³ to long tn/in³ = 0.00081 long tn/in³
• 51
through
75
kilograms per cubic decimeter
• 51 kg/dm³ to long tn/in³ = 0.00082 long tn/in³
• 52 kg/dm³ to long tn/in³ = 0.00084 long tn/in³
• 53 kg/dm³ to long tn/in³ = 0.00085 long tn/in³
• 54 kg/dm³ to long tn/in³ = 0.00087 long tn/in³
• 55 kg/dm³ to long tn/in³ = 0.00089 long tn/in³
• 56 kg/dm³ to long tn/in³ = 0.0009 long tn/in³
• 57 kg/dm³ to long tn/in³ = 0.00092 long tn/in³
• 58 kg/dm³ to long tn/in³ = 0.00094 long tn/in³
• 59 kg/dm³ to long tn/in³ = 0.00095 long tn/in³
• 60 kg/dm³ to long tn/in³ = 0.00097 long tn/in³
• 61 kg/dm³ to long tn/in³ = 0.00098 long tn/in³
• 62 kg/dm³ to long tn/in³ = 0.001 long tn/in³
• 63 kg/dm³ to long tn/in³ = 0.00102 long tn/in³
• 64 kg/dm³ to long tn/in³ = 0.00103 long tn/in³
• 65 kg/dm³ to long tn/in³ = 0.00105 long tn/in³
• 66 kg/dm³ to long tn/in³ = 0.00106 long tn/in³
• 67 kg/dm³ to long tn/in³ = 0.00108 long tn/in³
• 68 kg/dm³ to long tn/in³ = 0.0011 long tn/in³
• 69 kg/dm³ to long tn/in³ = 0.00111 long tn/in³
• 70 kg/dm³ to long tn/in³ = 0.00113 long tn/in³
• 71 kg/dm³ to long tn/in³ = 0.00115 long tn/in³
• 72 kg/dm³ to long tn/in³ = 0.00116 long tn/in³
• 73 kg/dm³ to long tn/in³ = 0.00118 long tn/in³
• 74 kg/dm³ to long tn/in³ = 0.00119 long tn/in³
• 75 kg/dm³ to long tn/in³ = 0.00121 long tn/in³
• 76
through
100
kilograms per cubic decimeter
• 76 kg/dm³ to long tn/in³ = 0.00123 long tn/in³
• 77 kg/dm³ to long tn/in³ = 0.00124 long tn/in³
• 78 kg/dm³ to long tn/in³ = 0.00126 long tn/in³
• 79 kg/dm³ to long tn/in³ = 0.00127 long tn/in³
• 80 kg/dm³ to long tn/in³ = 0.00129 long tn/in³
• 81 kg/dm³ to long tn/in³ = 0.00131 long tn/in³
• 82 kg/dm³ to long tn/in³ = 0.00132 long tn/in³
• 83 kg/dm³ to long tn/in³ = 0.00134 long tn/in³
• 84 kg/dm³ to long tn/in³ = 0.00135 long tn/in³
• 85 kg/dm³ to long tn/in³ = 0.00137 long tn/in³
• 86 kg/dm³ to long tn/in³ = 0.00139 long tn/in³
• 87 kg/dm³ to long tn/in³ = 0.0014 long tn/in³
• 88 kg/dm³ to long tn/in³ = 0.00142 long tn/in³
• 89 kg/dm³ to long tn/in³ = 0.00144 long tn/in³
• 90 kg/dm³ to long tn/in³ = 0.00145 long tn/in³
• 91 kg/dm³ to long tn/in³ = 0.00147 long tn/in³
• 92 kg/dm³ to long tn/in³ = 0.00148 long tn/in³
• 93 kg/dm³ to long tn/in³ = 0.0015 long tn/in³
• 94 kg/dm³ to long tn/in³ = 0.00152 long tn/in³
• 95 kg/dm³ to long tn/in³ = 0.00153 long tn/in³
• 96 kg/dm³ to long tn/in³ = 0.00155 long tn/in³
• 97 kg/dm³ to long tn/in³ = 0.00156 long tn/in³
• 98 kg/dm³ to long tn/in³ = 0.00158 long tn/in³
• 99 kg/dm³ to long tn/in³ = 0.0016 long tn/in³
• 100 kg/dm³ to long tn/in³ = 0.00161 long tn/in³
#### Foods, Nutrients and Calories
BROCCOLI CUTS, UPC: 60788001179 weigh(s) 91.93 gram per (metric cup) or 3.07 ounce per (US cup), and contain(s) 34 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
RED RASPBERRY FRUIT SPREAD, UPC: 051500141755 weigh(s) 321.23 gram per (metric cup) or 10.72 ounce per (US cup), and contain(s) 211 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
Foods high in Thiamin, foods low in Thiamin, and Recommended Dietary Allowances (RDAs) for Thiamin
#### Gravels, Substances and Oils
CaribSea, Marine, Aragonite, Bermuda Pink weighs 1 281.5 kg/m³ (80.00143 lb/ft³) with specific gravity of 1.2815 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Grain Wheat weighs 790 kg/m³ (49.31809 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-507, liquid (R507) with temperature in the range of -51.12°C (-60.016°F) to 60°C (140°F)
#### Weights and Measurements
The pennyweight per square yard surface density measurement unit is used to measure area in square yards in order to estimate weight or mass in pennyweights
The radiation absorbed dose is a measurement of radiation, in energy per unit of mass, absorbed by a specific object, such as human tissue.
lb/mm³ to lb/m³ conversion table, lb/mm³ to lb/m³ unit converter or convert between all units of density measurement.
#### Calculators
Online Food Calories and Nutrients Calculator | 2,912 | 7,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-43 | latest | en | 0.254806 |
http://docs.oracle.com/cd/B13789_01/olap.101/b10339/x_maxbytes005.htm | 1,469,409,224,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824201.28/warc/CC-MAIN-20160723071024-00088-ip-10-185-27-174.ec2.internal.warc.gz | 76,133,063 | 3,325 | Skip Headers
Oracle® OLAP DML Reference 10g Release 1 (10.1) Part Number B10339-02
## MIN
The MIN function calculates the smaller value of two expressions.
Return Value
DECIMAL
Syntax
MIN(expression1expression2)
Arguments
expression1
One expression to be compared.
expression2
The other expression to be compared.
Notes
Dimensions of the Result
Ordinarily, the dimensions of both the expressions you want to compare and the results of MIN are the same. When the dimensions of one expression are a subset of the other's dimensions, then the results of MIN are dimensioned by the larger set of dimensions. In any case, the results of MIN are dimensioned by the union of the dimensions of the two expressions.
Examples
Example 17-4 Calculating Whether Actual or Budget Values Are Smaller
Suppose, for each of the first six months of 1996, you want to find whether the `actual` value or the `budget` value is smaller for the line item Cost of Goods Sold (`Cogs`) in the `Sporting` division.
```LIMIT line TO 'Cogs'
LIMIT division TO 'Sporting'
LIMIT month TO 'Jan96' TO 'Jun96'
REPORT DOWN month actual budget MIN(actual budget)
```
The preceding statements produce the following output.
```DIVISION: SPORTING
--------------LINE--------------
--------------COGS--------------
MIN
(ACTUAL
MONTH ACTUAL BUDGET BUDGET)
-------------- ---------- ---------- ----------
Jan96 287,557.87 279,773.01 279,773.01
Feb96 315,298.82 323,981.56 315,298.82
Mar96 326,184.87 302,177.88 302,177.88
Apr96 394,544.27 386,100.82 386,100.82
May96 449,862.25 433,997.89 433,997.89
Jun96 457,347.55 448,042.45 448,042.45
``` | 448 | 1,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-30 | longest | en | 0.667948 |
https://testbook.com/question-answer/ab-c-bc-a-ca-b-is-equal-to--615aa4b8bdc4d39ad8a1ea48 | 1,642,520,912,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300934.87/warc/CC-MAIN-20220118152809-20220118182809-00504.warc.gz | 647,177,421 | 29,559 | a(b - c) + b(c - a) + c(a - b) is equal to
This question was previously asked in
HP JBT TET 2019 Official Paper
View all HP TET Papers >
1. ab + bc + ca
2. 2(ab + bc + ca)
3. None of these
Option 2 : 0
Detailed Solution
Calculation:
a(b - c) + b(c - a) + c(a - b)
⇒ ab - ac + bc - ab + ca - cb
⇒ ab - ac + bc - ab + ca - cb
⇒ 0
∴ The required answer will be 0. | 146 | 370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-05 | latest | en | 0.689594 |
http://www.askiitians.com/forums/Trigonometry/23/3808/iitjee-2000.htm | 1,484,704,341,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280133.2/warc/CC-MAIN-20170116095120-00177-ip-10-171-10-70.ec2.internal.warc.gz | 344,093,157 | 32,062 | Click to Chat
1800-2000-838
+91-120-4616500
CART 0
• 0
MY CART (5)
Use Coupon: CART20 and get 20% off on all online Study Material
ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R
There are no items in this cart.
Continue Shopping
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list
Get extra R 600 off
USE CODE: chait6
``` Consider the following statements in S and R : S : Both sin x and cos x are decreasing functions in the interval (π/2,π) R : If a differentiable function decreases in an interval (a, b), then its derivative also decreases in (a, b). Which of the following is true : (A) Both S and R are wrong (B) Both S and R are correct, but R is not the correct explanation of S. (C) S is correct and R is correct explanation for S. (D) S is correct and R is wrong.
```
7 years ago
Share
``` Hi,
S is correct and R is wrong.
Because f'(x) may be negative. But that does not say that f"(x) has to be negative too.
Regards
Rajat
```
7 years ago
# Other Related Questions on Trigonometry
sin^2 6°-sin^2 12°+sin^2 18°-sin^2 24°......15 solve it Urgent
Ajay, the complete qution isSolution is sin^2 6°-sin^2 12°+sin^2 18°-sin^2 24°..... upto 15 terms. sin 78°=0 sin 42°+sin 54°+ sin 66°+ + sin 18° sin 6°+ where )=0.5 (your required answer),...
Kumar 3 months ago
Not any people get my answer why. You can no give answer my question I am join this site
Vivek kumar 5 months ago
Hello If you want to get the solution quick you should post your question in clear manner. Its not clear what you wnat us to solve, and what does 15 at the end of question means?
Ajay 5 months ago
Solve: (sin theta+cosec theta)^2 + (cos theta +sec theta)^2- (tan^2 theta + cot ^2 theta)^2
What needs to be solved here ? The question is incomplete....................................................................
Ajay 6 months ago
i don’t know how to do this...............................................................................................
Saravanan 2 months ago
this is the question :: Solve: (sin theta+cosec theta)^2 + (cos theta +sec theta)^2 - (tan^2 theta + cot ^2 theta)
Naveen Shankar 6 months ago
solutions to Question no. 17,18 19 and 20 pleaseeeeeeeeeee
Let the feet of the altitudes on BC, AC, AB, be D,E,F resp. Let the orthocenter be H. The following can be proved easily: 1. HDCE and HFBD are cyclic quadrilaterals. Then chord HE subtends...
mycroft holmes one month ago
Draw which is Isoceles as OB = OC. Now which means . Let D, be the foot of the perp from O on BC ( which is also the midpoint of BC). Then OD = OC sin (OBC) = R cos A. Hence the required...
mycroft holmes one month ago
a cos A = b cos B 2R sin A cos A = 2R sin B cos B sin 2A = sin 2B Either A = B (isoceles or equilateral) or 2A = 180 o – 2B so that A+B = 90 o .(Right-angled)
mycroft holmes one month ago
If alpha is a real root of the equation ax 2 +bx+c and beta is a real root of equation -ax 2 +bx+c. Show that there exists a root gama of the equation (a/2)x 2 +bx+c which lies between alpha...
Ajay 6 months ago
Small Mistake in last para posting again..............................................................................................................
Ajay 6 months ago
We have Similarly, So if P(x) = a/2 x 2 +bx +c, then and are off opposite sign and hence there must exist a root between the two numbers.
mycroft holmes 6 months ago
In the listed image can you tell me how beta*gamma = 2 ….. . . .. ??
The value of gamma is still not correct, either printing mistake or you gave me wrong value. The correct value of gamma is below
Ajay 5 months ago
Thankyou so much............................. …......................................................................!
Anshuman Mohanty 5 months ago
Yes sorry..... . . . .it is not so clear.. ok the values are beta = α + α^2 + α^4 and gamma = α^3 + α^5 + α^7
Anshuman Mohanty 5 months ago
if |z - i| Options: a*) 14 b) 2 c) 28 d) None of the above
If |z-i| = ?? PLs complete the question
Nishant Vora one month ago
Got it! [z + 12 – 6 i ] can be rewritten as [ z – i + 12 – 5 i] => | z – i | and => |12 – 5 i | = sqrt ( 12^2 + 5^2) = 13......................(2) => | z + 12 – 6 i | => | z + 12 – 6 i |...
Divya one month ago
I tried posting the question several times, it kept cutting off the rest of the question. Here: If | z-1| Options: a*) 14 b) 2 c) 28 d) None of the above
Divya one month ago
View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: R 15,000
• View Details
Get extra R 6,000 off
USE CODE: chait6
Get extra R 600 off
USE CODE: chait6
More Questions On Trigonometry
Post Question | 1,386 | 4,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-04 | longest | en | 0.826295 |
http://www.cfd-online.com/Forums/ansys/97047-movement-piston-cylinder-fluent.html | 1,438,072,960,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042981753.21/warc/CC-MAIN-20150728002301-00018-ip-10-236-191-2.ec2.internal.warc.gz | 353,104,091 | 17,911 | # Movement of piston in a cylinder with fluent?
Register Blogs Members List Search Today's Posts Mark Forums Read
February 7, 2012, 20:18 Movement of piston in a cylinder with fluent? #1 New Member Andy Join Date: Feb 2012 Posts: 5 Rep Power: 5 I'm starting to build a model of a piston (say 45mm diameter, 20mm length) in a sealed cylinder of air (50mm diameter 100mm length). This piston moves up and down (+ and minus y) with a sinusoidal velocity. At the top of the cylinder there is a constant temperature of say 100 degrees centigrade and at the bottom 25 degrees centigrade. As the piston moves downward, the air is forced to the top of the cylinder, heating up and as the piston moves back upward, the air moves back down to the bottom. My question is, how would I start to go about modeling this in fluent? I've never modelled something with motion before and a tutorial would be great. The setup in my head would be... -geometry,cylinder of fluid constructed with a missing pocket 45mmx20mm, representing the piston. -energy equation enabled so heat transfer was active -no initial fluid velocity (piston would be at bottom dead center) -top and bottom zones set at constant temperature -cylinder wall set to 0 heat flux -viscous model selected depending on approx reynolds number But where might I find the options to translate the pocket in the continuum? Thanks in advance
February 10, 2012, 12:03 #2 Super Moderator Sijal Ahmed Memon (turboenginner@gmail.com) Join Date: Mar 2009 Location: Islamabad Pakistan Posts: 3,914 Blog Entries: 6 Rep Power: 38 Check this link Need help with meshing Diesel IC engine tsram90 likes this.
September 18, 2013, 09:09
#3
New Member
Rajendra Kumar
Join Date: Jun 2013
Posts: 2
Rep Power: 0
Quote:
Originally Posted by Whitworth I'm starting to build a model of a piston (say 45mm diameter, 20mm length) in a sealed cylinder of air (50mm diameter 100mm length). This piston moves up and down (+ and minus y) with a sinusoidal velocity. At the top of the cylinder there is a constant temperature of say 100 degrees centigrade and at the bottom 25 degrees centigrade. As the piston moves downward, the air is forced to the top of the cylinder, heating up and as the piston moves back upward, the air moves back down to the bottom. My question is, how would I start to go about modeling this in fluent? I've never modelled something with motion before and a tutorial would be great. The setup in my head would be... -geometry,cylinder of fluid constructed with a missing pocket 45mmx20mm, representing the piston. -energy equation enabled so heat transfer was active -no initial fluid velocity (piston would be at bottom dead center) -top and bottom zones set at constant temperature -cylinder wall set to 0 heat flux -viscous model selected depending on approx reynolds number But where might I find the options to translate the pocket in the continuum? Thanks in advance
Dear Whitworth,
i have also written UDF for moving wall, but the movement of wall is always +y and reaches in original position y=0 but not in -y direction. so how can i make movement of wall in + to - ?
September 19, 2013, 12:27 Tutorial : Cold Flow Simulation Inside an SI Engine #4 New Member Arash zareinezhad Join Date: Mar 2013 Location: Iran Posts: 11 Rep Power: 4 Dear Whitworth & rajendra9424 this file can help any one that want to simulate any kind of cylider with valve motion and injection! if you need more help just call me download file of Cold Flow Simulation Inside an SI Engine here or here amin_gls likes this.
March 3, 2015, 11:05 #5 New Member Vivek M Join Date: Mar 2015 Posts: 2 Rep Power: 0 I downloaded the file and tried to run the simulation. but after a certain number of iterations i got an access violation error. how can i rectify that. pls help me . Thanks in Advance
March 4, 2015, 00:50 Fluent Access Violation error after a specific iteration number #6 New Member Arash zareinezhad Join Date: Mar 2013 Location: Iran Posts: 11 Rep Power: 4
Thread Tools Display Modes Linear Mode
Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules
Similar Threads Thread Thread Starter Forum Replies Last Post Paul Reichl Main CFD Forum 16 December 8, 2012 16:21 mohammad CFX 2 January 8, 2012 20:39 opehterinar81 FLUENT 0 August 19, 2011 11:55 mosman Main CFD Forum 0 August 7, 2011 01:40 Michele Spinolo FLUENT 6 June 27, 2002 12:57
All times are GMT -4. The time now is 04:42. | 1,182 | 4,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2015-32 | longest | en | 0.88416 |
https://community.smartsheet.com/discussion/102047/can-i-over-ride-a-column-formula-for-some-row | 1,716,737,345,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058956.26/warc/CC-MAIN-20240526135546-20240526165546-00281.warc.gz | 148,255,070 | 105,512 | # Can I over ride a column formula for some row?
Options
✭✭✭
I have a grid to track periodic review of documents. Most documents must be reviewed every year, but some can be reviewed every two years.
My column formula is [Effective Date]@row + 365.
Can I put [Effective Date]@row + 730 in the rows where that would be true?
Thank you,
Lynn
• ✭✭✭✭✭
Options
Add a checkbox column called "Two Years" and change the formula to
=[Effective Date]@row + IF([Two Years]@row = 1, 730, 365)
Then check the box on any of the documents that are reviewed every two years.
• ✭✭✭
Options
Thank you so much! What does "=1" do in the formula?
• ✭✭✭✭✭
Options
"=1" is checking to see if the check box is checked.
Checked = 1
Unchecked = 0
• ✭✭✭✭✭
Options
Add a checkbox column called "Two Years" and change the formula to
=[Effective Date]@row + IF([Two Years]@row = 1, 730, 365)
Then check the box on any of the documents that are reviewed every two years.
• ✭✭✭✭✭✭
Options
I hope you're well and safe!
Unfortunately, it's not possible now, but it's an excellent idea!
Please submit this as a Product Feedback or Idea (If it hasn't been added already) when you have a moment.
Here's a possible workaround or workarounds
• Add a so-called helper column and update the formula so that it overrides the standard value if it's populated.
Would that work/help?
I hope that helps!
Be safe, and have a fantastic week!
Best,
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
Did my post(s) help or answer your question or solve your problem? Please support the Community by marking it Insightful/Vote Up, Awesome, or/and as the accepted answer. It will make it easier for others to find a solution or help to answer!
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:andree@workbold.com | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
• ✭✭✭
Options
Thank you so much! What does "=1" do in the formula?
• ✭✭✭✭✭✭
Options
Happy to help!
Remember! Did my post(s) help or answer your question or solve your problem? Please support the Community by marking it Insightful/Vote Up/Awesome or/and as the accepted answer. It will make it easier for others to find a solution or help to answer!
SMARTSHEET EXPERT CONSULTANT & PARTNER
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
W: www.workbold.com | E:andree@workbold.com | P: +46 (0) - 72 - 510 99 35
Feel free to contact me for help with Smartsheet, integrations, general workflow advice, or anything else.
• ✭✭✭✭✭ | 762 | 2,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.894827 |
https://mathexamination.com/class/spheroid.php | 1,618,511,103,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038087714.38/warc/CC-MAIN-20210415160727-20210415190727-00438.warc.gz | 488,040,453 | 6,981 | ## Do My Spheroid Class
A "Spheroid Class" QE" is a standard mathematical term for a generalized continuous expression which is used to resolve differential formulas and has options which are regular. In differential Class resolving, a Spheroid function, or "quad" is utilized.
The Spheroid Class in Class kind can be expressed as: Q( x) = -kx2, where Q( x) are the Spheroid Class and it is a crucial term. The q part of the Class is the Spheroid consistent, whereas the x part is the Spheroid function.
There are four Spheroid functions with proper solution: K4, K7, K3, and L4. We will now take a look at these Spheroid functions and how they are resolved.
K4 - The K part of a Spheroid Class is the Spheroid function. This Spheroid function can likewise be written in partial portions such as: (x2 - y2)/( x+ y). To fix for K4 we multiply it by the appropriate Spheroid function: k( x) = x2, y2, or x-y.
K7 - The K7 Spheroid Class has a service of the form: x4y2 - y4x3 = 0. The Spheroid function is then multiplied by x to get: x2 + y2 = 0. We then need to increase the Spheroid function with k to get: k( x) = x2 and y2.
K3 - The Spheroid function Class is K3 + K2 = 0. We then increase by k for K3.
K3( t) - The Spheroid function equationis K3( t) + K2( t). We increase by k for K3( t). Now we increase by the Spheroid function which gives: K2( t) = K( t) times k.
The Spheroid function is also called "K4" because of the initials of the letters K and 4. K indicates Spheroid, and the word "quad" is pronounced as "kah-rab".
The Spheroid Class is one of the primary approaches of resolving differential equations. In the Spheroid function Class, the Spheroid function is first multiplied by the proper Spheroid function, which will offer the Spheroid function.
The Spheroid function is then divided by the Spheroid function which will divide the Spheroid function into a genuine part and an imaginary part. This offers the Spheroid term.
Lastly, the Spheroid term will be divided by the numerator and the denominator to get the quotient. We are entrusted to the right hand side and the term "q".
The Spheroid Class is an essential principle to comprehend when resolving a differential Class. The Spheroid function is just one technique to fix a Spheroid Class. The approaches for fixing Spheroid equations consist of: particular worth decomposition, factorization, optimum algorithm, mathematical service or the Spheroid function approximation.
## Hire Someone To Do Your Spheroid Class
If you would like to end up being knowledgeable about the Quartic Class, then you require to very first start by checking out the online Quartic page. This page will reveal you how to utilize the Class by using your keyboard. The explanation will likewise reveal you how to create your own algebra equations to help you study for your classes.
Prior to you can comprehend how to study for a Spheroid Class, you should first comprehend the use of your keyboard. You will find out how to click the function keys on your keyboard, as well as how to type the letters. There are three rows of function keys on your keyboard. Each row has four functions: Alt, F1, F2, and F3.
By pressing Alt and F2, you can increase and divide the worth by another number, such as the number 6. By pressing Alt and F3, you can utilize the 3rd power.
When you press Alt and F3, you will type in the number you are attempting to increase and divide. To multiply a number by itself, you will push Alt and X, where X is the number you want to multiply. When you push Alt and F3, you will key in the number you are attempting to divide.
This works the exact same with the number 6, other than you will just key in the two digits that are 6 apart. Finally, when you push Alt and F3, you will utilize the 4th power. However, when you push Alt and F4, you will utilize the real power that you have actually discovered to be the most proper for your issue.
By using the Alt and F function keys, you can increase, divide, and after that use the formula for the 3rd power. If you require to multiply an odd variety of x's, then you will need to get in an even number.
This is not the case if you are attempting to do something complex, such as multiplying two even numbers. For instance, if you want to increase an odd variety of x's, then you will require to go into odd numbers. This is particularly true if you are attempting to figure out the answer of a Spheroid Class.
If you want to convert an odd number into an even number, then you will require to push Alt and F4. If you do not know how to multiply by numbers by themselves, then you will require to use the letters x, a b, c, and d.
While you can multiply and divide by utilize of the numbers, they are a lot easier to utilize when you can look at the power tables for the numbers. You will need to do some research when you initially begin to use the numbers, however after a while, it will be second nature. After you have actually created your own algebra formulas, you will be able to develop your own reproduction tables.
The Spheroid Solution is not the only way to fix Spheroid equations. It is important to find out about trigonometry, which uses the Pythagorean theorem, and then utilize Spheroid formulas to fix problems. With this method, you can know about angles and how to solve problems without having to take another algebra class.
It is necessary to try and type as quickly as possible, due to the fact that typing will help you understand about the speed you are typing. This will help you write your responses quicker.
## Hire Someone To Take My Spheroid Class
A Spheroid Class is a generalization of a direct Class. For instance, when you plug in x=a+b for a given Class, you get the value of x. When you plug in x=a for the Class y=c, you get the values of x and y, which give you an outcome of c. By using this basic principle to all the formulas that we have tried, we can now resolve Spheroid formulas for all the worths of x, and we can do it rapidly and efficiently.
There are numerous online resources offered that provide complimentary or affordable Spheroid formulas to resolve for all the values of x, consisting of the cost of time for you to be able to make the most of their Spheroid Class assignment help service. These resources usually do not require a membership fee or any sort of financial investment.
The answers supplied are the result of complex-variable Spheroid formulas that have actually been resolved. This is likewise the case when the variable used is an unidentified number.
The Spheroid Class is a term that is an extension of a direct Class. One benefit of using Spheroid formulas is that they are more general than the direct formulas. They are much easier to resolve for all the worths of x.
When the variable used in the Spheroid Class is of the form x=a+b, it is easier to resolve the Spheroid Class due to the fact that there are no unknowns. As a result, there are fewer points on the line specified by x and a continuous variable.
For a right-angle triangle whose base indicate the right and whose hypotenuse indicate the left, the right-angle tangent and curve chart will form a Spheroid Class. This Class has one unknown that can be found with the Spheroid formula. For a Spheroid Class, the point on the line specified by the x variable and a continuous term are called the axis.
The presence of such an axis is called the vertex. Because the axis, vertex, and tangent, in a Spheroid Class, are an offered, we can find all the worths of x and they will sum to the given worths. This is accomplished when we use the Spheroid formula.
The factor of being a constant aspect is called the system of equations in Spheroid equations. This is often called the central Class.
Spheroid formulas can be solved for other values of x. One method to solve Spheroid equations for other values of x is to divide the x variable into its factor part.
If the variable is offered as a positive number, it can be divided into its factor parts to get the regular part of the variable. This variable has a magnitude that is equal to the part of the x variable that is a consistent. In such a case, the formula is a third-order Spheroid Class.
If the variable x is unfavorable, it can be divided into the exact same part of the x variable to get the part of the x variable that is increased by the denominator. In such a case, the formula is a second-order Spheroid Class.
Service help service in fixing Spheroid formulas. When using an online service for fixing Spheroid equations, the Class will be solved immediately. | 2,002 | 8,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-17 | latest | en | 0.925288 |
https://www.mail-archive.com/everything-list@googlegroups.com/msg34678.html | 1,521,292,887,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645069.15/warc/CC-MAIN-20180317120247-20180317140247-00134.warc.gz | 844,055,928 | 5,563 | # Re: Math-> Computation-> Mind -> Geometry -> Space -> Matter
```
On 13 Jan 2013, at 10:46, Alberto G. Corona wrote:```
```
```
```I Bruno.
```
I wanted to put geometry in the chain because materialists seems to base their firm belief in the fact that space is both in mathematics, in the reality and in the mind, so space it is the firm thing where "real" things are located. I try to show that space is just our mental representation of a mathematical reality in R3 where information with survival value is presented and "colored" . This information is the matter. and therefore space and matter is only on the mind.
```
```
That can be locally correct, but is part of what I want an explanation. Geometry, topology, analysis *and* physics should emerge from the arithmetical (notably from the "view from inside. geometry is tricky because we have a qualia for the space of dimension 3 (and lower), but none for higher dimension, and I still don't know if this is a necessity or if it is contingent. Can we hardwired a machine so that he could imagine, and have qualia, for higher than 3 dimensional space?
```
```
```
```
Both chains can be alternative descriptions of the same cosmology (basically). since Arithmetic + computation unfold the set of all structures,
```
```
Only the subjective structure. The "objective structure of those subjective structure" is beyond arithmetic. There is sort of "Skolem paradox". With comp, arithmetic got "inside views", and the content of those views are bigger than arithmetic.
```
```
including the ones with good properties of simplicity etc. for biology. But there is an introduction of consciousness in your chain that is lacking in the one I propose.
```
Consciousness is mind in the first person perspective.
```
```
```
I'm conscious that mine is incomplete since the mind (or consciousness in your case) appears as a derivative and this is not so, since existence properly seen, is not possible without consciousness and therefore it must be more at the beginning of the chain by definition.
```
```
Here I disagree,if only methodologically. Consciousness is too much interesting to be taken as an assumption. Computer science suggest an explantion of consciousness in term of the truth that machine cannot avoid, despite they remain unjustifiable. It is really the coupling consciousness/material-realities which emerges from the addition and multiplication of natural numbers.
```
```
```
```
A better chain would be, with (<->) in the two first steps since the mind or in your case consciousness is the selector of existence.
```
```
Like in the UD Argument. There is no "magic" involved. I assume comp, and derive from it.
```
```
in your case, I think that consciousness would "cause-back" Arithmetic and computation:
```
Exactly: cause back, but not at the same logical state.
```
```Math<-> Computation<-> Mind -> Geometry -> Space -> Matter
```
We have only dreams, strictly speaking, and we must justifies in detail why we can share some of them.
```
Bruno
```
```
2013/1/13 Bruno Marchal <marc...@ulb.ac.be>
On 12 Jan 2013, at 13:48, Alberto G. Corona wrote:
```
Space and time may be a perception of the mind in the Kantian sense. I donĀ“t find that space must be independent of the mind. space and time may be the way we perceive a space-time manifold which is pure mathematical and nothing else. Maybe we can see space out there and we can think on geometrical figures in space (not algebraically) because we have space-mode rasoning on the mind, not because space is pre-existent to the mind, neither because space is something in mathematics, because space is described in math without gemetry.
```
```
And may be that the autopoietic computation, in the forms of natural selection, life and mind are trajectories in the space-time manifold, which, when looked closely form outside space-time, they are nothing but fortunate collisions of particle trajectories and molecules so that entropy stay controlled along these lines, with no reason but fortunate manifold structure and fortunate initial conditions. But looked from inside it appears to have phenomena like matter space, causality, termodinamic irreversibility, time, minds etc.
```
```
OK. My point is that if we assume computationalism it is necessarily so, and constructively so, so making that hypothesis testable.
```
We have the logical entaiment:
```
Arithmetic -> computations -> consciousness -> sharable dreams -> physical reality/matter -> human biology -> human consciousness.
```
```
It is a generalization of "natural selection" operating from arithmetical truth, and in which the physical reality is itself the result of a self-selection events (the global first person indeterminacy).
```
This generalizes both Darwin and Everett, somehow.
Bruno
http://iridia.ulb.ac.be/~marchal/
--
```
You received this message because you are subscribed to the Google Groups "Everything List" group.
```To post to this group, send email to everything-list@googlegroups.com.
```
To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/everything-list?hl=en .
```
--
Alberto.
--
```
You received this message because you are subscribed to the Google Groups "Everything List" group.
```To post to this group, send email to everything-list@googlegroups.com.
```
To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/everything-list?hl=en .
```
http://iridia.ulb.ac.be/~marchal/
--
You received this message because you are subscribed to the Google Groups
"Everything List" group.
To post to this group, send email to everything-list@googlegroups.com.
To unsubscribe from this group, send email to
For more options, visit this group at
``` | 1,328 | 5,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-13 | latest | en | 0.948861 |
https://captsschaudhari.com/2021/07/31/watchkeeping-art-of-navigation/ | 1,685,432,481,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645417.33/warc/CC-MAIN-20230530063958-20230530093958-00230.warc.gz | 185,231,866 | 24,454 | Q. How do you monitor the position being on the course line, without plotting it?
This is done by parallel indexing technique. While coasting, on every course a target must be identified for doing the monitoring by parallel indexing. On chart, the distance to pass a target is found by finding the nearest distance to pass, by drawing, an arc from the nearest tip of the target on course line.
On the radar screen, an independent EBL is drawn parallel to the ship’s course tangenting the VRM circle of the distance at which to pass the shore object. As the ship moves, the VRM must move ahead with the shore object maintaining tangency on the EBL.
The target moving shipwards of the EBL would indicate that the ship is drifting sideways towards the shore.
The target moving on the non-shipward of the EBL would indicate that the ship is drifting sideways away from the shore.
Q. In the middle of ocean, how would you find out the ship’s approximate latitude? You just have an old almanac.
The declination of the Sun usually remains constant with respect to the calendar dates. This is because the tropical period of revolution is taken into consideration when making calendar. Thus, if the ship’s latitude is around 20oN, and the declination for the day is say 15oS, the Sun at meridian passage would bear south and the altitude will be around 55o. Thus, if the altitude is 54o55’ at merpass, the MZD will be 35o05’ and the observed altitude would be 35o05’ – 15o = 20o05’N.
Q. How can the station pointer be used for finding out the ship’s position in coastal waters?
The horizontal angle between the lights is also the angular separation between the arms of a station pointer. This way if the tips of the arm are placed on the respective light houses on the chart. The centre of the central disc of station pointer is the position of the ship.
Q. You just have a harbor chart, and the only navigational equipment you have is the sextant. Can you find the position?
The sextant can be used to find the horizontal angles between the lights. The horizontal angle principle is used to get two position circles. The intersection is the position.
Q. How do you get the position circle?
Suppose the horizontal angle between one of the sets of light houses is θ, the lines are drawn to the shipward side of line joining the two light houses. The angle subtended should be 90-θ at both the lighthouses. To make an isosceles triangle. The point of intersection is also a circumcentre to draw the position circle.
Q. What is the fastest way to find the compass error while in harbor or coasting close to land?
A quick way of finding out compass error is by taking a transit bearing. A duty officer must be prepared to take such bearing before hand during pilotage, etc.
Q. Is there any other terrestrial method of finding the compass error?
In a harbor, while alongside a berth, ensuring that the ship is parallel to the wharf, the direction of wharf and the ship’s head must be same, thus, giving compass error. While approaching or leaving a port, the leading lights also can give the compass error.
Q. What is the main purpose of the leading lights?
The main purpose of the leading lights is to detect a cross drift which may drift the vessel out of channel. Thus, while approaching or leaving these lights must appear in the same vertical line.
(You may also visit my youtube videos @captsschaudhari.com) | 751 | 3,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-23 | longest | en | 0.934311 |
http://www.ck12.org/book/CK-12-Middle-School-Math-Concepts-Grade-8/r12/section/5.9/ | 1,490,275,303,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186895.51/warc/CC-MAIN-20170322212946-00597-ip-10-233-31-227.ec2.internal.warc.gz | 468,150,363 | 44,026 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 5.9: Use the Percent Equation to Find the Base, b
Difficulty Level: At Grade Created by: CK-12
Estimated5 minsto complete
%
Progress
Practice Percent Equation to Find the Base b
MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated5 minsto complete
%
Estimated5 minsto complete
%
MEMORY METER
This indicates how strong in your memory this concept is
Have you ever enjoyed autumn? Take a look at this dilemma involving trees and their leaves.
By mid-September, 50% of the trees lose their leaves. If 850 trees in a grove lost their leaves, how many trees are there in all?
Use the percent equation to solve this problem. You will learn all that you need to know in this Concept.
### Guidance
Did you know that you can use the percent equation to solve percent problems? Take a look at how you can move from the proportion to the percent equation.
ab=p100\begin{align*}\frac{a}{b}=\frac{p}{100}\end{align*}
When we solve the proportion ab=p100\begin{align*}\frac{a}{b}=\frac{p}{100}\end{align*}, we use cross products to find the missing variable. However, even if we leave it in terms of the variables, we can still use cross multiply.
ab100aaa=p100=pb=pb100=.01pb\begin{align*}\frac{a}{b} &= \frac{p}{100}\\ 100 a &= pb\\ a &= \frac{pb}{100}\\ a &= .01pb\end{align*}
If we change the percent to a decimal by moving the decimal point two places to the left, then there is no need to multiply p\begin{align*}p\end{align*} by .01 as we will have already accounted for the coefficient of .01 by moving the decimal point.
Okay, let’s go through it again. Look at what we just wrote.
We wrote the same thing we just didn’t include values. The variables stayed and we multiplied them.
The key is that if we change the percent to a decimal, then all we have to do is to multiply it by the base and we will be able to figure out the value of a\begin{align*}a\end{align*}.
This situation showed us how to use go from a proportion to the percent equation when solving for part a.
Sometimes, you will know the percent and a part of the ratio, or part a\begin{align*}a\end{align*}, but you will need to find the whole or the base, b\begin{align*}b\end{align*}. When this happens, you can use the same key words as before and simply figure out the base by using the percent equation. Let’s look at one like this.
78 is 65% of what number?
Here we know that the word “is” means equals. The numbers may be in a different location, but just pay attention to the key words and you will know what to do. Notice that we have been given the percent and we are missing the “of what number” that is the value of the base. Let’s write the equation.
78=65%b\begin{align*}78 = 65\% b\end{align*}
To work with the 65%, it makes sense to convert it to a decimal. We do this by dropping the percent sign and moving the decimal two places to the left.
78=.65b\begin{align*}78 = .65b\end{align*}
Now we can solve it for the value of b\begin{align*}b\end{align*}. Divide both sides of the equation by .65.
78.65120=.65b.65=b\begin{align*}\frac{78}{.65} &= \frac{.65b}{.65}\\ 120 &= b\end{align*}
Here is another one.
11 is 77% of what number?
Once again, pay attention to the key words. You can see that we are once again going to be looking for the value of the base. Let’s write the equation.
11=77%b\begin{align*}11 = 77 \% b\end{align*}
Convert the percent to a decimal and solve.
1111.7714.28=.77b=b=b\begin{align*}11 &= .77b\\ \frac{11}{.77} &= b\\ 14.28 &= b\end{align*}
In this problem, you could round to the nearest hundredths place as we did here. Sometimes, you may be asked to round to the nearest tenths place. In that case, the answer would have been 14.3.
Solve each problem using the percent equation.
#### Example A
10 is 50% of what number?
Solution: 20\begin{align*}20\end{align*}
#### Example B
45 is 20% of what number?
Solution: 225\begin{align*}225\end{align*}
#### Example C
68 is 40% of what number?
Solution: 170\begin{align*}170\end{align*}
Now let's go back to the dilemma from the beginning of the Concept.
Let’s start by breaking apart this problem. We have a percent, so we know that we won’t be looking for the percent. We know that 850 trees in a grove lost their leaves, but we don’t know the total number of trees in the grove. The total could be thought of as the whole and this is the base. We are going to be looking for the base.
Let’s write the equation.
850=.50b\begin{align*}850 = .50b\end{align*}
Now we solve by dividing both sides of the equation by .50.
850.501700=.50b.50=b\begin{align*}\frac{850}{.50} &= \frac{.50b}{.50}\\ 1700 &= b\end{align*}
There are 1700 trees in the grove.
### Vocabulary
Percent
a part of a whole out of 100.
### Guided Practice
Here is one for you to try on your own.
25 is 60% of what number?
Solution
First, let's write down the percent equation.
100a=pb\begin{align*}100a = pb\end{align*}
In this problem, we are solving for the base. Let's fill in the values that we know.
100(25)=60b\begin{align*}100(25) = 60b\end{align*}
2500=60b\begin{align*}2500 = 60b\end{align*}
Now we divide to solve for b\begin{align*}b\end{align*}.
250060=b\begin{align*}\frac{2500}{60} = b\end{align*}
41.6=42\begin{align*}41.6 = 42\end{align*}
Our answer is 41.6\begin{align*}41.6\end{align*} or 42\begin{align*}42\end{align*}.
### Practice
Directions: Solve each percent problem. You may round your answers to the nearest tenth when necessary.
1. 23 is 9% of what number?
2. 10 is 35% of what number?
3. 580 is 82% of what number?
4. 58 is 8% of what number?
5. 58 is 80% of what number?
6. 11 is 82% of what number?
7. 33 is 2% of what number?
8. 14 is 9% of what number?
9. 50 is 67% of what number?
10. 33 is 45% of what number?
11. 40 is 80% of what number?
12. 68 is 99% of what number?
13. 78 is 55% of what number?
14. 16 is 12% of what number?
15. 1450 is 80% of what number?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Decimal In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths).
Equation An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs.
Show Hide Details
Description
Difficulty Level:
Authors:
Tags:
Subjects:
## Concept Nodes:
Date Created:
Jan 23, 2013
Aug 11, 2016
Image Detail
Sizes: Medium | Original
MAT.ARI.764.L.2 | 2,004 | 6,853 | {"found_math": true, "script_math_tex": 26, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2017-13 | longest | en | 0.867976 |
http://mathhelpforum.com/calculus/208853-infimum-integral-open-set.html | 1,481,423,527,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543782.28/warc/CC-MAIN-20161202170903-00426-ip-10-31-129-80.ec2.internal.warc.gz | 177,111,494 | 10,668 | # Thread: Infimum of integral of open set
1. ## Infimum of integral of open set
I have already done part a and b. Part a is easy, for part b, i let the anti-derivative of f to be k(t)+c and arrive at the answer that F(f)= 1/2+ 2*k(1/2) - k(1). But i don't know how to do the next part, can anyone give me a hint? the question c ask me to show that the infimum of F is 0 and it is never attained on A.
2. ## Re: Infimum of integral of open set
Hi kanezila,
In part b, to prove that $F$ is continuous at $f$ we need to show that for $\epsilon>0$ there is $\delta>0$ such that whenever $\|f-g\|_{\infty}<\delta$, we have $|F(f)-F(g)|<\epsilon.$ Perhaps looking at an antiderivative will go somewhere, but this is probably best done by using the definition.
To prove part c looking at the following sequence of functions $\{f_{n}\}$ in $A$, where $f_{n}(x)=0$ for $x\in [0,1/2]$, $f_{n}(x)=(n+2)(x-1/2)$ for $x\in [1/2, 1/2+1/(n+2)]$ and $f_{n}(x)=1$ for $x\in [1/2+1/(n+2), 1].$ Then we should be able to show that $F(f_{n})=\frac{1}{2}(\frac{1}{n+2}).$ Since this goes to 0 as n goes to infinity the infimum must be 0, because each of the $f_{n}$'s belong to $A.$
Does this clear things up? Good luck!
3. ## Re: Infimum of integral of open set
thank you very much for this, i have been trying to do this question for a whole day already.
4. ## Re: Infimum of integral of open set
could you tell me how to prove the first one???many thanks | 482 | 1,447 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2016-50 | longest | en | 0.915914 |
http://mathhelpforum.com/algebra/193099-3x-y-4-2-z.html | 1,526,895,743,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863972.16/warc/CC-MAIN-20180521082806-20180521102806-00099.warc.gz | 190,448,931 | 8,638 | # Thread: 3x + y = 4(2-z)
2. ## Re: 3x + y = 4(2-z)
Originally Posted by ingyaningya
$\displaystyle x = \frac{4(2-z) -y}{3}$
$\displaystyle y = 4(2-z)-3x$
$\displaystyle z = 2-\frac{3x+y}{4}$
What do you exactly need solve?
For "x"
4. ## Re: 3x + y = 4(2-z)
Originally Posted by ingyaningya
3x + y = 4(2-z)...Solve for x.
Subtract y from both sides. By doing so, you will isolate the term 3x.
3x = 4(2 - z) - y
We now divide both sides by the coefficient of x, which is the number 3.
x = [4(2 - z) - y]/3
By the way, you can use the distributive rule in terms of 4(2 - z) to simplify further if needed. | 235 | 616 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-22 | latest | en | 0.818492 |
https://archive.softwareheritage.org/browse/content/sha1_git:780667a7c0b9a82732b34ae7fe52cb42d6c3924a/?path=9f4acbc2ed8a1db0ead5aa29ca043125a2c56313/user_guide/tensor_basics.rst | 1,720,912,259,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00634.warc.gz | 94,600,063 | 10,407 | tensor_basics.rst
Tensor basics
=============
Creating a tensor
-----------------
A tensor is nothing more than a multi-dimensional array.
Let's take for this example the tensor :math:\tilde X defined by its frontal slices:
.. math::
X_1 =
\left[
\begin{matrix}
0 & 2 & 4 & 6\\
8 & 10 & 12 & 14\\
16 & 18 & 20 & 22
\end{matrix}
\right]
\text{and}
X_2 =
\left[
\begin{matrix}
1 & 3 & 5 & 7\\
9 & 11 & 13 & 15\\
17 & 19 & 21 & 23
\end{matrix}
\right]
In Python, this array can be expressed as a numpy array::
>>> import numpy as np
>>> import tensorly as tl
>>> X = tl.tensor(np.arange(24).reshape((3, 4, 2)))
You can view the frontal slices by fixing the last axis::
>>> X[..., 0]
array([[ 0, 2, 4, 6],
[ 8, 10, 12, 14],
[16, 18, 20, 22]])
>>> X[..., 1]
array([[ 1, 3, 5, 7],
[ 9, 11, 13, 15],
[17, 19, 21, 23]])
Unfolding
---------
Also called **matrization**, **unfolding** a tensor is done by reading the element in a given way as to obtain a matrix instead of a tensor.
For a tensor of size :math:(I_1, I_2, \cdots, I_n), the k-mode unfolding of this tensor will be of size :math:(I_k, I_1 \times \cdots \times I_{k-1} \times I_{k+1} \cdots \times I_n).
.. important::
In tensorly we use an unfolding different from the classical one as defined in [1]_ for better performance.
Given a tensor :math:\tilde X \in \mathbb{R}^{I_1 \times I_2 \times \cdots \times I_N}, the
mode-n unfolding of :math:\tilde X is a matrix :math:\mathbf{X}_{[n]} \in \mathbb{R}^{I_n, I_M},
with :math:M = \prod_{\substack{k=1,\\k \neq n}}^N I_k and is defined by
the mapping from element :math:(i_1, i_2, \cdots, i_N) to :math:(i_n, j), with
.. math::
j = \sum_{\substack{k=1,\\k \neq n}}^N i_k \times \prod_{m=k+1}^N I_m.
.. warning::
Traditionally, mode-1 unfolding denotes the unfolding along the first dimension.
However, to be consistent with the Python indexing that always starts at zero,
in tensorly, unfolding also starts at zero!
Therefore unfold(tensor, 0) will unfold said tensor along its first dimension!
For instance, using the :math:\tilde X previously defined, the 0-mode unfolding of :math:\tilde X:
.. math::
\tilde X_{[0]} =
\left[ \begin{matrix}
0 & 1 & 2 & 3 & 4 & 5 & 6 & 7\\
8 & 9 & 10 & 11 & 12 & 13 & 14 & 15\\
16 & 17 & 18 & 19 & 20 & 21 & 22 & 23\\
\end{matrix} \right]
The 1-mode unfolding is given by:
.. math::
\tilde X_{[1]} =
\left[ \begin{matrix}
0 & 1 & 8 & 9 & 16 & 17\\
2 & 3 & 10 & 11 & 18 & 19\\
4 & 5 & 12 & 13 & 20 & 21\\
6 & 7 & 14 & 15 & 22 & 23\\
\end{matrix} \right]
Finally, the 2-mode unfolding is the unfolding along the last axis:
.. math::
\tilde X_{[2]} =
\left[ \begin{matrix}
0 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 & 20 & 22\\
1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 & 17 & 19 & 21 & 23\\
\end{matrix} \right]
In tensorly:
.. code-block:: python
>>> from tensorly import unfold
>>> unfold(X, 0) # mode-1 unfolding
array([[ 0, 1, 2, 3, 4, 5, 6, 7],
[ 8, 9, 10, 11, 12, 13, 14, 15],
[16, 17, 18, 19, 20, 21, 22, 23]])
>>> unfold(X, 1) # mode-2 unfolding
array([[ 0, 1, 8, 9, 16, 17],
[ 2, 3, 10, 11, 18, 19],
[ 4, 5, 12, 13, 20, 21],
[ 6, 7, 14, 15, 22, 23]])
>>> unfold(X, 2) # mode-3 unfolding
array([[ 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22],
[ 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23]])
Folding
-------
You can **fold** an unfolded tensor back from matrix to full tensor using the :func:tensorly.base.fold function.
.. code-block:: python
>>> from tensorly import fold
>>> unfolding = unfold(X, 1)
>>> original_shape = X.shape
>>> fold(unfolding, 1, original_shape)
array([[[ 0, 1],
[ 2, 3],
[ 4, 5],
[ 6, 7]],
[[ 8, 9],
[10, 11],
[12, 13],
[14, 15]],
[[16, 17],
[18, 19],
[20, 21],
[22, 23]]])
References
----------
.. [1] T.G.Kolda and B.W.Bader, "Tensor Decompositions and Applications",
SIAM REVIEW, vol. 51, n. 3, pp. 455-500, 2009. | 1,570 | 3,853 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-30 | latest | en | 0.698568 |
https://prezi.com/mlzgwjpa4hpw/second-grade-mathematics-lesson-plan/ | 1,506,359,500,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818692236.58/warc/CC-MAIN-20170925164022-20170925184022-00205.warc.gz | 702,926,828 | 22,646 | ### Present Remotely
Send the link below via email or IM
• Invited audience members will follow you as you navigate and present
• People invited to a presentation do not need a Prezi account
• This link expires 10 minutes after you close the presentation
Do you really want to delete this prezi?
Neither you, nor the coeditors you shared it with will be able to recover it again.
You can change this under Settings & Account at any time.
# Second Grade Mathematics Lesson Plan
Whole numbers and place value
by
## Chelsea Atkin
on 24 September 2012
Report abuse
#### Transcript of Second Grade Mathematics Lesson Plan
How are numbers organized? Place Value with Whole Numbers By: Chelsea Atkin Objective 1
Identify and represent place
value in numbers up to 100
using base ten blocks. Standard 1
Students will acquire
number sense with whole
numbers and identify place value
in whole numbers. Are all positive numbers
starting with 0. Whole Numbers Place Value The value of where the digit is in
the number, such as ones, tens,
hundreds, etc. Now let's practice with the place value song! And we get... Now let's solve this together 100 + 40 + 5 = 145 (hundreds) (tens) (ones) Click on the link below to get started Try this on your own http://www.free-training-tutorial.com/place-value/clickthedigit.html (n.d.). Retrieved from http://www.free-training-tutorial.com/place-value/clickthedigit.html APA References Place value math song: ones, tens, hundreds. (2011) | 345 | 1,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-39 | longest | en | 0.829867 |
https://www.daniweb.com/programming/software-development/threads/351990/to-find-a-largest-sector-in-a-given-input-string-of-1-s-and-0-s | 1,670,435,928,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711200.6/warc/CC-MAIN-20221207153419-20221207183419-00662.warc.gz | 762,076,480 | 15,934 | Hello,
Help me in writing a logic for the program:
User input:
11000
01100
00101
10001
01011
Two 1's are said to be connected if they are adjacent to each other horizontally, vertically & diagonally.
Need to find the largest sector of 1's in the above input string.
For example: In the above input string the largest sector is of size 5.
Thank you
## All 4 Replies
What is your current logic to the problem? I can give you a hint, use a search and iterate through the area until no more connected 1's is found. Don't forget to keep track of where you have already visited. The method could be recursive (easy to implement) or iterative (a bit longer).
Start by figuring out what the steps are at a high level, then figure out how to execute them. Executing them may, and probably will, involve decomposing them into steps which you'll then re-examine in the same way until you have methods which are trivial to execute. Write the trivial methods.
Test.
If results pass tests, done. Else, find and fix bugs.
Start by figuring out what the steps are at a high level, then figure out how to execute them. Executing them may, and probably will, involve decomposing them into steps which you'll then re-examine in the same way until you have methods which are trivial to execute. Write the trivial methods.
Test.
If results pass tests, done. Else, find and fix bugs.
I need to read a file for input like:
Sample Input
``````2
11000
01100
00101
10001
01011
1011
1010
``````
Sample Output
``````5
3
``````
Sample Input 2 indicates there are two grids.
`````` FileReader fr = new FileReader("input.txt");
String s;
while( (s = br.readLine()) != null)
{
System.out.println(s.length());
}
``````
How should I proceed further now?
Get the data
Solve the problem somehow for however many grids are submitted.
Report the result.
That middle step needs a little work.
You know you're going to have to solve the problem for one gird, so maybe you should work on that. Solve that, and you should be able to extend it to multiple grids without much trouble.
I'll give you this much for free: it'll look a little like this
private int solve(Grid g) {}
where Grid might be a 2D array or a custom object or whatever structure makes it easiest for you to solve it. Return your solution as an int. When you do multiple grids, you'll stuff that in an array or an ArrayList or whatever makes most sense to you at that point. Don't worry about that now, though, just return the int so you can use it later.
But now you have your solve method to break down.
Be a part of the DaniWeb community
We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, learning, and sharing knowledge. | 650 | 2,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-49 | latest | en | 0.946615 |
https://testbook.com/question-answer/find-the-smallest-number-that-can-be-expressed-as--5fe9e5d08c14b207b3c51388 | 1,642,327,007,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299852.23/warc/CC-MAIN-20220116093137-20220116123137-00189.warc.gz | 639,613,102 | 30,034 | # Find the smallest number that can be expressed as a sum of two cubes in two different ways.
1. 91
2. 341
3. 855
4. 1729
Option 4 : 1729
## Detailed Solution
Calculation:
1729 = 1728 + 1 = 123 + 13
1729 = 1000 + 729 = 103 + 9
The number 1729 is known as Hardy – Ramanujan Number.
FACT:
This is a story about one of India’s great mathematical geniuses, S. Ramanujan.
Once another famous mathematician Prof. G.H. Hardy came to visit him in a taxi whose number was 1729. While talking to Ramanujan, Hardy described this number
“a dull number”. Ramanujan quickly pointed out that 1729 was indeed interesting. He said it is the smallest number that can be expressed as a sum of two cubes in two different ways: | 211 | 717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-05 | latest | en | 0.918135 |
http://dreaminnet.com/error-bars/adding-error-bars-to-bar-graphs-excel.php | 1,545,140,558,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829399.59/warc/CC-MAIN-20181218123521-20181218145521-00341.warc.gz | 81,392,598 | 5,142 | Home > Error Bars > Adding Error Bars To Bar Graphs Excel
# Adding Error Bars To Bar Graphs Excel
## Contents
Top of Page Change the error amount options On a 2-D area, bar, column, line, xy (scatter), or bubble chart, click the error bars, the data point, or the data series Choose the style of the error bar. I have written a new article about Error Bars in Excel 2007. All X error bars in a series, and all Y error bars in a series, are formatted the same. this content
There are other ways to use error bars to embellish Excel charts, as listed at the end of this article. Please help feeling frustrated. Apple disclaims any and all liability for the acts, omissions and conduct of any third parties in connection with or related to your use of the site. The resulting Y error bars or X error bars are the same size and do not vary with each data point.
## How To Add Error Bars To Bar Graph In Excel 2010
Line chart showing error bars with Standard deviation(s) of 1.3 If you need to specify your own error formula, select Custom and then click the Specify Value button to This option Uses this equation Where Standard Error s = series number i = point number in series s m = number of series for point y in chart n = lgines 203,670 views 4:01 Standard Deviation Error Bars Excel 2010 - Duration: 1:31. Excel for Beginners Excel Essentials Excel for Business - Learn Excel Online Filed Under: Business, Students Top Udemy Courses: Top Java Courses Top Python Courses Top Excel Courses Learn Excel With
Images were taken using Excel 2013 on the Windows 7 OS. Change Fixed Value to Custom. This displays the Chart Tools, adding the Design, Layout, and Format tabs. Vertical Error Bars In Excel Doug H 92,901 views 4:18 Graphing Means and Standard Deviations with Excel - Duration: 9:13.
Note: The direction of the error bars depends on the chart type of your chart. Sign in 48 Loading... Click Close. Scatter charts can show both horizontal and vertical error bars.
You can define a dynamic range that contains error bar values, akin to defining dynamic ranges for the X and Y values of a chart series in a dynamic chart. How To Insert Error Bars In Excel Mac Sign in to add this video to a playlist. At -195 degrees, the energy values (shown in blue diamonds) all hover around 0 joules. Set style of error bars, including color, width, and transparency Select data for error bars.
## How To Add Error Bars To Bar Graph In Excel 2013
The formula used for error bar calculation will appear in the Set Column Values dialog. Get More Information You can also remove error bars. How To Add Error Bars To Bar Graph In Excel 2010 Note: The direction of the error bars depends on the chart type of your chart. Bar Graph With Error Bars Excel Mac I just switched my spreadsheet to Numbers, and now it all makes sense...!
Loading... http://dreaminnet.com/error-bars/adding-error-bars-to-bar-graphs-in-excel-2007.php However when I follow these steps..."Select the range with the averages and Chart it. Every time I change the cell range, it affects every column. Nestor Matthews 12,819 views 14:12 Add Error Bars to a Line Chart - Duration: 4:18. Error Bars Line Graphs
However, you can also use More Error Bars Options…. I appreciate any help. Click the Error Bars tab and choose Positive and Negative. http://dreaminnet.com/error-bars/adding-error-bars-to-excel-graphs.php This distribution of data values is often represented by showing a single data point, representing the mean value of the data, and error bars to represent the overall distribution of the
It's worth it if you use charts a lot.Jerry Helpful (0) Reply options Link to this post by williamhltdudu, williamhltdudu Aug 13, 2011 12:05 AM in response to Alexa M Level Standard Deviation Bar Graphs If you click a second time on just the X or Y bars, just that set will be selected and formatted. To add error bars to a selected data point or data series, click the data point or data series that you want, or do the following to select it from a
## This way the unique standard error value is associated with each mean.
This shows our chart with positive and negative X and Y error bars, using a percentage of 12%. All postings and use of the content on this site are subject to the Apple Support Communities Terms of Use. To use custom values to determine the error amount, click Custom, and then do the following: Click Specify Value. How To Add Error Bars In Excel 2013 Loading...
If your missing error bar shows as just a black line at the top of the column, this might be the case. First click the line in the graph so it is highlighted. A series of error bars can be used as above to generate a custom set of Gridlines in a chart. check my blog There are several ways to enter values: fixed values, a percentage of the point's value, a number of standard deviations, the standard error of the plotted points, and custom values.
When error bars are added to a data plot, the error data is output to a new column on the source worksheet. There are two common ways you can statistically describe uncertainty in your measurements. Once you have calculated the mean for the -195 values, then copy this formula into the cells C87, etc. If you want different formatting for individual error bars, such as different color error bar lines, you will have to add a series for each format you want.
Jim Ham 403,222 views 12:21 How to make a bar graph in Excel (Scientific data) - Duration: 4:01. Where do I find the Chart Inspector? Customizing error bars The Plot details dialog provides customization controls for error bars in both 2D and 3D graphs. When you are done, click OK.
Alexa M Level 1 (0 points) Q: Adding Standard Deviation Bars to graphs Hello, I searched to find the answer to my question, but when I tried a few of the Result: Note: if you add error bars to a scatter chart, Excel also adds horizontal error bars. OriginPro What's new in latest version Product literature SHOWCASE Applications User Case Studies Graph Gallery Animation Gallery 3D Function Gallery FEATURES 2D&3D Graphing Peak Analysis Curve Fitting Statistics Signal Processing Key Top of Page Change the display of error bars On a 2-D area, bar, column, line, xy (scatter), or bubble chart, click the error bars, the data point, or the data
Skip navigation UploadSign inSearch Loading... Press DELETE. How can we improve it? Error bars are a useful and flexible tool that help add detail to an Excel chart.
Do one of the following: On the Layout tab, in the Analysis group, click Error Bars, and then click None. Show more Language: English Content location: United States Restricted Mode: Off History Help Loading... Reply sana says: June 21, 2016 at 7:46 am I am trying to add Error Bars in data based on standard deviation in Excel 2007-2010. And as Alexa wrote it still selects both bars at the same time, it is impossible only to select one at at time!Thank you for you quick replyLouise Helpful (0) Reply | 1,576 | 6,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-51 | latest | en | 0.785573 |
https://deepoceanpowerphilippines.com/qa/quick-answer-how-zero-gravity-is-created.html | 1,623,506,596,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487584018.1/warc/CC-MAIN-20210612132637-20210612162637-00219.warc.gz | 198,290,618 | 9,481 | # Quick Answer: How Zero Gravity Is Created?
## What happens when gravity is zero?
A lack of gravity would eventually take its toll on our very planet, writes Masters.
“Earth itself would most likely break apart into chunks and float off into space.” …
Without the force of gravity to hold it together, the intense pressures at its core would cause it to burst open in a titanic explosion..
## Is Earth’s gravity decreasing?
“The Earth’s gravity field changes from one month to the next mostly due to the mass of water moving around on the surface,” said Watkins. … “Most models assume that the total mass of the ocean is constant — that there is no water being added to it or taken away.
## Can we create microgravity on Earth?
There are very few ways to simulate microgravity on Earth; besides the drop towers, microgravity research takes place in underwater neutral buoyancy simulators and in the sky in the C-9 astronaut training aircraft that is able to achieve about 20 seconds of near-weightlessness as it soars and dips.
## Does NASA have a zero gravity simulator?
But we do use several facilities to recreate the weightless, or microgravity, conditions of orbit. One of them, NASA Glenn’s Zero Gravity Research Facility, is the largest of its kind in the United States. … As the experiments fall, they are virtually weightless.
## At what height gravity is zero?
Near the surface of the Earth (sea level), gravity decreases with height such that linear extrapolation would give zero gravity at a height of one half of the Earth’s radius – (9.8 m. s−2 per 3,200 km.)
## How much does it cost to do a zero gravity flight?
For one seat on a Zero-G flight, it’ll cost \$5,400 plus 5 percent tax. The package includes breakfast, lunch, professional photos – and seven to eight minutes of weightlessness.
## Does gravity exist in a vacuum?
Yes, gravity does exist in a vacuum. A vacuum does not need to be completely devoid of matter, it just needs to have a lower pressure than the area around it.
## Who invented zero gravity?
Sir Isaac Newton, an English mathematician and physicist, discovered gravity. To help explain this concept better, we can take an example of a block of lead in free fall on planet x. The block is said to be in a state of weightlessness even though it is being pulled down by the planet’s gravity.
## How does zero gravity affect the human body?
One of the major effects of weightlessness that is more long-term is the loss of muscle and bone mass. In the absence of gravity there is no weight load on the back and leg muscles, so they begin to weaken and shrink. … Even destruction and construction processes of bones change when in space.
## What would happen if gravity stopped for 1 second?
When gravity disappears for 1 second the outwards force balanced by the gravity would be released causing a massive explosion.
## What does no gravity feel like?
“The feeling is completely different from being on a roller coaster. It is more like motionlessness than movement. I feel great in zero gravity either floating in place or flying through space. As long as you are in rational control of your movements, zero gravity is the realization of a dream.
## Is it possible to shield a body from gravitational field?
Is it possible to shield a body from gravitational effects? Solution : No, it is not possible to shield a body from gravitational effact because the gravitational force between any two bodies is independent of the intervening medium or the presence of other bosied in their way.
## What planet has no gravity?
Considering that Mercury is the closest object to the sun, you can see that space is a very empty place (at least as far as planets go – in terms of particles and fields, space is not empty). For this reason, falling in space looks like floating.
## Is it possible to create anti gravity?
Aside from the long-running Anti Gravity column in Scientific American, however, there is no such thing as antigravity. … Only way out in deep space, beyond the domain of any planets or stars, can you truly escape gravity. As of yet, no technology exists to neutralize the pull of gravity.
## Is there a zero gravity room on earth?
The Zero Gravity Research Facility is NASA’s premier facility for ground based microgravity research, and the largest facility of its kind in the world. The Zero-G facility is one of two drop towers located at the NASA site in Brook Park, Ohio. The Zero-G facility has been operational since 1966.
## Does zero gravity exist?
No Zero Gravity Contrary to popular belief, there’s no such thing as zero gravity. Weightlessness and zero gravity are two different things. The earth’s gravity keeps the moon in orbit. And astronauts are generally much closer to earth than the moon is, which means that the earth’s pull on them has to be much stronger.
## Why gravity at the Centre of Earth is zero?
If you are at the center of the earth, gravity is zero because all the mass around you is pulling “up” (every direction there is up!). … If the earth were about 36,000 km in diameter with the same mass and length-of-day then the gravity at the equator would be zero. This is the altitude of geostationary orbits. | 1,096 | 5,218 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-25 | latest | en | 0.949509 |
http://perplexus.info/show.php?pid=3166 | 1,720,943,131,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00644.warc.gz | 29,316,720 | 4,677 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
The soccer results (Posted on 2005-06-10)
After the close of a soccer season, with four teams (named A, B, C and D), the final table was:
```TEAM Won Drawn Lost Goals Goals Points
For Against
A 3 0 0 7 1 9
B 1 1 1 2 3 4
C 1 1 1 3 3 4
D 0 0 3 1 6 0```
Knowing that the four teams played against each other one time, and that team A beat team B 3x0, can you deduce the results of the other five games?
See The Solution Submitted by pcbouhid Rating: 3.0000 (5 votes)
Subject Author Date Puzzle Thoughts K Sengupta 2023-06-12 09:58:57 Another solution? ignatious 2005-10-02 06:29:21 Working ALison 2005-07-15 20:07:16 No Subject scott 2005-07-05 13:17:10 Final Scores Rupesh Khandelwal 2005-06-14 13:13:18 re: the solution! pcbouhid 2005-06-13 12:41:19 re: Yawn pcbouhid 2005-06-13 12:39:36 the solution! daniel 2005-06-13 11:45:44 Yawn Charley 2005-06-12 17:11:47 correcting my previous comment pcbouhid 2005-06-11 23:31:41 re: Solution - Carl pcbouhid 2005-06-11 23:30:42 Solution Carl 2005-06-10 23:41:04 re(2): Goals vrs. Points Penny 2005-06-10 17:36:17 re: Score pcbouhid 2005-06-10 16:19:49 Score john 2005-06-10 15:42:33 Solution Penny 2005-06-10 15:02:44 My Solution Jeff Little 2005-06-10 13:37:41 re: Goals vrs. Points Hugo 2005-06-10 13:21:49 Goals vrs. Points Penny 2005-06-10 13:16:33 Spoiler Juggler 2005-06-10 12:56:19
Search: Search body:
Forums (0) | 607 | 1,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-30 | latest | en | 0.695321 |
http://hsquizbowl.org/forums/viewtopic.php?p=28104 | 1,548,075,473,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583792338.50/warc/CC-MAIN-20190121111139-20190121133139-00096.warc.gz | 104,920,210 | 42,889 | Biology in the modern QB era.
This is the holding pen for the best threads containing quiz bowl talk.
recfreq
Wakka
Posts: 167
Joined: Mon Nov 07, 2005 10:11 pm
Location: Japan.
Biology in the modern QB era.
After hearing some rather uneventful biology questions over the last couple of years, I think I can safely say that biology question-writing in general lags behind writing in the physical sciences and math. Take the case of physics, one can find topics coming up in each of the sophmore-junior undergrad level courses in theoretical mechanics (Hamiltonian), E & M (vector potential), particle physics (Higgs boson), etc, but very seldom do you find comparable questions at the same level in the biology curriculum, like developmental bio, immunobio, genetics, etc. When questions do come up in those areas, they lack the clues that provide us with knowledge at the level of those undergrad courses.
I\'ll use examples from ACF fall, but be warned that these are already very good questions. Take a physics question:
\"In 2001, researchers at Lawrence Berkeley Labs discovered it in an aluminum matrix doped with molten lead. A mathematical description of this Wiener process typically begins with the Langevin equation and the Stokes flow for a sphere. Jean Perrin provided evidence for the most famous formulation of this process, for which, in the limit of small times, the rms displacement is proportional to the time squared. In the large time limit, it is a diffusive process which can be described by the Fokker-Planck equation and it has a diffusive constant proportional to the temperature. Observed by its namesake in pollen grains suspended in water, and explained by Einstein using the kinetic theory of gases, FTP name this random motion of small particles in a fluid.
ANSWER: Brownian motion\"
This is something I\'d very much want to hear if I\'ve taken a course in this. It mentions Langevin molec dynamics, Wiener process, and Perrin who ever he is, and Fokker-Planck, with some very detailed clues that I\'d love to hear as a nonspecialist, b/c I feel like I could learn something just from reading this question, or if not, at least look it up.
By contrast, here\'s a biology question from the same packet.
\"Defects in these objects can cause Kearn-Sayre Syndrome or Luft Disease. They contain porins to allow for passive diffusion of selected molecules through their outer membranes. They provide evidence for Margulis’ endosymbiotic theory and are important to evolutionary and genetic studies due to their derivation solely from the mother. They allow some ribosomes in their matrix to allow for processes such as chemiosmosis and the citric acid cycle. FTP, name these organelles whose primary function is to convert organic materials into ATP, the so-called “powerhouses” of the cell.
ANSWER: mitochondria\"
I don\'t mean to pick on whoever wrote this, b/c it is a very good question, but just to pt out the difference in level of sophistication b/t this question and questions in the physical sciences, note 1st that its only really challenging clue is about diseases, and not really ones that mitochondria are really famous for being associated with (I\'d say Leber\'s optic neuropathy is probly more famous), but still the clue is fine, just not the kind that I\'d love to read about if I\'m interested in the mitochondria. Instead, I\'d rather know about say, the petite mutation in yeast, or its intron-lacking DNA coding for ND1 to ND5 of alpha F1 ATPases, or NADH-CoQ reductase. The porin and endosymbiosis clues just don\'t strike me with same force as the Fokker-Planck, not to mention that they are always used (sometimes also for chloroplast).
In general, I\'ve found myself just enjoying listening to the physics and math questions more b/c they have more depth and asks about more interesting things. I remember ACF nats last year being ridiculously hard in the physical sciences, whereas all the bio TUs were pretty darn easy off the end of the 1st sentence or beginning of the 2nd. There just appears to be a trend against writing interesting questions on, say the pentose-phosphate shunt, or neurulation, or proteoglycan, or homologous recombination, or RNA interference. Those are not grad topics; I learned them in sophmore-junior level courses, and they are reasonable to ask, just as reasonable as Franck-Condon and Stern-Gerlach. Given that biology depts are the most diverse in the country (MCB, IB, PB in Berkeley, and something like 5 diff ones at UCLA--btw another proof that biochemistry is chemistry: our Chemistry & Biochemistry dept, also, they both study structure), it\'s hard to fathom not getting a more diverse set of questions and answers.
Finally, I just want to ask if medicine is biology. If the clues are clinical, i.e. based on symptoms and effects, I don\'t see them as being biological b/c there\'s no relevant _academic_ clues. If the clues are physiological, or based on molecular genetics, I\'d be ok with it. But in general, as the USC WIT packets will attest, diseases are easy ways to dodge the bio distribution, b/c you just copy from a clinical handbook. It\'s also not very interesting. Bio players, let us know how you feel, but I don\'t need another question on gonerhea. I know what it is, and it\'s not biology. (It\'s also not really taught until med school.) In an effort to mimick the interest of the physical sciences and math crowd, we should strive for more mechanistic, molecular, physiological clues that tell us about the underlying processes involved instead of relying on memorizing viruses and diseases. I think those outside the field would prefer hearing such questions, just as I\'d prefer hearing about Liouville\'s thm and parity.
Ray Luo, UCLA.
QuizBowlRonin
Wakka
Posts: 182
Joined: Fri May 30, 2003 2:21 am
Location: San Mateo, CA
Contact:
Re: Biology in the modern QB era.
recfreq wrote:After hearing some rather uneventful biology questions over the last couple of years, I think I can safely say that biology question-writing in general lags behind writing in the physical sciences and math. Take the case of physics, one can find topics coming up in each of the sophmore-junior undergrad level courses in theoretical mechanics (Hamiltonian), E & M (vector potential), particle physics (Higgs boson), etc, but very seldom do you find comparable questions at the same level in the biology curriculum, like developmental bio, immunobio, genetics, etc. When questions do come up in those areas, they lack the clues that provide us with knowledge at the level of those undergrad courses.
<snip>
In general, I've found myself just enjoying listening to the physics and math questions more b/c they have more depth and asks about more interesting things. I remember ACF nats last year being ridiculously hard in the physical sciences, whereas all the bio TUs were pretty darn easy off the end of the 1st sentence or beginning of the 2nd. There just appears to be a trend against writing interesting questions on, say the pentose-phosphate shunt, or neurulation, or proteoglycan, or homologous recombination, or RNA interference. Those are not grad topics; I learned them in sophmore-junior level courses, and they are reasonable to ask, just as reasonable as Franck-Condon and Stern-Gerlach. Given that biology depts are the most diverse in the country (MCB, IB, PB in Berkeley, and something like 5 diff ones at UCLA--btw another proof that biochemistry is chemistry: our Chemistry & Biochemistry dept, also, they both study structure), it's hard to fathom not getting a more diverse set of questions and answers.
- Let me preface my thoughts by stating that my philosophy and preferences with regards to science questions, and even writing for ACF in general, can be seen as soft. My approach to writing science questions, and that which has been directed by the head tournament editors while I’ve been editing for ACF (Roger, Ezequiel, and Andrew), is not to go overboard with difficulty, and allow most teams, even those consisting of humanities players, to get:
• - The tossup on the giveaway
- At least 10 points on bonuses if the team knows anything about the given topic
- I consider last year’s the editorship of the ACF Nationals science distribution a failure on my part especially on non-biology topics. Many of the top specialists in science were unable to get questions correct early in the question, when I expect those specialists to get those questions, and many questions even on the giveaways stumped those with some science knowledge. When this occurs, I as the editor have failed to differentiate those who know their subject areas from those that don’t, for the obvious reason that when nobody gets the question right, it is impossible to show who’s better in that subject area. If anything, the science, even at ACF Nationals, should be easier - I should be seeing folks who I expect know the most on particular topics make thirty points on the bonuses and get tossups early. I was terribly disappointed by the fact in the finals round that Selene, Subash, Susan, and Seth did not get tossups that I was sure that one of the three would get on the first line – this is not a reflection of their lack of skill as it is a reflection of poor editing and clue-writing on my part. (This was not exclusive to the playoffs.) Because I wasn’t able to oversee the writing of real-life gettable (as opposed to theoretically should buzz in on the first clue) questions, I may have cost Chicago a chance at the championship (most questions went dead for Michigan).
• - Therefore, tossups like Franck-Condon or RNA interference, while written in a pyramidal manner consistent with ACF values, do not successfully differentiate teams that know their material versus those that do not. There is a difference between knowing what a topic is about and answering the question correctly.
- Certainly I personally would like to see whole tournaments full of organic chemistry and molecular biology questions, but it’s hard for me to see that doing so will educate people of all skill levels, entertain players of all skill levels, and therefore I must conclude that it is not good for quiz bowl. Such subject matter is better brought up as the hard part of a three part bonus, as IMHO should all canon expansion or infrequently asked difficult topics.
- For some then, biology will remain a ho-hum, uninteresting part of the distribution, at least as long as I’m editing it. However, I'd rather not write tossups where specialists and only specialists will know the answers, and where specialists and only specialists will get any points on bonuses.
- As for topic choice, I cannot speak for other tournaments, I can only speak for ACF Nationals. Here was this past year’s distribution:
A copy of just the biology distribution can be obtained from me at quizbowlronin@gmail.com.
I should suggest that while there are questions that are medically related, that most (but lamentably not all) questions contain clues which relate to biochemistry, microbiology, anatomy, and physiology. Questions which relate to the drug therapy of a certain disease or the deviation in plasma sodium levels or data which is entirely descriptive and has no relationship to analytical science obviously shouldn't be written.
As far as the rest of the distribution, it just is a fact of life that we're going to continue to hear questions on plants and other non-mammalian organisms. Folks, if it's in Campbell, it's kosher - otherwise we might as well determine that quiz bowl biology is actually quiz bowl human biology, and also throw out much of the first year college biology course curriculum or high school biology curriculum.
recfreq wrote:Finally, I just want to ask if medicine is biology. If the clues are clinical, i.e. based on symptoms and effects, I don't see them as being biological b/c there's no relevant _academic_ clues. If the clues are physiological, or based on molecular genetics, I'd be ok with it. But in general, as the USC WIT packets will attest, diseases are easy ways to dodge the bio distribution, b/c you just copy from a clinical handbook. It's also not very interesting. Bio players, let us know how you feel, but I don't need another question on gonerhea. I know what it is, and it's not biology. (It's also not really taught until med school.) In an effort to mimick the interest of the physical sciences and math crowd, we should strive for more mechanistic, molecular, physiological clues that tell us about the underlying processes involved instead of relying on memorizing viruses and diseases. I think those outside the field would prefer hearing such questions, just as I'd prefer hearing about Liouville's thm and parity.
More on this topic later, after more experiments.
Last edited by QuizBowlRonin on Wed Nov 23, 2005 3:23 am, edited 4 times in total.
Jason Paik
retired
Washington University 1998-2002
University of Alabama School of Medicine 2002-2011
Residency in Internal Medicine, Stanford Hospital and Clinics 2011-2013
Fellowship in Hematology and Oncology, Stanford Cancer Institute 2013-2018
First Chairman
Auron
Posts: 3875
Joined: Sat Apr 19, 2003 8:21 pm
Location: Fairfax VA
Contact:
I'll let JP discuss the issues of biology and medicine editing for qb.
As for the diversity of question-writing for biology, I agree there is a lot of diversity in topics covered in "biology". On the other hand, I don't know how many evolutionary-developmental biologists there are there. I'd enjoy a question that is disciplinarily "biogeophysical chemistry" if such a thing exists (and actually I'm pretty sure there is at least one such area)... but then we're really hitting a very esoteric portion of science.
Maybe it is the nature in which people are forced to learn biology compared to the physical sciences. When it comes to learning taxa or recognizing principles with infection, there are mechanisms that one cannot really "apply" when it comes to an undergraduate education. I'm sure one could learn how to run a western blot (and you've probably seen many questions already written on WB's) but that sounds much more boring than some unsolved problem in mathematics... because it is.
Medicine should be covered under science, but obviously it should not predominate the biology portion of any syllabus (at least acknowledge botany and zoology). I guess I would prefer 1/1 clinical science and 1/1 basic science in biomedical topics to be fair. But one could go on writing an entire packet on biomedical sciences if you are dedicated to learning the literature.
Granted we don't ask many questions on pharmacy or nursing concepts, but I leave that to a different discussion.
Emil Thomas Chuck, Ph.D.
Founder, PACE
Facebook junkie and unofficial advisor to aspiring health professionals in quiz bowl
---
Pimping Green Tea Ginger Ale (Canada Dry)
grapesmoker
Sin
Posts: 6360
Joined: Sat Oct 25, 2003 5:23 pm
Location: Pittsburgh, PA
Contact:
Actually, I think the mitochondria question is not very good. The Margulis/endosymbiosis clue is way too early. Of course, for all I know, the answer could be chloroplasts, but if I was playing against a team with any biology knowledge and they hadn't gotten it yet, I would be buzzing then. I would put that clue towards the end of the question since it's something well-known to non-biologists.
I think one of the factors that contributed to this gap of which Ray speaks is the fact that some of the most active science writers on the circuit are physicists. Certainly Seth Teitler and myself are far better qualified to write physics than biology. When I have to write a bio question, I almost invariably err on the side of something I've heard of, just because I have no idea if anything else is gettable.
I would love to write more sophisticated questions, but I am afraid of botching it. If Ray (or others) could recommend to me some good sources for such questions, I'll do my best to come up with more of them in future packets.
Jerry Vinokurov
ex-LJHS, ex-Berkeley, ex-Brown, sorta-ex-CMU
code ape, loud voice, general nuissance
Mr. Kwalter
Tidus
Posts: 617
Joined: Tue Nov 18, 2003 1:48 am
Location: Houston, TX
I guess I would prefer 1/1 clinical science and 1/1 basic science in biomedical topics to be fair.
Rather than creating a 2/2 "biology" distribution, how about lumping clinical biology in with "miscellaneous science?" You have 1/1 bio, 1/1 chem, 1/1 physics, then 2/2 whose subject is at the writer's/editor's discretion; why not just say you have to have at least 1/1 non-clinical bio? It's just an idea.
vandyhawk
Tidus
Posts: 584
Joined: Sat Dec 13, 2003 3:42 am
Location: Seattle
As a biomedical engineer, I've taken a boatload of classes with bio in the title somewhere, but I think I've let myself fall in the trap of only writing about stuff that has come up before. It's sometimes hard for me to judge what would be gettable by someone who has only taken, say, intro bio, so like Jerry said, I tend to default to easier, more canonical stuff. That's not to say I don't try to include interesting clues, but I think I need to try harder at that.
The mitochondria question above is pretty bad. As soon as I heard "porins" and "outer membrane" I'd buzz. I honestly feel like taxonomy questions are the worst bio questions, though. Sure, a lot of people learn that info in high school, but does anyone actually cover that in college level courses anymore? I know I've never come across it here. It seems like non-science people who are forced to write science can come up with a taxonomy question pretty easily, and maybe asnwer some too, but I just don't see the value. Obviously taxonomy won't go away, but multiple tossups or bonuses in the subject in a given tournament is just too much.
As far as clinical questions, I don't mind some clinical clues in a tossup about an organ or disease or whatever, but there should be other info (e.g. molecular, structual aspects) as well, and I don't think answers should be allowed that one would only really learn in med school. In some packet from the archive I looked over recently (maybe from BOB?), there was a bonus on cancer drugs. I've taken advanced cancer biology, and I would've been able to get 10 points on the bonus for Taxol. I had heard of another answer, Gleevec, though would not have been able to pull it out. That kind of question is just unacceptable unless it's a tournament for med students. Now asking about p53, though, is a welcome trend, and mabye at ACF Nats this year I can throw in some more related things since those kinds of pathways are being taught to undergrads now too.
I think I've heard questions on RNAi and proteoglycan before at ACF Nats, but I'm not positive on that. I know I've heard a Western blot tossup a few years ago at ACF regs, and too many electrophoresis ones. I'd love to hear more questions on newer experimental techniques (RNAi, RT-PCR, quantum dots, ELISA), as well as some different biochem areas like gluconeogenesis that don't get the same coverage as glycolysis and the electron transport chain.
As far as separating clinical from basic science in the bio distribution, I don't know if that's the best idea. Often, a good question will combine aspects of both areas, and if people start trying to write truly "clinical" questions, that could lead to even more med school-ish questions, which I don't think anyone wants. At least, I don't think you guys want me to crack open our "clinical correlations" lectures from when I took physiology from the med school. A tournament editor can just try to make sure there's a good balance throughout the set.
Those are my thoughts at the moment. I'll be interested to see what Jason has to say.
Deviant Insider
Auron
Posts: 4439
Joined: Sun Jun 13, 2004 6:08 am
Location: Chicagoland
One possible reason for the difference between biology and physics questions has to do with their relationship to studies of the history of science or the philosophy of science. Brownian motion sometimes comes up as a topic outside of physics because Albert Einstein is such an interesting figure. Other physics topics from the early 20th century (which have nothing to do with Brownian motion) are used as the archetypal example of a scientific revolution. Topics from the last fifty years are important to some philosophers.
Mitochondria, on the other hand, to the best of my knowledge is a topic only useful to people studying biology.
David Reinstein
Head Writer and Editor for Scobol Solo and Masonics (Illinois), TD for New Trier Scobol Solo and New Trier Varsity, Writer for NAQT (2011-2017), IHSSBCA Board Member, IHSSBCA Chair (2004-2014), PACE Member, PACE President (2016-2018), New Trier Coach (1994-2011)
Susan
Forums Staff: Administrator
Posts: 1809
Joined: Fri Aug 15, 2003 12:43 am
A few comments:
-I don't see Gleevec as an unacceptable answer for the hard part of a bonus. Granted, I'm in a cancer biology program at Chicago, where Janet Rowley, whose work on the Philadelphia chromosome led to the development of Gleevec, works, so I may be biased here. I encountered Gleevec in several undergrad classes (cancer biology and pharmacology), and it's regularly mentioned by anyone talking about targeted therapies, modern drug development, and (sometimes) pharmacogenetics. It's arguable that the development of Gleevec is the most important story in the last decade of pharmacology.
-In part, at least, the reason people have been unwilling to write a lot of biology questions on challenging or otherwise new (to quizbowl) topics is that, when we've tried to do that before, other people have bitched about it. (I recall Selene Koo's ACF Nationals tossup on noted compound (mentioned in six of my undergrad classes) cisplatin being met with pearl-clutching.)
-I think a lot of the problem of weak bio distributions can be chalked up to the fact that very few people editing tournament sets are into biology. The tournaments I've attended where I was most impressed by the quality of the biology questions were Roger Bhan's ACF Nationals (2003) and Subash Maddipoti's ACF Regionals (2003), both written by players with substantial backgrounds in biology.
-To those who hate taxonomy questions, how do you propose addressing natural history/biogeophysical sciences/paleo while using askable topics? These topics are a part of biology, too.
-There are certainly questions being asked about newer experimental techniques; I know I've both heard and written questions on RNAi and ELISA. It's nice to see these sorts of things (also, p53, Holliday junctions, etc.--can tossups on Rad51 be far behind?), but on the other hand, while I've heard p53 come up more over the past few years, I can't say I've heard any great questions on it. I'd rather hear good questions than novel answers (though ideally both should be possible).
-For those looking for good biology writing sources, I'll point you towards Pubmed's Bookshelf, and AWAY from Wikipedia.
Last edited by Susan on Tue Nov 22, 2005 4:38 pm, edited 1 time in total.
First Chairman
Auron
Posts: 3875
Joined: Sat Apr 19, 2003 8:21 pm
Location: Fairfax VA
Contact:
I would point out some of the neater, newer stuff is published with Science and Nature's News & Views sections. I would warn anyone that you do need to have a very solid basis in your science before throwing something up as a tossup there. But I would suggest that (as a study hint) any of the molecules of the year in the last 3-4 years could be really good fodder for questions in the college circuit.
But I think the dearth of biology/medical writers/editors who are knowledgeable about quiz bowl questions is an issue. Of course, most of us tend to be extremely busy [most of us are pre-med or med]. :) What I do worry about is how far away from basic biology we are moving in writing the more difficult stuff. It is a challenge to write a very good "fundamental" biology question, and as someone who has dealt with these issues before, I recognize how hard it can be.
Emil Thomas Chuck, Ph.D.
Founder, PACE
Facebook junkie and unofficial advisor to aspiring health professionals in quiz bowl
---
Pimping Green Tea Ginger Ale (Canada Dry)
vandyhawk
Tidus
Posts: 584
Joined: Sat Dec 13, 2003 3:42 am
Location: Seattle
I don't see Gleevec as an unacceptable answer for the hard part of a bonus
Neither do I, but having it as the medium level answer with Taxol as the "easy" part is a bit much for anything but possibly ACF Nats.
To those who hate taxonomy questions, how do you propose addressing natural history/biogeophysical sciences/paleo while using askable topics?
Perhaps I was a bit harsh on taxonomy - it certainly does have its place, but what I meant to get across is that there are often too many cop-out "name the class common to these organisms" or similarly unimaginative or poorly written questions throughout a tournament. Fewer and better would be appreciated.
recfreq
Wakka
Posts: 167
Joined: Mon Nov 07, 2005 10:11 pm
Location: Japan.
E.T. Chuck wrote:It is a challenge to write a very good "fundamental" biology question, and as someone who has dealt with these issues before, I recognize how hard it can be.
I don't think it's actually that hard. There's so much info out there. One easy for nonspecialists to write a lead-in that makes sense is just to search for a review in pubmed and skim the 1st 2 pages. Then use your canonical bio text book, although (replying to Jerry) I'd try to use the more subject specific books like Kandel's Principles of Neural Science for neurobio, or Albert's Molec Cell Bio for cellbio, or Wolpert's Developmental Bio for develbio, etc. This would be akin to using say Jackson's Electrodynamics book or Kittle's Thermal Physics book, not for the giveaways, but for what you'd put in the beginning and middle of these questions. And as opposed to some of the math and physics, the bio books are _very very_ easy to read, almost like a novel about how people discovered such and such factors by doing this and then that, so while people have an excuse (in my opinion) to write a bad math TU, they really shouldn't be writing a bad bio question, as the subject is not very dfficult conceptually.
Just to show how easy it is to write a good, in depth bio question that is still very accessible, I'll point out the following TU, which I wrote just by reading section 4.5 from Lodish's Molec Cell Bio book (5ed), pages 125 to 130. I used nothing else. If you read that 5 pages, you'd be able to write this question:
"It can be regulated by phosphorylation of a serine residue on eIF2, which inhibits exchange of bound GDP for GTP, disrupting formation of the i ternary complex. This process is initiated by eIF4E binding to 7-methylguanylate, and eIF4A scanning for the Kozak sequence surrounding the start site, recognition of which leads to eIF2 and eIF5 GTP hydrolysis, the latter recruiting the eIF6-bound 60S subunit. Hydrolysis of EF1-alpha-GTP ejects unacylated adaptors from the E site and positions the aminoacyl-tRNA in the A site close to that in the P site, but only if its anticodon matches the mRNA codon. FTP name this process by which the ribosome makes proteins from messenger RNA, taking place after transcription.
:. translation"
I think we can agree that this is accessible; there's nothing beyond sophmore bio majors stuff here. There's also little work involved. BTW if you read Ch4 from Lodish, which is actually just an introductory review of all of nucleic acids, DNA transcription, translation, replication, viruses, and whatever else of importance they talk about later in the book, you're on your way to writing good molec bio, at least on the DNA domain. Notice some of the key terms in there: Kozak seq, which should probly be a little earlier now that I think about it; 7-mG; eIFs and EFs (elongation factor); then later A site, P site of ribosome. If you just mentioned a good number of these, your question is automatically good (I didn't even refer to Shine-Delgarno, which comes up a few times, so there are many variations to be considered). And the lead-in is still nontrivial even for experts to nail down, even if she (like most good mobiologists) would know it as soon as she hears Kozak (I guess the eIF2 might be a bit early, but people'd still have to parse that to the right answer space). I wrote Davison-Germer (which was ok if not spectacular right, Jerry?) just by reading a few pages from an undergrad quantum book, so math and physics people should be able to do the same with a few pages of Lodish.
Personally, I don't like lab techniques or specific ailments as much, just as I don't like Atwood machine or questions about building bridges. If one could relate the technique or disease back to the core biological concepts, it'd be much better. E.g. trisomy or Turner's or other genetic diseases could have clues about genetic implications and physiological concepts. Techniques like micro RNAs could provide implications for the role of RNA in evolutionary terms, etc. (BTW if you write an organelle question, make sure you have enough new stuff to put in there, b/c those types of questions get very old very quickly, e.g if you do Golgi, try talking about KDEL receptors and AP proteins, not just mention them, but tell us what they do.)
BTW I'm also looking forward to Jason's post. I just want to say before hand that I didn't imply that the editing was bad. In fact, I remember the uracil question from ACF nats being one of the best ever, esp with the base clue on tRNA. I just thought that in terms of difficulty, the other sciences may have overshot the bio, and made bio seem more at the regionals level. But the questions themselves were all good. I did wish that there'd be a few less diseases that are just clinical things that one would pull out of a medical manual (I don't know if even beginning med students'd really memorize it), but I've already mention this to Jason. I was referring more to the general state of writing in biology, which I felt was somewhat low, given that the only really pretty good question I've heard this entire fall was "glutamate." A lot of them were (except some at WIT) ok, just nothing I'd remember for the rest of my life, and not much learning took place for me.
Ray Luo, UCLA.
recfreq
Wakka
Posts: 167
Joined: Mon Nov 07, 2005 10:11 pm
Location: Japan.
grapesmoker wrote:I would love to write more sophisticated questions, but I am afraid of botching it. If Ray (or others) could recommend to me some good sources for such questions, I'll do my best to come up with more of them in future packets.
Jerry, I think you give yourself less credit than you deserve. I remember an angiogenesis question that was quite good out of your hands. But in terms of topics, I don't want to force anything down people's throats, but there're a lot of things in cell bio that everybody takes (like the MCB 102 course at Cal) (I mean _everybody_, including physiologists, chem, biomed, IB, nursing, you name it) that don't get asked, and some of them go into considerable detail. Books like Lehninger, Lodish, and Alberts will point those out, esp when there're multiple sections devoted to them. Then there're stuff that less people take like development, cancer, and evolution, but may be we'll take this a step at a time and put them in bonuses.
Also, I will endeavor to write physics and math to the best of my ability, even though sometimes they can be comical (I recall the rayleigh scattering I tried to send to someone once), and any comments you may have on my deficiencies in that regard is welcomed, and I'll thank you in advance.
Finally, I don't think we should strive to be inherently difficult, just to step out of the high school bio we keep asking for the last few years. As long as there're in depth clues at the soph-jun bio major level at the beginning, I could care less how easy the giveaway becomes, be it Hardy Weinberg or gastrulation.
Ray Luo, UCLA.
recfreq
Wakka
Posts: 167
Joined: Mon Nov 07, 2005 10:11 pm
Location: Japan.
myamphigory wrote:(I recall Selene Koo's ACF Nationals tossup on noted compound (mentioned in six of my undergrad classes) cisplatin being met with pearl-clutching.)
Yeah, I submitted the Anfinsen expt that won him the Nobel. Oh, and Eric's suggestion is good for me personally, but I don't think I can enforce that on other people; they'd just complain that medicine is bio (I disagree, but I think that's the prevailing view).
Ray Luo, UCLA.
Matt Weiner
Sin
Posts: 8413
Joined: Fri Apr 11, 2003 8:34 pm
Location: Richmond, VA
I'm inspired by some parts of this thread to remind people about the pitfalls of writing on new topics. Please do so by means of bonus parts. I'm sure that cisplatin is very important, but if most rooms at the tournament couldn't answer the tossup, then it's by definition too hard for a tossup. No amount of "you SHOULD know this" or "if there was a biology major on every team you'd find this easier" can contradict that truism.
Making the clues in existing tossups more real and putting things like cisplatin in second or third bonus parts in place of diseases or other disliked biology questions is great, and I fully encourage it and will try to follow the advice in this thread when writing my own science questions, but we can't lose sight of the need to write tossups that teams can answer with their current makeup.
Matt Weiner
Founder of hsquizbowl.org
grapesmoker
Sin
Posts: 6360
Joined: Sat Oct 25, 2003 5:23 pm
Location: Pittsburgh, PA
Contact:
I'm not sure how useful the medicine/bio distinction is, personally. If I wrote a tossup on a lab technique that is used in physics labs, I wouldn't say that's not a physics question.
My other question is about mid-level clues. I take Ray's word for it that the question he presented is quality. Does it have mid-level clues? I thought the Brownian motion question that I wrote did; it started with a very specific formulation, proceeded to mention things like diffusive processes and RMS displacement, and ended with the pollen-grains. Anyway, it's just a thought. Most of the things in the question don't mean much to me but I'm not a biologist so that's as it should be.
I've found as I've written questions over the years that after you take enough science, whatever your specific subject, you develop an intuition for what is more or less obvious. It doesn't necessarily translate into knowing more about that subject, but you can learn to write good science questions that way.
Jerry Vinokurov
ex-LJHS, ex-Berkeley, ex-Brown, sorta-ex-CMU
code ape, loud voice, general nuissance
cvdwightw
Auron
Posts: 3447
Joined: Tue May 13, 2003 12:46 am
Location: Southern CA
Contact:
recfreq wrote:It can be regulated by phosphorylation of a serine residue on eIF2,
At this point I think anyone with a minor grasp of biology should be realizing this is some molecular biology or biochemical process; anyone with a better grasp would probably start thinking about the true answer as a prime candidate for the right one at the mention of "eIF"
recfreq wrote: which inhibits exchange of bound GDP for GTP, disrupting formation of the i ternary complex. This process is initiated by eIF4E binding to 7-methylguanylate, and eIF4A scanning for the Kozak sequence surrounding the start site, recognition of which leads to eIF2 and eIF5 GTP hydrolysis, the latter recruiting the eIF6-bound 60S subunit.
Depending on how familiar you are with biology, this is either additional meaningless lead-in or a narrowing of answers. As Ray pointed out, anyone who knows and remembers molecular biology should be able to buzz correctly off of "Kozak sequence" while the "60S subunit" implies a process taking place in the ribosome, from which someone could make a logical guess.
recfreq wrote:Hydrolysis of EF1-alpha-GTP ejects unacylated adaptors from the E site and positions the aminoacyl-tRNA in the A site close to that in the P site, but only if its anticodon matches the mRNA codon.
E, A, and P sites are all well known to bio players. Any bio player who hasn't gotten it at that point is probably sitting. The introduction of "tRNA" and "codon" should start to make this accessible, or at least guessable, to non-bio players.
recfreq finally wrote: FTP name this process by which the ribosome makes proteins from messenger RNA, taking place after transcription.
:. translation
By the end of this question even most novice teams should be able to get this question.
So to answer Jerry's question, there are definitely mid-level clues in the question, although they probably don't seem that way to non-biologists.
grapesmoker
Sin
Posts: 6360
Joined: Sat Oct 25, 2003 5:23 pm
Location: Pittsburgh, PA
Contact:
cvdwightw wrote: So to answer Jerry's question, there are definitely mid-level clues in the question, although they probably don't seem that way to non-biologists.
Thanks, that's basically what I was looking for. This has been very educational for me : )
Jerry Vinokurov
ex-LJHS, ex-Berkeley, ex-Brown, sorta-ex-CMU
code ape, loud voice, general nuissance
mattreece
Lulu
Posts: 40
Joined: Fri Feb 11, 2005 2:06 am
Is the situation for biology and physics questions really so different? Some will recall that I was criticized for writing a tossup with the answer "S-matrix," which is a fundamental concept that should be introduced in any undergrad quantum mechanics class. It was pretty much universally agreed that this question was too difficult. But now we're seeing various biology terms defended (at least partly) on the basis of their being taught in undergrad classes.
I'm not really sure what the correct standard is for science questions. I think we can all agree that pyramidal questions with very well-known answers are acceptable at all levels of competition. For tournaments like ACF Nationals, these should start with very difficult clues.
On the other hand it's nice to have some questions with somewhat more obscure answers. The question is how much obscurity should be allowed, and how one can get some sense of how obscure an answer is. I was surprised to learn that almost no one could answer a question on the S-matrix. Perhaps if I had searched old tournaments I would have realized that it has not been asked, despite its importance. I think it's an unfortunate but inevitable consequence of the way the circuit works that relatively minor concepts in some fields can be asked frequently while major concepts in other fields are off-limits. Like it or not, there is a canon, and a writer must deal with this. In some fields it's more problematic than others. For instance, in psychology it's simply a fact that psychoanalysis is well-established in the canon, so in sticking to it one is less likely to write questions that are deemed too obscure. On the other hand it's a bit ridiculous from the point of view of what is important in the field of psychology itself.
I guess what I'm getting at is that this discussion has a wider scope than just biology questions. In many subjects (especially, I would argue, the sciences, since the canon was largely established by non-scientists), there are conflicts between the importance of ideas and their place in the quizbowl canon. This can make it very hard to tell if a question will be acceptable to a quizbowl audience.
QuizBowlRonin
Wakka
Posts: 182
Joined: Fri May 30, 2003 2:21 am
Location: San Mateo, CA
Contact:
As for the diversity of question-writing for biology, I agree there is a lot of diversity in topics covered in "biology". On the other hand, I don't know how many evolutionary-developmental biologists there are there. I'd enjoy a question that is disciplinarily "biogeophysical chemistry" if such a thing exists (and actually I'm pretty sure there is at least one such area)... but then we're really hitting a very esoteric portion of science.
Maybe it is the nature in which people are forced to learn biology compared to the physical sciences. When it comes to learning taxa or recognizing principles with infection, there are mechanisms that one cannot really "apply" when it comes to an undergraduate education. I'm sure one could learn how to run a western blot (and you've probably seen many questions already written on WB's) but that sounds much more boring than some unsolved problem in mathematics... because it is.
Wesley Matthews and I had a discussion when going to ACF Fall about the nature of earth science questions, and found that he was exasperated that questions covered in the first year curriculum of earth science received poor receptions at tournaments. Besides proving the point that just because it’s in the first year curriculum of any subject does not mean it deserves to have questions written for quiz bowl, it shows that there are lacunae in quiz bowl where because writers and editors just aren’t interested in certain topics, people don’t ask questions about them, even though such topics are basic. The question becomes whether we should write questions on topics nobody cares about, or write very hard questions on advanced topics that some, but only very few know about. Which is more deserving, a basic topic that deserves educating the player about, or an esoteric topic that, while guaranteed to be known by some experts, is something that the non-specialist should have no business knowing?
Medicine should be covered under science, but obviously it should not predominate the biology portion of any syllabus (at least acknowledge botany and zoology). I guess I would prefer 1/1 clinical science and 1/1 basic science in biomedical topics to be fair. But one could go on writing an entire packet on biomedical sciences if you are dedicated to learning the literature.
Granted we don't ask many questions on pharmacy or nursing concepts, but I leave that to a different discussion.
The kernel of my point to be stated above is that there is a false dichonomy between medicine and biology.
Jason Paik
retired
Washington University 1998-2002
University of Alabama School of Medicine 2002-2011
Residency in Internal Medicine, Stanford Hospital and Clinics 2011-2013
Fellowship in Hematology and Oncology, Stanford Cancer Institute 2013-2018
QuizBowlRonin
Wakka
Posts: 182
Joined: Fri May 30, 2003 2:21 am
Location: San Mateo, CA
Contact:
grapesmoker wrote:Actually, I think the mitochondria question is not very good. The Margulis/endosymbiosis clue is way too early. Of course, for all I know, the answer could be chloroplasts, but if I was playing against a team with any biology knowledge and they hadn't gotten it yet, I would be buzzing then. I would put that clue towards the end of the question since it's something well-known to non-biologists.
Seth sent me the biology distribution a few hours before it was due, and I had a lookover. I apologize if I didn't catch a few things and put in additional clues when appropriate - I just didn't have any of the books or the time to do some editing.
I think one of the factors that contributed to this gap of which Ray speaks is the fact that some of the most active science writers on the circuit are physicists. Certainly Seth Teitler and myself are far better qualified to write physics than biology. When I have to write a bio question, I almost invariably err on the side of something I've heard of, just because I have no idea if anything else is gettable.
I would love to write more sophisticated questions, but I am afraid of botching it. If Ray (or others) could recommend to me some good sources for such questions, I'll do my best to come up with more of them in future packets.
As suggested, Pubmed bookshelf is good, but descriptions of esoteric topics might be too difficult for some players without a basic molecular biology background. Otherwise, Campbell for basic biology, Stryer or Voet for biochemistry, Lodish or Alberts for molecular and cell biology. Physiology books vary, but Guyton is good, and for anatomy Moore and Persaud.
Folks are welcome to e-mail me at quizbowlronin@gmail.com if they need explanations of certain topics (please, don't abuse this service!).
Jason Paik
retired
Washington University 1998-2002
University of Alabama School of Medicine 2002-2011
Residency in Internal Medicine, Stanford Hospital and Clinics 2011-2013
Fellowship in Hematology and Oncology, Stanford Cancer Institute 2013-2018
QuizBowlRonin
Wakka
Posts: 182
Joined: Fri May 30, 2003 2:21 am
Location: San Mateo, CA
Contact:
vandyhawk wrote:As a biomedical engineer, I've taken a boatload of classes with bio in the title somewhere, but I think I've let myself fall in the trap of only writing about stuff that has come up before. It's sometimes hard for me to judge what would be gettable by someone who has only taken, say, intro bio, so like Jerry said, I tend to default to easier, more canonical stuff. That's not to say I don't try to include interesting clues, but I think I need to try harder at that.
The mitochondria question above is pretty bad. As soon as I heard "porins" and "outer membrane" I'd buzz. I honestly feel like taxonomy questions are the worst bio questions, though. Sure, a lot of people learn that info in high school, but does anyone actually cover that in college level courses anymore? I know I've never come across it here. It seems like non-science people who are forced to write science can come up with a taxonomy question pretty easily, and maybe asnwer some too, but I just don't see the value. Obviously taxonomy won't go away, but multiple tossups or bonuses in the subject in a given tournament is just too much.
As far as clinical questions, I don't mind some clinical clues in a tossup about an organ or disease or whatever, but there should be other info (e.g. molecular, structual aspects) as well, and I don't think answers should be allowed that one would only really learn in med school. In some packet from the archive I looked over recently (maybe from BOB?), there was a bonus on cancer drugs. I've taken advanced cancer biology, and I would've been able to get 10 points on the bonus for Taxol. I had heard of another answer, Gleevec, though would not have been able to pull it out. That kind of question is just unacceptable unless it's a tournament for med students. Now asking about p53, though, is a welcome trend, and mabye at ACF Nats this year I can throw in some more related things since those kinds of pathways are being taught to undergrads now too.
I think I've heard questions on RNAi and proteoglycan before at ACF Nats, but I'm not positive on that. I know I've heard a Western blot tossup a few years ago at ACF regs, and too many electrophoresis ones. I'd love to hear more questions on newer experimental techniques (RNAi, RT-PCR, quantum dots, ELISA), as well as some different biochem areas like gluconeogenesis that don't get the same coverage as glycolysis and the electron transport chain.
As far as separating clinical from basic science in the bio distribution, I don't know if that's the best idea. Often, a good question will combine aspects of both areas, and if people start trying to write truly "clinical" questions, that could lead to even more med school-ish questions, which I don't think anyone wants. At least, I don't think you guys want me to crack open our "clinical correlations" lectures from when I took physiology from the med school. A tournament editor can just try to make sure there's a good balance throughout the set.
Those are my thoughts at the moment. I'll be interested to see what Jason has to say.
Really, I appreciate a whole lot what you have to say. I'll do my best when it comes time to edit this years ACF Nationals to keep what you're suggesting in mind. However, please try to remember that I'm not only serving the specialist constituency.
Jason Paik
retired
Washington University 1998-2002
University of Alabama School of Medicine 2002-2011
Residency in Internal Medicine, Stanford Hospital and Clinics 2011-2013
Fellowship in Hematology and Oncology, Stanford Cancer Institute 2013-2018
QuizBowlRonin
Wakka
Posts: 182
Joined: Fri May 30, 2003 2:21 am
Location: San Mateo, CA
Contact:
myamphigory wrote:A few comments:
-I don't see Gleevec as an unacceptable answer for the hard part of a bonus. Granted, I'm in a cancer biology program at Chicago, where Janet Rowley, whose work on the Philadelphia chromosome led to the development of Gleevec, works, so I may be biased here. I encountered Gleevec in several undergrad classes (cancer biology and pharmacology), and it's regularly mentioned by anyone talking about targeted therapies, modern drug development, and (sometimes) pharmacogenetics. It's arguable that the development of Gleevec is the most important story in the last decade of pharmacology.
Agreed, but I don't think Gleevec is tossupworthy just yet.
-In part, at least, the reason people have been unwilling to write a lot of biology questions on challenging or otherwise new (to quizbowl) topics is that, when we've tried to do that before, other people have bitched about it. (I recall Selene Koo's ACF Nationals tossup on noted compound (mentioned in six of my undergrad classes) cisplatin being met with pearl-clutching.)
As Matt Reece suggests, this is not a problem exclusive to the biology part of the distribution. Finding out that a tossup answer is something that both teams have never heard of ever just isn't any fun, regardless of intrinsic value of the topic asked.
<snip>
-To those who hate taxonomy questions, how do you propose addressing natural history/biogeophysical sciences/paleo while using askable topics? These topics are a part of biology, too.
Agreed.
-There are certainly questions being asked about newer experimental techniques; I know I've both heard and written questions on RNAi and ELISA. It's nice to see these sorts of things (also, p53, Holliday junctions, etc.--can tossups on Rad51 be far behind?), but on the other hand, while I've heard p53 come up more over the past few years, I can't say I've heard any great questions on it. I'd rather hear good questions than novel answers (though ideally both should be possible).
I know that I've written on p53 in every ACF Nationals since 2002, but I imagine is now the time to expand our horizons a little bit? Rb perhaps?
I too, work in DNA damage response but I unfortunately can't tell you much about what Rad51 does. Perhaps we should start with BRCA1/2?
Last edited by QuizBowlRonin on Wed Nov 23, 2005 4:42 am, edited 1 time in total.
Jason Paik
retired
Washington University 1998-2002
University of Alabama School of Medicine 2002-2011
Residency in Internal Medicine, Stanford Hospital and Clinics 2011-2013
Fellowship in Hematology and Oncology, Stanford Cancer Institute 2013-2018
recfreq
Wakka
Posts: 167
Joined: Mon Nov 07, 2005 10:11 pm
Location: Japan.
Wo, hold on a minute. I'm not talking about specialist topics; don't subject us to the p75 NGF receptor or RSK, either. I was just talking about very basic stuff in very basic mobio courses, e.g. we spent a day and a half on pentose phosphate shunt. I don't know what S matrix is, but I suspect it isn't as important relative to quantum physics as pentose phosphate shunt relative to biochemistry. If it's that important, I guess I just haven't done the reading; 7/8 of my quantum book is unread, and I'll get back to you when I see the phrase "S matrix." I just think it's time to stretch it just a bit, e.g. we studied different types of transposons for a while, and stuff like LINEs and SINEs and LTRs, all with different properties, could very well make good introductory questions, if nothing else at least on bonuses. Lipid synthesis, quantitative genetics, signal transduction components, etc would all make good clues and questions, better than exhausting every amino acid or vitamin _again_. Moreover, this is stuff we should have learned, and people are learning.
No, I don't want people to write specialist stuff either. I don't want to hear more about p23, p46, p294, p1300, whatever, unless it's important (actually, something 1300 is important, I heard, I just don't remember the letter). But stuff we all had to learn in intro mobio, and found in standard texts like Lehninger and Lodish: that's askable. (Foraging the fun clues, that's the fun part.)
Ray Luo, UCLA.
recfreq
Wakka
Posts: 167
Joined: Mon Nov 07, 2005 10:11 pm
Location: Japan.
QuizBowlRonin wrote:
myamphigory wrote:I too, work in DNA damage response but I unfortunately can't tell you much about what Rad51 does. Perhaps we should start with BRCA1/2?
How about types of DNA repair? The mismatch repair type, the nucleotide excision repair type, the SOS type, add in some descriptions, plug in some diseases like xeroderma pigmentosum, throw in some proteins like UvrA, MutH/L/S, some helicases, whatever, I think we have a TU. We had to memorize this stuff for the intro mobio, so it should be accessible to bio majors.
I envy your research work in terms of QB compliance. I work on VMAT transporters in mushroom bodies of Drosophila (at least for these two quarters), and there's no way I'll ever subject you to any of the subject matter from my research (I promise), except in a very very tangential manner. On the other hand, I like flies more than humans, so maybe I'm not so envious. Have a good one.
Ray Luo, UCLA.
recfreq
Wakka
Posts: 167
Joined: Mon Nov 07, 2005 10:11 pm
Location: Japan.
Re: Biology in the modern QB era.
QuizBowlRonin wrote:I was terribly disappointed by the fact in the finals round that Selene, Subash, Susan, and Seth did not get tossups that I was sure that one of the three would get on the first line – this is not a reflection of their lack of skill as it is a reflection of poor editing and clue-writing on my part.
Jason, I don't recall them not getting the bio, but to be honest, (may be I should have brought this up when it occurred) hepB, Addison, tibia, tetanus, jaundice, and asthma sound like a lot of med. I'm sure all but one or two of those got answered. I'd just envisage one or two evolution-ecology, a bit more physiology, and may be just one question on development, one on neurobiology wouldn't hurt to diversify, but sure, it's all fair game.
QuizBowlRonin wrote:Therefore, tossups like Franck-Condon or RNA interference, while written in a pyramidal manner consistent with ACF values, do not successfully differentiate teams that know their material versus those that do not.
I guess one can argue that we'd not need difficult questions as long as the clues for questions with easy answers have difficult enough clues at the beginning, and I think that ACF nats had that, but I do think that biology != medicine also brings up embarrasing things like me not knowing WTF tibia is. In general, though, isn't it good to have a mix of questions at the nats level of both difficult and easy questions, to probe the breadth of certain players as well. E.g. one could easily write a good, clue-laden question on the glyoxylate cycle--there're at least 4 pages in Stryer. Testing for that knowledge is different from testing whether you could get Calvin cycle really early, so I don't know if it's a good idea to abandon it altogether at the nats level. (At the fall level, yes, let's ban it.)
Thanks for the thoughts, I wasn't trying to be critical of ACF nats in any way. I just found it to be consistent with a trend of having really really easy bio (HS level in most tournies), combined with significantly harder physics and math. Perhaps other people can point out the same trends in other sciences. I think CS is pretty even, other than the fact that people keep thinking programming syntax or unix keyword is computer science. I've found chem to be b/t bio and physics, or just about.
Finally to address the knowable vs. gettable, I think if we go by a standard group of people, it's hard to go wrong. If everybody except one or two is against S matrix (sorry for using this example, Matt), then the majority'd likely rule. I'm just banking on the fact that since most bio people have taken biochem, that pentose-phosphate would still be knowable by the majority of the target audience. Now, whether it's gettable even if you know it, that sounds to me like QB skill, or at least QB communication. If you know the answer and still can't get it, due to language or whatever, then you might not be interpreting the ques correctly, or the ques can be confusing you. But if you just can't say it despite knowing it, then that's just a failure to give the answer despite buzzing in and knowing it, and it's a neg just like any other neg, i.e. it's your skills.
Sorry, I think I really should stop writing at this point.
Ray Luo, UCLA.
grapesmoker
Sin
Posts: 6360
Joined: Sat Oct 25, 2003 5:23 pm
Location: Pittsburgh, PA
Contact:
recfreq wrote: I don't know what S matrix is, but I suspect it isn't as important relative to quantum physics as pentose phosphate shunt relative to biochemistry. If it's that important, I guess I just haven't done the reading; 7/8 of my quantum book is unread, and I'll get back to you when I see the phrase "S matrix."
The S is for "scattering." It's basically a matrix that connects before and after states for some quantum process. I think it was a fine question; it's just that in my class the professor never gave it a name (perhaps he called it the transfer matrix or something, I seem to remember that name). Once I realized what it was, I figured out that it was a basic concept from quantum that I had apparently never learned the name of. But it is important, probably as important in quantum as the pentose-phosphate shunt is in biochemistry, since it's the basic descriptive ingredient of any scattering process.
Jerry Vinokurov
ex-LJHS, ex-Berkeley, ex-Brown, sorta-ex-CMU
code ape, loud voice, general nuissance
setht
Auron
Posts: 1133
Joined: Mon Oct 18, 2004 2:41 pm
Location: Columbus, Ohio
There are many things I want to respond to in this thread; I guess I'll just wade in and go backwards, for the hell of it.
grapesmoker wrote:The S is for "scattering." It's basically a matrix that connects before and after states for some quantum process. I think it was a fine question; it's just that in my class the professor never gave it a name (perhaps he called it the transfer matrix or something, I seem to remember that name).
This question (and other ACF Nationals 2005 science questions) were discussed in the aftermath of ACF Nats last spring. The thread is still available on the second page of the College area archives.
In that thread, it was noted that the S matrix and transfer matrix are somewhat different things: the transfer matrix connects amplitudes to the left and right of a potential barrier, the S matrix connects incoming and outgoing amplitudes (incoming and outgoing with respect to a potential barrier). Various people (myself, Jerry, Jordan Boyd-Graber) shared heart-warming tales of learning/not really learning about this topic in upper-level quantum mechanics classes, some dude with screen name kws popped in with his thoughts on introductory quantum mechanics classes covering simple Hartree-Fock theory, and Jerry stated that the only reason he would have been able to get the S matrix question was that he was working at the time in a accelerator physics group.
The physical science--in fact, all of the non-bio science at ACF Nationals 2005 (from what I can remember) was harder than the biological science. Through the first 14 or so rounds, I felt that the bio was about the same level, or slightly easier than the rest of the questions (other sciences, and humanities). For the finals rounds, I felt that the bio was still comparable with or somewhat easier than pretty much everything else, while the non-bio science jumped to a significantly harder level than pretty much everything else. Was I fascinated by all the new science material that was coming up? I can't say that I was. I was playing on a team where all four players (Subash, Susan, Selene, and myself) had fairly good science knowledge in various areas. I would guess that that's one of the best science teams that's ever been assembled in terms of breadth and depth of coverage (we didn't know anything about CS, but that was about it for science holes), we were playing a Michigan team with decent but not great science coverage, and we did not derive any advantage from our knowledge (in the first game, we got 0 science tossups, Michigan got 1; in the second game, we got 1 and Michigan got 2; we did not do particularly well on science bonus questions). Any interest I had in learning about, say, the Rackett equation was pretty much entirely lost in feeling frustrated and useless--it really felt like there was little point having non-bio science players for those matches, since the humanities tossups were much more open to contention.
Ray, be careful what you wish for. Encouraging people to write bio tossups with harder answers will probably result in many more questions you will enjoy and find interesting, but it can also result in a couple rounds' worth of questions that leave you scratching your head, which can be frustrating, especially if losing out on the chance to convert your knowledge into points costs you a few key games.
I prefer asking questions with well-known answers (Orion or Orion Nebula rather than the Becklin-Neugebauer Object [yes, this was submitted once to a tournament I was partly editing]) and inserting new material as clues to well-known answers, or using the occasional hard/new answer as one part of a bonus where the other parts are gettable. I think it's important to emphasize the need to keep bonus questions from getting too crazy--it's usually fine to introduce some new, very hard answer as one part of a bonus at, say, ACF Nationals, but if you don't put something there for teams without super-specialized knowledge, it's just not a good bonus. An example that I feel went overboard, taken from the 6th Editor packet of ACF Nationals 2005: there was a bonus on galactic astronomy with parts on "de Vaucouleurs law," "distance modulus," and "Holmberg radius." These are, at best, topics that might appear in advanced undergraduate galactic astronomy courses, but there's a good chance the only people with any shot at getting points on this bonus are graduate astronomy students with a galactic astronomy class under their belts. Even then, it's far from a guaranteed 30 (among other things, de Vaucouleurs and Holmberg are both empirically-fitted quantities that have pretty much passed by the wayside in favor of better-fitting quantities). I think we can do a better job differentiating between teams than "Do you have one or more grad students studying this particular area?"
I really want to encourage people to work on improving questions and expanding the canon (including bio questions) by working on clues and bonus question selection.
-Seth
setht
Auron
Posts: 1133
Joined: Mon Oct 18, 2004 2:41 pm
Location: Columbus, Ohio
Re: Biology in the modern QB era.
recfreq wrote:Jason, I don't recall them not getting the bio
We didn't, according to my notes. I believe the bio tossups in the two final rounds were on asthma and hyphae; I think we negged asthma and got beat to hyphae (I don't remember whether we lost a buzzer race or got beat by a good early buzz). I don't think we felt these tossups were unfair, too hard, or otherwise objectionably (I could be wrong, but I don't remember Susan or Selene objecting). I think hyphae was one of three science tossups converted in two rounds.
recfreq wrote:In general, though, isn't it good to have a mix of questions at the nats level of both difficult and easy questions, to probe the breadth of certain players as well. E.g. one could easily write a good, clue-laden question on the glyoxylate cycle--there're at least 4 pages in Stryer. Testing for that knowledge is different from testing whether you could get Calvin cycle really early, so I don't know if it's a good idea to abandon it altogether at the nats level. (At the fall level, yes, let's ban it.)
A quick look on the internet gives me the impression that the glyoxylate cycle has some links to the Krebs cycle and acetyl-CoA, both of which seem much better-known. I could be wrong about this, but let's suppose for a moment that I'm not. Is there really anything much better about writing a tossup on the glyoxylate cycle than writing a tossup on acetyl-CoA with some clues about its role in the glyoxylate cycle? One possible objection would be that an acetyl-CoA tossup would presumably eventually move to clues about acetyl-CoA that have little or nothing to do with the glyoxylate cycle, while the glyoxylate cycle tossup would focus its full attention on exploring the minutiae of the glyoxylate cycle. However, it seems to me that a writer ought to be able to take pretty much any clue for a "glyoxylate cycle" tossup and convert it to a clue for an "acetyl-CoA" tossup, so unless you feel there's a strong need for tossups with lots and lots of glyoxylate cycle clues, or a need for tossups that fewer teams will answer at the end, I don't see the point of writing on glyoxylate cycle rather than acetyl-CoA. I also think you're much more likely to get teams interested in new information about a topic they've heard of then in new information on a topic they've never heard of.
recfreq wrote:I'm just banking on the fact that since most bio people have taken biochem, that pentose-phosphate would still be knowable by the majority of the target audience.
I'm not sure I believe the majority of the teams at most tournaments have a bio person. Also, is there a difference between "the number of teams that have a bio person" and "the number of teams that have a bio person that has taken a biochem class"? If a biochem class is part of the typical first-year curriculum in bio programs across the land, then the answer is no. But if biochem classes don't pop up until, say, the 3rd year, then the relevant question is "how many teams have an advanced bio player," and at this point I am confident that the majority of the teams at pretty much every tournament (including ACF Nationals) do not fit the bill.
-Seth
selene
Kimahri
Posts: 4
Joined: Tue Nov 08, 2005 6:39 pm
Location: Chicago, IL
I was going to reply to this thread yesterday, before the thread had approximately doubled in size. I had several things in mind to say, but now I'm no longer sure if they've already been discussed. I'll say them anyway, and hope that I'm not being too redundant.
- I feel like a lot of the complaints about the relative lack of difficulty of bio questions compared to that of physical science questions is not necessarily so much an issue of "the state of question writing in physical science is so much higher/more advanced" than disparities in the goals of editors of these categories. For example, ACF Nationals last year seemed in general to be plagued by questions that were both challenging and not very accessible (I like to believe that the goal of ACF is to write challenging but accessible questions, but that's the topic of another thread, I suppose). On the other hand, the biology questions continued to be both fairly challenging and accessible throughout the tournament. This unfortunately meant that the biology questions were significantly more gettable than the other questions, including the physical science questions, which could lead one to complain that the biology questions were "too easy," and thus perhaps "too boring/uninteresting" (by a gross oversimplification of Ray's argument). Similarly, with the ACF Fall set this year, the physical science questions seemed to have more difficult pyrimidality than the biological science questions; the "mitochondrion" question is *way* more gettable in the middle than the "Brownian motion" question. This isn't necessarily an issue of "the state of biology question writing is not advanced enough," so much as two editors having different ideas of what "ACF Fall difficulty" means.
- Please please don't write questions based on the first one or two pages of PubMed articles. I work in a biological science lab. The only papers I look up regularly on PubMed are those related to my research. Writing tossup leadins from PubMed will help perhaps one person in the whole nation (the one doing research in the field) to get the tossup; for everyone else, it's just more tossup verbiage to listen to before getting to the actual potentially gettable part of the question. It'll make tossups longer, but not actually more substantive.
- I'd also like to reiterate what people have already said, about how it's better to expand the canon with a bonus part or tossup clue, rather than a tossup answer. Yes, I wrote the "cisplatin" tossup, and it was not well-received. It's not one of the best question-writing decisions I've ever made. That was also the ACF Nationals that featured the "Gilman reagents" tossup that I believe no one in the whole tournament got (I didn't write that one). Now, there may be some off chance that someone will find the tossup interesting despite not getting it. Mostly, though, it just leads to pained silence and groans of frustration. Also, we don't want to start making questions gettable only by specialists (as would, I think, start becoming the case if the science writers wrote significant numbers of questions on stuff they learned in their second or third year of undergraduate coursework, which is already starting to get into the realm of "specialized," since the people taking those classes are mostly majors in the field), at least not for an invitational or ACF Fall. I'd rather hear challenging questions on easy topics, than "easy" questions on challenging topics.
- Instituting some kind of arbitrary separation between medicine and biology is not a good idea. Just keep in mind that you can write a biology question that doesn't involve medicine in any way. To some extent, a medically-related biology question is more widely interesting than a biochemistry question that rambles on and on about making some polysyllabic molecule from another polysyllabic molecule, since it ties in with things going wrong, and, well, maybe you know someone with the disease! Thinking that most biology questions need a medical component, though, results in a rather dull question-writing style in which every biology question starts with "It malfunctions in x disease and y disease blah blah blah." Bringing up something that Ray mentioned in his initial post, I learned about gonorrhea in high school sex ed. Granted, I learned about it again in med school. OK, in some sense it isn't "biology" because I didn't learn about it in a class that existed in the "biology" department. I don't think that kind of logic is reasonable, though.
I'm afraid I may have rambled for too long without saying anything new or interesting, so I'll stop here for now. I may actually think of something new and/or interesting to say later, but don't hold your breath.
Selene
mattreece
Lulu
Posts: 40
Joined: Fri Feb 11, 2005 2:06 am
Mostly I agree with what Seth and Selene have said. Harder questions on answerable things are good, questions on completely unknown things are bad. Simple enough, in principle.
But in the interests of helping to calibrate what is gettable and what is not, I have some questions.
selene wrote:Similarly, with the ACF Fall set this year, the physical science questions seemed to have more difficult pyrimidality than the biological science questions; the "mitochondrion" question is *way* more gettable in the middle than the "Brownian motion" question.
Is this really true? Yes, the mitochondria question gives away the answer pretty early even to a non-specialist. But Perrin is mentioned really early in the Brownian motion question, and I think of Perrin as almost a giveaway for Brownian motion. This isn't because I'm in any way an expert on Brownian motion, by the way; for one thing, I'm not, and for another, I think for me this is quizbowl knowledge possibly dating back to high school. I thought it was fairly well-known. Maybe I'm wrong.
Next, since several people have been saying that the physical sciences questions at ACF Nationals 2005 were much harder than the biological science questions: to what extent was this true within physics? I am aware of the complaints about "S-matrix" and "fluctuation-dissipation theorem," and I admit those were poor choices for a tossup. Were there a lot of other examples, or did these two just stand out? I want to make sure I don't cause similar problems at future tournaments.
recfreq
Wakka
Posts: 167
Joined: Mon Nov 07, 2005 10:11 pm
Location: Japan.
selene wrote:That was also the ACF Nationals that featured the "Gilman reagents" tossup that I believe no one in the whole tournament got (I didn't write that one). Now, there may be some off chance that someone will find the tossup interesting despite not getting it.
I don't recall seeing Gilman reagent, but have certainly gotten it somewhere, practice or otherwise. I thought the o-chem from ACF nats was solid: Birch reduction, Hunsdiecker rxn, Canizzaro rxn, etc alls seem at about the nats level when you consider what is asked at the regs level, so I was quite happy. But there's no real change in answer space for the biology, when I expected something a bit harder; still I was fine with it b/c it had more difficult clues (not just somewhat difficult, but ACF nats level difficulty) at the beginning. (And also, we shouldn't be 30ing a glycolysis bonus at nats.) It's fine just to have more difficult clues, I just think there're niches for a few things here or there that _really_ are very well known that gets ignored, akin say to the Casimir effect or Josephson junction for physics (e.g. there're more than just Krebs, Calvin, and glycolysis). b/c people don't write on them, we think they're hard, when in fact they're more than well represented in standard bio curricula comparable to the physics curricula. (And not even doing borderline stuff here.)
OK, I don't think I"m saying anything new either, but thanks for all the discussion, everybody.
Ray Luo, UCLA.
recfreq
Wakka
Posts: 167
Joined: Mon Nov 07, 2005 10:11 pm
Location: Japan.
Re: Biology in the modern QB era.
setht wrote:I don't see the point of writing on glyoxylate cycle rather than acetyl-CoA. I also think you're much more likely to get teams interested in new information about a topic they've heard of then in new information on a topic they've never heard of.
acetyl coA is involved in _everything_, including beta oxidation, lipid synthesis, etc, whereas glyoxylate cycle is it's own process. It'd be like writing glucose vs. glycolysis, they're just different things, even though one happens to appear in the other. E.g. glyoxylate cycle uses diff enzymes, some of which are analogous to Krebs, but some not, and converts Krebs intermediates into carbohydrates, basically another shunt. Also, we have questions on both author and book she's primarily known for, right? So given glyoxylate is semi-well-known, I don't see inherently why it shouldn't come up as TU answer, but only at, say, ACF nats. But I _do_ think for usual tournaments, it's better to have the easier thing come up, and in this case, either acetyl coA or Krebs.
setht wrote:But if biochem classes don't pop up until, say, the 3rd year, then the relevant question is "how many teams have an advanced bio player," and at this point I am confident that the majority of the teams at pretty much every tournament (including ACF Nationals) do not fit the bill.
At UCLA, people take it their 1st two years. At Berkeley, it varies, but I had both sophomore and junior classmates. But I think that applies to every other discipline like physics and math. How many people have taken analysis before junior year? Probly not many, still "compact," "closed," "metric space," "Stone-Wierstrauss," etc come up, and even at regs level. In fact, for QB terms the amt of training you require to get bio questions is the _least_ among all the major sciences, you just need to get through HS. (This is a plug for new QB players to study bio, b/c you'll catch on faster.) The amt of training required for physics is just greater (quantum, solid state, particle). Bio is such that if you took the intro bio and intro mobio, you could become one of the better bio players out there.
But anyhow, Seth, I am being very careful what I wish for. I'd rather get stumped on a mobio question than not hearing anything beyond the usual Calvin cycle Krebs cycle stuff. May be this is b/c I found the Hunsdiecker and Birch amusing from nats, so anything remotely approaching those things would be cool for me. I still can't get over the greatness of the glutamate question, though, and if you're responsible, I'd just like to give you a big warm hug. It even mentioned the loop pore structure and stuff, though skipped some of the TM terminology. Having sat through a kainate receptor (another glutamate receptor) lecture during SfN, I asked myself why I waited so long on that question, and then I realized that I got to hear most of the wonderfulness. Given that no one complained of its difficulty, I'd just like to label the ACF fall "glutamate" question as the quintessential model for all future bio questions, what ever that means, b/c it had most of what you'd look for in a bio question: the mechanism, the structure, applications, research, etc.
Ray Luo, UCLA.
setht
Auron
Posts: 1133
Joined: Mon Oct 18, 2004 2:41 pm
Location: Columbus, Ohio
mattreece wrote:Next, since several people have been saying that the physical sciences questions at ACF Nationals 2005 were much harder than the biological science questions: to what extent was this true within physics? I am aware of the complaints about "S-matrix" and "fluctuation-dissipation theorem," and I admit those were poor choices for a tossup. Were there a lot of other examples, or did these two just stand out? I want to make sure I don't cause similar problems at future tournaments.
I don't have a very clear memory of much of the tournament. From what I do remember, the science tossups for the first 14 rounds or so seemed generally fine--I'm guessing the non-bio science tossups were a bit harder than the bio tossups, but I don't think we were struggling to convert science tossups. Those last two rounds just had a sudden concentration of significantly harder non-bio science tossups (not just the physics, either). I really don't remember how we did on science bonus questions throughout the tournament, but I'm told we actually had reasonabe conversion rates in most of the physical sciences (with the possible exception of chemistry) for the first 14 rounds. I don't remember too much about the science bonus questions in the 2 final rounds, other than a pair of really hard astro and earth science bonus questions (one of them being the de Vaucouleurs/distance modulus/Holmberg radius bonus I alluded to earlier).
My general impression was that the bio set was pretty acceptable and appropriate for the audience. I think the non-bio science for the first 14 rounds followed the trend of the rest of the questions: pretty good, pyramidal tossups on reasonably gettable topics, with bonus questions that were too hard (I remember Andrew noting afterward that if he'd had a bit more time, he would have gone through and changed 10-12 hard bonus parts per round; certainly the science bonus questions weren't the only ones with this problem, but I also think they didn't avoid the problem [with the probable exception of biology]). I don't know if Andrew asked for harder questions for the finals packets; the humanities seemed to ramp up a little bit, while the non-bio science jumped up quite a bit (again, we weren't struggling to convert science tossups in the first 14 rounds--I don't remember any match prior to the final rounds where 2 or 3 science tossups went dead [which happened in both final games], and I'm pretty certain we didn't go 0 for 4 on the science tossups in any round prior to the first final match).
Those were my impressions, hopefully you'll get feedback from more people.
-Seth
setht
Auron
Posts: 1133
Joined: Mon Oct 18, 2004 2:41 pm
Location: Columbus, Ohio
Re: Biology in the modern QB era.
recfreq wrote:acetyl coA is involved in _everything_, including beta oxidation, lipid synthesis, etc, whereas glyoxylate cycle is it's own process. It'd be like writing glucose vs. glycolysis, they're just different things, even though one happens to appear in the other. E.g. glyoxylate cycle uses diff enzymes, some of which are analogous to Krebs, but some not, and converts Krebs intermediates into carbohydrates, basically another shunt. Also, we have questions on both author and book she's primarily known for, right? So given glyoxylate is semi-well-known, I don't see inherently why it shouldn't come up as TU answer, but only at, say, ACF nats. But I _do_ think for usual tournaments, it's better to have the easier thing come up, and in this case, either acetyl coA or Krebs.
If acetyl CoA really is involved in everything, all the more reason to prefer writing a tossup on it to writing a tossup on the glyoxylate cycle (unless that has similarly wide-ranging importance). Also, I don't think I believe your premise that glyoxylate cycle is semi-well-known (unless you mean that in the sense of "the glyoxylate cycle is semi-well-known to people who have completed multiple years' worth of coursework in bio/biochem," which is clearly very different from "the glyoxylate cycle is semi-well-known in the quizbowl community").
Having thrown that out there, I'll ask: suppose you come up with a juicy clue about the glyoxylate cycle; if you can think of a way to reword things so as to use it as a clue towards an acetyl CoA tossup, why would that be at all worse than using the clue in a glyoxylate cycle tossup? A player that knows about the glyoxylate cycle will beat players that don't to your acetyl CoA tossup, just as they would beat players to your glyoxylate cycle tossup. The difference between the two is that, in matches where both teams have never heard of the glyoxylate cycle, the acetyl CoA tossup has a chance of being converted. I think a fairly large fraction of players on the circuit have heard of acetyl CoA but not the glyoxylate cycle (which sounds like the sort of thing people generally study after multiple years of bio/biochem coursework).
recfreq wrote:I still can't get over the greatness of the glutamate question, though, and if you're responsible, I'd just like to give you a big warm hug.
The credit for that tossup lies entirely with some team and/or Sudheer. I'm glad you enjoyed the question, and I'll ask: if all the bio tossups at ACF Nationals were of similar quality, but on similarly widely-known topics, would that be such a bad thing? Personally, I'd rather see writers strive to write great ACF Nationals-level tossups on "the Earth's mantle" or "electric field" than on "the D'' layer" or "Jefimenko's equations."
-Seth
recfreq
Wakka
Posts: 167
Joined: Mon Nov 07, 2005 10:11 pm
Location: Japan.
Seth, I agree with you, not much would be better than an in-depth question on acetylcholine, or gastrulation. (I mean glyoxylate is somewhat known in the qb community.) But a couple of arguments why glyoxylate would still work at the ACF nats or regs level:
1. Actually, glyoxylate cycle is covered in basic biochem metabolism section, usually after Krebs and before light and dark reactions of photosynthesis (although that could be switched around). I remember a whole discussion section just about glyoxylate cycle, and I don't think people go on to study it in any future classes, except possibly for those doing research in the area. I think everybody who took say the Cal MCB 102 (Biochem and Mobio in one semester) will have heard of it and discussed it by the time the class is over. It's also in its own section in Lehninger, Stryer, etc. Competent bio majors will be all over it (aside: I remember this is what Raj Bhan told us once about o-chem).
2. I agree that such an acetyl coA question will be very good, but that doesn't mean we should never write the glyoxylate ques. Just b/c I could put a cool Crime and Punishment clue in the beg of my Dostoyevsky TU doesn't mean I couldn't also ask Crime and Punishment in its own TU _as long as I had enough for a good ques_. What I'm contending is that there's _a lot_ on glyoxylate just from no more than a few pages of Lehninger (akin to the way I wrote that translation ques from Lodish, just a few pages). I think you'll agree that the translation ques was good enough, and I think you can do the same with glyoxylate. Also, there's too much on glyoxylate cycle to just summarize in a sentence in front of the acetyl coA (glyoxisome, btw are named this way, you could bring in peroxisomes and various other processes). And also, how about "I already wrote an acetyl coA and I want to write something else."
But anyways, I think we're just arguing about not very much here. I just think that like you, I'd like to hear in-depth questions on basic things (much like your philosophy on writing the myth tourney--BTW we'll try to read that at some pt so Charles and Dwight can give you feedback), but I'd also like to hear a few questions testing players' breadth _once in a while_ in regs and nats tournies, much as we like asking Jean Rhys once in a while even though we'd like to ask Dickens and James most of the time, and Hunsdiecker in addition to Wittig. (I'll point you to my social science packet from TTGT11 for the confirmation of my belief in that easy-but-in-depth philosophy.)
BTW happy thanksgiving.
Ray Luo, UCLA.
cvdwightw
Auron
Posts: 3447
Joined: Tue May 13, 2003 12:46 am
Location: Southern CA
Contact:
Two things:
(1) At UCLA no one can take biochemistry until at least their second year. I am a third year and will finally be taking it next quarter.
(2) Had I been paying attention during the Brownian motion tossup (I still have a tendency to miss hearing important clues from time to time), I would have gotten it at "Wiener process," but only because I've read ahead in my probability textbook. This begs the question: is it good to have people answering biology questions based on things they learned outside biology (also e.g. Selene's example with "gonorrhea")?
grapesmoker
Sin
Posts: 6360
Joined: Sat Oct 25, 2003 5:23 pm
Location: Pittsburgh, PA
Contact:
cvdwightw wrote: is it good to have people answering biology questions based on things they learned outside biology (also e.g. Selene's example with "gonorrhea")?
Sure it is. Why not? After all, it's not like you're answering based on a "sounds like" clue or something stupid; you're answering based on real knowledge that you acquired in another context. Doesn't seem like there's anything wrong with that. In fact, it's good.
Jerry Vinokurov
ex-LJHS, ex-Berkeley, ex-Brown, sorta-ex-CMU
code ape, loud voice, general nuissance
QuizBowlRonin
Wakka
Posts: 182
Joined: Fri May 30, 2003 2:21 am
Location: San Mateo, CA
Contact:
grapesmoker wrote:
cvdwightw wrote: is it good to have people answering biology questions based on things they learned outside biology (also e.g. Selene's example with "gonorrhea")?
Sure it is. Why not? After all, it's not like you're answering based on a "sounds like" clue or something stupid; you're answering based on real knowledge that you acquired in another context. Doesn't seem like there's anything wrong with that. In fact, it's good.
Despite our best efforts, fakery will always be a part of quizbowl.
Jason Paik
retired
Washington University 1998-2002
University of Alabama School of Medicine 2002-2011
Residency in Internal Medicine, Stanford Hospital and Clinics 2011-2013
Fellowship in Hematology and Oncology, Stanford Cancer Institute 2013-2018 | 20,418 | 88,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-04 | latest | en | 0.919476 |
https://www.autoitscript.com/forum/topic/209732-if-condition-with-evalstring-not-working/ | 1,716,924,347,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059148.63/warc/CC-MAIN-20240528185253-20240528215253-00211.warc.gz | 555,099,001 | 18,247 | Jump to content
# If Condition with Eval(String) not working
## Recommended Posts
Hi all.
I have a 2^8 conditions to check in order to sharpen result.
TO "simplify it I used this method:
First I created a function that, based on which checkbox is select returns the "IF" string
```Func _GenerateConditions(\$mode)
Local \$condition = "Number(\$array_assets[\$n][14]) = 1"
If \$mode = 1 Then ; long
If GUICtrlRead(\$CB_A_Ichi) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][4]) >= Number(\$long_greenred) AND Number(\$array_assets[\$n][5]) > Number(\$long_TKvsKJ)"
If GUICtrlRead(\$CB_A_MACD) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][6]) = Number(\$long_MACD) AND Number(\$array_assets[\$n][7]) = Number(\$long_Histo)"
If GUICtrlRead(\$CB_A_RSI) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][11]) >= Number(\$RSI)"
If GUICtrlRead(\$CB_A_MA) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][8]) = Number(\$MA)"
If GUICtrlRead(\$CB_A_EMA) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][9]) = Number(\$EMA)"
If GUICtrlRead(\$CB_A_Mom) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][10]) = Number(\$Momentum)"
If GUICtrlRead(\$CB_A_Cross) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][13]) = Number(\$Crossover)"
If GUICtrlRead(\$CB_A_BB) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][12]) = Number(\$BollingerBands)"
Else ; short
If GUICtrlRead(\$CB_A_Ichi) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][4]) <= Number(\$short_greenred) AND Number(\$array_assets[\$n][5]) < Number(\$short_TKvsKJ)"
If GUICtrlRead(\$CB_A_MACD) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][6]) = Number(\$short_MACD) AND Number(\$array_assets[\$n][7]) = Number(\$short_Histo)"
If GUICtrlRead(\$CB_A_RSI) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][7]) >= Number(\$short_RSI)"
If GUICtrlRead(\$CB_A_MA) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][8]) = Number(\$short_MA)"
If GUICtrlRead(\$CB_A_EMA) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][9]) = Number(\$short_EMA)"
If GUICtrlRead(\$CB_A_Mom) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][10]) = Number(\$short_Mom)"
If GUICtrlRead(\$CB_A_Cross) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][13]) = Number(\$short_Crossover)"
If GUICtrlRead(\$CB_A_BB) = \$GUI_CHECKED Then \$condition &= " AND Number(\$array_assets[\$n][12]) = Number(\$short_Bollinger)"
EndIf
Return \$condition
EndFunc```
Then I call it in my loop:
```\$sConditions = _GenerateConditions(\$iTradeWallet)
__CW(\$sConditions)
For \$n = 0 To UBound(\$array_assets) - 1
If Eval(\$sConditions) Then
_send(\$array_assets[\$n][3] & " good to be added!", \$green)
_ArrayAdd(\$aRemoteAssets, \$array_assets[\$n][3])
EndIf
Next```
when I write to console \$conditions it properly shows the condition itself but using EVAL() seems not passing it to IF cycle so I never can enter the IF
Just to clarify: \$array_asset is a 2D array containing:
```; x,0 = id
; x,1 = Exchange
; x,2 = Currency
; x,3 = Asset
; x,4 = GreenRedStreaks
; x,5 = TKRatio
; x,6 = MACDTrend
; x,7 = HistoTren
; x,8 = MA
; x,9 = EMA
; x,10 = Mom
; x,11 = RSI
; x,12 = Bollinger
; x,13 = Crossover
; x,14 = Volume
; x,15 = MarginEnabled
; x,16 = LastUpdate```
Any suggestion to avoid a 2^8 If...then
Marco
##### Share on other sites
I think that Eval is not what you want, from the helpfileReturn the value of the variable defined by a string. You are not passing a variable to Eval, so it fails with 0, and your If check fails
I think what you want is Execute: https://www.autoitscript.com/autoit3/docs/functions/Execute.htm
Give Execute a read and try, and see if that works.
We ought not to misbehave, but we should look as though we could.
##### Share on other sites
Perfect! Exactly what I needed!!!
Thanks @mistersquirrle!
## Create an account or sign in to comment
You need to be a member in order to leave a comment
## Create an account
Sign up for a new account in our community. It's easy!
Register a new account
## Sign in
Already have an account? Sign in here.
Sign In Now
• ### Recently Browsing 0 members
• No registered users viewing this page.
×
• Wiki
• Back
• #### Beta
• Git
• FAQ
×
• Create New... | 1,343 | 4,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.395016 |
https://www.mathmadeeasy.co/contact-me | 1,723,389,915,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00820.warc.gz | 681,875,928 | 154,467 | top of page
## Discrete Mathematics Unveiled!
MULTINOMIAL THEOREM VIDEO
Watch this video to learn about the Multinomial theorem, multinomial coefficients and applications of multinomial theorem in problems.
Click to watch video.
​
​
RECURRENCE RELATIONS -HOMOGENEOUS AND NON HOMOGENEOUS
This gives you a very basic overview of what is a Recurrence Relation. You will learn 2 types of recurrence relations, namely homogeneous and non homogeneous recurrence relations and how to find the general solution and specific terms. This lesson teaches you about first order recurrence relations.A number of problems are fully solved and explained. A useful lesson in Discrete Mathematics. Useful for College Math and Engineering Math. I
Click to watch video.
PRINCIPLE OF INCLUSION AND EXCLUSION
Learn about the Principle of Inclusion and Exclusion and its applications in problems. This is a part of Discrete Mathematics/College Math. .I have tried to simplify this concept for you through problems. 2 versions of the Principle of Inclusion and Exclusion are presented here, The use of permutations and combinations is also illustrated in problem 3.
Click to watch video.
​
​
bottom of page | 267 | 1,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-33 | latest | en | 0.900681 |
http://www.mathworks.com/matlabcentral/fileexchange/40831-carpet-plot-toolkit/content/CarpetPlots/carpetconvert.m | 1,427,560,022,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131297622.30/warc/CC-MAIN-20150323172137-00268-ip-10-168-14-71.ec2.internal.warc.gz | 661,655,165 | 6,944 | Code covered by the BSD License
Carpet Plot Toolkit
Rob McDonald (view profile)
A set of routines to make carpet plotting easier.
[xc, yc]=carpetconvert(x1, x2, y, offset, x1p, x2p)
```function [xc, yc]=carpetconvert(x1, x2, y, offset, x1p, x2p)
%CARPETCONVERT Converts points into carpet plot coordinates.
% [xc, yc]=carpetconvert(x1, x2, y, offset, x1p, x2p) converts the points
% x1p, x2p into carpet plot coordinates defined by (x1, x2, y, offset)
% as described in CARPET.
%
% The converted points are returned in [xc, yc].
%
% Rob McDonald
% ramcdona@calpoly.edu
% 19 February 2013 v. 1.0
% Calculate carpet plot cheater axis
xc = x1p + x2p * offset;
% Interpolate contours to carpet plot y-axis
yc = interp2(x1,x2,y,x1p,x2p);
``` | 253 | 757 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2015-14 | longest | en | 0.706743 |
https://www.arxiv-vanity.com/papers/1708.04203/ | 1,611,293,628,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529128.47/warc/CC-MAIN-20210122051338-20210122081338-00519.warc.gz | 681,960,767 | 38,267 | # On the Separation of a Polyhedron from Its Single-Part Mold††thanks: This work has been supported in part by the Israel Science Foundation (grant no. 825/15), by the Blavatnik Computer Science Research Fund, and by the Hermann Minkowski–Minerva Center for Geometry at Tel Aviv University.
Prosenjit Bose School of Computer Science, Carleton University, Dan Halperin The Blavatnik School of Computer Science, Tel Aviv University, , . Shahar Shamai
###### Abstract
Casting is a manufacturing process where liquid material is poured into a mold having the shape of a desired product. After the material solidifies, the product is pulled out of the mold. We study the case in which the mold is made of a single part and the object to be produced is a three-dimensional polyhedron. Objects that can be produced this way are called castable with a single-part mold. A direction in which the object can be removed without breaking the mold is called a valid pull-out direction. We give an algorithm that decides whether a given polyhedron with facets is castable with a single-part mold, and if so indicates how to orient the polyhedron in the mold and a direction in which the product can be pulled out without breaking the mold. Our algorithm runs in time. The best previously known algorithm for this problem runs in time.
## I Introduction
Casting is a widely-used manufacturing process, where liquid material is poured into a cavity inside a mold, which has the shape of a desired product. After the material solidifies, the product is taken out of the mold. Typically a mold is used to manufacture numerous copies of a product, in which case we need to ensure that the solidified product can be separated from its mold without breaking it.
The problems that we study here belong to the larger topic termed Movable Separability of Sets; see Toussaint [1]. Problems in this area are often exceedingly challenging from a combinatorial- and computational-geometry point of view (see, e.g., [2]). At the same time solutions to these problems are needed in mold design [3], assembly planning [4], and 3D printing to mention a few application areas.
We focus in this paper on a fairly basic movable-separability question. We are given a polyhedron in with facets. We do not make any particular assumptions about the polyhedron beyond that it is a closed regular set, namely it does not have dangling edges or facets. The mold is box-shaped and the cavity has the shape of such that one of ’s facets is the top facet of the cavity. See Fig. 1 for an illustration in 2D. Once the top facet has been determined, we detect whether there is a direction in which the solidified object can be pulled out of the mold without colliding with the mold. Such a direction is a valid pull-out direction, the corresponding top facet is valid, and the polyhedron is castable. When we restrict the pull-out direction to be the outer normal of the top facet, this problem is known to be solvable in linear time [5].
We address two problems: .
All Facets Single Direction (AllFSD):
Determine which facets of can serve as a valid top facet and for each such facet indicate one valid pull-out direction.
All Facets All Directions (AllFAD):
Same as above but for each valid facet indicate all the valid pull-out directions.
Why would anyone bother to solve AllFAD and not be satisfied with AllFSD? First, a solution is more stable if there is a continuum of directions rather than a single direction of separation. Second, we use the availability of many possible directions to optimize other criteria.
Previous results The best previous algorithms that we are aware of [6, Chapter 4] solve the AllFAD and AllFSD problems by solving the following two easier problems times — handling each candidate facet to be a top facet separately:
Single Facets Single Direction (SingleFSD):
Determine whether a specific given facet of a polyhedron can serve as a valid top facet and if so, indicate one direction in which can be pulled out of the mold with as its top facet.
Single Facets All Directions (SingleFAD):
Same as above but indicate all the directions in which can be pulled out of the mold with as its top facet.
The existing algorithm for SingleFSD takes time and the existing algorithm for SingleFAD takes time. This means that it takes time to solve AllFSD and time to solve AllFAD. Both algorithms, as well as the algorithms that we present below use linear storage space.
Contribution Our contribution in this paper is an -time algorithm for the AllFSD problem and an -time algorithm for the AllFAD problem. We also present an -time solution to the AllFAD problem when the input polyhedron is restricted to be convex. Additionally, we prove that for any polyhedron there can be up to six valid top facets.
The efficiency of our solution makes it a good candidate for implementation, which will enable product designers and engineers to quickly verify the castability of their design. Indeed we are currently developing a robust implementation of our solution. Although some of the components that our algorithms use are non-trivial, most of them are available in the CGAL library111cgal.org.
We presented a preliminary and partial version of this paper at the informal EUROCG meeting222eurocg2016.usi.ch, which does not publish proceedings. Here we extend our original work by presenting an -time algorithm for the AllFSD problem and showing that the maximum number of valid orientations for a polyhedron is six.
## Ii Preliminary Analysis
Instead of considering all of the top facets as possible candidates and running a separate algorithm for each of them, as in [6, Chapter 4], we start by finding a small set of candidate top facets (up to 12 facets). Then we solve SingleFSD/singleFAD [6, Chapter 4] for each of the candidates.
In order to find the possible top facets we will consider an arrangement of great circles on the unit sphere . An arrangement of curves on the sphere is a subdivision of the sphere into vertices, edges, and faces as induced by the given curves: Vertices are intersection points of curves, edges are maximal portions of a curve not intersected by any other curve, and faces are maximal portions of the sphere not intersected by any curve; see, e.g., [7, 8]. Each point on represents a direction in —the direction of the vector from the center of to . We will use the terms points and directions on interchangeably.
Let be the facets of the given polyhedron . Let be the normal to the facet pointing into the polyhedron.
We will describe a mold by the pair , which should be interpreted as follows. The top facet of the mold is . To achieve this, the polyhedron needs to be rotated by a rotation matrix such that becomes the top facet. We apply the same rotation matrix to to obtain a pull-out direction . (In other words, for convenience during our analysis, we use the vector is relative to the given, original orientation of the polyhedron. Once we determine that the pair is valid we still need to rotate both such that becomes the top facet and is a valid pull out direction in the new orientation.)
###### Observation 1
The pair represents a valid mold and pull-out direction if and only if for each point in the polyhedron, the ray that starts at with direction :
(i) intersects , and
(ii) does not intersects . (It may partialy overlap333This corresponds to allowing to be moved out while sliding in contact with . .)
###### Lemma 1
The pair represents a valid mold and pull-out direction if and only if
(i)
(ii)
###### Proof.
For the case where is the top facet of , this fact is proved in Lemma 4.1 in [6]. It remains to notice that the Conditions (i) and (ii) are invariant under rotation. They hold in ’s given orientation if and only if they hold when is rotated such that becomes the top facet. ∎
###### Definition 1
A valid pair is a pair that obeys the conditions of Lemma 1.
###### Definition 2
A facet will be called a valid top facet if there exists a vector for which the pair is a valid pair.
Let be the set of all valid top facets. Notice that the actual pull-out direction is , where is the matrix that rotates such that becomes the top facet, or equivalently such that points vertically downwards.
Denote by := the closed hemisphere of directions on for which , and by or the open complement hemisphere. For a set of facets X we denote by the set . Let . Let denote the boundary great circle of , and let . Consider the arrangement on , namely the subdivision of induced by the great circles in . See Fig. 2 for an illustration.
###### Definition 3
The depth of a point on is the number of hemispheres in in which is contained.
###### Observation 2
All the points in any fixed cell of the arrangement have the same depth.
###### Theorem 2
In our setting there cannot be a face of zero depth in . Similarly, in Lemma 1, Condition (i) follows from Condition (ii).
###### Proof.
Assume, for the sake of contradiction that there exists a face of zero depth in the arrangement . let be a direction in this face. We assumed that for each facet , , or equivalently, for each facet , the direction is pointing into the polyhedron. If we choose an arbitrary point inside the polyhedron (not on its boundary) and go from it in the direction we will eventually leave the polyhedron through some facet , meaning that must be pointing out of the polyhedron from . Contradiction.
Notice that in order to avoid special degenerate cases, like crossing the polyhedron in an edge, we choose the arbitrary point such that it does not lie on any of the planes spanned by the direction and an edge of the polyhedron. ∎
###### Theorem 3
The polyhedron is castable with a single-part mold if and only if the arrangement contains a point of depth . A cell of depth in , which is covered by the hemisphere , represents a mold whose top facet is and each point in represents the valid pull-out direction , where is the orthonormal matrix that rotates to point vertically down (in the negative direction).
###### Proof.
Let be a cell of depth in covered by , and let be a point in . We establish that the pair is a valid pair by verifying that the conditions of Lemma 1 above hold for it.
It remains to show that no point in any cell of different depth can represent a valid pull-out direction for any top facet. Consider a cell of depth greater than and a point in it. Let be the index set of the hemispheres that cover : . One of the facets must serve as the top facet for Condition (i) of Lemma 1 to hold. But then for each of the remaining facets Condition (ii) of the lemma is violated. Finally, as shown in Theorem 2, no cell can have depth zero. ∎
Our goal is to find all the facets in (for which covers a depth cell in ). In Section III we give an algorithm that finds a set of up to 12 facets in which is contained and show that .
###### Definition 4
A covering-set is a set of open hemispheres such that the union of all the hemispheres in covers the entire unit sphere.
###### Theorem 4
For each covering-set , .
###### Proof.
Let be a covering-set. Assume, for the sake of contradiction that there exists some facet for which . By Theorem 3, a facet is a valid top facet if and only if there exists a depth cell in which is covered by (and only by ). We know that each point in the unit sphere is covered by some hemisphere , therefore some hemisphere covers a point in this cell. By the definition of a cell, if some point in the cell is contained in then the entire cell is contained in . We also assumed that covers this cell, this means that this cell is of depth 2, at least. Contradiction. ∎
Our next step is to give an algorithm that finds, in linear time, a set of open hemispheres in whose union covers the entire unit sphere and whose size is bounded by a constant (specifically, 12).
## Iii Finding A Covering-Set
In order to find a covering-set, we handle each of the upper open hemisphere, the equator, and the lower open hemisphere separately. We find a covering set for each of them, and merge the three sets. We start by finding a covering set for the upper open hemisphere. In order to do so, we will define and prove the following claims.
###### Definition 5
The half-plane of a hemisphere with respect to the upper hemisphere is the central projection of the intersection of and the upper hemisphere onto the plane . [6, Chapter 4]. See Fig. 3.
Remark. The “half-plane” of the lower hemisphere with respect to the upper hemisphere is empty and the “half-plane” of the upper hemisphere with respect to the upper hemisphere is the entire plane .
###### Lemma 5
If the union of a set of half-planes, , is the entire plane then there exists a set such that the union of the half-planes in is the entire plane and .
###### Proof.
Helly’s theorem [9] states that given a collection of convex subsets of where , if the intersection of any objects in this collection is nonempty, then the entire collection has a nonempty intersection. The contrapositive of this theorem is that if the intersection of the entire collection is empty then there exists a subset of size such that its intersection is empty.
In our case ,
By Helly’s theorem there exist in . By that we learn that . ∎
This subset can be computed in linear time with linear space using a version [10] of Megiddo’s LP algorithm [11], or in linear expected time and constant space using Seidel’s randomized incremental algorithm [12].
We can now find a covering set of size three for the upper open hemisphere in linear time, as follows. If some hemisphere in is the upper open hemisphere return it and stop; we are done. Otherwise, transform each hemisphere in into a half-plane on , as described earlier. Use the procedure we have just described to find three half-planes that cover the entire plane in linear time. A covering set for the lower open hemisphere can be found analogously.
###### Observation 3
Any open hemisphere on is covered by up to three hemispheres in .
Now we also need to find a covering set for the equator. We can run a similar algorithm for the 2D case—splitting the equator into a left circle, a right circle and two points, then finding a covering set of size six at most. Finally, we merge the three covering sets and get a covering-set of size 12 for .
A simpler way of finding a covering-set for , is to choose four open hemispheres (any four hemispheres, not necessarily from ) that cover . Now we find a covering set of size three for each of these hemispheres, as above, Now we find a covering set of size three for each of these hemispheres. Putting all together, we obtain a covering-set for of total size 12.
We continue by proving the existence of a covering-set for of size six.
###### Theorem 6
There exists a covering-set of size six for .
###### Proof.
Let be some great circle on that does not contain a vertex of and is not contained in . is composed of parts of cells in that are partially contained in the two open hemispheres defined by . This means that any covering-set of an open hemisphere defined by , covers as well.
By that and by Observation 3, the covering-set of is of size six at most. ∎
Remark. This is only proof of existence. This version of the paper will not provide an algorithm to compute a covering-set of size six. However, a linear time algorithm will be provided in the full version of the paper that will be uploaded to the arXiv444arxiv.org.
###### Theorem 7
The number of valid top facets for any polyhedron is at most, six.
###### Proof.
As we showed in Theorem 6, there exists a covering-set of size six, and by Theorem 4, we know that the number of valid top facets in a polyhedron is bounded by the size of any covering-set. Thus, the number of valid top facets is bounded by six.
Notice that this is tight—all the six facets of a parallelepiped are valid top facets. ∎
## Iv Algorithm
The algorithms for both the AllFAD and AllFSD problems, are very similar. Find a covering-set of constant size, as discussed before. For each facet in the covering-set solve SingleFSD/SingleFAD [6, Chapter 4]. Now return all the pull-out directions for each valid top facet in the covering-set and stop.
We end up with a linear time algorithm for AllFSD and an time algorithm for AllFAD.
Algorithm 1 uses the function handleSingleFacet(,), which for a given polyhedron and a facet, solves SingleFAD in time or SingleFSD in linear time [6, Chapter 4].
Remark. When solving the AllFAD variant, one may wish to run the AllFSD algorithm first in order to end up with a linear-time algorithm in case that the polyhedron is not castable.
Remark. AllFAD for a polygon can be found in linear time and constant space using similar technics. The details will be given in the full version of the paper.
## V Casting Convex Polyhedra
In this section we show how to determine the castability of a convex polyhedron more efficiently, still solving the AllFAD problem for this case. We start with a couple of simple observations, which will be needed later.
###### Observation 4
Let be a convex polygon in the plane whose edges are given in cyclic order , say clockwise. Given a direction in the plane, all the edges of whose inner-facing normals have a non-negative scalar product with form a consecutive subchain of .
Let be a plane in that is parallel to the direction , intersects some convex polyhedron, , but does not intersect any of its vertices. Let be the convex polygon, which is the non-empty intersection of the convex polyhedron with . For each that intersects with we denote by the edge of . Let denote the inner facing normal of in .
If , then .
###### Proof.
Let be a vector in direction and length for some small . indicates whether the direction points into or out of when it starts from a point on [6, Chapter 4], i.e, if () for each point , then is inside (outside of) the polyhedron.
After the intersection with , each point on must still fulfill these conditions, since for each point , is in and therefore it is in the polygon if and only if it is in the original polyhedron.
We say that two facets of the input polyhedron are neighbors if their closures intersect at an edge. Denote by the set of neighbors of the facet , by the cardinality of this set, and by the index set of the facets in . The efficient AllFAD algorithm for convex polyhedra stems from the following observation.
###### Lemma 9
For a convex polyhedron, the pair represents a valid mold and pull-out direction if and only if
(i)
(ii) .
###### Proof.
The only difference between Lemma 1 and the current lemma is the set of facets on which Condition (ii) is tested: here it is only on the neighboring facets of . We argue that if the condition holds for the neighboring facets it will hold for all facets other than . Assume, for a contradiction that it holds for all neighboring facets, and there is a non-neighboring facet for which it does not hold. Let be a segment connecting a point inside and a point inside . Let be a plane containing and parallel to . If intersects a vertex of then we move s (and ) slightly parallel to itself such that it does not cross any vertex of . Let , and as above let for every facet of that intersects . Let and be the two neighboring edges to in — notice that and are the intersection of with two neighboring facets of . By Lemma 8 the scalar product of with both and is non-negative and with both and is negative. However, these form an intertwining cyclic subsequence of edges of : , in contradiction with Observation 4. This means that the inner-facing normal of has non-negative scalar product with and the same holds for the inner-facing normal of . ∎
###### Theorem 10
Given a convex polyhedron and a specific top facet , it is possible to find all the pull-out directions in time.
###### Proof.
Without loss of generality let’s assume that is horizontal (parallel to the -plane) and points vertically downwards. We wish to find all the pull-out directions for which the conditions of Lemma 9 hold. This can be done in time by centrally projecting the inner closed hemisphere for each facet in onto the plane (as in555Each facet is represented by a half-plane on containing all the directions with a positive component such that Definition 5), obtaining a set of half-planes and intersecting the half-planes created (see [6, Chapter 4]). We wish to reduce the running time of this procedure to time. Luckily, these half-planes are sorted by their slope since the slope of the half-plane created by on is equal to the slope of the edge on created by (again, and are parallel). The edges of a convex polygon are ordered by their slope. The intersection of a given set of half-planes which are sorted according to the slope of their bounding lines, can be computed in time [13]. ∎
The algorithm proceeds by fixing a facet , computing its edge-neighboring facets and computing the set of allowable directions using Theorem 10. We repeat the procedure for each facet of . Notice that is in fact the number of edges on the boundary of . The overall cost of time over all candidate top facets is time by Euler’s formula [14, Chapter 13], In summary,
###### Theorem 11
Given a convex polyhedron with facets, we can solve the AllFAD problem for in time .
## Vi Conclusion and Further Work
We have described efficient solutions to determine the castability of a polyhedron with a single-part mold. Our algorithms are an order of magnitude faster than the best previously known algorithms for these problems.
We outline several directions for further research:
(i) Our focus here was on separability with a single translation. Can one devise efficient algorithms in case we are allowed to pull out the polyhedron from its cast using arbitrary motion? This problem is already interesting for separating a planar polygon from its planar cast. The latter can be resolved by considering the motion planning problem for a robot (the polygon) among obstacles (the mold). This approach however would yield a near-quartic time solution [15]. Thus, the goal here is to take advantage of the special structure of the problem to obtain a more efficient solution.
(ii) There are many interesting problems when the mold is made of two or more parts, and we hope that our novel observations here may open the door to more efficient algorithms for these more complicated casting problems.
As mentioned, we are also developing a robust implementation of our algorithms, based on tools from the CGAL library.
## Appendix A Maximum Number Of Orientations
We present one more result concerning the relation between the castability of a polyhedron and the castability of its convex hull (Theorem 12). We show that every valid pair (see Definition 1) of , induces a corresponding valid pair for . This means that the set of valid pairs of the convex hull of is a super-set of the valid pairs of . We use Theorem 12 to give an alternative proof (See Theorem 7 for the original proof) that a polyhedron has at most six valid top facets.
###### Theorem 12
Given a polyhedron , one of its valid top facets, and a pull-out direction , ’s convex hull , can be pulled out through (the convex hull of ) with direction .
###### Proof.
Assume, for the sake of contradiction, that is not castable in direction through . This means that for some point one of the conditions of Observation 1 does not hold. is convex, and therefore a ray that starts in cannot intersect two facets of . This means that any point in that fulfills the first condition must fulfill the second condition as well. Thus, the first condition must not hold for . Let be a ray in direction , and be the locus of all the points (not necessarily in ) for which the first condition holds. (when represents Minkowski sums [16]). Every point in holds the first condition of Observation 1, therefore and since is convex, . Contradiction. ∎
###### Theorem 13
Given a polyhedron and a top facet , there can be at most one other top facet that does not share an edge with .
We omit the proof for lack of space.
###### Theorem 14
A polyhedron can have up to six valid top facets.
###### Proof.
By Theorem 12, it is sufficient to show this theorem for a convex polyhedron . It is known that the dual graph of a convex polyhedron is a planar graph [17]. We wish to consider a subgraph of ’s dual graph, which will include only vertices that represent a valid top facet and all of the edges between these vertices; we donate this graph by . Of course must be planar, by Euler’s formula [14, Chapter 13], (as long as ). By Theorem 13, each vertex is must be connected to all the other vertices in except maybe one, .
Merging the two inequalities gives us , solving this will result in
Want to hear about new tools we're making? Sign up to our mailing list for occasional updates.
If you find a rendering bug, file an issue on GitHub. Or, have a go at fixing it yourself – the renderer is open source!
For everything else, email us at [email protected]. | 5,529 | 25,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-04 | latest | en | 0.930957 |
https://gmatclub.com/forum/in-newtonian-laws-of-motion-there-is-a-condition-and-3668.html?sort_by_oldest=true | 1,511,206,022,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806124.52/warc/CC-MAIN-20171120183849-20171120203849-00098.warc.gz | 609,474,340 | 42,823 | It is currently 20 Nov 2017, 12:27
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# In Newtonian laws of motion, <there is a condition and
Author Message
CEO
Joined: 15 Aug 2003
Posts: 3452
Kudos [?]: 924 [0], given: 781
In Newtonian laws of motion, [#permalink]
### Show Tags
13 Dec 2003, 09:08
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 100% (00:47) wrong based on 2 sessions
### HideShow timer Statistics
In Newtonian laws of motion, <there is a condition
and it's converse regarding bodies at rest and bodies in motion>.
A) there is a condition and it's converse regarding bodies at rest and bodies in motion
B) there is a condition and its converse regarding bodies at rest and bodies in motion
C) there are a condition and its converse regarding bodies at rest and bodies in motion
D) there are a condition and it's converse regarding bodies at rest and bodies in motion
E) there has been a condition and its converse regarding bodies at
rest and bodies in motion
Kudos [?]: 924 [0], given: 781
Manager
Joined: 29 Aug 2003
Posts: 241
Kudos [?]: 31 [0], given: 0
Location: MI
### Show Tags
13 Dec 2003, 09:32
IMO, D is the best answer.
"are" is the right usage for "a condition" AND "converse".
"it's" correctly conveys that the "converse" is of "a condition".
Official ans ?
Kudos [?]: 31 [0], given: 0
Director
Joined: 13 Nov 2003
Posts: 957
Kudos [?]: 178 [0], given: 0
Location: Florida
### Show Tags
13 Dec 2003, 23:05
'm strangled between B and D
Kudos [?]: 178 [0], given: 0
Manager
Joined: 12 Oct 2003
Posts: 247
Kudos [?]: 7 [0], given: 0
Location: USA
### Show Tags
14 Dec 2003, 06:16
actually I was thinking between A and C ... did I wake up too early today?
official ans?
Kudos [?]: 7 [0], given: 0
CEO
Joined: 15 Aug 2003
Posts: 3452
Kudos [?]: 924 [0], given: 781
### Show Tags
14 Dec 2003, 20:15
akamai, thoughts
Kudos [?]: 924 [0], given: 781
Manager
Joined: 29 Aug 2003
Posts: 241
Kudos [?]: 31 [0], given: 0
Location: MI
### Show Tags
15 Dec 2003, 07:37
My bad, the possesive form is "its" not " it's ".
Thus, C is the best answer.
Kudos [?]: 31 [0], given: 0
Intern
Joined: 29 Aug 2003
Posts: 47
Kudos [?]: 2 [0], given: 0
Location: Detroit, MI
### Show Tags
15 Dec 2003, 12:04
"its" is the possessive of "it", as "his" is the possessive form of "him"
it's is the short form of "it is" or "it has". it's does not indicate possession.
I like C also. Official answer?
Kudos [?]: 2 [0], given: 0
Senior Manager
Joined: 11 Nov 2003
Posts: 356
Kudos [?]: 5 [0], given: 0
Location: Illinois
### Show Tags
17 Dec 2003, 16:27
Guys,
Am I missing here something? Many of you have chosen C. Is the word "are" appropriate in C? It says " there are a condition". How can we use "are" with article "a". Furthermore, "condition" is singular.
I was thinking of B as the best answer choice. Please let me know if I am missing anything. Thanks
Kudos [?]: 5 [0], given: 0
SVP
Joined: 03 Feb 2003
Posts: 1603
Kudos [?]: 308 [0], given: 0
### Show Tags
18 Dec 2003, 04:44
gmatblast wrote:
Guys,
Am I missing here something? Many of you have chosen C. Is the word "are" appropriate in C? It says " there are a condition". How can we use "are" with article "a". Furthermore, "condition" is singular.
I was thinking of B as the best answer choice. Please let me know if I am missing anything. Thanks
there are a CONDITION and its CONVERSE...
there are a wife and her husband in the room.
Kudos [?]: 308 [0], given: 0
18 Dec 2003, 04:44
Display posts from previous: Sort by
# In Newtonian laws of motion, <there is a condition and
Moderators: GMATNinjaTwo, GMATNinja
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,340 | 4,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-47 | latest | en | 0.909045 |
https://fr.mathworks.com/matlabcentral/answers/495647-string-to-categorical-array | 1,579,490,005,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250597230.18/warc/CC-MAIN-20200120023523-20200120051523-00295.warc.gz | 463,307,451 | 19,659 | MATLAB Answers
# String to Categorical array
4 views (last 30 days)
Naina on 9 Dec 2019
Commented: Stephan on 9 Dec 2019
Hi
I have an array of x=[zeros(1,120) ones(1,240)];
I want to convert it into a categorical array of 'Z' of size 1x120 and 'O' of size 1x240.
How to do that.?
Thanks in advance.
#### 0 Comments
Sign in to comment.
### Accepted Answer
Stephan on 9 Dec 2019
Edited: Stephan on 9 Dec 2019
x = [zeros(1,120), ones(1,240)];
res = categorical(x, [0, 1], {'Z', 'O'});
#### 2 Comments
Naina on 9 Dec 2019
Thanks a lot Stephen, it works perfectly.
I have one more question:
I have a cell array S=1x512, where each cell is of size 1200x65.
I need to change the cell size of 1200x65--> 1x78000.
And i can't figure out how to do that.? ;(
Stephan on 9 Dec 2019
S_new = cell(1,512);
for k = 1:512
S_new{1,k} = reshape(S{1,k},1,[]);
end
Sign in to comment.
### More Answers (0)
Sign in to answer this question. | 320 | 928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-05 | latest | en | 0.866266 |
https://communities.sas.com/t5/Base-SAS-Programming/Finding-max-across-columns-summation/td-p/389978?nobounce | 1,521,886,105,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257650188.31/warc/CC-MAIN-20180324093251-20180324113251-00313.warc.gz | 538,248,789 | 29,464 | Finding max across columns summation
Solved
Super Contributor
Posts: 409
Finding max across columns summation
Hi, I'm trying to figure out how I can sum across columns and check to see which is highest (or sort them).
Here's what I mean.
I have the data
Acct Prov Field_1 Field_2 Msg1 Msg2 Msg3 ..... Msg999
123 ON A 123 1 1 0 0
234 ON B 124 0 1 1 1
235 BC A 154 1 1 1 0
256 YK D 146 0 0 1 1
938 MB F 156 0 1 1 1
Pretend I have 100,000 records, I want to see by addition which message pops up to most from Msg1 to Msg999 (may be some missing like Msg 903, etc. is not a columns.
for Example Msg1 : 234 times
Msg2 : 3484 times,
Msg3 : 23847 times,
Msg4 : 1 time
and so on...
Thanks.
Accepted Solutions
Solution
08-28-2017 09:23 AM
Super User
Posts: 22,041
Re: Finding max across columns summation
Use STACKODS in combination with PROC MEANS:
``````proc means data=fake stackods sum mean;
var msg1-msg100;
ods output summary=want;
run;
proc sort data=want;
by descending sum;
run;``````
All Replies
Super User
Posts: 12,394
Re: Finding max across columns summation
proc means data=yourdataset sum;
var msg: ;
run;
the msg: the : denotes a variable list of all variables whose names start with msg.
One of the very nice things of 0/1 coded variables is that the sum = count of values = 1. The mean gives you a percentage if values=1.
Super User
Posts: 6,214
Re: Finding max across columns summation
With so many variables, you probably want the machine to find the largest values. For example:
proc summary data=have;
var msg: ;
output out=sums (keep=msg: ) sum=;
run;
data want;
set sums;
array counts {*} msg: ;
do i=1 to dim(counts);
variable = vname(counts{i});
count = counts{i};
output;
end;
keep variable count;
run;
proc sort data=want;
by descending count;
run;
proc print data=want;
var variable count;
run;
Solution
08-28-2017 09:23 AM
Super User
Posts: 22,041
Re: Finding max across columns summation
Use STACKODS in combination with PROC MEANS:
``````proc means data=fake stackods sum mean;
var msg1-msg100;
ods output summary=want;
run;
proc sort data=want;
by descending sum;
run;``````
☑ This topic is solved. | 719 | 2,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-13 | longest | en | 0.589488 |
https://www.tickertape.in/blog/time-weighted-rate-of-return/ | 1,702,077,962,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100779.51/warc/CC-MAIN-20231208212357-20231209002357-00874.warc.gz | 1,135,592,284 | 34,752 | Last Updated on May 24, 2022 by Aradhana Gotur
Time-weighted rate of return refers to the quantum of returns an investor can get from his investment for a particular period. It is different from Compounded Annual Growth Rate (CAGR) and is a popular metric for performance measurement in investments such as in mutual funds. When there are many withdrawals or deposits within the investment horizon, calculating the rate of returns may be difficult. This is where the time-weighted rate of return comes to the rescue. This article details this concept and helps you understand the importance of the time-weighted rate of return and how to calculate it.
## What is a time-weighted rate of return?
The time-weighted rate of return is a technique for estimating an investment portfolio’s compound growth rate. When many transactions are happening in the investment during the tenure, it may not give the correct picture to average out the peaks and troughs and simply look at the rate of return. What we need is a measure that will consider these changes and derive the compounded growth rate.
The time-weighted rate of return divides a portfolio’s return into sub-periods or intervals depending on the investments and withdrawals made. It aggregates the returns of all the sub-periods to create the rate for the whole period.
The yearly rate of return, which is the proportion of profit or loss earned from an investment over a certain period is not the same as the time-weighted rate of return. One drawback of the rate of return is that it doesn’t account for changes in cash inflows and outflows. As a result, the approach eliminates the distorting effects of cash inflows and outflows on growth rates.
Thus, a time-weighted rate of return is more beneficial for public investment managers and fund managers dealing with public securities.
## What does the time-weighted rate of return tell you?
The contributions and withdrawals made over time determine how much money will be generated on a portfolio
The rate of return for each cash flow-changing sub-period or interval is calculated using the time-weighted return. By isolating the returns that have cash flow changes, the result is more accurate than simply taking the starting and ending balances of a fund. All cash dividends are assumed to be reinvested in the portfolio for calculating the time-weighted rate of return.
When there is external cash flow, such as a deposit or a withdrawal, which would indicate the start of a new sub-period, daily portfolio valuations are required. Sub-periods must also be the same for comparing the returns of different portfolios or investments, and then the calculations are completed by geometrically linking these periods.
As investment managers who deal in publicly traded securities do not typically have control over fund investors’ cash flows, the time-weighted rate of return is a well-known performance metric for these sorts of funds.
## What are the factors considered to calculate the time-weighted rate of return?
When calculating the time-weighted rate of return, keep the following in mind:
• To compare various investment portfolios, sub-periods or intervals must be similar.
• Following a deposit or redemption, an investment valuation is necessary to indicate the start of a new sub-period.
• All returns must be considered to be reinvested in a portfolio.
## How to calculate the time-weighted rate of return?
The basic formula for calculating the time-weighted rate of return:
TWR = (ending value – beginning value) / beginning value
For example, assume Mr A invested Rs. 10,000 in an ETF on a particular date. After 25 days his investment was valued at Rs. 11,000. Now, TWR would be calculated as:
TWR = (11,000 – 10,000)/10,000 = 0.1%
Now if multiple periods are involved, then it can be calculated as:
TWR = [(1+m1)*(1+m2)*…..*(1+mN)-1]
where,
m1 = rate of return from 1st period
mN = rate of return from nth period
## Importance of time-weighted rate of return
Calculating the rate of return for assets with numerous withdrawals and deposits can be difficult, which is where the time-weighted rate of return comes in handy.
A large number of investments and redemptions distorts the rate of return across the whole investment term. That being stated, one cannot simply subtract the amount at the beginning from the balance at the end because the latter does not account for cash flows. The rate of return for each period’s investment or withdrawal made is calculated using the time-weighted return.
## Time-weighted rate of return vs rate of return
The rate of return is the net profit or loss on a venture over a certain period, which is expressed as a percentage of the original investment cost. The rate of return calculations ignores the cash flow variations in the portfolio.
On the other hand, TWRR takes all deposits and withdrawals into account when calculating the rate of return. For portfolios such as funds with smaller but more frequent contributions or withdrawals, the TWRR algorithm is more straightforward and does not allow for the distorting effects of the cash flow.
### Summary
The time-weighted return (TWR) helps remove the impact of money inflows and outflows on growth rates. It is more compact than the rate of return to measure a fund’s value. It is always better to consult your financial planner to understand the nuances of various tools and technologies used to measure returns.
Latest posts by Ayushi Mishra (see all) | 1,125 | 5,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | latest | en | 0.944962 |
https://yamol.tw/exam-109+%E5%8F%B0%E7%81%A3%E7%B6%9C%E5%90%88%E5%A4%A7%E5%AD%B8%E7%B3%BB%E7%B5%B1%E5%AD%B8%E5%A3%AB%E7%8F%AD%E8%BD%89%E5%AD%B8%E7%94%9F%E8%81%AF%E5%90%88%E6%8B%9B%E7%94%9F%E8%A9%A6%E9%A1%8C%EF%BC%9A%E5%BE%AE%E7%A9%8D%E5%88%86A-96944.htm | 1,618,405,274,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077818.23/warc/CC-MAIN-20210414125133-20210414155133-00613.warc.gz | 1,220,782,726 | 32,623 | # 109 年 - 109 台灣綜合大學系統學士班轉學生聯合招生試題：微積分A#96944
【非選題】
1.
1. (10 pts) Evaluate the following limits if they exist.
【題組】
(a)
1F
【非選題】
2.【題組】
(b)
1F
【非選題】
3.2. (10 pts) A curve in ℝ2 is given parametrically by
x=t2+2t+3
y=t4-3t3
for all t＞0. Find at the point (6,-2).
1F
x = t²+2t+3 ->6 (t-1)(t+3)=0
y = t⁴-3t³ ->-2 (t-1)(t³-2t²-2t-2)=0
1 -3 0 0 2 ]
1 -2 -2 -2 ] 1
———————
1 -2 -2 -2 |_0
—————————————————
dy dy/dt 4t³-9t² (t代入1) -5
— = ———— = ——————— = —— = -5/4
dx dx/dt 2t+2 (t代入1) 4
【非選題】
4.
3. (10 pts) Let x＞0 and △ABC be a triangle whose side lengths are . Choose a point P on , and a point Q on , and a point R on so that
Let f(x) be the area of △PQR . Find the critical point and the minimum of f(x).
1F
Critical Point x = 1
f(1) = 1.5
【非選題】
5.
4. (10 pts) Find the radius of the convergence of the power series
50
【站僕】摩檸Morning：有沒有達人提供一下答案?
【非選題】
6.
5.(10 pts) Evaluate the improper integral
50
【站僕】摩檸Morning：有沒有達人提供一下答案?
【非選題】
7.
6. (10 pts) Let g :(0,∞)- be a twice differentiable function. Assume that
g(I)=I, g'(I)=3,g"(I)=-4.
Define a real valued function h on
50
【站僕】摩檸Morning：有沒有達人提供一下答案?
【非選題】
8.7.(10 pts) Let S be the surface defined by the equation
xcos(xy)+z2y4 -7xz =1
and P(0,1,1) be a point on S . Find an equation that defines the tangent plane to S at p and a parametric equation of the normal line to S at P.
50
【站僕】摩檸Morning：有沒有達人提供一下答案?
【非選題】
9.
8. (10 pts) Evaluate the double integral
50
【站僕】摩檸Morning：有沒有達人提供一下答案?
【非選題】
10.
9. (10 pts) Let C be the curve in defined by the parametric equation
x(t)=cos(t),y(t)=sin(t),z(t)=t
for 0 ≤ t ≤ a. Suppose that the arc length of C is . Evaluate the line integal of the vector field F=
50
【站僕】摩檸Morning：有沒有達人提供一下答案?
【非選題】
11.
10. (10 pts) Find the flux of the vector field F on defined by
through the surface S = oriented with upward pointing normal vector field.
50
【站僕】摩檸Morning：有沒有達人提供一下答案? | 843 | 1,967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-17 | latest | en | 0.434599 |
https://forum.bubble.io/t/saving-a-list-of-numbers/284985 | 1,702,191,711,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00868.warc.gz | 294,378,151 | 9,455 | # Saving a list of numbers
I want to put a large list of numbers into the database to reference it later. Say the range from 0-10,000. Is there anyway to do that? I have a data type with a single field for list of numbers and I just want one entry for that data type and to not have to manually type 10,001 numbers
Whatās the purpose of this large list of numbers?
I want to pull random numbers from it but also Iām just curious now if it can be automated even if thereās a better way to do rng
There are much easier (and less costly) ways to generate random numbers than thatā¦
But if you really want to create a list of numbers in the database, then just run a recursive backend workflow 10,000 times, to create one database entry each time.
1 Like
Thats terrifying
Good learning exercise to learn how to use recursive workflows i guess.
Ok well thanks for both lol. Good to know a backend workflow could do it but also Iāll try to figure out better ways.
If you are looking to generate a random number between 1 and 10000 you can use the :calculate formula operator and select āgenerate random stringā. Specify 4 as the number of characters and only check the āuse numbersā checkbox. Then use the :converted to number operator to convert it to an int type which you can use as a random number.
It will be between 0-9999 so you could add 1 to make it between 1 and 10000, or just keep as is.
Let us know if you are using the list of numbers for something else as well. But if its just for generating a random number then the above method is much simpler
That sounds good can I use it to do lists of random numbers without it being clunky? I was trying to set up a lotto. Tickets with idk say 100 numbers from a possible 10,000. The way I was going to do with was search for(database item with 10,000 numbers in a list) then randomise it and list the first 100 numbers.
As @nico.dicagno suggests, the simplest, and most Bubble way to do this is with the generate random string function (4 characters, numbers only), converted to Number.
That will give you a random number between 0 and 9,999, and adding 1 to the result will give you a random number between 1 and 10,000
Thereās no simple way to adapt that to give a number between 0 and 10,000 in vanilla Bubble, so if you really need that then using some simple JavaScript will be the easiest way.
Hereās an example of both methods:
Random Number (bubbleapps.io)
2 Likes
Create a Data Thing called Lotto Extraction.
Add a field for āWinning Numbersā as a list of numbers
Create a recursive backend workflow that uses the above mentioned method to generate a random number - Add it to the Winning Numbers field - reschedule the recursive workflow only if āThis Lotto Extraction - Winning Numbers - count - < desired winning numbersā (Desired winning numbers can be a hard coded static value or even a recursive workflow input)
Bubble lists cannot have duplicates when using the āaddā operator, so this self-handles the unlikely but possible scenario of a winning number being picked multiple times by the random number generator.
At the end you will have a list of winning numbers in the Lotto Extraction. And it will be saved to the database for future use.
1 Like
@GaryIreland whatās your app name, I can give you a plugin for it
Itās blowing my mind a little that a recursive workflow generating 100 random numbers every time someone takes a ticket is less costly on servers than randomly sorting a list of numbers already in the database.
Iāll definitely take you guys word for it but damn lol.
So yeah thank you I think that solves it. I can use the same method you described to pick the winning numbers to make the tickets.
While I have you guys here Iām struggling with doing the workflow to select the winning tickets(Iām actually morally certain Iāve solved this before and itās annoying). So I have a field for list of winning tickets and Iām trying to filter the total list of tickets for ones with numbers matching. But the search field for numbers is expecting a single number not a list so I get contains/doesnāt contain as an operator and canāt get it to filter for tickets with more than one winning number
Plugin is much easier, it can return the list you want easily in one action with some Javascript, what is your app name
ultimatelottery, itās not on a paid plan Iām just sandboxing trying to figure out the lottery elements before maybe nesting it in a bigger gambling app
okay click āAdd pluginā and it should be at the top of the list
1 Like
Thank you! lemme check it out
To be completely fair, using Javascript to generate the list in one step is the best approach. Itās just not very bubble. It takes a few lines of very simple code and you would avoid having to run the recursive workflow.
2 Likes
That plugin is super handy @tylerboodman thank you.
Still canāt figure out how to filter a search of all tickets so itāll show me the winners. The ticket has a field with list of numbers but for some reason in this search the field seems to expect only one number.
Sounds like you assigned the users lotto numbers, now you need to pick a certain number of lotto numbers from everyoneās pool of lotto numbers? Or maybe Iām misunderstanding your structure.
The users all have tickets with 100 numbers. The draw has 100 winning numbers. I want to put all the tickets with matching numbers on the draws list of jackpot winners. Then also populate other lists with less matches. | 1,290 | 5,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-50 | latest | en | 0.936831 |
https://fliphtml5.com/viso/gijq/Titanic_activities_2_-_Homeschool_Helper_Online/4/ | 1,713,859,579,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818468.34/warc/CC-MAIN-20240423064231-20240423094231-00275.warc.gz | 231,965,325 | 19,048 | Enjoying your free trial? Only 9 days left! Upgrade Now
Create a Flipbook Now
Read more
TitanicTitanicTitanic Assignment I – Language Arts / History 1. Name three or more famous passengers 2. Choose 1 person from above and ... Read More
Titanic activities 2 - Homeschool Helper Online
Publications:
Followers:
Follow
Publications
Read Text Version
More from
P:04
TitanicAssignment V - MathThe Titanic set sail on its famous journey at 12:00 noon on April 10, 1912.At 2:20am Monday April 15, 1912 the Titanic sank into the sea.1. How long was the Titanic at sail from the time it first set sail until it finally sank?2. There were 2227 passengers on board the Titanic at the time that it sank. 14 of the lifeboats held 65 passengers each. 14 X 65 = ?4 were collapsible, which held 47 people each. 4 X 47 =2 which held 40 people each. 2 X 40 = ?How many people would all of these lifeboats hold?3. When the Titanic left Queenstown, Ireland, it had 2208 people on board. How many people wouldhave been without a lifeboat if all the boats were filled to capacity?4. Only 705 people survived the sinking of the Titanic. How many more people could have been saved?5. How many more lifeboats were needed to save all people aboard the Titanic?
Create a Flipbook Now
Explore more | 332 | 1,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-18 | latest | en | 0.966768 |
https://www.physicsforums.com/threads/calculate-the-final-temperature-of-the-ice-cream.197381/ | 1,680,094,375,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00149.warc.gz | 1,030,497,940 | 15,448 | # Calculate the final temperature of the ice cream
karr
## Homework Statement
5L of liquid nitrogen @ -205 degree celsius and 1L of liquid ice cream @ 15 degree celsius - they are both mixed. upon mixture nitrogen evaporates and ice cream hardens or freezes. Assuming only 14% of the energy gained by the nitrogen in the process of warming to room temperature (25c) was absorbed from the cream mixture, calculate the final temperature of the ice cream.
## Homework Equations
Info for Nitrogen:
density = .807g/ml
boiling point N = -195.8c
melting point = -210c
delta H vaporation=199.1 J/g
specific heat of liquid N=1.04j/g*c
Info for Ice cream
density- 1.032g/ml
melting point=-.540c
boiling point=100.17c
specific heat liquid=3.93J/g*c
specific heat solid=2.042J/g*c
(c= degree celsius)
## The Attempt at a Solution
I know that they both have to reach a equilibrium and i can carry out the problem from there but the 14% and room temp are throwing me off, how do i use that info?
T(final) = (Mn*S*Tinitial n - Mc*S*Tinitial c)/(-Mn*S+Mc*S)
s=specific heat
Mn= mass of nitrogen
T=temp
c=ice cream
is there anything wrong that I am doing in this equation? please reply ASAP
thnx
Last edited:
eli64
wicked problem
from what I can see there are 2 parts:
1. calculate delta H for the nitrogen as it warms up = delta H (absorbed)
this involves 3 steps
a) from -205c to boiling pt N2 \
b) heat of vap of N2
c) from boiling pt to room temp
this is total delta H for N2 = a + b + c
2. since the N2 got only 14% of its energy from the cream mixture, then (this might be a leap) 14% of the total delta H for N2 would be the energy that the cream released as it became ice-cream
so that 14% of total delta H for N2 = delta H of ice-cream
this has 2 steps
d. delta H (liquid at 25C--> solid at mp)
e.delta H (solid at mp --> s at final t)
delta H ice-cream = d + e
you should be able to get d) from your data, you have the delta H of ice-cream, solve for e and final T
usually there is a term for delta H fusion (melting) and that is not given here, but since ice-cream is not a pure-substance, it may not be needed
p.s. watch your signs as you work this out
Last edited:
karr
srry i forgot to post the delta H fusion of ice cream is 85.272J/g
eli64
srry i forgot to post the delta H fusion of ice cream is 85.272J/g
well there you go then, add that into the delta H of the icecream
karr
thank you so much, u r great! | 703 | 2,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-14 | latest | en | 0.910644 |
https://arbital.obormot.net/page/equivalence_relation.html | 1,701,970,851,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100677.45/warc/CC-MAIN-20231207153748-20231207183748-00013.warc.gz | 128,252,943 | 3,362 | # Equivalence relation
https://arbital.com/p/equivalence_relation
by Dylan Hendrickson Jul 5 2016 updated Jul 7 2016
A relation that allows you to partition a set into equivalence classes.
[summary: An equivalence relation is a binary Relation that is reflexive, [symmetric_relation symmetric], and transitive. You can think of it as a way to say when two elements of a Set are "the same" or "equivalent," despite not being literally the same element. It can be used to [set_partition partition] a set into equivalence classes.]
An equivalence relation is a binary Relation $\sim$ on a Set $S$ that can be used to say whether [element_of_a_set elements] of $S$ are equivalent.
An equivalence relation satisfies the following properties:
1. For any $x \in S$, $x \sim x$ (the reflexive property).
2. For any $x,y \in S$, if $x \sim y$ then $y \sim x$ (the [symmetric_relation symmetric] property).
3. For any $x,y,z \in S$, if $x \sim y$ and $y \sim z$ then $x \sim z$ (the transitive property).
Intuitively, any element is equivalent to itself, equivalence isn't directional, and if two elements are equivalent to the same third element then they're equivalent.
## Equivalence classes
Whenever we have a set $S$ with an equivalence relation $\sim$, we can divide $S$ into equivalence classes, i.e. sets of mutually equivalent elements. The equivalence class of some $x \in S$ is the set of elements of $S$ equivalent to $x$, written $[x]=\{y \in S \mid x \sim y\}$, and we say that $x$ is a "representative" of $[x]$. We call the set of equivalence classes $S/\sim = \{[x] \mid x \in S\}$.
From the definition of an equivalence relation, it's easy to show that $x \in [x]$ and $[x]=[y]$ if and only if $x \sim y$.
If you have a set already [set_partition partitioned] into subsets ($S$ is the [-disjoint_union] of the elements of a collection $A$), you can go the other way and define a relation with two elements related whenever they're in the same subset ($x \sim y$ when there is some $U \in A$ with $x,y \in U$). Then this is an equivalence relation, and the partition of the set is the set of equivalence classes under this relation (that is, $[x] \in A$ and $A=S/\sim$).
## Defining functions on equivalence classes
Suppose you have a Function $f: S \to U$ and want to define a corresponding function $f^*: S/\sim \to U$, where $U$ can be any set. You define $f^*([x])$ to be $f(x)$. This could be a problem; what if $x \sim y$ but $f(x) \neq f(y)$? Then $f^*([x])=f^*([y])$ wouldn't be well-defined. Whenever you define a function on equivalence classes in terms of representatives, you have to make sure the definition doesn't depend on which representative you happen to pick. In fact, one way you might arrive at an equivalence relation is to say that $x \sim y$ whenever $f(x)=f(y)$.
If you have a function $f: S \to S$ and want to define $f^*: S/\sim \to S/\sim$ by $f^*([x])=[f(x)]$, you have to verify that whenever $x \sim y$, $[f(x)]=[f(y)]$, equivalently $f(x) \sim f(y)$.
# Examples
Consider the integers with the relation $x \sim y$ if $n|x-y$, for some $n \in \mathbb N$. This is the integers mod $n$. The [-addition] and [-multiplication] operations can be inherited from the integers, so it makes sense to talk about addition and multiplication mod $n$.
The real numbers can be defined as equivalence classes of Cauchy sequences.
Isomorphism is an equivalence relation (unless the objects form a [-proper_class]). | 945 | 3,455 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-50 | latest | en | 0.842974 |
https://discussions.unity.com/t/calculate-pitch-90-to-90-from-euler-angles/229081 | 1,709,034,417,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474674.35/warc/CC-MAIN-20240227085429-20240227115429-00641.warc.gz | 203,753,803 | 5,483 | # Calculate Pitch (-90 to 90) from Euler Angles
I have some trouble converting euler angles to pitch angles for aviation that ranges from -90 (pointing to -Y) and 90 (pointing to +Y). Therefore if exceeding those bounds the final pitch should be substracted again. I have made a sketch to visualize what I want to achieve (grey angles):
I have this code for pitch, but it goes only from 0 to 90.
``````pitch = Vector3.Angle(new Vector3(transform.forward.x, 0, transform.forward.z), transform.forward );
``````
Use SignedAngle.
To get the pitch angle all you need is this:
``````float pitch = Mathf.Asin(transform.forward.y) * Mathf.Rad2Deg;
`````` | 164 | 653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-10 | latest | en | 0.90816 |
http://mathhelpforum.com/algebra/8515-math-help-print.html | 1,527,345,548,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867417.75/warc/CC-MAIN-20180526131802-20180526151802-00323.warc.gz | 179,104,681 | 3,353 | # math help
• Dec 6th 2006, 02:48 PM
fancyface
math help
solve by addition method indicate whether it is independent, inconsistent or depnedent,
x+3(y-1)=11
2(x-y+8y=28 :confused:
• Dec 6th 2006, 04:39 PM
topsquark
Quote:
Originally Posted by fancyface
solve by addition method indicate whether it is independent, inconsistent or depnedent,
x+3(y-1)=11
2(x-y+8y=28 :confused:
Need a little help with that last equation. You're missing a ")."
-Dan
• Dec 7th 2006, 04:16 AM
topsquark
From a PM from fancyface:
This is the message:
x+3(y-1)=11
2(x-y)+8y=28
-Dan
• Dec 7th 2006, 04:19 AM
topsquark
Quote:
Originally Posted by topsquark
From a PM from fancyface:
This is the message:
x+3(y-1)=11
2(x-y)+8y=28
-Dan
I'm going to simplify these equations a touch:
$\displaystyle x + 3y = 14$
$\displaystyle 2x + 6y = 28$
Note that if we multiply the top equation by 2 we get:
$\displaystyle 2x + 6y = 28$
which is just the same as the second equation.
This means we have an infinite number of solutions to x and y, all lying on the line $\displaystyle y = -\frac{1}{3}x + \frac{14}{3}$. You'd have to look up the definitions of dependent and independent. But I can say that the system is NOT inconsistent since we have at least one solution.
-Dan
• Dec 7th 2006, 08:41 AM
AfterShock
Quote:
Originally Posted by topsquark
I'm going to simplify these equations a touch:
$\displaystyle x + 3y = 14$
$\displaystyle 2x + 6y = 28$
Note that if we multiply the top equation by 2 we get:
$\displaystyle 2x + 6y = 28$
which is just the same as the second equation.
This means we have an infinite number of solutions to x and y, all lying on the line $\displaystyle y = -\frac{1}{3}x + \frac{14}{3}$. You'd have to look up the definitions of dependent and independent. But I can say that the system is NOT inconsistent since we have at least one solution.
-Dan
x = 14
y = 0
Note the gausselim (echelon form) is x + 3y = 14 // -5y = 0;
I agree with topsquark when he says that the system is not inconsistent (and thus consistent), since we have at least one solution. In fact, we have exactly one solution. Setting up the matrix of coefficients for the above and then augmenting it, we can clearly see when row reducing it that there are no pivots in the augmented column, thus it is consistent. Further, we know it is linearly INDEPENDENT since there are no free variables and the columns span R^2.
• Dec 7th 2006, 11:27 AM
topsquark
Quote:
Originally Posted by AfterShock
In fact, we have exactly one solution.
As I mentioned previously we have an infinite number of solutions. All points on the line $\displaystyle y = -\frac{1}{3}x + \frac{14}{3}$ will work.
-Dan
• Dec 7th 2006, 11:36 AM
AfterShock
Quote:
Originally Posted by topsquark
As I mentioned previously we have an infinite number of solutions. All points on the line $\displaystyle y = -\frac{1}{3}x + \frac{14}{3}$ will work.
-Dan
Indeed, I believe I was looking at the homogeneous system. | 888 | 2,964 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2018-22 | latest | en | 0.899511 |
https://www.plati.market/itm/80-one-of-the-mathematical-pendulum-makes-n1-20-osci/1826680?lang=en-US | 1,550,943,769,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249508792.98/warc/CC-MAIN-20190223162938-20190223184938-00062.warc.gz | 888,643,769 | 11,568 | # 80. One of the mathematical pendulum makes N1 = 20 osci
Sold: 0
Content: fiz033_80.doc 37,5 kB
# Description
80. One of the mathematical pendulum makes N1 = 20 oscillations, the other for the same time made N2 = 12 oscillations. Determine the lengths of the two pendulums, if the difference of their lengths Δl = 16 cm. | 97 | 325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-09 | latest | en | 0.806823 |
http://jwilson.coe.uga.edu/EMAT6680/Horst/derivativequotient/derivativequotient.html | 1,416,767,675,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400379636.59/warc/CC-MAIN-20141119123259-00051-ip-10-235-23-156.ec2.internal.warc.gz | 191,002,494 | 1,456 | # Rules of Differentiation: The Quotient Rule
### Erin Horst
Now that we have explored derivatives, we can now progress to the rules of differentiation. These rules are simply formulas that instruct the learner how to compute derivatives depending on a given function.
This next rule is more complex than the sum/difference rule or the power rule, but is very useful. Let's say we have a function h(x) that is the quotient of two functions h(x)=f(x)/g(x), then to find the derivative of h(x), we use the following rule
Consider the following graphs of functions and their respective derivatives as illustrations of the rule. It is highly suggested that you verify the derivative.
The following graph illustrates the function and its derivative .
The following graph illustrates the function and its derivative
Return | 164 | 823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2014-49 | longest | en | 0.951672 |
https://discusstest.codechef.com/t/c00k0ff-editorial/17093 | 1,701,320,029,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100164.87/warc/CC-MAIN-20231130031610-20231130061610-00287.warc.gz | 246,045,948 | 7,672 | # C00K0FF - Editorial
Practice
Contest
Author: Hanlin Ren
Tester: Jakub Safin
Editorialist: Jakub Safin
CAKEWALK
none
### PROBLEM
Determine if it’s possible to select 5 problems with required difficulties for a Cook-Off, given a list of N problems with known difficulties.
### QUICK EXPLANATION
Straightforward - check the given conditions.
### EXPLANATION
As the easiest problem in this contest, there isn’t much to be said about its solution.
Notice that there are no overlaps among the difficulties required for the 5 problems, so we can just independently check for each problem if at least one of the required difficulties for it appears among the input strings.
We don’t even need to remember all the input strings - it’s enough to remember how many times each of them appeared in the input. That makes checking the required conditions even easier.
Time complexity: O(N). Memory complexity: O(1).
### CHALLENGE
Too easy? You can think about how this problem could be generalised. An arbitrary number of difficulties, arbitrary lists of allowed difficulties for each spot, upper or lower bounds on the total number of used problems of each difficulty, etc…
### AUTHOR’S AND TESTER’S SOLUTIONS
Setter’s solution
Tester’s solution
#include <bits/stdc++.h>
// iostream is too mainstream
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include <time.h>
#define dibs reserve
#define OVER9000 1234567890
#define ALL_THE(CAKE,LIE) for(auto LIE =CAKE.begin(); LIE != CAKE.end(); LIE++)
#define tisic 47
#define soclose 1e-8
#define chocolate win
// so much chocolate
#define patkan 9
#define ff first
#define ss second
#define abs(x) ((x < 0)?-(x):x)
#define uint unsigned int
#define dbl long double
#define pi 3.14159265358979323846
using namespace std;
// mylittledoge
``````typedef long long cat;
#ifdef DONLINE_JUDGE
// palindromic tree is better than splay tree!
#define lld I64d
#endif
int main() {
cin.sync_with_stdio(0);
cin.tie(0);
cout << fixed << setprecision(10);
int T;
cin >> T;
string diff[] ={
"cakewalk",
"simple",
"easy",
"easy-medium",
"medium",
"medium-hard",
"hard"
};
for(int t =0; t < T; t++) {
int cnt[] ={0,0,0,0,0,0,0};
int N;
cin >> N;
for(int i =0; i < N; i++) {
string s;
cin >> s;
for(int j =0; j < 7; j++) if(diff[j] == s) cnt[j]++;
}
if(cnt[0] && cnt[1] && cnt[2] && cnt[3]|cnt[4] && cnt[5]|cnt[6]) cout << "Yes\n";
else cout << "No\n";
}
return 0;}
// look at my code
// my code is amazing``````
Would have been more amazing if you’d had shared only solution link instead of copy pasting the entire code.
I did this question using set
if set size was 5 then print yes
else print no
here is the link to my solution
https://www.codechef.com/viewsolution/15560862
simple
medium-hard
medium
hard
easy
easy-medium
cakewalk
The problem statement says we should check for exactly one cakewalk, one simple,one easy
But the editorial solution works for the cases even when there is more than one cakewalk or simple or easy…!
1 Like
The contest must contain exactly one. You have to choose contest problems from the list given. You have to make sure that from the list of problems, you choose exactly one of them. Hope that clears!
It is because we then have a choice for selecting the question and we have atleast one of them . In the contest we have to choose exactly one . It doesn’t matter if that question is the only one in the pool or one among many .
Why would you just copy and paste the tester’s solution?
https://www.codechef.com/viewsolution/15827238
heres my code … could somebody help me get out of it…
@youknowwho96 yes the language of the problem is saying exactly one for ck , si and esy
but solution is accepting for these>1 also
1 Like | 991 | 3,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-50 | latest | en | 0.740447 |
https://convertoctopus.com/55-4-miles-to-decimeters | 1,632,773,754,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00654.warc.gz | 226,610,867 | 7,936 | ## Conversion formula
The conversion factor from miles to decimeters is 16093.44, which means that 1 mile is equal to 16093.44 decimeters:
1 mi = 16093.44 dm
To convert 55.4 miles into decimeters we have to multiply 55.4 by the conversion factor in order to get the length amount from miles to decimeters. We can also form a simple proportion to calculate the result:
1 mi → 16093.44 dm
55.4 mi → L(dm)
Solve the above proportion to obtain the length L in decimeters:
L(dm) = 55.4 mi × 16093.44 dm
L(dm) = 891576.576 dm
The final result is:
55.4 mi → 891576.576 dm
We conclude that 55.4 miles is equivalent to 891576.576 decimeters:
55.4 miles = 891576.576 decimeters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 decimeter is equal to 1.1216086502479E-6 × 55.4 miles.
Another way is saying that 55.4 miles is equal to 1 ÷ 1.1216086502479E-6 decimeters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that fifty-five point four miles is approximately eight hundred ninety-one thousand five hundred seventy-six point five seven six decimeters:
55.4 mi ≅ 891576.576 dm
An alternative is also that one decimeter is approximately zero times fifty-five point four miles.
## Conversion table
### miles to decimeters chart
For quick reference purposes, below is the conversion table you can use to convert from miles to decimeters
miles (mi) decimeters (dm)
56.4 miles 907670.016 decimeters
57.4 miles 923763.456 decimeters
58.4 miles 939856.896 decimeters
59.4 miles 955950.336 decimeters
60.4 miles 972043.776 decimeters
61.4 miles 988137.216 decimeters
62.4 miles 1004230.656 decimeters
63.4 miles 1020324.096 decimeters
64.4 miles 1036417.536 decimeters
65.4 miles 1052510.976 decimeters | 523 | 1,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-39 | latest | en | 0.797516 |
http://foxymath.com/BasicD041.aspx | 1,611,407,292,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538082.57/warc/CC-MAIN-20210123125715-20210123155715-00138.warc.gz | 42,332,157 | 7,855 | Basic Division Lesson 4: Number 4
Here's an easy example:
12
÷
4
=
3
÷
=
divided by equals
We start with twelve boxes. We divide these twelve boxes by four. This means: how many groups of four can we make with the twelve boxes that we have? Answer: 3.
Here's another example:
32
÷
4
=
8
÷
=
divided equals
We start with thirty-two boxes. We divide these thirty-two boxes by four. This means: how many groups of four can we make with the thirty-two boxes that we have? Answer: 6.
EX21TextBox1 EX21TextBox2 EX21TextBox3 EX21TextBox4 EX21TextBox5 EX21TextBox6
Try this problem.
44
÷
4
=
÷
=
divided equals
We start with forty-four boxes. We divide these forty-four boxes by four. This means: how many groups of four can we make with the forty-four boxes that we have? Answer: 11.
EX31TextBox1 EX31TextBox2 EX31TextBox3 EX31TextBox4 EX31TextBox5 EX31TextBox6 EX31TextBox7 EX31TextBox8 EX31TextBox9 EX31TextBox10 EX31TextBox11
In order to do basic division, you must memorize some equations but fear not: FoxyMath will help you! When you are finished with this lesson, you will know the following equations:
4 ÷ 4 = 1
8 ÷ 4 = 2
12 ÷ 4 = 3
16 ÷ 4 = 4
20 ÷ 4 = 5
24 ÷ 4 = 6
28 ÷ 4 = 7
32 ÷ 4 = 8
36 ÷ 4 = 9
40 ÷ 4 = 10
44 ÷ 4 = 11
48 ÷ 4 = 12
The following exercise will help you learn the above equations and what they mean. Please take your time: take whatever time you need to learn the material. | 485 | 1,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2021-04 | latest | en | 0.791858 |
https://www.iosnoops.com/appinfo/go-to-supermarket-1-preschool-math-for-iphone-and-ipad/552920445 | 1,621,371,227,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991514.63/warc/CC-MAIN-20210518191530-20210518221530-00518.warc.gz | 801,864,954 | 8,239 | (36)
Released: Aug 27, 2012
Version: 1.2
Size: 61 MB
Get notified when this app is on sale or goes free [privacy policy]
Price History
Ranking - Education (iPhone)
Go To Supermarket 1-Preschool Math
Wang Hongting
★★★\$2.99→Free★★★
★★New and Noteworthy in China Education★★
One dollar, five dollar, ten dollar, oh, how can I count the money clearly? I'm going to the supermarket to buy candy, and I need three dollars. How to pay for it? Don’t worry! Follow me to the Little Mouse Supermarket. Let’s play a game about money and shopping. Soon you will be able to buy your favorite products happily.
Do you like playing games? If you do, come to Little Mouse House and play with Betty and Pink! The fun game here will make you feel more confident. Ready? Give it a try! Math can be so much fun!
This is a math game designed for kids by a mother with over 20 years of experience in early childhood education. The enlightenment game applies the calculation way of counting to help kids learn the comprehensive calculation of one dollar, five dollar and ten dollar and simulation shopping. As long as the kids can count, they can easily master this game.
The game includes 80 questions divided into 8 checkpoints in 4 stages:
Stage 1. Counting and calculating of one dollar
Stage 2. Counting and calculating of five dollar and one dollar
Stage 3. Counting and calculating of ten dollar and one dollar
Stage 4. Counting and calculating of ten dollar, five dollar and one dollar
Playful and childish pictures together with a fun and easy, step-by-step learning model help kids grasp the simple counting calculation method. The supermarket simulation shopping game helps kids experience the fun of life and game. The advanced game is added calculations of continuous addition and continuous subtract.
There are a lot of ways to learn calculation. It’s very helpful for training kids’ thinking, reaction, analysis and operability to let them study math with a variety of calculation methods. Learning by games can help kids maintain interest in math. We have summarized some simple and flexible ways of learning maths and designed dedicated learning games:
1. Number composition method: help kids master the composition of numbers and be able to calculate quickly - Delicious Numbers 1 (released);
2. Addition and subtraction of real objects: make kids calculate real objects corresponding to numbers- Delicious Numbers 2 (released);
3. Counting calculation method: help kids use counting to calculate directly- Go to the Supermarket (released);
4. Ascending and descending method: help kids learn calculation by direct ascending or descending of numbers - Fruit Express(Coming soon);
5. Vertical calculation method: help kids learn calculation by combining abacus principle with vertical form- Fruit Acrobatic Troupe (Coming soon);
After learning through a series of math games, your kids will become more and more interested in math and their computing ability and speed will be quickly improved.
What's new in version 1.2
Add shopping mod, and add way of learning through counting cartoon conins。
Our verdict: Good
• + Go To Supermarket 1-Preschool Math is a highly rated app (5-star).
Similar Apps
Prepware Private Pilot
Universal App - Designed for iPhone and iPad
Developer: ASA || Version: 1.53
HOMER Learn & Grow
Universal App - Designed for iPhone and iPad
Developer: Homer || Version: 9.1.3
Baby Shark Best Kids Songs
Universal App - Designed for iPhone and iPad
Developer: SmartStudy || Version: 49.7.5 | 772 | 3,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-21 | longest | en | 0.934186 |
https://www.manchester.ac.uk/study/undergraduate/courses/2024/03927/meng-mechatronic-engineering/course-details/EEEN20282 | 1,709,349,948,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00578.warc.gz | 873,180,981 | 15,428 | Don't just imagine your future at University, experience it first-hand!
Step into the shoes of an undergraduate student and join us for our upcoming on-campus Discover Days in Science, Engineering and Fashion. These days are tailored exclusively for Year 12s who are interested in taking their academic journey to the next level. Find your favourite subject now!
# MEng Mechatronic Engineering / Course details
Year of entry: 2024
## Course unit details:Applied Mechanics & Industrial Robotics
Unit code EEEN20282 20 Level 2 Semester 2 No
### Overview
Syllabus
• Industrial robotics; terminology, classification, workspace, wrists and end-effectors.
• Description of the position of a rigid body in Robotics. Euler angles, roll, pitch and yaw angles, transformations. Forward and inverse kinematics for articulated manipulator, the Denavit-Hartenber (DH) convention.
• Robot velocity. Linear and angular velocities. General motion of a rigid body. Linear velocity of a point attached to a moving frame. Jacobian. Singularities.
• Robot motion. Force, moment, and torque. Newton-Euler formulation. Equation of motion. Lagrange formulation. Centre of Mass. Moment of inertia. Kinetic and potential energy of an n-link robot.
• Control of an independent joint model. Applying control theory (path planning and PID control) to industrial robotics.
• Introduction to materials; review of physical concepts, structure and its influence on properties, relative cost.
• Properties of materials; Young’s modulus elasticity, shear modulus of elasticity, Hooke’s law, Poisson’s ratio, yield strength, ultimate tensile strength, brittle and ductile materials.
• Mechanical structures under load; tensile, compressive, shear, thermal, stress, strain, strain energy, torsion, first moment of area (centroid), polar moment of inertia (shafts), second moment of area (beams).
• Deflection of structures under load; bending of beams, torque twist relation, stress transformations (Mohr’s Circle), buckling instability (Euler strut), forces in and analysis of structures, free body diagrams.
• Failure of materials. Stress concentration, toughness, fatigue failures, ultimate strength, corrosion.
### Pre/co-requisites
Unit title Unit code Requirement type Description
Energy Transport and Conversion EEEN10212 Pre-Requisite Compulsory
Machines, Drives & Power Electronics EEEN20212 Co-Requisite Compulsory
Signals and Systems EEEN20131 Pre-Requisite Compulsory
Control Systems I EEEN20252 Co-Requisite Compulsory
Mathematics 1E1 for EEE MATH19681 Pre-Requisite Compulsory
Mathematics 1E2 MATH19682 Pre-Requisite Compulsory
Mathematics 2E1 MATH29681 Pre-Requisite Compulsory
### Aims
This course unit detail provides the framework for delivery in 2020/21 and may be subject to change due to any additional Covid-19 impact. Please see Blackboard / course unit related emails for any further updates.
The unit aims to:
This course unit introduces industrial robot configuration, the mathematical methods used to define their movement, and the mechanical techniques required to describe their dynamical behaviour.
An industrial robot arm is an example of a mechatronic system; it combines the required mechanical structure, actuators, sensors and control to enable an end-effector to be moved from one position to another.
The influence of the materials and structures used in the design of the mechanical components of industrial robots will also be studied.
### Learning outcomes
All of the following Intended Learning Outcomes are developed and assessed. On the successful completion of the course, students will be able to:
ILO 1 Apply homogenous transformations to find the position and orientation of the end-effector ILO 2 Identify the DH parameters of a robot. ILO 3 Use trigonometry to find the inverse kinematics of 3-joints robots. ILO 4 Analyse the movement of a robotics system by constructing its Jacobian. ILO 5 Apply Euler-Lagrange equations to a simple mechanical system and planar robotics systems. ILO 6 Identify the influence of material properties on their strength in terms of stress, strain and elasticity. ILO 7 Analyse a mechanical system with respect to moment and force equilibrium. ILO 8 Calculate the stress and strain on mechanical components when subject to tensile, compressive and torsional forces. ILO 9 Calculate the deflection of mechanical components when subject to tensile, compressive and torsional forces. ILO 10 Identify potential failure points in mechanical components due to loading and environmental conditions.
### Teaching and learning methods
Theoretical knowledge is delivered in lectures and applied during practical session.
Method Weight
Other 30%
Written exam 70%
### Feedback methods
.
1. Hutchinson S, Vidyasagar M (Mathukumalli), eds. Extracts. In: Robot Modeling and Control . Hoboken, N.J: John Wiley & Sons; 2006:478 p.¿: https://contentstore.cla.co.uk/secure/link?id=3a922d45-e10a-e611-80bd-0cc47a6bddeb.
2. Corke PI. Robotics, Vision and Control¿: Fundamental Algorithms in MATLAB® . Second, completely revised, extended and updated edition. Cham, Switzerland: Springer; 2017. doi:10.1007/978-3-319-54413-7
3. Siciliano B author. Robotics¿: Modelling, Planning and Control . (Sciavicco L author., Villani L author., Oriolo G author., eds.). London: Springer London; 2009. doi:10.1007/978-1-84628-642-1
4. Siciliano B, Khatib O, eds. Springer Handbook of Robotics . 2nd edition. Berlin: Springer; 2016. doi:10.1007/978-3-319-32552-1
5. Mechanics of Materials – JM Gere et al (many versions available, fifth onwards recommended).
6. PP Benham el al. Mechanics of Engineering Materials – (two versions).
7. RR Craig. Mechanics of Materials
8. DR Askeland et al. The Science and Engineering of Materials – (many versions).
### Study hours
Scheduled activity hours
Lectures 40
Practical classes & workshops 15
Tutorials 8
Independent study hours
Independent study 137
### Teaching staff
Staff member Role | 1,368 | 5,982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-10 | latest | en | 0.810371 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.