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http://www.amosweb.com/cgi-bin/awb_nav.pl?s=wpd&c=dsp&k=total+utility+curve | 1,726,840,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652278.82/warc/CC-MAIN-20240920122604-20240920152604-00602.warc.gz | 32,585,574 | 9,332 | Friday September 20, 2024
AmosWEB means Economics with a Touch of Whimsy!
PERFECT COMPETITION, LONG-RUN PRODUCTION ANALYSIS: In the long run, a perfectly competitive firm adjusts plant size, or the quantity of capital, to maximize long-run profit. In addition, the entry and exit of firms into and out of a perfectly competitive market guarantees that each perfectly competitive firm earns nothing more or less than a normal profit. As a perfectly competitive industry reacts to changes in demand, it traces out positive, negative, or horizontal long-run supply curve due to increasing, decreasing, or constant cost.
TOTAL UTILITY CURVE:
A curve illustrating the relation between the total utility obtained from consuming a good and the quantity of the good consumed. The shape of the total utility curve, increasing at a decreasing rate, reflects the law of diminishing marginal utility. The reason for this is that slope of the total utility curve is marginal utility, meaning the total utility curve can be use to derive the marginal utility curve.
The total utility curve is a cornerstone for consumer demand theory and utility analysis. It graphically captures the overall relation between the utility generated from a good and the quantity of the good consumed. This relation can them be used to derive marginal utility, which is critical to an understanding of market demand and the law of demand.
Roller Coaster Utility
To set the stage for the total utility curve, consider the table of total utility values presented here. This table reports the total utility generated when Edgar Millbottom, Shady Valley's resident roller coaster aficionado, spends the day riding the Monster Loop Death Plunge roller coaster at the local Shady Valley Amusement Park. Edgar takes 8 separate rides on the Monster Loop Death Plunge roller coaster, graciously recording the total utility he accumulates after each ride.
The task at hand is to plot these values into a graph. The graph below is ideally suited for the construction of the total utility curve for Edgar's 8 rides on the Monster Loop Death Plunge roller coaster.
Deriving Total Utility Curve
First, a few words about the existing graph.
• The joined axes at the right present the plotting area.
• The horizontal axis measures quantity, the number of rides on the roller coaster. This ranges from 0 to 8.
• The vertical axis measures total utility, in utils, ranging from 0 to 40.
The next step is relatively easy. Click the "Plot" button to plot points indicating the total utility associated with each ride on the roller coaster.
• If there are no rides, then total utility is 0 utils.
• After one ride, total utility 11 utils.
• After two rides, total utility 20 utils.
• After three ride, total utility 27 utils.
• And on it goes...
If the quantity consumed is continuously divisible, rather than coming in discrete amounts, then these nine points represent nine of an infinite number that capture the relation between total utility and quantity. To illustrate the other possible points, click the [Curve] button. The resulting line is the total utility curve.
Consider these three observations about the curve.
1. The curve begins at the origin. If Edgar does not ride, then he receives no utility.
2. The total utility curve is shaped something like a hill. It increases for the first 6 rides, then decreases for the rides 7 and 8. In general, Edgar's utility increases when he consumes a larger quantity.
3. The total utility curve reaches a peak of 36 utils at 6 rides. This is the number of rides that maximizes Edgar's utility.
In closing, take note of the slope of the total utility curve. It rises, flattens, then turns negative. The slope of this total utility curve is marginal utility. The shape of the total utility curve means that marginal utility is high for small quantities, declines, then becomes negative as the quantity consumed increases. This decline in marginal utility and the flattening slope of the total utility curve reflect the law of diminishing marginal utility.
<= TOTAL UTILITY TOTAL VARIABLE COST =>
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Send comments or questions to: WebMaster | 1,110 | 5,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.893077 |
https://discuss.codecademy.com/t/dont-understand-how-value-is-retrieved-8-13/25920 | 1,534,715,000,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215393.63/warc/CC-MAIN-20180819204348-20180819224348-00711.warc.gz | 622,544,318 | 4,788 | # Don't understand how value is retrieved (8/13)
#1
The solution to this is:
prices = {
"banana": 4,
"apple": 2,
"orange": 1.5,
"pear": 3
}
stock = {
"banana": 6,
"apple": 0,
"orange": 32,
"pear": 15
}
for key in prices:
print key
print "price: %s" % prices[key]
print "stock: %s" % stock[key]
I understand how the loop retrieves a key such as banana, but what I don't understand is how prices [key] then retrieves its value such as 1.5. Should it not have to be prices [value] to retrieve that? If anyone can help me clarify my misunderstanding it would be appreciated.
Why does it not like what is printed for orange?
#2
Hello, @terminator7t!
Dictionaries sure abit hard to understand at first, since we are used with lists where the index to access a position is an integer.
The thing about dictionaries is: it's a "composition" of key-value pairs and due to that, to get the value of a certain part of your dictionary, all you'll need to do is:
``````print dictName[keyName] #Will bring back the position keyName
#at that key-value pair``````
The "key" is what we use to bring back the value on it. You can think of it like a key to a room, where the key of your dictionary is the door's key and the value is the room.
In order to enter that room, you'll need to use a key for it and that's how dictionaries work.
``````for dictionaryKey in dictionaryName:
print dictionaryKey #Will print the element at the left side of the double points :
print dictionaryName[dictionaryKey] #Will access that position and bring back
#the value at the right side of the double points :``````
Q: So I can't call the value without using the dictionary name with the brackets?
A: Not that I know.
Hope my explanation was clear, if not, let me know and I'll find another way to explain!
#3
Hi g4be,
That has really helped me. Thanks!
#4
No problem, Terminator! Glad to help! | 494 | 1,877 | {"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} | 3.078125 | 3 | CC-MAIN-2018-34 | latest | en | 0.931479 |
https://techfindings.net/archives/tag/stable-diffusion | 1,709,555,563,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476442.30/warc/CC-MAIN-20240304101406-20240304131406-00178.warc.gz | 544,330,551 | 12,440 | # Tag Archives: Stable Diffusion
## Stable Diffusion – Playing with parameters
It is fun to make images with Stable Diffusion, but it is also frustrating when the result is not what you expect and it takes long time generate new pictures.
I have been playing with the cpu-only-branch of Stable Diffusion on a linux computer with an 8th Generation Core i7 CPU and 16GB of RAM and here comes some findings.
## Basic prompt and parameters
I wanted to generate a useful picture for my Dungeons & Dragons game. So, as a somewhat qualified start I did:
fantasy art of village house, cliff, well, town square, market and storm, in the style of greg rutkowski
I used
• seed=2 (bacause I did not like seed=1)
• Sample Steps=10
• Guide=7.5
• Sample Model=Euler Ancestral
• Resolution=512×512
• The 1.4 model (the small one)
Not very far from default settings. My performance is about 10s per sample step, thus this picture took 1m40s to generate:
This is the unmodified 512×512 picture. Below I will publish smaller/scaled picture but unless otherwise mentioned they are all generated as 512×512. This picture was not so far from what I had in mind, but I don’t see any well or market, or town square.
## Sample Methods
I generated exactly the same thing, only changing the Sample Method parameter:
Three of the sample methods took (roughly) twice the time (200% in the name above). I can at least draw the conclusion that the sampling method is not just a mathematical detail but something that actually affects the output quite much.
## Sampling Steps
Next thing was to try different number of sampling steps, from 2 to 99:
## Guide
There is a guide parameter (how strongly the image should follow the prompt) and that is not a very obvious parameter. For this purpose i used 30 Sampling Steps and tried a few guide values (0-15 are allowed values):
To my amateur eye, guide seems to be mostly about contrast and sharpness. I do not see that the pictures resembles my prompt more or less.
## Resolution
I generated 6 images using different resolutions. Sampling Steps is now 20.
To my surprise the lower than 512×512 came out ok, I have had very bad results at lower resolutions below. It is obvious that changing the resolutions creates a different picture, like with a different seed with the same prompt. The smaller pictures are faster and the larger slower to generate (as indicated by the %), and the largest image caused my 16GB computer to use its swap (but I think something else was swapped out). My conclusion is that you can not generate many pictures a low resolution, and then regenerate the ones you want with higher resolution and the same seed (there are probably other ways to upscale).
## Image type
So far all images have been “fantasy art”. I tried a few alternatives with 20 Sampling Steps:
This changes much. The disposition is similar but the architecture is entirely different. What if I like a drawing with the roof style of fantasy art?
## Artists
So far I have been using Greg Rutkowski for everyting (at first opportunity I will buy a collection of Greg Rutkowskis work when I find one – so far I have not found any). How about different artists:
## Dropping Keywords
So far I have not seen much of wells and markets in my pictures. What about dropping those keywords from the prompt?
## Model Choice
There is a 1.4-model to download, and a larger (full) version. What is the difference. I tried three prompts (all fantasy art in the style of greg rutkowski):
• old well in medieval village
• medieval village on cliff
• medieval village under cliff
Conclusion here is that the result is slightly different depending on model, but it is not like it makes a huge difference when it comes to quality and preference.
## Trying to get a well
Not giving up on getting a picture of a well I made 9 pictures, using different seeds and the prompt:
• fantasy art old well in medieval village, greg rutkowski
None of them contains a well as I think of a well. If I do an image search on Google I get plenty of what I want. Perhaps Stable Diffusion does not know what a well looks like, or perhaps this is what fantasy art and/or Greg Rutkowski would draw wells as.
## Conclusion
I did this because I thought I could learn something and I did. Perhaps you learnt something from reading about my results. It is obviously possible to get cool pictures, but what if you want something specific? The prompt is important, but if you are playing with the wrong parameters you may be wasting you time.
## Stable Diffusion CPU-only
I spent much time trying to install Stable Diffusion on an Intel NUC Hades Canyon with Core i7 (8th Generation) and an AMD RX Vega (4GB), with no success. 4GB is tricky. AMD is trickier.
I gave up on my NUC and installed on my laptop with Windows, GeForce GTX 1650. That worked, and a typical image (512×512 and 20 samples) takes about 3 minutes to generate.
For practical reasons I wanted to run Stable Diffusion on my Linux NUC anyway, so I decided to give a CPU-only version of stable diffusion a try (stable-diffusion-cpuonly). It was a pretty easy install, and to my surprise generation is basically as fast as on my GeForce GTX 1650. I have 16GB of RAM and that works fine for 512×512. I think 8GB would be too little and as usual, lower resolutions than 512×512 generates very bad output for me.
So when you read “stable diffusion requires Nvidia GPU with at least 4GB of RAM”, for simple hobby purposes any computer with 16GB of RAM will be fine.
## Stable Diffusion, GeForce GTX 1650 4GB, Windows 10
I have been trying to get Stable Diffusion working. My Linux workstation has an AMD GPU and that did not work well. I have Dell laptop with a GeForce GTX 1650, with 4GB video RAM, running Windows 10, and I managed to get Stable Diffusion working as expected.
I used this guide.
The key successfactors for this computer are:
• Use an “optimized” branch, that allows you to generate 512×512 with only 4GB VRAM
• Dont generate smaller than 512×512 (at least not 256×256)
• Check [x] full_precision
I wasted a lot of time with non-optimized versions of Stable Diffusion, where I had to go to 256×256 to generate anything, and that anything was always garbage.
It takes about 3 minutes to generate an image with default settings. | 1,450 | 6,317 | {"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 | longest | en | 0.921317 |
http://www.physicsforums.com/showthread.php?p=3777331 | 1,371,663,979,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709000375/warc/CC-MAIN-20130516125640-00029-ip-10-60-113-184.ec2.internal.warc.gz | 635,811,113 | 8,527 | ## Solid cylinder with nonuniform volume charge density?
1. The problem statement, all variables and given/known data
A long, non conducting, solid cylinder of radius 4.2 cm has a nonuniform volume charge density ρ = Ar^2, a function of the radial distance r from the cylinder axis. A = 2.5 µC/m5.
(a) What is the magnitude of the electric field at a radial distance of 3.2 cm from the axis of the cylinder?
(b) What is the magnitude of the electric field at a radial distance of 5.2 cm from the axis of the cylinder?
2. Relevant equations
e0 Ø = charge enclosed ; e0 is the permittivity constant 8.85e-12 and Ø can be the flux
e0 E(2π r L) = λ L ; λ is the linear charge density
3. The attempt at a solution
I thought that I could just integrate Ar^2 twice, which would give me λ. Then I could multiply by the height of the cylinder, L. After I did all that I arrived at this:
E=(Ar^3) / (24e0) with e0 being the permittivity constant.
The correct form would have me do this though:
charge enclosed = ∫ 2π r L (Ar^2) dr ; from 0 to r
= 2π L A ∫ r^3 dr ; from 0 to r
If you integrate that and solve for E you get this:
E=(Ar^3) / (4e0) <----- This gives the correct answer.
Can someone explain to me why the method I'm using wrong? Is there some constant I am missing in my method? Or is it just wrong?
And this is also my first post here on Physics Forums? Please tell me if I am doing something wrong so I can correct it in any future post.
PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
Recognitions:
Homework Help
welcome to physics forums! Or at least, congrats on your first post on physics forums :)
Quote by LumiKnight I thought that I could just integrate Ar^2 twice, which would give me λ.
This is what you should be doing. How exactly did you integrate?
I would do this: ∫ A r^2 dr = A ∫ r^2 dr That would give me this: (1/3)A r^3 Integrating again treating A as a constant would then give me this: (1/3)A ∫ r^3 dr = (1/12)A r^4 I could then set λ = (1/12)A r^4 , and I could substitute λ into the equation: e0 E(2π r L) = λ L E = (A r^3) / (2π*e0*12) I forgot to divide by 2π. But doing that just makes the denominator bigger.
Recognitions:
Homework Help
## Solid cylinder with nonuniform volume charge density?
The integration is not correct. Remember this rule? $dxdy=r \ dr \ d \theta$
Quote by BruceW The integration is not correct. Remember this rule? $dxdy=r \ dr \ d \theta$
No. I don't think I've ever seen that.
Recognitions: Homework Help hmm. I don't think you can do this question without using that rule, so I'm surprised you haven't seen it before. It is essentially an area integral using the cylindrical coordinate system (where the area is perpendicular to the z axis). You could also directly integrate using dxdy, but it would be much more complicated. Anyway, if you use that rule (remember that r is inside the integral, along with the function), then you should get the right answer for lambda.
Tags charge, charge density, cylinder, electric field
Similar discussions for: Solid cylinder with nonuniform volume charge density? Thread Forum Replies Classical Physics 9 Introductory Physics Homework 5 Introductory Physics Homework 1 Introductory Physics Homework 6 Introductory Physics Homework 3 | 895 | 3,464 | {"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} | 3.625 | 4 | CC-MAIN-2013-20 | latest | en | 0.92353 |
https://fresherbell.com/quizdiscuss/quantitative-aptitude/the-average-marks-of-four-subjects-is-120-if-33-was-misread | 1,716,341,515,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058525.14/warc/CC-MAIN-20240522005126-20240522035126-00248.warc.gz | 228,029,633 | 9,822 | # Quiz Discussion
The average marks of four subjects is 120. If 33 was misread as 13 during the calculation, what will be the correct average?
Course Name: Quantitative Aptitude
• 1] 122
• 2] 120
• 3] 125
• 4] 121
##### Solution
No Solution Present Yet
#### Top 5 Similar Quiz - Based On AI&ML
Quiz Recommendation System API Link - https://fresherbell-quiz-api.herokuapp.com/fresherbell_quiz_api
# Quiz
1
Discuss
A shop of electronic goods is closed on Monday. The average daily sales for remaining six days of a week is Rs. 15,640 and the average sale of Tuesday to Saturday is Rs. 14,124. The sales on Sunday is -
• 1] Rs. 20,188
• 2] Data inadequate
• 3] Rs. 23,220
• 4] Rs. 21,704
##### Solution
2
Discuss
The mean of 100 observations was calculated as 40. It was founder later on that one of the observation was misread as 83 instead of 53. The correct mean is :
• 1] 39
• 2] 39.7
• 3] 40.3
• 4] 42.7
##### Solution
3
Discuss
Find the average of all prime numbers between 30 and 50.
• 1]
37.8
• 2]
38.9
• 3]
39.8
• 4]
42.8
##### Solution
4
Discuss
40% of the employees in a factory are workers. All the remaining employees are executives. The annual income of each worker is Rs. 390. The annual income of each executive is Rs. 420. What is the average annual income of all the employees in the factory together?
• 1] 390
• 2] 405
• 3] 408
• 4] 415
##### Solution
5
Discuss
In Arun's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Arun and he thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?
• 1] 67 kg.
• 2] 68 kg.
• 3] 69 kg.
• 4] Data inadequate
##### Solution
6
Discuss
The average of 6 numbers is 7. The average of three numbers of them is 5. What will be the average of remaining numbers?
• 1] 15
• 2] 30
• 3] 9
• 4] 42
##### Solution
7
Discuss
Four years ago, the average age of A and B was 18 years. At present the average age of A, B, and C is 24 years. What would be the age of C after 8 years?
• 1] 25 years
• 2] 28 years
• 3] 32 years
• 4] 36 years
##### Solution
8
Discuss
The average of 25 results is 18. The average of first 12 of those is 14 and the average of last 12 is 17. What is the 13th result?
• 1] 74
• 2] 75
• 3] 69
• 4] 78
##### Solution
9
Discuss
In a primary school the average weight of male students is 65.9 kg and the average weight of female students is 57 kg. If the average weight of all the students ( both male and female ) is 60.3 kg and the number of male students in the school is 66, what is the number of female students in the school?
• 1] 162
• 2] 168
• 3] 180
• 4] 112
##### Solution
10
Discuss
A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in the month of June beginning with a Sunday?
• 1] 276
• 2] 280
• 3] 285
• 4] 250
# Quiz | 967 | 3,030 | {"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} | 3.546875 | 4 | CC-MAIN-2024-22 | latest | en | 0.93749 |
https://www.sanfoundry.com/heat-transfer-operations-questions-answers-boiling-curve-properties/ | 1,685,669,467,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648245.63/warc/CC-MAIN-20230602003804-20230602033804-00187.warc.gz | 1,058,695,812 | 22,098 | Heat Transfer Operations Questions and Answers – Boilers – Boiling Curve Properties
«
»
This set of Heat Transfer Operations Multiple Choice Questions & Answers (MCQs) focuses on “Boilers – Boiling Curve Properties”.
1. The diameter of the bubble that gets detached from the heating surface during boiling can be found by which one of the following relations?
a) Stanton’s Relation
b) Fritz Relation
c) Fourier Theory
d) Nusselt’s Theory
Explanation: The relation was given by Fritz, as given below-
D = Cd β [2 σ/g (ϸL – ϸV )]½, which is used to calculate the departing bubble diameter by knowing the values of surface tension and density.
2. Which one of the following conditions does not influence Nucleate pool boiling?
a) Pressure
b) Liquid properties
c) Heat capacity of fluid
d) Material of heating
Explanation: The factors that determine the Nucleate boiling region are –
• Pressure
• Liquid properties
• Material of heating
3. The Fritz equation to calculate the bubble diameter is given by –
D = Cd β [2 σ/g (ϸL – ϸV )]½
What is the value of the constant Cd for water bubbles at 1atm?
a) 0.0148
b) 0.003
c) 0.0027
d) 0.00148
Explanation: The equation to the bubble diameter for water bubbles at 1atm is
D = 0.0148 β [2 σ/g ( ϸL – ϸV )]½
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4. Which one of the following is the correct representation of Fritz equation which is used to calculate bubble diameter in bubble pool boiling?
a) D = Cd β [2 σ/g (pt – pv)]½
b) D = Cd β [σ/g (pt – pv)]½
c) D = Cd β [2 σ/g (pt – pv)]2
d) D = Cd β [2 σ/g (pt – pv)]
Explanation: The relation was given by Fritz, as given below-
D = Cd β [2 σ/g (ϸL – ϸV )]½, which is used to calculate the departing bubble diameter by knowing the values of surface tension and density along with the parameters Cd and β.
5. What is the significance of the term “ σ ” in the given Fritz Equation used to calculate bubble departure diameter?
D = Cd β [2 σ/g (PL – PV )]½
a) Surface Tension
b) Contact Angle
c) Liquid density
d) Gas density
Explanation: The relation was given by Fritz, as given below-
D = Cd β [2 σ/g (PL – PV )]½, which is used to calculate the departing bubble diameter by knowing the values of surface tension σ and density difference ϸL – ϸV along with the parameters Cd and β.
6. The height of the superheated layer in nucleate boiling is __________ the diameter of the departing bubble.
a) Greater than
b) Less than
c) Equal to
d) Much greater than
Explanation: The superheated region represented by the dotted lines is less than the diameter of bubble.
7. The size of the bubble on nucleation _________ as it rises up.
a) Increases
b) Decreases
c) Remains same
d) Breaks to many bubbles
Explanation: The size increases due to the increasing vapour volume and also other bubbles coalesce into one another hence its size increases.
8. The correct representation of the bubble when it is about to depart is best represented by the diagram-
a)
b)
c)
d)
Explanation: The Superheated region is always below the bubble diameter otherwise the bubble would further rise in size till it crosses the superheated region.
9. With increase in temperature of boiling operation, the heat transfer flux ___________
a) Decreases
b) Increases
c) Increases at first, then decreases
d) Increases at first, then decreases and later again increases
Explanation: The boiling curve can be represented as –
Where the flux first increases then decreases.
10. Which one of the following correctly represents the axes of the boiling curve?
a) Y: Heat flux and X: Temperature
b) X: Heat transfer coefficient and Y: Temperature
c) X: Heat flux and Y: Temperature
d) Y: Heat flux and X: Temperature difference
Explanation: The graph is plotted with the axes as Y: Heat flux and X: Temperature
11. The increase in heat flux after the Leiden Frost Point is due to ___________
b) Convection
c) Conduction
d) Flash
Explanation: The final region of heat flux growth/rise after the inefficient film boiling is called the radiation boiling region. Here the fluid is not in contact with the wall rather it is the radiation from the heating surface that is causing the temperature rise.
12. The Leiden Frost effect is observed in _________ of liquids.
a) Boiling
b) Cooling
c) Freezing
Explanation: The region of minima of heat flux from the surface to the liquid after boiling the water/fluid is known as Leiden Frost Effect of the stable film boiling region because here exists a film that does not allow heat transfer.
13. When the temperature of water is at about 90-100 degrees more than its boiling point, then what is the name of this type of boiling?
a) Nucleate
b) Stable Film boiling
c) Partially stable film boiling
Explanation: This region is called the unstable region because the film has not yet formed completely and flickers from place to place, the heat flux further drops on increasing temperature.
14. The temperature of operation of nucleate boiling is ___________ film boiling.
a) More than
b) Less than
c) Equal to
d) Difficult to be compared than
Explanation: The region of nucleate boiling occurs before Critical flux whereas that of film boiling occurs after critical flux region.
15. Which one of the following statement is not correct about nucleate boiling?
a) The temperature of the surface is greater than the saturation temperature of the liquid
b) The heat flux is greater than film boiling
c) The temperature is greater than film boiling
d) Bubbles are created by the expansion of entrapped vapour in the heating surface
Explanation: The region of nucleate boiling occurs before Critical flux whereas that of film boiling occurs after critical flux region. Hence the temperature of nucleate boiling is below film boiling.
Sanfoundry Global Education & Learning Series – Heat Transfer Operations.
To practice all areas of Heat Transfer Operations, here is complete set of 1000+ Multiple Choice Questions and Answers. | 1,423 | 5,983 | {"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} | 3.359375 | 3 | CC-MAIN-2023-23 | longest | en | 0.860784 |
https://www.eductify.com/de/mathematik-ubungen/42/algebraische%C2%A0ausdrucke | 1,603,675,668,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890108.60/warc/CC-MAIN-20201026002022-20201026032022-00357.warc.gz | 704,644,255 | 7,945 | Mathematik Übungen
−b−3b−2a−7a=?
2p−p−(6+3q−p+q−3)=?
4(b+1)+3b(b+1)=?
(a+2b)3=?
# Factoring (easy)
xy5–xy=xy(xy4–1)
a5+5a+4a+20 = a(a+5)+4(a+5) = (a+4)(a+5)
# Factoring (difficult)
8x2+30x+25= 8x2+10x+20x+25= 2x(4x+5)+5(4x+5)= (2x+5)(4x+5)
1/a–1/a2=?
(6x+10x)/2 =? | 180 | 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} | 3.984375 | 4 | CC-MAIN-2020-45 | longest | en | 0.315897 |
http://mathhelpforum.com/trigonometry/94526-prove.html | 1,516,513,108,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890187.52/warc/CC-MAIN-20180121040927-20180121060927-00439.warc.gz | 223,037,535 | 11,896 | 1. ## Prove that
If Tn = sin^nA+cos^nA
Prove that
T3-T5 = T5-T7
...T1.......T3
plzzzz explain by covering all the steps as i belong to 10th standard ......
2. Originally Posted by Sanjana Das
If Tn = sin^nA+cos^nA
Prove that
T3-T5 = T5-T7
...T1.......T3
plzzzz explain by covering all the steps as i belong to 10th standard ......
= $\frac{T_5-T_7}{T_3-T_5}$
$\frac{sin^5A+cos^5A-sin^7A-cos^7A}{sin^3A+cos^3A-sin^5A-cos^5A}$
= $\frac{sin^5A-sin^7A+cos^5Asin^2A}{sin^3A-sin^5A+cos^3Asin^2A}$
= $\frac{sin^3A-sin^5A+cos^5A}{sinA-sin^3A+cos^3A}$
= $\frac{sin^3Acos^2A+cos^5A}{sinAcos^2A+cos^3A}$
= $\frac{sin^3A+cos^3A}{sinA+cosA}$
= $\frac{T_3}{T_1}$
3. (T3 - T5) = sin^3A + cos^3A -sin^5A - cos^5A
= sin^3A - sin^5A + cos^3A - cos^5A
= sin^3A( 1 - sin^2A) + cos^3A ( 1- cos^2A)
= sin^3A*cos^2A + cos^3A*sin^2A
= sin^2A*cos^2A(sinA + cosA)
If you divide this by T1 you get sin^2A*cos^2A.
Proceed in the same manner in right hand side.
4. He3llo, Sanjana Das!
Another approach . . .
If $T_n \:=\:\sin^n\!A+\cos^n\!A$
prove that: . $\frac{T_3-T_5}{T_1}\:=\:\frac{T_5 - T_7}{T_3}$
Left side:
. . $\frac{T_3 - T_5}{T_1} \;=\;\frac{(\sin^3\!A + \cos^3\!A) - (\sin^5\!A + \cos^5\!A)}{\sin A + \cos A} \;=$ $\frac{(\sin^3\!A - \sin^5\!A) - (\cos^3\!A - \cos^5\!A)}{\sin A + \cos A}$
. . . . . . . $= \;\frac{\sin^3\!A(1 - \sin^2\!A) + \cos^3\!A(1 - \cos^2\!A)}{\sin A+\cos A} \;=\;\frac{\sin^3\!A\cos^2\!A + \cos^3\!A\sin^2\!A}{\sin A+\cos A}$
. . . . . . . $=\;\frac{\sin^2\!A\cos^2\!A\cdot(\sin A + \cos A)}{\sin A + \cos A} \;=\;\sin^2\!A\cos^2\!A$
Right side:
. . $\frac{T_5-T_7}{T_3} \;=\;\frac{(\sin^5\!A+\cos^5\!A) - (\sin^7\!A + \cos^7\!A)}{\sin^3\!A + \cos^3\!A} \;= \;\frac{(\sin^5\! - \sin^7\!A) + (\cos^5\!A - \cos^7\!A)}{\sin^3\!A + \cos^3\!A}$
. . . . . . . $= \;\frac{\sin^5\!A(1 - \sin^2\!A) + \cos^5\!A( 1 - \cos^2\!A)}{\sin^3\!A + \cos^3\!A} \;= \;\frac{\sin^5\!A\cos^2\!A + \cos^5\!A\sin^2\!A}{\sin^3\!A + \cos^3\!A}$
. . . . . . . $= \;\frac{\sin^2\!A\cos^2\!A\cdot(\sin^3\!A + \cos^3\!A)}{\sin^3\!A + \cos^3\!A} \;=\;\sin^2\!A\cos^2\!A$
Therefore: . $\frac{T_3-T_5}{T_1} \;=\;\frac{T_5-T_7}{T_3}$
Edit: Too slow again . . . Sa-ri-ga-ma beat me to it!
.
5. Noticed this:
. . $\frac{T_n - T_{n+2}}{T_{n-2}} \;=\;\frac{(\sin^n\!A + \cos^n\!A) - (\sin^{n+2}\!A + \cos^{n+2}\!A)}{\sin^{n-2}\!A + \cos^{n-2}\!A} \;=$ $\frac{(\sin^n\!A - \sin^{n+2}\!A) - (\cos^n\!A - \cos^{n+2}\!A)}{\sin^{n-2}\!A + \cos^{n-2}\!A}$
. . . . . . . $= \;\frac{\sin^n\!A(1 - \sin^2\!A) + \cos^n\!A(1 - \cos^2\!A)}{\sin^{n-2}\!A + \cos^{n-2}\!A} \;=\;\frac{\sin^n\!A\cos^2\!A + \cos^n\!A\sin^2\!A}{\sin^{n-2}\!A + \cos^{n-2}\!A}$
. . . . . . . $=\;\frac{\sin^2\!A\cos^2\!A\cdot(\sin^{n-2}\!A + \cos^{n-2}\!A)}{\sin^{n-2}\!A + \cos^{n-2}\!A} \;=\;\sin^2\!A\cos^2\!A$
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# if tn=sin^n cos^n find 2T6-3T4 1
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FP1 - Hardest chapter?
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1. Just curious on what is the most difficult chapter in FP1 (edexcel).
2. well, depends what you find most tricky
CH1: Lots of content, but pretty easy to get the hang of
CH2: Short chapter, very easy.
CH3: Generally easy once you've practiced, however some of the exam questions are horrendous.
CH4: Lots of content, but simple once you've learnt it
CH5: Looks hard, but is actually quite simple.
CH6: A case of committing several methods to memory.
3. I would say the matrices chapter
Mainly because I think it fits too much in without expecting you to learn why it works
4. (Original post by syNK)
Just curious on what is the most difficult chapter in FP1 (edexcel).
Matrix algebra and Proof since they're both things which you previously wouldnt have at all learnt in pure maths and also because, as the above poster has mentioned, there's little focus on why methods are carried out. I remember learning things for the sake of the exam as opposed to why things work the way they do.
5. (Original post by syNK)
Just curious on what is the most difficult chapter in FP1 (edexcel).
I'd say proof - if you can't see how to do it, you're screwed. At least with the others you just follow a method.
I personally found matrices really easy. Once I'd learnt how they worked, it was just like algebra.
Chapter 3 can be tough for people who have relied on learning methods the whole time, and don't have a proper understanding of maths.
Overall, I think FP1 is more about understanding what you're doing and why, rather than following set methods. Often there are exam questions which just double check that you understand the principles behind what you've learnt, which some people don't.
Just my two pennies worth.
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In computer science, tornado codes are a class of erasure codes that support error correction and have fast encoding and decoding algorithms. Software-based implementations of tornado codes are about 100 times faster on small lengths and about 10,000 times faster on larger lengths than software-based Reed-Solomon erasure codes while having only slightly worse overhead.
Tornado codes are fixed rate, near optimal erasure correcting codes that use sparse bipartite graphs to trade encoding and decoding speed for reception overhead. Since the introduction of Tornado codes, many other similar codes have emerged, most notably Online codes, LT codes and Raptor codes.
Rough Overview
Tornado codes work via xor (exclusive or). Xor operates on binary values, 1s and 0s. A xor B is 1 if A and B have different values and 0 if A and B have the same values. If you are given (A xor B) and A, you can determine the value for B. (Note that A xor B xor A is equal to B.) Similarly, if you are given (A xor B) and B, you can determine the value for A. This extends to multiple values, so given (A xor B xor C xor D) and any 3 of the values, the missing value can be recovered.
The xor of a number of binary variables is called their "parity" and this is often used in error detection and correction. Tornado codes use it for error correction. They use another checksum (like CRC-32 or MD5) for error detection.
The Tornado code algorithm starts with the sender breaking an input file or message into equal sized blocks of bytes. Let's call these blocks A [1] through A [N] . The sender records the index of each block and computes a checksums for the block and its index. (These will be used to determine if a block has been damaged during transmission and therefore needs to be recovered.) The sender also calculates some parity blocks, B [1] through B [K] . Each of these parity blocks holds the parity for a subset of the input blocks A [1] through A [N] . The sizes of and composition of these subsets is key to the speed and success of this algorithm. For each parity block, the sender records the indices of the input blocks and a checksum for the parity block and its input indices.
The sender now sends the input and parity blocks (with their indices and checksums) to the receiver. During this transmission, some of the blocks may be corrupted.
The receiver uses the checksums to identify the bad blocks and discards them. The receiver is now left with a subset of the input blocks and some parity blocks. As long as the receiver has received N + C blocks (where C is some constant), it is highly probable that the receiver can recover the file. Now, each parity block is associated with a subset of input blocks and for most parity blocks there may be multiple input blocks missing from its subset. However, given the sizes of the random subsets, it is highly likely that there exists one parity block that is missing only one of its input blocks. Using the xor operation described above, that missing input block can be recovered. Once it is recovered, a parity block that was previously missing two input blocks may now be missing just one and that one can now be recovered. This process continues - input blocks being recovered and more parity blocks being available to recover missing blocks - until the entire input file or message is recovered.
The beauty of the algorithm comes from determining the sizes and composition of the subsets. On average, the sizes are low - making it very fast to create them and fast to recover the file. However, occasionally they are large - covering most of the input blocks - so that any missing block can be recovered.
Exact Algorithm
"Need details on the exact formula for computing the size of subsets, given a value for C and expected loss rate. Also need big O() notation for algorithm and comparison for best of Reed-Solomon implementation. Source code is always great."
The subsets for each parity block are generated in a semi-random fashion. The algorithm assigned for each input block a number of parity blocks it will be present in. The algorithm also assigns for each parity block the number of input blocks that will be in it. Then randomly parity blocks with remaining slots are assigned input blocks with remaining slots. This is called a random degree-constrained bipartite graph. This method generates parity blocks with random subsets of input blocks with a specific distribution that allows the input file or message to be recovered with very high probability.
Patent Issues
Tornado codes and many similar codes (LT codes, Online codes) are patented inside the United States of America and cannot be used with out the patent holder's permission.
Citations
Michael Luby created Tornado codes and has good descriptions on his website. [http://www.icsi.berkeley.edu/~luby/]
Probable patent citation from Michael Luby's CV: "Message Encoding and Transmission System and Method for Multilevel Data Redundancy", Andres Albanese, Michael Luby, Johannes Blomer and Je Edmonds, U.S. Patent No. 5,617,541, Issued April 1, 1997. Serial Number 08/361,802; 12-21-94, Assignment recorded February 27, 1995, Reel 7364, Frames 685-689.
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### Look at other dictionaries:
• Tornado — Tornado, WV U.S. Census Designated Place in West Virginia Population (2000): 1111 Housing Units (2000): 437 Land area (2000): 3.595035 sq. miles (9.311097 sq. km) Water area (2000): 0.084938 sq. miles (0.219989 sq. km) Total area (2000): 3.679973 … StarDict's U.S. Gazetteer Places
• Tornado, WV — U.S. Census Designated Place in West Virginia Population (2000): 1111 Housing Units (2000): 437 Land area (2000): 3.595035 sq. miles (9.311097 sq. km) Water area (2000): 0.084938 sq. miles (0.219989 sq. km) Total area (2000): 3.679973 sq. miles… … StarDict's U.S. Gazetteer Places
• Tornado, West Virginia — Infobox Settlement official name = Tornado, West Virginia settlement type = CDP nickname = motto = imagesize = image caption = image mapsize = 250x200px map caption = Location of Tornado, West Virginia mapsize1 = map caption1 = subdivision type … Wikipedia
• Tornado Alley — For the book by William S. Burroughs, see Tornado Alley (book). Tornado Alley is a colloquial term most often used in reference to the area of the United States in which tornadoes are most frequent. Although an official location is not defined,… … Wikipedia
• Tornado watch — See Severe weather terminology for a comprehensive article on related weather terms. A tornado watch (SAME code: TOA; sometimes referred to as a red box by meteorologists and storm chasers) is issued when weather conditions are favorable for the… … Wikipedia
• Tornado Gundam — The Tornado Gundams are fictional weapons from the Gundam metaseries and were introduced in SD Gundam GX.AppearanceThe Tornado Gundams are the initial units given to the player upon first starting the game. As the name suggests, they are Gundam… … Wikipedia
• Building code — Code Violation: This concrete block wall is penetrated by cable trays and cables. The hole should be firestopped to restore the fire resistance rating of the wall. Instead, it is filled with flammable polyurethane foam. A building code, or… … Wikipedia
• LT code — In computer science, LT codes (Luby Transform codes) are the first class of practical fountain codes that are near optimal erasure correcting codes invented by Michael Luby in 1998 and published in 2002. [http://ieeexplore.ieee.org/xpl/freeabs… … Wikipedia
• Erasure code — In information theory, an erasure code is a forward error correction (FEC) code for the binary erasure channel, which transforms a message of k symbols into a longer message (code word) with n symbols such that the original message can be… … Wikipedia
• List of computer technology code names — Following is a list of code names that have been used to identify computer hardware and software products while in development. In some cases, the code name became the completed product s name, but most of these code names are no longer used once … Wikipedia | 1,827 | 8,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-45 | latest | en | 0.863326 |
http://www.docstoc.com/docs/126373680/Kinetic-Lab | 1,386,399,429,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163053669/warc/CC-MAIN-20131204131733-00029-ip-10-33-133-15.ec2.internal.warc.gz | 312,779,917 | 18,942 | # Kinetic Lab
Document Sample
``` Studying the Kinetics of a Reaction
In this experiment we will study the kinetics of a chemical reaction. The reaction is called a
"clock" reaction because of the means of observing the reaction rate. The reaction involves the
oxidation of iodine ion by the bromate ion in the presence of an acid:
6 I1- (aq) + BrO31-(aq) + 6 H1+(aq) ----> 3 I2 (aq) + 3 H20 (l) + Br1- (aq) (1)
The reaction is somewhat slow at room temperature. Its rate depends on the concentration of
the reactants and on the temperature. The rate law is a mathematical expression that relates the
reaction rate to the concentrations of reactants. If we express the rate of reaction as the rate of
decrease in concentration of bromate ion, the rate law has the form:
a b c
Rate = - ∆[BrO31-] /∆t = k[I1-] [BrO31-] [H1+]
where the square brackets refer to the molar concentration of the indicated species. The rate is
equal to the change in concentration of the bromate ion, divided by the change in time for the
reaction to occur. The term "k" is the rate constant for the equation, and changes as temperature
changes. The exponents a, b, and c are called the "orders" of the reaction with respect to the
indicated substance and show how the concentration of each substance affects the rate of
reaction. Adding the exponents gives the order of the overall reaction.
One purpose of the experiment is to determine the total rate law for the process. To do this we
must measure the rate, evaluate the rate constant, k, and determine the order of the reaction for
each reactant, the values of a, b, and c. Also, we will see the effect a catalyst has on the rate.
To find the rate of the reaction we need some way of measuring the rate at which one of the
reactants is used up, or the rate at which one of the products is formed. The method that we will
use is based on the rate at which iodine forms. If thiosulfate ions are added to the solution they
react with iodine as it forms in this way:
I2(aq) + 2 S2O32-(aq) ----> 2 I1-(aq) + S4O62-(aq) (2)
Reaction (1) is somewhat slow. Reaction (2) proceeds extremely rapidly, so that as quickly as
iodine is produced in reaction (1), it is consumed in reaction (2). Reaction (2) continues until all
of the thiosulfate is used up. After that, iodine begins to increase in concentration in solution. If
some starch is present, iodine will react with the starch to form a deep blue-colored complex that
is rapidly apparent.
Carrying out reaction (1) in the presence of thiosulfate ion and starch produces a chemical
"clock." When the thiosulfate is consumed, the solution turns blue.
In all of our reactions we will used the same quantity of thiosulfate ion. The blue color appears
when all the thiosulfate is used up. An examination of equations (1) and (2) shows that 6 moles
of S2O32- are needed to react with the I2 formed from 1 mole of BrO31-. Knowing the
amount of thiosulfate used allows the calculation of the amount of I2 that is formed, and also the
amount of BrO31-that has reacted at the time of the color change. The reaction rate is expressed
as the decrease in concentration of BrO31-ion divided by the time it takes for the blue color to
appear.
The experiment is designed so that the amounts of the reactants that are consumed are small in
comparison with the total quantities present. This means that the concentration of reactants is
almost unchanged during the reaction, and therefore the reaction rate is almost constant during
this time.
The experiment is designed using a microscale procedure. Only 12 drops of reactants delivered
from capillary droppers will be used for each measurement. The steps involved are as follows:
1. Determine the volume of a drop of solution. This must be done so that the number of
moles of thiosulfate ion can be found, and so the amount of bromate ions that react can be
calculated.
2. Fine the order of the reaction for each of the reactants, and determine the rate constant.
We will do this by carrying out an experiment at specific concentrations of each of the
reactants and measuring the reaction rate. Then we will change the concentration of one
reactant and observe how the reaction rate changes. This will be repeated for each reactant.
This data allows the calculation of the order of each reactant. Once the orders are known, the
value of the rate constant can be calculated.
3. In the last part of the experiment we will observe the effect of a catalyst on the rate.
Chemicals
Potassium iodide, KI, 0.010 M Sodium thiosulfate, Na2S2O3, 0.0010 M
Potassium bromate, KBrO3, 0.040 M Hydrochloric acid, HCl, 0.10 M
Starch solution, 2% Copper (II) nitrate, Cu(NO3)2, 0.1 M
Baking Soda, NaHCO3, Solid
Equipment
Beral pipets with capillary tips Trough for hot and cold water bath
Thermometer Microplate, 12-well (well plate)
Beaker, 10-mL or 50-mL Toothpicks for stirring
Sensitive balance Cotton swabs
Procedure
Safety Alert - Hydrochloric acid is hazardous to skin and eyes. Wash off spills with
lots of water. Neutralize spills on counter top with baking soda. Wear Chemical Splash
Goggles and a Chemical-Resistant Apron.
1. Find the Volume of a Drop of Solution.
Obtain micro tip Beral pipets. Then place a small beaker on a sensitive balance and find it's
mass. Holding the dropper vertically, deliver 5 drops of water into the beaker, and find the total
mass. Add an additional 5 drops of water, again determine the mass. Deliver 5 more drops and
again find the mass. Record your data in a table in your notebook. See the Data and
Calculations section for help in setting up your notebook.
Assume that the density of each of the dilute solutions that will be used is the same as that of
water, 1.00 g/mL. Calculate the volume of 1 drop of each solution.
2. Determine the Reaction rate and Calculate the Rate Law.
The table that follows shows the reagent quantities to be used in carrying out the reactions
needed.
Because we don't want the reaction to start until we are ready, be sure the KBrO3 solution is the
last solution added. It is important to use care in measuring out the solutions. Since the total
solution volume is quite small, even one extra drop can cause a substantial change in
concentrations.
It is necessary to use consistently good technique to obtain reproductive data. Hold droppers
vertically and be sure no air bubbles are introduced. Since such small quantities of reagents are
used, it is very easy to repeat measurements. Practice your technique by carrying out the first
experiment at least three times (more, if necessary) until your values are reproducible.
Calculations of the orders of reactants are all based on the values obtained for the first
experiment, so be sure to get reproducible data from the beginning. All other experiments should
be carried out at least twice.
A study of Table 1 shows that all experiments contain the same total number of drops of
solution. Only one drop of sodium thiosulfate, Na2S2O3, and one drop of starch solution are
added to each well. In experiments 1, 2, and 3, the concentration of potassium iodide, KI, is
gradually increased while all other volumes remain constant. Experiments 1, 4 and 5 have an
increasing concentration of potassium bromate, KBrO3. Experiments 1, 6 and 7 show an
increase in the concentration of hydrochloric acid, HCl. Experiment 8 will be a test to see if
calculated orders of reactants agree with experimental values.
Table 1. Reagent Quantities for Experiments
Experiment KI H2O HCl Starch Na2S2O3 KBrO3
Number 0.010M 0.10M 2% 0.0010M 0.040M
1 2 drops 4 drops 2 drops 1 drop 1 drop 2 drops
2 4 drops 2 drops 2 drops 1 drop 1 drop 2 drops
3 6 drops 0 drops 2 drops 1 drop 1 drop 2 drops
4 2 drops 2 drops 2 drops 1 drop 1 drop 4 drops
5 2 drops 0 drops 2 drops 1 drop 1 drop 6 drops
6 2 drops 2 drops 4 drops 1 drop 1 drop 2 drops
7 2 drops 0 drops 6 drops 1 drop 1 drop 2 drops
8 3 drops 1 drops 3 drops 1 drop 1 drop 3 drops
Measure out the drops of solutions required for experiment 1 in one of the wells of a 12-well
strip. Be sure to add KBrO3 last. Stir the mixture thoroughly with a toothpick. This is very
important, because it is impossible to achieve good mixing in the small well without stirring.
Begin timing the reaction as soon as the KBrO3 is added. Record the time required for the first
blue color to appear. Repeat the measurements until consistently reproducible values are
obtained. Record the room temperature as the temperature of these reactions.
Empty the well plate, rinse with water and shake to dry the wells. Use detergent and a cotton
swab, if necessary, to be sure the wells are clean and dry for each experiment. Carry out the
experiments with solution volumes described in Experiments 2 through 8.
3. Observe the Effect of a Catalyst on the Rate
Repeat the procedure of Experiment 1, but this time add 1 drop of 0.10 M copper(II) nitrate
solution, Cu(NO3)2, and only 3 drops of water to the mixture. The total volume will still be 12
drops. Record the reaction time.
Disposal and Cleanup
The solutions can be washed down the drain with a 20-fold excess of water. Clean the well
plates with detergent and water using a cotton swab.
Data and Calculations
1. Find the volume of a drop of Solution.
Set up a table like the one below in your laboratory notebook in which you can record the
necessary data and calculated values. Show a sample of each type of calculation.
(a)Mass of empty beaker ___ Average mass of 1 drop of water ___
(b)Mass of beaker plus 5 drops water ___ (d)Mass of beaker = 15 drops water ___
(b)-(a)Mass of first 5 drops ___ (d)-(c)Mass of 3rd 5 drops water ___
Average mass of 1 drop of water ___ Average mass of one drop of water ___
(c)Mass of beaker plus 10 drops water ___ Average of masses of 1 drop of water___
(c)-(b)Mass of 2nd 5 drops water ___ Volume of 1 drop: ___
2. Determine the Reaction Rate
In your lab notebook, set up the table similar to the one below.
Trial 1 Trial 2 Trial 3 Trial 4 Avg time Temp Rxn Rate [I-]o [BrO3-]o [H+]o
________________________________________________________________________
Exp 1 ________________________________________________________________________
Exp 2 ________________________________________________________________________
Exp 3 ________________________________________________________________________
Exp 4 ________________________________________________________________________
Exp 5 ________________________________________________________________________
Exp 6 ________________________________________________________________________
Exp 7 ________________________________________________________________________
Exp 8 ________________________________________________________________________
2. A) Determine the moles of S2O32- ion that were used in each reaction
In each reaction there is 1 drop of 0.0010 M Na2S2O3 solution. Since you know the
molarity of the solution, and the volume (liters) of one drop, you can calculate the moles
of S2O32- used. (Molarity = moles / liter) It is the same for all eight reactions.
2. B) Determine the moles of BrO31- used in each reaction
Since the rate will be expressed as -∆[BrO3-]/∆t, we will find the moles of BrO31- that
were used before we stopped timing. This represent the change in moles of BrO31- over
time. The blue color begins to appear when all of the thiosulfate ion is consumed.
Examine equations (1) and (2) on the front page and determine the ratio of S2O32- ions to
BrO31- ions. Use this ratio, and the moles of S2O32- you just calculated to determine the
moles of BrO31- ions that reacted. Again, it is the same for each reaction.
2. C) Determine the change in molarity of the BrO31- ion
Since you've just determine the moles of BrO31- used, you can easily calculate the
molarity using the equation molarity = moles / liter. Consider all of the solutions that had
that many moles of BrO31- react, had a total of 12 drops! This is the value of -∆[BrO31-] ,
and it is the same for all reactions.
3. Calculate the Rate for each of the 8 reactions
The rate of each reaction can be found by dividing -∆[BrO31-] that was just found, by the
number of seconds required for each reaction to take place. Write these values in the data
table.
4. Calculate the Initial Concentrations
Use the formula for diluting a solution: (M1) (V1) = (M2) (V2) Where the volume is
measured in drops. This should be calculated before the lab. Because I said so. Write
these in the data table.
5. Calculate the Order of Each Reactant
Just like the problems we've been doing. Baby! Look for two experiments where the
concentration of only one of the reactants is different and see how that effects the rate.
6. Find the Rate Constant
Using the newly found rate law (differential), then plug in the values for each experiment
and find the average k.
Experiment 8 is a check on your data. Substitute the concentrations of the reactants for
this experiment into the rate law that you determined and calculate the rate of reaction.
Then determine the percent error between the calculated rate with the actual measured
rate for experiment 8?
Discussion Questions:
1) Why does reaction rate change as concentration changes?
2) Explain the procedure used to find the rate law.
3) Comment on the agreement between measured and calculated rates for Experiment 8.
4) Why does reaction rate change as the temperature changes?
5) Differentiate between reaction rate and specific rate constant.
6) Comment on the effect of a catalyst. How does this effect the activation energy?
7) How could you improve on the results of this lab?
```
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AngeI:
Chem <.<
3 months ago
AngeI:
A solution is prepared by dissolving 3.5g of sodium hydroxide in water and adding water until the total volume of the solution is 2.5L. What are the OH- and H30+ concentrations?
3 months ago
Vocaloid:
convert g of NaOH to moles (divide by molar mass) divide by liters to get molarity. this is conc. of OH since NaOH dissociated into Na + OH to get H3O+, remember that [H3O+][OH-] = 10 ^ (-14)
3 months ago
AngeI:
I'll try and solve this later and tell you how it goes, I can't rn. Thank you though <3
3 months ago
AngeI:
So 3.5 divided by 39.997 right? Then divide by 2.5 (L)?
3 months ago
AngeI:
I got H30=-.0349? I dont think it should be negative e.e
3 months ago | 231 | 724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-35 | latest | en | 0.945449 |
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The best answer from the choices is the owner lost \$100. The \$100 bill that was stolen was then given back to the owner. What the owner loses is the \$70 worth of goods and the \$30 in change, which makes for a total of \$70 + \$30 = \$100. The owner has lost \$100.
Technically, the owner lost \$30 plus the value, V, of the \$70 of goods. Since stores typically sell goods at a markup, the value may be less than \$70. But in the case of a loss leader, the owner may have lost more than \$70.
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## The Teachers Sunglasses Riddle
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## The Tree And The Teacher Riddle
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## Teacher's Dessert Riddle
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## The Teaching Cyclops Riddle
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## Teachers And Cashiers Riddle
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## The Long Island Iced Tea
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TTwo guys walk into a bar. They both order the special, Long Island Iced Tea, which is poured from the same pitcher. One of them drinks his drink slowly and one of them drinks it fast. They walk outside and suddenly the guy who drank his Long Island Tea slowly drops to the ground and dies. They order an autospy and find that he died from poison. How could this be if the other guy drank the same tea and didnt even get sick?
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## Engineers And Teachers Riddle
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One minds the train, the other trains the mind.
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Solved: 56% | 794 | 3,151 | {"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-2019-35 | latest | en | 0.916755 |
https://discuss.codechef.com/t/problem-candy123/102413 | 1,659,902,090,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570692.22/warc/CC-MAIN-20220807181008-20220807211008-00433.warc.gz | 224,561,511 | 3,952 | # Problem candy123
int main()
{
int t, a, b;
// a & b are limits of limik and bob.
//
cin >> t;
while (t–)
{
cin >> a >> b;
int bob = 2, limak = 1, count1 = 0, count2 = 0;
if (a == 1 && b == 1)
{
cout << “Limak” << endl;
}
else
{
`````` for (int i = 3; i <= a; i = i + 2)
{
limak += i;
//limik is a variable to show the no of candies eaten by limik. as well as bob variable is used for bob.
if (limak > a)
{
break;
}
else
count1++;
//count1 and count2 are variables to count the chance of eating means how many times they have eaten candies .
}
for (int j = 4; j <= b; j = j + 2)
{
bob += j;
if (bob > b)
{
break;
}
else
count2++;
}
if (count2 >= count1)
cout << "Bob" << endl;
else if (count2 < count1)
cout << "Limak" << endl;
}
}
return 0;
``````
}
Your code is failing some testcases like when a = 2 and b = 1, your code prints Bob as the winner which is incorrect.
I have modified your code to give AC. | 317 | 922 | {"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-2022-33 | latest | en | 0.820434 |
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# Since 1986when the Department of Labor began to allow
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Since 1986when the Department of Labor began to allow [#permalink]
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Since 1986，when the Department of Labor began to allow investment officers’ fees to be based on how the funds they manage perform, several corporations began paying their investment advisers a small basic fee，with a contract promising higher fees if the managers perform well．
(A) investment officers’ fees to be based on how the funds they manage perform，several corporations began
(B) investment officers’ fees to be based on the performance of the funds they manage，several corporations began
(C) that fees of investment officers be based on how the funds they manage perform，several corporations have begun
(D) fees of investment officers to be based on the performance of the funds they manage，several corporations have begun
(E) that investment officers’ fees be based on the performance of the funds they manage，several corporations began
OA:d
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30 Jul 2009, 03:27
Agree with (D). This one is the least awkward. I think here is a problem with style rather than with grammar.
It's better to use present perfect ('have begun') because we have 'since' at the beginning.
So we have (C) and (D) left. (C) ruled out because it begins with 'that'.
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### Show Tags
30 Jul 2009, 04:38
I too think D is correct
"fees of investment officers " is preferred to "investment officer's fee"
Hence A,B, E out.
between C and D, C uses subjunctive mood. Subjunctive implies uncertainty. but here the law is already in practice from 1986, so no uncertainty. hence C is out.
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### Show Tags
30 Jul 2009, 05:43
sudiptak wrote:
I too think D is correct
"fees of investment officers " is preferred to "investment officer's fee"
Hence A,B, E out.
I do not agree to above reason. As Denis said, present perfect is required: A, B, E are out
sudiptak wrote:
between C and D, C uses subjunctive mood. Subjunctive implies uncertainty. but here the law is already in practice from 1986, so no uncertainty. hence C is out.
Not sure whether subjunctive is correct or not.
I think D is better for following reasons:
1) In C, began to allow that - it seems awkward to me (D is simple and neat). Can someone qualify the usage of 'to allow that"?
2)
In D, based on the performance of the funds they manage - simple noun phrase
In C, based on how the funds they manage perform - Clause with subject and verb
also note that the above segment is the part of modifier when the Department ...,
* Recommended ==> Usage of verbs in the modifiers should be minimum
----
Please correct me wherever I am wrong!
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31 Jul 2009, 14:19
Agreeing with Sudeepe ; its not about the subjunctive but about the precise construction
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Re: Department of Labor [#permalink] 31 Jul 2009, 14:19
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Boiling point / evaporation
Propane and butane differ mainly in their different boiling points, i.e. the temperature point at which evaporation begins.
Propane above - 42 ° C Butane above + - 0 ° C The boiling point is at at
These boiling points mean that at a temperature below -42 °, propane remains liquid. Butane remains liquid at temperatures below + - 0 ° C.
Since the gas can and must only be used from the gas phase and never from the liquid phase for the use of the devices, the boiling point is a decisive value.
Theoretically, propane could be transported in an open vessel at temperatures below –42 ° C. But theory and practice are vastly different and, of course, they play a major role in safety.
The evaporation starts when the temperature rises. One could use water as an example. If water is heated, evaporation starts at + 100 ° C. With propane this point is just at (minus) –42 ° C and with butane at + -0 ° C. The heat is supplied to the liquid gas from the outside through the container wall. Even at sub-zero temperatures, the ambient temperature is still a source of heat with propane, but not with butane.
The boiling point, which influences evaporation, also plays a major role in the pressure in the pressure vessel / liquefied gas cylinder. Again the water for comparison. When you bring it to a boil, the pressure gets higher.
The pressure conditions for liquid gas in relation to the temperature can be read from a vapor pressure curve. The temperature can therefore be determined using the pressure.
It is therefore very important to know that liquid gas expands when it is heated and, at the same time, the pressure in a container is increased if there is no possibility of expansion in the container. Regardless of the amount in a container. This explains that the filling quantity in a container cannot be measured using a pressure measurement.
A pressure measurement would only show two things:
1. the ambient temperature, due to the different pressure when the temperature changes
2. whether there is propane or butane in the container, due to the different pressure of the two gases.
The filling quantity cannot be determined by measuring the pressure.
However, this also means that a gas phase must always be present in addition to the liquid phase in a container so that the possibly increasing pressure in the gas phase can be absorbed when the temperature increases. Otherwise the pressure vessel could burst.
The gas phase in a pressure vessel is therefore an essential safety aspect when using it. The gas phase is always above the liquid phase. The pressure is therefore absorbed by the gas phase when the ambient temperature rises. In plain language it means that when filling, the pressure vessel / gas cylinder must never be completely filled so that a gas phase remains.
In general, it can be said that the gas container may only be filled to 80% with the liquid phase.
Since it is not possible to determine the filling quantity by measuring the pressure, filling only takes place according to weight, i.e. gravimetrically. Especially since you cannot see through the pressure vessel / gas container. At the approved filling stations, filling is always carried out using a scale and then a checkweigher.
The easiest way for the user / operator of a system to determine the quantity in a liquid gas cylinder is with a so-called bag scale. The tare weight that is stated on the nameplate is subtracted from the total weight.
Devices that determine the amount of content with an ultrasonic measurement have also recently appeared on the market:
A Self-filling of gas bottles with unauthorized transfer hoses that are offered on the market is not only prohibited for the aforementioned reasons, but also absolutely dangerous to the public. Anyone who does something like this is obviously not clear about the scope of his action. Unfortunately, there are always some who think they know better. However, accidents prove the opposite. According to the Product Liability Act, the seller of such an article is also liable for damage if this article has been delivered to unauthorized persons / non-specialists.
Examples of incorrect filling:
1. Incorrect filling of blue GAZ bottles with propane without a safety gas cushion immediately after the fall of the Wall in the former GDR on a boat led to the spring pressure valve being blown out on several butane gas bottles when the temperature rose. Apart from the noise, no major damage was incurred by chance. Only the mast of the sailboat was pierced by a valve.
2. in an accident in Spain in 1973, the tank of a tanker truck burst due to overfilling with 3.8 t. The safety gas cushion was missing here. When filling at dawn, all safety conditions were clearly ignored. In the midday heat, the container could no longer withstand the pressure and burst. Unfortunately, the gas ignited on the motor vehicle and rolled over a campsite as a burning roller, with a devastating effect.
Unfortunately, the connection threads of the German gas cylinders are not identical to the connections of foreign gas cylinders. Almost every country has its own ideas here. A thread standardization only in the EU is currently hardly to be expected or feasible. When filling German propane bottles abroad, it can therefore be very helpful to take so-called intermediate pieces with you on the trip:
With these respective spacers, the threads can then be bridged by the foreign filling stations if the filling station is not set up for German users.
Conversely, there are also intermediate pieces on the market to possibly connect a foreign gas cylinder to a German system / pressure regulator:
However, there is a growing trend towards safety in other countries too, and with it new filling regulations that are not always beneficial for travelers. Sometimes you come across real problems when you want to fill them, which are also handled differently.
So the pressure in a container is very important in terms of safety.
In the case of propane, a change in temperature results in the following approximate values for pressure:
at 1 ° C increase in temperature - pressure increase = 0.3 bar i.e. at a temperature increase of 10 ° C - pressure increase = 3.0 bar thus extrapolated at an ambient temperature of + 20 ° C - pressure = 7.0 bar + 30 ° C - pressure = 10.0 bar + 40 ° C - pressure = 13.0 bar
It turns out that these pressures are not a problem if they rise due to solar radiation. Unless the solar radiation is intensified by glass or mirrors or other circumstances. Of course, this information cannot be used for the temperature conditions in warmer zones. This is why butane is mainly used there, because the pressure with butane is significantly lower due to the higher boiling point:
Butane + 20 ° C - pressure = 1.2 bar
This lower pressure, due to the higher boiling point, is the reason why the blue butane GAZ bottles can be thinner-walled. Therefore it should never be filled with propane.
In plastic gas lighters there is always butane because of the low pressure. When it's cold, the lighter only works with the heat generated in your pocket.
Serious accidents can occur if the gas cushion / gas phase was not taken into account when filling.
Here are the values at more irregular Filling:
at 1 ° C increase in temperature - pressure increase = 7.0 bar i.e. every 10 ° C increase in temperature - pressure increase = 70.0 bar
The gas containers such as propellant gas tanks and gas cylinders, which are approved according to the pressure vessel ordinance, would hardly withstand the pressure load if filled incorrectly. The containers would burst at a pressure of around 100 bar. Of course not, as is often falsely claimed, to explode. The liquid gas could only explode if an ignition source were present at the same time after the bursting in order to ignite the entire quantity at once.
Compared to other countries, every German propane gas cylinder has a safety extraction valve. This valve is set in such a way that it blows off at approx. 30 - 35 bar in the event of an uncontrolled increase in pressure. The safety valve to be screwed in to use a butane GAZ bottle blows off at approx. 14 bar. After the pressure has stabilized back to normal, the safety valve would close again.
However, the safety valve is not intended to blow off if an incorrect filling takes place and could not respond so quickly. The function of this safety valve is given when the pressure rises slowly and also not when the pressure rises, where the container would suddenly be exposed to a fire. In practice it has happened that gas cylinders were left in the fire during a fire. The strength of the steel jacket decreases sharply in the event of a fire, while the pressure in the bottle increases at the same time. At the weakest point, the steel jacket tears open and the gas escapes. There is an additional fire, of course, but no explosion. The steel bottle remains as a flat sheet metal or if the gas bottle tears open on the bottom, it can happen that the filling acts like a propellant and the bottle flies around.
As already stated, the gas phase in every container not only plays a very important role in terms of safety and filling, but the "fuel" gas can be taken from the gas phase for combustion purposes, i.e. to operate the devices. | 1,933 | 9,406 | {"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-2021-39 | latest | en | 0.919289 |
https://physics.stackexchange.com/questions/290762/does-resistivity-of-material-change-under-strain | 1,656,520,782,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00168.warc.gz | 525,313,084 | 65,652 | # Does resistivity of material change under strain
As I am reading the document on physics of strain gauge, I encounter the following paragraph that I do not understand. How could resistivity change (delta rho) under strain/elongation? I always thought resistivity is a constant.
• Under strain or elongation, the inter-atomic distances change, and this potentially could change the "effective-mass" of the conduction electrons.
– NaOH
Nov 4, 2016 at 8:47
• All well explained here in terms of changes in the dimensions of the strain gauge. en.m.wikipedia.org/wiki/Strain_gauge Nov 4, 2016 at 8:53
I always thought resistivity is a constant.
No, resistivity does change under strain. This phenomenon is called piezoresistivity, and it is, in fact, the main physical phenomenon on which strain gauges are based.
For metallic conductors the relative change in resistivity is approximately proportional to the change in volume of the conductors [1],
$$\frac{\Delta\rho}{\rho} = C\frac{\Delta V}{V},$$
where $$C$$ is called Bridgman's constant. For many metals its value is around unity.
A (too) simplified way to understand the above formula is to refer to the Drude's model of conductivity. According to this model the conductivity is given by
$$\sigma = \frac {ne^{2}\tau }{m},$$
where $$n$$ is the electron density, $$e$$ the elementary charge and $$\tau$$ is the relaxation time, a time constant related to the dynamics of the electrons within the crystal. Roughly assuming that the electron density is given by a constant number of electrons $$N$$ divided by the sample volume $$V$$ yields
$$\sigma = \frac {Ne^{2}\tau }{mV},$$
from which one obtains
$$\frac{\Delta\rho}{\rho} = \frac{\Delta\sigma}{\sigma} = \frac{\Delta V}{V}.$$
As I wrote, this model is very simplified and the proportionality coefficient between the relative change of resistance and that of volume is not one, but close to one for many alloys.
For semiconductors things are more complicated, and the effect is anisotropic. The explanation lies in the change of the energy bands caused by the strain. The relative size of the effect is much greater than the case of metallic conductors. Thus, semiconductor strain gauges have a much greater sensitivity with respect to the metallic ones, but are less stable and accurate.
It is worth noting that the piezoresistivity effect causes a change in the temperature coefficient of conductors when they are mechanically constrained, because the resistivity becomes affected by the thermally-induced strain of the supporting element. Interestingly, this effect can be exploited to build resistors with very low temperature coefficients, of the order or below $$10^{-6}/\mathrm{K}$$.
[1] R. Pallàs-Areny, Sensors and signal conditioning, John G. Webster, 2001. | 654 | 2,790 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-27 | longest | en | 0.928351 |
http://help.com/post/228220-would-a-73-be-considered-a-c | 1,387,611,746,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345775355/warc/CC-MAIN-20131218054935-00043-ip-10-33-133-15.ec2.internal.warc.gz | 90,604,369 | 15,724 | # would a 73% be considered a C-? - Help.com
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## would a 73% be considered a C-?
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depends on the individual teacher’s grading scale.
Nope.. Its A
80%-100% A+
70%-79% A
60%-69% B
50%-59% C
40%-49% D
33%-39% F
This is the system where I live
definately not c-
Maybe a c+ or B but not an A
My high school’s grading scale would consider that a C- or a C at the most.
Letter Percent
A+ 95–100
A 90-94
A− 85-89
B+ 80–84
B 75-79
B− 70-74
C+ 65-69
C 60–64
C− 55–59
D 50–54
E or UN or I 0–49 (temporary)
F 0–49 (permanent)
Kuvri (yodaluv12) wrote:
My high school’s grading scale would consider that a C- or a C at the most.
wow you go to a tough school or else they ask easy questions
yoda where do you live?
And I used to go to an easy school… right? The grading system i told you is what we used in our school…
Our’s was:
97-100 A+
94-96 A
90-93 A-
87-89 B+
84-86 B
80-83 B-
77-79 C+
74-76 C
70-73 C-
67-69 D+
64-66 D
60-63 D-
0-59 F
I live in the US. This has been the standard at all 6 schools I’ve gone to in the US. Where do you guys live?
The marks I gave were for B.C.
Oh it must be different by country because I’ve lived in various cities in the US and it’s all like that.
In Quebec,
A+ 97−100
A 93-96
A- 88−92
B+ 84−87
B 80−83
B- 76−79
C+ 71−75
C 67−70
C- 63−66
D+ 58−62
D 54−57
D- 50−53
F 0−49
I’ll go to Quebec for Higher studies…
Kuvri (yodaluv12) wrote:
Oh it must be different by country because I’ve lived in various cities in the US and it’s all like that.
Probably, its different…
Where are you from, qash?
oh nvm it didn’t show up till now.
Kuvri (yodaluv12) wrote:
oh nvm it didn’t show up till now.
for the U.S.
its the following:
A 90-100 3.5-4.0
B 80-89 2.5-3.49
C 70-79 1.5-2.49
D 60-69 1.0-1.49
F 0 - 59 0.0
I m from South Asia.. I’ll go to Canada for Higher studies
spiratec9 wrote:
for the U.S.its the following:Grade Percentage GPA valueA 90-100 3.5-4.0B 80-89 2.5-3.49C 70-79 1.5-2.49D 60-69 1.0-1.49F 0 - 59 0.0
Yep that’s to calculate the GPA but most schools break it down further with plusses and minuses. So for example, a 97% would equal a 3.88 GPA.
Just F is failing.
I always wonder if you go in for a brain operation
and your doctor got an “B” in brain surgery.
Does this mean he only knows how to do 3/4 of the work???
spiratec9 wrote:
I always wonder if you go in for a brain operationand your doctor got an “B” in brain surgery.Does this mean he only knows how to do 3/4 of the work???
He can hardly find something he dosn’t know and say excuse me I have to go
look that up.??
spiratec9 wrote:
I always wonder if you go in for a brain operation
and your doctor got an “B” in brain surgery.
Does this mean he only knows how to do 3/4 of the work???
Hehe well it would depend what country he got the B in. In the US he’d know 85% of the information. In BC, he’d only know 75%. Even scarier, if he got the B in South Asia, he’d only know 60%!
Operation on your brain knowing only a little over half the stuff.
Thrilling.
In BC can you graduate if you get all C’s or C-’s?
yes you can.We need garbage collectors and bridge designers.
Ah lol. Here you can go to a community college or occasionally a state college with all C-’s. Most of our trashmen either dropped out or were in the D-F range.
c- is knowing 2/3 of the stuff.
Other 1/3 you don’t need.
one garbage can is the same as the other.You call them “trash” cans.
haha I got a C in calculus. That means I knew 75% of the information, and since then I’ve used 0% of it!
See I told you more than 1/3 is junk.
Try performing a integration on a trash can.
We joke about trash guys.Many of them make big bucks.
Same here! State jobs pay really well. Also painting lines on the road and laying down traffic cones. Even the people who plant grass between lanes get a good paycheck!
yes,but look how hard it is to paint a line its so straight.
Requires math.
They must have paid attention in math to do it so perfectly lol!
Aw ya beat me to it :)
yes until they get to intersection “higher math”
haha maybe I should just quit and paint lines instead :P
yes but don’t use thoses boring colors.How about lime green?
and purple as well. I’ll mix it up, don’t worry.
wait,if its too nice you will look at the road and crash>
qASH wrote:
80%-100% A+
70%-79% A
60%-69% B
50%-59% C
40%-49% D
33%-39% F
This is the system where I live
Wow… no one told me that. My school graded in percents, no letters, and before that, we were in 1234 system (4 is an A, 3 is a B) and I had a move in the middle, so no one ever really explained how they changed, you know?
Help me with: I’m bored.
then is 0, F?
Laina1312 wrote:
qASH wrote:
80%-100% A+
70%-79% A
60%-69% B
50%-59% C
40%-49% D
33%-39% F
This is the system where I live
Wow… no one told me that. My school graded in percents, no letters, and before that, we were in 1234 system (4 is an A, 3 is a B) and I had a move in the middle, so no one ever really explained how they changed, you know?
I never got that low, but I think 1 is the lowest you can go, but maybe. It was elementary school, what can I say? *shrug*
Help me with: I’m bored.
I have never heard of that system..
qASH wrote:
I have never heard of that system..
Ontario.
Help me with: I’m bored.
Laina1312 wrote:
I never got that low, but I think 1 is the lowest you can go, but maybe. It was elementary school, what can I say? *shrug*
then what a student will get if he/she fails?
in the US, 0-0.99 is an F on the 1-4 scale.
qASH wrote:
Laina1312 wrote:
I never got that low, but I think 1 is the lowest you can go, but maybe. It was elementary school, what can I say? *shrug*
then what a student will get if he/she fails?
I dunno, I just know my old report cards all have 1234s on them.
Help me with: I’m bored.
MIne is
A 100-80
B 79-60
C 59-40
D 39-20
F 19-0
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# NCERT Revision Notes for Class 3 Maths Chapter 3 Give and Take
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## NCERT Class 3 Maths Chapter 3 Give and Take Revision Notes
Class 3 Maths Give and Take is a fundamental chapter for the students. This chapter focuses on basic concepts of addition and subtraction in story methods. Students of Class 3 will be able to learn and understand all concepts of this chapter from the revision notes for Chapter 3 Give and Take prepared by our subject matter experts.
In these revision notes, the elaborate description of various concepts as given in the CBSE Class 3 Mathematics Chapter 3 are covered. With these notes, students can easily comprehend the mathematical principles and solve the sums given in the exercises. Hence, using the Give and Take revision notes will help you score better in the exams. Download the NCERT Revision Notes for Class 3 Maths Chapter 3 Give and Take for free from Vedantu.
Also, check CBSE Class 3 Maths revision notes for other chapters:
Last updated date: 02nd Oct 2023
Total views: 119.7k
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## Access Class 3 Mathematics Chapter 3 - Give and Take Notes
• The addition is a mathematical operation that is used to add numbers.
• Addends are the numbers that are added, i.e., the numbers to which the addition operation is applied.
• The sum of the given numbers is the result of adding the addends.
• The addition symbol is made up of one horizontal and one vertical line. It is also known as the plus sign.
• There can be more than two addends. Example:
4 + 5 + 7 =16 has three addends.
## How to Add Two Numbers?
• Start by adding the digits in one’s place. (5 + 2 = 7)
• Then add the digits in the tens place. (3 + 6 = 9)
• Finally, add the digit in the hundreds place. (2 + 5 = 7)
• 235 + 562 = 797
Let’s understand this with an example
Addition of Two Digit Numbers with Carryover
• Digits at one’s place are added first i.e. 7 + 4 = 11 which is 1 ten and 1 one.
• 1 ten is carried over to ten’s side.
• Now digits at tens place are added along with carryover i.e., 2 + 1 +1 = 4 tens.
• Hence we get the final answer as 41.
## Addition by Breaking Large Numbers According to Place Value
• We can break large numbers according to their place value so that we can solve tough addition problems mentally.
Example 1 : 45 + 67
= 40 + 5 + 60 + 7
= (40 + 60) + (5 + 7)
= 100 + 12
= 112
• Example 2 : 345 + 25
= 300 + 40 + 5 + 20 + 5
= (300) + (40 + 20) + (5 + 5)
= 300 + 60 + 10
= 370
## What is Subtraction?
• When we subtract one number from another it is known as subtraction.
• Subtraction is represented by the symbol (-). This symbol is also called the minus sign
• It is used to calculate the difference between two numbers. Example: 9 - 5 = 4
## How to Subtract Two Numbers?
• Start by subtracting the digits in one’s place. (4 - 1 =3)
• Then subtract the digits in the tens place. (5 - 2 =3)
• Finally, subtract the digits in the hundreds place. (8 - 6 =2)
• 854 - 621 = 233
Subtraction of Two Numbers
## Word Problems Based on Addition
• When objects from two or more collections are combined.
Example: Arya has 20 lemon sweets and 14 orange sweets. What is the total number of sweets Arya has?
In this, we will add 20 and 14 to find the total number of sweets.
Now , 20 + 14 = 34
Arya has 34 sweets.
• When there is an increase in the number.
Example: Ajay has 14 stamps. His friend gave him 23 stamps. How many stamps does Ajay have in all?
In this, we will add 14 and 23 to find a total number of stamps.
Now , 14 + 23 = 37
Ajay has 37 stamps.
• In addition to problems, the following keywords are used:
Sum, total, more, in all, added, all together.
Example:
• 25 and 40 more are 65 ( 25 + 40 = 65)
• The sum of 19 and 44 is 63 (19 + 44 = 63)
## Word Problems Based on Subtraction
• Partitioning: When something is taken away, removed, or given away.
Example: Rita has 100 chocolates out of which few were given away to poor children. She was left with 70 chocolates, can you guess how many were given away to poor children?
If we subtract 70 from 100 we get the number of chocolates given away to poor children.
100 - 70 = 30.
Hence 30 chocolates were given away to poor children.
• Reducing: When we have to find how much is left or how much is removed.
Example: Ajay had Rs 1000. He spent Rs 200 on groceries. Calculate the amount of money he has now.
If we subtract 200 from 1000 we get the amount of money he is left with.
1000 - 200 = 800
Hence, Ajay is left with Rs 800.
• Comparison: When we have to compare two quantities (more than/ less than)
Example: Rahul has 80 candies and Aman has 45 candies. Which person has more candies and by how much.
Clearly, Rahul has more candies than Aman.
To find how many more candies Rahul has we will subtract 45 from 80
80 - 45 = 35
Hence Rahul has 35 more candies than Aman.
• In subtraction problems, the following keywords are used:
take away, how many more, how many less, how many left, reducing, greater, smaller, less than
Example :
• 3 less than 34 is 31 ( 34 - 3 = 31)
• Reducing 100 by 34 gives 66 ( 100 -34 = 66)
## Practice Questions
Question 1. Solve the following
• 67 + 9 + 8
• 80 + 150
• 345 - 20
• 84
• 230
• 325
Question 2. A teacher noted the choice of fruits of her students in the following table.
Students Mangoes Strawberries Girls 40 15 Boys 60 10 Total
Complete the above table and answer the following questions:
• How many students in the school like strawberries?
• How many students in the school like mangoes?
Students Mangoes Strawberries Girls 40 15 Boys 60 10 Total 100 25
• 25
• 100
## Importance of NCERT Class 3 Chapter 3 Maths Give and Take Revision Notes
Here is a list of reasons why Give and Take Class 3 Maths revision notes are the ideal study resource for your exam preparation.
• The sums of this chapter have been explained in a simple language. Students can read and understand the methods of solving the sums easily from these revision notes.
• Understand the use of words in the sums of this chapter. You will be able to understand what is asked in a sum instantly using these revision notes so, you can solve the sums of this chapter easily. For instance, you will be able to comprehend what reducing 98 by 34 gives and will be easily able to solve similar sums in no time.
• Learn how to solve all types of sums covered in the chapter with the help of these revision notes.
• The objective of introducing this chapter in the curriculum will be clear to the students. They will learn the basic concepts of addition and subtraction and use them to solve arithmetic problems efficiently.
Students will imbibe the concept of addition and subtraction in arithmetic. The explanation of the sums will help them realize the applications of these concepts. They will be able to understand the meaning of ‘23 less than 89 is’ and solve all sums covered in the chapter with the help of these revision notes easily.
## Key Features of Revision Notes for Class 3 Maths Chapter 3 Give and Take
CBSE Class 3 students will be benefitted by using the revision notes for Maths Chapter 3 Give and Take in the following ways.
• Top-quality revision notes from subject experts with the highest precision.
• In accordance with CBSE Class 3 Mathematics curriculum.
• Easy explanations for the conceptual development of the young minds.
• Ensuring a complete understanding of the basic principles of addition and subtraction.
• Provides an easy way to revise the fundamental concepts of the chapter before the exam.
• Covers basic tips from the experts to solve the exercise problems for better scores in the exams.
## Conclusion
Vedantu's NCERT Revision Notes for Class 3 Maths Chapter 3, "Give and Take," offer a comprehensive and engaging resource for young learners. These well-structured notes provide a simplified understanding of the chapter's concepts, fostering an enjoyable learning experience. By combining concise explanations with colorful illustrations, Vedantu makes learning math an exciting journey for Class 3 students. The notes emphasize the importance of sharing and caring, instilling essential values alongside mathematical knowledge. With Vedantu's user-friendly approach, children can grasp fundamental mathematical principles easily. Overall, these revision notes serve as an invaluable aid, empowering students to excel in their studies while cultivating essential life skills for their holistic development.
## FAQs on NCERT Revision Notes for Class 3 Maths Chapter 3 Give and Take
1. Are the revision notes for Class 3 Maths Chapter 3 PDF free?
You can download the revision notes PDF for free from the official website and mobile app of Vedantu and use it as a reference for your exam preparation.
2. How can I prepare the concepts for Chapter 3 Maths before an exam?
By learning the simpler description of the fundamental principles in these revision notes, you can thoroughly prepare this chapter before your Maths exam.
3. What are the tips to understand the sums in Chapter 3 Maths Class 3?
Follow the Solutions and revision notes of the arithmetic problems given on Vedantu for a good understanding of this chapter. You will find a pattern to understand how to solve these questions with the help of the study materials provided on Vedantu.
4. How can students effectively use the NCERT Revision Notes for exam preparation?
Students can effectively use the NCERT Revision Notes by regularly reviewing them, practising related problems, and comparing their understanding with the notes. They can also use the notes as a quick reference during revision sessions.
5. Are the NCERT Revision Notes for Class 3 Maths Chapter 3 "Give and Take" available online?
Yes, NCERT Revision Notes for Class 3 Maths Chapter 3 "Give and Take" may be available on various educational websites and platforms that offer study materials aligned with the CBSE curriculum. | 2,398 | 10,059 | {"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} | 4.6875 | 5 | CC-MAIN-2023-40 | longest | en | 0.9054 |
http://csimsoft.com/help/mesh_generation/meshing_schemes/traditional/hole.htm | 1,519,437,511,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815034.13/warc/CC-MAIN-20180224013638-20180224033638-00780.warc.gz | 82,064,487 | 3,599 | # Hole
Applies to: Annular Surfaces
Summary: Useful on annular surfaces to produce a "polar coordinate" type mesh (with the singularity removed).
To create a polar coordinate-like mesh
1. On the Command Panel, click on Mesh and then Surface.
2. Click on the Mesh action button.
3. Enter the values for Select Surface. This can also be done using the Pick Widget function.
4. Select Hole from the drop-down menu.
5. Optionally, click on Advanced to further specify the settings.
6. Click on Apply Scheme and then Mesh.
Surface <surface_id_range> Scheme Hole [Rad_intervals <int>] [Bias <double>] [Pair Node <id> With Node <id>]
Discussion:
A polar coordinate-like mesh with the singularity removed is produced with this scheme. The azimuthal coordinate lines will be of constant radius (unlike scheme map) The number of intervals in the azimuthal direction is controlled by setting the number of intervals on the inner and outer bounding loops of the surface (the number of intervals must be the same on each loop). The number of intervals in the radial direction is controlled by the user input, rad_intervals (default is one).
A bias may be put on the mesh in the radial direction via the input parameter bias. The default bias of 0 gives a uniform grading, a bias less than zero gives smaller radial intervals near the inner loop, and a bias greater than zero gives smaller radial intervals near the outer loop.
The correspondence between mesh nodes on the inner and outer boundaries is controlled with the pair node "<loop node-id> with node <loop node-id>" construct. One id on the inner loop and one id on the outer loop should be given to connect the two nodes by a radial mesh line. Not choosing this option may result in sub-optimal node pairings with possible negative Jacobians. To use this option, mesh the inner and outer curve loops and then determine the mesh node ids.
Figure 1. Example of Hole Scheme | 411 | 1,926 | {"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-2018-09 | latest | en | 0.794302 |
http://tutoringsanjose.net/?tag=chemistry-2 | 1,656,374,868,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103344783.24/warc/CC-MAIN-20220627225823-20220628015823-00365.warc.gz | 54,780,160 | 15,532 | ## SI unit prefixes
Even with scientific notation, very large numbers and very small numbers can be awkward to work with. That is why scientists have developed a set of standard prefixes that are attached to a unit to indicate multiples of that unit. Each of the prefixes represents a particular power of ten. There are currently 20 prefixes in use:
Prefixes greater than 1
Factor Name symbol 1024 yotta Y 1021 zetta Z 1018 exa E 1015 peta P 1012 tera T 109 giga G 106 mega M 103 kilo k 102 hecto h 101 deka da
Prefixes less than 1
Factor Name symbol 10-1 deci d 10-2 centi c 10-3 milli m 10-6 micro μ 10-9 nano n 10-12 pico p 10-15 femto f 10-18 atto a 10-21 zepto z 10-24 yocto y
Do you have to memorize all of these? No. If you are a high school chemistry or physics student, you should know the prefixes in blue above. If you are an AP chemistry or AP physics student, you should also learn the prefixes in red. If you plan to be a scientist or an engineer, it wouldn’t hurt you to learn the rest, but the extreme prefixes are not so common.
Why do the prefixes skip three powers of 10? This allows us to use numbers that are always between 1 and 999. This is easiest to explain with an example. Let’s say we are measuring very small electrical currents. The following table shows how we would state the values without scientific notation, with scientific notation, and with prefixes. You can see how convenient this prefix method is:
Current Scientific notation Prefix notation 0.50 A 5.0 x 10-1 A 500 mA 0.050 A 5.0 x 10-2 A 50 mA 0.0050 A 5.0 x 10-3 A 5 mA 0.00050 A 5.0 x 10-4 A 500 μA 0.000050 A 5.0 x 10-5 A 50 μA 0.0000050 A 5.0 x 10-6 A 5 μA 0.00000050 A 5.0 x 10-7 A 500 nA 0.000000050 A 5.0 x 10-8 A 50 nA 0.0000000050 A 5.0 x 10-9 A 5 nA 0.00000000050 A 5.0 x 10-10 A 500 pA
## How to use ICE charts
When you are given the initial concentrations of the components in a reaction and you want to know the final concentrations at equilibrium, an ICE chart can be very useful. The ICE chart (some teachers call it an OCE chart or an IΔE chart) is a tool that helps you set up the proper algebraic expression in order to calculate the concentration change of each compound. It’s based on a very simple concept: the initial concentration of a compound plus the change in concentration of that compound must equal its final (equilibrium) concentration. To set up an ICE chart, follow these easy rules:
1. Write out the balanced equilibrium equation.
2. On the first line (I), put down the initial concentrations of each compound. If a compound is not in the initial mix, write 0. Be sure that you calculate the molarity, and not just the number of moles.
3. Pick a direction for the reaction to proceed. It’s usually pretty obvious, but if you aren’t sure which way the reaction will go, just choose one direction. It doesn’t matter if you guess wrong.
4. On the second line (C), use a variable to write the change in concentration for each compound using the balanced equation as a guide. For each compound that is consumed, write –Ax, where A is the coefficient of the compound in the balanced reaction. For each compound that is being formed, write +Ax. See the example below.
5. On the third line (E), write the sum of the first two lines. This must be the final concentration of each compound.
6. Use the equilibrium law and the value of Kc to write an expression with the equilibrium concentrations. Then solve this equation for x.
7. Use this value to determine the equilibrium concentrations of each compound. [Note: if the value of x is negative, this means you guessed wrong in step 3, and the reaction actually proceeds in the opposite direction. You will still get the correct final values of each compound if you add or subtract this negative value in this step.]
Example 1: 1.0 mole of H2 and 1.0 mole of I2 are placed in a 2.0 L flask. The following reaction takes place at 400°C: H2 (g) + I2 (g) → 2 HI (g). What are the concentrations of the three compounds at equilibrium? The Kc for the formation of hydrogen iodide (HI) from hydrogen and iodine gases is 64.1 at 400°C.
Solution: Start by writing the equation and the initial concentrations. Don’t forget to divide by the volume to find the molarities of all the compounds.
Next, put in the changes in concentration in terms of x. This reaction must go to the right, and we get two moles of HI for each mole of the reactants:
Then add the two columns to calculate to equilibrium concentrations:
Finally, write the equilibrium expression and solve:
$K_C=64.1= \dfrac{(2x)^2}{(0.5-x)^2}$
$\sqrt{64.1} = \sqrt{\dfrac{(2x)^2}{(0.5-x)^2}}$
$8.0= \dfrac{2x}{0.5-x}$
$4-8x=2x$
$4=10x$
$x=0.4$
$\therefore [ \text{H}_2]=[ \text{I}_2]=0.1 \, M; \, [ \text{HI}]=0.80 \, M$
Note that if you are given the change in concentration or the initial and final concentrations of a compound, you do not need to use an ICE chart. You can calculate the equilibrium values with a little bit of arithmetic. See the following example:
Example 2: The reaction in Example 1 is repeated at a different temperature with the same initial concentrations, and the final concentration of HI is determined to be 0.60 M. Determine the value of Kc at this new temperature.
Solution: To solve for Kc, we need to know the final concentrations of all three components. We are given the final value of HI, and knowing that we produced 0.60 M HI means that we must have lost 0.30 M of each of the reactants (use the balanced reaction to determine this). Since we started with 0.50 M of each, the final concentrations of each reactant must be 0.20 M (just subtract the numbers). So the equilibrium expression gives us:
$K_C= \dfrac{(0.6)^2}{(0.2)(0.2)}=9.0$
## Balancing redox reactions in basic solutions
When balancing redox reactions, you must start by finding the number of electrons transferred in each half reaction. You usually can’t get the charges to balance if you don’t equate the electrons. In this post, I show you how to balance a redox reaction in a basic solution.
Example : Iodide ions are oxidized in a basic solution by permanganate ions to produce iodine and Mn2+ ions. Show the balanced reaction.
Solution: I’ve made this example a little harder than in my other redox reaction posts because we first need to figure out from the example what the half reactions look like. This is not as complicated as it sounds. Just follow the English and turn it into chemical formulas. Our problem should therefore look like this:
I + MnO4 → I2 + Mn2+
Now comes the tricky part. It turns out it isn’t very easy to balance a redox reaction in basic solutions by adding OH ions, because when we balance the hydrogen atoms, we throw off the balance of the oxygen atoms. So we employ a clever technique. We first balance the reaction as though it were an acidic solution, then we change the solution from acidic to basic. This is a five-step process, and you’ll notice the first four steps are identical to the process for balancing a redox reaction in an acidic solution:
1. Determine the number of electrons transferred in each half reaction, and balance the electrons.
2. Balance the oxygen atoms by adding water molecules.
3. Balance the hydrogen atoms by adding H+ ions.
4. Check you answer by verifying that the charges are balanced.
5. Cancel out the H+ ions by adding an equal number of OH ions to both sides, which changes the H+ ions into water.
Let’s see how this works on our example. The iodine atom goes from an oxidation number of 1- on the left to 0 on the right, so each iodine atom loses one electron. Since there are two iodine atoms in I2, we think of this half-reaction as losing two electrons. The manganese goes from an oxidation number of 7+ in the permanganate ion to an oxidation number of 2+. It has gained five electrons. To balance the electrons, we need a total transfer of ten electrons. Multiply the iodine compounds by five and the manganese compounds by two. (Note that the iodide ion was multiplied by 10 rather than 5 to balance the number of iodine atoms):
10 I + 2 MnO4 → 5 I2 + 2 Mn2+
Now we balance the oxygen atoms. There are 8 oxygen atoms on the left and none on the right, so we add 8 water molecules to the right side:
10 I + 2 MnO4 → 5 I2 + 2 Mn2+ + 8 H2O
Next we balance the hydrogen atoms by adding 16 H+ ions to the left (we pretend we have an acidic solution):
10 I + 2 MnO4 + 16 H+ → 5 I2 + 2 Mn2+ + 8 H2O
The equation should be fully balanced, and we check our work by verifying the charges balance. 10- (the iodide ions), 2- (the permanganate ions) and 16+ (the hydrogen ions) gives a net charge on the left of 4+. On the right side, we also have a total charge of 4+.
Finally, we are ready to change the reaction from acidic to basic. We cancel out the 16 H+ ions by adding 16 OH ions to both sides:
10 I + 2 MnO4 + 16 H+ + 16 OH → 5 I2 + 2 Mn2+ + 8 H2O + 16 OH
Because H+ and OH ions react to form water molecules, we simplify the left side:
10 I + 2 MnO4 + 16 H2→ 5 I2 + 2 Mn2+ + 8 H2O + 16 OH
Our last step is to simplify the equation by subtracting the 8 extra water molecules from the right side. We have to subtract the same number from the left side to keep the reaction balanced:
10 I + 2 MnO4 + 8 H2O → 5 I2 + 2 Mn2+ + 16 OH
(Look for other tips on this website if you need help with balancing simple redox reactions or redox reactions in acidic solutions.)
## Balancing redox reactions in acidic solutions
When balancing redox reactions, you must start by finding the number of electrons transferred in each half reaction. You usually can’t get the charges to balance if you don’t equate the electrons. In this post, I show you how to balance a redox reaction in an acidic solution.
Example 1: Balance the following reaction:
MnO4 + NO + H+ → Mn2+ + H2O + NO3
Solution: To balance reactions like this, use the following four-step procedure:
1. Determine the number of electrons transferred in each half reaction, and balance the electrons.
2. Balance the oxygen atoms by adding water molecules.
3. Balance the hydrogen atoms by adding H+ ions.
4. Check you answer by verifying that the charges are balanced.
In this reaction, the Mn atom goes from an oxidation number of 7+ in the permanganate ion to an oxidation number of 2+. It has gained five electrons. The nitrogen atom goes from an oxidation number of 2+ in the nitric oxide to 5+ in the nitrate ion. It has lost three electrons. To balance the electrons, we need to make the total number of electrons transferred equal 15 (because that’s the least common multiple of three and five!). We multiply the Mn ions by 3 and the N compounds by 5:
3 MnO4 + 5 NO + H+ → 3 Mn2+ + H2O + 5 NO3
The next step is to balance the oxygen atoms. There are 17 oxygen atoms on the left side of the reaction and 15 on the right side (before you include the water). Therefore we need two water molecules on the right:
3 MnO4 + 5 NO + H+ → 3 Mn2+ + 2 H2O + 5 NO3
Finally, we balance the hydrogen atoms (there are four of them in the water molecule on the right) by adding 4 hydrogen ions on the left:
3 MnO4 + 5 NO + 4 H+ → 3 Mn2+ + 2 H2O + 5 NO3
We check our work by verifying the charges balance. 3- (the permanganate ions) and 4+ (the hydrogen ions) gives a net charge on the left of 1+. 6+ (the manganese ions) and 5- (the nitrate ions) on the right also gives a net charge of 1+. We have succeeded!
Example 2: Complete and balance the following reaction. Assume it takes place in an acidic solution:
Cr2O72 + HSO3 → SO42 + Cr3+
Solution: In this reaction, the Cr atoms go from an oxidation number of 6+ in the dichromate ion to an oxidation number of 3+ in the chromate ion. Therefore, each chromium atom gains three electrons. But since there are two chromium atoms in the dichromate ion, we must think of this half-reaction as gaining six electrons overall. The sulfur atom goes from an oxidation number of 4+ to 6+, so it loses two electrons.
In order to balance the electrons, we need a net transfer of six electrons in each half-reaction. We multiply the ions with the sulfur atoms by three. Note that we also multiply the chromium ion on the right by two to balance the number of Cr atoms:
Cr2O72 + 3 HSO3 → 3 SO42 + 2 Cr3+
The next step is to balance the oxygen atoms. There are 16 oxygen atoms on the left and 12 on the right, so we add 4 water molecules to the right side:
Cr2O72 + 3 HSO3 → 3 SO42 + 2 Cr3+ + 4 H2O
Finally, we balance the hydrogen atoms by adding H+ ions. There are three H atoms on the left and 8 H atoms on the right, so we need five H+ ions on the left:
5 H+ + Cr2O72 + 3 HSO3 → 3 SO42 + 2 Cr3+ + 4 H2O
We check our answer by checking the charge on each side. The left side has charges of 5+ (the H ions), 2- (the dichromate ion) and 3- (the hydrogen sulfite ions) for a net of zero. The right side has charges of 6- (the sulfate ions) and 6+ (the chromate ions) also for a net of zero.
(Look for other tips on this website if you need help with balancing simple redox reactions or redox reactions in basic solutions.)
## Balancing redox reactions– the basics
When balancing redox reactions, you must start by finding the number of electrons transferred in each half reaction. You usually can’t get the charges to balance if you don’t equate the electrons. In this post, I show you how the process works when both the oxidizing agent and the reducing agent are single elements.
Example (a very simple example): Balance the following reaction:
… Mg(s) + … Al3+(aq) → … Mg2+(aq) + … Al(s)
Solution: At first glance, it may appear that the reaction is already balanced because there is one magnesium atom and one aluminum atom on each side. But notice that the charges do not balance. Because aluminum gains three electrons in this reaction, but magnesium atoms lose only two electrons, you will need more magnesiums than aluminums. To find the right ratio, determine the least common multiple of 2 and 3, which is 6. Then multiply the magnesiums by 3 and the aluminums by 2 so that there is a total transfer of 6 electrons in this reaction:
3 Mg(s) + 2 Al3+(aq) → 3 Mg2+(aq) + 2 Al(s)
The charges balance, as required.
(Look for other tips on this website if you need help with balancing redox reactions in acidic or basic solutions.)
## Balancing simple chemical reactions
When balancing equations, start with the atoms that show up in only one compound on each side. Figure out those coefficients first, and then work back and forth from the reactants side to the products side to find the coefficients of each of the other compounds.
Example 1: Balance the following equation:
… C5H12 + … O2 → … CO2 + … H2O
Solution: This reaction showing the combustion of pentane is not too complicated. Notice that carbon appears in only one compound on each side, so it’s a good place to start. The five carbon atoms on the left need to be balanced by five carbon atoms on the right:
1 C5H12 + … O2 → 5 CO2 + … H2O
There are 12 hydrogen atoms on the left, so balance these on the right:
1 C5H12 + … O2 5 CO2 + 6 H2O
Finally, count up all the oxygen atoms on the right (there are 16 of them!) and balance with the O2 on the left:
1 C5H128 O2 5 CO2 + 6 H2O
Example 2: Balance the following equation:
… C6H14 + … O2 → … CO2 + … H2O
Solution: Changing the pentane to hexane has made this equation more complicated. Let’s see why. First, let’s balance the carbon and hydrogen as before:
1 C6H14 + … O2 6 CO27 H2O
There are now 19 oxygen atoms on the right side. How do you balance this on the left side? Some teachers will let you use fractions for your coefficients, and 19/2 is the correct fraction in this case. But most teachers want you to use only whole numbers for the coefficients. If you encounter a fraction, you need to multiply the entire equation by the denominator to finish the problem. Here, we multiply the equation above by 2:
2 C6H14 + … O2 12 CO2 + 14 H2O
Notice there are now 38 oxygen atoms on the right side, and the last step is simple:
2 C6H14 + 19 O2 12 CO2 + 14 H2O
Here’s a trick you can apply when the equation is a double replacement reaction:
Example 3: Balance the following reaction:
… Na2SO4 + … Al(NO3)3 → … NaNO3 + … Al2(SO4)3
Solution: Let’s start by balancing the aluminum:
… Na2SO4 + 2 Al(NO3)3 → … NaNO3 + 1 Al2(SO4)3
Now instead of balancing the sulfur atoms, we can save time by balancing the sulfate ions instead. We treat them as a group rather than as separate atoms. There are three sulfates on the right side, so we need three on the left:
3 Na2SO4 + 2 Al(NO3)3 → … NaNO3 + 1 Al2(SO4)3
Note that you can only balance polyatomic ions like this when the ions do not change from reactant side to product side. Finally, you can see there are six sodium atoms on the left (or, six nitrate ions), so the final answer is:
3 Na2SO4 + 2 Al(NO3)3 6 NaNO3 + 1 Al2(SO4)3
## Figuring out conversion factors
A conversion factor expresses the relationship between the unit in the numerator and the unit in the denominator. The quantities are actually the same, but the numbers are different because the units are different. Any time you can express a fixed relationship between two quantities, you can make two conversion factors out of that relationship. (Flipping the original conversion factor upside down gives you the second one!)
Examples
• 2.54 cm = 1 in. so the conversion factor is:
$\dfrac {2.54 \, cm}{1 \, in} \text{ or } \dfrac{1 \, in}{2.54 \, cm}$
• 1.00 cal = 4.184 J, so the conversion factor is:
$\dfrac {1.00 \, cal}{4.184 \, J} \text{ or } \dfrac{4.184 \, J}{1.00 \, cal}$
• 1.00 mole O2 has a mass of 15.999 g, so the conversion factor is:
$\dfrac {15.999 \, g \text{ O}_2}{1.00 \, mol \text{ O}_2} \text{ or } \dfrac{1.00 \, mol \text{ O}_2}{15.999 \, g \text{ O}_2}$
• In the production of ammonia, N2 + 3 H2 → 2 NH3, two moles of ammonia are produced for every 3 mol of H2 consumed, so the conversion factor is:
$\dfrac {2 \, mol \text{ NH}_3}{3 \, mol \text{ H}_2} \text{ or } \dfrac{3 \, mol \text{ H}_2}{2 \, mol \text{ NH}_3}$
## Ideal Gas Law vs. Combined Gas Law
When you are doing gas law problems, your first decision should be whether to use the Ideal gas law or the combined gas law. How do you know which to use?
The Ideal Gas Law is used when you have one set of conditions for your system. Then if you know any three of the values for pressure, volume, moles and temperature, you can solve for the unknown quantity. The Combined Gas Law is used when you have two sets of conditions. In other words, you know the initial pressure, volume and temperature, and two of those quantities change to a new set of conditions. Then you can calculate the new value of the third quantity.
Example 1: What is the volume of a container that holds 0.561 mol of a gas at 1.15 atm at 27°C?
Answer: This problem describes a single set of conditions, so use the Ideal Gas Law:
$PV=nRT$
$(1.15 \, atm)(V)=(0.561 \, mol) \! \left ( 0.0821 \, \frac{\ell \cdot atm}{mol \cdot K} \right ) \! (300 \, K)$
$V=12.0 \, L$
Example 2: A 3.00 L container holds an ideal gas at STP. What will the new pressure be if the volume is increased to 3.75 L and the temperature is increased to 500 K?
Answer: This problem describes an initial condition and a new set of conditions, so use the combined gas law:
$\dfrac{P_1V_1}{T_1}= \dfrac{P_2V_2}{T_2}$
$\dfrac{(1.00 \, atm)(3.00 \, L)}{273 \, K}= \dfrac{(P)(3.75 \, L)}{500 \, K}$
$P=1.47 \, atm$
Blue Taste Theme created by Jabox | 5,330 | 19,508 | {"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": 18, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.820045 |
https://stat.ethz.ch/pipermail/r-help/2021-September/472226.html | 1,716,748,752,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00504.warc.gz | 454,555,990 | 2,525 | # [R] Fastest way to extract rows of smaller matrix many times by index to make larger matrix? and multiply columsn of matrix by vector
Jeff Newmiller jdnewm|| @end|ng |rom dcn@d@v|@@c@@u@
Tue Sep 14 09:49:00 CEST 2021
```That is about as fast as it can be done. However you may be able to avoid doing it at all if you fold V2 into a matrix instead. Did you mean to use matrix multiplication in your calculation of M3?
On September 13, 2021 11:48:48 PM PDT, nevil amos <nevil.amos using gmail.com> wrote:
>Hi is there a faster way to "extract" rows of a matrix many times to for a
>longer matrix based in a vector or for indices than M[ V, ]
>
>I need to "expand" ( rather than subset) a matrix M of 10-100,000 rows x
>~50 columns to produce a matrix with a greater number (10^6-10^8) of rows
>using a vector V containing the 10^6 -10^8 values that are the indices of
>the rows of M. the output matrix M2 is then multiplied by another vector V2
>With the same length as V.
>
>Is there a faster way to achieve these calculations (which are by far the
>slowest portion of a function looped 1000s of times? than the standard M2
><- M[ V, ] and M3<-M2*V2, the two calculations are taking a similar time,
>Matrix M also changes for each loop.
>
>
>M<-matrix(runif(50*10000,0,100),nrow=10000,ncol=50)
>x = 10^7
>V<-sample(1:10000,x,replace=T)
>V2<-(sample(c(1,NA),x,replace=T))
>print<-(microbenchmark(
>M2<-M[V,],
>M3<-M2*V2,
>times=5,unit = "ms"))
>
>
>
>thanks for any suggestions
>
>Nevil Amos
>
> [[alternative HTML version deleted]]
>
>______________________________________________
>R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
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Chem Differential Eq HW Solutions Fall 2011 103
# Chem Differential Eq HW Solutions Fall 2011 103 - Fourier...
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Section 6.6 The Biharmonic Operator 103 Exercises 6.6 1. u xxyy = 0, u xxxx = 4!, u yyyy = - 4!, 4 u = 0. 5. Express v in Cartesian coordinates as follows: v = r 2 cos(2 θ )(1 - r 2 ) = r 2 [cos 2 θ - sin 2 θ ](1 - r 2 ) = ( x 2 - y 2 ) ( 1 - ( x 2 + y 2 ) ) . Let u = x 2 - y 2 . Then u is harmonic and so v is biharmonic by Example 1, with A = 1, D = 1, B = C = 0. 7. Write v = r 2 · r n cos and let u = r n cos . Then u is harmonic (use the Laplacian in polar coordinates to check this last assertion) and so v is biharmonic, by Example 1 with A = 1 and B = C = D = 0. 9. Write v = ar 2 ln r + br 2 + c ln r + d = φ + ψ , where φ = [ ar 2 + c ] ln r and ψ = br 2 + d . From Example 1, it follows that ψ is biharmonic. Also, ln r is harmonic (check the Laplacian in polar coordinates) and so, by Example 1, φ is biharmonic. Consequently, v is biharmonic, being the sum of two biharmonic functions. 13. We follow the method of Theorem 1, as illustrated by Example 2. First, solve the Dirichlet problem 2 w = 0, w (1 ) = cos 2 θ , for 0 r < 1, 0 θ 2 π . The solution in this case is w ( r,θ ) = r 2 cos 2 θ . (This is a simple application of the method of Section 4.4, since the boundary function is already given by its
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Unformatted text preview: Fourier series.) We now consider a second Dirichlet problem on the unit disk with boundary values v (1 , θ ) = 1 2 ( w r (1 ,θ )-g ( θ ) ) . Since g ( θ ) = 0 and w r ( r, θ ) = 2 r cos 2 θ , it follows that v (1 , θ ) = cos 2 θ . The solution of th Dirichlet problem in v is v ( r, θ ) = r 2 cos 2 θ . Thus the solution of biharmonic problem is u ( r, θ ) = (1-r 2 ) r 2 cos 2 θ + r 2 cos 2 θ = 2 r 2 cos 2 θ-r 4 cos 2 θ. This can be veri±ed directly by plugging into the biharmonic equation and the boundary conditions. 17. u (1 , 0) = 0 implies that w = 0 and so v (1 , θ ) =-g ( θ ) 2 . So v ( r, θ ) =-1 2 ± a + ∞ ² n =1 r n (cos nθ + b n sin nθ ) ³ , where a n and b n are the Fourier coefficients of g . Finally, u ( r,θ ) = (1-r 2 ) v ( r,θ ) =-1 2 (1-r 2 ) ´ a + ∞ ² n =1 r n (cos nθ + b n sin nθ ) µ ....
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https://www.isixsigma.com/topic/central-composite-designs-and-box-behnken-designs/ | 1,611,299,236,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529128.47/warc/CC-MAIN-20210122051338-20210122081338-00577.warc.gz | 824,587,172 | 44,159 | # central composite designs and Box-Behnken designs
Six Sigma – iSixSigma Forums Old Forums General central composite designs and Box-Behnken designs
Viewing 5 posts - 1 through 5 (of 5 total)
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• #49838
vikranth
Member
Dear all,
When you are doing DOE using response surface methods, how do you chose between Central Composite designs and Box-Behnken designs? Please eloberate.
Please advise if there is any good book which talks about DOE in detail.(something on the lines of DOE for dummies).
Best Regards,
Vikranth.
0
#171074
George4
Participant
A good question for real experts
0
#171086
Vallee
Participant
Clearly stated, I have not used these tools in DOE before; however it looks similar to matrix and vector rotation I did (many) years ago in factor analysis. The designs on which way to rotate your matrix takes a lot of practice and conceptual understanding. Below is a simple link with examples of your question but I can not validate its correctness. Post a question to Robert Butler and he may be able to help you.
http://www-rohan.sdsu.edu/doc/matlab/toolbox/stats/doe5.html
HF Chris Vallee
0
#171089
Robert Butler
Participant
Box-Behnken designs are much more efficient that 3**k factorial designs. They do not contain any corner points in the design space which may or may not be an advantage. They are nearly orthogonal Res V designs and they estimate all linear effects, all quadratic effects, and all linear 2 way interactions. The last is probably their biggest drawback if you dont want all 2 ways there isnt any easy way to reduce the design to take this desire into account. As with most designs the slight non-orthogonality isnt an issue.
Central composite designs are much more flexible with respect to the issue of 2 way interactions. They are comprised of a standard 2**k factorial, center points, and axial points. Because their core is a 2**k factorial you have the option of running a full factorial at the center or, if you dont desire information on some or all of the 2 way interactions you can run the core as a fractional factorial design. What this means is that, if all of the 2way interactions are not of interest, the CCD will be more efficient than the Box-Behnken.
The axial (sometimes called star points) are outside of the bounds of the design space box defined by the factorial part of the design. They are chosen to produce rotatability. All this means is that the predicted response is capable of being estimated with equal variance regardless of the direction from the center of the design space. From the standpoint of practice this means the axial points will correspond to your design space minimum and maximum and the min/max points of the factorial will be something greater or less than the overall minimum and maximum. CCDs are also slightly non-orthogonal but again, it doesnt matter that much.
If you cant run factor levels at 5 distinct values (axial+, 1, 0, -1, axial-) then you can collapse the axial values to 1 and run a central composite face design. This design wont have rotatability and will have a degree less orthogonality for the quadratic terms but, again, the issue is more of a theoretical concern than a practical one.
As for books, give the breadth of what you are asking there isnt any one single source of which Im aware. One book which I think does a very good job of describing the various design options is Understanding Industrial Designed Experiments 4th edition Schmidt and Launsby. Chapter 3 in particular is probably the best general overview of the field Ive read.
0
#171099
vikranth
Member
Thanks Mr.Butler.
It is very nice of you to answer the question.
Thanks for the rest of the people who have actively participated in the discussion..
Regards,
Vikranth.
0
Viewing 5 posts - 1 through 5 (of 5 total)
The forum ‘General’ is closed to new topics and replies. | 928 | 3,913 | {"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} | 2.578125 | 3 | CC-MAIN-2021-04 | longest | en | 0.897773 |
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Word Problems For Mixed Addition And Subtraction Word Problems 1 Division Word Problems Addition Words Math Word Problems | 983 | 5,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-30 | latest | en | 0.91843 |
http://de.metamath.org/mpegif/pm4.71d.html | 1,597,416,528,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739328.66/warc/CC-MAIN-20200814130401-20200814160401-00027.warc.gz | 31,011,234 | 3,457 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > pm4.71d Structured version Unicode version
Theorem pm4.71d 634
Description: Deduction converting an implication to a biconditional with conjunction. Deduction from Theorem *4.71 of [WhiteheadRussell] p. 120. (Contributed by Mario Carneiro, 25-Dec-2016.)
Hypothesis
Ref Expression
pm4.71rd.1
Assertion
Ref Expression
pm4.71d
Proof of Theorem pm4.71d
StepHypRef Expression
1 pm4.71rd.1 . 2
2 pm4.71 630 . 2
31, 2sylib 198 1
Colors of variables: wff setvar class Syntax hints: wi 4 wb 186 wa 369 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 This theorem depends on definitions: df-bi 187 df-an 371 This theorem is referenced by: difin2 3714 resopab2 5144 fcnvres 5747 resoprab2 6382 psgnran 16866 efgcpbllemb 17099 cndis 20087 cnindis 20088 cnpdis 20089 blpnf 21194 dscopn 21388 itgcn 22543 limcnlp 22576 nb3gra2nb 24884 usg2wlkeq 25137 clwwlkn2 25204 rngosn3 25855 1stpreima 27982 fsumcvg4 28398 mbfmcnt 28729 topdifinffinlem 31277 ptrest 31433 isidlc 31707 dih1 34319 lzunuz 35075
Copyright terms: Public domain W3C validator | 527 | 1,221 | {"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-2020-34 | latest | en | 0.30401 |
https://quant.stackexchange.com/questions/74783/questions-about-the-replicating-portfolio-in-the-binomial-model | 1,708,862,557,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474595.59/warc/CC-MAIN-20240225103506-20240225133506-00467.warc.gz | 467,467,062 | 41,033 | # Questions about the replicating portfolio in the binomial model
I'm starting to teach myself quantitative finance and I've got several questions (marked in bold) regarding the replicating portfolio of a security in the binomial model. I'm following, among others, the classical book "Stochastic Calculus for Finance I: The Binomial Asset Pricing Model".
First, I'll start with some notation so that there is no confusion. As always, let $$d$$, $$u$$, $$r$$ be the down-factor, up-factor, and risk-free interest rate, respectively, verifying that $$d < 1 + r < u$$. Then, if $$V_{n}$$, $$0 \leq n \leq N$$, is the value at time $$n$$ of a security that has a unique payoff $$V_N$$ at maturity, we know that the discounted process $$\dfrac{V_n}{(1+r)^n}$$ is a martingale under the risk-neutral probability measure $$\widetilde{\mathbb{P}}$$, where the probability of heads is $$p := \dfrac{1+r-d}{u-d}$$, and thus we can easily compute the value of each $$V_n$$ via
$$V_{n} = \widetilde{E}_{n}\Big(\frac{V_{N}}{(1+r)^{N-n}}\Big).$$
If $$d = d_{n}$$, $$u = u_{n}$$, and $$r = r_{n}$$ are not constant numbers but an adapted stochastic process, the result is the same but now the discounted process is given by $$\dfrac{V_n}{(1+r_{0})·\dots·(1+r_{n-1})}$$, and the risk-neutral pricing formula still holds true provided that
$$\widetilde{\mathbb{P}}(w_{n+1} = H|w_{1},\dots,w_{n}) := p_{n} := \dfrac{1+r_n-d_n}{u_n-d_n},$$
$$\widetilde{\mathbb{P}}(w_{n+1} = T|w_{1},\dots,w_{n}) = 1- p_{n}.$$
To show the previous result, one usually constructs the following portfolio: suppose that $$V$$ is, for instance, an European call. Start with $$X_{0}$$ wealth, buy $$\Delta_{0}$$ shares of the underlying, and invest (or borrow) the remaining money at the risk-free rate $$r$$. At time $$1$$, sell the portfolio and reinvest the money doing the same strategy. At time $$n+1$$, the value of the replicating portfolio is given by
$$X_{n+1} = \Delta_{n}S_{n+1} + (1+r)(X_{n}-\Delta_{n}S_{n}).$$
Now that the notation is clear, my first question is the following:
1) Do we need to replicate the derivative security using the underlying security $$S$$? I'm aware of the hedging benefits of combining a derivative and its underlying, but since here the goal is to construct a replicating portfolio, could it be constructed trading with another security? What are the advantages of using the underlying over the rest of securities? The only benefit I see is that you only need to model the prices of one stock.
Now suppose that we want to price zero-coupon bonds using the binomial model. Assume that the interest rates form an stochastic adapted process, in such a way that 1 dollar invested at time $$n$$ yields $$(1+r_{n})$$ at time $$n+1$$. Let $$B_{n,m}$$ be the value at time $$n$$ of a zero-coupon bond that pays $$1$$ dollar at time $$m$$. Since the risk-neutral pricing formula also applies here, we can easily conclude that $$B_{n,m} = \widetilde{E}_{n}\Big(\frac{1}{(1+r_{n})·\dots·(1+r_{m-1})}\Big).$$ However, here goes another question:
2) How would one construct a replicating portfolio in this case? In what securities does it make sense to trade?
Finally, I see that in the book that I mention at the beginning, a portfolio process is constructed by trading in the zero-coupon bonds and the money market via the following equation:
where $$\Delta_{n,m}$$ is the number of zero-coupon bonds of maturity $$m$$ held by the investor between times $$n$$ and $$n+1$$. I understand that this portfolio process, properly discounted, is a martingale, and hence there can't be arbitrage when trading in the zero-coupon bonds and the money market. My final question is:
3) How is this formula related to the proof that $$B_{n,m} = \widetilde{E}_{n}\big(\frac{1}{(1+r_{n})·\dots·(1+r_{m-1})}\big)$$?
If some of my questions are not clear enough, please let me know. Thanks a lot!
Here are some answers, hope it helps:
Regarding question 1, Do we need to replicate the derivative security using the underlying security $$S$$? I think the answer is related to your other question: What are the advantages of using the underlying over the rest of securities? The reason is that it is the underlying for which the derivative has a known sensitivity. Imagine you'd like to hedge a derivative on $$S$$ using an underlying $$H$$. How would you do it? How is an increase in value of the derivative linked to the performance or evolution of $$H$$? In the case of $$S$$ you know (or can estimate) that, is just the partial derivative (assuming no jumps).
2) How would one construct a replicating portfolio in this case? In what securities does it make sense to trade? Since these are adapted processes (the rate from $$t$$ to $$t + \Delta t$$ is known t and it is suppossed to be a rate at which any market player can capitalize, then you can just replicate it by buying zero coupon bonds or via a bank account.
Question 3 is related to this, and just comes from the compounded capitalization/borrowing formula. If the rate from $$t_0$$ to $$t_1$$ is $$r_0$$, the price of a zero coupon bond paying 1 unit at time $$t_1$$ is
$$\dfrac{1}{1 + r_0 (t_1 - t_0)}$$,
now for a ZCB paying at time $$t_2$$, if the rate from $$t_1$$ to $$t_2$$ is $$r_1$$, following the same argument
$$\dfrac{1}{(1 + r_0 (t_1 - t_0))(1 + r_1 (t_2 - t_1))},$$
and it follows directly. Note that in the book they're missing the $$\Delta t$$ term (maybe it is just taken into account in the $$r$$ factor).
• Thank you so much for your time! But I still can't see why I can't use another stock $H$ different from the underlying $S$, at least in the binomial model. If you start with $X_{0}$ wealth and follow the self-financing strategy defined above, replacing $S$ with $H$, I can impose X_{N} = V_{N} at maturity and determine all the $\Delta$'s and $X's$ so that the long or short position in V is hedged, I can't see any mathematical obstruction, at least in the binomial model. Mar 1, 2023 at 19:26
• As for questions 2 and 3, If interest rates are deterministic, your argument is obviously true, but if not, the answer should be the expected value under the risk-neutral measure of the quantities you mention. In order to construct the replicating strategy, I can't see why the bonds & money market strategy that I describe is useful, since you need to know the $B_{n,m}$'s, and that is precisely what you want to find. Mar 1, 2023 at 19:34
• Trading in the stock and money market isn't useful, since the delta's would be 0 and you'd be left with $X_{n} = (1+r_{n-1})\dots(1+r_0)X_{0}$, but that can't replicate a constant payoff unless the interest rates are constant. I guess the answer is found by trading in the stocks and derivatives markets. Mar 1, 2023 at 19:38 | 1,872 | 6,750 | {"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": 59, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-10 | longest | en | 0.85984 |
https://www.jiskha.com/display.cgi?id=1369703629 | 1,516,687,888,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891750.87/warc/CC-MAIN-20180123052242-20180123072242-00095.warc.gz | 909,916,067 | 3,614 | # math
posted by .
Amir's class has 24 students. There are 15 boys in the class. Whitch of the following represents the ratio of girls to boys?
• math -
9/15 = 3/5
• math -
3/5
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There are 5 girls more than boys in a class. if 2 boys joined the class the ratio of girls to boys is 5:4. find the total number of student in the class. | 488 | 1,741 | {"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} | 3.546875 | 4 | CC-MAIN-2018-05 | latest | en | 0.947253 |
https://www.experts-exchange.com/questions/28862799/Excel-exclude-list-of-values-from-all-areas-of-workbook.html | 1,544,776,785,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825495.60/warc/CC-MAIN-20181214070839-20181214092339-00634.warc.gz | 902,279,889 | 15,184 | # Excel exclude list of values from all areas of workbook
Greetings,
I have a table of 300+ email addresses. I want to do a FIND in the cell B2:B350 in cells C2:C50
I want to check if the value of each cell is unique in the workbook and return some kind of indicator.
Thank you!!!
JohnD
LVL 1
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Commented:
JohnD,
I'm assuming your questions have two parts..
1. If you want to see whether values of C2:C50 exists in cells B2:B350 then use this formula in D2 and drag it down...
``````=Countif(B\$2:B\$350,C2)
``````
Now drag this down till D50 and for any values which exist it will give you 1 or how many times it exists if it doesn't exist it will give you a zero as answer...
2. Assuming you don't have any blank values from B2 to B350..You can use this formula at any cell it will tell you whether it has duplicates or all values are unique..
``````=IF(COUNTA(B2:B350)<>SUMPRODUCT(1/COUNTIF(B2:B350,B2:B350)),"Has Duplicates","Unique")
``````
Saurabh...
PMAuthor Commented:
I posed my question poorly; my apologies
I want to make sure the values in Column B are unique throughout the workbook. Is there a formula which I can place in Column C which checks for each value in Column B across the whole workbook?
Commented:
Which are the columns or range you want to check for duplicate???
Assuming if you want to check columns D TO E then you can use..
=countif(D:E,B2)
This formula need to be applied in C2..Now if the count is more that means this value exists more then once IN Column D and E.. You can play around with this formula as in range and drag this down to check how many times values exists..
Saurabh...
PMAuthor Commented:
Thank you; I have 8 tables on 8 separate worksheets which need to be checked for duplicates
Commented:
You can do something like this...
``````=COUNTIF(Table1,b2)+COUNTIF(Table2,b2)+COUNTIF(Table3,b2)+COUNTIF(Table4,b2)+COUNTIF(Table5,b2)+COUNTIF(Table6,b2)+COUNTIF(Table7,b2)+COUNTIF(Table8,b2)
``````
Now anytime when the value is more then 1 you know its a duplicate..and i have assumed your table name starts with table1 and goes till table8..
Saurabh...
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https://wiki.analog.com/university/tools/pluto/users/receiver_sensitivity?rev=1529694597 | 1,631,910,417,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00254.warc.gz | 664,477,393 | 12,269 | # Analog Devices Wiki
This version is outdated by a newer approved version.This version (22 Jun 2018 21:10) was approved by Larry Hawkins.The Previously approved version (10 Mar 2017 22:46) is available.
This is an old revision of the document!
Receiver sensitivity is a measure of the ability of a receiver to demodulate and get information from a signal. Most people quantify sensitivity as the lowest signal power level from which we can get useful information.
Since the above definition uses “demodulate”, you should immediately understand, this is a meaningless specification for a pure radio. There is no demodulation in the base design, and we pass samples around. The actual receiver sensitivity will depend on channel bandwidth, temperature, modulation scheme, how robust the demodulator is, something that we just don't control (that's all up to you).
What is specified, and measured, is noise:
• Tx noise: ≤−157 dBm/Hz noise floor
• Rx noise figure : 2.5 dB @ Maximum Rx gain
• Integrated Phase Noise : 0.3 ° rms, 100 Hz to 100 MHz
These noise numbers can be used to calculate the min received power to decode something.
## Rx Noise Figure
The IEEE Standard definition of noise figure, states that:
The noise factor, at a specified input frequency, is defined as the ratio of:
• (1) the total noise power per unit bandwidth available at the output port when noise temperature of the input termination is standard (290 K) to
• (2) that portion of (1) engendered at the input frequency by the input termination.
• NF = Noise Figure
• k = Boltzmann’s Constant (1.38 x 10-23 joules/K),
• = temperature in Kelvin (in this case 290 K)
• B = bandwidth of the device
• G = gain of the device
This is measured using test equiment, as specified in Agilent Fundamentals of RF and Microwave Noise Figure Measurements.
university/tools/pluto/users/receiver_sensitivity.1529694597.txt.gz · Last modified: 22 Jun 2018 21:09 by Larry Hawkins
| 456 | 1,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-39 | latest | en | 0.883168 |
http://docu.openrepgrid.org/elements_distances_standardized.html | 1,493,179,040,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121153.91/warc/CC-MAIN-20170423031201-00138-ip-10-145-167-34.ec2.internal.warc.gz | 101,699,994 | 11,297 | ### Introduction
As a similarity measure in grids different types of Minkowski metrics, especially the Euclidean and city-block metric are frequently used. The Euclidean distance is the sum of squared differences between the ratings on two different elements. They are, however, no standardized measure. The distances strongly depend on the number of constructs and the rating range. The figure below demonstrates this fact. Note how the distance changes although the rating pattern remains identical.
In order to be able to compare distances across grids of different size and rating range a standardization is desireable. Also, the notion of significance of a distance, i.e. a distance which is unusually big, is easier with a standard reference measure. Different suggestions have been made in the literature of how to standardize Euclidean interelement distances (Hartmann, 1992; Heckmann, 2012; Slater, 1977). The three variants will be briefly discussed and the corresponing R-Code is demonstrated.
### Slater distances (1977)
#### Description
The first suggestion to standardization was made by Slater (1977). He essentially calculated an expected average Euclidean distance $$U$$ for the case if the ratings are randomly distributed. To standardize the grids he suggested to divide the matrix of Euclidean distances $$E$$ by this unit of expected distance $$U$$. The Slater standardization thus is the division of the Euclidean distances by the distance expected on average. Hence, distances bigger than 1 are greater than expected, distances smaller than 1 are smaller than expected.
#### R-Code
The function distanceSlater calculates Slater distances for a grid.
distanceSlater(boeker)
##########################
Distances between elements
##########################
Distance method: Slater (standardized Euclidean)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
(1) self 1 1.03 0.75 0.69 0.87 1.19 0.80 1.03 0.99 0.59 1.79 0.58 0.55 0.64 0.54
(2) ideal self 2 1.11 0.78 1.06 1.31 1.07 0.97 1.14 0.97 1.56 1.22 1.21 1.24 1.25
(3) mother 3 0.73 0.53 0.95 0.55 0.81 0.64 0.67 1.58 0.69 0.83 0.77 0.69
(4) father 4 0.63 1.15 0.65 0.90 0.83 0.69 1.66 0.84 0.91 0.98 0.89
(5) kurt 5 0.89 0.57 0.79 0.57 0.72 1.51 0.79 0.93 0.87 0.83
(6) karl 6 0.94 0.74 0.66 1.09 0.97 1.17 1.22 1.08 1.15
(7) george 7 0.92 0.65 0.81 1.51 0.73 0.91 0.92 0.78
(8) martin 8 0.68 0.80 1.27 1.09 1.10 1.01 1.07
(9) elizabeth 9 0.87 1.31 1.00 1.13 1.03 0.98
(10) therapist 10 1.74 0.65 0.63 0.69 0.65
(11) irene 11 1.83 1.86 1.72 1.84
(12) childhood self 12 0.43 0.50 0.34
(13) self before illness 13 0.43 0.41
(14) self with delusion 14 0.45
(15) self as dreamer 15
Note that Slater distances cannot be compared across grids with a different number of constructs (see Hartmann, 1992).
You can save the results and define the way they are displayed using the print method. For example we could display distances only within certain boundaries, using the cutoff values .8 and 1.2 to indicate very big or small distances as suggested by Norris and Makhlouf-Norris (1976).
d <- distanceSlater(boeker)
print(d, cutoffs=c(.8, 1.2))
##########################
Distances between elements
##########################
Distance method: Slater (standardized Euclidean)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
(1) self 1 0.75 0.69 0.80 0.59 1.79 0.58 0.55 0.64 0.54
(2) ideal self 2 0.78 1.31 1.56 1.22 1.21 1.24 1.25
(3) mother 3 0.73 0.53 0.55 0.64 0.67 1.58 0.69 0.77 0.69
(4) father 4 0.63 0.65 0.69 1.66
(5) kurt 5 0.57 0.79 0.57 0.72 1.51 0.79
(6) karl 6 0.74 0.66 1.22
(7) george 7 0.65 1.51 0.73 0.78
(8) martin 8 0.68 0.80 1.27
(9) elizabeth 9 1.31
(10) therapist 10 1.74 0.65 0.63 0.69 0.65
(11) irene 11 1.83 1.86 1.72 1.84
(12) childhood self 12 0.43 0.50 0.34
(13) self before illness 13 0.43 0.41
(14) self with delusion 14 0.45
(15) self as dreamer 15
Note that Slater distances cannot be compared across grids with a different number of constructs (see Hartmann, 1992).
#### Calculation
Let $$G$$ be the raw grid matrix and $$D$$ be the grid matrix centered around the construct means, with $$d_{ij} =g_{..} - g_{ij}$$, where $$g_{..}$$ is the mean of the construct. Further, let
$P=D^TD \qquad \text{and} \qquad S=tr\;P$
The Euclidean distances results in:
$(\sum{ (d_{ij} - d_{ik} )^2})^{1/2}$
$\Leftrightarrow (\sum{ (d_{ij}^2 + d_{ik}^2 - 2d_{ij}d_{ik})})^{1/2}$
$\Leftrightarrow (\sum{ d_{ij}^2 } + \sum{d_{ik}^2} - 2\sum{d_{ij}d_{ik} })^{1/2}$
$\Leftrightarrow (S_j + S_k - 2P_{jk})^{1/2}$
For the standardization, Slater proposes to use the expected Euclidean distance between a random pair of elements taken from the grid. The average for $$S_j$$ and $$S_k$$ would then be $$S_{avg} = S/m$$ where $$m$$ is the number of elements in the grid. The average of the off-line diagonals of $$P$$ is $$S/m(m-1)$$ (see Slater, 1951, for a proof). Inserted into the formula above it gives the following expected average euclidean distance $$U$$ which is outputted as unit of expected distance in Slater’s INGRID program.
$U = (2S/(m-1))^{1/2}$
The calculated euclidean distances are then divided by $$U$$, the unit of expected distance to form the matrix of standardized element distances $$E_{std}$$, with
$E_{std} = E/U$
### Hartmann distances (1992)
#### Description
Hartmann (1992) showed in a Monte Carlo study that Slater distances (see above) based on random grids, for which Slater coined the expression quasis, have a skewed distribution, a mean and a standard deviation depending on the number of constructs elicited. Hence, the distances cannot be compared across grids with a different number of constructs. As a remedy he suggested a linear transformation (z-transformation) of the Slater distance values which take into account their estimated (or alternatively expected) mean and their standard deviation to standardize them. Hartmann distances represent a more accurate version of Slater distances. Note that Hartmann distances are multiplied by -1 to allow an interpretation similar to correlation coefficients: negative Hartmann values represent an above average dissimilarity (i.e. a big Slater distance) and positive values represent an above average similarity (i.e. a small Slater distance).
The Hartmann distance is calculated as follows (Hartmann, 1992, p. 49).
$D = -1 \frac{D_{slater} - M_c}{sd_c}$
Where $$D_{slater}$$ denotes the Slater distances of the grid, $$M_c$$ the sample distribution’s mean value and $$sd_c$$ the sample distributions’s standard deviation.
#### R-Code
The function distanceHartmann calculates Hartmann distances. The function can be operated in two ways. The default option (method="paper") uses precalculated mean and standard deviations (as e.g. given in Hartmann (1992)) for the standardization.
distanceHartmann(boeker)
##########################
Distances between elements
##########################
Distance method: Hartmann (standardized Slater distances)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
(1) self 1 -0.28 1.58 1.92 0.80 -1.33 1.20 -0.29 -0.04 2.62 -5.24 2.66 2.87 2.28 2.89
(2) ideal self 2 -0.78 1.36 -0.47 -2.09 -0.56 0.12 -1.02 0.12 -3.69 -1.50 -1.45 -1.63 -1.71
(3) mother 3 1.70 2.99 0.22 2.82 1.15 2.27 2.09 -3.84 1.91 1.06 1.44 1.92
(4) father 4 2.31 -1.04 2.23 0.55 1.00 1.92 -4.39 0.96 0.50 0.08 0.63
(5) kurt 5 0.63 2.72 1.27 2.69 1.74 -3.37 1.30 0.35 0.79 1.01
(6) karl 6 0.29 1.63 2.14 -0.66 0.10 -1.21 -1.53 -0.60 -1.04
(7) george 7 0.45 2.19 1.17 -3.39 1.70 0.54 0.42 1.35
(8) martin 8 2.03 1.22 -1.85 -0.67 -0.73 -0.13 -0.53
(9) elizabeth 9 0.76 -2.07 -0.08 -0.91 -0.29 0.05
(10) therapist 10 -4.91 2.20 2.35 1.97 2.22
(11) irene 11 -5.47 -5.65 -4.79 -5.52
(12) childhood self 12 3.66 3.16 4.22
(13) self before illness 13 3.60 3.79
(14) self with delusion 14 3.52
(15) self as dreamer 15
For calculation the parameters from Hartmann (1992) were used. Use 'method=new' or method='simulate' for a more accurate version.
The second option (method="simulate") is to simulate the distribution of distances based on the size and scale range of the grid under investigation. A distribution of Slater distances is derived using quasis and used for the Hartmann standardization instead of the precalculated values. The following simulation is based on reps=1000 quasis.
h <- distanceHartmann(boeker, method="simulate", reps=1000)
h
##########################
Distances between elements
##########################
Distance method: Hartmann (standardized Slater distances)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
(1) self 1 -0.28 1.56 1.90 0.79 -1.32 1.19 -0.29 -0.04 2.59 -5.19 2.63 2.84 2.25 2.86
(2) ideal self 2 -0.77 1.35 -0.47 -2.07 -0.56 0.12 -1.01 0.12 -3.65 -1.48 -1.43 -1.62 -1.70
(3) mother 3 1.68 2.96 0.21 2.79 1.14 2.25 2.07 -3.80 1.89 1.04 1.42 1.90
(4) father 4 2.28 -1.04 2.20 0.54 0.99 1.90 -4.35 0.95 0.49 0.08 0.62
(5) kurt 5 0.62 2.69 1.25 2.66 1.72 -3.34 1.28 0.34 0.78 0.99
(6) karl 6 0.28 1.61 2.11 -0.65 0.09 -1.20 -1.51 -0.59 -1.03
(7) george 7 0.44 2.17 1.15 -3.36 1.68 0.53 0.41 1.33
(8) martin 8 2.01 1.20 -1.83 -0.67 -0.73 -0.13 -0.52
(9) elizabeth 9 0.75 -2.05 -0.08 -0.90 -0.28 0.05
(10) therapist 10 -4.86 2.18 2.32 1.95 2.19
(11) irene 11 -5.41 -5.59 -4.74 -5.46
(12) childhood self 12 3.62 3.12 4.18
(13) self before illness 13 3.56 3.75
(14) self with delusion 14 3.48
(15) self as dreamer 15
If the results are saved, there are a couple of options for printing the object (see ?print.hdistance).
print(d, p=c(.05, .95))
##########################
Distances between elements
##########################
Distance method: Slater (standardized Euclidean)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
(1) self 1 1.03 0.75 0.69 0.87 1.19 0.80 1.03 0.99 0.59 1.79 0.58 0.55 0.64 0.54
(2) ideal self 2 1.11 0.78 1.06 1.31 1.07 0.97 1.14 0.97 1.56 1.22 1.21 1.24 1.25
(3) mother 3 0.73 0.53 0.95 0.55 0.81 0.64 0.67 1.58 0.69 0.83 0.77 0.69
(4) father 4 0.63 1.15 0.65 0.90 0.83 0.69 1.66 0.84 0.91 0.98 0.89
(5) kurt 5 0.89 0.57 0.79 0.57 0.72 1.51 0.79 0.93 0.87 0.83
(6) karl 6 0.94 0.74 0.66 1.09 0.97 1.17 1.22 1.08 1.15
(7) george 7 0.92 0.65 0.81 1.51 0.73 0.91 0.92 0.78
(8) martin 8 0.68 0.80 1.27 1.09 1.10 1.01 1.07
(9) elizabeth 9 0.87 1.31 1.00 1.13 1.03 0.98
(10) therapist 10 1.74 0.65 0.63 0.69 0.65
(11) irene 11 1.83 1.86 1.72 1.84
(12) childhood self 12 0.43 0.50 0.34
(13) self before illness 13 0.43 0.41
(14) self with delusion 14 0.45
(15) self as dreamer 15
Note that Slater distances cannot be compared across grids with a different number of constructs (see Hartmann, 1992).
### Heckmann’s approach (2012)
#### Description
Hartmann (1992) suggested a transformation of Slater (1977) distances to make them independent from the size of a grid. Hartmann distances are supposed to yield stable cutoff values used to determine ‘significance’ of inter-element distances. It can be shown that Hartmann distances are still affected by grid parameters like size and the range of the rating scale used (Heckmann, 2012). The function distanceNormalize applies a Box-Cox (1964) transformation to the Hartmann distances in order to remove the skew of the Hartmann distance distribution. The normalized values show to have more stable and nearly symmetric cutoffs (quantiles) and better properties for comparison across grids of different size and scale range.
#### R-Code
The function distanceNormalize will return Slater, Hartmann or power transformed Hartmann distances (Heckmann, 2012) if prompted. It is also possible to return the quantiles of the sample distribution and only the element distances consideres ‘significant’ according to the quantiles defined.
n <- distanceNormalized(boeker)
n
##########################
Distances between elements
##########################
Distance method: Power transformed Hartmann distances
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
(1) self 1 -0.28 1.56 1.90 0.79 -1.32 1.19 -0.29 -0.04 2.59 -5.18 2.63 2.84 2.25 2.86
(2) ideal self 2 -0.77 1.35 -0.47 -2.07 -0.56 0.12 -1.01 0.12 -3.65 -1.48 -1.43 -1.62 -1.70
(3) mother 3 1.68 2.96 0.21 2.79 1.14 2.25 2.07 -3.80 1.89 1.04 1.42 1.90
(4) father 4 2.28 -1.04 2.20 0.54 0.99 1.90 -4.34 0.94 0.49 0.08 0.62
(5) kurt 5 0.62 2.69 1.25 2.66 1.72 -3.34 1.28 0.34 0.78 0.99
(6) karl 6 0.28 1.61 2.11 -0.65 0.09 -1.19 -1.51 -0.59 -1.03
(7) george 7 0.44 2.17 1.15 -3.36 1.68 0.53 0.41 1.33
(8) martin 8 2.01 1.20 -1.83 -0.67 -0.73 -0.13 -0.52
(9) elizabeth 9 0.75 -2.05 -0.08 -0.90 -0.28 0.05
(10) therapist 10 -4.86 2.18 2.32 1.95 2.19
(11) irene 11 -5.41 -5.59 -4.74 -5.46
(12) childhood self 12 3.62 3.12 4.17
(13) self before illness 13 3.56 3.75
(14) self with delusion 14 3.48
(15) self as dreamer 15
#### Calculation
The form of normalization applied by Hartmann (1992) does not account for skewness or kurtosis. Here, a form of normalization - a power transformation - is explored that takes into account these higher moments of the distribution. For this purpose Hartmann values are transformed using the ‘’Box-Cox’’ family of transformations (Box & Cox, 1964). The transformation is defined as
$Y_i^{\lambda}= \left\{ \begin{matrix} \frac{(Y_i + c)^\lambda - 1}{\lambda} & \mbox{for }\lambda \neq 0 \\ ln(Y_i + c) & \mbox{for }\lambda = 0 \end{matrix} \right.$
As the transformation requires values $$\ge 0$$ a constant $$c$$ is added to derive positive values only. For the present transformation $$c$$ is defined as the minimum Hartmann distances from the quasis distribution. In order to derive at a transformation that resembles the normal distribution as close as possible, an optimal $$\lambda$$ is searched by selecting a $$\lambda$$ that maximizes the correlation between the quantiles of the transformed values $$Y_i^\lambda$$ and the standard normal distribution. As a last step, the power transformed values $$Y_i^\lambda$$ are z-transformed to remove the arbitrary scaling resulting from the Box-Cox transformation yielding $$Y_i^P$$.
$$Y_{i}^P = \frac{Y^{\lambda}_i - \overline Y^{\lambda}}{\sigma_{Y^{\lambda}}}$$
### Literature
Box, G. E. P., & Cox, D. R. (1964). An Analysis of Transformations. Journal of the Royal Statistical Society. Series B (Methodological), 26(2), 211–252. Retrieved from http://www.jstor.org/stable/2984418
Hartmann, A. (1992). Element comparisons in repertory grid technique: Results and consequences of a monte carlo study. International Journal of Personal Construct Psychology, 5(1), 41–56. doi:10.1080/08936039208404940
Heckmann, M. (2012, July). Standardizing inter-element distances in grids – A revision of Hartmann’s distances. Talk held at the 11th Biennial Conference of the European Personal Construct Association (EPCA), Dublin, Irland.
Norris, H., & Makhlouf-Norris, F. (1976). The measurement of self-identity. In P. Slater (Ed.), The measurement of intrapersonal space by grid technique: Explorations of intrapersonal space (Vol. 1, pp. 79–92). London: Wiley & Sons.
Slater, P. (1951). The transformation of a matrix of negative correlations. British Journal of Statistical Psychology, 6, 101–106.
Slater, P. (1977). The measurement of intrapersonal space by grid technique: Dimensions of intrapersonal space (Vol. 2). London: Wiley & Sons. | 6,098 | 20,376 | {"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": 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} | 3.34375 | 3 | CC-MAIN-2017-17 | latest | en | 0.870106 |
https://ucscphysicsdemo.sites.ucsc.edu/physics-5c6c-demos/lenzs-law-jumping-ring/ | 1,582,903,766,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147234.52/warc/CC-MAIN-20200228135132-20200228165132-00118.warc.gz | 583,761,450 | 10,024 | # Lenz’s Law – Jumping Ring
Figure 1
Figure 2
Figure 3
A coil with an iron core and two aluminum rings are used to demonstrate electromagnetic induction and Lenz’s Law. One of the rings has a slit cut in it, and the other does not (see Figure 3). The rings are placed around the core. When the apparatus is turned on, the solid ring is ejected into the air. The ring with the slit remains. The effect is enhanced if the ring is cooled with liquid nitrogen.
If the button is held down and the ring is dropped from the top of the iron bar, it will remain suspended as shown in Figure 2.
The apparatus also includes a light bulb attached to a small copper coil which shines when the coil is placed around the core and the button is pressed.
Materials:
• Iron core solenoid with push button(see figure 1)
• Solid aluminum ring
• Aluminum ring with slit cut in it(see figure 3)
• Small copper coil with light bulb attached
• Liquid nitrogen for cooling(optional)
Demo:
Place the two aluminum rings around the iron core. Plug the apparatus in and press the button to send current through the coil. The solid ring will eject into the air, while the slitted ring will remain unaffected.
Now remove the rings. Press and hold the button and drop the solid ring from the top of the iron bar. Notice that the ring becomes suspended at an equilibrium position until we release the button, shutting off current flow.
Remove the solid ring again. Place the copper coil with the attached light bulb around the iron bar. Press the button and observe the light bulb shine while we hold the button down.
DO NOT hold the button down for longer than five seconds, as this can heat up the metal or wire significantly and cause burns.
Explanation:
Figure 4 (source)
Faraday’s Law states that any change in a magnetic field about a coil of wire generates an voltage, or emf, about the wire. When we turn our solenoid on, it produces a magnetic field much like a current through a straight wire produces a magnetic field described by the right-hand rule. Using the right hand rule, you can use your thumb to represent the direction of the current and curl your fingers around your imaginary wire. The direction your fingers curl describes the shape of the magnetic field lines in a circle about the wire(see figure 4). A current through a solenoid also produces a magnetic field which can be visualized using figure 5. The strength of this magnetic field depends on the length of the coil, l, and the number of times the wire is looped, n.
Figure 5 (source)
By turning our solenoid on, the magnetic field around the coil is formed and the change in magnetic environment creates an emf in our aluminum ring or copper wire. This emf induces a current in our aluminum ring that flows in the opposite direction as the current through the solenoid.
We can describe the induced emf mathematically, where:
• B = External magnetic field due to solenoid = μnlsolenoid
• A = Area of coil
• Iring = Current through our ring or copper coil
• Φ = BA = Magnetic flux
• n = Number of turns in coil
• ΔΦ/Δt = Change in magnetic flux over time
Using these definitions, the induced emf in our aluminum ring or copper coil is:
$emf = \cfrac{n\Delta\phi}{\Delta t}$
Lenz’s Law states that an induced current will always flow in the opposite direction of that which produced it. In other words, the emf induced by the solenoid creates a current in the ring or wire that flows in the opposite direction of the solenoid’s current.
These opposite currents repel one another, as they each create a magnetic force pointing in the opposite direction of one another, causing the aluminum ring to be ejected away from the solenoid. The slitted ring does not eject because an induced current cannot complete a loop around the ring, and remains unaffected by this force. In addition, the copper coil with the light bulb attached is too heavy to be ejected, but we can still observe the induced emf and current via the lit bulb.
Figure 6
When we drop the aluminum ring from above the iron core, the ring stays suspended at some equilibrium position. This equilibrium position is where the magnetic force due to the opposing currents and force due to gravity are exactly equal and opposite, resulting in a ring suspended by these two forces.
$F_{magnetic} = -F_{gravity}$
Notes:
• DO NOT touch the ring or copper wire while the apparatus is on.
• The use of liquid nitrogen is not needed to produce significant effects, however can be used to make predictions and compare the effects at room temperature versus when cooled.
Written by Lydia Seymour | 1,007 | 4,620 | {"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": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | longest | en | 0.910683 |
http://convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=acre-foot&To=section+modulus | 1,653,732,206,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00471.warc.gz | 13,025,625 | 3,855 | Partner with ConvertIt.com
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Conversion Result: ```acre-foot = 1233.48183754752 volume (volume) ``` Related Measurements: Try converting from "acre-foot" to balthazar, board foot, coomb, cord foot (of wood), dram fluid (fluid dram), dry pint, fifth, gill, last, magnum, minim, oz fluid (fluid ounce), pipe, pony, rehoboam, tablespoon, UK bushel (British bushel), UK oz fluid (British fluid ounce), wine arroba (Spanish wine arroba), wine bottle, or any combination of units which equate to "length cubed" and represent capacity, section modulus, static moment of area, or volume. Sample Conversions: acre-foot = 35,003.25 bushel (dry bushel), 5,873.03 chetvert (Russian chetvert), 8,750.81 coomb, 333,671,862.86 dram fluid (fluid dram), 1,089 freight ton, 10,427,245.71 gill, 407,314.29 jeroboam, 27,805,988.57 jigger, 437.54 last, 162,925.71 methuselah, 65,170.29 nebuchadnezzar, 10,427,245.71 noggin, 41,708,982.86 oz fluid (fluid ounce), 140,012.98 peck (dry peck), 2,606,811.43 pint (fluid pint), 217,234.29 rehoboam, 11,667.75 sack, 17,501.62 strike, 6,951,497.14 tea cup, 875.08 wey.
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Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 509 | 1,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-21 | latest | en | 0.691956 |
http://www.mapleprimes.com/products/MaplePlayer | 1,462,505,715,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861718132.40/warc/CC-MAIN-20160428164158-00086-ip-10-239-7-51.ec2.internal.warc.gz | 664,307,716 | 26,099 | # MaplePlayer Questions and PostsMaplePlayer Questions and Posts Feed
These are Posts and Questions associated with the product, MaplePlayer
### Weyl scalars from null tetrad...
April 07 2016
0 7
hello dear,
I was intersted in finding Weyl scalars for a given metric, would be really helpful if anyone can give me an example of how to find a Weyl Scalar for a given metric or set of null tetrads using Debever formalism in Tensor package(more specifically, how to define h in debever formalism for a given set of null tetrad or for a given metric).
Regards,
Suresh
### How to use .mws in Maple Player?...
February 24 2016
2 6
Just starting some work on numerical engineering solutions by going through a workbook, which includes a Maple Document on CD (with one index.mw and several .mws files).
As I don't have acces to a Maple system, I downloaded Maple Player. However - Maple Player will only accept .mw and .mwz files(at least that shows up on 'FILE>OPEN'); so trying to open an .mws file results in - NOTHING. In your forum I found some information that .mwz is a nonpublic Maple format !?...... Now I don't see a way to work with PLAYER. Isn't PLAYER meant for exactly that scenario - sharing MAPLE worksheets ?
So - what to do ?
Thanks for some hint,
GW.G
### Set plot zoom factor using command...
February 21 2016
4 7
Hi,
Can anyone please suggest a way to set the zoom factor for a 3D plot from the plot commands or using DocumentTools or in some other way that doesn't involve the interactive tools?
What I'm trying to achieve is the following:
I'm developing MapleCloud worksheets for a course I'm teaching and want to give my students several 3D plots to interact with through their web browser. When I make the plots, Maple determines the scaling so that boxed axes with labels will fit in the area of the plot component. However, I'm making plots with axes=none that look better at a higher zoom factor (the corners of the unseen boxed axes would be out of the plot area but all the plot components still fit) and I'm keen to create the plots that look like this if possible.
My ideal solution would be for there to be a zoomfactor option for the plots[display] command or to be able to use some code like SetProperty("Plot0",zoomfactor,1.25) but I haven't been able to find a way to achieve this. The closest I found was setting the viewpoint option, but that locked the view, preventing rotation, and I want to just set the initial view. Is there some other method I haven't found?
Alex
### Use of Maple Player...
December 24 2015
1 1
I downloaded Maple Player, and I don't know how can I use it resolve Maths problems. Kindly, could you explain to me the procedure I shall follow for the reason to use it?
### Displacement in elasticity...
December 12 2015
0 3
Hi
I've this equation ,what is displacement,
U := sum(C[i]*lambda[i]*h^2*M*(C[13]+C[55])*exp(Z*lambda[i]*M)*cos(M*x)/C[11], i = 1 .. 4);
But when I used the data ,I do not get the right result
the datas are
h := .25;
M := Pi/L; L := 1;
E[3] := 1; E[1] := 40; E[2] := 1;
1
V[13] := .25; V[31] := .25; V[12] := .25; V[23] := .25; V[32] := .25; V[21] := .25;
G[44] := .2; G[55] := .2; G[66] := .2;
G[13] := .5;
C[55] := G[13];
Omega := -2*V[13]*V[21]*V[32]-V[12]*V[21]-V[13]*V[31]-V[23]*V[32]+1;
C[11] := (-V[23]*V[32]+1)*E[1]/Omega;
C[12] := (V[13]*V[32]+V[12])*E[2]/Omega;
C[33] := (1-V[12]*V[12])*E[3]/Omega;
C[1] := -.2596702449; C[2] := -2.411545729; C[3] := -0.6051254981e-1; C[4] := .4085627399;
lambda[1] := .6891428810; lambda[2] := -.6891428810; lambda[3] := 9.177422410; lambda[4] := -9.177422410;
could you help me plase
### Use of Maple Player...
October 19 2015
0 2
i have an application of emmbeded components i will like to know if i can create a stand alone exececutable using maple player
### How to write to a file in MAPLE PLAYER?...
August 31 2015
0 2
I have been unable to successfully open a text file for writing in MAPLEPLAYER, whether it is requested in the startup code section or via a document tools button. Is there a way to write to a file in MAPLEPLAYER? It returns a "no write access" statement.
The write works perfectly in MAPLE 2015.1
MRB
### How to use my module in Maple Player ?...
July 24 2015
2 9
I have created and saved a MAPLE module in an .mla archive. The module contains three procedures A, B, C, where
A calls, B and C.
Once the module library has been loaded, A acccepts inputs and generates outputs.
Is it possible to create a MAPLE player worksheet which calls the module and share it with a Maple Player (only) user, so that they can then supply the inputs and observe the outputs from A using the Maple Player programme components?
Can anyone help?
MRB
### Lock some contents for viewing...
June 23 2015
1 2
Maple 18 and MapleNet 2015.
Show/Hide Contents allows one to hide certain elements of the worksheet. Is there a way settings there (or somewhere) can be locked so that another user is prevented from seeing certain elements of the worksheet?
Rationale: As an example: I'd like my students to use Maple Player to interact with a worksheet, using Maple Component GUI elemnts. I do not want them to see all the code behind that, and in fact explicitly want to rule them seeing some function definitions. I can hide "input, output" when I create the document, but under "View" in Maple Player, the intrepid student could always unhide that and see the code.
### MaplePlayer support Maplets?...
June 23 2015
1 1
I'm using Maple Player 2105, Maple 18 and exploring using Maple/MapleNet/MaplePlayer for education.
Is the following correct? In order for a my students to access a Maplet, they would need a full version of Maple installed on their system (Maple Player would not be enough) I couldnt get Maplets to work with Maple Player, but maybe that's just me.
(I do not have the \$ at the moment to go the MapleNet route)
Greg
### Difference in Explore between Maple and MaplePlaye...
June 19 2015
2 1
My code works in Maple (18.02), not Maple Player (2015.1). Much headache ensued.
I've isolated the problem in a toy form:
Defn := proc (scaleby)
whattype(scaleby);
end;
Explore(Defn(p), parameters = [p = [1, 10, 100]]);
In Maple will return "integer" for any value of p selected in the combobox.
In Maple Player returns "string" for any value of p selected in the combobox
Is there something I'm missing here? Is Maple Player not backwards compatible to versions of Maple prior to 2015?
My understanding is that Maple Code should export without headache to Maple Player. Is this a solid understanding?
Thanks for helping out a newbie!
### MaplePlayer App...
December 09 2014
1 0
I have downloaded MaplePlayer on my iPad. I see that it comes with a number of "pre-recorded" programs that it can run. Is there a way to have it run a Maple worksheet that I have written? If so, how do I use this app to open an existing worksheet?
Also, are there any plans for this app on the Android platform any time soon?
### Voting is now open for the next Möbius App Challen...
April 15 2014 Maple MapleNet MaplePlayer
3
1
Voting is now open for the next individual prize to be awarded as part of the Möbius App Challenge. The winner will receive an Xbox One Prize Pack!
Here are the finalist Apps:
Note that, if you ever have any problems viewing Apps in your browser, or simply want to work offline, you can always download a Möbius App and view it in Maple or the free Maple Player. To download a Möbius App, follow the link to the App and then click on the Download button near the top left of the page.
You can vote for your favorite through our Facebook page or, if you’re not on Facebook, send an email with your vote to Mobius-Project@maplesoft.com.
And remember, we are now accepting entries for the next quarterly prize. You could win a Music Prize Pack, including the 64GB 5th Generation Apple iPod Touch, Sennheiser In-Ear Noise Cancelling Headphones and the Bose SoundLink Bluetooth Speaker III! See the Möbuis App Challenge for details.
Voting closes April 25th, 2014.
### can i read worksheets with a .mws extention with m...
March 17 2014
1 2
I took a calculus 1 class in 2002, so i have many maple worksheets i would like to view on my new dell venue 8 pro. can the player read the .mws ext ? Donald Altringer
ps. I have maple 8 on my laptop but not on th tablet and i don't have a way of installing it
### Maple Player app for iPad crashes when I run an ap...
January 29 2014
1 3
I'm trying to create a small app for Maple Player to graph solids of revolution. When I run the app in the desktop version of Maple T.A it works, but when I move it into the Maple Player App it crashes. Any ideas?
1 2 3 Page 1 of 3
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# Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
Anonymous Coward
User ID: 39532440
United States
05/08/2013 07:45 PM
Report Abusive Post
Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
The number divided by the sum of it's digits always = 37
111, 1+1+1 = 3 , 111/3 = 37
222, 2+2+2 = 6 , 222/6 = 37
333, 3+3+3 = 9 , 333/9 = 37
444, 4+4+4 = 12, 444/12 = 37
555, 5+5+5 = 15, 555/15 = 37
666, 6+6+6 = 18, 666/18 = 37
777, 7+7+7 = 21, 777/21 = 37
888, 8+8+8 = 24, 888/24 = 37
999, 9+9+9 = 27, 999/27 = 37
9 x 37 = 333
Anonymous Coward
User ID: 39533915
United States
05/08/2013 07:55 PM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
Check that out!
Anonymous Coward (OP)
User ID: 39532440
United States
05/08/2013 07:58 PM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
" Abraham fell facedown; he laughed and said to himself, "Will a son be born to a man a hundred years old? Will Sarah bear a child at the age of ninety?"
Genesis 17:17
100/90 = 1.111111111111111
Enim1
User ID: 1089842
United States
05/08/2013 07:59 PM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
Cool!
Enim1
Anonymous Coward
User ID: 8811543
United States
05/08/2013 08:01 PM
Report Abusive Post
Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
The number divided by the sum of it's digits always = 37
111, 1+1+1 = 3 , 111/3 = 37
222, 2+2+2 = 6 , 222/6 = 37
333, 3+3+3 = 9 , 333/9 = 37
444, 4+4+4 = 12, 444/12 = 37
555, 5+5+5 = 15, 555/15 = 37
666, 6+6+6 = 18, 666/18 = 37
777, 7+7+7 = 21, 777/21 = 37
888, 8+8+8 = 24, 888/24 = 37
999, 9+9+9 = 27, 999/27 = 37
9 x 37 = 333
Quoting: Anonymous Coward 39532440
noice!
Anonymous Coward (OP)
User ID: 39532440
United States
05/08/2013 08:13 PM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
Curiously, 37 is called a " centered hexagonal number " .
It's curious because when light passes through hexagonal ice crystals in the air, it makes the common 22 degree halo we see around the sun and moon.
Anonymous Coward (OP)
User ID: 39532440
United States
05/08/2013 08:35 PM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
Now to dig a little deeper....
111+222+333+444+555+666+777+888+999 = 4995
4+9+9+5 = 27
Magic number 27
Thoth ( To the Egyptians ) and Hermes Trismegistus ( to the Greeks ), gambled with the Moon for 1/72nd of each day's light, to give us the 365 day year.
The moon is 27% the size of the Earth
72/27 = 2.666666666666667
One half of the moon's radius is approximately 540 miles.
144,000/540 = 2.666666666666667
Incidentally, the word " beast " in Hebrew, does not have a gematria value of 666.
It has the value of 540
The ancient Jewish Midrash said for 36 hours, in the beginning, was a " different light ", and on the 4th day, light was created.
4 days = 96 hours
96/36 = 2.666666666666667
should we go deeper into the numbers ?
Anonymous Coward (OP)
User ID: 39532440
United States
05/08/2013 08:59 PM
Report Abusive Post
Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
The number divided by the sum of it's digits always = 37
111, 1+1+1 = 3 , 111/3 = 37
222, 2+2+2 = 6 , 222/6 = 37
333, 3+3+3 = 9 , 333/9 = 37
444, 4+4+4 = 12, 444/12 = 37
555, 5+5+5 = 15, 555/15 = 37
666, 6+6+6 = 18, 666/18 = 37
777, 7+7+7 = 21, 777/21 = 37
888, 8+8+8 = 24, 888/24 = 37
999, 9+9+9 = 27, 999/27 = 37
9 x 37 = 333
Quoting: Anonymous Coward 39532440
37/27 = 1.37037037037037
137 is the number of the Fine Structure Constant ( Alpha ) and the Golden Angle !
That's pretty cool
Anonymous Coward
User ID: 29852642
Belgium
05/08/2013 09:25 PM
Report Abusive Post
Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
The number divided by the sum of it's digits always = 37
111, 1+1+1 = 3 , 111/3 = 37
222, 2+2+2 = 6 , 222/6 = 37
333, 3+3+3 = 9 , 333/9 = 37
444, 4+4+4 = 12, 444/12 = 37
555, 5+5+5 = 15, 555/15 = 37
666, 6+6+6 = 18, 666/18 = 37
777, 7+7+7 = 21, 777/21 = 37
888, 8+8+8 = 24, 888/24 = 37
999, 9+9+9 = 27, 999/27 = 37
9 x 37 = 333
Quoting: Anonymous Coward 39532440
I suspect this relationship doesn't hold only for these triple digit numbers of which there are only nine.
If you multiply 37 with 4, you get 148 (which is situated between 111 and 222), and the sum of 1+4+8 is 13 (further reduced to 4 as always with the gematria method).
Thus 148, 1+4+8 = 4, 148/4 = 37
and I am sure this rule can be generalized beyond triple digit numbers.
Anonymous Coward
User ID: 39537653
Brazil
05/08/2013 09:29 PM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
37 is the key?
not 42?
Anonymous Coward
User ID: 35325670
United States
05/08/2013 09:30 PM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
Cool story bro.
Anonymous Coward (OP)
User ID: 39532440
United States
05/08/2013 09:35 PM
Report Abusive Post
Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
The number divided by the sum of it's digits always = 37
111, 1+1+1 = 3 , 111/3 = 37
222, 2+2+2 = 6 , 222/6 = 37
333, 3+3+3 = 9 , 333/9 = 37
444, 4+4+4 = 12, 444/12 = 37
555, 5+5+5 = 15, 555/15 = 37
666, 6+6+6 = 18, 666/18 = 37
777, 7+7+7 = 21, 777/21 = 37
888, 8+8+8 = 24, 888/24 = 37
999, 9+9+9 = 27, 999/27 = 37
9 x 37 = 333
Quoting: Anonymous Coward 39532440
I suspect this relationship doesn't hold only for these triple digit numbers of which there are only nine.
If you multiply 37 with 4, you get 148 (which is situated between 111 and 222), and the sum of 1+4+8 is 13 (further reduced to 4 as always with the gematria method).
Thus 148, 1+4+8 = 4, 148/4 = 37
and I am sure this rule can be generalized beyond triple digit numbers.
Quoting: Anonymous Coward 29852642
Wouldn't you have to have some numbers other then 1-9 for your statement to be true ?
101010, for example, is not a repeating string.
bigD111
User ID: 39191150
United States
05/08/2013 09:40 PM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
Yes, and 37=10 which is fulfillment! All who see the numbers are lost tribes of Israel! Some are the 144,000.
deplorably republican
rlc3764
User ID: 39509893
Philippines
05/08/2013 09:43 PM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
37?
37=heart of wisdom
37 HOLOGRAPHIC BEERATIG SET
CUBE3 + 37(KEY) = CIBE4
December 25 = 12.25 ( 1+2 = 3, 2+5 = 7) 12+25 = 37
MDAS = tool of numbers
M=13th letter
D=4
A=1
S=19
MDAS = 13+4+1+19 = 37
37 = 3 SEVENS = 777
21 = 7+7+7 = +777+
37 X (7+7+7) = 777
Anonymous Coward
User ID: 26300532
United States
05/08/2013 09:45 PM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
OP - Very Interesting......have you heard of Ivan Panin?
Anonymous Coward
User ID: 29852642
Belgium
05/08/2013 09:52 PM
Report Abusive Post
Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
The number divided by the sum of it's digits always = 37
111, 1+1+1 = 3 , 111/3 = 37
222, 2+2+2 = 6 , 222/6 = 37
333, 3+3+3 = 9 , 333/9 = 37
444, 4+4+4 = 12, 444/12 = 37
555, 5+5+5 = 15, 555/15 = 37
666, 6+6+6 = 18, 666/18 = 37
777, 7+7+7 = 21, 777/21 = 37
888, 8+8+8 = 24, 888/24 = 37
999, 9+9+9 = 27, 999/27 = 37
9 x 37 = 333
Quoting: Anonymous Coward 39532440
I suspect this relationship doesn't hold only for these triple digit numbers of which there are only nine.
If you multiply 37 with 4, you get 148 (which is situated between 111 and 222), and the sum of 1+4+8 is 13 (further reduced to 4 as always with the gematria method).
Thus 148, 1+4+8 = 4, 148/4 = 37
and I am sure this rule can be generalized beyond triple digit numbers.
Quoting: Anonymous Coward 29852642
Wouldn't you have to have some numbers other then 1-9 for your statement to be true ?
101010, for example, is not a repeating string.
Quoting: Anonymous Coward 39532440
There is obviously one number which, used as a divider on 101010, will give 37 as a result. Offhand, it may or may not be an integer but a quick calculation will show it to be 2730.
So obviously this will hold true for the whole series of integers and doubtlessz even all natural numbers.
Gematrically, the sum of the digits composing the number you are dividing will always fit the sum of the digits composing the divider. Those repetitive triple digits are apparently only special cases where the sum of the digits will be the actual divider, without further reducing gematrically.
Anonymous Coward
User ID: 29852642
Belgium
05/08/2013 10:02 PM
Report Abusive Post
Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
20553.5/555.5 = 37
and 2+5+5+3+5 = 2
and 5+5+5+5 = 2
See ? Gematrically they're the same even if they are neither repetitive nor integers.
Anonymous Coward (OP)
User ID: 39532440
United States
05/08/2013 10:02 PM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
Here's the video I learned this from, a great Youtube channel called " Numberphile ".
Anonymous Coward (OP)
User ID: 39532440
United States
05/08/2013 10:08 PM
Report Abusive Post
Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
20553.5/555.5 = 37
and 2+5+5+3+5 = 2
and 5+5+5+5 = 2
See ? Gematrically they're the same even if they are neither repetitive nor integers.
Quoting: Anonymous Coward 29852642
I understand what you are saying, but this isn't only true for these numbers ?
If there are only 9 single digits, 1-9 and only 9 triple digit repeating strings that can be written with them, 111...999
Also, when you say " Gematrically ", there are quite a few methods of gematria. Which one would this be that you are referring to ?
Anonymous Coward
User ID: 33294695
Ireland
05/08/2013 10:09 PM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
Do you know how many days old the President was when Benghazi happened?
If you read between the lines, you will see the beast exposed and it's imminent downfall
And it all began on that day
Anonymous Coward (OP)
User ID: 39532440
United States
05/08/2013 10:15 PM
Report Abusive Post
Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
Do you know how many days old the President was when Benghazi happened?
If you read between the lines, you will see the beast exposed and it's imminent downfall
And it all began on that day
Quoting: Anonymous Coward 33294695
Isn't that called " Apophenia " ?
Anonymous Coward
User ID: 33294695
Ireland
05/08/2013 10:24 PM
Report Abusive Post
Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
Do you know how many days old the President was when Benghazi happened?
If you read between the lines, you will see the beast exposed and it's imminent downfall
And it all began on that day
Quoting: Anonymous Coward 33294695
Isn't that called " Apophenia " ?
Quoting: Anonymous Coward 39532440
Call it whatever you like
It cannot be debunked
It's a fact
Thread: The Pope Verified The Information I Gave You By Resigning
Smoking Weed and diving Down Rabbit Holes can be quite fun and sometimes dangerous
Anonymous Coward (OP)
User ID: 39532440
United States
05/08/2013 10:37 PM
Report Abusive Post
Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
Do you know how many days old the President was when Benghazi happened?
If you read between the lines, you will see the beast exposed and it's imminent downfall
And it all began on that day
Quoting: Anonymous Coward 33294695
Isn't that called " Apophenia " ?
Quoting: Anonymous Coward 39532440
Call it whatever you like
It cannot be debunked
It's a fact
Thread: The Pope Verified The Information I Gave You By Resigning
Smoking Weed and diving Down Rabbit Holes can be quite fun and sometimes dangerous
Quoting: Anonymous Coward 33294695
It's also a fact that every president since I have been alive ( 40 years so far ) has been called the " antichrist " by some group of religious nuts.
They have all been wrong.
Anonymous Coward
User ID: 33294695
Ireland
05/08/2013 10:51 PM
Report Abusive Post
Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
It's also a fact that every president since I have been alive ( 40 years so far ) has been called the " antichrist " by some group of religious nuts.
They have all been wrong
Quoting: Anonymous Coward 39532440
The Oval Office produces the eggs, the Seat is the Uterus
And out of her comes The One
All US Presidents past, present and future are one of the eggs
But which one makes the breakthrough into the light?
I recall seeing a thread around here somewhere asking who the egg was sitting in the chair in the Stonecutters secret room in the Simpsons
Council on Foreign Relations
User ID: 38982770
United States
05/09/2013 12:10 AM
Report Abusive Post
Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
37 is a prime number.
here is a list of all prime #'s
2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939.
there aren't really a lot of them.
Interestingly, the signal SETI sends into space is based on prime numbers.
The cardinality of 1679 was chosen because it is a semiprime (the product of two prime numbers), to be arranged rectangularly as 73 rows by 23 columns. The alternative arrangement, 23 rows by 73 columns, produces jumbled nonsense (File:AreciboMessageShifted.svg). The message forms the image shown on the right, or its inverse, when translated into graphics characters and spaces.[4]
Read up on the Arecibo message, the message that SETI sends into space,
This is a response, sent back in time to earth
Last Edited by Protagonist on 05/09/2013 12:35 AM
SevenThunders
User ID: 15851599
United States
05/09/2013 12:15 AM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
The number divided by the sum of it's digits always = 37
111, 1+1+1 = 3 , 111/3 = 37
222, 2+2+2 = 6 , 222/6 = 37
333, 3+3+3 = 9 , 333/9 = 37
444, 4+4+4 = 12, 444/12 = 37
555, 5+5+5 = 15, 555/15 = 37
666, 6+6+6 = 18, 666/18 = 37
777, 7+7+7 = 21, 777/21 = 37
888, 8+8+8 = 24, 888/24 = 37
999, 9+9+9 = 27, 999/27 = 37
9 x 37 = 333
Quoting: Anonymous Coward 39532440
this is trivial to prove actually. Behold
xxx = x * 111
x + x + x = 3x
xxx / (3x) = x * 111 / (3x) = 111/3 = 37
QED
Vision Thing
User ID: 36175629
United States
05/09/2013 12:32 AM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
Curiously, 37 is called a " centered hexagonal number " .
It's curious because when light passes through hexagonal ice crystals in the air, it makes the common 22 degree halo we see around the sun and moon.
Quoting: Anonymous Coward 39532440
If you keep talking like that I'm gonna have an acid flashback
Anonymous Coward
User ID: 33867892
Australia
05/09/2013 12:57 AM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
Now to dig a little deeper....
111+222+333+444+555+666+777+888+999 = 4995
4+9+9+5 = 27
Magic number 27
Thoth ( To the Egyptians ) and Hermes Trismegistus ( to the Greeks ), gambled with the Moon for 1/72nd of each day's light, to give us the 365 day year.
The moon is 27% the size of the Earth
72/27 = 2.666666666666667
One half of the moon's radius is approximately 540 miles.
144,000/540 = 2.666666666666667
Incidentally, the word " beast " in Hebrew, does not have a gematria value of 666.
It has the value of 540
The ancient Jewish Midrash said for 36 hours, in the beginning, was a " different light ", and on the 4th day, light was created.
4 days = 96 hours
96/36 = 2.666666666666667
should we go deeper into the numbers ?
Quoting: Anonymous Coward 39532440
and most really stupid cunts are convinced its all just a collosal accident.
Anonymous Coward
User ID: 11853442
United States
05/09/2013 01:04 AM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
The number divided by the sum of it's digits always = 37
that's creepy
Quoting: Anonymous Coward 39532440
Anonymous Coward
User ID: 18790409
United States
05/09/2013 01:34 AM
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Re: Did you know the relationship between all triple digit repeating strings like 111, 666 and 888
I'm not even supposed to be here today XD
Anonymous Coward
User ID: 39509893
Philippines
05/09/2013 01:56 AM
Report Abusive Post | 5,845 | 17,666 | {"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} | 3.453125 | 3 | CC-MAIN-2017-26 | latest | en | 0.8466 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/845/2/n/a/ | 1,638,426,820,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361169.72/warc/CC-MAIN-20211202054457-20211202084457-00375.warc.gz | 917,066,228 | 57,202 | # Properties
Label 845.2.n.a Level $845$ Weight $2$ Character orbit 845.n Analytic conductor $6.747$ Analytic rank $1$ Dimension $4$ CM no Inner twists $4$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$845 = 5 \cdot 13^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 845.n (of order $$6$$, degree $$2$$, not minimal)
## Newform invariants
Self dual: no Analytic conductor: $$6.74735897080$$ Analytic rank: $$1$$ Dimension: $$4$$ Relative dimension: $$2$$ over $$\Q(\zeta_{6})$$ Coefficient field: $$\Q(\zeta_{12})$$ Defining polynomial: $$x^{4} - x^{2} + 1$$ Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 65) Sato-Tate group: $\mathrm{SU}(2)[C_{6}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{12}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + \zeta_{12} q^{2} -2 \zeta_{12} q^{3} -\zeta_{12}^{2} q^{4} + ( -2 - \zeta_{12}^{3} ) q^{5} -2 \zeta_{12}^{2} q^{6} -3 \zeta_{12}^{3} q^{8} + \zeta_{12}^{2} q^{9} +O(q^{10})$$ $$q + \zeta_{12} q^{2} -2 \zeta_{12} q^{3} -\zeta_{12}^{2} q^{4} + ( -2 - \zeta_{12}^{3} ) q^{5} -2 \zeta_{12}^{2} q^{6} -3 \zeta_{12}^{3} q^{8} + \zeta_{12}^{2} q^{9} + ( 1 - 2 \zeta_{12} - \zeta_{12}^{2} ) q^{10} + ( -2 + 2 \zeta_{12}^{2} ) q^{11} + 2 \zeta_{12}^{3} q^{12} + ( -2 + 4 \zeta_{12} + 2 \zeta_{12}^{2} ) q^{15} + ( 1 - \zeta_{12}^{2} ) q^{16} + \zeta_{12}^{3} q^{18} -6 \zeta_{12}^{2} q^{19} + ( -\zeta_{12} + 2 \zeta_{12}^{2} + \zeta_{12}^{3} ) q^{20} + ( -2 \zeta_{12} + 2 \zeta_{12}^{3} ) q^{22} + 6 \zeta_{12} q^{23} + ( -6 + 6 \zeta_{12}^{2} ) q^{24} + ( 3 + 4 \zeta_{12}^{3} ) q^{25} + 4 \zeta_{12}^{3} q^{27} + ( -6 + 6 \zeta_{12}^{2} ) q^{29} + ( -2 \zeta_{12} + 4 \zeta_{12}^{2} + 2 \zeta_{12}^{3} ) q^{30} -6 q^{31} + ( -5 \zeta_{12} + 5 \zeta_{12}^{3} ) q^{32} + ( 4 \zeta_{12} - 4 \zeta_{12}^{3} ) q^{33} + ( 1 - \zeta_{12}^{2} ) q^{36} + 6 \zeta_{12} q^{37} -6 \zeta_{12}^{3} q^{38} + ( -3 + 6 \zeta_{12}^{3} ) q^{40} + ( -8 + 8 \zeta_{12}^{2} ) q^{41} + ( 6 \zeta_{12} - 6 \zeta_{12}^{3} ) q^{43} + 2 q^{44} + ( \zeta_{12} - 2 \zeta_{12}^{2} - \zeta_{12}^{3} ) q^{45} + 6 \zeta_{12}^{2} q^{46} + 8 \zeta_{12}^{3} q^{47} + ( -2 \zeta_{12} + 2 \zeta_{12}^{3} ) q^{48} + ( -7 + 7 \zeta_{12}^{2} ) q^{49} + ( -4 + 3 \zeta_{12} + 4 \zeta_{12}^{2} ) q^{50} -12 \zeta_{12}^{3} q^{53} + ( -4 + 4 \zeta_{12}^{2} ) q^{54} + ( 4 + 2 \zeta_{12} - 4 \zeta_{12}^{2} ) q^{55} + 12 \zeta_{12}^{3} q^{57} + ( -6 \zeta_{12} + 6 \zeta_{12}^{3} ) q^{58} -2 \zeta_{12}^{2} q^{59} + ( 2 - 4 \zeta_{12}^{3} ) q^{60} -6 \zeta_{12}^{2} q^{61} -6 \zeta_{12} q^{62} -7 q^{64} + 4 q^{66} -12 \zeta_{12} q^{67} -12 \zeta_{12}^{2} q^{69} -2 \zeta_{12}^{2} q^{71} + ( 3 \zeta_{12} - 3 \zeta_{12}^{3} ) q^{72} + 6 \zeta_{12}^{3} q^{73} + 6 \zeta_{12}^{2} q^{74} + ( 8 - 6 \zeta_{12} - 8 \zeta_{12}^{2} ) q^{75} + ( -6 + 6 \zeta_{12}^{2} ) q^{76} + ( -2 - \zeta_{12} + 2 \zeta_{12}^{2} ) q^{80} + ( 11 - 11 \zeta_{12}^{2} ) q^{81} + ( -8 \zeta_{12} + 8 \zeta_{12}^{3} ) q^{82} -4 \zeta_{12}^{3} q^{83} + 6 q^{86} + ( 12 \zeta_{12} - 12 \zeta_{12}^{3} ) q^{87} + 6 \zeta_{12} q^{88} + ( 8 - 8 \zeta_{12}^{2} ) q^{89} + ( 1 - 2 \zeta_{12}^{3} ) q^{90} -6 \zeta_{12}^{3} q^{92} + 12 \zeta_{12} q^{93} + ( -8 + 8 \zeta_{12}^{2} ) q^{94} + ( -6 \zeta_{12} + 12 \zeta_{12}^{2} + 6 \zeta_{12}^{3} ) q^{95} + 10 q^{96} + ( -6 \zeta_{12} + 6 \zeta_{12}^{3} ) q^{97} + ( -7 \zeta_{12} + 7 \zeta_{12}^{3} ) q^{98} -2 q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4 q - 2 q^{4} - 8 q^{5} - 4 q^{6} + 2 q^{9} + O(q^{10})$$ $$4 q - 2 q^{4} - 8 q^{5} - 4 q^{6} + 2 q^{9} + 2 q^{10} - 4 q^{11} - 4 q^{15} + 2 q^{16} - 12 q^{19} + 4 q^{20} - 12 q^{24} + 12 q^{25} - 12 q^{29} + 8 q^{30} - 24 q^{31} + 2 q^{36} - 12 q^{40} - 16 q^{41} + 8 q^{44} - 4 q^{45} + 12 q^{46} - 14 q^{49} - 8 q^{50} - 8 q^{54} + 8 q^{55} - 4 q^{59} + 8 q^{60} - 12 q^{61} - 28 q^{64} + 16 q^{66} - 24 q^{69} - 4 q^{71} + 12 q^{74} + 16 q^{75} - 12 q^{76} - 4 q^{80} + 22 q^{81} + 24 q^{86} + 16 q^{89} + 4 q^{90} - 16 q^{94} + 24 q^{95} + 40 q^{96} - 8 q^{99} + O(q^{100})$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/845\mathbb{Z}\right)^\times$$.
$$n$$ $$171$$ $$677$$ $$\chi(n)$$ $$-\zeta_{12}^{2}$$ $$-1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
484.1
−0.866025 + 0.500000i 0.866025 − 0.500000i −0.866025 − 0.500000i 0.866025 + 0.500000i
−0.866025 + 0.500000i 1.73205 1.00000i −0.500000 + 0.866025i −2.00000 1.00000i −1.00000 + 1.73205i 0 3.00000i 0.500000 0.866025i 2.23205 0.133975i
484.2 0.866025 0.500000i −1.73205 + 1.00000i −0.500000 + 0.866025i −2.00000 + 1.00000i −1.00000 + 1.73205i 0 3.00000i 0.500000 0.866025i −1.23205 + 1.86603i
529.1 −0.866025 0.500000i 1.73205 + 1.00000i −0.500000 0.866025i −2.00000 + 1.00000i −1.00000 1.73205i 0 3.00000i 0.500000 + 0.866025i 2.23205 + 0.133975i
529.2 0.866025 + 0.500000i −1.73205 1.00000i −0.500000 0.866025i −2.00000 1.00000i −1.00000 1.73205i 0 3.00000i 0.500000 + 0.866025i −1.23205 1.86603i
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
5.b even 2 1 inner
13.c even 3 1 inner
65.n even 6 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 845.2.n.a 4
5.b even 2 1 inner 845.2.n.a 4
13.b even 2 1 845.2.n.b 4
13.c even 3 1 845.2.b.a 2
13.c even 3 1 inner 845.2.n.a 4
13.d odd 4 1 845.2.l.a 4
13.d odd 4 1 845.2.l.b 4
13.e even 6 1 845.2.b.b 2
13.e even 6 1 845.2.n.b 4
13.f odd 12 1 65.2.d.a 2
13.f odd 12 1 65.2.d.b yes 2
13.f odd 12 1 845.2.l.a 4
13.f odd 12 1 845.2.l.b 4
39.k even 12 1 585.2.h.b 2
39.k even 12 1 585.2.h.c 2
52.l even 12 1 1040.2.f.a 2
52.l even 12 1 1040.2.f.b 2
65.d even 2 1 845.2.n.b 4
65.g odd 4 1 845.2.l.a 4
65.g odd 4 1 845.2.l.b 4
65.l even 6 1 845.2.b.b 2
65.l even 6 1 845.2.n.b 4
65.n even 6 1 845.2.b.a 2
65.n even 6 1 inner 845.2.n.a 4
65.o even 12 1 325.2.c.b 2
65.o even 12 1 325.2.c.e 2
65.q odd 12 1 4225.2.a.e 1
65.q odd 12 1 4225.2.a.m 1
65.r odd 12 1 4225.2.a.h 1
65.r odd 12 1 4225.2.a.k 1
65.s odd 12 1 65.2.d.a 2
65.s odd 12 1 65.2.d.b yes 2
65.s odd 12 1 845.2.l.a 4
65.s odd 12 1 845.2.l.b 4
65.t even 12 1 325.2.c.b 2
65.t even 12 1 325.2.c.e 2
195.bh even 12 1 585.2.h.b 2
195.bh even 12 1 585.2.h.c 2
260.bc even 12 1 1040.2.f.a 2
260.bc even 12 1 1040.2.f.b 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
65.2.d.a 2 13.f odd 12 1
65.2.d.a 2 65.s odd 12 1
65.2.d.b yes 2 13.f odd 12 1
65.2.d.b yes 2 65.s odd 12 1
325.2.c.b 2 65.o even 12 1
325.2.c.b 2 65.t even 12 1
325.2.c.e 2 65.o even 12 1
325.2.c.e 2 65.t even 12 1
585.2.h.b 2 39.k even 12 1
585.2.h.b 2 195.bh even 12 1
585.2.h.c 2 39.k even 12 1
585.2.h.c 2 195.bh even 12 1
845.2.b.a 2 13.c even 3 1
845.2.b.a 2 65.n even 6 1
845.2.b.b 2 13.e even 6 1
845.2.b.b 2 65.l even 6 1
845.2.l.a 4 13.d odd 4 1
845.2.l.a 4 13.f odd 12 1
845.2.l.a 4 65.g odd 4 1
845.2.l.a 4 65.s odd 12 1
845.2.l.b 4 13.d odd 4 1
845.2.l.b 4 13.f odd 12 1
845.2.l.b 4 65.g odd 4 1
845.2.l.b 4 65.s odd 12 1
845.2.n.a 4 1.a even 1 1 trivial
845.2.n.a 4 5.b even 2 1 inner
845.2.n.a 4 13.c even 3 1 inner
845.2.n.a 4 65.n even 6 1 inner
845.2.n.b 4 13.b even 2 1
845.2.n.b 4 13.e even 6 1
845.2.n.b 4 65.d even 2 1
845.2.n.b 4 65.l even 6 1
1040.2.f.a 2 52.l even 12 1
1040.2.f.a 2 260.bc even 12 1
1040.2.f.b 2 52.l even 12 1
1040.2.f.b 2 260.bc even 12 1
4225.2.a.e 1 65.q odd 12 1
4225.2.a.h 1 65.r odd 12 1
4225.2.a.k 1 65.r odd 12 1
4225.2.a.m 1 65.q odd 12 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(845, [\chi])$$:
$$T_{2}^{4} - T_{2}^{2} + 1$$ $$T_{11}^{2} + 2 T_{11} + 4$$
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$1 - T^{2} + T^{4}$$
$3$ $$16 - 4 T^{2} + T^{4}$$
$5$ $$( 5 + 4 T + T^{2} )^{2}$$
$7$ $$T^{4}$$
$11$ $$( 4 + 2 T + T^{2} )^{2}$$
$13$ $$T^{4}$$
$17$ $$T^{4}$$
$19$ $$( 36 + 6 T + T^{2} )^{2}$$
$23$ $$1296 - 36 T^{2} + T^{4}$$
$29$ $$( 36 + 6 T + T^{2} )^{2}$$
$31$ $$( 6 + T )^{4}$$
$37$ $$1296 - 36 T^{2} + T^{4}$$
$41$ $$( 64 + 8 T + T^{2} )^{2}$$
$43$ $$1296 - 36 T^{2} + T^{4}$$
$47$ $$( 64 + T^{2} )^{2}$$
$53$ $$( 144 + T^{2} )^{2}$$
$59$ $$( 4 + 2 T + T^{2} )^{2}$$
$61$ $$( 36 + 6 T + T^{2} )^{2}$$
$67$ $$20736 - 144 T^{2} + T^{4}$$
$71$ $$( 4 + 2 T + T^{2} )^{2}$$
$73$ $$( 36 + T^{2} )^{2}$$
$79$ $$T^{4}$$
$83$ $$( 16 + T^{2} )^{2}$$
$89$ $$( 64 - 8 T + T^{2} )^{2}$$
$97$ $$1296 - 36 T^{2} + T^{4}$$ | 4,836 | 8,968 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-49 | latest | en | 0.417374 |
https://search.r-project.org/CRAN/refmans/BDEsize/html/plots.Block.html | 1,719,326,469,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00176.warc.gz | 434,363,659 | 2,179 | plots.Block {BDEsize} R Documentation
Diagnosis Graphs for the number of Blocks of Randomized Complete Block Design
Description
This function produces graphs between the sample size, power and the detectable standardized effect size of randomized complete block design.
Usage
plots.Block(factor.lev, interaction = FALSE, delta_type = 1, delta = c(1, 0, 1),
deltao = NULL, alpha = 0.05, beta = 0.2, type = 1, maxsize = 1000)
Arguments
factor.lev vector of the numbers of levels for each factor. interaction specifies whether two-way interaction effects are included in a model with the main effects. When interaction = TRUE, two-way interaction effects are include in a model. delta_type specifies the type of standardized effect size: 1 for standard deviation type and 2 for range type. delta vector of effect sizes: delta[1] for main effects, delta[2] for two-way interaction effects, and delta[3] for standard deviation of noise. When interaction=FALSE, delta[2] is 0. deltao the minimal detectable standardized effect size for power vs the number of blocks plot when type = 3. alpha Type I error. beta Type II error. type graph type: 1 for Power vs Delta plot, 2 for Delta vs the Number of Blocks plot, and 3 for Power vs the Number of Blocks plot. maxsize tolerance for the number of blocks.
Details
In a randomized complete block design (without replications), the optimal number of blocks need to be determined. This function produces graph between Number of Block, power 1-beta and the detectable standardized effect size delta of randomized complete block design. According to type, it displays plot of Power vs Delta, Delta vs Number of Blocks, or Power vs Number of Blocks.
Value
plot of Power vs Delta, Delta vs Number of Blocks, or Power vs Number of Blocks according to type.
Examples
# plot of Power vs Delta for randomized complete block design
# with 2 factors without the interaction effects
plots.Block(factor.lev=c(2, 2), interaction=FALSE,
delta_type=1, delta=c(1, 0, 1), alpha=0.05, beta=0.2, type=1)
# plot of Power vs Number of Blocks for randomized complete block design
# with 2 factors with the interaction effects
plots.Block(factor.lev=c(2, 3), interaction=TRUE,
delta_type=1, delta=c(1, 1, 1), deltao=1.5, alpha=0.05, beta=0.2, type=3)
[Package BDEsize version 1.6 Index] | 570 | 2,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-26 | latest | en | 0.804977 |
https://www.spoj.com/problems/VPL2_BC/ | 1,550,704,517,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247496855.63/warc/CC-MAIN-20190220230820-20190221012820-00109.warc.gz | 996,115,175 | 9,933 | ## VPL2_BC - Peter Quest
no tags
Is Peter’s birthday! And his friends are preparing a big party, however, Peter is obsessed with a variation of the famous game Minesweeper, moreover, Peter hates losing, so if any of you or your friends can beat him in the game, he will get angry and will attend the party that everyone is organizing for him.
This new mode of Minesweeper consists in building the gameboard given the mines, so, if the matrix has size of (2x2) and there is a mine in the position (0,0) of the matrix, the resultant gameboard will be *1 11 Your task is to beat Peter and celebrate his birthday before its too late! Please have in consideration that:
• The cell (i,j) where there is a mine will be denoted as ’*’
• The cell (i,j) where there is no mine will be denoted as ’-’
• The cell (i,j) where there is N mines adjacent to it, will be denoted as a number from 1 to 8 (depending on the number of adjacencies)
Input
The first line contains an integer T , which specifies the number of test cases. Then, will follow the descriptions of T test cases.
For each test case, you will have a line with three numbers N , M , K, denoting, respectively, the size of the matrix, formed by N columns and M rows, and K mines. Then, K lines will follow, each containing two numbers Ki and Kj denoting the position where the mine is in the board.
Output
For each test case you should print the string "Scenario #i:" where i represents the test case you are analyzing (starting from 1), then, in the next line, you shall print the gameboard as specified.
INPUT OUTPUT 3 3 2 2 0 0 1 1 3 3 3 0 0 1 1 2 2 8 8 10 0 1 5 0 2 5 4 5 2 6 5 6 6 6 5 7 6 7 7 7 Scenario #1: *2 2* 11 Scenario #2: *21 2*2 12* Scenario #3: 1*1----- 111-1221 ----1**1 ----2331 11--1*32 *1--13** 11---2** -----13*
• 1 ≤ T ≤ 10
• 1 ≤ N ,M ≤ 8
• 1 ≤ K ≤ min(8,N *M )
• 1 ≤ T ≤ 10
• 1 ≤ N ,M ≤ 128
• 1 ≤ K ≤ min(128,N *M )
• 1 ≤ T ≤ 10
• 1 ≤ N ,M ≤ 1024
• 1 ≤ K ≤ min(1024,N *M ) | 629 | 1,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-09 | latest | en | 0.90329 |
https://www.jiskha.com/search?query=a+800+gram+ball+at+the+end+of+a+cord+is+whirled+in+an+almost+horizontal+circle+of+radius+1.50+meters.+its+tangential+speed+in+the+circle+is+10+meters+per+second.+What+must+be+the+tension+in+the+cord%3F | 1,624,293,639,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488286726.71/warc/CC-MAIN-20210621151134-20210621181134-00475.warc.gz | 783,993,200 | 10,545 | # a 800 gram ball at the end of a cord is whirled in an almost horizontal circle of radius 1.50 meters. its tangential speed in the circle is 10 meters per second. What must be the tension in the cord?
38,921 results
1. ## Physics
A 1.0-kg ball on the end of a string is whirled at a constant speed of 2.0 m/s in a horizontal circle of radius 1.5 m. What is the work done by the centripetal force during one revolution? I did: F=m(v^2)/r F=1(2^2)/1.5=2.7 J The answer that was provided
2. ## physics
A 0.55- ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.5 on a frictionless horizontal surface.If the cord will break when the tension in it exceeds 95 , what is the maximum speed the ball can have?
3. ## physics
A 0.40kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 2.1m on a frictionless horizontal surface.If the cord will break when the tension in it exceeds 80N , what is the maximum speed the ball can have?
4. ## Physics
A sphere of mass 3.0X10^-4 kg is suspended from a cord. A steady horizontal breeze pushes the sphere so that the cord makes a constant angle of 37 degrees with the vertical. find a) the magnitude of that push and b) the tension of the cord. I looked
5. ## Physics
The hammer throw is a track-and-field event in which a 7.30-kg ball (the hammer) is whirled around in a circle several times and released. It then moves upward on the familiar curved path of projectile motion and eventually returns to the ground some
6. ## Physics
A stone with mass 0.8kg is attached to one end of a string 0.9m long . The string will break if its tension exceeds 600N. The stone is whirled in a horizontal circle ,the other end of the string remains fixed. Find the maximum speed , the stone can attain
7. ## Physics - HELP
A 0.40 kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.5 m on a frictionless horizontal surface. The cord will break when the tension in it exceeds 61 N. What is the maximum speed the ball can have?
8. ## physics
A mass 5kg is whirled in a horizontal circle @ 1 end of a string 50cm long,the other end being fixed. If the string when hanging vertically will just support a load of 200kg mass without breaking,find the maximum whirling speed in revolution per second and
9. ## physics
a .45kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.3m on a frictionless horizontal surface. if the cord will break when the tension in it exceeds 75 N, what is the maximum speed the ball can have?
10. ## physics
A ball of mass 200 grams is whirled in a circle at the end of a strong 100 cm long rope whose breaking strength is 10 N. Neglecting gravity, the maximum speed of the ball is approximately.
11. ## physics
A 0.50 kg ball that is tied to the end of a 1.4 m light cord is revolved in a horizontal plane with the cord making a 30° angle with the vertical. (b) If the ball is revolved so that its speed is 4.0 m/s, what angle does the cord make with the vertical?
12. ## Physics
Imagine that you swing about your head a ball attached to the end of a string. The ball moves at a constant speed in a horizontal circle. Part A If the mass of the ball is 0.160kg and you supply a tension force of 14.5N to the string, what angle would the
13. ## physics
A 0.80-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.7 m on a frictionless horizontal surface.If the cord will break when the tension in it exceeds 90 N, what is the maximum speed the ball can have? Express your
14. ## Physics
A tether ball of mass 0.25 kg is attached to a vertical pole by a cord 1.4 m long. Assume the cord attaches to the center of the ball. The cord makes an angle of 20° with the vertical. (a) What is the tension in the cord? (b) What is the speed of the
15. ## Physics
A small lead ball, attached to a 1.50-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 0.75 m
16. ## AP physics B
A phone cord is 3.26 m long. The cord has a mass of 0.224 kg. A transverse wave pulse is produced by plucking one end of the taut cord. The pulse makes four trips down and back along the cord in 0.865 s. What is the tension in the cord? Answer in units of
17. ## Physics
A small wooden block with mass 0.800kg is suspended from the lower end of a light cord that is 1.52m long. The block is initially at rest. A bullet with mass 0.0126kg is fired at the block with a horizontal velocity v0. The bullet strikes the block and
18. ## physics
A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless
19. ## physics help!
A small 520-gram ball on the end of a thin, light rod is rotated in a horizontal circle of radius 1.5 m. Calculate the moment of inertia of the ball about the center of the circle. Then calculate the torque needed to keep the ball rotating at constant
20. ## Physics
A 0.40-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.3 m on a frictionless surface. If the cord will break when the tension in it exceeds 60 N, what is the maximum speed the ball can have? How would your answer be
21. ## physics
Amass of 0.5 kg is whirled at 120 rpm in a horizontal circle at the end of a string 2 meters long. Neglecting the force due gravity, find: (a) the angular velocity (b) the centripetal acceleration.
22. ## physics
a ball of weight 200 grams,is tied to the end of a cord and whirled in a horizontal circle of radius 0.6 m.if the ball makes five complete revolutions in 2 seconds,what will be the linear speed of the ball?
23. ## college physics
A stone with a mass of 0.800 is attached to one end of a string 0.800 long. The string will break if its tension exceeds 60.0 . The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. Find the
24. ## Dynamics of rigid bodies
A body of mass 300g is whirled in a horizontal circle by means of a 60cm long cord. If the cord can sustain a tension of 400N before, find the max. number of revolution per minute which the body can have
25. ## physics
The hammer throw is a track-and-field event in which a 7.30-kg ball (the hammer) is whirled around in a circle several times and released. It then moves upward on the familiar curved path of projectile motion and eventually returns to the ground some
26. ## physics
An object weighting 300 N is suspended by means of two cords. (There is a cord on the ceiling with is at 45 degrees. The second cord is attached to the wall and is horizontal. These two cords are attached to a cord which has the 300N object on it.) The
27. ## physics
A small ball of mass 65 g is suspended from a string of length 53 cm and whirled in a circle lying in the horizontal plane.If the string makes an angle of 28◦ with the vertical, find the centripetal force experienced by the ball. The acceleration of
28. ## Physics
A ball is fastened to one end of a 30cm string, and the other end is held fixed to a support. the ball whirls in a horizontal circle. Find the speed of the ball in its circular path if the string makes an angle of 30 degrees to the vertical.
29. ## physics
A 0.2 kg mass hangs at the end of a wire. what is the tension in the wire if the ball is whirled around in a horizontal circle with tangential velocity of 5 m/sec
30. ## dynamics
a 2kg ball revolves in a horizontal circle as shown at a constant speed of 1.5m/s. Knowing that L=600mm, determine (a) the angle that the cord with the vertical (b) tension in the cord.
31. ## physics
A phone cord is 4.25 m long. The cord has a mass of 0.200 kg. A transverse wave pulse is produced by plucking one end of the taut cord. The pulse makes four trips down and back along the cord in 0.780 s. What is the tension in the cord?
32. ## Physics
A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless
33. ## physics
A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless
34. ## Physics
A stationary arrangement of two crayon boxes and three cords. Box A has a mass of 11.0kg and is on a ramp at angle è=30.0 degrees; box B has a mass of 7.00 kg and hangs on a cord. The cord connected to Box A is parallel to the ramp which is frictionless
35. ## Physic
A 55-kg female bungee jumper fastens one end of the cord (made of elastic material) to her ankle and the other end to a bridge. Then she jumps off the bridge. As the cord is stretching, it exerts an elastic force directed up on her. Calculate her
36. ## physics
a ball of weight 200 grams,is tied to the end of a cord and whirled in a horizontal circle of radius 0.6 m.if the ball makes five complete revolutions in 2 seconds,what will be the linear speed of the ball?
37. ## Physics
A 0.40-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 2.0-m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 75-N, what is the maximum speed the ball can have?
38. ## physics
a .45 kg ball attached to the end of a horizontal cord is rotated in a circle of radius 1.3 m on a frictionless horizontal surface. if the cord will break when the tension in it exceeds 75N what is the maximum speed the ball can have?
39. ## Physics
A 0.45-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.3-m on a frictionless horizontal surface. If the maximum speed of the ball is 14.7 m/s, how long does it take the ball to make on rotation if it is moving at its
40. ## Physics
A tether ball of mass 0.25 kg is attached to a vertical pole by a cord 1.4 m long. Assume the cord attaches to the center of the ball. The cord makes an angle of 20° with the vertical. (a) What is the tension in the cord? (b) What is the speed of the
41. ## PHYSICS!!
A ball on the end of a string is whirled around in a horizontal circle of radius 0.386 m. The plane of the circle is 1.10 m above the ground. The string breaks and the ball lands 2.00 m (horizontally) away from the point on the ground directly beneath the
42. ## Physics
A ball on the end of a string is whirled around in a horizontal circle of radius 0.486 m. The plane of the circle is 1.83 m above the ground. The string breaks and the ball lands 2.39 m away from the point on the ground directly beneath the ball’s
43. ## Physics
A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless
Suppose a cord is a heavy rope of mass 1.0 kg. Calculate the acceleration of each box and the tension at each end of the cord, using the free-body diagrams. Assume the cord doesn't sag. (m1 = 8.5 kg , m2 = 12.5 kg and FP = 36.0 N)
45. ## Physics
Although a mass on the end of a string can be whirled around in a horizontal circle, (plane of the mass is horizontal), is it possible for the string itself to be horizontal? Explain and justify your answer. I know that it is impossible, because it just
46. ## physics
a mass of 5 kg is whirled in a horizontal circle at one end of a string 50 cm long. the other end being fixed. if the string when hanging vertically will just support a load of 200 kg mass without breaking , find the maximum whirling speed in m/s
47. ## physics
A ball of mass 0.750 kg is attached to the end of a 1.20m long cord that can withstand a maximum tension of 50.0 N. If swung in a vertical circle, what is the maximum velocity the ball can achieve before the string breaks?
48. ## Physics
A uniform rope with length L and mass m is held at one end and whirled in a horizontal circle with angular velocity w. You can ignore the force of gravity on the rope. (a) At a point on the rope a distance r from the end that is held, what is the tension
49. ## physics
A sphere of mass 3.3 10-4 kg is suspended from a cord. A steady horizontal breeze pushes the sphere so that the cord makes an angle of 41° with the vertical when at rest (a) Find the magnitude of that push. (b) Find the tension in the cord.
50. ## Physics
a 800 gram ball at the end of a cord is whirled in an almost horizontal circle of radius 1.50 meters. its tangential speed in the circle is 10 meters per second. What must be the tension in the cord? What angle does the cord make with the horizontal?
51. ## Physics
a 800 gram ball at the end of a cord is whirled in an almost horizontal circle of radius 1.50 meters. its tangential speed in the circle is 10 meters per second. What must be the tension in the cord? What angle does the cord make with the horizontal?
52. ## Physics
a 800 gram ball at the end of a cord is whirled in an almost horizontal circle of radius 1.50 meters. its tangential speed in the circle is 10 meters per second. What must be the tension in the cord? What angle does the cord make with the horizontal?
53. ## Physics
A 0.4kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.3 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 65 N, what is the maximum rpm(revolution per minute) the ball can
54. ## Physics Circular Motion
A 0.40-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 2.0 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 75 N, what is the maximum speed the ball can have? May someone
55. ## physics
A 0.50 kg ball that is tied to the end of a 1.1 m light cord is revolved in a horizontal plane with the cord making a 30° angle, with the vertical (a) Determine the ball's speed. (b) If, instead, the ball is revolved so that its speed is 4.0 m/s, what
56. ## physics
A 0.50 kg ball that is tied to the end of a 1.2 m light cord is revolved in a horizontal plane with the cord making a 30° angle, with the vertical
57. ## physics
A 0.50 kg ball that is tied to the end of a 1.2 m light cord is revolved in a horizontal plane with the cord making a 30° angle, with the vertical
58. ## physics
A 0.50 kg ball that is tied to the end of a 1.4 m light cord is revolved in a horizontal plane with the cord making a 30° angle with the vertical. (b) If the ball is revolved so that its speed is 4.0 m/s, what angle does the cord make with the vertical?
59. ## physics
A 0.50 kg ball that is tied to the end of a 1.4 m light cord is revolved in a horizontal plane with the cord making a 30° angle with the vertical. determine the balls speed.
60. ## Kinetic energy
steel ball of mass 0.310 kg is fastened to a cord that is 86.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal, as shown in the figure. At the bottom of its path, the ball strikes a 3.10 kg steel block initially at
61. ## physics
A steel ball of mass 0.300 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal. At the bottom of its path, the ball strikes a 2.50 kg steel block initially at rest on a frictionless
62. ## PHYSICS
A metal ball tied to the end of a string 1.8 m long is whirled in a horizontal circle at 4 m/s. What is the mass of the ball if the tension on the string is 30 N?
63. ## PHYSICS
a metal ball tied to the end of a string 1.8 m long is whirled in a horizontal circle at 4 m/s. what is the mass of the ball if the tension on the string is 30 N ?
64. ## Physics
A metal ball tied to the end of a string 1.8 m long is whirled in a horizontal circle at 4 m/s. What is the mass of the ball if the tension on the string is 30 N? What formula/s should I use?
65. ## Physics
A small ball is attached to one end of a spring that has an unstrained length of 0.202 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.18 m/s. The spring remains nearly parallel to the ground
66. ## Physics
A small ball is attached to one end of a spring that has an unstrained length of 0.202 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.18 m/s. The spring remains nearly parallel to the ground
67. ## Physics
A 0.75-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.6 m on a frictionless horizontal surface.
68. ## physics homework
0.2kg mass hangs at the end of the wire. what is the tension if the ball is then whirled around in a horizontal circle with tangential velocity of 5m/s
69. ## physics
32.6 gram mass is whirled on a cord 131 cm long at a speed of 6 m/s. What is the centripetal acceleration of the mass in MKS units ?
70. ## Physics
A ball on the end of a string is whirled around in a horizontal circle of radius 0.300m. The plane of the circle is 1.19m above the ground. The string breaks and the ball lands 1.91m (horizontally) away from the point on the ground directly beneath the
71. ## physics
A mass M of 2.95 kg is attached to the end of a string whose length is 0.420 m. The mass slides without friction on a horizontal surface as indicated in the diagram. If the string can withstand a maximum tension of 127.5 N, what is the maximum tangential
72. ## Physics
A ball on the end of a string is whirled around in a horizontal circle of radius 0.486 m. The plane of the circle is 1.83 m above the ground. The string breaks and the ball lands 2.39 m away from the point on the ground directly beneath the ball’s
73. ## Physics
A ball on the end of a string is whirled around in a horizontal circle of radius 0.33 m. The plane of the circle is 1.21 m above the ground. The string breaks and the ball lands 2.22 m away from the point on the ground directly beneath the ball’s
74. ## physics
A ball on the end of a string is whirled around in a horizontal circle of radius 0.33 m. The plane of the circle is 1.26 m above the ground. The string breaks and the ball lands 2.4 m away from the point on the ground directly beneath the ball’s location
75. ## physics
A ball on the end of a string is whirled around in a horizontal circle of radius 0.467 m. The plane of the circle is 1.24 m above the ground. The string breaks and the ball lands 2.49 m away from the point on the ground directly beneath the ball’s
76. ## physics
A ball on the end of a string is whirled around in a horizontal circle of radius 0.22 m. The plane of the circle is 1.69 m above the ground. The string breaks and the ball lands 2.68 m away from the point on the ground directly beneath the ball’s
77. ## Physics
A phone cord is 2.85 m long. The cord has a mass of 0.254 kg. A transverse wave pulse is produced by plucking one end of the taut cord. The pulse makes four trips down and back along the cord in 0.758 s. What is the tension in the cord?
78. ## physics
A phone cord is 4.75 m long. The cord has a mass of 0.200 kg. A transverse wave pulse is produced by plucking one end of the taut cord. The pulse makes four trips down and back along the cord in 0.900 s. What is the tension in the cord?
79. ## physics
A phone cord is 3.55 m long. The cord has a mass of 0.200 kg. A transverse wave pulse is produced by plucking one end of the taut cord. The pulse makes four trips down and back along the cord in 0.775 s. What is the tension in the cord?
80. ## physics
A phone cord is 4.75 m long. The cord has a mass of 0.200 kg. A transverse wave pulse is produced by plucking one end of the taut cord. The pulse makes four trips down and back along the cord in 0.900 s. What is the tension in the cord?
81. ## physics
A phone cord is 3.55 m long. The cord has a mass of 0.200 kg. A transverse wave pulse is produced by plucking one end of the taut cord. The pulse makes four trips down and back along the cord in 0.775 s. What is the tension in the cord?
82. ## Physics
A 0.2kg ball is attached to a rope and is whirled in a horizontal circle. Calculate the maximum speed of the ball if the length of the rope is 0.8m and its breaking strength is 0.9N
83. ## physics
I have put a 100 gram weight ball in a plastic bottle and weight of this bottle is 20 gram .Now I shake this bottle in horizontal direction towards 5 centimeter back and forth with the applying force of 1 newton .So what will be the kinetic energy of this
84. ## physics
Tetherball is a game played by kids. The equipment consists of a volleyball on a string, with the other end of the string tied to the top of a post. Kids hit the ball back and forth around the post. Consider a volleyball of total mass 200 g attached to the
85. ## physic
A telephone cord is 6.20 m long and has a mass of 0.220 kg. A transverse pulse is produced by plucking one end of the taut cord. The pulse makes four trips down and back along the cord in 0.830 s. What is the tension in the cord?
86. ## PHYSICS
A wheel in the shape of a uniform disk of radius R and mass mp is mounted on a frictionless horizontal axis. The wheel has moment of inertia about the center of mass Icm=(1/2)mpR2 . A massless cord is wrapped around the wheel and one end of the cord is
87. ## Physics Classical Mechanics`
A wheel in the shape of a uniform disk of radius R and mass mp is mounted on a frictionless horizontal axis. The wheel has moment of inertia about the center of mass Icm=(1/2)mpR2 . A massless cord is wrapped around the wheel and one end of the cord is
88. ## phy111
A ball of mass 50 g tied to the end of a 50 cm inextensible string is whirled around in a vertical circle. Find the tension in the string when the ball is at the top of the circle. Take g=10m s2
89. ## physics
Tetherball is a game played by kids. The equipment consists of a volleyball on a string, with the other end of the string tied to the top of a post. Kids hit the ball back and forth around the post. Consider a volleyball of total mass 200 g attached to the
90. ## Physics
A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless
91. ## physics
A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless
92. ## physics
An object of mass 0.50 kg on the end of a string is whirled round in a horizontal circle of radius 2.0m with a constant speed of 10m/s.Find its angular velocity and tension in the string( string is horizontal)answer should be 25.5 N
93. ## physics
A non-uniform bar of weight W = 20.0 N is suspended at rest in a horizontal position by two massless cords. The angle phi between cord 2 and the vertical is equal to 32.5o. The bar has a length L = 6.5 m and the distance of its center of gravity from the
94. ## physics
a 10 kg mass is attached to one end of a string with a lenght of 5m and whirled in a horizontal circle overheadd at a rate of 1 revolution per second what is tension in the sring
95. ## physics
A ball of mass 21.7 g is attached to a cord of length 0.61 m and rotates in a vertical circle. What is the minimum speed the ball must have at the top of the circle so the cord does not become slack? The acceleration of gravity is 9.8 m/s . Answer in units
96. ## Physics
a small body of mass m revolving in a horizontal circle with velocity V of a constant magnitude at the end of a cord of length L. As the body swings around its path, the cord sweeps over the surface of a cone. The cord makes an angle è with the vertical,
97. ## physics 1403
10kg mass is attached to a 5kg mass by a 20cm cord. The 5 kg mass is attached by a 30cm cord to a motor that rotates the entire mass/cord system 5rev/s in the horizontal plane. What is the tension in each cord?
98. ## Physics
A 275-gram ball bounces off a wall. The ball stays in a horizontal plane throughout the motion. The ball strikes the wall with a speed of 4.50 m/s at an angle of 𝜃 = 35.0o relative to the wall and rebounds with the same speed at the same angle. If the
99. ## physics
The figure shows a stationary arrangement of two crayon boxes and three cords. Box A has a mass of 12.6 kg and is on a ramp at angle θ = 28.9o; box B has a mass of 6.06 kg and hangs on a cord. The cord connected to box A is parallel to the ramp, which is
100. ## physics
A sphere of mass 3.3 10-4 kg is suspended from a cord. A steady horizontal breeze pushes the sphere so that the cord makes an angle of 41° with the vertical when at rest (a) Find the magnitude of that push. (b) Find the tension in the cord. | 6,655 | 25,445 | {"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} | 3.53125 | 4 | CC-MAIN-2021-25 | latest | en | 0.86974 |
https://www.coursehero.com/file/6309257/Excel-15/ | 1,513,132,799,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948520218.49/warc/CC-MAIN-20171213011024-20171213031024-00641.warc.gz | 755,294,526 | 46,185 | # Excel-15 - Exercise 15 5 Manufacturing Costs 1 and 2 Direct...
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Exercise 15 - 5 Manufacturing Costs 1. and 2. Direct Materials: Oak wood \$50,000 Furniture parts (wheels, locks, etc.) 5,500 Total \$55,500 Direct labor: Payroll - production line employees' wages \$45,500 Total 45,500 Manufacturing overhead: Miscellaneous supplies (glue, saw blades, varnish, etc.) \$10,000 Payroll - plant manager's salary 25,000 Building depreciation (0.75 * \$28,000) 21,000 Maintenance - plant and equipment 5,000 Utilities (0.75 * \$16,000) 12,000 Total 73,000 Period costs: Payroll - administrative salaries \$100,000 Building depreciation (0.25 * \$28,000) 7,000 Utilities (0.25 * \$16,000) 4,000 Income taxes 8,500 Total 119,500 Total costs for the year \$293,500
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Exercise 15 - 11 C-V-P Analysis 1. \$12,000,000 fixed costs / ( \$80 ticket price - \$1 ticket production cost) = 151,899 tickets (rounded up) 2. \$80,000,000 fixed costs / ( \$80 ticket price - \$1 ticket production cost) = 1,012,659 tickets (rounded up)
Exercise 15 - 15 Product Costing and C-V-P Analysis 1. 2 hours * \$120 =
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Excel-15 - Exercise 15 5 Manufacturing Costs 1 and 2 Direct...
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Ask a homework question - tutors are online | 464 | 1,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-51 | latest | en | 0.786662 |
https://www.teachtasticiep.com/iep-goal/iep-goal-to-solve-subtraction-word-problems-up-to-two-digits | 1,713,914,764,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818835.29/warc/CC-MAIN-20240423223805-20240424013805-00150.warc.gz | 895,996,734 | 164,166 | top of page
Learning Standard
### 2.OA.A.1
Use addition and subtraction within 100 to solve one- and two-step word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.
Target IEP Goal
By (date), when given exercises with subtraction, two digits, the student will solve subtraction word problems (up to two digits), improving operations and algebraic thinking skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
Teach Tastic IEP goals written to be SMART: Specific, Measurable, Attainable, Results-Oriented, and Time-Bound.
IEP Goal Objectives
1
Subtract Two-Digit Numbers - With Regrouping
By (date), when given exercises with subtraction, two digits, the student will subtract two-digit numbers - with regrouping, improving number and operations in base ten skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
2
Subtract Two-Digit Numbers With Regrouping (using Models)
By (date), when given exercises with subtraction strategies, two digits, the student will subtract two-digit numbers with regrouping (using models), improving number and operations in base ten skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
3
Subtract a One-Digit Number from a Two-Digit Number (using models, With Regrouping)
By (date), when given exercises with subtraction strategies, two digits, the student will subtract a one-digit number from a two-digit number (using models, with regrouping), improving number and operations in base ten skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
4
Use Models to Subtract a One-Digit Number from a Two-Digit Number Without Regrouping
By (date), when given exercises with subtraction strategies, two digits, the student will use models to subtract a one-digit number from a two-digit number without regrouping, improving number and operations in base ten skills from 0/10 work samples out of ten consecutive trials to 8/10 work samples in ten consecutive trials.
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codingsimplifylife.blogspot.com | 1,632,854,584,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060882.17/warc/CC-MAIN-20210928184203-20210928214203-00668.warc.gz | 230,748,817 | 17,339 | ## 2016年3月12日 星期六
### [Codeforces 631E]Product Sum
Definitions:
$SUM[i]=\sum_{k=1}^{i}A_{k}$
For each element, $A_{i}$, how does the characteristic of the array change if we move $A_{i}$ to position $j$?
It changes $A_{i}(j-i)+SUM[i]-SUM[j]$!
Enumerate $i$ and $j$, we get an $O(N^{2})$ solution.
How to speed up?
Treat $j$ as coordinate $x$.
Then we can treat $SUM[j]$ as a function $f(x)$, $A_{i}*j+SUM[i]-A_{i}*i$ as a line $ax+b$ ($a=A_{i},b=SUM[i]-A_{i}*i$)
What we should find is $\max\{(ax+b)-f(x)\}$, where $f(x)$ is a fixed function, $ax+b$ is a line depend on which $A_{i}$ we choose.
Now we have a new way to enumerate: $x$
Enumerate $x$, and find the line whose $ax+b$ is maximum.
Actually, we can use $deque$ to maintain the $convex\ hull$ of these lines.
For each $x$, we just find the corresponding point on the $convex\ hull$, $(x,y)$, $y$ is what we want to get (maximum $ax+b$).
On this picture, black straight lines are $ax+b$ we want to pick to form the $convex\ hull$, the green one.
How to build the $convex\ null$?
Well, this is a long story... Learn it!
(keywords: $monotone\ queue$,$slope\ optimization$)
(hints: sort the lines by their slopes, $a$)
By enumerating $x$ in increasing order, we can reach linear time complexity with the help of monotonicity of $x$, get the maximum $ax+b$ for every $x$.
Update maximum $ax+b-SUM[x]$, this is how much we can increase the characteristic of the array.
Initial characteristic of the array is $\sum_{i=1}^{N}A_{i}*i$
Time complexity：$O(N\log N)$ (remember that $sorting$ was required)
p.s. Actually, we can avoid $sorting$ to reach $O(N)$, that is, only allow element move toward one direction, then these $ax+b$ become rays (instead of lines), enumerate $x$ in that directions, and dynamic maintain the $convex\ hull$. Please remenber that it's required to do the same thing on another direction.
p.s. For $O(N)$ implementation, you can see this.
code:
#include<cstdio>
#include<cassert>
#include<algorithm>
using namespace std;
typedef long long LL;
void getmax(LL &a,const LL &b){if(b>a)a=b;}
struct Deq
{
int DATA[200000],L,R;
void Clear(){L=0,R=-1;}
void PushBack(const int v){DATA[++R]=v;}
void PopBack(){R--;}
void PopFront(){L++;}
int Front(const int i)const{return DATA[L+i];}
int Back(const int i)const{return DATA[R-i];}
int Size()const{return R-L+1;}
}DEQ;
int N;
LL A[200001],SUM[200001];
LL GetA(const int i){return A[i];}
LL GetB(const int i){return SUM[i]-A[i]*i;}
double GetX(const int i1,const int i2){return (double)(GetB(i1)-GetB(i2))/(GetA(i2)-GetA(i1));}
bool NeedPopBack(const int i)
{
if(DEQ.Size()==0)return false;
if(GetA(i)==GetA(DEQ.Back(0)))return GetB(i)>=GetB(DEQ.Back(0));
if(DEQ.Size()==1)return false;
return GetX(DEQ.Back(0),i)<=GetX(DEQ.Back(1),DEQ.Back(0));
}
void BuildConvexHull()
{
vector<int>order;
for(int i=1;i<=N;i++)order.push_back(i);
sort(order.begin(),order.end(),[](const int a,const int b)->bool{return A[a]<A[b];});
DEQ.Clear();
for(const int i:order)
{
while(NeedPopBack(i))DEQ.PopBack();
DEQ.PushBack(i);
}
}
LL Solve()
{
LL change=0LL;
for(int i=0;i<=N;i++)
{
while(DEQ.Size()>=2&&GetX(DEQ.Front(0),DEQ.Front(1))<=i)DEQ.PopFront();
// printf("i=%d,DEQ.Front()=%d,a=%I64d,b=%I64d\n",i,DEQ.Front(0),GetA(DEQ.Front(0)),GetB(DEQ.Front(0)));
getmax(change,GetA(DEQ.Front(0))*i+GetB(DEQ.Front(0))-SUM[i]);
}
LL initial_cost=0LL;
for(int i=1;i<=N;i++)initial_cost+=A[i]*i;
// printf("initial_cost=%I64d,change=%I64d\n",initial_cost,change);
return initial_cost+change;
}
int main()
{
// freopen("in.txt","r",stdin);
while(scanf("%d",&N)==1)
{
SUM[0]=0LL;
for(int i=1;i<=N;i++)scanf("%I64d",&A[i]),SUM[i]=SUM[i-1]+A[i];
BuildConvexHull();
printf("%I64d\n",Solve());
}
return 0;
} | 1,256 | 3,720 | {"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} | 3.875 | 4 | CC-MAIN-2021-39 | latest | en | 0.712561 |
http://www.r-bloggers.com/le-monde-puzzle-825/ | 1,469,535,218,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824853.47/warc/CC-MAIN-20160723071024-00240-ip-10-185-27-174.ec2.internal.warc.gz | 663,038,563 | 17,213 | # Le Monde puzzle [#825]
June 18, 2013
By
(This article was first published on Xi'an's Og » R, and kindly contributed to R-bloggers)
Yet another puzzle which first part does not require R programming, even though it is a programming question in essence:
Given five real numbers x1,…,x5, what is the minimal number of pairwise comparisons needed to rank them? Given 33 real numbers, what is the minimal number of pairwise comparisons required to find the three largest ones?
I do not see a way out of considering the first question as the quickest possible sorting of a real sample. Using either quicksort or heapsort, I achieve sorting the 5 numbers in exactly 6 comparisons for any order of the initial sample. (Now, there may be an even faster way based on comparing partial sums first… I just do not see how!) Update: Oops! I realised I made my reasoning based on a reasonable case, the correct answer is indeed 7!!!
For the second part, let us start from the remark that 32 comparisons are needed to find the largest number, then at most 31 for the second largest, and at most 30 for the third largest (since we can take advantage of the partial ordering resulting from the determination of the largest number). This is poor. If I instead use a heap algorithm, I need O(n log{n}) comparisons to build this binary tree whose parents are always larger than their siblings, as in the above example. (I can produce a sort of heap structure, although non-binary, in an average 16×2.5=40 steps. And a maximum 16×3=48 steps.) The resulting tree provides the largest number (100 in the above example) and at least the second largest number (36 in the above). To get the third largest number, I first need a comparison between the one-before-last terms of the heap (19 vs. 36 in the above), and one or two extra comparisons (25 vs. 19 and maybe 25 vs. 1 in the above). (This would induce an average 1.5 extra comparison and a maximum 2 extra comparisons, resulting in a total of 41.5 average and 49.5 maximum comparisons with my sub-optimal heap construction.) Once again, using comparisons of sums may help in speeding up the process, for instance comparing numbers by groups of 3, but I did not pursue this solution…
If instead I try to adapt quicksort to this problem, I can have a dynamic pivot that always keep at most two terms above it, providing the three numbers as a finale result. Here is an R code to check its performances:
```quick3=function(x){
comp=0
i=1
lower=upper=NULL
pivot=x[1]
for (i in 2:length(x)){
if (x[i]<pivot){ lower=c(lower,x[i])
}else{
upper=c(upper,x[i])
if (length(upper)>1) comp=comp+1}
comp=comp+1
if (length(upper)==3){
pivot=min(upper)
upper=sort(upper)[-1]
}}
if (length(upper)<3) upper=c(pivot,upper)
list(top=upper,cost=comp)
}
```
When running this R code on 10⁴ random sequences of 33 terms, I obtained the following statistics, I obtained the following statistics on the computing costs
```</p>
<p style="text-align: justify;">> summary(costs)
Min. 1st Qu. Median Mean 3rd Qu. Max.
32.00 36.00 38.00 37.89 40.00 49.00
```
and the associated histogram represented above. Interestingly, the minimum is the number of comparisons needed to produce the maximum!
Reading the solution in Le Monde in the train to London and Bayes 250, I discovered that the author suggests a 7 comparison solution in the first case [compare A and B, C and D = 2 comparisons; if A>B and C>D, compare A and C and get, say A>C>D = 1 comparison; insert E in this ordering by comparing with C and then A or D = 2 comparisons, obtaining e.g. A>E>C>D; conclude by inserting B by first comparing with C then with D or E = 2 comparisons] and a 41 comparison solution in the second case. He is (or I am!) I was clearly mistaken in the first case while the quick3 algorithm does 41 or less most of the time (90%) but not always.
Filed under: Books, Kids, pictures, R Tagged: heapsort, Le Monde, mathematical puzzle, pracma, quicksort, R
To leave a comment for the author, please follow the link and comment on their blog: Xi'an's Og » R.
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##### objective Ques (148 results)
41)
A can complete a certain work in 30 days. He started the work. After 4 days, B joined him and the whole work was completed in 10 days from the beginning. B alone can complete one-third of the same work in:
SSC CHSL 2021
A)
6 days
B)
3 days
C)
2 days
D)
4 days
42)
A can complete a work in 20 days, while B can complete the same work in 25 days. Both worked together for 10 days and then C alone completed the remaining work in 10 days. In how many days will A, B and C together complete the same work?
SSC CHSL 2021
A)
12 days
B)
8 days
C)
5 days
D)
10 days
43)
A can do 60% of a work in 18 days, while B can do 40% of the same work in 10 days. Both work together for 10 days only. C completed the remaining work in 4 days. In how many days will C complete the same work?
SSC CHSL 2021
A)
18
B)
25
C)
15
D)
12
44)
Sachin alone can complete a piece of work for Rs. 8,500 in $$8\frac{1}{2}$$ days. But with the help of Vishnu, the work is completed in 6 days. The share to be paid to Vishnu is:
SSC CHSL 2021
A)
Rs. 2,400
B)
Rs. 2,000
C)
Rs. 2,500
D)
Rs. 3,200
45)
Geeta can assemble a toy in 10 minutes, whereas Sudha can assemble the same toy in 15 minutes. If they work together, how much time will they take to assemble 60 toys?
SSC CHSL 2021
A)
7 hours 30 minutes
B)
7 hours
C)
5 hours
D)
6 hours
46)
A can do a piece of work in 12 days. He worked for 6 days and left, and then B finished it in 10 days. If both work together, then in how many days will they finish the same work?
SSC CHSL 2021
A)
7
B)
9
C)
9.5
D)
7.5
47)
Suman and Lata working together can complete a task in 8 days. If Lata can complete the same task in 12 days, then how many days will Suman take to complete the same task?
SSC CHSL 2021
A)
15
B)
24
C)
10
D)
18
48)
40 persons take 6 days to complete a certain task, working 10 hours a day. How many hours a day will be sufficient for 30 persons to complete the same task in 10 days?
SSC CHSL 2021
A)
8
B)
10
C)
9
D)
6
49)
A can complete a task in 6 days, while B can complete the same task in 12 days. How many days will they take to complete the same task if they work together?
SSC CHSL 2021
A)
3
B)
5
C)
$$4\frac{1}{2}$$
D)
4
50)
A can complete one-sixth of a work in 5 days and B can complete one-fourth of the same work in 15 days. In how many days, will both working together complete the work?
SSC CHSL 2021
A)
25 days
B)
20 days
C)
10 days
D)
12 days
showing 41 - 50 results of 148 results
Full Syllabus in 1 Video [Contains Everything basic to advance] | 844 | 2,693 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-18 | latest | en | 0.876036 |
http://www.jiskha.com/display.cgi?id=1408493288 | 1,498,294,417,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320243.11/warc/CC-MAIN-20170624082900-20170624102900-00519.warc.gz | 587,708,194 | 3,993 | # math
posted by .
58. Let GCF (x,y) represent the greatest common factor of x any y. If p is a positive even integer less than 11, for what value of p does GCF (p^2,81) have the greatest value?
• math -
I'd make that 6, since 81 = 3^4.
p^2 must have some powers of 3 in it, else GCF(p^2,81) = 1.
GCF(36,81) = 9
9 would be better, but we needed an even integer.
• math -
Is it 4?
• math -
hmmm. Is it 4?
Is 4^2 = 36?
Does 4 have 3 as a factor?
Better read what I wrote, and this time pay attention. | 176 | 508 | {"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-2017-26 | latest | en | 0.949821 |
http://hvac-talk.com/vbb/printthread.php?t=73692&pp=13&page=1 | 1,508,263,106,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822145.14/warc/CC-MAIN-20171017163022-20171017183022-00653.warc.gz | 168,061,437 | 3,499 | # Milli Volts
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• 04-03-2005, 05:39 PM
tonymjr
What would you set your multi meter to to check for 750mv with no actual mv scale?
• 04-03-2005, 05:45 PM
Wild Leg
milli = thousandths of something.
There are 1000 millivolts in one volt.
• 04-03-2005, 05:49 PM
tonymjr
ac and dc scales
• 04-03-2005, 05:52 PM
Wild Leg
Which do you want to measure?
ac or dc?
• 04-03-2005, 05:53 PM
tonymjr
well it's a thermopile for a gas boiler
• 04-03-2005, 05:56 PM
Wild Leg
That narrows it down.
A thermopile is a big thermocouple.
Have you tried your meter on it yet?
• 04-03-2005, 05:59 PM
tonymjr
so what does a thermocouple generate ac or dc? I always thought dc
• 04-03-2005, 06:01 PM
Wild Leg
That's a good question, and the one you could have started this thread with.
Another name for a thermopile is a pilot generator.
Generators produce DC.
• 04-03-2005, 08:29 PM
thecomfortzone44
Yes and you will be darned lucky to get 750 mv off that power pile.
• 04-03-2005, 08:35 PM
tinknocker service tech
it is not worth it to check
replace it and clean the pilot burner at the same time
• 04-03-2005, 08:42 PM
me75006
0.00075v = 750uv microvolts
0.75v = 750mv millivolts
1000v = 1kv kilovolts
• 04-03-2005, 10:32 PM
ct2
BWAL2:lol@
0.00075v = 750uv microvolts
0.75v = 750mv millivolts
1000v = 1kv kilovolts
Hey maybee you could use that in class sometime.
Bwal2 is one of the most respected educators on this site
• 04-03-2005, 11:26 PM
Wild Leg
Quote:
Originally posted by ct2
Bwal2 is one of the most respected educators on this site
Now THAT's a first!
Careful, ct2, you'll give me a swelled head. :D
( I do use that sort of thing in class. Last I heard, math was a necessary skill.)
My motivation for posting & reading on this forum is the same as everybody else's.
I learn something EVERY time I log on, because I don't know it all.
[Edited by bwal2 on 04-03-2005 at 11:43 PM]
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last | 705 | 2,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-43 | latest | en | 0.842817 |
http://faculty.kfupm.edu.sa/ES/oncel/StudyQuestionsforMidTermofGeol581.htm | 1,544,757,538,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825349.51/warc/CC-MAIN-20181214022947-20181214044447-00594.warc.gz | 91,890,361 | 2,543 | MID-TERM STUDY QUESTIONS FOR GEOL581
Define
• wavelength of a wave
• frequency of a wave
• period of a wave
• amplitude of a wave
Give a formula which relates wavelength and frequency.
Distinguish between a seismograph and a seismogram.
Explain in very simple terms how a seismograph works.
What are the 4 principle types of waves produced by an earthquake?
Which of these are body waves?
Which are surface waves?
Which waves are the fastest? Gives equations for the velocities of P-waves and S-waves.
What is the most important factor in determining velocity in rocks?
What are SV and SH waves?
Why can't S-waves travel through liquid (give a mathematical proof)?
Show mathematically why P-waves are faster than S-waves.
Describe the motion of a particle of rock as each of the four principle seismic waves passes.
What is the effect of depth of focus on the production of surface waves?
What happens when a seismic wave meets a surface of discontinuity within the Earth?
What is the relation between the angle of reflection and the angle of incidence?
Give Snell's Law.
What happens to a wave when it meets a surface of discontinuity at the critical angle?
Be able to trace the paths of earthquake waves such as PS, PcS, PKP, PKIKP, pPcP, etc.
List some reflected and/or refracted waves which might be mistaken on seismograms for surface waves and for aftershocks.
Explain how the epicenter of an earthquake can be located.
How do we know that the upper mantle is partially molten?
At what depths do major discontinuities exist within the Earth? What are the names of these discontinuities?
Name 2 minor discontinuities in the Earth.
What are the two methods commonly used in seismic prospecting?
Which is most often used? Which gives the most information?
What sources of energy are most often used in seismic exploration?
Describe in detail how refraction profiles would be produced, and what information could be obtained from them.
What kind of time/distance graph would be produced where velocity increases continuously with depth?
What kind of graph would be produced where velocity first increases continuously with depth and then abruptly changes?
What kind of graph would be obtained if a seismic profile was taken perpendicular to the strike of a vertical fault?
How can the throw of the fault be determined?
List 3 factors which help determine seismic wave velocity.
How does one recognize a reflecting horizon on seismograms?
How can the depth to a reflecting horizon be determined?
How can the dip of a reflecting horizon be determined?
What is seismic/earthquake tomography? What information can it provide?
List and discuss possible geophysical properties which may be indicators of forthcoming earthquakes.
Hooke's Law.
• Define
• stress
• strain
• Young's modulus
• Poisson's ratio
• bulk modulus
• compressibility
• modulus of rigidity
• shear modulus
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updated October 30-2004 | 632 | 2,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-51 | latest | en | 0.942554 |
https://www.jiskha.com/display.cgi?id=1259123199 | 1,516,181,128,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886860.29/warc/CC-MAIN-20180117082758-20180117102758-00219.warc.gz | 889,438,646 | 3,889 | # math
posted by .
how do you use the property of one to write 3 fractions equivalent to 5/9
• math -
(5/9)(4/4) =
(5/9)(2/2) =
(5/9)(765/765) =
• math -
15424 idk
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More Similar Questions | 579 | 2,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-05 | latest | en | 0.907332 |
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Test: Light, Sound and Force - 2
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10 Questions MCQ Test Science for Class 3 | Test: Light, Sound and Force - 2
Test: Light, Sound and Force - 2 for Class 3 2022 is part of Science for Class 3 preparation. The Test: Light, Sound and Force - 2 questions and answers have been prepared according to the Class 3 exam syllabus.The Test: Light, Sound and Force - 2 MCQs are made for Class 3 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Light, Sound and Force - 2 below.
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Test: Light, Sound and Force - 2 - Question 1
Roy wants to make a list of actions that show friction produces heat. Which of the following could he include in his list? (a) Rubbing hands together (b) Scratching a match stick against the box (c) Sawing wood (d) Drilling a hole into metal sheet (e) Rubbing off some pencil marks with an eraser
Detailed Solution for Test: Light, Sound and Force - 2 - Question 1
As all these actions involve two surfaces rubbing against each other, thus force of friction comes into action which produces heat.
Test: Light, Sound and Force - 2 - Question 2
Select the forces that the women X and Y in the diagram need to overcome to carry their groceries.
Detailed Solution for Test: Light, Sound and Force - 2 - Question 2
Woman 'X' needs to overcome friction to slide the trolley smoothly. Woman 'Y' needs to hold up the shopping bags against gravity,
Test: Light, Sound and Force - 2 - Question 3
Nidhi was asked to draw the shadow of a tumbler formed on the screen behind it, when it is kept in different manners at the same position and light is shone on it from the same direction. She drew three shadows as shown here. Read the following statements regarding it and select the INCORRECT ones. (i) Shadow 1 is formed when the tumbler is kept horizontally. (ii) Shadow 2 is formed when the tumbler is kept vertically upright. (iii) Shadow 3 is incorrect because the given tumbler cannot cast a rectangular shadow.
Detailed Solution for Test: Light, Sound and Force - 2 - Question 3
A tumbler or any object may cast differently shaped shadows depending upon the angle of light source and the manner in which the object is kept.
Test: Light, Sound and Force - 2 - Question 4
Q can hear the sound made by P in the shortest time, when the medium is a _______.
Detailed Solution for Test: Light, Sound and Force - 2 - Question 4
Sound waves travel fastest in solid medium.
Test: Light, Sound and Force - 2 - Question 5
Refer to the given figure and select the correct option.
Detailed Solution for Test: Light, Sound and Force - 2 - Question 5
When a pushing force is applied, the object moves farther from the source of force. In case of opposing forces the object moves towards the weaker force. In case both the forces are equal, object shows no movement at all.
Test: Light, Sound and Force - 2 - Question 6
A torch is shone on a wooden prism and a beach ball as shown below.
What will be the shape of shadow observed on the screen?
Test: Light, Sound and Force - 2 - Question 7
The above figures show the force required to move a 100 g object over different surfaces. The friction is greatest at ____.
Detailed Solution for Test: Light, Sound and Force - 2 - Question 7
Friction is a force or resistance offered by a surface to the movement of object in contact. As it is shown, that on surface W, the force required to move an object is maximum, this shows that the opposing force of friction is provided maximum by surface W.
Test: Light, Sound and Force - 2 - Question 8
Sandhya made a musical instrument as shown in the figure using large wooden box, a broom handle and a length of string.
She tightened the string by pulling harder on the broom handle. What kind of sound will she hear now?
Test: Light, Sound and Force - 2 - Question 9
The girl a is putting more efforts to pull the sack bag as compared to the girl b. Girl a has to work more against a force which also _______.
Detailed Solution for Test: Light, Sound and Force - 2 - Question 9
Wheel cart reduces friction between sack bag and floor making it easier to pull the bag. Friction is present between a surface and object put on it and resists its motion. Friction is the force which prevents us from slipping when we walk.
Test: Light, Sound and Force - 2 - Question 10
Ashish shone a torch at a ball. The shadow of the ball is projected onto a screen as shown.
What could Ashish do to get a smaller shadow of the ball on the screen?
1. Move the ball closer to the screen.
2. Move the torch further away from the ball.
3. Move the screen further away from the ball.
Detailed Solution for Test: Light, Sound and Force - 2 - Question 10
By increasing the distance between light source and object the size of the shadow can be reduced. The distance between them can be increased either by moving ball closer to the screen or by moving torch further away from the ball.
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16 videos|43 docs|30 tests | 1,450 | 6,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-49 | latest | en | 0.925017 |
http://www.unitconverters.net/length/centimeter-to-rod-us-survey.htm | 1,487,858,709,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171171.24/warc/CC-MAIN-20170219104611-00137-ip-10-171-10-108.ec2.internal.warc.gz | 647,883,382 | 3,798 | Home / Length Conversion / Convert Centimeter to Rod (US Survey)
# Convert Centimeter to Rod (US Survey)
Please provide values below to convert centimeter [cm] to rod (US survey) [rd], or vice versa.
From: centimeter To: rod (US survey)
### Centimeter to Rod (US Survey) Conversion Table
Centimeter [cm]Rod (US Survey) [rd]
0.01 cm1.9883838383838E-5 rd
0.1 cm0.00019883838383838 rd
1 cm0.0019883838383838 rd
2 cm0.0039767676767677 rd
3 cm0.0059651515151515 rd
5 cm0.0099419191919192 rd
10 cm0.019883838383838 rd
20 cm0.039767676767677 rd
50 cm0.099419191919192 rd
100 cm0.19883838383838 rd
1000 cm1.9883838383838 rd
### How to Convert Centimeter to Rod (US Survey)
1 cm = 0.0019883838383838 rd
1 rd = 502.92100584201 cm
Example: convert 15 cm to rd:
15 cm = 15 × 0.0019883838383838 rd = 0.029825757575758 rd | 292 | 816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-09 | latest | en | 0.298679 |
http://www.programmingwithbasics.com/2015/03/multiplication-of-two-no-without-using.html | 1,516,587,136,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890947.53/warc/CC-MAIN-20180122014544-20180122034544-00087.warc.gz | 544,210,947 | 23,707 | # C++ Program To Multiply Two Number Without Using Asterisk
Problem:- Write a C++ program to find multiplication of two number without using Asterisk ( * ) Symbol.
Logic:- There are two methods we can use for solving this problem, this is a very important question for interview purpose because we never think about how we can multiply two number without using Asterisk Sign( * ), we just take two value and use the Asterisk Sign( * ) and put between two variables. We will take an example and through an example, we can try to solve the problem and find out the number of the way we can solve the problem.
For solving a mathematical equation there is a VEDIC Math( Old Indian Math ), By the help of VEDIC Math, we can solve large equation is second for an example if you want to check a Number is divisible by 11 or not How you can check in a second and remember number may be very large. For an example, we take a number 161051 is divisible by 11 or not. you can check the demo of VEDIC math, C Program To Check Number Is Divisible By 11 Or Not Using (VEDIC MATH).
Now come to the problem there is two way, So let's take an example suppose we want to multiply 20 and 10 there are two methods you can follow.
You can direct multiple 20 * 10 = 200 this is a very famous Traditional way for calculating or solving multiplication but here one problem occurring in problem statement specially mention that you can not use an Asterisk Sign(*) So you have to Follow another method.
Method 2:-
So here we go we know that multiplication is an addition of one number to another number of (First / Second) number of times. Don't get you let's take an example we want to multiply 20 * 10 so we can add 20 into 10 times like 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 = 200 or we can add 10 into 20 times answer will be same like 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 = 200. Here i am using method 2.
Solution:-
#include<iostream>
using namespace std;
int main ()
{
long a,b,i,temp=0;
cout<<"Enter Two numbers for multiplication : \n";
cin>>a>>b;
for(i=1;i<=b;i++)
{
temp=temp+a;
}
cout<<endl<<"Result are : "<<temp<<endl;
}
Output:- | 580 | 2,198 | {"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} | 3.671875 | 4 | CC-MAIN-2018-05 | latest | en | 0.866942 |
https://www.fodors.com/community/europe/i-need-the-metric-system-for-dummies-please-help-me-spanish-or-catalan-672560/ | 1,558,333,568,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255773.51/warc/CC-MAIN-20190520061847-20190520083847-00521.warc.gz | 781,380,815 | 26,010 | > >
>
Jan 22nd, 2007, 06:04 AM
#1
Original Poster
Join Date: Nov 2003
Posts: 4,874
I hate to confess this, but we didn't order meat or cheese from the deli in grocery stores in Italy because we didn't know how much we needed in kilograms.
I found an on-line conversion chart, so now I know that a pound of meat or cheese is .4535 kg, a half pound is .226 and a quarter pound is .113. Darn...I was hoping it would go the other way, so I could just hold up two fingers or something when ordering.
I'd love to be able to sample cured meats in Barcelona, but my Spanish is bad, and my Catalan is non-existent. Are there some common words to use when ordering, e.g. a half or a quarter kilo, or am I better off just writing down the amount I need?
Jan 22nd, 2007, 06:17 AM
#2
Join Date: Jun 2006
Posts: 17,549
You might be better off if you expressed those less-than-one-Kilo weights in terms of GRAMS (such as that 113 which equals about a 1/4 pound).
1000 Grams = one Kilo = 2.2 pounds
Jan 22nd, 2007, 06:22 AM
#3
Join Date: Jun 2004
Posts: 12,492
you can also order by the slice (loncha).
dos lonchas gordas, tres lonchas FINAS.. and they weigh it.
that is what i do when calculating exact sandwiches for a party.
gordas/finas/muy finas, muy gordas
thick/thin
Jan 22nd, 2007, 06:24 AM
#4
Join Date: Jan 2003
Posts: 2,568
If you can remember that 500kg is a bit more than a pound, you should be able to go from there. I've always used that as my guide and have had no trouble -- you know the numbers 1-10 and 100 in the local language so you are all set.
Jan 22nd, 2007, 06:34 AM
#5
Join Date: Jun 2006
Posts: 17,549
500KG is a LOT more than a pound...you must mean 500 GRAMS
Jan 22nd, 2007, 06:39 AM
#6
Join Date: Jan 2003
Posts: 4,326
Or just go to a supermarket where meat and cheese are sold pre-packed!
Jan 22nd, 2007, 06:44 AM
#7
Original Poster
Join Date: Nov 2003
Posts: 4,874
Thanks, everyone. Alec, that's what we did in Italy and that's what I want to avoid in Spain. Our choices were quite limited because we were metric impaired.
Jan 22nd, 2007, 06:48 AM
#8
Join Date: Jun 2004
Posts: 19,000
Which begs the question - are you metric impaired, or simply math challenged? If the latter, then I suggest investing \$2 in a pocket calculator. Maybe one with metric conversions printed on the back?
Jan 22nd, 2007, 06:50 AM
#9
Join Date: Jan 2003
Posts: 2,568
Thanks, Dukey, for catching my typo. Of course if you just say "two hundred" the vendor will know that you mean grams and won't give you that many kilos.
Jan 22nd, 2007, 06:55 AM
#10
Original Poster
Join Date: Nov 2003
Posts: 4,874
Robespierre, the issue is knowing how to say what I want...do I say a half a kilogram, or 500 grams? Which is more common? How should I order so that I am most likely to get what I want (e.g. writing it down?)....I sure don't want to get 500 kg when I meant 500 grams!
Jan 22nd, 2007, 06:58 AM
#11
Join Date: Jan 2003
Posts: 13,194
Occasionally (perhaps more often in restaurants, referring to the price of a steak, or a lobster), you might see "per etto", referring to "ettogrammi" (hectograms, in English), so, in a deli... 1, 2, 3 or 4 "etto" is slightly under under 1/4, 1/2, 3/4 or a full pound.
And as indicated above, pointing, and saying "cento, due centi, tre centi", etc., is pretty unambiguous - - since grams is the only unit that makes sense when pointing at a 8 kg ham.
Best wishes,
Rex
Jan 22nd, 2007, 07:00 AM
#12
Join Date: Jun 2004
Posts: 19,000
What Rex said.
If the language is the issue, then calculate what you want and round it up to the nearest convenient number. Then show the calculator to the vendor.
For example, if you're buying an expensive spice* and the answer is 84g., then round up to 100g. For cheese, maybe the nearest 250g (about ½ lb.).
* Did you know that saffron is the most expensive item in most grocery stores (on a \$/pound basis)? It's typically in the range of \$2-3 for 1/8 of an ounce.
Jan 22nd, 2007, 11:50 AM
#13
Join Date: Jan 2003
Posts: 74,166
Hi MP
>do I say a half a kilogram, or 500 grams?
If you want about 2 oz of something, on a piece of paper write "50 gram", 4 oz is "100 gram", etc.
Jan 22nd, 2007, 11:52 AM
#14
Original Poster
Join Date: Nov 2003
Posts: 4,874
Okay, now I'm going to really show my ignorance, but are the words "gram" and "kilogram" the same in all western European languages?
Jan 22nd, 2007, 11:58 AM
#15
Join Date: Feb 2004
Posts: 23,074
Yes, a kilogram is the same weight around the world. Same with a gram.
Jan 22nd, 2007, 12:00 PM
#16
Join Date: Jun 2004
Posts: 19,000
I think mp was asking about the terminology. I don't have a definitive answer, but in French and German, the words are cognate with English.
Jan 22nd, 2007, 12:00 PM
#17
Original Poster
Join Date: Nov 2003
Posts: 4,874
Not the weight - the word. Is the Spanish word for "kilogram" still "kilogram"?
Jan 22nd, 2007, 12:01 PM
#18
Join Date: Apr 2005
Posts: 2,385
A friend of mine tells a funny story about this. While traveling in Germany as an 18 year old, he and his friend went in to order at a deli counter. They both were large boys with very healthy appetites, so they ordered "zwei gram" of lunch meat, hoping that it wouldn't be TOO much. The counter attendant just about fell over laughing - and tried to explain that two grams would be about two specks!
Jan 22nd, 2007, 12:12 PM
#19
Join Date: Feb 2005
Posts: 9,017
German is Kilogramm ;-)
Jan 22nd, 2007, 12:17 PM
#20
Join Date: Feb 2004
Posts: 23,074
Fun things one can find online:
Italian: grammo
Greek: γραμμάριο
French: gramme
German: Gramm
Portugese: grama
Spanish: gramo
Russian: грамм
etc...
But I am pretty sure you won't have problem at any store, saying "gram"! | 1,812 | 5,718 | {"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-2019-22 | latest | en | 0.948167 |
https://openwetware.org/wiki/MATLAB_calculations | 1,603,512,963,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881640.29/warc/CC-MAIN-20201024022853-20201024052853-00637.warc.gz | 474,320,712 | 5,875 | # MATLAB calculations
```%% First I define fixed constants
R=6.75e-2; % in m
muo=4*pi*(1e-7);
muB=9.274e-24; %using J
h=6.626e-34; %using J
N=320;
%% The data is then used to calculate the B-field through the Helmholtz
%coils and the sample
%SMALL
datS=[90.00 1.839;
85.00 1.766;
80.00 1.685;
75.00 1.567;
70.00 1.477;];
fqS=(1e6)*datS(:,1)';
iS=datS(:,2)';
BS=((4/5)^(3/2))*((muo*N*iS)/R);
%MEDIUM
datM=[70.00 1.335
65.00 1.253
60.00 1.171
55.00 1.071
50.00 0.972
45.00 0.898
40.00 0.790
35.00 0.690
30.00 0.591
25.00 0.492];
fqM=(1e6)*datM(:,1)';
iM=datM(:,2)';
BM=((4/5)^(3/2))*((muo*N*iM)/R);
%Large
datL=[30.00 0.581
26.00 0.472
22.00 0.427
18.00 0.355
14.00 0.265];
fqL=(1e6)*datL(:,1)';
iL=datL(:,2)';
BL=((4/5)^(3/2))*((muo*N*iL)/R);
%% Now I calculate g_s for each data pair and then from linear least square
% fits for the 3 data sets
gSmall=(h*fqS)./(muB*BS)
gSmean=mean(gSmall)
gMed=(h*fqM)./(muB*BM)
gMmean=mean(gMed)
gLarge=(h*fqL)./(muB*BL)
gLmean=mean(gLarge)
gTot=(gSmean+gMmean+gLmean)/3
Sfit=polyfit(BS,(h*fqS)/muB,1)
Sval=polyval(Sfit,BS)
Mfit=polyfit(BM,(h*fqM)/muB,1)
Mval=polyval(Mfit,BM)
Lfit=polyfit(BL,(h*fqL)/muB,1)
Lval=polyval(Lfit,BL)
AvgFit=(Sfit(1)+Mfit(1)+Lfit(1))/3
figure(1)
plot(BS,Sval,BS,(h*fqS)/muB,'r*');title('B vs.(h*v)/muB for small coil'),xlabel('T'),ylabel('T')
saveas(figure(1),'smallcoil.jpg')
figure(2)
plot(BM,Mval,BM,(h*fqM)/muB,'r*');title('B vs.(h*v)/muB for medium coil'),xlabel('T'),ylabel('T')
saveas(figure(2),'mediumcoil.jpg')
figure(3)
plot(BL,Lval,BL,(h*fqL)/muB,'r*');title('B vs.(h*v)/muB for large coil'),xlabel('T'),ylabel('T')
saveas(figure(3),'largecoil.jpg')
sdmS=sqrt(sum((gSmall-gSmean).^2)*(1/(length(gSmall)*(length(gSmall)-1))))
sdmM=sqrt(sum((gMed-gMmean).^2)*(1/(length(gMed)*(length(gMed)-1))))
sdmL=sqrt(sum((gLarge-gLmean).^2)*(1/(length(gLarge)*(length(gLarge)-1))))
sdmTot=(sdmS+sdmM+sdmL)/3
``` | 893 | 1,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-45 | latest | en | 0.34901 |
https://www.numbersaplenty.com/2446365562 | 1,701,345,169,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00647.warc.gz | 1,049,227,687 | 2,961 | Search a number
2446365562 = 21223182781
BaseRepresentation
bin1001000111010000…
…1001001101111010
320022111021020122221
42101310021031322
520002232144222
61042430034254
7114423523054
oct22164111572
96274236587
102446365562
111045a00723
1258334898a
132ccaa09b4
14192c8c8d4
15e4b840c7
hex91d0937a
2446365562 has 4 divisors (see below), whose sum is σ = 3669548346. Its totient is φ = 1223182780.
The previous prime is 2446365553. The next prime is 2446365599. The reversal of 2446365562 is 2655636442.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in only one way, i.e., 2427434361 + 18931201 = 49269^2 + 4351^2 .
It is an unprimeable number.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 611591389 + ... + 611591392.
Almost surely, 22446365562 is an apocalyptic number.
2446365562 is a deficient number, since it is larger than the sum of its proper divisors (1223182784).
2446365562 is a wasteful number, since it uses less digits than its factorization.
2446365562 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1223182783.
The product of its digits is 1036800, while the sum is 43.
The square root of 2446365562 is about 49460.7476894557. The cubic root of 2446365562 is about 1347.4328097912.
The spelling of 2446365562 in words is "two billion, four hundred forty-six million, three hundred sixty-five thousand, five hundred sixty-two".
Divisors: 1 2 1223182781 2446365562 | 497 | 1,546 | {"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 | 0.822466 |
https://community-forums.domo.com/main/discussion/61106/cy-vs-py-and-multiline-chart | 1,721,393,552,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514902.63/warc/CC-MAIN-20240719105029-20240719135029-00556.warc.gz | 159,313,571 | 99,994 | # CY vs PY and Multiline Chart
Options
Member
Hi All,
I am trying to build a multiline chart that would have 2 Y-axis line chart.
Y1 Axis to show Cumulative Line of YTD Sales vs PY to Date Sales and
Y2 Axis to show YTD Average Sales vs PY to Date average sales.
I have tried all the available line charts and seems not to be the one as per the picture attached.
Also, what's the best approach to calculate previous year to date in DOMO. Watched the brilliant video by @jaeW_at_Onyx on YouTube and seems a little confusing. Could some help please with some direction?
## Best Answer
• Coach
edited August 2023 Answer ✓
Options
I guess it all depends on your specific needs, if showing the average lines is optional, you could go with the default Period Over Period Running Total chart, selecting Month To Date and comparing against 1 Year ago, which would give you something really similar to the image you shared, and has the flexibility that you could change to see Quarter to Date or other time periods with ease.
Working with Beastmodes adds a lot more flexibility on what you can show, but it can be tricky to control the date range shown, so it is a matter of what works best for you. If you go the beast mode route, you'll need to do a line chart with 4 different beast modes, one per line and work around controlling what will be shown on the X axis, doable, but might require some trial an error.
## Answers
• Coach
Options
For previous year to date, I do:
"Case when Date >= DATE_SUB(CurrentDate, interval 1 year) and Year(Date) = Year(DATE_SUB(CurrentDate, interval 1 year) ) then Value End"
Find the year value of today's date 1 year ago and only selects dates less than or equal to today 1 year ago.
• Member
Options
Thanks @trafalger for the response. The beast mode calculation you gave above logically makes sense however, it pulls the same amounts for the current year figures. what could be going on here?
• Coach
edited August 2023 Answer ✓
Options
I guess it all depends on your specific needs, if showing the average lines is optional, you could go with the default Period Over Period Running Total chart, selecting Month To Date and comparing against 1 Year ago, which would give you something really similar to the image you shared, and has the flexibility that you could change to see Quarter to Date or other time periods with ease.
Working with Beastmodes adds a lot more flexibility on what you can show, but it can be tricky to control the date range shown, so it is a matter of what works best for you. If you go the beast mode route, you'll need to do a line chart with 4 different beast modes, one per line and work around controlling what will be shown on the X axis, doable, but might require some trial an error. | 622 | 2,765 | {"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-2024-30 | latest | en | 0.932196 |
https://www.safalta.com/blog/reasoning-ability-quiz-17-march-2021?src=guide | 1,685,318,542,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644571.22/warc/CC-MAIN-20230528214404-20230529004404-00006.warc.gz | 1,081,394,082 | 104,881 | # Reasoning Ability Quiz: 17 March 2021
Reasoning Ability Quiz: 17 March 2021 Published by: Anmol Bhatnagar Updated Wed, 17 Mar 2021 02:50 PM IST
Reasoning Ability Quiz: 17 March 2021
Q: 1-5) Study the following arrangement carefully and answer the questions given below.
Eight members are seating in a meeting, namely, Tarun, Sachin, Rahul, Arun, Pradeep, Mohit, Lalit, and Kunal are sitting around a square table in which four of them are sitting at the corner facing outside and the remaining four are sitting at the middle side of the table facing inside. There are five are male and three of them are female.
Lalit and kunal are sitting opposite to each other. Pradeep sits third to the right of Sachin. None of the females are sitting immediately neighbor of Rahul. Sachin does not sits at the corner.
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Arun sits third to the left of Mohit. Mohit sits at one of the corners and she is sitting immediate right of female. The one how is forth right of sachin belongs to male. Only one person sits between Arun and Rahul. Rahul is not the immediate neighbor of Mohit. kunal is not an immediate neighbor of Rahul. No two female is sitting diagonally opposite to each other. Sachin is not a female.
Q-1) Four of the following five are alike in a certain way and hence form a group. Which is the odd that does not belong to the group?
A) Lalit
B) Arun
C) Tarun
D) Rahul
Q-2) Who sits second to the right of Sachin ?
A) Tarun
B) Arun
C) Lalit
E) None of these
Q-3) Who sits in the front of Tarun ?
A) Kunal
C) Rahul
D) Arun
E) Sachin
Q-4) Four of the following five are alike in a certain way and hence form a group. Which is the odd that does not belong to the group?
A) Kunal
B) Mohit
C) Lalit
E) Arun
Q-5) Who sits to the third left of Rahul?
A) Mohit
B) Sachin
C) Tarun
D) Kunal
E) None of the above
Q: 6-8) Study the following arrangement carefully and answer the questions given below.
Point N is 10m east of M. Point P is 10m west of D. L is 20m north of M. D is 10m west of Q. D is 10m south of N.
Q-6) L is in which direction with respect to P ?
A) South
B) North
C) North west
D) South west
E) North east
Q-7) What is the total distance between P and Q ?
A) 10m
B) 20m
C) 30m
D) 40 m
E) None of these
Q-8) N is in which direction with respect to Q ?
A) South
B) North
C) Northwest
D) Southwest
E) Northeast
Q: 9) In a certain code “SOLVED” is written as “DEVLOS”, then how MENTION can be written as ?
A)NOTINEM
B)TIONMEN
C)NOITNEM
D)NOTIENM
Q-10) In a certain code “ ACTOR ” is written as “BDUPS”, then how REASONING can be written as ?
A)SFBOPTJOH
B) SFBPTJOHO
C) SFPBTHJOH
D) SFBTPOJOH
Solution:
Q: 1-5)
Q-1) C) Tarun
Q-2) B) Arun
Q-3) D) Arun
Q-4) E) Arun
Q-5) D) Kunal
Q:6-8)
Q-6) B) North
Q-7) B) 20m
Q-8) C) Northwest
Q-9) Ans – c) NOITNEM
Q-10) ANS- D) SFBTPOJOH
Solution:
A(+1) = B ; C(+1) =D ; T(+1) =U ; O(+1) =P ; R(+1) =S
Similarly,
R(+1) = S ; E(+1) =F ; A(+1) =B ; S(+1) =T ; O(+1) =P ; N(+1) =O ; I(+1) =J ; N(+1) =O ; G(+1)=H
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Now at just ₹ 1499 ₹ 599975% off | 1,266 | 3,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-23 | longest | en | 0.934648 |
http://www.solutioninn.com/1-kevin-jones-is-single-and-recently-graduated-from-law | 1,503,352,130,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109670.98/warc/CC-MAIN-20170821211752-20170821231752-00078.warc.gz | 693,618,202 | 8,509 | # Question: 1 Kevin Jones is single and recently graduated from law
1. Kevin Jones is single and recently graduated from law school. He earns \$9000 per month, an awesome salary for someone only 26 years old. He also has \$1400 withheld for federal income tax, \$540 for state income taxes, \$688 for Medicare and Social Security taxes, and \$230 for health insurance every month. Kevin has outstanding student loans of almost \$80,000 on which he pays about \$900 per month and a 0% auto loan payment of \$300 on a Ford Fusion Hybrid he purchased new during law school. He is considering taking out a loan to buy a Kawasaki motorcycle.
(a) What is Kevin’s debt payments-to-disposable income ratio?
2. Carmen and Juan Montoya have just finished putting their three daughters through college. As empty-nesters, they are considering purchasing a vacation home on a nearby lake because prices have dropped in recent years. The house might also serve as a retirement home once they retire in 12 years. The Montoyas’ net worth is \$283,000 including their home worth about \$265,000 on which they currently owe \$143,000 for their first mortgage. Their outstanding debts in addition to their mortgage include \$12,500 on one car loan, \$13,700 on a second car loan, and a \$25,500 second mortgage on their home taken out to help pay for college expenses.
(a) Calculate the Montoyas’ debt-to-equity ratio.
(b) Advise them as to the wisdom of borrowing for a vacation home at this time.
3. Chelsea Menken recently graduated with a degree in food science and now works for a major consumer foods company earning \$20K per year with about \$36,000 in take-home pay. She rents an apartment for \$1040 per month. While in school, she accumulated about \$38,000 in student loan debt on which she pays \$385 per month. During her last fall semester in school, she had an internship in a city about 100 miles from her campus. She used her credit card for her extra expenses and has a current debt on the account of \$8000. She has been making the minimum payment on the account of about \$320. She has assets of \$14,000.
(a) Calculate Chelsea’s debt payment-to-disposable income and debt-service-to-income ratios.
(b) Calculate Chelsea’s debt-to-equity ratio.
(c) Comment on Chelsea’s debt situation and her use of student loans and credit cards while in college.
View Solution:
Sales22
Views604 | 531 | 2,385 | {"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-2017-34 | latest | en | 0.977649 |
https://resources.quizalize.com/view/quiz/g2-math-term-1-lq-2-9502ba57-3261-4b04-9154-32ae3ac420f0 | 1,716,388,764,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058557.6/warc/CC-MAIN-20240522132433-20240522162433-00248.warc.gz | 422,682,654 | 17,914 | # G2 Math Term 1 LQ #2
## Quiz by Esther San Juan
### Our brand new solo games combine with your quiz, on the same screen
Correct quiz answers unlock more play!
15 questions
• Q1
I. Multiple Choice
Base ten blocks can represent larger numbers.
B. False
A. True
300s
• Q2
I. Multiple Choice
A flat represents _________________.
b. one thousand
a. one hundred
300s
• Q3
I. Multiple Choice
One long represents ______________________.
b. one thousand
a. ten
300s
• Q4
I. Multiple Choice
A square represents _____________________.
b. one
a. five
300s
• Q5
I. Multiple Choice
Another way to represent numbers is to use tens and ones.
a. Yes
b. No
300s
• Q6
Choose the correct amount or number being represented by the following models.
2555
3777
300s
• Q7
Choose the correct amount or number being represented by the following models.
456
123
300s
• Q8
Choose the correct amount or number being represented by the following models.
1253
1324
300s
• Q9
Choose the correct amount or number being represented by the following models.
123
350
300s
• Q10
Choose the correct amount or number being represented by the following models.
3363
3456
300s
• Q11
Choose the correct amount or number being represented by the following models.
1387
2287
300s
• Q12
Choose the correct amount or number being represented by the following models.
506
304
300s
• Q13
Choose the correct amount or number being represented by the following models.
145
135
300s
• Q14
Choose the correct amount or number being represented by the following models.
6455
8255
300s
• Q15
Choose the correct amount or number being represented by the following models.
5884
3794
300s
Teachers give this quiz to your class | 468 | 1,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-22 | latest | en | 0.770933 |
http://www.zhitov.ru/widgets/so/credit/ | 1,579,976,346,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251678287.60/warc/CC-MAIN-20200125161753-20200125190753-00041.warc.gz | 294,055,497 | 7,465 | # Calculator Loan
dheer Loan bilihii
dulsaarka
lacagaha magdhowga
lacagaha kala duwan
## Calculator Loan
calculator The xisaabin doonaa qadarka lacagta, deynta iyo kharashka credit.
## lacagaha kala duwan
Marka hab kala duwan ee bixinta daymaha lacagta amaahda ah oo amaah ah ugu qaybsan yahay saamiyada loo siman yahay. saamiyada Kuwani koobnaan bulk oo ka mid ah lacagta bil kasta. Qaybta haraaga ah wakiil xiiso on dheelitirka la bixin on amaahda. Sidaas bil ka bil lacagta hoos size.
In habkan of bixinta amaahda waxaa jira cillado.
Chief iyaga ka mid ah waa xaqiiqada ah in si amaah ah kula bixinta daymaha macnaheedu waa ka sii adag tahay in la helo.
Bank tirinta lacagta ugu badan ee amaahda ku saleysan in amaahdaana waa inay awoodaan inay bixiyaan lacagta wajigeeda hore. Taas macnaheedu waa in ay u helaan amaah noocan oo kale ah waxaa loo baahan yahay in ay leeyihiin dakhli ku filan oo sare. Xaaladaha qaarkood, waxa ku caawin kara damiin damiin ama co-deynsanaya jiidashada.
khasaare kale waa in qeybtii hore ee muddada lacagta amaahdaana si gaar ah u daran. Haddii aynu ka hadlayno amaah weyn u noqon kartaa rar culus for amaahdaana. Laakiin mustaqbalka khasaaro this noqon kartaa sharaf. Sicir iyo hoos lacagaha dulsaarka dhigay yar culaab.
## lacagaha magdhowga
At hab magdhow ah bixinta daymaha amaah wuxuu u qaybsan yahay qaybo siman ee ma aha oo kaliya xaddiga daynta, laakiin sidoo kale danaha waxa on muddo ah amaah dhan. Sidaas darteed, wax amaahdaana bixinaya qaaraan siman ugu weyn mudada lacag bixinta. Maanta, habkan of lacagta loo isticmaalo by ugu bangiyo ganacsi. Faa'iida weyn ee macaashka waa in cadadka lacag amaah ah uu noqon doono ka badan in nidaamka kala. In lacagta magdhowga waxaa dheer habka loo maleeyo in qeybtii hore ee xilli amaah ah aad bixiso inta dulsaarka amaahda. Laakiin wadarta weyn ee daynta muddo this hadhay ku dhawaad Sucuudiga. | 717 | 1,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-05 | latest | en | 0.154012 |
http://nrich.maths.org/public/leg.php?group_id=4&code=-351 | 1,502,970,258,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103270.12/warc/CC-MAIN-20170817111816-20170817131816-00696.warc.gz | 315,336,690 | 4,912 | # Search by Topic
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### There are 6 results
Broad Topics > Calculations and Numerical Methods > Approximate solution by numerical methods.
### Root Hunter
##### Stage: 5 Challenge Level:
In this short problem, try to find the location of the roots of some unusual functions by finding where they change sign.
### Equation Attack
##### Stage: 5 Challenge Level:
The equation a^x + b^x = 1 can be solved algebraically in special cases but in general it can only be solved by numerical methods.
##### Stage: 4 Challenge Level:
A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?
### The Fastest Cyclist
##### Stage: 4 Challenge Level:
Andy is desperate to reach John o'Groats first. Can you devise a winning race plan?
### Archimedes Numerical Roots
##### Stage: 5 Challenge Level:
How did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?
### Farey Approximation
##### Stage: 5 Short Challenge Level:
Use Farey sequences to obtain rational approximations to irrational numbers. | 278 | 1,247 | {"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-2017-34 | latest | en | 0.878346 |
https://www.mercurynews.com/2021/04/22/bridge-april-22-2021/ | 1,620,411,476,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988802.93/warc/CC-MAIN-20210507181103-20210507211103-00599.warc.gz | 951,174,269 | 42,414 | # Bridge: April 22, 2021
PUBLISHED: | UPDATED:
I asked my friend the English professor whether he had any colleagues in the math department.
“One just got back from the annual Fibonacci conference,” the prof said.
“Did he enjoy it?”
“He had a great time,” the prof replied. “He said it was as good as the last two put together.”
Good declarer play can mean adding one chance to another. At today’s slam, South took the king of diamonds and let the jack of clubs ride hopefully. Alas, East took the queen. South then had four club tricks, two diamonds, four hearts and one spade, but no 12th trick.
SECOND CHANCE
South could get a second chance. He leads a heart to dummy at Trick Two and returns a spade to his nine. West takes the king and leads another diamond, and South wins and cashes the ace of spades.
When the queen falls from East, South has four spades, four hearts, two diamonds and two clubs. If nothing good happened in spades (South also succeeds if East has the K-Q), South could try for five club tricks.
DAILY QUESTION
You hold: S J 10 8 7 2 H K J 9 D Q 7 C A K 7. Your partner opens one diamond, you respond one spade and he bids two clubs. What do you say?
ANSWER: Based on what you know at this point, a bid of 3NT would be reasonable and would be the choice of many players. Still, partner’s hand is not well defined, and he could have None, A 3 2, A K J 9 6 3, Q 10 5 4. Bid two hearts, a forcing “fourth-suit” action, and let partner continue to describe his hand.
South dealer
N-S vulnerable
NORTH
S J 10 8 7 2
H K J 9
D Q 7
C A K 7
WEST
S K 6 5 3
H 8 6
D 10 9 8 5
C 6 3 2
EAST
S Q 4
H 10 7 5 4
D J 6 4 3 2
C Q 5
SOUTH
S A 9
H A Q 3 2
D A K
C J 10 9 8 4
South West North East
1 C Pass 1 S Pass
2 H Pass 3 C Pass
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## ROC Curve and Performance Metrics
This topic describes the performance metrics for classification, including the receiver operating characteristic (ROC) curve and the area under a ROC curve (AUC), and introduces the Statistics and Machine Learning Toolbox™ object `rocmetrics`, which you can use to compute performance metrics for binary and multiclass classification problems.
### Introduction to ROC Curve
After training a classification model, such as `ClassificationNaiveBayes` or `ClassificationEnsemble`, you can examine the performance of the algorithm on a specific test data set. A common approach is to compute a gross measure of performance, such as quadratic loss or accuracy, averaged over the entire test data set. You can inspect the classifier performance more closely by plotting a ROC curve and computing performance metrics. For example, you can find the threshold that maximizes the classification accuracy, or assess how the classifier performs in the regions of high sensitivity and high specificity.
#### Receiver Operating Characteristic (ROC) Curve
A ROC curve shows the true positive rate (TPR, or sensitivity) versus the false positive rate (FPR, or 1-specificity) for different thresholds of classification scores.
Each point on a ROC curve corresponds to a pair of TPR and FPR values for a specific threshold value. You can find different pairs of TPR and FPR values by varying the threshold value, and then create a ROC curve using the pairs.
For a multiclass classification problem, you can use the one-versus-all coding design and find a ROC curve for each class. The one-versus-all coding design treats a multiclass classification problem as a set of binary classification problems, and assumes one class as positive and the rest as negative in each binary problem.
A binary classifier typically classifies an observation into a class that yields a larger score, which corresponds to a positive adjusted score for a one-versus-all binary classification problem. That is, a classifier typically uses 0 as a threshold and determines whether an observation is positive or negative. For example, if an adjusted score for an observation is 0.2, then the classifier with a threshold value of 0 assigns the observation to the positive class. You can find a pair of TPR and FPR values by applying the threshold value to all observations, and use the pair as a single point on a ROC curve. Now, assume you use a new threshold value of 0.25. Then, the classifier with a threshold value of 0.25 assigns the observation with an adjusted score of 0.2 to the negative class. By applying the new threshold to all observations, you can find a new pair of TPR and FPR values and have a new point on the a ROC curve. By repeating this process for various threshold values, you find pairs of TPR and FPR values and create a ROC curve using the pairs.
#### Area Under ROC Curve (AUC)
The area under a ROC curve (AUC) corresponds to the integral of a ROC curve (TPR values) with respect to FPR from `FPR` = `0` to `FPR` = `1`.
The AUC provides an aggregate performance measure across all possible thresholds. The AUC values are in the range `0` to `1`, and larger AUC values indicate better classifier performance.
• A perfect classifier always correctly assigns positive class observations to the positive class and has a true positive rate of `1` for any threshold values. Therefore, the line passing through `[0,0]`, `[0,1]`, and `[1,1]` represents the perfect classifier, and the AUC value is `1`.
• A random classifier returns random score values and has the same values for the false positive rate and true positive rate for any threshold values. Therefore, the ROC curve for the random classifier lies on the diagonal line, and the AUC value is `0.5`.
### Performance Curve with MATLAB
You can compute a ROC curve and other performance curves by creating a `rocmetrics` object. The `rocmetrics` object supports both binary and multiclass classification problems and provides the following object functions:
• `plot` — Plot ROC or other classifier performance curves. `plot` returns a `ROCCurve` graphics object for each curve. You can modify the properties of the objects to control the appearance of each curve. For details, see ROCCurve Properties.
• `average` — Compute performance metrics for an average ROC curve for multiclass problems.
• `addMetrics` — Compute additional classification performance metrics.
You can also compute the confidence intervals of performance curves by providing cross-validated inputs or by bootstrapping the input data.
After training a classifier, use a performance curve to evaluate the classifier performance on test data. Various measures such as mean squared error, classification error, or exponential loss can summarize the predictive power of a classifier in a single number. However, a performance curve offers more information because it lets you explore the classifier performance across a range of thresholds on the classification scores.
#### Plot ROC Curve for Binary Classifier
Compute the performance metrics (FPR and TPR) for a binary classification problem by creating a `rocmetrics` object, and plot a ROC curve by using `plot` function.
Load the `ionosphere` data set. This data set has 34 predictors (`X`) and 351 binary responses (`Y`) for radar returns, either bad (`'b'`) or good (`'g'`).
`load ionosphere`
Partition the data into training and test sets. Use approximately 80% of the observations to train a support vector machine (SVM) model, and 20% of the observations to test the performance of the trained model on new data. Partition the data using `cvpartition`.
```rng("default") % For reproducibility of the partition c = cvpartition(Y,Holdout=0.20); trainingIndices = training(c); % Indices for the training set testIndices = test(c); % Indices for the test set XTrain = X(trainingIndices,:); YTrain = Y(trainingIndices); XTest = X(testIndices,:); YTest = Y(testIndices);```
Train an SVM classification model.
`Mdl = fitcsvm(XTrain,YTrain);`
Compute the classification scores for the test set.
```[~,Scores] = predict(Mdl,XTest); size(Scores)```
```ans = 1×2 70 2 ```
The output `Scores` is a matrix of size `70`-by-`2`. The column order of `Scores` follows the class order in `Mdl`. Display the class order stored in `Mdl.ClassNames`.
`Mdl.ClassNames`
```ans = 2x1 cell {'b'} {'g'} ```
Create a `rocmetrics` object by using the true labels in `YTest` and the classification scores in `Scores`. Specify the column order of `Scores` using `Mdl.ClassNames`.
`rocObj = rocmetrics(YTest,Scores,Mdl.ClassNames);`
`rocObj` is a `rocmetrics` object that stores the AUC values and performance metrics for each class in the `AUC` and `Metrics` properties. Display the `AUC` property.
`rocObj.AUC`
```ans = 1×2 0.8587 0.8587 ```
For a binary classification problem, the AUC values are equal to each other.
The table in `Metrics` contains the performance metric values for both classes, vertically concatenated according to the class order. Find the rows for the first class in the table, and display the first eight rows.
```idx = strcmp(rocObj.Metrics.ClassName,Mdl.ClassNames(1)); head(rocObj.Metrics(idx,:))```
``` ClassName Threshold FalsePositiveRate TruePositiveRate _________ _________ _________________ ________________ {'b'} 15.545 0 0 {'b'} 15.545 0 0.04 {'b'} 15.105 0 0.08 {'b'} 11.424 0 0.16 {'b'} 10.077 0 0.2 {'b'} 9.9716 0 0.24 {'b'} 9.9417 0 0.28 {'b'} 9.0338 0 0.32 ```
Plot the ROC curve for each class by using the `plot` function.
`plot(rocObj)`
For each class, the `plot` function plots a ROC curve and displays a filled circle marker at the model operating point. The legend displays the class name and AUC value for each curve.
Note that you do not need to examine ROC curves for both classes in a binary classification problem. The two ROC curves are symmetric, and the AUC values are identical. A TPR of one class is a true negative rate (TNR) of the other class, and TNR is 1-FPR. Therefore, a plot of TPR versus FPR for one class is the same as a plot of 1-FPR versus 1-TPR for the other class.
Plot the ROC curve for the first class only by specifying the `ClassNames` name-value argument.
`plot(rocObj,ClassNames=Mdl.ClassNames(1))`
#### Plot ROC Curves for Multiclass Classifier
Compute the performance metrics (FPR and TPR) for a multiclass classification problem by creating a `rocmetrics` object, and plot a ROC curve for each class by using the `plot` function. Specify the `AverageROCType` name-value argument of `plot` to create the average ROC curve for the multiclass problem.
Load the `fisheriris` data set. The matrix `meas` contains flower measurements for 150 different flowers. The vector `species` lists the species for each flower. `species` contains three distinct flower names.
`load fisheriris`
Train a classification tree that classifies observations into one of the three labels. Cross-validate the model using 10-fold cross-validation.
```rng("default") % For reproducibility Mdl = fitctree(meas,species,Crossval="on");```
Compute the classification scores for validation-fold observations.
```[~,Scores] = kfoldPredict(Mdl); size(Scores)```
```ans = 1×2 150 3 ```
The output `Scores` is a matrix of size `150`-by-`3`. The column order of `Scores` follows the class order in `Mdl`. Display the class order stored in `Mdl.ClassNames`.
`Mdl.ClassNames`
```ans = 3x1 cell {'setosa' } {'versicolor'} {'virginica' } ```
Create a `rocmetrics` object by using the true labels in `species` and the classification scores in `Scores`. Specify the column order of `Scores` using `Mdl.ClassNames`.
`rocObj = rocmetrics(species,Scores,Mdl.ClassNames);`
`rocObj` is a `rocmetrics` object that stores the AUC values and performance metrics for each class in the `AUC` and `Metrics` properties. Display the `AUC` property.
`rocObj.AUC`
```ans = 1×3 1.0000 0.9636 0.9636 ```
The table in `Metrics` contains the performance metric values for all three classes, vertically concatenated according to the class order. Find and display the rows for the second class in the table.
```idx = strcmp(rocObj.Metrics.ClassName,Mdl.ClassNames(2)); rocObj.Metrics(idx,:)```
```ans=13×4 table ClassName Threshold FalsePositiveRate TruePositiveRate ______________ _________ _________________ ________________ {'versicolor'} 1 0 0 {'versicolor'} 1 0.01 0.7 {'versicolor'} 0.95455 0.02 0.8 {'versicolor'} 0.91304 0.03 0.9 {'versicolor'} -0.2 0.04 0.9 {'versicolor'} -0.33333 0.06 0.9 {'versicolor'} -0.6 0.08 0.9 {'versicolor'} -0.86957 0.12 0.92 {'versicolor'} -0.91111 0.16 0.96 {'versicolor'} -0.95122 0.31 0.96 {'versicolor'} -0.95238 0.38 0.98 {'versicolor'} -0.95349 0.44 0.98 {'versicolor'} -1 1 1 ```
Plot the ROC curve for each class. Specify `AverageROCType="micro"` to compute the performance metrics for the average ROC curve using the micro-averaging method.
`plot(rocObj,AverageROCType="micro")`
The filled circle markers indicate the model operating points. The legend displays the class name and AUC value for each curve.
### ROC Curve for Multiclass Classification
For a multiclass classifier, the `rocmetrics` function computes the performance metrics of a one-versus-all ROC curve for each class, and the `average` function computes the metrics for an average of the ROC curves. You can use the `plot` function to plot a ROC curve for each class and the average ROC curve.
#### One-Versus-All (OVA) Coding Design
The one-versus-all (OVA) coding design reduces a multiclass classification problem to a set of binary classification problems. In this coding design, each binary classification treats one class as positive and the rest of the classes as negative. `rocmetrics` uses the OVA coding design for multiclass classification and evaluates the performance on each class by using the binary classification that the class is positive.
For example, the OVA coding design for three classes formulates three binary classifications:
Each row corresponds to a class, and each column corresponds to a binary classification problem. The first binary classification assumes that class 1 is a positive class and the rest of the classes are negative. `rocmetrics` evaluates the performance on the first class by using the first binary classification problem.
`rocmetrics` applies the OVA coding design to a binary classification problem as well if you specify classification scores as a two-column matrix. `rocmetrics` formulates two one-versus-all binary classification problems each of which treats one class as a positive class and the other class as a negative class, and `rocmetrics` finds two ROC curves. You can use one of them to evaluate the binary classification problem.
#### Average of Performance Metrics
You can compute metrics for an average ROC curve by using the `average` function. Alternatively, you can use the `plot` function to compute the metrics and plot the average ROC curve. For examples, see Find Average ROC Curve (example for `average`) and Plot Average ROC Curve for Multiclass Classifier (example for `plot`).
`average` and `plot` support three algorithms for computing the average false positive rate (FPR) and average true positive rate (TPR) to find the average ROC curve:
• Micro-averaging — The software combines all one-versus-all binary classification problems into one binary classification problem and computes the average performance metrics as follows:
1. Convert the values in the `Labels` property of a `rocmetrics` object to logical values where logical `1` (`true`) indicates a positive class for each binary problem.
2. Stack the converted vectors of labels, one vector from each binary problem, into a single vector.
3. Convert the matrix that contains the adjusted values of the classification scores (the `Scores` property) into a vector by stacking the columns of the matrix.
4. Compute the components of the confusion matrix for the combined binary problem for each threshold (each distinct value of adjusted scores). A confusion matrix contains the number of instances for true positive (TP), false negative (FN), false positive (FP), and true negative (TN).
5. Compute the average FPR and TPR based on the components of the confusion matrix.
• Macro-averaging — The software computes the average values for FPR and TPR by averaging the values of all one-versus-all binary classification problems.
The software uses three metrics—threshold, FPR, and TPR—to compute the average values as follows:
1. Determine a fixed metric. If you specify `FixedMetric` of `rocmetrics` as `"FalsePositiveRate"` or `"TruePositiveRate"`, then the function holds the specified metric fixed. Otherwise, the function holds the threshold values fixed.
2. Find all distinct values in the `Metrics` property for the fixed metric.
3. Find the corresponding values for the other two metrics for each binary problem.
4. Average the FPR and TPR values of all binary problems.
• Weighted macro-averaging — The software computes the weighted average values for FPR and TPR using the macro-averaging algorithm and using the prior class probabilities (the `Prior` property) as weights.
### Performance Metrics
The `rocmetrics` object supports these built-in performance metrics:
• Number of true positives (TP)
• Number of false negatives (FN)
• Number of false positives (FP)
• Number of true negatives (TN)
• Sum of TP and FP
• Rate of positive predictions (RPP)
• Rate of negative predictions (RNP)
• Accuracy
• True positive rate (TPR), recall, or sensitivity
• False negative rate (FNR), or miss rate
• False positive rate (FPR), fallout, or 1-specificity
• True negative rate (TNR), or specificity
• Positive predictive value (PPV), or precision
• Negative predictive value (NPR)
• Expected cost
`rocmetrics` also supports a custom metric specified as a function handle. For details, see the `AdditionalMetrics` name-value argument of the `rocmetrics` function.
`rocmetrics` computes performance metric values for various thresholds for each one-versus-all binary classification problem using a confusion matrix, scale vector, and misclassification cost matrix. Each performance metric is a function of a confusion matrix and scale vector. The expected cost is also a function of the misclassification cost matrix, as is a custom metric.
• Confusion matrix — A confusion matrix contains the number of instances for true positive (TP), false negative (FN), false positive (FP), and true negative (TN). `rocmetrics` computes confusion matrices for various threshold values for each binary problem.
• Scale vector — A scale vector is defined by the prior class probabilities and the number of classes in true labels. `rocmetrics` finds the probabilities and number of classes for each binary problem from the prior class probabilities specified by the `Prior` name-value argument and the true labels specified by the `Labels` input argument.
• Misclassification cost matrix — `rocmetrics` converts the misclassification cost matrix specified by the `Cost` name-value argument to the values for each binary problem.
By default, `rocmetrics` uses all distinct adjusted score values as threshold values for each binary problem. For more details on threshold values, see Thresholds, Fixed Metric, and Fixed Metric Values.
#### Confusion Matrix
A confusion matrix is defined as
`$\left[\begin{array}{cc}TP& FN\\ FP& TN\end{array}\right],$`
where
• `P` stands for "positive".
• `N` stands for "negative".
• `T` stands for "true".
• `F` stands for "false".
For example, the first row of the confusion matrix defines how the classifier identifies instances of the positive class: TP is the count of correctly identified positive instances, and FN is the count of positive instances misidentified as negative.
`rocmetrics` computes confusion matrices for various threshold values for each one-versus-all binary classification. The one-versus-all binary classification model classifies an observation into a positive class if the score for the observation is greater than or equal to the threshold value.
#### Prior Class Probabilities
By default, `rocmetrics` uses empirical probabilities, which are class frequencies in the true labels.
`rocmetrics` normalizes the `1`-by-K prior probability vector π to a `1`-by-`2` vector for each one-versus-all binary classification, where K is the number of classes.
The prior probabilities for the kth binary classification in which the positive class is the kth class is $\left[{\pi }_{k},1-{\pi }_{k}\right]$, where πk is the prior probability for class k in the multiclass problem.
#### Scale Vector
`rocmetrics` defines a scale vector sk of size `2`-by-`1` for each one-versus-all binary classification problem:
`${\text{s}}_{k}=\frac{1}{{\pi }_{k}N+\left(1-{\pi }_{k}\right)P}\left[\begin{array}{c}{\pi }_{k}N\\ \left(1-{\pi }_{k}\right)P\end{array}\right],$`
where P and N represent the total instances of positive class and negative class, respectively. That is, P is the sum of TP and FN, and N is the sum of FP and TN. sk(1) (first element of sk) and sk(2) (second element of sk) are the scales for the positive class (kth class) and negative class (the rest), respectively.
`rocmetrics` applies the scale values as multiplicative factors to the counts from the corresponding class. That is, the function multiplies counts from the positive class by sk(1) and counts from the negative class by sk(2). For example, to compute the positive predictive value (```PPV = TP/(TP+FP)```) for the kth binary problem, `rocmetrics` scales `PPV` as follows:
`$PPV=\frac{{s}_{k}\left(1\right)\cdot TP}{{s}_{k}\left(1\right)\cdot TP+{s}_{k}\left(2\right)\cdot FP}.$`
#### Misclassification Cost Matrix
By default, `rocmetrics` uses a K-by-K cost matrix C, where C(i,j) = `1` if i ~= j, and C(i,j) = `0` if i = j. C(i,j) is the cost of classifying a point into class j if its true class is i (that is, the rows correspond to the true class and the columns correspond to the predicted class).
`rocmetrics` normalizes the K-by-K cost matrix C to a `2`-by-`2` matrix for each one-versus-all binary classification:
`${C}_{k}=\left[\begin{array}{cc}0& {\text{cost}}_{k}\left(N|P\right)\\ {\text{cost}}_{k}\left(P|N\right)& 0\end{array}\right].$`
Ck is the cost matrix for the kth binary classification in which the positive class is the kth class, where `costk(N|P)` is the cost of misclassifying a positive class as a negative class, and `costk(P|N)` is the cost of misclassifying a negative class as a positive class.
For class k, let πk+ and πk- be K-by-`1` vectors with the following values:
πki+ and πki- are the ith elements of πk+ and πk-, respectively.
The cost of classifying a positive-class (class k) observation into the negative class (the rest) is
`${\text{cost}}_{k}\left(N|P\right)={\left({\pi }_{k}^{+}\right)}^{\prime }C{\pi }_{k}^{-}.$`
Similarly, the cost of classifying a negative-class observation into the positive class is
`${\text{cost}}_{k}\left(P|N\right)={\left({\pi }_{k}^{-}\right)}^{\prime }C{\pi }_{k}^{+}.$`
### Classification Scores and Thresholds
The `rocmetrics` function determines threshold values from the input classification scores or the `FixedMetricValues` name-value argument.
#### Classification Score Input for `rocmetrics`
`rocmetrics` accepts classification scores (`Scores`) in a matrix of size n-by-K or a vector of length n, where n is the number of observations and K is the number classes. For cross-validated data, `Scores` can be a cell array of vectors or a cell array of matrices.
• Matrix of size n-by-K — Specify `Scores` using the second output argument of the `predict` function of a classification model object (such as `predict` of `ClassificationTree`). Each row of the output contains classification scores for an observation for all classes, and the column order of the output matches the class order in the `ClassNames` property of the classification model object. You can specify `Scores` as a matrix for both binary classification and multiclass classification problems.
If you use a matrix format, `rocmetrics` adjusts the classification scores for each class relative to the scores for the rest of the classes. Specifically, the adjusted score for a class given an observation is the difference between the score for the class and the maximum value of the scores for the rest of the classes. For more details, see Adjusted Scores for Multiclass Classification Problem.
• Vector of length n — Specify `Scores` using a vector when you have classification scores for one class only. A vector input is also suitable when you want to use a different type of adjusted scores for a multiclass problem. As an example, consider a problem with three classes, `A`, `B`, and `C`. If you want to compute a performance curve for separating classes `A` and `B`, with `C` ignored, you need to address the ambiguity in selecting `A` over `B`. You can use the score ratio `s(A)/s(B)` or score difference `s(A)–s(B)` and pass the vector to `rocmetrics`; this approach can depend on the nature of the scores and their normalization.
You can use `rocmetrics` with any classifier or any function that returns a numeric score for an instance of input data.
• A high score returned by a classifier for a given instance and class signifies that the instance is likely from the respective class.
• A low score signifies that the instance is not likely from the respective class.
For some classifiers, you can interpret the score as the posterior probability of observing an instance of a class given an observation. An example of such a score is the fraction of observations for a certain class in a leaf of a decision tree. In this case, scores fall into the range from 0 to 1, and scores from all classes add up to 1. Other functions can return scores ranging between minus and plus infinity, without any obvious mapping from the score to the posterior class probability.
`rocmetrics` does not impose any requirements on the input score range. Because of this lack of normalization, you can use `rocmetrics` to process scores returned by any classification, regression, or fit functions. `rocmetrics` does not make any assumptions about the nature of input scores.
`rocmetrics` is intended for use with classifiers that return scores, not those that return only predicted classes. Consider a classifier that returns only classification labels, 0 or 1, for data with two classes. In this case, the performance curve reduces to a single point because the software can split classified instances into positive and negative categories in one way only.
#### Adjusted Scores for Multiclass Classification Problem
For each class, `rocmetrics` adjusts the classification scores (input argument `Scores` of `rocmetrics`) relative to the scores for the rest of the classes if you specify `Scores` as a matrix. Specifically, the adjusted score for a class given an observation is the difference between the score for the class and the maximum value of the scores for the rest of the classes.
For example, if you have [s1,s2,s3] in a row of `Scores` for a classification problem with three classes, the adjusted score values are [s1-`max`(s2,s3),s2-`max`(s1,s3),s3-`max`(s1,s2)].
`rocmetrics` computes the performance metrics using the adjusted score values for each class.
For a binary classification problem, you can specify `Scores` as a two-column matrix or a column vector. Using a two-column matrix is a simpler option because the `predict` function of a classification object returns classification scores as a matrix, which you can pass to `rocmetrics`. If you pass scores in a two-column matrix, `rocmetrics` adjusts scores in the same way that it adjusts scores for multiclass classification, and it computes performance metrics for both classes. You can use the metric values for one of the two classes to evaluate the binary classification problem. The metric values for a class returned by `rocmetrics` when you pass a two-column matrix are equivalent to the metric values returned by `rocmetrics` when you specify classification scores for the class as a column vector.
#### Model Operating Point
The model operating point represents the FPR and TPR corresponding to the typical threshold value.
The typical threshold value depends on the input format of the `Scores` argument (classification scores) specified when you create a `rocmetrics` object:
• If you specify `Scores` as a matrix, `rocmetrics` assumes that the values in `Scores` are the scores for a multiclass classification problem and uses adjusted score values. A multiclass classification model classifies an observation into a class that yields the largest score, which corresponds to a nonnegative score in the adjusted scores. Therefore, the threshold value is `0`.
• If you specify `Scores` as a column vector, `rocmetrics` assumes that the values in `Scores` are posterior probabilities of the class specified in `ClassNames`. A binary classification model classifies an observation into a class that yields a higher posterior probability, that is, a posterior probability greater than `0.5`. Therefore, the threshold value is `0.5`.
For a binary classification problem, you can specify `Scores` as a two-column matrix or a column vector. However, if the classification scores are not posterior probabilities, you must specify `Scores` as a matrix. A binary classifier classifies an observation into a class that yields a larger score, which is equivalent to a class that yields a nonnegative adjusted score. Therefore, if you specify `Scores` as a matrix for a binary classifier, `rocmetrics` can find a correct model operating point using the same scheme that it applies to a multiclass classifier. If you specify classification scores that are not posterior probabilities as a vector, `rocmetrics` cannot identify a correct model operating point because it always uses `0.5` as a threshold for the model operating point.
The `plot` function displays a filled circle marker at the model operating point for each ROC curve (see `ShowModelOperatingPoint`). The function chooses a point corresponding to the typical threshold value. If the curve does not have a data point for the typical threshold value, the function finds a point that has the smallest threshold value greater than the typical threshold. The point on the curve indicates identical performance to the performance of the typical threshold value.
For an example, see Find Model Operating Point and Optimal Operating Point.
#### Thresholds, Fixed Metric, and Fixed Metric Values
`rocmetrics` finds the ROC curves and other metric values that correspond to the fixed values (`FixedMetricValues` name-value argument) of the fixed metric (`FixedMetric` name-value argument), and stores the values in the `Metrics` property as a table.
The default `FixedMetric` value is `"Thresholds"`, and the default `FixedMetricValues` value is `"all"`. For each class, `rocmetrics` uses all distinct adjusted score values as threshold values, computes the components of the confusion matrix for each threshold value, and then computes performance metrics using the confusion matrix components.
If you use the default `FixedMetricValues` value (`"all"`), specifying a nondefault `FixedMetric` value does not change the software behavior unless you specify to compute confidence intervals. If `rocmetrics` computes confidence intervals, then it holds `FixedMetric` fixed at `FixedMetricValues` and computes confidence intervals for other metrics. For more details, see Pointwise Confidence Intervals.
If you specify a nondefault value for `FixedMetricValues`, `rocmetrics` finds the threshold values corresponding to the specified fixed metric values (`FixedMetricValues` for `FixedMetric`) and computes other performance metric values using the threshold values.
• If you set the `UseNearestNeighbor` name-value argument to `false`, then `rocmetrics` uses the exact threshold values corresponding to the specified fixed metric values.
• If you set `UseNearestNeighbor` to `true`, then among the adjusted scores, `rocmetrics` finds a value that is the nearest to the threshold value corresponding to each specified fixed metric value.
The `Metrics` property includes an additional threshold value that replicates the largest threshold value for each class so that a ROC curve starts from the origin `(0,0)`. The additional threshold value represents the reject-all threshold, for which `TP` = `FP` = `0` (no positive instances, that is, zero true positive instances and zero false positive instances).
Another special threshold in `Metrics` is the accept-all threshold, which is the smallest threshold value for which `TN` = `FN` = `0` (no negative instances, that is, zero true negative instances and zero false negative instances).
Note that the positive predictive value (`PPV = TP/(TP+FP)`) is `NaN` for the reject-all threshold, and the negative predictive value (`NPV = TN/(TN+FN)`) is `NaN` for the accept-all threshold.
#### NaN Score Values
`rocmetrics` processes `NaN` values in the classification score input (`Scores`) in one of two ways:
• If you specify `NaNFlag="omitnan"` (default), then `rocmetrics` discards rows with `NaN` scores.
• If you specify `NaNFlag="includenan"`, then `rocmetrics` adds the instances of `NaN` scores to false classification counts in the respective class for each one-versus-all binary classification. That is, for any threshold, the software counts instances with `NaN` scores from the positive class as false negative (FN), and counts instances with `NaN` scores from the negative class as false positive (FP). The software computes the metrics corresponding to a threshold of `1` by setting the number of true positive (TP) instances to zero and setting the number of true negative (TN) instances to the total count minus the `NaN` count in the negative class.
Consider an example with two rows in the positive class and two rows in the negative class, each pair having a `NaN` score:
True Class LabelClassification Score
Negative0.2
Negative`NaN`
Positive0.7
Positive`NaN`
If you discard rows with `NaN` scores (`NaNFlag="omitnan"`), then as the score threshold varies, `rocmetrics` computes performance metrics as shown in the following table. For example, a threshold of 0.5 corresponds to the middle row where `rocmetrics` classifies rows 1 and 3 correctly and omits rows 2 and 4.
Threshold`TP``FN``FP``TN`
10101
0.51001
01010
If you add rows with `NaN` scores to the false category in their respective classes (`NaNFlag="includenan"`), `rocmetrics` computes performance metrics as shown in the following table. For example, a threshold of 0.5 corresponds to the middle row where `rocmetrics` counts rows 2 and 4 as incorrectly classified. Notice that only the `FN` and `FP` columns differ between these two tables.
Threshold`TP``FN``FP``TN`
10211
0.51111
01120
### Pointwise Confidence Intervals
`rocmetrics` computes pointwise confidence intervals for the performance metrics, including the AUC values and score thresholds, by using either bootstrap samples or cross-validated data. The object stores the values in the `Metrics` and `AUC` properties.
You cannot specify both options. If you specify a custom metric in `AdditionalMetrics`, you must use bootstrap to compute confidence intervals. `rocmetrics` does not support cross-validation for a custom metric.
`rocmetrics` holds `FixedMetric` (threshold, FPR, TPR, or a metric specified in `AdditionalMetrics`) fixed at `FixedMetricValues` and computes the confidence intervals on AUC and other metrics for the points corresponding to the values in `FixedMetricValues`.
• Threshold averaging (TA) (when `FixedMetric` is `"Thresholds"` (default)) — `rocmetrics` estimates confidence intervals for performance metrics at fixed threshold values. The function takes samples at the fixed thresholds and averages the corresponding metric values.
• Vertical averaging (VA) (when `FixedMetric` is a performance metric) — `rocmetrics` estimates confidence intervals for thresholds and other performance metrics at the fixed metric values. The function takes samples at the fixed metric values and averages the corresponding threshold and metric values.
The function estimates confidence intervals for the AUC value only when `FixedMetric` is `"Thresholds"`, `"FalsePositiveRate"`, or `"TruePositiveRate"`.
## References
[1] Fawcett, T. “ROC Graphs: Notes and Practical Considerations for Researchers”, Machine Learning 31, no. 1 (2004): 1–38.
[2] Zweig, M., and G. Campbell. “Receiver-Operating Characteristic (ROC) Plots: A Fundamental Evaluation Tool in Clinical Medicine.” Clinical Chemistry 39, no. 4 (1993): 561–577 .
[3] Davis, J., and M. Goadrich. “The Relationship Between Precision-Recall and ROC Curves.” Proceedings of ICML ’06, 2006, pp. 233–240.
[4] Moskowitz, C. S., and M. S. Pepe. “Quantifying and Comparing the Predictive Accuracy of Continuous Prognostic Factors for Binary Outcomes.” Biostatistics 5, no. 1 (2004): 113–27.
[5] Huang, Y., M. S. Pepe, and Z. Feng. “Evaluating the Predictiveness of a Continuous Marker.” U. Washington Biostatistics Paper Series, 2006, 250–61.
[6] Briggs, W. M., and R. Zaretzki. “The Skill Plot: A Graphical Technique for Evaluating Continuous Diagnostic Tests.” Biometrics 64, no. 1 (2008): 250–256.
[7] Bettinger, R. “Cost-Sensitive Classifier Selection Using the ROC Convex Hull Method.” SAS Institute, 2003. | 8,186 | 35,594 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "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-2023-14 | longest | en | 0.847358 |
https://www.astronomyclub.xyz/spherical-aberration-2/info-xye.html | 1,566,099,323,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313589.19/warc/CC-MAIN-20190818022816-20190818044816-00500.warc.gz | 743,103,340 | 10,241 | Info
3.2 Characteristic Function and Seidel (3rd order) aberrations 59 Now it can easily be shown that the vertex curvature is c = 72 = n 1 ' (3-6)
where e is the eccentricity. Then
;z" = >2+ 8>+iCs«2 y'+■■■ <">
A hyperbola gives the same result except for alternating negative signs:
Considering now the three equations (3-3), (3-7) and (3-8) and following Schwarzschild [3-1], they can all be written from (3-6) in the form:
Z =2y2 + C3 (1 — e2)y4 + 1j(1 — e2)2y' + ... (3-9)
or, introducing the Schwarzschild (conic) constant bs [3-1] as bs = —e2 (3-10)
c c3 c5
z = 2y2 + 8 (1 + bs)y4 + 16(1 + bs)2y' + ••• (3-11)
To the third order (second term), all surfaces have conic sections, uniquely defined by bs as follows:
bs = —1, e = 1, parabola — 1 <bs < 0, 0 < e < 1, ellipse bs < —1, e > 1, hyperbola
A further case is of interest, as it can be useful in optical systems:
bs > 0, e imaginary, oblate spheroid (minor axis of an ellipse) (3-13)
The formulation of Eqs- (3-11) and (3-12) is the essential basis of aberration theory for telescopes.
3.2 Characteristic Function and Seidel (3rd order) aberrations: aberration theory of basic telescope forms
3.2.1 The Characteristic Function of Hamilton
In 1833, Hamilton [3-2] published one of the most profound and elegant analyses in the history of geometrical optics: the Characteristic Function- Based solely on the property of symmetry of a centered optical system about its axis, he deduced the general form of the aberration function in terms of three fundamental parameters: the aperture radius (p) (normalized to 1 at the edge), the field radius (a) (normalized to 1 at the edge), and the azimuth angle of the plane containing the ray and principal ray in the image forming wavefront. Following Hopkins [3.3], we will define these normalized parameters as p, a and 0, the dashes of p',a' and 0', denoting the image space, being omitted for simplicity. Hamilton showed that, because of symmetry, these parameters can only appear in the general aberration function in the forms
The aberration function must take the form, expressed as wavefront aberration:
W (a,p,0) = okoo + (0^20 p2 + ikn ap cos 0 + 2kooa2)
It is easily shown that some of these terms must be zero because of the definitions of Gaussian optics. All terms 2k00a2, 4k00a4 ... are zero; also the constant term okoo must be zero with the normal definition of the wavefront. The first terms that remain are then
W(a,p,0) = ok2op2 + ikiiap cos 0 + ok40p4 + ik3iap3 cos 0 (3 15)
This can be expressed as the general function
W(a, p, 0) = E(i+„)k(m+„),„a(!+")p(m+") cos" 0 , (3.16)
in which l, m are even positive integers or zero, and n is any positive integer or zero. However, the terms with m = n =0
are excluded if the Gaussian conditions are defined as zero. In telescope optics, this is not, in general, the case for the latter two of these conditions.
The type of aberration depends on the functions of p and cos 0, whereas the function of a shows how the effect varies in the field.
Table 3.1 shows the first, third and fifth order aberrations of this general function.
m n = 0 n = 1 n = 2 n = 3 Order
Telescopes Mastery
Through this ebook, you are going to learn what you will need to know all about the telescopes that can provide a fun and rewarding hobby for you and your family!
Get My Free Ebook | 966 | 3,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-35 | longest | en | 0.872749 |
https://brainly.com/question/146766 | 1,484,644,887,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279650.31/warc/CC-MAIN-20170116095119-00526-ip-10-171-10-70.ec2.internal.warc.gz | 805,484,327 | 8,428 | 2014-10-09T17:32:01-04:00
The easiest way to do this is to write the decimals with equal amount of places after the dot and you'll see it more clearly:
1.09 = 1.090
1.901 = 1.901
1.9 = 1.900
1.19 = 1.190
So from the least to the greatest:
1.090
1.190
1.900
1.901
In a form you gave at the beginning:
1.09 < 1.19 < 1.9 < 1.901 | 132 | 330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-04 | latest | en | 0.870184 |
http://unizor.blogspot.com/2020/08/ac-capacitors-unizorcom-physics4teens.html | 1,600,551,409,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192887.19/warc/CC-MAIN-20200919204805-20200919234805-00307.warc.gz | 138,326,189 | 11,148 | ## Saturday, August 15, 2020
### AC Capacitors: UNIZOR.COM - Physics4Teens - Electromagnetism - AC Induction
Notes to a video lecture on http://www.unizor.com
Alternating Current and Capacitors
Let's start with a clear illustration of what we will discuss in this lecture.
The main idea is that alternating current (AC) goes through a capacitor, while direct current (DC) does not.
To explain this fact, consider how electricity is produced in both cases DC and AC.
For DC case we will use a battery that uses some chemical reaction to
separate some outer layer ("free") electrons from their atoms and moves
them toward a battery's negative pole, where excess of electrons will be
present, leaving deficiency of electrons on the positive pole.
If we connect the poles of this battery to two plates of a capacitor,
the excess of electrons from the negative pole of a battery will travel
to one plate, while the deficiency of electrons will be on the other
plates. It will continue until we reach a saturation of electrons on one
plate and their absence on another. The electromotive power of the
battery could not push electrons anymore to one plate, because their
repulsive force will be too strong. And that's it. Electrons will not
jump from one plate of a capacitor to another (unless the EMF of a
battery is so strong that it breaks the isolation between the plates of a
capacitor, which defeats the purpose), the circuit is not closed, the
current will not go through.
In AC case the generated electromotive force (EMF) is variable in magnitude and direction.
It starts with separating "free" electrons from their atoms, gradually
increasing the concentration of electrons on one pole and on the
capacitor's plate connected to it.
At the same time deficiency of electrons is observed on the other pole and a capacitor's plate connected to it.
That lasts for some short period of time (half a period), when the
generated EMF gets weaker and goes down to zero. This causes the excess
electrons from one capacitor plate to go back to a generator and
compensate the deficiency of electrons on the other plate. This ends
half a period of EMF oscillation.
On the next half a period the EMF changes the sign, electrons will be
accumulated on the plate, where previously we observed their deficiency.
On the plate where we had excess of electrons during the previous half a
period we will have their deficiency.
So, during the full cycle of two half-periods electrons move to one
plate of a capacitor, then back to a generator and to another plate of a
capacitor. Then the cycle repeats itself, electrons move within a
circuit with a capacitor back and forth, thus facilitating the
alternating current in this circuit.
Let's compare the alternating current in a case of a closed circuit without a capacitor with a circuit that has a capacitor.
We start in both cases with a device that generates alternating sinusoidal EMF
U(t) = Umax·sin(ωt)
where
U(t) is the generated EMF as a function of time t,
Umax is the peak value of EMF,
ω is the angular speed of rotation of a device generating the alternating EMF.
In a closed circuit with an alternating EMF and no capacitors, according to the Ohm's Law, the electric current I(t), as a function of time t, will be proportional to an EMF and will alternate synchronously with it
I(t) = Imax·sin(ωt)
where
Imax is the peak value of electric current in a circuit.
In case of a capacitor being a part of a circuit the situation is more complex.
As we know, the amount of electricity Q(t) accumulated in the capacitor is proportional to voltage U(t) applied to its plates, and capacity of a capacitor C
(see lecture "Electric Fields" - "Capacitors" in this course) is the
constant proportionality factor that depends on a type of a capacitor
C = Q(t)/U(t)
Therefore,
Q(t) = C·U(t) = C·Umax·sin(ωt)
Knowing the amount of electricity Q(t) accumulated in a capacitor as a function of time t, we can determine the electric current I(t) in a circuit, which is a rate of change (that is, derivative by time) of the amount of electricity
I(t) = dQ(t)/dt =
= C·Umax·ω·cos(ωt) =
= Imax·cos(ωt) =
= Imax·sin(ωt+π/2)
where
Imax = C·Umax·ω
Incidentally, expression which is called reactance, plays for a capacitor similar role to that of a resistance because, using this reactance, the formula for a peak value of the electric current in a circuit with a capacitor resembles the Ohm's Law
Imax = Umax/XC
What's most important in the formula I(t)=Imax·sin(ωt+π/2)
and the most important property of a capacitor in an AC circuit is
that, while an EMF in a circuit oscillates with angular speed ω, the electric current oscillates with the same angular speed ω as EMF, but its period is shifted in time by π/2 relative to EMF.
This is a very important capacitor's property used in AC motors, which will be a subject of the next lecture.
XC = 1/(ω·C), | 1,167 | 4,915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-40 | latest | en | 0.92085 |
http://www.mrperezonlinemathtutor.com/A2/3_3_Matrices_Systems_Eq_Aug_Matrix.html | 1,371,674,247,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709135115/warc/CC-MAIN-20130516125855-00021-ip-10-60-113-184.ec2.internal.warc.gz | 484,771,562 | 3,978 | # Matrices and Systems of Equations
### Fortunately for you this lesson takes a great deal in addressing all these questions. It uses your natural ability to follow the flow of events by highlighting each step with colors, and keeping track of what has been done. You will really find it easy to understand and apply to more problems you may later encounter.
Lesson's Content
Lesson In PDF Format (no animations)
Lesson's Glossary
Determinant for matrix 2x2:
matrix
has determinant
Determiant for matrix 3x3:
matrix
has determinant
Equivalent system: A system of equations having the same solution set as another system.
Graph of an equation in two variables: All the points that may be graph from the solution set of the equation.
Graph of a function: The solution set graphed for the function in the given domain.
Graph of a number: The location of a point paired with a number in the number line.
Graph of an ordered pair: The location in the coordinate plane associated with an ordered pair of real numbers.
Linear equation: Any equation with all exponents = 1 regardless of the form the equation is represented.
Linear equation in two variables: All equivalent equations to the one in the form of ax + by = c, where a, b, and c are in the set of the real numbers and a and b can't be zero at the same time. The graph is a straight line.
Linear function: A function of the form: f(x) = mx + b.
Matrix: Rectangular arrangement of numbers in rows and columns that uses large brackets in order to define the matrix. A matrix size is defined by number of rows and number of columns.
Ordered pair: In a coordinate plane is the location of a point.
Slope of a line:The measure of how steep a line is. The change in y (rise) divided by the change in x (run). Slope = m
Slope-intercept form of an equation: The equation of a line in the form y = mx + b, where m represents the slope, and be represents the y-intercept.
Solution of an equation in two variables: Any ordered pair of real numbers that makes the sentence true.
Solution of a system of two equations in two variables: Ordered pair that when replaced in the equations produces a true statement for both equations.
Solve a system of two equations in x and y: Finding all ordered pairs (x,y) which make both equations in the system true.
Standard form of a linear equation: ax + by = c, where a, b, and c are integers and a and b are not both zero.
Didn't you find what you were looking for? Do your search here! | 568 | 2,517 | {"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} | 4.8125 | 5 | CC-MAIN-2013-20 | latest | en | 0.911176 |
https://www.stat.math.ethz.ch/pipermail/r-help/2018-March/451728.html | 1,670,453,466,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711221.94/warc/CC-MAIN-20221207221727-20221208011727-00695.warc.gz | 1,034,915,767 | 2,346 | # [R] adjusted values
Cristiano Alessandro cri.alessandro at gmail.com
Thu Mar 22 16:43:27 CET 2018
```Hi all,
I am fitting a linear mixed model with lme4 in R. The model has a single
factor (des_days) with 4 levels (-1,1,14,48), and I am using random
intercept and slopes.
Fixed effects: data ~ des_days
Value Std.Error DF t-value p-value
(Intercept) 0.8274313 0.007937938 962 104.23757 0.0000
des_days1 -0.0026322 0.007443294 962 -0.35363 0.7237
des_days14 -0.0011319 0.006635512 962 -0.17058 0.8646
des_days48 0.0112579 0.005452614 962 2.06469 0.0392
I can clearly use the previous results to compare the estimations of each
"des_day" to the intercept, using the provided t-statistics. Alternatively,
I could use post-hoc tests (z-statistics):
> ph_conditional <- c("des_days1 = 0",
"des_days14 = 0",
"des_days48 = 0");
> lev.ph <- glht(lev.lm, linfct = ph_conditional);
> summary(lev.ph)
Simultaneous Tests for General Linear Hypotheses
Fit: lme.formula(fixed = data ~ des_days, data = data_red_trf, random
= ~des_days |
ratID, method = "ML", na.action = na.omit, control = lCtr)
Linear Hypotheses:
Estimate Std. Error z value Pr(>|z|)
des_days1 == 0 -0.002632 0.007428 -0.354 0.971
des_days14 == 0 -0.001132 0.006622 -0.171 0.996
des_days48 == 0 0.011258 0.005441 2.069 0.101
(Adjusted p values reported -- single-step method)
The p-values of the coefficient estimates and those of the post-hoc tests
differ because the latter are adjusted with Bonferroni correction. I wonder
whether there is any form of correction in the coefficient estimated of the
LMM, and which p-values are more appropriate to use.
Thanks
Cristiano
[[alternative HTML version deleted]]
``` | 579 | 1,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-49 | latest | en | 0.662821 |
https://number.academy/28847 | 1,652,680,546,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662509990.19/warc/CC-MAIN-20220516041337-20220516071337-00733.warc.gz | 513,591,733 | 12,031 | # Number 28847
Number 28,847 spell 🔊, write in words: twenty-eight thousand, eight hundred and forty-seven . Ordinal number 28847th is said 🔊 and write: twenty-eight thousand, eight hundred and forty-seventh. The meaning of number 28847 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 28847. What is 28847 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 28847.
## What is 28,847 in other units
The decimal (Arabic) number 28847 converted to a Roman number is (X)(X)(V)MMMDCCCXLVII. Roman and decimal number conversions.
#### Weight conversion
28847 kilograms (kg) = 63596.1 pounds (lbs)
28847 pounds (lbs) = 13084.9 kilograms (kg)
#### Length conversion
28847 kilometers (km) equals to 17925 miles (mi).
28847 miles (mi) equals to 46425 kilometers (km).
28847 meters (m) equals to 94642 feet (ft).
28847 feet (ft) equals 8793 meters (m).
28847 centimeters (cm) equals to 11357.1 inches (in).
28847 inches (in) equals to 73271.4 centimeters (cm).
#### Temperature conversion
28847° Fahrenheit (°F) equals to 16008.3° Celsius (°C)
28847° Celsius (°C) equals to 51956.6° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
28847 seconds equals to 8 hours, 47 seconds
28847 minutes equals to 2 weeks, 6 days, 47 minutes
### Codes and images of the number 28847
Number 28847 morse code: ..--- ---.. ---.. ....- --...
Sign language for number 28847:
Number 28847 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Mathematics of no. 28847
### Multiplications
#### Multiplication table of 28847
28847 multiplied by two equals 57694 (28847 x 2 = 57694).
28847 multiplied by three equals 86541 (28847 x 3 = 86541).
28847 multiplied by four equals 115388 (28847 x 4 = 115388).
28847 multiplied by five equals 144235 (28847 x 5 = 144235).
28847 multiplied by six equals 173082 (28847 x 6 = 173082).
28847 multiplied by seven equals 201929 (28847 x 7 = 201929).
28847 multiplied by eight equals 230776 (28847 x 8 = 230776).
28847 multiplied by nine equals 259623 (28847 x 9 = 259623).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 28847
Half of 28847 is 14423,5 (28847 / 2 = 14423,5 = 14423 1/2).
One third of 28847 is 9615,6667 (28847 / 3 = 9615,6667 = 9615 2/3).
One quarter of 28847 is 7211,75 (28847 / 4 = 7211,75 = 7211 3/4).
One fifth of 28847 is 5769,4 (28847 / 5 = 5769,4 = 5769 2/5).
One sixth of 28847 is 4807,8333 (28847 / 6 = 4807,8333 = 4807 5/6).
One seventh of 28847 is 4121 (28847 / 7 = 4121).
One eighth of 28847 is 3605,875 (28847 / 8 = 3605,875 = 3605 7/8).
One ninth of 28847 is 3205,2222 (28847 / 9 = 3205,2222 = 3205 2/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
28847
#### Is Prime?
The number 28847 is not a prime number. The closest prime numbers are 28843, 28859.
#### Factorization and factors (dividers)
The prime factors of 28847 are 7 * 13 * 317
The factors of 28847 are 1 , 7 , 13 , 91 , 317 , 2219 , 4121 , 28847
Total factors 8.
Sum of factors 35616 (6769).
#### Powers
The second power of 288472 is 832.149.409.
The third power of 288473 is 24.005.014.001.423.
#### Roots
The square root √28847 is 169,844046.
The cube root of 328847 is 30,669043.
#### Logarithms
The natural logarithm of No. ln 28847 = loge 28847 = 10,269761.
The logarithm to base 10 of No. log10 28847 = 4,460101.
The Napierian logarithm of No. log1/e 28847 = -10,269761.
### Trigonometric functions
The cosine of 28847 is 0,624539.
The sine of 28847 is 0,780993.
The tangent of 28847 is 1,250511.
### Properties of the number 28847
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 28847 in Computer Science
Code typeCode value
28847 Number of bytes28.2KB
Unix timeUnix time 28847 is equal to Thursday Jan. 1, 1970, 8:47 a.m. GMT
IPv4, IPv6Number 28847 internet address in dotted format v4 0.0.112.175, v6 ::70af
28847 Decimal = 111000010101111 Binary
28847 Decimal = 1110120102 Ternary
28847 Decimal = 70257 Octal
28847 Decimal = 70AF Hexadecimal (0x70af hex)
28847 BASE64Mjg4NDc=
28847 MD5e4825224ef50eac17bcdf3576f66784e
28847 SHA1f87f16c8c6dd89194a5dcb457450e4024868b5da
28847 SHA22430ac9843c8b8a2b864e089d52c7322b10af59aab6e4d0acc7abf7703
28847 SHA384a03dbf5420510e45b7570dde1b3344d81da293a939f575996591d73f39ab8367d49a934d507ae627cd5522b0da1e7683
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## Numerology 28847
### Character frequency in number 28847
Character (importance) frequency for numerology.
Character: Frequency: 2 1 8 2 4 1 7 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 28847, the numbers 2+8+8+4+7 = 2+9 = 1+1 = 2 are added and the meaning of the number 2 is sought.
## Interesting facts about the number 28847
### Asteroids
• (28847) 2000 JT50 is asteroid number 28847. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 5/9/2000.
## Number 28,847 in other languages
How to say or write the number twenty-eight thousand, eight hundred and forty-seven in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 28.847) veintiocho mil ochocientos cuarenta y siete German: 🔊 (Anzahl 28.847) achtundzwanzigtausendachthundertsiebenundvierzig French: 🔊 (nombre 28 847) vingt-huit mille huit cent quarante-sept Portuguese: 🔊 (número 28 847) vinte e oito mil, oitocentos e quarenta e sete Chinese: 🔊 (数 28 847) 二万八千八百四十七 Arabian: 🔊 (عدد 28,847) ثمانية و عشرون ألفاً و ثمانمائة و سبعة و أربعون Czech: 🔊 (číslo 28 847) dvacet osm tisíc osmset čtyřicet sedm Korean: 🔊 (번호 28,847) 이만 팔천팔백사십칠 Danish: 🔊 (nummer 28 847) otteogtyvetusinde og ottehundrede og syvogfyrre Dutch: 🔊 (nummer 28 847) achtentwintigduizendachthonderdzevenenveertig Japanese: 🔊 (数 28,847) 二万八千八百四十七 Indonesian: 🔊 (jumlah 28.847) dua puluh delapan ribu delapan ratus empat puluh tujuh Italian: 🔊 (numero 28 847) ventottomilaottocentoquarantasette Norwegian: 🔊 (nummer 28 847) tjue-åtte tusen, åtte hundre og førti-syv Polish: 🔊 (liczba 28 847) dwadzieścia osiem tysięcy osiemset czterdzieści siedem Russian: 🔊 (номер 28 847) двадцать восемь тысяч восемьсот сорок семь Turkish: 🔊 (numara 28,847) yirmisekizbinsekizyüzkırkyedi Thai: 🔊 (จำนวน 28 847) สองหมื่นแปดพันแปดร้อยสี่สิบเจ็ด Ukrainian: 🔊 (номер 28 847) двадцять вiсiм тисяч вiсiмсот сорок сiм Vietnamese: 🔊 (con số 28.847) hai mươi tám nghìn tám trăm bốn mươi bảy Other languages ...
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## Comment
If you know something interesting about the number 28847 or any natural number (positive integer) please write us here or on facebook. | 2,473 | 7,097 | {"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} | 2.875 | 3 | CC-MAIN-2022-21 | latest | en | 0.684405 |
http://mathhelpforum.com/calculus/168068-help-series.html | 1,529,372,032,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861641.66/warc/CC-MAIN-20180619002120-20180619022120-00340.warc.gz | 201,523,924 | 9,890 | # Thread: Help with a series
1. ## Help with a series
I have the series (from 1 to infinity) of (1-(1/e^n))^n. I can't seem to find a test to help me establish convergence or divergence.
Thanks for your help!
Bret
2. What is the limit at infinity for (1-(1/e^n))^n?
If it is not zero, then it diverges.
3. $\displaystyle $\mathop {\lim }\limits_{n \to \infty } {(1 - \frac{1}{{{e^n}}})^n} = \mathop {\lim }\limits_{n \to \infty } {e^{\ln (1 - \frac{1}{{{e^n}}}) \cdot n}} = {e^{\mathop {\lim }\limits_{n \to \infty } \frac{{ - {n^2}}}{{{e^n} - 1}}}} = {e^{\mathop {\lim }\limits_{n \to \infty } \frac{{ - 2}}{{{e^n}}}}} = 1$$
4. Thanks everyone. I get really embarassed when I forget basic things. Cool explanation Pranas. Thanks also snowtea. | 264 | 750 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-26 | latest | en | 0.766924 |
https://crosswordfiend.com/2015/08/12/thursday-august-13-2015/ | 1,723,230,774,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640768597.52/warc/CC-MAIN-20240809173246-20240809203246-00223.warc.gz | 131,598,332 | 45,646 | # Thursday, August 13, 2015
NYT 7:20 (Derek)
LAT 4:31 (Gareth)
BEQ 7:45 (Ben)
### Jim Hilger’s New York Times crossword – Derek’s write-up
Trading with Doug Peterson on the Thursday NYT; Doug will be back to blog the Friday puzzle. He said to me earlier that he hoped I would get a fun one, and he was right! The instructions to the puzzle state: “When this puzzle is done, four squares will remain empty. Which ones and why are for you to determine. The grid shows “which ones,” and the “why” is that there are eight answers that all end in “BREAK.” Here are the pertinent clues/answers, with an underscrore where the empty grid square is:
• 17A [12:05 t0 12:20 p.m., maybe] SHORT L_UNCH (BREAK)
• 34A [Bit of misfortune] B_AD (BREAK)
• 36A [A couple weeks off partying in Florida, say] SPR_ING VACATION (BREAK)
• 6D [Daring escape] JA_IL (BREAK)
• 22D [It might include the line “You’re listening to WABC”] STA_TION (BREAK)
• 23D [Turning point in a tennis match, maybe] SER_VICE (BREAK)
• 46D [Lover’s hurt] HEA_RT (BREAK)
Pretty clever. Puzzle started off a tad difficult, especially since there is an element of slight fear knowing you’re going to have empty squares! I got the clue from STATION BREAK. I thought at first it was STATION ID, but that didn’t fit. Then the middle answer had to something along the lines of spring “break,” so from there it was just looking for answers that could be a phrase ending in break. Minor quibbles: seemed like an overuse of “say” and “maybe” in some of the thematic clues. And is “spring break vacation” ever said? I would think most people just say “spring break.” I said my quibbles were minor. Decent fill in this puzzle. I liked OLD BATDC AREA (I think I’ve seen this often recently), SHAVETAILSORVISENNUI (one of my favorite words!), and JINX. Some crosswordese used (AZOAAREAGAR, and DOST to name a few), but not to an annoying degree. I didn’t know 4D [Sword: Fr.] FER
All in all a fun puzzle, which is what Doug it would be! 4.1 stars, mainly for the fresh idea. Here’s hoping Doug gets an equally fun puzzle for Friday!
### Brendan Emmett Quigley’s website crossword – “ATOP” — Ben’s Review
I’m going to partially blame my slower time on this week’s puzzle to Lollapuzzoola, as the wide range of creative puzzles at the tournament led to me overthinking what the trick was to this week’s theme answers. I was looking for answers to be atop one another when I should have considering the title was a hint that I should be looking to change A TO P in the theme answers:
• 16A: Continuous change of a nothing worker? — PEON FLUX
• 19A: Cleaned up after a drive-by? — SWEPT BULLETS
• 35A: Skip the surveillance? — GO WITHOUT SPYING
• 49A: Bronco tamer who avoided mass layoffs? — BUSTER KEPT ON
• 55A: The tater to rule them all? — KING SPUD
It was a mixed week for my music references, as I immediately knew that 33A‘s “‘Tiny Bubbles’ entertainer” was DON HO (thanks, that one episode of The Mole!), but couldn’t remember Frank Zappa’s “Burnt WEENY Sandwich” (40A) without a few of the crossing clues. By the time I got to it in the downs, I had enough crossings to realize that singer 5D was looking for (“He died on ‘The Day the Music Died'”) was VALENS, so it’s a mixed bag all around.
Lollapuzzoola was also on my brain when I hit 44D. “Low-fare bus” didn’t resolve to BOLT (too few letters), but then I remembered the JITNEY as well and it all made sense. Similarly the “versatile white bean” at 63A wasn’t going to be a CANNELINI (too many letters this time), so SOY fit the bill. It was also nice to see Cheri OTERI get a call out in 46D for something other than her work on other frequent crossword filler SNL. I’ve seen a lot of ALI Larter (known for her work in the “Resident Evil” movies, if 27D is any indication) in puzzles lately – must be something going on in the puzzle hivemind.
This week’s puzzle is a step down from last week’s fantastic puzzle, but it was a fun solve nonetheless. A solid Thursday from BEQ.
3.5/5 stars
### Mark Bickham’s LA Times crossword – Gareth’s write-up
LA Times
150813
It’s “word that completes” theme time today with Mark Bickham. Today’s word is HOLDING and it completes the second part of 5 theme answers. The revealer; [Type of Tax…], WITHHOLDING; is on the bland side, but HOLDING itself is more interesting in the phrases that it forms. HOLDING court, HOLDING hands, HOLDING pattern, HOLDING cell, and HOLDING back make for a more interesting and idiomatic set than most.
The answers themselves are mostly solid, with two gems – the spoken word BERIGHTBACK and LOOKMANOHANDS. Listed, we have:
• [“Hang on for a sec”], BERIGHTBACK
• [It can be white or red], BLOODCELL
• [Way with words?], SPEECHPATTERN
• [Cry from a daring biker], LOOKMANOHANDS. I spent two days unconscious and permanently lost the hearing in my right ear in a bike accident at age 8, so that cuts a little close to home…
• [French Open surface], CLAYCOURT
• [Type of tax … or how the ends of the other five longest across answers might be seen?], WITHHOLDING
The medical subtheme of BLOODCELL continues with two of the punchier non-theme answer: a LABTEST and WEBMD – a mixed blessing of the internet age. Along with 6 themers, Mr. Bickham manages two long downs in SHARONSTONE and BEEFSAUSAGE. DKNY and THEOC are also fun choices among the medium-length fill.
For a densely themed crossword, there are not too many casualties: RESEEK is not the most “in the language” “re” word there is. There are also several short abbreviations, but mostly of the common and in the language variety – no foul in my book.
Clue commentary:
• [Casino conveniences], ATMS. Convenient for whom?
• [Literary breakdown], GENRES. Simple, elegant clue.
• Jeans appliqué], IRONON. Bonus points for “appliqué”!
Technically a well put together puzzle.
3.5 Stars
Gareth
### Alan Arbesfeld’s CrosSynergy/Washington Post crossword, “Double O Seven”—Ade’s write-up
CrosSynergy/Washington Post crossword solution, 08.13.15: “Double O Seven”
Hello once again! Hope all is well, and my apologies for the late post. Today’s crossword puzzle, brought to us by Mr. Alan Arbesfeld has a grand total of seven themes, and each of them are two-word entries in which each of the words start with the letter “O.” Funny enough, I was having a conversation with a TV producer/director and a cameraman from the French sports station RDS, and the producer and I came up with nickname for the cameraman, who was known as someone that was very charming to the ladies. We said he was “James Bond.”
• OVAL OFFICE (17A: [It’s part of the West Wing])
• OPTS OUT (21A: [Chooses not to participate])
• OLIVE OIL (28A: [Italian dressing ingredient])
• OSAGE ORANGE (34A: [Fruit of the mulberry family])
• OLE OLSEN (47A: [“Hellzapoppin'” costar])
• OLD ONES (53A: [Well-worn witticisms])
• OPEN ORDERS (58A: [Market authorizations awaiting execution])
This is probably only the second or third time I’ve ever seen POOR BOY spelled out in reference to the sandwich (1D: [New Orleans hero sandwich (var.)]). I’ve had a couple of Po’boy sandwiches and I can’t say that I’ve been blown away. Must have not gotten the sandwich from the right place. Nice to be in Quebec and be greeted with a French expression, ET VOILÀ, in a grid (2D: [Magician’s dramatic cry]). Oh, and if you’re not a sports fan, then you might not know that SAFECO, after last night, is the sight of the latest no-hitter in Major League baseball (5D: [______ Field (Seattle Mariners stadium]). Hisashi Iwakuma of the Seattle Mariners pulled off the trick last night, no-hitting the Baltimore Orioles in a 3-0 win. Iwakuma is the second Japanese pitcher to throw a no-hitter in Major League Baseball, as Hideo Nomo threw two in his career, including one of the most impossible feats in the history of sports: throwing a no-hitter at EXTREMELY hitter-friendly Coors Field, which he did on Sept. 17, 1996. Don’t believe me?…
“Sports will make you smarter” moment of the day: EATONS (48D: [2012 Olympic decathlon gold medalist and “Goldfinger” actress Shirley]) – Of the two EATONS referenced, I’ll concentrate on Ashton, the American who not only won the decathlon at the 2012 Olympics on London, but, during the Olympic Trials leading up to those games, set the world record for most points in a decathlon, with 9,039. He also holds the world record for most points accumulated in an indoor heptathlon, with 6,645.
It’s TGIV tomorrow!! (Thank goodness it’s vendredi!) See you on Friday!
Take care!
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### 8 Responses to Thursday, August 13, 2015
1. Avg Solvr says:
I think you mean SPR_ING (BREAK) VACATION. Liked the theme.
2. huda says:
Never saw the note, so I was flying blind for quite a while and tumble to it at STATION as well, and then HEART soon after.
Never heard of SHAVETAILS so it took a lot of crosses.
I too really liked the theme.
Across Lite forced me to write BREAK in the empty squares before I got the happy pencil…
• CY Hollander says:
Never saw the note
I wish I hadn’t. These tricky themes are much better without the hint. I wish they wouldn’t print them over the puzzle, where they’re hard to miss.
I read [Beldam] as [Bedlam] and was quite surprised when HAG/SHAVETAILS was correct. (I almost opted for SLAVETAILS since they were rookies.)
This is one where I wished I hadn’t read the note first, I think the hint gave too much away for a Thursday.
• Derek Allen says:
I thought the hint was too much as well. As mentioned, a rebus with BREAK in the square would work, but I think I get the idea of including an actual word “break” in the grid. Still a fun solve.
• CY Hollander says:
This is one where I wished I hadn’t read the note first, I think the hint gave too much away for a Thursday.
100% agree. It’s such a waste of a fresh twist to tell you exactly what it is in advance.
• Gareth says:
I never read the notes. It’s easy not to in Across Lite. I like my puzzle themes abstruse…
4. Howard B says:
Loved the theme, fill was rather tough; SHAVETAILS, ORVIS and BREDA may be rather different fill, but they just created additional blockages and detours here, instead of enjoyment on discovery. None of them were on my radar, but were not very interesting after working through every crossing of each answer there.
N.B. – I didn’t read the hint, so that did throw me for a bit – but also added to the enjoyment. It’s nice to have the note available if you want it though. | 2,804 | 10,530 | {"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-2024-33 | latest | en | 0.929161 |
https://www.jiskha.com/display.cgi?id=1332802367 | 1,503,236,733,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106754.4/warc/CC-MAIN-20170820131332-20170820151332-00050.warc.gz | 921,510,805 | 3,679 | Math problem
posted by .
subtract simplify if possible i am confused
(4-z)/(z-6)-(6z-5)/(6-z)
• Math problem -
You have
(4-z)/(z-6) + (5-6z)/(z-6)
= (4-z+5-6z)/(z-6)
= (9-7z)/(z-6)
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More Similar Questions | 569 | 1,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-34 | latest | en | 0.798355 |
https://math.stackexchange.com/questions/1260880/understanding-proof-that-sequence-x-n-1n-is-not-convergent | 1,643,179,352,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304915.53/warc/CC-MAIN-20220126041016-20220126071016-00113.warc.gz | 426,107,190 | 34,525 | # Understanding proof that sequence $x_n =( -1)^{n}$ is not convergent
To Understanding proof that sequence $x_n =( -1)^{n}$ is not convergent .
Here goes the proof :-
Firstly i assume that sequence converges to x .i.e -1 and 1 lies between $x$-$\epsilon$ and $x+$$\epsilon$ . The length of interval is $2 \epsilon$ which is greater than $2$ (as can be seen by drawing number line ) . so this happens for $\epsilon$ greater than 1 .Because since $2\epsilon$ can only accommodate -1 and 1 if it is greater than 2 i.e if epsilon is greater than 1 .How do i deal with other case when epsilon is less than 1
Thanks
• This reminds me of a joke..."let $\epsilon$ be less than 0".
– user223391
May 1 '15 at 17:10
• It is easy to conclude that this sequence is not a Cauchy sequence. Hence it is not convergent. May 1 '15 at 17:12
• Have you learned about Cauchy sequences yet? May 1 '15 at 17:12
• @Sloan not yet ,no May 1 '15 at 17:13
• I gave a strong hint. The case when $\epsilon<1$ should be the easier case to prove--and you really don't need cases... just make $\epsilon$ very small. May 1 '15 at 17:13
The definition of a limit is that for any $\epsilon>0$ there exists an $N$ so that if $n\geq N$ then $|x_{n}-x|<\epsilon$. Now, you just need to argue that if $|-1 - x|<\epsilon$ then $|1-x|>\epsilon$ which is a contradiction (to the assumption that $x_n\to x$). Take $\epsilon$ as small as lets you make your argument (smaller than $1$).
• if mod of -1 -x is less than $\epsilon$ ,then how is mod 1 -x less than $\epsilon$ May 1 '15 at 17:19
• @K.Dutta, If $|-1-x|<\epsilon$ then $|1-x|>\epsilon$ is the point... we don't want both to be smaller than $\epsilon$ (that's how we get the contradiction). May 1 '15 at 18:11
• How come "$|-1 - x|<\epsilon$ then $|1-x|>\epsilon$"? All I can come up with is $|1 + x|<\epsilon$. Mar 7 '16 at 2:42
• @Tim, it helps to think of what $|-1-x|$ means... it is the distance from $x$ to $-1$. If that distance is small, then $x$ cannot also be close to $1$. And, $|1-x|$ is the distance from $x$ to $1$. Basically, all this is saying is that you cannot be simultaneously close to $1$ and $-1$. Mar 7 '16 at 13:51
You can ignore tha case $\epsilon>1$ right away. Convergence says that for every $\epsilon>0$ we have $|x_n-x|<\epsilon$ for all $n>n_0$ (with $n_0$ dependig on $\epsilon$). Specifically, it should hold for $\epsilon=\frac1{42}$. Since it doesn't, we do not have convergence. | 775 | 2,436 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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-2022-05 | latest | en | 0.897304 |
http://www.physicsforums.com/showthread.php?t=504269 | 1,368,986,627,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697917013/warc/CC-MAIN-20130516095157-00008-ip-10-60-113-184.ec2.internal.warc.gz | 643,552,383 | 7,056 | ## electricity and magnetism
PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
Recognitions: Homework Help You're trying to calculate the current induced by the movement of the rod. That current is 0 when the rod is at equilibrium, because the rod wouldn't be moving. Even when it's not at equilibrium, the induced current is usually so small as to be negligible compared to the current supplied by the power source. So you just need to consider the force applied by the magnetic field on the rod, due to the existing current I. There's no need to consider induction.
OK.... As I try your idea I get into the following dilemma: which way is V? The movement of the rod or the movement of the current? Unfortunately neither one of those options creates a force upwards using the right hand rule... \: | 189 | 904 | {"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-2013-20 | latest | en | 0.932774 |
http://mathsblog.co.uk/2007/08/02/maths-worksheets-partition-into-thhtu-s4/ | 1,521,366,222,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645604.18/warc/CC-MAIN-20180318091225-20180318111225-00171.warc.gz | 190,108,564 | 9,621 | Posted by Peter on 2nd August 2007
# Maths worksheets: Partition into ThHTU. Y4
A good understanding of place value is crucial for successful mental arithmetic and progress with written methods of the four rules. The value of a digit depends on its place in the number. For example: the 2 in 1 234 is worth 200 whilst the 2 in 4 321 is worth only 20. Our whole number system is based around this idea but don’t be surprised if your children find it difficult – many do! It seems quite an old fashioned idea to place the letters HTU to stand for hundreds, tens and units, above a number or sum, but it does help with this understanding and in reading numbers out loud.
In year 4 children are expected to read numbers up to the thousands. Many books separate the thousands from the hundreds by a very small gap and when we write the number in words we place a comma between the thousands and the rest of the number. The reason for this is to help with reading the number.
4201-04 Partition into ThHTU (pg 2) | 233 | 1,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-13 | longest | en | 0.94577 |
https://hilbertthm90.wordpress.com/2009/11/03/the-next-inequality/ | 1,500,633,806,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423769.10/warc/CC-MAIN-20170721102310-20170721122310-00646.warc.gz | 679,823,560 | 39,644 | # The Next Inequality
Considering it has been at least a post removed, I’ll bring us back to our situation. We have a local Noetherian ring $(R, \frak{m})$. Our notation is that $\delta(R)$ is the least number of generators of an $\frak{m}$-primary ideal (which was shown to be independent of choice of ideal here). The goal for the day is to show that $d(R)\geq \dim R$.
Suppose $\frak{q}$ is $\frak{m}$-primary. We’ll prove something more general. Let $M$ be a finitely generated $R$-module, $x\in R$ a non-zero divisor in $M$ and $M'=M/xM$. Then the claim is that $\deg\chi_q^{M'}\leq \deg\chi_q^M -1$.
Since $x$ is not a zero-divisor, we have an iso as $R$-modules: $xM\cong M$. Define $N=xM$. Now take $N_n=N\cap \frak{q}^nM$. Since $\frak{q}^nM$ is a stable $\frak{q}$-filtration of $M$, by Artin-Rees we get that $(N_n)$ is a stable $\frak{q}$-filtration of $N$.
For each $n$ we have $0\to N/N_n \to M/\frak{q}^nM\to M'/\frak{q}^nM'\to 0$ exact.
Thus we get $l(N/N_n)-l(M/\frak{q}^nM)+l(M'/\frak{q}^nM')=0$. So if we let $g(n)=l(N/N_n)$, we get for large $n$: $g(n)-\chi_q^M(n)+\chi_q^{M'}(n)=0$.
But $(N_n)$ is also a stable $\frak{q}$-filtration of $M$, since $N\cong M$. We already showed that the degree and leading coefficient of $g(n)$ depends only on $M$ and $\frak{q}$ and not on the filtration. Thus $g(n)$ and $\chi_q^M(n)$ have the same degree and leading coefficient, so the highest powers kill eachother which gives $\deg\chi_q^{M'}\leq \deg \chi_q^M-1$.
In particular, we will need that $R$ as an $R$-module gives us $d(R/(x))\leq d(R)-1$.
Now we prove the goal for today. For simplicity, let $d=d(R)$. We will induct on $d$. The base case gives that $l(R/\frak{m}^n)$ is constant for large $n$. In particular, there is some $N$ such that $\frak{m}^n=\frak{m}^{n+1}$ for all $n>N$. But we are local, so $\frak{m}=J(R)$ and hence by Nakayama, $\frak{m}^n=0$. Thus for any prime ideal $\frak{p}$, we have $\frak{m}^k\subset \frak{p}$ for some $k$, so take radicals to get $\frak{m}=\frak{p}$. Thus there is only one prime ideal and we actually have an Artinian ring and hence have $\dim R=0$.
Now suppose $d>0$ and the result holds for $\leq d-1$. Let $p_0\subset p_1\subset \cdots \subset p_r$ be a chain of primes. Choose $x\in p_1\setminus p_0$. Define $R'=R/p_0$ and $\overline{x}$ be the image of $x$ in $R'$.
Note that since $R'$ is an integral domain, and $\overline{x}$ is not 0, it is not a zero-divisor. So we use our first proof from today to get that $d(R'/(\overline{x}))\leq d(R')-1$.
Let $\frak{m}'$ be the maximal ideal of $R'$. Then $R'/\frak{m}'$ is the image of $R/\frak{m}$, so $l(R/\frak{m}^n)\geq l(R'/\frak{m}'^n)$ which is precisely $d(R)\geq d(R')$. Plugging this into the above inequality gives $d(R'/(\overline{x}))\leq d(A)-1=d-1$.
So by the inductive hypothesis, $\dim(R'/\overline{x})\leq d-1$. Take our original prime chain. The images form a chain $\overline{p}_1, \ldots , \overline{p}_r$ in $R'/(\overline{x})$. Thus $r-1\leq d-1\Rightarrow r\leq d$. Since the chain was arbitrary, $\dim R\leq d(R)$.
A nice corollary here is that the dimension of any Noetherian local ring is finite. Another similar corollary is that in any Noetherian ring (drop the local) the height of a prime ideal is finite (and hence primes satisfy the DCC), since $ht(p)=\dim A_p$ which is local Noetherian. | 1,204 | 3,348 | {"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": 80, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.82193 |
https://ch.mathworks.com/matlabcentral/answers/1742655-im-currently-still-learning-how-to-do-matlab-why-do-i-getting-index-in-position-1-exceeds-array-boun | 1,708,540,483,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473524.88/warc/CC-MAIN-20240221170215-20240221200215-00091.warc.gz | 183,230,907 | 26,552 | # im currently still learning how to do Matlab why do i getting Index in position 1 exceeds array bounds. Index must not exceed 6? (line 51) Rob.plot(y(i,:));
5 views (last 30 days)
Timothy Ambun on 17 Jun 2022
Commented: Timothy Ambun on 18 Jun 2022
clear all;
% Robot Kinematics
% Robot dimension
L1 = 0.4; L2 =0.5; L3 = 0.4; d1 = 0.05;
%DH Parameters
L(1) = Link([0 L1 0 1.5708])
L(2) = Link([0 0 L2 0 ])
L(3) = Link([0 0 L3 0 ])
Rob.name = ('Group7')
P = transl(2.338e-06, 0.5365, 1.036)
P1= transl(0.1, 0.3, -0.3)
P2= transl(0.4, -0.2, 0.1)
P3= transl(-0.4, 0.4, -0.3)
P4= transl(-0.3,-0.3, 0.2)
q0=[0 0 0];
%Inverse
I=Rob.ikine(P,'q0',[0 0 0],'mask',[1 1 1 0 0 0]);
I1=Rob.ikine(P1,'q0',[0 0 0],'mask',[1 1 1 0 0 0]);
I2=Rob.ikine(P2,'q0',[0 0 0],'mask',[1 1 1 0 0 0]);
I3=Rob.ikine(P3,'q0',[0 0 0],'mask',[1 1 1 0 0 0]);
I4=Rob.ikine(P4,'q0',[0 0 0],'mask',[1 1 1 0 0 0]);
%Trajectory
t=0:0.04:0.2;
y=jtraj(q0,I,t);
y1=jtraj(I,I1,t);
y2=jtraj(I1,I2,t);
y3=jtraj(I2,I3,t);
y4=jtraj(I3,I4,t);
y5=jtraj(I4,q0,t);
hold on
atj=zeros(4,4);
view(3)
for i = 1:1:51
atj=Rob.fkine(y(1,:));
jta=transl(atj);
JTA(i,:) = jta;
jta=JTA;
plot2(jta(i,:),'r.')
Rob.plot(y(i,:));
plot2 (JTA,'b')
end
for i= 1:1:51
atj1=Rob.fkine(y1(1,:));
jta=transl(atj1);
JTA1(i,:) = jta;
jta=JTA1;
plot2(jta(i,:),'r.')
Rob.plot(y1(i,:))
plot2 (JTA1,'b')
end
for i= 1:1:51
atj2=Rob.fkine(y2(1,:));
jta=transl(atj2);
JTA2(i,:) = jta;
jta=JTA2;
plot2(jta(i,:),'r.')
Rob.plot(y1(i,:))
plot2 (JTA2,'b')
end
for i= 1:1:51
atj3=Rob.fkine(y3(1,:));
jta=transl(atj3);
JTA3(i,:) = jta;
jta=JTA3;
plot2(jta(i,:),'r.')
Rob.plot(y3(i,:))
plot2 (JTA3,'b')
end
for i= 1:1:51
atj4=Rob.fkine(y4(1,:));
jta=transl(atj4);
JTA4(i,:) = jta;
jta=JTA4;
plot2(jta(i,:),'r.')
Rob.plot(y4(i,:))
plot2 (JTA4,'b')
end
for i= 1:1:51
atj5=Rob.fkine(y5(1,:));
jta=transl(atj5);
JTA5(i,:) = jta;
jta=JTA5;
plot2(jta(i,:),'r.')
Rob.plot(y5(i,:))
plot2 (JTA5,'b')
end
Geoff Hayes on 17 Jun 2022
@Timothy Ambun - the error message is telling you that there are only 6 elements in y so
for i = 1:1:51
atj=Rob.fkine(y(1,:));
jta=transl(atj);
JTA(i,:) = jta;
jta=JTA;
plot2(jta(i,:),'r.')
Rob.plot(y(i,:));
plot2 (JTA,'b')
end
since i iterates from 1 to 51, then when i is 7, you will observe this error. You should be able to use the MATLAB debugger to confirm that y only has six items.
Also, why have you chosen 51? How does that relate to anything?
Timothy Ambun on 18 Jun 2022
okay thank you, i chose 51 because i copied the code from my lecturer and he can run it well. But i have change it to 6 and it work well. Thanks for you explaination.
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Start Hunting! | 1,249 | 2,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-10 | latest | en | 0.240788 |
http://www.lofoya.com/Solved/18/two-vessels-a-and-b-contain-spirit-and-water-mixed-in-the-ratio-5-2 | 1,513,197,336,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948530841.24/warc/CC-MAIN-20171213201654-20171213221654-00185.warc.gz | 399,342,142 | 14,332 | # Moderate Alligations or Mixtures Solved QuestionAptitude Discussion
Q. Two vessels A and B contain spirit and water mixed in the ratio 5:2 and 7:6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel c containing spirit and water in the ratio 8:5?
✖ A. 1 : 7 ✖ B. 2 : 9 ✔ C. 7 : 9 ✖ D. 3 : 8
Solution:
Option(C) is correct
Spirit in 1 litre mix of A = 5/7 litre.
Spirit in 1 litre mix of B = 7/13 litre.
Spirit in 1 litre mix of C = 8/13 litre.
By rule of alligation we have required ratio X:Y
X : Y
5/7 7/13
\ /
(8/13)
/ \
(1/13) : (9/91)
7 9
Therefore required ratio = 1/13 : 9/91
= 7:9
## (1) Comment(s)
Gokul
()
frnds i dont know how to calculate mean value | 268 | 795 | {"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} | 3.828125 | 4 | CC-MAIN-2017-51 | longest | en | 0.682232 |
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0 Members and 2 Guests are viewing this topic.
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##### Re: Starting my project :Tube cell
« Reply #8 on: February 19, 2009, 10:48:39 am »
stevie !!!!! mathhhhhhh stevie MATH!!! hahah sorry but you just took the diameter....you just included the left side spacing and the right side spacing.....you need to take the radius!!!! NOT THE DIAMETERRRRR.....your math is right for the diameter now just divide it by 2!
(http://i2.photobucket.com/albums/y2/kinesisfilms/TUBESSSSS.jpg)
Hi Kinesis,
well, As you state it like this, it looks like i done something dumn...but....
But the facts as i see are different:
http://www.mathgoodies.com/lessons/vol2/circumference.html
Diameter = 2 times the radius
WO1992007861A1.pdf page 9 and page 14
Rod of 0.5 inch outer diameter = 0.5 x 25.4 = 12.7mm
Tube of 0.75 inside diameter = 0.75 x 25.4 = 19.05mm
Total gap = 6.35mm
Devided by 2 = 3.175mm
Gapspace is accoording to this doc of Stan 3mm.
In fact, i have those tubes here on my desk.......and they prove the same.
HOWEVER.....
On page 9 of WO 198901270A1 Stan descripes an example:
The outside cyilinder was 0.75 inch in inside diameter; the inner cylinder was 0.5 inch in outer diameter.
Spacing from the outside of the innersurface of the inside cylinder to the inner surface of the outside cylinder was 0.0625 inch.
That 0.0625 is 1.5mm.........
Well, i eat my shoe, if i know what that Meyer person wants to tell....
br
Steve
« Last Edit: February 19, 2009, 12:27:06 pm by stevie1001 »
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##### Re: Starting my project :Tube cell
« Reply #9 on: February 19, 2009, 18:36:05 pm »
If ya want to get technical about it Then
Outer Tube OD 3/4 0.7500 19.0500
Outer Wall 0.0354 0.9000
Take your 19.0500 OD in diameter Tube and subtract 0.9000 for the left wall, and subtract 0.9000 for the right wall. This leaves you with 17.25 Mili Meters
Inner Tube OD 1/2 0.5000 12.7000
Take 12.7000 Mili Meters and subtract it from 17.25 you now have 4.55 Mili Meters Free Space.
Since you have a gap on both sides of your tube then divide 4.55 by 2 to get 2.275 Mili Meters.
Stans gap was 2.2 Mili Meters.
I have researched stans tube cell for 2 years, here is some of my work. http://waterfuelcell.org/phpBB2/viewtopic.php?t=1194&postdays=0&postorder=asc&&start=0&sid=6f22f4d6958be2075163fd047f8cfd4e
Being the fact that i researched the wrong tube gap, i hate to see you repeat my mistake, its not cheap.
Also, Dynodon, you stood 2 foot from stans fuel cell, you seen that the gap was not 1.5 or 1.6 Mili Meters, 2 Mili Meters is easy to determin from 1.5 Mili Meters if you're use to looking at tube gaps.
You also know that stans cell was not 3/4 in diameter but 1 inch, the tube gap is clearly bigger than 2 Mili Meters, i know without a doubt you know this!
Listen to who you want on your tube selection, i have over 2 thousand dollars invested in stans tube cells and 2 years of nothing but tinkering to boot, do what you want.
I'll tell you this, you want beat my production of .600 ML of gas per Minute using a 1.6 mm gap because your tubes want produce any more. Thats 10 cc's per second for all 12 of my tubes, or .83 cc's a second per tube where .83 cc's x 12 tubes is 9.96 cc's a second or close to 600 cc's per min.
My tubes was 6 inches tall and 1 inch in diameter,,, and im telling you now I dont care how long you make your tubes your not going to beat my production with that gap!
Your after atleast .800 cc's a min per tube where .800 x nine tubes is 7.2 Liters Per min, Cram that up your tubes!!!
Dont like my math,, open up the test of evaluation and do your own math... I've seen stans gap.
Hi Resonanceking,
Thank you very much for guiding me in the tube jungle.. Yes it can be expensive to buy wrong type of tubes..
I think you are saying we are not into ordinary electrolysis here, so thats why the gap must be bigger...
so a gap size on 1,5 mm or less will produce more gas but use more amp with a "dead short condition" and we want to split water with only voltage pulses so maybe i will go for a gap in the 2-3 mm range..
Your fuelcells looks great and is a great inspiration so thanks to you for pictures and visdom.
Best regards
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##### Re: Starting my project :Tube cell
« Reply #10 on: February 19, 2009, 19:00:10 pm »
Hi,
I do not want to sound like granddade, but as far as I can do math, is the tubeset of Stan with the VIC as follows:
ROD of 0.5 inch outerdiameter
Tube of 0.75 inch inner diameter.
That leaves a gap of 3mm...
As far as my and Hydrocars/resonanceking, we know that with normal electrolysis, 3mm is too big.
Max gap is 1.5mm
I suggest you go with Resonancekings advises.
br
Steve
Hi Steve,
You dont sound like a granddade LOL... I am open to all point of veiws.
I think i will try a bigger gap than i was intend to and see if i get lucky.
Thank you for your help
Best regards
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##### Re: Starting my project :Tube cell
« Reply #11 on: February 19, 2009, 19:09:26 pm »
Thank you to Dynodon, Kinesisfilms,Dankie and the rest..
Please feel free to discuss tubes so much you want here.. It is interesting and even funny.
Best regards
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##### Re: Starting my project :Tube cell
« Reply #12 on: February 19, 2009, 20:57:38 pm »
I've been thinking about this, and I think I might just use some SS rod, .75" dia, 3-4" length, and drill a hole in it to so specific size, then take several other 3-4" long rods, and machine them down on the lathe to various diameters, to give various spacings, from .025" to .125" and see what happens, then the center rod will be solid... which I think would be just as good if not better, considering the nature of a conductor is that all the charge lies on the surface... no reason to have an inner surface at all. And then to change the spacing I can just switch out the center rod. Stan made a variable spacing cell... but this would be easy enough.
Comments on that approach?
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##### Re: Starting my project :Tube cell
« Reply #13 on: February 19, 2009, 22:18:08 pm »
here's about a 2.5 mm gap where i used about 28 volts applied to the cell.
I just cant see a small gap doing what this gap can, 2.5 mm is all i had so thats the reason for that..
i'm not sure if its 2.5, more like 3.5,, its a big gap i have here. Outter tube 1" inner tube 1/2, so its pretty big. To big. anyways thought id share what it can do.
You are on the right track
There is no doubt in my mind anymore, this is over 1,5 mm!!!.... more like your 2,5 mm.. Stan still surprises me... this is one important piece...
And inner tubes looks thicker than the outer tubes on this picture...
« Last Edit: February 19, 2009, 22:44:46 pm by WaytoGo »
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##### Re: Starting my project :Tube cell
« Reply #14 on: February 20, 2009, 01:26:52 am »
Since The patent stated .35 instead of .035 doesn't mean it was a typo, zero's are invalid and false numbers thats usually void.
It's quite obvious stan did not use a wall thickness of (7 MM, HENCE (typo .35)) but .035 Which is around 1/32 of an inch which is half the size of 1.5 Mili Meters, Thats quite thin isn't it, almost paper thin!
so at first it wasn't a typo when i said it was.....then after being proven wrong now it's a typo.....THANK YOU FOR AGREEING WITH ME THE SECOND TIME AROUND......make up your damn mind.
YES IF IT IS A TYPO THEN YOUR SUBTRACTING SKILLS ARE VERIFIED.....
here is the actual sizing of the alternator tubes....with the TYPO like i said.....and which now you have realized.
(http://i2.photobucket.com/albums/y2/kinesisfilms/ALTERNATORTUBES.jpg)
once again like i said the only way it will work is if it is a typo.....but nope HYDROCARZ DOESN'T KNOW ABOUT SIGNIFICANT FIGURES.....ZEROES DON'T MEAN ANYTHING AND CAN BE DROPPED.....pfffft.
oh wait now YOU Realize it......the above quotes are classic.
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##### Re: Starting my project :Tube cell
« Reply #15 on: February 20, 2009, 01:38:07 am »
WO1992007861A1.pdf page 9 and page 14
Rod of 0.5 inch outer diameter = 0.5 x 25.4 = 12.7mm
Tube of 0.75 inside diameter = 0.75 x 25.4 = 19.05mm
Total gap = 6.35mm
Devided by 2 = 3.175mm
HAHAH STEVIE YOU DID IT AGAIN!!!! REALLY IS THIS HAPPENING??? ALRIGHT LET'S END THIS ONCE AND FOR ALL
take the .5 inches diameter multiply it by 25.4 to get 12.7mm
take the .75x 25.4 = 19.05mm......
divide 12.7mm by 2....6.35mm
divide 19.05mm by 2...9.525
now subtract......the wall space from the 9.525-1.5875=7.9375
subtract 7.9375-6.35= 1.5875mm
STEVIE IS IS SIMPLE ALGEBRA.....IT'S CALLED ORDER OF OPERATIONS.....YOU CAN'T SUBTRACT OR ADD BEFORE MULTIPLYING AND DIVIDING.....ORDER OF OPERATIONS.....THIS IS 5TH GRADE STUFF......
BUT SERIOUSLY THIS IS SUCH A COMMON MISTAKE!.....
no worries stevie you just messed up with order of operations.
ANYONE ELSE STILL QUESTION THE TUBE CELL SIZE?...i will gladly help you out.
THIS IS A GREAT POST THANK YOU STEVIE THIS WAS PROBABLY A COMMON PROBLEM WITH PEOPLE! | 2,770 | 9,367 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2023-06 | latest | en | 0.847554 |
https://www.assignmenthelptutors.com/finite-difference-method-nonlinear-ode.html | 1,610,999,498,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703515235.25/warc/CC-MAIN-20210118185230-20210118215230-00308.warc.gz | 668,666,575 | 8,879 | # Finite Difference Method–Nonlinear Ode
If we once more consider the heat in a metal bar of length L but this time consider the effect of radiation as well as conduction then the steady state equation has the form
uxx− d(u4− u4b) = −g(x),
Where ub is temperature of the background d incorporates a coefficient of radiation as well as g(x) is the heat source.
If we again restore the continuous problem by its discrete approximation then we get:
(ui+1− 2ui+ ui−1)/ h2− d(u4i− u4b) = −gi= −g(xi).
This equation is nonlinear in the unknowns therefore we no longer have a system of linear equations to solve however a system of nonlinear equations. One way to resolve these equations would be by the multivariable Newton method. In its place we introduce another interactive method.
Implementing the Relaxation Method:
In the following program we resolve the finite difference equations with the boundary conditions:
u(0) = 0 and u(L) = 0. We let L = 4, n = 4, d = 1, and g(x) = sin(πx/4).
Notice that the vector u forever contains the current estimate of the values of u.
L = 4; n = 4;
h = L/n; hh = h^2/3;
u0 = 0; uL = 0; ub = .5; ub4 = ub^4;
x = 0:h:L; g = sin(pi*x/4); u = zeros(1,n+1);
steps = 4;
u(1)=u0; u(n+1)=uL;
for j = 1:steps
u(2:n) = (u(3:n+1)+u(2:n)+u(1:n-1))/3 + hh*(-u(2:n).^4+ub4+g(2:n));
end
plot(x,u)
If you run this program with the specified n and steps the result will not seem reasonable.
We are able to plot the initial guess by adding the command plot(x,u) right before the for loop. We can as well plot successive iterations by moving the last plot(x,u) before the end. We can experiment as well as see if the iteration is converging. Try a variety of values of steps and n to produce a good plot. You will notice so as to this method converges quite slowly. In particular as we raise n we need to increase steps like n2 that is if n is large then steps needs to be really large.
Relaxation Method for Nonlinear Finite Differences:
We are able to rewrite equation as:
ui+1− 2ui+ ui−1 = h2d(u4i− u4b) − h2gi.
From this we are able to solve for uiin terms of the other quantities:
2ui= ui+1 + ui−1− h2d(u4i− u4b) + h2gi.
After that we add uito both sides of the equation to obtain:
3ui= ui+1 + ui+ ui−1− h2d(u4i− u4b) + h2gi,
And afterwards divide by 3 to get:
ui=1/3(ui+1 + ui+ ui−1) −h2/3(d(u4i− u4b) + gi)
Now for the major idea. We will start with an initial guess for the value of uifor each i which we can represent as a vector u0. Subsequently we will use the above equation to get better estimates, u1, u2, . . . , and expect that they converge to the correct answer.
If we let:
uj= (uj0, uj1, uj2, . . . , ujn−1, ujn)
Denote the jth approximation afterwards we can obtain that j + 1st estimate from the formula:
Notice that gi and ub don’t change. In the resulting equation we have ui at every successive step depending on its previous value as well as the equation itself.
Assignmenthelptutors.com is dedicated to outstanding service with:
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When you need the best Finite Difference Method – Nonlinear ODE online assistance, our team is here for you. We provide you 24/7 support teams dedicated to your success, Assignmenthelptutors.com experts having experience and degrees in their fields Finite Difference Method – Nonlinear ODE programming work and degrees, and even easy safe methods for reaching our experts when you need them. Assignmenthelptutors.com Finite Difference Method – Nonlinear ODE homework assignments do not have to be the worst experience of your school career, Assignmenthelptutors.com is here to provide you with the highest quality Finite Difference Method – Nonlinear ODE assistance every time. | 1,070 | 3,943 | {"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} | 3.8125 | 4 | CC-MAIN-2021-04 | latest | en | 0.918573 |
https://mathematica.stackexchange.com/questions/268875/eigenvalues-arent-given-to-be-imaginary | 1,721,835,054,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518304.14/warc/CC-MAIN-20240724140819-20240724170819-00615.warc.gz | 332,878,484 | 39,467 | # Eigenvalues aren't given to be imaginary
So I have this matrix I am working with that looks like $$S=\begin{pmatrix}0 & -a & b\\ a & 0 & -c\\ -b & c& 0 \end{pmatrix}$$ and upon asking mathematica for the eigenvalues I get Eigenvlaues[S] $$=0, \sqrt{-a^2+-b^2+-c^2},-\sqrt{-a^2+-b^2+-c^2}$$. Yes the eigenvalues are imaginary, but why doesn't it simplify it to be imaginary, and how can I force it to replace those minus sines and slap an $$i$$ in there instead? When I go to find the Eigenvectors does it recognize the terms I have as imaginary eigenvalues and simplify as such, or do I need to force it somehow.
• People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find the meta Q&A, How to copy code from Mathematica so it looks good on this site, helpful. -- Also, keep in mind a, b, and c are treated as complex numbers. -- Finally, something like this might work: PowerExpand[Sqrt[-a^2]] Commented May 30, 2022 at 22:59
• Oh wow I didn't know that! Okay thank you for the heads up. Could I just make the assumption in the code that a,b,c are all real?
– Joey
Commented May 30, 2022 at 23:02
• Try ComplexExpand. Commented May 30, 2022 at 23:04
• Compare: Simplify[Sqrt[-a^2 - b^2], a > 0 && b > 0], Simplify[Sqrt[-a^2 - b^2], a >= 0 && b >= 0], Simplify[Sqrt[-a^2 - b^2], {a, b} \[Element] Reals], Simplify[Sqrt[-a^2 - b^2], a != 0 && b != 0 && {a, b} \[Element] Reals] -- That the case a == b == 0 seems important here looks like a weakness of Mathematica. It probably does not check whether the expression under the radical can change sign or not, probably for efficiency's sake. Commented May 30, 2022 at 23:14
• Mma seems to want to treat 0 as a non-imaginary number and keep it separate from the (nonzero) imaginary numbers: Assuming[a^2 + b^2 >= 0, FullSimplify@PiecewiseExpand@ComplexExpand[Sqrt[-a^2 - b^2]] ] Commented May 30, 2022 at 23:28 | 628 | 2,021 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-30 | latest | en | 0.88299 |
http://kidport.com/Grade6/Science/Lesson_11___Things_That_Float.html | 1,624,042,392,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487640324.35/warc/CC-MAIN-20210618165643-20210618195643-00264.warc.gz | 32,545,479 | 4,669 | << Back to Lessons Index
# 6th Grade Science / Lesson 11 - Things That Float
Things That Float What will we be learning today? In this lesson, we are going to learn all about why do things float.
why do things float. Have you ever dived deep under the water while swimming? If so you know that you feel more pressure on your ears as you go deeper. Pressure is the weight or force on a given area. The weight, in this case, comes from the air and water above you.
When any object is submerged in a fluid, the fluid "pushes" in on the object. The push is greater at the bottom than at the top, so the fluid actually pushes the object toward the surface. This push is called the buoyant force.
Any submerged object displaces fluid (moves fluid out of the way). The photograph on the next slide shows how the water level rises when a lead mass is submerged in water. If you look at the weight readings on the spring scale, you will see that the reading is less when the lead mass is submerged-the buoyant force has made it seem lighter. It turns out that the amount of buoyant force equals the weight of the fluid displaced by the lead mass.
An object seems to lose weight when it is placed underwater. This is because the water pushes the object upward. The push is called the buoyant force. The unit of force is the newton (N).
When an object first begins sinking into a fluid, little fluid is displaced. There is little buoyant force, and the object keeps sinking. However, some objects may displace a weight of fluid equal to their own weight before they sink all the way in. These objects float on the fluid. Other objects, though, cannot displace enough water to float. Even when fully submerged, these objects have more weight than buoyant force, and they sink to the bottom.
Let's Review What We Learned Today! Now complete your PDA! | 406 | 1,852 | {"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-2021-25 | latest | en | 0.946638 |
https://www.aqua-calc.com/one-to-one/density/grain-per-us-quart/grain-per-cubic-inch/1 | 1,582,206,451,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144979.91/warc/CC-MAIN-20200220131529-20200220161529-00032.warc.gz | 648,623,331 | 8,662 | # 1 grain per US quart [gr/US qt] in grains per cubic inch
## gr/US qt to gr/in³ unit converter of density
1 grain per US quart [gr/qt] = 0.02 grain per cubic inch [gr/in³]
### grains per US quart to grains per cubic inch density conversion cards
• 1
through
25
grains per US quart
• 1 gr/qt to gr/in³ = 0.02 gr/in³
• 2 gr/qt to gr/in³ = 0.03 gr/in³
• 3 gr/qt to gr/in³ = 0.05 gr/in³
• 4 gr/qt to gr/in³ = 0.07 gr/in³
• 5 gr/qt to gr/in³ = 0.09 gr/in³
• 6 gr/qt to gr/in³ = 0.1 gr/in³
• 7 gr/qt to gr/in³ = 0.12 gr/in³
• 8 gr/qt to gr/in³ = 0.14 gr/in³
• 9 gr/qt to gr/in³ = 0.16 gr/in³
• 10 gr/qt to gr/in³ = 0.17 gr/in³
• 11 gr/qt to gr/in³ = 0.19 gr/in³
• 12 gr/qt to gr/in³ = 0.21 gr/in³
• 13 gr/qt to gr/in³ = 0.23 gr/in³
• 14 gr/qt to gr/in³ = 0.24 gr/in³
• 15 gr/qt to gr/in³ = 0.26 gr/in³
• 16 gr/qt to gr/in³ = 0.28 gr/in³
• 17 gr/qt to gr/in³ = 0.29 gr/in³
• 18 gr/qt to gr/in³ = 0.31 gr/in³
• 19 gr/qt to gr/in³ = 0.33 gr/in³
• 20 gr/qt to gr/in³ = 0.35 gr/in³
• 21 gr/qt to gr/in³ = 0.36 gr/in³
• 22 gr/qt to gr/in³ = 0.38 gr/in³
• 23 gr/qt to gr/in³ = 0.4 gr/in³
• 24 gr/qt to gr/in³ = 0.42 gr/in³
• 25 gr/qt to gr/in³ = 0.43 gr/in³
• 26
through
50
grains per US quart
• 26 gr/qt to gr/in³ = 0.45 gr/in³
• 27 gr/qt to gr/in³ = 0.47 gr/in³
• 28 gr/qt to gr/in³ = 0.48 gr/in³
• 29 gr/qt to gr/in³ = 0.5 gr/in³
• 30 gr/qt to gr/in³ = 0.52 gr/in³
• 31 gr/qt to gr/in³ = 0.54 gr/in³
• 32 gr/qt to gr/in³ = 0.55 gr/in³
• 33 gr/qt to gr/in³ = 0.57 gr/in³
• 34 gr/qt to gr/in³ = 0.59 gr/in³
• 35 gr/qt to gr/in³ = 0.61 gr/in³
• 36 gr/qt to gr/in³ = 0.62 gr/in³
• 37 gr/qt to gr/in³ = 0.64 gr/in³
• 38 gr/qt to gr/in³ = 0.66 gr/in³
• 39 gr/qt to gr/in³ = 0.68 gr/in³
• 40 gr/qt to gr/in³ = 0.69 gr/in³
• 41 gr/qt to gr/in³ = 0.71 gr/in³
• 42 gr/qt to gr/in³ = 0.73 gr/in³
• 43 gr/qt to gr/in³ = 0.74 gr/in³
• 44 gr/qt to gr/in³ = 0.76 gr/in³
• 45 gr/qt to gr/in³ = 0.78 gr/in³
• 46 gr/qt to gr/in³ = 0.8 gr/in³
• 47 gr/qt to gr/in³ = 0.81 gr/in³
• 48 gr/qt to gr/in³ = 0.83 gr/in³
• 49 gr/qt to gr/in³ = 0.85 gr/in³
• 50 gr/qt to gr/in³ = 0.87 gr/in³
• 51
through
75
grains per US quart
• 51 gr/qt to gr/in³ = 0.88 gr/in³
• 52 gr/qt to gr/in³ = 0.9 gr/in³
• 53 gr/qt to gr/in³ = 0.92 gr/in³
• 54 gr/qt to gr/in³ = 0.94 gr/in³
• 55 gr/qt to gr/in³ = 0.95 gr/in³
• 56 gr/qt to gr/in³ = 0.97 gr/in³
• 57 gr/qt to gr/in³ = 0.99 gr/in³
• 58 gr/qt to gr/in³ = 1 gr/in³
• 59 gr/qt to gr/in³ = 1.02 gr/in³
• 60 gr/qt to gr/in³ = 1.04 gr/in³
• 61 gr/qt to gr/in³ = 1.06 gr/in³
• 62 gr/qt to gr/in³ = 1.07 gr/in³
• 63 gr/qt to gr/in³ = 1.09 gr/in³
• 64 gr/qt to gr/in³ = 1.11 gr/in³
• 65 gr/qt to gr/in³ = 1.13 gr/in³
• 66 gr/qt to gr/in³ = 1.14 gr/in³
• 67 gr/qt to gr/in³ = 1.16 gr/in³
• 68 gr/qt to gr/in³ = 1.18 gr/in³
• 69 gr/qt to gr/in³ = 1.19 gr/in³
• 70 gr/qt to gr/in³ = 1.21 gr/in³
• 71 gr/qt to gr/in³ = 1.23 gr/in³
• 72 gr/qt to gr/in³ = 1.25 gr/in³
• 73 gr/qt to gr/in³ = 1.26 gr/in³
• 74 gr/qt to gr/in³ = 1.28 gr/in³
• 75 gr/qt to gr/in³ = 1.3 gr/in³
• 76
through
100
grains per US quart
• 76 gr/qt to gr/in³ = 1.32 gr/in³
• 77 gr/qt to gr/in³ = 1.33 gr/in³
• 78 gr/qt to gr/in³ = 1.35 gr/in³
• 79 gr/qt to gr/in³ = 1.37 gr/in³
• 80 gr/qt to gr/in³ = 1.39 gr/in³
• 81 gr/qt to gr/in³ = 1.4 gr/in³
• 82 gr/qt to gr/in³ = 1.42 gr/in³
• 83 gr/qt to gr/in³ = 1.44 gr/in³
• 84 gr/qt to gr/in³ = 1.45 gr/in³
• 85 gr/qt to gr/in³ = 1.47 gr/in³
• 86 gr/qt to gr/in³ = 1.49 gr/in³
• 87 gr/qt to gr/in³ = 1.51 gr/in³
• 88 gr/qt to gr/in³ = 1.52 gr/in³
• 89 gr/qt to gr/in³ = 1.54 gr/in³
• 90 gr/qt to gr/in³ = 1.56 gr/in³
• 91 gr/qt to gr/in³ = 1.58 gr/in³
• 92 gr/qt to gr/in³ = 1.59 gr/in³
• 93 gr/qt to gr/in³ = 1.61 gr/in³
• 94 gr/qt to gr/in³ = 1.63 gr/in³
• 95 gr/qt to gr/in³ = 1.65 gr/in³
• 96 gr/qt to gr/in³ = 1.66 gr/in³
• 97 gr/qt to gr/in³ = 1.68 gr/in³
• 98 gr/qt to gr/in³ = 1.7 gr/in³
• 99 gr/qt to gr/in³ = 1.71 gr/in³
• 100 gr/qt to gr/in³ = 1.73 gr/in³
• gr/qt stands for gr/US qt
#### Foods, Nutrients and Calories
RAUSCH, NOUMEA PREMIUM MILK CHOCOLATE, UPC: 858335002066 contain(s) 591 calories per 100 grams or ≈3.527 ounces [ price ]
LA FE, WHITE TUNA IN OLIVE OIL, UPC: 023545206744 weigh(s) 232.47 gram per (metric cup) or 7.76 ounce per (US cup), and contain(s) 109 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
Foods high in Calcium, Ca, foods low in Calcium, Ca, and Recommended Dietary Allowances (RDAs) for Calcium
#### Gravels, Substances and Oils
CaribSea, Freshwater, Instant Aquarium, Kon Tiki weighs 1 601.85 kg/m³ (100.00023 lb/ft³) with specific gravity of 1.60185 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 ]
Vitamin E [C29H50O2] weighs 950 kg/m³ (59.30656 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-404A, liquid (R404A) with temperature in the range of -51.12°C (-60.016°F) to 68.34°C (155.012°F)
#### Weights and Measurements
The stone per square inch surface density measurement unit is used to measure area in square inches in order to estimate weight or mass in stones
The units of data measurement were introduced to manage and operate digital information.
oz t/l to t/yd³ conversion table, oz t/l to t/yd³ unit converter or convert between all units of density measurement.
#### Calculators
Calculate volume of a spherical segment and its surface area | 2,377 | 5,575 | {"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-2020-10 | latest | en | 0.480449 |
http://www.acadblock.com/mechanics/w-p-e-d3vk/ | 1,726,143,021,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00504.warc.gz | 33,511,189 | 7,240 | # w.p & e
Block A is of mass 30kg & block B is of mass 5kg. if μ kinetic friction between the inclined & block A is 0.2, determine the speed of block A after it moves 1m down the plane, starting from rest.
62
Lokesh Verma ·
m=mass of A
M=mass of B
F=frictional force
first step: Find the constraint equation
2T.Sa-TSb=0
so 2A-B=0
A=10 is given. Thus, B=20
Now the FBD of A block
mg sin 37 - 2T - F = maA
mg cos 37 = N = 30.g.4/5 = 24g
Tension is internal force.. So work done by tension is 0
Work by friction is F.s = μN.10 = 2.24=48g
Potential Energy change will be -mg.10+Mg20 = -200g
Increase of Kinetic Energy+increase of Potential Energy + energy Loss=0
KE change = 200-48 = 152
1/2mVa2+1/2MVb2 = 152
30. V2 + 5. 4 V2 = 304
V2 = 152/25 =
i may have done some mistake sindhu.. cos did this one in too much jaldi jaldi.. But pls check the method! | 318 | 865 | {"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-2024-38 | latest | en | 0.824182 |
https://shop.proudtobeprimary.com/products/february-math-spiral-review-worksheets-for-kindergarten | 1,726,045,839,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00572.warc.gz | 497,280,509 | 103,993 | # February Math Spiral Review Worksheets for Kindergarten
Sale price\$5.00
Kindergarten Math Mats - Spiral math review for February. This printable and digital resource (Google Slides) includes 20 different Kindergarten math spiral review mats for winter and Valentine's Day. These math worksheets are perfect for February morning work, math journals, and math centers. Each Math Mat has 4 different spiral math skills and standards on it for students to practice and review.
This resource is comprehensive and provides spiral math practice for the month of February.
This Kindergarten math morning work includes questions and activities on
• Subtraction within 10
• Teen numbers
• Addition & Subtraction Word problems
• 2D and 3D geometry and shapes
• Measurement
• Time to the hour and half hour
• Numbers to 20
• Graphing
• Money
• Patterns
• One to One Correspondence
• Numerical Order
• Comparing Numbers
• Skip Counting by 10s
• Tally Marks
• Number writing
• … and more!
The math spiral review activities align with the Common Core standards for kindergarten. This is perfect for kindergarten, but could also be used as a review or extra practice in 1st grade.
Three Kindergarten Math Spiral Review Resources in one:
1. PDF - Open the PDF file and print! This file is NOT interactive or digital.
2. GOOGLE SLIDES - You will receive every mat digitally enhanced in Google Slides with interactive components (preformatted text boxes, moveable pieces).
3. POWERPOINT – The Google Slides version has been downloaded and is included as an interactive PPT file.
This Kindergarten math morning work includes:
• Subtraction word problems
• Addition math fact practice with fingers, counters, number lines, 10 frames, and pictures.
• Subtraction math fact practice with fingers, counters, number lines, 10 frames, and pictures.
• 2D Geometry - draw, identify, and describe shapes
• 3D Geometry – identify and match shapes with real life objects
• Measurement - counting number of units, describing which object is tallest/shortest/longest/heaviest/lightest, and drawing objects that are taller/shorter/longer/heaviest/lightest.
• Time to the hour and half hour - add hands on clocks & reading clocks
• Numbers to 20 - fill in missing numbers and complete number charts
• Writing numbers to 50
• Place value to 20 - counting ones and tens blocks
• Greater than, less than, and equal to
• Number order to 20
• Tally marks
• Skip Counting by 10's
• Patterning - create and complete different types of patterns with shapes and letters
• Money - identify and count coins (provided in Canadian on additional pages)
4 ways to use Math Mats in your classroom!
✅ Print on card stock & laminate. Use with whiteboard markers during your math centers!
✅ Photocopy with the booklet cover and staple into a booklet. Have your students complete 1 mat a day for morning work or for extra practice during math workshop!
✅ Shrink the pages and have your students use the mats in their Interactive Math Notebooks!
✅ Photocopy and place into a math folder to practice each day.
Check out a detailed blog post about Math Mats HERE!
Look at the PREVIEW for a breakdown of what a Math Mat looks like and how it can be used.
Upon purchase, you can instantly download a ZIP file for each month that includes a printable PDF, details for using with Google, and interactive PPT file.
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© Copyright Proud to be Primary, Elyse Rycroft. All rights reserved. Permission is granted to copy pages specifically designed for student or teacher use by the original purchaser or licensee. This product is licensed for personal classroom use ONLY unless multiple licenses are purchased. The reproduction, alteration, adaptation, copying, or sale of any part of this product is strictly prohibited. Read the full Terms and Conditions HERE.
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(0) | 941 | 4,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-38 | latest | en | 0.820607 |
https://www.numerade.com/questions/how-many-sets-of-quantum-numbers-are-possible-for-a-hydrogen-atom-for-which-a-n1-b-n2-c-n3-d-n4-and-/ | 1,620,570,312,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988986.98/warc/CC-MAIN-20210509122756-20210509152756-00123.warc.gz | 973,940,900 | 31,455 | Our Discord hit 10K members! 🎉 Meet students and ask top educators your questions.Join Here!
# How many sets of quantum numbers are possible for a hydrogen atom for which (a) $n=1,$ (b) $n=2,$ (c) $n=3$ , (d) $n=4,$ and (e) $n=5$ ?
## By the above process, we see that in $n=1,$ there are 2 elements; in $n=2,$ there are $8 ;$ in $n=3$ , there are 18 elements and so on, which indicates that the formula for finding the electron $=2 n^{2} .$
Quantum Physics
Atomic Physics
### Discussion
You must be signed in to discuss.
LB
##### Marshall S.
University of Washington
##### Farnaz M.
Simon Fraser University
##### Jared E.
University of Winnipeg
### Video Transcript
the number of possible quantum states, which I just used. The member sign here That pound sign is equal to two times in squared. So for part A when N is equal to one, the number of states is equal to two times one squared, which is equal to two. Your part B you have in is equal to two. So the number of states is going to be equal to two times two squared where two squared is four times two is equal to eight. So the number of states is equal to eight on the for part C we have in is equal to three. So the number of states is gonna be two times three squared three squared is nine claims to is 18. So the number of states is 18. Making box it in is their solution for part C.
University of Kansas
#### Topics
Quantum Physics
Atomic Physics
LB
##### Marshall S.
University of Washington
##### Farnaz M.
Simon Fraser University
##### Jared E.
University of Winnipeg | 398 | 1,559 | {"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} | 3.453125 | 3 | CC-MAIN-2021-21 | latest | en | 0.919644 |
https://betterlesson.com/common_core/browse/34/ccss-math-content-k-md-a-2-directly-compare-two-objects-with-a-measurable-attribute-in-common-to-see-which-object-has-more-of-le | 1,532,118,891,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591831.57/warc/CC-MAIN-20180720193850-20180720213850-00179.warc.gz | 604,704,117 | 22,241 | ## Directly compare two objects with a measurable attribute in common, to see which object has "more of"/"less of" the attribute, and describe the difference. For example, directly compare the heights of two children and describe one child as taller/shorter.
26 Lesson(s)
### Measurement- Comparing Length Day 2
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### Measurement- Comparing Length Day 1
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### How Do We Measure? Comparing Weight Day 1
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### Measurement- Comparing Length Day 3
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### Measurement- Comparing Height Day 1
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### Measurement- Comparing Height Day 2
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### The First Day of Kindergarten - Scary and Strange? More or Less!
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### Guppy and the Goldfish
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Kindergarten Science » Unit: Let's go fishing!
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### Grandma's Button Box - Sorting
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### Math Centers - Sorting
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### The Gingerbread Man Loose in the School- Sorting
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### How Long? A Fishy "Tail"
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### Sorting By Shape Buttons
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### Sorting By Size
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### Wow! That's Heavy!
Kindergarten Math » Unit: How Does It All Measure Up?
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Common Core Math | 964 | 4,011 | {"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} | 3.8125 | 4 | CC-MAIN-2018-30 | latest | en | 0.624261 |
https://www.doorsteptutor.com/Exams/IMO/Class-3/Questions/Topic-Computational-Operations-1/Part-9.html | 1,527,409,007,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868132.80/warc/CC-MAIN-20180527072151-20180527092151-00077.warc.gz | 729,741,667 | 14,534 | Computational Operations (IMO- Mathematics Olympiad (SOF) Class 3): Questions 54 - 60 of 64
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Question number: 54
» Computational Operations
MCQ▾
Question
Each of these plates has exactly 12 pieces of chocolate in it. How many pieces of chocolate are there in all?
Choices
Choice (4) Response
a.
16
b.
36
c.
24
d.
18
Question number: 55
» Computational Operations
MCQ▾
Question
What is the missing number in the box?
Choices
Choice (4) Response
a.
4
b.
5
c.
9
d.
6
Question number: 56
» Computational Operations
MCQ▾
Question
What is the missing number in the box?
Choices
Choice (4) Response
a.
5
b.
6
c.
9
d.
4
Question number: 57
» Computational Operations
MCQ▾
Question
How many 100’s, 10’s and 1’s should be added to give largest 3 digit number?
Choices
Choice (4) Response
a.
1,0, 9
b.
9,9, 9
c.
1,0, 0
d.
9,0, 1
Question number: 58
» Computational Operations
MCQ▾
Question
Nisha to order rulers for the 58 students in her class.
1. Each student needs one ruler.
2. One ruler is equal to 6 box.
Which of the following can be used to find how many boxes Nisha needs?
Choices
Choice (4) Response
a.
b.
c.
d.
Question number: 59
» Computational Operations
Appeared in Year: 2012
MCQ▾
Question
The missing number is ________.
Choices
Choice (4) Response
a.
3
b.
5
c.
6
d.
7
Question number: 60
» Computational Operations
Appeared in Year: 2012
MCQ▾
Question
The value of is ________.
Choices
Choice (4) Response
a.
198
b.
112
c.
124
d.
120
f Page | 528 | 1,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-22 | latest | en | 0.748732 |
http://list.seqfan.eu/pipermail/seqfan/2009-February/052563.html | 1,686,308,245,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00714.warc.gz | 27,541,855 | 1,978 | [seqfan] Re: Additive combination of unique cyclotomic polynomials
Maximilian Hasler maximilian.hasler at gmail.com
Sat Feb 7 03:55:55 CET 2009
```> - For x^n, what is the minimum integer A such that x^n can
> be expressed using only some subset of the first A
> cyclotomic polynomials?
> The second sequence seems to begin (with n=0):
>
> 4,4,3,8,8,10,12,...
I get
x^5 = -polcyclo(1)+polcyclo(2)-polcyclo(3)+polcyclo(7)-polcyclo(8)-polcyclo(9)
and
x^6 = polcyclo(1)-polcyclo(2)+polcyclo(3)-polcyclo(5)+polcyclo(8)+polcyclo(9)
so 10,12 should read 9,9 - or did I get something wrong ?
Maximilian
```
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Decision making is an important aspect of the Paper F5 syllabus, and questions on this topic will be common. The range of possible questions is considerable, but this article will focus on only one: linear programming
The ideas presented in this article are based on a simple example. Suppose a profit-seeking firm has two constraints: labour, limited to 16,000 hours, and materials, limited to 15,000kg. The firm manufactures and sells two products, X and Y. To make X, the firm uses 3kg of material and four hours of labour, whereas to make Y, the firm uses 5kg of material and four hours of labour. The contributions made by each product are \$30 for X and \$40 for Y. The cost of materials is normally \$8 per kg, and the labour rate is \$10 per hour.
The first step in any linear programming problem is to produce the equations for constraints and the contribution function, which should not be difficult at this level.
In our example, the materials constraint will be 3X + 5Y ≤ 15,000, and the labour constraint will be 4X + 4Y ≤ 16,000. You should not forget the non-negativity constraint, if needed, of X,Y ≥ 0.
The contribution function is 30X + 40Y = C
Figure 1: Optimal production plan
Plotting the resulting graph (Figure 1, the optimal production plan) will show that by pushing out the contribution function, the optimal solution will be at point B – the intersection of materials and labour constraints.
The optimal point is X = 2,500 and Y = 1,500, which generates \$135,000 in contribution. Check this for yourself (see Working 1). The ability to solve simultaneous equations is assumed in this article.
The point of this calculation is to provide management with a target production plan in order to maximise contribution and therefore profit. However, things can change and, in particular, constraints can relax or tighten. Management needs to know the financial implications of such changes. For example, if new materials are offered, how much should be paid for them? And how much should be bought? These dynamics are important.
Suppose the shadow price of materials is \$5 per kg (this is verifiable by calculation – see Working 2). The important point is, what does this mean? If management is offered more materials it should be prepared to pay no more than \$5 per kg over the normal price. Paying less than \$13 (\$5 + \$8) per kg to obtain more materials will make the firm better off financially. Paying more than \$13 per kg would render it worse off in terms of contribution gained. Management needs to understand this.
There may, of course, be a good reason to buy ‘expensive’ extra materials (those costing more than \$13 per kg). It might enable the business to satisfy the demands of an important customer who might, in turn, buy more products later. The firm might have to meet a contractual obligation, and so paying ‘too much’ for more materials might be justifiable if it will prevent a penalty on the contract. The cost of this is rarely included in shadow price calculations. Equally, it might be that ‘cheap’ material, priced at under \$13 per kg, is not attractive. Quality is a factor, as is reliability of supply. Accountants should recognise that ‘price’ is not everything.
Students need to realise that as you buy more materials, then that constraint relaxes and so its line on the graph moves outwards and away from the origin. Eventually, the materials line will be totally outside the labour line on the graph and the point at which this happens is the point at which the business will cease to find buying more materials attractive (point D on the graph). Labour would then become the only constraint.
We need to find out how many materials are needed at point D on the graph, the point at which 4,000 units of Y are produced. To make 4,000 units of Y we need 20,000kg of materials. Consequently, the maximum amount of extra material required is 5,000kg (20,000 – 15,000). Note: Although interpretation is important at this level, there will still be marks available for the basic calculations.
#### WORKINGS
Working 1:
The optimal point is at point B, which is at the intersection of:
3X + 5Y = 15,000 and
4X + 4Y = 16,000
Multiplying the first equation by four and the second by three we get:
12X + 20Y = 60,000
12X + 12Y = 48,000
The difference in the two equations is:
8Y = 12,000, or Y = 1,500
Substituting Y = 1,500 in any of the above equations will give us the X value:
3X + 5 (1,500) = 15,000
3X = 7,500
X = 2,500
The contribution gained is (2,500 x 30) + (1,500 x 40) = \$135,000
Working 2: Shadow price of materials
To find this we relax the material constraint by 1kg and resolve as follows:
3X + 5Y = 15,001 and
4X + 4Y = 16,000
Again, multiplying by four for the first equation and by three for the second produces:
12X + 20Y = 60,004
12X + 12Y = 48,000
8Y = 12,004
Y = 1,500.5
Substituting Y = 1,500.5 in any of the above equations will give us X:
3X + 5 (1,500.5) = 15,001
3X = 7,498.5
X = 2,499.5
The new level of contribution is: (2,499.5 x 30) + (1,500 x 40) = \$135,005
The increase in contribution from the original optimal is the shadow price:
142,505 – 142,500 = \$5 per kg.
Geoff Cordwell is former examiner for Paper F5
Last updated: 9 Sep 2013 | 1,375 | 5,389 | {"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-2014-23 | longest | en | 0.891533 |
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```IIT JEE 2009 Physics Paper2 Code 1 Solutions
MATRIX MATCH TYPE
10. Column I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits S1 and S2. In each of these cases S1P0 = S2P0, S1P1 - S2P1 = λ/4 and S1P2 - S2P2 = λ/3, where l is the wavelength of the light used. In the cases B, C and D a transparent sheet of refractive index m and thickness t is pasted on slit S2. The thickness of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by δ(P) and the intensity by I(P). Match each situation given in Column I with the statement(s) in Column II valid for that situation.
Column I
Column II
(A)
(p)
δ(P0) = 0
(B)
(µ - 1)t = λ/4
(q)
δ(P1) = 0
(C)
(µ - 1)t = λ/2
(r)
I(P1) = 0
(D)
(µ - 1)t = 3λ/4
(s)
I(P0) > I(P1)
(t)
I(P2) > I(P1)
Sol. (A ->p, s); (B -> q); (C -> t); (D -> r, s, t)
<< Back || Next >>
``` | 452 | 1,336 | {"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} | 2.59375 | 3 | CC-MAIN-2018-09 | latest | en | 0.769815 |
https://www.airmilescalculator.com/distance/ykh-to-hek/ | 1,709,266,350,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00141.warc.gz | 603,752,617 | 23,538 | # How far is Heihe from Yingkou?
The distance between Yingkou (Yingkou Lanqi Airport) and Heihe (Heihe Aihui Airport) is 707 miles / 1138 kilometers / 614 nautical miles.
The driving distance from Yingkou (YKH) to Heihe (HEK) is 789 miles / 1269 kilometers, and travel time by car is about 14 hours 41 minutes.
707
Miles
1138
Kilometers
614
Nautical miles
1 h 50 min
125 kg
## Distance from Yingkou to Heihe
There are several ways to calculate the distance from Yingkou to Heihe. Here are two standard methods:
Vincenty's formula (applied above)
• 706.893 miles
• 1137.634 kilometers
• 614.273 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 706.983 miles
• 1137.778 kilometers
• 614.351 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).
## How long does it take to fly from Yingkou to Heihe?
The estimated flight time from Yingkou Lanqi Airport to Heihe Aihui Airport is 1 hour and 50 minutes.
## Flight carbon footprint between Yingkou Lanqi Airport (YKH) and Heihe Aihui Airport (HEK)
On average, flying from Yingkou to Heihe generates about 125 kg of CO2 per passenger, and 125 kilograms equals 276 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Yingkou to Heihe
See the map of the shortest flight path between Yingkou Lanqi Airport (YKH) and Heihe Aihui Airport (HEK).
## Airport information
Origin Yingkou Lanqi Airport
City: Yingkou
Country: China
IATA Code: YKH
ICAO Code: ZYYK
Coordinates: 40°32′33″N, 122°21′30″E
Destination Heihe Aihui Airport
City: Heihe
Country: China
IATA Code: HEK
ICAO Code: ZYHE
Coordinates: 50°10′17″N, 127°18′31″E | 543 | 1,934 | {"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} | 2.734375 | 3 | CC-MAIN-2024-10 | latest | en | 0.801489 |
http://playzona.net/Washington/how-to-calculate-error-in-an-experiment.html | 1,544,994,458,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827992.73/warc/CC-MAIN-20181216191351-20181216213351-00015.warc.gz | 225,834,450 | 8,489 | Address 2805 N Market St, Spokane, WA 99207 (509) 328-9872 http://www.modernofficeequip.com
# how to calculate error in an experiment Hooper, Washington
WiedergabelisteWarteschlangeWiedergabelisteWarteschlange Alle entfernenBeenden Wird geladen... The mean is given by the following. After he recovered his composure, Gauss made a histogram of the results of a particular measurement and discovered the famous Gaussian or bell-shaped curve. Understanding Experimental Error Mr.
Computable Document Format Computation-powered interactive documents. Pugh and G.H. Get the best of About Education in your inbox. It is important you drop any negative sign since you cannot have a negative error.
The definition of is as follows. Since the correction is usually very small, it will practically never affect the error of precision, which is also small. Repeating the measurement gives identical results. The same measurement in centimeters would be 42.8 cm and still be a three significant figure number.
The density of water at 4 degrees Celsius is 1.0 g/mL is an accepted value. This fact gives us a key for understanding what to do about random errors. Services Technical Services Corporate Consulting For Customers Online Store Product Registration Product Downloads Service Plans Benefits Support Support FAQ Customer Service Contact Support Learning Wolfram Language Documentation Wolfram Language Introductory Book Thus, all the significant figures presented to the right of 11.28 for that data point really aren't significant.
Of course, for most experiments the assumption of a Gaussian distribution is only an approximation. Next, the sum is divided by the number of measurements, and the rule for division of quantities allows the calculation of the error in the result (i.e., the error of the Calculating Experimental Error So how do you judge how close you came to duplicating the correct data in an experiment? In science it is important that you express exactly what you mean so that others looking at your work know exactly what you meant.
You might also enjoy: Sign up There was an error. In[4]:= In[5]:= Out[5]= We then normalize the distribution so the maximum value is close to the maximum number in the histogram and plot the result. Using a better voltmeter, of course, gives a better result. The function AdjustSignificantFigures will adjust the volume data.
Another example is AC noise causing the needle of a voltmeter to fluctuate. Not only have you made a more accurate determination of the value, you also have a set of data that will allow you to estimate the uncertainty in your measurement. For example, one could perform very precise but inaccurate timing with a high-quality pendulum clock that had the pendulum set at not quite the right length. In the case that the error in each measurement has the same value, the result of applying these rules for propagation of errors can be summarized as a theorem.
You would find different lengths if you measured at different points on the table. What is her experimental error? Chemistry Chemistry 101 - Introduction to Chemistry Chemistry Tests and Quizzes Chemistry Demonstrations, Chemistry Experiments, Chemistry Labs & Chemistry Projects Periodic Table and the Elements Chemistry Disciplines - Chemical Engineering and Another similar way of thinking about the errors is that in an abstract linear error space, the errors span the space.
Tutoring Platform DIY Create your own inquiry space and share it with your students or other teachers powered by Graasp. Clearly, if the errors in the inputs are random, they will cancel each other at least some of the time. The accepted value is the measurement that scientists throughout the world accept as true. The correct data has already been determined in a research lab - the correct data is called the "accepted value".
Limitations imposed by the precision of your measuring apparatus, and the uncertainty in interpolating between the smallest divisions. In[43]:= Out[43]= The above number implies that there is meaning in the one-hundred-millionth part of a centimeter. In[1]:= In[2]:= In[3]:= We use a standard Mathematica package to generate a Probability Distribution Function (PDF) of such a "Gaussian" or "normal" distribution. Nonetheless, our experience is that for beginners an iterative approach to this material works best.
Not too bad. Another way of saying the same thing is that the observed spread of values in this example is not accounted for by the reading error. Nearly all of the graphics are created in Adobe Illustrator, Fireworks and Photoshop. Wolfram Universal Deployment System Instant deployment across cloud, desktop, mobile, and more.
Experimental error (also known as Percent Error) is the percentage you missed the accepted value in the experiment. When you subtract (Step #1) round your answer to the correct number of significant figures. Similarly for many experiments in the biological and life sciences, the experimenter worries most about increasing the precision of his/her measurements. In[25]:= Out[25]//OutputForm=Data[{{789.7, 2.2}, {790.8, 2.3}, {791.2, 2.3}, {792.6, 2.4}, {791.8, 2.5}, {792.2, 2.5}, {794.7, 2.6}, {794., 2.6}, {794.4, 2.7}, {795.3, 2.8}, {796.4, 2.8}}]Data[{{789.7, 2.2}, {790.8, 2.3}, {791.2, 2.3}, {792.6, 2.4}, {791.8,
The correct procedure to do this is to combine errors in quadrature, which is the square root of the sum of the squares. Here is his data: Mass of Aluminum: 18.36 grams Volume of Aluminum: 6.87 mL Density: 18.36 grams / 6.87 mL = 2.672489 g/mL = 2.67 g/mL Accepted Value for the Density Divide this difference (between the experimental value and the accepted value) by the accepted value. Zero error is as close as you can get - you cannot have a -2 % error.
In[8]:= Out[8]= Consider the first of the volume data: {11.28156820762763, 0.031}. A Washington D.C. The accepted value is the measurement that scientists throughout the world accept as true. The particular micrometer used had scale divisions every 0.001 cm.
Water freezes at 0 degrees Celsius is an accepted value. The answer to this depends on the skill of the experimenter in identifying and eliminating all systematic errors. Maybe we are unlucky enough to make a valid measurement that lies ten standard deviations from the population mean. You get a friend to try it and she gets the same result.
Chapter 7 deals further with this case. In this case the meaning of "most", however, is vague and depends on the optimism/conservatism of the experimenter who assigned the error. An important and sometimes difficult question is whether the reading error of an instrument is "distributed randomly". Education All Solutions for Education Web & Software Authoring & Publishing Interface Development Software Engineering Web Development Finance, Statistics & Business Analysis Actuarial Sciences Bioinformatics Data Science Econometrics Financial Risk Management
We form lists of the results of the measurements. In[28]:= Out[28]//OutputForm=Datum[{70, 0.04}]Datum[{70, 0.04}] Just as for Data, the StandardForm typesetting of Datum uses ±. It is important to emphasize that the whole topic of rejection of measurements is awkward. Still others, often incorrectly, throw out any data that appear to be incorrect. | 1,611 | 7,296 | {"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-2018-51 | latest | en | 0.880289 |
https://www.hackmath.net/en/word-math-problems/reason?page_num=30 | 1,571,524,133,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986700435.69/warc/CC-MAIN-20191019214624-20191020002124-00396.warc.gz | 764,939,962 | 7,230 | # Reason - math word problems - page 30
1. Magic number
The number 135 split to two addends so that one addend was 30 greater than 2/5 the addend.
2. Big number
hat is the remainder when dividing number 10 to 47 - 111 by number 9?
3. 3 cats
3 cats eat 3 mice in 3 days. How many mice will eat 10 cats in 10 days?
4. Remainder
A is an arbitrary integer that gives remainder 1 in the division with 6. B is an arbitrary integer that gives remainder 2 the division by. What makes remainder in division by 3 product of numbers A x B ?
5. Family 8
Father is 38 years old, daughter 12, son 14. How many years will father have as many years as his children together?
6. Flowers
The flower has six flowers and each flower has a number. These are the numbers: 20,40,39,28,8,9. What number will be in the middle of the flower so that the numbers come from the flowers when we subtract and add?
7. Unknown numbers
The sum of two consecutive natural numbers and their triple is 92. Find these numbers.
8. Clubhouse
There were only chairs and table in the clubhouse. Each chair had four legs, and the table was triple. Scouts came to the clubhouse. Everyone sat on their chair, two chairs were left unoccupied, and the number of legs in the room was 101. How many chairs w
9. Suzan
Susan's age will be after 12 years four times as much as twelve years ago. How old is Susan now?
10. Soaps
Each box has the same number of soaps. A quarter of all boxes contain only white soaps, and in each of the remaining 120 boxes there are always half the white soaps and half the green. White soaps total 1200. (a) the number of all soap boxes; (b) the sma
11. Summands
Find two summands of the number 42, so that its product is minimized.
12. One hundred stamps
A hundred letter stamps cost a hundred crowns. Its costs are four levels - twenty tenths , one crown, two-crown and five-crown. How many are each type of stamps? How many does the problem have solutions?
13. Warmer weather
This morning it was -6 °C. What temperature did the thermometer show yesterday if it was three times warmer?
14. House numbering
The residential house has three entrances numbered even numbers, successive immediately behind. The sum of the two numbers on the outside entrances is 68. Calculate the middle of these three numbers.
15. Digit sum
The digit sum of the two-digit number is nine. When we turn figures and multiply by the original two-digit number, we get the number 2430. What is the original two-digit number?
Several pieces of bread were in the store. Half of them plus a half of bread bought a cookbook. Of the remaining bread, half of it plus half of the bread was bought by Mr. Novák. The last bread was bought by Ms. Small. How many breads were in the store at.
17. Clouds
Approximately at what height is the cloud we see under an angle of 26°10' and see the Sun at an angle of 29°15' and the shade of the cloud is 92 meters away from us?
18. Ľé sweets
There are 20 sweets in the bag. Some are chocolate, other coconuts, and the remaining marzipan. Chocolate is 4 times more than coconut. Marzipan's less than chocolate. How much is in a bag of coconut sweets?
19. Train delay
The train should be in the station for 10 minutes. It still has 32 km left. What is the expected delay if the train travels in 2 minutes 3 km except for the last section where it is 2 km in 5 minutes.
20. Prove
Prove that k1 and k2 is the equations of two circles. Find the equation of the line that passes through the centers of these circles. k1: x2+y2+2x+4y+1=0 k2: x2+y2-8x+6y+9=0
Do you have an interesting mathematical word problem that you can't solve it? Enter it, and we can try to solve it.
To this e-mail address, we will reply solution; solved examples are also published here. Please enter e-mail correctly and check whether you don't have a full mailbox. | 982 | 3,823 | {"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} | 3.84375 | 4 | CC-MAIN-2019-43 | longest | en | 0.961983 |
http://projectpokemon.org/wiki/index.php?title=Pokemon_Save_Structure&diff=prev&oldid=3840 | 1,560,822,429,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998600.48/warc/CC-MAIN-20190618003227-20190618025227-00343.warc.gz | 143,582,468 | 22,654 | # This Page Is Currently Under Construction And Will Be Available Shortly, Please Visit Reserve Copy Page
Pokémon in the games Pokémon Diamond, Pearl, and Platinum are stored in a 136 byte or 236 byte structures. All unencrypted values are stored in little-endian format. The game encrypts the data when it is stored in the save file. Party Pokémon have additional values to hold current battle stats and conditions. The information below describes the Pokémon data format.
## Checksum
The checksum serves two purposes:
1. It validates the data after decryption, and
2. It serves as the encryption key for the data.
The checksum is calculated in three steps:
1. Split the unencrypted data from offsets 0x08 to 0x87 into two-byte words,
2. Take the sum of the words, and
3. Truncate the sum to 16 bits.
## Block Shuffling
The 128 bytes of Pokémon data are split into four 32-byte blocks for shuffling. The blocks are shuffled according to a shift value derived from the personality value. Given the personality value pv, the expression yielding the shift value is:
((pv >> 0xD) & 0x1F) % 24
The right shifting (pv >> 0xD) is equivalent to a division of 8192.
To shuffle the blocks, take the four blocks of unencrypted data, A, B, C, and D. The blocks shall be rearranged in the encrypted data according to the Block Order column of the following table. (To unshuffle, use the Inverse column.)
Shift Value (decimal) Block Order Inverse
00 ABCD ABCD
01 ABDC ABDC
02 ACBD ACBD
06 BACD BACD
09 BCDA DABC
11 BDCA DACB
15 CBDA DBAC
16 CDAB CDAB
17 CDBA DCAB
18 DABC BCDA
19 DACB BDCA
20 DBAC CBDA
21 DBCA DBCA
22 DCAB CDBA
23 DCBA DCBA
## Encryption
The encryption uses the pseudorandom number generator (PRNG), a linear congruential generator. Elements of the PRNG can be described with the recursive function:
X[n+1] = (0x41C64E6D * X[n] + 0x6073)
To decrypt the data, given a function rand() which returns the upper 16 bits of consecutive results of the above given function:
1. Seed the PRNG with the checksum (let X[n] be the checksum).
2. Sequentially, for each 2-byte word Y from 0x08 to 0x87, apply the transformation: unencryptedByte = Y xor rand()
3. Unshuffle the blocks using the block shuffling algorithm above.
To encrypt the data:
1. Shuffle the blocks using the block shuffling algorithm above.
2. Seed the PRNG with the checksum (let X[n] be the checksum),
3. Sequentially, for each 2-byte word Y from 0x08 to 0x87, apply the transformation: unencryptedByte = Y xor rand()
## Unencrypted bytes
Offset Contents
0x00-0x03 Personality value (Also known as the PID)
0x04-0x05 Unused
0x06-0x07 Checksum
## Encrypted bytes
### Block A
Offset Contents
0x08-0x09 National Pokédex ID
0x0A-0x0B Held Item
0x0C-0x0D OT ID
0x0E-0x0F OT Secret ID
0x10-0x13 Experience points
0x14 Friendship (Steps to Hatch if an egg)
0x15 Ability
0x16 Markings
0x17 Original Language
0x18 HP Effort Value
0x19 Attack Effort Value
0x1A Defense Effort Value
0x1B Speed Effort Value
0x1C SP Attack Effort Value
0x1D SP Defense Effort Value
0x1E Cool Contest Value
0x1F Beauty Contest Value
0x20 Cute Contest Value
0x21 Smart Contest Value
0x22 Tough Contest Value
0x23 Sheen Contest Value
0x24-0x25 Sinnoh Ribbon Set 1
0x26-0x27 Sinnoh Ribbon Set 2
### Block B
Offset Contents
0x28-0x29 Move 1 ID
0x2A-0x2B Move 2 ID
0x2C-0x2D Move 3 ID
0x2E-0x2F Move 4 ID
0x30 Move 1 Current PP
0x31 Move 2 Current PP
0x32 Move 3 Current PP
0x33 Move 4 Current PP
0x34-0x37 Move PP Ups
0x38-0x3B Bits 0-29 - Individual Values<br>
HP ( [0-31] << 0 )<br> Attack ( [0-31] << 5 )<br> Defense ( [0-31] << 10 )<br> Speed ( [0-31] << 15 )<br> SP Attack ( [0-31] << 20 )<br> SP Defense ( [0-31] << 25 )<br> Bit 30 - IsEgg Flag<br>Bit 31 - IsNicknamed Flag
0x3C-0x3D Hoenn Ribbon Set 1
0x3E-0x3F Hoenn Ribbon Set 2
0x40 Bit 0 - Fateful Encounter Flag<br>Bit 1 - Female<br>Bit 2 - Genderless<br>Bit 3-7 - Alternate Forms
0x41 Shiny Leaves (HGSS)<br>Bit 5 - Leaf Crown<br>Bits 0–4 - Leaves A–E (bit 0 is leftmost)
0x42-0x43 Unused
0x44-0x45 Egg Location (Platinum)
0x46-0x47 Met at Location (Platinum)
### Block C
Offset Contents
0x48-0x5D Nickname
0x5E Unused
0x5F Hometown
0x60-0x61 Sinnoh Ribbon Set 3
0x62-0x63 Sinnoh Ribbon Set 4
0x64-0x67 Unused
### Block D
Offset Contents
0x68-0x77 OT Name
0x7B-0x7D Date Met
0x7E-0x7F Egg Location (Diamond/Pearl)
0x80-0x81 Met At Location (Diamond/Pearl)
0x82 Pokérus
0x83 Poké Ball
0x84 Bit 0-6 - Met At Level<br>Bit 7 - Female OT Gender
0x85 Encounter Type
0x86 HGSS Poké Ball
0x87 Unused
## Battle Stats
The battle stats are encrypted in the same manner, however the seed is not the checksum, it is the PID.
### Encrypted Bytes
Offset Contents
0x88 Bits 0-2 - Asleep (0-7 rounds)<br>Bit 3 - Poisoned<br>Bit 4 - Burned<br>Bit 5 - Frozen<br>Bit 6 - Paralyzed<br>Bit 7 - Toxic
0x89 Unknown - Flags - Max Value 0xF0
0x8A-0x8B Unknown
0x8C Level
0x8D Capsule Index (Seals)
0x8E-0x8F Current HP
0x90-0x91 Max HP
0x92-0x93 Attack
0x94-0x95 Defense
0x96-0x97 Speed
0x98-0x99 Special Attack
0x9A-0x9B Special Defense
0x9C-0xD3 Unknown - Contains Trash Data
0xD4-0xEB Seal Coordinates
## Original Language
Value Language
0x1 日本語 (Japan)
0x2 English (US/UK/AU)
0x3 Français (France/Québec)
0x4 Italiano (Italy)
0x5 Deutsch (Germany)
0x7 Español (Spain/Latin Americas)
0x8 한국어 (South Korea)
Bit Marking
0x01 Circle
0x02 Triangle
0x04 Square
0x08 Heart
0x10 Star
0x20 Diamond
## Alternate Forms
Alternate forms are stored at offset 0x40 of the Pokemon structure. Each form follows the pattern of index << 3 where index starts at zero for the main form.
### Unown's Forms
Bit Form Bit Form Bit Form Bit Form
0x00 A 0x38 H 0x70 O 0xA8 V
0x08 B 0x40 I 0x78 P 0xB0 W
0x10 C 0x48 J 0x80 Q 0xB8 X
0x18 D 0x50 K 0x88 R 0xC0 Y
0x20 E 0x58 L 0x90 S 0xC8 Z
0x28 F 0x60 M 0x98 T 0xD0 !
0x30 G 0x68 N 0xA0 U 0xD8 ?
Bit Form
0x00 Normal
0x08 Attack
0x10 Defense
0x18 Speed
Bit Form
0x00 Plant
0x08 Sandy
0x10 Trash
Bit Form
0x00 West
0x08 East
### Rotom's Forms
Rotom's alternate forms will only be displayed in Pokemon Platinum. Rotom may be traded between Diamond, Pearl, and Platinum without losing the alternate form.
Bit Form
0x00 Normal
0x08 Heat
0x10 Wash
0x18 Frost
0x20 Fan
0x28 Cut
### Giratina's Forms
Giratina's Origin form will only be displayed in Pokemon Platinum.
Bit Form
0x00 Altered
0x08 Origin
### Shaymin's Forms
Shaymin's Sky form will only be displayed in Pokemon Platinum.
Bit Form
0x00 Land
0x08 Sky
### Arceus' Forms
Bit Form Bit Form Bit Form
0x00 Normal 0x30 Insect 0x60 Zap
0x08 Fist 0x38 Spooky 0x68 Mind
0x10 Sky 0x40 Iron 0x70 Icicle
0x18 Toxic 0x48 Flame 0x78 Draco
0x20 Earth 0x50 Splash 0x80 Dread
0x28 Stone 0x58 Meadow 0x4C Curse
## Encounter Types
Value Country
0x0 Pal Park, Egg, Hatched, Special Event
0x2 Tall Grass
0x4 Dialga/Palkia In-Game Event
0x5 Cave, Hall of Origin
0x7 Surfing, Fishing
0x9 Building
0xA Great Marsh (Safari Zone)
## Ribbons
Diamond, Pearl, and Platinum store the ribbon information as bitfields in 1-byte words. Given below are the bytewise representation of the ribbon bitfields.
### Sinnoh Ribbon Set 1
Bit Ribbon
0x24 & 0x01 Sinnoh Champ Ribbon
0x24 & 0x02 Ability Ribbon
0x24 & 0x04 Great Ability Ribbon
0x24 & 0x08 Double Ability Ribbon
0x24 & 0x10 Multi Ability Ribbon
0x24 & 0x20 Pair Ability Ribbon
0x24 & 0x40 World Ability Ribbon
0x25 & 0x01 Shock Ribbon
0x25 & 0x02 Downcast Ribbon
0x25 & 0x04 Careless Ribbon
0x25 & 0x08 Relax Ribbon
0x25 & 0x10 Snooze Ribbon
0x25 & 0x20 Smile Ribbon
0x25 & 0x40 Gorgeous Ribbon
0x25 & 0x80 Royal Ribbon
### Sinnoh Ribbon Set 2
Bit Ribbon
0x26 & 0x01 Gorgeous Royal Ribbon
0x26 & 0x02 Footprint Ribbon
0x26 & 0x04 Record Ribbon
0x26 & 0x08 History Ribbon
0x26 & 0x10 Legend Ribbon
0x26 & 0x20 Red Ribbon
0x26 & 0x40 Green Ribbon
0x26 & 0x80 Blue Ribbon
0x27 & 0x01 Festival Ribbon
0x27 & 0x02 Carnival Ribbon
0x27 & 0x04 Classic Ribbon
0x27 & 0x08 Premier Ribbon
### Sinnoh Ribbon Set 3
Bit Ribbon
0x60 & 0x01 Cool Ribbon
0x60 & 0x02 Cool Ribbon Great
0x60 & 0x04 Cool Ribbon Ultra
0x60 & 0x08 Cool Ribbon Master
0x60 & 0x10 Beauty Ribbon
0x60 & 0x20 Beauty Ribbon Great
0x60 & 0x40 Beauty Ribbon Ultra
0x60 & 0x80 Beauty Ribbon Master
0x61 & 0x01 Cute Ribbon
0x61 & 0x02 Cute Ribbon Great
0x61 & 0x04 Cute Ribbon Ultra
0x61 & 0x08 Cute Ribbon Master
0x61 & 0x10 Smart Ribbon
0x61 & 0x20 Smart Ribbon Great
0x61 & 0x40 Smart Ribbon Ultra
0x61 & 0x80 Smart Ribbon Master
### Sinnoh Ribbon Set 4
Bit Ribbon
0x62 & 0x01 Tough Ribbon
0x62 & 0x02 Tough Ribbon Great
0x62 & 0x04 Tough Ribbon Ultra
0x62 & 0x08 Tough Ribbon Master
### Hoenn Ribbon Set 1
Bit Ribbon
0x3C & 0x01 Cool Ribbon
0x3C & 0x02 Cool Ribbon Super
0x3C & 0x04 Cool Ribbon Hyper
0x3C & 0x08 Cool Ribbon Master
0x3C & 0x10 Beauty Ribbon
0x3C & 0x20 Beauty Ribbon Super
0x3C & 0x40 Beauty Ribbon Hyper
0x3C & 0x80 Beauty Ribbon Master
0x3D & 0x01 Cute Ribbon
0x3D & 0x02 Cute Ribbon Super
0x3D & 0x04 Cute Ribbon Hyper
0x3D & 0x08 Cute Ribbon Master
0x3D & 0x10 Smart Ribbon
0x3D & 0x20 Smart Ribbon Super
0x3D & 0x40 Smart Ribbon Hyper
0x3D & 0x80 Smart Ribbon Master
### Hoenn Ribbon Set 2
Bit Ribbon
0x3F & 0x01 Tough Ribbon
0x3F & 0x02 Tough Ribbon Super
0x3F & 0x04 Tough Ribbon Hyper
0x3F & 0x08 Tough Ribbon Master
0x3F & 0x10 Champion Ribbon
0x3F & 0x20 Winning Ribbon
0x3F & 0x40 Victory Ribbon
0x3F & 0x80 Artist Ribbon
0x3E & 0x01 Effort Ribbon
0x3E & 0x02 Marine Ribbon
0x3E & 0x04 Land Ribbon
0x3E & 0x08 Sky Ribbon
0x3E & 0x10 Country Ribbon
0x3E & 0x20 National Ribbon
0x3E & 0x40 Earth Ribbon
0x3E & 0x80 World Ribbon
## Save File Location
The party Pokémon are stored in the save file beginning at offset 0x00098 for the first small block, and 0x40098 for the second small block.
In the first small save block, the first Daycare Pokémon is located at 0x0141C, and the second is located at 0x01508. In the second small block, the first Daycare Pokémon is located at 0x4141C, and the second is located at 0x41508.
In the first small save block, the Pal Park Pokémon are stored beginning at 0x0BA28. In the second small save block, the Pal Park Pokémon are stored beginning at 0x4BA28.
Each party, Daycare, and Pal Park Pokémon is 236 bytes in size.
The PC storage Pokémon are stored in the save file from Box 1 to Box 18. The offset starts at 0x0C104 for the first big block and 0x4C104 for the second big block. Each PC stored Pokémon is 136 bytes in size. | 3,673 | 10,397 | {"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-2019-26 | latest | en | 0.775099 |
https://nrich.maths.org/865 | 1,596,836,816,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737225.57/warc/CC-MAIN-20200807202502-20200807232502-00205.warc.gz | 454,089,172 | 4,841 | ### Is There a Theorem?
Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?
### The Old Goats
A rectangular field has two posts with a ring on top of each post. There are two quarrelsome goats and plenty of ropes which you can tie to their collars. How can you secure them so they can't fight each other but can reach every corner of the field?
### Rolling Triangle
The triangle ABC is equilateral. The arc AB has centre C, the arc BC has centre A and the arc CA has centre B. Explain how and why this shape can roll along between two parallel tracks.
# Turning Triangles
##### Age 11 to 14 Challenge Level:
A triangle ABC resting on a horizontal line is "rolled" along the line. Describe the paths of each of the vertices and the relationships between them and the original triangle.
Describe what happens if the triangle is not equilateral and explain the outcomes you see.
Here are two examples for you.
Right Angle Triangle:
Equilateral Triangle:
You might also like to look at the related problem How Far Does it Move? | 244 | 1,146 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-34 | latest | en | 0.937961 |
https://math.stackexchange.com/questions/1785361/diagonal-operators-on-infinite-dimensional-hilbert-spaces | 1,638,037,475,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00058.warc.gz | 472,393,557 | 35,486 | # Diagonal operators on infinite dimensional Hilbert spaces
the following is a short question regarding a theorem from a quantum mechanics book I am working through but the question is a mathematical one.
There is a theorem which states:
Theorem: The eigenstates of a Hermitian operator defines a complete set of mutually orthonormal basis states. The operator is diagonal in this eigenbasis with its diagonal elements equal to the eigenvalues.
Question:
• For a finite dimensional space we can describe the Hermitian operator as a Hermitian matrix, it is then clear that a diagonal operator is then a diagonal matrix. When considering an operator on an infinite dimensional Hilbert space, what is a diagonal operator defined as? Could it be described as an infinite diagonal matrix. If so, is it required that this is a special case where the spectrum is discrete?
Thanks.
• In the infinite dimensional case, this usually means the operator, say $A$, is unitarily equivalent to a multiplication operator on some $L^{2}$-space. In general, $A$ cannot be described by an $\infty\times\infty$ matrix. May 14 '16 at 19:41
If you're working strictly within the context of an inner product space or Hilbert space, the statement you have learned is false. Even if you could identify certain "states" that you would call "eigenstates," those objects are not vectors in the space. They are some unknown, unspecified object in some unknown and extended space.
For example, start with the position operator $M$ on an interval $[0,1]$. This operator is multiplication by $x$, meaning $(Mf)(x)=xf(x)$. This is a perfectly legitimate selfadjoint linear operator on the Hilbert space $L^2[0,1]$, but it has no eigenvectors, meaning that there are no non-trivial functions $f \in L^2[0,1]$ for which $Mf = \lambda f$, regardless of the choice of $\lambda$. The operator $M$ has "continuous spectrum" but no eigenvectors or eigenvalues. There just are no such "eigenstate" objects.
Physicists typically introduce the $\delta$ function into such a space (which is logically inconsistent) and claim that $(M \delta_{y})(x) = x\delta(x-y)=y\delta_{y}$. For many reasons, on many levels, that is Mathematical non-sense, and cannot be salvaged within $L^2[0,1]$ in a logically consistent way. The axioms of Quantum use Hilbert spaces, but $\delta$ functions cannot live in spaces such as $L^2[0,1]$ because two functions that are equal a.e. in $L^2[0,1]$ are identical, which means that pointwise values have no meaning for elements of $L^2[0,1]$. Dirac knew this, but he liked the intuition.
John von Neumann, who was a contemporary of P.A.M. Dirac, created consistent ways of dealing with the selfadjoint operators of Physics through the Spectral Theorem, and this was available to Dirac at the time; but Dirac chose an intuitive presentation over logical consistency. And it wasn't because rigorous Mathematics was unavailable at the time to handle everything; the correct Math was there.
All of the Hilbert spaces of Quantum must be separable, meaning that the space contains a countable dense subset. The reason for this has to do with constructibility, and being able to bootstrap to general answers through finite approximations. $L^2[0,1]$ is a separable space. One of the consequences of having a separable space, is that an orthonormal basis of such a space is always finite or countably infinite. You cannot have mutually orthogonal objects that are indexed by an interval of the real line, for example. That's impossible in separable spaces. And, if you have a selfadjoint operator $M$ on a Hilbert space, and you have $Mf=\lambda f$ and $Mg = \mu g$ for $\lambda\ne \mu$, then $(f,g)=0$ must hold. So, a selfadjoint operator on a Hilbert space that is allowed in Quantum Mechanics cannot have more than a countable number of actual eigenvalues. That's a consequence of correct axiomatic systems for Quantum. You can have continuous spectrum, but that's different than having a continuum of eigenvalues. If you enlarge the space to allow for such things, then you lose the ability to approximate in a finite way, which disconnects the theory from a setting where finite approximation makes sense.
Sorry to disappoint you. The theorem you have--as stated--is not true. Correct spectral theory gives you essentially the same thing, but not exactly that.
• is there a way to recognize that a normal operator on a separable Hilbert space is (or is not) of the form $T x = \sum_{n=1}^\infty \lambda_n \langle e_n,x\rangle e_n$ for some orthonormal basis $(e_n)$ ? if say $T$ is compact, it automatically has a SVD, and hence if it is normal it has this form, right ? May 14 '16 at 21:26
• @user1952009 : Operators with a continuous spectrum would not be compact. You can have a bounded position observable on a bounded interval, but it is not compact. Compact resolvents make more sense, such as for a Hamiltonian. May 14 '16 at 21:31
• "and you say $\delta_x$ is not so consistent". the operator $P_x : f \to \langle \delta_x, f \rangle$ with $\|f\|^2 = \int_{X} |P_x f|^2 d\mu$ is a definition of a Hilbert space space of functions. if all the $P_x$ are bounded then we are probably in $l^2(X)$, but if they are not then $P_x$ can be defined as some sort of weak-* limit of a sequence of operators of $H^*$ ? is there a problem with those considerations ? May 14 '16 at 21:44
• @user1952009 : There are all kinds of extended structures designed to salvage the Dirac idea in some way for cases of continuous spectrum, but you can't do this in a separable Hilbert space. Sometimes it's just better to start doing correct Math, live without the "ideals," but have a consistent system that really isn't any more difficult to deal with than the kludged-up alternative. May 14 '16 at 21:50 | 1,407 | 5,785 | {"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} | 3.5 | 4 | CC-MAIN-2021-49 | latest | en | 0.915531 |
https://www.unitconverters.net/volume/ton-register-to-acre-inch.htm | 1,566,200,051,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314696.33/warc/CC-MAIN-20190819073232-20190819095232-00000.warc.gz | 1,022,045,435 | 4,455 | Home / Volume Conversion / Convert Ton Register to Acre-inch
# Convert Ton Register to Acre-inch
Please provide values below to convert ton register [ton reg] to acre-inch [ac*in], or vice versa.
From: ton register To: acre-inch
### Ton Register to Acre-inch Conversion Table
Ton Register [ton Reg]Acre-inch [ac*in]
0.01 ton reg0.0002754821 ac*in
0.1 ton reg0.0027548209 ac*in
1 ton reg0.0275482094 ac*in
2 ton reg0.0550964187 ac*in
3 ton reg0.0826446281 ac*in
5 ton reg0.1377410468 ac*in
10 ton reg0.2754820937 ac*in
20 ton reg0.5509641873 ac*in
50 ton reg1.3774104683 ac*in
100 ton reg2.7548209366 ac*in
1000 ton reg27.5482093664 ac*in
### How to Convert Ton Register to Acre-inch
1 ton reg = 0.0275482094 ac*in
1 ac*in = 36.3 ton reg
Example: convert 15 ton reg to ac*in:
15 ton reg = 15 × 0.0275482094 ac*in = 0.4132231405 ac*in | 306 | 841 | {"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-2019-35 | longest | en | 0.330697 |
https://cbseexpert.com/class-12-physics-chapter-6-case-study-question-electromagnetic-induction-pdf-download/ | 1,726,795,495,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00086.warc.gz | 145,340,808 | 23,032 | # Class 12 Physics Chapter 6 Case Study Question Electromagnetic Induction PDF Download
In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Download of CBSE Class 12 Physics Chapter 6 Electromagnetic Induction Case Study and Passage Based Questions with Answers were Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Physics Electromagnetic Induction to know their preparation level.
In CBSE Class 12 Physics Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.
# Electromagnetic Induction Case Study Questions With answers
Here, we have provided case-based/passage-based questions for Class 12 Physics Chapter 6 Electromagnetic Induction
Case Study/Passage Based Questions
Question: 1
Currents can be induced not only in conducting coils but also in conducting sheets or blocks. Current is induced in solid metallic masses when the magnetic flux threading through them changes. Such currents flow in the form of irregularly shaped loops throughout the body of the metal. These currents look like eddies or whirlpools in water so they are known as eddy currents. Eddy currents have both undesirable effects and practically useful applications. For example, it causes unnecessary heating and wastage of power in electric motors, dynamos, and in the cores of transformers.
(I) The working of speedometers of trains is based on
(ii) Identify the wrong statement
(iii) Which of the following is the best method to reduce eddy currents?
(iv) The direction of eddy currents is given by
(v) Eddy currents can be used to heat localized tissues of the human body. This branch of medical therapy is called
Question: 2
Self Induction. When a current I flows through a coil, flux linked with it is φ = LI, where L is a constant known as self-inductance of the coil.
Any change in current sets up an induced emf in the coil. Thus, the self-inductance of a coil is the induced emf set up in it when the current passing through it changes at the unit rate. It is a measure of the opposition to the growth or the decay of current flowing through the coil. Also, the value of self-inductance depends on the number of turns in the solenoid, its area of cross-section, and the permeability of its core material.
(i) The inductance in a coil plays the same role as
(a) inertia in mechanics
(b) energy in mechanics
(c) momentum in mechanics
(d) force in mechanics
(ii) A current of 2.5 A flows through a coil of inductance 5 H. The magnetic flux linked with the coil is
(a) 0.5 Wb
(b) 12.5 Wb
(c) zero
(d) 2 Wb
The inductance L of a solenoid depends upon its radius R as
(a) L ∝ R
(b) L ∝ 1/R
(c) L ∝ R2
(d) L ∝ R3
(iv) The unit of self-inductance is
(a) Weber ampere
(b) Weber-1 ampere
(c) Ohm second
(v) The induced emf in a coil of 10 henry inductance in which current varies from 9 A to 4 A in 0.2
second is
(a) 200 V
(b) 250 V
(c) 300 V
(d) 350 V | 792 | 3,195 | {"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-2024-38 | latest | en | 0.927123 |
https://apk-dl.com/color-puzzle/com.liquid.androidgamei.colores.app/ | 1,521,589,113,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647545.84/warc/CC-MAIN-20180320224824-20180321004824-00576.warc.gz | 524,082,041 | 22,701 | # Color Puzzle. 1.1 APK
1.1 / September 15, 2014
(5.0/5) (1)
## Description
In this fun puzzle game the goal is to makethe falling lights match the cubes at the bottom. By adding andsubtracting the primary colors change the falling lights color andgain points! -Prism Light
## App Information Color Puzzle.
• App Name
Color Puzzle.
• Package Name
com.liquid.androidgamei.colores.app
• Updated
September 15, 2014
• File Size
918k
• Requires Android
Android 4.0 and up
• Version
1.1
• Developer
Liquid Ink.
• Installs
1 - 5
• Price
\$0.99
• Category
Casual
• Developer
Email rrcarrion4jc@gmail.com
4931 Town N Country Blvd.
Algebra Slope . 3.2 APK
Discontinued, the new app can befoundhere:https://www.google.com/url?q=https://play.google.com/store/apps/details?id=com.liquidink.calculushelperFeatures in the new app include:-Slope Intercept Form-Point Slope Form-General Form-y intercept-x intercept-midpoint between two points-distance between two points
Color Puzzle 1.1 APK
In this free fun puzzle game the goal istomake the falling lights match the cubes at the bottom. Byaddingand subtracting the primary colors change the falling lightscolorand gain points! To unlock the full features buy the proversionhttps://play.google.com/store/apps/details?id=com.liquid.androidgamei.colores.app-PrismLight
Parabola Solver 1.0 APK
Parabola Solver can help you quicklyfindout all the properties of a parabola(quadratic equation). Alltheanswers are in forms that teachers accept!Features include:-Standard Form-Vertex Form-Find the equation from three points-Zeroes-Y intercept-Focus-Directrix-Domain-Range
Algebra Slope. 1.1 APK
Algebra Slope can help youquicklyfind out all the properties of a linear equation (line). Alltheanswers are in forms that teachers accept!Features include:-Slope Intercept Form-Point Slope Form-General Form-y intercept-x intercept-midpoint between two points-distance between two points
Color Puzzle. 1.1 APK
In this fun puzzle game the goal is to makethe falling lights match the cubes at the bottom. By adding andsubtracting the primary colors change the falling lights color andgain points! -Prism Light
Algebra Slope (Pro) 1.1 APK
Discontinued, the new app can befoundhere:https://www.google.com/url?q=https://play.google.com/store/apps/details?id=com.liquidink.calculushelperFeatures in the new app include:-Slope Intercept Form-Point Slope Form-General Form-y intercept-x intercept-midpoint between two points-distance between two points
Algebra Slope 1.4 APK
Algebra Slope can help you quickly findoutall the properties of a linear equation (line). All the answersarein forms that teachers accept!Features include:-Slope Intercept Form-Point Slope Form-General Form-y intercept-x intercept-midpoint between two points-distance between two points | 692 | 2,781 | {"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-2018-13 | latest | en | 0.75628 |
http://www.ma.huji.ac.il/~razk/iWeb/My_Site/Teaching.html | 1,552,939,705,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201672.12/warc/CC-MAIN-20190318191656-20190318213656-00468.warc.gz | 321,127,132 | 4,393 | Teaching
Below are links to lecture notes written while teaching at the Hebrew University. You are welcome to use them, but be aware that they are not meant to replace comprehensive textbooks. In particular, they may contain errors. Please let me know if you find any mistake.
1. Measure Theory (80517, Fall 2018)
2. 1.Introduction [Chapter1.pdf]
3. 2.Measure Spaces [Chapter2.pdf]
4. 3.Integration [Chapter3.pdf]
5. 4.Differentiation [Chapter4.pdf]
6. 5.Lp-Spaces [Chapter5.pdf]
8. 7.Some Notes on Topology [Topology.pdf]
This appendix contains a summary of definitions and facts in topology. It is offered as a quick reference for students did not take this course, however, it is by no means a substitute for a comprehensive course.
9. Probability (80420, Fall 2016)
10. 1.Preface [Preface.pdf]
11. 2.Notions in Set Theory [Chapter0.pdf]
12. 3.Probability Spaces [Chapter1.pdf]
13. 4.Conditional Probability [Chapter2.pdf]
14. 5.Discrete Random Variables [Chapter3.pdf]
15. 6.Expectation [Chapter4.pdf]
16. 7.Inequalities [Chapter5.pdf]
17. 8.Continuous Random Variables [Chapter6.pdf]
18. 9.Limit Theorems [Chapter7.pdf]
19. 10. Sequences and Series [Appendix.pdf]
This appendix contains a summary of definitions and facts concerning sequences and series. It is offered as a quick reference for students who might not have learned throughly these subjects.
20. Calculus (80131, Fall 2015)
21. 1.Preface
22. 2.Real Numbers [Chapter2.pdf]
23. 3.Sequences [Chapter3.pdf]
24. 4.Functions [Chapter4.pdf]
25. 5.Derivatives [Chapter5.pdf]
26. Mathematical Methods for Physicists (Fall 2014)
27. 1.Functions of One Variable [Chapter4.pdf]
28. 2.Derivatives and Integrals [Chapter5.pdf]
29. 3.Taylor Polynomials and Taylor Series [Chapter6.pdf]
30. 4.Indefinite Integrals [Chapter7.pdf]
31. 5.Curves in Rn [Chapter8.pdf]
32. 6.Functions of Several Variables [Chapter9.pdf]
33. Basic Notions in Functional Analysis (Fall 2013)
34. 1.Hilbert Spaces [Hilbert.pdf]
35. 2.Banach Spaces [Banach.pdf]
36. 3.Topological Vector Spaces [TVS.pdf]
37. Calculus for Engineers (Spring 2013) [Calculus.pdf]
38. ODEs (Spring 2012) [ODEs.pdf]
Chapter on asymptotic methods [ODEs2.pdf]
39. Mechanics (Spring 2008) [Mechanics.pdf]
40. Asymptotic Methods (Spring 2008) [Asymptotics.pdf] | 659 | 2,301 | {"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-2019-13 | latest | en | 0.745445 |
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