url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
values | snapshot_type stringclasses 2
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value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.techieindoor.com/go-how-to-get-the-arc-tangent-of-y-x/ | 1,719,292,467,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00619.warc.gz | 896,736,831 | 29,330 | # Go – How to get the arc tangent of y/x
Here, We will learn how to get the arc tangent of y/x in go. We can get it by using Atan2() function in math package in go golang.
Function prototype:
`func Atan2(y, x float64) float64`
Return value:
`Atan2() function in math package returnsthe arc tangent of y/x by using the signs ofthe two to determine the quadrant of thereturn value.`
Example with code:
```package main
import (
"fmt"
"math"
)
func main() {
no := math.Atan2(1, 2)
fmt.Printf("%.2f\n", no)
no = math.Atan2(0, 0)
fmt.Printf("%.2f\n", no)
no = math.Atan2(1, 4)
fmt.Printf("%.2f\n", no)
no = math.Atan2(3, 4)
fmt.Printf("%.2f\n", no)
no = math.Atan2(5, 2)
fmt.Printf("%.2f\n", no)
}
```
Output:
`0.460.000.240.641.19`
`https://www.techieindoor.com/go-lang-tutorial/`
`https://golang.org/doc/https://golang.org/pkg/` | 280 | 842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.564117 |
http://www.dbfinteractive.com/forum/index.php?topic=733.msg9335 | 1,386,962,386,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164981569/warc/CC-MAIN-20131204134941-00089-ip-10-33-133-15.ec2.internal.warc.gz | 306,406,939 | 7,997 | ### Author Topic: Looking for White noise / tv static algorithms (Read 1079 times)
0 Members and 1 Guest are viewing this topic.
• Atari ST
• Posts: 105
• Karma: 3
##### Looking for White noise / tv static algorithms
« on: October 09, 2006 »
Wondering if anyone has any pseudo code or theories for coding white noise aka. tv static.
- Black background.
- random White/grey pixels?
#### benny!
• Senior Member
• Posts: 4372
• Karma: 227
• in this place forever!
##### Re: Looking for White noise / tv static algorithms
« Reply #1 on: October 09, 2006 »
I only know the famous tiny ptc test code by gaffer.
Code: [Select]
`//// TinyPTC by Gaffer// www.gaffer.org/tinyptc//#include "tinyptc.h"#define WIDTH 320#define HEIGHT 200#define SIZE WIDTH*HEIGHT static int noise;static int carry;static int index;static int seed = 0x12345;static int pixel[SIZE];int main(){ if (!ptc_open("test",WIDTH,HEIGHT)) return 1; while (1) { for (index=0; index<SIZE; index++) { noise = seed; noise >>= 3; noise ^= seed; carry = noise & 1; noise >>= 1; seed >>= 1; seed |= (carry << 30); noise &= 0xFF; pixel[index] = (noise<<16) | (noise<<8) | noise; } ptc_update(pixel); }}`
And btw. If you want this fx as a screensaver you can dl it from my page :
http://www.weltenkonstrukteur.de/?site=works&prd=TVNoise
« Last Edit: October 09, 2006 by benny! »
[ mycroBLOG - POUET :: whatever keeps us longing - for another breath of air - is getting rare ]
Challenge Trophies Won:
#### Shockwave
• good/evil
• Founder Member
• Posts: 17297
• Karma: 489
• evil/good
##### Re: Looking for White noise / tv static algorithms
« Reply #2 on: October 09, 2006 »
I've found that the effect works best with only a few different colours, say 3 greys and black.
Shockwave ^ Codigos
Challenge Trophies Won:
• Atari ST
• Posts: 105
• Karma: 3
##### Re: Looking for White noise / tv static algorithms
« Reply #3 on: October 09, 2006 »
Thanks for the feedback. I'm try to convert the code above to BlitzMax but am stuck on 1 or two lines:
noise >>= 3;
noise ^= seed;
What is >>= and ^= ? Is this c++ ?
#### Jim
• Founder Member
• Posts: 5257
• Karma: 394
##### Re: Looking for White noise / tv static algorithms
« Reply #4 on: October 09, 2006 »
It's C.
>> is Shr
^ is Xor
So
noise>>=3 is
noise = noise Shr 3
and
noise^=seed is
noise=noise Xor seed
Basically though what's being done here is a random number generator.
Both Blitz and C have one built in so it's not really necessary to write your own.
In Blitz use
noise=Rand(255)
in C use
#include <stdio.h>
noise=rand()&0xff
and in freebasic do
#include "crt.bi"
noise=rand() and &hff
You might want to make sure it's totally random too. You need to seed the random number generator.
In the sample posted that means change 'seed' to be something else.
In Blitz you do
SeedRnd MilliSecs()
In C you do
#include <time.h>
srand(time(NULL));
Jim
Challenge Trophies Won: | 888 | 3,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2013-48 | latest | en | 0.509564 |
https://www.tensorflow.org/api_docs/python/tf/keras/metrics/Hinge?hl=nb-NO | 1,638,957,096,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363465.47/warc/CC-MAIN-20211208083545-20211208113545-00340.warc.gz | 1,028,989,960 | 60,336 | Help protect the Great Barrier Reef with TensorFlow on Kaggle
# tf.keras.metrics.Hinge
Computes the hinge metric between `y_true` and `y_pred`.
`y_true` values are expected to be -1 or 1. If binary (0 or 1) labels are provided we will convert them to -1 or 1.
`name` (Optional) string name of the metric instance.
`dtype` (Optional) data type of the metric result.
#### Standalone usage:
````m = tf.keras.metrics.Hinge()`
`m.update_state([[0, 1], [0, 0]], [[0.6, 0.4], [0.4, 0.6]])`
`m.result().numpy()`
`1.3`
```
````m.reset_state()`
`m.update_state([[0, 1], [0, 0]], [[0.6, 0.4], [0.4, 0.6]],`
` sample_weight=[1, 0])`
`m.result().numpy()`
`1.1`
```
Usage with `compile()` API:
``````model.compile(optimizer='sgd', loss='mse', metrics=[tf.keras.metrics.Hinge()])
``````
## Methods
### `merge_state`
View source
Merges the state from one or more metrics.
This method can be used by distributed systems to merge the state computed by different metric instances. Typically the state will be stored in the form of the metric's weights. For example, a tf.keras.metrics.Mean metric contains a list of two weight values: a total and a count. If there were two instances of a tf.keras.metrics.Accuracy that each independently aggregated partial state for an overall accuracy calculation, these two metric's states could be combined as follows:
````m1 = tf.keras.metrics.Accuracy()`
`_ = m1.update_state([[1], [2]], [[0], [2]])`
```
````m2 = tf.keras.metrics.Accuracy()`
`_ = m2.update_state([[3], [4]], [[3], [4]])`
```
````m2.merge_state([m1])`
`m2.result().numpy()`
`0.75`
```
Args
`metrics` an iterable of metrics. The metrics must have compatible state.
Raises
`ValueError` If the provided iterable does not contain metrics matching the metric's required specifications.
### `reset_state`
View source
Resets all of the metric state variables.
This function is called between epochs/steps, when a metric is evaluated during training.
### `result`
View source
Computes and returns the metric value tensor.
Result computation is an idempotent operation that simply calculates the metric value using the state variables.
### `update_state`
View source
Accumulates metric statistics.
For sparse categorical metrics, the shapes of `y_true` and `y_pred` are different.
Args
`y_true` Ground truth label values. shape = `[batch_size, d0, .. dN-1]` or shape = `[batch_size, d0, .. dN-1, 1]`.
`y_pred` The predicted probability values. shape = `[batch_size, d0, .. dN]`.
`sample_weight` Optional `sample_weight` acts as a coefficient for the metric. If a scalar is provided, then the metric is simply scaled by the given value. If `sample_weight` is a tensor of size `[batch_size]`, then the metric for each sample of the batch is rescaled by the corresponding element in the `sample_weight` vector. If the shape of `s` | 766 | 2,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-49 | latest | en | 0.701777 |
http://www.cfd-online.com/Forums/cfx/25260-energy-imbalance-turbulence.html | 1,438,240,879,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042987135.9/warc/CC-MAIN-20150728002307-00254-ip-10-236-191-2.ec2.internal.warc.gz | 366,169,908 | 17,190 | # Energy imbalance and turbulence
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February 19, 2008, 22:00 Energy imbalance and turbulence #1 yunhee Guest Posts: n/a Hello, I want to ask a few questions as follows: 1)How are T energy imbalance and H energy imbalance? (My system consists of steel block and cooling water which passes the inside of the block. By the way, the domain for water has H-energy imbalance option as well as various options (e.g., P-mass imbalance and etc) while the block domain has the only T-energy imbalance in cfx-solver. 2) How can I interpret the imbalance from monitor in cfx-solver? (As I explained above, my model has two domains of water and block. So, '%imbalance=domain imbalance/maximum over all domain' is needed for my case (i.e., multidomains)? If yes, maximum over all domain means maximum imbalance over all domain? Am I right?) 3) In the block, there are two different sizes of cross-secitonal area diameter (hole diameter) for water to pass. For example, the water passes the inlet hole with big cross-sec diameter first and later the water is divided according to two small holes (this cross section is very small). For a certain water flow rate, the water is therfore turbulent at the inlet hole and simultaneously laminor at the small hole. In this case, how do I have to set simulation options in cfx-pre in order to cover both water's turbulance and laminar flow at different locations? Thank you so much.
February 20, 2008, 17:39 Re: Energy imbalance and turbulence #2 Glenn Horrocks Guest Posts: n/a Hi, 1) CFX solves an enthalpy equation in the fluid regions but a temperature equation in the solid regions. The imbalance is error in the conservation of that quantity. Read the documentation for more information. 2) No, the balances are calculated from the entire solution space. It is not a maximum imbalance, but a total imbalance of the entire solution. Again this is described in the documentation. 3) Relaminarisation of flows is not modelled in CFX. However, I have had some success in modelling flows like this using a k-w based turbulence model (such as SST) as these models allow the turbulence intensity to go to zero and predict a laminar-like state with no turbulent viscosity. It's not perfect but good enough for most engineering applications. The turbulence transition model cannot help you as it only predicts laminar to turbulent transitions, not the other way. Glenn Horrocks
February 20, 2008, 18:22 Re: Energy imbalance and turbulence #3 yunhee Guest Posts: n/a Hi, Glenn. Thank you for your kind answer. If you don't mind, I want to ask additional questions. 1) If I set the default domain-> fluid models->turbulence model option: k-Epsilon (in cfx-pre), then this option can cover both laminar and turbulent cases ( I mean it covers all ranges of reynolds number)? 2) From the energy imbalance graph presented in monitoring window (i.e., plot for energy imbalance in cfx-solver), the unit of variable value (y value) is the percentage? OR unitless? Could you please let me know answers? Thanks a lot. Sincerely,
February 21, 2008, 19:16 Re: Energy imbalance and turbulence #4 Glenn Horrocks Guest Posts: n/a Hi, 1) The k-e model does not handle low turbulence levels well. The k-w family of turbulence models do. Have a look at "Turbulence Modeling for CFD" by Wilcox for details. 2) The imbalances displayed in Solver manager are percent. Glenn Horrocks
February 21, 2008, 20:02 Re: Energy imbalance and turbulence #5 yunhee Guest Posts: n/a Dear Glenn, Thank you for your kind answer again. Glenn, I am not sure still about the meaning of y value for the imbalance plot displayed in Solver manager... If y is zero, does it mean energy in = energy out? In my opinion, y is residual value for imbalance or energy equation measurment . So, if y is graudually approaching to 0 (or equal to zero), it means that residual is zero and thus the solution is convering. But this convergence does not gurantee the energy conservation (i.e., energy in = energy out). Because the solution convergence can be also guranteed in case of energy inconservation... If you don't mind, could you please give me an answer? It might be the last question... Thanks again.
February 24, 2008, 18:19 Re: Energy imbalance and turbulence #6 Glenn Horrocks Guest Posts: n/a Hi, If the imbalances are zero that means the energy equation has fully balanced. That is, integrated across the entire simulation domain Energy in - energy stored = energy out. This is another form of convergence criterion. The residuals used by CFX as a default are another. It is quite possible for the residuals to have converged well but the imbalances to not be converged; and also it is quite possible for the residuals to not have converged but the imbalances have converged. Feel free to ask as many questions as you wish, that is why the forum is here! Glenn Horrocks
February 25, 2008, 10:56 Re: Energy imbalance and turbulence #7 yunhee Guest Posts: n/a Thank you so much for your kind answer again, Glenn. You are a treasure in this forum! Thanks again. Have a great day~!
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Contact Us - CFD Online - Top | 1,413 | 5,859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-32 | longest | en | 0.913762 |
https://www.univerkov.com/determine-the-acceleration-with-which-a-sled-is-rolling-down-a-slide-7-m-high-and-17-m-long/ | 1,701,471,195,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00842.warc.gz | 1,173,473,148 | 6,336 | # Determine the acceleration with which a sled is rolling down a slide 7 m high and 17 m long
Determine the acceleration with which a sled is rolling down a slide 7 m high and 17 m long. The coefficient of friction of the sleigh on the snow is 0.3.
h = 7 m.
L = 17 m.
μ = 0.3.
g = 10 m / s2.
a -?
m * a = Ft + Ftr + N, where Ft is the force of gravity that acts on the sled, Ftr is the friction force of the sled against the slide, N is the reaction force of the slide to the sled.
ОХ: m * a = m * g * sinα – Ftr.
OU: 0 = – m * g * cosα + N.
a = (m * g * sinα – Ftr) / m.
N = m * g * cosα.
Ftr = μ * N = μ * m * g * cosα.
a = (m * g * sinα – μ * m * g * cosα) / m = g * (sinα – μ * cosα).
sinα = h / L = 7 m / 17 m = 0.41.
cosα = √ (1 – sin2α).
cosα = √ (1 – (0.41) 2) = 0.91.
a = 10 m / s2 * (0.41 – 0.3 * 0.91) = 1.37 m / s2.
Answer: the sled will move with acceleration a = 1.37 m / s2.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 410 | 1,218 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2023-50 | latest | en | 0.85925 |
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# Problem 1198. Handle to an array of functions
Solution 189859
Submitted on 11 Jan 2013 by Tomasz
• Size: 24
• This is the leading solution.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% f{1}=@(x) x^2; f{2}=@(x) x+3; f{3}=@(x) x/2; g=cf(f); x=[1 2 3]; y_correct = [2 3.5 6]; assert(isequal(g(x),y_correct))
ans = @real
2 Pass
%% f{1}=@(x) x^0.5; f{2}=@(x) x-1; f{3}=@(x) x^2; f{4}=@(x) x/3; g=cf(f); x=[ 16 49 100]; y_correct = [3 12 27]; assert(isequal(g(x),y_correct))
ans = @real
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Start Hunting! | 279 | 749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-45 | latest | en | 0.644958 |
https://handleonlineclass.com/2021/05/27/exercise-deductive-inductive-reasoning-in-chapter-3-you-have-read/ | 1,653,574,998,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662606992.69/warc/CC-MAIN-20220526131456-20220526161456-00687.warc.gz | 343,050,153 | 22,026 | # Exercise: Deductive & Inductive Reasoning In Chapter 3, you have read
In Chapter 3, you have read about the two types of human reasoning, “deductive” and “inductive.” Deductive reasoning relies on logical relationships between claims and tells us, given what we know or assume to be true, what must necessarily be true. Consider the following example of a deductive argument in the form of modus ponens:
Premise 1: If it is raining outside, then I have an umbrella.
Premise 2: It is raining outside.
Conclusion: Therefore, I have an umbrella.
Kris Barton & Barbara G. Tucker: Inductive Reasoning ______________________________________________________
For this assignment, you will be demonstrating your understanding of deductive and inductive reasoning. First, create your own example of each of the three deductive argument forms discussed on pp. 45-46: modus pones, modus tollens, and hypothetical syllogism. I gave an example of one, modus ponens, above, about it raining outside and me having an umbrella. That’s exactly what I’m asking you to do. Don’t use my example or an example from the book or internet. Just think of your own example to show you can create the argument forms. It is only necessary to create one example of each argument. Please state the argument so that the logical form is apparent rather than discussing your argument in a passage. You may use this format:
1. premise
2. premise
3. conclusion
Your examples should look just like my example of modus ponens above, only with your own original content. To show your understanding of inductive reasoning, answer the following using the reading in the textbook and the link provided above:
1. What is inductive reasoning?
2. How does inductive reasoning differ from deductive reasoning?
3. What is a generalization?
4. What is causal reasoning?
5. What is sign reasoning?
6. What is analogical reasoning?
7. Describe a situation where you have used one of the types of inductive reasoning that you defined in questions 2-5. To be clear, your submission to this dropobox should include:
1) one original example for each of the three deductive argument forms, and | 506 | 2,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.873989 |
https://blog.reduza.com.br/en/common-denominators-worksheet.html | 1,709,232,875,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474852.83/warc/CC-MAIN-20240229170737-20240229200737-00088.warc.gz | 129,806,469 | 7,616 | # Common Denominators Worksheet
Common Denominators Worksheet - Web results for common denominators worksheets 3,200 + results sort by: Web there are generally four methods that can be used for comparing fractions. Web a way to address this is to learn how to compare fractions and find common denominators so that two fractions can be. Web this printable offers students more practice subtracting fractions with common denominators and reducing them. Web the worksheets in this section deal with adding fractions with like denominators, which are an easy first step to dealing. Each file linked below contains 10 worksheets. First is to use common denominators. Reading fractions and matching to their words. Web common denominator worksheet worksheet 3, level 5 put these over a common denominator : The common denominator is the term used if the denominator of two or more fractions is.
Web this is a free lesson about finding a common denominator in fraction addition. Web results for common denominators worksheets 3,200 + results sort by: A fraction represents a part of a whole or any number of equal parts. Develop addition skills with simple fractions having common denominators. Web what is a common denominator? When the denominators of two or more fractions are the same, they are common. Web writing fractions from a numerator and denominator.
Common Denominators Worksheet - The common denominator has to be a multiple. Web this page has 50 common denominator worksheets, sorted into 5 different levels of difficulty. Web results for common denominators worksheets 3,200 + results sort by: The common denominator is the term used if the denominator of two or more fractions is. Question 1 4 9 and 10 27:. Web this is a free lesson about finding a common denominator in fraction addition. First is to use common denominators. Web common denominator worksheet worksheet 1, level 1 put these over a common denominator : Web common denominator worksheet worksheet 3, level 5 put these over a common denominator : Each file linked below contains 10 worksheets.
Question 1 3 5 and 4 7:. Web results for common denominators worksheets 3,200 + results sort by: Web there are generally four methods that can be used for comparing fractions. Web common denominator worksheet help your pupils develop and enhance their knowledge and understanding of fractions using this. The common denominator has to be a multiple.
## When The Denominators Of Two Or More Fractions Are The Same, They Are Common.
Reading fractions and matching to their words. Each file linked below contains 10 worksheets. Web a way to address this is to learn how to compare fractions and find common denominators so that two fractions can be. Web with common denominators fraction addition.
## This Activity Teaches Students How To Find The Common Denominator Of Two Fractions With Unlike.
Web this is a free lesson about finding a common denominator in fraction addition. Question 1 4 9 and 10 27:. Develop addition skills with simple fractions having common denominators. Web this page has 50 common denominator worksheets, sorted into 5 different levels of difficulty.
## Web Writing Fractions From A Numerator And Denominator.
Web common denominator worksheet worksheet 1, level 1 put these over a common denominator : A fraction represents a part of a whole or any number of equal parts. Web this printable offers students more practice subtracting fractions with common denominators and reducing them. Web this page has printable worksheets on finding the least common denominator, lcd, or lowest common denominator.
## First Is To Use Common Denominators.
Web each worksheet has a variety of fractions all with a common (same) denominator. Web common denominator worksheet worksheet 3, level 5 put these over a common denominator : Question 1 3 5 and 4 7:. Web the worksheets in this section deal with adding fractions with like denominators, which are an easy first step to dealing. | 794 | 3,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-10 | latest | en | 0.908738 |
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Using this rule, the sentence "Jack and Jill went up the hill" is changed to "Etar tzn Evmm gkzc dw cuk uvmm". The Hill Cipher: A Cryptosystem Using Linear Algebra Robyn N. The Playfair cipher uses a 5 by 5 table containing a keyword or phrase. The ciphers in this book (except for the RSA cipher in the last chapter) are all centuries old, and modern computers now have the computational power to hack their encrypted messages. 3) Only 26 letters in the system [using MOD 26]. Let m be a positive integer. Please leave any questions or comments in the comments section below. Ciphers Polybius - 2 examples found. The most complex in the Hill cipher is to create a sufficiently large inverse matrix. SQGKC PQTYJ using the Hill cipher with the inverse key calculations and the result. All of these are examples of symmetric encryption. INTRODUCTION: The study of the advanced Hill cipher [1], which depends mainly upon the concept of an involutory matrix (a matrix which is equal to its inverse), has roused the interest of researchers in the areas of cryptography and image cryptography. • Caesar cipher. ciphers,firstdescribedbyLesterHill[4]in1929. For example, let's assume the key is 'point'. Hill in 1929, the Hill cipher is a polygraphic substitution cipher based on linear algebra. The Hill cipher The Playfair cipher is a polygraphic cipher; it enciphers more than one letter at a time. 3) Only 26 letters in the system [using MOD 26]. If the keyword is as long as the plaintext, for example, a previously agreed upon sample of text, the cipher is unbreakable if a new key is used for every message. In this post, I have shown how RSA works, I will follow this up L1 with another post explaining why it works. 20 -25 & practice encryption/decryption, key strength discussion. Upper case and lower case characters are treated equally). Hill in 1929, it was the first polygraphic cipher in which it was practical (though barely) to operate on more than three symbols at once. A must be invertible mod 26. The missing E in DESP A RATLY from K3. Symmetric ciphers use symmetric algorithms to encrypt and decrypt data. How do you find inverse of matrix in hill cipher place the 4x4 identity matrix on the right and adjoined. Pigpen Cipher is used extensively in Freemason documentation. According to definition in wikipedia, in classical cryptography, the Hill cipher is a polygraphic substitution cipher based on linear algebra. redThe Hill Cipher - Computer Science and Engineering Nov 16, 2004 The Hill Cipher. We shall use Vigenère Table. Once the Hill Cipher is chosen as our encryption algorithm, the next task is to create a matrix to encrypt the text. Hill who introduced them in two papers. Well, to be honest I am not a expert on the topics related to Encryption, Cryptography or Advanced Encryption Standard. • Block ciphers are substitution ciphers (and not transpositions). The scheme was invented in 1854 by Charles Wheatstone, but bears the name of Lord Playfair who promoted the use of the cipher. 2 thoughts on “ Rail Fence Cipher Program in C and C++[Encryption & Decryption] ” Ashokkumar July 22, 2018. Hill's Cipher Lester S. To encipher a message, first the plaintext is broken into blocks of n letters which are converted to numbers, where A=0, B=1, C=2 Y=24, Z=25 (so each character is assigned to a number which is usually from the range of 00-25 for the characters A-Z. The method described above can solve a 4 by 4 Hill cipher in about 10 seconds, with no known cribs. Examples JavaScript Example of a Hill Cipher [2] This is a JavaScript implementation of a Hill Cipher. Vigenere Cipher is a method of encrypting alphabetic text. Viswambari# Department of Computer Science, Nehru Memorial College, Puthanampatti, Trichy, India Abstract Hill cipher encryption is one of the polygraph cipher of classical encryption in which if the encryption key matrix called key matrix is not chosen properly,. (AFPDRFPBA – Caesared tip. for the determinant there is usually a formula, such as: a x d - b x c However, for the Hill Cipher I am completely lost. See Table A, repeated below, for quick. Note that all vowels in the English alphabet map to even numbers using the standard Hill cipher encoding! Conversely, solving the system modulo 13 tells you the fourth letter of the unknown plaintext (up to rot13). Here is known plaintext and resulting cipher. In this algorithm, we consider all possible states from the current state and then pick the best one as successor , unlike in the simple hill climbing technique. , has the same structure like an affine Hill cipher. For this example we will use a 3x3 matrix. In order for a matrix to be used in this equation, the matrix must be invertible. 1 (441 ratings) Course Ratings are calculated from individual students’ ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. ” It is not discussed why interweaving and interlacing strengthen the Hill cipher. Solving the linear system modulo 2 gives you the parity of the second and third letters of the unknown plaintext. You need 97 characters to do the Hill cypher, so the idea was to temporarily bring in a couple of printable characters (which no one is likely to use on input) that have codes greater than 128, map them to 127 and 128 to do the cypher, and then afterwards convert them back. June 28-30, 2006. The missing E in DESP A RATLY from K3. I have done the following: a) found the inverse of K:. Submitted by Monika Sharma, on January 08, 2020. Contribute to tstevens/Hill-Cipher development by creating an account on GitHub. When Knuckles arrives at Pumpkin Hill, he's surprised to learn he can fly (well, glide). It also make use of Modulo Arithmetic (like the Affine Cipher). Invented by Lester S. And it is, of. This cipher was invented in 1929 by Laster S. The rail fence cipher (sometimes called zigzag cipher) is a transposition cipher that jumbles up the order of the letters of a message using a basic algorithm. These implement command-line DES and IDEA encryption algorithms. Abraham Sinkov and Todd Feil. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. A new image encryption technique that combines Elliptic Curve Cryptosystem with Hill Cipher (ECCHC) has been proposed in this paper to convert Hill cipher from symmetric technique to asymmetric one and increase its security and efficiency and resist the hackers. Cryptography is a cipher, hashing, encoding and learning tool for all ages. I have a project to make an encryption and decryption for string input in java. RC4 is a stream cipher, meaning that it encrypts one byte at a time. They are not selected or validated by us and can contain inappropriate terms or ideas. The Vigenère Cipher Encryption and Decryption. Decryption is. The Encoding Matrix Is Typically Kept Private To Keep The Original. In classical cryptography, the Hill cipher is a polygraphic substitution cipher based on linear algebra. The case here is restricted to 2x2 case of the hill cipher for now, it may be expanded to 3x3 later. Cut out the 2 circles leaving the inner circle NOT HOLLOW. Ciphers Polybius - 2 examples found. , the E is already missing in the original plaintext. It was the first cipher that was able to operate on 3 symbols at once. 11 Data Encryption Standard 23 2. Let's look at a simple example of a classical cipher based on substitution. However, I have been tasked with finding the solution without a known chunk of plaintext. The idea is to take m linear combinations of the m alphabetic characters in one. Post navigation. Hill cipher was the first polygraphic cipher. This tool solves monoalphabetic substitution ciphers, also known as cryptograms. 14 ÷ 10 = 1 R 4 14 Mod 10 = 4 24 Mod 10 = 4 Modulus Examples. In G cipher, A becomes G, B becomes H and so on. Also if you don't want to use Cramer's Formula for it (using the determinant) you can just do linear operations and row-reduce it (if you have ever seen that technique before, or want to learn it now). Suppose "M" is the plaintext and the key is given as 4, then you get the Ciphertext as the letter "Q". A stream cipher is an encryption algorithm that encrypts 1 bit or byte of plaintext at a time. About this tool. Please leave any questions or comments in the comments section below. The cipher accomplishes this using uses a text string (for example, a word) as a key, which …. Thus, a matrix of order 32 is generated which means the length of the quantum shift register is 5 qubits. +1 equals A+B It is Related to the Caesar Cipher. In classical cryptography, the Hill cipher is a polygraphic substitution cipher based on linear algebra. Decipher the message MWALO LIAIW WTGBH JNTAK QZJKA ADAWS SKOKU AYARN CSODN IIAES OOKJY B using the Hill cipher with the 2 23 inverse key Show your calculations and the result. But, I will not your question unanswered and share with you what I know of the topic. If the keyword was longer than the 4 letters needed, we would only take the first 4 letters, and if it was shorter, we would fill it up with the alphabet in order (much like a Mixed Alphabet). a 7!D b 7!E x 7!A To decrypt, shift the letters to the left by 3 and wrap around. Visit Stack Exchange. 13 Blowfish 28 2. Encryption: To encrypt a message using the Hill cipher. Evening •History of cryptology Read TCB Intro, pgs. Generation of Key Matrix for Hill Cipher using Magic Rectangle K. Please leave any questions or comments in the comments section below. GitHub is home to over 40 million developers working together to host and review code, manage projects, and build software together. Rude or colloquial translations are usually marked in red or orange. Even though it is a type of classical and historical cryptography method, it has a special place in my heart because of strong math background and easy adaptation. More complex ciphers like the polyalpha-betic Vigenere cipher, are harder to solve and the solution by hand takes much more time. Hill cipher is one of the techniques to convert a plain text into ciphertext and vice versa. 2 Hill Cipher The Hill cipher (HC) algorithm is one of the famous and known symmetric key algorithms in the field of cryptography. Indeed, the protection of sensitive communications has. The algorithm as follows. So, before going further, we should throw some light on the matrix. This is a simplified implementation of the cipher, using a 2x2 matrix. Hill in 1929, it was the first polygraphic cipher in which it was practical (though barely) to operate on more than three symbols at once. Find answers to pascal code hill cipher 4 x 4? from the expert community at Experts Exchange. Hill Cipher. The idea is to take m linear combinations of the m alphabetic characters in one. Also if you don't want to use Cramer's Formula for it (using the determinant) you can just do linear operations and row-reduce it (if you have ever seen that technique before, or want to learn it now). To decode the message, the person has to be aware which cipher has been used. Hill s cipher machine, from figure 4 of the patent In classical cryptography, the Hill cipher is a polygraphic substitution cipher based on linear algebra. This is in contrast to a substitution cipher, in which the plaintext letters are replaced by letters from another alphabet (or by different letters from the same alphabet). Polygraphic Substitution Ciphers Encrypts letters in groups Frequency analysis more difficult Hill Ciphers Polygraphic substitution cipher Uses matrices to encrypt and decrypt Uses modular arithmetic (Mod 26) Modular Arithmetic For a Mod b, divide a by b and take the remainder. (ABS, traction control, brake assist) and, for off-roading, Hill Descent Control. This cipher was invented in 1929 by Laster S. Hill Cipher decryption has been added with a given decryption matrix. Because of this, the cipher has a significantly more mathematical nature than some of the others. Take for example the Hill or matrix cryptosystem as implemented in HillCryptosystem. The Hill Cipher uses an area of mathematics called Linear Algebra, and in particular requires the user to have an elementary understanding of matrices. The Caesar cipher is one example of a substitution cipher. Globe Magazine Jubilee Jim Fisk and the great Civil War score In 1865, a failed stockbroker tries to pull off one of the boldest financial schemes in American history: the original big short. Part C: Hill Cipher Given the key displayed below 4x4 key. Similar to the Keyword Cipher, the Playfair cipher also uses a keyword. Hill Cipher Please email your Mathematica file to [email protected] com Abstract – Hill cipher is a polygraphic substitution cipher based on linear algebra. The cipher is named after the ancient Roman general Julius Caesar, due to its use in military affairs and private communications. Characters of the plain text. [3], which was broken by Keliher [4] using a known/chosen-plaintext attack similar to the linear algebraic attack on the original Hill Cipher. The main drawback of Hill Cipher is selecting the correct encryption key matrix for encryption. To counter charges that his system was too complicated for day to day use, Hill constructed a cipher machine for his system using a series of geared wheels and. For this example we will use a 3x3 matrix. 100% Working codes and genuine output. Also Read: Caesar Cipher in Java. ciphers,firstdescribedbyLesterHill[4]in1929. Caesar cipher technique was founded by Julius caesar. The scheme was invented in 1854 by Charles Wheatstone , but bears the name of Lord Playfair for promoting its use. Invented by Lester S. Caesar Cipher ; Mono- alphabetic Cipher; Playfair Cipher; Hill Cipher; Polyalphabetic Cipher; The Caesar Cipher involves replacing each letter of the alphabet with the letter - standing places down or up according to the key given. If they key is easy to guess, it is easy for an attacker to decrypt the encrypted data and possibly create fake messages herself. HILL CIPHER In classical cryptography, the Hill cipher is a polygraphic substitution cipher based on linear algebra. the proposed cipher. Text test - a simple way of seeing what characters are passed over in a form. mastermathmentor. The restore command restores the cipher's key to a previously stored state. E orts to redeem the Hill Cipher are not without merit. Learn Hill Cipher with 3×3 Matrix Multiplicative Inverse Example By pnp. THE HILL CIPHER, RINGS, AND SOME LINEAR ALGEBRA MATH 195 Usually, in linear algebra we take matrices with entries in the real numbers or some other field, but this is not strictly required. Romantic Tantalizers Cipher as an Improved Version of Hill Cipher Andru Putra Twinanda Informatics Engineering, School of Electrical Engineering and Informatics, Institut Teknologi Bandung e-mail: nd[email protected] 1-14 • Monoalphabetic substitution ciphers with spaces • Finish classwork and complete examples, pgs. charAt(i)]); } return cipher. These are ciphers where each letter of the clear text is replaced by a corresponding letter of the cipher alphabet. Plain Text is used Column. It is scalable in the number of bits and in the number of rounds. C p key= +( )mod26. Hill in 1929, it was the first polygraphic cipher in which it was practical (though barely) to operate on more than three symbols at once. 10 Rotor Machines 22 2. Louis CSE571S ©2011 Raj Jain Summary 1. Directions: Type or paste ciphertext into cipher box. This course covers the complete syllabus of Cryptography and Network security for GATE examination. INTRODUCTION: The study of the advanced Hill cipher [1], which depends mainly upon the concept of an involutory matrix (a matrix which is equal to its inverse), has roused the interest of researchers in the areas of cryptography and image cryptography. These numbers will form the key (top row, bottom row. Polybius extracted from open source projects. Gronsfeld ciphers can be solved as well through the Vigenère tool. Next you would create a message, have 1-26 represents letter A-Z, respectively with 27 meaning a space and place the numerical value into the matrix. The actual field construction shall match the approved plans. That is, generate a random $$m$$ x $$m$$ matrix to be used as a block permutation, where $$m$$ is the block length of this Hill cipher. As time progressed, the study of cryptography began to involve higher level mathematics. Also Read: Java Vigenere Cipher. The Hill cipher was the first cipher purely based on mathematics (linear algebra). Caesar Cipher is the earliest known substitution cipher. To encrypt a message, each n block of letters will be multiplied by the n*n matrix, against modulus 26. I have the message, "Which wristwatches are swiss wristwatches. LIMITATIONS. • Atbash cipher. The cipher is based on the Vigenère cipher, with the main difference that also numbers can be encoded. thanks in advance!!. Testing text messages using the hill cipher algorithm successfully carried out in accordance with the flow or the steps so as to produce a ciphertext in the form of randomization of the letters of the alphabet. In Hill Cipher algorithm to encrypt plaintext block of size n, we need key matrix ( with entries are between 10,p included, but the determinant must be relatively prime to , each entry in the plaintext block is. Encipher In order to encrypt a message using the Hill cipher, the sender and receiver must first agree upon a key matrix A of size n x n. In this example above, t he fundamental pieces of information about the Hill cipher are given. For example Key matrix. Invented by Lester S. and in this manner got its name. First, you need to assign two numbers to each letter in the alphabet and also assign numbers to space,. It is very attractive. What is plaintext and ciphertext? plaintext is the input message given by user. Encryption has been used for many thousands of years. Kamal Hossain[1], Md. The Atbash Cipher has been added. In this algorithm, we consider all possible states from the current state and then pick the best one as successor , unlike in the simple hill climbing technique. Symmetric and Asymmetric Encryption. Invented by Lester S. The question should indicate that they are to compute the decryption matrix using the Hill Cipher and provide the key. If $$n$$ is the size of the cryptosystem alphabet, then there are $$n^{m^2}$$ possible keys. Each letter is first encoded as a number. S A K N O X A O J X 3. The example here is restricted to a 2x2 case of the Hill cipher. RC4 (Ron’s Code) is a symmetric key encryption algorithm. JavaScript Example of the Hill Cipher [2] This is a JavaScript implementation of the Hill Cipher. Hill cipher decryption needs the matrix and the alphabet used. 15 Conclusion 31. Cryptanalysis is the process of breaking the cipher and discovering the meaning of the message. " You convert everything to upper case and write it without spaces. You can not see it directly on the screenshot from the movie [6] and also not in the movie itself. In the Hill cipher each letter corresponds to one unique number. Hill cipher - Example of 3x3 matrices - Decryption part. In this particular example let's say the digraphs with the most frequencies are RH and NI. CE ciphers given in The Cryptogram are all solvable by pencil and paper methods, although computers and other mechanical aids are often used to assist. The Vigenere cipher consists of using several Caesar ciphers in sequence with different shift values. Pick a plain text message (three words, no spaces; meetmetonight, for example—or use something else!), go through the detailed steps of the Hill cipher explained in the textbook (Chapter 2 and any supporting. redThe Hill Cipher - Computer Science and Engineering Nov 16, 2004 The Hill Cipher. Indeed, the protection of sensitive communications has. 29 was chosen because there are 29 characters and it is also a prime number, which solves some. It is not a true cipher, but just a way to conceal your secret text within plain sight. An example is a modifled Hill Cipher due to Sastry et al. 7 Novel Modification to the Algorithm 18 2. Furthermore, due to its linear nature, the basic Hill cipher succumbs to known-plaintext attacks [4][9]. Plain Text is used Column. Ivplyprr th pw clhoic pozc. Prime numbers play important roles in various encryption schemes. For example, a king is about 13 inches tall and a rook is about 6. The first step is to create a matrix using the. Also Read: Hill cipher code in python. I use "LibreOffice Calc" and "Sage Mathematics" for Linux. This makes block ciphers popular today. Hill Cipher 4x4, with a 4x4 square matrix, is used as a symmetrical algorithm, which has a relatively fast processing time. Hope that this will help to understand the concept Monoalphabetic cipher algorithm. GitHub Gist: instantly share code, notes, and snippets. Write a MATLAB function decrypt(c,A,b,L) which decrypts a given coded message block c of length L, supposing it was coded using a Hill cipher with matrix A and vector b. Decryption In order to decrypt, we turn the ciphertext back into a vector, then simply multiply by the inverse matrix of the key matrix (IFKVIVVMI in letters). Simple ciphers Simple encryption algorithms, which were invented long before first computers, are based on substitution and transposition of single plaintext characters. Breaking a Hill Cipher - ciphertext-only. Hill cipher is a polygraphic substitution cipher based on linear algebra. However, a main drawback of this algorithm is that it encrypts identical plaintext blocks to identical ciphertext blocks and cannot encrypt images that contain large areas of a single color. If the G cipher is used, then A becomes G, B becomes H, C becomes I. the Hill cipher scheme. As can be seen, this extension has some interesting consequences for cryptography. The missing E in DESP A RATLY from K3. Hill in 1929, it was the first polygraphic cipher in which it was practical (though barely) to operate on more than three symbols at once. Hill cipher. Input Expected output Actual output 1 2 1 2 4 4 3 6 9 4 8 16 problem in coding for inverse the key matrix used for encryption and decryption in Hill cipher This tell us. All pieces are moved with a hook on the end of the stick. Sayers's Have His Carcase (1932). Hill, an American mathematician. The technique encrypts pairs of letters (bigrams or digrams), instead of single letters as in the simple. Since is a non-abelian group and matrix multiplication is not commutative, then the multiplication of Hill Cipher keys with RGB image. xls But 100 x 100 would be just as easy to perform. n = 26, k = 3. For example. Meet other local Jeep owners to discuss styles, performance, customizations, and off-roading. Hill in 1929, it was the first polygraphic cipher in which it was…. Learn Hill Cipher with 3×3 Matrix Multiplicative Inverse Example By pnp. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, How to find the key matrix of a 2x2 Hill Cipher? Ask Question Asked 1 year, 11 months ago. • To counter charges that his system was too complicated for day to day use, Hill constructed a cipher machine for his system using a series of geared wheels and chains. The cipher is based on the Vigenère cipher, with the main difference that also numbers can be encoded. Sefik Serengil December 4, 2018 January 5, 2020 Cryptography. The inverse of matrix K for example is (1/det(K)) * adjoint(K), where det(K) <> 0. The Clinton Bus Tour 1992 Roadrunner a Ford E350 4x4 The newer WHCA Roadrunners are electronic platforms built into a Ford E350 4x4 Econoline with an extended cab platform was adapted in 1992 the van was ordered with a 460 V-8 engine with two generators one under the hood as well as one in the rear both ran off the vehicles gas tank. Hill's Cipher Lester S. Rolf-Dieter Reiss∗ and Ulf Cormann. Please report examples to be edited or not to be displayed. In classical cryptography, a permutation cipher is a transposition cipher in which the key is a permutation. In the following example A is a 2 x 2 matrix and the message will be enciphered in blocks of 2 characters. Input Expected output Actual output 1 2 1 2 4 4 3 6 9 4 8 16. These are ciphers where each letter of the clear text is replaced by a corresponding letter of the cipher alphabet. Well, to be honest I am not a expert on the topics related to Encryption, Cryptography or Advanced Encryption Standard. When creating the matrix, use numbers under 26 (representing letters in the english alphabet). Secret Codes & Number Games 17 SYMBOL CIPHERS – SAY IT WITH PIGGIES Small changes to the shape, color or composition of an object might not be perceptible to a casual viewer. Today, we will discuss yet another substitution technique – Hill Cipher which is far better than monoalphabetic cipher. In a Hill cipher encryption the plaintext message is broken up into blocks of length according to the matrix chosen. Many kinds of polygraphic ciphers have been devised. mastermathmentor. JavaScript Example of the Hill Cipher [2] This is a JavaScript implementation of the Hill Cipher. Read an input file to a byte array and write the encrypted/decrypted byte array to an output file accordingly. The applications of algebra in cryptography is a lot and hill cipher is just an example of it. This project for my Linear Algebra class is about cryptography. According to definition in wikipedia, in classical cryptography, the Hill cipher is a polygraphic substitution cipher based on linear algebra. The Double Transposition Cipher Back to Crack the Ciphers This was one of the most secure hand ciphers used in the Second World War. Note that all vowels in the English alphabet map to even numbers using the standard Hill cipher encoding! Conversely, solving the system modulo 13 tells you the fourth letter of the unknown plaintext (up to rot13). I know with other matrices, e. First we write the plaintext in a block of reasonable size for the plaintext. find C,C++,JAVA programs with output images. The Hill Cipher. Determine the number of different (good) keys there are for a 2 * 2 Hill cipher without counting them one by one, using the following steps: a. Solving the linear system modulo 2 gives you the parity of the second and third letters of the unknown plaintext. charAt(i)][message. Tried to use the 4x4 matrix as the Hill cipher key, without success. , pulling rather than pushing on the building. 7 Novel Modification to the Algorithm 18 2. charAt(i)][message. Similar to the Keyword Cipher, the Playfair cipher also uses a keyword. OUTPUT Enter plain-text: meet Enter block size of matrix: 2 Enter key Matrix 3 1 5 2 Encrypted Text is: OQ FG Enter key Inverse Matrix: 2 -1 -5 3 Decrypted Text is: ME ET. Example: S n Abelian Groups Commutativity Most nite abelian groups are also rings Rings Closed under a second operation is commutative is associative Multiplicative identity, 1 Multiplication distributes over addition Example: Z m Fields is commutative Multiplicative inverse for all non-zero elements Example: Z p, prime p. The program will take tw. edu by 4pm on Monday. The Hill Cipher uses matrix multiplication to encrypt a message. 307 In the most general terms possible, an encryption system must combine two ele- ments: some information--called the key-- known only to the authorized communi- cants, and an algorithm which operates on this key and the message (plaintext) to produce the cipher. The Hill Cipher encrypts blocks of letters simultaneously. A random key within the key space of this Hill cipher. Mathematical Cryptography - Crack The Code 4. The first row of this table has the 26 English letters. Hill Cipher Please email your Mathematica file to [email protected] We have explored three simple substitution ciphers that generated ciphertext C from plaintext p by means of an arithmetic operation modulo 26. The key space is the set of all invertible matrices over Z 29. 15 Conclusion 31. A polyalphabetic cipher is any cipher based on substitution, using multiple substitution alphabets. In a Hill cipher encryption the plaintext message is broken up into blocks of length according to the matrix chosen. Pick a plain text message (three words, no spaces; meetmetonight , for example—or use something else!), go through the detailed steps of the Hill cipher explained in the textbook (Chapter 2 and any supporting material), and encrypt it. - bmml bfobhear sunday, october 14, 12. In a substitution cipher, a letter such as A or T, is transposed into some other. As a small example, a 10 x 10 involutory matrix Mod 41 in the Picture. "5 17 4 15" (without quotes). Hill's cipher machine, from figure 4 of the patent In classical cryptography , the Hill cipher is a polygraphic substitution cipher based on linear algebra. For example, this would mean that in a Caesar cipher shift of three: A would become D; B would become E; C would become F etc. We can read Plaintext and we can not read Ciphertext because it is encrypted code. A block cipher is capable of encrypting a single fixed-sized block of data; and, by the evidence around us, apparently it is easier to build good block ciphers than stream ciphers. Start at about 3% incline, and try to hold your pace. HILL CIPHER In classical cryptography, the Hill cipher is a polygraphic substitution cipher based on linear algebra. Simple ciphers Simple encryption algorithms, which were invented long before first computers, are based on substitution and transposition of single plaintext characters. Also if you don't want to use Cramer's Formula for it (using the determinant) you can just do linear operations and row-reduce it (if you have ever seen that technique before, or want to learn it now). ?,-0: 1: 2: 3: 4: 5: 6. For example, one can define a similar game for measuring the security of a block cipher-based encryption algorithm, and then try to show (through a reduction argument) that the probability of an adversary winning this new game is not much more than P E (A) for some A. Assume Plaintext = P, Ciphertext = C, and the Key = K. HILL CIPHER Hill cipher is a classical cipher which is based on linear algebra. We illustrate the method with the help of examples and show that the suggested modifications in Hill Cipher do not make it significantly stronger. The Hill Cipher uses matrix multiplication to encrypt a message. 7) Remark: Many modern cryptosystems (DES, AES, RSA) are also block ciphers. The process of encryption and. There are still surviving examples of letters written using the cipher from the 16th Century. Hill Cipher Polygraphic Substitution Cipher. Jadi, secara. The encryption of the original text is done using the Vigenère square or Vigenère table. The Caesar shift cipher, named because it was used by Julius Caesar himself, is actually 26 different ciphers, one for each letter of the alphabet. The Hill cipher • The first cipher devised to make use of algebraic methods was one created by mathematician Lester Hill in 1929. Berbeda dengan yang lain meski bisa diperpanjang untuk mengerjakan blok huruf berukuran berbeda. HILL CIPHER Hill cipher is a classical cipher which is based on linear algebra. According to these results Hill Cipher with randomized approach proved that it has better encryption quality compared to Hill image cipher. Sequences of numbers are preceded by a 'Q' and a character that implies how many numbers will follow. Register Now! It is Free Math Help Boards We are an online community that gives free mathematics help any time of the day about any problem, no matter what the level. "5 17 4 15" (without quotes). 1) Write a C++ console program in three parts. Hill Cipher Please email your Mathematica file to [email protected] 7 Novel Modification to the Algorithm 18 2. To encrypt a message, each block of n letters (considered as an n-component vector) is multiplied by an. An Efficient Modification to Playfair Cipher 1 Md. The output from cipherProc cget -key can be passed back into the restore command to implement a crude save feature. Recall that the Playfair cipher enciphers digraphs - two-letter blocks. It is a poly-alphabetic cipher based on linear algebra. THE HILL CIPHER, RINGS, AND SOME LINEAR ALGEBRA MATH 195 Usually, in linear algebra we take matrices with entries in the real numbers or some other field, but this is not strictly required. As an example you can crack the following cipher text with this tool: Altd hlbe tg lrncmwxpo kpxs evl ztrsuicp qptspf. As stated before, the Hill Cipher is an encryption algorithm that uses polygraphic substitution ciphers based. 2 Forapolygraphicsubstitution,changing justoneortwoplaintextletterscancompletelychangethecorrespondingciphertext!That. It is a famous Polygram and classical ciphering algorithm based on matrix transformation. 223 In this set of exercises, using matrices to encode and decode messages is examined. , and ? or !. (Hill Cipher –Authors’ Contribution) 17 2. I Transposition ciphers. The Hill Cipher uses an area of mathematics called Linear Algebra, and in particular requires the user to have an elementary understanding of matrices. Complications also. Cipher is also called as Cypher. Hope that this will help to understand the concept Monoalphabetic cipher algorithm. In the example below, we want to transmit the word HELLO which is stored on the plain text tape. Hill's cipher machine, from figure 4 of the patent In classical cryptography , the Hill cipher is a polygraphic substitution cipher based on linear algebra. For example. The Double Transposition Cipher Back to Crack the Ciphers This was one of the most secure hand ciphers used in the Second World War. In this method, matrices and matrix multiplication have been used to combine the plaintext. An online, on-the-fly Baconian cipher encoder/decoder. [3], which was broken by Keliher [4] using a known/chosen-plaintext attack similar to the linear algebraic attack on the original Hill Cipher. Multi-letter ciphers •Multi-letter ciphers work by substituting a group of letters (2, 3 or more at a time) by another group of letters (usually the same length) -The Playfair cipher uses square diagrams to substitute digrams of the plaintext -The Hill Cipher uses matrix operations to substitute letter sequences, n at a time, where n is a. The most complex in the Hill cipher is to create a sufficiently large inverse matrix. Multiplication of Matrices Adding and Subtracting Matrices - ChiliMath. For example, this would mean that in a Caesar cipher shift of three: A would become D; B would become E; C would become F etc. It then asks to use the Hill Cipher to show the calculations and the plain text when I decipher the same encrypted message "KCFL". The Rot Cipher is when you take a letter and put it back or fourth to equal a different letter. The example here is restricted to a 2x2 case of the Hill cipher. Keywords: Keywords: Hill cipher, matrix, eigenvalue, diagonal matrix, exponentiation, pseudo-random number, dynamic key, image encryption 1 Introduction The Hill cipher (HC) [1] and [2] is a well-known symmetric cryptosystem that multiplies a plaintext vector by a key matrix to get the ciphertext. A 2-Dimensional Hill Cipher Example. Hill cipher (matrix based); exams will involve 3x3 or 4x4 matrices. instead of ‘26’, as is the case with Hill cipher. The matrix is called a cipher key. In this method, matrices and matrix multiplication have been used to combine the plaintext. • Combine the stream with the plaintext to produce the ciphertext (typically by XOR) = ⊕ Example of Stream Encryption Key Ciphertext Stream Plaintext Example of. Simple G Code Example Mill – G code Programming for Beginners. 'Prepare for examinations and take any number of courses from various topics on Unacademy Hill cipher - Example of 3x3 matrices - Encryption part. A substitution cipher is a type of encryption where characters or units of text are replaced by others in order to encrypt a text sequence. HARDWARE AND SOFTWARE REQUIREMENT: 1. Substitution Solver. The Hill cipher is considered to be the first polygraphic cipher in which it is practical to work on more than three symbols at once. That is, for each plain-text character we have a corresponding substitution known as a key letter. decrpytion We will now decrypt the ciphertext "SYICHOLER" using the keyword "alphabet" and a 3x3 matrix. 12 Example: Playfair Cipher Program file for this chapter: This project investigates a cipher that is somewhat more complicated than the simple substitution cipher of Chapter 11. Java & C Programming Projects for $30 -$250. Hill cipher - Example of 3x3 matrices - Encryption part. Hill Cipher Description The hill cipher is a polygraphic substitution cipher based on linear algebra modular mathematics. Hill in 1929, it was the first polygraphic cipher in which it was practical (though barely) to operate on more than three symbols at once. First, you need to assign two numbers to each letter in the alphabet and also assign numbers to space,. For a recap of how the Hill cipher works, see here. Start MATLAB from there. Up vote 0 down vote favorite. Rude or colloquial translations are usually marked in red or orange. The inverse of matrix K for example is (1/det(K)) * adjoint(K), where det(K) <> 0. As a running example, we use the ciphertext below; it is the encryption, by a randomly chosen substitution cipher, of the first two sentences in Section 1. [3], which was broken by Keliher [4] using a known/chosen-plaintext attack similar to the linear algebraic attack on the original Hill Cipher. Note that (a) the windward pressure varies along the height of the building, while the leeward pressure is assumed to be constant; and that (b) the leeward pressure is negative, i. Example: Encipher Hill cipher. As can be seen, this extension has some interesting consequences for cryptography. 14:24 mins. I really appreciate if you have sample source code or function method for Hill cipher in java that I may use in my project. Problem 1 Download the various MATLAB sample files and save them in a directory where you want to store your whole project. Vignere's cipher - Encryption & decryption. Grade-1, first quality ceramic tile for commercial and residential wall and top use. To decrypt hill ciphertext, compute the matrix inverse modulo 26 (where 26 is the alphabet length), requiring the matrix to be invertible. These should be run at a slower pace and slightly lower incline than the intervals described above. Hill in 1929, it was the first polygraphic cipher in which it was practical (though barely) to operate on more than three symbols at once. But, I will not your question unanswered and share with you what I know of the topic. Recall that the Playfair cipher enciphers digraphs - two-letter blocks. It is scalable in the number of bits and in the number of rounds. Code: Output: Enter 3×3 matrix for key (It should be inversible): 6 24 1 13 16 10 20 17 15 Enter a 3 letter string: act Encrypted string is: poh Inverse Matrix is: -626351520 -647356092 Read more…. Details on picking a key. Hill used matrices and matrix multiplication to mix up the plaintext. • To counter charges that his system was too complicated for day to day use, Hill constructed a cipher machine for his system using a series of geared wheels and chains. The Hill cipher The Playfair cipher is a polygraphic cipher; it enciphers more than one letter at a time. This is the method used in the "Cryptograms" often found in puzzle books or. append(square[k. Meanwhile, the operations performed in modern encryption algorithms are usually similar but they affect single bits and bytes. 11 Data Encryption Standard 23 2. It also uses matrices and matrix multiplication to form a ciphertext from a plain text and vice versa. Support this blog on Patreon! Hill cipher is a kind of a block cipher method. 13 Blowfish 28 2. Caesar Cipher ; Mono- alphabetic Cipher; Playfair Cipher; Hill Cipher; Polyalphabetic Cipher; The Caesar Cipher involves replacing each letter of the alphabet with the letter - standing places down or up according to the key given. The Hill cipher is based on linear algebra and overcomes the frequency distribution problem of the Caesar cipher that was previously discussed. Hill cipher in python. In this paper, we have used an improved version Hill Cipher method which uses an orthogonal matrix as its key matrix. All separated beams shall receive full depth solid blocking at 24" on center, maximum spacing. " You convert everything to upper case and write it without spaces. please share the program to do the same as soon as possible. Pick a plain text message (three words, no spaces; meetmetonight, for example—or use something else!), go through the detailed steps of the Hill cipher explained in the textbook (Chapter 2 and any supporting material), and encrypt it. The plaintext is written in a grid, and then read off following the route chosen. Hill Cipher. It is a polygraphic substitution cipher that depends on linear algebra. To decode the message, the person has to be aware which cipher has been used. In Hill Cipher Index of Ceaser Cipher is used. The Hill cipher was developed by Lester Hill & introduced in an article published in 1929. Hill cipher was the first polygraphic cipher. The Vigenère cipher is probably the best-known example of a polyalphabetic cipher, though it is a simplified special case. 'Prepare for examinations and take any number of courses from various topics on Unacademy Hill cipher - Example of 3x3 matrices - Encryption part. The m = 2 Hill Cipher. Hill in 1929, it was the first polygraphic cipher in which it was practical (though barely) to operate on more than three symbols at once. Application to Hill Cipher¶. harmonize with or adjust to; "key one's actions to the voters' prevailing attitude" any of 24 major or minor diatonic scales that provide the tonal framework for a piece of music. Hill created the Hill cipher, which uses matrix manipulation. Vigenere Cipher is a method of encrypting alphabetic text. The Caesar shift cipher, named because it was used by Julius Caesar himself, is actually 26 different ciphers, one for each letter of the alphabet. Upper case and lower case characters are treated equally). See Table A, repeated below, for quick. A cipher was present for each letter of the alphabet, for example ROT1 is one of the ciphers. 2x2 Hill is a simple cipher based on linear algebra, see this link. Teams will decode encrypted messages using cryptanalysis techniques, or show skill with advanced ciphers by encrypting or decrypting a message. If the effort. Why? • Consider DES, with 64 bit block cipher. Hill cipher requires a matrix based. Wheatstone and Baron Playfair of St. Until 1996 export from the U. Given a message encoded with a shift/rotation cipher, such as rot13, this recipe recovers the most probable plain text for the message. Application to Hill Cipher¶. 1) A 4x4 encryption matrix. Practical - 5 Write a program to find hamming distance. Hill in 1929, it was the first polygraphic cipher in which it was practical (though barely) to operate on more than three symbols at once. A random key within the key space of this Hill cipher. For encryption we write the message diagonally in zigzag form in a matrix having total rows = key and total columns = message length. This will of course exclude a letter, but in this cipher, the letters I and J are combined to represent one letter. We can read Plaintext and we can not read Ciphertext because it is encrypted code. You can try to crack it with guessing a known plaintext inside the cipher and then reverse the encoding matrix, like it is shown here: Practical Cryptography You should be able to solve the matrix equation if you have n^2 text, means for a 4x4 mat. First, you need to assign two numbers to each letter in the alphabet and also assign numbers to space,. The Hill Cipher: A Cryptosystem Using Linear Algebra Robyn N. Generation of Key Matrix for Hill Cipher using Magic Rectangle K. ElGamal Elliptic Curve is used as an asymmetric algorithm, where the key used for the encryption and decryption process is a different key, so no key distribution is needed. Search hill cipher, 300 result(s) found modern h_p hill ips upfc hi this code involves upfc with modern h_p hill ips codes results are obtained for time vs speed,power oscillation damping controller with consideration of its closed-loop stability, dynamic tracking optimality and robustness against to variation of the power system operating. Rabiul Islam 1 Department of Computer Science & Engineering, University of Inf ormation T echnology and Sciences,. 12 International Data Encryption Algorithm 26 2. Basically Hill cipher is a cryptography algorithm to encrypt and decrypt data to ensure data security. Ideal for a class that wants to discover rather than wash, rinse and repeat :-). You must write a program in either Java or C that encrypts the alphabetic letters in a file using the Hill cipher where the Hill matrix can be any size from 2 x 2 up to 9 x 9. Hill cipher decryption needs the matrix and the alphabet used. More complex ciphers like the polyalpha-betic Vigenere cipher, are harder to solve and the solution by hand takes much more time. A Vigenère cipher builds on this method by using multiple Caesar ciphers at different points in the message; this article shows you how to use it. Imagined by Lester S. As his search moves from abandoned camping grounds to more sinister subterranean settings and his grip on reality loosens. Next we look at our table (where a space is replaced with a #):. The CB band plan consists of 40 channels where a few half-step frequencies are commonly used in remote-control aircraft and cars. The user must be able to choose J = I or no Q in the alphabet. Header file, driver file and implementation file. A polyalphabetic cipher is any cipher based on substitution, using multiple substitution alphabets. It is scalable in the number of bits and in the number of rounds. The method described above can solve a 4 by 4 Hill cipher in about 10 seconds, with no known cribs. 8 Poly-Alphabetic Cipher 21 2. The working is shown. Users can change the entries a, b, c, and d in the encryption matrix [a b; c d] using. Partial-Size Key Ciphers • Actual ciphers cannot use full size keys, as the size is large. In order for a matrix to be used in this equation, the matrix must be invertible. For example, the Data Encryption Standard (DES) encryption algorithm is considered highly insecure; messages encrypted using DES have been decrypted by brute force within a single day by machines such as the Electronic Frontier Foundation's (EFF) Deep […]. Hey I'm taking the Hardvard CS50 course through ItunesU and I'm working on writing a code that will encrypt a message using the vigenere cipher which uses the equation Ci=(Pi+Kj)%26 where P is plaintext and K is the word to encrypt by. It also uses matrices and matrix multiplication to form a ciphertext from a plain text and vice versa. Let's look at a simple example of a classical cipher based on substitution. Wind pressure is deployed on the "windward" and "leeward" sides of the building as shown in Fig. The working is shown. It is an example of a block cipher, because it encrypts blocks of characters of plaintext simultaneously. Assuming the language being dealt with is English, then the letter 'E' would normally be the most frequent letter, or in a mono-alphabetic cipher some other letter would be assigned the same frequency as 'E'. 20-21 • Practice cracking Caesar shift ciphers. instead of ‘26’, as is the case with Hill cipher. Posts about Hill Cipher in Algorithm Program in C written by samir. Part C: Hill Cipher Given the key displayed below 4x4 key. For example, one can define a similar game for measuring the security of a block cipher-based encryption algorithm, and then try to show (through a reduction argument) that the probability of an adversary winning this new game is not much more than P E (A) for some A. But it is not true for. The longer the keyword, the more secure the cipher. The cipher is named after the ancient Roman general Julius Caesar, due to its use in military affairs and private communications. The app allows the user to change the encryption matrix and encipher a message. Rail fence cipher, Columnar cipher are some examples of the transposition cipher. • Combine the stream with the plaintext to produce the ciphertext (typically by XOR) = ⊕ Example of Stream Encryption Key Ciphertext Stream Plaintext Example of. even with arbitrarily many examples of said encryption being used , it would be theoretically impossible to break without just brute forcing it. The matrix is called a cipher key. For example Key matrix. Caesar cipher: The Caesar cipher is an additive cipher. What is the inverse of the 3x3 matrix (mod 26) when, A = [ 4 9 15 15 17 6 24 0 17 ] (Don't need to know how, just need the inverse). Suppose "M" is the plaintext and the key is given as 4, then you get the Ciphertext as the letter "Q". Hill in 1929, it was the first polygraphic cipher in which it was practical (though barely) to operate on more than three symbols at once. Below is an example of a Playfair cipher, solved by Lord Peter Wimsey in Dorothy L. It is an example of a block cipher, because it encrypts blocks of characters of plaintext simultaneously. Caesar cipher, Hill cipher, mono-alphabetic cipher are some examples of the substitution cipher. The main drawback of Hill Cipher is selecting the correct encryption key matrix for encryption. If $$n$$ is the size of the cryptosystem alphabet, then there are $$n^{m^2}$$ possible keys. Quality is at the heart of everything we do at CARiD, so whatever your project, our brand name products and qualified experts will ensure success. The Hill cipher The Playfair cipher is a polygraphic cipher; it enciphers more than one letter at a time. References Bauer, F. com Abstract - Hill cipher is a polygraphic substitution cipher based on linear algebra. Pigpen gravestone. All of these are examples of symmetric encryption. ) I Homophonic ciphers. Developed in 1987 by Ronald Rivest, it is used in SSL and many applications such as Lotus Notes and Oracle Secure SQL. In Hill cipher, each letter is represented by a number modulo 26. Luckily for you though, its very simple. Example: Using the example matrix, compute the inverse matrix : \begin. In cryptography, a substitution cipher is a method of encrypting by which units of plaintext are replaced with ciphertext, according to a fixed system; the "units" may be single letters (the most common), pairs of letters, triplets of letters, mixtures of the above, and so forth. The question should indicate that they are to compute the decryption matrix using the Hill Cipher and provide the key. Hill cipher is a type of monoalphabetic polygraphic substitution cipher. For example, if n=5 then k=2, 3, 4. These numbers will form the key (top row, bottom row). These should be run at a slower pace and slightly lower incline than the intervals described above. C minor, for one. The following discussion assumes an elementary knowledge of matrices. Looking for the latest tile trends? Shop our new products section to see the newest tile products added to our collection. Invented by Lester S. The structure of the Playfair cipher is a 5×5 grid of the English alphabet. In this article, we are going to learn three Cryptography Techniques: Vigenére Cipher, Playfair Cipher, and Hill Cipher. , the E is already missing in the original plaintext. As an example of this, look at the following drawings of a cartoon pig. Imagined by Lester S. 15 Conclusion 31. 14 Decipher the message YITJP GWJOW FAQTQ XCSMA ETSQU SQAPU SQGKC PQTYJ using the Hill cipher with the inverse key. The Playfair Cipher was popularised by Lyon Playfair, but it was invented by Charles Wheatstone, one of the pioneers of the telegraph. +1 equals A+B It is Related to the Caesar Cipher. A Step by Step Hill Cipher Example. Next we look at our table (where a space is replaced with a #):. Mathematical Cryptography - Crack The Code 4. C code to Encrypt & Decrypt Message using Substitution Cipher Here, we have given C program to implement Substitution Cipher to encrypt and decrypt a given message. ciphers,firstdescribedbyLesterHill[4]in1929. The technique encrypts pairs of letters (bigrams or digrams), instead of single letters as in the simple. First, you need to assign two numbers to each letter in the alphabet and also assign numbers to space,. Hill's cipher is a cipher with a two part key, a multiplier $$m$$ which is a square $$n\times n$$ matrix and a shift $$s$$ which is a vector with $$n$$ entries; typically all the arithmetic is done modulo 26. But based on the fact, that you can see the beginning characters of the next lines and assuming that each line has the same amount of characters, the E can not be part of the plaintext, i. Invented by Lester S. Download it now and see much more!. the proposed cipher. The Hill cipher has achieved Shannon's diffusion, and an n-dimensional Hill cipher can diffuse fully across n symbols at once. 14:24 mins. 14 ÷ 10 = 1 R 4 14 Mod 10 = 4 24 Mod 10 = 4 Modulus Examples. The Caesar shift cipher, named because it was used by Julius Caesar himself, is actually 26 different ciphers, one for each letter of the alphabet. redThe Hill Cipher - Computer Science and Engineering Nov 16, 2004 The Hill Cipher. Similar to the Hill cip her the affine Hill cipher is polygraphic cipher, encrypting/decrypting 𝑚𝑚 letters at a time. Plaintext. The app allows the user to change the encryption matrix and encipher a message. As an example, let's encrypt the message 'Meet me at the Hammersmith Bridge tonight'. However, I have been tasked with finding the solution without a known chunk of plaintext. regulate the musical pitch of. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Our emphasis is on making explanations easy to understand in order to further the general interest in cryptography and cryptanalysis. 1 #21 and 23 and What Makes a Matrix Invertible Clicker questions in 2.
8va7tzgll7, yorxg7dmshdhy, rpxmi9ku6485, q6cgbibha5idh, akm1uxskxg, jdgvx49ocf90sam, qhh65czhxzrs, 2y502d2x4s, y8x6ntne64, id9s77ownv4pfb, b4jd4r4tn2ks, ouikf0r7v2eyrh4, 4coc402ow7rmp, 3rporwl7g3e, 2gklhk455e, regt0p2sakc, ozwo3hq9kuj, aku3q1npzdj, mtjgfzurc6tcib1, 3in5wrbwgmb, btm0hflghirj, eird5x4brv2igk, jnvujjz62nxx7t2, qqzdwh6yoo7, eu9cdyg69y, ioss4edj7zoe, oi3fwnd890cy | 12,355 | 54,864 | {"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} | 2.9375 | 3 | CC-MAIN-2020-24 | latest | en | 0.909788 |
https://www.nagwa.com/en/explainers/386178283437/ | 1,702,228,418,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102612.80/warc/CC-MAIN-20231210155147-20231210185147-00164.warc.gz | 972,323,321 | 16,525 | Lesson Explainer: Percentage Change | Nagwa Lesson Explainer: Percentage Change | Nagwa
# Lesson Explainer: Percentage Change Mathematics • 7th Grade
In this explainer, we will learn how to apply percentage change and work out the percentage increase or decrease.
You should already be familiar with proportions and percentages.
We are here considering a quantity, called the original quantity, that undergoes a change (i.e., it either increases or decreases).
We want to express the amount of change (either the increase or the decrease) as a percentage of the original quantity. This means that we are comparing the increase or the decrease to the original quantity as we do in any proportion that compares a part to a whole. (The only difference is that the increase here is not a part of the original quantity. The rest of the mathematical comparison goes exactly the same way.)
### How To: Finding the Percentage of Change
If we are given the original quantity and the quantity after change, we first need to find the amount by which the quantity has either increased or decreased, the increase or decrease. For this, we simply take the absolute value of the difference between the quantity after change and the original quantity:
Then, we compare the increase or decrease to the original quantity as a percentage:
Combining the two steps in one gives
Let us look at the first example to check our understanding of the percentage of change.
### Example 1: Increasing a Quantity by a Given Percentage
Increase 45 by 12%.
Here, the original quantity is 45, and we want to increase it by 12%. So, we need to find 12% of 45. This is given by multiplying 45 by 12%, or . We find that 12% of 45 is
We have found the value of the increase. Then the final quantity is
In the previous example, we have found that when we increase a quantity by , the final quantity is then which can be rewritten as The same is true if a quantity is decreased by ; it is given by which can be rewritten as
Now we are going to look at examples where we are given the quantity after the change (the increase or decrease) and the percent of change, and we need to find the original quantity.
### Example 2: Calculating the Percentage of Change
The costs of two different video games decreased by \$9. The original cost of the first game was \$49, and the original cost of the second one was \$43. Which game had the greater percent of decrease?
We see that the decrease was the same in absolute value: \$9. But as the original prices are different, if we express the decrease as a percentage of the original price, they will be different. Without performing the calculation, we see that the same amount (\$9) is a greater fraction of the smaller original price (\$43).
Our answer is, the second game had the greater percent of decrease.
### Example 3: Finding the Original Quantity Knowing Its Value after a given Percentage of Change
A bottle of hand lotion is on sale for \$5.01. If this price represents a 28% discount from the original price, determine the original price to the nearest cent.
Here we have the price of the bottle after the discount, which means that this price is the price after change. The bottle is on sale with a 28% discount. This means that the price has been reduced by 28%. This can be summarized with this diagram.
Therefore, the sale price (\$5.01) is of the original price.
We can see on the diagram that dividing 5.01 into 72 equal shares will give the value of 1% of the original price. The original price is then 100 times this amount: .
We can also write an equation that describes that 72% of the original price is the sale price: , where is the original price.
We have .
Dividing both sides of the equation by 0.72, we find that
Therefore, the original price was \$6.96.
### Example 4: Finding a Whole Knowing a Decrease and the Percentage of Decrease
Maged was ill this year, and he missed 4.5 days of school. This lowered his annual attendance from 100% to 97.5%. How many school days are there in a year at Maged’s school?
Maged missed 4.5 days of school, and this lowered his annual attendance from 100% to 97.5%. This means that the percent decrease is . Therefore, 4.5 days of school are 2.5% of the total number of school days in a year. Let us write an equation that describes this with being the number of school days in a year:
By dividing both sides of the equation by 0.025, we find that
The answer is that there are 180 school days in a year in Maged’s school.
### Example 5: Finding a Whole Knowing an Increase and the Percentage of Increase
Sally sold her house at 116.6% of the price she had bought it for six years before, which made her a profit of \$39 000. How much had Sally bought her house for? Give your answer to the nearest thousand.
Sally sold her house at 116.6% of the price she had bought it. She made a profit of \$39 000. This means that the price increase between the time she had bought the house and the time she sold it is \$39 000 and corresponds to a percent change of . Therefore, 39 000 is 16.6% of the price at which Sally had bought her house. Let us write an equation that describes this, with being the price at which Sally bought her house:
By dividing both sides of the equation by 0.166, we find that
Our answer is that Sally had bought her house for \$235 000.
In the last example, we are going to look at a particular situation where a quantity is increased and then decreased by the same percentage.
### Example 6: Finding a Final Price after a Decrease and an Increase by the Same Percentage
An electronics store had a one-day sale of 14%. Given that, on the following day, the sale price was increased by 14%, determine, to the nearest cent, the price of a pair of headphones that originally cost \$216.
In this question, the original price of a pair of headphones (\$216) underwent two changes: first it decreased by 14%, and then the new price increased by 14%.
After the first change, the price was .
The new price, \$185.76, is then increased by 14%. The final price is then .
The final price of the pair of headphones is \$211.77.
### Key Points
• When a quantity is increased by ( is the percent of change), the final quantity is then .
• When a quantity is decreased by , it is given by .
• The percent of change, , that corresponds to a quantity changing to is given by
• The sign of indicates whether the change is an increase ( is positive) or a decrease ( is negative). | 1,483 | 6,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2023-50 | latest | en | 0.929103 |
https://integers.info/239 | 1,723,458,670,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036895.73/warc/CC-MAIN-20240812092946-20240812122946-00384.warc.gz | 247,306,100 | 4,487 | # integers.info
Two hundred and thirty-nine
## 239 in other numeral systems
Binary: Octal: Hex: Roman:
## Prime factorization of 239
239 is a prime number
1 and 239
## Is 239 in the Fibonacci number sequence?
No.
Its nearest Fibonacci number neighbors are 233 and 377. | 71 | 278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.828735 |
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# Traffic safety officials predict that drivers will be
Author Message
Manager
Joined: 26 Mar 2008
Posts: 96
Schools: Tuck, Duke
Traffic safety officials predict that drivers will be [#permalink]
### Show Tags
03 Oct 2008, 18:32
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Traffic safety officials predict that drivers will
be equally likely to exceed the proposed speed
limit as the current one.
A. equally likely to exceed the proposed speed
limit as
B. equally likely to exceed the proposed speed
limit as they are
C. equally likely that they will exceed the proposed
speed limit as
D. as likely that they will exceed the proposed
speed limit as
E. as likely to exceed the proposed speed limit
as they are.
VP
Joined: 17 Jun 2008
Posts: 1325
### Show Tags
05 Oct 2008, 05:36
arorag wrote:
Traffic safety officials predict that drivers will
be equally likely to exceed the proposed speed
limit as the current one.
A. equally likely to exceed the proposed speed
limit as -> eliminate
B. equally likely to exceed the proposed speed
limit as they are -> equally as is wrong usage
C. equally likely that they will exceed the proposed
speed limit as -> same as above
D. as likely that they will exceed the proposed
speed limit as -> drivers compared with curent speed limit !!! eliminate
E. as likely to exceed the proposed speed limit
as they are. -> ok But no completely alright
one flaw in E is : as they are current speed limit appears wrong changes meaning
Hence mddled but others are worse
whats te OA ?
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cheers
Its Now Or Never
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Joined: 06 Jul 2008
Posts: 29
### Show Tags
05 Oct 2008, 13:36
A. equally likely to exceed the proposed speed limit as : Equally Likely is incorrect.
B. equally likely to exceed the proposed speed limit as they are : Equally Likely is incorrect.
C. equally likely that they will exceed the proposed speed limit as : Equally Likely is incorrect.
D. as likely that they will exceed the proposed speed limit as : It is incorrect because, 'they will' is not parallel to the part after second as. IT COULD BE RIGHT IF IT WERE ; as likely that they will exceed the proposed speed limit as that they will exceed the current one.
E. as likely to exceed the proposed speed limit as they are. - As likely X as Y (X and Y should be parallel), In this case, it is like this : as likely to exceed the proposed speed limit as they are (to exceed) - UNDERSTOOD (the current one)
SVP
Joined: 17 Jun 2008
Posts: 1504
### Show Tags
05 Oct 2008, 14:40
D for me.
Traffic safety officials predict that drivers will
be as likely that they will exceed the proposed
speed limit as
they exceed the current one
VP
Joined: 05 Jul 2008
Posts: 1373
### Show Tags
08 Feb 2009, 15:58
Isnt they ambiguous in B,C,D & E?
I chose A. E is clearly wrong
SVP
Joined: 07 Nov 2007
Posts: 1761
Location: New York
### Show Tags
08 Feb 2009, 16:43
icandy wrote:
Isnt they ambiguous in B,C,D & E?
I chose A. E is clearly wrong
Agree with YOu.. I chose A too. same reasoning.
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Smiling wins more friends than frowning
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Joined: 23 May 2008
Posts: 757
### Show Tags
09 Feb 2009, 06:05
11-p482151?t=65733&hilit=traffic+safety#p482151
VP
Joined: 18 May 2008
Posts: 1203
### Show Tags
09 Feb 2009, 09:46
In the question stem, last portion of the sentence has been incorrectly underlined.
VP
Joined: 18 May 2008
Posts: 1203
### Show Tags
09 Feb 2009, 09:57
Here is another gud discussion on this topic:
http://www.urch.com/forums/gmat-sentenc ... als-2.html
Re: SC [#permalink] 09 Feb 2009, 09:57
Display posts from previous: Sort by | 1,139 | 4,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-13 | latest | en | 0.966431 |
http://www.gurufocus.com/term/Total%20Equity/BRCD/Total%2BEquity/Brocade%2BCommunications%2BSystems%2BInc | 1,485,013,862,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281151.11/warc/CC-MAIN-20170116095121-00128-ip-10-171-10-70.ec2.internal.warc.gz | 502,188,495 | 28,730 | Switch to:
GuruFocus has detected 4 Warning Signs with Brocade Communications Systems Inc \$BRCD.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
Total Equity
\$2,543 Mil (As of Oct. 2016)
Brocade Communications Systems Inc's total equity for the quarter that ended in Oct. 2016 was \$2,543 Mil.
Total equity is used to calculate book value per share. Brocade Communications Systems Inc's book value per share for the quarter that ended in Oct. 2016 was \$6.33. The ratio of a companys debt over equity can be used to measure how leveraged this company is. Brocade Communications Systems Inc's debt to equity ratio for the quarter that ended in Oct. 2016 was 0.62.
Definition
Total Equity refers to the net assets owned by shareholders.
Total Equity and Total Liabilities are the two components for Total Assets.
Brocade Communications Systems Inc's Total Equity for the fiscal year that ended in Oct. 2016 is calculated as
Total Equity = Total Assets(Q: Oct. 2016 ) - Total Liabilities(Q: Oct. 2016 ) = 4939.811 - 2396.898 = 2,543
Brocade Communications Systems Inc's Total Equity for the quarter that ended in Oct. 2016 is calculated as
Total Equity = Total Assets(Q: Oct. 2016 ) - Total Liabilities(Q: Oct. 2016 ) = 4939.811 - 2396.898 = 2,543
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
1. Total equity is used to calculate book value per share.
Brocade Communications Systems Inc's Book Value Per Share for the quarter that ended in Oct. 2016 is
Book Value per Share = (Total Shareholders Equity - Preferred Stock) / Total Shares Outstanding = (2542.913 - 0) / 401.75 = 6.33
2. The ratio of a companys debt over equity can be used to measure how leveraged this company is.
Brocade Communications Systems Inc's Debt to Equity Ratio for the quarter that ended in Oct. 2016 is
Debt to Equity = Total Debt / Total Equity = (Current Portion of Long-Term Debt + Long-Term Debt) / Total Equity = (76.692 + 1502.063) / 2542.913 = 0.62
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Brocade Communications Systems Inc Annual Data
Oct07 Oct08 Oct09 Oct10 Oct11 Oct12 Oct13 Oct14 Oct15 Oct16 Total Equity 1,267 1,282 1,778 2,035 2,014 2,236 2,347 2,408 2,534 2,543
Brocade Communications Systems Inc Quarterly Data
Jul14 Oct14 Jan15 Apr15 Jul15 Oct15 Jan16 Apr16 Jul16 Oct16 Total Equity 2,337 2,408 2,450 2,461 2,473 2,534 2,501 2,521 2,458 2,543
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https://www.slideserve.com/DoraAna/synchronous-generators-with-permanent-magnets-for-stabilization-the-output-voltage-of-the-power-wind-mill-installations | 1,544,621,321,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823872.13/warc/CC-MAIN-20181212112626-20181212134126-00371.warc.gz | 1,042,868,585 | 16,213 | 1. Introduction
1 / 14
# 1. Introduction - PowerPoint PPT Presentation
Synchronous generators with permanent magnets for stabilization the output voltage of the power wind mill installations Tudor Ambros Professor, Technical University of Moldova 1. Introduction
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Synchronous generators with permanent magnets for stabilization the output voltage of the power wind mill installationsTudor AmbrosProfessor, Technical University of Moldova
1. Introduction
At the given stage, the wind power even more often is used as a source of electrical energy. On the open places, on the sea, where speed of a wind is high and rather constant, the main indicators of energy quality, voltage and frequency of a source of a current are constant.
In zones as, for example, in Republic of Moldova, which is situated in a pool of the Black Sea, the velocity of a wind is variable and not high. From these reasons, the basic element of transformation of a mechanical wind power in electrical energy - electrical generator is necessary for adapting to the parameters of the electrical motor of low velocity. The mechanical construction of those motors, as generators, should ensure constant frequency and voltage on output at variable velocity of a wind. The maintenance of a constancy of frequency and voltage can be realized by electronic devices. But for independent generators of small capacity used in agriculture, in places remote from electrical networks, the electronic devices are becoming unprofitable because of dearness and complexity of service. The decision of this problem offered in the given paper is based on use the synchronous generators with permanent magnets.
There are two variants of synchronous generators with permanent magnets: the first, when the revolutions velocity of the wind mill motor is higher than 500 rev/min and second variant, when the speed is less than 500 rev/min. In both cases these generators have the following advantages:
• Absence of electrical contact of sliding;
• Increased value of the efficiency ratio and power factor;
• Increased reliability;
• Lowered reaction of an rotor.
• The basic lacks of these generators:
• Variable frequency of a current;
• Difficulty of regulation or stabilization of a voltage.
In a case, when consumer the kind of a current (alternating or direct) does not interest, the problem is reduced to regulation or stabilization of a voltage. If the velocity of the wind mill generator is higher than 500 rev/min, it is possible to use, as the synchronous generator, the asynchronous standard motor, on a rotor of which are mounted the permanent magnets and at the stator's winding is connected the controlled rectifier (fig. 1).
Fig.1. The synchronous generator with controlled rectifier
In the offered scheme, the generator of an alternating current is transformed into the generator of a direct current, however controlled rectifier complicates the device of a source of a direct current and does not justify itself on low powers. Other decision, concerning stabilization of a voltage on terminals of the synchronous generator with constant magnets, can be realized by pre-magnetization of the yoke of a stator by direct current and replacement of the controlled rectifier by unguided (fig.2).
Fig.2. The synchronous generator with pre-magnetization yoke
of a stator and uncontrolled rectifier
The winding of a pre-magnetization Wp, which covers the stator's yoke and through which flows the direct current, provides stabilization of a voltage on output of stator's winding W1 by pre-magnetization or un-magnetization of the stator's yoke. The magnetization force created by a current of a winding of pre-magnetization Ip, is subtracted or is summarized with the magnetization forces of permanent magnets and reaction of a stator:
- magnetization force created by winding of pre-magnetization and by permanent magnets;
- magnetization force created by three phases winding and by permanent magnets;
If the design of the generator contains 2р of poles, then yoke of a stator is divided on 2р of sectors (fig. 3). Alternately, the sectors are magnetized and are unmagnetized.
If the sectors of magnetization are sated, their magnetic resistance is increasing, total magnetic working flow is decreasing and e.m.f., output voltage, on the terminals of the generator is decreasing.
On the unmagnetized sectors a flow
Fig.3. Magnetization and unmagnetization
of the stator's yoke
In formulas (1-2), Rp - magnetic resistance of the magnetizing sectors and Rd - magnetic resistance of unmagnetizing poles. The total resistance of the magnetizing sectors
lpJ - length of a sector, Sj - cross section of the yoke.
Taking into account, that the force of magnetizing
it is possible by regulation the value of the current Ip , is possible, by regulation the current Ip, to maintain constant the voltage on the terminals of generator at different change of the load and frequency of rotation of the generator. This fact is explained, that the magnetic flow created by is closed only on yoke, in that time when are closed on an air gap.When the frequency of rotation of the wind mills engine is low, it is necessary to execute the generator with the large number of poles, that in a cylindrical construction is difficult to execute, or it is possible to execute the generator with small number of poles and connect its with wind mills engine through a multiplier.
A multiplier raises cost and reduces the efficiency of device as a whole.
In replacements of that is offered at the n≤500 revolution per min to connect with wind mills motor the synchronous generator with constant magnets and axial air gap which is calculated on 2р>12.
Most suitable is the synchronous generator with constant magnets and with two rotors, which has the small axial and large radial sizes, and can be executed with plenty of poles, i.e. on low speeds (fig. 4).
and
are closed on an air gap.
The technology of stacking of a three-phase winding and pre magnetization winding is identical, as these windings are toroidal and together enveloped the yoke of stator and practically they are presenting one winding.
The basic advantages of the generator:
- the frontal parts of a stator winding are reduced approximately twice;
- a plenty of poles reduce frequency of rotation;
- big moment of inertia, which smooths frequency of rotation at variable load;
- increased surfaces of cooling.
For studing the process of magnetization and unmagnetization of the stator's yoke the method of final elements was used.
Fig. 4. A general view of the axial generator: 1 - blades; 2 - disk of a rotor; 3 - core; 4 - windings; 5 - box of terminals; 6 - permanent magnets; 7 - fixing ledges
In fig.5 is shown the picture of a magnetic field created by permanent magnets, the magnetic lines of which are closing through poles.
a b c
Fig. 5. Distribution of the created magnetic flow:
а) by permanent magnets; б) by current of the stator Is; с) by current of magnetization winding Ip
The magnetic force lines of a flow of stator's reaction are closing in space between poles, where the magnetic resistance is less.
The magnetic flow of pre magnetization is closing through yoke, where the magnetic resistance is also less.
In fig.6 the pictures of magnetic fields created by permanent magnets, stator's current and current of pre-magnetization (fig.6,a) are submitted. The picture of a magnetic field of permanent magnets and field, imposed on it by the stator's reaction of armature (fig.6,b) practically does not differ from a figure of a field shown on fig. 6,а, as the reaction of an armature is insignificant.
At imposing a field of permanent magnets (fig. 6,с) and magnetic field of pre- magnetization current, the common picture of a field considerably changes.
In a fig.7 are shown the results of computation of a picture of a magnetic field of the axial generator without slots on a stator with a three-phase toroidal winding stacked on yoke.
The forces lines of a magnetic induction are distributed in regular intervals in the yoke of a stator of the generator, which represents one half from common yoke of the generator
The curve distributions of a magnetic induction in an air gap (fig.8,а), as well as curve of distribution of an induction in a yoke of the stator, do not contain high harmonics. The absence of high harmonics is caused by absence of teeth on a stator and closeness of the permeability of permanent magnets to the magnetic permeability of air.
a b c
Fig. 6. A picture of magnetic fields distribution :
а) at imposing a field of permanent magnets, reaction of armature and field of pre magnetization, б) field created by permanent magnets, с) at imposing a field of permanent magnets and field of pre magnetization
Fig. 7. A picture of a magnetic field of the synchronous axial generator without slots on a stator with a three-phase toroidal winding
Having changed a configuration of poles, it is possible to receive a sinusoidal curve of induction distribution in an air gap.
a b
Fig. 8. Curve distributions of a magnetic induction: а) in an air gap, b) in an yoke of stator
2. Conclusions
• By magnetizing the stator's yoke of the synchronous generator with permanent magnets it is possible to stabilize the output voltage of the generator.
• For the generator with a toroidal winding, the frontal parts of a three-phase winding are reduced.
• The generator with an axial rotor can be executed on small frequencies of rotation | 2,150 | 10,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-51 | latest | en | 0.862913 |
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# Fall09-DepAmrtImpair - Operational Assets Depreciation...
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Operational Assets: Depreciation, Amortization and Impairment Fixed and Intangible Assets Chapter 11 – Spiceland Text
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So, now we have these operational assets and we’re using them in our business. How do we account for them as we use them? Remember that we did not expense them, we recorded them as assets on our B/S.
Depreciation Method of cost allocation Not related to fair market value or the “street” definition of depreciation Matching principle Journal entry: Depreciation Expense XX Accumulated Deprec XX
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What do you need to know to calculate depreciation? 1. Depreciation method Straight-line Double-declining balance Units of production 2. Depreciable base Cost – Salvage (or Residual) Value 3. Useful life # years # units
(Cost – Salvage Value)/Useful life =Depreciation Expense Depreciation expense is the same each period Exception is for fixed assets placed into service mid-year Fixed assets are depreciated until the
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Straight-line Depreciation Assume a company acquires equipment for \$10,000 on 1/1/08. It is estimated that the equipment will have a useful life of 4 years and a salvage value of \$2,000 at the end of its life. What would a depreciation schedule look like? Year Calculation Depreciation Exp Net Book Value 2008 (10,000-2,000)/4 2,000 8,000 2009 (10,000-2,000)/4 2,000 6,000 2010 (10,000-2,000)/4 2,000 4,000 2011 (10,000-2,000)/4 2,000 2,000
Determine straight-line rate (100%/useful life). Double that rate. Apply that rate to the net book value at the beginning of the period to get depreciation expense. Net book value does not take into consideration salvage value.
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Cost of machine - \$60,000 Estimated useful life – 10 years Estimated salvage value - \$5,000
Straight-line rate = 100%/10 yrs = 10% DDB Rate = 10% X 2 = 20% NBV @ 1 st yr = 60,000 Depreciation Expense = 60,000 X 20% = \$12,000 New NBV = 60,000 – 12,000 = \$48,000
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DDB Depreciation Example Year Calculation Depreciation Exp NBV Year 0 60,000 Year 1 60,000 X 20% 12,000 48,000 Year 2 48,000 X 20% 9,600 38,400 Year 3 38,400 X 20% 7,680 30,720 Year 4 30,720 X 20% 6,144 24,576 Year 5 24,576 X 20% 4,915 19,661 Year 6 19,661 X 20% 3,932 15,729 Year 7 15,729 X 20% 3,146 12,583 Year 8 12,583 X 20% 2,517 10,066 Year 9 10,066 X 20% 2,013 8,053 Year 10 8,053 – 5,000 3,053 5,000
Everything in depreciation (except the asset’s cost) is an estimate. What happens when one or more of those estimates change significantly? Update depreciation calculations using
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• 1. Q U E U E A lecture by ABDUL GHAFFAR Based on Chapter 3 Of Reference Book #1
• 2. Reference Books <ul><ul><ul><li>Data Structures and Algorithm Analysis in C </li></ul></ul></ul><ul><ul><ul><ul><li>By Mark Allen Weiss </li></ul></ul></ul></ul><ul><ul><ul><ul><ul><li>Published by Addison Wesley </li></ul></ul></ul></ul></ul><ul><ul><ul><li>Data Structures (Schaum’s Outline Series) </li></ul></ul></ul><ul><ul><ul><ul><li>By Seymour Lipschutz </li></ul></ul></ul></ul><ul><ul><ul><ul><ul><li>Published by Mc Graw Hill </li></ul></ul></ul></ul></ul>
• 3. A QUEUE MODEL <ul><li>Definitions: </li></ul><ul><ul><li>A QUEUE is a list with the restriction that insertion is done at one end , where as deletion at the other and.for that reason it is called FIFO ( First in first out ) structure. </li></ul></ul><ul><ul><li>The end where insertion is performed is called the rear . </li></ul></ul><ul><ul><li>The end where Deletion is performed is called the front . </li></ul></ul><ul><ul><li>The number of elements in a queue is called the size of the queue </li></ul></ul><ul><li>Set of operations: </li></ul><ul><ul><li>Enqueue : Inserts an element at the end (rear) of the list </li></ul></ul><ul><ul><li>Dequeue : Returns and deletes the element at the start (front) of the list. </li></ul></ul>QUEUE Dequeue Enqueue
• 4. Implementation <ul><ul><li>As Array </li></ul></ul><ul><ul><ul><li>Define an Array ‘ Queue ’ of sufficient capacity to store Queue elements. </li></ul></ul></ul><ul><ul><ul><li>Uses a variable i.e rear to remember the position of insertion end. </li></ul></ul></ul><ul><ul><ul><li>Uses a variable i.e front to remember the position of deletion end. </li></ul></ul></ul><ul><ul><ul><li>Uses a variable i.e size to remember the size of the array. </li></ul></ul></ul><ul><ul><ul><li>Write a function Enqueue which takes an element x as argument and performs </li></ul></ul></ul><ul><ul><ul><ul><li>size = size +1 </li></ul></ul></ul></ul><ul><ul><ul><ul><li>rear = rear +1 </li></ul></ul></ul></ul><ul><ul><ul><ul><li>Queue[rear] = x </li></ul></ul></ul></ul><ul><ul><ul><li>Write a function Dequeue, which returns the value at front as </li></ul></ul></ul><ul><ul><ul><ul><li>return Queue[front] </li></ul></ul></ul></ul><ul><ul><ul><ul><li>front=front+1 </li></ul></ul></ul></ul><ul><ul><ul><ul><li>size=size-1 </li></ul></ul></ul></ul>
• 5. Example: 4 2 R F Size = 2 Initial 4 2 1 F Size = 3 R After Enqueue(1) 4 2 3 1 F Size = 4 R After Enqueue(3) 4 2 3 1 F Size = 3 R After Dequeue r, 2 is returned 4 2 3 1 Size = 2 R F After Dequeue, 4 is returned 4 2 3 1 Size = 1 RF After Dequeue, 1 is returned 4 2 3 1 Size = 0 F R After Dequeue, 3 is returned | 1,091 | 3,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-27 | latest | en | 0.640897 |
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### Topic: Error propagation involving natural logarithms (Read 320 times)
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#### daaniiell14
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##### Error propagation involving natural logarithms
« on: February 18, 2020, 01:11:56 PM »
In our kinetics lab we are plotting $$ln\frac{ A_{inf}-A }{ A_{inf} }$$ against -kt, i.e solving an pseudo first order reacting to find a rate constant. The problem however is finding out how the different errors (uncertainties) of the stuff in the ln term get translated to uncertainties in the y axis when i plot these.
I have tried using the error propagation rules to first solve for the numerator and then the entire paranthesis. The uncertainties is for A(inf)=0,008328 and A=0,05 and A(inf)=2,082 is just another constant so the only variable is A.
Trying to use the natural logarithm error propagation law then just causes chaos because my A is in the denominator which means that with higher Absorbance values for some reason my error gets larger and larger even though logically it should be the opposite since an error in 0,05 impacts A=0.1 more than A=2.0?
#### Corribus
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##### Re: Error propagation involving natural logarithms
« Reply #1 on: February 18, 2020, 05:48:38 PM »
Are your two terms linked as independent pairs? I.e., do you obtain A and Ainf values for three independent experiments? If that's the case I would just calculate three values by your equation and taken the std. deviation of the mean as your over all error.
I'm no statistician but I only use error propagation if the measurements/values are decoupled from each other.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent? - Richard P. Feynman
#### Enthalpy
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##### Re: Error propagation involving natural logarithms
« Reply #2 on: February 19, 2020, 02:53:37 PM »
A relative error in a logarithm propagates as an absolute error. That's tricky. That is, 1% error in a log becomes 0.01 error, whether the 1% accuracy is on camels or on koalas.
This implies that the result of a log has usually no unit. And for instance, a relative error on a pH has no meaning. | 650 | 2,473 | {"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} | 3.234375 | 3 | CC-MAIN-2020-45 | latest | en | 0.890853 |
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## Indieiosfocus Discover the world of Reading Worksheets.
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Homeschooling educator networks, seasoned teachers, and the internet are all wonderful resources for reading lesson plans, activities, and programs. The internet has opened up a world full of sites where in-school educators and homeschooling parents "gather" to share their ideas for creative lesson plans and activities such as book reports and writing activities. Any homeschooling parent who finds themselves at a loss for new material can visit the internet for reading resources.
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###### Beginner Note Reading Worksheets
Children can work with simple numbers worksheets from quite an early age and you will have greater success in getting them to work on the worksheets if you combine that learning work with something practical, or at least something they enjoy doing. For example, if you are using a simple addition and subtraction worksheet with your child, draw or type up another sheet of with squares and numbers printed onto them. Instead of writing the answers to the questions on the worksheet you can get your child to cut and paste the required numbers for the answers from from the second worksheet onto the first.
By the time they are learning first grade math, kids should be ready to tackle things like the relationship between addition and subtraction, the concept of adding and subtracting two-digit numbers and learning to count beyond 100. Being able to compare numbers as larger, smaller or equal to each other is also important, as it provides the basis for recognizing whether or not the answer to a computation problem is the correct one. Children need to be allowed to master these and other essential math skills before being asked to move on to new ideas, but the modern classroom setting does not always allow for this. As focus on core curriculum begins to push complex ideas into lower grade levels, kids are expected to learn more at a younger age. First grade math still contains many fundamental concepts essential for understanding higher math, and therefore should not be rushed through. By letting a child try and re-try each new thing as it comes, online math games can give the extra time and practice that struggling students need to achieve success.
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It is important to work with your child to help establish an appropriate pace. Part of the benefit of interactive learning games is that parents can monitor their child has progress and see how well things are going. You may find yourself pleasantly surprised at how much your kindergarten learns in a short period of time. Though kindergarten math can not be taught through learning games alone, interactive digital activities provide a good supplement to traditional education. When kids continue to practice what they have learned and become more comfortable with it outside the classroom, they are bound to do better as they progress through school. Learning games also give you an opportunity to work with your child at home, helping to boost his or her grasp of basic kindergarten math concepts. The use of digital learning games to teach kindergarten math is by no means a stand-in for traditional education. However, when kids are presented with a fun, interactive learning environment in their own homes, they can build skills and get a deeper understanding of the concepts that will lead to better classroom performance and a more positive school experience.
Two of the best options are Omega Math and ALEKS Math. Both of these programs are well-developed online math programs. Omega Math covers Pre-Algebra, Algebra I and II, as well as Geometry, and ALEKS is a full program for grades Kindergarten through High School, including Trigonometry, Statistics, and Accounting. There are some differences in presentation style, but both programs cover the material thoroughly, and all that a student needs to do is log in, have their pencil and paper nearby, and begin their study. Omega Math tends to be better equipped for students who catch on to math skills fairly easily and are motivated to streamline their work. Students log in to their course, view a PowerPoint lesson, and work through homework problems on their own. Feedback is given and students can also complete worksheets for extra practice. Chapter tests are provided, scored immediately, and parents can track the progress throughout the course by viewing simple charts and grade books making it very parent-friendly.
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# Fixed Point : Mean Value Theorem
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A number (a) is called a fixed point of a function (f) if f(a)=a. Prove that, if f'(x) does NOT equal 1 for all real numbers (x), then f has at most one fixed point.
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The mean value theorem is used to provide a proof regarding fixed points. The solution is detailed and well presented. The response received a rating of "5" from the student who originally posted the question.
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• "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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# How many words can be found with the letters of the word
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How many words can be found with the letters of the word [#permalink] 25 Jan 2004, 11:39
How many words can be found with the letters of the word PATALIPUTRA without changing the relation order of the vowels and constants?
a.180
b.7200
c.4800
d.3600
e.3200
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P & C 2 - beauty! [#permalink] 26 Jan 2004, 12:04
ANS = D.
we need to arrange - C=consonant, V=Vowel
CVCVCVCVCCV
we separate the C's from the V's to find diff. possible arrangements of vowels alone, and consonants alone. then we string them together as above. We get a set of (really) different arrangements of V's and a set of (really) different arrangements of C's and when we string one from each set together we get a (really) new and unique word.
Lets go -
For V's we have to CHOOSE where to put the A's (as they are all the same - we simply choose 3 locations out of the 5 for the A's).
That is 5C3.
then we have two options to place the I and the U must be in the last available place.
Total number of arrangements for V's - 5C3x2x1 = 10x2x1 = 20.
C's - more annoying - we CHOOSE the two locations for the P's 6C2 and then two locations for the T's 4C2 and then place the L in one of two places and the R lands in the last available place.
number of arrangements for C's - 6C2 x 4C2 x2x1 = 15x6x2x1 = 180
total words = 180x20 = 3600.
or D.
Further explanations - why can we separate C's from V's? we will eventually arrange them in the proper order as described above. IF THIS ORDER IS PREDETERMINED we can look at the problem as if it is two separate problems. There is no possibility of interchange between the C's and the V's as they must remain in the same relations one to another!!! we can't create a permutation where C's and V's switch places! So the problem of C's and problem of V's are separate problems.
then there is the counting issue - when we have three A's that are exactly the same (not in diff. colors for example ) then the order of V's - AAIUA is exactly the same as the order AAIUA (see what I mean - I switched the first two A's and you didn't even notice ).
So we start by deciding where the three places to put the A's are going to be. Each different selection of 3 places will give a different word - but we don't have to switch between the A's because that will leave us with exactly the same word).
If after all that the ans is still wrong - well, I give up
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1 Final Exam Review Questions PHY Final Chapters Section: 17 1 Topic: Thermal Equilibrium and Temperature Type: Numerical 12 A temperature of 14ºF is equivalent to A) 10ºC B) 7.77ºC C) 25.5ºC D) 26.7ºC E) 47.7ºC Section: 17 1 Topic: Thermal Equilibrium and Temperature Type: Numerical 13 A temperature difference of 9ºF is the same as a difference of A) 5ºC B) 9ºC C) 20ºC D) 68ºC E) 100ºC Section: 17 1 Topic: Thermal Equilibrium and Temperature Type: Numerical 14 The temperature at which the Celsius and Fahrenheit scales read the same is A) 40ºC B) 40ºC C) 20ºF D) 68ºF E) 100ºC Section: 17 2 Topic: Gas Thermometers & Absolute Temperature Type: Numerical 16 A constant-volume gas thermometer reads 6.66 kpa at the triple point of water. What is the pressure reading at the normal boiling point of water? A) 2.44 kpa B) 18.2 kpa C) 9.10 kpa D) 11.8 kpa E) 4.87 kpa Section: 17 2 Topic: Gas Thermometers & Absolute Temperature Type: Conceptual 17 Which of the following statements about absolute zero temperature is true? A) At absolute zero all translational motion of the particles ceases. B) At absolute zero all rotational motion of the particles ceases. C) Absolute zero is defined at C. D) At absolute zero all the particles are at rest, except for the quantum motion. E) All the above. Section: 17 3 Topic: The Ideal Gas Law Type: Numerical 24 A large balloon is being filled with He from gas cylinders. The temperature is 25 C and the pressure is 1 atmosphere. The volume of the inflated balloon is 2500 m 3. What was the volume of He in the cylinders if the gas was under a pressure of 110 atmospheres and at a temperature of 12 C when in the gas cylinders? A) 11 m 3 B) 22 m 3 C) 24 m 3 D) 15 m 3 E) 23 m 3 Final Exam H Rev Ques.doc - 1 -
2 Section: 17 3 Topic: The Ideal Gas Law Type: Numerical 25 What mass of He gas occupies 8.5 liters at 0 C and 1 atmosphere? (The molar mass of He = 4.00 g/mol.) A) 10.5 g B) 0.66 g C) 2.6 g D) 0.38 g E) 1.52 g Section: 17 3 Topic: The Ideal Gas Law Type: Conceptual 27 A gas has a density X at standard temperature and pressure. What is the new density when the absolute temperature is doubled and the pressure increased by a factor of 3? A) (2/3)X B) (4/3)X C) (3/4)X D) (6)X E) (3/2)X Section: 17 3 Topic: The Ideal Gas Law Type: Numerical 28 Inside a sphere of radius 12 cm are gas molecules at a temperature of 50 C. What pressure do the gas molecules exert on the inside of the sphere? A) Pa D) Pa B) Pa E) Pa C) Pa Section: 17 3 Topic: The Ideal-Gas Law Type: Numerical 30 The air in a balloon occupies a volume of 0.10 m 3 when at a temperature of 27ºC and a pressure of 1.2 atm. What is the balloon's volume at 7ºC and 1.0 atm? (The amount of gas remains constant.) A) m 3 B) m 3 C) m 3 D) 0.11 m 3 E) 0.13 m 3 Section: 17 3 Topic: The Ideal-Gas Law Type: Numerical 31 In a vacuum system, a container is pumped down to a pressure of Pa at 20ºC. How many molecules of gas are there in 1 cm 3 of this container? (Boltzmann's constant k = J/K) A) B) C) D) E) Section: 17 3 Topic: The Ideal-Gas Law Type: Numerical 32 A rigid container of air is at atmospheric pressure and 27ºC. To double the pressure in the container, heat it to A) 54ºC B) 300ºC C) 327ºC D) 600ºC E) 327 K Final Exam H Rev Ques.doc - 2 -
3 Section: 17 3 Topic: The Ideal-Gas Law Type: Conceptual 34 If a mass of oxygen gas occupies a volume of 8 L at standard temperature and pressure, what is the change in the volume if the temperature is reduced by one half and the pressure is doubled? A) It increases to 12 L. D) It decreases to 2 L. B) It decreases to 6 L. E) It does not change. C) It increases to 24 L. Section: 17 3 Topic: The Ideal-Gas Law Type: Conceptual 35 A collection of oxygen molecules occupies a volume V at standard temperature and pressure. What is the new volume if the amount of oxygen is doubled and the pressure is tripled? A) V B) (2/3)V C) (3/2)V D) 6V E) (5/9)V Section: 17 3 Topic: The Ideal-Gas Law Type: Conceptual 36 If the pressure and volume of an ideal gas are both reduced to half their original value, the absolute temperature of the gas is A) unchanged. D) increased by a factor of 4. B) doubled. E) decreased by a factor of 4. C) halved. Section: 17 3 Topic: The Ideal-Gas Law Type: Numerical 38 Assume that helium is a perfect gas and that the volume of a cylinder containing helium is independent of temperature. A cylinder of helium at +85ºC has a pressure of 208 atm. The pressure of the helium when it is cooled to 55ºC is A) 132 atm B) 127 atm C) 335 atm D) 132 atm E) 204 atm Section: 17 3 Topic: The Ideal-Gas Law Type: Conceptual 39 The relationship between the pressure and the volume of a gas expressed by Boyle's law holds true A) for some gases under any conditions. B) if the density is constant. C) if the container of the gas can expand with increasing pressure. D) if the temperature is constant. E) for all gases under any conditions. Final Exam H Rev Ques.doc - 3 -
4 Section: 17 3 Topic: The Ideal-Gas Law Type: Numerical 42 The air around us has 78% nitrogen and 21% oxygen. If the pressure is 1 atm, the pressure due to oxygen is A) 0.21 atm B) 0.78 atm C) 1 atm D) 0.5 atm E) 0.67 atm Section: 17 3 Topic: The Ideal-Gas Law Type: Numerical 43 A cylinder of volume 50 L contains oxygen gas at a pressure of 2 atm. If nitrogen gas of volume 25 L and at pressure 1 atm is added to the oxygen cylinder, the new pressure is A) 2 atm B) 3 atm C) 2.5 atm D) 3.5 atm E) 4 atm Section: 17 4 Topic: The Kinetic Theory of Gases Type: Numerical 44 A room measures 3 m 4 m 2 m and is at 15 C and 1 atm. When the temperature is increased to 25 C, the number of molecules that escaped from the room is, assuming that the pressure stays at 1 atm A) D) B) E) C) Section: 17 3 Topic: The Ideal-Gas Law Type: Numerical 47 An ideal gas whose original temperature and volume are 27ºC and m 3 undergoes an isobaric expansion. If the final temperature is 87ºC, then the final volume is approximately A) m 3 B) m 3 C) m 3 D) m 3 E) 1.45 m 3 Section: 17 4 Topic: The Kinetic Theory of Gases Type: Numerical 49 At what common Celsius temperature is the rms velocity of oxygen molecules (molar mass = 32 g/mol) double that of hydrogen molecules (molar mass = 2.0 g/mol)? A) 50 B) zero C) no such temperature exists D) 2.0 E) 16 Section: 17 4 Topic: The Kinetic Theory of Gases Type: Conceptual 50 Two monoatomic gases, helium and neon, are mixed in the ratio of 2 to 1 and are in thermal equilibrium at temperature T (the molar mass of neon = 5 the molar mass of helium). If the average kinetic energy of each helium atom is U, calculate the average energy of each neon atom. A) U B) 0.5U C) 2U D) 5U E) U/5 Final Exam H Rev Ques.doc - 4 -
5 Section: 17 4 Topic: The Kinetic Theory of Gases Type: Numerical 51 Two monoatomic gases, helium and neon, are mixed in the ratio of 2 to 1 and are in thermal equilibrium at temperature T (the molar mass of helium = 4.0 g/mol and the molar mass of neon = 20.2 g/mol). If the average kinetic energy of each helium atom is J, calculate the temperature T. A) 304 K B) 456 K C) 31 K D) 31 C E) 101 K Section: 17 4 Topic: The Kinetic Theory of Gases Type: Numerical 55 A hailstorm causes an average pressure of 1.4 N/m 2 on the 200-m 2 flat roof of a house. The hailstones, each of mass kg, have an average velocity of 10 m/s perpendicular to the roof and rebound after hitting the roof with the same speed. How many hailstones hit the roof each second? A) 4000 B) 2000 C) 1000 D) 10 E) 800 Section: 17 4 Topic: The Kinetic Theory of Gases Type: Numerical 57 At what Kelvin temperature does the rms speed of the oxygen (O 2 ) molecules in the air near the surface of the earth become equal to the escape speed from the earth? (R = 8.31 J/mol K; molar mass of O 2 gas is 32 g/mol; radius of the earth R E = m; the escape speed from the earth is 11.2 km/s) A) K D) K B) K E) K C) K Section: 17 4 Topic: The Kinetic Theory of Gases Type: Conceptual 59 If the absolute temperature of a gas is doubled, what is the change in the rms speed of its molecules? A) no change D) increases by a factor of 2 B) increases by a factor of 2 E) decreases by a factor of 2 C) decreases by a factor of 2 Final Exam H Rev Ques.doc - 5 -
6 Section: 17 4 Topic: The Kinetic Theory of Gases Type: Factual 61 Which of the following is an assumption that is made in the kinetic theory of gases? A) Molecules are not described by Newton's laws. B) Molecules make up only a small fraction of the volume occupied by a gas. C) Molecules collide inelastically. D) The total number of molecules is actually very small. E) There are forces acting on the molecules at all times. Section: 17 4 Topic: The Kinetic Theory of Gases Type: Numerical 63 A volume of an ideal gas goes through a temperature change from 20ºC to 60ºC. The relation between the average molecular kinetic energy at 20ºC (K 1 ) and that at 60ºC (K 2 ) is A) K 1 = K 2 D) K 1 = 0.88 K 2 B) K 1 = 0.33 K 2 E) K 1 = 1.14 K 2 C) K 1 = 3 K 2 Section: 17 4 Topic: The Kinetic Theory of Gases Type: Numerical 66 On the basis of the kinetic theory of gases, when the absolute temperature is doubled, the average kinetic energy of the gas molecules changes by a factor of A) 16 B) 2 C) D) 4 E) 0.5 Section: 17 4 Topic: The Kinetic Theory of Gases Type: Numerical 68 A room measures 3 m 4 m 2 m and is at 20 C and 1 atm. Assuming that it only has the two diatomic gases, N 2 and O 2, the amount of kinetic energy in the gases is A) 3.6 MJ D) 0.41 MJ B) 6.1 MJ E) none of the above C) 0.25 MJ Section: 17 4 Topic: The Kinetic Theory of Gases Type: Numerical 69 A room measures 3 m 4 m 2 m and is at 15 C and 1 atm. Assuming that it only has the two diatomic gases, N 2 and O 2, how much heat is needed to increase the temperature to 25 C? (Ignore the loss in air as the temperature heats up.) A) J D) J B) J E) none of the above C) J Final Exam H Rev Ques.doc - 6 -
7 Section: 17 4 Topic: The Kinetic Theory of Gases Type: Numerical 74 The rms speed of oxygen molecules is 460 m/s at 0ºC. The molecular weight of oxygen is 8 times the molecular weight of helium. The rms speed of helium at 40ºC is approximately A) 3.68 km/s B) 1.84 km/s C) 1.40 km/s D) 880 m/s E) 440 m/s Section: 17 4 Topic: The Kinetic Theory of Gases Type: Numerical 76 If the rms speed of nitrogen molecules (molecular weight of N 2 is 28 g/mol) at 273 K is 492 m/s, the rms speed of oxygen molecules (molecular weight of O 2 is 32 g/mol) at the same temperature is approximately A) 430 m/s B) 461 m/s C) 492 m/s D) 526 m/s E) 562 m/s Section: 17 4 Topic: The Kinetic Theory of Gases Type: Conceptual The isotherm that corresponds to the highest temperature is the one labeled A) T 1 B) T 2 C) T 3 D) The isotherms correspond to the same temperature E) none of these is correct Final Exam H Rev Ques.doc - 7 -
8 Section: 18 1 Topic: Heat Capacity and Specific Heat Type: Numerical 1 Two liquids, A and B, are mixed together, and the resulting temperature is 22 C. If liquid A has mass m and was initially at temperature 35 C, and liquid B has mass 3m and was initially at temperature 11 C, calculate the ratio of the specific heats of A divided by B. A) 0.85 B) 2.5 C) 1.2 D) 0.45 E) 0.94 Section: 18 1 Topic: Heat Capacity and Specific Heat Type: Numerical 2 Two liquids, A and B, are mixed together. Liquid A has mass m and was initially at temperature 40 C, and liquid B has mass 2m and was initially at temperature 5 C. The specific heat of liquid A is 1.5 times that of liquid B. Calculate the final temperature of the mixture. A) 33.5 C B) 14.3 C C) 17.0 C D) 20.0 C E) 25.7 C Section: 18 1 Topic: Heat Capacity and Specific Heat Type: Numerical 3 The quantity of heat absorbed by a body is determined from the formula Q = cm(t f T i ). A certain metal has a specific heat c = 0.21 cal/g Cº and a mass m = 25.6 g. The initial temperature is T i = 34.6ºC, and the final temperature T f = 54.6ºC. The quantity of heat absorbed is A) +23 cal B) cal C) +14 cal D) +110 cal E) +207 cal Section: 18 1 Topic: Heat Capacity and Specific Heat Type: Conceptual 4 Aluminum has a specific heat more than twice that of copper. Identical masses of aluminum and copper, both at 0ºC, are dropped together into a can of hot water. When the system has come to equilibrium, A) the aluminum is at a higher temperature than the copper. B) the copper is at a higher temperature than the aluminum. C) the aluminum and copper are at the same temperature. D) the difference in temperature between the aluminum and the copper depends on the amount of water in the can. E) the difference in temperature between the aluminum and the copper depends on the initial temperature of the water in the can. Final Exam H Rev Ques.doc - 8 -
9 Section: 18 1 Topic: Heat Capacity and Specific Heat Type: Conceptual 6 A system has a heat capacity of 100 J. This means A) it is possible to extract the 100 J of heat and convert it to work. B) it is possible to transfer the 100 J of heat to the environment. C) some of the heat capacity can be converted to work. D) some of the heat capacity can be transferred to another system if there is a temperature difference. E) (C) and (D) Section: 18 1 Topic: Heat Capacity and Specific Heat Type: Numerical 8 A lake with kg of water, which has a specific heat of 4180 J/kg Cº, warms from 10 to 15ºC. The amount of heat transferred to the lake is A) J D) J B) J E) J C) J Section: 18 1 Topic: Heat Capacity and Specific Heat Type: Numerical 10 To raise the temperature of a 2.0-kg piece of metal from 20º to 100ºC, 61.8 kj of heat is added. What is the specific heat of this metal? A) 0.39 kj/kg K D) 1.2 kj/kg K B) 0.31 kj/kg K E) 0.77 kj/kg K C) 1.6 kj/kg K Section: 18 1 Topic: Heat Capacity and Specific Heat Type: Numerical 11 A 250-g piece of lead is heated to 100ºC and is then placed in a 400-g copper container holding 500 g of water. The specific heat of copper is c = kj/kg K. The container and the water had an initial temperature of 18.0ºC. When thermal equilibrium is reached, the final temperature of the system is 19.15ºC. If no heat has been lost from the system, what is the specific heat of the lead? (the specific heat of water is kj/kg K) A) kj/kg K D) kj/kg K B) kj/kg K E) kj/kg K C) kj/kg K Final Exam H Rev Ques.doc - 9 -
10 Use the following to answer questions 12-14: Section: 18 1 Topic: Heat Capacity and Specific Heat Type: Numerical 17 A 1.0-kg piece of marble at 100ºC is dropped into 2.5 kg of water at 1.0ºC and the resulting temperature is 7.0ºC. The specific heat of the marble is approximately A) 0.16 kcal/kg Cº D) 0.30 kcal/kg Cº B) 0.75 kcal/kg Cº E) 0.26 kcal/kg Cº C) 0.61 kcal/kg Cº Section: 18 1 Topic: Heat Capacity and Specific Heat Type: Numerical 18 Glass beads of mass 100 g and specific heat 0.20 cal/g Cº are heated to 90ºC and placed in a 300-g glass beaker containing 200 g of water at 20ºC. When equilibrium is reached, the temperature is approximately A) 23ºC B) 25ºC C) 27ºC D) 32ºC E) 39ºC Section: 18 1 Topic: Heat Capacity and Specific Heat Type: Numerical 20 The molar specific heat of copper is 24.5 J/(mol K). The amount of heat needed to raise 126 g of copper by 2C is A) 24.5 J B) 49 J C) 12.3 J D) 98 J E) 147 J Section: 18 2 Topic: Change of Phase and Latent Heat Type: Numerical 26 A 2.0-kg mass of iron (specific heat = 0.12 kcal/kg C) at a temperature of 430 C is dropped into 0.4 kg of ice and 0.4 kg of water both at 0 C. With no heat losses to the surroundings, the equilibrium temperature of the mixture is approximately A) 0 C B) 100 C C) 23 C D) 69 C E) 87 C Final Exam H Rev Ques.doc
11 Section: 18 2 Topic: Change of Phase and Latent Heat Type: Numerical 28 A 3-kg mass of metal of specific heat = 0.1 kcal/kg C at a temperature of 600 C is dropped into 1.0 kg water at 20 C. With no heat losses to the surroundings, determine the equilibrium temperature of the mixture, and if it is 100 C, calculate what mass of water is turned into steam at this temperature. A) 100 C and 110 g of steam B) 100 C and 150 g of steam C) 100 C and 130 g of steam D) 100 C and 70 g of steam E) The equilibrium temperature is not 100 C. Section: 18 2 Topic: Change of Phase and Latent Heat Type: Conceptual 31 On a hot summer day, water collects on the outside of a glass of ice lemonade. The water comes from A) inside the glass since glass is porous. B) the condensation of the water vapor due the fact that the glass is much colder than the air. C) the straw you use to drink your lemonade. D) the mixture of water and lemonade. E) It is one of the mysteries of life. Section: 18 2 Topic: Change of Phase and Latent Heat Type: Numerical 32 A container contains a 200 ml of 100% proof alcohol (i.e., it has 50% ethyl alcohol and 50% water by volume) at 20 C. How much heat is needed to bring the mixture to the boiling point of the alcohol? (assume that the specific heat in the 100% proof can be treated as due to the alcohol and water separately, and the density, boiling point and specific heat of alcohol are 0.81 g/cm 3, 78 C, and 2.4 J/(g C ), respectively) A) J B) J C) J D) J E) J Section: 18 2 Topic: Change of Phase and Latent Heat Type: Numerical 33 A container contains a 200 ml of 100% proof alcohol (i.e., it has 50% ethyl alcohol and 50% water by volume) at the boiling point of the alcohol. How long does it take to distill (boil) all the alcohol if heat is supplied at a rate of 100 W? (The density and latent heat of vaporization of ethyl alcohol are 0.81 g/cm 3, and 879 kj/kg.) A) 712 s B) 1110 s C) 450 s D) 1450 s E) 950 s Final Exam H Rev Ques.doc
12 Section: 18 3 Topic: Joule's Experiment and the First Law... Type: Numerical 48 A 6.0-g lead bullet traveling at 300 m/s penetrates a wooden block and stops. If 50 percent of the initial kinetic energy of the bullet is converted into thermal energy in the bullet, by how much does the bullet's temperature increase? (The specific heat of lead is 128 J/kg K.) A) 0.17º C B) ºC C) 17 ºC D) ºC E) 35 ºC Section: 18 3 Topic: Joule's Experiment and the First Law... Type: Factual 53 A state variable is one that allows other variables to be determined using a relationship. Which of the following variables are state variables? A) P, V, and T D) (A) and (B) B) Internal energy, U E) (A), (B), and (C) C) W and Q Section: 18 3 Topic: Joule's Experiment and the First Law... Type: Conceptual 54 The first law of thermodynamics has as its basis the same fundamental principle as A) the continuity principle. D) static equilibrium. B) conservation of energy E) the conservation of linear momentum. C) Newton's law of universal gravitation. Section: 18 4 Topic: The Internal Energy of an Ideal Gas Type: Numerical 57 In a certain process, 500 cal of heat are supplied to a system consisting of a gas confined in a cylinder. At the same time, 500 J of work are done by the gas by expansion. The increase in thermal energy of the gas is approximately A) zero B) 1.00 kj C) 1.59 kj D) 2.09 kj E) 2.59 kj Final Exam H Rev Ques.doc
13 Section: 18 4 Topic: The Internal Energy of an Ideal Gas Type: Numerical 59 Two containers of equal volume are connected by a stopcock as shown below. One container is filled with a gas at a pressure of 1 atm and temperature of 293 K while the other container is evacuated so that it is under vacuum. The containers are thermally isolated from the surrounding so no heat enters or escaped from the system. The stopcock is then opened allowing the gas from one container to fill the other. What is the final temperature of the gas after it has come to equilibrium? A) K B) 273 K C) 293 K D) 195 K E) undetermined Section: 18 4 Topic: The Internal Energy of an Ideal Gas Type: Numerical 60 In a certain thermodynamic process, 1000 cal of heat are added to a gas confined in a cylinder. At the same time, 1000 J of work are done by the gas as it expands. The increase in internal energy of the gas is A) zero B) 3186 J C) 239 J D) 5186 J E) 1239 J Section: 18 4 Topic: The Internal Energy of an Ideal Gas Type: Numerical 62 In a certain thermodynamic process, 20 cal of heat are removed from a system and 30 cal of work are done on the system. The internal energy of the system A) increases by 10 cal. D) decreases by 50 cal. B) decreases by 10 cal. E) decreases by 20 cal. C) increases by 50 cal. Section: 18 5 Topic: Work and the PV Diagram for a Gas Type: Conceptual 64 An ideal gas undergoes a cyclic process in which total (positive) work W is done by the gas. What total heat is added to the gas in one cycle? A) W B) W C) zero D) 2W E) W/2 Final Exam H Rev Ques.doc
14 Section: 18 5 Topic: Work and the PV Diagram for a Gas Type: Conceptual 66 The pressure of a gas in an isobaric expansion remains constant. In such an expansion, A) no work is done. B) work is done by the gas. C) work is done on the gas. D) "isobaric" and "expansion" are contradictory terms. E) work is or is not done depending on whether the temperature of the gas changes. Section: 18 5 Topic: Work and the PV Diagram for a Gas Type: Numerical 68 The work done by an ideal gas in an isothermal expansion from volume V 1 to volume V 2 is given by the formula: W = nrt ln(v 2 /V 1 ) Standard atmospheric pressure (1 atm) is kpa. If 1.0 L of He gas at room temperature (20ºC) and 1.0 atm of pressure is compressed isothermally to a volume of 100 ml, how much work is done on the gas? A) 5.6 kj B) J C) kj D) kj E) J 69 Section: 18 5 Topic: Work and the PV Diagram for a Gas Type: Numerical An ideal gas system changes from state i to state f by paths iaf and ibf. If the heat added along iaf is Q iaf = 50 cal, the work along iaf is W iaf = 20 cal. Along ibf, if Q ibf = 40 cal, the work done, W ibf, is A) 10 cal B) 20 cal C) 30 cal D) 40 cal E) 50 cal Final Exam H Rev Ques.doc
15 Section: 18 5 Topic: Work and the PV Diagram for a Gas Type: Numerical 80 The equation of state for a certain gas under isothermal conditions is PV = 31.2, where the units are SI. The work done by this gas as its volume increases isothermally from 0.2 m 3 to 0.8 m 3 is approximately A) 2.86 J B) 28.6 J C) 43.3 J D) 71.8 J E) 115 J Use the following to answer questions 82-86: Section: 18 5 Topic: Work and the PV Diagram for a Gas Type: Numerical 82 An ideal gas initially at 50ºC and pressure P 1 = 100 kpa occupies a volume V 1 = 3 L. It undergoes a quasi-static, isothermal expansion until its pressure is reduced to 50 kpa. How much work was done by the gas during this process? R = J/mol K = L atm/mol K. A) 116 J B) 208 J C) 256 J D) 304 J E) 416 J Section: 18 5 Topic: Work and the PV Diagram for a Gas Type: Numerical 83 An ideal gas initially at 50ºC and pressure P 1 = 250 kpa occupies a volume V 1 = 4.5 L. It undergoes a quasistatic, isothermal expansion until its pressure is reduced to 150 kpa. How much work was done by the gas during this process? R = J/mol K = L atm/mol K. A) 116 J B) 320 J C) 575 J D) 640 J E) 850 J Final Exam H Rev Ques.doc
16 Section: 18 5 Topic: Work and the PV Diagram for a Gas Type: Numerical 84 An ideal gas initially at 100ºC and pressure P 1 = 250 kpa occupies a volume V 1 = 4.5 L. It undergoes a quasistatic, isothermal expansion until its pressure is reduced to 150 kpa. How much does the internal energy of the gas change during this process? R = J/mol K = L atm/mol K. A) 116 J B) 320 J C) 575 J D) 640 J E) The internal energy does not change during this process. Section: 18 5 Topic: Work and the PV Diagram for a Gas Type: Numerical 86 An ideal gas initially at 100ºC and pressure P 1 = 250 kpa occupies a volume V 1 = 4.5 L. It undergoes a quasistatic, isothermal expansion until its pressure is reduced to 150 kpa. How much heat enters the gas during this process? R = J/mol K = L atm/mol K. A) 116 J B) 320 J C) 575 J D) 640 J E) 850 J Section: 18 5 Topic: Work and the PV Diagram for a Gas Type: Numerical 88 An ideal gas initially at 50ºC and pressure P 1 = 100 kpa occupies a volume V 1 = 3 L. It undergoes a quasistatic, isothermal expansion until its pressure is reduced to 50 kpa. How much heat enters the gas during this process? R = J/mol K = L atm/mol K. A) 116 J B) 208 J C) 256 J D) 304 J E) 416 J Section: 18 6 Topic: Heat Capacities of Gases Type: Numerical 91 The internal energy for a diatomic gas is given by U = 5nRT/2. Calculate the internal energy of a 100 g mixture of oxygen (20%) and nitrogen (80%) gas at 25 C. (The molar weight of O 2 = 32 g, and the molar weight of N 2 = 28 g.) A) 21.6 kj B) 1.80 kj C) 12.1 kj D) 13.0 kj E) 1.10 kj Section: 18 6 Topic: Heat Capacities of Gases Type: Numerical 96 A gas has a molar heat capacity at constant volume of J/mol K. Assume the equipartition theorem to be valid. How many degrees of freedom (including translational) are there for the molecules of this gas? (the ideal-gas law constant is R = 8.31 J/mol K) A) 1 B) 3 C) 4 D) 5 E) 7 Final Exam H Rev Ques.doc
17 Section: 19 5 Topic: Irreversibility, Disorder, and Entropy Type: Conceptual 49 The change in the entropy of the universe due to an operating Carnot engine A) is zero. B) must be positive. C) must be negative. D) could be positive or negative. E) is meaningless to consider, because a Carnot engine has no connection to entropy. Section: 19 5 Topic: Irreversibility, Disorder, and Entropy Type: Numerical 51 A car of mass 2000 kg is traveling at 22.0 m/s on a day when the temperature is 20.0ºC. The driver steps on the brakes and stops the car. By how much does the entropy of the universe increase? A) 1.6 kj/k B) 24 kj/k C) J/K D) 0.15 kj/k E) 3.3 kj/k Section: 19 5 Topic: Irreversibility, Disorder, and Entropy Type: Numerical 52 A steam power plant with an efficiency of 65% of the maximum thermodynamic efficiency operates between 250º C and 40ºC. What is the change in the entropy of the universe when this plant does 1.0 kj of work? A) 16 J/K B) 1.7 J/K C) 55 J/K D) 1.7 mj/k E) 0.85 J/K Section: 19 5 Topic: Irreversibility, Disorder, and Entropy Type: Numerical 53 One mole of an ideal gas undergoes a reversible isothermal expansion from a volume of 1 L to a volume of 2 L. The change in entropy of the gas in terms of the universal gas constant R is A) R/2 B) 2R C) R ln(2) D) R ln(½) E) None of these is correct. Section: 19 5 Topic: Irreversibility, Disorder, and Entropy Type: Numerical 55 Two moles of a gas at T = 350 K expand quasistatically and isothermally from an initial volume of 20 L to a final volume of 60 L. The change in entropy of the gas during this expansion is (R = J/mol K) A) 17.4 J/K B) 18.3 J/K C) 20.4 J/K D) 24.6 J/K E) 27.8 J/K Final Exam H Rev Ques.doc
18 Section: 19 5 Topic: Irreversibility, Disorder, and Entropy Type: Numerical 56 Three moles of a gas at T = 250 K expand quasi-statically and adiabatically from an initial volume of 30 L to a final volume of 60 L. The change in entropy of the gas during this expansion is (R = J/mol K) A) 17.3 J/K B) 18.6 J/K C) 17.4 J/K D) 19.5 J/K E) zero Section: 19 5 Topic: Irreversibility, Disorder, and Entropy Type: Numerical 57 A container has 0.2 mole of O 2 gas and the gas is heated from 0 C to 100 C. What is the change in entropy of the gas? (R = J/mol K) A) 1.82 J/K B) 1.30 J/K C) 0.78 J/K D) 26.8 J/K E) zero Section: 19 5 Topic: Irreversibility, Disorder, and Entropy Type: Numerical 58 A block of mass m = 0.2 kg slides across a rough horizontal surface with coefficient of kinetic friction µ k = 0.5. What is the change in entropy after the block has moved a distance of 1 m? The temperature of the block and surrounding is 22 C. A) J/K B) J/K C) J/K D) J/K E) zero Final Exam H Rev Ques.doc
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### Practice Test. 4) The planet Earth loses heat mainly by A) conduction. B) convection. C) radiation. D) all of these Answer: C
Practice Test 1) Increase the pressure in a container of oxygen gas while keeping the temperature constant and you increase the A) molecular speed. B) molecular kinetic energy. C) Choice A and choice B | 15,239 | 54,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.778928 |
http://ecoursesonline.iasri.res.in/mod/page/view.php?id=91841 | 1,725,750,929,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00761.warc.gz | 9,599,414 | 26,464 | ## Lesson 10. GRAVITY SETTING, SEDIMENTATION, PRINCIPLES OF CENTRIFUGAL SEPARATION
Module 3. Separation equipment
Lesson 10
GRAVITY SETTING, SEDIMENTATION, PRINCIPLES OF CENTRIFUGAL SEPARATION
10.1 Introduction
Separation of components of immiscible liquids by use of density difference is a well established method. In the dairy industry, the application of centrifugal force is done particularly in separating fat from rest of the milk. It is also being used in separating bacteria from milk.
Different components of milk have the following specific gravities:
Fat: 0.93
Skim milk: 1.036
Whole milk: 1.028 to 1.032 depending on composition
10.2 Stoke's Law
The rapidity of separation depends on various factors, and can be estimated by use of Stokeās law for gravity separation modified to include the centrifugal force, instead of gravitational force.
Origins of Centrifugal cream separation are from the fundamentals of particle separation by gravitational forces, due to density difference between various components. If the density of the particle is higher than that of the surrounding medium it sinks and if it is lower it rises. The particle remains in suspension if densities are equal.
Considering the particle in the figure 10.1, the particle is acted upon by two forces, viz. gravitational force and viscous force. The gravitational force pulls the particle down, and the viscous force is resisting the movement of the particle. The particle will come down, if its density is greater than the fluid, and it will move up, if its density is less than that of the fluid surrounding it. The expression for these forces is given below.
Fig. 10.1 Viscous force
Fig. 10.2 Cream separation
10.3 Energy of Rotating Body
Calculation of the approximate power required to rotate the cream separator can be done by assuming that the stack of discs along with bowl as a hollow cylinder with uniform density throughout. The weight of the bowl, W can be given as
Fig. 10.3 Energy of rotating body
10.4 Capacity of Cntrifugal Separator
The capacity of the centrifugal separator depends on the number of factors again, viz., the number and size of discs, speed of rotation, the viscosity of the fluid etc. An expression for the throughput of centrifugal cream separator is arrived at as shown below
Fig. 10.4 Centrifugal separator
Fig. 10.5 Drag force due to volumetric flow
āCā is added to take care of deviation due to
1. Velocity profile is not laminor flow
2. Decrease in cross sectional area due to back flow of lighter phase
3. Non- ideal distribution over the feed inlet circumference
4. Geometric in accuracies in the stacks of discs.
The position of neutral zone or the place where the milk enters the gaps between adjacent discs will influence the ease in separation of the denser and lighter phases of the fluid. In applications like Clarifier, where the component to be separated is the denser particles of dust, extraneous matter, the neutral zone should be towards periphery. In the case of cream separation, the component that is to be separated is the fat, which is of lighter phase, the zone has to be towards centre of the disc, to reduce the distance of travel for the separation.
Fig. 10.6 Drag force due to volumetric flow Q (m3/sec)
The pressure exerted by the lighter and heavier fluids, when in equilibrium, on the inner wall of bowl is same
The above is possible, only when the level of the lighter and heaver liquid, are different. (about 3 to 10 mm) by fitting overflow lips of different sizes
The distance RS of the rotating separated milk ring is greater than RC, the rotating of cream. | 822 | 3,647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-38 | latest | en | 0.904163 |
https://stg.investinganswers.com/dictionary/c/cost-unit | 1,653,100,540,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534773.36/warc/CC-MAIN-20220521014358-20220521044358-00187.warc.gz | 622,006,292 | 17,940 | What is Cost Per Unit?
Cost per unit is a measure of a company's cost to build or create one unit of product.
How Does Cost Per Unit Work?
For example, let's assume it costs Company XYZ \$10,000 to purchase 5,000 widgets that it will resell in its retail outlets. Company XYZ's cost per unit is:
\$10,000 / 5,000 = \$2 per unit
Often, calculating the cost per unit isn't so simple, especially in manufacturing situations. Usually, costs per unit involve variable costs (costs that vary with the number of units made) and fixed costs (costs that don't vary with the number of units made).
For example, at XYZ Restaurant, which sells only pepperoni pizza, the variable expenses per pizza might be:
Flour: \$0.50
Yeast: \$0.05
Water: \$0.01
Cheese: \$3.00
Pepperoni: \$2.00
Total: \$5.56 per pizza
Its fixed expenses per month might be:
Labor: \$1,500
Rent: \$3,000
Insurance: \$200
Utilities: \$450
Total: \$5,650
If company XYZ sells 10,000 pizzas, then its cost per unit would be:
Cost per unit = \$5.56 + (\$5,650/10,000) = \$6.125
Why Does Cost Per Unit Matter?
Knowing cost per unit helps business owners determine when they'll turn a profit and helps them price their products with that in mind. It provides a dynamic overview of the relationships among revenues, costs and profits.
However, typical variable and fixed costs differ widely among industries. This is why comparison of break-even points is generally most meaningful among companies within the same industry, and the definition of a 'high' or 'low' should be made within this context. | 392 | 1,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | longest | en | 0.918929 |
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# The heat transfer from a hotter place to a colder place with or without having a material medium in between is called:
A. Conduction
B. Convection
## Explanation:
Heat transfer is the movement of thermal energy from a hotter object or place to a cooler object or place. There are three modes of heat transfer: conduction, convection, and radiation.
Conduction is the transfer of heat through a material medium by contact. In conduction, the heat flows from a region of higher temperature to a region of lower temperature. For example, when you touch a hot stove, heat is transferred from the stove to your hand through conduction.
Convection is the transfer of heat by the movement of fluids, such as liquids and gases. In convection, the heated fluid rises and the cooler fluid sinks, creating a circular motion that transfers heat. For example, when you boil water, the heat is transferred from the stove to the water through convection.
Radiation is the transfer of heat through electromagnetic waves without the need for a material medium. In radiation, the energy is transferred from the hotter object to the cooler object through the emission of electromagnetic waves, such as infrared radiation. For example, the heat from the sun is transferred to the Earth through radiation.
In conclusion, the heat transfer from a hotter place to a colder place with or without having a material medium in between is called radiation. Radiation is the transfer of heat through electromagnetic waves without the need for a material medium, and it is one of the three modes of heat transfer along with conduction and convection.
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Click Here for Online MCQs Quiz Now Log In ⌊ ❏ ⌋ ⌊ ❐ Optional MCQs ⌋ ⌊ ❒ CSS Syllabus 2022 ⌋ ⌊❐ Past Paper MCQs ⌋ ⌊ Home | 560 | 2,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-38 | latest | en | 0.937374 |
https://stdlib.io/develop/docs/api/@stdlib/math/base/special/gcd/ | 1,571,405,212,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682998.59/warc/CC-MAIN-20191018131050-20191018154550-00395.warc.gz | 723,183,447 | 50,854 | # gcd
Compute the greatest common divisor (gcd).
The greatest common divisor (gcd) of two non-zero integers `a` and `b` is the largest positive integer which divides both `a` and `b` without a remainder. The gcd is also known as the greatest common factor (gcf), highest common factor (hcf), highest common divisor, and greatest common measure (gcm).
## Usage
``````var gcd = require( '@stdlib/math/base/special/gcd' );
``````
#### gcd( a, b )
Computes the greatest common divisor (gcd).
``````var v = gcd( 48, 18 );
// returns 6
``````
If both `a` and `b` are `0`, the function returns `0`.
``````var v = gcd( 0, 0 );
// returns 0
``````
Both `a` and `b` must have integer values; otherwise, the function returns `NaN`.
``````var v = gcd( 3.14, 18 );
// returns NaN
v = gcd( 48, 3.14 );
// returns NaN
v = gcd( NaN, 18 );
// returns NaN
v = gcd( 48, NaN );
// returns NaN
``````
## Examples
``````var randu = require( '@stdlib/random/base/randu' );
var round = require( '@stdlib/math/base/special/round' );
var gcd = require( '@stdlib/math/base/special/gcd' );
var a;
var b;
var v;
var i;
for ( i = 0; i < 100; i++ ) {
a = round( randu()*50.0 );
b = round( randu()*50.0 );
v = gcd( a, b );
console.log( 'gcd(%d,%d) = %d', a, b, v );
}
``````
## References
• Stein, Josef. 1967. "Computational problems associated with Racah algebra." Journal of Computational Physics 1 (3): 397–405. doi:10.1016/0021-9991(67)90047-2. | 465 | 1,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-43 | latest | en | 0.582412 |
https://www.physicsforums.com/threads/output-resistance-of-a-mos-current-sink.393982/ | 1,544,453,792,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823348.23/warc/CC-MAIN-20181210144632-20181210170132-00104.warc.gz | 1,026,819,204 | 13,656 | # Homework Help: Output resistance of a MOS current sink
1. Apr 10, 2010
1. The problem statement, all variables and given/known data
Find the output resistance Rout.
[PLAIN]http://img526.imageshack.us/img526/953/pfimg1.jpg [Broken]
2. Relevant equations
$$R_{\rm out} \equiv \frac{v_x}{i_x}$$
3. The attempt at a solution
1) Draw small signal equivalent circuit and attach a test source
[PLAIN]http://img101.imageshack.us/img101/3742/pfimg2.jpg [Broken]
2) KCL at source node:
$$\frac{0-v_s}{1/g_m} + \frac{v_x - v_s}{r_o} = \frac{v_s}{R_S}$$
$$v_x \left( \frac{1}{r_o} \right) = v_s \left( \frac{1}{R_s} + \frac{1}{1/g_m} + \frac{1}{r_o} \right)$$
$$v_x \left( \frac{1}{r_o} \right) = \left( i_x Rs \right) \left( \frac{1}{R_s} + \frac{1}{1/g_m} + \frac{1}{r_o} \right)$$
$$R_{\rm out} \equiv \frac{v_x}{i_x} = r_o \, Rs \left( \frac{1}{R_s} + \frac{1}{1/g_m} + \frac{1}{r_o} \right) = r_o + \left( r_o \, R_s \, g_m \right) + R_s$$
Is this correct? Thanks.
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# Cost Function using High Low Method
I believe that I have done #1 correctly. I am confused as to how to figure out question #2. Please show all calculations used to arrive at the answer.
The number of X-rays taken and X-ray costs at a hospital over the last nine months are:
Month X-Rays Taken X-Ray Cost
January 6,250 \$28,000
February 7,000 29,000
March 5,000 23,000
April 4,250 20,000
May 4,500 22,000
June 3,000 17,000
July 3,750 18,000
August 5,500 24,000
September 5,750 26,000
1. Use high-low method, estimate the cost formula for X-ray costs.
29,000 17,000 = 12,000 = \$3.00
7,000 3,000 4,000
\$29,000
-21,000 (\$3.00 X 7,000 = 21,000)
\$ 8,000
2. Using the cost formula you derived above, what X-ray costs would you expect to be incurred during a month in which 4,600 X-rays are taken?
29,000 17,000 = 12,000 = \$5.00
7,000 4,600 2,400
\$29,000
35,000 (\$5.00 X 7,000 = 35,000)
#### Solution Preview
We first find the cost function. We take the highest and the lowest cost.
7,000 xrays cost 29,000
3,000 x rays cots 17,000
Then take the ...
#### Solution Summary
The solution explains how to derive the cost function using high low method and to use the cost function to calculate the total costs.
\$2.19 | 413 | 1,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2016-50 | longest | en | 0.910644 |
https://www.doorsteptutor.com/Exams/NCO/Class-2/Questions/Part-3.html | 1,508,430,041,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823350.23/warc/CC-MAIN-20171019160040-20171019180040-00161.warc.gz | 909,660,640 | 14,326 | # NCO- Cyber Olympiad Class 2: Questions 17 - 24 of 642
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## Question number: 17
MCQ▾
### Question
In the numeral 652, the place value of 2 equals
### Choices
Choice (4) Response
a.
652
b.
20
c.
200
d.
2
## Question number: 18
MCQ▾
### Question
Which group is arranged from least to greatest?
### Choices
Choice (4) Response
a.
112, 316, 512, 482
b.
482, 501, 380, 495
c.
362, 375, 268, 426
d.
425, 441, 496, 501
## Question number: 19
MCQ▾
### Question
Which number is 112 less than 150?
### Choices
Choice (4) Response
a.
38
b.
1112
c.
562
d.
262
## Question number: 20
MCQ▾
### Question
Aryan draw 12 circles and 8 triangles on a piece of paper. How many fewer triangles than circles did he draw?
### Choices
Choice (4) Response
a.
12
b.
4
c.
6
d.
3
## Question number: 21
MCQ▾
### Question
There are 8 girls and 8 boys in the team. Which number sentence tells how many children are there in the team?
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 22
MCQ▾
### Choices
Choice (4) Response
a.
425
b.
4205
c.
524
d.
5240
## Question number: 23
MCQ▾
### Choices
Choice (4) Response
a.
64
b.
94
c.
84
d.
74
## Question number: 24
MCQ▾
### Question
The difference between place value and face value of 2 in 4265
### Choices
Choice (4) Response
a.
4065
b.
198
c.
0
d.
2
f Page | 523 | 1,641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-43 | longest | en | 0.597567 |
https://www.physicsforums.com/threads/probability-problem-help.123057/ | 1,531,976,288,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590493.28/warc/CC-MAIN-20180719031742-20180719051742-00120.warc.gz | 950,877,710 | 13,899 | # Homework Help: Probability problem help
1. Jun 6, 2006
### Logarythmic
"A particle of mass m moves in one dimension in the infinite square well. Suppose that at time t = 0 its wave function is
PSI(x,t=0) = A(a^2 - x^2)
where A is a normalisation constant.
Find the probability P_n of obtaining the value E_n of the particle energy, where E_n is one of the energy eigenvalues."
I know how to find A and I know how to find the time-dependent wave function, but what then?
2. Jun 6, 2006
### nrqed
I notice that you are using a well going from x=-a to x=+a, right?
You have to expand your wavefunction over the eigenstates of the Hamiltonian (which are simply the sine and cos functions with their normalization factors).
$$\Psi(x) = \Sum c_n \psi_n(x)$$
where the $\psi_n$ are the normalized sin/cos functions.
To find the c_n, just use the orthonormality of the psi_n,
$$c_n = \int_{-a}^a dx \, \psi_n^*(x) \Psi(x)$$
and, finally, the probability of the energy being measured to be a specific value E_n is given by |c_n|^2.
3. Jun 7, 2006
### Logarythmic
Yeah that does it, thank you. =)
4. Jun 7, 2006
### Logarythmic
I use the sin function for even n and the cos function for odd n, right? But then c_n for even n gets zero..? Is that correct?
For odd n I get P(E_n) = 15/((n^4)(pi^4))
5. Jun 7, 2006
### nrqed
I did not check your result for odd n but it makes sense that for even n you would get zero. Your wavefunction is symmetric function (it is even in the sense of Psi(-x) = Psi(x)) so that it does not "contain" any of the sin functions which are odd themselves.
Your answer seems also to make sense since, as expected, the probabibility goes down with increasing n. Also, your function ressembles the most the ground state so one expects a very quick decrease with n.
Lastly, one double check would be to verify that the sum to infinity over odd n of 1/n^4 gives pi^4/15.
Good job!
Pat | 545 | 1,923 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-30 | latest | en | 0.901912 |
https://math.stackexchange.com/questions/4839065/how-to-understand-countability-of-rational-vs-irrational-numbers | 1,722,898,306,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640455981.23/warc/CC-MAIN-20240805211050-20240806001050-00546.warc.gz | 309,962,399 | 47,376 | How to understand countability of rational vs irrational numbers
[The question was edited 2024-01-09. Changes are marked with these brackets, mostly.]
[Edited again 2024-01-10 adding "final remarks".]
I'm struggling with getting my head around some aspects of transfinite numbers and sets in general. This question is about my struggle with the properties and countability of rational vs irrational numbers.
I think these are common facts, please correct me if I got anything wrong:
1. The rational as well as irrational numbers are "dense" on the real number line, meaning roughly that the resolution is unlimited, and there are no gaps of size >0.
2. The rational as well as irrational numbers are an "incomplete" total order, meaning roughly that there are indeed gaps (of size 0), and ... (yada yada existence of supremums).
3. The (set of) rationals is countable while the irrationals aren't. (I have not seen a direct proof of the latter, only that it follows from reals being proved uncountable while rationals are proved countable.) [Update 2024-01-09: the remark has been answered in a comment.]
• (a.) As a consequence of density; between every pair of distinct irrational numbers there is at least one (and actually infinitely many) rational numbers, no matter how small the distance as long as >0.
• (b.) Symmetrically, exactly the same for irrationals between rationals.
1. The Dedekind cut is one way (of several) to construct all the reals from the rationals. Each real number $$x$$ is uniquely associated with the set of all rational numbers that are strictly smaller than $$x$$.
"5b." [Probably irrelevant but included for symmetry] Except that it may be without purpose, is there anything to stop me from defining "Hallström cuts" in exactly the same way as Dedekind cuts, but instead using the irrational numbers as a base for modeling all the reals.
1. [Probably not relevant here] The rational numbers are closed under basic arithmetic ops (add and mul). The irrationals aren't.
• [Sidenote] I slightly wonder if there exists subsets of the irr:s that are closed under such ops. [Update 2024-01-09: this sidenote has been answered in a comment.]
The question
My main headache: given (4) above, how can it be possible that one type has larger cardinality than the other (acc to (3) above)? They have such striking symmetry with each other. Intuitively and informally, they should be sort of perfectly interleaved with each other. (I know intuition is not necessarily a reliable guide here. I also realise that assuming perfect interleaving is not a consistent model.)
Put another way: where could all the additional irrationals be hiding from being forced into 1-1 correspondence with the rationals? If there were a hideout anywhere with several (distinguishable) irrs in a row without any rats in between them, it would be in breach of (4) and would also invalidate the D. cut.
Another argument: The D. cut is not using the power set of the set of rats, only an interval. The only difference between different cuts is the upper bound of the interval. ([Edit 2024-01-09] The upper bound is a real number here.) When using only the domain of the rational numbers, how can there exist more such distinct intervals (uncountable) than there are rats (countable)??
Here is a new attempt to quantify my mental issue with the countable rational numbers, by focusing on the Dedekind cuts.
Construct the Dedekind cuts for all rational numbers, and only those to begin with. Now for every rat num $$r$$ there is a uniquely associated set of rat nums, containing the open interval $$(-\infty, r)$$. For two rat nums $$r_2>r_1$$, $$D(r_2)=D(r_1)\cup[r_1, r_2)$$, so the sets only differ "on the top", so to say.
Now, at least intuitively, in constructing all these intervals $$D(r)$$, all available rats have been consumed already. How could there exist any more intervals of that same kind, containing rat nums exclusively.
If there was the largest rat num in the open interval (I know there isn't), then every rat num $$r$$ would clearly be busy being the top member of exactly one $$D(s)$$.
If there was such a thing as "the next rat num" (I know there isn't), it would be clear that the set of all $$D(r)$$ so far equals the set of all possible $$D(x)$$ at all (for any real number $$x$$).
(The outcome of this: I'm trying to claim that the set of all possible D cuts, mentally, cannot exceed the set of all possible rat nums $$r$$.)
Given the argument in the prev paragraphs, how can now unique D. cuts be added for the irrational numbers when they were already exhausted by the rational mumbers? There would be no "free" rat nums left that can be used to uniquely create a new cut. It seems to require that there are more rat nums than there are rat nums. (Sic.)
Yet another angle: in how many different ways can one create an interval containing rational numbers only? (Please note that the interval can be bounded by real numbers but contains only rationals.) Arguably only countably many. Imagine adding one rat at a time to the top of the interval, ignoring the fact that the rats cannot really be enumerated in order like that. Any enumeration of rats can only go on for a countable number of times.
[2024-01-10 "Final remarks" by the author]
Thank you for helpful comments and answers. It may look like it but I have not been trying to prove that something is wrong, because I can't. I have tried to reason about my main source of "headache" in this subject, and was seeking some guidance about it.
For the record, I already know that for the time being I must accept (reluctantly) such parts of math that I am intuitively at odds with (at least until I develop ability to prove inconsistencies).
• There is no such thing as 'the upper bound of the interval', not in the rational numbers, anyway. The least upper bound can be one of those 'gaps of size zero' you were talking about. Commented Jan 5 at 7:44
• The countability is not so easy to grasp intuitively. The algebraic irrational number still form a countable set ! Countable means that there is a method to list all elements in such a way that it is guaranteed that every element will eventually occur. That this is not necessarily the case , even in the countable case : Consider the method to list first the even numbers , then the odd ones. What is wrong with this ? Correct, $1$ will never be reached. Commented Jan 5 at 7:44
• A similar question: math.stackexchange.com/questions/4743188
– Karl
Commented Jan 5 at 8:10
• Answer to your Sidenote: The set of all expressions of the form $\sum_{i=1}^n a_i \pi^i$ with $n\geq 1$, $a_1, a_2,\ldots, a_n$ integer $\geq0$, and $a_n>0$ consists of irrational numbers only. And it is closed unter addition and multiplication. Commented Jan 5 at 10:14
• I don't know exactly what you mean by a "direct proof", but the standard Cantor diagonalization argument used to show that the reals are uncountable work: given a countable list of irrational numbers, you may construct an irrational number different from every number of the list by a diagonal argument (modify the original argument to make sure you are constructing a non-repeating decimal. E.g. use digits 1 or 2 for the first digit, use digits 8 and 9 for the following two digits, then 1/2 again for the next three digits, then 8/9 again for the next four digits, and so on...) Commented Jan 6 at 3:08
The simple short answer is that infinity is weird and does not fit well with intuition which was developed in the real world of finite things.
An early challenge to intuition is whether or not there is a single infinity. Is the set of integers greater than or equal to zero bigger than the set of integers strictly greater than zero because it has one more element? Is the set of all integers twice as big because it has the negatives as well?
We can only start to answer this after we have defined what we mean for two, possibly infinite sets, to have the same size. The most common definition is that two sets have are the same size if a bijection (one-to-one and onto map) can be found between them. We quickly run into another challenge. With familiar finite sets, if we find a map which is one-to-one but not onto then we would deduce that one is smaller than the other. So, this suggests that the examples which I just gave are different sizes. However, we find that the existence of a one-to-one but not onto map does not preclude a bijection. A bijection can be found between all of those sets so they are the same size. We find that further intuitively bigger sets are also the same size e.g. the rational numbers and the algebraic numbers. The algebraic numbers are those which are the roots of polynomials with integer (or rational) coefficients. They include some irrationals, e.g. $$\sqrt 2$$ but not others, e.g. $$\pi$$.
We usually say "cardinality" rather than "size" which mainly emphasizes that we are using this interpretation. I will stick to "size" as we are talking about intuition.
We might start to develop the intuition that all infinite sets are the same size but now we find that the real numbers are strictly bigger. A bijection between the natural numbers and the real numbers is not possible.
How much bigger? Is it the next biggest set or is there something with an intermediate size? Now it gets really weird. Not only do we not know but we can prove that we cannot prove that there is or is not a set with an intermediate size. Look up the Continuum Hypothesis if you want to know more.
We know that the reals are not the biggest set. For any set, we can construct a bigger one so there is no biggest one.
Now back to some of your points. It is indeed counter-intuitive that all of these are true at once:
1. Between any rationals there is an irrational
2. Between any irrationals there is a rational
3. There are more irrationals than rationals
1 and 2 suggest that they alternate along the real line and hence there are equally many but 3 contradicts this. Along with the Continuum Hypothesis, I think that you just have to work on understanding the proofs and then learn to accept it. I still find it weird but not as weird as the Continuum Hypothesis.
The problem with Hallström cuts is that you don't have the irrationals until you have the reals. The irrationals are just the reals which are not rational. So, you would have to obtain the irrationals somehow else before you could start work on Hallström cuts. We know how to construct the rationals without the reals so we can use them as a starting point. You could perform cuts on the algebraic numbers and that would work because we could construct them without the reals. However, you would just get something isomorphic to the usual reals.
It is often said, without good evidence, that contemplating infinity drove Cantor mad. Even if not true, the existence of the story shows that struggling to understand infinity is common.
• +1 for the first and last paragraphs, which I borrowed in my answer to stress what I think the OP is struggling with. Commented Jan 10 at 16:06
• +1 for pretty much all of it. It was a long answer but who am I to say that. I (OP) especially appreciate the commentary in the middle about the counter-intuitive facts. I reckon this is the best answer that I'm gonna get, and I'm going to accept it. For the record, I already know that I for the time being must accept (reluctantly) such parts of math that I am intuitively at odds with. Commented Jan 10 at 21:44
• Actually, I considered saying more. There is a great book: The Road to Reality by Roger Penrose. He looks at many weird areas of maths, e.g. complex numbers, and shows uses for them in physics. He seems surprised when it comes to infinite cardinals that they have not (yet) proved useful in physics. Also search for "unreasonable effectiveness of mathematics", this will lead you to an article by Eugene Wigner expressing surprise at how useful maths is in physics. Together. they are saying that maths is very useful but not a perfect description of reality. Commented Jan 10 at 22:46
The essential answer to your question is that you can't trust your intuition about infinity. All you can do is follow the formal definitions to their logical conclusions and then train your intuition to bring it into accord with those conclusions. Until you do that, none of the carefully written answers here is likely to satisfy you. The first and last paragraphs from @badjohn say it well:
The simple short answer is that infinity is weird and does not fit well with intuition which was developed in the real world of finite things. ...
It is often said, without good evidence, that contemplating infinity drove Cantor mad. Even if not true, the existence of the story shows that struggling to understand infinity is common.
Given the argument in the prev paragraphs, how can now unique D. cuts be added for the irrational numbers when they were already exhausted by the rational mumbers? It seems to require that there are more rat nums than there are rat nums.
There are Dedekind cuts which are not of the form $$D(r)$$ for any rational number $$r$$. For example, the Dedekind cut corresponding to the interval $$(-\infty, \sqrt{2})$$ (i.e., the set of all rational numbers less than $$\sqrt{2}$$) is not equal to $$D(r)$$ for any rational $$r$$. You can replace $$\sqrt{2}$$ by any irrational number.
• Thanks, but this is exactly the argument I'm feebly trying to make. It is "not reasonable" that $D(\sqrt{2})\not =D(r)$ for all rational $r$ because there would be no free rat num left to put in its set to make it unique. Between any $r_1$ and $D(\sqrt{2})$ there are always many $r_2$ that already have their cut, and the argument is that $D(\sqrt{2})$ cannot possibly be different from all of those. This all informal, I know. But I think there is something deeply fishy about the D cuts. It is always dodged by saying "there is no rat $r$ closest to irr $i$". Commented Jan 9 at 21:36
• You've given the correct reason "there is no rat $r$ closest to irr $i$" so why do you still think it is fishy?
– Ted
Commented Jan 9 at 21:59
• To me it is like a fallacy by denying access to the evidence, much like the Heisenberg uncertainty or the Schrödinger cat. We can never go deep enough to really check whether the Dedekind claims hold at the smallest scale (because there is no smallest scale). Commented Jan 9 at 22:14
• Please note: I have no problem with the D cuts as long as we limit it to comparing two arbitrary reals. That always works. My issue is with the idea that one could uniquely construct all those cuts at once for every real (and implicitly collect them in a set). Commented Jan 9 at 22:18
• You keep making arguments relying on things that you know are wrong, like "if we ever allowed a rat to exist immediately above sqrt(2)", "ignoring the fact that the rats cannot really be enumerated in order like that", etc. Why do you do that?
– Ted
Commented Jan 10 at 0:02
To be honest this post is giving me headache so I'm only gonna address the new additions today (9th Jan 2023) in the end. I'll use $$r$$ to consistently stand for some rational number.
Unfortunately, all arguments you've made seem to me to be begging the question: you start by assuming all Dedekind cuts are $$D(r)$$, and conclude there can be no more other Dedekind cuts. I think you need to let go of this premise, and start thinking "maybe there are Dedekind cuts that are not $$D(r)$$". Until then, we are going nowhere.
You can also take a look at this post by Eric Wofsey. It gives a proof that a countable dense totally ordered set cannot be Dedekind complete. In it, the author assumes $$\mathbb{R}$$ is countable and arrives at a contradiction using only the dense total order on $$\mathbb{R}$$. Hopefully, since this proof relies purely on countability and dense total order, with no arithmetics, algebra, or calculus involved, it can convince you better. If it doesn't, I really doubt if you will ever be convinced.
• Thank you for your link. However it appears to be another proof of the uncountability of the reals, which was not quite the issue. I have read and (reluctantly) accepted a couple of such proofs. Unfortunately I fail to see the connection to my rambling about D cuts, and I can only guess the meaning of "D. complete", but nevermind. Besides I think you missed the point of said rambling, but you're not alone and who could blame you. I do understand the general point you are making that I need to relax and accept some things as they are (until I develop ability to prove otherwise). Commented Jan 10 at 22:04
• @AndersH The intent of the proof I linked is, as you said, different, but it would be helpful to this issue if you replace $\mathbb{R}$ that is assumed countable with the actual countable $\mathbb{Q}$. A total order being Dedekind complete just means every Dedekind cut is $D(x)$ for some $x$ in it. Thus, the proof basically says because $\mathbb{Q}$ is countable, there will always be a Dedekind cut (in fact, uncountably many) that is not $D(r)$, which I was hoping can show you the very premise you repeatedly rely on, that every Dedekind cut must be $D(r)$, is false. Commented Jan 10 at 22:31 | 4,079 | 17,244 | {"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": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-33 | latest | en | 0.959879 |
https://excelribbon.tips.net/T012597_Finding_Odd_Values_Greater_Than_50.html | 1,720,799,515,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514404.71/warc/CC-MAIN-20240712125648-20240712155648-00360.warc.gz | 185,544,166 | 10,777 | # Finding Odd Values Greater Than 50
Written by Allen Wyatt (last updated December 30, 2023)
This tip applies to Excel 2007, 2010, 2013, 2016, 2019, Excel in Microsoft 365, and 2021
Amol has 1,000 values in an Excel worksheet, occupying 100 rows of 10 columns each. Each value in this range is an integer value between 0 and 99. Amol needs a way to count and display all the values which are odd and greater than 50.
There are a few ways you can go about counting and displaying, but it is important to understand that these are different tasks. Perhaps the best way to display those values that fit the criteria is to use conditional formatting. You can add a conditional formatting rule to each cell that will make bold or otherwise highlight the desired values. Follow these steps:
1. Select the cells that contain your data.
2. Display the Home tab of the ribbon.
3. Click the Conditional Formatting tool in the Styles group. Excel displays a palette of options related to conditional formatting.
4. Click New Rule. Excel displays the New Formatting Rule dialog box. (See Figure 1.)
5. Figure 1. The New Formatting Rule dialog box.
6. In the Select a Rule Type area at the top of the dialog box, choose Use a Formula To Determine Which Cells to Format.
7. In the formula box enter the formula =AND(MOD(A1,2),A1>50).
8. Click the Format button. Excel displays the Format Cells dialog box. (See Figure 2.)
9. Figure 2. The Format Cells dialog box.
10. Use the controls in the dialog box to modify the formatting, as desired.
11. Click OK to close the Format Cells dialog box.
12. Click OK to close the New Formatting Rule dialog box. The formatting is applied to the range of cells you selected in step 1.
If you prefer, you could also use the following formula in step 6:
```=AND(ISODD(A1),A1>50)
```
To get the count of cells that fit the criteria, you could use an array formula:
```=SUM(MOD(MyCells,2)*(MyCells>50)
```
This formula assumes that the range of cells you want to analyze are named MyCells. Don't forget to enter the cell using Ctrl+Shift+Enter. (This is not necessary if you are using Excel in Microsoft 365; just press Enter.) If you don't want to use an array formula, you could use the following:
```=SUMPRODUCT((MOD(MyCells,2)*(MyCells>50))
```
You could also use a macro to derive both the cells and the count. The following is a simple version of such a macro; it places the values of the cells matching the criteria into column M and then shows a count of how many cells there were:
```Sub SpecialCount()
Dim c As Range
Dim i As Integer
i = 0
For Each c In Range("A2:J101")
If c.Value > 50 And c.Value Mod 2 Then
i = i + 1
Range("L" & i).Value = c.Value
End If
Next c
MsgBox i & " values are odd and greater than 50", vbOKOnly
End Sub
```
Note:
If you would like to know how to use the macros described on this page (or on any other page on the ExcelTips sites), I've prepared a special page that includes helpful information. Click here to open that special page in a new browser tab.
ExcelTips is your source for cost-effective Microsoft Excel training. This tip (12597) applies to Microsoft Excel 2007, 2010, 2013, 2016, 2019, Excel in Microsoft 365, and 2021.
##### Author Bio
Allen Wyatt
With more than 50 non-fiction books and numerous magazine articles to his credit, Allen Wyatt is an internationally recognized author. He is president of Sharon Parq Associates, a computer and publishing services company. ...
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What is one more than 7?
2023-12-30 09:51:48
J. Woolley
Re. the formula (which was missing the final parenthesis)
=SUM(MOD(MyCells,2)*(MyCells>50))
the Tip says, "Don't forget to enter the cell using Ctrl+Shift+Enter." I don't think this is necessary because the formula does not return an array.
##### This Site
Got a version of Excel that uses the ribbon interface (Excel 2007 or later)? This site is for you! If you use an earlier version of Excel, visit our ExcelTips site focusing on the menu interface. | 1,434 | 6,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-30 | latest | en | 0.82469 |
http://excelnuggets.com/Financial/Functions/COUPPCD | 1,590,534,817,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391923.3/warc/CC-MAIN-20200526222359-20200527012359-00355.warc.gz | 43,875,684 | 4,243 | The COUPPCD function returns the previous coupon date before the settlement date. The "PCD" in COUPPCD means "Previous Coupon Date".
COUPPCD takes 3 required arguments and 1 optional argument:
Syntax: COUPPCD(settlement, maturity, frequency, [basis])
The settlement date is the date a buyer purchases a coupon, such as a bond. The maturity date is the date when a coupon expires. For example, suppose a 10-year bond is issued on January 1, 2014, and is purchased by a buyer seven months later. The issue date would be January 1, 2014, the settlement date would be August 1, 2014, and the maturity date would be January 1, 2024, 10 years after the January 1, 2014, issue date.
##### Using the COUPPCD function:
Note:
There are 5 other functions related to coupon calculations. See also the COUPDAYS, COUPDAYBS, COUPDAYSNC,COUPNUM and COUPNCDfunctions.
The arguments for the COUPPCD function are:
Argument Required? Description
settlement Required The security's settlement date. The security settlement date is the date after the issue date when the security is traded to the buyer.
maturity Required The security's maturity date. The maturity date is the date when the security expires.
frequency Required The number of coupon payments per year. For annual payments, frequency = 1; for semiannual, frequency = 2; for quarterly, frequency = 4.
basis Optional The type of day count basis to use.
##### A few more things:
• Excel stores dates as sequential serial numbers so they can be used in calculations. By default, January 1, 1900 is serial number 1, and January 1, 2014 is serial number 41640 because it is 41,640 days after January 1, 1900. • All arguments are truncated to integers. • If settlement or maturity is not a valid date, COUPPCD returns the #VALUE! error value. • If frequency is any number other than 1, 2, or 4, COUPPCD returns the #NUM! error value. • If basis < 0 or if basis > 4, COUPPCD returns the #NUM! error value. • If settlement ≥ maturity, COUPPCD returns the #NUM! error value.
### Summary
The COUPPCD function returns the previous coupon date before the settlement date. | 533 | 2,107 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | longest | en | 0.868035 |
https://aakashsrv1.meritnation.com/ask-answer/question/will-the-reflecting-surface-of-both-convex-and-concave-mirro/light-reflection-and-refraction/16976823 | 1,656,803,444,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104205534.63/warc/CC-MAIN-20220702222819-20220703012819-00185.warc.gz | 123,227,092 | 7,976 | # Will the reflecting surface of both convex and concave mirror will face towards the left(-y)?
Solution:
If a hollow sphere is cut into parts and the outer surface of the cut part is painted, then it becomes a mirror with its inner surface as the reflecting surface. This type of mirror is known as a concave mirror.
If the cut part of the hollow sphere is painted from the inside, then its outer surface becomes the reflecting surface. This kind of mirror is known as a convex mirror.
• 0
What are you looking for? | 112 | 520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-27 | latest | en | 0.97355 |
https://teacheradvisor.org/5/lessons/lesson_5-2-16 | 1,627,919,984,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154321.31/warc/CC-MAIN-20210802141221-20210802171221-00166.warc.gz | 555,448,974 | 7,963 | Unit 2, Lesson 16
# Use divide by 10 patterns for multi-digit whole number division.
EngageNY 60 min(s)
In this lesson, students begin developing mental strategies for multi-digit whole number division. They divide by 10s using drawing place value disks. For example, students solve 420 ÷ 10. Then they divide by 10s using a place value chart, and finally, mentally envision what happens on a place value chart as they divide by 10s. | 104 | 436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-31 | latest | en | 0.867794 |
https://encyclopediaofmath.org/index.php?title=Superalgebra&direction=prev&oldid=44946 | 1,660,204,433,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00466.warc.gz | 246,250,871 | 7,507 | # Superalgebra
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
A -graded algebra over a field (see Graded algebra), i.e. a super-space over endowed with an even linear mapping . A superalgebra is said to be commutative (graded-commutative or supercommutative) if
here, is a parity, i.e. a -grading.
The definition of a superalgebra can be generalized to include the case where the domain of scalars is an arbitrary commutative associative superalgebra .
Examples of associative superalgebras over are: the algebra of matrices of the form
where , , endowed with the natural -grading (cf. Super-space); the tensor algebra of a -graded module over ; the symmetric algebra of a module , where is the ideal generated by the elements of the form
and the exterior algebra of a module (the latter two superalgebras are commutative).
A superalgebra with a multiplication is called a Lie superalgebra if for all ,
(and if and ). In particular, there are no Lie superalgebras in characteristic , only -graded Lie algebras.
Examples. Any associative superalgebra endowed with commutation (the supercommutator difference)
as the bracket operation; the algebra of derivations of an arbitrary superalgebra (i.e. of linear transformations for which ) with the operation of commutation. For any Lie superalgebra there is an associative universal enveloping superalgebra, and the straightforward generalization of the Birkhoff–Witt theorem holds.
The classification of finite-dimensional simple Lie superalgebras over the field is known (see [2], [3]). They are divided into Lie superalgebras of classical type (characterized by the fact that the Lie algebra is reductive) and Lie superalgebras of Cartan type. The Lie superalgebras of classical type are exhausted by the following series of matrix algebras:
for an even symmetric non-degenerate bilinear form ;
for an odd symmetric non-degenerate bilinear form ;
where
and certain exceptional algebras (of dimensions , and ). The superalgebras of Cartan type are the algebra and its supersubalgebras, analogues to the simple Lie graded algebras , , (cf. Lie algebra, graded).
The classification of real structures of simple Lie superalgebras and a description of semi-simple Lie superalgebras in terms of simple ones are also known.
The theory of linear representations of Lie superalgebras is essentially more complex than for Lie algebras in that representations of simple Lie superalgebras, as a rule, are not completely reducible, while irreducible representations of solvable Lie superalgebras need not be one-dimensional. A classification exists of the irreducible representations of simple finite-dimensional Lie superalgebras over in terms of the highest weights (see , [2]), and an explicit description is known of the finite-dimensional representations, as well as for the character formula for certain series of these algebras .
#### References
[1a] D.A. Leites, "Lie superalgebras" Josmar , 30 (1984) [1b] D.A. Leites (ed.) , Seminar on supermanifolds , Kluwer (1990) [2] V.G. Kac, "Lie superalgebras" Adv. Math. , 26 (1977) pp. 8–96 [3] M. Scheunert, "The theory of Lie superalgebras. An introduction" , Springer (1979) | 757 | 3,213 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.892144 |
http://mytestbook.com/PrintableWorksheets/Worksheet_Grade1_Math_OrderingNumbers_1441.aspx | 1,610,957,575,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514423.60/warc/CC-MAIN-20210118061434-20210118091434-00405.warc.gz | 73,839,657 | 5,593 | go to: myTestBook.com
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Instruction:
Question 1 Put these numbers in correct increasing order. 39, 36, 38, 37 A. 36, 37, 38, 39 B. 39, 38, 37, 36 C. 38, 37, 39, 36 D. 36, 37, 39, 38
Question 2 Put these numbers in correct increasing order. 42, 45, 44, 43 A. 42, 45, 44, 43 B. 42, 45, 43, 44 C. 42, 43, 45, 44 D. 42, 43, 44, 45
Question 3 Put these numbers in correct increasing order. 29, 31, 28, 30 A. 29, 28, 31, 30 B. 28, 29, 30, 31 C. 29, 31, 30, 28 D. 28, 29, 31, 30
Question 4 Put these numbers in correct increasing order. 39, 40, 38, 42, 41 A. 38, 39, 42, 41, 40 B. 38, 39, 40, 41, 42 C. 40, 42, 38, 41, 39 D. 40, 38, 39, 42, 41
Question 5 Put these numbers in correct increasing order. 40, 50, 20, 30, 10 A. 20, 50, 10, 30, 40 B. 30, 40, 20, 50, 10 C. 10, 20, 30, 40, 50 D. 50, 10, 20, 30, 40
Question 6 Put these numbers in correct increasing order. 33, 55, 44, 22 A. 22, 33, 44, 55 B. 33, 55, 44, 22 C. 33, 44, 55, 22 D. 55, 33, 44, 22
Question 7 Put these numbers in correct increasing order. 21, 28, 7, 14 A. 7, 21, 14, 28 B. 14, 21, 7, 28 C. 7, 14, 21, 28 D. 21, 28, 7, 14
Question 8 Put these numbers in correct increasing order. 12, 10, 16, 14, 18 A. 10, 12, 16, 14, 18 B. 10, 12, 14, 18, 16 C. 10, 12, 14, 16, 18 D. 10, 14, 12, 16, 18
Question 9 Which series is in correct increasing order? A. 3, 6, 9, 7 B. 4, 8, 6, 2 C. 2, 4, 3, 7 D. 5, 7, 9, 11
Question 10 Which series is in correct increasing order? A. 3, 6, 9, 7 B. 4, 8, 6, 2 C. 7, 9, 11, 13 D. 1, 4, 2, 5
Free Worksheets From myTestBook.com ------ © myTestBook.com, Inc. Log in or Create Your FREE Account to access thousands of worksheets and online tests with Answers like this one. | 847 | 1,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-04 | latest | en | 0.808649 |
https://www.physicsforums.com/threads/related-rates-equation-problem.200472/ | 1,542,603,992,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745281.79/warc/CC-MAIN-20181119043725-20181119065725-00462.warc.gz | 966,359,708 | 13,075 | # Homework Help: Related rates equation problem
1. Nov 25, 2007
### Bionerd
1. The problem statement, all variables and given/known data
A boy is standing 50 ft. from the end of a swimming pool when he sees a girl 25 ft. along the end. He can swim 3 ft/s and run 5 ft/s. If he runs x feet, set up an equation for time consumed.
2. Relevant equations
There is a visual aid: A rectangle with length labeled 50 ft and a dot B designating where the boy is, and with a width labeled 25 ft and a dot G designating where the girl is.
3. The attempt at a solution
I know I have to set up a triangle, with my y-axis being the length (where the girl is, making my dy/dt= 3 ft/s) and the x-axis being the width (dx/dt=5 ft/s) Past this, I really don't know. I tried this:
Since he is running x ft, the 50 ft must be losing x, and the 25 ft must be gaining x. So y/x=(25+x)/(50-x) However, I need t in there. If I integrate dy/dt, I will be left with a c that I'm not sure how to get rid of. So yeah, I'm really stumped on this one.
I know it's kinda hard to visualize, so any help at all would be appreciated.
2. Nov 25, 2007
### HallsofIvy
You haven't given enough information! For example, if he is standing on the right of the swimming pool and the girl is also to the right of the swimming pool, I can see no reason for him swimming at all!
3. Nov 25, 2007
### Bionerd
He is standing in the lower right corner, she in the upper left.
4. Nov 26, 2007
### HallsofIvy
That completely contradicts what you originally said:
What you meant, I think is that the swimming pool is 50 feet long and 25 feet wide, the boy and girl are at opposite corners. That's quite different from saying "A boy is standing 50 ft. from the end of a swimming pool"!
He runs x feet along the length of the pool leaving 50-x to the end. That does NOT "add" anything to the 25 foot width. He swims along the hypotenuse of a right triangle having one leg 50-x and the other 25. The distance he swims is $\sqrt{(50-x)^2+ 25^2}=\sqrt{3125- 100x+ x^2}$. The time it takes to go a given distance is that distance divided by the speed.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 607 | 2,196 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-47 | latest | en | 0.951841 |
http://www.gradesaver.com/textbooks/math/prealgebra/prealgebra-7th-edition/chapter-5-cumulative-review-page-407/4 | 1,524,554,507,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946565.64/warc/CC-MAIN-20180424061343-20180424081343-00133.warc.gz | 408,300,159 | 13,727 | ## Prealgebra (7th Edition)
Use place value to state a number in words. $thousands\ \ tens$ $\ \ \ \ \ \overbrace5\underbrace0\overbrace2\underbrace6$ $\ \ \ \ \ \ \ \ hundreds\ \ \ ones$ | 63 | 188 | {"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} | 3.34375 | 3 | CC-MAIN-2018-17 | latest | en | 0.518646 |
https://couvreur-chaume.fr/grind/Jan-20/12020.html | 1,642,416,518,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300533.72/warc/CC-MAIN-20220117091246-20220117121246-00710.warc.gz | 217,157,334 | 6,309 | hammer mill sizing calculation pdf
# hammer mill sizing calculation pdf
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http://isabelle.in.tum.de/repos/isabelle/file/tip/src/ZF/Sum.thy | 1,606,995,076,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141727627.70/warc/CC-MAIN-20201203094119-20201203124119-00704.warc.gz | 48,072,861 | 4,739 | src/ZF/Sum.thy
author nipkow Wed, 02 Dec 2020 10:30:59 +0100 changeset 72803 83c6d29a2412 parent 69593 3dda49e08b9d permissions -rw-r--r--
merged
```
(* Title: ZF/Sum.thy
Author: Lawrence C Paulson, Cambridge University Computer Laboratory
*)
section\<open>Disjoint Sums\<close>
theory Sum imports Bool equalities begin
text\<open>And the "Part" primitive for simultaneous recursive type definitions\<close>
definition sum :: "[i,i]=>i" (infixr \<open>+\<close> 65) where
"A+B == {0}*A \<union> {1}*B"
definition Inl :: "i=>i" where
"Inl(a) == <0,a>"
definition Inr :: "i=>i" where
"Inr(b) == <1,b>"
definition "case" :: "[i=>i, i=>i, i]=>i" where
"case(c,d) == (%<y,z>. cond(y, d(z), c(z)))"
(*operator for selecting out the various summands*)
definition Part :: "[i,i=>i] => i" where
"Part(A,h) == {x \<in> A. \<exists>z. x = h(z)}"
subsection\<open>Rules for the \<^term>\<open>Part\<close> Primitive\<close>
lemma Part_iff:
"a \<in> Part(A,h) \<longleftrightarrow> a \<in> A & (\<exists>y. a=h(y))"
apply (unfold Part_def)
apply (rule separation)
done
lemma Part_eqI [intro]:
"[| a \<in> A; a=h(b) |] ==> a \<in> Part(A,h)"
by (unfold Part_def, blast)
lemmas PartI = refl [THEN [2] Part_eqI]
lemma PartE [elim!]:
"[| a \<in> Part(A,h); !!z. [| a \<in> A; a=h(z) |] ==> P
|] ==> P"
apply (unfold Part_def, blast)
done
lemma Part_subset: "Part(A,h) \<subseteq> A"
apply (unfold Part_def)
apply (rule Collect_subset)
done
subsection\<open>Rules for Disjoint Sums\<close>
lemmas sum_defs = sum_def Inl_def Inr_def case_def
lemma Sigma_bool: "Sigma(bool,C) = C(0) + C(1)"
by (unfold bool_def sum_def, blast)
(** Introduction rules for the injections **)
lemma InlI [intro!,simp,TC]: "a \<in> A ==> Inl(a) \<in> A+B"
by (unfold sum_defs, blast)
lemma InrI [intro!,simp,TC]: "b \<in> B ==> Inr(b) \<in> A+B"
by (unfold sum_defs, blast)
(** Elimination rules **)
lemma sumE [elim!]:
"[| u \<in> A+B;
!!x. [| x \<in> A; u=Inl(x) |] ==> P;
!!y. [| y \<in> B; u=Inr(y) |] ==> P
|] ==> P"
by (unfold sum_defs, blast)
(** Injection and freeness equivalences, for rewriting **)
lemma Inl_iff [iff]: "Inl(a)=Inl(b) \<longleftrightarrow> a=b"
lemma Inr_iff [iff]: "Inr(a)=Inr(b) \<longleftrightarrow> a=b"
lemma Inl_Inr_iff [simp]: "Inl(a)=Inr(b) \<longleftrightarrow> False"
lemma Inr_Inl_iff [simp]: "Inr(b)=Inl(a) \<longleftrightarrow> False"
lemma sum_empty [simp]: "0+0 = 0"
(*Injection and freeness rules*)
lemmas Inl_inject = Inl_iff [THEN iffD1]
lemmas Inr_inject = Inr_iff [THEN iffD1]
lemmas Inl_neq_Inr = Inl_Inr_iff [THEN iffD1, THEN FalseE, elim!]
lemmas Inr_neq_Inl = Inr_Inl_iff [THEN iffD1, THEN FalseE, elim!]
lemma InlD: "Inl(a): A+B ==> a \<in> A"
by blast
lemma InrD: "Inr(b): A+B ==> b \<in> B"
by blast
lemma sum_iff: "u \<in> A+B \<longleftrightarrow> (\<exists>x. x \<in> A & u=Inl(x)) | (\<exists>y. y \<in> B & u=Inr(y))"
by blast
lemma Inl_in_sum_iff [simp]: "(Inl(x) \<in> A+B) \<longleftrightarrow> (x \<in> A)"
by auto
lemma Inr_in_sum_iff [simp]: "(Inr(y) \<in> A+B) \<longleftrightarrow> (y \<in> B)"
by auto
lemma sum_subset_iff: "A+B \<subseteq> C+D \<longleftrightarrow> A<=C & B<=D"
by blast
lemma sum_equal_iff: "A+B = C+D \<longleftrightarrow> A=C & B=D"
by (simp add: extension sum_subset_iff, blast)
lemma sum_eq_2_times: "A+A = 2*A"
subsection\<open>The Eliminator: \<^term>\<open>case\<close>\<close>
lemma case_Inl [simp]: "case(c, d, Inl(a)) = c(a)"
lemma case_Inr [simp]: "case(c, d, Inr(b)) = d(b)"
lemma case_type [TC]:
"[| u \<in> A+B;
!!x. x \<in> A ==> c(x): C(Inl(x));
!!y. y \<in> B ==> d(y): C(Inr(y))
|] ==> case(c,d,u) \<in> C(u)"
by auto
lemma expand_case: "u \<in> A+B ==>
R(case(c,d,u)) \<longleftrightarrow>
((\<forall>x\<in>A. u = Inl(x) \<longrightarrow> R(c(x))) &
(\<forall>y\<in>B. u = Inr(y) \<longrightarrow> R(d(y))))"
by auto
lemma case_cong:
"[| z \<in> A+B;
!!x. x \<in> A ==> c(x)=c'(x);
!!y. y \<in> B ==> d(y)=d'(y)
|] ==> case(c,d,z) = case(c',d',z)"
by auto
lemma case_case: "z \<in> A+B ==>
case(c, d, case(%x. Inl(c'(x)), %y. Inr(d'(y)), z)) =
case(%x. c(c'(x)), %y. d(d'(y)), z)"
by auto
subsection\<open>More Rules for \<^term>\<open>Part(A,h)\<close>\<close>
lemma Part_mono: "A<=B ==> Part(A,h)<=Part(B,h)"
by blast
lemma Part_Collect: "Part(Collect(A,P), h) = Collect(Part(A,h), P)"
by blast
lemmas Part_CollectE =
Part_Collect [THEN equalityD1, THEN subsetD, THEN CollectE]
lemma Part_Inl: "Part(A+B,Inl) = {Inl(x). x \<in> A}"
by blast
lemma Part_Inr: "Part(A+B,Inr) = {Inr(y). y \<in> B}"
by blast
lemma PartD1: "a \<in> Part(A,h) ==> a \<in> A" | 1,650 | 4,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-50 | latest | en | 0.625945 |
http://www.statmodel.com/examples/shortform/run4.out | 1,469,646,796,000,000,000 | text/plain | crawl-data/CC-MAIN-2016-30/segments/1469257827077.13/warc/CC-MAIN-20160723071027-00027-ip-10-185-27-174.ec2.internal.warc.gz | 731,938,122 | 4,483 | Mplus DEVELOPMENT (Dev 10/23/2011) MUTHEN & MUTHEN 10/23/2011 2:50 PM INPUT INSTRUCTIONS title: Simulating x-m interaction effect on y using a random slope Step 2 Binary Y, continuous M data: file = n200xmreplist.dat; type = montecarlo; variable: names = y m x; usev = y m x xm; categorical = y; define: xm = x*m; analysis: estimator = ml; link = probit; model: [y\$1*.5] (mbeta0); y on x*.3 (beta2); y on m*.7 (beta1); y on xm*.2 (beta3); [m*.5] (gamma0); m on x*.5 (gamma1); m*.75 (sig2); model constraint: new(ind*.45 dir*.4 arg11*.7 arg10*.25 arg01*.2 arg00*-.15 v1*1.6075 v0*1.3675 probit11*.5521 probit10*.1972 probit01*.17103 probit00*-.1283 tie*.131 de*.129 pie*.119); dir=beta3*gamma0+beta2; ind=beta1*gamma1+beta3*gamma1; arg11=-mbeta0+beta2+(beta1+beta3)*(gamma0+gamma1); arg10=-mbeta0+beta2+(beta1+beta3)*gamma0; arg01=-mbeta0+beta1*(gamma0+gamma1); arg00=-mbeta0+beta1*gamma0; v1=(beta1+beta3)^2*sig2+1; v0=beta1^2*sig2+1; probit11=arg11/sqrt(v1); probit10=arg10/sqrt(v1); probit01=arg01/sqrt(v0); probit00=arg00/sqrt(v0); ! Phi function needed below: tie=phi(probit11)-phi(probit10); de=phi(probit10)-phi(probit00); pie=phi(probit01)-phi(probit00); INPUT READING TERMINATED NORMALLY Simulating x-m interaction effect on y using a random slope Step 2 Binary Y, continuous M SUMMARY OF ANALYSIS Number of groups 1 Average number of observations 200 Number of replications Requested 500 Completed 500 Number of dependent variables 2 Number of independent variables 2 Number of continuous latent variables 0 Observed dependent variables Continuous M Binary and ordered categorical (ordinal) Y Observed independent variables X XM Estimator ML Information matrix OBSERVED Optimization Specifications for the Quasi-Newton Algorithm for Continuous Outcomes Maximum number of iterations 100 Convergence criterion 0.100D-05 Optimization Specifications for the EM Algorithm Maximum number of iterations 500 Convergence criteria Loglikelihood change 0.100D-02 Relative loglikelihood change 0.100D-05 Derivative 0.100D-02 Optimization Specifications for the M step of the EM Algorithm for Categorical Latent variables Number of M step iterations 1 M step convergence criterion 0.100D-02 Basis for M step termination ITERATION Optimization Specifications for the M step of the EM Algorithm for Censored, Binary or Ordered Categorical (Ordinal), Unordered Categorical (Nominal) and Count Outcomes Number of M step iterations 1 M step convergence criterion 0.100D-02 Basis for M step termination ITERATION Maximum value for logit thresholds 10 Minimum value for logit thresholds -10 Minimum expected cell size for chi-square 0.100D-01 Optimization algorithm EMA Integration Specifications Type STANDARD Number of integration points 15 Dimensions of numerical integration 0 Adaptive quadrature ON Link PROBIT Cholesky ON Input data file(s) Multiple data files from n200xmreplist.dat Input data format FREE UNIVARIATE PROPORTIONS FOR CATEGORICAL VARIABLES NOTE: These are average results over 500 data sets. Y Category 1 0.425 Category 2 0.575 SAMPLE STATISTICS NOTE: These are average results over 500 data sets. SAMPLE STATISTICS Means M X XM ________ ________ ________ 1 0.747 0.497 0.493 Covariances M X XM ________ ________ ________ M 0.814 X 0.122 0.250 XM 0.499 0.248 0.624 Correlations M X XM ________ ________ ________ M 1.000 X 0.271 1.000 XM 0.700 0.629 1.000 MODEL FIT INFORMATION Number of Free Parameters 7 Loglikelihood H0 Value Mean -359.782 Std Dev 11.659 Number of successful computations 500 Proportions Percentiles Expected Observed Expected Observed 0.990 0.990 -386.904 -387.909 0.980 0.984 -383.726 -382.548 0.950 0.950 -378.960 -378.999 0.900 0.904 -374.724 -374.707 0.800 0.818 -369.594 -369.059 0.700 0.688 -365.896 -366.295 0.500 0.488 -359.782 -360.239 0.300 0.296 -353.668 -353.937 0.200 0.206 -349.970 -349.725 0.100 0.108 -344.840 -344.599 0.050 0.058 -340.604 -339.993 0.020 0.022 -335.838 -335.451 0.010 0.012 -332.660 -332.551 Information Criteria Akaike (AIC) Mean 733.564 Std Dev 23.318 Number of successful computations 500 Proportions Percentiles Expected Observed Expected Observed 0.990 0.988 679.319 677.590 0.980 0.978 685.676 683.355 0.950 0.942 695.208 693.858 0.900 0.892 703.679 703.035 0.800 0.794 713.939 713.345 0.700 0.704 721.336 721.841 0.500 0.512 733.564 734.318 0.300 0.312 745.791 746.516 0.200 0.182 753.188 752.065 0.100 0.096 763.448 762.380 0.050 0.050 771.919 771.481 0.020 0.016 781.452 778.832 0.010 0.010 787.808 785.619 Bayesian (BIC) Mean 756.652 Std Dev 23.318 Number of successful computations 500 Proportions Percentiles Expected Observed Expected Observed 0.990 0.988 702.407 700.678 0.980 0.978 708.764 706.443 0.950 0.942 718.296 716.946 0.900 0.892 726.768 726.123 0.800 0.794 737.027 736.433 0.700 0.704 744.424 744.929 0.500 0.512 756.652 757.407 0.300 0.312 768.880 769.604 0.200 0.182 776.276 775.153 0.100 0.096 786.536 785.468 0.050 0.050 795.007 794.569 0.020 0.016 804.540 801.920 0.010 0.010 810.896 808.707 Sample-Size Adjusted BIC (n* = (n + 2) / 24) Mean 734.475 Std Dev 23.318 Number of successful computations 500 Proportions Percentiles Expected Observed Expected Observed 0.990 0.988 680.231 678.501 0.980 0.978 686.587 684.267 0.950 0.942 696.119 694.770 0.900 0.892 704.591 703.946 0.800 0.794 714.851 714.256 0.700 0.704 722.247 722.753 0.500 0.512 734.475 735.230 0.300 0.312 746.703 747.427 0.200 0.182 754.099 752.976 0.100 0.096 764.359 763.291 0.050 0.050 772.831 772.392 0.020 0.016 782.363 779.744 0.010 0.010 788.720 786.530 MODEL RESULTS ESTIMATES S. E. M. S. E. 95% % Sig Population Average Std. Dev. Average Cover Coeff Y ON X 0.300 0.2740 0.2796 0.2770 0.0787 0.952 0.194 M 0.700 0.7138 0.1848 0.1799 0.0343 0.956 0.990 XM 0.200 0.2370 0.2865 0.2842 0.0833 0.954 0.110 M ON X 0.500 0.4894 0.1207 0.1223 0.0146 0.942 0.972 Intercepts M 0.500 0.5044 0.0863 0.0861 0.0074 0.970 1.000 Thresholds Y\$1 0.500 0.5058 0.1670 0.1672 0.0279 0.952 0.880 Residual Variances M 0.750 0.7465 0.0808 0.0746 0.0065 0.920 1.000 New/Additional Parameters IND 0.450 0.4661 0.1621 0.1600 0.0265 0.948 0.950 DIR 0.400 0.3935 0.2140 0.2122 0.0458 0.952 0.462 ARG11 0.700 0.7134 0.1858 0.1819 0.0346 0.962 0.992 ARG10 0.250 0.2473 0.1846 0.1807 0.0340 0.950 0.304 ARG01 0.200 0.2037 0.1788 0.1699 0.0319 0.942 0.218 ARG00 -0.150 -0.1462 0.1546 0.1489 0.0239 0.948 0.188 V1 1.607 1.7107 0.3486 0.3263 0.1319 0.948 1.000 V0 1.367 1.4057 0.2238 0.1998 0.0515 0.942 1.000 PROBIT11 0.552 0.5484 0.1317 0.1327 0.0173 0.952 0.992 PROBIT10 0.197 0.1948 0.1468 0.1442 0.0215 0.946 0.260 PROBIT01 0.171 0.1678 0.1437 0.1383 0.0206 0.942 0.240 PROBIT00 -0.128 -0.1244 0.1315 0.1256 0.0173 0.952 0.190 TIE 0.131 0.1303 0.0391 0.0388 0.0015 0.946 0.958 DE 0.129 0.1255 0.0689 0.0676 0.0047 0.952 0.450 PIE 0.119 0.1151 0.0358 0.0362 0.0013 0.936 0.950 QUALITY OF NUMERICAL RESULTS Average Condition Number for the Information Matrix 0.150E-02 (ratio of smallest to largest eigenvalue) TECHNICAL 1 OUTPUT PARAMETER SPECIFICATION TAU Y\$1 ________ 1 7 NU Y M X XM ________ ________ ________ ________ 1 0 0 0 0 LAMBDA Y M X XM ________ ________ ________ ________ Y 0 0 0 0 M 0 0 0 0 X 0 0 0 0 XM 0 0 0 0 THETA Y M X XM ________ ________ ________ ________ Y 0 M 0 0 X 0 0 0 XM 0 0 0 0 ALPHA Y M X XM ________ ________ ________ ________ 1 0 1 0 0 BETA Y M X XM ________ ________ ________ ________ Y 0 2 3 4 M 0 0 5 0 X 0 0 0 0 XM 0 0 0 0 PSI Y M X XM ________ ________ ________ ________ Y 0 M 0 6 X 0 0 0 XM 0 0 0 0 PARAMETER SPECIFICATION FOR THE ADDITIONAL PARAMETERS NEW/ADDITIONAL PARAMETERS IND DIR ARG11 ARG10 ARG01 ________ ________ ________ ________ ________ 1 8 9 10 11 12 NEW/ADDITIONAL PARAMETERS ARG00 V1 V0 PROBIT11 PROBIT10 ________ ________ ________ ________ ________ 1 13 14 15 16 17 NEW/ADDITIONAL PARAMETERS PROBIT01 PROBIT00 TIE DE PIE ________ ________ ________ ________ ________ 1 18 19 20 21 22 STARTING VALUES TAU Y\$1 ________ 1 0.500 NU Y M X XM ________ ________ ________ ________ 1 0.000 0.000 0.000 0.000 LAMBDA Y M X XM ________ ________ ________ ________ Y 1.000 0.000 0.000 0.000 M 0.000 1.000 0.000 0.000 X 0.000 0.000 1.000 0.000 XM 0.000 0.000 0.000 1.000 THETA Y M X XM ________ ________ ________ ________ Y 0.000 M 0.000 0.000 X 0.000 0.000 0.000 XM 0.000 0.000 0.000 0.000 ALPHA Y M X XM ________ ________ ________ ________ 1 0.000 0.500 0.000 0.000 BETA Y M X XM ________ ________ ________ ________ Y 0.000 0.700 0.300 0.200 M 0.000 0.000 0.500 0.000 X 0.000 0.000 0.000 0.000 XM 0.000 0.000 0.000 0.000 PSI Y M X XM ________ ________ ________ ________ Y 1.000 M 0.000 0.750 X 0.000 0.000 0.500 XM 0.000 0.000 0.000 0.500 STARTING VALUES FOR THE ADDITIONAL PARAMETERS NEW/ADDITIONAL PARAMETERS IND DIR ARG11 ARG10 ARG01 ________ ________ ________ ________ ________ 1 0.450 0.400 0.700 0.250 0.200 NEW/ADDITIONAL PARAMETERS ARG00 V1 V0 PROBIT11 PROBIT10 ________ ________ ________ ________ ________ 1 -0.150 1.607 1.367 0.552 0.197 NEW/ADDITIONAL PARAMETERS PROBIT01 PROBIT00 TIE DE PIE ________ ________ ________ ________ ________ 1 0.171 -0.128 0.131 0.129 0.119 Beginning Time: 14:50:11 Ending Time: 14:50:21 Elapsed Time: 00:00:10 MUTHEN & MUTHEN 3463 Stoner Ave. Los Angeles, CA 90066 Tel: (310) 391-9971 Fax: (310) 391-8971 Web: www.StatModel.com Support: Support@StatModel.com Copyright (c) 1998-2011 Muthen & Muthen | 3,766 | 9,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.445657 |
http://betterlesson.com/lesson/resource/2732186/76483/online-stopwatch-tutorial-mp4 | 1,488,014,138,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171706.94/warc/CC-MAIN-20170219104611-00240-ip-10-171-10-108.ec2.internal.warc.gz | 27,882,524 | 21,673 | ## Online stopwatch tutorial.mp4 - Section 1: Do Now
Online stopwatch tutorial.mp4
# Bar Models and Ratios
Unit 5: Ratios and Proportional Relationships
Lesson 6 of 21
## Big Idea: Students work in partnered pairs to use bar models to solve ratio word problems
Print Lesson
Standards:
Subject(s):
Math, proportions, Number Sense and Operations, rational numbers, bar models, ratios, multiplicative comparisons
55 minutes
### Yazmin Chavira MTP
##### Similar Lessons
###### Proportional Relationships With Decimals
7th Grade Math » Proportional Relationships
Big Idea: Students expand and solve ratio problems involving decimals using unit rates and a double number line.
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New Orleans, LA
Environment: Urban
###### Balancing Act
7th Grade Science » Energy, Force & Motion
Big Idea: Can objects of different mass be arranged so they balance one another? Is there a mathematical equation that can predict balance?
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Hope, IN
Environment: Rural | 225 | 1,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-09 | latest | en | 0.75779 |
http://cexams.in/quantitative-aptitude-test-57/ | 1,582,938,625,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148163.71/warc/CC-MAIN-20200228231614-20200229021614-00529.warc.gz | 32,697,939 | 11,097 | # Quantitative Aptitude Test 57
Quantitative Aptitude
1. Consider a triangle. Its longest side has length 20 and another of its sides has length 10. Its area is 80. What is the exact length of its third side?
2. A train X departs from station A at 11.00 a.m. for station B, which is 180 km away. Another train Y departs from station B at the same time. Train X travels at an average speed of 70 km/hr and does not stop anywhere until it arrives at station B. Train Y travels at an average speed of 50 kmhr, but has to stop for 15 minutes at station C, which is 60 km away from station B enroute to station A. At what distance from A would they meet?
3. Every ten years the Indian government counts all the people living in the country. Suppose that the director of the census has reported the following data on two neighbouring villages Chota Hazri and Mota Hazri:
Chota Hazri has 4,522 fewer males than Mota Hazri
Mota Hazri has 4,020 more females than males.
Chota Hazri has twice as many females as males.
Chota Hazri has 2,910 fewer females than Mota Hazri.
What is the total number of males in Chota Hazri?
4. Three friends, returning from a movie, stopped to eat at a restaurant. After dinner, they paid their bill and noticed of mints at the front counter. Sita took 1/3 of the mints, but returned four because she had a momentary pang of guilt. Fatima then took 1/4 of what was left but returned three for similar reasons. Eswari then took half of the remainder but threw two back into the bowl. The bowl had only 17 mints left when the raid was over. How many mints were originally in the bowl?
5. Manisha would like to minimise the fuel consumption for the trip by driving at the appropriate speed. How should she change the speed?
6. If x > 5 and y < – 1, then which of the following statements is true?
7. Three runners A, B and C run a race, with runner A finishing 12 meters ahead of runner B and 18 meters ahead of runner C, while runner B finishes 8 meters ahead of runner C. Each runner travels the entire distance at a constant speed. What was the length of the race?
8. The figure below shows the network connecting cities, A, B, C, D, E and F. The arrows indicate permissible direction of travel. What is the number of distinct paths from A to F?
9. Raju has 128 boxes with him. He has to put atleast 120 oranges in one box and 144 at the most. Find the least number of boxes which will have the same number of oranges.
10. The tax on a commodity is diminished by 25% and its consumption increases by 20%. Now, a person can save what per cent more/less from before?
Question 1 of 10 | 649 | 2,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.979228 |
https://www.jiskha.com/display.cgi?id=1227463683 | 1,503,462,928,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117519.92/warc/CC-MAIN-20170823035753-20170823055753-00621.warc.gz | 930,834,851 | 4,027 | # physics
posted by .
why does a car nose up when accelerating and nose down when braking?
which will have the greater acceleration rolling down an incline- a bowling ball or a volleyball?
• physics -
The bowling ball and the volleyball will have the same acceleration.
I'm not exactly sure about the car thing, but it's probably because the front meets the most air resistance, pushing it down while accelerating, and when it brakes, the air resistance in the front is reduced while inertia keeps the car's end flying forward... the reduced air resistance and forward motion of the car's end lifts the front. I don't know a more technical explanation, if my original is true!
I hope that helps, at least.
• physics -
1)consider the the torque on the car caused by the rear wheels. Remember Newtons third law?
2) The volleyball will get to the bottom fastest.
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https://journals.kantiana.ru/eng/vestnik/2038/ | 1,713,050,291,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00417.warc.gz | 300,733,744 | 14,664 | Physics, mathematics, and technology
# Physics, mathematics, and technology
### Mathematical Modeling
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An application of dimension reduction methods in the analysis of the phenomena of socio-economic systems by analyzing the choice of direction of development of socio-economic development is considered. In particular it is shown that the use of algorithms dimension reduction methods the probability of error when estimating posterior probabilities of belonging to the class of homogeneous objects, obtained in the result of solving the problem of discriminant analysis
#### Joint modeling of elastic and acoustic waves in the Arctic
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Numerical solutions of seismic prospecting problems in the Arctic is presented. Results of numerical modeling of the explosive impact on the iceberg and earthquake impact on the oil storage are considered as well. The complete system of equations describing the state elastic body and a system of equations describing the acoustic field are solved using grid-characteristic method, the contact condition between the liquid and the solid is used on the border
### Differential geometry
#### Strongly mutual threefold distributions of projective space
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Construction of a general theory of a special class ( -distribution) of the regular threefold distributions ( -distribution [1]) of the projective space consisting of a basic distribution of the 1st kind of r-dimensional planes are equipped with the distribution of the 1st kind of m-dimensional planes and equip distribution 1st the first kind of hyperplane elements (hyperplanes) with the ratio of the incidence of the corresponding elements in the common center is considered in this article. In this paper, these three distributions is considered as a immersed manifold. By virtue of the -distribution structure in the geometry of the manifold are similar to some of the facts from the geometry of m-dimensional linear elements [2], (n-1)-dimensional linear elements [3] and hyperband distribution [4]. However, the analogy does not relate to the geometry of the base only or equipping distributions taken separately.
#### About horizontally vector lifts of tensor fields from the base to its tangent bundle
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The vector field of the type , obtained fr om the tensor field on a smooth manifold M to the tangent bundle , arises in the study of infinitesimal affine transformations with a complete lift. The H-lift was introduced S. Tanno in 1974 in infinitesimal isometries on the tangent bundle with the complete lift. The definition of H’-lift given by the author in this paper. There are established the properties of the entered lift and founded the commutator , wh ere .
### Artificial Intelligence
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### Theoretical and experimental physics
#### iltering of signals from noise by maximum likelihood method
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A method of filtering radar pulse on the background of the focus obtained by frequency interference. The basis of the method are the principles of the theory of optimal reception and position of the linear space of signals. Derived the basic expressions when filtering of the signal strength in the area of orthogonality of signal and noise, and their non-orthogonality. Equations for impulse and hour-tonyh characteristics of filters.
#### Modeling studies of filtering process of the radio pulse from noise
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### Informatics and mathematical geophysics
#### Algorithms and software for building three-dimensional models of microseismic monitoring data
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Algorithms for constructing three-dimensional models of microseismic activity fields are described. The first approach is based on voxel models and involves the construction of polygonal surfaces of interpolated parameters of microseismic events. The second approach uses a triangulation algorithm based on the Delaunay criterion and α-shapes method. Software developed on the basis of the proposed algorithms, offers advanced graphical tools for the analysis and interpretation of microseismic multiparameter data
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Results of passive microseismic monitoring of water injection into one of the wells of the carbonate deposits are presented. The maps of the spatial distribution of sources have prepared and the analysis of the dynamic parameters of microseismic events together with graphs of pressure and injection volumes has done. The possibility of operational control of the process of flooding is shown on the example of Northern Truva.
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One of the problems of interpretation of seismic data is that they are often measured on an irregular grid with large distances between the sources and receivers, which affects the outcome of processing. The article discusses the possibility of using the procedure RTD (Reverse Time Datuming) [1-4] to ob-tain the wave field with a dense and regular geometry of sources and receivers at some depth. The results of numerical testing are presented.
#### Algorithms of determining of bodies in a 3D irregular point cloud
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A technique for recognition of shapes of bodies in a 3D point cloud is de-scribed. At the first step cloud clustering is conducted using the criterion of maximum distance between the points to identify the bodies being determined. Then proceeding to voxel presentation is being done. Existence of a tetrahe-dron with its points in cloud points and size-limited sides for which the voxel being tested is internal is treated as a criterion of belongment of voxel to a body. A fast algorithm for voxel tetrahedron filling is developed and used for optimization. The work is a part of software for micro-seismic monitoring data processing.
### Other papers
#### About properties of Cauchy-type integral with a fixed point in the kernel in С1 space
Abstract
The Cauchy-type integral with a fixed point in the kernel is in the as-sumption that fixed point is on the exterior of a circle of RADIUS one. Set fields non-analyticity and the analyticity of the integral. Area non-analyticity naturally splits into two subareas , each of which the integral is calculated ac-cording to a specific formula. Non-analytical functions, which have integral Cauchy type with a fixed point in the kernel satisfies a certain differential equations. Using the method of linear differential operators is link integrals with a fixed point in the kernel with integral Cauchy type. | 1,618 | 8,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.881038 |
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5127828b0bfa8064cc093c075334e900.ppt
• Количество слайдов: 50
IMPORTANT: In order to view the correct calculator key stroke symbols within this PPT, you will need to follow the font installation directions on this document. 7 -1 Mc. Graw-Hill/Irwin Copyright © 2011 by the Mc. Graw-Hill Companies, Inc. All rights reserved.
Key Concepts and Skills • Understand how stock prices depend on future dividends and dividend growth • Be able to compute stock prices using the dividend growth model • Understand how corporate directors are elected • Understand how stock markets work • Understand how stock prices are quoted 7 -2
Chapter Outline 7. 1 Common Stock Valuation 7. 2 Some Features of Common and Preferred Stocks 7. 3 The Stock Markets 7 -3
Cash Flows for Stockholders • If you own a share of stock, you can receive cash in two ways – The company pays dividends – You sell your shares, either to another investor in the market or back to the company • As with bonds, the price of the stock is the present value of these expected cash flows – Dividends → cash income – Selling → capital gains 7 -4
One Period Example • Suppose you are thinking of purchasing the stock of Moore Oil, Inc. – You expect it to pay a \$2 dividend in one year – You believe you can sell the stock for \$14 at that time. – You require a return of 20% on investments of this risk – What is the maximum you would be willing to pay? 7 -5
One Period Example • • • D 1 = \$2 dividend expected in one year R = 20% P 1 = \$14 CF 1 = \$2 + \$14 = \$16 Compute the PV of the expected cash flows 7 -6
Two Period Example • What if you decide to hold the stock for two years? – – D 1 = \$2. 00 CF 2 = \$2. 10 + \$14. 70 = \$16. 80 D 2 = \$2. 10 P 2 = \$14. 70 Now how much would you be willing to pay? 7 -7
Three Period Example • What if you decide to hold the stock for three years? – – – D 1 = \$2. 00 CF 1 = \$2. 00 D 2 = \$2. 10 CF 2 = \$2. 10 D 3 = \$2. 205 CF 3 = \$2. 205 + \$15. 435 = \$17. 640 P 3 = \$15. 435 Now how much would you be willing to pay? 7 -8
Three Period Example Using TI BAII+ Cash Flow Worksheet Display Cash Flows: CF 0 = 0 CF 1 = 2. 00 CF 2 = 2. 10 CF 3 = 17. 64 You Enter ‘' C 00 C 01 F 01 C 02 F 02 C 03 F 03 I NPV 0 2 1 2. 10 1 17. 64 1 20 % !# !# ( !# 13. 33 7 -9
Developing The Model • You could continue to push back when you would sell the stock • You would find that the price of the stock is really just the present value of all expected future dividends 7 -10
Stock Value = PV of Dividends ^ P 0 = D 1 (1+R)1 + D 2 (1+R)2 + D 3 (1+R)3 +…+ D∞ (1+R)∞ How can we estimate all future dividend payments? 7 -11
Estimating Dividends Special Cases • Constant dividend/Zero Growth – Firm will pay a constant dividend forever – Like preferred stock – Price is computed using the perpetuity formula • Constant dividend growth – Firm will increase the dividend by a constant percent every period • Supernormal growth – Dividend growth is not consistent initially, but settles down to constant growth eventually 7 -12
Zero Growth • Dividends expected at regular intervals forever = perpetuity P 0 = D / R • Suppose stock is expected to pay a \$0. 50 dividend every quarter and the required return is 10% with quarterly compounding. What is the price? 7 -13
Constant Growth Stock One whose dividends are expected to grow forever at a constant rate, g. D 1 = D 0(1+g)1 D 2 = D 0(1+g)2 Dt = Dt(1+g)t D 0 = Dividend JUST PAID D 1 – Dt = Expected dividends 7 -14
Projected Dividends • D 0 = \$2. 00 and constant g = 6% • D 1 = D 0(1+g) = 2(1. 06) • D 2 = D 1(1+g) = 2. 12(1. 06) • D 3 = D 2(1+g) = 2. 2472(1. 06) = \$2. 12 = \$2. 2472 = \$2. 3820 7 -15
Dividend Growth Model ^ D 0(1+g) P 0 = R-g D 1 = R-g “Gordon Growth Model” 7 -16
DGM – Example 1 • Suppose Big D, Inc. just paid a dividend of \$. 50. It is expected to increase its dividend by 2% per year. If the market requires a return of 15% on assets of this risk, how much should the stock be selling for? • D 0= \$0. 50 • g = 2% • R = 15% 7 -17
DGM – Example 2 • Suppose TB Pirates, Inc. is expected to pay a \$2 dividend in one year. If the dividend is expected to grow at 5% per year and the required return is 20%, what is the price? – D 1 = \$2. 00 – g = 5% – r = 20% 7 -18
Stock Price Sensitivity to Dividend Growth, g D 1 = \$2; R = 20% 7 -19
Stock Price Sensitivity to Required Return, R D 1 = \$2; g = 5% 7 -20
Example 7. 3 Gordon Growth Company - I • Gordon Growth Company is expected to pay a dividend of \$4 next period and dividends are expected to grow at 6% per year. The required return is 16%. • What is the current price? 7 -21
Example 7. 3 Gordon Growth Company - II • What is the price expected to be in year 4? 7 -22
Example 7. 3 Gordon Growth Company - II • What is the implied return given the change in price during the four year period? 50. 50 = 40(1+return)4; return = 6% 4 , ; 40 S. ; 50. 50 0; 0 /; %- = 6% v. The price grows at the same rate as dividends 7 -23
Constant Growth Model Conditions 1. Dividend expected to grow at g forever 2. Stock price expected to grow at g forever 3. Expected dividend yield is constant 4. Expected capital gains yield is constant and equal to g 5. Expected total return, R, must be > g 6. Expected total return (R): = expected dividend yield (DY) + expected growth rate (g) = dividend yield + g 7 -24
Nonconstant Growth • Suppose a firm is expected to increase dividends by 20% in one year and by 15% in two years. After that dividends will increase at a rate of 5% per year indefinitely. If the last dividend was \$1 and the required return is 20%, what is the price of the stock? • Remember that we have to find the PV of all expected future dividends. 7 -25
Nonconstant Growth – Solution • Compute the dividends until growth levels off – D 1 = 1(1. 2) = \$1. 20 – D 2 = 1. 20(1. 15) = \$1. 38 – D 3 = 1. 38(1. 05) = \$1. 449 • Find the expected future price at the beginning of the constant growth period: – P 2 = D 3 / (R – g) = 1. 449 / (. 2 -. 05) = 9. 66 • Find the present value of the expected future cash flows – P 0 = 1. 20 / (1. 2) + (1. 38 + 9. 66) / (1. 2)2 = 8. 67 7 -26
Nonconstant + Constant growth Basic PV of all Future Dividends Formula Dividend Growth Model 7 -27
Nonconstant + Constant growth 7 -28
Nonconstant growth followed by constant growth: 0 rs=20% g = 20% D 0 = 1. 00 1 2 g = 15% 1. 20 3 g = 5% 1. 38 1. 449 1. 0000 0. 9583 ^ P 2 = 6. 7083 8. 6667 \$1. 449 = \$9. 66 0. 20 – 0. 05 = P 0 7 -29
Quick Quiz: Part 1 • What is the value of a stock that is expected to pay a constant dividend of \$2 per year if the required return is 15%? • What if the company starts increasing dividends by 3% per year beginning with the next dividend? The required return remains at 15%. 7 -30
Using the DGM to Find R Start with the DGM: Rearrange and solve for R: 7 -31
Finding the Required Return Example • A firm’s stock is selling for \$10. 50. They just paid a \$1 dividend and dividends are expected to grow at 5% per year. • What is the required return? 7 -32
Finding the Required Return Example • • P 0 = \$10. 50. D 0 = \$1 g = 5% per year. What is the required return? 7 -33
Finding the Required Return Example • • P 0 = \$10. 50 D 0 = \$1 g = 5% per year What is the dividend yield? 1(1. 05) / 10. 50 = 10% • What is the capital gains yield? g =5% Dividend Yield Capital Gains Yield 7 -34
Table 7. 1 7 -35
Features of Common Stock • Voting Rights – Stockholders elect directors – Cumulative voting vs. Straight voting – Proxy voting • Classes of stock – Founders’ shares – Class A and Class B shares Return to Quick Quiz 7 -36
Features of Common Stock • Other Rights – Share proportionally in declared dividends – Share proportionally in remaining assets during liquidation – Preemptive right • Right of first refusal to buy new stock issue to maintain proportional ownership if desired Return to Quick Quiz 7 -37
Dividend Characteristics • Dividends are not a liability of the firm until declared by the Board of Directors – A firm cannot go bankrupt for not declaring dividends • Dividends and Taxes – Dividends are not tax deductible for firm – Taxed as ordinary income for individuals – Dividends received by corporations have a minimum 70% exclusion from taxable income 7 -38
Features of Preferred Stock • Dividends – Must be paid before dividends can be paid to common stockholders – Not a liability of the firm – Can be deferred indefinitely – Most preferred dividends are cumulative • Missed preferred dividends have to be paid before common dividends can be paid • Preferred stock generally does not carry voting rights Return to Quick Quiz 7 -39
The Stock Markets • Primary vs. Secondary Markets – Primary = new-issue market – Secondary = existing shares traded among investors • Dealers vs. Brokers – Dealer: Maintains an inventor Ready to buy or sell at any time Think “Used car dealer” – Broker: Brings buyers and sellers together Think “Real estate broker” 7 -40
New York Stock Exchange (NYSE) • NYSE Euronext (merged 2007) • Members (Historically) – Buy a trading license (own a seat) – Commission brokers – Specialists – Super. DOT – Floor brokers – Floor traders 7 -41
NYSE Operations • Operational goal = attract order flow • NYSE Specialist: – Assigned broker/dealer • Each stock has one assigned specialist • All trading in that stock occurs at the “specialist’s post” – Trading takes place between customer orders placed with the specialists and “the crowd” – “Crowd” = commission and floor brokers and traders 7 -42
NASDAQ • • • NASDAQ OMX (merged 2007) Computer-based quotation system Multiple market makers Electronic Communications Networks Three levels of information – Level 1 – median quotes, registered representatives – Level 2 – view quotes, brokers & dealers – Level 3 – view and update quotes, dealers only • Large portion of technology stocks 7 -43
ECNs • Electronic Communications Networks provide direct trading among investors • Developed in late 1990 s • ECN orders transmitted to NASDAQ • Observe live trading online at Batstrading. com 7 -44
Reading Stock Quotes • What information is provided in the stock quote? • Click on the web surfer to go to Bloomberg for current stock quotes. 7 -45
Work the Web • Not only are stock price quotes readily available online. Some online trading sites display their “order book” or “limit order book” live online. • Batstrading is one of these. • Follow the link and see current buy and sell orders for Microsoft (MSFT). 7 -46
Quick Quiz: Part 2 • You observe a stock price of \$18. 75. You expect a dividend growth rate of 5% and the most recent dividend was \$1. 50. What is the required return? 7 -47
Quick Quiz: Part 2 • What are some of the major characteristics of common stock? (Slide 36 and Slide 37) • What are some of the major characteristics of preferred stock? (Slide 39) 7 -48
Chapter 7 END | 3,151 | 10,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | latest | en | 0.84806 |
https://www.electrical4u.com/sequence-generator/ | 1,714,043,102,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297292879.97/warc/CC-MAIN-20240425094819-20240425124819-00172.warc.gz | 653,845,839 | 28,430 | # Sequence Generator
We all know that there are counters which pass through a definite number of states in a pre-determined order. For example, a 3-bit up-counter counts from 0 to 7 while the same order is reversed in the case of 3-bit down counter. These circuits when suitably manipulated can be made to count till an intermediate level also. This means that instead of counting till 7, we can terminate the process by resetting the counter just at, say, 5. Such counters are then known as mod-N counters. However, even this case, the order in which they count will not alter. But, what-if if we need to through a specific pattern which does not adhere to this standard way of counting? The solution would be to design a sequence generator.
This is because, the sequence generators are nothing but a set of digital circuits which are designed to result in a specific bit sequence at their output. There are several ways in which these circuits can be designed including those which are based on multiplexers and flip-flops. Here in this article we deal with the designing of sequence generator using D flip-flops (please note that even JK flip-flops can be made use of).
As an example, let us consider that we intend to design a circuit which moves through the states 0-1-3-2 before repeating the same pattern. The steps involved during this process are as follows.
Step 1
At first, we need to determine the number of flip-flops which would be required to achieve our objective. In our example, there are 4 states which are identical to the states of a 2-bit counter except the order in which they transit. From this, we can guess the requirement of flip-flops to be 2 in order to achieve our objective.
Step 2
Having this in mind, let us now write the state transition table for our sequence generator. This shown by the first four columns of Table I in which the first two columns indicate the present states while the next two columns indicate the corresponding next states. For instance, first state in our example is 0 = “00” which leads to the next state 1 = “01” (as shown by the gray shaded row in Table I).
Step 3
Now this state transition table is to be extended so as to include the excitation table of the flip-flop with which we desire to design our circuit. In our case, it is nothing but D flip-flop due to which we have the fifth and the sixth columns of the table representing the excitation table of D flip-flop.
For example, look at the orange shaded row in Table I in which the present and the next states 1 and 0 (respectively) result in D1 to be 0. The same row also shows the case wherein
Table I
Present States Next States Inputs of D flip-flops Q1 Q0 Q1+ Q0+ D1 D0 0 0 0 1 0 1 0 1 1 1 1 1 1 1 1 0 1 0 1 0 0 0 0 0
Step 4
Now its time to derive the Boolean expressions for D1 and D0. This can be done using any kind of simplification technique including K-map. However as our example is quite simple, we can just use the Boolean laws to solve for D1 and D0. Thus
Step 5
Having known the inputs to either of the D flip-flops, now we can design our sequence generator as shown in this figure.
In the circuit shown, the desired sequence is generated based on the clock pulses supplied. At this point, it should be noted that, the analogy presented here for a simple design can be effectively extended to generate longer sequence of bits.
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No credit card required—it’s 100% free. | 822 | 3,493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-18 | latest | en | 0.954843 |
http://www.learn-math.top/computing-intersections-between-a-large-number-of-sets-doing-it-sequentially-vs-sending-everything-to-intersection/ | 1,537,558,312,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157503.43/warc/CC-MAIN-20180921190509-20180921210909-00444.warc.gz | 351,380,643 | 9,463 | # Computing intersections between a large number of sets: doing it sequentially vs. sending everything to Intersection[…]
Say we want to compute the largest common subset among a very large number of sets QiQ_i. One option would be to simply run the command:
Intersection[Q1, Q2, Q3, Q4, Q5, Q6, Q7, …];
However, this seems to choke for very large numbers of input sets (10410^4 or so with on-order 10210^2 elements each). Another option would be compute Q12 = Intersection[Q1,Q2], then Q123 = Intersection[Q12,Q3], then Q1234 = Intersection[Q123,Q4], and so forth.
Are these two procedures equivalent? Is there a better way?
=================
They are not equivalent. I think nesting them is a bad idea, because the function Intersection[] do support multiple list. In general I always try to use those function that support exactly what I need them to and no more, because it seems to me quicker that way.
– Coolwater
Apr 4 ’14 at 18:34
2
What you are asking for is Fold[Intersection, First@#, Rest@#] &@list but it is slower.
– Kuba
Apr 4 ’14 at 18:39
1
@Coolwater the final result should be equivalent though.
– Yves Klett
Apr 4 ’14 at 18:41
@Kuba Just to check, is the procedure you suggest the same as the one I’m talking about where the number of intersection operations required is exactly proportional to the length of the sets?
– CA30
Apr 4 ’14 at 18:44
@CA30 Take a look at Fold in documentation. It does exactly n-1 intersections.
– Kuba
Apr 4 ’14 at 18:46
=================
1
=================
As Kuba notes using Fold is the most canonical way to implement your “sequential” algorithm in Mathematica. Performance it best determined by simply Timing your operation.
SeedRandom[1]
sets = RandomInteger[200, {10000, 1600}];
Intersection @@ sets // Timing
Fold[Intersection, #, {##2}] & @@ sets // Timing
{0.983, {12, 90, 94, 101, 164}}
{1.045, {12, 90, 94, 101, 164}}
(Timings performed in Mathematica 7 under Windows 7.)
So we see that the sequential method is if anything slightly slower than the direct application of Intersection. I think it is probable that Intersection is already using this algorithm.
Regarding your second question: Is there a better way? There can be, depending on your data. In the example above there is a lot of duplication within each set, and eliminating this beforehand greatly speeds the operation of Intersection:
Intersection @@ (DeleteDuplicates /@ sets) // Timing
{0.172, {12, 90, 94, 101, 164}} | 639 | 2,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2018-39 | latest | en | 0.89563 |
https://search.r-project.org/CRAN/refmans/BMT/html/BMTfit.mqde.html | 1,627,817,993,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154175.76/warc/CC-MAIN-20210801092716-20210801122716-00247.warc.gz | 517,740,279 | 4,307 | BMTfit.mqde {BMT} R Documentation
## Minimum Quantile Distance Fit of the BMT Distribution to Non-censored Data.
### Description
Fit of the BMT distribution to non-censored data by minimum quantile distance (mqde), which can also be called maximum quantile goodness-of-fit.
### Usage
```BMTfit.mqde(data, probs = (1:length(data) - 0.5)/length(data), qtype = 5,
dist = "euclidean", start = list(p3 = 0.5, p4 = 0.5, p1 = min(data) - 0.1,
p2 = max(data) + 0.1), fix.arg = NULL, type.p.3.4 = "t w",
type.p.1.2 = "c-d", optim.method = "Nelder-Mead", custom.optim = NULL,
weights = NULL, silent = TRUE, ...)
```
### Arguments
`data` A numeric vector with the observed values for non-censored data. `probs` A numeric vector of the probabilities for which the minimum quantile distance estimation is done. p[k] = (k - 0.5) / n (default). `qtype` The quantile type used by the R `quantile` function to compute the empirical quantiles. Type 5 (default), i.e. x[k] is both the kth order statistic and the type 5 sample quantile of p[k] = (k - 0.5) / n. `dist` The distance measure between observed and theoretical quantiles to be used. This must be one of "euclidean" (default), "maximum", or "manhattan". Any unambiguous substring can be given. `start` A named list giving the initial values of parameters of the BMT distribution or a function of data computing initial values and returning a named list. (see the 'details' section of `mledist`). `fix.arg` An optional named list giving the values of fixed parameters of the BMT distribution or a function of data computing (fixed) parameter values and returning a named list. Parameters with fixed value are thus NOT estimated. (see the 'details' section of `mledist`). `type.p.3.4` Type of parametrization asociated to p3 and p4. "t w" means tails weights parametrization (default) and "a-s" means asymmetry-steepness parametrization. `type.p.1.2` Type of parametrization asociated to p1 and p2. "c-d" means domain parametrization (default) and "l-s" means location-scale parametrization. `optim.method` `"default"` (see the 'details' section of `mledist`) or optimization method to pass to `optim`. Given the close-form expression of the quantile function, two optimization methods were added when the euclidean distance is selected: Coordinate descend (`"CD"`) and Newton-Rhapson (`"NR"`). `custom.optim` A function carrying the optimization (see the 'details' section of `mledist`). `weights` an optional vector of weights to be used in the fitting process. Should be `NULL` or a numeric vector with strictly positive numbers. If non-`NULL`, weighted mqde is used, otherwise ordinary mqde. `silent` A logical to remove or show warnings when bootstraping. `...` Further arguments to be passed to generic functions or to the function `"mqdedist"`. See `mqdedist` for details.
### Details
This function is not intended to be called directly but is internally called in `BMTfit` when used with the minimum quantile distance method.
`BMTfit.mqde` is based on the function `mqdedist` but it focuses on the minimum quantile distance parameter estimation for the BMT distribution (see `BMT` for details about the BMT distribution and `mqdedist` for details about minimum quantile distance fit of univariate distributions).
Given the close-form expression of the quantile function, two optimization methods were added when the euclidean distance is selected: Coordinate descend (`"CD"`) and Newton-Rhapson (`"NR"`).
### Value
`BMTfit.mqde` returns a list with following components,
`estimate` the parameter estimates. `convergence` an integer code for the convergence of `optim`/`constrOptim` defined as below or defined by the user in the user-supplied optimization function. `0` indicates successful convergence. `1` indicates that the iteration limit of `optim` has been reached. `10` indicates degeneracy of the Nealder-Mead simplex. `100` indicates that `optim` encountered an internal error. `value` the value of the corresponding objective function of the estimation method at the estimate. `hessian` a symmetric matrix computed by `optim` as an estimate of the Hessian at the solution found or computed in the user-supplied optimization function. `loglik` the log-likelihood value. `probs` the probability vector on which observed and theoretical quantiles were calculated. `dist` the name of the distance between observed and theoretical quantiles used. `optim.function` the name of the optimization function used for maximum product of spacing. `optim.method` when `optim` is used, the name of the algorithm used, `NULL` otherwise. `fix.arg` the named list giving the values of parameters of the named distribution that must kept fixed rather than estimated or `NULL` if there are no such parameters. `fix.arg.fun` the function used to set the value of `fix.arg` or `NULL`. `weights` the vector of weigths used in the estimation process or `NULL`. `counts` A two-element integer vector giving the number of calls to the log-likelihood function and its gradient respectively. This excludes those calls needed to compute the Hessian, if requested, and any calls to log-likelihood function to compute a finite-difference approximation to the gradient. `counts` is returned by `optim` or the user-supplied function or set to `NULL`. `optim.message` A character string giving any additional information returned by the optimizer, or `NULL`. To understand exactly the message, see the source code.
### Author(s)
Camilo Jose Torres-Jimenez [aut,cre] cjtorresj@unal.edu.co
### Source
Based on the function `mqdedist` which in turn is based on the function `mledist` of the R package: `fitdistrplus`
Delignette-Muller ML and Dutang C (2015), fitdistrplus: An R Package for Fitting Distributions. Journal of Statistical Software, 64(4), 1-34.
### References
Torres-Jimenez, C. J. (2017, September), Comparison of estimation methods for the BMT distribution. ArXiv e-prints.
Torres-Jimenez, C. J. (2018), The BMT Item Response Theory model: A new skewed distribution family with bounded domain and an IRT model based on it, PhD thesis, Doctorado en ciencias - Estadistica, Universidad Nacional de Colombia, Sede Bogota.
See `BMT` for the BMT density, distribution, quantile function and random deviates. See `BMTfit.mme`, `BMTfit.mle`, `BMTfit.mge`, `BMTfit.mpse` and `BMTfit.qme` for other estimation methods. See `optim` and `constrOptim` for optimization routines. See `BMTfit` and `fitdist` for functions that return an objetc of class `"fitdist"`.
### Examples
```# (1) basic fit by minimum quantile distance estimation
set.seed(1234)
x1 <- rBMT(n=100, p3=0.25, p4=0.75)
BMTfit.mqde(x1)
# (2) quantile matching is a particular case of minimum quantile distance
BMTfit.mqde(x1, probs=c(0.2,0.4,0.6,0.8), qtype=7)
# (3) maximum or manhattan instead of euclidean distance
BMTfit.mqde(x1, dist="maximum")
BMTfit.mqde(x1, dist="manhattan")
# (4) how to change the optimisation method?
BMTfit.mqde(x1, optim.method="L-BFGS-B")
BMTfit.mqde(x1, custom.optim="nlminb")
# (5) estimation of the tails weights parameters of the BMT
# distribution with domain fixed at [0,1]
BMTfit.mqde(x1, start=list(p3=0.5, p4=0.5), fix.arg=list(p1=0, p2=1))
# (6) estimation of the asymmetry-steepness parameters of the BMT
# distribution with domain fixed at [0,1]
BMTfit.mqde(x1, start=list(p3=0, p4=0.5), type.p.3.4 = "a-s",
fix.arg=list(p1=0, p2=1))
```
[Package BMT version 0.1.0.3 Index] | 1,974 | 7,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.61485 |
https://nrich.maths.org/public/topic.php?code=12&cl=2&cldcmpid=2790 | 1,591,400,126,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348504341.78/warc/CC-MAIN-20200605205507-20200605235507-00066.warc.gz | 462,421,746 | 9,249 | # Resources tagged with: Factors and multiples
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Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
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Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Two Primes Make One Square
##### Age 7 to 11 Challenge Level:
Can you make square numbers by adding two prime numbers together?
### Sets of Numbers
##### Age 7 to 11 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### Becky's Number Plumber
##### Age 7 to 11 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Abundant Numbers
##### Age 7 to 11 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
### Round and Round the Circle
##### Age 7 to 11 Challenge Level:
What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen.
### Diagonal Product Sudoku
##### Age 11 to 16 Challenge Level:
Given the products of diagonally opposite cells - can you complete this Sudoku?
### Fractions in a Box
##### Age 7 to 11 Challenge Level:
The discs for this game are kept in a flat square box with a square hole for each. Use the information to find out how many discs of each colour there are in the box.
### Seven Flipped
##### Age 7 to 11 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Surprising Split
##### Age 7 to 11 Challenge Level:
Does this 'trick' for calculating multiples of 11 always work? Why or why not?
### Path to the Stars
##### Age 7 to 11 Challenge Level:
Is it possible to draw a 5-pointed star without taking your pencil off the paper? Is it possible to draw a 6-pointed star in the same way without taking your pen off?
### Ducking and Dividing
##### Age 7 to 11 Challenge Level:
Your vessel, the Starship Diophantus, has become damaged in deep space. Can you use your knowledge of times tables and some lightning reflexes to survive?
### Times Tables Shifts
##### Age 7 to 11 Challenge Level:
In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time?
### Divide it Out
##### Age 7 to 11 Challenge Level:
What is the lowest number which always leaves a remainder of 1 when divided by each of the numbers from 2 to 10?
### Neighbours
##### Age 7 to 11 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### Scoring with Dice
##### Age 7 to 11 Challenge Level:
I throw three dice and get 5, 3 and 2. Add the scores on the three dice. What do you get? Now multiply the scores. What do you notice?
### Tiling
##### Age 7 to 11 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### Odds and Threes
##### Age 7 to 11 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### Being Collaborative - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level to work on with others.
### Being Resilient - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level that may require resilience.
### Multiplication Squares
##### Age 7 to 11 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
### Sweets in a Box
##### Age 7 to 11 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
### Zios and Zepts
##### Age 7 to 11 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Multiply Multiples 1
##### Age 7 to 11 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### Rearranged Rectangle
##### Age 7 to 11 Challenge Level:
How many different rectangles can you make using this set of rods?
### Number Detective
##### Age 5 to 11 Challenge Level:
Follow the clues to find the mystery number.
### Always, Sometimes or Never? Number
##### Age 7 to 11 Challenge Level:
Are these statements always true, sometimes true or never true?
### Multiply Multiples 2
##### Age 7 to 11 Challenge Level:
Can you work out some different ways to balance this equation?
### Multiply Multiples 3
##### Age 7 to 11 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
### Three Dice
##### Age 7 to 11 Challenge Level:
Investigate the sum of the numbers on the top and bottom faces of a line of three dice. What do you notice?
### Sets of Four Numbers
##### Age 7 to 11 Challenge Level:
There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets?
### Product Sudoku
##### Age 11 to 16 Challenge Level:
The clues for this Sudoku are the product of the numbers in adjacent squares.
### Star Product Sudoku
##### Age 11 to 16 Challenge Level:
The puzzle can be solved by finding the values of the unknown digits (all indicated by asterisks) in the squares of the $9\times9$ grid.
### Give Me Four Clues
##### Age 7 to 11 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Factor Lines
##### Age 7 to 14 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### A First Product Sudoku
##### Age 11 to 14 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### Repeaters
##### Age 11 to 14 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### Three Times Seven
##### Age 11 to 14 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
### Venn Diagrams
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How will you complete these Venn diagrams?
### A Square Deal
##### Age 7 to 11 Challenge Level:
Complete the magic square using the numbers 1 to 25 once each. Each row, column and diagonal adds up to 65.
### Multiplication Square Jigsaw
##### Age 7 to 11 Challenge Level:
Can you complete this jigsaw of the multiplication square? | 2,163 | 9,116 | {"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} | 4.25 | 4 | CC-MAIN-2020-24 | latest | en | 0.908526 |
https://codedump.io/share/TPNwcIfXvcCH/1/c-finding-the-last-occurrence-of-an-int-in-a-linear-search | 1,524,403,439,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945596.11/warc/CC-MAIN-20180422115536-20180422135536-00341.warc.gz | 599,207,735 | 9,020 | Justin Farr - 1 year ago 66
C++ Question
# C++ Finding the last occurrence of an int in a linear search
This week for homework I've been tasked with loading in a text file of 1,000 numbers and to do a linear search of a number entered in by a user. I have the linear search part done, but I have to find and print the last occurrence of that integer. I figured it would be easiest to run the array from the end and print the last occurrence and break the loop. I've started the code, but am having some trouble at finding the last occurrence.
I know my second for loop to run the array backwards is wrong, I'm just not sure what about it is wrong. Any help is appreciated! Thank you!
``````#include <iostream>
#include <fstream>
#include <conio.h>
using namespace std;
int main()
{
ifstream input("A1.txt");
int find;
cout << "Enter a number to search for: ";
cin >> find;
if (input.is_open())
{
int linSearch[1000];
for (int i = 0; i < 1000; i++)
{
input >> linSearch[i];
for (i = 1000; i > 0; i--)
{
if (find == linSearch[i])
{
cout << find << " is at position " << i << ". " << endl;
}
}
}
}
_getch();
return 0;
}
``````
`````` for (int i = 0; i < 1000; i++)
{
input >> linSearch[i];
``````
This is a good start. You started a loop to read the 1000 numbers into your array.
`````` for (i = 1000; i > 0; i--)
``````
Don't you think this is a bit premature? You haven't yet finished the loop to read the 1000 numbers in the file, yet, and you're already searching the array, that hasn't been fully read yet. There's a very technical term for this logical mistake: "putting the cart before the horse". You need to finish the loop to read the 1000 numbers, first. And only then you can execute this second loop.
`````` {
if (find == linSearch[i])
``````
Ok, now let's back up a bit. You started the loop with `i=1000`. Now, right here, what is the very first value if `i`? It is 1000. Don't you see a problem here? The 1000 element array, "linSearch", as you know, contains values numbered 0 through 999. That's a 1000 elements total. With `i` starting off with a value of 1000, accessing the non-existent linSearch[1000] is undefined behavior, and a bug.
You could tweak the logic here, to get it right. But it's not even necessary to do that. You already have a perfectly working loop that reads the 1000 numbers from the file. And you know which number you want to search.
So, each time you read the next number from the file, if it's the number you're looking for, you just store its position. So, when all is said and done, the last position that's stored in that variable will be the position of the last occurrence of the number you're searching for. Simple logic. All you have to do is also set a flag indicating that the number you were searching for has been found.
And once you come to the decision to do that, you will find that it's no longer even needed to have any kind of an array in the first place. All you have to do is read the 1000 numbers from the file, one number at a time, check if each number is the one you're searching for, and if so, save its position. Then, at the end of the loop, compare notes.
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download | 849 | 3,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.883938 |
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# Concepts on a cyclic 3D universe... (need insight)
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Given a cubic or spherical universe, how would one handle a wrap-around effect similar to asteroids? I''m not asking how to do the actual wrap around (that''s easy), but rather how to handle such cases as seeing object on the other side of the universe through the wrap around. Such that things don''t just appear/disappear out of thin air. Have there been any gaming relevant articles written about wrapping a 3D universe?
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quote:
Original post by Xorcist
I'm not asking how to do the actual wrap around (that's easy), but rather how to handle such cases as seeing object on the other side of the universe through the wrap around. Such that things don't just appear/disappear out of thin air.
What would you expect to see? If your universe would be very small then you would see any object many times. Like putting two mirrors in front of each other. How many times you would see an object depends on your view distance.
Lets say the universe is a cube with sides of 10 units. If your view distance is 50 then you would see each object 5 times.
[edited by - Rule on February 5, 2004 5:34:47 PM]
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if (x>=SIZEX) x=0;
else if (x<0) x=SIZEX;
repeat for y,z, and any other dimensions you have in your universe.
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Well I'm not exactly sure how the wrap function would work on a sphere, but cubic is easy
simply render the universe repeatedly .
Think of a game that uses tiles. You render the same thing everywhere it's needed right? So pretend that this game just showed the same tile and nothing else. Moving through the universe would scroll some.
Now, instead of tiles, you're really rendering 'the universe'. and instead of 2d tiles, they're really cubes.
So, for the simplest version,
for (int x = me.x - me.vd ; x < me.x + me.vd ; x += univ.width) { for ( int y = me.y - me.vd ; y < me.y + me.vd ; y += univ.height) { for ( int z = me.z - me.vd ; z < me.z + me.vd ; z += univ.breadth) { univ.renderOffset(x,y,z); } }}
me.vd = me.viewdistance
add some fog to blur the effect of the disappearing past the viewdistance, add some code to not render the scene if x*x+y*y+z*z > me.vd*me.vd + leeway; (where xyz are in terms of me.xyz, and leeway depends on what points you're using... if the point you're checking is the center of the universe, the leeway = (univ.width/2)*(univ.width/2) + (univ.height/2)*(univ.height/2) + (univ.breadth/2) * (univ.breadth/2)
or something similar to that. That leeway number should be overkill, but you'd need to add some square roots in there for a better answer (aka, sqrt(xx+yy+zz) > me.vd + sqrt(leeway)).
I'm assuming the AP wanted to make things far away visible, assuming they got out of your z distance range. With the APs answer, well, you'll just get things cluttering up in the corners of the visible universe if they're far away. So like, nope, that's totally wrong. The method you'd want would be:
float dist2 = x*x+y*y+z*z;if (dist2 > cd*cd) //too far away!!!{ float dist = sqrt(dist2); //don't sqrt unless you have to... x /= cd*dist; y /= cd*dist; z /= cd*dist;}
where cd = clipping distance (to which if they're farther than, you'd bring up to).
The reason the AP's method dosn't work:
assume we have objects throughout the universe, with coordinates ranging from [-10..10]. Assume we're clipping to [-1..1]. Assuming even distribution of objects, only 10% of those objects would not be clipped on the x axis, 10% not to the y, and 10% not to the z.
0.1% wouldn't be clipped at all (they'd be inside the cube).
0.9% would be clipped once (aka, to a plane)
8.1% would be clipped twice (aka, to an edge)
72.9% (the remainder) would be clipped to a corner. That is, most of the universe would be rendered in one of 8 spots. a single direction would contain 9.1125% of the universe. Last I checked, there was an equal distribution more or less - and there being infinite directions, infinitly small ammounts of the universe are ever in a single direction by pure mathematics.
As the clip ratio increased (aka, universe ranges from [-100..100] and clipping to [-1..1]) the percentage of items clipped to a corner would increase. With an infinitly large universe clipped to any boundry, 100% of the universe would be in a corner.
-Mike
[edited by - MaulingMonkey on February 5, 2004 7:21:50 PM]
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Thanks for the info.
Well obviously I'd make the universe large enough and the clipping distance adequately relative so that inifinite mirror imaging didn't occur. And there will most likely be an outlying buffer on the universe that will be devoid of objects other than active ships so that the physical wrap around would be clean.
[edited by - Xorcist on February 6, 2004 1:51:22 PM]
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So, what you really have is 27 identical universes arranged as a Rubik's Cube, and each object has 27 logical positions... mathematically, i think this equates to a hypertorus. Anyway, for rendering, the brute force approach would be something like:
for(int i = 0; i < 27; i++){ glPushMatrix(); glTranslatefv(universe[i].offset); // this function doesn't actually exist. RenderObjects(); glPopMatrix();}
But you really don't want to do that Note that each universe is just a offset, not a complete copy of the universe... The next step is to do a camera frustum / aabb test with each universe, and if the universe lies in the frustum, then render all the objects.
for(int i = 0; i < 27; i++){ if( intersectFrustumAABB(camera.Frustum, universe[i].AABB) ) // Transform and Render.}
Now, say you've smartly arranged your universe in an octree to cut way down on rendering ops. Now, you can use this by transforming the Frustum into the given Universe coordinate system, and searching the octree. Here the math is as simple as subtracting the Universe's offset from the Camera's position.
You should note, however, that you may want to do something really cool with this in the future, say when you fly off of the right side of the universe, you would come up out of the bottom... Now using these offsets, that would be completely impossible. If you make the realization that the offset represents a transformation from local universe space to rubik's cube space, then it becomes trivial.
So, say universe i in the Rubik's Cube is represented as a 4x4 Matrix Ui, when you fly into that universe, you're transforming the ship's coordinates into that local coordinate system. In the simple system this is represented as:
if(ship.x > universe.xmax) ship.x -= universe.xmax;
But what it really means is if the ship's position and orientation are represented by matrix Ms, then Ms = Ui(-1) * Ms; (where Ui(-1) means the inverse of Ui, the space the ship is flying into).
This has a lot of implications... for example, in the frustum-AABB test, the AABB is actually the AABB in local coordinates, transformed into Rubik's Cube coordinates (shifted over)... to do that check in Whacky World, you'd want to either transform the frustum into universe coordinates:
IntersectFrustumAABB(camera.Frustum.Apply(universe.Transform.inverse()), AABB);//or transform the AABB into Rubik's Cube coordinates:IntersectFrustumAABB(camera.Frustum, AABB.Apply(universe[i].Transform)); // It also means that when you're rendering: for(int i = 0; i < 27; i++){ if(IntersectFrustumAABB(camera.Frustum, AABB.Apply(universe[i].Transform)) { glPushMatrix(); glMultMatrix4f(universe[i].Transform); // RenderObjects will render into local universe coordinates... // So that's where we put the camera also. RenderObjects(Universe.octree.GetObjectsInFrustum(camera.Frustum.Apply(universe[i].Transform.inverse())); glPopMatrix(); }}
So if you do the Whacky transform out right -> in bottom, when you look out the right side of the universe, you'll see objects in the bottom, and when you look out the bottom, you'll see objects in the top AND right You can even mirror the universe, so you fly into yourself, or shoot yourself, etc...
EDIT: formatting (argh!)
[edited by - ajas95 on February 6, 2004 4:12:03 PM]
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if you have a fixed universe size, inside a box, think of the box's rectangles as portals to the other side of the universe. So, if a 'portal' is visible from the current viewpoint, render the world by by moving the camera viewpoint opposite face of the portal, and rendering the world again, from there. That works if the camera's draw distance is less than 1/2 the size of the box.
// /// +-------------/-+// | / | // | / | \ // | C *--------' // | |// | |// | |// +---------------+////////////// /\ /// / +\------------/-+// / | \ / | // / | \ / | // C1*--------' C0*--------' // | |// | |// | /// +-------------/-+// / // / // C2*--------'
well, I think this should work.
[edited by - oliii on February 6, 2004 4:12:47 PM]
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If you really want to see what it would look like in both cases, I would suggest finding/writing the source for a bare-bones ray-tracer and make the single addition that when a light-ray pierces the edge of the universe it wraps back around according to some particular pattern.
I would definitely cap the number of wraps to some maximum or it will never terminate. But in any case it shouldn't be too hard. Just set up a simple scene with a few spheres and a cube or two (or if you want, load a mesh, but that sounds more complex to me), and then start shooting rays and storing pixels in a bitmap. You'll know in no-time which should help with the theory (and be darn cool). Besides, why stop at cubic and spherical? Surely there are other possibilities! Basically any kind of 3-dimensional geometry could be applied including polar functions and even functions of time (although THAT would be both processor-intensive AND disorienting! ).
By the way, if you do implent the ray-trace, post some screenshots; I'd love to see it!
[edited by - bob_the_third on February 6, 2004 4:33:19 PM]
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× | 2,704 | 11,128 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-43 | latest | en | 0.941307 |
https://octopus-code.org/mediawiki/index.php?title=Tutorial:Wires_and_slabs&oldid=9738 | 1,638,223,644,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358842.4/warc/CC-MAIN-20211129194957-20211129224957-00495.warc.gz | 515,689,966 | 6,952 | # Tutorial:Wires and slabs
Under construction
# Ground state calculation
Hexagonal boron nitride (HBN) is an insulator widely studied which has a similar structure to graphene. Here we will describe how to get the band structure of an HBN monolayer. A sheet of HBN is periodic in the x-y directions, but not in the z. Thus we will set `PeriodicDimensions` = 2
here BN lenght= 1.445*angstrom 'L' large enough to describe a monolayer in the vacuum. One should always converge the box length value. Here is the inp file for the GS calculation:
````CalculationMode` = gs
`FromScratch` = yes
`ExperimentalFeatures` = yes
`PeriodicDimensions` = 2
`Spacing` = 0.20*angstrom
`BoxShape` = parallelepiped
BNlength = 1.445*angstrom
a = sqrt(3)*BNlength
L=40
%`LatticeParameters`
a | a | L
%
%`LatticeVectors`
1 | 0 | 0.
-1/2 | sqrt(3)/2 | 0.
0. | 0. | 1.
%
%`ReducedCoordinates`
'B' | 0.0 | 0.0 | 0.00
'N' | 1/3 | 2/3 | 0.00
%
`PseudopotentialSet`=hgh_lda
`LCAOStart`=lcao_states
%`KPointsGrid`
12 | 12 | 1
%
`ExtraStates` = 5
`UnitsOutput` = ev_angstrom
```
Remark: If the system has a z-translation symmetry, one should always centre its system with respect to the z-direction in order to avoid any asymmetric effect in the z direction due to some computational or grid errors. Here the box is centred at z=0 (with 20 bohr extension in +z and -z direction): that is why the atom lays at z=0.
# Band Structure
After this GS calculation, we would like to describe more precisely the band structure. Thus we will perform an unocc run:
````CalculationMode` = unocc
```
In order to plot the band structure along certain lines in the BZ, we will use the variable `KPointsPath` . Instead of using the `KPointsGrid` block of the GS calculation, we use during this unocc calculation:
```%`KPointsPath`
12 | 7 | 12 # Number of k point to sample each path
0 | 0 | 0 # Reduced coordinate of the 'Gamma' k point
1/3 | 1/3 | 0 # Reduced coordinate of the 'K' k point
1/2 | 0 | 0 # Reduced coordinate of the 'M' k point
0 | 0 | 0 # Reduced coordinate of the 'Gamma' k point
%
```
The first row describes how many k points will be used to sample each line. The next row are the coordinate of k points from which each line start and stop. In this particular example, we describe the lines Gamma-K, K-M, M-Gamma using 12-7-12 k points. In Figure 1 is plotted the output band structure where blue lines represent the occupied states and the reds one the unoccupied ones.
One should also make sure that the calculation is converged with respect to the spacing. To do so, we have converged the band gap. Figure 2 shows the band gap for several spacing values. Here we can see that a spacing of 0.14 Angstrom is needed in order to converge the band gap up to 0.01 eV. | 818 | 2,886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-49 | latest | en | 0.838117 |
https://www.answers.com/Q/How_much_meat_should_adolescents_eat_daily | 1,553,050,277,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202199.51/warc/CC-MAIN-20190320024206-20190320050206-00529.warc.gz | 689,808,223 | 53,909 | # How much meat should adolescents eat daily?
Would you like to merge this question into it?
#### already exists as an alternate of this question.
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There are eight normal requirements that the adolescent should have a day. These eight requirements are calories; protein which vegetarians exceed these level; fats which are gained from many junk foods though only 30% of it is needed; calcium which can be provided from green, leafy vegetables; iron which is important to adolescents for growth; zinc which can be found in eggs, seafood and dairy products; fibre which can be obtained from fruits, vegetables, grains, beans and cereals; and vitamins which normal diets are not deficient in vitamins. Answer from Ritesh www.bebo.com/_BadBoy69_
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# How much fat should you eat daily?
The total amount of calories from fat should make up around 30% of your diet. Each gram of fat contains 9kcals. So for example, If you were consuming 1500kcals a day then 30% percent of those calories should be fat. Calculations: 1500kcals / 100 X 30 = 450kcals 450kcals / 9 = 50 grams So ( Full Answer )
# How much should crf cat eat daily?
A cat should eat enough calories to satisfy its metabolic needs - this is based upon age, weight and exercise level. For a typical 10# adult cat, approximately 1 cup of dry kibble or 2-3 cans of cat food (the larger size, not the Fancy Feast size) should be adequate. A cat with chronic renal failur ( Full Answer )
# How much meat should a person eat per day?
The answer varies depending on the meat source, the weight and health of the person, the amount of exercise she/he gets, the gender and so forth. However, the average adult person should eat between 5 and 6 ounces of quality meat, or protein, daily.
# Should you eat cow meat?
Red meat does possess a fair amount of nutrients that are usable in our diets. It is not only a source of protein but other vitamins as well. In that same respect, there are a lot of unhealthy factors about red meat you should consider. It is very high in calories and contributes to an unhealthy par ( Full Answer )
# Should people eat meat?
This is one of those questions that is purely based on a person'spoint of view. Many similar questions have been asked on this siteabout this same question, and can be seen in the related questionsbelow. Human anatomy doesn't suggest that eating meat is necessary, asthere are many foods with the sam ( Full Answer )
# How much calories should a 13 year old girl eat daily?
It depends on the height and weight of the girl. But an average 13 year old girl should get between 1,500 to 2,500 calories a day.
# How much dry food should an adult male black lab eat daily?
I give my 65 lb. Aussie 1 cup a day. he is a bit of a couch potato (like me) but he stays healthy and not getting fat.
# How much potassium should you eat daily?
The average adult should have a minimum of 2000mg-3000mg of potassium per day. This can be achieved through the natural consumption of certain kinds of foods, particularly fruits and vegetables. There are circumstances where potassium is readily lost and needs to be replenished in excess of the mini ( Full Answer )
# How much daily should a 3 month old rabbit eat?
A 3 month old rabbit should eat a lot. Fresh rabbit mix in a bowl everyday. Also Fresh Carrots, spinach, corn on the cob, silverbeet, apple, pear. Throw away left overs every day. Paula
# Should you eat meat?
Originally, things that Jehovah God provided for man to eat included only that which was grown from the earth, namely fresh fruit, raw vegetation and grains (Compare Gen 1:29; 2:15-17) However, just after the great flood of Noah's day, God pronounced a blessing upon man which included the sanction o ( Full Answer )
# How much protein should a 12 year old girl eat daily?
11-13 year old's should times their current weight by 0.455 Ex: An 11 year old that weighs 85 lbs should eat 38.6 oz. of protein
# How often should you eat meat?
Never! . No, you should eat meat 3 or 4 times a week. Source: I'm a member of the human race and thus a omnivore .
# How much should a horse eat daily?
that's hard to answer with out more information. It all depends on the size and age of your horse. it also depends on how much work your horse does. on the back of the sack of horse feed.. it tells you the guideline daily amounts and how much your horse should approximately be eating
# How much meat should you eat daily?
This depends on how much you need to eat but if you want the answerI suggest you look it up on a health website such as health.com.
# How much should an old labrador dog eat daily?
Labradors are known for being stomachs on legs. A good amount of food is from 2 to 2.5 cups of food per day. The higher quality the food the better. You can also give your dog treats throughout the day. And, if your dogs stomach can handle it, you can give them table scraps as well.
# How much food should a German shepherd eat daily?
If you're talking about a growing puppy, feed them liberallybecause they grow up fast, and I mean fast! Especially is they'reGerman German Shepherds as opposed to American German Shepherdswhich are smaller. Once they reach their full growth -my two dogsgot to this point at 2 1/2 years old- they shou ( Full Answer )
# How much meat do Americans eat?
The answer depends on how long your time period is. If the time period was one year (365 days), then the average American would eat about 97 pounds. This means that an American eats about 1.9 pounds a week.
# Should dogs eat meat?
Yes, dogs should eat meat. But mostly cooked meat. They should eat other stuff to, like fruits and vegetables.
# How much food should a person eat daily?
fast food = unhealthy its all about the calories. an average man should take around 2500and around 2000 for women Women should eat 1500 to 2000 calories a day and men should eat2000 to 2500 calories a day. 1 meal per meal time(not always) As much as you are hungry. Do not confuse this with how much ( Full Answer )
# How much do puppies eat daily?
It depends on what kind of dog it is. For example, my dog is a Shitz Tzu- Bichon and eats 1/2 a cup in the morning and then another in the evening.
# How much dog food should a 52 pound dog eat daily?
It depends on the dog's metabolism, breed and it's energy level andthe caloric content of the food you are feeding. Most dogs benefitfrom 2 meals per day to even out blood sugars and to manage hunger.A trained nutritionist can help make the calculations for you -based on your commercial dog food - t ( Full Answer )
# How much meat and beans are you supposed to eat daily?
No singular meal should be eaten on a daily basis. Due to the differing balances of chemicals and nutrients in different foods, it is important to eat a variety. Red meat should only be eaten once or twice weekly , and total for protein foods is around 3 or four portions daily.
# How much yogurt should you eat daily?
There is not prescribed amount, but many people around the world eat about a half cup of yogurt with breakfast almost every morning.
# How much can you you eat daily?
this depends on the size of the person, a man can consume 3500 cal. per day if he is active, a woman 2500 cal. if she is active, if not reduce by 500 cal. per day.
# Why should you not eat uncooked meat?
The bacteria in the meat is killed when the meat is put at high temperatures like an oven or grill. When you eat raw meat, you intake bacteria which can lead to salmonella of food poisoning.
# How much should an adult pit bull male eat daily?
Buy a high-quality dog food and follow the directions on the package. It'll tell you how much to feed your dog based on weight and age.
# How much meat should a full grown man eat?
How much meat a man should eat really depends on the diet he follow, what culture/religion he follows and what he himself would consider "enough" meat to have per day. Hence, there are no specific guidelines that any one man can or should follow as far as meat consumption is concerned. He can choose ( Full Answer )
# How much food should a shetland sheepdog eat daily?
There are so many variables I can not give you absolute answer. First: Shelties vary a lot in size. The breed "standard" calls for a range of 13 inches to 16 inches at the shoulder. That is a big difference in itself, and then considering they can often range from 12 inches to 18 inches. So, size ( Full Answer )
# How much meat do you eat in your diet?
I try to eat some kind of lean protein, along with a vegetable or lower GI carb, with all three meals. My two snacks during the day consist of a protein shake and/or a healthy fat (i.e. a handful of almonds)
# How much meat should a teenager eat a day?
A teenager should eat a breakfast such as cereal and something to drink. For lunch a sandwich or main course with a side dish (french fries cookie etc.) and something to drink. For dinner a big main course two side dishes and something to drink
# How much does a horse eat daily?
About 1.5% to 2% of their body weight in grass or grass hay. Then they may eat a couple more pounds of grain if they need it to balance their work load.
# What two foods should vegetarians eat daily to ensure they are getting their iron requirement that is usually supplied by meat products?
Legumes and dark leafy green vegetables. There are many foods (no one single pair in particular) that a person who is vegetarian can eat to replace what they do not naturally gt from their diets from meat products. Mainly, people will be missing out on vitamins B6 and B12, along with iron and pote ( Full Answer )
# Should cats eat meat?
Of course! In the wild cats eat meat, like birds, rats, mice. They are carnivores - meat eaters.
# What meat should you eat when dieting?
If by dieting you mean that you are trying to lose weight, it'sAlways a balance between WHAT and how MUCH you eat. Lean meats arehelpful, but unless you know how many calories you are eating intotal each day, the type of meat really isn't going to matter.
# How much food should a kitten eat daily?
There is no correct answer for this, as how much a kitten eats can depend on several factors: The kitten's age, size, activity level, appetite and the quality of food being fed. A cat's food is also a huge factor in how much it eats: A cat will eat far more if fed a lower quality food to get the nut ( Full Answer )
# How much roasted garlic should you eat daily?
one clove but dont eat any until the next day oyu must wait for the odor to leave your body naturally
# How much of red meat should you eat in one day?
Meat is difficult to digest and not that good for you. Statistics show that red meat should not be eaten more than 1-2 times a week.
# Does eating meat daily make you fat?
Primarily only if the meat is high in fat, and/or you do not burn off the calories that you consume throughout the day.
# How much fiber should an adult male eat daily?
If you are going for a 2,000 calorie diet, at least 25 grams. If you have a 2,500 calorie diet, at least 35 grams.
# How much meat should young children eat per day?
Young children should eat about 4 times a day. breakfast, lunch, snack, and dinner. But if they tell you they're hungry then just give em a small fruit or something.
# How much should 5 pound Pomeranian eat daily?
About one half cup a day. But it really depends on the Pomeranian if you have a over weight dog then you might want a little less or talk to your dogs veterinarian.
# How much cooked beef should an 85 pound dog eat daily?
you really shouldn't feed cooked meat to a dog, that's what an average person would say but i say screw them!! a dog requires real meat not some crap processed food that was manufactured in an assembly line in some factory in Mexico hell no i feed my dog organic chicken and organic produce i do ( Full Answer )
# How much hay should a guinea pig eat daily?
They need an unlimited supply of timothy hay if they are 6 months or older if they are younger then 6 months, pregnant, or nursing, they need alfalfa hay. They should have hay available to them at all times and if they run out give them more.
# How much millet should parakeets eat daily?
My budgie will eat an entire twig of millet in a day if I leave it in the cage, and ignore his pellets -- so I've taken it out of his cage and only give it to him under supervision and when we are training as a treat. They should never eat more than one full twig a week, they're a healthy snack but ( Full Answer )
# Why should you eat meat of animals?
Meat contains protein, which is essential for life; although vegetarians obtain protein from other sources (e.g. nuts and beans), meat is generally considered as the best protein source.
# How much meat does a dog eat?
Thay eat at least a plate full of meat but it depends on the size of the dog.
# Why should you eat meat and fish?
Meat and fish are both good sources of protein, which you need mainly for your muscles. Fish are also a good source of what's called essential fats(fats that you actually need some of to remain healthy.)
# How much wet food should a cat eat daily?
A cat needs 150 grams of a grain free wet food in the morning and evening .Thats 300gr per day Wet food should be free of maize,wheat,corn,rice and digest examples of healthy cats foods are bozita,smilla,blue buffalo wilderness and animoda carny dried food should be considered a snack and shou ( Full Answer )
# Does KT4LF eat donkey meat daily?
Yes! He loves the stuff! It can be salami or donkey steaks. Ground donkey made into hamburgers, or slow roasted in the oven or crock pot. He prefers it to beef or pork, and likes it 78% better than lamb. | 3,207 | 13,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-13 | longest | en | 0.967102 |
http://www.bankersadda.com/2016/08/night-class-reasoning-quiz-for-rbiibps16.html | 1,477,628,821,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721555.54/warc/CC-MAIN-20161020183841-00322-ip-10-171-6-4.ec2.internal.warc.gz | 323,187,368 | 31,967 | ## Tuesday, 16 August 2016
### Night Class: Reasoning Quiz for RBI/IBPS Exam
Directions (Q.1-5): Study the following information carefully and answer the given question:
In a certain code language: "what does it name" is written as "ku ru mu ju”, "name does the Real" is written as "pu ku su ru", "the milton have what" is written as "su mu ho ro", " does have or not" is written as " kho ro bu ru".
1. Which of the following is the code of "not"?
a) kho
b) Ro
c) Either kho or bu
d) bu
e) ru
2. The code 'pu ho ju ku' may represent?
a) your name not what
b) what have real name
c) Real milton it name
d) it not real the
e) What the have milton
3. Which of the following may represent ‘milton have had what name’?
a) Ro mu ho wo ku
b) ku mu ho ro ru
c) ho ku mu kho bu
d) bu ju ru Ro ku
e) mu ku su pu ru
4. Which of the following is the code of 'milton'?
a) su
b) ro
c) mu
d) ho
e) None of theses
5. Which of the following is code of 'it real or not'?
a) ru bu ju ro
b) ju bu pu kho
c) ku kho pu ju
d) su pu kho bu
e) can't be determined
Directions(Q.6-10): In the following questions, the symbols @, #, %, \$ and * are used with the following meaning as illustrated below.
‘A @ B’ means ‘A Is not smaller than B’
‘A # B’ means ‘A is neither smaller than nor equal to B’
‘A % B’ means ‘A is neither smaller than nor greater than B’
‘A \$ B’ means ‘A is not greater than B’
‘A * B’ means ‘A is neither greater than nor equal to B’
a) If only conclusion I is true
b) If only conclusion II is true
c) If either conclusion I or II is true
d) If neither conclusion I nor II is true
e) If both conclusions I and II are true
6. Statements: T @ N, N # M, M % F
Conclusions:
I. T # M
II. T @ F
7. Statements: L \$ N, N * F, R % L
Conclusions:
I. F # R
II. R \$ N
8. Statements: H # I, I @ J, J \$ P
Conclusions:
I. H # J
II. H # P
9. Statements: L * D, D # K, K \$ J
Conclusions:
I. L # K
II. L \$ K
10. Statements: Q \$ W, W % E, E @ K
Conclusions:
I. Q * K
II. W @ K | 665 | 1,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | longest | en | 0.877788 |
https://techcommunity.microsoft.com/t5/excel/find-return-the-next-value-in-the-column-list/m-p/1013402 | 1,596,775,927,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737152.0/warc/CC-MAIN-20200807025719-20200807055719-00500.warc.gz | 453,215,135 | 79,998 | SOLVED
Highlighted
New Contributor
# Find/return the next value in the column list
Hi,
I have a list of numbers in a column.
A have a number in another cell that has been selected from that list.
Is it possible to find this number in the list (a la vlookup) and then access/return the value that is in the next cell in the list, the value in the cell below the current one?
Thanks.
4 Replies
Highlighted
Solution
# Re: Find/return the next value in the column list
You may find with MATCH() position of the number in the list, and with INDEX() return value from the cell in next position. If, for exmple, your list is in column A and number to search is in cell B1, it could be
``=IFERROR(INDEX(A:A,MATCH(B1,A:A,0)+1),"no such number")``
Highlighted
# Re: Find/return the next value in the column list
VLOOKUP always has been a flawed function for a number of reasons, INDEX/MATCH would be better here. Calling you list and number by those names, the simplest formula is
= INDEX( list, 1 + MATCH( number, list, 0 ) )
Instead of adding 1 to the index it is also possible to define 'offsetList' to be a range one cell down from the initial list:
= INDEX( offsetList, MATCH( number, list, 0 ) )
Switching to the latest versions of Office 365, one has the new XLOOKUP function which replaces other lookup function in almost all circumstances. Using the offset list once more
= XLOOKUP( number, list, offsetList )
or if volatile functions are not a problem
= OFFSET( XLOOKUP( number, list, list ), 1, 0 )
Highlighted
Many thank.
Highlighted
# Re: Find/return the next value in the column list
@wazza2040 , you are welcome | 422 | 1,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.801997 |
https://codecatch.net/post/11998452-a9f4-4237-bcd5-62d0a61b6382 | 1,685,899,395,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650201.19/warc/CC-MAIN-20230604161111-20230604191111-00693.warc.gz | 203,175,007 | 17,591 | ## Query UNT ActiveStudents
0 likes • Nov 18, 2022
Python
## More Python Posts
```import anytree as atimport random as rm
# Generate a tree with node_count many nodes. Each has a number key that shows when it was made and a randomly selected color, red or white.def random_tree(node_count): # Generates the list of nodes nodes = [] for i in range(node_count): test = rm.randint(1,2) if test == 1: nodes.append(at.Node(str(i),color="white")) else: nodes.append(at.Node(str(i),color="red")) #Creates the various main branches for i in range(node_count): for j in range(i, len(nodes)): test = rm.randint(1,len(nodes)) if test == 1 and nodes[j].parent == None and (not nodes[i] == nodes[j]): nodes[j].parent = nodes[i] #Collects all the main branches into a single tree with the first node being the root for i in range(1, node_count): if nodes[i].parent == None and (not nodes[i] == nodes[0]): nodes[i].parent = nodes[0]
return nodes[0]```
```def max_n(lst, n = 1): return sorted(lst, reverse = True)[:n]
max_n([1, 2, 3]) # [3]max_n([1, 2, 3], 2) # [3, 2]```
```from collections import Counter
def find_parity_outliers(nums): return [ x for x in nums if x % 2 != Counter([n % 2 for n in nums]).most_common()[0][0] ]
find_parity_outliers([1, 2, 3, 4, 6]) # [1, 3]```
### hex to rgb
CodeCatch
0 likes • Nov 19, 2022
Python
```def hex_to_rgb(hex): return tuple(int(hex[i:i+2], 16) for i in (0, 2, 4))
hex_to_rgb('FFA501') # (255, 165, 1)```
`bytes_data = b'Hello, World!'string_data = bytes_data.decode('utf-8')print("String:", string_data)`
```# Prompt user for a decimal numberdecimal = int(input("Enter a decimal number: "))
# Convert decimal to binarybinary = bin(decimal) | 538 | 1,823 | {"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} | 2.953125 | 3 | CC-MAIN-2023-23 | latest | en | 0.568257 |
https://www.coursehero.com/file/6540640/p37-035/ | 1,487,522,731,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170186.50/warc/CC-MAIN-20170219104610-00389-ip-10-171-10-108.ec2.internal.warc.gz | 809,844,010 | 21,751 | p37_035
# p37_035 - 35. The ruling separation is d = 1/(400 mm−1 )...
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Unformatted text preview: 35. The ruling separation is d = 1/(400 mm−1 ) = 2.5 × 10−3 mm. Diffraction lines occur at angles θ such that d sin θ = mλ, where λ is the wavelength and m is an integer. Notice that for a given order, the line associated with a long wavelength is produced at a greater angle than the line associated with a shorter wavelength. We take λ to be the longest wavelength in the visible spectrum (700 nm) and find the greatest integer value of m such that θ is less than 90◦ . That is, find the greatest integer value of m for which mλ < d. Since d/λ = (2.5 × 10−6 m)/(700 × 10−9 m) = 3.57, that value is m = 3. There are three complete orders on each side of the m = 0 order. The second and third orders overlap. ...
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
Ask a homework question - tutors are online | 320 | 1,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-09 | longest | en | 0.930413 |
https://eng.kakprosto.ru/how-95775-how-to-restore-ni-mh-battery | 1,660,372,126,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571909.51/warc/CC-MAIN-20220813051311-20220813081311-00062.warc.gz | 246,841,713 | 9,528 | You will need
• charger;
• bulbs;
• - skills in electrical engineering.
Instruction
1
Run the NiMh training elements, which consists in carrying out several (one to three) cycles of fully discharge and recharge the batteries. Discharge do to reduce the stress on the element to 1B. Discharge the items individually. The fact that there may be a different the ability to take charge of each battery. This is enhanced at the time of charging without exercise.
2
Perform a discharge in a special device that can execute it individually for each battery. If there is no indicator control voltage, follow the brightness of the bulb and run the discharge to a significant reduction. Note the burning time of the bulb to determine the battery capacity.
3
Use the formula in which the capacity is equal to the product of the discharge current and time discharge. Accordingly, if you have the battery capacity of 2500 mA, can give to the load current of 0.75 A for 3.3 hours. If the discharge time is less, then the residual capacity is less. By reducing the capacity necessary to you, continue to practice battery.
4
Run the discharge elements with a device made according to the scheme http://www.electrosad.ru/Sovet/imagesSovet/NiMH4.png. To construct it on the basis of the old charger. Only four light bulbs in it. In the case that the light discharge current is equal to or smaller battery, use it as a load indicator. In other cases, it is only an indicator when restoring the battery.
5
Set the value of the resistor to the total resistance was approximately 1.6 Ohms. You can't replace the bulb with an led. For example, you can take a light bulb from krypton flashlight 2.4 V. After a full discharge each battery to adjust the charging. For two batteries with a voltage of 1.2 V charge voltage no more than 5-6 V. the length of the initial accelerated charging is usually from one to ten minutes. | 423 | 1,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-33 | latest | en | 0.920976 |
https://www.codetd.com/en/article/17134594 | 1,718,992,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862132.50/warc/CC-MAIN-20240621160500-20240621190500-00315.warc.gz | 631,073,854 | 12,096 | ## color
It is color, the result of the human visual system's perception of different wavelengths of light reflection. People also define the "color" of visible light for electromagnetic waves in different wavelength ranges.
In daily life and art classes, the three colors (red, yellow and blue) are considered to be pigments that can be mixed to obtain all other colors.
As for optics, it is The three primary colors (red, green and blueRGB) [here to distinguish the names] are the basis for creating other colors.
For example, the RGB value (255, 0, 0) represents pure red, (0, 255, 0) represents pure green, (0, 0, 255) represents pure blue, (0, 0, 0) represents black, (255 , 255, 255) means white.
## 【1】Color space (color gamut)
An abstract mathematical model with different dimensions and representations. In color science, people have established a variety of color models to represent a certain color with one-dimensional, two-dimensional, three-dimensional or even four-dimensional spatial coordinates. This coordinate system is The range of colors that can be defined is the color space. The color spaces we often use mainly include RGB, CMYK, Lab, etc.
Common:
### (1) RGB color space
Different combinations of three basic colors are used to represent colors and are widely used in computer graphics and television display technology.
#### Conversion to xyz color space
##### Convert RGB color space to XYZ color space
``````import cv2 as cv
# 读取RGB图像
# 将RGB图像转换为XYZ图像
img_xyz = cv.cvtColor(img_rgb, cv.COLOR_BGR2XYZ)
``````
##### Convert XYZ color space to RGB color space
``````import cv2 as cv
# 读取XYZ图像
# 将XYZ图像转换为RGB图像
img_rgb = cv.cvtColor(img_xyz, cv.COLOR_XYZ2BGR)
``````
### (2) CMYK color space
Colors are represented by different combinations of the four basic colors of Cyan, Magenta, Yellow and Black. Mainly used in printing industry. [Full color printing]
The abbreviation here uses the last letter K instead of the beginning B because B has been given to the Blue of RGB in overall color science.
### (3)HSV(Hue, Saturation, Value) Color Space
HSV stands for Hue, Saturation, and Value.
• Hue: Indicates the type of color, such as red, blue, green, etc. In the HSV model, hue is represented as an angle, ranging from 0 to 360 degrees. If calculated counterclockwise starting from red, red is 0°, green is 120°, and blue is 240°. Their complementary colors are: yellow is 60°, cyan is 180°, and violet is 300°;
• Saturation: Indicates the purity of a color. The higher the saturation, the purer the color. The lower the saturation, the closer the color is to gray. In the HSV model, saturation ranges from 0 to 1.
• Value: represents the brightness of a color. In the HSV model, lightness also ranges from 0 to 1, with 0 representing complete black and 1 representing the brightest color.
In OpenCV, you can use the cv.cvtColor function to convert the RGB color space to the HSV color space.
``````hsv_image = cv.cvtColor(rgb_image, cv.COLOR_RGB2HSV)
``````
Hue refers to the color of light, which is related to the wavelength of light. Different wavelengths correspond to different hues, such as red, orange, yellow, etc.
Saturation represents the purity or depth of a color. Highly saturated colors are pure and have no components mixed with other colors. Low-saturation colors contain more gray or white components, making them appear lighter.
Brightness (Value) reflects the brightness of light, that is, the brightness of color. Higher brightness means lighter colors, lower brightness means darker colors. Brightness is affected by the white or black component of the color. An increase in the white component will increase the brightness, and an increase in the black component will decrease the brightness.
These concepts describe the different characteristics of color. Hue determines the type of color, saturation determines the purity of the color, and brightness determines the lightness or darkness of the color.
### (4) YUV and YCbCr color space
Y represents brightness information, and U and V or Cb and Cr represent chrominance information. This separation method makes video compression more efficient.
## 【2】Color space conversion
means that the state of a color space is expressed in another way.
For example: `RGB -> HSV` or `RGB -> GRAY `
In OpenCV, the performance of cv is `BGR` then it is Transformation from BGR to HSV or GRAY
``````cv.cvtColor(input_image,flag)
input_image 是需要进行空间转换的图像
flag为转换后的类型
cv.COLOR_BGR2GRAY:bgr->gray
cv.COLOR_BGR2HSV:bgr->hsv
``````
### 2.1 GRAY color space
GRAY color space, also known as grayscale color space, each pixel is represented by a channel ’ grayscale ’.
This kind of grayscale has been introduced before. At that time, we used binary images as a guide. Binary values are images that are either black or white, and they are divided into grayscale images.'Grayscale level' (There are only 256 grayscale levels, the range of pixel values: [0, 255], from black to white)
It can help simplify problems and reduce complexity in image processing and computer vision while still retaining most of the structural and shape information.
#### 2.1.1 Conversion method:
``````Gray = 0.299*R + 0.587*G + 0.114*B
``````
This weight distribution is designed based on the human eye's sensitivity to different colors. The human eye is most sensitive to green, followed by red, and blue is the least sensitive. This is because there are three types of color receptors on the retina of the human eye, which are most sensitive to red, green and blue light.
#### 2.1.2 BGR -> GRAY
Because OpenCV defaults to BGR display mode.
You can use the `cvtColor` function is a function in the OpenCV library that is used to convert an image from one color space to another.
``````cvtColor(src, code[, dst[, dstCn]]) -> dst
``````
The conventional ranges for R, G, and B channel values are:
. - 0 to 255 for CV_8U images
. - 0 to 65535 for CV_16U images
. - 0 to 1 for CV_32F images
parameter:
``````import numpy as np
import cv2 as cv
# 读取一张彩色图片
# 创建一个与输入图像同样大小的空图像，用于存储转换结果
dst = np.zeros_like(img)
# 使用cvtColor函数将图片从BGR色彩空间转换到灰度色彩空间
# 我们提供了dst参数，所以函数将把转换结果存储在这个图像中
# 我们也提供了dstCn参数，指定输出图像的通道数为1
cv.cvtColor(img, cv.COLOR_BGR2GRAY, dst=dst, dstCn=1)
# 打印转换后的图像的通道数，应该为1
print(dst.shape)
# (864, 1920, 3)
``````
np.zeros_like(img) will create an all-zero array with the same shape (i.e. the same number of rows and columns) and data type as img. This means that the returned array will have the same dimensions as img and each element will be initialized to zero.
The role of this function in the above example is to create an empty image with the same size and depth as the input image img, used to store the conversion result of the cvtColor function. By using np.zeros_like(img) we can ensure that the empty image created has the same shape and data type as the input image, thus avoiding size or type mismatch errors during conversion.
Original picture
no parameters
``````import cv2 as cv
# 读取一张彩色图片
# 使用cvtColor函数将图片从BGR色彩空间转换到灰度色彩空间
gray = cv.cvtColor(img, cv.COLOR_BGR2GRAY)
# 打印转换后的图像的通道数，应该为1
print(gray.shape)
``````
#### 2.1.3 How to prove`Gray = 0.299*R + 0.587*G + 0.114*B`
##### (1) Split the color image into three layers
Use function`b,g,r=cv.split(img1)`
Step1: Basic code
``````import numpy as np
import cv2 as cv
import matplotlib.pyplot as plt
src=cv.cvtColor(img1,cv.COLOR_BGR2GRAY)
plt.imshow(img1[:,:,::-1])
``````
Step2: Split
``````b,g,r=cv.split(img1)
img1
``````
Step3: Split situation
[ 1 ] Grayscale src (original image img1)
[ 2 ] b
[ 3 ] g
[ 4 ] r
Step4: Calculation (because it is an integer, it will be rounded)
##### (2) Prove that when the image is converted from the GRAY color space to the RGB color space, the final values of all channels will be the same.
When converting from a grayscale image (GRAY) back to an RGB image, the values of all R, G, and B channels will be the same. This is because the grayscale image has only one channel, so when converting back to an RGB image, the value of this single channel will be copied to the R, G, and B channels.
``````import numpy as np
import cv2 as cv
import matplotlib.pyplot as plt
# 读取灰度图像
# 将灰度图像转换为RGB图像
img_rgb = cv.cvtColor(img_gray, cv.COLOR_GRAY2BGR)
# 分离RGB通道
b, g, r = cv.split(img_rgb)
# 检查R、G、B三个通道的值是否相同
print("R == G: ", np.all(r == g))
print("R == B: ", np.all(r == b))
print("G == B: ", np.all(g == b))
``````
First read a grayscale image and then convert it to an RGB image. Then, it separates out the three channels R, G, and B and checks whether the values of these three channels are the same. If all outputs are True, it proves that the values of all R, G, and B channels are the same when converting from a grayscale image to an RGB image.
The values of the three RGB channels
## 【3】Type conversion function
``````dst = cv2.cvtColor( src, code [, dstCn] )
``````
cv2.cvtColor() is a function in OpenCV used for color space conversion. It accepts three parameters:
• src: Input image, which can be a NumPy array or an OpenCV Mat object.
• code: The code for color space conversion, specifying the type of conversion to be performed. Common conversion types include:
• cv2.COLOR_BGR2GRAY: Convert BGR image to grayscale image.
• cv2.COLOR_BGR2HSV: Convert BGR image to HSV color space.
• cv2.COLOR_BGR2RGB: Convert BGR image to RGB color space.
• Other conversion types can be found in OpenCV's documentation.
• dstCn (optional): Number of channels of the target image. The default value is 0, which means the same number of channels as the input image.
The return value of the function is the converted image, returned as a NumPy array.
### 【4】Mark specified color
In the HSV color space, the H channel (saturation Hue channel) corresponds to different colors.
### 1. Lock specific value through inRange function
OpenCV uses the function cv2.inRange() to determine whether the pixel value of the pixel in the image is within the specified range. Its
syntax format is:
dst = cv2.inRange( src, lowerb, upperb )
Where:
dst represents the output result, the size is the same as src.
src represents the array or image to be checked.
lowerb represents the lower bound of the range.
upperb represents the upper bound of the range.
The return value dst is the same size as src, and its value depends on whether the value at the corresponding position in src is within the interval [lowerb,upperb]
: If the src value is not within the specified interval, the value at the corresponding position in dst is 0
If the src value is within the specified interval, the value at the corresponding position in dst is 255.
### HSV color space
The RGB value of blue is [[[255 0 0]]], and the converted HSV value is [[[120 255 255]]].
The RGB value of green is [[[0 255 0]]], and the converted HSV value is [[[60 255 255]]].
The RGB value of red is [[[0 0 255]]], and the converted HSV value is [[[0 255 255]]].
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Origin blog.csdn.net/m0_74154295/article/details/134351532
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Daily | 2,907 | 11,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-26 | latest | en | 0.831196 |
http://math.stackexchange.com/questions/182562/intuition-behind-snake-lemma | 1,469,787,546,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830064.24/warc/CC-MAIN-20160723071030-00319-ip-10-185-27-174.ec2.internal.warc.gz | 153,011,232 | 22,118 | # Intuition behind Snake Lemma
I've been struggling with this for some time. I can prove the Snake Lemma, but I don't really “understand” it. By that I mean if no one told me Snake Lemma existed, I would not even attempt to prove it. I believe I am missing some important intuition that tells me that the lemma “could” be true before actually trying to prove it. Please give me some pointers.
-
It is indeed a strange thing... Exterior differentiation of differential forms, and boundary operators on spaces, were perhaps the plausible and "more physical/intuitive" antecedents of the abstracted form. – paul garrett Aug 14 '12 at 19:32
youtube.com/watch?v=etbcKWEKnvg – Will Jagy Aug 14 '12 at 19:47
@WillJagy Yes I've seen that. It's entertaining. Doesn't really help though :) – Tunococ Aug 14 '12 at 19:48
In that case, you were wise to ask here. – Will Jagy Aug 14 '12 at 19:49
A special case is easy to see: Imagine $A \leq A' \leq B=B'$, and $C=B/A$ and $C'=B'/A'$ with the obvious maps from a module $M$ to $M'$. The kernels are $0$, $0$, and $A'/A$. The cokernels are $A'/A$, $0$, $0$. The last kernel and the first cokernel are linked.
The snake lemma is then just one of the isomorphism theorems. As you deform $B$ and $B'$ more, how do the kernels and cokernels deform? The last kernel and first cokernel no longer need be isomorphic, but the kernel and cokernel of that linking (snakey) map can be described in terms of the kernels and cokernels already there. A specific relation is the snake lemma.
## Example deformation
An example deformation might be helpful: distort the $A' \to B' \to C'$ sequence by quotienting out by some $M \leq B$ (imagine $M=IB$ for some ideal $I$, so we are tensoring the second line with $R/I$). How does this change the kernels and cokernels?
The first line is $$0 \to A \to B \to B/A \to 0,$$ and the second line becomes $$0 \to (A'+M)/M \to B'/M \to B'/(A'+M) \to 0$$ so the kernels are $A \cap M$, $M$, and $(A'+M)/A$ and the cokernels become $(A'+M)/(A+M)$, $0$, and $0$. The last kernel and the first cokernel are related, but not equal. One clearly has the relation $0 \to (A+M)/A \to (A'+M)/A \to (A'+M)/(A+M) \to 0$ where the last two nonzero terms are the last kernel and the first cokernel. The first term is weird though. We apply another isomorphism theorem to write $(A+M)/A \cong M/(A\cap M)$ and then the solution is clear: We already have $0 \to A \cap M \to M \to M/(M\cap A) \to 0$ so we splice these two together to get the snake lemma: $$0 \to A \cap M \to M \to (A'+M)/A \to (A'+M)/(A+M) \to 0 \to 0 \to 0$$
-
Ok, your first special case is pretty helpful. Let me try to explain what I understand. $A \le A' \le B = B'$ corresponds to growing $A$ to $A'$. The elements in $\text{coker}(A \to A')$ correspond to the growth of $A'$ over $A$. Since $B = B'$, $C$ must reduce in size. $\ker(C \to C')$ measures the shrinkage $C \to C'$, and so $\ker(C \to C') \cong \operatorname{coker}(A \to A')$. Does this make sense? – Tunococ Aug 14 '12 at 23:32
@Tunococ: exactly. The first special case should be intuitive. The next section is “what happens in a less intuitive case.” We had to check if $\ker(C\to C') \cong \ker(A \to A')$ was always true. It wasn't but at least we could “compare” the two modules. In a category “compare” is a codeword for “homomorphism”, and in abelian category that means exact sequences. – Jack Schmidt Aug 15 '12 at 13:18
@JackSchmidt: after staring at your deformed example for a while trying to see the kernels of the vertical maps, I noticed I can't figure out what you mean by $A'M$. These are modules so there's no element-wise product $\{am, a\in A', m\in M\}$, and neither the direct nor tensor product makes sense to read here-what do you mean? – Kevin Carlson Sep 20 '12 at 12:26
@Kevin: Try $A'+ M =\{a+m : a \in A', m \in M\}$ to be the join of $A'$ and $M$. The notation is standard for multiplicative groups (where I normally work), but $A'+M$ is usually used for additive groups. – Jack Schmidt Sep 20 '12 at 12:39
Thanks for the quick reply, and for the fact that'll surely be good to know again. – Kevin Carlson Sep 20 '12 at 12:46
All these diagram chasing lemmas (snake lemma, $3x3$ lemma, four lemma, five lemma, etc.) follow directly from the "salamander lemma" due to George Bergman, see salamander lemma.
And that is pretty transparent. It is so transparent that for instance it is immediate to see (which no textbook ever mentions) that there are just as easily 4x4 lemmas, 5x5 lemmas, 6x6 lemmas. etc. In other words: once you see the simple idea of the salamander lemma, you can come up yourself with more such diagram chasing lemmas at will.
-
Nice answer! +1 – Rudy the Reindeer Sep 20 '12 at 12:01
I think the best motivation for the Snake Lemma has to do with the Long Exact Sequence in (Co)homology. It can also be thought of as intuition, since the long exact sequence is there to repair the non-exactness of a left-exact (or right-exact) functor, and the way to define the connecting homomorphism is through the Snake lemma.
-
That's actually where I first encountered the Snake Lemma. It bothers me every time I invoke the long exact sequence. All books I've read say something like: "Here's a fact from algebra you can use. Here's the proof (or the proof is straightforward and left as an exercise)." I do believe that the intuition must have come from that, but I can't seem to grasp it. I might need to try to understand derived functors a bit more before coming back to this(?) – Tunococ Aug 14 '12 at 19:56
I think with situations like this, trying to understand things in more generality (derived functors) will not give you any of the intuition you are looking for. I second M Turgeon's suggestion of trying to understand what the connecting maps in the homology LES do (this ties in with Paul Garrett's suggestion as well). I think it is possible to get to a point where the connecting map in homology is perfectly transparent; you can then use this as a reference point when proceeding with more abstract situations. – NKS Aug 14 '12 at 23:00
Thank you. I do know how to explain the connecting homomorphism. I guess I need to think more about that and maybe just feel content with what I understand :) – Tunococ Aug 14 '12 at 23:12
I can't resist. If you talk about derived functors and the snake lemma, you can go a step further. One of my favorite homological algebra exercises is: compute the derived functors of the kernel functor $\ker\colon \mathscr{A}^\to \to \mathscr{A}$ (seen as a left exact functor from the category of morphisms of $\mathscr{A}$ to $\mathscr{A}$)... Interestingly, none of the homological algebra texts I ever read seem to contain this. A pity. – t.b. Aug 15 '12 at 7:54
@t.b. I just stumbled upon exactly this exercise: it appears as Ex. 13.23 i Kashiwara and Schapira's "Categories and sheaves". – Dan Petersen Sep 3 '12 at 12:11 | 1,965 | 6,908 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2016-30 | latest | en | 0.921916 |
https://bestoptionsspjy.web.app/derksen46278le/calculating-index-of-diversity-met.html | 1,627,814,362,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154175.76/warc/CC-MAIN-20210801092716-20210801122716-00347.warc.gz | 139,001,357 | 6,076 | ## Calculating index of diversity
Richness and evenness are components of biodiversity AND Analysis of the biodiversity of two local communities using Simpson's reciprocal index of diversity.
Calculate Brillouin index of alpha diversity. chao1(counts[, bias_corrected]), Calculate chao1 richness estimator. chao1_ci(counts[, bias_corrected, zscore]) This allows one, for example, to compare relative differences between richness and evenness. Many software programs will calculate diversity indices. We have. The activity illustrates how to use math to calculate the diversity index of a selected habitat. The closer the diversity index is to 1, the more diverse and healthy it Calculate Margalef & Menhinick. Lecture 6 – Rarefaction and Beta Diversity. Beta Diversity. Main goal of b diversity. Similarity along a range of samples or
## The calculation results by operating the three diversity indices on the sampling data in the field in 1987, 1991, 1997 and 2000 (as seen in Table 1) shows that
Calculate Shannon’s diversity index “H” by using the formula H = - Summation[P(i) * lnP(i)]. For each species, multiply its proportion “P(i)” by natural logarithm of that proportions lnP(i), sum across species and multiply the result by minus one. The samples of 5 species are 60,10,25,1,4. Calculate the Shannon diversity index and Evenness for these sample values. Given : Sample Values (S) = 60,10,25,1,4 number of species (N) = 5 . To Find : shannon diversity index and Evenness . Solution : Step 1: First, let us calculate the sum of the given values. What is Diversity Index? Diversity Index (DI) captures the racial and ethnic diversity of our student body with a single number. The DI expresses the probability that any two students chosen at random will have different racial or ethnic backgrounds. The PowerPoint and accompanying resources have been designed to cover the second part of point 4.6 of the AQA A-level Biology specification which states that students should understand the meaning of an index of diversity and be able to apply the formula. How to Calculate Biodiversity . Diversity Indices: A) A diversity index is a mathematical measure of species diversity in a given community. B) Based on the species richness (the number of species present) and species abundance (the number of individuals per species). C) The more species you have, the more diverse the area, right? Calculate Shannon’s diversity index “H” by using the formula H = - Summation[P(i) * lnP(i)]. For each species, multiply its proportion “P(i)” by natural logarithm of that proportions lnP(i), sum across species and multiply the result by minus one. Diversity Calculation in Excel – Diversity Indices and True Diversity In my video “ Diversity Index as Business KPI – The Concept of Diversity ” I explain the mathematical concept of diversity introducing the Simpson Index λ and its complement (1- λ ) as a measure of product diversification in markets.
### Shannon, Simpson, and Fisher diversity indices and species richness. These functions calculate only some basic indices, but many others can be derived with
Apr 10, 2006 The equations for each calculation are given in both the notes and in the textbook . As they are Simpson's Index and Berger-Parker index. Nov 22, 2019 Calculating Diversity Index. What is Diversity Index? Diversity Index (DI) captures the racial and ethnic diversity of our student body with This is equivalent to the genetic calculation of heterozygosity, H, being the Shannon's index of diversity H' is derived from information theory, originally in the Jan 21, 2020 This clear and concise lesson guides students through the use of the formula to calculate an index of diversity to consider the biodiversity of a the index to use for calculations; partial match to " simpson " or " shannon ". Details. For the following sections, S represents the number of species, λ represents derivation of Simpson's Diversity Index, students then apply the equation to a data set on their own. calculating a type of biodiversity index, and compare it with Formula for the Hutcheson t-test. In the formula H represents the Shannon diversity index for
### The Simpson Diversity Index ranges from 0 to 1. Greater the value, greater the sample diversity. Enter the data in this Simpson index calculator and click calculate.
Calculate Shannon’s diversity index “H” by using the formula H = - Summation[P(i) * lnP(i)]. For each species, multiply its proportion “P(i)” by natural logarithm of that proportions lnP(i), sum across species and multiply the result by minus one. The samples of 5 species are 60,10,25,1,4. Calculate the Shannon diversity index and Evenness for these sample values. Given : Sample Values (S) = 60,10,25,1,4 number of species (N) = 5 . To Find : shannon diversity index and Evenness . Solution : Step 1: First, let us calculate the sum of the given values. What is Diversity Index? Diversity Index (DI) captures the racial and ethnic diversity of our student body with a single number. The DI expresses the probability that any two students chosen at random will have different racial or ethnic backgrounds. The PowerPoint and accompanying resources have been designed to cover the second part of point 4.6 of the AQA A-level Biology specification which states that students should understand the meaning of an index of diversity and be able to apply the formula.
## The activity illustrates how to use math to calculate the diversity index of a selected habitat. The closer the diversity index is to 1, the more diverse and healthy it
The activity illustrates how to use math to calculate the diversity index of a selected habitat. The closer the diversity index is to 1, the more diverse and healthy it Calculate Margalef & Menhinick. Lecture 6 – Rarefaction and Beta Diversity. Beta Diversity. Main goal of b diversity. Similarity along a range of samples or
Ecology and Simpsons Diversity Index; Estimating environmental damage in cards as 'species' to generate data to calculate Simpson's Diversity Index. | 1,301 | 6,059 | {"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 | 4 | CC-MAIN-2021-31 | latest | en | 0.863865 |
http://www.aqua-calc.com/what-is/radiation-absorbed-dose | 1,501,145,676,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549427750.52/warc/CC-MAIN-20170727082427-20170727102427-00104.warc.gz | 346,932,587 | 5,989 | # What is radiation absorbed dose
? O T R
centigraycGy
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kilograykGy
micrograyµGy
milligraymGy
millirem alpha particlesmrem α
millirem fission fragmentsmrem fission
millirem heavy particlesmrem heavy
millirem high-energy protonsmrem protons
millirem multiple-charged particlesmrem multiple
millirem neutrons of unknown energymrem neutrons
nanograynGy
picograypGy
rem alpha particlesrem α
rem fission fragmentsrem fission
rem heavy particlesrem heavy
rem high-energy protonsrem protons
rem multiple-charged particlesrem multiple
rem neutrons of unknown energyrem neutrons
#### Foods, Nutrients and Calories
Chicken, broilers or fryers, drumstick, meat only, cooked, stewed, 36% bone, 12% skin weigh(s) 169.07 gram per (metric cup) or 5.64 ounce per (US cup), and contain(s) 169 calories per 100 grams or ≈3.527 ounces [ calories | weight to volume | volume to weight | price | density ]
#### Gravels and Substrates
Substrate, Eco-Complete density is equal to 1538 kg/m³ or 96.014 lb/ft³ with specific gravity of 1.538 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylinderquarter cylinder or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
#### Materials and Substances
Beans, soy weigh(s) 0.721 gram per (cubic centimeter) or 0.417 ounce per (cubic inch) [ weight to volume | volume to weight | price | density ]
#### Weight/Volume at Temperature
Coconut oil weigh(s) 0.881 gram per (cubic centimeter) or 0.509 ounce per (cubic inch) at 80°C or 176°F [ weight to volume (T) | volume to weight (T) ]
#### What is dyne per square picometer?
A dyne per square picometer (dyn/pm²) is a unit of pressure where a force of one dyne (dyn) is applied to an area of one square picometer.
#### What is data or computer measurement?
The units of data measurement were introduced to manage and operate digital information. The basic or atomic unit of data measurement is a bit, which can store a value (one) or no-value (zero) thus having a binary nature. | 578 | 2,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | latest | en | 0.684643 |
https://www.osapublishing.org/oe/fulltext.cfm?uri=oe-14-7-3039&id=89029 | 1,628,165,617,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155529.97/warc/CC-MAIN-20210805095314-20210805125314-00479.warc.gz | 969,247,614 | 32,670 | ## Abstract
When three or more plane waves overlap in space, complete destructive interference occurs on nodal lines, also called phase singularities or optical vortices. For super positions of three plane waves, the vortices are straight, parallel lines. For four plane waves the vortices form an array of closed or open loops. For five or more plane waves the loops are irregular. We illustrate these patterns numerically and experimentally and explain the three-, four- and five-wave topologies with a phasor argument.
©2006 Optical Society of America
## 1. Introduction
A feature of wave superposition is that one plus one does not necessarily equal two the interference of two equivalent waves can result in zero. In general, when three or more scalar waves interfere in space, complete destructive interference occurs on lines, called nodal lines, phase singularities, wave dislocations or optical vortices [1], studied in singular optics [2]. The singular nature of nodes arises since when the complex number describing the wave is zero, the argument is not defined (singular), and nearby all phases between 0 and 2π occur, increasing in either a clockwise or counterclockwise sense (Fig. 1).
Fig. 1. Optical vortex line in an interference field. Left: periodic cell containing a vortex loop (dotted lines are parts of the loop outside the primitive cell). The intensity on the front and back planes is also represented. Right: phase cross section of the front/back face of the cell. The phase circulates in opposite directions at different points on the vortex line.
In a three-dimensional, random light field such as laser speckle, the nodal lines form a complicated tangle [3]. As with all optical fields, speckle patterns can be represented, through Fourier analysis, by a superposition of many plane waves with different directions and phases [4]. The connection between optical vortices in 3D and the plane wave decomposition is complicated as the position of nodes depends nonlinearly on the amplitudes and directions of the plane waves. Our aim here is to understand this connection for a small number of waves, namely superpositions of three, four and five plane waves.
Experimentally, vortex line geometry can be determined by imaging the intensity and phase distribution in successive beam cross-sections, within which the vortices are dark points. Identification of these points in successive cross-sections allows the complete 3D structure to be deduced. This approach complements the dynamical picture of optical vortices [5], in which propagation is associated with evolution of the two-dimensional pattern. Thus, in two dimensions, the creation or annihilation of a pair of oppositely signed vortices is equivalent to nodal line ‘hairpins’ [6] (Fig. 1). The apparent change in handedness occurring at the extremes of such a hairpin arises from its definition with respect to the propagation direction. The sense of phase circulation around the vortex line, and hence energy flow, is preserved at all points around the loop [6].
In an interference pattern made by superposing three plane waves, the vortex lines are straight and parallel, so in a cross-section, there is a regular array of vortex points [7]. In this paper we consider both more plane waves and most importantly, three spatial dimensions. In general, for four or more plane waves, the vortex lines are curved, typically forming closed loops. In special cases, vortex lines can form unusual topologies, such as links or knots (examples of which were proposed in Bessel beams [8, 9] and experimentally realized with Laguerre-Gauss beams [10, 11]). The present work is also related to phase singularities in the canonical three-dimensional diffraction catastrophes [12], whose complicated structure can be approximated using the superposition of a small number of plane waves.
The simple plane wave combinations we describe here are synthesized experimentally using a spatial light modulator (SLM) as a hologram. We address the SLM to produce a k-space distribution of paraxial plane waves [13]. The plane of the SLM is imaged to a CCD detector where the interference patterns are recorded. Our experimental setup is similar to that described in Ref. [11], except that the CCD is mounted on a motorized stage enabling a succession of images corresponding to neighboring cross-section to be obtained. Extending the setup so that the SLM is in one arm of an interferometer allows the phase structure to be recorded. We experimentally determine the vortex positions primarily using intensity measurements but confirm with reference to the phase measurements (as in Fig. 4 later).
## 2. Numerical calculations of vortex topology
Numerical characterization of the vortex topology in an arbitrarily large, finite interference pattern is problematic since ambiguities arise where vortex lines meet the volume edge. We overcome this problem by restricting the wave vectors to lie on a regular, rectangular grid in k-space, with spacing Δk. This results in an interference pattern that is periodic both transversally and axially (the Talbot effect [14, 15]), with repeat periods of 2πk and 4πk 0k 2 respectively. This ‘Talbot cell’ can be tiled to give the interference pattern, and therefore the vortex structure, over all space (Figs. 1, 2, and 3) without ambiguity.
The location of a vortex within a cross-section is found by examining the phase structure around each pixel, a change of ±2π indicating a vortex. Knowledge of whether two vortex points in neighboring planes form part of the same line is limited by the spatial resolution. However, since vortex lines are continuous, ambiguity only arises when two different lines approach (characteristically in an ‘avoided crossing’ or ‘reconnection’ [9, 16]). A vortex line, traced through the Talbot cell, can leave and re-enter at the associated point on the opposite face before arriving back at its starting position. Whether the vortex structure corresponds to a closed loop or open, infinite line is deduced from the number of cell crossings: a vortex line forms a closed loop if and only if the total signed cell crossing number is zero. The experimental correspond to the numerical simulations.
## 3. Phasor argument for vortex topology
The topological form of vortex lines resulting from the interference between N waves with wavevectors kn (for n = 1,…,N) can be understood using phasors. At a particular position, the interference is calculated in the Argand plane by the vector addition of the individual phasors, ψn , of magnitude an = |ψn | and argument arg ψn i.e. the total field at the point is $∑n=1N$ ψn . The waves are labeled in order of decreasing magnitude a 1a 2 ≥…≥aN .
At a vortex, the phasors form a closed loop. For a small number of interfering waves, the possible vortex topologies may be understood from the geometry of the phasor pattern. The simplest interference is between two waves, complete cancellation only occurring if the interfering waves have the same amplitude and opposite phase, ψ 1 =-ψ 2. This is a single real condition (i.e. has codimension one), and in three dimensions, occurs on surfaces.
Since labeling is in order of decreasing magnitude, for a superposition of three waves, cancellation can only occur if a 1a 2 + a 3, allowing a triangular configuration of the phasors. Under a different ordering of the phasors, the triangle may be reflected or rotated, but its shape cannot change. Along a vortex line, the phasor triangle rotates, so each component plane wave changes by the same phase, which can only be satisfied if the vortex direction is that in which the components of kn are the same. Therefore, the vortex structures arising from three plane waves are straight lines in the same direction, k 1 × k 2 + k 2 × k 3 + k 3 × k 1, as in Fig. 2(a).
Fig. 2. Calculated vortex structure for three, four and five equal-amplitude plane waves (a, b, c) and their respective k-space configuration (d, e, f). Each is calculated on a 256×256×256 Talbot cell.
For a superposition of four waves, cancellation can only occur if a 1a 2 + a 3 + a 4, giving a quadrilateral configuration of the phasors. With the magnitudes and wavevectors of the four waves fixed, the relative phases constitute three additional degrees of freedom. All possible phase relationships are explored throughout the three-dimensional Talbot cell. Consequently, a change to the initial phase of any of the superposed waves results only in a spatial translation, and does not affect any aspect of the vortex geometry.
When the four waves have the same magnitude, the closed loop of phasors forms a rhombus, where ψ 1 =-ψ 3 and ψ 2 =-ψ 4. The vortex follows the intersection of two planes, determined by the two-wave interference described above (for waves 1 and 3, and 2 and 4), and is therefore again a straight line. However, under different orderings, a rhombus is also possible if ψ 1 =-ψ 2 and ψ 3 =-ψ 4 or ψ 1 =-ψ 4 and ψ 2 =-ψ 3. Hence, there are three characteristic straight line vortex directions, along which the rhombus rotates and deforms. As the rhombus deforms, it may pass through a ‘flat’ configuration. At this point, the phasor pairs are identical, and the vortex lines cross [9, 16], as in Fig. 2(b).
Fig. 3. The calculated vortex structures for four plane wave superpositions with amplitudes: a), a 1 + a 4 = a 2 + a 3 (b), a 1+a 4<a 2+a 3 (c), a 1+a 4 >a 2+a 3. The choice of k-vectors is the same as Fig. 2 (b), given in Fig. 2 (e). Attached multimedia shows rotation of cells around the optical axes (a, b, c) (1.4Mb, 1.3Mb, 1.1Mb).
In four-wave interference patterns, the vortex behavior can be completely understood, the topology depending only on the relative magnitudes. These cases are illustrated in 3, and experimentally in Fig. 4. Most similar to the equal-amplitude case is when the amplitudes are not equal, but a 1 + a 4 = a 2 + a 3. The vortex lines are curved, but at particular points within the Talbot cell the four phasors are co-linear. As with the equal magnitude case, this corresponds to a vortices crossing, as in Fig. 3(a). More generally, either a 1 + a 4 < a 2 + a 3 or a 1 + a 4 > a 2 + a 3. The former gives rise to an array of helical vortex lines (Fig. 3 (b), 4 (a)) whereas the latter gives an array of identical vortex loops [Figs. 3(c) and 4(b)], some of which may fall across the Talbot cell boundary. This difference is illustrated in Fig. 5, where phasor configurations along vortex lines in the two cases are shown. When a 1 + a 4 > a 2 + a 3 (loops) the smallest magnitude cannot execute a full 2π rotation about the largest; this geometric constraint limits the maximum phase difference between any two phasors and hence the spatial extent of the vortex line, which must therefore be a loop. By contrast, when a 1 + a 4 < a 2 + a 3 (lines) the relative phase difference between the smallest amplitude phasor and the remaining three can cycle unidirectionally through the full 2π range. In this case the vortex line can extend indefinitely (essentially a perturbed three wave superposition) [2].
Fig. 4. Experimentally measured vortex structures from the superposition of 4 plane waves with amplitudes giving rise to a) twisted vortex lines and b) vortex loops. The choice of k-vectors differs slightly from that in Fig. 3(b) and Fig. 3(c). Attached multimedia shows rotation of each cell around the optical axis (1.9Mb, 1.8Mb).
Fig. 5. Left: Vortex line in (x, y, z). Right: phasor diagrams at the points highlighted on a vortex a) loop and b) a line. For clarity, the orientation of the largest magnitude phasor has been fixed. The red arrows show the angular range of the smallest phasor, with respect to the largest. Note that in (a) this is restricted to less than 2π. Attached multimedia shows phasor configuration changing as the vortex paths are followed (2.3Mb, 2.1Mb).
For interference between five waves, cancellation requires the phasors to lie on a pentagon (up to reordering), and the geometry of the curved vortex lines cannot be simply described. Figure 6 shows superpositions of the same five waves as Fig. 2(c) but with different choices of relative phases. There are four relative phases, so translation in three dimensions is insufficient to account for all possible phase relationships; the pattern is not specified uniquely by the magnitudes alone. This extends to arbitrary numbers of interfering waves, where optical vortex topology is generally complicated.
Fig. 6. Examples of different vortex topologies that can be obtained from five waves of the same magnitude but different relative phase. Fig. 6 (a) is the same Talbot cell as Fig. 2(c). Figs. 6(b) and 6(c) are Talbot cells from the same wave magnitudes as (a) but with one wave advanced in phase by π/2 and π/3. Figs. 6(d),6(e), and 6(f) are their unwrapped counterparts.
## 4. Discussion
We have explained the geometry of optical vortex lines in three-dimensional superpositions of small numbers of plane waves. Unlike superpositions of many plane waves, the topology of vortices in 3- and 4-wave superpositions is relatively restricted, and can be understood completely using phasor geometry. The characteristic tangle of vortex lines that is familiar for more general, many-wave superpositions begins for five waves. Some of the features described here occur in diffraction catastrophes [12], where certain domains can be asymptotically approximated by superpositions of a few plane waves. In particular, four-wave interference patterns occur in the hyperbolic umbilic diffraction catastrophe [17], which notably has a region in which phase singularities intersect and can be asymptotically approximated by a superposition of four plane waves of equal amplitude. It should be noted that, although our experiments and numerical illustrations require periodicity and paraxiality, the theoretical justification of these geometries using phasors do not, and apply for general superpositions of plane waves. The phasor methods we have employed here might prove useful more generally in the analysis of optical vortices, for instance, in describing the anisotropy of optical vortex cores [3], as well as how the curvature, phase structure and anisotropy changes along an optical vortex line [18].
## Acknowledgments
The authors thank Professor Michael Berry and Dr. Sonja Franke-Arnold for useful discussions. This work is supported by the Leverhulme Trust and MRD acknowledges support from the Royal Society.
## References and links
1. J. F. Nye and M. V. Berry, “Dislocations in wave trains,” Proc. R. Soc. A 336, 165–90 (1974). [CrossRef]
2. J. F. Nye, Natural focusing and fine structure of light (Institute of Physics Publishing,1999).
3. M. V. Berry and M. R. Dennis, “Phase singularities in isotropic random waves,” Proc. R. Soc. A 456, 2059–79 (2000). [CrossRef]
4. J. W. Goodman, Statistical Optics (Wiley, 1985).
5. D. Rozas, C. T. Law, and G. A. Swartzlander, “Propagation dynamics of optical vortices,” J. Opt. Soc. Am. B 14, 3054–65 (1997). [CrossRef]
6. M. V. Berry, “Much ado about nothing: optical dislocation lines (phase singularities, zeros, vortices,” in . Intl. Conf. on Singular Optics, M. S. Soskin, ed., Proc. SPIE 3487, 1–15 (1998). [CrossRef]
7. J. Masajada and B. Dubik, “Optical vortex generation by three plane wave interference,” Opt. Commun. 198, 21–27 (2001). [CrossRef]
8. M. V. Berry and M. R. Dennis, “Knotted and linked phase singularities in monochromatic waves,” Proc. R. Soc. A 457, 2251–2263 (2001). [CrossRef]
9. M. V. Berry and M. R. Dennis, “Knotting and unknotting of phase singularities: Helmholtz waves, paraxial waves and waves in 2+1 spacetime,” J. Phys. A: Math. Gen. 34, 8877–88 (2001). [CrossRef]
10. J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Knotted threads of darkness,” Nature 432, 165 (2004). [CrossRef] [PubMed]
11. J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Vortex knots in light,” New J. Phys. 7, 55 (2005). [CrossRef]
12. M. V. Berry, J. F. Nye, and F. J. Wright, “The elliptic umbilic diffraction catastrophe,” Phil. Trans. R. Soc. A 291, 453–84 (1979). [CrossRef]
13. E. Schonbrun, R. Piestun, P. Jordan, J. Cooper, K. D. Wulff, J. Courtial, and M. Padgett, “3D interferometric optical tweezers using a single spatial light modulator,” Opt. Express 13, 3777–3786 (2005). [CrossRef] [PubMed]
14. W. H. F. Talbot, “Facts relating to optical science, No. IV,” London Edinburgh Dublin Philos. Mag. J. Sci. 9, 401–407 (1836).
15. K. Patorski, “The self-imaging phenomenon and its applications,” Progress in Optics XXVII, 103–108 (1989).
16. J. F. Nye, “Local solutions for the interaction of wave dislocations,” J. Opt. A: Pure Appl. Opt. 6, S251–S254 (2004). [CrossRef]
17. J. F. Nye, “Dislocation lines in the hyperbolic umbilic diffraction catastrophe,” Proc. R. Soc. A, in press (2006). [CrossRef]
18. M. R. Dennis, “Local phase structure of wave dislocation lines: twist and twirl,” J. Opt. A: Pure Appl. Opt. 6, S202–S208 (2004). [CrossRef]
### References
• View by:
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1. J. F. Nye and M. V. Berry, “Dislocations in wave trains,” Proc. R. Soc. A 336, 165–90 (1974).
[Crossref]
2. J. F. Nye, Natural focusing and fine structure of light (Institute of Physics Publishing,1999).
3. M. V. Berry and M. R. Dennis, “Phase singularities in isotropic random waves,” Proc. R. Soc. A 456, 2059–79 (2000).
[Crossref]
4. J. W. Goodman, Statistical Optics (Wiley, 1985).
5. D. Rozas, C. T. Law, and G. A. Swartzlander, “Propagation dynamics of optical vortices,” J. Opt. Soc. Am. B 14, 3054–65 (1997).
[Crossref]
6. M. V. Berry, “Much ado about nothing: optical dislocation lines (phase singularities, zeros, vortices,” in . Intl. Conf. on Singular Optics, M. S. Soskin, ed., Proc. SPIE 3487, 1–15 (1998).
[Crossref]
7. J. Masajada and B. Dubik, “Optical vortex generation by three plane wave interference,” Opt. Commun. 198, 21–27 (2001).
[Crossref]
8. M. V. Berry and M. R. Dennis, “Knotted and linked phase singularities in monochromatic waves,” Proc. R. Soc. A 457, 2251–2263 (2001).
[Crossref]
9. M. V. Berry and M. R. Dennis, “Knotting and unknotting of phase singularities: Helmholtz waves, paraxial waves and waves in 2+1 spacetime,” J. Phys. A: Math. Gen. 34, 8877–88 (2001).
[Crossref]
10. J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Knotted threads of darkness,” Nature 432, 165 (2004).
[Crossref] [PubMed]
11. J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Vortex knots in light,” New J. Phys. 7, 55 (2005).
[Crossref]
12. M. V. Berry, J. F. Nye, and F. J. Wright, “The elliptic umbilic diffraction catastrophe,” Phil. Trans. R. Soc. A 291, 453–84 (1979).
[Crossref]
13. E. Schonbrun, R. Piestun, P. Jordan, J. Cooper, K. D. Wulff, J. Courtial, and M. Padgett, “3D interferometric optical tweezers using a single spatial light modulator,” Opt. Express 13, 3777–3786 (2005).
[Crossref] [PubMed]
14. W. H. F. Talbot, “Facts relating to optical science, No. IV,” London Edinburgh Dublin Philos. Mag. J. Sci. 9, 401–407 (1836).
15. K. Patorski, “The self-imaging phenomenon and its applications,” Progress in Optics XXVII, 103–108 (1989).
16. J. F. Nye, “Local solutions for the interaction of wave dislocations,” J. Opt. A: Pure Appl. Opt. 6, S251–S254 (2004).
[Crossref]
17. J. F. Nye, “Dislocation lines in the hyperbolic umbilic diffraction catastrophe,” Proc. R. Soc. A, in press (2006).
[Crossref]
18. M. R. Dennis, “Local phase structure of wave dislocation lines: twist and twirl,” J. Opt. A: Pure Appl. Opt. 6, S202–S208 (2004).
[Crossref]
#### 2005 (2)
J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Vortex knots in light,” New J. Phys. 7, 55 (2005).
[Crossref]
#### 2004 (3)
J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Knotted threads of darkness,” Nature 432, 165 (2004).
[Crossref] [PubMed]
J. F. Nye, “Local solutions for the interaction of wave dislocations,” J. Opt. A: Pure Appl. Opt. 6, S251–S254 (2004).
[Crossref]
M. R. Dennis, “Local phase structure of wave dislocation lines: twist and twirl,” J. Opt. A: Pure Appl. Opt. 6, S202–S208 (2004).
[Crossref]
#### 2001 (3)
J. Masajada and B. Dubik, “Optical vortex generation by three plane wave interference,” Opt. Commun. 198, 21–27 (2001).
[Crossref]
M. V. Berry and M. R. Dennis, “Knotted and linked phase singularities in monochromatic waves,” Proc. R. Soc. A 457, 2251–2263 (2001).
[Crossref]
M. V. Berry and M. R. Dennis, “Knotting and unknotting of phase singularities: Helmholtz waves, paraxial waves and waves in 2+1 spacetime,” J. Phys. A: Math. Gen. 34, 8877–88 (2001).
[Crossref]
#### 2000 (1)
M. V. Berry and M. R. Dennis, “Phase singularities in isotropic random waves,” Proc. R. Soc. A 456, 2059–79 (2000).
[Crossref]
#### 1998 (1)
M. V. Berry, “Much ado about nothing: optical dislocation lines (phase singularities, zeros, vortices,” in . Intl. Conf. on Singular Optics, M. S. Soskin, ed., Proc. SPIE 3487, 1–15 (1998).
[Crossref]
#### 1989 (1)
K. Patorski, “The self-imaging phenomenon and its applications,” Progress in Optics XXVII, 103–108 (1989).
#### 1979 (1)
M. V. Berry, J. F. Nye, and F. J. Wright, “The elliptic umbilic diffraction catastrophe,” Phil. Trans. R. Soc. A 291, 453–84 (1979).
[Crossref]
#### 1974 (1)
J. F. Nye and M. V. Berry, “Dislocations in wave trains,” Proc. R. Soc. A 336, 165–90 (1974).
[Crossref]
#### 1836 (1)
W. H. F. Talbot, “Facts relating to optical science, No. IV,” London Edinburgh Dublin Philos. Mag. J. Sci. 9, 401–407 (1836).
#### Berry, M. V.
M. V. Berry and M. R. Dennis, “Knotted and linked phase singularities in monochromatic waves,” Proc. R. Soc. A 457, 2251–2263 (2001).
[Crossref]
M. V. Berry and M. R. Dennis, “Knotting and unknotting of phase singularities: Helmholtz waves, paraxial waves and waves in 2+1 spacetime,” J. Phys. A: Math. Gen. 34, 8877–88 (2001).
[Crossref]
M. V. Berry and M. R. Dennis, “Phase singularities in isotropic random waves,” Proc. R. Soc. A 456, 2059–79 (2000).
[Crossref]
M. V. Berry, “Much ado about nothing: optical dislocation lines (phase singularities, zeros, vortices,” in . Intl. Conf. on Singular Optics, M. S. Soskin, ed., Proc. SPIE 3487, 1–15 (1998).
[Crossref]
M. V. Berry, J. F. Nye, and F. J. Wright, “The elliptic umbilic diffraction catastrophe,” Phil. Trans. R. Soc. A 291, 453–84 (1979).
[Crossref]
J. F. Nye and M. V. Berry, “Dislocations in wave trains,” Proc. R. Soc. A 336, 165–90 (1974).
[Crossref]
#### Courtial, J.
J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Vortex knots in light,” New J. Phys. 7, 55 (2005).
[Crossref]
J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Knotted threads of darkness,” Nature 432, 165 (2004).
[Crossref] [PubMed]
#### Dennis, M. R.
J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Vortex knots in light,” New J. Phys. 7, 55 (2005).
[Crossref]
J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Knotted threads of darkness,” Nature 432, 165 (2004).
[Crossref] [PubMed]
M. R. Dennis, “Local phase structure of wave dislocation lines: twist and twirl,” J. Opt. A: Pure Appl. Opt. 6, S202–S208 (2004).
[Crossref]
M. V. Berry and M. R. Dennis, “Knotting and unknotting of phase singularities: Helmholtz waves, paraxial waves and waves in 2+1 spacetime,” J. Phys. A: Math. Gen. 34, 8877–88 (2001).
[Crossref]
M. V. Berry and M. R. Dennis, “Knotted and linked phase singularities in monochromatic waves,” Proc. R. Soc. A 457, 2251–2263 (2001).
[Crossref]
M. V. Berry and M. R. Dennis, “Phase singularities in isotropic random waves,” Proc. R. Soc. A 456, 2059–79 (2000).
[Crossref]
#### Dubik, B.
J. Masajada and B. Dubik, “Optical vortex generation by three plane wave interference,” Opt. Commun. 198, 21–27 (2001).
[Crossref]
#### Goodman, J. W.
J. W. Goodman, Statistical Optics (Wiley, 1985).
#### Leach, J.
J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Vortex knots in light,” New J. Phys. 7, 55 (2005).
[Crossref]
J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Knotted threads of darkness,” Nature 432, 165 (2004).
[Crossref] [PubMed]
#### Masajada, J.
J. Masajada and B. Dubik, “Optical vortex generation by three plane wave interference,” Opt. Commun. 198, 21–27 (2001).
[Crossref]
#### Nye, J. F.
J. F. Nye, “Local solutions for the interaction of wave dislocations,” J. Opt. A: Pure Appl. Opt. 6, S251–S254 (2004).
[Crossref]
M. V. Berry, J. F. Nye, and F. J. Wright, “The elliptic umbilic diffraction catastrophe,” Phil. Trans. R. Soc. A 291, 453–84 (1979).
[Crossref]
J. F. Nye and M. V. Berry, “Dislocations in wave trains,” Proc. R. Soc. A 336, 165–90 (1974).
[Crossref]
J. F. Nye, Natural focusing and fine structure of light (Institute of Physics Publishing,1999).
J. F. Nye, “Dislocation lines in the hyperbolic umbilic diffraction catastrophe,” Proc. R. Soc. A, in press (2006).
[Crossref]
#### Padgett, M. J.
J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Vortex knots in light,” New J. Phys. 7, 55 (2005).
[Crossref]
J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Knotted threads of darkness,” Nature 432, 165 (2004).
[Crossref] [PubMed]
#### Patorski, K.
K. Patorski, “The self-imaging phenomenon and its applications,” Progress in Optics XXVII, 103–108 (1989).
#### Talbot, W. H. F.
W. H. F. Talbot, “Facts relating to optical science, No. IV,” London Edinburgh Dublin Philos. Mag. J. Sci. 9, 401–407 (1836).
#### Wright, F. J.
M. V. Berry, J. F. Nye, and F. J. Wright, “The elliptic umbilic diffraction catastrophe,” Phil. Trans. R. Soc. A 291, 453–84 (1979).
[Crossref]
#### J. Opt. A: Pure Appl. Opt. (2)
J. F. Nye, “Local solutions for the interaction of wave dislocations,” J. Opt. A: Pure Appl. Opt. 6, S251–S254 (2004).
[Crossref]
M. R. Dennis, “Local phase structure of wave dislocation lines: twist and twirl,” J. Opt. A: Pure Appl. Opt. 6, S202–S208 (2004).
[Crossref]
#### J. Phys. A: Math. Gen. (1)
M. V. Berry and M. R. Dennis, “Knotting and unknotting of phase singularities: Helmholtz waves, paraxial waves and waves in 2+1 spacetime,” J. Phys. A: Math. Gen. 34, 8877–88 (2001).
[Crossref]
#### London Edinburgh Dublin Philos. Mag. J. Sci. (1)
W. H. F. Talbot, “Facts relating to optical science, No. IV,” London Edinburgh Dublin Philos. Mag. J. Sci. 9, 401–407 (1836).
#### Nature (1)
J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Knotted threads of darkness,” Nature 432, 165 (2004).
[Crossref] [PubMed]
#### New J. Phys. (1)
J. Leach, M. R. Dennis, J. Courtial, and M. J. Padgett, “Vortex knots in light,” New J. Phys. 7, 55 (2005).
[Crossref]
#### Opt. Commun. (1)
J. Masajada and B. Dubik, “Optical vortex generation by three plane wave interference,” Opt. Commun. 198, 21–27 (2001).
[Crossref]
#### Phil. Trans. R. Soc. A (1)
M. V. Berry, J. F. Nye, and F. J. Wright, “The elliptic umbilic diffraction catastrophe,” Phil. Trans. R. Soc. A 291, 453–84 (1979).
[Crossref]
#### Proc. R. Soc. A (3)
M. V. Berry and M. R. Dennis, “Knotted and linked phase singularities in monochromatic waves,” Proc. R. Soc. A 457, 2251–2263 (2001).
[Crossref]
J. F. Nye and M. V. Berry, “Dislocations in wave trains,” Proc. R. Soc. A 336, 165–90 (1974).
[Crossref]
M. V. Berry and M. R. Dennis, “Phase singularities in isotropic random waves,” Proc. R. Soc. A 456, 2059–79 (2000).
[Crossref]
#### Proc. SPIE (1)
M. V. Berry, “Much ado about nothing: optical dislocation lines (phase singularities, zeros, vortices,” in . Intl. Conf. on Singular Optics, M. S. Soskin, ed., Proc. SPIE 3487, 1–15 (1998).
[Crossref]
#### Progress in Optics (1)
K. Patorski, “The self-imaging phenomenon and its applications,” Progress in Optics XXVII, 103–108 (1989).
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J. F. Nye, “Dislocation lines in the hyperbolic umbilic diffraction catastrophe,” Proc. R. Soc. A, in press (2006).
[Crossref]
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### Supplementary Material (7)
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### Figures (6)
Fig. 1. Optical vortex line in an interference field. Left: periodic cell containing a vortex loop (dotted lines are parts of the loop outside the primitive cell). The intensity on the front and back planes is also represented. Right: phase cross section of the front/back face of the cell. The phase circulates in opposite directions at different points on the vortex line.
Fig. 2. Calculated vortex structure for three, four and five equal-amplitude plane waves (a, b, c) and their respective k-space configuration (d, e, f). Each is calculated on a 256×256×256 Talbot cell.
Fig. 3. The calculated vortex structures for four plane wave superpositions with amplitudes: a), a 1 + a 4 = a 2 + a 3 (b), a 1+a 4<a 2+a 3 (c), a 1+a 4 >a 2+a 3. The choice of k-vectors is the same as Fig. 2 (b), given in Fig. 2 (e). Attached multimedia shows rotation of cells around the optical axes (a, b, c) (1.4Mb, 1.3Mb, 1.1Mb).
Fig. 4. Experimentally measured vortex structures from the superposition of 4 plane waves with amplitudes giving rise to a) twisted vortex lines and b) vortex loops. The choice of k-vectors differs slightly from that in Fig. 3(b) and Fig. 3(c). Attached multimedia shows rotation of each cell around the optical axis (1.9Mb, 1.8Mb).
Fig. 5. Left: Vortex line in (x, y, z). Right: phasor diagrams at the points highlighted on a vortex a) loop and b) a line. For clarity, the orientation of the largest magnitude phasor has been fixed. The red arrows show the angular range of the smallest phasor, with respect to the largest. Note that in (a) this is restricted to less than 2π. Attached multimedia shows phasor configuration changing as the vortex paths are followed (2.3Mb, 2.1Mb).
Fig. 6. Examples of different vortex topologies that can be obtained from five waves of the same magnitude but different relative phase. Fig. 6 (a) is the same Talbot cell as Fig. 2(c). Figs. 6(b) and 6(c) are Talbot cells from the same wave magnitudes as (a) but with one wave advanced in phase by π/2 and π/3. Figs. 6(d),6(e), and 6(f) are their unwrapped counterparts. | 8,807 | 30,527 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-31 | latest | en | 0.877087 |
https://testbook.com/question-answer/an-exterior-angle-of-a-triangle-is-100-if-o--60091b7128da4995ec0c8c1e | 1,627,470,717,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153709.26/warc/CC-MAIN-20210728092200-20210728122200-00153.warc.gz | 571,585,271 | 20,563 | # An exterior angle of a triangle is 100º. If one of the opposite interior angles is 30º, then the other two angles of the triangle are
Free Practice With Testbook Mock Tests
## Options:
1. 40º, 110º
2. 70º, 80º
3. 60º, 90º
4. 50º, 100º
### Correct Answer: Option 2 (Solution Below)
This question was previously asked in
Official Paper 10: Tripura TET 2016 Paper 1
## Solution:
Given:
An exterior angle of a triangle is 100º
one of the opposite interior angles is 30º
Concept used:
The sum of all the angles of the triangle is 180°
Linear pair: The sum of the angle on the straight line is 180°
Calculation:
The angle on a line is 180º
⇒ c = 180º - 100° = 80°
⇒ a + b + c = 180°
⇒ 30° + b + 80° = 180°
⇒ b = 180° - 110° = 70°
∴ The other two angles of the triangle are 70°, 80° | 276 | 803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-31 | latest | en | 0.643821 |
http://iq-inteligencja.pl/ronnies-penrith-gvrw/0978f7-harley-davidson-street-glide-cvo | 1,632,158,846,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057083.64/warc/CC-MAIN-20210920161518-20210920191518-00492.warc.gz | 32,431,888 | 12,477 | {\displaystyle s>a} f Extreme Value Theorem If a function f(x) is continuous on the closed interval [a, b], then f(x) has an abosolute maximum and minimum on the interval [a, b]. : e say, which is greater than is bounded on so that We focus now to the analysis via GPD and the possible way to estimate VaR and ES. M ) Thus, these distributions are important in statistics. s − a In calculus, the extreme value theorem states that if a real-valued function $${\displaystyle f}$$ is continuous on the closed interval $${\displaystyle [a,b]}$$, then $${\displaystyle f}$$ must attain a maximum and a minimum, each at least once. This defines a sequence . is bounded, the Bolzano–Weierstrass theorem implies that there exists a convergent subsequence 1 x {\displaystyle x} , {\displaystyle [a,e]} , Thus {\displaystyle U_{\alpha _{1}},\ldots ,U_{\alpha _{n}}} Let f be continuous on the closed interval [a,b]. These three distributions are also known as type I, II and III extreme value distributions. The extreme value theorem cannot be applied to the functions in graphs (d) and (f) because neither of these functions is continuous over a closed, bounded interval. {\displaystyle U_{\alpha }} ) We first prove the boundedness theorem, which is a step in the proof of the extreme value theorem. The Rayleigh distribution method uses a direct calculation, based on the spectral moments of all the data. in {\displaystyle [a,b],} {\displaystyle b} The Heine–Borel theorem asserts that a subset of the real line is compact if and only if it is both closed and bounded. a in We conclude that EVT is an useful complemen t to traditional VaR methods. then all points between a f | f ) f δ {\displaystyle s+\delta \in L} interval , then has both a {\displaystyle M[a,e] [ 2 This is usually stated in short as "every open cover of M . which overlaps f maximum and a minimum on That is, there exists a point x1 in I, such that f(x1) >= f(x) for all x in I. a {\displaystyle f(x_{n})>n} L {\displaystyle [s-\delta ,s+\delta ]} f ( Here we want to review briefly the most common EVT approaches and models and look into some applications. iii) bounded . , f so that , [a,b]. Thevenin’s Theorem Basic Formula Electric Circuits; Thevenin’s Theorem Basic Formula Electric Circuits. ) , hence there exists {\displaystyle a} This means that for all Suppose the contrary viz. ) ] | A continuous real function on a closed interval has a maximum and a minimum, This article is about the calculus concept. {\displaystyle [a,b]} x {\displaystyle K} + a 2 in {\displaystyle s} x {\displaystyle f} d δ M is sequentially continuous at Fermat’s Theorem If fx has a relative (or local) extrema at x c, then x c is a critical point of fx . , we know that The function has an absolute maximum over $$[0,4]$$ but does not have an absolute minimum. It is used in mathematics to prove the existence of relative extrema, i.e. Inhaltsverzeichnis . M in {\displaystyle e} Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. a ( is continuous on . . Therefore, there must be a point x in [a, b] such that f(x) = M. ∎, In the setting of non-standard calculus, let N be an infinite hyperinteger. t The Extreme Value Theorem guarantees the existence of a maximum and minimum value of a continuous function on a closed bounded interval. a . = s [ {\displaystyle f} f can be chosen such that for all d . , such that K > ( a − x + ( x {\displaystyle f(a)-1} 0 Proof of the Extreme Value Theorem Theorem: If f is a continuous function defined on a closed interval [a;b], then the function attains its maximum value at some point c contained in the interval. α {\displaystyle m} U Thus, before we set off to find an absolute extremum on some interval, make sure that the function is continuous on that interval, otherwise we may be hunting for something that does not exist. point. R . {\displaystyle \delta >0} max. attains its supremum, or in other words f on the interval V i f {\displaystyle x} f {\displaystyle [a,s+\delta ]} f • P(Xu) = 1 - [1+g(x-m)/s]^(-1/g) for g <> 0 1 - exp[-(x-m)/s] for g = 0 • Parameters: – m = location – s = spread – g = shape – u = threshold. {\displaystyle [a,e]} s Mean value is easily distorted by extreme values/outliers. m is compact, then e ≥ x L {\displaystyle M-d/2} That is, there exists a point x1 in I, such that f(x1) >= f(x) for all x in I. We have seen that they can occur at the end points or in the open interval . i {\displaystyle f} [ {\displaystyle s} )} converges to f(d). i . b {\displaystyle \delta >0} α is bounded on {\displaystyle f(x)} If we then take the limit as $$n$$ goes to infinity we should get the average function value. ⊂ f Closed interval domain, … We must therefore have That is, there exist numbers x1 and x2 in [a,b] such that: The extreme value theorem does not indicate the value of… The Extreme value theorem states that if a function is continuous on a closed interval [a,b], then the function must have a maximum and a minimum on the interval. is another point in and consider the following two cases : (1) Visit my website: http://bit.ly/1zBPlvmSubscribe on YouTube: http://bit.ly/1vWiRxWHello, welcome to TheTrevTutor. s k ] Hence, by the transfer principle, there is a hyperinteger i0 such that 0 ≤ i0 ≤ N and V Therefore, f attains its supremum M at d. ∎. δ of points , hence there exists We call these the minimum and maximum cases, respectively. ( {\displaystyle [a,b]} ≤ ] , M B = U ] e where ( {\displaystyle x} M is bounded above by Extreme value distributions arise as limiting distributions for maximums or minimums (extreme values) of a sample of independent, identically distributed random variables, as the sample size increases.Thus, these distributions are important in probability and mathematical statistics. L + , then this theorem implies that ( is bounded on this interval. must attain a maximum and a minimum, each at least once. {\displaystyle B} ) {\displaystyle e} The GEV distribution unites the Gumbel, Fréchet and Weibull distributions into a single family to allow a continuous range of possible shapes. is bounded on f This is restrictive if the algorithm is used over a long time and possibly encounters samples from unknown new classes. it follows that the image must also ( f , , − [ > L In this paper we apply Univariate Extreme Value Theory to model extreme market riskfortheASX-AllOrdinaries(Australian)indexandtheS&P-500(USA)Index. , Let us call it , there exists an say, which is greater than and + N ( which overlaps ] − {\displaystyle f(x)\leq M-d_{2}} {\displaystyle d_{n_{k}}} Note that for this example the maximum and minimum both occur at critical points of the function. x In calculus, the general Leibniz rule, named after Gottfried Wilhelm Leibniz, generalizes the product rule (which is also known as "Leibniz's rule"). s increases from ] Given topological spaces {\displaystyle M[a,x]} / we can deduce that a K x 2 1. R . For the statistical concept, see, Functions to which the theorem does not apply, Generalization to metric and topological spaces, Alternative proof of the extreme value theorem, Learn how and when to remove this template message, compact space#Functions and compact spaces, "The Boundedness and Extreme–Value Theorems", http://mizar.org/version/current/html/weierstr.html#T15, Regiomontanus' angle maximization problem, List of integrals of exponential functions, List of integrals of hyperbolic functions, List of integrals of inverse hyperbolic functions, List of integrals of inverse trigonometric functions, List of integrals of irrational functions, List of integrals of logarithmic functions, List of integrals of trigonometric functions, https://en.wikipedia.org/w/index.php?title=Extreme_value_theorem&oldid=1000573202, Short description is different from Wikidata, Articles lacking in-text citations from June 2012, Articles with unsourced statements from June 2011, Creative Commons Attribution-ShareAlike License. This means that . ) {\displaystyle x} {\displaystyle x_{n}\in [a,b]} In the proof of the boundedness theorem, the upper semi-continuity of f at x only implies that the limit superior of the subsequence {f(xnk)} is bounded above by f(x) < ∞, but that is enough to obtain the contradiction. . {\displaystyle M-d/2} History. We note that ] s {\displaystyle [a,x]} f {\displaystyle d_{1}} In most traditional textbooks this section comes before the sections containing the First and Second Derivative Tests because many of the proofs in those sections need the Mean Value Theorem. The extreme value type I distribution has two forms. s is closed and bounded for any compact set , a {\displaystyle x} [ {\displaystyle [s-\delta ,s+\delta ]} so that all these points belong to is said to be compact if it has the following property: from every collection of open sets is continuous on the right at δ There has been rapid development over the last decades in both theory and applications. {\displaystyle f^{*}(x_{i_{0}})\geq f^{*}(x_{i})} 1. {\displaystyle M} (2) then we are done. , the existence of the lower bound and the result for the minimum of x is continuous on say, belonging to also belong to The absolute maximum is shown in red and the absolute minimumis in blue. ) Extreme Value Theory for Time Series using Peak-Over-Threshold method - Gianluca Rosso (2015) 3 () = ( | O) (23) Now, we can consider that one of the best way to analyze the peak of our time series is the POT method. interval I=[a,b]. This relation allows to study Hitting Time Statistics with tools from Extreme Value Theory, and vice versa. inf in a − x ] Unlimited random practice problems and answers with built-in Step-by-step solutions. + = on the interval {\displaystyle s} {\displaystyle s} {\displaystyle [s-\delta ,s]} ) 3.3 Increasing and Decreasing Functions. {\displaystyle d_{1}} ⊃ {\displaystyle L} Among all ellipses enclosing a fixed area there is one with a smallest perimeter. {\displaystyle s 0 (Frechet, heavy tailed) type III with ξ < 0 (Weibull, bounded) =exp− s+�� − . on the interval ) s = b { \displaystyle x } total ozone data do not adequately address the structure the! Each fails to attain a maximum and a minimum on the interval using the first derivative and guidleines! Which is a non-empty interval, then has both a maximum on the same interval is similarly. Distribution is also compact very large literature written during last years are also as! 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To attain a maximum and minimum both occur at critical points of the point we are seeking i.e | may! Method we suggest to refer to the supremum only if it is used to thing! Identifies candidates for local Extreme-Value points 1 tool for creating Demonstrations and anything technical forecasts derived EVT... Very small or very large literature written extreme value theorem formula last years use the derivative to determine intervals on which value! To show that There must be a finite maximum value on$ [ a, b $. Value type I distribution is chosen: on a closed interval [ 0 d! And answers with built-in step-by-step solutions over a long Time and possibly samples! For example, a metric space has the Heine–Borel theorem asserts that a is. Address the structure of the extreme value theorem: [ 2 ] ;! < M { \displaystyle a } 0.1 1 10 100 turns out that multi-period VaR forecasts distribution method a... Learn the extreme value theorem: [ 2 ] s2is a long-term average value of the theorem ''! 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Values can be a very small or very large literature written during last.... Maximum over \ ( n\ ) goes to infinity we should get the average function value is... Are -3.7, 1.07 cite | improve this question | follow | asked may '15! Local extremum at a boundary point -coordinate of the subsequence converges to the.... See from the above that s > a { \displaystyle s } completes... Converges to the very large literature written during last years refer to the supremum German market. Very rare or extreme events Formula Electric Circuits answers with built-in step-by-step solutions the list isn ’ t,... We then take the limit as \ ( [ 0,4 ] \ ) but not... Ecosystems, etc show that s > a { \displaystyle s > a { \displaystyle s.! A maximum and minimum both occur at the proof that $f$ attains its minimum the! The extrema on a extreme value theorem formula interval can be shown to preserve compactness: [ 2 ] standard! Be shown to preserve compactness: [ 2 ] the structure of the function must. 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Fact find an extreme value Theory, and vice versa described in the usual sense {... Moments of all the data which extreme value type I distribution is chosen.! ( this Generalised Pareto distribution German hog market the result was also discovered later by Weierstrass in 1860 known type., Normal Curvature at a Regular point of a maximum and a minimum, this article is about the of... Which can distort the mean value theorem. I distribution has two forms is therefore fundamental to develop algorithms to... Distribution unites the Gumbel distribution use continuity to show that s > a { \displaystyle }. Or very large value which can distort the mean value theorem. b } we can fact. Tells us that we can deduce that s > a { \displaystyle f } is interval... 4-7: the mean the usual sense that the image below shows continuous. And thus the VaR forecasts learn the extreme value theorem tells us that we in! \Displaystyle L } is a step in the open interval, then f is above! Time statistics with tools from extreme value theorem. ( a ) =M } then we are done |! Application of EVT is an useful complemen extreme value theorem formula to traditional VaR methods adequately... Extremum on an open interval, then the extremum occurs at a Regular of! By b { \displaystyle s } is the point where the function 15 January 2021, at 18:15 to thing! Rare or extreme events ) indexandtheS & P-500 ( USA ) Index true for an upper semicontinuous function via and... Find an extreme value Theory to model extreme market riskfortheASX-AllOrdinaries ( Australian ) &. Global extremum occurs at a critical point a way to estimate VaR and ES its supremum type. Theory, and vice versa ) of total ozone data do not adequately address the structure of the extreme distributions... Of s { \displaystyle s > a } to estimate VaR and ES identifies! A permanent a metric space has the Heine–Borel theorem asserts that a function s = b \displaystyle. | 4,539 | 19,169 | {"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": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-39 | latest | en | 0.901795 |
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matlab
Plane 1.1. The bases of a trapezoid are $a$ and $b$. Find the length of the segment that the diagonals of the trapezoid intersept on the trapezoid’s midline.
Solutions
1.1. a) Let $P$ and $Q$ be the midpoints of $A B$ and $C D$; let $K$ and $L$ be the intersection points of $P Q$ with the diagonals $A C$ and $B D$, respectively. Then $P L=\frac{a}{2}$ and $P K=\frac{1}{2} b$ and so $K L=P L-P K=\frac{1}{2}(a-b)$.
b) Take point $F$ on $A D$ such that $B F | C D$. Let $E$ be the intersection point of $M N$ with $B F$. Then
$$M N=M E+E N=$$
$$\frac{q \cdot A F}{p+q}+b=\frac{q(a-b)+(p+q) b}{p+q}=\frac{q a+p b}{p+q}$$
1.2. Prove that the midpoints of the sides of an arbitrary quadrilateral are vertices of a parallelogram. For what quadrilaterals this parallelogram is a rectangle, a rhombus, a square?
1.2. Consider quadrilateral $A B C D$. Let $K, L, M$ and $N$ be the midpoints of sides $A B$, $B C, C D$ and $D A$, respectively. Then $K L=M N=\frac{1}{2} A C$ and $K L | M N$, that is $K L M N$ is a parallelogram. It becomes clear now that $K L M N$ is a rectangle if the diagonals $A C$ and $B D$ are perpendicular, a rhombus if $A C=B D$, and a square if $A C$ and $B D$ are of equal length and perpendicular to each other.
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https://www.physicsforums.com/threads/relative-motion-with-vectors.859358/ | 1,508,192,244,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820466.2/warc/CC-MAIN-20171016214209-20171016234209-00182.warc.gz | 1,219,872,332 | 17,061 | # Relative Motion With Vectors
Tags:
1. Feb 25, 2016
1. The problem statement, all variables and given/known data
A ball rolls with a velocity of 14 mm/s [W] on a board that is being pulled [E 60o N] at 20.0 mm/s. What is the velocity of the ball relative to the floor?
b = ball
x = board
f = floor
2. Relevant equations
V_bf = V_bx + V_xf
3. The attempt at a solution
I tried to work through this question by drawing it on a Cartesian grid and by drawing out the vectors. I used trigonometry on the triangle to figure out the component of V_xf horizontally, and then plugging V_xf into the V_bf = V_bx + V_xf equation. The answer I got was 4 mm/s W, but is was incorrect. I am not sure where I went wrong.
2. Feb 25, 2016
### haruspex
Strangely, neither am I. Why might that be?
3. Feb 28, 2016
### ehild
The velocity vector has both westward and northward components.
Do you mean 60° East of North as the direction of velocity of the board?
4. Feb 29, 2016
### HallsofIvy
Staff Emeritus
60 degrees east of north would be written as N60E. E60N is not a standard notation ("standard notation" goes from N or S first) but I would interpret it as "60 degrees north of east" which would be the same as "30 degrees east of north" or N30E. If you want someone to explain "what you did wrong", you will have to tell us what you did! | 384 | 1,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-43 | longest | en | 0.960192 |
http://mymathforum.com/algebra/39146-maximum-value.html | 1,569,253,111,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514577363.98/warc/CC-MAIN-20190923150847-20190923172847-00172.warc.gz | 136,320,512 | 8,769 | My Math Forum maximum value
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October 28th, 2013, 02:57 AM #1 Member Joined: Nov 2012 Posts: 61 Thanks: 0 maximum value If a,b are natural numbers such that $2013 +a^2= b^2,$ then the maximum possible value of ab is
October 28th, 2013, 03:31 AM #2 Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: maximum value Here is an 'intuitive' approach. The difference between consecutive integer squares is in arithmetic progression of the odd integers. $1^2 \ - \ 0^2 \ = \ 1 \\ 2^2 \ - \ 1^2 \ = \ 3 \\ 3^2 \ - \ 2^2 \ = \ 5 \\ 4^2 \ - \ 3^2 \ = \ 7$ . . . $(n + 1)^2 \ - \ n^2 \= \ 2n \ + \ 1$ So we want 2n + 1 = 2013 which makes n = 1006 $2013 \ + \ 1006 ^2 \= \ 1007^2$ ab = 1006*1007 = 1013042 This is the maximum since it gives $2013 \= \ b^2 \ - \ a^2$ $2013 \= \ (b \ + \ a)(b \ - \ a)$ $2013 \= \ (1007 \ + \ 1006)(1007 \ - \ 1006)$ $2013 \= \ (2013)(1)$ WE can't do better than that because there are no integer squares between consecutive integer squares.
November 8th, 2013, 02:29 AM #3 Member Joined: Nov 2012 Posts: 61 Thanks: 0 Re: maximum value Can you tell what will be the minimum value?
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Contact - Home - Forums - Cryptocurrency Forum - Top | 584 | 1,666 | {"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": 8, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2019-39 | latest | en | 0.814709 |
https://www.datanumen.com/ai-story-generator/the-boy-and-the-chessboard-a-magical-journey-in-3d-world/ | 1,713,583,615,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817474.31/warc/CC-MAIN-20240420025340-20240420055340-00888.warc.gz | 665,676,850 | 27,334 | # The Boy and The Chessboard: A Magical Journey in 3D World
## Section 1: Introduction
Jack, a nimble and sharp-minded 10-year-old boy, lived in a small town, far removed from the hustle and bustle of the city. His humble abode was nestled between scenic landscapes and vast green fields. However, what intrigued Jack more than this scenic beauty was a game – the game of chess.
Chess was not merely a game for Jack, it was a mystery, a problem to be solved, a world to be explored. Jack was endowed with a sharp intellect, a keen sense for patterns and an innate gift for mathematics. His world revolved around equations, patterns, numbers, and variables.
Jack would often spend hours engrossed in a challenging game of chess with his grandfather, a retired army general and a proficient chess player. His grandfather was his mentor, who recognized Jack’s unique talent early on and started honing it. He introduced Jack to diverse strategy games, logical reasoning puzzles, but nothing fascinated Jack more than the chessboard and its sixty-four black and white squares. Jack was captivated by the strategic depth, the myriad possibilities, the unusual movements of the chess pieces, and the riveting endgames.
Among all chess pieces, Jack held a special fascination for the knight, a piece that moves in an unconventional pattern compared to others. His interest sparked even more when his grandfather introduced him to the “Knight’s Tour,” a sequence where the knight visits each square on the chessboard exactly once. This strategic puzzle was the beginning of an extraordinary journey for Jack into the magical world of geometry and patterns.
## Section 2: Discovery of Tessellations
On a typical afternoon, Jack was engrossed in his usual knight tour exercises. With every move he made, he meticulously traced the knight’s path, marking each stop with a dedicated colored pencil. Somewhere in those rhythmic moves, Jack noticed an uncanny pattern unfolding – each time the knight made its complete tour, it left behind an intricate pattern on the chessboard.
These patterns were not random, nor chaotic. They were tessellations – repeating patterns of shapes perfectly fitting together without gaps or overlaps. Observing these tessellations brought about a whole new perspective for Jack on the chessboard. It was no longer just a checkered board of sixty-four squares, it had transformed into a canvas where each knight’s move was a stroke of paint filling it with beautiful tessellations.
Intrigued and excited with this revelation, Jack delved deeper into understanding and decoding the mystery behind these changing patterns. Equipped with a powerful analytical mind and a strong sense for patterns, he looked up books, read online articles, and extensively analyzed his numerous played games. He tirelessly sought to comprehend the inner workings of these tessellations.
Hours turned into days and days into weeks, as Jack furiously worked with purpose, his mind preoccupied with the bewitching geometry of the chessboard. Each exercise filled the chessboard with dynamic tessellations, driving Jack further into a world of patterns and geometry that emerged from his beloved chessboard.
## Section 3: Exploration of the Patterns
The patterns unfolding from the knight’s tour occupied Jack’s mind day in and day out. He would spend dozens of hours tracing the paths, trying to decipher the tessellations, and exploring the large array of possible arrangements. Each knight’s tour was equivalent to an exciting treasure hunt for him, where hidden within the black and white squares lied a panorama of geometric patterns.
After numerous trials with different start and end points, Jack found that the tessellations weren’t just random by-effects. They were the tangible impressions of the knight’s journey that bore magical meaning. To the innocent yet wise mind of Jack, these patterns seemed to form a kind of coded language, a secret key to an extraordinary reality – a dimension which was not visible to the naked eye.
One day, while quietly observing the patterns he had traced over several games, an idea germinated in his head. He realized that while each square was a point of visit, the journey of the knight between these points could be visualized in three dimensions. The concept of 2D and 3D came rushing into his mind and he imagined a new world, where the chessboard extended not just horizontally but also vertically.
This epiphany marked a turning point in Jack’s experiments. The patterns weren’t just patterns anymore – they were a portal, a gateway to an extraordinary dimensional reality. The knight’s tour turned from a simple chess puzzle into a voyage through this newly discovered, magical dimension.
## Section 4: A New Dimension
Armed with the realization of a new dimension hidden within the chessboard, Jack was ecstatic. Using the sheer power of his imagination and the mathematical acumen at his disposal, he began to explore this untrod dimensional reality. He sketched out the movements of the knight, resembling spiraling staircases and towering constructions, penning down the three-dimensional path of the knight on paper.
The knight’s prancing became dramatic plunges into depth or exhilarating ascents into the skies. Each square in the knight’s path became an intricate entity of space, having not only length and breadth but also height. The chessboard was no longer just a mere checkered board of sixty-four squares; it had transformed into an endless cube of countless geometric possibilities.
Jack found himself unlocking dimension after dimension as he maneuvered the knight across this newfound reality. Each game, each knight’s tour was an exciting expedition into the heart of this thrilling maze of endless chess cubes. The complexity of the game escalated, and with it, Jack’s fascination and engagement.
Indeed, Jack was no longer just playing chess – he was creating art, architecting spatial structures of magnificent beauty. The two-dimensional domain of the chessboard had become a multi-dimensional labyrinth, teeming with unlimited patterns, endless pathways and countless geometric wonders. The transformation was nothing short of magical. Jack had indeed found a whole new world within the confines of the sixty-four squares.
## Section 5: Commencement of an Extraordinary Journey
Filled with the exhilaration of discovery and the eagerness of a pioneer voyager, Jack set forth on his exploration journey through this amazing new dimension. The familiar checkered chessboard had morphed into an uncharted territory – a 3D realm of infinite possibilities. His heart brimming with child-like wonder, Jack ventured courageously into this magical world.
Everything around him was imbued with an air of fantasy. The chess pieces were no longer constrained to their flat squares, they floated freely in this space, defying the laws of gravity. They created mysterious pathways, weaving an intricate tapestry of complex patterns that painted a stunning landscape before Jack’s eyes. It was like stepping into a fantastical world straight out of a fairy tale – full of excitement, myriad twists and turns, and unimaginable wonders that surpassed the ordinary chessboard.
Jack’s analytical mind worked in tandem with his active imagination, constantly translating the complex 2D operations into three dimensional plots. Each knight’s tour became a sensory experience, manipulating dimensions with deftness and creating marvels which were hitherto unknown.
This extraordinary journey was by no means a fleeting intrigue. It symbolized the commencement of a whole new phase in Jack’s life. It was the beginning of Jack’s love affair with dimensions – the dawn of his journey into the fascinating world of geometry and spatial understanding. And every step forward was a mesmerizing dance between his creativity and his brilliantly analytical mind.
## Section 6: Conclusion
It was indeed an extraordinary journey for a 10-year-old boy, who transformed a simple game of chess into a voyage through an enchanting dimensional reality. Jack’s discovery was a testament to his sharp intellect, analytical mindset and unquenchable curiosity.
Through his fascinating expedition, he not only uncovered the beauty of mathematics and patterns but also the elusive concept of 3D geometry, all encoded within the simplicity of a chessboard. The impression of knight’s movements took him far beyond the confines of the checkered board into a spatial spectacle full of geometrical complexities.
Jack’s tale illustrates a profound truth – extraordinary can be found within the ordinary for those who dare to look closely, to question, to explore. What seemed like a simple game was actually a trove of mathematical and geometrical wonders. This discovery spoke volumes about the immense potential and magic that resides within seemingly mundane things.
Moreover, this eye-opening journey was just the beginning of many more explorations and discoveries to follow. It instilled a sense of wonder and curiosity in Jack’s youthful heart and fueled his passion for continual learning. His love for patterns, geometry, dimensions, and the very essence of space deepened, setting a firm foundation for his exciting journey ahead in the world of mathematics and beyond. | 1,800 | 9,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.96401 |
https://brightside.me/articles/8-puzzles-that-might-keep-you-up-all-night-524960/?show_all_comments | 1,721,569,231,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517701.96/warc/CC-MAIN-20240721121510-20240721151510-00840.warc.gz | 119,664,672 | 95,071 | # 7 Puzzles That Might Keep You Up All Night
Quizzes
2 years ago
Some research has proven that solving riddles and puzzles is exceptionally beneficial for your brain. It sharpens your mind, improves your memory, and teaches you to pay attention to small details. And it’s simply fun. Enjoy these riddles with a little twist.
We at Bight Side love riddles. Read all the information, the answer is right there. In some puzzles we deliberately included more details than necessary, you just have to fish out what you need. Good luck.
## 1. Candle mystery
Once an old man and his 3 daughters lived in a village. One day the old man got very sick so Death came to take his spirit. The oldest daughter asked Death to let her father live for 2 years and Death agreed. 2 years later Death returned, the second daughter appealed and Death agreed to grant the old man one more year.
A year later Death returned and now the youngest daughter came with a candle and asked Death to allow her father to live until the candle in her hand burned out. Death agreed. As the youngest daughter planned, Death never returned.
Question. How did she plan it?
The youngest daughter blew out the candle after Death left and the candle never burned out.
Paul was standing on a 60-ft ladder. He slipped from the ladder and fell to the ground. But he didn’t get badly injured.
Question. How is this possible?
he was only on the bottom rung
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Paul was standing on the bottom rung of the ladder.
## 3. Coins mystery
You have 2 buckets of water, inside the 1st bucket the temperature of the water is 25ºC, inside the 2nd bucket the water is 25ºF. You drop a coin into each bucket from the same height and they hit water at exactly the same time.
Question. Which coin touches the bottom first?
The coin in the 1st bucket. At 25ºC water is liquid, while at 25ºF it turns into ice.
## 4. Grass mystery
In a field, there is a patch of magic grass, which doubles in quantity every day. If on day one it’s one patch of grass, on day 2 there will be 2 patches and so on. It takes 10 days for the grass to cover the entire field.
Question. How long will it take for it to cover half of the field?
The grass doubles in quantity every day, so on the 9th day it will be half the quantity it is on the 10th day. Hence the half of the patch will be covered on the 9th day.
## 5. An apple mystery
There are 10 people in a room and each person can see the entire room and everyone else. You want to place an apple so all but one person can see it.
Question. Where would you place the apple in the room?
Place the apple on anyone’s head.
## 6. Car vs car
There are 2 cars, the blue car is going 60mph and the red car is going 40mph. They both started at the same time however they still cross each other at some point.
Question. How is that possible?
They are travelling in opposite directions.
## 7. Apple tree mystery
2 friends Ann and Mary went for a bush walk. Suddenly heavy rain started, the girls saw an abandoned house and decided to get shelter in the house. When the rain stopped, the friends were dry but very hungry. So, they went different ways to find fruit.
Ann found an apple tree but when she ate an apple she turned into the tree. Half an hour later Mary got there and she heard someone say: “Your friend turned into a tree, you must find her to bring her back. You only have one chance. If you get it wrong, your friend will remain a tree forever.”
Question. How can Mary find out which tree is her friend?
Since it rained half an hour back, all the trees in the forest must be wet. The only tree that is not wet is Mary’s friend Ann.
Have you cracked them all? Tell us about it in the comments and share these puzzles with your friends so we can all have some fun.
I disagree with the last one. There are 3 men, each one carrying the log on his shoulder. The weight gets distributed by proportion, but you have to admit that end ones are not positioned at the edge of the log, but they have a certain portion "hanging". If that portion is equal to half distance to the middle guy, they are all carrying the same. Buy the way the picture looks, it is not far from the truth.
1
1
Shut
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Why are you so rude?
???
🐱
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3 years ago
Wrong article
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- | 1,136 | 4,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.882889 |
https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_15&diff=prev&oldid=153615 | 1,632,574,769,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057622.15/warc/CC-MAIN-20210925112158-20210925142158-00088.warc.gz | 158,590,292 | 15,643 | # Difference between revisions of "2021 AIME II Problems/Problem 15"
## Problem
Let $f(n)$ and $g(n)$ be functions satisfying $$f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}$$and $$g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}$$for positive integers $n$. Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$.
## Solution 1
Consider what happens when we try to calculate $f(n)$ where n is not a square. If $k^2 for (positive) integer k, recursively calculating the value of the function gives us $f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n$. Note that this formula also returns the correct value when $n=(k+1)^2$, but not when $n=k^2$. Thus $f(n)=k^2+3k+2-n$ for $k^2.
If $2 \mid (k+1)^2-n$, $g(n)$ returns the same value as $f(n)$. This is because the recursion once again stops at $(k+1)^2$. We seek a case in which $f(n), so obviously this is not what we want. We want $(k+1)^2,n$ to have a different parity, or $n, k$ have the same parity. When this is the case, $g(n)$ instead returns $(k+2)^2-n+g((k+2)^2)=k^2+5k+6-n$.
Write $7f(n)=4g(n)$, which simplifies to $3k^2+k-10=3n$. Notice that we want the $LHS$ expression to be divisible by 3; as a result, $k \equiv 1 \pmod{3}$. We also want n to be strictly greater than $k^2$, so $k-10>0, k>10$. The LHS expression is always even (why?), so to ensure that k and n share the same parity, k should be even. Then the least k that satisfies these requirements is $k=16$, giving $n=258$.
Indeed - if we check our answer, it works. Therefore, the answer is $\boxed{258}$.
-Ross Gao
## Solution 2 (More Variables)
We restrict $n$ in which $k^2 for some positive integer $k,$ or $$n=(k+1)^2-p\hspace{40mm}(1)$$ for some nonnegative integer $p.$ By observations, we get \begin{align*} f\left((k+1)^2\right)&=k+1, \\ f\left((k+1)^2-1\right)&=k+2, \\ f\left((k+1)^2-2\right)&=k+3, \\ &\cdots \\ f\bigl(\phantom{ }\underbrace{(k+1)^2-p}_{n}\phantom{ }\bigr)&=k+p+1. \hspace{19mm}(2) \\ \end{align*} If $n$ and $(k+1)^2$ have the same parity, then starting with $g\left((k+1)^2\right)=k+1,$ we get $g(n)=f(n)$ by a similar process. This contradicts the precondition $\frac{f(n)}{g(n)} = \frac{4}{7}.$ Therefore, $n$ and $(k+1)^2$ must have different parities, from which $n$ and $(k+2)^2$ must have the same parity.
Along with the earlier restriction, we obtain $k^2 or $$n=(k+2)^2-2q\hspace{38.25mm}(3)$$ for some positive integer $q.$ By observations, we get \begin{align*} g\left((k+2)^2\right)&=k+2, \\ g\left((k+2)^2-2\right)&=k+4, \\ g\left((k+2)^2-4\right)&=k+6, \\ &\cdots \\ g\bigl(\phantom{ }\underbrace{(k+2)^2-2q}_{n}\phantom{ }\bigr)&=k+2q+2. \hspace{15.5mm}(4) \\ \end{align*} By $(2)$ and $(4),$ we have $$\frac{f(n)}{g(n)}=\frac{k+p+1}{k+2q+2}=\frac{4}{7}. \hspace{27mm}(5)$$ From $(1)$ and $(3),$ equating the expressions for $n$ gives $(k+1)^2-p=(k+2)^2-2q.$ Solving for $k$ produces $$k=\frac{2q-p-3}{2}. \hspace{41.25mm}(6)$$ We substitute $(6)$ into $(5),$ then simplify, cross-multiply, and rearrange: \begin{align*} \frac{\tfrac{2q-p-3}{2}+p+1}{\tfrac{2q-p-3}{2}+2q+2}&=\frac{4}{7} \\ \frac{p+2q-1}{-p+6q+1}&=\frac{4}{7} \\ 7p+14q-7&=-4p+24q+4 \\ 11p-11&=10q \\ 11(p-1)&=10q. \hspace{29mm}(7) \end{align*} Since $\gcd(11,10)=1,$ we know that $p-1$ must be divisible by $10,$ and $q$ must be divisible by $11.$ Furthermore, rewriting the restriction $k^2 in terms of $p$ and $q$ gives \begin{align*} k^2&<\underbrace{(k+2)^2-2q}_{\text{by }(3)} \\ q&<2k+2 &\hspace{5.5mm}(8) \\ q&<2\biggl(\phantom{ }\underbrace{\frac{2q-p-3}{2}}_{\text{by }(6)}\phantom{ }\biggr)+2 \\ p+1& Combining $(8)$ and $(9),$ we get the compound inequality $$p+1 Note that if we start with $k^2<\underbrace{(k+1)^2-p}_{\text{by }(1)}$ instead, then we get $p<2k+1$ and $p both of which agree with $(10).$ In order to minimize $n,$ we should minimize $k.$ By $(10),$ we should minimize $p$ and $q.$ From $(7),$ we construct the following table: $\[\begin{array}{c|c|c} & & \\ [-2.5ex] \boldsymbol{p} & \boldsymbol{q} & \textbf{Satisfies }\boldsymbol{(9)?} \\ [0.5ex] \hline & & \\ [-2ex] 11 & 11 & \\ 21 & 22 & \\ 31 & 33 & \checkmark \\ \geq41 & \geq44 & \checkmark \\ \end{array}$$ We have $(p,q)=(31,33).$ Substituting this result into $(6)$ produces $k=16.$ Finally, substituting everything into either $(1)$ or $(3)$ generates $n=\boxed{258}.$
Remark
We can verify that $$\frac{f(258)}{g(258)}=\frac{31+\overbrace{f(289)}^{17}}{2\cdot33+\underbrace{f(324)}_{18}}=\frac{48}{84}=\frac47.$$
~MRENTHUSIASM | 1,810 | 4,625 | {"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": 93, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2021-39 | latest | en | 0.666564 |
https://www.codecademy.com/courses/learn-javascript-control-flow/lessons/control-flow/exercises/comparison-operators?action=lesson_resume&course_redirect=introduction-to-javascript | 1,656,382,377,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103347800.25/warc/CC-MAIN-20220628020322-20220628050322-00113.warc.gz | 760,893,863 | 34,022 | Learn
When writing conditional statements, sometimes we need to use different types of operators to compare values. These operators are called comparison operators.
Here is a list of some handy comparison operators and their syntax:
• Less than: `<`
• Greater than: `>`
• Less than or equal to: `<=`
• Greater than or equal to: `>=`
• Is equal to: `===`
• Is not equal to: `!==`
Comparison operators compare the value on the left with the value on the right. For instance:
``10 < 12 // Evaluates to true``
It can be helpful to think of comparison statements as questions. When the answer is “yes”, the statement evaluates to `true`, and when the answer is “no”, the statement evaluates to `false`. The code above would be asking: is 10 less than 12? Yes! So `10 < 12` evaluates to `true`.
We can also use comparison operators on different data types like strings:
``'apples' === 'oranges' // false``
In the example above, we’re using the identity operator (`===`) to check if the string `'apples'` is the same as the string `'oranges'`. Since the two strings are not the same, the comparison statement evaluates to `false`.
All comparison statements evaluate to either `true` or `false` and are made up of:
• Two values that will be compared.
• An operator that separates the values and compares them accordingly (`>`, `<`, `<=`,`>=`,`===`,`!==`).
Let’s practice using these comparison operators!
### Instructions
1.
Using `let`, create a variable named `hungerLevel` and set it equal to `7`.
2.
Write an `if...else` statement using a comparison operator. The condition should check if `hungerLevel` is greater than `7`. If so, the conditional statement should log, `'Time to eat!'`. Otherwise, it should log `'We can eat later!'`.
After you press run, play around with the condition by tweaking the comparison of `hungerLevel` by using different operators such as `<=`,`>=`,`>`, and `<`. | 450 | 1,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-27 | latest | en | 0.867062 |
https://forum.math.toronto.edu/index.php?PHPSESSID=33b0kvfl3cs01p0mtsrfql0vk1&action=printpage;topic=1977.0 | 1,624,485,916,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488540235.72/warc/CC-MAIN-20210623195636-20210623225636-00308.warc.gz | 246,655,518 | 2,428 | # Toronto Math Forum
## MAT244--2019F => MAT244--Test & Quizzes => Quiz-3 => Topic started by: Yingyingz on October 11, 2019, 02:00:00 PM
Title: TUT5103 Quiz3
Post by: Yingyingz on October 11, 2019, 02:00:00 PM
Find the solution of the given initial problem.
$$y^{\prime \prime}+y^{\prime}-2 y=0, y(0)=1, y^{\prime}(0)=1$$
We assume $y=e^{r t}$, then $r$ must be a root of the characteristic equation:
$$\begin{array}{l}{r^{2}+r-2=0} \\ \therefore (r-1)(r+2)=0 \\ {r_{1}=1 \quad r_{2}=-2}\end{array}$$
$\therefore$ the general solution is $y=c_{1} e^{t}+c_{2} e^{-2 t}$
$$\begin{array}{l}{y^{\prime}=c_{1} e^{t}-2 c_{2} e^{-2 t}} \\ {\operatorname{plug} \text { in } y(0)=1, y^{\prime}(0)=1}\end{array}$$
$$\left\{\begin{array}{l}{C_{1}+C_{2}=1\qquad {{\small 1}}} \\ {C_{1}-2 C_{2}=1\quad~~ {{\small 2}}}\end{array}\right.$$
${{\small 1}}-{{\small 2}}$ , we get
$$\begin{array}{r}{3 C_{2}=0} \\ {C_{2}=0}\end{array}$$
$$\therefore C_{1}=1$$
$\therefore$ the solution is $y=e^{t}$ | 427 | 986 | {"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} | 4.28125 | 4 | CC-MAIN-2021-25 | latest | en | 0.650141 |
https://www.physicsforums.com/threads/cantilever-beam-failure-as-a-function-of-force-applied-to-free-end.649596/ | 1,519,498,873,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815918.89/warc/CC-MAIN-20180224172043-20180224192043-00549.warc.gz | 923,132,243 | 16,076 | # Cantilever Beam Failure as a function of Force Applied to Free End
1. Nov 4, 2012
### seuss
Hello,
First post, go easy.
Trying to do some basic calculations on how to find when a steel cantilever beam will yield when a load is applied to the free end of the beam. I have attached my best MS Paint diagram of the problem. Assume the left end is clamped and essentially a fixed end. Failure is when the "beam" permanently deforms (yields)
E - Modulus of Elasticity 200 GPa 29000 ksi
G - Shear Modulus 80 GPa
I - (1/12)(25mm)(4mm)^3 133.33 mm^4
L - 60 mm
I have done typical axial stress-strain problems in undergrad, and I have done bending moment, shear, deflection, diagrams of beams, but never analyzed a beam in this way for failure.
I think if calculate the shear stress at O as a function of the load applied at the other end, L, and find when that shear exceeds the shear stress for yielding, that might be the answer I'm looking for. BUT, how do I determine the yield point? I know it's going to be very clear once I get the problem framed correctly, but I'm not connecting all the dots.
Any validity to this line of thinking?
This isn't a homework problem, it's for a project I'm working on.
The application is a latch securing between a door and a frame with by a rotating arm (cam).
Need some preliminary numbers. Help me get started! Thanks!
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6.9 KB
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1,256
2. Nov 4, 2012
### SteamKing
Staff Emeritus
What makes you think the beam is going to fail in shear before the bending stress
at the clamped end reached yield?
3. Nov 4, 2012
### AlephZero
For beam bending, the usual way to exceed the yield stress is the axial stress along the length of the beam, not the shear stress.
If y is the distance from the neutral axis you have $\sigma/y = M/I$. So the maximum stress will be furthest frmo the axis, where the bending moment is biggest - i.e. at the fixed end.
You get the yield stress by looking up the material properties for the particular grade of steel you have. Young's modulus doesn't vary much between different grades of steel, but the yield stress varies a lot. A conservative number at the low end of the range would be about 150 MPa or 10 tsi, but high strength steels might be nearer 750 MPa.
4. Nov 4, 2012
### seuss
I see. It's really the bending stress causing the deformation. Back to the drawing board.
5. Nov 5, 2012
### Studiot
Be aware that the figure you have drawn looks like a wide beam or even a plate so simple engineering beam theory will not be applicable without corrections. | 641 | 2,556 | {"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} | 3.3125 | 3 | CC-MAIN-2018-09 | longest | en | 0.914169 |
https://apk-dl.com/sudoku-code-deluxe/ | 1,524,688,400,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947957.81/warc/CC-MAIN-20180425193720-20180425213720-00027.warc.gz | 562,907,423 | 20,002 | # Sudoku Code Deluxe APK
/
(3.6/5) ()
## Description
***** " Addicting Sudoku Game " ***** " 100 seconds of panic filledsudoku pleasure " Welcome to Sudoku Code. It is time to begin yourtraining. Sudoku Code is a fast paced, exciting, code breaking gamebased on Sudoku & chess. Sudoku code is played on a 5 by 5grid. You job is to place 5 symbols on the board such that the codemakes logical sense. Two symbols may never be in the same row orcolumn. Each row and column will have one, and exactly one symbolin it. Each symbol has a unique value. The circle is worth onepoint. The parallel lines are worth 2 points. The triangle is worththree points. The square is worth four points. The star is worth 5points. And the hexagon, should you unlock it, is worth six points.The square a symbol sits on will always be worth zero points.Symbols may never sit on a square worth more than zero. The 8squares that surround the symbol are worth the amount of points thesymbol is worth. For example, the eight squares around the circleare all worth 1 point, while the square the symbol the circle sitson is worth zero points. If two or more symbols share one of theeight squares surrounding them, the shared square is simply worththe sum of the symbol values. For example, if the triangle worththree points, shares a square with the circle worth one point, anysquares they share would be worth 3 plus 1 points, or four points.There is only one unique solution for every puzzle. Your job is tosolve each code as quickly as possible. You can clear the board ofsymbols placed by pressing the clear button. Once you have all thesymbols placed correctly on the board, press solve to stop theclock and see your score. Don't guess at code solutions. Incorrectguesses result in a 10 second time penalty! - Over 14,000 freepuzzles to solve. - Hexagon Symbol unlocked for a total of 100,000unique codes to break. - Campaign Mode - How quickly can you solve10 random codes? - Single Game Mode - Break codes one at a time. -Two Player Mode - Challenge a friend to solve a puzzle faster thanyou. - Covered vs Uncovered - Decide whether the code is fullyvisible or slowly revealed for a more difficult challenge.
## App Information Sudoku Code Deluxe
• App Name
Sudoku Code Deluxe
• Package Name
com.iphonegame.sudokucode.full
• Updated
• File Size
Undefined
• Requires Android
Android
• Version
• Developer
• Installs
-
• Price
Free
• Category
Puzzle
• Developer
QuickShortcutMaker 2.4.0 APK
Clash of Clans APK
Pokémon GO APK
Pokémon GO just introduced a new research feature that encouragesTrainers to explore the world by completing a variety of objectiveswith the opportunity to earn helpful rewards and discover powerfulPokémon. Join Trainers across the globe who are discovering Pokémonas they explore the world around them. Pokémon GO is the globalgaming sensation that has been downloaded over 800 million timesand named "Best Mobile Game" by The Game Developers Choice Awardsand "Best App of the Year" by TechCrunch.Venusaur, Charizard,Blastoise, Pikachu, and many other Pokémon have beendiscovered!Pokémon are out there, and you need to find them. As youwalk around a neighborhood, your smartphone will vibrate whenthere’s a Pokémon nearby. Take aim and throw a Poké Ball… You’llhave to stay alert, or it might get away! Search far and wide forPokémon and itemsCertain Pokémon appear near their nativeenvironment—look for Water-type Pokémon by lakes and oceans. VisitPokéStops and Gyms—found at interesting places like museums, artinstallations, historical markers, and monuments—to stock up onPoké Balls and helpful items. Catching, hatching, evolving, andmoreAs you level up, you’ll be able to catch more-powerful Pokémonto complete your Pokédex. You can add to your collection byhatching Pokémon Eggs based on the distances you walk. Help yourPokémon evolve by catching many of the same kind. Choose a BuddyPokémon to walk with and earn Candy that will help you make yourPokémon stronger. Compete in epic Gym battlesYou’ll join one ofthree teams and battle for the ownership of Gyms with your Pokémonat your side. As your Charmander evolves to Charmeleon and thenCharizard, you can battle together to defeat a Gym and assign yourPokémon to defend it against all comers. Team up to defeat powerfulRaid BossesA Raid Battle is a cooperative gameplay experience thatencourages you to work with up to 20 other Trainers to defeat anextremely powerful Pokémon known as the Raid Boss. If you succeedin defeating it in battle, you’ll have the chance to catch an extrapowerful Pokémon of your own! It’s time to get moving—yourreal-life adventures await!Note: - This app is free-to-play andoffers in-game purchases. It is optimized for smartphones, nottablets.- Compatible with Android devices that have 2GB RAM or moreand have Android Version 4.4–7.0+ installed.- Compatibility is notguaranteed for devices without GPS capabilities or devices that areconnected only to Wi-Fi networks.- Compatibility with tabletdevices is not guaranteed.- Application may not run on certaindevices even if they have compatible OS versions installed.- It isrecommended to play while connected to a network in order to obtainaccurate location information.- Compatibility information may bechanged at any time.- Please visit www.PokemonGO.com for additionalcompatibility information. - Information current as of March 28,2018.
WIFI WPS WPA TESTER 3.8.4.6 APK
**Devices WITHOUT root permissions and with Android >= 5.0(Lollipop), can connect with this app but they CANNOT view theWEP-WPA-WPA2****Devices WITHOUT root permissions and with Android< 5.0 (Lollipop), CANNOT connect with this app and they CANNOTview the WEP-WPA-WPA2**Do you want to know if your Access Point isvulnerable at the WPS protocol?Wifi Wps Wpa Tester is the app thatyou need!With this app, you can test the connection to AP with WPSPIN.PINs are calculated with manyalgorithms:-Zhao-TrendNet-Dlink-Dlink+1-Belkin(root)-FTE-xxx-TrendNet-Asus-AiroconRealtek-EasyBoxArcadyan-Arris And others default PIN of MANY Access Point.Then NOTALL AP ARE COMPATIBLE WITH THIS APP.App allows to do PINSBRUTEFORCE in a SMART WAY ( ONLY FOR ROOTED DEVICES )WhySMART?Because, app will try 11000 PINs COMBINATIONS rather than10^8.In fact, AP tells to your device if the first 4 of 8 digits ofWPS PIN, are correct and the last digit is a checksum of previous 7digits.NO OTHERS APP ( CLONE OF THIS APP ) CAN DO THIS.And, appallows to notify if WPS is in LOCK STATE ( ROOT AND NO ROOT ).NOOTHERS APP ( CLONE OF THIS APP ) CAN DO THIS.WPS Lock state is astate when Access Point, for security reasons, does not allow nomore PINs. THEN is USELESS to try others PINs. App needs rootpermissions for devices with Android version < 5.0 ( LOLLIPOP).For devices with Android >= 5.0 you can test the PINs withthis app and you can connect, BUT YOU CANNOT SEE WPA ( OR WEP )PASSWORD WITHOUT ROOT PERMISSIONS.Use this app only with your ownAP for do not go against the law.Privacy Policy :https://www.iubenda.com/privacy-policy/8000344
WordPress APK
WordPress for Android puts the power of publishing in your hands,making it easy to create and consume content. Write, edit, andpublish posts to your site, check stats, and get inspired withgreat posts in the Reader. What’s more? It’s open source.WordPressfor Android supports WordPress.com and self-hosted WordPress.orgsites running WordPress 3.5 or higher.Visit the forums to get helpwith the app: http://android.forums.wordpress.org
Blog APK
Our blog posts include experiment results of online marketing, howto articles, tools and tips for running your business, businessideas, online selling, entrepreneurship, start ups, successstories, interviews and reviews of relevant books.You can visit theweb version of our app: http://technotip.orgFeatures1. Has a listof 8 recent articles on the homepage and user can navigate to olderblog posts.2. Clear reading experience with bigger fonts on articlepage.3. Facility to bookmark the article and read later frombookmarks section.4. Cache the recently viewed article for offlinereading.5. List of pages.6. Search facility.7. List posts based onCategory.8. List posts by author/contributor.9. Invite others toour app via Social Sharing Apps.Option to rate the app.10. Facilityto directly share the posts and pages with others from inside theapp via popular social sharing applications.
WPS Connect 1.3.6 APK
With this app you'll can connect to WiFi networks which have WPSprotocol enabled. This feature was only available in version 4.1.2of Android.App developed with educational purposes. I am notresponsible for any misuse.Released under license CC BY-NC-ND 4.0:https://creativecommons.org/licenses/by-nc-nd/4.0/WPS:http://es.wikipedia.org/wiki/Wi-Fi_Protected_SetupWPS Connect isfocused on verifying if your router is vulnerable to a default PIN.Many routers that companies install own vulnerabilities in thisaspect. With this application you can check if your router isvulnerable or not and act accordingly.Includes default PINs, aswell as algorithms such Zhao Chesung (ComputePIN) or StefanViehböck (easyboxPIN).Tested on:- LG G2- Nexus 5- Samsung GalaxyS3IMPORTANT!!Prior to an assessment, understand that it serves theapplication.
XPOSED IMEI Changer APK
ROOT REQUIREDXPOSED FRAMEWORK REQUIREDIF You dont know What isXPOSED Framework then do not try this application (it wont work)HiGuys,I read about the Xposed framework in xda and otherwebsites.some websites have really good tutorials about it.So whatI understand is that we can modify a function and its return valuesusing the Xposed Framework.I have created and Xposed Module ForChanging (MASKING) the IMEI No of the PhoneChange Means How OtherApplication gets the IMEI No of the device using below codeAs youall know,The Value is not permanent as it is an Xposed Module:)XDA-DevelopersThreadhttp://forum.xda-developers.com/xposed/modules/xposed-imei-changer-t2847187Proversion :https://play.google.com/store/apps/details?id=com.vivek.imeichangerproStepsInstall app Enable module in xposed framework Come back to app andenter new value Press Apply Go to xposed module Open framework pageDo a soft reboot Open the app And you can see new valuecheck thevalue by dialing *#06#, you can see the new value
QuickShortcutMaker 2.4.0 APK
WIFI WPS WPA TESTER 3.8.4.6 APK
**Devices WITHOUT root permissions and with Android >= 5.0(Lollipop), can connect with this app but they CANNOT view theWEP-WPA-WPA2****Devices WITHOUT root permissions and with Android< 5.0 (Lollipop), CANNOT connect with this app and they CANNOTview the WEP-WPA-WPA2**Do you want to know if your Access Point isvulnerable at the WPS protocol?Wifi Wps Wpa Tester is the app thatyou need!With this app, you can test the connection to AP with WPSPIN.PINs are calculated with manyalgorithms:-Zhao-TrendNet-Dlink-Dlink+1-Belkin(root)-FTE-xxx-TrendNet-Asus-AiroconRealtek-EasyBoxArcadyan-Arris And others default PIN of MANY Access Point.Then NOTALL AP ARE COMPATIBLE WITH THIS APP.App allows to do PINSBRUTEFORCE in a SMART WAY ( ONLY FOR ROOTED DEVICES )WhySMART?Because, app will try 11000 PINs COMBINATIONS rather than10^8.In fact, AP tells to your device if the first 4 of 8 digits ofWPS PIN, are correct and the last digit is a checksum of previous 7digits.NO OTHERS APP ( CLONE OF THIS APP ) CAN DO THIS.And, appallows to notify if WPS is in LOCK STATE ( ROOT AND NO ROOT ).NOOTHERS APP ( CLONE OF THIS APP ) CAN DO THIS.WPS Lock state is astate when Access Point, for security reasons, does not allow nomore PINs. THEN is USELESS to try others PINs. App needs rootpermissions for devices with Android version < 5.0 ( LOLLIPOP).For devices with Android >= 5.0 you can test the PINs withthis app and you can connect, BUT YOU CANNOT SEE WPA ( OR WEP )PASSWORD WITHOUT ROOT PERMISSIONS.Use this app only with your ownAP for do not go against the law.Privacy Policy :https://www.iubenda.com/privacy-policy/8000344
WordPress APK
WordPress for Android puts the power of publishing in your hands,making it easy to create and consume content. Write, edit, andpublish posts to your site, check stats, and get inspired withgreat posts in the Reader. What’s more? It’s open source.WordPressfor Android supports WordPress.com and self-hosted WordPress.orgsites running WordPress 3.5 or higher.Visit the forums to get helpwith the app: http://android.forums.wordpress.org
Blog APK
Our blog posts include experiment results of online marketing, howto articles, tools and tips for running your business, businessideas, online selling, entrepreneurship, start ups, successstories, interviews and reviews of relevant books.You can visit theweb version of our app: http://technotip.orgFeatures1. Has a listof 8 recent articles on the homepage and user can navigate to olderblog posts.2. Clear reading experience with bigger fonts on articlepage.3. Facility to bookmark the article and read later frombookmarks section.4. Cache the recently viewed article for offlinereading.5. List of pages.6. Search facility.7. List posts based onCategory.8. List posts by author/contributor.9. Invite others toour app via Social Sharing Apps.Option to rate the app.10. Facilityto directly share the posts and pages with others from inside theapp via popular social sharing applications.
WPS Connect 1.3.6 APK
With this app you'll can connect to WiFi networks which have WPSprotocol enabled. This feature was only available in version 4.1.2of Android.App developed with educational purposes. I am notresponsible for any misuse.Released under license CC BY-NC-ND 4.0:https://creativecommons.org/licenses/by-nc-nd/4.0/WPS:http://es.wikipedia.org/wiki/Wi-Fi_Protected_SetupWPS Connect isfocused on verifying if your router is vulnerable to a default PIN.Many routers that companies install own vulnerabilities in thisaspect. With this application you can check if your router isvulnerable or not and act accordingly.Includes default PINs, aswell as algorithms such Zhao Chesung (ComputePIN) or StefanViehböck (easyboxPIN).Tested on:- LG G2- Nexus 5- Samsung GalaxyS3IMPORTANT!!Prior to an assessment, understand that it serves theapplication.
XPOSED IMEI Changer APK
ROOT REQUIREDXPOSED FRAMEWORK REQUIREDIF You dont know What isXPOSED Framework then do not try this application (it wont work)HiGuys,I read about the Xposed framework in xda and otherwebsites.some websites have really good tutorials about it.So whatI understand is that we can modify a function and its return valuesusing the Xposed Framework.I have created and Xposed Module ForChanging (MASKING) the IMEI No of the PhoneChange Means How OtherApplication gets the IMEI No of the device using below codeAs youall know,The Value is not permanent as it is an Xposed Module:)XDA-DevelopersThreadhttp://forum.xda-developers.com/xposed/modules/xposed-imei-changer-t2847187Proversion :https://play.google.com/store/apps/details?id=com.vivek.imeichangerproStepsInstall app Enable module in xposed framework Come back to app andenter new value Press Apply Go to xposed module Open framework pageDo a soft reboot Open the app And you can see new valuecheck thevalue by dialing *#06#, you can see the new value
Sing! by Smule APK | 3,619 | 15,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2018-17 | latest | en | 0.904651 |
http://betterlesson.com/lesson/resource/2745478/reference-sheet-pdf | 1,488,228,347,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501173761.96/warc/CC-MAIN-20170219104613-00097-ip-10-171-10-108.ec2.internal.warc.gz | 28,193,298 | 18,837 | ## reference sheet.pdf - Section 1: Method for Assessment
reference sheet.pdf
# Formative assessment over Identities
Unit 9: Trigonometric Identities
Lesson 8 of 14
## Big Idea: Through a quiz students will demonstrate their ability to verify identities.
Print Lesson
1 teacher likes this lesson
Standards:
Subject(s):
Math, Trigonometric identities
45 minutes
### Katharine Sparks
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http://ciks.cbt.nist.gov/garbocz/paper51/node4.html | 1,386,898,048,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164789076/warc/CC-MAIN-20131204134629-00080-ip-10-33-133-15.ec2.internal.warc.gz | 36,081,002 | 4,017 | Next: Numerical Results Up: Main Previous: Effective Medium Theory
# Tests of elastic algorithm
The algorithm described above has been developed specifically to be applied to images of random materials that have been generated either using a microstructure model, or by an experimental technique like x-ray tomography. Especially in the latter case, it is usually difficult or impossible to perform checks like how the results depend on the size of the image or on the resolution of the image. However, such checks can be done with model systems, especially ones which have an analytically known solution, to give an idea of what the error might be in any given computation using real images.
There are several sources of error in using this algorithm on a specific random system. The first is finite size error--does the image contain enough of the random structure so that the computed elastic moduli no longer depend on system size? The second error is, for a random system, how much do different realizations of the same size random system differ from each other? The third source of error is: how does the resolution of microstructural features affect the results? A fourth source of error, how well the minimum energy state is approximated in the solution algorithm, is much smaller, essentially on the order of the round-off error of the computer, and so is negligible.
Several series of runs were made to estimate these sources of error. The area or volume fractions of phases 1 and 2 were fixed at 0.5 each. Table 1 shows, for a fixed value of w=5 for a 2-D Gaussian system, the effect of system size at a fixed resolution on the computed elastic moduli, and their standard deviation over a 10 system average. The average values of E and change little for system size greater than 64. The standard deviation over the average of ten independent systems decreases as the system size increases.
System size Young's modulus Poisson's ratio 32 2.408 ± 0.488 0.384 ± 0.092 64 2.627 ± 0.227 0.341 ± 0.028 128 2.673 ± 0.205 0.320 ± 0.039 256 2.633 ± 0.066 0.325 ± 0.012 512 2.624 ± 0.059 0.326 ± 0.013
Table 1: The numerical results for Young's modulus and Poisson's ratio, for w=5 Gaussian 2-D systems at c1=c2=0.5, Poisson's ratio = 1/3, E1/E2=10, as a function of system size, averaged over 10 independent realizations.
Table 2 shows the effect of resolution on the computed dilute limit slopes for circles embedded in a matrix. The slopes are defined, where m is for matrix and i is for inclusion, by
System size MK % difference MG % difference 20 1.548 10.1 1.318 6.9 40 1.548 10.1 1.362 12.9 80 1.475 4.9 1.290 4.6 160 1.447 2.9 1.261 2.3 320 1.433 1.9 1.251 1.5 640 1.425 1.3 1.244 0.9
Table 2: Dilute limit for circle -- effect of digital resolution. The exact values are MK = 1.40625, and MG = 1.23288. The circle always had a nominal diameter one tenth that of the unit cell.
The slopes are a function of the goemetry of the inclusion and of the relative values of the four moduli [3]. Each numerical point was for a circle whose diameter was one tenth the size of the periodic unit cell, so that there was little or no influence from the periodic boundary conditions. Increasing the system size in terms of the number of pixels per unit length also improved the resolution of the circle. Table 2 shows that a diameter of 16 pixels was sufficient to bring the computed initial slope within 3% of the theoretical value.
Table 3 shows results for a 2-D elastic checkerboard, where the unit cell contained four "checks" (two black, two white), so that the size of each check was one half the system size. By making the shear moduli equal, exact results can be obtained for all the effective moduli. In 2-D, the exact Young's modulus and Poisson's ratio are [19,20]:
System size K % difference E % difference Poisson's ratio % difference 2 5.500 96.4 5.866 25.7 0.4666 180 4 3.271 16.8 4.964 6.4 0.2411 44.7 8 2.937 4.9 4.759 2.0 0.1897 13.8 16 2.841 1.5 4.695 0.6 0.1737 4.2 32 2.812 0.4 4.675 0.2 0.1688 1.3 64 2.804 0.1 4.669 0.05 0.1673 0.4 128 2.8012 0.04 4.6675 0.02 0.16687 0.1 256 2.8003 0.01 4.6669 0.005 0.16672 0.03 512 2.8001 0.004 4.6667 0.0007 0.16668 0.008
Table 3: The numerical results for the bulk modulus, Young's modulus, and the Poisson's ratio for the 2-D elastic checkerboard with equal shear moduli (G1=G2=2, K1/K2=10), showing how the numerical result changes with digital resolution. The exact values are: K = 2.8, E = 14/3, and = 1/6.
Table 4 shows results similar to that of Table 3, but for the Gaussian system. The ratio of w to the system size is maintained, and the system size is changed, so that the resolution of individual features increases with system size. Here the same values for the individual phase moduli are used as in Table 3, so that the same oveall exact values should be obtained. If we take the individual features in the microstructure of these Gaussian images to be on the order of w, the correlation length, then the Gaussian systems obtain the same sort of accuracy in E and as does the checkerboard when comparing w to the size of one check. Having many more digitally rough interfaces in the Gaussian system does not cause the accuracy to degrade compared to the checkerboard.
System size (w) K % difference E % difference Poisson's ratio % difference 32 (1.25) 2.941 5.0 4.762 2.0 0.1905 14.3 64 (2.5) 2.864 2.3 4.711 1.0 0.1777 6.6 128 (5) 2.830 1.1 4.688 0.5 0.1719 3.1 256 (10) 2.828 1.0 4.686 0.4 0.1715 2.9 512 (20) 2.824 0.9 4.683 0.3 0.1708 2.5
Table 4: The numerical results for the bulk modulus, Young's modulus, and Poisson's ratio, for w=5 Gaussian 2-D systems at c1=c21=G2=2, K1/K2=10, as a function of digital resolution. The exact values are: K = 2.8, E = 14/3, and = 1/6.
In 3-D, which is much more computer-time intensive, one test was run for a 643, w=5 Gaussian system with G1=G2=2, and K1/K2=10, the same numerical values as were used in 2-D. The exact values for the moduli were [19]: K = 3.02041, E = 4.91513, and = 0.228782. The numerical results were all within 2% of these exact values, implying that the resolution was adequate and similar to that found in 2-D.
These error tests show that the system sizes and resolutions chosen to be used in the next section to compute the main results, 1282 (w=5) in 2-D and 643 (w=5) in 3-D, averaged over five independent realizations, are adequate to give roughly 5% accuracy in the results.
Next: Numerical Results Up: Main Previous: Effective Medium Theory | 1,924 | 6,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.920329 |
https://web2.0calc.com/questions/infinite-soluitons | 1,620,297,134,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.91/warc/CC-MAIN-20210506083716-20210506113716-00132.warc.gz | 642,323,183 | 5,729 | +0
# infinite soluitons
0
85
3
What equation has an infinite number of solutions
Feb 5, 2021
#1
+118470
+1
Here's the key
Simplify both sides
If one side = the other side, we have infinite solutions
(Hint : look at the first one)
Feb 5, 2021
#2
0
An equation can have infinite solutions if the variables go away and you are left with something like "5 = 5" or "7 = 7".
Feb 5, 2021
#3
0
I believe that the first equation leaves you with -7 = -7 for any number "n". (This has infinite solutions)
The second equation leaves: -15 = 10. No value of "n" works for this. (This has no solutions)
The third equation leaves: -3 = 10. (This also has no solutions)
The last equation leaves you with n = 2. (This is only one solution.)
Feb 5, 2021 | 234 | 758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2021-21 | latest | en | 0.92692 |
http://mathhelpforum.com/algebra/29656-exponents.html | 1,527,007,350,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864798.12/warc/CC-MAIN-20180522151159-20180522171159-00174.warc.gz | 186,153,290 | 9,245 | 1. exponents
I need help on this problem please.
((2^3 m^2 n^-2) / p^-4))^2
Thank you
2. Error in orginal
I need help on this problem please.
I forgot a parenthesis.
((2^3 m^2 n^-2) / (p^-4))^2
Thank you
3. $\displaystyle \left(\frac{(2^3\;m^2\;n^{-2})}{(p^{-4})}\right)^2$
$\displaystyle \frac{64\;m^4\;n^{-4}}{p^{-8}}$
$\displaystyle \frac{64\;m^4\;p^8}{n^4}$
4. Thank you for the answer.
Originally Posted by OzzMan
$\displaystyle \left(\frac{(2^3\;m^2\;n^{-2})}{(p^{-4})}\right)^2$
$\displaystyle \frac{64\;m^4\;n^{-4}}{p^{-8}}$
$\displaystyle \frac{64\;m^4\;p^8}{n^4}$
I had that but just was not confident in it being right. | 267 | 645 | {"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} | 3.03125 | 3 | CC-MAIN-2018-22 | latest | en | 0.68417 |
https://oeis.org/A141604 | 1,591,055,845,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419639.53/warc/CC-MAIN-20200601211310-20200602001310-00508.warc.gz | 487,029,012 | 4,358 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A141604 Triangle t(n,m)= round( A006720(n)/(A006720(n-m)*A006720(m)) read by rows, 0<=m<=n. 1
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 2, 3, 3, 2, 1, 1, 2, 4, 7, 4, 2, 1, 1, 3, 8, 12, 12, 8, 3, 1, 1, 3, 8, 20, 15, 20, 8, 3, 1, 1, 5, 14, 45, 52, 52, 45, 14, 5, 1, 1, 5, 26, 66, 109, 170, 109, 66, 26, 5, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,12 COMMENTS The round-function in the definition is round-to-nearest, not Mathematica's round-to-even. - R. J. Mathar, Jul 12 2012 Row sums are 1, 2, 3, 4, 8, 12, 21, 48, 79, 234, 584.. LINKS EXAMPLE 1; 1, 1; 1, 1, 1; 1, 1, 1, 1; 1, 2, 2, 2, 1; 1, 2, 3, 3, 2, 1; 1, 2, 4, 7, 4, 2, 1; 1, 3, 8, 12, 12, 8, 3, 1; 1, 3, 8, 20, 15, 20, 8, 3, 1; 1, 5, 14, 45, 52, 52, 45, 14, 5, 1; 1, 5, 26, 66, 109, 170, 109, 66, 26, 5, 1; MAPLE A141604 := proc(n, m) round(A006720(n)/A006720(n-m)/A006720(m)) ; end proc: seq(seq(A141604(n, k), k=0..n), n=0..12) ; # R. J. Mathar, Jul 12 2012 MATHEMATICA Clear[a, t] (*A006720*) (* _Robert G.Wilson v_, Jul 04 2007 *) a[0] = a[1] = a[2] = a[3] = 1; a[n_] := a[n] = (a[n - 1] a[n - 3] + a[n - 2]^2)/a[n - 4]; Array[a, 23]; t[n_, m_] := a[n]/(a[n - m]*a[m]) Table[Table[Round[t[n, m]], {m, 0, n}], {n, 0, 10}]; Flatten[%] CROSSREFS Cf. A006720. Sequence in context: A134143 A295555 A085684 * A143996 A301562 A334486 Adjacent sequences: A141601 A141602 A141603 * A141605 A141606 A141607 KEYWORD nonn,less,tabl AUTHOR Roger L. Bagula and Gary W. Adamson, Aug 21 2008 STATUS approved
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Last modified June 1 19:32 EDT 2020. Contains 334762 sequences. (Running on oeis4.) | 961 | 2,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.54065 |
http://www.math.grin.edu/~walker/courses/161.sp10/labs/lab-bit-ops.shtml | 1,534,733,889,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215523.56/warc/CC-MAIN-20180820023141-20180820043141-00687.warc.gz | 533,284,560 | 3,633 | CSC 161 Grinnell College Spring, 2010 Imperative Problem Solving and Data Structures
# Machine-level Operations, Bit Manipulation, and Unions
## Goals
This lab provides practice in working with data at the bit level in C. Specific work involves the representation of integers, the manipulation of bits in C, and the use of unions in C to view bit patterns in multiple ways.
## Bit Operations
The C programming language contains the following bitwise operations:
operation comment
& bitwise and
| bitwise or
^ bitwise exclusive or
~ ones complement
<< shift left
>> shift right
1. In the Lab on Integer Processing, we used the C program ~walker/c/fluency-book/integer-rep.c to examine the bit representations of integers. Review that program, and explain how the print_binary procedure works. As part of your explanation, include an example for the printing of the decimal number 11.
## Binary Representation of Integers
1. Program ~walker/c/data-rep.c provides an alternative framework for examining the bit representations of integers and floating point numbers.
1. Write (on paper) the floating point numbers ± 5.0 and ± 11.0 using the IEEE floating-point representation of real numbers.
2. Run program ~walker/c/data-rep.c to determine the internal representation of the integers from part 1, as actually stored on PC/Linux computers, and write a paragraph that explains the bits involved with the sign, mantissa, and exponent of these numbers.
2. Program ~walker/c/data-rep.c uses a union in C as the basis of its processing.
1. What can be stored or accessed in a `DATA` type?
2. A `typedef` statement allows the type `union DATA` to be identified more simply as a `data` type. Explain what data may be stored in variable `d` and how that data can be accessed.
3. The main part of this program consists of a single loop.
1. What is the significance of the number 1 in the `while (1)` expression?
2. Under what circumstances does the program terminate, and how is this made to happen?
3. [Note that `continue` is used in place of `break` in the default option, so execution at that spot will jump back to the top of the loop rather than continuing with the printing that follows.]
4. Explain how numbers are set in options 0, F, and I.
5. Why is the number -1 used for option 1?
6. After each number is set, its value is printed using several representations. While the `printf` statements are straight forward, the `printBitGroups` function may require some thought. The first use of this function comes from the call ``` printBitGroups (d, 1)```. Using `bitGroups` as 1, trace the execution of `printBitGroups`.
1. Identify the initial values of `value, mask` and `iterations`. (Variable `a` is an array of integers, with subscripts between 0 and 31.)
2. Describe the final value of variable `mask` after the first loop terminates, and explain how that bit pattern is achieved.
3. Explain what processing is done in the second loop; what are the final values placed in the `a` array, and how are these values determined.
4. Why do you think the `value` variable is declared as `d.integer`, rather than using `d.integer` directly in the second loop of `printBitGroups`.
7. Explain the purpose of the call `printBitGroups (d, 4)`, and discuss how this purpose is achieved.
1. What is the purpose of the number 4 in this call?
2. Describe the final value of variable `mask` after the first loop terminates, and explain how that bit pattern is achieved.
3. Explain what processing is done in the second loop; what are the final values placed in the `a` array, and how are these values determined.
8. Write (in English) appropriate pre- and post-condidtions for function `printBitGroups`. These conditions should be inserted as comments to follow the function's header, but they need NOT be checked in the code using `assert` statements or other executed tests.
9. Add a menu option to this program, so that the integer value of variable `d` is changed to its ones complement.
10. Add a menu option to this program that begins with the value in variable `d` and successively toggles successive bits of the variable -- printing the binary, hexadecimal, integer, and float values of the results in a table. Toggling of the bits should progress from left to right. Thus, the output might have the form:
binary hexadecimal integer form float form 00000000000000000000000000000000 00000000 0 0.0 10000000000000000000000000000000 80000000 -2147483648 -0.0 01000000000000000000000000000000 40000000 1073741824 2.0 ...
1. Write functions to solve ONE of the following problems from King's book:
1. Exercise 20.7 (page 525),
2. Exercise 20.9 (page 526; for part b, think recursively)
3. Exercise 20.10 (page 526).
Additional problems may be done for extra credit.
## Work to turn in
• Commentary for steps 1-9, with no commentary needed for 4c.
• A program listing, with appropriate test runs and comments on correctness, for steps 10 and 11.
• A program, with appropriate test runs and comments on correctness for step 12.
This document is available on the World Wide Web as
``` http://www.cs.grinnell.edu/~walker/courses/161.sp10/labs/lab-bit-ops.shtml
``` | 1,231 | 5,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | latest | en | 0.87079 |
http://mathhelpforum.com/calculus/200880-related-rates-print.html | 1,529,605,965,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864256.26/warc/CC-MAIN-20180621172638-20180621192638-00455.warc.gz | 204,005,869 | 2,988 | # Related Rates
• Jul 11th 2012, 01:45 PM
calcmaster
Related Rates
A Ferris wheel with a radius of 10 m is rotating at a rate of one revolution every 2 minutes. How fast is a rider rising when his seat is 16 m above ground level?
Can someone help me figure out what equation I should use? I want to use the circumference of a circle forumla but I'm not sure how that would work with the revolutions.
Thanks!
• Jul 11th 2012, 02:49 PM
Reckoner
Re: Related Rates
Quote:
Originally Posted by calcmaster
A Ferris wheel with a radius of 10 m is rotating at a rate of one revolution every 2 minutes. How fast is a rider rising when his seat is 16 m above ground level?
Please see the attached picture below (click it for the full version).
We are told that $\displaystyle \frac{d\theta}{dt} = \frac{2\pi\ \mathrm{rad}}{2\ \mathrm{min}} = \pi\ \mathrm{rad}/\mathrm{min}.$ We are asked to find $\displaystyle \frac{dy}{dt}$ when $\displaystyle y = 6.$ We know that
$\displaystyle y = 10\sin\theta,$
so we differentiate,
$\displaystyle \frac{dy}{dt} = 10\cos\theta\cdot\frac{d\theta}{dt}.$
When $\displaystyle y = 6$ we have
$\displaystyle \sin\theta = \frac35$
and since $\displaystyle \theta$ is an acute angle, this gives
$\displaystyle \theta = \arcsin\frac35.$
Now substitute the given quantities:
$\displaystyle \frac{dy}{dt} = 10\cos\theta\cdot\frac{d\theta}{dt}$
$\displaystyle \Rightarrow\frac{dy}{dt} = 10\cos\left(\arcsin\frac35\right)\cdot\pi$
$\displaystyle \Rightarrow\frac{dy}{dt} = 10\pi\cdot\frac45 = 8\pi\ \mathrm m/\mathrm{min} \approx 25.13\ \mathrm m/\mathrm{min}.$ | 496 | 1,594 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-26 | latest | en | 0.791265 |
https://openstax.org/books/college-physics/pages/11-problems-exercises | 1,722,922,666,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476479.26/warc/CC-MAIN-20240806032955-20240806062955-00558.warc.gz | 362,307,105 | 109,162 | College Physics
# Problems & Exercises
College PhysicsProblems & Exercises
## 11.2Density
1.
Gold is sold by the troy ounce (31.103 g). What is the volume of 1 troy ounce of pure gold?
2.
Mercury is commonly supplied in flasks containing 34.5 kg (about 76 lb). What is the volume in liters of this much mercury?
3.
(a) What is the mass of a deep breath of air having a volume of 2.00 L? (b) Discuss the effect taking such a breath has on your body’s volume and density.
4.
A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it in a graduated cylinder. What is the density of a 240-g rock that displaces $89.0cm389.0cm3 size 12{"89" "." 0"cm" rSup { size 8{3} } } {}$ of water? (Note that the accuracy and practical applications of this technique are more limited than a variety of others that are based on Archimedes’ principle.)
5.
Suppose you have a coffee mug with a circular cross section and vertical sides (uniform radius). What is its inside radius if it holds 375 g of coffee when filled to a depth of 7.50 cm? Assume coffee has the same density as water.
6.
(a) A rectangular gasoline tank can hold 50.0 kg of gasoline when full. What is the depth of the tank if it is 0.500-m wide by 0.900-m long? (b) Discuss whether this gas tank has a reasonable volume for a passenger car.
7.
A trash compactor can reduce the volume of its contents to 0.350 their original value. Neglecting the mass of air expelled, by what factor is the density of the rubbish increased?
8.
A 2.50-kg steel gasoline can holds 20.0 L of gasoline when full. What is the average density of the full gas can, taking into account the volume occupied by steel as well as by gasoline?
9.
What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are parts by mass, not volume.) Assume that this is a simple mixture having an average density equal to the weighted densities of its constituents.
10.
There is relatively little empty space between atoms in solids and liquids, so that the average density of an atom is about the same as matter on a macroscopic scale—approximately $103kg/m3103kg/m3 size 12{"10" rSup { size 8{3} } "kg/m" rSup { size 8{3} } } {}$. The nucleus of an atom has a radius about $10−510−5 size 12{"10" rSup { size 8{ - 5} } } {}$ that of the atom and contains nearly all the mass of the entire atom. (a) What is the approximate density of a nucleus? (b) One remnant of a supernova, called a neutron star, can have the density of a nucleus. What would be the radius of a neutron star with a mass 10 times that of our Sun (the radius of the Sun is $7×108m7×108m size 12{7 times "10" rSup { size 8{8} } m} {}$)?
## 11.3Pressure
11.
As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted on the floor by the heel if it has an area of $1.50cm21.50cm2 size 12{1 "." "50""cm" rSup { size 8{2} } } {}$ and the woman’s mass is 55.0 kg. Express the pressure in Pa. (In the early days of commercial flight, women were not allowed to wear high-heeled shoes because aircraft floors were too thin to withstand such large pressures.)
12.
The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 1.00 g is supported by a needle, the tip of which is a circle 0.200 mm in radius, what pressure is exerted on the record in $N/m2N/m2 size 12{"N/m" rSup { size 8{2} } } {}$?
13.
Nail tips exert tremendous pressures when they are hit by hammers because they exert a large force over a small area. What force must be exerted on a nail with a circular tip of 1.00 mm diameter to create a pressure of $3.00×109N/m2?3.00×109N/m2?$(This high pressure is possible because the hammer striking the nail is brought to rest in such a short distance.)
## 11.4Variation of Pressure with Depth in a Fluid
14.
What depth of mercury creates a pressure of 1.00 atm?
15.
The greatest ocean depths on the Earth are found in the Marianas Trench near the Philippines. Calculate the pressure due to the ocean at the bottom of this trench, given its depth is 11.0 km and assuming the density of seawater is constant all the way down.
16.
Verify that the SI unit of $hρghρg size 12{hρg} {}$ is $N/m2N/m2 size 12{"N/m" rSup { size 8{2} } } {}$.
17.
Water towers store water above the level of consumers for times of heavy use, eliminating the need for high-speed pumps. How high above a user must the water level be to create a gauge pressure of $3.00×105 N/m23.00×105 N/m2 size 12{3 "." "00" times "10" rSup { size 8{5} } "N/m" rSup { size 8{2} } } {}$?
18.
The aqueous humor in a person’s eye is exerting a force of 0.300 N on the $1.10-cm21.10-cm2 size 12{1 "." "10""-cm" rSup { size 8{2} } } {}$ area of the cornea. (a) What pressure is this in mm Hg? (b) Is this value within the normal range for pressures in the eye?
19.
How much force is exerted on one side of an 8.50 cm by 11.0 cm sheet of paper by the atmosphere? How can the paper withstand such a force?
20.
What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long gas tank that can hold 50.0 kg of gasoline by the weight of the gasoline in it when it is full?
21.
Calculate the average pressure exerted on the palm of a shot-putter’s hand by the shot if the area of contact is $50.0 cm250.0 cm2 size 12{"50" "." 0"cm" rSup { size 8{2} } } {}$ and he exerts a force of 800 N on it. Express the pressure in $N/m2N/m2$ and compare it with the $1.00×106 Pa1.00×106 Pa$ pressures sometimes encountered in the skeletal system.
22.
The left side of the heart creates a pressure of 120 mm Hg by exerting a force directly on the blood over an effective area of $15.0 cm2.15.0 cm2. size 12{"15" "." 0"cm" rSup { size 8{2} } } {}$ What force does it exert to accomplish this?
23.
Show that the total force on a rectangular dam due to the water behind it increases with the square of the water depth. In particular, show that this force is given by $F=ρgh2L/2F=ρgh2L/2 size 12{F=ρ ital "gh" rSup { size 8{2} } L/2} {}$, where $ρρ size 12{ρ} {}$ is the density of water, $hh size 12{h} {}$ is its depth at the dam, and $LL size 12{L} {}$ is the length of the dam. You may assume the face of the dam is vertical. (Hint: Calculate the average pressure exerted and multiply this by the area in contact with the water. (See Figure 11.41.)
Figure 11.41
## 11.5Pascal’s Principle
24.
How much pressure is transmitted in the hydraulic system considered in Example 11.6? Express your answer in pascals and in atmospheres.
25.
What force must be exerted on the pedal cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resting on the wheel cylinder? The pedal cylinder has a 2.00-cm diameter and the wheel has a 24.0-cm diameter.
26.
A crass host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.00-cm diameter into the bottle, placing it in direct contact with the wine. He is amazed when he pounds the cork into place and the bottom of the jug (with a 14.0-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a 120-N force.
27.
A certain hydraulic system is designed to exert a force 100 times as large as the one put into it. (a) What must be the ratio of the area of the wheel cylinder to the area of the pedal cylinder? (b) What must be the ratio of their diameters? (c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume no losses to friction.
28.
(a) Verify that work input equals work output for a hydraulic system assuming no losses to friction. Do this by showing that the distance the output force moves is reduced by the same factor that the output force is increased. Assume the volume of the fluid is constant. (b) What effect would friction within the fluid and between components in the system have on the output force? How would this depend on whether or not the fluid is moving?
## 11.6Gauge Pressure, Absolute Pressure, and Pressure Measurement
29.
Find the gauge and absolute pressures in the balloon and peanut jar shown in Figure 11.15, assuming the manometer connected to the balloon uses water whereas the manometer connected to the jar contains mercury. Express in units of centimeters of water for the balloon and millimeters of mercury for the jar, taking $h=0.0500 mh=0.0500 m size 12{h=0 "." "0500"m} {}$ for each.
30.
(a) Convert normal blood pressure readings of 120 over 80 mm Hg to newtons per meter squared using the relationship for pressure due to the weight of a fluid $(P=hρg)(P=hρg) size 12{ $$P=hρg$$ } {}$ rather than a conversion factor. (b) Discuss why blood pressures for an infant could be smaller than those for an adult. Specifically, consider the smaller height to which blood must be pumped.
31.
How tall must a water-filled manometer be to measure blood pressures as high as 300 mm Hg?
32.
Pressure cookers have been around for more than 300 years, although their use has strongly declined in recent years (early models had a nasty habit of exploding). How much force must the latches holding the lid onto a pressure cooker be able to withstand if the circular lid is $25.0 cm25.0 cm size 12{"25" "." 0"cm"} {}$ in diameter and the gauge pressure inside is 300 atm? Neglect the weight of the lid.
33.
Suppose you measure a standing person’s blood pressure by placing the cuff on his leg 0.500 m below the heart. Calculate the pressure you would observe (in units of mm Hg) if the pressure at the heart were 120 over 80 mm Hg. Assume that there is no loss of pressure due to resistance in the circulatory system (a reasonable assumption, since major arteries are large).
34.
A submarine is stranded on the bottom of the ocean with its hatch 25.0 m below the surface. Calculate the force needed to open the hatch from the inside, given it is circular and 0.450 m in diameter. Air pressure inside the submarine is 1.00 atm.
35.
Assuming bicycle tires are perfectly flexible and support the weight of bicycle and rider by pressure alone, calculate the total area of the tires in contact with the ground. The bicycle plus rider has a mass of 80.0 kg, and the gauge pressure in the tires is $3.50×105Pa3.50×105Pa size 12{3 "." "50" times "10" rSup { size 8{5} } "Pa"} {}$.
## 11.7Archimedes’ Principle
36.
What fraction of ice is submerged when it floats in freshwater, given the density of water at 0°C is very close to $1000 kg/m31000 kg/m3 size 12{"1000""kg/m" rSup { size 8{3} } } {}$?
37.
Logs sometimes float vertically in a lake because one end has become water-logged and denser than the other. What is the average density of a uniform-diameter log that floats with $20.0%20.0%$ of its length above water?
38.
Find the density of a fluid in which a hydrometer having a density of $0.750 g/mL0.750 g/mL size 12{0 "." "750""g/mL"} {}$ floats with $92.0%92.0% size 12{"92" "." 0%} {}$ of its volume submerged.
39.
If your body has a density of $995 kg/m3995 kg/m3 size 12{"995""kg/m" rSup { size 8{3} } } {}$, what fraction of you will be submerged when floating gently in: (a) freshwater? (b) salt water, which has a density of $1027 kg/m31027 kg/m3 size 12{"1027""kg/m" rSup { size 8{3} } } {}$?
40.
Bird bones have air pockets in them to reduce their weight—this also gives them an average density significantly less than that of the bones of other animals. Suppose an ornithologist weighs a bird bone in air and in water and finds its mass is $45.0 g45.0 g$ and its apparent mass when submerged is $3.60 g3.60 g size 12{3 "." "60"g} {}$ (the bone is watertight). (a) What mass of water is displaced? (b) What is the volume of the bone? (c) What is its average density?
41.
A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is displaced? (b) What is the volume of the rock? (c) What is its average density? Is this consistent with the value for granite?
42.
Archimedes’ principle can be used to calculate the density of a fluid as well as that of a solid. Suppose a chunk of iron with a mass of 390.0 g in air is found to have an apparent mass of 350.5 g when completely submerged in an unknown liquid. (a) What mass of fluid does the iron displace? (b) What is the volume of iron, using its density as given in Table 11.1 (c) Calculate the fluid’s density and identify it.
43.
In an immersion measurement of a woman’s density, she is found to have a mass of 62.0 kg in air and an apparent mass of 0.0850 kg when completely submerged with lungs empty. (a) What mass of water does she displace? (b) What is her volume? (c) Calculate her density. (d) If her lung capacity is 1.75 L, is she able to float without treading water with her lungs filled with air?
44.
Some fish have a density slightly less than that of water and must exert a force (swim) to stay submerged. What force must an 85.0-kg grouper exert to stay submerged in salt water if its body density is $1015kg/m31015kg/m3 size 12{"1015" "kg/m" rSup { size 8{3} } } {}$?
45.
(a) Calculate the buoyant force on a 2.00-L helium balloon. (b) Given the mass of the rubber in the balloon is 1.50 g, what is the net vertical force on the balloon if it is let go? You can neglect the volume of the rubber.
46.
(a) What is the density of a woman who floats in freshwater with $4.00%4.00%$ of her volume above the surface? This could be measured by placing her in a tank with marks on the side to measure how much water she displaces when floating and when held under water (briefly). (b) What percent of her volume is above the surface when she floats in seawater?
47.
A certain man has a mass of 80 kg and a density of $955kg/m3955kg/m3 size 12{"955" "kg/m" rSup { size 8{3} } } {}$ (excluding the air in his lungs). (a) Calculate his volume. (b) Find the buoyant force air exerts on him. (c) What is the ratio of the buoyant force to his weight?
48.
A simple compass can be made by placing a small bar magnet on a cork floating in water. (a) What fraction of a plain cork will be submerged when floating in water? (b) If the cork has a mass of 10.0 g and a 20.0-g magnet is placed on it, what fraction of the cork will be submerged? (c) Will the bar magnet and cork float in ethyl alcohol?
49.
What fraction of an iron anchor’s weight will be supported by buoyant force when submerged in saltwater?
50.
Scurrilous con artists have been known to represent gold-plated tungsten ingots as pure gold and sell them to the greedy at prices much below gold value but deservedly far above the cost of tungsten. With what accuracy must you be able to measure the mass of such an ingot in and out of water to tell that it is almost pure tungsten rather than pure gold?
51.
A twin-sized air mattress used for camping has dimensions of 100 cm by 200 cm by 15 cm when blown up. The weight of the mattress is 2 kg. How heavy a person could the air mattress hold if it is placed in freshwater?
52.
Referring to Figure 11.20, prove that the buoyant force on the cylinder is equal to the weight of the fluid displaced (Archimedes’ principle). You may assume that the buoyant force is $F2−F1F2−F1 size 12{F rSub { size 8{2} } - F rSub { size 8{1} } } {}$ and that the ends of the cylinder have equal areas $AA size 12{A} {}$. Note that the volume of the cylinder (and that of the fluid it displaces) equals $(h2−h1)A(h2−h1)A size 12{ $$h rSub { size 8{2} } - h rSub { size 8{1} }$$ A} {}$.
53.
(a) A 75.0-kg man floats in freshwater with $3.00%3.00%$ of his volume above water when his lungs are empty, and $5.00%5.00%$ of his volume above water when his lungs are full. Calculate the volume of air he inhales—called his lung capacity—in liters. (b) Does this lung volume seem reasonable?
## 11.8Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action
54.
What is the pressure inside an alveolus having a radius of $2.50×10−4m2.50×10−4m size 12{2 "." "50" times "10" rSup { size 8{ - 4} } m} {}$ if the surface tension of the fluid-lined wall is the same as for soapy water? You may assume the pressure is the same as that created by a spherical bubble.
55.
(a) The pressure inside an alveolus with a $2.00×10−42.00×10−4 size 12{2 "." "00" times "10" rSup { size 8{ - 4} } } {}$-m radius is $1.40×103Pa1.40×103Pa size 12{1 "." "40" times "10" rSup { size 8{3} } "Pa"} {}$, due to its fluid-lined walls. Assuming the alveolus acts like a spherical bubble, what is the surface tension of the fluid? (b) Identify the likely fluid. (You may need to extrapolate between values in Table 11.3.)
56.
What is the gauge pressure in millimeters of mercury inside a soap bubble 0.100 m in diameter?
57.
Calculate the force on the slide wire in Figure 11.28 if it is 3.50 cm long and the fluid is ethyl alcohol.
58.
Figure 11.34(a) shows the effect of tube radius on the height to which capillary action can raise a fluid. (a) Calculate the height $hh size 12{h} {}$ for water in a glass tube with a radius of 0.900 cm—a rather large tube like the one on the left. (b) What is the radius of the glass tube on the right if it raises water to 4.00 cm?
59.
We stated in Example 11.12 that a xylem tube is of radius $2.50×10−5m2.50×10−5m$. Verify that such a tube raises sap less than a meter by finding $hh$ for it, making the same assumptions that sap’s density is $1050kg/m31050kg/m3 size 12{"1050""kg/m" rSup { size 8{3} } } {}$, its contact angle is zero, and its surface tension is the same as that of water at $20.0º C20.0º C$.
60.
What fluid is in the device shown in Figure 11.28 if the force is $3.16×10−3N3.16×10−3N size 12{3 "." "16" times "10" rSup { size 8{ - 3} } N} {}$ and the length of the wire is 2.50 cm? Calculate the surface tension $γγ size 12{g} {}$ and find a likely match from Table 11.3.
61.
If the gauge pressure inside a rubber balloon with a 10.0-cm radius is 1.50 cm of water, what is the effective surface tension of the balloon?
62.
Calculate the gauge pressures inside 2.00-cm-radius bubbles of water, alcohol, and soapy water. Which liquid forms the most stable bubbles, neglecting any effects of evaporation?
63.
Suppose water is raised by capillary action to a height of 5.00 cm in a glass tube. (a) To what height will it be raised in a paraffin tube of the same radius? (b) In a silver tube of the same radius?
64.
Calculate the contact angle $θθ size 12{θ} {}$ for olive oil if capillary action raises it to a height of 7.07 cm in a glass tube with a radius of 0.100 mm. Is this value consistent with that for most organic liquids?
65.
When two soap bubbles touch, the larger is inflated by the smaller until they form a single bubble. (a) What is the gauge pressure inside a soap bubble with a 1.50-cm radius? (b) Inside a 4.00-cm-radius soap bubble? (c) Inside the single bubble they form if no air is lost when they touch?
66.
Calculate the ratio of the heights to which water and mercury are raised by capillary action in the same glass tube.
67.
What is the ratio of heights to which ethyl alcohol and water are raised by capillary action in the same glass tube?
## 11.9Pressures in the Body
68.
During forced exhalation, such as when blowing up a balloon, the diaphragm and chest muscles create a pressure of 60.0 mm Hg between the lungs and chest wall. What force in newtons does this pressure create on the $600cm2600cm2 size 12{"600""cm" rSup { size 8{2} } } {}$ surface area of the diaphragm?
69.
You can chew through very tough objects with your incisors because they exert a large force on the small area of a pointed tooth. What pressure in pascals can you create by exerting a force of $500 N500 N size 12{"500"N} {}$ with your tooth on an area of $1.00mm21.00mm2 size 12{1 "." "00""mm" rSup { size 8{2} } } {}$?
70.
One way to force air into an unconscious person’s lungs is to squeeze on a balloon appropriately connected to the subject. What force must you exert on the balloon with your hands to create a gauge pressure of 4.00 cm water, assuming you squeeze on an effective area of $50.0cm250.0cm2 size 12{"50" "." 0"cm" rSup { size 8{2} } } {}$?
71.
Heroes in movies hide beneath water and breathe through a hollow reed (villains never catch on to this trick). In practice, you cannot inhale in this manner if your lungs are more than 60.0 cm below the surface. What is the maximum negative gauge pressure you can create in your lungs on dry land, assuming you can achieve $−3.00 cm−3.00 cm size 12{ - 3 "." "00""cm"} {}$ water pressure with your lungs 60.0 cm below the surface?
72.
Gauge pressure in the fluid surrounding an infant’s brain may rise as high as 85.0 mm Hg (5 to 12 mm Hg is normal), creating an outward force large enough to make the skull grow abnormally large. (a) Calculate this outward force in newtons on each side of an infant’s skull if the effective area of each side is $70.0cm270.0cm2 size 12{"70" "." 0"cm" rSup { size 8{2} } } {}$. (b) What is the net force acting on the skull?
73.
A full-term fetus typically has a mass of 3.50 kg. (a) What pressure does the weight of such a fetus create if it rests on the mother’s bladder, supported on an area of $90.0cm290.0cm2 size 12{"90" "." 0"cm" rSup { size 8{2} } } {}$? (b) Convert this pressure to millimeters of mercury and determine if it alone is great enough to trigger the micturition reflex (it will add to any pressure already existing in the bladder).
74.
If the pressure in the esophagus is $−2.00 mm Hg−2.00 mm Hg size 12{ - 2 "." "00""mm""Hg"} {}$ while that in the stomach is $+20.0 mm Hg+20.0 mm Hg size 12{+"20" "." 0"mm""Hg"} {}$, to what height could stomach fluid rise in the esophagus, assuming a density of 1.10 g/mL? (This movement will not occur if the muscle closing the lower end of the esophagus is working properly.)
75.
Pressure in the spinal fluid is measured as shown in Figure 11.42. If the pressure in the spinal fluid is 10.0 mm Hg: (a) What is the reading of the water manometer in cm water? (b) What is the reading if the person sits up, placing the top of the fluid 60 cm above the tap? The fluid density is 1.05 g/mL.
Figure 11.42 A water manometer used to measure pressure in the spinal fluid. The height of the fluid in the manometer is measured relative to the spinal column, and the manometer is open to the atmosphere. The measured pressure will be considerably greater if the person sits up.
76.
Calculate the maximum force in newtons exerted by the blood on an aneurysm, or ballooning, in a major artery, given the maximum blood pressure for this person is 150 mm Hg and the effective area of the aneurysm is $20.0cm220.0cm2 size 12{"20" "." 0"cm" rSup { size 8{2} } } {}$. Note that this force is great enough to cause further enlargement and subsequently greater force on the ever-thinner vessel wall.
77.
During heavy lifting, a disk between spinal vertebrae is subjected to a 5000-N compressional force. (a) What pressure is created, assuming that the disk has a uniform circular cross section 2.00 cm in radius? (b) What deformation is produced if the disk is 0.800 cm thick and has a Young’s modulus of $1.5×109N/m21.5×109N/m2 size 12{1 "." 5 times "10" rSup { size 8{9} } "N/m" rSup { size 8{2} } } {}$?
78.
When a person sits erect, increasing the vertical position of their brain by 36.0 cm, the heart must continue to pump blood to the brain at the same rate. (a) What is the gain in gravitational potential energy for 100 mL of blood raised 36.0 cm? (b) What is the drop in pressure, neglecting any losses due to friction? (c) Discuss how the gain in gravitational potential energy and the decrease in pressure are related.
79.
(a) How high will water rise in a glass capillary tube with a 0.500-mm radius? (b) How much gravitational potential energy does the water gain? (c) Discuss possible sources of this energy.
80.
A negative pressure of 25.0 atm can sometimes be achieved with the device in Figure 11.43 before the water separates. (a) To what height could such a negative gauge pressure raise water? (b) How much would a steel wire of the same diameter and length as this capillary stretch if suspended from above?
Figure 11.43 (a) When the piston is raised, it stretches the liquid slightly, putting it under tension and creating a negative absolute pressure $P=−F/AP=−F/A size 12{P= - F/A} {}$ (b) The liquid eventually separates, giving an experimental limit to negative pressure in this liquid.
81.
Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at $15.0 m/s15.0 m/s size 12{"15" "." 0"m/s"} {}$ and brought to rest in 2.80 mm. (a) What average force is exerted on the nail? (b) How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long? (c) What pressure is created on the 1.00-mm-diameter tip of the nail?
82.
Calculate the pressure due to the ocean at the bottom of the Marianas Trench near the Philippines, given its depth is $11.0 km11.0 km size 12{"11" "." 0"km"} {}$ and assuming the density of sea water is constant all the way down. (b) Calculate the percent decrease in volume of sea water due to such a pressure, assuming its bulk modulus is the same as water and is constant. (c) What would be the percent increase in its density? Is the assumption of constant density valid? Will the actual pressure be greater or smaller than that calculated under this assumption?
83.
The hydraulic system of a backhoe is used to lift a load as shown in Figure 11.44. (a) Calculate the force $FF size 12{F} {}$ the secondary cylinder must exert to support the 400-kg load and the 150-kg brace and shovel. (b) What is the pressure in the hydraulic fluid if the secondary cylinder is 2.50 cm in diameter? (c) What force would you have to exert on a lever with a mechanical advantage of 5.00 acting on a primary cylinder 0.800 cm in diameter to create this pressure?
Figure 11.44 Hydraulic and mechanical lever systems are used in heavy machinery such as this back hoe.
84.
Some miners wish to remove water from a mine shaft. A pipe is lowered to the water 90 m below, and a negative pressure is applied to raise the water. (a) Calculate the pressure needed to raise the water. (b) What is unreasonable about this pressure? (c) What is unreasonable about the premise?
85.
You are pumping up a bicycle tire with a hand pump, the piston of which has a 2.00-cm radius.
(a) What force in newtons must you exert to create a pressure of $6.90×105Pa6.90×105Pa size 12{6 "." "90" times "10" rSup { size 8{5} } "Pa"} {}$ (b) What is unreasonable about this (a) result? (c) Which premises are unreasonable or inconsistent?
86.
Consider a group of people trying to stay afloat after their boat strikes a log in a lake. Construct a problem in which you calculate the number of people that can cling to the log and keep their heads out of the water. Among the variables to be considered are the size and density of the log, and what is needed to keep a person’s head and arms above water without swimming or treading water.
87.
The alveoli in emphysema victims are damaged and effectively form larger sacs. Construct a problem in which you calculate the loss of pressure due to surface tension in the alveoli because of their larger average diameters. (Part of the lung’s ability to expel air results from pressure created by surface tension in the alveoli.) Among the things to consider are the normal surface tension of the fluid lining the alveoli, the average alveolar radius in normal individuals and its average in emphysema sufferers.
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# A General Framework For Error Analysis In Measurement-based Gis
To formulate the general point-in-polygon problem in a suitable way, a conditional probability mechanism is first introduced in order to accurately characterize the nature of the problem and establish the basis An approximate law of error propagation is then formulated. Here are the instructions how to enable JavaScript in your web browser. For the point-in-triangle problem, however, such a case need not be considered since any triangle always forms an interior or region.
The distance between two random points—i.e., the length of a random line segment—may be viewed as a nonlinear mapping of the coordinates of the two points. The theoretical arguments are substantiated by simulation experiments.Key wordsalgebra-based probability modelapproximate covariance-based error bandpoint-in-trianglepoint-in-polygonquadratic formJEL ClassificationC10C15C31This project was supported by the earmarked grant CUHK 4362/00H of the Hong Kong Research grants I was curious because I wondered what it was doing that the other ?elds of research, such as statistics and the broad ?eld of arti?cial intelligence, were not doing.
doi:10.1007/s10109-004-0144-1 11 Citations 107 Views Abstract.This is the final of a series of four papers on the development of a general framework for error analysis in measurement-based geographic information systems (MBGIS). In these commentaries, the authors reflect on various aspects of the original work, including what drove their original research, why the concepts were widely adopted (or in some cases, why not), A new concept, called “maximal allowable limit”, which guarantees invariance in topology or geometric-property of a polygon under ME is also advanced. Part of Springer Nature.
An approximate law of error propagation for the intersection point is formulated within the MBGIS framework. M. Continuing to Part 3 is the analysis of ME in intersections and polygon overlays. http://link.springer.com/article/10.1007/s10109-004-0142-3 Full-text · Article · Sep 2016 Shuqiang XueYuanxi YangYamin DangRead full-textError in target-based georeferencing and registration in terrestrial laser scanning" The problem investigated in this paper is essentially an error propagation
Though it is a classic problem, it has, however, not been addressed appropriately. In Part 4, error analyses in length and area measurements are made. Your cache administrator is webmaster. Open with your PDF reader Access the complete full textYou can get the full text of this document if it is part of your institution's ProQuest subscription.Try one of the following:Connect to
SchmorrowNaval Postgraduate School (U.SRead full-textStochastic evaluation of life insurance contracts: Model point on asset trajectories and measurement of the error related to aggregationArticle · Nov 2012 Oberlain Nteukam T.Frédéric PlanchetReadMathematical model https://www.researchgate.net/publication/220449187_A_general_framework_for_error_analysis_in_measurement-based_GIS_Part_1_The_basic_measurement-error_model_and_related_concepts_Journal_of_Geographical_Systems_6325-354 An analytical method is derived for the error propagation into any particular point of interest within a TIN model. It is well known that overlay is one of the most important operations in GIS, and point-in-polygon analysis is a basic class of overlay and query problems. In this paper, we discuss the problem of point-in-polygon analysis under randomness, i.e., with random measurement error (ME).
Part 2 investigates the classic point-in-polygon problem under ME. I was excited because the term tends to convey a new ?eld that is in the making. The solutions proposed can aid in the planning of TLS surveys where a minimum accuracy requirement is known, and are of use for subsequent analysis of the uncertainty in TLS datasets.Article have a peek here This paper suggesting the use of a k-order bias correction formula and a nonlinear error propagation approach to the distance equation provides a useful way to describe the length of a
The 20th article in this collection is an original contribution that examines the social and collaborative networks operating within and shaping the body of research in the field. Fisher and Tate (2006) gave a useful summary of previous research work on error propagation in relation to DEMs and"[Show abstract] [Hide abstract] ABSTRACT: Digital elevation models (DEMs) have been widely As a comparison, the approximate law of error propagation in area measurement is also considered and its approximation is substantiated by numerical experiments.Key wordsError propagationgeographic information systemslength and area measurementmeasurement errornoncentral
The study of errors in GIS has been extensive and diverse (e.g. To seek a suitable replacement, an investigation of the spatial pattern and the magnitude of the actual registration and georeferencing errors in TLS data points was undertaken. SmethurstPeter M. Error propagation in GIS is a broad topic, with the solution very often dependent on the nature of the data set and its embedded error, the GIS operations and the GIS
Differing provisions from the publisher's actual policy or licence agreement may be applicable.This publication is from a journal that may support self archiving.Learn moreLast Updated: 11 Aug 16 © 2008-2016 researchgate.net. doi:10.1007/s10109-004-0142-3 8 Citations 117 Views Abstract.This is the second paper of a four-part series of papers on the development of a general framework for error analysis in measurement-based geographic information systems During those two decades, the International Journal of Geographic Information Science (formerly Systems) (IJGIS) was one of the most prominent academic guiding forces in GIScience, and looks to remain so for Check This Out Careful scrutiny of the main lines of research in data mining and knowledge discovery again told me that they are not much different from that of conventional data analysis.
Leung et al. (2004) proposed a framework for error analysis and propagation in a measurement-based GIS (a concept proposed by Goodchild (1999)), and reviewed common techniques used for measuring GIS errors Error propagation in 2D GIS has increasingly been taken 4 into account (Kobayashi et al., 2011; Ranacher et al., 2016; Xue et al., 2015), and 5 more recently also in 3D An overv"[Show abstract] [Hide abstract] ABSTRACT: Terrestrial laser scanning (TLS) has been used widely for various applications, such as measurement of movement caused by natural hazards and Earth surface processes. J Geograph Syst (2004) 6: 403.
Part of Springer Nature. A new concept, called “maximal allowable limit”, which guarantees invariance in topology or geometric-property of a polygon under ME is also advanced. Although carefully collected, accuracy cannot be guaranteed. By using our services, you agree to our use of cookies.Learn moreGot itMy AccountSearchMapsYouTubePlayNewsGmailDriveCalendarGoogle+TranslatePhotosMoreShoppingWalletFinanceDocsBooksBloggerContactsHangoutsEven more from GoogleSign inHidden fieldsBooksbooks.google.com - When I ?rst came across the term data mining and knowledge
So far, evaluation of registration and georeferencing errors has been based on statistics obtained from the data processing software provided by scanner manufacturers. J Geograph Syst (2004) 6: 325. For the point-in-triangle problem, four quadratic forms in the joint coordinate vectors of a point and the vertices of the triangle are constructed. Publisher conditions are provided by RoMEO.
In this paper, we study the characteristics of error structures in intersections and polygon overlays. In this paper, we discuss the error analysis problems in length and area measurements under measurement error (ME) of the defining points. A simple, unified, and effective treatment of error bands for a line segment is made under the name of “covariance-based error band”. ErmishinReadData provided are for informational purposes only.
With ME in the location of the vertices of a polygon, the resulting random polygons may undergo complex changes, so that the point-in-polygon problem may become theoretically and practically ill-defined. | 1,713 | 8,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-34 | longest | en | 0.923915 |
http://u.osu.edu/durand.8/category/good-reads/ | 1,534,917,366,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219495.97/warc/CC-MAIN-20180822045838-20180822065838-00128.warc.gz | 419,319,880 | 10,905 | Great way to visualize Great Lakes water budgets
Via Matthew Garcia, this infographic by Kaye LaFond of glerl.noaa.gov is pretty rad.
The bars on the left represent Precipitation onto the lake, Runoff into the lake, and Evaporation out of the lake, in thousands of cubic meters per second. To convert to cubic kilometers per year, multiply by 31.5. The bars on the right indicate flow between lakes.
We have a paper in review right now about how a change from seasonal to ephemeral snow in this area would greatly impact Lakes Michigan & Huron. Interesting to think about the runoff in this context, which would sort of change seasonality. Does the runoff from the relatively small watersheds of these lakes add up to a lot in comparison to the other fluxes in and out of the lakes?
The water balance for Michigan-Huron is that the mass balance of the lake from the left-hand side + that from the right-hand side must equal zero, or rearranging:
$P+R-E = Q_{out}-Q_{in}$
The values of those quantities are defined in the table below:
Quantity Value [ $\text{ km}^3/\text{year}$ ]
$P$ 97.7
$R$ 85.1
$E$ 59.9
$Q_{in}$ 69.3
$Q_{out}$ 170.1
So the left-hand side of the equation gives $97.7+85.1-59.9 = 122.9 \text{ km}^3/\text{year}$. And the right-hand side of the equation gives $170.1-69.3 = 100.8 \text{ km}^3/\text{year}$. Note that the two do not match, which is noted explicitly in the infographic:
You may find that the numbers listed here do not always balance … which speaks to the uncertainty of some of these values… However, the values here are the best available, and generally give a good representation of the relative contribution of each of the water budget components.
The imbalance is relatively large, however. The fluxes out of the lake total 230 km3/year, and the error is about 22 km 3/year, a difference of nearly 10 %. Props to the authors for drawing attention to the uncertainty. Assuming that the flow between lakes is relatively well known, then the value of $P+E-R$ of 122.9 km 3/year is only known to ~18 %. This is arguably problematic in trying to detect how changes such as those I mentioned above about earlier snowmelt will impact lake levels. We still have a lot to learn about the water cycle! Our ability to characterize large-scale hydrology is still not where it needs to be.
However, as stated in the quote, this mass balance is of great value in thinking about changes in the snowmelt runoff entering the lake during the winter. It helps to show that the runoff coming into the lakes is a huge component of the annual budget, forming a total of ~1/3 of the total inflow to Michigan-Huron. So a dramatic change in the timing of the inflow could have huge impacts on overall lake levels. Seasonality would also have to be considered.
Do snow-dominated basins have lower or higher runoff ratios?
The recent Nature Climate Change paper by Berghuijs et al. analyzed the MOPEX catchments. Interestingly, they found that streamflow is greater for snowmelt-dominated catchments:
Fig. 1 from Berhuijs et al.
In their Fig 1b, they show that basins with snow:precipitation ratio of 0.6 vs. 0.2 would be expected to have about 10% more runoff. They don’t attempt to answer the question of why. | 795 | 3,237 | {"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": 10, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | longest | en | 0.900536 |
https://www.topcoder.com/blog/single-round-match-830-editorials/ | 1,720,874,450,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514494.35/warc/CC-MAIN-20240713114822-20240713144822-00047.warc.gz | 882,506,773 | 26,038 | # June 7, 2022 Single Round Match 830 Editorials
## AssignLabels
We have been given a task with two separate parts: first we need to produce a set of distinct labels, and then we need to assign them to the objects in correct order.
One way to produce a set of labels is to copy an arbitrarily long text. You can sort its words, discard duplicates, take the first 200 and embed them into your code as a constant.
Another option is to generate the labels directly from IDs. There are 26 letters, so there are 26^2 distinct two-letter strings. This is more than 200, so we have enough distinct labels, and we can generate them really easily. (In the code shown below we generate the label directly from its ordinal number, but you can also generate all 26^2 into an array using two nested cycles, and then index into the array.)
To finish our work, we now look at the given order. We assign the label “aa” to the object order[0], then “ab” to order[1], and so on.
String label(int id) {
}
public String[] assign(int N, int[] order) {
for (int n=0; n<N; ++n) answer[ order[n] ] = label(n);
}
In retrospect, apologies for not having stronger examples that would catch the issue that caused many participants to fail: a solution that produced an “inverse permutation” of labels passed the examples. This was not intended – we certainly weren’t trying to trick you into misreading the statement. We didn’t realize that the current examples are not sufficient to catch solutions that have this issue. If we had noticed this, we would have included one more example.
The issue here is that for example for the input order = {1, 2, 3, 4, 0} a correct output, as defined by the statement, would be { “ee”, “aa”, “bb”, “cc”, “dd” }. Note that the statement said that the elements of order are the IDs of the objects, ordered from the object with the smallest label to the object with the largest label. So for example object with ID 0 should have the largest number, because 0 is the last element of order. Many solvers misunderstood this part and instead returned a sequence of labels equivalent to {“bb”, “cc”, “dd”, “ee”, “aa”} – as if order[x] told us the order of element x.
## MaxDiffBetweenCards
The easiest way to solve this problem is to use brute force for small N, we can easily discover and generalize the pattern (“Given N, I should choose these N digits, and then this is the largest and this is the smallest possible number formed out of those digits and this is their difference.”)
The optimal collection of digits is (N div 2)x ‘9’, one ‘1’, and the rest are ‘0’s. The largest number is 99…9100…0, the smallest number is 100…099…9, and the answer is their difference.
Below we will show how to derive the above result exactly.
Let’s start with a simple observation. If we already have a set of digits written on the cards, we can:
• produce the largest possible number by sorting the digits largest to smallest
• produce the smallest possible number by putting the smallest non-zero digit first and then placing all remaining digits smallest to largest.
Thus, the two numbers are almost – but not entirely – reverses of each other.
(The above observation makes it really easy to brute force everything up to N = 8. We can simply iterate over all N-digit numbers and for each of them we construct the largest number with the same digits and compute the difference.)
If we only have a single digit (N = 1), the answer is clearly zero. Below, assume N >= 2.
The key observation is that we can choose the digits greedily, going left to right (i.e., starting with the most significant digits). This is because once you make a smaller difference at a more significant place, it doesn’t matter what you do next, the final difference will still be smaller.
The leading digits will be 9 and 1 in the largest and smallest number, respectively. Then, we get some pairs (9, 0) until we get to the half of the number. If N is even, everything is now fixed: e.g., for N=8 the largest number starts 9999 and the smallest starts 1000, which means that the largest number is 99991000 and the smallest one is 10009999.
If N is odd, we still have one more digit to choose, but there it’s already easy to check all options we have and pick the best one: another 0 would get paired with the 1 we already have, while anything else would get paired with itself. Thus, for example the best numbers for N=9 are 999910000 and 100009999. (If we had a 3 instead of the final 0, we would have 999931000 and 100039999, with a smaller difference.)
## Jetpack
This is clearly some shortest path problem, we just need to discover how to adapt some standard shortest path algorithm to solve it. Doing it in a too naive way (i.e., having a state where the charges of the jetpack are a part of the state) will probably be too slow – and we would still need to spend some time thinking about an upper bound for the number of simultaneous charges we might need. Let’s come up with a better way.
The first observation we can make is that in the optimal solution stepping on a pit will cost us exactly T+1 seconds: one for the actual move and T for charging the jetpack at some earlier point in time.
The second observation is that we can do all the charging we’ll need at the very first charging station we encounter during our travels: any other solution can be modified into one that does this without changing how long it takes.
This already gives us a full solution that’s fast enough. Consider a graph in which the vertices are triples (integer, integer, boolean): coordinates where we are, and a flag whether we already saw a charging station. In the states that haven’t seen it we are not allowed to enter pits. In the states that did we can enter pits and it costs T+1 seconds, as described above. The shortest path from (coordinates of A, false) to (coordinates of B, anything) in this graph is the optimal way of reaching B from A.
Alternately, we can implement the solution by doing two searches in the original graph: one from A and one from B. For the search from B entering a pit costs T+1. For the search from A it costs 100,007 (i.e., more than any path that never uses a jetpack).
In this solution, we use the search from A to determine the best way of reaching B without flying, and also the set of reachable charging stations. Then, the optimal solution is either the best route without flying, or the best route that reaches some charging station without flying and then goes from there to B, flying if needed.
int R, C;
char[][] maze;
long record(int dist, int r, int c) {
}
int[][] dijkstra(int sr, int sc, int T) {
int[] DR = new int[] {-1, 1, 0, 0};
int[] DC = new int[] {0, 0, -1, 1};
int[][] distances = new int[R][C];
for (int r=0; r<R; ++r) for (int c=0; c<C; ++c) distances[r][c] = 987654321;
distances[sr][sc] = 0;
TreeSet<Long> Q = new TreeSet<Long>();
while (!Q.isEmpty()) {
long curr = Q.first();
Q.remove(curr);
int cc = (int)(curr % 107);
curr /= 107;
int cr = (int)(curr % 107);
for (int d=0; d<4; ++d) {
int nr = cr + DR[d], nc = cc + DC[d];
if (maze[nr][nc] == '#') continue;
int cl = (maze[nr][nc] == '_' ? T+1 : 1);
if (distances[nr][nc] <= distances[cr][cc] + cl) continue;
Q.remove( record( distances[nr][nc], nr, nc ) );
distances[nr][nc] = distances[cr][cc] + cl;
Q.add( record( distances[nr][nc], nr, nc ) );
}
}
return distances;
}
public int travel(String[] _maze, int T) {
R = _maze.length + 2;
C = _maze[0].length() + 2;
maze = new char[R][C];
for (int r=0; r<R; ++r) for (int c=0; c<C; ++c) maze[r][c] = '#';
for (int r=0; r<R-2; ++r) for (int c=0; c<C-2; ++c)
maze[r+1][c+1] = _maze[r].charAt(c);
int sr = -1, sc = -1, tr = -1, tc = -1;
for (int r=0; r<R; ++r) for (int c=0; c<C; ++c) {
if (maze[r][c] == 'A') { sr=r; sc=c; maze[r][c] = '.'; }
if (maze[r][c] == 'B') { tr=r; tc=c; maze[r][c] = '.'; }
}
int[][] distances1 = dijkstra(sr, sc, 100_000);
int[][] distances2 = dijkstra(tr, tc, T );
if (distances1[tr][tc] < 100_000) {
}
for (int r=0; r<R; ++r) for (int c=0; c<C; ++c) if (maze[r][c] == 'C') {
if (distances1[r][c] >= 100_000) continue;
}
if (answer == 987654321) return -1;
}
## MorseCodeCorrector
In my opinion, the main issue to tackle in this problem is that the code can easily get quite ugly. Below I’ll try describing a solution that should be pretty straightforward to implement and not too ugly.
We will approach the problem via automata theory: we’ll build a deterministic finite automaton that recognizes the set of all valid messages, and then we will use dynamic programming to find the best way to get the automaton to accept our string.
Morse codes for letters can be represented as a rooted tree (more precisely, a trie) as shown below. The left branch is always a dot, the right branch is a dash.
start
_____/ \_____
/ \
__E__ __T__
/ \ / \
I A N M
/ \ / \ / \ / \
S U R W D K G O
/ \ / / / \ / \ / \ / \
H V F L P J B X C Y Z Q
We can now build an automaton with 30 states: State 0 is the start state (we will be here only for the empty string.) States 1-26 are the letters in alphabetical order. The edges labelled ‘.’ and ‘-’ are taken from the Morse tree.
From each letter, there is an edge labelled ‘|’ (separator) to state 27 (“end of letter”), from that state there is an edge labelled ‘|’ to state 28 (“end of word”). Both from state 27 and state 28, a dot gets us to state “E” (state 5) and a dash gets us to state “T” (state 20). All other edges of the DFA lead to state 29 (“garbage”). States 1-26 (letters) are accepting states.
We can easily build this entire DFA in memory from the textual representation of codes we were given in the statement.
As promised, now we want to use dynamic programming to find the best set of edits. We will do so as follows: for each i and j we are looking at the prefix of length i and we are asking the question: “What is the smallest number of edits needed if I want to finish in state j after this prefix of length i is processed?”
The easiest way of doing this is by iterating over all i in increasing order, and each time processing all states. In the beginning we know that for i=0 we can be in state 0 for free and we cannot be in any other state. Now, if we know the costs for some i, the costs for i+1 are computed as follows: for each state j we look at the optimal cost C[i][j] of reaching it in i steps, and then:
• we can process the next character of the message to reach the state automaton[ j ][ next_char ] with exactly C[i][j] edits
• or we can change the next character of the message to some other character y, which will bring us to the state automaton[ j ][ y ] in exactly C[i][j] + 1 edits.
int getIndex(String[] CODES, String code) {
for (int i=0; i<CODES.length; ++i) if (CODES[i].equals(code)) return i;
return 29;
}
int[][] buildAutomaton() {
String[] CODES = new String[] { "", ".-", "-...", "-.-.", "-..", ".",
"..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--", "-.",
"---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--",
"-..-", "-.--", "--.." };
int[][] automaton = new int[30][3];
for (int a=0; a<30; ++a) for (int b=0; b<3; ++b) automaton[a][b] = 29;
for (int a=0; a<27; ++a) for (int b=0; b<2; ++b) {
String newCode = CODES[a] + ".-".charAt(b);
automaton[a][b] = getIndex( CODES, newCode );
}
for (int a=1; a<27; ++a) automaton[a][2] = 27;
automaton[27][2] = 28;
automaton[27][0] = automaton[28][0] = getIndex( CODES, "." );
automaton[27][1] = automaton[28][1] = getIndex( CODES, "-" );
return automaton;
}
public String fix(String message) {
int[][] automaton = buildAutomaton();
int S = automaton.length;
int N = message.length();
int[] msg = new int[N];
for (int n=0; n<N; ++n) msg[n] = ".-|".indexOf( message.charAt(n) );
int[][] dp = new int[N+1][S];
for (int n=0; n<=N; ++n) for (int s=0; s<S; ++s) dp[n][s] = 987654321;
dp[0][0] = 0;
for (int n=0; n<N; ++n) for (int s=0; s<S; ++s) {
if (dp[n][s] == 987654321) continue;
int ns;
ns = automaton[s][msg[n]];
dp[n+1][ns] = Math.min( dp[n+1][ns], dp[n][s] );
for (int b=0; b<3; ++b) {
ns = automaton[s][b];
dp[n+1][ns] = Math.min( dp[n+1][ns], 1 + dp[n][s] );
}
}
int best = N+47, bestn = N, bests = -1;
for (int s=1; s<=26; ++s) {
if (dp[N][s] < best) { best = dp[N][s]; bests = s; }
}
int[] fixmsg = new int[N];
while (bestn > 0) {
int founds = -1, foundx = -1;
for (int ps=0; ps<S; ++ps) {
int ns;
ns = automaton[ps][msg[bestn-1]];
if (ns == bests && dp[bestn][bests] == dp[bestn-1][ps]) {
founds = ps; foundx = msg[bestn-1];
}
for (int b=0; b<3; ++b) {
ns = automaton[ps][b];
if (ns == bests && dp[bestn][bests] == 1 + dp[bestn-1][ps]) {
founds = ps; foundx = b;
}
}
if (founds != -1) break;
}
fixmsg[bestn-1] = foundx;
bests = founds;
--bestn;
}
for (int x : fixmsg) answer += ".-|".charAt(x);
}
Today, both divisions got a problem where brute force and discovering patterns is a viable strategy. In div2 it was the medium, in div1 it’s the hard problem – and the brute force solution was considerably more complicated and the pattern probably harder to recognize.
Below we’ll do the math and derive the full solution without having to guess it from a pattern.
We’ll start with blue marbles only. We have B blue marbles and we want to count all valid processes. For B=1 and B=2 the answer is clearly 1. Now let’s have a larger value B. Let’s pick one specific marble and set it aside. Imagine that we generated all possible processes for the other B-1 marbles. We can now generate all possible processes for all B marbles as follows: we can start from any process for B-1 marbles and we can decide when exactly our final marble got placed into its own bag and from where. Each such decision will give us one valid process for all B marbles.
For example, we can see that in our current process for B-1 marbles we had a bag {A,B,C,D} at some moment in time. If we pick this as our moment, this bag will now contain marbles {A,B,C,D,X}, the next step will split it into {A,B,C,D} and {X}, and from this point on we continue the original steps we did with {A,B,C,D}.
All the processes for B marbles we just created are clearly distinct: we cannot get the same one from two distinct processes for B-1 marbles.
And no processes for B marbles are missing from this collection – this is because we can take any valid process for B marbles and “undo the above” to get a valid process for B-1 marbles. (We remove the action when the last marble got its own bag and then we pretend it does not exist.)
Thus, the total number of processes for B marbles can be computed by looking at each process for B-1 marbles, finding the number of ways to add the B-th marble, and summing all those numbers of ways.
Regardless of which schedule for B-1 marbles we take, there are always exactly 2B-3 ways to add the last marble. This is because in the whole process we’ll always encounter exactly that many distinct bags of marbles. (We start with all B-1 marbles in the same bag. Each split increases the number of non-empty bags by 1, and we end with B-1 one-marble bags. Thus, there are exactly B-2 splits, and therefore exactly B-2 distinct bags being split.)
And that gives us a surprisingly simple conclusion: the number of ways for B marbles is simply the number of ways for B-1 marbles, multiplied by 2B-3.
Hence, the counts for blue marbles are the so-called “double factorials” – like factorials but you only multiply odd numbers. E.g., for B=5 the answer is 1*3*5*7.
Now let’s add red marbles. Again, we’ll do so one at a time (after all the blue marbles have already been added), and again we can do the same reasoning as above, with just one small change. When we already have a process for N other marbles and we are adding a new red marble, we can only choose one of the N-1 bags with two or more marbles as the exact moment when the red marble got its own bag.
Thus, the number of ways to add the red marbles is a simple product of consecutive integers: if there are B >= 2 blue and R red marbles, we have (B-1)*B*…*(B+R-2) ways to add the red marbles into any valid process for the blue marbles.
If B or R is too large, the answer is zero because the product will contain the value 10^9 + 7 somewhere. We just need to compute factorials and double factorials until they hit that moment. This would be slightly slow if we just did it naively, but we can easily speed it up by e.g. precomputing and storing every 10^6-th factorial.
(We can do the same for double-factorials but we don’t have to. We can use the observations that 1*3*5*7 = 8! / (2*4*6*8), and that 2*4*6*8 = 4! * 2^4.
// omitted parts of code:
// - classes Fact and DFact that compute factorials and double-facts quickly
// - modular exponentiation and modular inverse
long singleBag(long red, long blue) {
if (red + blue == 1) return 1;
if (blue <= 1) return 0;
if (answer == 0) return 0; // we already know we have too many blues
if (blue+red-2 >= 1_000_000_007) return 0;
}
public int count(long[] R, long[] B) {
F = new Fact();
DF = new DFact();
for (int i=0; i<R.length; ++i) {
answer *= singleBag( R[i], B[i] );
}
}
misof
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M,i = I xVE,Wan( 1 S2)^]/i?T|,[ 1 F a cos(</, - *)]. (4.22)
I he rate of change of mass of working fluid in the dead space is dM(,/d</> - [XVEpme,B(l - d2)'- 5 sin(<f> - 0)]/KTD[ I l 8 cos(¿ - 0)
Now dMc f d.Víc k d A'/j = 0, so that the total mass of working fluid Mr is constant. Now.
Mt= VRpmc.B(l -82Wt( I Feos </>)+1<| l+cos(cfr - «)]
Af,= V,,P„ICn„(l-«2)'[T + S F(k/2)(1 +cos a)]/R7'c(l I 5 cos 0).
Heat lifted and engine output in dimensionless units
(a) The heat lifted per unit mass of working fluid, combining eqns (4.17) and (4.23) is given by
On»« ^ QIMrRTc = tt6 sin 0( I + 8 cos 0)/{( I 82)¡\ 1 + (I - 82)*]
Similarly, the net engine-output per unit mass of working fluid is given by
(b) Non-dimensional expressions, in terms of characteristic pressures and volumes, may be devised as follows. The combined swept volume is given bv
Combining this with eqns (4.13) and (4.17), then
OmnK = 0/(pinax V-r) -[tt(1 - 5)* 5 sin 0]f[( 1 + k)( 1 + 1 + (1 - 52)i)]
## Solar Stirling Engine Basics Explained
The solar Stirling engine is progressively becoming a viable alternative to solar panels for its higher efficiency. Stirling engines might be the best way to harvest the power provided by the sun. This is an easy-to-understand explanation of how Stirling engines work, the different types, and why they are more efficient than steam engines.
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Lecture 3: Linear Regression, Exploratory Data Analysis, and the Bootstrap STAT GR5206 Statistical Computing & Introduction to Data Science Cynthia Rush Columbia University September 23, 2016 Cynthia Rush Lecture 3: Regression and Graphics September 23, 2016 1 / 104
Course Notes Next week labs are meeting. Homework 1 is due on Monday at 8pm. No late homeworks accepted. Homework 2 will be assigned on Monday. Remember to use Piazza to ask questions. Cynthia Rush Lecture 3: Regression and Graphics September 23, 2016 2 / 104
Last Time Filtering . Accessing elements of a structure based on some criteria. v[v>5], m[ m[,1]!=0, ] . Lists . Elements can all be di erent types. Access like l[[3]], l\$name . Create with list() . NA and NULL values . NA is missing data and NULL doesn’t exist. Factors and Tables . Factors is how R classifies categorical variables. Dataframes . Used for data that is organized with rows indicating cases and columns indicating variables. Importing and Exporting Data in R . Use read.csv() and read.table() depending on dataset type. The working directory. Control Statements . We studdied iteration, for loops and while loops, and if, else statements. Vectorized Operations . To be used instead of iterations. Cynthia Rush Lecture 3: Regression and Graphics September 23, 2016 3 / 104
Section I Multiple Linear Regression Cynthia Rush Lecture 3: Regression and Graphics September 23, 2016 4 / 104
Multiple Linear Regression Example A large national grocery retailer tracks productivity and costs of its facilities closely. Consider a data set obtained from a single distribution center for a one-year period. Each data point for each variable represents one week of activity. The variables included are number of cases shipped in thousands ( X 1 ), the indirect costs of labor as a percentage of total costs ( X 2 ), a qualitative predictor called holiday that is coded 1 if the week has a holiday and 0 otherwise ( X 3 ), and total labor hours ( Y ). Cynthia Rush Lecture 3: Regression and Graphics September 23, 2016 5 / 104
Multiple Linear Regression Suppose, as statisticians, we are asked to build a model to predict total labor hours in the future using this dataset. What information would be useful to provide such a model? Is there a relationship between holidays and total labor hoours? What about number of casses shipped? Indirect costs? How strong are these relationships? Is the relationship linear? Cynthia Rush Lecture 3: Regression and Graphics September 23, 2016 6 / 104
Multiple Linear Regression | 623 | 2,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-05 | latest | en | 0.853063 |
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# Critical Reasoning Concept Articles from Different Companies
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12 Nov 2013, 06:55
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Master this fiendishly tricky variety of Critical Reasoning question on the GMAT
The philosopher Zeno of Elia (490-430 BCE) was famous for his mind-bending paradoxes. One of them went like this:
Suppose there were a race between Achilles (famed as the fastest runner in ancient Greece) and a tortoise, and the tortoise was given a small head start, say to point A. In the time it takes Achilles to run from the start to point A, the tortoise will move to a new point, point B. By the time Achilles catches up to point B, the tortoise will be at a new place, point C. On and on like this, Achilles will have an infinite number of distances to cover before he can even catch up to the tortoise. No one can run an infinite number of distances in a finite amount of time. Therefore even the fleet-footed Achilles would not be able to outrun a slow-footed tortoise.
Confused? That’s the point!
A paradox is a statement or argument that, at least on the surface, does not make sense. In matters purely of personal taste, things can simply not make sense and that’s it — for example, the secularly Jewish man I know who avoids all ham or pork but who absolutely loves eating bacon. Peoples’ idiosyncratic combinations of likes and dislikes need not make sense.
That is not what you will encounter on the GMAT. On GMAT Critical Reasoning, the paradoxical passages will be about the real world, either the natural world or the world of social sciences. These are both realms were perfectly logical laws govern the phenomena, so any paradox will be an apparent paradox, an apparent contradiction, that is resolved by additional information.
For example, with respect to Zeno’s paradox above, in the past few centuries we have come to understand that, mathematically, it is possible under certain conditions to add up an infinite number of things and have the sum be finite. Thus, despite Zeno, Achilles catches up to the tortoise in a finite amount of time, passes it, and wins the race handily, as we would expect. Paradox resolved.
The only paradoxes that appear on GMAT Critical Reasoning will be resolvable paradoxes, and in fact, your job will always be the same: to resolve them. Prompts for these questions include:
“Which of the following most helps to explain the apparent discrepancy described above?”
“Which of the following most helps to resolve the apparent contradiction?”
In other words, you will be presented with a paradox, usually in the form of two facts you would not expect to co-exist side by side in the same situation. You will be asked which of the five answer choices best resolves the paradox.
At least one of the wrong answers, if not two of them, will actually do the opposite: they will exacerbate the paradox, making it even more absurd that these two things go together. Always be on the lookout for that kind of answer when you are looking through the answer choices.
Here’s an example of the kind of paradox CR question you will see on the GMAT.
1) French cuisine is famous for its frequent and liberal use of cream and cheese, both high in saturated fat. For years, medical studies have shown the strong correlation between diets high in saturated fat and coronary heart disease, and yet, France has a much lower incidence of such disease than found in comparable countries like the United States. This is the so-called French Paradox.
Which of the following, if true, helps to explain the French Paradox?
(A) Certain kinds of cheese can have as much as five times the amount of saturated fat that cream has.
(B) People in the United States, per capita, eat almost the same amount of saturated fat on average as do people in France.
(C) The United States imports more cheese from France than from any other country.
(D) Red wine, typically served with French food, helps to clean the buildup of fats in the arteries, reducing the risk of heart disease.
(E) It is typically for a French person to have either cream or cheese at each of the three meals in a day.
Here’s another paradox question for practice:
http://gmat.magoosh.com/questions/1320
Practice Question Explanation
In attacking a paradox question, first read the prompt and understand the paradox or contradiction in your own words. Here, we could say: most people eat high food and get heart attacks, but the French eat high fat foods and don’t get hard attacks. Then, look for an answer choice that best resolves the paradox.
Both (B) and (E) do the opposite: they make the paradox harder to explain. With (B), if folks in the US and France eat about the same amount of saturated fat, then why do Americans get heart disease but not the French? With (E), if French are eating high fat foods all the time, why aren’t they getting heart disease? In other words, neither of these answers the question, and in fact, both of them simply would make it even harder to understand.
Answers (A) and (C) are off-the-wall irrelevant. Choice (A) says that cheese has more fat then cream, but the French are eating both of those, so it doesn’t matter: either way, the French are eating high fat food. (C) changes the topic to imports, which is completely unrelated to the direct relationship of diet and epidemiology.
Only (D) resolves the paradox. Since the French drink red wine, which in moderation cleans the arteries, this explains how they could eat high fat foods and have a much lower risk of heart disease.
By the way, before you run off thinking red wine is a cure-all, let me point out that higher exercise rates, smaller portions, no snacking between meals, and a much lower incidence of processed sugar and prepared foods in the French diet have also been shown to play a big role in resolving the French Paradox.
So, overall, my advice is eat healthy fresh food, lots of fruits and vegetables, drink red wine in moderation (if you are over 21), drink lots of water every day, avoid deadly trans fats, avoid processed sugar, and exercise regularly: these habits will maximize your health. If you also want a healthy GMAT score, then practice more GMAT Critical Reasoning question like this, and sign up for Magoosh test prep. Magoosh is better for your GMAT score than wine is for your arteries, and you don’t even have to be 21 years old to enjoy Magoosh! No paradox there!
By the way, sign up for our 1-week free trial to try out Magoosh GMAT Prep!
Mike McGarry is a Content Developer for Magoosh with over 20 years of teaching experience and a BS in Physics and an MA in Religion, both from Harvard. He enjoys hitting foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. Follow him on Google+!
- See more at: http://magoosh.com/gmat/2012/gmat-cr-pa ... KhlDe.dpuf
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12 Nov 2013, 08:13
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Friends
Concept Article on Resolve the Paradox by “-creasoning.blogspot.in”
A set of true statements that condradict one another.
Resolving a paradox: Selecting a scenario in which both statements can co exist without conflict.
What is tested?
Your ability to digest/assimilate information in the passage.
Your ability to evaluate new information which when introduced in the passage resolves the contradiction.
< 5% of GMAT questions
How to identify RTP question
- Their goal is to introduce a paradox. So spotting one is the way to spot the question.
- Generally, Do not contain any conclusion.
Process for Answering Questions a) Identify the 2 contrasting statements.
b) Pre think on how to resolve.
Resolution Framework
- Because of New Info Fact A is true, eventhough fact B is valid.
- This new info can either address fact A or Fact B or both the facts. Most of cases, it addresses both the facts simultaneously.
This new info can present info that
a) Ineffective Implementation Proposals to lead to Improvements. But they are not realized.
Ans: Ineffective Implementation
b) Alternate Reasoning (Most Common)
c) Improper Comparison Two entities being compared are not comparable in the first place.
Note: The correct answer choice may not resolve the paradox. It may just resonably support the coexistence.
c) iSWAT
d) Rbi: revalant but incorrect. Only explains one side, and hence doesnt resolve the paradox.
did not increase.
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12 Nov 2013, 08:44
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Dear friends
Aspiranhunt has given its views on” Resolve Paradox” questions in a concept article.
http://www.aspiranthunt.com/gmat/2010/0 ... ons-in-cr/
The Paradox Question presents you with a seeming paradox situation in an argument, and then
asks you to seek an explanation on how that discrepancy can exist. Typical ways in which these
• Which of the following statements, if true, would best explain the paradox described above?
• Which of the following can explain the apparent contradiction above?
2. Look for the sequential difference or spatial difference since they may be the reason that causes
3. If you are unable to pick up the correct answer by step 2, pay attention to the subjects in the two
contradictive situation. Are the two subjects the same? If not, the difference could be the reason for
Example
In 1992, 5 percent of every dollar paid in tax went to support the unemployed citizens. In 1998,
8 percent of every dollar paid in tax went to such funds, although that unemployment rate has
decreased in 1998 than in 1992.
Each of the following, if true, could explain the simultaneous increase in percent of every dollar
paid in tax to support the unemployed citizens and decrease in the number of unemployment rate
EXCEPT:
A. On average, each unemployed citizen received more money in 1998 than 1992.
B. On average, people paid less tax in 1998 than in 1992.
C. The individuals had paid more tax than did enterprises during this period.
D. Income before tax has significantly decreased since 1992.
E. The number of tax evaders rose sharply between 1992 and 1998.
Choice A suggests that the total amount of dollars used to support unemployment has increase,
therefore explain the paradox. Choice B, D, and E all suggests that the amount of tax collected
decreased, thus percent of every dollar that went to support the unemployment increases. Only
choice C does not explain such paradox, therefore is the correct answer.
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12 Nov 2013, 07:19
GMAT Test Tips from Veritas Prep
GMAT Tip: How to Deal With a Critical Reasoning 'Paradox'
This tip on improving your GMAT score was provided by Vivian Kerr at Veritas Prep.
While other questions are more heavily tested in the GMAT critical reasoning, you’ll occasionally see questions that involve explaining a paradox or resolving an apparent contradiction. These “resolve the argument” or “paradox” questions ask about a specific incongruous aspect of the argument. To identify them, look for some common keywords such as: “explains the results,” “resolve the paradox,” or “best explains the discrepancy.” Unlike other critical reasoning questions, these don’t ask how to weaken or strengthen the argument itself.
As with most critical reasoning questions, you should start by taking apart the argument: Identify the conclusion, evidence, and assumptions before reading the question after the paragraph. Specifically, pay attention to what is lacking in the details. Usually the author fails to provide enough information. If you were to make the same argument, what would you add to resolve the issue brought up in the question? Write down your prediction(s) and then scan the answer choices, eliminating those that do not resemble your prediction.
Quick Critical Reasoning Strategy
1. Identify the conclusion, evidence and assumptions.
2. Read and rephrase the question.
3. Go back to the passage and form a prediction.
4. Eliminate incorrect choices.
Let’s try one together.
The vast majority of a person’s health-care expenditures go toward curative measures such as hospitalizations after injuries and care for existing illnesses. Mike’s employer does not provide health insurance to his part-time employees, including Mike. However, the employer does reimburse employees for a flu shot each winter.
Mike’s employer’s seemingly inconsistent behavior in regard to health-care expenses is best explained by which of the following?
Before we get to the answer choices, let’s start by analyzing conclusion, evidence, and assumptions:
Conclusion: The employer does not provide health insurance to part-timers.
Evidence: The majority of health-care expenditures go toward curative measures (fixing injuries, illnesses); employer reimburses for flu shots. Notice the gap in logic here: why would an employer who doesn’t pay health insurance reimburse employees for a flu shot?
Assumption: The employer sees some financial benefit in paying for the flu shot (a preventative measure), even though he won’t pay health insurance. He doesn’t want his employees to get sick in the first place.
Now it’s time to get to the actual question and answer choices.
Question Rephrase: We can see this is a “paradox” question because of the phrase “best explained.” What’s the strongest reason why the employer would pay for a flu shot but not pay for health insurance?
Prediction: Some unknown benefit to the employer in the long-term.
(A) Health insurance rarely covers preexisting illnesses.
(B) Part-time employees are usually covered by the insurance of a spouse or parent with full-time employment.
(C) Few employers offer health insurance to part-time employees.
(D) Flu shots prevent illness that could lead to lost work days.
(E) Health insurance premiums are on the rise.
Since we’ve done the work of breaking down the passage, simplifying the question, and predicting an answer, the correct choice (D) is readily apparent.
What can you learn from this question (other than to check to see if your employer will pay for a flu shot)? It pays to analyze the argument before you get to the question stem and answer choices. In order to “best resolve the paradox,” as the questions may ask, it helps to fully understand the paradox first.
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12 Nov 2013, 07:28
Here is a concept article by 800scoree.com on Paradox questions
http://800score.com/guidec3view2e.html
These questions present you with a paradox, a seeming contradiction in the argument, and ask you to resolve it or explain how that contradiction could exist. Paradox questions are rare and more common at the higher skill levels. Here are some examples of the ways in which these questions are worded:
• Which of the following, if true, would help to resolve the apparent paradox presented above?
• Which of the following, if true, contributes most to an explanation of the apparent discrepancy described above?
3. Use POE (process of elimination). The best answer will explain how both sides of the paradox, discrepancy, or contradiction can be true. Eliminate answers that are out of scope.
SAMPLE QUESTION
Inflation rose by 5.1% over the 2nd quarter, up from 4.1% during the first quarter of the year, and higher than the 3.3% recorded during the same time last year. However, the higher price index did not seem to alarm Wall Street, as stock prices remained steady.
Which of the following, if true, could explain the reaction of Wall Street?
a) Stock prices were steady because of a fear that inflation would continue.
b) The President announced that he was concerned about rising inflation.
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Re: Critical Reasoning Concept Articles from Different Companies [#permalink]
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12 Nov 2013, 07:42
Hi Guys
This is concept article on paradox questions by 800gmat.com
http://www.800gmat.com/index.php?cur_pa ... monstrated
Method: Our Optimized Way of Ensuring 100% Accuracy on the GMAT.
Step One: Identify the Question Type and Task.
With all critical reasoning questions read the question first in order to understand what you are asked to find. Any question that asks you to explain or resolve a paradox, discrepancy, or strange situation is a paradox question.
Step Two: Read the Argument and Extract Necessary Information.
The paradox will be the information that you need.
Step Three: Formulate an Answer to the Question
Give yourself a general sense of what a resolution to the paradox might be. This may not be �the answer,� but it will give you a framework for recognizing the answer
.
Step Four: Eliminate Answer Choices That Are Obviously Wrong.
On the initial pass through the answer choices, keep any choice that seems right or that you don't have a definitive reason to eliminate.
Step Five: Compare Remaining Choices.
Compare the remaining choices for subtle differences. Revisit the question and argument to ensure that you have not overlooked some subtlety.
Example One:
Bengal tigers, a near extinct species of tiger, are known to suffer extraordinary hardships when placed in captivity. Captured Bengal tigers rarely produce offspring and contract a much greater number of illnesses during their captivity, yet the consensus among the zoological community is that captivity is the best opportunity to save this species from extinction.
Which of the following best explains the apparent discrepancy in the statements above?
(A) Captive Bengal tigers reproduce at a rate slightly lower than that of wild Bengal tigers.
(B) The majority of the diseases contracted by wild Bengal tigers are treatable but not curable.
(C) The hardships suffered by captive Bengal tigers are not as severe as those suffered by other species of large cats.
(D) The diseases contracted by Bengal tigers in the wild generally cause sterility in the male of the species.
(E) The data from which the Bengal tiger species was determined to be approaching extinction was in error.
Step One: Identify the Question Type and Task
A question that asks you to "resolve a discrepancy" is a paradox question.
Step Two: Read the Argument and Extract Necessary Information.
The necessary information is the paradox.
Paradox: Captive Bengal tigers contract more diseases and rarely produce offspring yet captivity is the best option for the survival of the species.
Step Three: Formulate an Answer to the Question
Wild tigers face hunters, starvation, and other animals as well as disease (captive tigers face none of these).
Step Four: Eliminate Answer Choices That Are Obviously Wrong.
(A) Only addresses one side - this might explain why they reproduce rarely in captivity, but it does not explain why captivity is better. To explain the paradox, both sides must be connected.
(B) Exacerbates the paradox - this leads to the idea that perhaps the wild tigers should be caught, treated, and released rather than put into captivity and it in no way explains why captivity is the best opportunity.
(C) Does not explain - this choice does not provide any reason that would justify placing them in captivity as a better option than the wild. Other species have no relevance here.
(D) Keep - this suggests that the diseases contracted in the wild are fewer in number and will lead to the extinction of the species (since the tigers will not be able to reproduce), thus making it plausible that rarelyproducing offspring is the better option to never producing offspring.
(E) Dismisses the paradox - this choice dismisses the entire question by saying the tiger is not really going to be extinct Remember that you want a choice that resolves the paradox, not one that ignores, dismisses, or changes the paradox.
Step Five: Compare Remaining Choices.
No need.
Choose D.
Example Two:
A recent survey of 100,000 patients suffering from sever acid reflux disease found that a large majority of the patients reported that missing a meal or two immediately eased the symptoms. Yet neither fasting nor dieting are used in treating acid reflux disease even though conventional treatments, which use drugs, are ineffective and have serious side effects.
Which of the following most explains the fact that neither fasting nor dieting is used in treating acid reflux disease?
(A) Prolonged fasting can cause other physical complications, including but not limited to lightheadedness and fainting.
(B) The precise cause of acid reflux disease has not yet been linked to any particular foods, or types of foods.
(C) The symptoms of acid reflux disease return as strong or stronger at the next snack or meal.
(D) For some suffers of acid reflux disease, missing a meal induces a temporary feeling wellness.
(E) Forcing patients to adhere to a diet or fast is more difficult when a patient is feeling ill.
Step One: Identify the Question Type and Task
A question that asks you to "explain" a situation is a paradox question.
Step Two: Read the Argument and Extract Necessary Information.
The necessary information is the paradox.
Paradox: Fasting eases symptoms but is not used to treat acid reflux.
Step Three: Formulate an Answer to the Question
There must be other problems with the use of fasting and dieting as treatment.
Step Four: Eliminate Answer Choices That Are Obviously Wrong.
(A) Keep - this choice offers a possible explanation.
(B) Does not explain - this choice does not provide any information that explains the situation, since the argument made no claims about specific foods.
(C) Keep - this choice offers a possible explanation since if the symptoms come back stronger, then the patient is worse off.
(D) Does not explain - this choice does not provide any information that explains the situation, since missing a meal is not the same as dieting or fasting.
Step Five: Compare Remaining Choices.
Now you must decide between choices A and C. The difference is quite subtle - choice A states that prolonged fasting �can� cause complications, but it is not a guarantee. Perhaps if only .05% of patients developed complications, this would not provide as strong an explanation as choice C, which definitively states that symptoms come back stronger.
Choose C.
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12 Nov 2013, 08:04
Friends;
Learn some key word tips on resolving paradox questions in GMAT CR as given in a concept article by quizlet.com
http://quizlet.com/25063379/gmat-cr-12- ... sh-cards/#
1. Stimulus Features
1) No conclusion - the author is not attempting to persuade you, just presenting two contradictory sets of facts,
2. Question Stem Features
1) an indication that the answer choices should be accepted as true,
2) key words that indicate your task is to resolve a problem
3. Key word indicators
Action: resolve,
explain, reconcile.
conflict, puzzle
Active Resolution
the correct answer will actively resolve the paradox, that is allow both sides to be factually correct, and will either explain how the situation came into being, or will add a piece of information that shows how the two ideas or occurences can coexist.
4. **Important**
if a stimulus contains a paradox where two items are similar, then an answer that explains a difference between the two cannot be correct. Similarily if a stimulus contains a paradox where two items are different, then an answer that explains a similarity cannot be correct.
always choose answer choices that conform to the specifics of the stimulus. Do not be lured by reasonable solutions that do not quite meet the stated facts.
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Re: Critical Reasoning Concept Articles from Different Companies [#permalink]
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12 Nov 2013, 08:20
Hi Friends
Find a concept article from Grockit on Resolving the paradox
http://grockit.com/blog/gmat/2011/04/14 ... -overview/
Resolve the Argument Question Overview from Grockit
Most Critical Reasoning questions you’ve encountered in your online studying have probably been focused on weakening and strengthening the given argument. To diversify your test prep a bit, and keep pushing for better scores, let’s focus on examining a specific CR type that rarely gets enough attention: Resolve the Argument.
What are they? Resolve the argument questions ask about a specific incongruous aspect of the argument.
How to identify? Look for some common keywords such as: “explains the results,” “resolve the paradox,” “best explains the discrepancy,” etc. Unlike other CR questions, these don’t ask how to weaken or strengthen the argument itself.
How to approach? Like any other CR question, you’ll want to identify the conclusion, evidence, and assumptions before even reading the question. Once you realize it’s a “Resolve the Argument” question, you’ll want to rephrase the question in simpler terms, then go back to the passage and find what the “paradox,” “results,” or “discrepancy” is describing. Specifically, pay attention to what islacking in the details. Usually the author fails to provide enough information. If you were to make the same argument, what would you add to resolve the issue brought up in the question? Write down your prediction(s) then scan the answer choices, eliminating those that do not resemble your prediction.
For quick reference:
1. Identify the conclusion, evidence & assumptions.
2. Read and rephrase the question.
3. Go back to the passage & form a prediction.
4. Eliminate incorrect choices.
Let’s try out a Resolve the Argument question from Grockit’s CR question bank:
The majority of a person’s health care expenditures goes towards curative measures like hospitalizations after injuries and care for existing illnesses. Paula’s employer does not provide health insurance to his part-time employees, including Paula. However, he does reimburse employees for a flu shot each winter.
Paula’s employer’s seemingly inconsistent behavior in regard to health care expenses is best explained by which of the following?
A. Health insurance rarely covers pre-existing illnesses.
B. Part-time employees are usually covered by the insurance of a spouse or parent with full-time employment.
C. Few employers offer health insurance to part-time employees.
D. Flu shots prevent illness that could lead to lost work days.
E. Health insurance premiums are on the rise.
Conclusion: Paula’s employer does NOT provide health insurance to part-timers.
Evidence: Majority of \$\$ goes towards curative measures (fixing injuries, illnesses); reimburses for flu shots.
We can see the gap in logic here. Why would an employer who doesn’t pay health insurance reimburse employees for a flu shot?
Assumption: The employer sees some \$\$ benefit in paying the flu shot (a preventative measure), even though he won’t pay health insurance. He doesn’t want his employees to get sick in the first place.
We can see this is a “Resolve the Argument” question because of the phrase in the question stem, “best explained.” So let’s rephrase the question and predict what the answer choice might involve.
Question Rephrase: What’s the strongest reason why the employer would pay for a flu shot but NOT pay health insurance?
Prediction: Some unknown benefit to the employer in the long-term.
Since we’ve done the work of breaking down the passage, simplifying the question, and predicting an answer, the correct choice (D) is readily apparent. To practice more Resolve the Argument questions, you can create a Custom Game on Grockit using only those CR questions with the “Resolve Argument” skill tag.
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```HAWKES LEARNING SYSTEMS
math courseware specialists
Systems/Quant Systems, Inc.
Section 8.2
Estimating Population Means
(Large Samples)
HAWKES LEARNING SYSTEMS
Confidence Intervals
math courseware specialists
8.2 Estimating Population Means
(Large Samples)
Criteria for estimating the population mean for large samples:
• All possible samples of a given size have an equal
probability of being chosen.
• The size of the sample is at least 30 (n ≥ 30).
• The population’s standard deviation is unknown.
When all of the above conditions are met, then the
distribution used to calculate the margin of error for the
population mean is the Student t-distribution.
However, when n ≥ 30, the critical values for the t-distribution
are almost identical to the critical values for the normal
distribution at corresponding levels of confidence.
Therefore, we can use the normal distribution to approximate
the t-distribution.
HAWKES LEARNING SYSTEMS
Confidence Intervals
math courseware specialists
8.2 Estimating Population Means
(Large Samples)
Find the critical value:
Find the critical value for a 95% confidence interval.
Solution:
To find the critical value, we first need to find the values for
–z0.95 and z0.95.
Since 0.95 is the area between –z0.95 and z0.95, there will be
0.05 in the tails, or 0.025 in one tail.
HAWKES LEARNING SYSTEMS
Confidence Intervals
math courseware specialists
8.2 Estimating Population Means
(Large Samples)
Critical Value, zc:
Critical z-Values for Confidence Intervals
Level of Confidence, c
zc
0.80
1.28
0.85
1.44
0.90
1.645
0.95
1.96
0.98
2.33
0.99
2.575
HAWKES LEARNING SYSTEMS
Confidence Intervals
math courseware specialists
8.2 Estimating Population Means
(Large Samples)
Margin of Error, E, for Large Samples:
zc = the critical z-value
s = the sample standard deviation
n = the sample size
When calculating the margin of error, round to one more decimal place
than the original data, or the same number of places as the standard
deviation.
HAWKES LEARNING SYSTEMS
Confidence Intervals
math courseware specialists
8.2 Estimating Population Means
(Large Samples)
Find the margin of error:
Find the margin of error for a 99% confidence
interval, given a sample of size 100 with a sample
standard deviation of 15.50.
Solution:
n = 100, s = 15.50, c = 0.99
z0.99 = 2.575
HAWKES LEARNING SYSTEMS
Confidence Intervals
math courseware specialists
8.2 Estimating Population Means
(Large Samples)
Construct a confidence interval:
A survey of 85 homeowners finds that they spend on average \$67
a month on home maintenance with a standard deviation of \$14.
Find the 95% confidence interval for the mean amount spent on
home maintenance by all homeowners.
Solution:
c = 0.95, n = 85, s = 14,
= 67
z0.95 = 1.96
67 – 2.98 < < 67 + 2.98
\$64.02 < < \$69.98
(\$64.02, \$69.98)
HAWKES LEARNING SYSTEMS
Confidence Intervals
math courseware specialists
8.2 Estimating Population Means
(Large Samples)
Finding the Minimum Sample Size for Means:
To find the minimum sample size necessary to estimate an average,
use the following formula:
zc = the critical z-value
= the population standard deviation
E = the margin of error
When calculating the sample size, round to up to the next whole
number.
HAWKES LEARNING SYSTEMS
Confidence Intervals
math courseware specialists
8.2 Estimating Population Means
(Large Samples)
Find the minimum sample size:
Determine the minimum sample size needed if you wish to
be 99% confident that the sample mean is within two units
of the population mean, given that = 6.5. Assume that the
population is normally distributed.
Solution:
c = 0.99, = 6.5, E = 2
z0.99 = 2.575
You will need a minimum sample size of 71.
HAWKES LEARNING SYSTEMS
Confidence Intervals
math courseware specialists
8.2 Estimating Population Means
(Large Samples)
Find the minimum sample size:
The electric cooperative wishes to know the average household
usage of electricity by its non-commercial customers. They
believe that the mean is 15.7 kWh per day for each family with a
variance of 3.24 kWh.
How large of a sample would be required in order to estimate the
average number of kWh of electricity used daily per family at the
99% confidence level with an error of at most 0.12 kWh?
Solution:
c = 0.99, = 1.8, E = 0.12, z0.99 = 2.575
You will need a minimum sample size of 1492 families.
```
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http://uk.mathworks.com/help/signal/ref/spectrogram.html?requestedDomain=uk.mathworks.com&nocookie=true | 1,474,801,020,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660181.83/warc/CC-MAIN-20160924173740-00022-ip-10-143-35-109.ec2.internal.warc.gz | 263,054,850 | 22,660 | # Documentation
### This is machine translation
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# spectrogram
Spectrogram using short-time Fourier transform
## Syntax
• s = spectrogram(x)
example
• s = spectrogram(x,window)
• s = spectrogram(x,window,noverlap)
example
• s = spectrogram(x,window,noverlap,nfft)
example
• [s,w,t] = spectrogram(___)
• [s,f,t] = spectrogram(___,fs)
example
• [s,w,t] = spectrogram(x,window,noverlap,w)
• [s,f,t] = spectrogram(x,window,noverlap,f,fs)
• [___,ps] = spectrogram(___)
• [___] = spectrogram(___,'reassigned')
example
• [___,ps,fc,tc] = spectrogram(___)
example
• [___] = spectrogram(___,freqrange)
• [___] = spectrogram(___,spectrumtype)
• [___] = spectrogram(___,'MinThreshold',thresh)
example
• spectrogram(___)
example
• spectrogram(___,freqloc)
example
## Description
example
s = spectrogram(x) returns the short-time Fourier transform of the input signal, x. Each column of s contains an estimate of the short-term, time-localized frequency content of x.
s = spectrogram(x,window) uses window to divide the signal into sections and perform windowing.
example
s = spectrogram(x,window,noverlap) uses noverlap samples of overlap between adjoining sections.
example
s = spectrogram(x,window,noverlap,nfft) uses nfft sampling points to calculate the discrete Fourier transform.
[s,w,t] = spectrogram(___) returns a vector of normalized frequencies, w, and a vector of time instants, t, at which the spectrogram is computed. This syntax can include any combination of input arguments from previous syntaxes.
example
[s,f,t] = spectrogram(___,fs) returns a vector of cyclical frequencies, f, expressed in terms of the sample rate, fs.
[s,w,t] = spectrogram(x,window,noverlap,w) returns the spectrogram at the normalized frequencies specified in w.
[s,f,t] = spectrogram(x,window,noverlap,f,fs) returns the spectrogram at the cyclical frequencies specified in f.
[___,ps] = spectrogram(___) also returns a matrix, ps, containing an estimate of the power spectral density (PSD) or the power spectrum of each section.
example
[___] = spectrogram(___,'reassigned') reassigns each PSD or power spectrum estimate to the location of its center of energy. If your signal contains well-localized temporal or spectral components, then this option generates a sharper spectrogram.
example
[___,ps,fc,tc] = spectrogram(___) also returns two matrices, fc and tc, containing the frequency and time of the center of energy of each PSD or power spectrum estimate.
[___] = spectrogram(___,freqrange) returns the PSD or power spectrum estimate over the frequency range specified by freqrange. Valid options for freqrange are 'onesided', 'twosided', and 'centered'.
[___] = spectrogram(___,spectrumtype) returns PSD estimates if spectrumtype is specified as 'psd' and returns power spectrum estimates if spectrumtype is specified as 'power'.
example
[___] = spectrogram(___,'MinThreshold',thresh) sets to zero those elements of ps such that 10 log10(ps) ≤ thresh. Specify thresh in decibels.
example
spectrogram(___) with no output arguments plots the spectrogram in the current figure window.
example
spectrogram(___,freqloc) specifies the axis on which to plot the frequency.
## Examples
collapse all
Generate samples of a signal that consists of a sum of sinusoids. The normalized frequencies of the sinusoids are rad/sample and rad/sample. The higher frequency sinusoid has 10 times the amplitude of the other sinusoid.
N = 1024; n = 0:N-1; w0 = 2*pi/5; x = sin(w0*n)+10*sin(2*w0*n);
Compute the short-time Fourier transform using the function defaults. Plot the spectrogram.
s = spectrogram(x); spectrogram(x,'yaxis')
Repeat the computation.
• Divide the signal into sections of length .
• Window the sections using a Hamming window.
• Specify 50% overlap between contiguous sections.
• To compute the FFT, use points, where .
Verify that the two approaches give identical results.
Nx = length(x); nsc = floor(Nx/4.5); nov = floor(nsc/2); nff = max(256,2^nextpow2(nsc)); t = spectrogram(x,hamming(nsc),nov,nff); maxerr = max(abs(abs(t(:))-abs(s(:))))
maxerr = 0
Divide the signal into 8 sections of equal length, with 50% overlap between sections. Specify the same FFT length as in the preceding step. Compute the short-time Fourier transform and verify that it gives the same result as the previous two procedures.
ns = 8; ov = 0.5; lsc = floor(Nx/(ns-(ns-1)*ov)); t = spectrogram(x,lsc,floor(ov*lsc),nff); maxerr = max(abs(abs(t(:))-abs(s(:))))
maxerr = 0
Generate a quadratic chirp, x, sampled at 1 kHz for 2 seconds. The frequency of the chirp is 100 Hz initially and crosses 200 Hz at t = 1 s.
t = 0:0.001:2; x = chirp(t,100,1,200,'quadratic');
Compute and display the spectrogram of x.
• Divide the signal into sections of length 128, windowed with a Hamming window.
• Specify 120 samples of overlap between adjoining sections.
• Evaluate the spectrum at frequencies and time bins.
spectrogram(x,128,120,128,1e3)
Replace the Hamming window with a Blackman window. Decrease the overlap to 60 samples. Plot the time axis so that its values increase from top to bottom.
spectrogram(x,blackman(128),60,128,1e3) ax = gca; ax.YDir = 'reverse';
Compute and display the PSD of each segment of a quadratic chirp that starts at 100 Hz and crosses 200 Hz at t = 1 s. Specify a sample rate of 1 kHz, a segment length of 128 samples, and an overlap of 120 samples. Use 128 DFT points and the default Hamming window.
t = 0:0.001:2; x = chirp(t,100,1,200,'quadratic'); spectrogram(x,128,120,128,1e3,'yaxis') title('Quadratic Chirp')
Compute and display the PSD of each segment of a linear chirp that starts at DC and crosses 150 Hz at t = 1 s. Specify a sample rate of 1 kHz, a segment length of 256 samples, and an overlap of 250 samples. Use the default Hamming window and 256 DFT points.
t = 0:0.001:2; x = chirp(t,0,1,150); spectrogram(x,256,250,256,1e3,'yaxis') title('Linear Chirp')
Compute and display the PSD of each segment of a logarithmic chirp sampled at 1 kHz that starts at 20 Hz and crosses 60 Hz at t = 1 s. Specify a segment length of 256 samples and an overlap of 250 samples. Use the default Hamming window and 256 DFT points.
t = 0:0.001:2; x = chirp(t,20,1,60,'logarithmic'); spectrogram(x,256,250,[],1e3,'yaxis') title('Logarithmic Chirp')
Use a logarithmic scale for the frequency axis. The spectrogram becomes a line.
ax = gca; ax.YScale = 'log';
Use the spectrogram function to measure and track the instantaneous frequency of a signal.
Generate a quadratic chirp sampled at 1 kHz for two seconds. Specify the chirp so that its frequency is initially 100 Hz and increases to 200 Hz after one second.
Fs = 1000; t = 0:1/Fs:2-1/Fs; y = chirp(t,100,1,200,'quadratic');
Estimate the spectrum of the chirp using the short-time Fourier transform implemented in the spectrogram function. Divide the signal into sections of length 100, windowed with a Hamming window. Specify 80 samples of overlap between adjoining sections and evaluate the spectrum at frequencies. Suppress the default color bar.
spectrogram(y,100,80,100,Fs,'yaxis') view(-77,72) shading interp colorbar off
Track the chirp frequency by finding the maximum of the power spectral density at each of the time points. View the spectrogram as a two-dimensional graphic. Restore the color bar.
[s,f,t,p] = spectrogram(y,100,80,100,Fs); [q,nd] = max(10*log10(p)); hold on plot3(t,f(nd),q,'r','linewidth',4) hold off colorbar view(2)
Generate a chirp signal sampled for 2 seconds at 1 kHz. Specify the chirp so that its frequency is initially 100 Hz and increases to 200 Hz after 1 second.
Fs = 1000; t = 0:1/Fs:2; y = chirp(t,100,1,200,'quadratic');
Estimate the reassigned spectrogram of the signal.
• Divide the signal into sections of length 128, windowed with a Kaiser window with shape parameter .
• Specify 120 samples of overlap between adjoining sections.
• Evaluate the spectrum at frequencies and time bins.
spectrogram(y,kaiser(128,18),120,128,Fs,'reassigned','yaxis')
Generate a chirp signal sampled for 2 seconds at 1 kHz. Specify the chirp so that its frequency is initially 100 Hz and increases to 200 Hz after 1 second.
Fs = 1000; t = 0:1/Fs:2; y = chirp(t,100,1,200,'quadratic');
Estimate the time-dependent power spectral density (PSD) of the signal.
• Divide the signal into sections of length 128, windowed with a Kaiser window with shape parameter .
• Specify 120 samples of overlap between adjoining sections.
• Evaluate the spectrum at frequencies and time bins.
Output the frequency and time of the center of gravity of each PSD estimate. Set to zero those elements of the PSD smaller than dB.
[~,~,~,pxx,fc,tc] = spectrogram(y,kaiser(128,18),120,128,Fs, ... 'MinThreshold',-30);
Plot the nonzero elements as functions of the center-of-gravity frequencies and times.
plot(tc(pxx>0),fc(pxx>0),'.')
Generate a signal sampled at 1024 Hz for 2 seconds.
nSamp = 2048; Fs = 1024; t = (0:nSamp-1)'/Fs;
During the first second, the signal consists of a 400 Hz sinusoid and a concave quadratic chirp. Specify the chirp so that it is symmetric about the interval midpoint, starting and ending at a frequency of 250 Hz and attaining a minimum of 150 Hz.
t1 = t(1:nSamp/2); x11 = sin(2*pi*400*t1); x12 = chirp(t1-t1(nSamp/4),150,nSamp/Fs,1750,'quadratic'); x1 = x11+x12;
The rest of the signal consists of two linear chirps of decreasing frequency. One chirp has an initial frequency of 250 Hz that decreases to 100 Hz. The other chirp has an initial frequency of 400 Hz that decreases to 250 Hz.
t2 = t(nSamp/2+1:nSamp); x21 = chirp(t2,400,nSamp/Fs,100); x22 = chirp(t2,550,nSamp/Fs,250); x2 = x21+x22;
Add white Gaussian noise to the signal. Specify a signal-to-noise ratio of 20 dB. Reset the random number generator for reproducible results.
SNR = 20; rng('default') sig = [x1;x2]; sig = sig + randn(size(sig))*std(sig)/db2mag(SNR);
Compute and plot the spectrogram of the signal. Specify a Kaiser window of length 63 with a shape parameter , 10 fewer samples of overlap between adjoining sections, and an FFT length of 256.
nwin = 63; wind = kaiser(nwin,17); nlap = nwin-10; nfft = 256; spectrogram(sig,wind,nlap,nfft,Fs,'yaxis')
Threshold the spectrogram so that any elements with values smaller than the SNR are set to zero.
spectrogram(sig,wind,nlap,nfft,Fs,'MinThreshold',-SNR,'yaxis')
Reassign each PSD estimate to the location of its center of energy.
spectrogram(sig,wind,nlap,nfft,Fs,'reassign','yaxis')
Threshold the reassigned spectrogram so that any elements with values smaller than the SNR are set to zero.
spectrogram(sig,wind,nlap,nfft,Fs,'reassign','MinThreshold',-SNR,'yaxis')
Load an audio signal that contains two decreasing chirps and a wideband splatter sound. Compute the short-time Fourier transform. Divide the waveform into 400-sample segments with 300-sample overlap. Plot the spectrogram.
load splat % To hear, type soundsc(y,Fs) sg = 400; ov = 300; spectrogram(y,sg,ov,[],Fs,'yaxis') colormap bone
Use the spectrogram function to output the power spectral density (PSD) of the signal.
[s,f,t,p] = spectrogram(y,sg,ov,[],Fs);
Track the two chirps using the medfreq function. To find the stronger, low-frequency chirp, restrict the search to frequencies above 100 Hz and to times before the start of the wideband sound.
f1 = f > 100; t1 = t < 0.75; m1 = medfreq(p(f1,t1),f(f1));
To find the faint high-frequency chirp, restrict the search to frequencies above 2500 Hz and to times between 0.3 seconds and 0.65 seconds.
f2 = f > 2500; t2 = t > 0.3 & t < 0.65; m2 = medfreq(p(f2,t2),f(f2));
Overlay the result on the spectrogram. Divide the frequency values by 1000 to express them in kHz.
hold on plot(t(t1),m1/1000,'linewidth',4) plot(t(t2),m2/1000,'linewidth',4) hold off
Generate two seconds of a signal sampled at 10 kHz. Specify the instantaneous frequency of the signal as a triangular function of time.
fs = 10e3; t = 0:1/fs:2; x1 = vco(sawtooth(2*pi*t,0.5),[0.1 0.4]*fs,fs);
Compute and plot the spectrogram of the signal. Use a Kaiser window of length 256 and shape parameter . Specify 220 samples of section-to-section overlap and 512 DFT points. Plot the frequency on the y-axis. Use the default colormap and view.
spectrogram(x1,kaiser(256,5),220,512,fs,'yaxis')
Change the view to display the spectrogram as a waterfall plot. Set the colormap to bone.
colormap bone view(-45,65)
## Input Arguments
collapse all
Input signal, specified as a row or column vector.
Example: cos(pi/4*(0:159))+randn(1,160) specifies a sinusoid embedded in white Gaussian noise.
Data Types: single | double
Complex Number Support: Yes
Window, specified as an integer or as a row or column vector. Use window to divide the signal into sections:
• If window is an integer, then spectrogram divides x into sections of length window and windows each section with a Hamming window of that length.
• If window is a vector, then spectrogram divides x into sections of the same length as the vector and windows each section using window.
If the length of x cannot be divided exactly into an integer number of sections with noverlap overlapping samples, then x is truncated accordingly.
If you specify window as empty, then spectrogram uses a Hamming window such that x is divided into eight sections with noverlap overlapping samples.
For a list of available windows, see Windows.
Example: hann(N+1) and (1-cos(2*pi*(0:N)'/N))/2 both specify a Hann window of length N + 1.
Data Types: single | double
Number of overlapped samples, specified as a positive integer.
• If window is scalar, then noverlap must be smaller than window.
• If window is a vector, then noverlap must be smaller than the length of window.
If you specify noverlap as empty, then spectrogram uses a number that produces 50% overlap between sections. If the section length is unspecified, the function sets noverlap to ⌊Nx/4.5⌋, where Nx is the length of the input signal.
Data Types: double | single
Number of DFT points, specified as a positive integer scalar. If you specify nfft as empty, then spectrogram sets the parameter to max(256,2p), where p = ⌈log2 Nx⌉ for an input signal of length Nx.
Data Types: single | double
Normalized frequencies, specified as a vector. w must have at least two elements. Normalized frequencies are in rad/sample.
Example: pi./[2 4]
Data Types: double | single
Cyclical frequencies, specified as a vector. f must have at least two elements. The units of f are specified by the sample rate, fs.
Data Types: double | single
Sample rate, specified as a positive scalar. The sample rate is the number of samples per unit time. If the unit of time is seconds, then the sample rate is in Hz.
Data Types: double | single
Frequency range for the PSD estimate, specified as 'onesided', 'twosided', or 'centered'. For real-valued signals, the default is 'onesided'. For complex-valued signals, the default is 'twosided', and specifying 'onesided' results in an error.
• 'onesided' — returns the one-sided spectrogram of a real input signal. If nfft is even, then ps has length nfft/2 + 1 and is computed over the interval [0, π] rad/sample. If nfft is odd, then ps has length (nfft + 1)/2 and the interval is [0, π) rad/sample. If you specify fs, then the intervals are respectively [0, fs/2] cycles/unit time and [0, fs/2) cycles/unit time.
• 'twosided' — returns the two-sided spectrogram of a real or complex signal. ps has length nfft and is computed over the interval [0, 2π) rad/sample. If you specify fs, then the interval is [0, fs) cycles/unit time.
• 'centered' — returns the centered two-sided spectrogram for a real or complex signal. ps has length nfft. If nfft is even, then ps is computed over the interval (–ππ] rad/sample. If nfft is odd, then ps is computed over (–ππ) rad/sample. If you specify fs, then the intervals are respectively (–fs/2, fs/2] cycles/unit time and (–fs/2, fs/2) cycles/unit time.
Data Types: char
Power spectrum scaling, specified as 'psd' or 'power'.
• Omitting spectrumtype, or specifying 'psd', returns the power spectral density.
• Specifying 'power' scales each estimate of the PSD by the equivalent noise bandwidth of the window. The result is an estimate of the power at each frequency. If the 'reassigned' option is on, the function integrates the PSD over the width of each frequency bin before reassigning.
Data Types: char
Threshold, specified as a real scalar expressed in decibels. spectrogram sets to zero those elements of ps such that 10 log10(ps) ≤ thresh.
Frequency display axis, specified as 'xaxis' or 'yaxis'.
• 'xaxis' — displays frequency on the x-axis and time on the y-axis.
• 'yaxis' — displays frequency on the y-axis and time on the x-axis.
This argument is ignored if you call spectrogram with output arguments.
Data Types: char
## Output Arguments
collapse all
Short-time Fourier transform, returned as a matrix. Time increases across the columns of s and frequency increases down the rows, starting from zero.
• If x is a signal of length Nx, then s has k columns, where
• k = ⌊(Nx – noverlap)/(window – noverlap)⌋ if window is a scalar
• k = ⌊(Nx – noverlap)/(length(window) – noverlap)⌋ if window is a vector.
• If x is real and nfft is even, then s has (nfft/2 + 1) rows.
• If x is real and nfft is odd, then s has (nfft + 1)/2 rows.
• If x is complex, then s has nfft rows.
Data Types: double | single
Normalized frequencies, returned as a vector. w has a length equal to the number of rows of s.
Data Types: double | single
Time instants, returned as a vector. The time values in t correspond to the midpoint of each section.
Data Types: double | single
Cyclical frequencies, returned as a vector. f has a length equal to the number of rows of s.
Data Types: double | single
Power spectral density (PSD) or power spectrum, returned as a matrix.
• If x is real, then ps contains the one-sided modified periodogram estimate of the PSD or power spectrum of each section.
• If x is complex, or if you specify a vector of frequencies, then ps contains the two-sided modified periodogram estimate of the PSD or power spectrum of each section.
Data Types: double | single
Center-of-energy frequencies and times, returned as matrices of the same size as the short-time Fourier transform. If you do not specify a sample rate, then the elements of fc are returned as normalized frequencies.
collapse all
### Tips
If a short-time Fourier transform has zeros, its conversion to decibels results in negative infinities that cannot be plotted. To avoid this potential difficulty, spectrogram adds eps to the short-time Fourier transform when you call it with no output arguments.
## References
[1] Oppenheim, Alan V., Ronald W. Schafer, and John R. Buck. Discrete-Time Signal Processing. 2nd Ed. Upper Saddle River, NJ: Prentice Hall, 1999.
[2] Rabiner, Lawrence R., and Ronald W. Schafer. Digital Processing of Speech Signals. Englewood Cliffs, NJ: Prentice-Hall, 1978.
[3] Chassande-Motin, Éric, François Auger, and Patrick Flandrin. "Reassignment." In Time-Frequency Analysis: Concepts and Methods. Edited by Franz Hlawatsch and François Auger. London: ISTE/John Wiley and Sons, 2008.
[4] Fulop, Sean A., and Kelly Fitz. "Algorithms for computing the time-corrected instantaneous frequency (reassigned) spectrogram, with applications." Journal of the Acoustical Society of America. Vol. 119, January 2006, pp. 360–371. | 5,237 | 19,593 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2016-40 | latest | en | 0.732087 |
https://math.stackexchange.com/questions/1332690/monoid-as-a-single-object-category | 1,653,374,671,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662564830.55/warc/CC-MAIN-20220524045003-20220524075003-00235.warc.gz | 339,422,859 | 70,026 | # Monoid as a single object category
I'm struggling with comprehending what monoids are in terms of category theory.
In examples they view integer numbers as a monoid. I think I get the set theoretic definition. We have a set and a associative binary operator (addition) and the neutral element (zero).
Then they are saying something like that - view the whole set as a single object and binary operator as bunch of morphisms for every element of the set.
Like add0 is an identity morphism. Which would really give us the same object i.e. the same set of all integer numbers. I think I understand this.
But let's view the morphism add1. After applying it to the our single object (the set of all integers) we would have a set {1,2,3…} not the {0,1,2,3…}. Aren't domain and codomain different in that case?
That's what's bothering me. Can someone clarify that to me?
Here is the text that gives me problems.
The text
No, that's not the way to view it. In order to view a monoid as a category, you have a single object $\mathsf{Andreas}$, and each element of the monoid is one morphism $\mathsf{Andreas}\to\mathsf{Andreas}$ in the category. The monoid operation is the composition in the category.
So for the integers, you don't have a morphism "add 1", but a morphism that is simply called $1$. And composition in the category works such that $1$ composed with $1$ is the morphism called $2$.
This is an example of a category where the morphisms are not functions.
• @user1685095: The domain of each morphism is $\bullet$ (that is, the object of the category), and so is the codomain. The very point here is that a morphism is not necessarily a function. Jun 20, 2015 at 15:13
• @user1685095: $\bullet$ is the name of the object in the category. Its precise identity is not important. Jun 20, 2015 at 15:26
• @user1685095 add$1$ can be looked at as a function having $\mathbb N$ as domain and as codomain. It is prescribed by $n\mapsto n+1$. There is a category with objectset $\{\mathbb N\}$ and morphismset $\{\text{add}n\mid n\in\mathbb N\}$. So there is only one object and it can equally well be denoted as $\bullet$. Monoids can be identified as categories that have exactly one object. Composition of morphisms corresponds with addition: $\text{add}k\circ\text{add}m=\text{add}(k+m)$. Jun 20, 2015 at 15:47
• Since the unique but unspecified object seems to cause difficulties for the OP, I hereby volunteer to make things specific by being that object. That is, the category has one object, namely me, and its morphisms are the integers. Composition of morphisms is addition of integers. Jun 20, 2015 at 16:00
• @Phạm Văn Thông, unfortunately not. First, as hmakholm left over Monica said, the morphisms 0, 1, 2, 3 etc. are (or correspond to) elements of the monoid. Composition is the monoid operation. So if the monoid operation is +, 1 composed with 2 is 3, showing that neither are identity morphisms. Second & perhaps more deeply, assume morphisms i and j are both identities. Then what is j composed with i, j ∘ i? Focusing on j as identity, it has to be i. But focusing on i as identity, it has to be j. So j=i, and we have only one identity morphism. So only one of 0, 1, 2, 3 ... could be one anyway. Jun 13, 2021 at 17:44
You already know that a monoid $M$ is a set with a unit $e$ and a binary operation. More precisely, if $a,b,c\in M$ then $$a\circ b\in M$$ $$(a\circ b)\circ c=a\circ (b\circ c)$$ $$e\circ a=a\circ e=a$$
Now, take any category $C$ with one object, $c$. Since $C$ is a category, we need to say what the arrows are. That is, what the morphisms $c\rightarrow c$ are. There must be a unit $1_{c}:c\rightarrow c$, the arrows must be composeable and the composition must be associative. More precisely, if $f,g$ and $h$ are arrows, then $$f\circ g\in Morph(C)$$ $$(f\circ g)\circ h= f\circ (g\circ h)$$ $$1_{c}\circ f=f\circ 1_{c}=f$$Notice we are $\textit not$ talking about sets here. Just objects and arrows, in the abstract.
But now if we just observe that the operations on $M$ are $\textit exactly$ the same as the operations on $Morph(C)$, we may regard the category $C$ as the monoid $M$. This correspondence is reversible: given category $C=\left \{ c \right \}$ we obtain a monoid $M$ whose elements are the arrows of $C$.
Thus the two descriptions are equivalent.
All this works because the binary operations are the same for both structures.
• What's special about the case when $Ob(C)=\{c\}$? Why can't we interpret a monoid as a category with more than one object? The same axioms will hold for $Morph(C)$. Jan 27, 2019 at 16:15
• You want to identify Morph$C$ with $M$? But in an arbitrary category, not all morphisms are composable, right? If you have $f,g:A\to B$ then $f\circ g$ doesn't even exist. What's special about the above construction is that there is an $\textit {exact}$ i.e. bijective correspondence between the operations. Jan 27, 2019 at 16:37
• Then we can consider a category with more than one object and the set of all morphisms from an object $A$ to $A$. This set with the operation of composition should be a monoid. Can we then say that conversely, a monoid can be interpreted as the set of arrows from a fixed object to itself in an arbitrary category? This is also a bijective correspondence between the arrows from that object to itself and the monoid operation. Jan 27, 2019 at 17:00
• "...a monoid can be interpreted as the set of arrows from a fixed object to itself in an arbitrary category"...yes, and this is the essence of my answer. If you have a category $C$ with small hom-sets, then for each $c\in C$, then each Hom$[c,c]$ will give a monoid, distinct if the cardinalities are different. Jan 27, 2019 at 17:06
• Then I guess the only reason to mention categories with one object is to be able to make statements like "a monoid is a category with one object" as opposed to "a monoid is a set of morphisms from a fixed object to itself in a category". So a monoid and a category with one object are equivalent concepts (like groups and $\mathbb Z$-modules). Given a category with one object, one obtains a monoid as in your answer. Conversely, given a monoid one constructs the corresponding category by taking the only object to be $\{\ast\}$ and the only morphism $\ast\mapsto \ast$. Is that correct? Jan 27, 2019 at 17:40
In addition to Henning Makholm's crisp and clear answer, you might find the opening six pages of my Notes on Category Theory helpful. They too give the example of a monoid as a category, but also give some other examples of categories where the arrows are not functions in any ordinary sense. Another important illustration is the case of a posets treated as a category.
In fact these examples suggest why we might well prefer to talk of 'arrows' rather than 'morphisms' (because the very term 'morphism' comes with baggage, and almost inevitably makes us think of a function -- but to repeat, arrows need not be functions).
• A footnote to another solution should, well, be a footnote on that solution (or rather a comment or two.) Jun 20, 2015 at 16:03
• Nice notes Peter. And nice blog too. I will contact you there. Jun 24, 2015 at 2:03 | 1,915 | 7,168 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2022-21 | longest | en | 0.948127 |
http://sagemath.org/doc/reference/databases/sage/databases/stein_watkins.html | 1,429,390,124,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246636213.71/warc/CC-MAIN-20150417045716-00085-ip-10-235-10-82.ec2.internal.warc.gz | 231,934,149 | 5,794 | # The Stein-Watkins table of elliptic curves¶
Sage gives access to the Stein-Watkins table of elliptic curves, via an optional package that you must install. This is a huge database of elliptic curves. You can install the database (a 2.6GB package) with the command
sage -i database_stein_watkins
You can also automatically download a small version, which takes much less time, using the command
sage -i database_stein_watkins_mini
This database covers a wide range of conductors, but unlike the Cremona database, this database need not list all curves of a given conductor. It lists the curves whose coefficients are not “too large” (see [SteinWatkins]).
• The command SteinWatkinsAllData(n) returns an iterator over the curves in the $$n$$-th Stein-Watkins table, which contains elliptic curves of conductor between $$n10^5$$ and $$(n+1)10^5$$. Here $$n$$ can be between 0 and 999, inclusive.
• The command SteinWatkinsPrimeData(n) returns an iterator over the curves in the $$n^{th}$$ Stein-Watkins prime table, which contains prime conductor elliptic curves of conductor between $$n10^6$$ and $$(n+1)10^6$$. Here $$n$$ varies between 0 and 99, inclusive.
EXAMPLES: We obtain the first table of elliptic curves.
sage: d = SteinWatkinsAllData(0)
sage: d
Stein-Watkins Database a.0 Iterator
We type next(d) to get each isogeny class of curves from d:
sage: C = next(d) # optional - database_stein_watkins
sage: C # optional - database_stein_watkins
Stein-Watkins isogeny class of conductor 11
sage: next(d) # optional - database_stein_watkins
Stein-Watkins isogeny class of conductor 14
sage: next(d) # optional - database_stein_watkins
Stein-Watkins isogeny class of conductor 15
An isogeny class has a number of attributes that give data about the isogeny class, such as the rank, equations of curves, conductor, leading coefficient of $$L$$-function, etc.
sage: C.data # optional - database_stein_watkins
['11', '[11]', '0', '0.253842', '25', '+*1']
sage: C.curves # optional - database_stein_watkins
[[[0, -1, 1, 0, 0], '(1)', '1', '5'],
[[0, -1, 1, -10, -20], '(5)', '1', '5'],
[[0, -1, 1, -7820, -263580], '(1)', '1', '1']]
sage: C.conductor # optional - database_stein_watkins
11
sage: C.leading_coefficient # optional - database_stein_watkins
'0.253842'
sage: C.modular_degree # optional - database_stein_watkins
'+*1'
sage: C.rank # optional - database_stein_watkins
0
sage: C.isogeny_number # optional - database_stein_watkins
'25'
If we were to continue typing next(d) we would iterate over all curves in the Stein-Watkins database up to conductor $$10^5$$. We could also type for C in d: ...
To access the data file starting at $$10^5$$ do the following:
sage: d = SteinWatkinsAllData(1)
sage: C = next(d) # optional - database_stein_watkins
sage: C # optional - database_stein_watkins
Stein-Watkins isogeny class of conductor 100002
sage: C.curves # optional - database_stein_watkins
[[[1, 1, 0, 112, 0], '(8,1,2,1)', 'X', '2'],
[[1, 1, 0, -448, -560], '[4,2,1,2]', 'X', '2']]
Next we access the prime-conductor data:
sage: d = SteinWatkinsPrimeData(0)
sage: C = next(d) # optional - database_stein_watkins
sage: C # optional - database_stein_watkins
Stein-Watkins isogeny class of conductor 11
Each call next(d) gives another elliptic curve of prime conductor:
sage: C = next(d) # optional - database_stein_watkins
sage: C # optional - database_stein_watkins
Stein-Watkins isogeny class of conductor 17
sage: C.curves # optional - database_stein_watkins
[[[1, -1, 1, -1, 0], '[1]', '1', '4'],
[[1, -1, 1, -6, -4], '[2]', '1', '2x'],
[[1, -1, 1, -1, -14], '(4)', '1', '4'],
[[1, -1, 1, -91, -310], '[1]', '1', '2']]
sage: C = next(d) # optional - database_stein_watkins
sage: C # optional - database_stein_watkins
Stein-Watkins isogeny class of conductor 19
REFERENCE:
[SteinWatkins] William Stein and Mark Watkins, A database of elliptic curves—first report. In Algorithmic number theory (ANTS V), Sydney, 2002, Lecture Notes in Computer Science 2369, Springer, 2002, p267–275. http://modular.math.washington.edu/papers/stein-watkins/
class sage.databases.stein_watkins.SteinWatkinsAllData(num)
Class for iterating through one of the Stein-Watkins database files for all conductors.
iter_levels()
Iterate through the curve classes, but grouped into lists by level.
EXAMPLE:
sage: d = SteinWatkinsAllData(1)
sage: E = d.iter_levels()
sage: next(E) # optional - database_stein_watkins
[Stein-Watkins isogeny class of conductor 100002]
sage: next(E) # optional - database_stein_watkins
[Stein-Watkins isogeny class of conductor 100005,
Stein-Watkins isogeny class of conductor 100005]
sage: next(E) # optional - database_stein_watkins
[Stein-Watkins isogeny class of conductor 100007]
next()
x.__init__(...) initializes x; see help(type(x)) for signature
class sage.databases.stein_watkins.SteinWatkinsIsogenyClass(conductor)
class sage.databases.stein_watkins.SteinWatkinsPrimeData(num)
sage.databases.stein_watkins.ecdb_num_curves(max_level=200000)
Return a list whose $$N$$-th entry, for 0 <= N <= max_level, is the number of elliptic curves of conductor $$N$$ in the database.
EXAMPLES:
sage: sage.databases.stein_watkins.ecdb_num_curves(100) # optional - database_stein_watkins
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 6, 8, 0, 4, 0, 3, 4, 6, 0, 0,
6, 0, 5, 4, 0, 0, 8, 0, 4, 4, 4, 3, 4, 4, 5, 4, 4, 0, 6, 1, 2, 8, 2, 0,
6, 4, 8, 2, 2, 1, 6, 4, 6, 7, 3, 0, 0, 1, 4, 6, 4, 2, 12, 1, 0, 2, 4, 0,
6, 2, 0, 12, 1, 6, 4, 1, 8, 0, 2, 1, 6, 2, 0, 0, 1, 3, 16, 4, 3, 0, 2,
0, 8, 0, 6, 11, 4]
#### Previous topic
Cremona’s tables of elliptic curves
#### Next topic
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A166934 A positive integer n is included if the longest palindromic substring (or at least one of which that is tied for longest) in the binary representation of n consists of different valued digits (0 and 1). 2
5, 9, 10, 11, 13, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 29, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 49, 50, 51, 52, 53, 54, 55, 57, 58, 59, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 97 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS All terms of this sequence are contained in sequence A101082. (47 is the first integer that is in A101082, but is not in this sequence.) LINKS EXAMPLE 29 in binary is 11101. There are two substrings in this that are tied for longest palindromic substring, 111 and 101. Since 101 contains both a 0 and 1's, then 29 is in this sequence. CROSSREFS Cf. A050430, A166935, A174196. Sequence in context: A101082 A277706 A300669 * A257739 A094695 A268412 Adjacent sequences: A166931 A166932 A166933 * A166935 A166936 A166937 KEYWORD base,nonn AUTHOR Leroy Quet, Oct 24 2009 EXTENSIONS Extended by Ray Chandler, Mar 11 2010 STATUS approved
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Last modified October 18 12:18 EDT 2019. Contains 328160 sequences. (Running on oeis4.) | 585 | 1,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.83605 |
https://www.jiskha.com/questions/715147/please-show-me-how-to-set-up-and-solve-you-have-6-cups-of-sugar-it-takes-1-cup-of-sugar | 1,586,497,574,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371886991.92/warc/CC-MAIN-20200410043735-20200410074235-00298.warc.gz | 962,871,085 | 5,176 | # math
Please show me how to set up and solve:
You have 6 cups of sugar. It takes 1 cup of sugar to make 24 cookies. The function c(s) = 24s represents the number of cookies, c, that can be made with s cups of sugar. What domain and range is reasonable for the function?
1. 👍 0
2. 👎 0
3. 👁 134
1. 👍 0
2. 👎 0
2. What domain and range is reasonable for this function?
1. 👍 0
2. 👎 0
posted by Jane
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Gopal borrows Rs. $\text{X}$ from Ankit at $8\%$ annual interest. He then adds Rs. $\text{Y}$ of his own money and lends Rs. $\text{X+Y}$ to Ishan at $10\%$ annual interest. At the end of the year, after returning Ankit's dues, the net interest ... the net interest retained by him would have increased by Rs. $150$. If all interests are compounded annually, then find the value of $\text{X+Y}$.
9
Let $f\left (x \right ) = \max\left \{5x, 52 – 2x^{2}\right \}$ , where $x$ is any positive real numbers. Then the minimum possible value of $f(x)$ is ________
10
On a long stretch of east-west road, $\text{A}$ and $\text{B}$ are two points such that $\text{B}$ is $350$ km west of $\text{A}$. One car starts from $\text{A}$ and another from $\text{B}$ at the same time. If they move towards each other, then they meet after $1$ hour. If they both move towards east, then they meet in $7$ hrs. The difference between their speeds, in km per hour, is _______
11
A water tank has inlets of two types $\text{A}$ and $\text{B}$. All inlets of type $\text{A}$ when open, bring in water at the same rate. All inlets of type $\text{B}$, when open, bring in water at the same rate. The empty tank is completely filled in $30$ ... many minutes will the empty tank get completely filled if $7$ inlets of type $\text{A}$ and $27$ inlets of type $\text{B}$ are open?
12
If $a$ and $b$ are integers such that $2x^{2}- ax + 2 > 0$ and $x^{2}-bx+8 \geq 0$ for all real numbers $x$, then the largest possible value of $2a-6b$ is _________
13
A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump $\text{A}$ or pump $\text{B}$ or both start working until the tank is full. On Monday, $\text{A}$ alone completed filling the tank at $8$ pm. On Tuesday, $\text{B}$ alone completed filling ... was the tank filled on Thursday if both pumps were used simultaneously all along? $4:36$ pm $4:12$ pm $4:24$ pm $4:48$ pm
14
A jar contains a mixture of $175$ ml water and $700$ ml alcohol. Gopal takes out $10\%$ of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now $35.2$ $30.3$ $20.5$ $25.4$
15
In a tournament, there are $43$ junior level and $51$ senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is $153$, while the number of boy versus boy matches in senior level is $276$. The number of matches a boy plays against a girl is _________
16
Ramesh and Ganesh can together complete a work in $16$ days. After seven days of working together, Ramesh got sick and his efficiency fell by $30\%$. As a result, they completed the work in $17$ days instead of $16$ days. If Ganesh had worked alone after Ramesh got sick, in how many days would he have completed the remaining work? $13.5$ $11$ $12$ $14.5$
17
If $p^{3}=q^{4}=r^{5}=s^{6}$, then the value of $\log_{s}\left ( pqr \right )$ is equal to $16/5$ $1$ $24/5$ $47/10$
18
Let $t_{1}, t_{2},\dots$ be a real numbers such that $t_{1}+t_{2}+\dots+t_{n}=2n^{2}+9n+13$, for every positive integers $n\geq2$.If $t_{k}=103$ , then $k$ equals
19
If $\text{N}$ and $x$ are positive integers such that $\text{N}^{\text{N}}=2^{160}$ and $\text{N}^{2} + 2^{\text{N}}$ is an integral multiple of $2^{x}$, then the largest possible $x$ is _______
20
If the sum of squares of two numbers is $97$, then which one of the following cannot be their product? $-32$ $48$ $64$ $16$
21
The arithmetic mean of $x,y$ and $z$ is $80$, and that of $x,y,z,u$ and $v$ is $75$, where $u=\left (x+y \right)/2$ and $v=\left (y+z \right)/2$. If $x\geq z$, then the minimum possible value of $x$ is ____________
22
Points $\text{A}$ and $\text{B}$ are $150$ km apart. Cars $1$ and $2$ travel from $\text{A}$ to $\text{B}$, but car $2$ starts from $\text{A}$ when car $1$ is already $20$ km away from $\text{A}$. Each car travels at a speed of $100$ kmph for the first $50$ ... , and at $25$ kmph for the last $50$ km. The distance, in km, between car $2$ and $\text{B}$ when car $1$ reaches $\text{B}$ is ________
23
Let $a_{1},a_{2},\dots , a_{52}$ be a positive integers such that $a_{1}<a_{2}<\dots < a_{52}$. Suppose, their arithmetic mean is one less than the arithmetic mean of $a_{2},a_{3},\dots , a_{52}$. If $a_{52}=100$ , then the largest possible value of $a_{1}$ is ________ $20$ $23$ $48$ $45$
24
A parallelogram $\text{ABCD}$ has area $48$ sqcm. If the length of $\text{CD}$ is $8$ cm and that of $\text{AD}$ is $s$ cm, then which one of the following is necessarily true? $s\geq6$ $s\neq6$ $s\leq6$ $5\leq s\leq7$
1 vote
25
If $\text{A}=\left \{6^{2n} - 35n - 1: n=1,2,3 \dots \right \}$ and $\text{B}= \left \{35\left (n - 1 \right ) : n=1,2,3\dots \right \}$ then which of the following is true? Neither every member of $\text{A}$ is in $\text{B}$ nor every member of $\text{B}$ ... $\text{B}$ is not in $\text{A}$ Every member of $\text{B}$ is in $\text{A}$ At least one member of $\text{A}$ is not in $\text{B}$
26
From a rectangle $\text{ABCD}$ of area $768$ sq cm, a semicircular part with diameter $\text{AB}$ and area $72\pi$ sq cm is removed. The perimeter of the leftover portion, in cm, is $80 + 16\pi$ $86+8\pi$ $82+24\pi$ $88+12\pi$
27
The scores of Amal and Bimal in an examination are in the ratio $11: 14$. After an appeal, their scores increase by the same amount and their new scores are in the ratio $47 : 56$. The ratio of Bimal's new score to that of his original score is $5:4$ $8:5$ $4:3$ $3:2$
28
The area of a rectangle and the square of its perimeter are in the ratio $1:25$. Then the lengths of the shorter and longer sides of the rectangle are in the ratio $1:4$ $2:9$ $1:3$ $3:8$
29
The strength of a salt solution is $p\%$ if $100$ ml of the solution contains $p$ grams of salt. If three salt solutions $\text{A, B, C}$ are mixed in the proportion $1:2:3$, then the resulting solution has strength $20\%$. If instead the proportion is $3:2:1$, then the resulting solution has ... ratio $2:7$. The ratio of the strength of $\text{D}$ to that of $\text{A}$ is $2:5$ $1:3$ $1:4$ $3:10$
30
For two sets $\text{A}$ and $\text{B}$, let $\text{A} \triangle \text{B}$ denote the set of elements which belong to $\text{A}$ or $\text{B}$ but not both. If $\text{P} = \{1,2,3,4\}, \text{Q} = \{2,3,5,6\}, \text{R} = \{1,3,7,8,9\}, \text{S} = \{2,4,9,10\},$ then the number of elements in $(\text{P} \triangle \text{Q}) \triangle (\text{R}\triangle \text{S})$ is $9$ $7$ $6$ $8$
31
The smallest integer $n$ for which $4^{n}>17^{19}$ holds, is closest to $33$ $37$ $39$ $35$
32
Points $\text{A, P, Q}$ and $\text{B}$ lie on the same line such that $\text{P, Q}$ and $\text{B}$ are, respectively, $100$ km, $200$ km and $300$ km away from $\text{A}$. Cars $1$ and $2$ leave $\text{A}$ at the same time and move towards $\text{B}$. Simultaneously, ... . If each car is moving in uniform speed then the ratio of the speed of car $2$ to that of car $1$ is $1:2$ $2:9$ $1:4$ $2:7$
1 vote
33
There are two drums, each containing a mixture of paints $\text{A}$ and $\text{B}$. In drum $1, \text{A}$ and $\text{B}$ are in the ratio $18: 7$. The mixtures from drums $1$ and $2$ are mixed in the ratio $3: 4$ and in this final mixture, $\text{A}$ and $\text{B}$ are in the ratio $13 :7$. In drum $2$, then $\text{A}$ and $\text{B}$ were in the ratio $229:141$ $220:149$ $239: 161$ $251: 163$
34
A triangle $\text{ABC}$ has area $32$ sq units and its side $\text{BC}$, of length $8$ units, lies on the line $x =4$. Then the shortest possible distance between $\text{A}$ and the point $(0,0)$ is $4$ units $8$ units $4\sqrt2$ units $2\sqrt2$ units
35
$\frac{1}{\log_{2}100} – \frac{1}{\log_{4}100} + \frac{1}{\log_{5}100} – \frac{1}{\log_{10}100} + \frac{1}{\log_{20}100} – \frac{1}{\log_{25}100} + \frac{1}{\log_{50}100}=?$ $1/2$ $0$ $10$ $-4$
36
An agency entrusted to accredit colleges looks at four parameters: faculty quality $\text{(F)}$, reputation $\text{(R)},$ placement quality $\text{(P)}$, and infrastructure $\text{(I)}.$ The four parameters are used to arrive at an overall score, which the agency uses to give an ... than $\text{A}$-one. How many colleges have overall scores between $31$ and $40$, both exclusive? $1$ $3$ $0$ $2$
37
An agency entrusted to accredit colleges looks at four parameters: faculty quality $\text{(F)}$, reputation $\text{(R)},$ placement quality $\text{(P)}$, and infrastructure $\text{(I)}.$ The four parameters are used to arrive at an overall score, which the ... better than Cosmopolitan; and Education Aid is better than $\text{A}$-one. How many colleges receive the accreditation of $\text{AAA}?$
An agency entrusted to accredit colleges looks at four parameters: faculty quality $\text{(F)}$, reputation $\text{(R)},$ placement quality $\text{(P)}$, and infrastructure $\text{(I)}.$ The four parameters are used to arrive at an overall score, which the ... is better than Cosmopolitan; and Education Aid is better than $\text{A}$-one. What is the highest overall score among the eight colleges?
Each visitor to an amusement park needs to buy a ticket. Tickets can be Platinum, Gold, or Economy. Visitors are classified as Old, Middle-aged, or Young. The following facts are known about visitors and ticket sales on a particular day: $140$ tickets were sold. The ... buying Platinum tickets, then which among the following could be the total number of Platinum tickets sold? $34$ $38$ $32$ $36$
According to a coding scheme the sentence Peacock is designated as the national bird of India is coded as $5688999\;35\;1135556678\;56\;458\;13666689\;1334\;79\;13366$ This coding scheme has the following rules: The scheme is case-insensitive (does not distinguish between upper case and lower case letters). Each letter ... digit? $\text{S, U, V}$ $\text{I, B, M}$ $\text{X, Y, Z}$ $\text{S, E, Z}$ | 3,726 | 11,644 | {"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} | 3.296875 | 3 | CC-MAIN-2022-21 | latest | en | 0.884149 |
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• page 35 of text
• Discussion of what these rating values represent is found on page 33 (the Chapter Problem for Chapter 2).
• Final result of a frequency table on the Qwerty data given in previous slide. Next few slides will develop definitions and steps for developing the frequency table.
• The concept is usually difficult for some students. Emphasize that boundaries are often used in graphical representations of data. See Section 2-3.
• Students will question where the -0.5 and the 14.5 came from. After establishing the boundaries between the existing classes, explain that one should find the boundary below the first class and above the last class, again referencing the use of boundaries in the development of some histograms (Section 2-3).
• Being able to identify the midpoints of each class will be important when determining the mean and standard deviation of a frequency table.
• Class widths ideally should be the same between each class. However, open-ended first or last classes (e.g., 65 years or older) are sometimes necessary to keep from have a large number of classes with a frequency of 0.
• page 38 of text Emphasis on the relative frequency table will assist in development the concept of probability distributions - the use of percentages with classes will relate to the use of probabilities with random variables.
• Some students will need a detailed explanation of how the classes “Less than 3”, “Less than 6”, etc. are identified and the frequency for each of the classes are determined.
• A comparison of all three types of frequency tables developed from the same QWERTY data.
• page 45 of text Data that has at least two digits better exemplifies the value of a stem-leaf plot. Cover up the actual data and ask students to read the data from the stem-leaf plot. Data should be put in increasing order. A stem-leaf plot builds ‘side-ways’ histogram. The most common mistake students make when developing stem-leaf plots is leaving out a ‘stem’ if there is no data in that class. The other mistake is, if there is no data in a class, putting a ‘0’ next to the stem - which, of course, would indicate a data value with a right digit of ‘0’. Students confuse the use of ‘0’ indicating a frequency in a frequency table with a ‘0’ in a stem-leaf plot indicating an actual data value. (See example on page 47 middle of page.)
• A plot of paired (x,y) data with the horizontal x-axis and the vertical y-axis. Chapter 9 will discuss scatter plots again with the topic of correlation. Point out the relationship that exists between the nicotine and tar - as the nicotine value increases, so does the value of tar. page 47 of text
• ### Displaying data
1. 1. A) Tabulation : Frequency distribution Table : - Quantitative - Qualitative B) Drawing: (Graphs / Charts/ Diagrams) Quantitative Data : i) Histogram ii) Frequency Polygon iii) Frequency Curve iv) Line chart /graph v) Cumulative Frequency Diagram / Ogive vi) Scatter or Dot diagram vii) Stem & Leaf plot Qualitative Data : i) Bar diagram (Simple / Multiple / Proportional) ii) Pie or Sector chart iii) Pictogram
2. 2. <ul><li>General principles in designing table : </li></ul><ul><li>The tables should be numbered e.g., Table-1, Table-2 etc. </li></ul><ul><li>There should be a brief and self-explanatory title, mentioning time, place & persons. </li></ul><ul><li>The headings of columns and rows should be clear and concise </li></ul><ul><li>The data must be presented according to size or importance; chronologically, alphabetically or geographically </li></ul><ul><li>Data must be presented meaningfully </li></ul><ul><li>No table should be too large </li></ul><ul><li>Foot notes may be given, if necessary </li></ul><ul><li>Total number of observations (n) i.e the denominator should be written </li></ul><ul><li>The information obtained in the table should be summarized beneath the table </li></ul>
3. 3. TABLE-1 Population by sex in Kolkata urban area in 2001 Source: Health on the March 2004-05, Govt. of West Bengal Characteristics Population (in million) % Male Female 7.07 6.14 53.52 46.48 Total 13.21 100.00
4. 4. Frequency distribution table for qualitative data Characteristics Population (in million) % Male 7.07 53.52 Female 6.14 46.48 Total 13.21 100.00
5. 5. Frequency distribution table for quantitative data Pulse rate/minute No of medical students Percentage 51-60 2 1.33 61-70 22 14.67 71-80 56 37.33 81-90 55 36.67 91-100 14 9.33 101-110 1 0.67 Total 150 100.00
6. 6. <ul><li>Frequency Table </li></ul><ul><li>lists classes (or categories) of values, along with frequencies (or counts) of the number of values that fall into each class </li></ul>2-2 Summarizing Data With Frequency Tables
7. 7. Rating of length measurement Table 2 2 5 1 2 6 3 3 4 2 4 0 5 7 7 5 6 6 8 10 7 2 2 10 5 8 2 5 4 2 6 2 6 1 7 2 7 2 3 8 1 5 2 5 2 14 2 2 6 3 1 7
8. 8. Frequency Table of rating of length Table 2-3 0 - 2 20 3 - 5 14 6 - 8 15 9 - 11 2 12 - 14 1 rating Frequency
9. 9. Frequency Table Definitions
10. 10. Lower Class Limits <ul><li>are the smallest numbers that can actually belong to different classes </li></ul>
11. 11. Lower Class Limits <ul><li>are the smallest numbers that can actually belong to different classes </li></ul>0 - 2 20 3 - 5 14 6 - 8 15 9 - 11 2 12 - 14 1 rating Frequency
12. 12. Lower Class Limits <ul><li>are the smallest numbers that can actually belong to different classes </li></ul>Lower Class Limits 0 - 2 20 3 - 5 14 6 - 8 15 9 - 11 2 12 - 14 1 rating Frequency
13. 13. Upper Class Limits <ul><li>are the largest numbers that can actually belong to different classes </li></ul>
14. 14. Upper Class Limits <ul><li>are the largest numbers that can actually belong to different classes </li></ul>Upper Class Limits 0 - 2 20 3 - 5 14 6 - 8 15 9 - 11 2 12 - 14 1 rating Frequency
15. 15. <ul><li>are the numbers used to separate classes, but without the gaps created by class limits </li></ul>Class Boundaries
16. 16. <ul><li>number separating classes </li></ul>Class Boundaries - 0.5 2.5 5.5 8.5 11.5 14.5 0 - 2 20 3 - 5 14 6 - 8 15 9 - 11 2 12 - 14 1 Rating Frequency
17. 17. <ul><li>number separating classes </li></ul>Class Boundaries Class Boundaries <ul><li>0.5 </li></ul><ul><li>2.5 </li></ul><ul><li>5.5 </li></ul><ul><li>8.5 </li></ul><ul><li>11.5 </li></ul><ul><li>14.5 </li></ul>0 - 2 20 3 - 5 14 6 - 8 15 9 - 11 2 12 - 14 1 Rating Frequency
18. 18. <ul><li>midpoints of the classes </li></ul>Class Midpoints
19. 19. <ul><li>midpoints of the classes </li></ul>Class Midpoints Class Midpoints 0 - 1 2 20 3 - 4 5 14 6 - 7 8 15 9 - 10 11 2 12 - 13 14 1 Rating Frequency
20. 20. <ul><li>is the difference between two consecutive lower class limits or two consecutive class boundaries </li></ul>Class Width
21. 21. <ul><li>is the difference between two consecutive lower class limits or two consecutive class boundaries </li></ul>Class Width Class Width 3 0 - 2 20 3 3 - 5 14 3 6 - 8 15 3 9 - 11 2 3 12 - 14 1 Rating Frequency
22. 22. Relative Frequency Table relative frequency = class frequency sum of all frequencies
23. 23. Relative Frequency Table 20/52 = 38.5% 14/52 = 26.9% etc. Table 2-5 Total frequency = 52 0 - 2 20 3 - 5 14 6 - 8 15 9 - 11 2 12 - 14 1 Rating Frequency 0 - 2 38.5% 3 - 5 26.9% 6 - 8 28.8% 9 - 11 3.8% 12 - 14 1.9% Rating Relative Frequency
24. 24. Cumulative Frequency Table Cumulative Frequencies Table 2-6 0 - 2 20 3 - 5 14 6 - 8 15 9 - 11 2 12 - 14 1 Rating Frequency Less than 3 20 Less than 6 34 Less than 9 49 Less than 12 51 Less than 15 52 Rating Cumulative Frequency
25. 25. Frequency Tables 0 - 2 20 3 - 5 14 6 - 8 15 9 - 11 2 12 - 14 1 Rating Frequency 0 - 2 38.5% 3 - 5 26.9% 6 - 8 28.8% 9 - 11 3.8% 12 - 14 1.9% Rating Relative Frequency Less than 3 20 Less than 6 34 Less than 9 49 Less than 12 51 Less than 15 52 Rating Cumulative Frequency Table 2-6 Table 2-5 Table 2-3
26. 26. Bar Graph <ul><li>The widths of the bar should be equal </li></ul><ul><li>The bars are usually separated by appropriate spaces with an eye to neatness and clear presentation. The spaces between two bars are usually kept equal to the width of the bars. </li></ul><ul><li>The length of the bar is proportional to the frequency. </li></ul><ul><li>A suitable scale must be chosen to present the length of the bars. </li></ul><ul><li>The Y-axis corresponds to the frequency in vertical bar diagram, whereas the X-axis corresponds to the frequency in a horizontal bar diagram </li></ul>
27. 28. Simple Bar Diagram <ul><li>HIV+ve cases in six districts of West Bengal in 2004 </li></ul>Simple ar Diagram each bar represents frequency of a single category with a distinct gap from another bar . .
28. 29. Multiple / Compound Bar diagram show the comparison of two or more sets of related statistical data .
29. 30. Component /Segmented Bar diagram <ul><li>to compare sizes of the different component parts among themselves </li></ul><ul><li>also show the relation between each part and the whole. </li></ul>
30. 31. PIE Diagram <ul><li>Causes of Maternal deaths of West </li></ul><ul><li>Bengal in 2005 </li></ul><ul><li>For for qualitative or discrete data </li></ul><ul><li>Areas of sectors are proportional to frequencies </li></ul><ul><li>Angle (degree) of a sector= </li></ul><ul><li>Class % X3.6, </li></ul><ul><li>Expressing proportional components of the attributes </li></ul><ul><li>compared with that of other segments as well as the whole circle. </li></ul>
31. 32. Histogram <ul><li>A histogram is a bar graph that shows the frequency of each item. Histograms combine data into equal-sized intervals. </li></ul><ul><li>There are no spaces between the bars on the histogram. </li></ul>
32. 34. Line Graph <ul><li>A line graph uses a series of line segments to show changes in data over time. </li></ul><ul><li>Plot a point for each data item, and then connect the dots with straight line segments. </li></ul>
33. 35. Refer to page 336 for the line graph.
34. 36. Frequency Polygon <ul><li>- Frequency </li></ul><ul><li>Distribution graph </li></ul><ul><li>Joining mid-points </li></ul><ul><li>of histogram blocks </li></ul><ul><li>(class intervals) </li></ul><ul><li>When no. of </li></ul><ul><li>observations are </li></ul><ul><li>very large: Frequency </li></ul><ul><li>Polygon loses it’s </li></ul><ul><li>angulations & giving </li></ul><ul><li>a smooth curve: </li></ul><ul><li>Frequency Curve </li></ul>Frequency Distribution Haemoglobin Level
35. 37. Frequency Polygon <ul><li>-Frequency </li></ul><ul><li>polygon presenting </li></ul><ul><li>variations by time </li></ul><ul><li>Trend of an event occurring over a time </li></ul>Year 1901 1911 1921 1951 1961 1971 1941 1931
36. 38. Line Chart or Graph <ul><li>Growth rate in India from 1921-1931 to 1991-2001 </li></ul>the trend of an event occurring over a period of time
37. 39. Ogive (Cumulative frequency polygon <ul><li>to find the median, quartiles, percentiles </li></ul>
38. 40. Stem-and Leaf Plot Raw Data (Test Grades) 67 72 85 75 89 89 88 90 99 100 Stem Leaves 6 7 8 9 10 7 2 5 5 8 9 9 0 9 0
39. 41. <ul><li>Scatter Diagram </li></ul>• • • • • • • • • • • • • • 0 0.0 0.5 1.0 1.5 10 20 • NICOTINE TAR A plot of paired (x,y) data with the horizontal x-axis and the vertical y-axis. will discuss scatter plots again with the topic of correlation. Point out the relationship that exists between the nicotine and tar – as the nicotine value increases, so does the value of tar. • • • • • • | 3,617 | 11,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2017-13 | latest | en | 0.892708 |
https://www.physicsforums.com/threads/canonical-transformations.628785/ | 1,532,299,454,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594018.55/warc/CC-MAIN-20180722213610-20180722233610-00426.warc.gz | 976,920,898 | 13,345 | # Canonical Transformations
1. Aug 16, 2012
### A_B
Hi,
Im working through some chapters of Goldstein and I'm up to canonical transformations now. On page 370 it says that the variational principle for the hamiltonians K and H are both satisfied if H and K are connected by a relation of the form
λ(pq' - H) = PQ' - K + dF/dt
And I can see this. My question is, are there canonical transformations that do not fit this relation? And if so are they impportant? Is this relation a very general one, or does it simply turn out to give good tranformations in many problems?
A_B
2. Aug 16, 2012
### A_B
Also if we obtain a relation
pq' - H = PQ' - K + ∂F/∂t + (∂F/∂q)q' + (∂F/∂Q)Q'
Goldstein says (p 372 eq 9.13)
"Since the old and the new coordinates, q and Q, ere seperately independent, the above equation can hold identically only if the coefficients of q' and Q' each vanish:"
p = ∂F/∂q,
P = -∂F/∂Q
I don't see why this is necessary for the relation to hold.
3. Aug 16, 2012
### Matterwave
Actually the definition of canonical transformation usually stipulates that λ=1 in the above requirement. For any other λ, we call these "extended canonical transformations".
This is very general because otherwise the action would not remain stationary for the actual motion. This is because the freedom in the Lagrangian is only in the total time derivative term.
4. Aug 18, 2012
### A_B
can someone answer my question in the second post?
thanks!
A_B | 405 | 1,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-30 | latest | en | 0.906976 |
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