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Home / Questions / A worker earns \$16 per hour at a plant and is told that only 5% of all workers make a high...
# A worker earns \$16 per hour at a plant and is told that only 5% of all workers make a higher wage The wage is assumed
A worker earns \$16 per hour at a plant and is told that only 5% of all workers make a higher wage. The wage is assumed to be normally distributed with the standard deviation of \$5 per hour.
a. What is the average wage for the plant workers per hour?
b. If a worker is selected at random, what is the probability that he/she earns at most \$10 per hour?
c. If two workers are selected at random, what is the probability that both earn at most \$10 per hour?
A worker earns \$16 per hour at a plant and is told that only 5% of all workers make a higher wage. Thewage is assumed to be normally distributed with the standard deviation of \$5 per hour.a. What...
Apr 23 2020 View more View Less | 231 | 956 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-39 | latest | en | 0.96035 |
https://mathsolver.microsoft.com/zh-Hant/solve-problem/q%20%3D%20%60frac%20%7B%20K%20(%202%20)%20(%203%20)%20%5E%20%7B%202%20%7D%20%7D%20%7B%208%20%7D | 1,721,443,760,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00665.warc.gz | 333,455,772 | 78,088 | ## 共享
q=\frac{K\times 2\times 9}{8}
q=\frac{K\times 18}{8}
q=K\times \frac{9}{4}
K\times \frac{9}{4}=q
\frac{9}{4}K=q
\frac{\frac{9}{4}K}{\frac{9}{4}}=\frac{q}{\frac{9}{4}}
K=\frac{q}{\frac{9}{4}}
K=\frac{4q}{9}
q 除以 \frac{9}{4} 的算法是將 q 乘以 \frac{9}{4} 的倒數。
q=\frac{K\times 2\times 9}{8}
q=\frac{K\times 18}{8}
q=K\times \frac{9}{4} | 190 | 341 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-30 | latest | en | 0.243651 |
https://www.scienceforums.net/topic/95813-prime-products-just-one-last-time/page/4/ | 1,709,159,496,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474746.1/warc/CC-MAIN-20240228211701-20240229001701-00707.warc.gz | 973,180,413 | 29,918 | # Prime Products just one last time
## Recommended Posts
Ok, so 40,000 people have seen this post. Does no one agree the equations show a pattern in factoring. I’m serious in my approach to this problem. I am not trying to deceive anyone. As my instructor in college would say this is a 5 second problem. Plug in N and x can only be a Prime factor.
I know it isn’t difficult solving knowing both x and N. But these equations show where x must be knowing only N.
Many have seen it but does anyone believe me? You see no one believes me that an amateur found a pattern to Prime numbers. And they would be correct. These equations and patterns are based on patterns in factoring. So if N was 10 an x of 2 would prove the equations true.
So has anyone actually tried what I proposed? Or, did it just sound impossible and not worth the effort?
I ask because I must not be explaining it correctly. Does anyone have any questions? This was the first time I wrote so much about a set of math problems. I have learned it can get out of control; not knowing how to describe something that isn’t completely finished.
So let me know good or bad what you think. Like in show business no comments is worse than criticism.
My question is, “Does no one see it?”
A pattern in Primes may be impossible, but a pattern in factoring is possible. And is important to reverse public key cryptography.
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• 1 month later...
• Replies 83
• Created
#### Posted Images
Compare to
Compare these 2 Wolfram Alpha links. There is a pattern here.
Easy knowing both x and n. But the pattern is there. There is more to the patterns I am working on.
I am compiling a Kindle book.
If you believe there is a pattern post here or check my profile.
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• 2 weeks later...
Ok so no response means that no one sees my pattern. I know my equations are cumbersome, but there is a pattern. In fact I have about 3 distinct equations that show patterns when solved. I know it isn’t a perfect solution where x is found knowing only n. You have to use test values of x to determine where n’s least common multiple is.
There is a pattern and I am trying to show that pattern of the least common multiple. I will post again a simplified explanation. My goal was to show the pattern before just plugging a chugging an answer. I will show this pattern, but it is labor intensive to organize all 4 equations.
I just wish I was getting more feedback. Over 5 years ago I posted to the Wolfram community. This problem is not designed for their boards. But I did receive some good feedback that knowing n and x the equation checked. However that is not useful. So as I refined my idea where the equation would show a pattern knowing only n. The only problem was it was such a complex nth degree polynomial, I could not solve it. I could only use it in a computer program to compare test values of x. I mean, I led me to work what was known about solving polynomials and if there was anyway to solve it graphically. But I still feel my equations though rugged, showed important patterns.
My equations had a range of error, but I now know this was due to the limitations of the calculator. But will my average coding skills I do not know how to program for over 100 digit numbers.
But in my future attempts to show my work, I will show my patterns and where you can learn more about them. But until then don’t just look at this equation of finding the least common multiple. Look at the equations with the factors known and see those factors as forming a pattern as they are multiplied together to form n. A pattern in Prime numbers sounds impossible. But that is not what my equation does. It takes n and finds it factors according to an equation that will solve for the least common multiple. I remember a previous comment that solving a Prime pattern was not serious. I argue that it may be impossible, put finding an equation describing the least common multiple is a serious endeavor.
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• 3 weeks later...
(x^2 * (N^2 + 2 * x^3)) / N^2
x^2 = ((N^4*x^2*N^2*x^5) +x^8) / N^4
((N^4*x^2 + 2*N^2) * x^5) / (N^4*x^2)
(((N^4/x) + 2*N^2*x^2 + x^5)/N^3) / N) * x^3 = x^2
Where y = ((N^4/x) + 2*N^2*x^2 + x^5)/N^3)
Each of these 4 equations is separate. I just wanted to post and show just how fun plugging and chugging can be. It doesn’t mean every equation is a useful solution. But plugging and chugging is fun when working with series. It is a place where computers and automation of equations does prove useful. I recently read a journal entry where it described brute force as a way to prove or disprove. I am not worried about that debate, but I am interested in using computers to look for patterns.
I hope I typed these equations correctly. It is late so if you get any typos let me know the equations can be extremely hard not to create mass confusion. But I challenge you to test these equations and try your own creations. There are infinitely many, but I want one that is useful to describe the patterns of factorization. I don’t care if the equation simplifies to zero equals zero. I would prefer it didn’t but the quest for the useful equation begins.
Oh, N is the known product of 2 numbers x and y, where x is the smaller factor
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See attached PDF.
Here is a bunch of patterns when trying to substitute and find x knowing only N.
Again, a lot of zero equals zero; and a bunch of high degree hard to simplify polynomials.
But if you look at the fraction Mathematica has created (while knowing both x and N) you see there is a simplification of the polynomial. No guarantees, but it may just be possible to simplify the polynomial equation to make it way more useful.
For example, the equations and fractions are of the form: (N^4 * x) / x^5 = y
y^4= N^4 / x^4
So these fractions are:
(Something Simplified) / (x^5y)
The right combination might just simplify a large, cumbersome polynomial into something more useful. Or not. You be the judge. Download the PDF and test for yourself.
20190204VeryImportantPatterns20190330.pdf
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Reduced this polynomial:
{(((((N^4 / x) + 2 * (N^2 * x^2) + x^5) / n^3) * x^3) / N}
to
Sqrt{((N^4 / x^5 + 2 * N^2 / x^2 + x) * x^3) / (N4 / x^4)} =x
Need some math help. I reduced the polynomial. However, the result does not seem any easier to solve. The black is the original the red is reduced.
N is known. And x is to be solved. Can anyone solve this simplified polynomial?
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I hope someone responds. I am serious. And to show that I am serious I offer the equation as far as I can simplify.
[ x^4 + (2*x)/N^2 + x^10/N^4 ] ^(1/4) = x
The brackets are the 4th root of
We know N = 85, so solving for x should equal 5. It works when you know both N and x, but is this enough to solve for N?
This is my best attempt so far. So if you are curious please respond with your own polynomial simplifications.
Respond if you fill that I am serious and on the right path of solving the least common multiple.
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[ x^4 + (2*x)/N^2 + x^10/N^4 ] ^(1/4) = x
The brackets are the 4th root of
N = x * y , the product of 2 Prime numbers
x is the smaller Prime factor
Example N = 85, x = 5, and y = 17
I apologize for the blunder above. But you need to understand I literally tried thousands of equations.
So far this is my best. Check it for errors. Is anyone having any luck with other combinations?
x^3 + 2*x^4/N^2 + x^6/N^4 = x^2
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On 2/23/2018 at 1:19 PM, Trurl said:
Ok, this is my final post to this thread, unless someone asks a question.
On 7/24/2018 at 3:17 PM, Trurl said:
I don’t know if anyone believes me when I say this series is significant.
On 7/24/2018 at 3:17 PM, Trurl said:
So, if you think there is any meaning to my equations, please post me a message in this thread to let me know. I will respond with more information.
On 6/17/2019 at 7:15 PM, Trurl said:
Ok so no response means that no one sees my pattern.
!
Moderator Note
I'm afraid this is something only YOU see. If you find a way to persuade others, let me know and I can open this monologue up to discussion again.
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This topic is now closed to further replies.
× | 2,093 | 8,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-10 | latest | en | 0.960689 |
https://www.calltutors.com/Assignments/take-the-bertillon-measures-from-two-people-and-fill-the-first-table-in-centimetres | 1,603,257,231,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107875980.5/warc/CC-MAIN-20201021035155-20201021065155-00112.warc.gz | 646,026,402 | 18,492 | ## Take the Bertillon measures from two people and fill the first table (in centimetres).
### computer science
##### Description
Biometric Identification Francesc Serratosa Exercises of Part 1: Biometrics for Human Identification Week 1 1. Take the Bertillon measures from two people and fill the first table (in centimetres). Then, fill the second table with the Euclidean distances and decide which is the maximum threshold such that the four people are considered as different subjects. After that, suppose that Will West and William West were actually the same person as was thought at first. What are the minimum and maximum threshold values? Trait Will West William West Person 1 Person 2 1) Height (fig. 1) 178.5 177.5 2) Arm with (fig. 2) 187.0 188.0 3) Height seated (fig. 3) 91.2 91.3 4) Head width (fig. 4) 19.7 19.8 5) Head length (fig. 5) 15.8 15.9 6) Cheek width (fig. 6) 14.8 14.8 7) Right ear length (fig. 6) 6.6 6.5 8) Left foot length (fig. 7) 28.2 27.5 9) Third left finger length (fig. 8) 12.3 12.2 10) Small left finger length (fig. 8) 9.7 9.6 11) Left forearm length (fig. 9) 50.2 50.3 Distance Will West William West Person 1 Person 2 Will West 0 William West 0 Person 1 0 Person 2 0 2. Which is the percentage of revenues of the fingerprint trait with respect to the whole biometric features? And what is the second biometric characteristic with the highest revenues? 3. Revenues of biometric systems from 2007 to 2015 with respect to time can be easily modelled as a linear function. What is the equation of the line? If predictions hold, which value is estimated in 2020? Week 2 4. Explain the differences between the verification and recognition systems. When is the first used and when is the second used? The algorithms to compare biometric traits can be very expensive temporarily. For this reason, it is usual to use algorithms that deduce sub-optimal distances very fast but can fail to get the exact distance. In which cases is important to use a sub-optimal algorithm? 5. Explain in what consists the enrolment to a biometric system. Typically, in the enrolment systems, there is a responsible for verifying the data in a whole process of capturing biometric features. Why it is so important to verify that this process is done correctly? 6. What are the six processes related on the verification, identification and registration systems? 7. Describe a real application that uses an off-line system and another that uses an online system. 8. Explain the difference between the two possible types of queries: Positives and negatives. Describe a pair of real applications that are of two types | 681 | 2,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-45 | longest | en | 0.877431 |
http://uu.diva-portal.org/smash/record.jsf?pid=diva2%3A650350&c=5&searchType=SIMPLE&language=sv&query=&af=%5B%5D&aq=%5B%5B%7B%22journalId%22%3A%227592%22%7D%5D%5D&aq2=%5B%5B%5D%5D&aqe=%5B%5D&noOfRows=50&sortOrder=author_sort_asc&sortOrder2=title_sort_asc&onlyFullText=false&sf=all | 1,582,099,750,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144058.43/warc/CC-MAIN-20200219061325-20200219091325-00221.warc.gz | 145,842,799 | 16,749 | uu.seUppsala universitets publikationer
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A limit law of almost l-partite graphs
Uppsala universitet, Teknisk-naturvetenskapliga vetenskapsområdet, Matematisk-datavetenskapliga sektionen, Matematiska institutionen. (Mathematical logic)
2013 (Engelska)Ingår i: Journal of Symbolic Logic (JSL), ISSN 0022-4812, E-ISSN 1943-5886, Vol. 78, nr 3, s. 911-936Artikel i tidskrift (Refereegranskat) Published
##### Abstract [en]
For integers l >= 1, d >= 0 we study (undirected) graphs with vertices 1,..., n such that the vertices can be partitioned into l parts such that every vertex has at most d neighbours in its own part. The set of all such graphs is denoted P-n (l, d). We prove a labelled first-order limitlaw, i.e., for every first-order sentence phi, the proportion of graphs in P-n (l, d) that satisfy phi converges as n -> infinity. By combining this result with a result of Hundack, Promel and Steger [12] we also prove that if 1 <= s(1) <=...<= s(1) are integers, then Forb(A(I),s(1),...,s(l)) has alabelled first-order limit law, where Forb (A(I),s(1),...,s(l)) denotes the set of all graphs with vertices 1,..., n, for some n, in which there is no subgraph isomorphic to the complete (l + 1)-partite graph with parts of sizes 1, S-1,..., S-l. In the course of doing this we also prove that there exists a first-order formula depending only on l and d, such that the proportion of g e P (I, d) with the following property approaches 1 as n ->infinity: there is a unique partition of {1,..., n} into l parts such that every vertex has at most d neighbours in its own part, and this partition, viewed as an equivalence relation, is defined by xi.
##### Ort, förlag, år, upplaga, sidor
Association for Symbolic Logic , 2013. Vol. 78, nr 3, s. 911-936
##### Nyckelord [en]
Finite model theory, limit law, random graph, forbidden subgraph
##### Nationell ämneskategori
Algebra och logik
Matematik
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• rtf | 1,102 | 3,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-10 | latest | en | 0.452591 |
https://www.allmath.com/RadicalRootCalculator.php | 1,726,623,644,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00758.warc.gz | 604,350,776 | 11,766 | Put the number, write the root, and click calculate to find the root value using radical calculator
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Radical calculator is used to simplify radical expressions. A radical expression is any mathematical expression containing a square root, cube root, ... up to the nth root.
## What is Radical (Nth Root)?
In mathematics, the radical, specifically the nth root, refers to the operation that calculates the value that, when raised to the power of n, gives a given number. The symbol used to represent the radical is √, known as the radical sign /the square root symbol.
The term "radical" is often used to refer to the square root (), but it can also represent other roots, such as cube roots (), fourth roots (), and so on. The number "n" in the nth root indicates the degree of the root. For example, when n = 2, it represents the square root; when n = 3, it represents the cube root, and so on.
To calculate the nth root of a number, you raise that number to the power of 1/n. For instance, if you want to find the cube root of 8, you would solve it as follows:
`∛8 = 81/3 = 2`
In this case, raising 8 to the power of 1/3 gives you the value of 2, which is the cube root of 8.
## How to find the root value?
Example 1:
Determine the fourth root of the number 625.
Solution
It can be expressed in mathematical form
4√625
= (5 × 5 × 5 × 5)1/4
= (54)1/4
= 5
Example 2:
Determine the Sixth root of the number 64
Solution
In mathematical form, it can be expressed as:
6√64
= (64)1/6
= (2 × 2 × 2 × 2 × 2 × 2)1/6
= (26)1/6
= 2
Example 3:
Find the 4th root of the number 196
Solution
In mathematical form, it can be expressed as:
4√196
= (196)1/4
= (4 × 4 × 4 × 4)1/4
= (44)1/4
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https://www.blatygranitowe.net.pl/how-stem-and-leaf-separator-close/ | 1,653,415,027,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00426.warc.gz | 762,806,008 | 6,599 | Get a Quote
### How Stem And Leaf Separator Rampd
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Aug 30, 2018 Stem and Leaf Plot Example. Suppose that your class had the following test scores: 84, 62, 78, 75, 89, 90, 88, 83, 72, 91, and 90 and you wanted to see at a glance what features were present in the data. You would rewrite the list of scores in order and then use a stem-and-leaf plot. The stems are 6, 7, 8, and 9, corresponding to the tens place ...
### Stem-and-Leaf Plots Gizmo - ExploreLearning
Check out this Gizmo from @ExploreLearning! Build a data set and compare the line plot of the data set to the stem-and-leaf plot. Time's Up! As a guest, you can only use this Gizmo for 5 minutes a day. Sign up for a free Gizmos account and …
### Stem-and-Leaf Plots Gizmo : Lesson Info : ExploreLearning
Gizmo User from Arkansas. Download. Stem and Leaf Questionairre. Can be used before, during, or after the lesson. 5 short questions over stats. Best For: Math 7.
### Solved: How can I format a stem-and-leaf plot using SAS?
Jan 17, 2019 Seems to get close with the given data for the Stem and Leaf though the order is still smallest to largest upwards. If your data changes you may need to change the plot size. I think the default number of rows is based on the number of records and range of values.
### Stem And Vein Kratom
Jan 18, 2022 The products you find at all the vendors in America contain only the kratom leaf’s fat tissue. How Is Stem And Vein Kratom Used? As we mentioned earlier, the twigs, stems, and leaf vein networks are dried together and sent to manufacturing units. However, these manufacturers remove the stem and vein kratom from fat tissue and use the latter.
### Stem-and-Leaf Plots: How to Find Mean Median Mode
May 05, 2021 A stem-and-leaf plot is a type of plot that displays data by splitting up each value in a dataset into a stem and a leaf. This tutorial explains how to calculate the mean, median, and mode of a stem-and-leaf plot. Example: Mean, Median Mode of Stem-and-Leaf Plot
### Biogeographical patterns of biomass allocation in leaves
Nov 03, 2015 Relationship of forest biomass allocation to stand characteristics and environmental factors. The PCA demonstrated that 75.5%, 16.6% and 7.9% of variation in leaf, stem and root biomass in all ...
### How to Understand a Stem and Leaf Plot Diagram
Nov 04, 2019 If you've done it correctly, it should yield a stem-and-leaf plot graph that looks like the one in this section. To find the median, count all the days in the month, which in the case of June is 30. Divide 30 by two, yielding 15, count either up from the lowest temperature of 50 or down from the highest temperature of 87 until you get to the ...
### Stem-and-Leaf Graphs (Stemplots) Introduction to Statistics
One simple graph, the stem-and-leaf graph or stemplot, comes from the field of exploratory data analysis. It is a good choice when the data sets are small. To create the plot, divide each observation of data into a stem and a leaf. The leaf consists of a final significant digit. For example, $23$ has stem two and leaf three.
### Stem-and-Leaf Plots: Definition Examples - Statology
Sep 23, 2018 A stem-and-leaf plot displays data by splitting up each value in a dataset into a “stem” and a “leaf.” This tutorial explains how to create and interpret stem-and-leaf plots. How to Make a Stem-and-leaf Plot. The following two examples illustrate how to create a stem-and-leaf plot from scratch for a given dataset. Example 1
### Tobacco leaf separator - Monk-Austin International Inc.
Sep 28, 1994 What is claimed is: 1. An air separator for separating a mixture of tobacco leaf lamina and stem into lighter and heavier segments comprising: (a) a separation chamber having an upper end including a lighter segment discharge opening, a back wall including a leaf mixture inlet and a front wall including a heavier segment discharge opening;
### How to Make a Stem and Leaf Plot -
Teacher Sarah shows you how to make a stem and leaf graph to help you organize your information. Close Dialog. You are now leaving TVOKids.com. TVOKids doesn't have control over the new place you're about to visit, so please make sure you get your Parent or Guardian's permission first!
### Stem-and-Leaf Plots: Examples - Purplemath
This example has two lists of values. Since the values are similar, I can plot them all on one stem-and-leaf plot by drawing leaves on either side of the stem I will use the tens digits as the stem values, and the ones digits as the leaves. Since 9 (in the Econ 101 list) has no tens digit, the stem value will be 0 .
### Stem and Leaf Plot Maker Stem and Leaf Plot in Excel
To generate a plot: Open the Stem and Leaf Plot Template by clicking on QI Macros menu - Chart Templates - Stem and Leaf Plot. The template will open and contains some pre-populated sample data. Input or cut and paste your data into the yellow input area. Double click on the update plot button and QI Macros will sort the data and populate the ...
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<script type="text/javascript" src="https://v1.cnzz.com/z_stat.php?id=1281089964&web_id=1281089964"></script> | 1,773 | 7,524 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-21 | latest | en | 0.920405 |
http://at.metamath.org/ileuni/ax-1.html | 1,508,449,788,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823478.54/warc/CC-MAIN-20171019212946-20171019232946-00463.warc.gz | 31,058,878 | 4,023 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > ax-1 GIF version
Axiom ax-1 5
Description: Axiom Simp. Axiom A1 of [Margaris] p. 49. One of the 3 axioms of propositional calculus. The 3 axioms are also given as Definition 2.1 of [Hamilton] p. 28. This axiom is called Simp or "the principle of simplification" in Principia Mathematica (Theorem *2.02 of [WhiteheadRussell] p. 100) because "it enables us to pass from the joint assertion of φ and ψ to the assertion of φ simply." General remarks: Propositional calculus (axioms ax-1 5 through ax-3 7 and rule ax-mp 8) can be thought of as asserting formulas that are universally "true" when their variables are replaced by any combination of "true" and "false." Propositional calculus was first formalized by Frege in 1879, using as his axioms (in addition to rule ax-mp 8) the wffs ax-1 5, ax-2 6, pm2.04 75, con3 550, notnot2 718, and notnot1 541. Around 1930, Lukasiewicz simplified the system by eliminating the third (which follows from the first two, as you can see by looking at the proof of pm2.04 75) and replacing the last three with our ax-3 7. (Thanks to Ted Ulrich for this information.) The theorems of propositional calculus are also called tautologies. Tautologies can be proved very simply using truth tables, based on the true/false interpretation of propositional calculus. To do this, we assign all possible combinations of true and false to the wff variables and verify that the result (using the rules described in wi 4 and wn 3) always evaluates to true. This is called the semantic approach. Our approach is called the syntactic approach, in which everything is derived from axioms. A metatheorem called the Completeness Theorem for Propositional Calculus shows that the two approaches are equivalent and even provides an algorithm for automatically generating syntactic proofs from a truth table. Those proofs, however, tend to be long, since truth tables grow exponentially with the number of variables, and the much shorter proofs that we show here were found manually.
Assertion
Ref Expression
ax-1 (φ → (ψφ))
Detailed syntax breakdown of Axiom ax-1
StepHypRef Expression
1 wph . 2 wff φ
2 wps . . 3 wff ψ
32, 1wi 4 . 2 wff (ψφ)
41, 3wi 4 1 wff (φ → (ψφ))
Colors of variables: wff set class This axiom is referenced by: a1i 10 id 18 id1 19 a1d 21 a1dd 41 jarr 90 pm2.86i 91 pm2.86d 92 pm5.1im 161 biimt 229 pm5.4 237 pm4.45im 316 pm2.51 560 pm4.8 602 pm2.53 617 imorri 644 pm2.64 689 pm2.82 700 biort 714 oibabs 806 pm5.14 832 dedlem0a 887 oplem1 899 meredith 1250 a9wa9lem6 1410 a9wa9lem6OLD 1411 a9wa9lem7 1412 ax46to4 1450 ax467to4 1456 stdpc4 1619 ax11 1655 sbi2 1669 ax11v 1703 ax11eq 1810 ax11el 1811 ax11f 1812 ax11indi 1814 ax11indalem 1815 ax11inda2ALT 1816 ax11inda2 1817 moimv 1872 euim 1874
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# Vertex normals
This topic is 5458 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
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Yup, it''s me again with a normal question... I finaly perfected my face normal calculation code, but now i need vertex normals. Can any of give me a link or just tell me how to get vertex normals? Thanks!
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It''s easy. Just find the faces that share this vertex and average their normals, you''ll get your vertex normal.
##### Share on other sites
here you go:
CVector3 *vNormals = NULL; CVector3 sum, v1, v2, v3; t3DObject* pObject; int i, j, k, l; bool e; for (i = 0; i < pModel->numOfObjects; i++) { pObject = &pModel->pObject[i]; pObject->pNormals = new CVector3 [pObject->numOfVerts]; vNormals = new CVector3 [pObject->numOfFaces]; for (j = 0; j < pObject->numOfFaces; j++) { v1 = pObject->pVerts[pObject->pFaces[j].vertIndex[0]]; v2 = pObject->pVerts[pObject->pFaces[j].vertIndex[1]]; v3 = pObject->pVerts[pObject->pFaces[j].vertIndex[2]]; vNormals[j] = Normalize(Cross(v1 - v2, v1 - v3)); } for (j = 0; j < pObject->numOfVerts; j++) { sum = CVector3(0, 0, 0); for (l = 0; l < pObject->numOfFaces; l++) { e = false; for (k = 0; k < 3; k++) { if (pObject->pFaces[l].vertIndex[k] == j) e = true; } if (e) { sum = sum + vNormals[l]; } } sum = Normalize(sum); pObject->pNormals[j] = sum; } delete [] vNormals; }
It works with the data structures on www.gametutorials.com (for 3DS loading and vectors)
EDIT: now i realized that both the "Vertex normals" and "Face normals" thread were yours .
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Implement Newton’s First Law of Motion as a teaching principle with this packet: students (bodies at rest) need many hands-on activities (impressed forces) to learn (compelling change)! This collection of Physical Science Action Labs will give your students plenty of experience with motion. The labs
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Wheeee! Swing High! This packet includes a complete science fair project for a kindergarten first grade class. It could also be used as a simple 3-5 grade individual project. Using a simple pendulum, students discover the changes in number of swings based on the string length. It includes discuss
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This resource gives step by step directions for building a pendulum from plastic pipe. The pendulum is used for creating fantastic works of art while integrating science and math (STEAM). ✅Following these directions must be done by someone who has the tools and expertise to safely cut, drill, and
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Though Edgar Allan Poe is always a student favorite,“The Pit and the Pendulum” proves to be one of Poe’s most challenging short stories. I designed this lesson with engaging activities that not only help students to better understand the content, but also to make this text more enjoyable. I use t
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What variable would make the speed of a pendulum speed up? Pushing it harder?Adding weight? Increasing the length? How about all three variables? Your students will prove or disprove their hypothesis as they tests them out in this simple, yet engaging experiment. There is a math connection as the
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19 slides! The Pit and the Pendulum brought to you in Powerpoint!!! This is the perfect way to introduce this frightening story to your students!! Filled with a ton of information related to state standards, as well as images to spark the imagination! Included in this power point are the following:
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This 2 page worksheet has three parts. In the first part students match twelve examples of Figurative Language to their respective definitions; the examples are the ones E. A. Poe employs in "The Pit and the Pendulum:" personification, parallel structure, metaphor, oxymoron, irony, symbolism, alli
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Pendulum Inquiry and Lab including 3 experiments, graphing & conclusion questions Good for grades 4-8
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This lab is an excellent way for students to demonstrate their understanding of Potential and Kinetic Energy and practice basic algebra skills. In addition, the lab equipment is easy for the instructor to acquire and set up. This assessment covers a few key concepts in physics, such as transformatio
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Looking for a good physics lab to do with your students? This is it! This lab helps students understand the relationship between mass, length and the rate of swing. It also helps them practice with collecting data as well as graphing it (which we all know is crucial in a science class!) This lab can
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Poe's short stories can be challenging for students to comprehend. Use this great unit to help students' completely understand and appreciate this great story. Bundle includes: Introduction PPT (Background info, check quiz, and MORE) Reading Guide Images PPT After Activity (Summary, Analysis, &
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This 4 page, scantron-ready quiz has 50 questions: 38 multiple choice questions on plot and 12 matching questions on figurative language. For Part 2 the students match the example to a quoted phrase or clause, i.e., "... figures of fiends ..."-alliteration. This is a very good quiz to assess reading
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"Pit and the Pendulum" by Edgar Allan Poe ➡All our stories are Digital Ready. Students can highlight/annotate the text on a computer. ✅A 2nd organizer for teaching close/active reading/annotation is included. "The Pit and the Pendulum" by Edgar Allan Poe is part of our Short Story Annotation Ser
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This activity is meant to be used after reading Edgar Allan Poe’s “The Pit and the Pendulum.” It is a Jigsaw activity in which each student has different roles in a group. The groups explore different elements of the story - symbolism, themes, and emotions - and then present to each other. This i
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Use this reading guide in your classroom to help students' understand this challenging, yet intriguing story. Guide focuses on comprehension, context, character, and vocabulary. Guide explores Poe's use of suspense. Also, students must explain quotations as they relate to the story and context.
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10 open-ended comprehension and discussion questions for Poe's short story "The Pit and the Pendulum." Questions range in level of difficulty and assess various levels of Bloom's Taxonomy. Download includes a student handout with typed questions for easy distribution as well as a teacher answer key.
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Here is EAP's short story "The Pit and the Pendulum" complete with thought questions. I usually have my students complete this story in literature circles or as an entire class, as it is a somewhat difficult story to understand if you are reading alone. Enjoy :)
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With a focus on imagery, symbolism, comprehension, and mood, this 15 question multiple choice quiz can be used as a culminating assessment following the study of Poe's challenging short story. Before purchasing, consider that this quiz is included in The Pit and the Pendulum Bundle (\$5). RELATED
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Scientific Method and Pendulums- use the scientific method to learn about the concepts of variables, amplitude, bob weight and rod length with a simple, yet fascinating, object- the pendulum! With this product you will be able to teach your students how to set up and conduct an investigation using
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Covers plot, setting, symbolism, theme, and historical basis for the short story in a short review of the work.
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Looking for a nice activity to do with your students featuring Poe's story, Pit and Pendulum? This may be right for you! This 9 page packet has quite a bit to offer! Included is: -Pre and Post Reading printables -article coinciding with story -questions for discussion -Poe-ster project -Poe-ster g
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Here is everything you need to teach Edgar Allan Poe's "The Pit and the Pendulum" This lesson is part of my author study unit which is also available for sale. 12 pages including: - a link to a downloadable version of the story - background/introduction to time period - 2 page vocabulary chart - 2 p
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Welcome to your “The Pit and the Pendulum” Short Story Resource Package. This package is broken into five parts so that you can easily choose which parts you wish to use. Each section can be used independently within your own plans or together to form the basis of a full unit. Enjoy! 1. The first p
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The wordsearch for E. A. Poe's "The Pit and the Pendulum" has 50 words; the crossword puzzle has 35 clues and answers. Each contains terms, ideas, details and information from the short story. These can be used as a reading check," just for fun," extra credit, homework, or as a silent activity afte
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 2,032 | 8,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | latest | en | 0.905578 |
https://www.convert-measurement-units.com/convert+Metric+Pound+to+Picogram.php | 1,653,584,158,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662619221.81/warc/CC-MAIN-20220526162749-20220526192749-00115.warc.gz | 821,548,310 | 12,919 | Convert Metric Pound to pg (Metric Pound to Picogram)
## Metric Pound into Picogram
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Metric+Pound+to+Picogram.php
## How many Picogram make 1 Metric Pound?
1 Metric Pound = 500 000 000 000 000 Picogram [pg] - Measurement calculator that can be used to convert Metric Pound to Picogram, among others.
# Convert Metric Pound to Picogram (Metric Pound to pg):
1. Choose the right category from the selection list, in this case 'Mass / Weight'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Metric Pound'.
4. Finally choose the unit you want the value to be converted to, in this case 'Picogram [pg]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '864 Metric Pound'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Mass / Weight'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '69 Metric Pound to pg' or '16 Metric Pound into pg' or '9 Metric Pound -> Picogram' or '97 Metric Pound = pg' or '20 Metric Pound to Picogram' or '70 Metric Pound into Picogram'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(43 * 84) Metric Pound'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '864 Metric Pound + 2592 Picogram' or '48mm x 81cm x 10dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 1.397 098 258 891 3×1026. For this form of presentation, the number will be segmented into an exponent, here 26, and the actual number, here 1.397 098 258 891 3. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 1.397 098 258 891 3E+26. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 139 709 825 889 130 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 859 | 3,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-21 | latest | en | 0.848899 |
https://mathoverflow.net/questions/296623/what-is-the-extreme-value-distribution-for-the-kolmogorov-smirnov-d-statistic | 1,670,255,019,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711017.45/warc/CC-MAIN-20221205132617-20221205162617-00749.warc.gz | 430,030,614 | 26,578 | # What is the extreme value distribution for the Kolmogorov-Smirnov D statistic?
I occasionally find that I want to apply a K-S test in the context of unit-testing software that involves random behaviors. Unit testing with sampling statistics is a bit tricky because you want to minimize false-failures.
Extreme value distributions seem like one useful approach, since they make it possible to run a series of experiments, find a maximum D statisic between experimental results and an expected distribution, and measure the probability that D is an outlier using Extreme Value theory.
The idea is that I might run (n) K-S tests, comparing n pairs of samples. This will result in n D-statistic values; the maximum of these values will adhere to some variation of extreme value distribution. I could use the formula for this (or an approximation).
I have never found an extreme value distribution for the K-S D statistic. I suspect it at least adheres to the Weibull form of the EV distribution since its value has a finite maximum, but not even sure of that. I might do some empirical fitting but a more general formula would be even nicer.
UPDATE: Although I have posted an answer that works for a given sample size, it would also be interesting to derive a limiting cdf, that is identify $\alpha$ for the Weibull family of EVD: $$F(x) = e^{-(-(\frac{x - \mu}{\sigma}))^\alpha}$$
• The downvote doesn't seem exactly welcoming... Anyway, in case you didn't know it, there's also statistics.stackexchange (= Cross Validated) which might be more welcoming. If other users agree that would be a good move, then we could migrate it over. Apr 3, 2018 at 14:49
• @ToddTrimble Perhaps it is out of scope? Hard for me to tell; I see other questions about properties of KS and/or extreme value distributions in this forum.
– eje
Apr 3, 2018 at 17:45
• If you want a result for finite $n$ (instead of only an asymptotic result), you can use the Dvoretzky–Kiefer–Wolfowitz inequality to get an explicit (non-asymptotic) bound. Oct 31, 2018 at 13:44 | 484 | 2,043 | {"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} | 3.28125 | 3 | CC-MAIN-2022-49 | latest | en | 0.935789 |
https://hu.pinterest.com/cece09/properties-of-operations/ | 1,603,341,017,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878879.33/warc/CC-MAIN-20201022024236-20201022054236-00472.warc.gz | 363,752,194 | 60,882 | # Properties of Operations
## Collection by Carol Camp
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Throughout the Math Common Core for Grades 1 - 8, "properties of operations" is mentioned a lot.....that students should know how to use them to solve problems.
Multi-digit multiplication is a difficult concept to teach. Long gone are the days where we teach one method, such as long multiplication, and just *hope* that all of our students catch on and can use
Distributive Property of Multiplication Anchor Chart
Okay, raise your hand if you’re an elementary teacher and just the mention of the distributive property makes you break out in a cold sweat. Go ahead…no one can see you. Hmmmm, that’s what I thought. 🙂 The common core standards are all about the properties, and that’s very new to most elementary teachers. It’s […]
Vind de verliefde harten. De juf highlight ze. Het getal wat overblijft krijgt een rondje.
Looking for fun, new resources for teaching arrays in 2nd grade? This set of 20 task cards provides photographed examples of real life uses for arrays. Great for small group math centers, whole class scoot games, or math tubs!
A Hands-On introduction the Distributive Property of Multiplication. I love this!
Arrays activities 2nd grade to introduce students to multiplication fun ideas such as this one to paint repeated addition equations
Teaching kids to find the area of composite shapes- math unit for teaching area. Hands-on, engaging math activities to help kids understand how to find the area of composite shapes. \$
The DISTRIBUTIVE PROPERTY of multiplication can be tricky! This post gives you fun, engaging STRATEGIES and ACTIVITIES that students will love!
FREE Worksheet with answers, Bell Work, Guided Notes, Exit Quiz, Power Point, Lesson Plans and more to help teach the Distributive Property!
Apply properties of operations as strategies to add and subtract - commutative and associative (Common Core 1.OA.3) | Operations & Algebraic Thinking | Learnist
Okay, raise your hand if you’re an elementary teacher and just the mention of the distributive property makes you break out in a cold sweat. Go ahead…no one can see you. Hmmmm, that’s what I thought. 🙂 The common core standards are all about the properties, and that’s very new to most elementary teachers. It’s […]
Inverse means opposite or reverse. In math, addition and subtraction are inverse operations. When we add, we are putting together. When we subtract, we are taking apart. Multiplication and division are inverse operations. When we divide, we are separating a whole into equal groups. When we multiply, we are combining equal groups into a whole. … | 577 | 2,654 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2020-45 | latest | en | 0.84731 |
http://oeis.org/A254524 | 1,548,110,614,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583814455.32/warc/CC-MAIN-20190121213506-20190121235506-00287.warc.gz | 170,946,553 | 4,581 | This site is supported by donations to The OEIS Foundation.
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A254524 n is the a(n)-th positive integer having its digitsum. 12
1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 3, 3, 3, 3, 3, 3, 3, 3, 2, 1, 4, 4, 4, 4, 4, 4, 4, 3, 2, 1, 5, 5, 5, 5, 5, 5, 4, 3, 2, 1, 6, 6, 6, 6, 6, 5, 4, 3, 2, 1, 7, 7, 7, 7, 6, 5, 4, 3, 2, 1, 8, 8, 8, 7, 6, 5, 4, 3, 2, 1, 9, 9, 8, 7, 6, 5, 4, 3, 2, 1, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 10, 5 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,10 COMMENTS a(A051885(n)) = 1. - Reinhard Zumkeller, Oct 09 2015 Ordinal transform of A007953. - Antti Karttunen, May 20 2017 LINKS David A. Corneth, Table of n, a(n) for n = 1..10000 EXAMPLE 35 is the 4th positive integer having digitsum 8 (the others before are 8, 17 and 26) so a(35) = 4. MATHEMATICA c[n_, k_] := If[n >= k, Binomial[n, k], 0]; b[s_, q_, n_] := (s1 = q; If[s <= q*(n - 1), s1 = s + q; Sum[(-1)^i*c[q, i]*c[s1 - 1 - n*i, q - 1], {i, 0, q - 1}], 0]); a[n_] := (r = 1; v = IntegerDigits[n]; l = v[[-1]]; For[i = Length[v] - 1, i >= 1, i--, For[j = 1, j <= v[[i]], j++, r += b[l + j, Length[v] - i, 10]]; l += v[[i]]]; r); Table[a[n], {n, 1, 110}] (* Jean-François Alcover, Nov 14 2016, adapted from PARI *) PROG (PARI) \\This algorithm needs a modified binomial. C(n, k)=if(n>=k, binomial(n, k), 0) \\ways to roll s-q with q dice having sides 0 through n - 1. b(s, q, n)=if(s<=q*(n-1), s+=q; sum(i=0, q-1, (-1)^i*C(q, i)*C(s-1-n*i, q-1)), 0) \\main algorithm a(n)={r = 1; v=digits(n); l=v[#v]; forstep(i = #v-1, 1, -1, for(j=1, v[i], r+=b(l+j, #v-i, 10)); l+=v[i]); r} (Haskell) import Data.IntMap (empty, findWithDefault, insert) a254524 n = a254524_list !! (n-1) a254524_list = f 1 empty where f x m = y : f (x + 1) (insert q (y + 1) m) where y = findWithDefault 1 q m; q = a007953 x -- Reinhard Zumkeller, Oct 09 2015 CROSSREFS Cf. A007953, A051885, A069877, A143164, A263017, A263109, A263110. Cf. A286478 (analogous sequence for factorial base). Sequence in context: A160093 A259143 A035931 * A140438 A132211 A067441 Adjacent sequences: A254521 A254522 A254523 * A254525 A254526 A254527 KEYWORD nonn,base,look,nice AUTHOR David A. Corneth, Jan 31 2015 STATUS approved
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Last modified January 21 17:43 EST 2019. Contains 319350 sequences. (Running on oeis4.) | 1,191 | 2,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.481949 |
https://www.airmilescalculator.com/distance/ild-to-alc/ | 1,611,457,894,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703544403.51/warc/CC-MAIN-20210124013637-20210124043637-00258.warc.gz | 655,059,475 | 24,366 | # Distance between Lleida (ILD) and Alicante (ALC)
Flight distance from Lleida to Alicante (Lleida–Alguaire Airport – Alicante–Elche Airport) is 245 miles / 394 kilometers / 213 nautical miles. Estimated flight time is 57 minutes.
Driving distance from Lleida (ILD) to Alicante (ALC) is 331 miles / 533 kilometers and travel time by car is about 5 hours 46 minutes.
## Map of flight path and driving directions from Lleida to Alicante.
Shortest flight path between Lleida–Alguaire Airport (ILD) and Alicante–Elche Airport (ALC).
## How far is Alicante from Lleida?
There are several ways to calculate distances between Lleida and Alicante. Here are two common methods:
Vincenty's formula (applied above)
• 244.718 miles
• 393.835 kilometers
• 212.654 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 245.017 miles
• 394.317 kilometers
• 212.914 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Lleida–Alguaire Airport
City: Lleida
Country: Spain
IATA Code: ILD
ICAO Code: LEDA
Coordinates: 41°43′41″N, 0°32′6″E
B Alicante–Elche Airport
City: Alicante
Country: Spain
IATA Code: ALC
ICAO Code: LEAL
Coordinates: 38°16′55″N, 0°33′29″W
## Time difference and current local times
There is no time difference between Lleida and Alicante.
CET
CET
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 61 kg (135 pounds).
## Frequent Flyer Miles Calculator
Lleida (ILD) → Alicante (ALC).
Distance:
245
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
245
Round trip? | 501 | 1,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.783764 |
https://jeeneetqna.in/752/the-equivalent-resistance-between-and-for-mesh-shown-figure | 1,696,021,137,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510528.86/warc/CC-MAIN-20230929190403-20230929220403-00617.warc.gz | 360,208,471 | 9,824 | # The equivalent resistance between A and B for the mesh shown in the figure is
more_vert
The equivalent resistance between A and B for the mesh shown in the figure is
(1) 4.8 $\Omega$
(2) 7.2 $\Omega$
(3) 16 $\Omega$
(4) 30 $\Omega$
more_vert
verified
Ans: (3) 16 $\Omega$
Sol.
RAB = 16 $\Omega$
more_vert | 101 | 318 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-40 | latest | en | 0.716207 |
http://de.metamath.org/mpeuni/frgrancvvdeqlem6.html | 1,718,653,190,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861737.17/warc/CC-MAIN-20240617184943-20240617214943-00541.warc.gz | 8,194,844 | 6,209 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > frgrancvvdeqlem6 Structured version Visualization version GIF version
Theorem frgrancvvdeqlem6 26562
Description: Lemma 6 for frgrancvvdeq 26569. The mapping of neighbors to neighbors applied on a vertex is the intersection of the corresponding neighborhoods. (Contributed by Alexander van der Vekens, 23-Dec-2017.)
Hypotheses
Ref Expression
frgrancvvdeq.nx 𝐷 = (⟨𝑉, 𝐸⟩ Neighbors 𝑋)
frgrancvvdeq.ny 𝑁 = (⟨𝑉, 𝐸⟩ Neighbors 𝑌)
frgrancvvdeq.x (𝜑𝑋𝑉)
frgrancvvdeq.y (𝜑𝑌𝑉)
frgrancvvdeq.ne (𝜑𝑋𝑌)
frgrancvvdeq.xy (𝜑𝑌𝐷)
frgrancvvdeq.f (𝜑𝑉 FriendGrph 𝐸)
frgrancvvdeq.a 𝐴 = (𝑥𝐷 ↦ (𝑦𝑁 {𝑥, 𝑦} ∈ ran 𝐸))
Assertion
Ref Expression
frgrancvvdeqlem6 ((𝜑𝑥𝐷) → {(𝐴𝑥)} = ((⟨𝑉, 𝐸⟩ Neighbors 𝑥) ∩ 𝑁))
Distinct variable groups: 𝑦,𝐷,𝑥 𝑥,𝑉,𝑦 𝑥,𝐸,𝑦 𝑦,𝑌 𝜑,𝑦 𝑦,𝑁 𝑥,𝐷 𝑥,𝑁 𝜑,𝑥
Allowed substitution hints: 𝐴(𝑥,𝑦) 𝑋(𝑥,𝑦) 𝑌(𝑥)
Proof of Theorem frgrancvvdeqlem6
StepHypRef Expression
1 simpr 476 . . . 4 ((𝜑𝑥𝐷) → 𝑥𝐷)
2 riotaex 6515 . . . 4 (𝑦𝑁 {𝑥, 𝑦} ∈ ran 𝐸) ∈ V
3 frgrancvvdeq.a . . . . 5 𝐴 = (𝑥𝐷 ↦ (𝑦𝑁 {𝑥, 𝑦} ∈ ran 𝐸))
43fvmpt2 6200 . . . 4 ((𝑥𝐷 ∧ (𝑦𝑁 {𝑥, 𝑦} ∈ ran 𝐸) ∈ V) → (𝐴𝑥) = (𝑦𝑁 {𝑥, 𝑦} ∈ ran 𝐸))
51, 2, 4sylancl 693 . . 3 ((𝜑𝑥𝐷) → (𝐴𝑥) = (𝑦𝑁 {𝑥, 𝑦} ∈ ran 𝐸))
65sneqd 4137 . 2 ((𝜑𝑥𝐷) → {(𝐴𝑥)} = {(𝑦𝑁 {𝑥, 𝑦} ∈ ran 𝐸)})
7 frgrancvvdeq.nx . . 3 𝐷 = (⟨𝑉, 𝐸⟩ Neighbors 𝑋)
8 frgrancvvdeq.ny . . 3 𝑁 = (⟨𝑉, 𝐸⟩ Neighbors 𝑌)
9 frgrancvvdeq.x . . 3 (𝜑𝑋𝑉)
10 frgrancvvdeq.y . . 3 (𝜑𝑌𝑉)
11 frgrancvvdeq.ne . . 3 (𝜑𝑋𝑌)
12 frgrancvvdeq.xy . . 3 (𝜑𝑌𝐷)
13 frgrancvvdeq.f . . 3 (𝜑𝑉 FriendGrph 𝐸)
147, 8, 9, 10, 11, 12, 13, 3frgrancvvdeqlem4 26560 . 2 ((𝜑𝑥𝐷) → {(𝑦𝑁 {𝑥, 𝑦} ∈ ran 𝐸)} = ((⟨𝑉, 𝐸⟩ Neighbors 𝑥) ∩ 𝑁))
156, 14eqtrd 2644 1 ((𝜑𝑥𝐷) → {(𝐴𝑥)} = ((⟨𝑉, 𝐸⟩ Neighbors 𝑥) ∩ 𝑁))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 = wceq 1475 ∈ wcel 1977 ≠ wne 2780 ∉ wnel 2781 Vcvv 3173 ∩ cin 3539 {csn 4125 {cpr 4127 ⟨cop 4131 class class class wbr 4583 ↦ cmpt 4643 ran crn 5039 ‘cfv 5804 ℩crio 6510 (class class class)co 6549 Neighbors cnbgra 25946 FriendGrph cfrgra 26515 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-int 4411 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-om 6958 df-1st 7059 df-2nd 7060 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-1o 7447 df-er 7629 df-en 7842 df-dom 7843 df-sdom 7844 df-fin 7845 df-card 8648 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-nn 10898 df-2 10956 df-n0 11170 df-z 11255 df-uz 11564 df-fz 12198 df-hash 12980 df-usgra 25862 df-nbgra 25949 df-frgra 26516 This theorem is referenced by: frgrancvvdeqlem7 26563 frgrancvvdeqlemA 26564 frgrancvvdeqlemC 26566
Copyright terms: Public domain W3C validator | 2,582 | 4,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-26 | latest | en | 0.420719 |
https://fr.scribd.com/document/261374288/Chapter-1-lecture-1 | 1,566,544,844,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318011.89/warc/CC-MAIN-20190823062005-20190823084005-00067.warc.gz | 463,624,437 | 68,941 | Vous êtes sur la page 1sur 12
# Introduction to Fluid Mechanics
Chapter 1
Introduction
Pritchard
Main Topics
Definition of Fluid Mechanics
Definition of a Fluid
Basic Equations
Methods of Analysis
Dimensions and Units
Pritchard
Fluid Mechanics
What is Fluid Mechanics?
It is the study of fluids in motion or at rest
under the action of applied forces
Fluid Mechanics
is divided into three areas:
Fluid statics:
Study of fluids
at rest
Fluid kinematics:
Study of fluids
in motion
Fluid dynamics:
Study of effect of
forces on fluid motion
Pritchard
Fluid Mechanics
What are the approaches for solving fluid mechanics problems?
There are three approaches:
Analytical
Experimental and
Computational
Based on these, fluid mechanics is sometimes classified into:
Analytical Fluid Dynamics
Experimental Fluid Dynamics and
Computational Fluid Dynamics or CFD
Pritchard
Definition of a Fluid
When a shear stress is applied:
Fluids continuously deform
Solids deform or bend up to the elastic limit
Pritchard
Definition of a Fluid
Fluid is a substance that deforms continuously under the
application of a shear stress
## Fluids comprise liquids (e.g. water, alcohol, oil, mercury)
and gases (air, oxygen, helium)
## From fluid mechanics point of view, all matter consists
of only two states. What are they?
Note that a solid also deforms when a shear stress is applied,
but its deformation is not continuous
Pritchard
Basic Equations
Conservation of mass
The mass of a system is constant
Net force F ma
## The principle of angular momentum
The net torque equals the rate of change of angular momentum of the
system
## The first law of thermodynamics
The change in stored energy of a system equals the net energy added to
the system as heat and work
## The second law of thermodynamics
The entropy of any isolated system cannot decrease during any
## process between equilibrium states
Pritchard
Basic Equations
Not all the basic laws are always required to solve any one problem
In some problems, it is necessary to employ additional relations in
the form of equations of state or constitutive equations
o
## Appropriate boundary and initial conditions (e.g. no-slip) must also be
satisfied
Pritchard
Methods of Analysis
System and Control Volume
## The system is separated from its surrounding by its boundaries
Mass cannot pass through a system boundary
Everything external to this system is termed: surroundings
System boundaries can move and the shape of the boundary can deform
Pritchard
Methods of Analysis
System and Control Volume
Control Volume (CV):
An arbitrary volume in space
with a boundary
A boundary is called a control surface (CS)
CV and CS can move and deform
Mass can pass through a CV
Pritchard
Methods of Analysis
Differential or Integral Approach
Differential Approach
Considers infinitesimal elements of a system or a control volume
Provides differential equations defining the flow
Solutions provide detailed picture of the flow
## Differential approach is discussed in Chapter 5
Integral Approach
Applies basic laws to a finite system or control volume
Provides information on gross behavior of a fluid flow
o E.g., overall lift of a airplane wing
Integral approach is treated in Chapter 4
Pritchard
Methods of Analysis
Lagrangian or Eulerian System
Lagrangian System (after J. L. Lagrange)
Individual fluid particle is tracked as it moves through the flow
Tracking is based on particles position and time
It is not widely used in fluid mechanics. Why?
Because it is very difficult to track individual particles since fluids consist of
so many particles
## Eulerian System (after L. Euler)
Gives the details of the entire flow field at a given position as a function of time
At a given point in the flow, new particles are arriving all the time
Difficult to apply Newtons second law. Why?
Because his 2nd law applies only to particles of fixed mass but in this
case the mass may vary: F = ma
Pritchard | 866 | 3,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-35 | latest | en | 0.884651 |
https://www.calculatoratoz.com/en/tensile-stress-in-the-shaft-when-it-is-subjected-to-axial-tensile-force-calculator/Calc-13978 | 1,624,367,283,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488517820.68/warc/CC-MAIN-20210622124548-20210622154548-00526.warc.gz | 621,041,871 | 35,022 | 🔍
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Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force Solution
STEP 0: Pre-Calculation Summary
Formula Used
tensile_stress = 4*Axial Force/pi*Diameter of shaft
σ = 4*P/pi*d
This formula uses 1 Constants, 2 Variables
Constants Used
pi - Archimedes' constant Value Taken As 3.14159265358979323846264338327950288
Variables Used
Axial Force - The Axial Force value/ axial force (Measured in Newton)
Diameter of shaft - The Diameter of shaft is defined as the diameter of the hole in the iron laminations that contains the shaft. (Measured in Centimeter)
STEP 1: Convert Input(s) to Base Unit
Axial Force: 10 Newton --> 10 Newton No Conversion Required
Diameter of shaft: 10 Centimeter --> 0.1 Meter (Check conversion here)
STEP 2: Evaluate Formula
Substituting Input Values in Formula
σ = 4*P/pi*d --> 4*10/pi*0.1
Evaluating ... ...
σ = 1.27323954473516
STEP 3: Convert Result to Output's Unit
1.27323954473516 Pascal -->1.27323954473516 Newton per Square Meter (Check conversion here)
1.27323954473516 Newton per Square Meter <-- Tensile Stress
(Calculation completed in 00.031 seconds)
< 10+ Desgin of Shafts Calculators
Diameter of Shaft When Torsional Shear Stress in a Shaft is Given(Pure Torsion)
diameter_of_shaft = (16*Torsional Moment/pi*Torsional Shear Stress)^(1/3) Go
Torsional Moment When Torsional Shear Stress in a Shaft is Given(Pure Torsion)
torsional_moment = Torsional Shear Stress*pi*(Diameter of shaft^3)/16 Go
Torsional Shear Stress in a Shaft(Pure Torsion)
torsional_shear_stress = 16*Torsional Moment/(pi*Diameter of shaft^3) Go
Diameter of the Shaft When Bending Stress is Given(Pure Bending)
diameter_of_shaft = (32*Bending moment/pi*Bending Stress)^(1/3) Go
Bending Moment When Bending Stress is Given(Pure Bending)
bending_moment = (Bending Stress*pi*Diameter of shaft^3)/32 Go
Bending Stress in the Shaft (Pure Bending Moment)
bending_stress = 32*Bending moment/pi*Diameter of shaft^3 Go
Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force
tensile_stress = 4*Axial Force/pi*Diameter of shaft Go
Diameter of Shaft When Tensile Stress in the Shaft is Given
diameter_of_shaft = 4*Axial Force/pi*Tensile Stress Go
Axial Force When Tensile Stress in the Shaft is Given
axial_force = Tensile Stress*pi*Diameter of shaft/4 Go
Normal Stress When Both Bending and Torsional act on the Shaft
normal_stress = Bending Stress+Tensile Stress Go
Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force Formula
tensile_stress = 4*Axial Force/pi*Diameter of shaft
σ = 4*P/pi*d
Define Tensile Stress?
Tensile stress can be defined as the magnitude of force applied along an elastic rod, which is divided by the cross-sectional area of the rod in a direction perpendicular to the applied force. Tensile means the material is under tension and that there are forces acting on it trying to stretch the material.
How to Calculate Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force?
Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force calculator uses tensile_stress = 4*Axial Force/pi*Diameter of shaft to calculate the Tensile Stress, The Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force formula is defined as the magnitude of force applied along an elastic rod, which is divided by the cross-sectional area of the rod in a direction perpendicular to the applied force. . Tensile Stress and is denoted by σ symbol.
How to calculate Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force using this online calculator? To use this online calculator for Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force, enter Axial Force (P) and Diameter of shaft (d) and hit the calculate button. Here is how the Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force calculation can be explained with given input values -> 1.27324 = 4*10/pi*0.1.
FAQ
What is Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force?
The Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force formula is defined as the magnitude of force applied along an elastic rod, which is divided by the cross-sectional area of the rod in a direction perpendicular to the applied force. and is represented as σ = 4*P/pi*d or tensile_stress = 4*Axial Force/pi*Diameter of shaft. The Axial Force value/ axial force and The Diameter of shaft is defined as the diameter of the hole in the iron laminations that contains the shaft.
How to calculate Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force?
The Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force formula is defined as the magnitude of force applied along an elastic rod, which is divided by the cross-sectional area of the rod in a direction perpendicular to the applied force. is calculated using tensile_stress = 4*Axial Force/pi*Diameter of shaft. To calculate Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force, you need Axial Force (P) and Diameter of shaft (d). With our tool, you need to enter the respective value for Axial Force and Diameter of shaft and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Tensile Stress?
In this formula, Tensile Stress uses Axial Force and Diameter of shaft. We can use 10 other way(s) to calculate the same, which is/are as follows -
• tensile_stress = 4*Axial Force/pi*Diameter of shaft
• diameter_of_shaft = 4*Axial Force/pi*Tensile Stress
• axial_force = Tensile Stress*pi*Diameter of shaft/4
• bending_stress = 32*Bending moment/pi*Diameter of shaft^3
• bending_moment = (Bending Stress*pi*Diameter of shaft^3)/32
• diameter_of_shaft = (32*Bending moment/pi*Bending Stress)^(1/3)
• torsional_shear_stress = 16*Torsional Moment/(pi*Diameter of shaft^3)
• torsional_moment = Torsional Shear Stress*pi*(Diameter of shaft^3)/16
• diameter_of_shaft = (16*Torsional Moment/pi*Torsional Shear Stress)^(1/3)
• normal_stress = Bending Stress+Tensile Stress
Where is the Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force calculator used?
Among many, Tensile Stress in the Shaft When It is Subjected to Axial Tensile Force calculator is widely used in real life applications like {FormulaUses}. Here are few more real life examples -
{FormulaExamplesList}
Let Others Know | 1,769 | 6,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2021-25 | latest | en | 0.674286 |
http://lists.extropy.org/pipermail/extropy-chat/2007-June/036523.html | 1,696,177,258,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510903.85/warc/CC-MAIN-20231001141548-20231001171548-00432.warc.gz | 24,544,236 | 3,920 | # [ExI] Happy Solstice!
spike spike66 at comcast.net
Sat Jun 23 17:52:45 UTC 2007
```Cool Lee! Since you take into account the latitude and longitude, your next
digit of precision on your yardstick experiment comes from taking into
account the analemma. This website gives a reasonable explanation:
http://www.analemma.com/Pages/framesPage.html
The analemma is the result of the fact that the earth isn't in a circular
orbit around the sun, but rather an ellipse. This time of year when we are
at the aphelion, or farthest point from the sun, the apparent traverse of
the sun across the ecliptic is slightly less than than the usual ~ degree
per day. Mid winter it will traverse slightly more than the average.
By happy coincidence, the aphelion and perihelion almost correspond with the
soltices.
I am setting up an experiment to mark the pavement on my back patio
corresponding to the shadow of the peak of the house at exactly noon. Of
course most days at noon I would not be home, so it will take years to get
most of the calendar days marked. When I do, I will have a figure 8 shaped
calendar back there. Isaac will love it. Is this cool or what?
> -----Original Message-----
> From: extropy-chat-bounces at lists.extropy.org [mailto:extropy-chat-
> bounces at lists.extropy.org] On Behalf Of Lee Corbin
> Sent: Saturday, June 23, 2007 10:32 AM
> To: ExI chat list
> Subject: Re: [ExI] Happy Solstice!
>
> Happy Solstice to you too, Spike.
>
> I performed the appropriate ritual in the parking lot at work with the
> help of an assistant. We used a yardstick as plumb line, waited until
> 1pm, and measured the length of the shadow ('twas 9 inches).
>
> (Oddly, I noted that our shadows didn't seem to point exactly north
> until about 7 minutes after 1pm---1 pm, of course, because of the
> dratted daylight savings time.)
>
> So the angle generated by the sun and yardstick was arctan of 1/4,
> which is almost exactly 14 degrees. So (drawing a sanity-check
> diagram and) adding to the 23.5 degree axial tilt of the Earth, we
> got 37.5 for our latitude. Very nice for Santa Clara, California,
> no? (Google Earth gave 37.38 or something. Had to be pretty
> lucky given how much the wind was blowing the yardstick around.)
>
> But it got even better. My assistant pointed out that we were
> somewhere in the middle of the time zone, so substituting +8 hours
> from UCT (GMT in London) was only an approximation. Google
> suggested that the *longitude* of our parking lot is 121.99 W.
> My assistant's idea was
> that if we were, say, a bit to the west of the middle of our time zone,
> then we should have to wait a bit for the sun to get exactly overhead.
> Lo and behold once more! Two degrees (1.99, that is) yields eight
> minutes because the Earth turns one degree every four minutes (an
> hour is 1/24 of 360, or 15 degrees, so 60 minutes = 15 degrees is
> of course four minutes.)
>
> Delighted was I, since I had noted that we seemed to have to wait
> seven minutes for the sun to get the yardstick's shadow lined up right!
>
> You can do this any day this summer! See how many minutes before
> or after the hour you have to wait for a shadow to point exactly
> John Harrison's quest all the more poignant.
>
> Every time that ritual measurements like this are performed, the claims
> of the flat-Earthers are weakend a little more. :-)
>
> Lee
>
>
> ----- Original Message -----
> From: "spike" <spike66 at comcast.net>
> To: "'ExI chat list'" <extropy-chat at lists.extropy.org>
> Sent: Friday, June 22, 2007 7:56 AM
> Subject: Re: [ExI] Happy Solstice!
>
>
> >
> > Growing seasons would lag behind the solar seasons, which is the most
> > critical schedule to most societies. Those smart Celts, being astronomy
> > minded, would perhaps be more likely ignore the air temperature and note
> the
> > celestial cues.
> >
> > spike
> >
> >> -----Original Message-----
> >> From: extropy-chat-bounces at lists.extropy.org [mailto:extropy-chat-
> >> bounces at lists.extropy.org] On Behalf Of Alex Ramonsky
> >> Sent: Friday, June 22, 2007 2:53 AM
> >> To: ExI chat list
> >> Subject: Re: [ExI] Happy Solstice!
> >>
> >> Happy Solstice Amara : )
> >> ...Something that I have wondered for many years...Does anyone know why
> >> the Midsummer Solstice/Midwinter Solstice are called so, when (at least
> >> in the UK) they're considered to be the _beginning_ of summer/winter?
> >> The Celts treated them as the middle of the seasons...Is this one of
> >> those eccentric British things or is there genuine worldwide confusion?
> >> Best,
> >> AR
> >> **********
> >>
> >> Amara Graps wrote:
> >>
> >> >Happy June Solstice [1] to you Northerners and Southerners
> (hemispheres,
> >> >that is)!! Celebration time!
> >> >
> >> >Midsummer Night [2]
> >> >by Zinta Aistars
> >> >
> >> >One night each year, that longest night
> >> >be
> >> >
> >>
> >>
> >>
> >>
> >> _______________________________________________
> >> extropy-chat mailing list
> >> extropy-chat at lists.extropy.org
> >> http://lists.extropy.org/mailman/listinfo.cgi/extropy-chat
> >
> >
> > _______________________________________________
> > extropy-chat mailing list
> > extropy-chat at lists.extropy.org
> > http://lists.extropy.org/mailman/listinfo.cgi/extropy-chat
> >
> _______________________________________________
> extropy-chat mailing list
> extropy-chat at lists.extropy.org
> http://lists.extropy.org/mailman/listinfo.cgi/extropy-chat
``` | 1,475 | 5,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.908492 |
http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blob_plain;f=topics/_week5_system_F.mdwn;hb=57f69d438433032f3c302607a91a20a45617c686 | 1,610,754,890,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-04/segments/1610703497681.4/warc/CC-MAIN-20210115224908-20210116014908-00670.warc.gz | 60,937,661 | 1,131 | # System F and recursive types In the simply-typed lambda calculus, we write types like σ -> τ. This looks like logical implication. We'll take that resemblance seriously when we discuss the Curry-Howard correspondence. In the meantime, note that types respect modus ponens:
Expression Type Implication
----------------------------------
fn α -> β α ⊃ β
arg α α
------ ------ --------
fn arg β β
The implication in the right-hand column is modus ponens, of course. System F is usually attributed to Girard, but was independently proposed around the same time by Reynolds. It enhances the simply-typed lambda calculus with quantification over types. In System F, you can say things like Λ α (\x.x):(α -> α) This says that the identity function maps arguments of type α to results of type α, for any choice of α. So the Λ is a universal quantifier over types. | 198 | 939 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-04 | latest | en | 0.83772 |
https://thermodynamics-engineer.com/2013/06/27/nozzle-and-diffuser-exercise1/ | 1,685,590,005,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647525.11/warc/CC-MAIN-20230601010402-20230601040402-00115.warc.gz | 641,154,502 | 23,111 | # Nozzle and Diffuser (Exercise1)
Air flows through an adiabatic and horizontal placed pipe with changeable cross section. At the inlet side air flows with velocity cin=10 m/s under the pressure pin=1bar and the temperature ϑin=25°C. At the outlet side air flows out of the pipe with velocity cout=150 m/s under the pressure pin=0.8bar.
1) Determine whether this pipe is nozzle or diffuser without calculating the ratio of cross section of outlet and inlet Aout/Ain
2) Determine now the ratio of cross section of outlet and inlet Aout/Ain
Air can be considered as perfect gas with cp=1 J/K∙g and R=0.29 J/K∙g
Solution:
1)
The 1.Law of thermodynamics:Stationary → d/dτ=0
Without input and output (work) → ΣW=0
Horizontal placed → no change in potential energy → Δ(g·z)=0
Then we can obtain the simplified 1.Law: wobei hin=cp·Tin und hout=cp·Tout
Therefore the temperature of air at the outlet side is:
We now calculate the entropy change during this process:
Hence it is Nozzle.
2)
In order to determine Aout/Ain, we need additional the law of conservation of mass and the ideal gas law:
Ain>Aout , it is also proved that this pipe is a nozzle. | 311 | 1,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2023-23 | latest | en | 0.821293 |
https://rdrr.io/github/AEBilgrau/Bmisc/man/MultivariateSpecial.html | 1,550,297,868,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479885.8/warc/CC-MAIN-20190216045013-20190216071013-00107.warc.gz | 673,015,740 | 13,864 | # MultivariateSpecial: Multivariate special functions In AEBilgrau/Bmisc: Biostatistical miscellaneous functions
## Description
Special matematical functions related to the multivariate gamma function which appears in the Wishart and Inverse Wishart distributions. The interface and usage should mirror the Special mathematical functions from the base package.
## Usage
``` 1 2 3 4 5 6 7 8 9 10 11``` ```gammap(x, p = 1) digammap(x, p = 1) Dgammap(x, p = 1) psigammap(x, p = 1, deriv = 0) trigammap(x, p = 1, deriv = 0) lgammap(x, p = 1L) ```
## Arguments
`x` A numeric vector of values. The values of `x` should be strictly greater than `(p - 1)/2`. `p` A (single) integer giving the dimension. Default is `1`. `deriv` A (single) integer giving the order of the derivative. Default is `0`.
## Details
Warnings are thown when the functions are evaluated outside its domain. Run `Bmisc:::plotMultivariateSpecial()` for more info.
## Value
Returns a numeric vector with the same length as `x`. Similar to `gamma` and related functions `NaN`s are returned with warning when the function is evaluated outside its domain
`gammap` is the multivariate gamma function.
`digammap` is the multivariate digamma function.
`Dgammap` is the derivative of the multivariate gamma function.
`psigammap` is the n'th derivative log multivariate gamma function (i.e. multivariate polygamma).
`trigammap` is the 2'nd derivative log multivariate gamma function (i.e. the multivariate trigamma).
`lgammap` is the log multivariate gamma function.
## Note
For `lgammp`, warnings of the type ```"value out of range in 'lgamma'"``` is due to evaluation in the half integers below (p-1)/2.
## Author(s)
Anders Ellern Bilgrau <anders.ellern.bilgrau (at) gmail.com>
```1 2 3 4 5 6 7 8``` ```gammap(3.2, p = 1) gammap(c(0.2, 0.5, 0.9, 1.2, 1.5), p = 3) # x < (p-1)/2 ! lgammap(3.1, p = 2) digammap(2, p = 3) Dgammap(3.2, p = 1) # A plot of all special functions Bmisc:::plotMultivariateSpecial() ``` | 605 | 1,992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.677083 |
https://alison.com/topic/learn/117980/extreme-pathways-and-elementary-modes | 1,660,185,624,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571232.43/warc/CC-MAIN-20220811012302-20220811042302-00406.warc.gz | 112,407,592 | 43,361 | Alison's New App is now available on iOS and Android!
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Metabolic Engineering Prof. Amit Ghosh School of Energy Science and Engineering Indian Institute of Technology - Kharagpur Lecture - 23 Extreme Pathways and Elementary Modes Welcome to metabolic engineering course, today we are going to learn a new topic that is extreme pathways and elementary modes. These are 2 important things in the metabolic network where you can get the flux distribution because of the elementary pathway, extreme pathway and the elementary mode present in the network. (Refer Slide Time: 00:48)
So to start with, we will cover some of the concept today like linear basis for null space, the null space we already know S dot v = 0. So we want to calculate the linear basis for the null space and then followed by convex basis for the null space, which is basically the extreme pathway and the type of extreme pathways and then at the end we will learn about the elementary modes. (Refer Slide Time: 01:16)
So this is the constant base method which are available, you can actually install MATLAB or Python version of anaconda can be installed and where you can run this constraint based method. So far we learned about FBA flux variability analysis. Also we learn about for flux dependencies like robustness, phase planes, flux coupling. Today we are going to learn about the allowable solution among that we have the extreme pathways and elementary mode sampling we have already completed. So today we will cover these 2 topics the extreme pathways and the elementary modes to learn about the elementary modes and extreme pathways we start with the metabolic pathways. (Refer Slide Time: 02:01)
Historically, pathways include glycolysis, TCA cycle as shown in the left you can see that in early in the history of biochemistry enzymes are isolated from cell were shown to carry out
specific chemical reaction. So, in from biochemistry, you know the enzyme from the cell which actually carry out a specific biochemical reaction, it was then recognized that the product of one reaction was substrate of another. So, here you can see that the glucose and then glucose 6 phosphate and that glucose 6 phosphate become converted into fructose 6 phosphate and so on. So, each of the product become the reactant of the other reaction. And it was then recognize that the product bound reaction was subset of another and thus one could link different chemical transformation to form a series of reaction. So glycolysis is a series of reactions you can see to form a basic metabolic pathway. So this is where glycolysis is the basic metabolic pathway, TCA cycle is again another metabolic pathway which you already know which are present in most of the organism and so on. Like there are many pathways like TCA cycle, calvin cycle and then we have urea cycle. And the definition and biochemical function of such pathway have been taught to generation of life scientists in the classes previously in different courses, you have already learned about these biochemical pathways. With the advent of whole genome sequencing and the development of network reconstruction method which I have already learned, we can now piece together an entire network. So you can add these pathways into different metabolic pathway you can add to form a complete network for example, we have integrated the TCA cycle and the glycolysis together so, that they can function as a whole. And these pathways are linked reaction which can be grouped together based upon conceptual understanding of a functional role. So, on the functional basis, you have to connect them this pathway together and try to bring about a function as a whole. (Refer Slide Time: 04:39)
So, this is another example where you can the network based pathway definition. In the first one you can see here only the reaction with the advanced biochemical techniques years of research have led to the PCS characterization of the individual reactions. The individual reactions are characterized by the biochemistry in the lab for several years and as a result, the complete stoichiometry of many metabolic reactions can be characterized. So that because we have biochemistry that is why; you would not be able to make these stoichiometric matrix. But most of the reactions have been grouped into traditional pathway that is glycolysis, TCA cycle. So, in this section you can see that the individual reaction can be combined to form a pathway like glycolysis but they do not account for cofactors. So, there is no cofactor added or by product in a way that lends itself into a mathematical description with sequence. And annotated genome you can account we can take into account many metabolic reaction in organism to make a model subsequently network based pathway can be analyzed that account for a complete network. So, you can see that some of the reactions which are active which are highlighted in a black arrow and the reaction whose arrow are given in grey arrows are basically in active reaction. So, in a network, we can see this hormone only flux flow some of the reaction become active some of the reactions remain inactive depending on the regulation it has. So, the network was pathways where you can actually simulate and see which are the pathways which are the
functional state of the network? And the functional changes from one condition to the other condition that we have, we can actually verify using the flux calculation. (Refer Slide Time: 06:34)
So, then in this class we will learn about linear bases of a null space, the linear bases of a null space, the null space contain all steady state solution, flux solution and thus all balanced to use of the network. So, this null space is basically all steady state solution that is S dot v = 0. In the last class was also I told that, we took a rectangular box and then within the box we have a plane which cuts the box and that is the intersection of the plane and the box is basically this steady state solution space. And this steady state solution flux solution you can actually calculate by estimating the linear basis set. The linear basis set or the vector which are present which are actually spanning the steady state solution space. So, today we are going to see how we can calculate the linear basis set of a null space. (Refer Slide Time: 07:30)
So, the null space of the stoichiometric matrix for any matrix you can calculate the null space yesterday I told you that the null space you can calculate for stoichiometric matrix as well. So, the null space of S is defined as S dot v = 0 or V ss is basically the steady state flux. So, when S dot v = 0 then the solution for this equation you will get the flux for only steady state thus all the steady state flux distribution are found in the null space. So, whenever you calculate the null space there that is where we will get only the steady state flux distribution then the null space has a dimension of n, and the null space is spanned by n - r basis set the set of basis vectors are represented of a matrix that satisfy a SR is SR = 0. So, the null space the null space is actually have a spanned by n - r basis vector. So, R is the rank of the matrix and then the basis vector for that satisfy SR = 0. So, if S is a stoichiometric matrix and R basis vector which is a column vector, and then if you multiply what you get is basically that is equal to 0 and then you solve this equation. So, you only solve this equation you get it gets steady state fluxes, which is actually a linear combination of r that is the basis vector. So, ultimately if you know the basis vector that is r, then you would be able to get these steady state fluxes as a set of linear basis vector is not unique. But once the set is chosen, and then when you chose the weights, that is W i then it become unique the steady state flux become unique. So, the null space has a dimension of n - r, r is missing. So, n is the number of reaction where n is number of reaction and r is the rank of the
matrix. So, given a matrix of m into n dimension, you can calculate the basis vector and this vector is easy to calculate. You can use MATLAB to actually calculate the basis vector of a matrix any matrix can be decomposed into a basis vector and the dimension and the null space has a dimension of n - r, where r is the rank of the matrix. And the basis set is actually if there are r i is the number of basis set where if you multiply it with a weight like non negative number W i then you get this steady state solutions. So, the weight of a particular v ss is actually unique. Once you choose the weight then your steady state fluxes are unique. (Refer Slide Time: 10:42)
So, as I told the basis for the null space can be found in Mathematica and MATLAB without with one command. So, in one command, you just have null, within bracket S that is the matrix we want to calculate the basis vector, immediately MATLAB will give the basis vector. And also you can now use the command null for transpose of S. This way you can calculate the basis vector for any matrix and those basis vectors, the linear combination of your basis vector is basically the steady state fluxes. (Refer Slide Time: 11:18)
So, let us take an example so basis vector for the null of S so how do you calculate for this small network reaction network where we have 6 reactions. So 6 reactions you can see v 1, v 2, v 3, v 4, v 5, v 6, and we have 4 metabolites. So, let us calculate the basis vector for this small network. How do you calculate? So, it is very straightforward like you do S dot v = 0. So you have the S matrix here and the flux v if you do S dot v, what do we get? We will get 4 equations because there are 4 metabolites. So, we have 4 metabolites, that is why you have 4 equations. Now, you solve this equation and you get the flux out of it, we in the very well, how do you do that? So, it is very, by handy you can do just you write the flux vector v 1, v 2, v 3, v 4, v 5, v 6 then you write v 1 in terms of v 4 v 6, because then if that is v 1 = v 1 the first, if you do the matrix multiplication, what you will get v 1 = v 1, v 2 = v 4 - v 6 that is nothing but this one v 2 = v 4 - v 6. And then we can v 3 = v 4 - v 6. So v 3 = v 4 minus this equation, and then we have v 2 = v 4 - v 6, v 2 equal to and then we have the v 4 = v 4, v 5 = v 6, v 6 = v 6. So, v 5 = v 6 is basically nothing but this one and v 2 = v 3 is basically so if I know v 2 = v 3, that is why we have v 4 - v 6 before that can also be further decomposed into in a vector. So v 4 is nothing but 1, 1, 1, 1, 1, 0, 0, 0. Because we have 1, the coefficient of v 4 is 1 here before is 1 here, v 4 is 1 here, 1 here, and this is 0, 0, 0 got it so, v 4 coefficients is only 1, 1, 1, 1 but here it is v 6 v 6, that is why you have a coefficient of 0, 0. And similarly, for v 6, we have coefficient over here is 0, here it is
minus 1 here it is minus 1, here it is 0, and v 6. So, the coefficient over v 4 is 0 over here and the coefficient or v 6 is minus 1 here, here it is also minus 1 here v 4 is 0, v 6 is 1 and v 6 is 1. So, you have that this linear set of equation can be written as v = v 4 multiplied by a constant non negative value b 1 + v 6 into b 2. So now we can see that the your this is vector v 4 and v 6 when these 2 we have the basis vector. So that is basically b 1 so, b 1 is basically this one, and b 2 is this one. So b 1 and b 2 is basically your basis vector and v 4 and v 6 and can be any negative constant. So, it is become a linear combination so, your flux distribution v is basically nothing but the linear combination of the basis vector v 1 and v 2 multiplied by the constant value v 4 and v 6. So, if you draw these basis vectors on the network, we will see that these basis vectors the first basis vector b 1 that is 1, 1, 1, 1, 1 that is v 1, v 2, v 3, v 4 are actually carrying flux so, this which is nothing but this pathway. So, this is v 1 and this is v 2 this is v 3 and this is v 4. So, your first equation is basically 1, 1, 1, 1, 1, 1, 2 v 1, v 2, v 3, v 4, v 1, v 2, v 3, v 4 are actually carrying flux and the other b 2 is basically v 2, v 3 are actually v 5, v 6 are actually negative and it is going against the reaction. So, you are in the other basis vector which is v 5 and v 6, are positive so, these are positive these are positive. So, v 2, v 5, v 6 are positive, so, it is going in this direction this way. So, these basis vectors basically we have 2 basis vector for this network and these basis vectors are actually encompassing the entire flux solution. The solution space is actually can expand by these 2 basis vectors and these basis vectors you can immediately calculate using a MATLAB given a matrix you can and just with one command you can calculate the basis vector. Now, you can see the basis vector 2 is wrong, why it is wrong? Basis vector 1 is fine, because the reactions flux is along the direction of the basis vector, but if you look into basis vector b 2, it is going against the reaction so, is going in the opposite direction. So, for reaction v 3 and v 2 basically, we have the reaction which is actually going against the reaction. So, the basis vector v 2 is kind of not satisfied. (Refer Slide Time: 17:37)
So, how do you correct that? So, this is what do we have the particular solution where you have the basis vector 1 basis vector 2 and we multiplied by a constant weight like 2 and 1 then it becomes a particular solution. Similarly, if you see the flux distribution can be decomposed into 2 pathways, first this is pathway 1 and this is pathway 2 and then you have a weightage. So, if you put weightage 2 what do you have and the flux 2 which is entering from here which is distributed into 1, 1, 1 and then again it combined 1, 1 and become 2 so, it is fine. So, how do you change the basis set? They will be changing the basis it is also possible where you can multiply by a small 2 by 2 matrix. So, if you multiply it by a transformation matrix basis transformation matrix that is 1, 1, 0, 1 that is shown over here if you multiplied with the with the basis vector. So, this is the original basis vector and this is the transformation basis vector. So, if you multiply these 2 you get a new basis vector where the basis vector first basis vector remain the same only the second basis vector has changed where you can see that the negative sign the basis vector p 2 is basically shown over here. So, if you plot it what you will see, this is basis this is the first basis vector p 1 and this is the p 2. So, here you can see that, well the v 1 is carrying flux and then we then v 4 and v 5 carrying flux and v 4 and v 5 and then also and this is also very clear. So, this is the other basis with p 2. So, now, the basis vectors are correct because they are along the direction of the reaction.
This is none of the pathway is against the reactions. So, this is how you correct by using a basis transformation matrix, this one. (Refer Slide Time: 19:45)
So, now we will talk about the convex basis for the null space that is extreme pathway. So now we will come to the extreme but there is some idea you have got about the basis vector. How do you can calculate the basis vector for a matrix and then how basis vector looks in terms of pathway and now what is extreme pathway based on null space? Once you have the null space you will be able to calculate the extreme pathways that is EM extreme EP. (Refer Slide Time: 20:16)
So, extreme pathways is very important for that you have to do convex analysis what is convex analysis that this study of a system of linear equation and inequality the convex analysis actually deal with linear equation and also inequality. The converse analysis is used to study metabolic network while the linear equations are derived from the mass balance
involve linear equation and that you get from the mass balance equation that is S dot v = 0. The inequality are generated from thermodynamic information or the reversibility of the reaction that is a boundary condition that is, v greater than 0. From linear algebra null spaces defined which contains all the solution space to the set of linear homogeneous equation. So, once you have defined null space, then you will see that all your solution actually lie in this linear homogeneous equation. When we add inequality constant such as all variables must be positive the solutions become restricted to positive quadrant as we remain in the positive quadrant. Because you were I may because all the variables must be positive when you add inequalities and constantly make sure that all the variables are positive. So, the solution becomes restricted to positive quadrant. So, this is what we have the convex space when you apply the constraint that is the inequalities from thermodynamics and also apply the mass balance equations that is S dot v = 0 then you see becomes a convex cone and all your inequalities are positive that is why it remained in the positive quadrant. (Refer Slide Time: 21:58)
So, what is convexity? The definition of convex space is for every 2 point in this space the line connecting these 2 points should lie entirely within this space that it is solution space you are defining the convex space. So, any 2 point in the convex space if you connect it by a line and then if it lies within the space then it becomes then easily become the it will entirely lie within the space for example, the convex shapes are given over here and these are non convex space.
So, now we are in the right hand side. So, any 2 point if you draw in the convex space, you will see that if you connect this line, then you will lies within this sphere also in the cylinder also or in the rectangle as well you can see that any 2 point within this space if you connect it then it lies within that solution space. But if you non convex whereas, if you draw a line over here and you try to connect it, it is going outside the convex space, so, it is not actually not a non convex space it is a convex one. (Refer Slide Time: 23:00)
So, linear spaces and convex spaces, we have defined over here they distributed by both distribute by linear equation both have linear equation but in convex space we have the inequalities that is the difference between the linear space. The vector space defined by a set of linearly independent basis vector b. The vector space the in linear spaces are defined by the basis vector which I have already defined b i. Whereas, in convex volume polyhedral cone, you define a generating basis vector v which is basically an alpha p i linear combination of the basis vector, but your alpha is always positive see here that is the difference. So, in the linear space the w i goes from minus infinity to plus infinity, but whereas in convex where we have 0 to infinity that is the restrain you are the inequality you have a convex space where everything is positive. So, every point in the vector space is uniquely described by a linear combination of the basis vector whereas, in convex, the every point in the vector space is described by a non negative linear combination of the generating vector. So, this alpha is basically always positive and the
number of genetic vector may exceed the null space through these p i it can it can exceed the dimension of the null space and we have a unique set of generating vector. So, the number of basis vector is actually equal to the dimension of the null space in case of linear space and there are in finite number of bases that can be used to span the space whereas, there are a unique set of generating vector to represent the convex space. So, this is a different way in the convex space and the linear space. (Refer Slide Time: 24:49)
So, now the convex space generating convex generating vector must lie at the edge to avoid the use of negative weight. So, we can this is one of the outcome that you get from the convex space that the vector always lies at the edge. So, easy lies at the edge of the solution space and biochemically, this steady state solid solution can be represented by non negative linear combination of the convex steady state vector, which I have already told you. So, the state fluxes are always can be represented by a non negative linear combination of the convex basis vector. So p i is the basis vector alpha i is a non negative linear non negative parameter constant value that you multiply to get the steady state fluxes. (Refer Slide Time: 25:47)
So, the extreme pathway you can capture the phenotypic potential of the metabolic network. How do you capture the phenotypic potential? We will start with the network and then you create this stoichiometric matrix and that is the reactions are column and metabolites are rows and then you construct the metabolic analysis that is in calculating the basis vector. So, once you calculate the basis vector this is one basis vector another vector which actually enclosing the entire solution space. So, how many basis vector you have depend only and then based on that you can draw the solution space in this space. So, the extreme pathways have following characteristics as well, these are the following criteria these are the extreme paths which are given by the basis is vector. So, what are the properties for the extreme pathway they are unique and minimal set of basis vectors. So, they are unique and they have a minimum set of basis vector. So, all possible phenotypes can be represented by non negative linear combination of the elementary pathway was basically a linear combination, which I have already told all possible phenotype V ss is basically sigma of alpha i p i. So, this is the steady state flux distribution, this is the steady state flux distribution, these are basically different phenotypes. So, these all phenotypes, you can calculate all possible phenotypes can be calculated from nonlinear combination of the elementary pathways. So, you just change the alpha i, you get different phenotypes, that is what elementary pathway says and then they are represented by time variant properties of the metabolic network. So, the elementary pathway is basically not going to change, if we change the
condition it is the elementary pathway remain the same and they that is why they are time invariant. And they are there a computer program ever available to calculate the extreme pathway. So, as I told as in MATLAB or Mathematica, you can easily calculate the elementary pathway. (Refer Slide Time: 27:57)
So, the metabolic genotype and phenotype which is defined within the context of convex analysis so, metabolic genotype and phenotype you can define by doing convex analysis, like whereby if you do a convex analysis, you will be able to get the basis vector p 1, p 2, p 3, p 4, p 5 which actually encompasses the entire solution. So, the convex analysis can be related to cellular biology like convex cone, which is basically capability of the metabolic network in cellular biology. And then unique convex that is the basis vector are nothing but the basis vector you get from the convex analysis is nothing but the extreme pathways which actually describe different phenotypic state of a cell and then we have a particular solution that is basically a metabolic phenotype. So, for whatever solution you get from the convex analysis is basically a metabolic phenotype and the flux vector is nothing but the positive combination of the extreme pathway. So, the combination of p 1, p 2, p 3, p 4, p 5 you get the flux vector that is different phenotype of the cell can be described by extreme pathways. (Refer Slide Time: 29:15)
So, what are the different types of extreme pathways? (Refer Slide Time: 29:18)
So, there are 3 types of extreme pathways. The first one is a primary metabolic pathway and then we have the type 2 futile cycle only currency exchange fluxes are active and then type 3 we have the internal loops. So, in the type 1 you can see that the metabolite the substrate is entering the cell and then some product is coming out. So, primary type 1 metabolic pathway you can see that the exchange reactions are involved what is the difference between the main characteristic of type 1 pathway? That they have actually involves the type 1 extreme pathway involve the conversion of primary input into primary output and thus contains the exchange fluxes with the environment. So, these are the exchange fluxes, you can see that it is crossing the boundary and here also it is crossing the boundary. So, in both the cases we see that I say extreme is a
exchange reaction exchange fluxes and type 1 metabolic pathway is actually involving the exchange fluxes. And the type 2 extreme pathway is actually internal there is an internal exchange of metabolites such as and that is currency metabolite. So, only the currency metabolites are exchanged in a futile cycle. So, this is a cycle which is present inside the cell and then the currency metabolites are exchanged in this loop, the type 2 extreme pathway involve the internal exchange of currency metabolite that is ADP energy, NADH. And then we have type 3 pathway in type 3 extreme pathway, we have internal cycle. So, this is the internal cycle you can see we are here and then important characteristic of type 3 pathway is that it does not carry any fluxes. There is no there is 0 fluxes across all system boundaries. So, these are the extreme pathway is the type of extreme pathway. (Refer Slide Time: 31:35)
Then, we have the extreme pathways which are shown over here the type 1, type 2, type 3. In type 1 you can see the general fluxes are involved and the exchange fluxes are actually exchange fluxes are not equal to 0 in case of type 1. Whereas, in type 2 the exchange fluxes are 0 for primary exchange fluxes as per the currency metabolite exchange fluxes is not 0. Whereas in type 3, which; is basically internal cycle in the both exchange fluxes are 0, either primary or currency. And the pathway matrix P that is the matrix P is given by p 1, p 2, p 3, and so on. These are the extreme pathway that if you multiply it with S this stoichiometric matrix that is S dot P =
0 or P is the extreme pathways. So, these columns are the individual extreme pathways and the rows correspond to the fluxes. (Refer Slide Time: 32:45)
Now, we will calculate the extreme pathways for this metabolic network. So, these metabolic network what do you see is basically there are 7 internal fluxes v 1, v 2, v 3, v 4, v 5, v 6, v 7 and these are the 7 internal fluxes. And then we have 7 exchange fluxes. We have 4 exchange fluxes 1, 2, 3, 4 these are the 4 exchange fluxes. So, that is why how many metabolite we have in this network we have 5 metabolites. So, we have 11 internal flexes so total 11 internal fluxes, so, 7 + 4, 11, 7 internal fluxes and 4 exchange reaction that give rise to 11 fluxes and 5 metabolites. Now, you want to calculate this stoichiometric matrix. So, stoichiometric matrix will have a dimension of 5 by 11 matrix and the dimension of the null space of either you can calculate in a pen and paper for null space also, you can use the computer like MATLAB to calculate the null space if you calculate the null space. What you will get is basically the dimension of the null space will be 6 and the number of extreme pathway which is evaluated for these network is basically 8 how it is possible? So, we have p 1, p 2, p 3, p 4, p 5, p 6, p 7, p 8 these are the 8 extreme pathways. And out of that is 8 extreme pathways 6 are the basis vectors. So, the basis vector p 1, p 2, p 3, p 4, p 5, p 6 these are the basis vector which is basically p 1 is nothing but composed of reaction v 1, v 2, v 6 and p 2 is composed of v 1, v 4.
And this exchange metabolite and then we have p 3 which is composed of metabolites A, B, C and p 6 is composed of metabolites A, B, C, D, E and p 4 composed of metabolite A, B, C, D, E. So, these are the extreme pathway, which are present which you can calculate from the basis vector and those are plotted over here. And out of 6 basis vectors 7, 6 basis vectors are present in the extreme pathway, but the 2 pathway that is p 6, 7 and p 6, they are forming a futile loop so, this is a futile loop. So, actually not carrying and not involve in any particular metabolite formation, they are present in the network which is forming a loop inside the thermodynamic an invisible loop inside the network. So, mainly, the metabolite formation and flux distribution are represented by the null space basis vector that was shown over from p 1 to p 6. (Refer Slide Time: 35:55)
So, this is the extreme pathway analysis have been done in glycolysis where we have the primary metabolite and the currency metabolites. We have our primary metabolite and the currency metabolites and the type 1 pathways have been calculated for glycolysis. So, we have 2, type 1 pathway. So, what are those 2, type 1 pathway the first one is basically so, they this is the first day they were subset is entering and then subset is going out. So, this is one of the type 1 pathway and the other one is a subset is going in and then product is going out. So, this is another this is 1 this is 2 total type 1 pathway present is 2, 1 and 2, and they both are actually type 1 extreme pathways, then type 2 extreme pathways how many type 2 extreme pathways are present, there are only 1 type 2 extreme pathway. So, this is the
currency metabolite exchange you can see, so, that is why I said type 2 in pathway and type 3 extreme pathway there are 2 type 3 extreme pathway and this is 1 and this is 2. So, these are the type 3 extreme patterns are basically the futile cycle over which is shown over here and is also shown over here. So, a futile cycles are basically this one and also this one these 2 constraints of type 3 extreme pathway. This way you can decompose any metabolic network or part of the network into different types of extreme pathway and type 1, type 2, type 3. (Refer Slide Time: 37:50)
So, now we will discuss about the elementary mode and | 6,987 | 30,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.918011 |
https://code-research.eu/en/50-is-40-of-what-number.2152.html | 1,642,776,570,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303385.49/warc/CC-MAIN-20220121131830-20220121161830-00689.warc.gz | 223,449,791 | 11,407 | Pensick227
4
# 50 is 40% of what number?
$50-40\%\\ x-100\%\\ \\ 50 \cdot 100=40x\\ 40x=5000\\ x=125\\ Answer:Searched\ number\ is\ 125.$
$40\%x=50\\0,4x=50/:0,4\\x=125$ Answer: 50 is 40% of 125. I hope I helped :D | 108 | 216 | {"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} | 3.9375 | 4 | CC-MAIN-2022-05 | latest | en | 0.594322 |
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points
178
Hi,
I have a formula that looks for a specific word or variance of the word, is it possible to look for more than one word?
The formula i have is =COUNTIFS(Response1, FIND("shower", LOWER(@cell)) > 0)
This works for the word showers or shower which is perfect, however am i am able to count if the
Hi,
I have done a formula for Countifs
=COUNTIFS({July Range unders}, <0, {July Range Manager}, [A/M]1)
This works well and returns the correct data, however i now want the total sum of these entries so tried the following:
=SUMIFS({July Range unders}, <0, {July Range Manager}, [A/M]1)
But this is
Hi,
i have created a workspace which has some reports and other sheets on, this is all rolled up into a dashboard.
I have shared the workspace with colleagues, however when they click on a report it brings back a blank report with no data.
Hi, i am looking for some help please.
I have been reporting using today and looking back 30 days, however my business now wants me to report on the previous months data. so if in Mid June it looks at May's data
Is this possible to have as a formula?
Thank you
Dave
Hi,
I am working hard to get my head around formulas and have managed to get the correct return with the below formula, if greater than 14 days then return "Yes", if less than 14 days then return no. Where i am struggling is i also want to return "Yes" if the cell is blank.
current formula:
Hi,
I am looking for some help with a formula please?
I have a scheduled date which is static, Column name "1st Scheduled" i then have a completed date, Column name "Completed date"
I am trying to have the status column update to "Overdue" if no date is entered in the "Completed date" and today is
Hi,
I am trying to count the total in a range on another sheet if entries fall within a certain date range (30 days) and are a specific name "Missed Hours"
I have the following formula but cant get it to work:
=SUMIFS({Help Desk Range 1}, <=TODAY(), {Help Desk Range 1}, >=TODAY(-30), {Help Desk
Hi, i am looking for some help please.
I have a % complete column that i want to automate to 100% if a date in another column has a date in, but also return 0% if the date column is blank.
I feel this should be easy but am struggling and would appreciate any help.
Thank you as always
Hi I am trying to create a formula that count all entries by a certain person within the last 30 days. This is a cross sheet formula and seems to work well.
Now comes the however!
I have the below formula which does the calculation, however if a date is in the future it still counts it.
Hi All,
I have a sheet that has been populated via forms for the past 12 months. What i would like to do is use a count if formula but only when the date is this year.
my current formula is =COUNTIFS({Pure Gym Issue Type YTD}, "HR / Lateness or leaving early")
This works really well but counts | 711 | 2,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.94837 |
https://www.homeworkmarket.com/content/finance-question-help-0 | 1,579,459,589,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594705.17/warc/CC-MAIN-20200119180644-20200119204644-00313.warc.gz | 895,647,955 | 39,954 | # Finance Question: Help!
mmi089
A firm is all equity financed with 10,000 outstanding shares with a market value of \$20 each. Its net income was \$30,000 and it decides to pay a cash dividend of \$2000. Calculate the value of each share after the dividend payout.
a. \$22.8
b. \$20
c. \$19.8
d. not enough information
Grandma's Applesauce, Inc. has a 0.60 probability of a good year with operating cash flow of \$50,000 and 0.40 probability of a bad year with operating cash flow of \$30,000. The company has a debt of \$35,000 with 8 percent interest due next year. Assuming the company has no means of servicing its debt other than operations, and a 0% tax rate, which of the following is true?
a. Shareholders expected claim is \$12,200
b. Creditors expected claim is \$37,800
c. Creditors expected claim is \$35, 680
d. None of the above
The owners of a firm facing a high probability of bankruptcy prefer to invest in ____ projects, because ____.
a. safer; riskier projects make bankruptcy more likely
b. no new; the firm is likely to go bankrupt anyway
c. risky; the shareholders have little to lose and might win if successful
d. risky; creditors prefer taking a gamble rather than having the company default
You are trying to decide whether to accept or reject a one-year project. The project is estimated to generate \$5,000 in incremental gross profit, which includes \$200 in depreciation. Incremental SG&A expense is \$400. At a 35% tax rate, the after-tax incremental cash flow is:
a. \$2,990
b. \$3,190
c. \$3,250
d. \$3,510
What is the present value of a growing perpetuity that makes a paymet of \$100 in the first year, which thereafter grows at 3% per year? Apply a discount rate of 7%.
a. 2,000
b. 3,500
c. 2,500
d. 4,000
• Posted: 5 years ago
• Due:
• Budget: \$20 | 497 | 1,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-05 | latest | en | 0.925188 |
http://library.kiwix.org/wikipedia_en_computer_novid_2018-10/A/Hexadecagon.html | 1,558,878,909,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259177.93/warc/CC-MAIN-20190526125236-20190526151236-00264.warc.gz | 125,228,283 | 12,660 | Type Regular polygon
Edges and vertices 16
Schläfli symbol {16}, t{8}, tt{4}
Coxeter diagram
Symmetry group Dihedral (D16), order 2×16
Internal angle (degrees) 157.5°
Dual polygon Self
Properties Convex, cyclic, equilateral, isogonal, isotoxal
In mathematics, a hexadecagon (sometimes called a hexakaidecagon or 16-gon) is a sixteen-sided polygon.[1]
A regular hexadecagon is a hexadecagon in which all angles are equal and all sides are congruent. Its Schläfli symbol is {16} and can be constructed as a truncated octagon, t{8}, and a twice-truncated square tt{4}. A truncated hexadecagon, t{16}, is a triacontadigon, {32}.
### Construction
As 16 = 24 (a power of two), a regular hexadecagon is constructible using compass and straightedge: this was already known to ancient Greek mathematicians.[2]
at a given circumcircle
at a given side length, animation. (The construction is very similar to that of octagon at a given side length.)
## Measurements
Each angle of a regular hexadecagon is 157.5 degrees, and the total angle measure of any hexadecagon is 2520 degrees.
The area of a regular hexadecagon with edge length t is
Because the hexadecagon has a number of sides that is a power of two, its area can be computed in terms of the circumradius R by truncating Viète's formula:
Since the area of the circumcircle is the regular hexadecagon fills approximately 97.45% of its circumcircle.
## Symmetry
The 14 symmetries of a regular hexadecagon. Lines of reflections are blue through vertices, purple through edges, and gyration orders are given in the center. Vertices are colored by their symmetry position.
The regular hexadecagon has Dih16 symmetry, order 32. There are 4 dihedral subgroups: Dih8, Dih4, Dih2, and Dih1, and 5 cyclic subgroups: Z16, Z8, Z4, Z2, and Z1, the last implying no symmetry.
On the regular hexadecagon, there are 14 distinct symmetries. John Conway labels full symmetry as r32 and no symmetry is labeled a1. The dihedral symmetries are divided depending on whether they pass through vertices (d for diagonal) or edges (p for perpendiculars) Cyclic symmetries in the middle column are labeled as g for their central gyration orders.[3]
The most common high symmetry hexadecagons are d16, a isogonal hexadecagon constructed by eight mirrors can alternate long and short edges, and p16, an isotoxal hexadecagon constructed with equal edge lengths, but vertices alternating two different internal angles. These two forms are duals of each other and have half the symmetry order of the regular hexadecagon.
Each subgroup symmetry allows one or more degrees of freedom for irregular forms. Only the g16 subgroup has no degrees of freedom but can seen as directed edges.
## Dissection
16-cube projection 112 rhomb dissection
Regular
Isotoxal
Coxeter states that every zonogon (a 2m-gon whose opposite sides are parallel and of equal length) can be dissected into m(m-1)/2 parallelograms. [4] In particular this is true for regular polygons with evenly many sides, in which case the parallelograms are all rhombi. For the regular hexadecagon, m=8, and it can be divided into 28: 4 squares and 3 sets of 8 rhombs. This decomposition is based on a Petrie polygon projection of a 8-cube, with 28 of 1792 faces. The list enumerates the number of solutions as 1232944, including up to 16-fold rotations and chiral forms in reflection.
8-cube
{8}#{ } {83}#{ } {85}#{ }
A regular skew hexadecagon is seen as zig-zagging edges of a octagonal antiprism, a octagrammic antiprism, and a octagrammic crossed-antiprism.
A skew hexadecagon is a skew polygon with 24 vertices and edges but not existing on the same plane. The interior of such an hexadecagon is not generally defined. A skew zig-zag hexadecagon has vertices alternating between two parallel planes.
A regular skew hexadecagon is vertex-transitive with equal edge lengths. In 3-dimensions it will be a zig-zag skew hexadecagon and can be seen in the vertices and side edges of a octagonal antiprism with the same D8d, [2+,16] symmetry, order 32. The octagrammic antiprism, s{2,16/3} and octagrammic crossed-antiprism, s{2,16/5} also have regular skew octagons.
### Petrie polygons
The regular hexadecagon is the Petrie polygon for many higher-dimensional polytopes, shown in these skew orthogonal projections, including:
A15 B8 D9 2B2 (4D)
15-simplex
8-orthoplex
8-cube
611
161
8-8 duopyramid
8-8 duoprism
A hexadecagram is a 16-sided star polygon, represented by symbol {16/n}. There are three regular star polygons, {16/3}, {16/5}, {16/7}, using the same vertices, but connecting every third, fifth or seventh points. There are also three compounds: {16/2} is reduced to 2{8} as two octagons, {16/4} is reduced to 4{4} as four squares and {16/6} reduces to 2{8/3} as two octagrams, and finally {16/8} is reduced to 8{2} as eight digons.
Deeper truncations of the regular octagon and octagram can produce isogonal (vertex-transitive) intermediate hexadecagram forms with equally spaced vertices and two edge lengths.[5]
A truncated octagon is a hexadecagon, t{8}={16}. A quasitruncated octagon, inverted as {8/7}, is a hexadecagram: t{8/7}={16/7}. A truncated octagram {8/3} is a hexadecagram: t{8/3}={16/3} and a quasitruncated octagram, inverted as {8/5}, is a hexadecagram: t{8/5}={16/5}.
## In art
The hexadecagonal tower from Raphael's The Marriage of the Virgin
In the early 16th century, Raphael was the first to construct a perspective image of a regular hexadecagon: the tower in his painting The Marriage of the Virgin has 16 sides, elaborating on an eight-sided tower in a previous painting by Pietro Perugino.[6]
A hexadecagrammic pattern from the Alhambra
Hexadecagrams (16-sided star polygons) are included in the Girih patterns in the Alhambra.[7]
## Others
In the Philippines, in local carnivals (peryahan), Ferris Wheels with maximum of 16 seats or gondolas | 1,629 | 5,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-22 | latest | en | 0.9115 |
https://endevco.com/ask-the-experts/understanding-iepe-accelerometer-dynamic-range/ | 1,566,551,776,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318243.40/warc/CC-MAIN-20190823083811-20190823105811-00487.warc.gz | 447,747,266 | 13,374 | Home/Resources/Ask the experts/Understanding IEPE Accelerometer Dynamic Range
# Understanding IEPE Accelerometer Dynamic Range
### Question
Question: Please explain the concept of dynamic range, as it applies to an IEPE accelerometer? I cannot seem to find this information on Endevco® product data sheets.
IEPE accelerometer dynamic range can be defined as the range between lowest and highest levels that an accelerometer can measure. Accelerometer dynamic range can be defined in many ways; this article, however, will cover the most commonly used method for estimating transducer dynamic range. When making this estimation, we generally refer to a term known as residual noise, or resolution, to determine our minimum discernible signal and our maximum amplitude range. As dynamic range is expressed in terms of decibels (dB), the following equation is used to make this type of calculation:
• Dynamic Range = 20 log (range/mds)
Where:
• range = specified range in g’s (RMS)
• mds = minimum discernible signal, which is the residual noise x 3
As an example, let’s take an Endevco® model 65-10. This is an IEPE (offered by Meggitt as ISOTRON®) triaxial accelerometer with a 10 mV/g sensitivity and the following specifications:
• Range: ±500 g peak
• 500 X 0.707 = 354 g RMS
• Resolution: 0.0008 g RMS (800 µg)
• Bandwidth: 2 Hz to 10 kHz
Thus, the dynamic range of the Endevco® model 65-10 is 103.4 dB.
So, in this calculation, why did we multiply the residual noise level times three? This is because we are trying to determine a level where it is possible to see actual signal. If we simply used the noise level specification, we assume that if the signal level equals the noise level, we could see the signal, a concept or assumption which is incorrect.
In the above example, the range was converted to RMS, so that the terms are consistent.
In practice, when determining the minimum discernible signal, it is necessary to take into account the noise contribution of any additional amplification in the string. Noise improvement can also be realized by filtering out any out-of-band noise. | 477 | 2,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-35 | latest | en | 0.886472 |
https://programsandcourses.anu.edu.au/2015/course/math3029 | 1,713,804,299,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818312.80/warc/CC-MAIN-20240422144517-20240422174517-00558.warc.gz | 412,670,625 | 11,649 | • Offered by Department of Mathematics
• ANU College ANU Joint Colleges of Science
• Course subject Mathematics
• Areas of interest Mathematics
• Course convener
• Dr Boris Buchmann
• Mode of delivery In Person
• Offered in First Semester 2015
Probability Modelling with Applications (MATH3029)
The course introduces stochastic processes with a view towards applications in fields such as finance, insurance, risk management, and operations research. The aim is to provide mathematics students with basic knowledge of stochastic processes where practical rather than theoretical aspects are emphasized.
Probability Modelling and Applications provides a sound foundation to progress to honours and post-graduate courses emphasizing the theory of mathematical finance and stochastic analysis.
The course contains sufficient material for students to feel comfortable with Markov chains, Poisson processes, and Brownian motion, and the conceptual formulation of topics in continuous time finance, insurance and risk management, where these processes are applied. Also the concept of martingales, which is fundamental for understanding the modern option pricing theory of Black and Scholes, is introduced.
Note: This is an HPC. It continues the development of sophisticated mathematical and probabilistic techniques and their application begun in STAT2001(HPC)
## Learning Outcomes
Upon successful completion, students will have the knowledge and skills to:
On satisfying the requirements of this course, students will have the knowledge and skills to:
1. Explain the fundamental concepts of probability theory and its position in modern mathematics and applied contexts;
2. Demonstrate accurate and efficient use of probability theory techniques;
3. Demonstrate capacity for mathematical reasoning through analyzing, proving and explaining concepts from probability theory;
4. Apply problem-solving using probabilistic methods in various situations in mathematical finance.
## Indicative Assessment
Assessment will be based on:
• Weekly Assignments (50%; LO 1-4)
• Final examination (50%; LO 1-4)
The ANU uses Turnitin to enhance student citation and referencing techniques, and to assess assignment submissions as a component of the University's approach to managing Academic Integrity. While the use of Turnitin is not mandatory, the ANU highly recommends Turnitin is used by both teaching staff and students. For additional information regarding Turnitin please visit the ANU Online website.
This course will only be offered subject to staff availability
## Requisite and Incompatibility
To enrol in this course you must have successfully completed STAT2001 or MATH2320 or MATH3116.
## Fees
Tuition fees are for the academic year indicated at the top of the page.
If you are a domestic graduate coursework or international student you will be required to pay tuition fees. Tuition fees are indexed annually. Further information for domestic and international students about tuition and other fees can be found at Fees.
Student Contribution Band:
2
Unit value:
6 units
If you are an undergraduate student and have been offered a Commonwealth supported place, your fees are set by the Australian Government for each course. At ANU 1 EFTSL is 48 units (normally 8 x 6-unit courses). You can find your student contribution amount for each course at Fees. Where there is a unit range displayed for this course, not all unit options below may be available.
Units EFTSL
6.00 0.12500
## Course fees
Domestic fee paying students
Year Fee
2015 \$3096
International fee paying students
Year Fee
2015 \$4146
Note: Please note that fee information is for current year only.
## Offerings, Dates and Class Summary Links
ANU utilises MyTimetable to enable students to view the timetable for their enrolled courses, browse, then self-allocate to small teaching activities / tutorials so they can better plan their time. Find out more on the Timetable webpage.
The list of offerings for future years is indicative only.
Class summaries, if available, can be accessed by clicking on the View link for the relevant class number.
### First Semester
Class number Class start date Last day to enrol Census date Class end date Mode Of Delivery Class Summary
1808 16 Feb 2015 06 Mar 2015 31 Mar 2015 29 May 2015 In Person N/A | 870 | 4,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-18 | latest | en | 0.900947 |
http://theelementarymathmaniac.blogspot.com/2014/07/more-quick-and-fun-math-practice-for.html | 1,490,624,956,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189474.87/warc/CC-MAIN-20170322212949-00469-ip-10-233-31-227.ec2.internal.warc.gz | 359,397,762 | 24,172 | Tuesday, July 29, 2014
More Quick and Fun Math Practice For Little Kids
Last week, I started describing how I have been making busy bags to use in the classroom in conjunction with my math browse boxes to keep kids busy and learning in the classroom, at home or on the road.
You can check out last week's post about number sticks and read more about math busy bags and browse boxes here!
This week, I want to show you another simple to make but effective way to keep kids busy and practicing math!
Materials List
-Felt sheets: I used a full blue one as well as about 1/5 of a sheet of yellow, green and orange and a few scraps of a black sheet.
-Wiggle eyes: I included some round and some oval shaped ones. The sky is the limit here! They have all kinds of wiggle eyes available these days. I picked up a multi-pack with a lot of varieties for various projects.
-Permanent Marker
-Scissors
-Small scrap of cardstock or cardboard
-Small circular object
I simply traced the bottom of a round yogurt cut onto a scrap piece of cardstock. Then I cut this out and used it as a pattern for cutting out the felt. When I had 12 circles, I wrote the numbers 1-12 on them. The kid I was making this for has been working on ordering numbers up to 10 and was ready for a bit of a stretch. You could also put other numbers here. I will be making another set for some first graders who need practice pulling through decades. This one might only contain 8 circles and have the numbers from 57-64. If I was making one of these for second graders, it might contain numbers that pull through a century such as 297-306. If I wanted to make one for a third grader, I might use skip counting numbers to work on multiplication.
I have tried this out on a few kids and here is what I have discovered.
When given to a 5 year old, her first instinct was to line up the numbers in sequential order.
Another 5 year old put the numbers in order around the perimeter and used the black pieces and eyes to make a design in the middle.
Another kid put them in order in a long line and then made a worm out of them.
The entire task fits in a quart size Ziploc bag and is ready to be added to a browse box or taken out when a student needs an activity to keep them busy and learning.
After sitting with kids and showing them how to put the numbers in order and make a caterpillar, they were very successful. Without instruction, they did tend to focus on the numbers and putting them in order and that is the main goal for this activity anyway!
I will be sharing more ideas for busy bags and browse boxes for a variety of grades in the coming weeks so stay tuned! The easiest way to get updates is to like me on Facebook!
1 comment:
1. I think I remember something like this from my childhood!
~Lucy
Kids Math Teacher | 654 | 2,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-13 | longest | en | 0.961934 |
http://nrich.maths.org/524 | 1,503,571,687,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886133449.19/warc/CC-MAIN-20170824101532-20170824121532-00196.warc.gz | 288,410,219 | 4,760 | ### Pebbles
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!
### Have You Got It?
Can you explain the strategy for winning this game with any target?
# Repeaters
##### Stage: 3 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594).
Whatever digits you choose the number will always be divisible by 7 and by 11 and by 13, in each case without a remainder.
Can you explain why? | 197 | 792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-34 | latest | en | 0.920684 |
https://uk.mathworks.com/matlabcentral/cody/problems/190-great-circle-distance/solutions/69132 | 1,590,494,034,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390758.21/warc/CC-MAIN-20200526112939-20200526142939-00576.warc.gz | 586,874,463 | 15,718 | Cody
# Problem 190. Great Circle Distance
Solution 69132
Submitted on 2 Apr 2012 by Ben Petschel
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
%% assert(isequal(round(sphere_distance(100,10,50,-20,14)*10000)/10000,75.9097));
2 Pass
%% assert(isequal(round(sphere_distance(6371e3,-97.7430608,30.267153,-74.0244265,40.6081588)*10000)/10000,2426004.8394));
3 Pass
%% assert(isequal(round(sphere_distance(6371e3,-97.7430608,31.267153,-74.0244265,40.6081588)*10000)/10000,2364307.7819)); | 211 | 613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.713101 |
https://pyproveit.github.io/Prove-It/packages/proveit/physics/quantum/circuits/__pv_it/axioms/b071f55dc5c27621dae8ad34530b8021a6fdb6850/expr.html | 1,675,502,540,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500095.4/warc/CC-MAIN-20230204075436-20230204105436-00580.warc.gz | 490,355,986 | 5,593 | # from the theory of proveit.physics.quantum.circuits¶
In [1]:
import proveit
# Automation is not needed when building an expression:
proveit.defaults.automation = False # This will speed things up.
proveit.defaults.inline_pngs = False # Makes files smaller.
%load_expr # Load the stored expression as 'stored_expr'
# import Expression classes needed to build the expression
from proveit import A, B, S, k
from proveit.logic import Equals, Forall
from proveit.numbers import NaturalPos
from proveit.physics.quantum.circuits import Input, MultiQubitElem
In [2]:
# build up the expression from sub-expressions
expr = Forall(instance_param_or_params = [k], instance_expr = Forall(instance_param_or_params = [A, B, S], instance_expr = Equals(Equals(MultiQubitElem(element = Input(state = A, part = k), targets = S), MultiQubitElem(element = Input(state = B, part = k), targets = S)), Equals(A, B))), domain = NaturalPos)
expr:
In [3]:
# check that the built expression is the same as the stored expression
assert expr == stored_expr
assert expr._style_id == stored_expr._style_id
print("Passed sanity check: expr matches stored_expr")
Passed sanity check: expr matches stored_expr
In [4]:
# Show the LaTeX representation of the expression for convenience if you need it.
print(stored_expr.latex())
\forall_{k \in \mathbb{N}^+}~\left[\forall_{A, B, S}~\left(\left(\begin{array}{c} \Qcircuit@C=1em @R=.7em{
& \qin{A~\mbox{part}~k~\mbox{on}~S} & \qw
} \end{array} = \begin{array}{c} \Qcircuit@C=1em @R=.7em{
& \qin{B~\mbox{part}~k~\mbox{on}~S} & \qw
} \end{array}\right) = \left(A = B\right)\right)\right]
In [5]:
stored_expr.style_options()
namedescriptiondefaultcurrent valuerelated methods
with_wrappingIf 'True', wrap the Expression after the parametersNoneNone/False('with_wrapping',)
condition_wrappingWrap 'before' or 'after' the condition (or None).NoneNone/False('with_wrap_after_condition', 'with_wrap_before_condition')
wrap_paramsIf 'True', wraps every two parameters AND wraps the Expression after the parametersNoneNone/False('with_params',)
justificationjustify to the 'left', 'center', or 'right' in the array cellscentercenter('with_justification',)
In [6]:
# display the expression information
stored_expr.expr_info()
core typesub-expressionsexpression
0Operationoperator: 7
operand: 2
1ExprTuple2
2Lambdaparameter: 34
body: 4
3ExprTuple34
4Conditionalvalue: 5
condition: 6
5Operationoperator: 7
operand: 11
6Operationoperator: 9
operands: 10
7Literal
8ExprTuple11
9Literal
10ExprTuple34, 12
11Lambdaparameters: 13
body: 14
12Literal
13ExprTuple32, 33, 28
14Operationoperator: 19
operands: 15
15ExprTuple16, 17
16Operationoperator: 19
operands: 18
17Operationoperator: 19
operands: 20
18ExprTuple21, 22
19Literal
20ExprTuple32, 33
21Operationoperator: 24
operands: 23
22Operationoperator: 24
operands: 25
23NamedExprselement: 26
targets: 28
24Literal
25NamedExprselement: 27
targets: 28
26Operationoperator: 30
operands: 29
27Operationoperator: 30
operands: 31
28Variable
29NamedExprsstate: 32
part: 34
30Literal
31NamedExprsstate: 33
part: 34
32Variable
33Variable
34Variable | 935 | 3,096 | {"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} | 2.640625 | 3 | CC-MAIN-2023-06 | latest | en | 0.469963 |
http://www.justintools.com/unit-conversion/length.php?k1=astronomical-unit&k2=parsecs | 1,568,999,059,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574050.69/warc/CC-MAIN-20190920155311-20190920181311-00438.warc.gz | 277,042,240 | 19,610 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# Convert [Astronomical Unit] to [Parsecs], (au to pc)
## LENGTH
1 Astronomical Unit
= 4.8481368111E-6 Parsecs
*Select units, input value, then convert.
Embed to your site/blog Convert to scientific notation.
Category: length
Conversion: Astronomical Unit to Parsecs
The base unit for length is meters (SI Unit)
[Astronomical Unit] symbol/abbrevation: (au)
[Parsecs] symbol/abbrevation: (pc)
How to convert Astronomical Unit to Parsecs (au to pc)?
1 au = 4.8481368111E-6 pc.
1 x 4.8481368111E-6 pc = 4.8481368111E-6 Parsecs.
Always check the results; rounding errors may occur.
Definition:
The astronomical unit (symbol au, or ua) is a unit of length, roughly the distance from Earth to the Sun. However, that distance varies as Earth orbits the Sun, fr ..more definition+
A parsec (symbol: pc) is a unit of length used to measure large distances to objects outside our Solar System. One parsec is the distance at which one astronomical unit subt ..more definition+
In relation to the base unit of [length] => (meters), 1 Astronomical Unit (au) is equal to 149597870700 meters, while 1 Parsecs (pc) = 3.08567758149E+16 meters.
1 Astronomical Unit to common length units
1 au =149597870700 meters (m)
1 au =149597870.7 kilometers (km)
1 au =1.495978707E+13 centimeters (cm)
1 au =490806662402 feet (ft)
1 au =5.88967994882E+12 inches (in)
1 au =163602220801 yards (yd)
1 au =92955807.273 miles (mi)
1 au =1.5812056939E-5 light years (ly)
1 au =5.65409346319E+14 pixels (PX)
1 au =9.34986691875E+45 planck length (pl)
Astronomical Unit to Parsecs (table conversion)
1 au =4.8481368111E-6 pc
2 au =9.69627362219E-6 pc
3 au =1.45444104333E-5 pc
4 au =1.93925472444E-5 pc
5 au =2.42406840555E-5 pc
6 au =2.90888208666E-5 pc
7 au =3.39369576777E-5 pc
8 au =3.87850944888E-5 pc
9 au =4.36332312999E-5 pc
10 au =4.8481368111E-5 pc
20 au =9.69627362219E-5 pc
30 au =0.000145444104333 pc
40 au =0.000193925472444 pc
50 au =0.000242406840555 pc
60 au =0.000290888208666 pc
70 au =0.000339369576777 pc
80 au =0.000387850944888 pc
90 au =0.000436332312999 pc
100 au =0.00048481368111 pc
200 au =0.000969627362219 pc
300 au =0.00145444104333 pc
400 au =0.00193925472444 pc
500 au =0.00242406840555 pc
600 au =0.00290888208666 pc
700 au =0.00339369576777 pc
800 au =0.00387850944888 pc
900 au =0.00436332312999 pc
1000 au =0.0048481368111 pc
2000 au =0.00969627362219 pc
4000 au =0.0193925472444 pc
5000 au =0.0242406840555 pc
7500 au =0.0363610260832 pc
10000 au =0.048481368111 pc
25000 au =0.121203420277 pc
50000 au =0.242406840555 pc
100000 au =0.48481368111 pc
1000000 au =4.8481368111 pc
1000000000 au =4848.1368111 pc
(Astronomical Unit) to (Parsecs) conversions | 1,062 | 3,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.754015 |
https://coachenergyimage.com/nintendo-switch-ikflh/057ef6-failure-is-cumulative-meaning | 1,627,905,171,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154320.56/warc/CC-MAIN-20210802110046-20210802140046-00124.warc.gz | 185,620,835 | 5,017 | El-Nashar reported that there were no significant differences in the time to treatment failure between the two groups, with a 3-year cumulative failure rate of 9% in … Consider a grouped data set, where 4 failures are found in the interval 0-100 hours and 2 failures are found in the second interval 100-180 hours. Do you have any comments on this article? What does CFR stand for? Cumulative doses can have toxic effects on the heart, including heart failure, so the drug is only used for a limited period of 2 to 3 years. CFR abbreviation stands for Cumulative Failure Rate. It is a temporary detour, not a dead end. Thus new devices start life with high reliability and end with a high failure probability. These points are temporary moments or small snapshots in the larger scope of your month, year, or lifetime. The cumulative distribution function (CDF), also called the unreliability function or the probability of failure, is denoted by Q(t). Cumulative MTBF is the average time-between-failure from the beginning of the test (i.e., t = 0) up to time t. The instantaneous MTBF is the average time-between-failure in a given interval, dt. It is the integral of h(t) from 0 to t, or the area under the hazard function h(t) from 0 to t. MTTF is the average time to failure. (Also called the mean time to failure, expected time to failure, or average life.) The events in cumulative probability may be sequential, like coin tosses in a row, or they may be in a range. Define failure as only temporary. In the context of repairs over time, the value of the MCF can be thought of as the average number of repairs that each system will have undergone after a certain time. Cumulative incidence is calculated as the number of new events or cases of disease divided by the total number of individuals in the Cumulative incidence, in epidemiology, estimate of the risk that an individual will experience an event or develop a disease during a specified period of time. The failure probability, on the other hand, does the reverse. Failure is delay, not defeat. You should get a heart evaluation, including evaluation of left ventricular ejection fraction (LVEF), before starting this drug and before each dose administration. The time interval between 2 failures if the component is called the mean time between failures (MTBF) and is given by the first moment if the failure density function: Failure is something we can avoid only by saying nothing, doing nothing, and being nothing.” —Denis Waitley Click To Tweet Failure is a loss, but avoid looking at this loss as permanent. How to use cumulative in a sentence. The Mean Cumulative Function (MCF) is a cumulative history function that shows the cumulative number of recurrences of an event, such as repairs over time. What is the abbreviation for Cumulative Failure Rate? Cumulative definition is - increasing by successive additions. H(t) is the cumulative hazard function. MTTF = . For example, if you're observing a response with three categories, the cumulative probability for an observation with response 2 would be the probability that the predicted response is 1 OR 2. That's cumulative probability. By seeing failure as temporary, it will seem less overwhelming. What does cumulative damage on an iowa title mean 14.05.2020 what does cumulative damage on an iowa title mean Comments: A certificate of title shall not be issued for a motor vehicle unless a damage disclosure statement has been made by the transferor of the vehicle and is furnished with the application for certificate of title. It represents the probability that a brand new component will fail at or before a specified time. Failure happens at various points in our lives. If so send them to murray@omdec.com. | 792 | 3,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-31 | latest | en | 0.941538 |
https://testbook.com/question-answer/the-average-of-three-numbers-is-35-if-the-smalles--607a643d2cc381f5083b4b69 | 1,643,240,310,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305006.68/warc/CC-MAIN-20220126222652-20220127012652-00486.warc.gz | 586,156,398 | 29,581 | # The average of three numbers is 35. If the smallest number is half of the largest number and middle number is 15 less than the largest number. Find the smallest number.
1. 48
2. 56
3. 24
4. 12
Option 3 : 24
## Detailed Solution
Given
Average of three numbers = 35
Smallest number = largest number/2
Middle number = larger number - 15
Formula used
Average = Sum of all observation/ Number of observation
Calculation
Let the largest number is l
Smallest number is l/2
Middle number is (l - 15)
According to question
[l + (l/2) + (l - 15)]/3 = 35
⇒ (2l + l + 2l - 30)/6 = 35
⇒ 5l - 30 = 210
⇒ 5l = 240
⇒ l = 48
Smallest number = 48/2
⇒ 24
∴ Smallest number is 24 | 234 | 683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2022-05 | latest | en | 0.764889 |
https://electronics.stackexchange.com/questions/176173/why-do-we-short-the-input-when-calculating-the-output-impedance | 1,713,840,696,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00789.warc.gz | 197,196,171 | 39,355 | # Why do we short the input when calculating the output impedance?
I believe the title says everything about the question. My issue has to to do with the following: Consider a two-port given by:
$$V_1 = Z_{11}I_1 + Z_{12}I_2$$ $$V_2 = Z_{21}I_1 + Z_{22}I_2$$
Cleary by the equations above, one would expect that the measurement of the $Z_{ij}$ parameters to be performed while keeping one port at a time in open loop configuration to guarantee $I = 0$ for the "unwanted" current. However this seems not to be the case, as can be seen for instance in the derivation of the of the output resistance of a simple CE amplifier where the input is shorted to determine the output resistance. What am I missing here?
• The question is, do you want the output impedance as the circuit is used, or do you want the circuit-theoretical $Z_{22}$ of the circuit itself? Most people want the former. Jun 18, 2015 at 22:51
• Thanks a lot for clarifying. I was thinking along those lines but didn't find any material to confirm / deny it. Could you maybe post this as answer for me to accept (and upvote?)
– R.G.
Jun 18, 2015 at 23:19
The two test configurations tell you two different things.
If you short the input, you're testing the output impedance of the circuit as it is normally used. This is because you typically drive the circuit with a (relatively) low impedance source.
If you open the input, you get the $Z_{22}$ as defined in circuit theory as a characteristic of the circuit itself (without effects from any driving circuit).
The first option is often more practically useful.
If you want to solve for any of the Z parameters it is clear that you must assume Iac=0, which means Idc=const.
However, remember that there are other sets of parameters for describing the BJT:
• Y-parameters are all "short-circuit"-parameters (Vac=0, Vdc=const)
• h-parameters are "short-circuit-parameters" for h11 and h21 and "open-circuit-parameters" for h12 and h22.
And remember: "Short ciruit" means: DC conditions constant (no ac signal). | 507 | 2,035 | {"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} | 3.484375 | 3 | CC-MAIN-2024-18 | latest | en | 0.929103 |
https://forum.openoffice.org/en/forum/viewtopic.php?f=9&t=3871 | 1,606,646,419,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197593.33/warc/CC-MAIN-20201129093434-20201129123434-00597.warc.gz | 293,855,939 | 8,550 | ## Function for irregular date progression
### Function for irregular date progression
Hi,
I need to have a function in calc that does the following: Take a date an give the corresponding dates in 1 day, 1 week, 2 weeks, 1 month, 2 month, 6 month
For example:
1th January --> 2nd January, 8th January, 1th February, 1th March, 1th July
Is there any way I can do this with a function or would I need to record a macro or something like that?
Monetix
Posts: 3
Joined: Tue Mar 25, 2008 5:04 pm
### Re: Function for irregular date progression
I guess the results have to be in different cells. In that case, you should use the DATE() function with the YEAR(), MONTH() and DAY() functions. The formulas would look like that (assuming that initial date is in A1):
=DATE(YEAR(\$A\$1);MONTH(\$A\$1)+6;DAY(\$A\$1)) for the date + 6 months.
NB: I'm using the French version at work, I hope it's similar for the English.
Thanks to add '[Solved]' in your first post title (edit button) if your issue has been fixed.
LO 6.4.6 on Xubuntu 20.04 and Windows 10.
Hagar Delest
Moderator
Posts: 29066
Joined: Sun Oct 07, 2007 9:07 pm
Location: France
### Re: Function for irregular date progression
A1:A7
2008-01-01
2008-01-02
2008-01-08
2008-02-01
2008-03-01
2008-07-01
2009-01-01
A8: =DATE(YEAR(A1)+1;MONTH(A1);DAY(A1)))
copy down A8
Please, edit this topic's initial post and add "[Solved]" to the subject line if your problem has been solved.
Ubuntu 18.04, no OpenOffice, LibreOffice 6.4
Villeroy
Volunteer
Posts: 28825
Joined: Mon Oct 08, 2007 1:35 am
Location: Germany
### Re: Function for irregular date progression
1 day, 2 days, 2 weeks are easy - just add 1, 2, 7 14 respectively.
But... what date is one month after 31st January? Is it 28th February? (or 29th in a leap year?), or is it 3rd March? (or 2nd in a leap year?)
And then what is a month after 3rd February? Is it also 3rd March? That means either a month after 1st February (1st March) would come before a month after 31st January (3rd March), or you have 28th-31st January all mapping to 28th February.
I'd stick with 30, 60, 180 days or similar.
Hagar de l'Est wrote:I guess the results have to be in different cells. In that case, you should use the DATE() function with the YEAR(), MONTH() and DAY() functions. The formulas would look like that (assuming that initial date is in A1):
=DATE(YEAR(\$A\$1);MONTH(\$A\$1)+6;DAY(\$A\$1)) for the date + 6 months.
What happens if MONTH(\$A\$1)>6?
huw
Volunteer
Posts: 417
Joined: Wed Nov 21, 2007 1:57 pm
### Re: Function for irregular date progression
What happens if MONTH(\$A\$1)>6
Just give a try. DATE can calculate all kinds of date offsets from start of Gregorian Calendar (1582-10-15) until year 9956 by using all kind of offset values, including negative ones.
=DATE(2009;2;29) => 2009-03-01
=DATE(2009;-2;29) => 2008-10-29
Please, edit this topic's initial post and add "[Solved]" to the subject line if your problem has been solved.
Ubuntu 18.04, no OpenOffice, LibreOffice 6.4
Villeroy
Volunteer
Posts: 28825
Joined: Mon Oct 08, 2007 1:35 am
Location: Germany
### Re: Function for irregular date progression
Villeroy wrote:
What happens if MONTH(\$A\$1)>6
Just give a try. DATE can calculate all kinds of date offsets from start of Gregorian Calendar (1582-10-15) until year 9956 by using all kind of offset values, including negative ones.
=DATE(2009;2;29) => 2009-03-01
=DATE(2009;-2;29) => 2008-10-29
I see, thanks Villeroy (and sorry, Hagar). I did something silly when I tried it before (I just got errors).
But still, I don't see the resulting dates (back-tracking & repetition) as meaningful:
Code: Select all Expand viewCollapse view
`Date +1 month28/01/2007 28/02/200729/01/2007 01/03/200730/01/2007 02/03/200731/01/2007 03/03/200701/02/2007 01/03/200702/02/2007 02/03/200703/02/2007 03/03/200704/02/2007 04/03/2007`
huw
Volunteer
Posts: 417
Joined: Wed Nov 21, 2007 1:57 pm
### Re: Function for irregular date progression
That's the corner case of each end of month (with less than 31 days). It comes from the fact that dates are based on numbers. So that's a limitation of the tool. If it's not enough for the OP needs, then, the formula needs to be more complicated to introduce alternatives and handle these situations.
LO 6.4.6 on Xubuntu 20.04 and Windows 10.
Hagar Delest
Moderator
Posts: 29066
Joined: Sun Oct 07, 2007 9:07 pm
Location: France
### Re: Function for irregular date progression
My solution starts with 7 constant values, assuming that the following dates are (7 cells up)+1 year. This won't change the days and months, unless there is a leap day in the start sequence: 2008-02-29 -> 2009-03-01 -> [remains 1st March]
Please, edit this topic's initial post and add "[Solved]" to the subject line if your problem has been solved.
Ubuntu 18.04, no OpenOffice, LibreOffice 6.4
Villeroy
Volunteer
Posts: 28825
Joined: Mon Oct 08, 2007 1:35 am
Location: Germany
### Re: Function for irregular date progression
Hi Villeroy, you focused on the Year+1 but the OP needs more than this one (and does he really need the year+1 one ?). Or am I missing something?
Monetix wrote:Take a date an give the corresponding dates in 1 day, 1 week, 2 weeks, 1 month, 2 month, 6 month
For example:
1th January --> 2nd January, 8th January, 1th February, 1th March, 1th July
LO 6.4.6 on Xubuntu 20.04 and Windows 10.
Hagar Delest
Moderator
Posts: 29066
Joined: Sun Oct 07, 2007 9:07 pm
Location: France
### Re: Function for irregular date progression
Another proposal for any irregular sequence:
A1: 2008-01-01 (start date)
B1: Days (label)
C1: Months (label)
D1: Years (label)
Take the date from previous row and add the time units as specified in this row:
A2 =DATE(YEAR(\$A1)+\$D2;MONTH(\$A1)+\$C2;DAY(\$A1)+\$B2)
Take the date in A1 and add the time units as specified in this row:
A2 =DATE(YEAR(\$A\$1)+\$D2;MONTH(\$A\$1)+\$C2;DAY(\$A\$1)+\$B2)
Please, edit this topic's initial post and add "[Solved]" to the subject line if your problem has been solved.
Ubuntu 18.04, no OpenOffice, LibreOffice 6.4
Villeroy
Volunteer
Posts: 28825
Joined: Mon Oct 08, 2007 1:35 am
Location: Germany | 1,944 | 6,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-50 | latest | en | 0.926866 |
https://qforquestions.com/category/ncert-solutions/ | 1,653,675,352,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662675072.99/warc/CC-MAIN-20220527174336-20220527204336-00298.warc.gz | 514,001,553 | 7,779 | # Category: NCERT Solutions
## Probability Class 12 NCERT Solutions
Probability Class 12 NCERT Solutions Exercise 13.1 Solutions Exercise 13.2 Solutions Exercise 13.3 Solutions Exercise 13.4 Solutions Exercise 13.5 Solutions Miscellaneous Exercise Solutions Exercise 13.1 Solutions Exercise 13.2 Solutions …
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Application of Integrals Class 12 NCERT Solutions Exercise 8.1 Solutions Exercise 8.2 Solutions Miscellaneous Exercise Solutions Exercise 8.1 Solutions Exercise 8.2 Solutions Miscellaneous Exercise 8.3 Solutions
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Integrals Class 12 NCERT Solutions Exercise 7.1 Solutions Exercise 7.2 Solutions Exercise 7.3 Solutions Exercise 7.4 Solutions Exercise 7.5 Solutions Exercise 7.6 Solutions Exercise 7.7 Solutions Exercise 7.8 Solutions …
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Application of Derivatives Class 12 NCERT Solutions Exercise 6.1 Solutions Exercise 6.2 Solutions Exercise 6.3 Solutions Exercise 6.4 Solutions Exercise 6.5 Solutions Miscellaneous Exercise Solutions Exercise 6.1 Solutions Exercise …
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Continuity and Differentiability Class 12 NCERT Solutions Exercise 5.1 Solutions Exercise 5.2 Solutions Exercise 5.3 Solutions Exercise 5.4 Solutions Exercise 5.5 Solutions Exercise 5.6 Solutions Exercise 5.7 Solutions Exercise …
## Determinants Class 12 NCERT Solutions
Determinants Class 12 NCERT Solutions Exercise 4.1 Solutions Exercise 4.2 Solutions Exercise 4.3 Solutions Exercise 4.4 Solutions Exercise 4.5 Solutions Exercise 4.6 Solutions Miscellaneous Exercise Solutions Exercise 4.1 Solutions … | 641 | 2,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.670959 |
https://research.stlouisfed.org/fred2/series/LINC731INFO | 1,462,228,003,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860117783.16/warc/CC-MAIN-20160428161517-00207-ip-10-239-7-51.ec2.internal.warc.gz | 991,163,563 | 19,194 | # All Employees: Information in Lincoln, NE (MSA)
2016-03: 2.6 Thousands of Persons
Monthly, Seasonally Adjusted, LINC731INFO, Updated: 2016-04-15 5:37 PM CDT
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The data services of the Federal Reserve Bank of St. Louis include series that are seasonally adjusted. To make these adjustments, we use the X-12 Procedure of SAS to remove the seasonal component of the series so that non-seasonal trends can be analyzed. This procedure is based on the U.S. Bureau of the Census X-12-ARIMA Seasonal Adjustment Program. More information on this program can be found at http://www.census.gov/srd/www/x12a/.
The seasonal moving average function used is that of the Census Bureau’s X-11-ARIMA program. This includes a 3x3 moving average for the initial seasonal factors and a 3x5 moving average to calculate the final seasonal factors. The D11 function is also used to output the entire seasonally adjusted series that is displayed. For specific information on the SAS X-12 procedure, please visit their website: http://support.sas.com/documentation/cdl/en/etsug/60372/HTML/default/viewer.htm#etsug_x12_sect001.htm.
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Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Graph Data
Graph Image
Suggested Citation
``` Federal Reserve Bank of St. Louis, All Employees: Information in Lincoln, NE (MSA) [LINC731INFO], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/LINC731INFO, May 2, 2016. ```
Retrieving data.
Graph updated. | 616 | 2,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2016-18 | latest | en | 0.852192 |
https://discourse.processing.org/t/solved-p3d-ellipse-rotation-problem/5908 | 1,656,217,855,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037089.4/warc/CC-MAIN-20220626040948-20220626070948-00212.warc.gz | 255,022,746 | 5,897 | # [SOLVED] P3D ellipse rotation problem
Hello there fellow coders!
Let’s say I want to rotate an ellipse around it’s axis. I came up with the following code, but it is not doing what I want.
``````float alpha = 0;
int r = 200;
void setup()
{
size(500, 500, P3D);
}
void draw()
{
background(0);
pushMatrix();
translate(width/2, height/2, 0);
noFill();
stroke(255);
rotateY(alpha);
ellipse(0, 0, r, r);
popMatrix();
alpha += 0.01;
}
``````
For reference, my problem is the same as what this guy posted (no solution):
I am expecting this: http://i.imgur.com/T3qFwv7.gifv
but I’m getting this: http://i.imgur.com/twLn9vQ.gifv
So it’s obviously not rotating around its Y axis, and I can’t understand why that is. Setting a different ellipseMode doesn’t fix the issue.
Any idea would be greatly appreciated.
Cheers
1 Like
for this alone!
The problem is that the sketch you are drawing is 3D!
So the part of the ring that is far away from the camera looks smaller than the part that is closer.
It’s a pretty weird effect.
The fix is simple: Change your camera to ignore the Z distance. In short, call ortho():
``````float alpha = 0;
int r = 200;
void setup()
{
size(500, 500, P3D);
ortho();
}
void draw()
{
background(0);
pushMatrix();
translate(width/2, height/2, 0);
noFill();
stroke(255);
rotateY(alpha);
ellipse(0, 0, r, r);
popMatrix();
alpha += 0.01;
}
``````
You might also consider not rotating the ring at all, but drawing a 2D ellipse who’s width is constantly decreasing/increasing.
2 Likes
Fantastic, ortho() worked! Thanks a lot!
It had been driving me nuts
Yeah I knew about the width trick, but my project is more complex and involves rotations of multiple objects along X, Y and Z. I just simplified it to get straight to the point.
Thanks again @TfGuy44
Have a great day
Just want to add that you made me understand it was a perspective problem.
So I fiddled with that and was able to mitigate the weird effect without having to completely give up on perspective.
``````void setup()
{
size(500, 500, P3D);
float fov = PI/24.0;
float cameraZ = (height/2.0) / tan(fov/2.0);
perspective(fov, float(width)/float(height),
cameraZ/10.0, cameraZ*10.0);
}
``````
Using the code above and setting the object very far on the Z axis (around -3200), yields a decent result. Full code below:
``````float alpha = 0;
int r = 200;
void setup()
{
size(500, 500, P3D);
float fov = PI/24.0;
float cameraZ = (height/2.0) / tan(fov/2.0);
perspective(fov, float(width)/float(height),
cameraZ/10.0, cameraZ*10.0);
}
void draw()
{
background(0);
pushMatrix();
translate(width/2, height/2, -3200);
noFill();
stroke(255);
rotateY(alpha);
ellipse(0, 0, r, r);
popMatrix();
alpha += 0.01;
}
``````
1 Like
Thanks for sharing this perspective solution to your problem with the community - very helpful! | 810 | 2,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.890502 |
http://www.bikeforums.net/archive/index.php/t-505259.html | 1,394,441,295,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010707300/warc/CC-MAIN-20140305091147-00051-ip-10-183-142-35.ec2.internal.warc.gz | 248,953,634 | 5,511 | # Southern California - Happy Birthday, Grumpy Pig!!
Bikeforums.net is a forum about nothing but bikes. Our community can help you find information about hard-to-find and localized information like bicycle tours, specialties like where in your area to have your recumbent bike serviced, or what are the best bicycle tires and seats for the activities you use your bike for.
View Full Version : Happy Birthday, Grumpy Pig!!
Mo'Phat
01-24-09, 09:57 PM
(better late than never)
Happy birthday to you.
Happy birthday to you.
Happy birthday, Dear Frank.
Happy birthday to you.
Happy birthday to you
You live in a zoo
You look like a monkey
and you smell like one, too.
Indolent58
01-24-09, 10:21 PM
hbdgp
GP
01-24-09, 10:27 PM
Thanks, guys.
It was a pretty tough week.
http://i265.photobucket.com/albums/ii204/grumpypig/horsey.jpg
Indolent58
01-24-09, 10:29 PM
You even wear poser gear on a merry-go-round?
Extort
01-24-09, 10:35 PM
It proves he is 'HARD CORE'...
HB!
efficiency
01-25-09, 04:56 AM
Happy Birthday Frank!
alicestrong
01-25-09, 08:29 AM
Happy Birthday!!
:bday::bday::bday::bday:
01-25-09, 09:01 AM
H b!!
sojourn
01-25-09, 01:08 PM
Happy Birthday GP!
You look pretty young.......what are you, like 13??
01-25-09, 01:56 PM
Happy Birthday GP!
You look pretty young.......what are you, like 13??
No, he's like...70.
GP
01-25-09, 03:35 PM
Happy Birthday GP!
You look pretty young.......what are you, like 13??
No, he's like...70.
If you add 13 and 70 then divide by 1.66 you'll come up with my age. Hint: I got my AARP card in the mail last month.
01-25-09, 03:53 PM
Bo don't do math.
IanInSD
01-25-09, 04:09 PM
If you add 13 and 70 then divide by 1.66 you'll come up with my age. Hint: I got my AARP card in the mail last month.
Sweet! Senior discount at Applebees.
sojourn
01-25-09, 04:12 PM
If you add 13 and 70 then divide by 1.66 you'll come up with my age. Hint: I got my AARP card in the mail last month.
Yeah, so like I said......you're like....a LOT younger than me! | 641 | 2,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2014-10 | latest | en | 0.877185 |
https://mail.python.org/pipermail/numpy-discussion/2009-June/043289.html | 1,642,613,171,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301475.82/warc/CC-MAIN-20220119155216-20220119185216-00191.warc.gz | 448,646,071 | 3,357 | # [Numpy-discussion] second 2d fft gives the same result as fft+ifft
Béla MIHALIK bela.mihalik at gmail.com
Wed Jun 10 11:53:31 EDT 2009
```I made the transformations with scipy. It works with complex numbers.
The results are here:
http://server6.theimagehosting.com/image.php?img=second_fourier_scipy.png
Also i asked it from others... as i understand now: the second fourier
transformation of a function always gives back the mirror of the function
itself.
I didn't know it :-)
thanks,
bela
2009/6/9 David Goldsmith <d_l_goldsmith at yahoo.com>
>
> Sorry, I meant:
>
> Im(iFT(FT(f))) = Im(FT^2(f)), Re(iFT(FT(f))) != Re(FT^2(f))
>
> DG
>
> --- On Tue, 6/9/09, David Goldsmith <d_l_goldsmith at yahoo.com> wrote:
>
> > From: David Goldsmith <d_l_goldsmith at yahoo.com>
> > Subject: Re: [Numpy-discussion] second 2d fft gives the same result as
> fft+ifft
> > To: "Discussion of Numerical Python" <numpy-discussion at scipy.org>
> > Date: Tuesday, June 9, 2009, 11:01 AM
> >
> > --- On Tue, 6/9/09, Matthieu Brucher <matthieu.brucher at gmail.com>
> > wrote:
> >
> > > Hi,
> > >
> > > Is it really ?
> > > You only show the imaginary part of the FFT, so you
> > can't
> > > be sure of
> > > what you are saying.
> >
> > Indeed, is there not a "label" for a function f which
> > satisfies
> >
> > Im(iFFT(f)) = Im(FFT^2(f)),
> > Re(iFFT(f)) != Re(FFT^2(f))?
> >
> > (And similarly if Im and Re roles are reversed.)
> > Seems like the class of such functions (if any exist) might
> > have some interesting properties...
> >
> > DG
> >
> > > Don't forget that the only difference between FFT and
> > iFFT
> > > is (besides
> > > of teh scaling factor) a minus sign in the exponent.
> > >
> > > Matthieu
> > >
> > > 2009/6/9 bela <bela.mihalik at gmail.com>:
> > > >
> > > > I tried to calculate the second fourier
> > transformation
> > > of an image with the
> > > > following code below:
> > > >
> > > >
> > >
> > ---------------------------------------------------------------
> > > > import pylab
> > > > import numpy
> > > >
> > > > ### Create a simple image
> > > >
> > > > fx = numpy.zeros( 128**2
> > ).reshape(128,128).astype(
> > > numpy.float )
> > > >
> > > > for i in xrange(8):
> > > > for j in xrange(8):
> > > > fx[i*8+16][j*8+16] = 1.0
> > > >
> > > > ### Fourier Transformations
> > > >
> > > > Ffx = numpy.copy( numpy.fft.fft2( fx ).real )
> > # 1st
> > > fourier
> > > > FFfx = numpy.copy( numpy.fft.fft2( Ffx ).real )
> > #
> > > 2nd fourier
> > > > IFfx = numpy.copy( numpy.fft.ifft2( Ffx ).real )
> > #
> > > inverse fourier
> > > >
> > > > ### Display result
> > > >
> > > > pylab.figure( 1, figsize=(8,8), dpi=125 )
> > > >
> > > > pylab.subplot(221)
> > > > pylab.imshow( fx, cmap=pylab.cm.gray )
> > > > pylab.colorbar()
> > > > pylab.title( "fx" )
> > > >
> > > > pylab.subplot(222)
> > > > pylab.imshow( Ffx, cmap=pylab.cm.gray )
> > > > pylab.colorbar()
> > > > pylab.title( "Ffx" )
> > > >
> > > > pylab.subplot(223)
> > > > pylab.imshow( FFfx, cmap=pylab.cm.gray )
> > > > pylab.colorbar()
> > > > pylab.title( "FFfx" )
> > > >
> > > > pylab.subplot(224)
> > > > pylab.imshow( IFfx, cmap=pylab.cm.gray )
> > > > pylab.colorbar()
> > > > pylab.title( "IFfx" )
> > > >
> > > > pylab.show()
> > > >
> > >
> > ---------------------------------------------------------------
> > > >
> > > > On my computer FFfx is the same as IFfx..... but
> > why?
> > > >
> > > > I uploaded a screenshot about my result here:
> > > > http://server6.theimagehosting.com/image.php?img=second_fourier.png
> > > >
> > > > Bela
> > > >
> > > >
> > > > --
> > > > View this message in context:
> http://www.nabble.com/second-2d-fft-gives-the-same-result-as-fft%2Bifft-tp23945026p23945026.html
> > > > Sent from the Numpy-discussion mailing list
> > archive at
> > > Nabble.com.
> > > >
> > > > _______________________________________________
> > > > Numpy-discussion mailing list
> > > > Numpy-discussion at scipy.org
> > > > http://mail.scipy.org/mailman/listinfo/numpy-discussion
> > > >
> > >
> > >
> > >
> > > --
> > > Information System Engineer, Ph.D.
> > > Website: http://matthieu-brucher.developpez.com/
> > > Blogs: http://matt.eifelle.com and http://blog.developpez.com/?blog=92
> > > _______________________________________________
> > > Numpy-discussion mailing list
> > > Numpy-discussion at scipy.org
> > > http://mail.scipy.org/mailman/listinfo/numpy-discussion
> > >
> >
> >
> >
> > _______________________________________________
> > Numpy-discussion mailing list
> > Numpy-discussion at scipy.org
> > http://mail.scipy.org/mailman/listinfo/numpy-discussion
> >
>
>
>
> _______________________________________________
> Numpy-discussion mailing list
> Numpy-discussion at scipy.org
> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>
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``` | 1,583 | 4,987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.795898 |
https://encyclopediaofmath.org/wiki/Bayes,_Thomas | 1,721,793,783,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00498.warc.gz | 213,346,165 | 9,704 | # Bayes, Thomas
This article Thomas Bayes was adapted from an original article by D.V. Lindley, which appeared in StatProb: The Encyclopedia Sponsored by Statistics and Probability Societies. The original article ([http://statprob.com/encyclopedia/ThomasBAYES.html StatProb Source], Local Files: pdf | tex) is copyrighted by the author(s), the article has been donated to Encyclopedia of Mathematics, and its further issues are under Creative Commons Attribution Share-Alike License'. All pages from StatProb are contained in the Category StatProb.
Thomas BAYES
b.c. 1701 - d. 7 April 1761
Summary. The problem of passing from a population to the properties of a sample was one of the first studied in probability. Thomas Bayes, a nonconformist minister, was the first to solve the inverse problem of passage from sample to population, using ideas that are widely used today.
Thomas Bayes, born in London, the son of a nonconformist minister, spent most of his adult life in a similar position in Tunbridge Wells, England. He was educated at Edinburgh University and was elected a fellow of the Royal Society in 1742. During his lifetime he published a few mathematical papers, of which the best-known is a 1736 defence of Newton's ideas against an attack by Bishop Berkeley. He is today remembered for a paper that his friend Richard Price claimed to have found amongst his possessions after death. It appeared in the Society's Transactions in 1763 and has often been republished. Apart from these bare facts, surprisingly little is known of Bayes' life.
By the middle of the 18th century it was well-understood that if, to use modern terminology, in each of $n$ independent trials, the chance of success had the same value, $\theta$ say; then the probability of exactly $r$ successes was given by the binomial distribution $$P(r|\theta,n) = {n \choose r}\theta^r(1-\theta)^{n-r}.$$
Jacob Bernoulli had established the weak law of large numbers and de Moivre had found the normal approximation to the binomial. The passage from a known value of $\theta$ to the empirical observation of $r$ was therefore extensively appreciated. Bayes studied the inverse problem; what did the data $(r,n)$ say about the chance $\theta$? There already existed partial answers. For example, Arbuthnot had observed $r$ male, and $n-r$ female, births with $r$ considerably in excess of $1/2n$. He argued that, on the basis of the binomial with $\theta = 1/2$, a value of $r$ as high as this was so improbable that $\theta$ could not be $1/2$. That idea has been much extended into the modern form of a significance test and its associated $P$-value or significance level.
Bayes proceeded differently using the theorem that nowadays always bears his name, though it does not appear explicitly in the 1763 paper, $$P(A|B) = P(B|A)P(A)/P(B)$$
for events $A, B$ with $P(B)\neq 0$. The theorem permits the inversion of the events in $P(B|A)$ into $P(A|B)$. Applied when $A$ refers to $\theta$ and $B$ to the empirical $r$, we have $$P(\theta|r,n) \propto P(r|\theta,n)P(\theta|n).$$
(The missing constant of proportionality does not depend on $\theta$. It is $P(r|n)^{-1}$ but is most easily found by making the product integrate to one by multiplying by the constant). The result effects the passage from the binomial, on the right, to a probability statement about the change, on the left. It therefore becomes possible to pass from the data to a statement about what are probable, and what are improbable, values of the chance.
This elegantly and simply solves the problem, except for one difficulty. It requires a value for $P(\theta|n)$, a probability distribution for the chance before the result of the trials has been observed. It is usual to describe this as the prior distribution (prior, that is, to $r$) and the final result as the posterior distribution. Thus the theorem describes how your views of $\theta$ change, from prior to posterior, as a result of data $r$. Bayes discussed the choice of prior but his approach is ambiguous. He is usually supposed to have taken $P(\theta|n)$ uniform in (0,1) - the so-called Bayes's postulate - but an alternative reading suggests he took $P(r|n)$ to be uniform. Mathematically these lead to the same result.
Little notice was taken of the 1763 paper at the time. It was first appreciated by Laplace, in the early years of the next century, who used the ideas in his eclectic approach to probability. The theorem is of basic importance because it provides a solution to the general problem of inference or induction. Let $H$ be a universal hypothesis and $E$ empirical evidence bearing on $H$. A simple example might be $H$, all swans are white, and $E$ the observation of the colour of a swan. A more sophisticated one would have $H$ as Newton's laws and $E$ observation of the motions of the planets. In either case, $P(E|H)$ can be calculated. Bayes's theorem says $$P(H|E) \propto P(E|H)P(H),$$
expressing a view about the hypothesis, given the evidence, in terms of the known probability of the evidence, given the hypothesis, and the prior view about $H$. As more evidence supporting $H$ accrues, having large probability on $H$, so even the sceptic, with low $P(H)$, will become convinced, $P(H|E)$ will approach one and the hypothesis accepted. Many people, following Jeffreys, who extensively developed these ideas into a practicable scientific tool, hold that this provides a description of the scientific method. This view differs from that of Popper, who only admits refutation of a hypothesis and whose attitude to probability is regarded as unsound by supporters of Jeffrey's ideas.
Recently Bayes's theorem has been used as a means of processing evidence in a court of law. Let $G$ be the hypothesis that the defendant is truly guilty of the offence with which he or she has been charged and $E$ a new piece of evidence. Then applying the theorem both to $G$ and to $\bar G$, denoting innocence, $$\frac{P(G|E)}{P(\bar G|E)} = \frac{P(E|G)}{P(E|\bar G)}\;\frac{P(G)}{P(\bar G)}.$$
The expression on the left is the odds on guilt, given $E$; that on the right is the same odds without $E$. The remaining term is the likelihood ratio, being a comparison of the probability of the evidence, supposing guilt, to the same probability supposing innocence. Forensic scientists present evidence to the court in the form of a likelihood ratio. The court can then multiply their former (prior) odds by the likelihood ratio to obtain new (posterior) odds as a result of hearing the evidence.
Bayes's result has attained increased importance following work by Ramsey, de Finetti and Savage between 1925 and 1955, which demonstrated that our knowledge had to be based on probability and that our beliefs must obey the rules of the probability calculus, of which Bayes's is essentially the multiplication rule. In this view, the significance levels of the classical school are unsound because they do not express opinions about hypotheses like $H$, or parameters like $\theta$, in terms of direct probabilities of $H$ or $\theta$. The resulting methodology is called Bayesian statistics and has rather different procedures and results from those of the classical school. Bayesians regard probability as a measure of a person's belief, whereas the classical school only admits probability as a frequency concept.
Bayes's result is also central to modern ideas on decision-making under uncertainty. Suppose there is a choice to be made amongst a set ${d}$ of decisions in the presence of uncertainty about a parameter $\theta$. The work of Ramsey and others leads to the introduction of a utility function $u(d, \theta)$; describing the worth of decision $d$ when the parameter has the value $\theta$, and the choice of that $d$ which maximizes the expected utility $\sigma_{\theta}u(d,\theta)P(\theta)$. Additional evidence $E$ updates $P(\theta)$ to $P(\theta|E)$, by the theorem, and improves the decision-making. All this is a long way from Bayes's original problem and its resolution. He would doubtless be astonished were he to realize how his wonderful idea has been extended and his name used.
#### References
[1] The original paper appeared in The Philosophical Transactions of the Royal Society of London (1763) 53, 370-418. There is a reprint in Biometrika (1958) 45, 296-315. An illuminating commentary on it is provided by S.M. Stigler (1982) Thomas Bayes's Bayesian Inference. Journal of the Royal Statistical Society, Series A, 145, 250-258. The most complete biography is provided by A.W.F. Edwards in the latest edition of The Dictionary of National Biography. [2] Two recent books on modern Bayesion methods are A.O'Hagan (1994) Bayesian Inference. Vol.2B of Kendall's Advanced Theory of Statistics. Edward Arnold, London; John Wiley, New York. J.M. Bernardo $&$ A.F.M. Smith (1994) Bayesian Theory. John Wiley, Chichester. The latter is part of a forthcoming 3-volume work and has an extensive bibliography. The modern `classic' is B. de Finetti (1974/5) Theory of Probability. John Wiley, London, in 2 volumes, translated from the Italian. [3] C.G.G. Aitken (1995) Statistics and the Evaluation of Evidence for Forensic Scientists. John Wiley, Chichester, deals with legal applications. D.V. Lindley (1985) Making Decisions. John Wiley, London, extends Bayesian ideas to decision-making.
Reprinted with permission from Christopher Charles Heyde and Eugene William Seneta (Editors), Statisticians of the Centuries, Springer-Verlag Inc., New York, USA.
How to Cite This Entry:
Bayes, Thomas. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Bayes,_Thomas&oldid=52886 | 2,274 | 9,668 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.97169 |
https://leanprover-community.github.io/archive/stream/113489-new-members/topic/begginer.20matrix.20question.html | 1,621,203,116,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989914.60/warc/CC-MAIN-20210516201947-20210516231947-00495.warc.gz | 389,856,452 | 4,580 | ## Stream: new members
### Topic: begginer matrix question
#### Chris Birkbeck (Apr 06 2021 at 09:56):
Hello, I'm new here and have been playing around with matrices in Lean. First of all, sorry if this is not the correct stream to ask this in.
My problems are essentially with matrix coercion. I’ve seen that we there is the following notation for matrices and I would like to know how to use it for special_linear_group. Specifically, how do I take
import linear_algebra.matrix
import data.matrix.notation
def N : matrix (fin 2) (fin 2 ) ℤ:= ![![-1, 0], ![0, -1]]
and turn it into an element of
import linear_algebra.special_linear_group
def SL2Z := special_linear_group (fin 2) ℤ
I assume this wont be automatic as one has prove N has determinant 1, but I dont know how to get lean to take a 2x2 matrix of determinant 1 and think of it as an element of SL2Z.
Thank you
#### Johan Commelin (Apr 06 2021 at 10:00):
@Chris Birkbeck Welcome! This stream is fine (-;
#### Johan Commelin (Apr 06 2021 at 10:00):
You can turn a matrix into an element of SL2Z using \< _, _ \>. If you type that into VScode, the \< and \> will turn into funny unicode brackets.
#### Johan Commelin (Apr 06 2021 at 10:01):
The end result will be \<your_matrix, proof_that_det_eq_1\>
#### Chris Birkbeck (Apr 06 2021 at 10:10):
Ah that's what I needed. Thank you very much!
#### Johan Commelin (Apr 06 2021 at 10:13):
I guess the following also works:
def my_matrix : SL2Z :=
_
Now put your cursor on the _, and wait till a little :bulb: appears.
#### Johan Commelin (Apr 06 2021 at 10:13):
Then you can click it, or type Ctrl-..
#### Johan Commelin (Apr 06 2021 at 10:14):
Now choose generate skeleton (almost at the bottom of hte menu)
#### Chris Birkbeck (Apr 06 2021 at 13:25):
Thats really useful, thank you. Now, say I wanted to use this notation. I was trying to prove that N^2 is 1. I end up with something that looks like
⟨N, proof⟩ * ⟨N, proof⟩ = 1
Now, I can't seem to figure out how to simplify this. I guess I'm confused how one splits this into working with the matrices and proofs separetely (but maybe one doesnt want to do this and there is something better?).
#### Johan Commelin (Apr 06 2021 at 13:26):
@Chris Birkbeck So now there is hopefully an ext-lemma for SL2Z that says that for proving equalities it suffices to check equality of the matrices.
#### Johan Commelin (Apr 06 2021 at 13:26):
If it doesn't exist apply subtype.coe_injective hopefully works. (I'm just guessing.)
#### Johan Commelin (Apr 06 2021 at 13:28):
@Chris Birkbeck You might want to just work with special_linear_group, instead of making your SL2Z definition on top of it.
#### Johan Commelin (Apr 06 2021 at 13:29):
Then the ext tactic will help you further.
#### Johan Commelin (Apr 06 2021 at 13:29):
Which says that if all coefficients are equal, then you are done.
#### Johan Commelin (Apr 06 2021 at 13:30):
After ext, you do intros i j, and then fin_cases i; fin_cases j should give you 4 goals, one for each coefficient.
#### Johan Commelin (Apr 06 2021 at 13:30):
With a bit of luck, each of them can be closed with simp.
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Question
# A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 2 days B had to leave and A alone completed the remaining work. The total time taken for the work to be completed was
A
8 days
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12 days
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10 days
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15 days
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Solution
## The correct option is B 12 days (A + B)'s 1 day's work = 115 + 110 = 16 Work done by A and B in 2 days = 16 × 2 = 13 Remaining work = 1 - 13 = 23 Now, 115 work is done by A in 1 day. So, 23 work will be done by A in 15 × 23 = 10 days Hence, the total time taken = 10 + 2 = 12 days.
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Frequent Visitor
## Formula help, create new column with values from 2 columns
Hi Everyone.
Not really sure how to explain this properly, but I am trying to figure out a way to calculate all rows with one result based on the values in 2 other columns. Here is my example below.
So if any of the order numbers have a TRUE from my if statmenent I would like the new column to carry that true into all rows for that order number.
Order Work True Desired result 10 TRUE TRUE 10 TRUE TRUE 10 FALSE TRUE 20 FALSE FALSE 20 FALSE FALSE 20 FALSE FALSE 30 FALSE TRUE 30 FALSE TRUE 30 TRUE TRUE
Thanks
1 ACCEPTED SOLUTION
Community Champion
Hi @Dave403
``Desired result = if(CALCULATE(COUNT(Table_2[Work True]),ALLEXCEPT(Table_2,Table_2[Order]), Table_2[Work True] = "TRUE")>=1,"TRUE","FALSE")``
2 REPLIES 2
Community Champion
Hi @Dave403
``Desired result = if(CALCULATE(COUNT(Table_2[Work True]),ALLEXCEPT(Table_2,Table_2[Order]), Table_2[Work True] = "TRUE")>=1,"TRUE","FALSE")``
Frequent Visitor
Hi Pranit
Works perfectly, thanks for the help.
Dave
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Top Kudoed Authors | 449 | 1,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-33 | latest | en | 0.805407 |
https://instrumentationtools.com/humix/video/8833b745c5581860cb7e7294cf1d4084c2259b23b3dc28fa086ac2790fdccf40 | 1,675,125,179,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499831.97/warc/CC-MAIN-20230130232547-20230131022547-00053.warc.gz | 340,199,249 | 37,333 | ###### How to Calculate 4 to 20 mA from Percentage? - Instrumentation Formula
3K views · Aug 31, 2022
Visit Channel:
```In this video, we will learn how to calculate the equivalent 4 to 20 mA current signal from the 0 to 100 percentage.
The 4 to 20 mA and 0 to 100 % signals are
0% - - - 4 mA
25 % - - - 8 mA
50 % - - - 12 mA
75 % - - - 16 mA
100 % - - - 20 mA
4-20 mA into 0-100 % Calculation Formula
## Instrumentation Formula
1. Formula to Calculate 4-20 mA from Process Variable
https://instrumentationtools.com/formula-to-calculate-4-20-ma-from-process-variable/
2. Formula for Percentage to Process Variable Conversion
https://instrumentationtools.com/formula-for-percentage-to-process-variable/
3. Formula for Current 4-20ma to Percentage Conversion
https://instrumentationtools.com/formula-for-current-to-percentage-conversion/
4. Calculate Process Variable from 4-20mA using Fraction of Measurement
https://instrumentationtools.com/calculate-process-variable-from-4-20ma-using-fraction-of-measurement/
5. Formulas to calculate mA from PV, LRV and URV
https://instrumentationtools.com/calculate-process-variable-from-4-20ma-using-fraction-of-measurement/
#4to20mA #InstrumentationFormula #InstrumentFormulas``` | 326 | 1,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.633439 |
http://www.cfo.com/printable/article.cfm/14574893 | 1,529,425,353,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863100.8/warc/CC-MAIN-20180619154023-20180619174023-00032.warc.gz | 398,166,858 | 3,895 | # Spreadsheets: More than One Way to Skin a VLOOKUP
Use the one-two punch of these concatenated keystrokes to solve a VLOOKUP puzzle.
Bill Jelen, CFO.com | US
May 11, 2011
Editor's Note: To read more about spreadsheets, share a tip with other readers, or suggest a topic for Bill Jelen to cover in an upcoming column, click here to visit CFO.com's Spreadsheet Tips Page.
In last week's column, Greg asked about a multiconditional wizard that would generate something similar to a SUMIF. While that column explained the old Conditional Sum Wizard, reader Dominique B. writes: "It seems to me the reader's request was not fully answered, as he was looking for a multi-condition VLOOKUP, not a sum. My recommended solution in this situation is to add an extra column containing a composite text key built using a separator unlikely to appear in the actual data (e.g., a double underscore). In your example, this composite key would be 'C__Bob' and using VLOOKUP you can achieve what I understand Greg T. wanted. Thanks for you column, which brings out small Excel gems." For her suggestion, Dominique wins an autographed copy of Learn Excel from MrExcel from CFO and MrExcel.com.
There are many ways to solve most problems in Excel. The sum formula from last week's column would handle multiple matches, but would not return text results. The methods explained in this week's column (below) ignore multiple matches, but can return a text result. Which method you use depends on the shape of your data.
Say, for example, that you need to do a VLOOKUP to fill in the yellow range in Figure 1 (below).
Fig. 1
If you are allowed to change the original data set, you could insert a new column C (see Figure 2). Put this formula in C2 and copy down:
=A2&"__"&B2
Fig. 2
There is nothing magic about the double underscore; Dominique used it as it is unlikely to appear anywhere in the data in columns A or B. The formula for G3 is then =VLOOKUP(G\$2&"__"&\$F3,\$C\$2:\$D\$27,2,False). (See Figure 3.)
Fig. 3
If you are not allowed to change the original data set to add the concatenated key, and if the original data set is sorted by company, then you can use the method illustrated in Figure 4.
Fig. 4
CFO contributing editor Bill Jelen is the author of 32 books about Excel and the host of MrExcel.com. You can win a copy of one of his books if your question or suggestion is selected for a column. Post your question at the Community Center "Ask a Question" box on the right. Tune into Bill's next Excel Webcast, "Intro to VBA and Editing Recorded Code," on Wednesday, May 18, at 2:00 pm. | 634 | 2,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-26 | latest | en | 0.896097 |
https://www.physicsforums.com/threads/what-does-up-indice-paired-with-up-indice-mean-in-relativity-notation.848349/ | 1,722,783,151,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640404969.12/warc/CC-MAIN-20240804133418-20240804163418-00559.warc.gz | 743,563,422 | 20,247 | # What Does Up Indice Paired with Up Indice Mean in Relativity Notation?
• Joe D
In summary, this equation is trying to find the four functions that correspond to the values of beta for a given electric field and magnetic field.
Joe D
## Homework Statement
Find t(τ), x(τ), y(τ), z(τ) for constant electric field E=E(sinθx+cosθz) and B=Bz, and constant magnetic field where E,B, and θ are all constants.
I haven't seen this notation before, and I really just want to know what it means specifically. I know and up indice paired with a down indice essentially gives an inner product, but I don't really know what an up indice paired with an up indice gives me. So this is equation 12.32 in Jackson Electrodynamics that I'm trying to understand.
## Homework Equations
m(d2xα/dτ2)=(e/c)(∂αAβ-∂βAα)(dxβ/dτ)
## The Attempt at a Solution
Also, the 4 vector potential that I've already calculated is:
Aα=(-E(xsinθ+zcosθ),-By,0,0) where E, B and θ are all constants.
or Aα=(-E(xsinθ+zcosθ),0,Bx,0) I can't decide which one is more convenient.
I've attached an image of my best guess for what that equation means.
#### Attachments
• IMG_20151215_143708814_HDR.jpg
25.3 KB · Views: 380
Last edited:
I think its best to think of this way:
Since beta is a free index, you get one equation for each value of beta (so you get 4 equations). In each of those equations, alpha is summed over (it is up for A and down for x). So each one of those equations can now be solved for the four functions you need.
By the way for n free indicides in an equation in dimension d, you get d^n equations. In this case, d=4 and n=1
Last edited:
Brian T said:
I think its best to think of this way:
Since beta is a free index, you get one equation for each value of beta (so you get 4 equations). In each of those equations, alpha is summed over (it is up for A and down for x). So each one of those equations can now be solved for the four functions you need.
By the way for n free indicides in an equation in dimension d, you get d^n equations. In this case, d=4 and n=1
Just a small correction (I am sure it is a typo): it is alpha which is a free index and beta is summed over.
nrqed said:
Just a small correction (I am sure it is a typo): it is alpha which is a free index and beta is summed over.
Ohh thanks. Got them flipped xD
## What is relativity mathematical notation?
Relativity mathematical notation is a system of symbols and equations used to represent the concepts and principles of Einstein's theory of relativity. It allows for precise and concise communication of complex ideas in the field of physics.
## What is the difference between special relativity and general relativity?
Special relativity is a theory that describes the laws of physics in non-accelerating frames of reference, while general relativity is a theory that extends these laws to include accelerating frames of reference and the effects of gravity.
## What is the significance of the speed of light in relativity?
The speed of light, denoted by the symbol c, is a fundamental constant in relativity. It is the maximum speed at which all energy, matter, and information can travel. It also plays a crucial role in the equations of special relativity, including the famous equation E=mc^2.
## What is the role of tensors in relativity?
Tensors are mathematical objects used to describe the relationships between different quantities in space and time. In relativity, tensors are used to represent concepts such as energy, momentum, and curvature, and they allow for the precise formulation of Einstein's equations.
## How does relativity mathematical notation differ from classical mechanics notation?
Relativity mathematical notation differs from classical mechanics notation in several ways. For example, it includes the use of four-vectors and tensors to describe quantities in space and time, as well as the incorporation of the speed of light as a fundamental constant. It also differs in terms of the concepts and principles it represents, as relativity introduces new ideas such as time dilation and the curvature of spacetime.
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Question 8
# Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
Speed of sound in air = 346 m/s
Speed of sound in Aluminium = 6420 m/s
Let the length of the aluminium rod = d.
Therefore, time taken by the sound wave to reach the other end,
Therefore, time taken by sound wave to reach the other end,
The ratio of time taken by the sound wave in air and aluminium:
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# Parents-Shocked-Over-Violent-Math-Problems
page: 1
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posted on Mar, 2 2012 @ 05:21 AM
If you think ATS breeds Conspiracy Theorists. Wait till you get a load of these guys. These people are getting the kids minds primed and ready while they're still in 3rd grade. I never had any questions on tests like this when I was in school. What a weird World we live now where space aliens, cannibalism, mass murder and terrorism are part of our elementary school curriculum.
Read this article and the actual math test and tell me if you don't find it a bit odd. I suppose if it keeps the kids interested in math and in school it can't be 100% bad. But I don't know how good these kinds of questions are going to be for the mental health of our future generations.
Parents-Shocked-Over-Violent-Math-Problems
WASHINGTON, D.C. (WUSA) - Some DC parents are shocked over math problems that were sent home with third graders at the Trinidad Center City Public Charter School. One says the "bloodthirsty aliens then sucked the blood of 828 teachers and left them for dead." Another talks about cooking Africans, Americans and Indians in ovens to bake: "How many desperate people were in each oven?"
There are killings, kidnappings and deaths. Here, on his birthday, a boy named John swallows 585 marbles and died. Only nine friends came to his funeral.
Math Problem 1
Green aliens landed in Chicago and rounded up 1479 math teachers. The bloodthirsty aliens then sucked the blood of 828 teachers and left them for dead. The aliens tied up the rest of the teachers and marched them into 3 UFOs. If there were an equal number of poor math teachers in each UFO, how many teachers were in each UFO?
Math Problem 3
Human terrorists landed on Planet Xintaks and abducted Princess
Txsclotyl. The enraged aliens hurled 1420 asteroids at 7 terrorist training
camps on Earth. 482 asteroids landed harmlessly in the Pacific Ocean.
However, the rest of the asteroids smashed into each camp in equal
numbers. How many asteroids smashed into each camp
Math Problem 13
Brian, a brave member of a SWAT team in California, had a terribly busy
week last week. He had to work for 7 whole days. He killed 163 terrorists,
296 murderers and 206 arsonists. How many criminals did he kill on
average each day?
What are we teaching the kids these days?
edit on 2-3-2012 by ThirdRock69 because: (no reason given)
posted on Mar, 2 2012 @ 05:37 AM
WOW thats sneaky!
I certinly never had any questions like that at school, it was all apples oranges and random first names.
"timmy has 5 apples and jonny gives him the 3 apples he had, how many apples does timmy have now."
I think parents need to be paying attention to this, sounds almost unbelivable!
posted on Mar, 2 2012 @ 05:45 AM
Right that sounds more familiar.
How can these kids focus on math when they are reading these kinds of horrible acts.
Is it a way to further desensitize our youth to killing or to create more fear of terrorism and such?
posted on Mar, 2 2012 @ 05:55 AM
I found myself laughing at first, then had a strange compulsion to do the maths
Hmmmmmmmm back to school for me lol.
posted on Mar, 2 2012 @ 05:57 AM
Is this in America? Clicking on the link for the worksheet says it's in Singapore...
NVM I'm stupid...clearly says DC..carry on
edit on 2-3-2012 by WatcherMan because: (no reason given)
posted on Mar, 2 2012 @ 06:02 AM
I watched the clip and my question is what does any of this have to do with "violent video games" why even mention games at all? Strange math questions though, when it was mentioned the quiz came from a website my first thought was maybe the site was hacked and someone thought it was funny to replace the questions with these.
posted on Mar, 2 2012 @ 06:08 AM
In the follow up article. It says the math teacher got fired for this.
Another teacher is made to be the scapegoat it sounds like.
I would like to hear from the people who wrote and published this test for 3rd graders. What's their excuse?
Violent Math problems lead to teachers firing
WASHINGTON, D.C. (WUSA) -- The teacher who sent home those violence-laced math problems at Center City Public Charter School's Trinidad campus in Northeast has been fired as of Thursday.
Wheeler says instead of using the resources provided in the school for the mandated 20 homework problems on Monday, the third grade teacher downloaded them off the the 'HomeschoolingParadise.com' website. She says the teacher used "bad judgment".
Several parents have insisted that the teacher, who they say is a minister, would never have sent those problems home, unless someone made her do it.
"Not true", said Wheeler. She explained that the mistake was worthy of firing because parents trust the school to look at what they're putting in front of them children.
She said that even if a teacher were told to send those problems home, she or he should question it.
posted on Mar, 2 2012 @ 06:10 AM
I think what is happening here, is that the teachers, or whoever created the questions, are trying to be "down" with it, trying to speak to the kids in a language, that they (the teachers), think the kids can associate with. They think the kids play violent video games, hang out with gangs, wear the pants to low, etc etc, so the teachers try to be "cool" creating these questions.
I am not saying it right, because the questions are rather explicit, however, i feel that is what they are trying to do.
vvv
posted on Mar, 2 2012 @ 06:16 AM
I think what is happening here, is that the teachers, or whoever created the questions, are trying to be "down" with it, trying to speak to the kids in a language, that they (the teachers), think the kids can associate with. They think the kids play violent video games, hang out with gangs, wear the pants to low, etc etc, so the teachers try to be "cool" creating these questions.
I would agree with you if this were a high school test or even middle school. But this was a 3rd grade test. That's typically around 8 years old. Way to young to be into gangs and violence. If this was a movie script it would be rated R or PG-13 at least.
edit on 2-3-2012 by ThirdRock69 because: (no reason given)
posted on Mar, 2 2012 @ 06:16 AM
Originally posted by ThirdRock69
Math Problem 13
Brian, a brave member of a SWAT team in California, had a terribly busy
week last week. He had to work for 7 whole days. He killed 163 terrorists,
296 murderers and 206 arsonists. How many criminals did he kill on
average each day?
I always had problems with math like this and the basic problems are still there.
How many criminals did he kill. Well that depends. I would not classify a terrorist as a criminal. So we now have two viable answers. I could also ask if the 163 terrorists also murdered anyone or started fires, both likely events and therefore they could have been counted multiple times. This is police and government figures after all, they do like to stretch the math. How the hell is a third grade student supposed to answer.
The questions are too hard.
posted on Mar, 2 2012 @ 06:22 AM
Originally posted by ThirdRock69
In the follow up article. It says the math teacher got fired for this.
Good.
2
posted on Mar, 2 2012 @ 06:25 AM
Remember we are supposed to assume that any one who is labeled a terrorist is automatically a criminal even if they haven't been found guilty in a court of law yet.
So if we assume that all of them are criminals. This is a problems about averages.
Math Problem 13
Brian, a brave member of a SWAT team in California, had a terribly busy
week last week. He had to work for 7 whole days. He killed 163 terrorists,
296 murderers and 206 arsonists. How many criminals did he kill on
average each day?
Fisrt step is to add up all the criminals
163 + 296 + 206 = 665
Then divide by 7 days
665 divided 7 = 95 criminals per day
posted on Mar, 2 2012 @ 06:34 AM
No it's govt reporting. The murders and arson attacks were carried out by the terrorists so it is simply 163/7 = 23 and a bit.
posted on Mar, 2 2012 @ 06:41 AM
163/7 = 23 and a bit.
Well no wonder that Brian is working 7 days a week.
With numbers that low he's not meeting the quota.
posted on Mar, 2 2012 @ 06:48 AM
Oh boy, where is the school system heading to?
on a positive note we can say, that with kids learning math with aliens and terrorists, we will see alot of new ATS members soon after school is over.
Another one (slightly off topic) is the teacher who sent a student to detention, since the student repeatedly opposed the teachers claim that 'a kilometer is longer than a mile'.
He also writes, that the rest of the class was all agreeing...
Here s the link teachers note
posted on Mar, 2 2012 @ 06:49 AM
Originally posted by ThirdRock69
In the follow up article. It says the math teacher got fired for this.
Violent Math problems lead to teachers firing
WASHINGTON, D.C. (WUSA) -- The teacher who sent home those violence-laced math problems at Center City Public Charter School's Trinidad campus in Northeast has been fired as of Thursday.
SNIP
She said that even if a teacher were told to send those problems home, she or he should question it.
Read the final line in this NEWS ARTICLE 'Wrottem by Peggy Fox'.
Most disturbing...what school did Peggy Fox go to ...this is a sign of the times folks...the world is going to hell, one bad spelling at a time...btw I wonder why they have not interviewed the teacher further to find out WHO MADE YOU DO THIS (because she said she was 'made to'). Now, I know these questions have been downloaded of a homeschooling website, but the problems certainly look like they're designed to make genocide, kidnapping, torture, baking ppl in ovens, violence, aggression, death, acceptable in the heads of the kids, instead of the shock-creating, horrific acts they really are. Reminds of of that FEMA camp rumor.
note: did a whois, homeschooling-paradise.com is owned by some Malasian with an Indian name. tres bizarre.
posted on Mar, 2 2012 @ 07:24 AM
Originally posted by VreemdeVlieendeVoorwep
I think what is happening here, is that the teachers, or whoever created the questions, are trying to be "down" with it, trying to speak to the kids in a language, that they (the teachers), think the kids can associate with. They think the kids play violent video games, hang out with gangs, wear the pants to low, etc etc, so the teachers try to be "cool" creating these questions.
I am not saying it right, because the questions are rather explicit, however, i feel that is what they are trying to do.
vvv
Yet another completely implausible rationalization proposed by an ATS member. What's going on here? Does anybody think 3rd graders would play such sick video games? Does anybody think a teacher would care about being perceived as "down with" anything to impress 3rd graders? Or that 3rd graders would make a connection between a teachers' interests in vid games and a cool factor?
It's time to start considering what would have been incomprehensible last year. Just as the banks appear to be run by depraved luciferian thieves, the same is most likely true here.
There has been a 300% increase year over year in arrests of child molesters in public schools where I live. The sick f*cks have infected our schools from Secratery Arnie Duncan, right down to teachers in DC schools.
posted on Mar, 2 2012 @ 07:30 AM
The article was written by Peggy Fox but the person who made the comment was Beverley Wheeler the school's CEO.
www.wusa9.com...
"I was absolutely distressed ", says Dr. Beverley Wheeler, the CEO of Center City PCS which oversees it's Trinidad elementary school in NE.
That's another funny thing I never had in the public school system while I was growing up.
I never knew any public school that had a CEO. We all had principals who were at the top.
Now these schools are run like a business and not concerned with educating kids as much as they are indoctrinating them with propaganda to become future sheeple.
edit on 2-3-2012 by ThirdRock69 because: (no reason given)
posted on Mar, 2 2012 @ 07:32 AM
Yet another completely implausible rationalization proposed by an ATS member. What's going on here?
I will admit, i did not see that the test was for 3rd graders, and i apologize.
If this was for older kids though, i would stick to this as a possible explanation. Because honestly, there has to be an explanation for this, i cannot accept it that it was just because some teacher wanted fun.
However, i apologize for not noticing it was for 3rd graders, now calm down please. I made a mistake, and it was pointed out.
vvv
posted on Mar, 2 2012 @ 07:33 AM
1 too many
2 not enough
3 not enough
happy columbus day
new topics
top topics
12 | 3,105 | 12,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.936844 |
https://www.teacherspayteachers.com/Product/2nd-9-Weeks-Lesson-Plans-for-3rd-Grade-Math-Common-Core-1941886 | 1,511,595,427,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809695.97/warc/CC-MAIN-20171125071427-20171125091427-00536.warc.gz | 838,208,944 | 25,129 | Total:
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# (2nd 9 Weeks) Lesson Plans for 3rd Grade Math Common Core
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Product Description
You are purchasing 8 weeks of math lesson plans aligned to the common core standards. This is just for the lesson plans.
I sell the lesson plans and sheets together for \$7 a week or \$45 a nine weeks bundle.
In this bundle, the skills are in the following order:
Week 8: Skill - 3.OA.6 Understand division as an unknown factor problem
Week 9: Skill - 3.OA.7 Fluently Multiply and Divide within 100
Week 10: Skill - 3.OA.8 2 Step Word Problems - Addition & Subtraction
Week 11: Skill - 3.OA.9 Patterns - Function Machines
Week 12: Skill - 3.OA.8 2Step Word Problems - Multiply & Divide & 3.OA.9 Patterns - Input/Output
Week 13: Skill - 3.NBT.3 Multiply one-digit numbers by multiples of 10
Week 14: Skill - 3.MD.3 Graphs - Bar Graphs
Week 15: Skill - 3.MD.3 Graphs - Picture Graph
The lesson plan includes -
Common Core Standard
Essential Questions
I Can Statements
Vocabulary
Literature Connection
This is a detailed lesson plan that follows the following format
Day 1
Math Journal - Introduce strategy #1 (direct instruction)
Target Skill Lesson
Practice Skill Sheets (10 pages included - the front side is skill practice and the back side is a review of previously taught skills)
Day 2
Math Journal - Introduce strategy #2 (direct instruction)
Target Skill Lesson
Practice Skill Sheets (10 pages included - front side skill practice and the back side is a review of previously taught skills)
Day 3
Follows a guided math schedule: (no whole class instruction on this day)
+Journal Problem of the Day (multiple choice question aligned to EOG)(included)
+Struggling students work with teacher on weekly math skill sheet (included)
+advanced students work on enrichment activities (links included)
+ students work in center on foundation math facts
Day 4
Follows a guided math schedule: (no whole class instruction on this day)
+Journal Problem of the Day (multiple choice question aligned to EOG)(included)
+Struggling students work with teacher on weekly math skill sheet (included)
+advanced students work on enrichment activities (links included)
+ students work in center on foundation math facts
Day 5
Assessment (included)
Each weekly lesson plan focuses on 2 math strategies that can be added to a classroom chart for easy reference. The students also add the strategies to their math journal on Monday and Tuesday. The weekly strategies are listed below.
Week 8 - Use a Model - Number line to divide
Week 8 - Use a Model - Bar Model
Week 9 - Number Bonds
Week 9 - Understand the Multiplication Table
Week 10 - Paraphrasing
Week 10 - Visualize and Draw
Week 11 - Working Backwards
Week 11 - Look for a pattern
Week 12 - Input / Output Function tables
Week 12 - Guess and Check
Week 13 - Draw and Use Base Ten Blocks
Week 13 - Break Numbers Apart
Week 14 - Analyze - What do you notice?
Week 14 - Analyze - What does it not tell you?
Week 15 - Parts of a Picture Graph
Week 15 - Answering Test Questions
The plans are included using Microsoft Word so you can easily make changes.
School Year Worksheets & Lesson Plans for 3rd Grade Math Common Core aligned
Math Lesson Plans:
(1st 9 Weeks) Lesson Plans for 3rd Grade Math Common Core
(2nd 9 Weeks) Lesson Plans for 3rd Grade Math Common Core
(3rd 9 Weeks) Lesson Plans for 3rd Grade Math Common Core
(4th 9 Weeks) Lesson Plans for 3rd Grade Math Common Core
Math Lesson Plans & Practice Sheets:
Math Plans & Sheets (1st 9 Weeks) 3rd Grade Common Core
Math Plans & Sheets (2nd 9 Weeks) 3rd Grade Common Core
Math Plans & Sheets (3rd 9 Weeks) 3rd Grade Common Core
Math Plans & Sheets (4th 9 Weeks) 3rd Grade Common Core
Total Pages
40+
N/A
Teaching Duration
1 hour
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\$15.00 | 1,033 | 4,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-47 | latest | en | 0.880001 |
https://www.hackmath.net/en/math-problem/121?tag_id=38 | 1,611,563,408,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565376.63/warc/CC-MAIN-20210125061144-20210125091144-00148.warc.gz | 799,162,943 | 15,592 | # RT - hypotenuse and altitude
Right triangle BTG has hypotenuse g=117 m and altitude to g is 54 m.
How long are hypotenuse segments?
Correct result:
gb = 36 m
gt = 81 m
#### Solution:
$g_b+g_t=117 \ \\ g_b \cdot g_t = 54^2 \ \\ \ \\ g^2 -117g +2916 =0 \ \\ \ \\ a=1; b=-117; c=2916 \ \\ D = b^2 - 4ac = 117^2 - 4\cdot 1 \cdot 2916 = 2025 \ \\ D>0 \ \\ \ \\ g_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 117 \pm \sqrt{ 2025 } }{ 2 } \ \\ g_{1,2} = \dfrac{ 117 \pm 45 }{ 2 } \ \\ g_{1,2} = 58.5 \pm 22.5 \ \\ g_{1} = 81 \ \\ g_{2} = 36 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (g -81) (g -36) = 0 \ \\ \ \\ \ \\ g_b=36 \ \text{m}$
$g_t = 81 \ \text{m}$
Try calculation via our triangle calculator.
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Calculate the perimeter and the area of square with a diagonal length 30 cm. | 1,105 | 3,809 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.812705 |
https://acemypapers.com/net-present-value-help-please/ | 1,657,135,195,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104676086.90/warc/CC-MAIN-20220706182237-20220706212237-00173.warc.gz | 124,138,589 | 12,164 | # “Net Present Value” Help Please
Assignment: Chapter 14 LO 3
1. Cornerstone Exercise 14-23
Net Present Value
Holland, Inc., has just completed development of a new cell phone. The new product is expected to produce annual revenues of \$1,350,000. Producing the cell phone requires an investment in new equipment, costing \$1,440,000. The cell phone has a projected life cycle of five years. After five years, the equipment can be sold for \$180,000. Working capital is also expected to increase by \$180,000, which Holland will recover by the end of the new product’s life cycle. Annual cash operating expenses are estimated at \$810,000. The required rate of return is 8 percent.
1. Prepare a schedule of the projected annual cash flows.
Holland, Inc.
Annual cash flows
For Five Years
Years and Items
Cash Flow
Year 0
Equipment
\$ (1440000)
Working Capital
(180000)
Total
\$ (1620000)
Year 1-4
Revenues
\$ 1350000
Operating Expenses
(810000)
Total
\$ 540000
Year 5
Revenues
\$ 1350000
Operating Expenses
(810000)
Salvage
180000
Recovery of Working
180000
Total
\$ 900000
2. Calculate the NPV using only discount factors from Exhibit 14B-1. Round present value calculations and your final answer to the nearest whole dollar.
\$ _________________
3. Calculate the NPV using discount factors from both Exhibit 14B-1 and Exhibit 14B-2. Round present value calculations and your final answer to the nearest whole dollar.
\$ _________________
2. Exercise 14-28 Net Present Value Use the Exhibit 14B-1 and Exhibit 14B-2 to locate the present value of an annuity of \$1, which is the amount to be multiplied times the future annual cash flow amount. Each of the following scenarios is independent. Assume that all cash flows are after-tax cash flows. a. Southward Manufacturing is considering the purchase of a new welding system. The cash benefits will be \$400,000 per year. The system costs \$2,250,000 and will last 10 years. b. Kaylin Day is interested in investing in a women’s specialty shop. The cost of the investment is \$180,000. She estimates that the return from owning her own shop will be \$35,000 per year. She estimates that the shop will have a useful life of six years. c. Goates Company calculated the NPV of a project and found it to be \$21,300. The project’s life was estimated to be eight years. The required rate of return used for the NPV calculation was 10 percent. The project was expected to produce annual after-tax cash flows of \$45,000. 1. Compute the NPV for Southward Manufacturing, assuming a discount rate of 12 percent. Round to the nearest dollar. \$ _________________ Should the company buy the new welding system? __Yes__ 2. Conceptual Connection: Assuming a required rate of return of 8 percent, calculate the NPV for Kaylin Day’s investment. Round to the nearest dollar. \$ _________________ Should she invest? __No__ Calculate the NPV assuming the estimated return was \$45,000 per year. Round to the nearest dollar. \$ ________ Would this affect the decision? What does this tell you about your analysis? The input in the box below will not be graded, but may be reviewed and considered by your instructor. _________________ 3. What was the required investment for Goates Company’s project? Round to the nearest dollar. \$ _________________
3. Exercise 14-30
Net Present Value and Competing Projects
Follow the format shown in Exhibit 14B-1 and Exhibit 14B-2 as you complete the requirements below. You may use the attached spreadsheet to help you complete this activity, but you are not required to do so. You will find the spreadsheet by clicking on the link in the drop-down menu above.
Wilburton Hospital is investigating the possibility of investing in new dialysis equipment. Two local manufacturers of this equipment are being considered as sources of the equipment. After-tax cash inflows for the two competing projects are as follows:
Both projects require an initial investment of \$700,000. In both cases, assume that the equipment has a life of five years with no salvage value.
Round present value calculations and your final answers to the nearest dollar.
1. Assuming a discount rate of 12 percent, compute the net present value of each piece of equipment.
Puro equipment: \$ ______________ Briggs equipment: \$ ______________
2. A third option has surfaced for equipment purchased from an out-of-state supplier. The cost is also \$700,000, but this equipment will produce even cash flows over its five-year life. What must the annual cash flow be for this equipment to be selected over the other two? Assume a 12 percent discount rate.
\$ _________________
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Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result. | 1,343 | 5,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-27 | latest | en | 0.864485 |
abhinavgupta2812.wordpress.com | 1,607,067,124,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141735395.99/warc/CC-MAIN-20201204071014-20201204101014-00556.warc.gz | 186,124,103 | 39,009 | I have working pretty hard on multiple projects recently, and that’s why I haven’t got the time to document my experiments with Cubieboard 2, but rest assured a good number of blog posts are coming up on that front. However, I picked up this game from the internet called the 2048-tile game. Its a pretty neat (and addictive game) you can find the link here.
Here is the jist of what the game is – you have a 4*4 tile board and begins with 2 tiles with the number “2” on them. You can use the arrow keys to move all the tiles on the board in the 4 directions. Two tiles with the same number with combine to form a new tile with the next power of two, when they move in the direction you want them to.
Also, with every direction move, the game spawns a random “2” or “4” in one of the empty spaces on the board.
As an example, I have shown the starting frame, and the frames after the first move in the left directions and the second move in the downward direction, notice how the tiles of 2 combine to make a tile of 4. You win the game when you reach the 2048 tile; or if you run out of space to move (the more likely possibility). Its more difficult than it sounds, not to forget, the game is very addictive.
A lot of people I know, have played the game, but not many have succeeded in reaching to the 2048 tile (If github is to be believed, even the creator himself hasn’t finished the game yet.) The one thing I realized that was always messing up my game was these godforsaken tiles coming up from all directions and blocking by bigger tiles – and it forced me to take another route which took a lot of space on the board. So I figured that restricting the directions from which the tiles could block would be very crucial to the solution.
A simple way to do the above would be to restrict all your movement to the bottom side of the board and never pressing the up direction, just use the lefts and rights and downs to manoeuvre the tiles as much as you can. There will be a few instances where you will have to use the up direction thus causing a disruption if a 2 or 4 blocks your tiles – but its still possible to control single anomalies.
The second step that you need to take is to ensure that the you block the bottom left corner for the tile with the highest power of two, and the subsequent tiles in the bottom row arranged in the descending order. This is very important – because you will soon realize that when you do this, a single tile can sweep away all tiles taking you to the next higher power of two – making your path and hence your movements defined at any point of time – basically at any point in the game, you would be concentrating in making a power of two sufficient enough to trigger a chain reaction of geometric progressions.
A few pictures will make this clearer:
The above picture is an example of this strategy, I kept the all my movements devoid of the up button, and concentrated on keeping the highest tile in the corner with the bottom row in the descending order. If you notice the images, all it will take in say the first image is a tile of 32 to introduce the 256 tile, similarly you see the strategy being applied in the subsequent images as well.
It takes time, but once you have the flow going you know what particular tile to focus on next and once you do that, the entire game becomes a series of local problems.
And now for the winning set of tiles (Narcissistic ego boost ):
I have tried this strategy, and have reached the end goal three times. It seems like a strategy that is reproducible and seems logical that it will work. However, I am trying to see if the problem can be formalized in mathematical terms, and hence the solution would be a formal one as well.
It’s been a long time since I had/did something worth writing for the blog; never got around to hacking more on the Cubieboards. However, this morning I got a chance to get working on a Cubieboard 2! So got an SD card and was all prepared to get an OS up and running on it. But as luck would have had it, the SD card refused to do anything. Tried checking the dmesg log, and got this:
`mmc0: error -110 whilst initializing SD card `
Seemed more or less like a driver issue of some sort, so the first thing that I tried was running lspci and getting what the hardware actually is. This was my lspci output:
```abhinav@AsusK55V:/~ lspci
00:00.0 Host bridge: Intel Corporation Ivy Bridge DRAM Controller (rev 09)
00:01.0 PCI bridge: Intel Corporation Ivy Bridge PCI Express Root Port (rev 09)
00:02.0 VGA compatible controller: Intel Corporation Ivy Bridge Graphics Controller (rev 09)
00:14.0 USB controller: Intel Corporation Panther Point USB xHCI Host Controller (rev 04)
00:16.0 Communication controller: Intel Corporation Panther Point MEI Controller #1 (rev 04)
00:1a.0 USB controller: Intel Corporation Panther Point USB Enhanced Host Controller #2 (rev 04)
00:1b.0 Audio device: Intel Corporation Panther Point High Definition Audio Controller (rev 04)
00:1c.0 PCI bridge: Intel Corporation Panther Point PCI Express Root Port 1 (rev c4)
00:1c.1 PCI bridge: Intel Corporation Panther Point PCI Express Root Port 2 (rev c4)
00:1c.3 PCI bridge: Intel Corporation Panther Point PCI Express Root Port 4 (rev c4)
00:1d.0 USB controller: Intel Corporation Panther Point USB Enhanced Host Controller #1 (rev 04)
00:1f.0 ISA bridge: Intel Corporation Panther Point LPC Controller (rev 04)
00:1f.2 SATA controller: Intel Corporation Panther Point 6 port SATA Controller [AHCI mode] (rev 04)
00:1f.3 SMBus: Intel Corporation Panther Point SMBus Controller (rev 04)
01:00.0 VGA compatible controller: NVIDIA Corporation Device 1058 (rev ff)
03:00.0 Network controller: Atheros Communications Inc. AR9485 Wireless Network Adapter (rev 01)
04:00.0 Unassigned class [ff00]: Realtek Semiconductor Co., Ltd. Device 5289 (rev 01)
04:00.2 Ethernet controller: Realtek Semiconductor Co., Ltd. RTL8111/8168B PCI Express Gigabit Ethernet controller (rev 0a)
```
The device 04:00.0 seems to be unrecognised by the OS, so my guess was it was the SD card reader, and on googling the part number, turns out it was! And also that these drivers were not on the mainline kernel. The issue is faced by a lot of other brands of laptops as well, HP, Dell what not! On checking the Realtek’s driver download page here. However, they do not have a driver for 5289, the very device that my Asus has (But I still did not lose it on Realtek, I mean they have amazing WiFi drivers/hardware support). So I decided to look around, and luckily found a mirror where they still have the driver for 5289. In case you are going through the same issue, you can follow the steps here:
```abhinav@AsusK55V:/~ wget https://bugs.launchpad.net/bugs/971876/+attachment/2991730/+files/rts_bpp.tar.bz2
abhinav@AsusK55V:/~ tar jxf rts_bpp.tar.bz2
abhinav@AsusK55V:/~ cd rts_bpp
abhinav@AsusK55V:/~ make
abhinav@AsusK55V:/~ su -c 'make install ; modprobe rts_bpp ; depmod -a' ```
All hell broke loose after this, I start getting these compilation errors while running make:
```/home/abhinav/tmp/rts_pstor/rtsx.c:916:22: error: expected ‘=’, ‘,’, ‘;’, ‘asm’ or ‘__attribute__’ before ‘rtsx_probe’
/home/abhinav/tmp/rts_pstor/rtsx.c:1080:23: error: expected ‘=’, ‘,’, ‘;’, ‘asm’ or ‘__attribute__’ before ‘rtsx_remove’
/home/abhinav/tmp/rts_pstor/rtsx.c:1106:11: error: ‘rtsx_probe’ undeclared here (not in a function)
/home/abhinav/tmp/rts_pstor/rtsx.c:1107:2: error: implicit declaration of function ‘__devexit_p’ [-Werror=implicit-function-declaration]
/home/abhinav/tmp/rts_pstor/rtsx.c:1107:24: error: ‘rtsx_remove’ undeclared here (not in a function)
/home/abhinav/tmp/rts_pstor/rtsx.c:485:12: warning: ‘rtsx_control_thread’ defined but not used [-Wunused-function]
/home/abhinav/tmp/rts_pstor/rtsx.c:596:12: warning: ‘rtsx_polling_thread’ defined but not used [-Wunused-function]
/home/abhinav/tmp/rts_pstor/rtsx.c:745:13: warning: ‘quiesce_and_remove_host’ defined but not used [-Wunused-function]
/home/abhinav/tmp/rts_pstor/rtsx.c:780:13: warning: ‘release_everything’ defined but not used [-Wunused-function]
/home/abhinav/tmp/rts_pstor/rtsx.c:790:12: warning: ‘rtsx_scan_thread’ defined but not used [-Wunused-function]
/home/abhinav/tmp/rts_pstor/rtsx.c:816:13: warning: ‘rtsx_init_options’ defined but not used [-Wunused-function]
cc1: some warnings being treated as errors
make[2]: *** [/home/abhinav/tmp/rts_pstor/rtsx.o] Error 1
make[1]: *** [_module_/home/abhinav/tmp/rts_pstor] Error 2
make[1]: Leaving directory `/usr/src/linux-headers-3.8.0-26-generic'
make: *** [default] Error 2
```
After a bit of googling I realized that this error exists for kernel versions above 3.6.x only. Below that, it works like a charm (After this point I had completely lost it on Realtek, I mean write proper drivers for crying out loud!). So the solution as it turns out is to edit the driver file “rtsx.c” and remove all instances of the string “__devinit” “__devexit” and “__devexit_p” from it, leaving the rest of the code (and the rest of the line they occur in) intact. Meaning a line in the code which looked like :
` static void __devinit rtsx_remove(struct pci_dev *pci) `
Becomes:
` static void rtsx_remove(struct pci_dev *pci) *pci_id) `
After editing the file, I ran make again, the code compiled, and the kernel file was now ready. After this, I had to load the driver file onto the kernel using modprobe :
```abhinav@AsusK55V:/~ sudo modprobe rts_bpp
abhinav@AsusK55V:/~ sudo depmod -a ```
Now, in an ideal world, my SD card would be running fine and we’d have light sabers and death stars. However, my card reader still gave the same error, but apparently following the above method worked for most people. So I was pretty sure that the driver file wasn’t wrong. On running lspci again, it was clear what the problem was:
``` abhinav@AsusK55V:/~ lspci -v -s 04:00.0
04:00.0 Unassigned class [ff00]: Realtek Semiconductor Co., Ltd.
Device 5289 (rev 01)
Subsystem: ASUSTeK Computer Inc. Device 1457
Flags: bus master, fast devsel, latency 0, IRQ 16
Memory at f7800000 (32-bit, non-prefetchable) [size=64K]
Kernel driver in use: rtsx_pci
Kernel modules: rts_bpp, rtsx_pci
```
So as you can see, although our newly compiled driver for the SD card, “rts_bpp” is available for the OS to use, it still continues to use the old buggy “rtsx_pci”. So the obvious steps would be ideally to unload this module and use our rts_bpp module. So I run:
```abhinav@AsusK55V:/~ sudo modprobe -r rtsx_pci
```
Only to get this error:
```FATAL: Module still in use
```
Turns out that there are certain other hidden modules which use the rtsx_pci module, so you had to unload them first and then unload this godforesaken module:
```abhinav@AsusK55V:/~ lsmod|grep rtsx_pci
rtsx_pci 62705 2 rtsx_pci_sdmmc, rtsx_pci_ms
```
So as we see, the rtsx_pci module cannot be unloaded unless we unload the two modules that uses it, like this:
```abhinav@AsusK55V:/~ sudo modprobe -r rtsx_pci_sdmmc
abhinav@AsusK55V:/~ sudo modprobe -r rtsx_pci_ms
abhinav@AsusK55V:/~ sudo modprobe -r rtsx_pci
abhinav@AsusK55V:/~ sudo modprobe -r rts_bpp
abhinav@AsusK55V:/~ sudo depmod -a
abhinav@AsusK55V:/~ sudo modprobe rts_bpp
abhinav@AsusK55V:/~ sudo depmod -a
```
So essentially, I unloaded all the kernel modules, including the one I compiled, ran a depmod, loaded just the rts_bpp and ran a depmod again. After a few seconds, my SD card finally mounted!
However rebooting the laptop reconfigured the settings again, and I was stuck with the rtsx_pci module again. So after running the steps above again, I blacklisted the rtsx_pci module:
`abhinav@AsusK55V:/~ sudo nano /etc/modprobe.d/blacklist.conf `
by appending the file above with “blacklist rtsx_pci” in the end. After this you need to update the appropriate init files so that the setting sticks everytime you boot your system. You can do that by running:
`abhinav@AsusK55V:/~ sudo updateinitramfs -u `
Its annoying how an error like this takes the whole day, when I could have just gone on Windows and done it. But there is something about fixing a problem that makes you feel happy. Now, I shall get back to getting arch on my SD card for the new Cubieboard 2! Hopefully I get to write some posts about GPIO interrupts, something that I have wanted to experiment with for so long now!
I frequently use GitHub to maintain a lot of my personal projects, minor hacks and if I am upto it, contributing to other open source packages. It is like this free (almost) unlimited remote server where you can push all your code. The downside is, unless you pay them some money, all your code ends up being open to everyone. Now if for some reason, you don’t want the code to be public, you might have to buy private space on GitHub or maybe try and apply for the free educational repositories. (Or move to BitBucket, but again the free ones are limited).
However, there is a cheap, albeit risky alternative to that – You can use your Dropbox space (Or Google Drive – if you are a fan of Google reading through your stuff) as a free Git remote server! This is not a fool proof method and the chances for conflicts arising are high if multiple users are pushing code, so it is best recommended for personal work, where only one person is pushing code to the remote repo.
A lot of people have done this before, so this is old news, but I think its a decent hack and worth telling people about it.
## The Steps
Let’s move to the command line and set up our repository!
Step 1 : Move to your Dropbox folder and create a folder (Git in our example) where you will be creating your repositories and initialize a bare repository like this:
``` cd ~/Dropbox/Git
git init --bare dropboxrepo.git ```
This will create a folder with the name dropboxrepo.git and inside it, you can see that it has all the essentials for a git repository.
Guts of the bare repository
Now technically you are making this repository on your local machine, but since it is inside the Dropbox folder, you are syncing it to their servers as well, and can access the folder contents anywhere.
Step 2 : Now go the local repository, where you maintain the code that you want to push to your remote repository, and add the above folder as a remote like this:
` git remote add dropboxremote ~/Dropbox/Git/dropboxrepo.git `
Basically what you did here is add the path to your dropboxrepo.git folder as a remote alias. So now the alias dropboxremote will refer to this folder and will be identified as a remote repository.
Step 3 : Now that you have initialized a remote repo and added it to the remote alias in your local repo, you can now push your code like this:
``` git push -u dropboxremote master </code>
Counting objects: 6, done.
Delta compression using up to 4 threads.
Compressing objects: 100% (2/2), done.
Writing objects: 100% (6/6), 458 bytes, done.
Total 6 (delta 0), reused 0 (delta 0) Unpacking objects: 100% (6/6), done.
To /Users/abhinav/Dropbox/Git/dropboxrepo.git
* [new branch] master -> master
Branch master set up to track remote branch master from origin.```
After this, you should see the Dropbox spin sign for a while as it syncs all your files. However you will notice that after syncing, you don’t see your files in the folder but the same files you saw before (they are modified now though). But don’t worry all your files are synced. You can now follow the same steps as above and clone this repository on another machine and you will see all your committed code right there!
## Word of caution
As I said before, this is not exactly the best way to have your own private remote server (It’s free!) and it comes with its own dangers. Firstly, if you push or pull while the folder is still syncing you might encounter errors like this:
```error: unable to find ea2ae105b8955b3f73d79065bc52ee3126cd2e4f
error: refs/heads/master does not point to a valid object!
Your configuration specifies to merge with the ref 'master'
from the remote, but no such ref was fetched. ```
If you see this error, just wait for the syncing to finish and try again. If it still doesn’t work, you will have to delete your current remote repo and create a new one and try again.
The above issue occurs frequently if there are a lot of contributors pushing the code to the same remote repo. Of course you can be careful and avoid them, just don’t pull/push while syncing!
I recently started using this method to maintain some of my private projects and it works just fine!
Happy tinkering!
A dedicated Home Server is a useful thing to have if you’re a hacker. You can use it as a remote repository for all your code, for hosting your website, dumping all your media and accessing it with any other device and even printer sharing. This post gives a detailed procedure of making (in my opinion) a near perfect home server using a Cubieboard!
## Benefits of using a Cubieboard
Low Cost : For \$50, the Cubieboard is like the cheapest computer in the world. Complete with 1GB RAM, 4GB space, 10/100 Ethernet port, two USB ports, SDcard slot, camera interface, audio out, HDMI AND along with all this awesomeness, it also has a complete SATA port. It also comes with a nice glass box for protection.
Low Power : The low cost doesn’t just end there with the pricing, the average power consumption of the Cubieboard is 8W, which if compared to standard low power servers is 5 times less! So it just doesn’t cost less, it consumes less as well!
## Getting Started
You will need a functional operating system with SSH and Ethernet ports enabled on the Cubieboard before installing the server stack.In case you haven’t seen already, I have described how to do that in this post. Make sure you have followed the steps and installed a headless Debian on your Cubieboard before reading further. I have released the latest kernel images with fixed patches and SATA power enabled over here.
Now that the SATA is enabled, you can now easily connect a HDD, thus removing any space constraint on your server. (NOTE: The SATA works for a 2.5 inch drive only, for other HDD types, you will need external power source)
To ensure that the SATA interface works on boot, just type this command in your terminal (on the Cubieboard, of course!)
`echo "sw_ahci_platform" >> /etc/modules`
After connecting the Ethernet port, make sure you note down the IP of your server. You can do that by using the DHCP list or just doing a shell login with the TTL pins and running ifconfig, you will get all the network connection details including the IP.
Now that we have all the hardware and Operating System stuff settled, we consider our server stack options. One of the interesting options is the Napkin Server. It is based on JRuby, Sinatara and Neo4j graph database. The Napkin server offers a REST API with predefined services for storing and retrieving configuration information and collecting time series data from client devices. A plug-in model provides extensibility in HTTP handling and periodic processing of stored data. It is a very interesting project, you can refer to the wiki for Cubieboard installation here.
The second option would be to go for the more traditional LAMP stack. The installation is a very simple 3-step process.
#### Step 1: Installing Apache 2 and PHP 5
Apache is usually the go-to choice for most linux based servers. You can easily install Apache and configure it with PHP 5 with the package manager tool like this:
` root@cubie:~# apt-get install apache2 php5 libapache2-mod-php5 `
The folder holding all the php pages by default is located in /var/www/
To check if php is properly installed, just open a sample php file:
` root@cubie:~# vi /var/www/test_file.php `
and add the following php code
` {codecitation}<?php phpinfo();?>{/codecitation} `
After you have saved this file, you can test by opening a browser on a different machine and connect to the following link: http://%5Bcubieboard’s-ip%5D//test_file.php and you will see this :
#### Step 2 : Installing MySQL Database server
The next step would be installing the database backbone. Again, installing MySQL is an easy task with the package manager:
` root@cubie:~# apt-get install mysql-server mysql-client php5-mysql `
While installing, you will get be prompted to insert a root mysql password. You can add it there,but in case you forget you can even add it later like this:
``` mysql -u root
mysql> USE mysql;
mysql> FLUSH PRIVILEGES; ```
#### Step 3 : Installing PHPMyAdmin
PHPMyAdmin is an extremely easy-to-use web-tool to manage the databases on your server. It is a very handy tool to remotely manage databases on your server. To install it simply go:
` root@cubie:~# apt-get install phpmyadmin `
During the installation process, you will be prompted with a screen to choose your choice of web-server. In our case, it is Apache2
Once the installation is complete, to test it, open a browser on a different machine connected in the same network, and open the link : http://%5Bcubieboard’s-ip%5D/phpmyadmin
Doing so will result in this:
To enter, you must use the Cubieboard’s login details. Once you enter the correct values, it should lead you to a Dashboard, which is pretty self explanatory to use!
Now you have a fully functional web-server!! You can now start experimenting with your server, maybe by installing wordpress on it and using it as a host for your own website!
Happy tinkering!
In my last post, we saw how flexible and powerful the Cubieboard can be, with the correct knowledge of the pin multiplexing and choosing the pin functionality for your application. Given that you have 7 UART ports, and around 3 SPI ports at your disposal, you can only imagine the power that this gives you.
The company that produces the SoC on which the Cubieboard is based on, AllWinner, primarily wanted to develop this chip as a cheap alternative to Android enabled device manufacturers, such as phones, tablets and what not! And hence, the chipset and subsequently the development boards, are powerful, flexible and have ports designed for external device interfacing, including 2 camera interfaces. (However, only one can be active at one time, like front cam and back cam) But, what it makes for in these interfaces, it lacks in the number of ADC’s (Only one, for accelerometer. People argue that it has the ADC’s for the LCD screen, but involves hacking the driver files, interesting project if someone is willing me to pay for it) and PWM channels. Basically, you cannot collect data or move motors by just using the Cubieboard.
So, if you are planning to use the Cubieboard in your Embedded Design Project, you will sooner or later have the need to communicate with a daughter board, either for getting a bunch of sensor data, or performing some actuation. This post comes as a result of a similar need I had, where I had to trigger a bunch of OpenCV codes which depended on certain sensor values.
Since I could not interface my sensors on the Cubieboard, I decided to use an Arduino for measuring the sensor data and communicate the processed values to the Cubieboard via UART.
The Cubieboard has 8 UART ports in all, including the 4-pin TTL port. However, this 4 pin TTL by default is used in the following ways:
1. Console Messages (including bootup messages)
2. A getty so you can login via serial
So, I would suggest that you avoid using the TTL pins, since you would need to edit the boot and initab files in order to use them. Plus, this port comes in very handy for debugging purposes. So instead, we will configure the other GPIO pins for UART.
On the Arduino side, instead of connected pins 0 and 1 (The default UART pins) , I decided to use the SoftSerial library instead. Why? Because I wanted to hook up my Arduino to my PC via USB, whilst it communicates with the Cubieboard, for debugging purposes.
Now I will show the codes on the Arduino and the Cubieboard side respectively, after which you can safely wire up the proper pins on both sides and communicate seamlessly.
## Arduino side:
As mentioned before, I won’t be using the default serial pins for interfacing and instead use the SoftSerial library to configure pins 2 and 3 instead. However, the SoftSerial library does not support baudrates above 9600. We will see how that is actually a good thing for us later. Following code snippet sends a character to both serial ports if the ADC value is above a threshold, and prints whatever value it gets from CubieBoard to the PC:
```/*
Connects Arduino to Cubieboard
Arduino: SoftSerial
Raspberry Pi: GPIO UART
Based on Arduino SoftSerial example
*/
#include
SoftwareSerial mySerial(2, 3); // RX, TX
int pressure;
int THRESHOLD = 500;
void setup()
{
// Open serial communications to PC and wait for port to open:
Serial.begin(57600);
// set the data rate for the SoftwareSerial port to Cubieboard
mySerial.begin(9600);
// Set A1 as input sensor
pinMode(A1,INPUT);
}
void loop() // run over and over
{
// Measure the sensor value
// If pressure crosses the threshold, send character to both ports
if(pressure>THRESHOLD) {
Serial.write("Y");
mySerial.write("Y");
}
// If data is available on Cubieboard, print it to PC
if (mySerial.available())
}```
## Cubieboard side:
First order of business would be to configure a UART port of our choice on the Cubieboard. For this case, UART5 is an optimal choice, it has only the Rx and Tx ports, just like the Arduino side, thus not wasting any GPIOs unnecessarily.
Now we make changes in the fex file like mentioned in my previous post and note down the pins:
```[uart_para5]
uart_used = 1
uart_port = 5
uart_type = 2
uart_tx = port:PH06
uart_rx = port:PH07```
The pins are : Tx (Port H, pin 6) and Rx (Porn H, pin 7)
Make sure to compile the fex file to an appropriate bin file, otherwise the changes will not be seen.
Just to check all the UART ports that you have, you can just list by the following command:
`ls /dev/tty*`
OR
`dmesg|grep tty `
Now for the communication part, hackers tend to use Python due to the ease with which you can perform serial communication, thanks to the PySerial library. But in my case, since I had to trigger some OpenCV based codes, I decided to go the C++ way, since I could just include a header file in the main file and save a lot of unwanted mess.
Serial Initialization function
Although you can just open the Serial Port on any linux box as a file, you need to define some flow control and flag options for the desired communication terminal since this is an asynchronous communication and follows a proper handshaking protocol(If you need more clarification on this, please mention on the comments). So we will need some standard libraries for this purpose, you can add them like this:
```#include
#include <sys/ioctl.h>```
Now we initialize the serial port terminal options :
```/*Init function, takes serialport address and baudrate as input */
int serialport_init(const char* serialport, int baud)
{
struct termios toptions; //from the termios lib
int fd; //file descriptor```
Next step would be to open the port with the proper flags and initialize the input and output data speed equal to the input baudrate:
```//Open the port as a file
//O_RDWR = Open as read and write
//O_NOCTTY = TTY file not to be used for control purposes
fd = open(serialport, O_RDWR | O_NOCTTY)
//set the baudrate
speed_t brate = baud
//setting the input and output speed
cfsetispeed(&toptions, brate);
cfsetospeed(&toptions, brate);
```
Now that the serial port has been initialized with the correct options, you can just read it as a file and accordingly write the functions for them. For the complete code with the read and write functions, please refer to this link.
Now you can include this library in your main file and initialize the port like this:
` serialport_init("/dev/ttyS5", 9600); `
## Final Connections:
Now that we have configured the UART pins and written the appropriate code on both ends, time to hook them up. Following are the pin outs:
Cubieboard Arduino Rx – Port H, pin 7 Rx – pin 2 Tx – Port H, pin 6 Tx – Pin 3
The possibilities from here are endless, you can send commands from the Board to the Arduino and use them to control motors or actuators. In case you need further clarifications on any of these, feel free to drop a comment here!
NOTE: I mentioned before that the 9600 baudrate on the Arduino is actually a blessing in disguise for us, that is because Arduino supports only 2 pin UART, which is actually not at all ideal for a full duplex communication. Imagine this scenario, the Arduino sends some data and triggers a code on the Cubieboard, now any data that the Arduino sends while the Cubieboard is still executing the code will be lost because of the lack of proper data control pins. A quick fix around this would be to run a thread which accumulates the sensor data in a buffer on the board, or to time your code properly so that it finishes executing before the next data comes in
Happy Tinkering!
In the previous post I had mentioned how you can customize the way you use the Cubieboard’s pins by changing the script.bin file. This post will be expand on that thought further.
## Fex files, bin files and sunxi-tools
Every chipset has a human-readble file which sort of acts like a metadata file for the I/O pins of the board, this file is called the fex file. It is of no use to the chip itself. It needs to be parsed and converted into an equivalent binary file, named script.bin, like the one we saw in the previous post while installing our OS on the board.
Thankfully the tools for converting fex to binary and vice versa for the A10 boards already exist. You will need to get the source files for the same and compile them in order to use them. Run the following command to get the source files:
`home@user:~\$ git clone git://github.com/linux-sunxi/sunxi-tools.git `
This should clone a copy of the sunxi-tools repository in your folder. The tools provided here are exclusively for AllWinner A10 based boards. They won’t work for generic devices. Now that we have the source files, we can proceed to build the tools:
```home@user:~\$ cd sunxi-tools
home@user:~\$ make fex2bin
home@user:~\$ make bin2fex ```
This will spit out two executables, fex2bin and bin2fex. These executables do exactly what they are named. Next step would be to transfer both these executables to the folder where you want to keep your script.bin files.
`home@user:~\$ cp /path/to/sunxi-tools/fex2bin /path/to/script/files`
Now we use our existent script file, the one we downloaded from the git repo here, and get it’s equivalent fex file like this:
`home@user:~\$ ./bin2fex script.bin cubieboard.fex `
If it executes (and it should) successfully, it shows the following lines in the terminal:
```fexc-bin: script.bin: version: 0.1.2
fexc-bin: script.bin: size: 44104 (79 sections) ```
Now use your favourite text editor to open the cubieboard.fex file. You will see a lot of things that may seem gibberish at present.
Each line consists of a key = value pair combination under a [sectionheader]. All three [sectionheader], key and value are case-sensitive. For comments a semi-colon (;) is used and everything following a semi-colon is ignored.
## Configuring the GPIO pins in the fex file
Now since we intend to use some of the GPIO’s here, go to the [gpio-para] section header. You should see the following under this section header :
```gpio_used = 1
gpio_num = 4
gpio_pin_1 = port:PG00<1><default><default><default>
gpio_pin_2 = port:PB19<1><default><default><default>
gpio_pin_3 = port:PG02<0><default><default><default>
gpio_pin_4 = port:PG04<0><default><default><default>
```
The first line has the parameter gpio_used which is initialized to 1. This parameter basically denotes if the functionality described by the section header is enabled(1) or disabled(0). Thus in this case, it seems like the GPIO functionality is enabled for 4 pins (gpio_num). The properties of each pin are given in the subsequent lines.
Description of the GPIO configuration in the form:
port:<port><mux feature><pullup/down><drive capability><output level>
where:
<port> is the port to configure (ie. PH15)
<mux feature> is the function to configure the port for, mux 0 is as input, mux 1 as output and for 2-7 see A10/PIO for details.
<pullup/down> is 0 = disabled; 1 = pullup enabled; 2 = pulldown enabled (only valid when port is an input)
<drive capability> defines the output drive in mA, values are 0-3 corresponding to 10mA, 20mA, 30mA and 40mA.
<output level> sets the initial output level for the port; 0 = low; 1 = high (only valid for outputs)
The <pullup/down> <drive capability> and <output level> can be set to <default> which means don’t change. Any trailing <default> options can be omitted.
You can refer to the following image to see the I/O pins on the Cubieboard:
Cubieboard GPIO
## Configuring the UART ports
The Cubieboard supports a total of 8 UART ports, apart from the TTL connection provided. To configure them, you can check under the [uart_para\$] section header, where the \$ stands for a number from 0-7. These are the configuration parameters for UART1 (ttyS1):
```[uart_para1]
uart_used = 0
uart_port = 1
uart_type = 8
uart_tx = port:PA10<4><1><default><default>
uart_rx = port:PA11<4><1><default><default>
uart_rts = port:PA12<4><1><default><default>
uart_cts = port:PA13<4><1><default><default>
uart_dtr = port:PA14<4><1><default><default>
uart_dsr = port:PA15<4><1><default><default>
uart_dcd = port:PA16<4><1><default><default>
uart_ring = port:PA17<4><1><default><default>
```
Since the uart_used parameter here is initialized to zero, this particular UART (ttyS1) is disabled and cannot be used. To enable it, you need to set it’s value to 1 just like we discussed for gpio_used in the previous section. This particular UART allows for the full 8 pin connection (uart_type = pin connections allowed). In case you are not familiar with it, these are the functions of each of the pin:
TX Transmit Data RX Receive Data RTS Request To Send CTS Clear To Send DSR Data Set Ready DTR Data Terminal Ready DCD Data Carrier Detect RING Ring Indicator (Modem ring)
To interface this UART, you will just have to connect wires from the pins mentioned in the section header above to the appropriate pins on the slave device.
## Generating the modified script file
Once we are done making all the modifications and customization we want, we need to generate the chip-readable script.bin file. To do that, we shall use the second executable that was taken from sunxi-tools earlier, the fex2bin.
```home@user:~\$ ./fex2bin cubieboard.fex script.bin
```
Succesful execution of this command will generate the new script.bin file which you can now transfer to the appropriate folder of your bootable SD card. On booting up your Cubieboard again, you will be able to use the modified pins just as you configured them in the fex file. To use your newly configured GPIO and UART pins, read ahead!
GPIO: After booting up the Cubieboard, do the following to use the GPIOs:
```cubie@board:~\$ echo 2 > /sys/class/gpio/export
cubie@board:~\$ ls /sys/class/gpio
export gpio2_pb19 gpiochip1 unexport
```
Thus activating the second pin in your GPIO config (PB19) by generating a special file by the name gpio2_pb19. Similarly echoing 1,3 or 4 will activate the appropriate pins. You can then write/read values from these files thus fulfilling the I/O function.
UART : For UARTs, you need to make the proper connections and then use some serial program like minicom. Or you can write your own code in Python or C to do it. In my next post, I will be writing a C program to perform a UART communication with an Arduino board, this will enable your Cubieboard to make more complicated tasks.
A few weeks back, I had the chance of working with this new family of ARM based System on Chip (SoC) chipset from this company called Allwinner Technologies. My work involved designing a system around the A10 family of chipset. So to get myself acquainted with it and creating a first prototype, I decided to play around with the development boards based on it.
Hackberry and CubieBoard are the two most famous development boards based on the Allwinner A10 chipset. And from its specifications, I can assure you, they are far more powerful and customizable than Raspberry Pi and BeagleBoards. If you want an in detailed comparison between CubieBoard and Raspberry Pi, you can have a look at Alexandre’s blog.
As Alexandre has pointed out rightly in his blog, the CubieBoard community is very young and the support is as good as zero. Hence this post is going to be an in-detail procedure on how to install a customizable version of a headless Debian on the CubieBoard. And maybe later, I might have some posts on how to interface with daughter controllers or something like using the board to set up a dedicated server.
The CubieBoard comes with a stock Android ICS on its NAND flash memory. However, there is a provision to boot from SD card. Hence we create a boot-able SD card for using the OS of our choice. This involves four steps:
1. Partitioning the SD card (Preferred SD card size = 4GB)
2. Installing the bootloader
3. Installing the kernel files and the script.bin
4. Installing a proper root file-system
A lot of out-of-the-box images for SD cards exist, but if you use them, you won’t be able to customize the GPIOs for your specific purpose. Hence we will manually do each step and look at the flexibilities it provides. To keep the post simple and short, I won’t be discussing how to build the tool-chains and kernel images in this post and use the kernel images I created myself instead. I will have separate posts for building the kernel images and the pin binaries.
The first and foremost requirement is to have Linux on your host machine (Laptop/Desktop). I am not sure about Mac systems, but I am guessing that should be fine as well. But Windows is a strict no-no.
## Partitioning the SD card:
The first MB of the SD card contains the partition table along with the bootloader binaries for booting. The next partition contains the kernel Images and the boot.scr files. The biggest partition contains the rootfs which takes around more than 2GB space. Thus it is best to use a 4GB SD card.
Now assuming that you have installed the sfdisk tools and your card is connected to the host machine, we begin with checking the partition table and the physical characteristics of the SD card, such as bytes per sector. Here we assume that the card is named sdd in the /dev/ folder. To know the name of your SDcard, run the command
`home@user:~\$ sudo ls -la /dev/sd*`
This will list all the disk type devices connected to your machine and you can read the name from there. After running
`home@user:~\$ fdisk -l`
this is the typical 4GB SD card partition that you should see:
```Disk /dev/sdd: 3965 MB, 3965190144 bytes
122 heads, 62 sectors/track, 1023 cylinders, total 7744512 sectors
Units = sectors of 1 * 512 = 512 bytes
Sector size (logical/physical): 512 bytes / 512 bytes
I/O size (minimum/optimal): 512 bytes / 512 bytes
Disk identifier: 0xb4e92fdc
Device Boot Start End Blocks Id System
/dev/sdd1 2048 34815 16384 83 Linux
/dev/sdd2 34816 7744511 3854848 83 Linux
```
Notice how sdd1 starts at 2048 sector. As the byte per sectors is 512, 2048 sector means that the partition sdd1 starts at 1MB. Usually the first 1MB space in a memory device is reserved for the Master Boot Record (MBR), the partition table and in case the memory device is bootable, the Initial Program Loader (IPL) and the Secondary Program Loader (SPL), also known as the bootloader binaries. Resetting the first 1MB of the SDcard ‘formats’ it. to do that you need to run the following command:
`home@user:~\$ dd if=/dev/zero of=/dev/sdd bs=1M count=1`
This writes zeroes for the byte size (bs) of 1MB, starting from the first sector itself. Thus rewriting over the entire column and subsequently formatting the entire card. If you run
`home@user:~\$ fdisk -l`
now, it will show no partition table exists. Now we will use the sfdisk tools to make our partitions.
Most SD cards have similar properties as the one shown above, i.e bytes per sector is usually 512. Using this information, we calculate the number of sectors for our partitions and start from sector 1MB (Since the first MB is preserved for the above mentioned reasons). The following commands create a 64MB fat partitions and the rest is used as an ext4 partition.
```home@user:~\$ sfdisk -uS /dev/sdd
2048,131072,c #Here it tells you to define sdd1
131073,,L #Here it tells you to define sdd2. Being the last partition,
just press enter twice to exit ```
Now we format these partition appropriately by running the following commands:
```home@user:~\$ mkfs.vfat /dev/sdd1
home@user:~\$ mkfs.ext4 /dev/sdd2
home@user:~\$ sync```
## Installing the kernel image and filesystem on the card:
Now that we have the card partitioned according to our needs, its time to set up the card by putting the appropriate files in their respective partitions. I have prepared three versions of kernel files and the bootloaders for the CubieBoard board. You can get these files from here.
A brief explanation of the files in this repository:
1. Bootloader : This directory contains the bootloader files for the CubieBoard
2. kernel_sunxi_3_* : These directories are the different kernel versions
1. lib : This directory contains the firmware for that specific kernel version
2. uImage : The actual kernel image file
3. script.bin : The files that specifies the functionality of each pin.
So the flexibility provided here is that you can make your own script files and hence specify every pin’s function. Furthermore, you can write your own patches for the kernel and make custom images depending on what you desire. How to do generate custom files for these will be shown in later posts.
We should now have 2 partitions. One 64M in vfat format, and another in ext4. We also need to make this bootable, so lets blat in the IPL and SPL to the SD at the appropriate places. These files are in the directory named “bootloader” in the link given above.
Write spl ->8k and then write uBoot ->32k on the SD card.
```home@user:~\$ dd if=sunxi-spl.bin of=/dev/sdd bs=1024 seek=8
home@user:~\$ dd if=u-boot.bin of=/dev/sdd bs=1024 seek=32```
Now create a folder on your linux machine and mount both the partitions there. Do the following assuming that you are the superuser (sudo su):
```home@user:~\$ mkdir /mnt
home@user:~\$ mkdir /mnt/boot
home@user:~\$ mkdir /mnt/rootfs
home@user:~\$ mount -t auto /dev/sdd1 /mnt/boot
home@user:~\$ mount -t auto /dev/sdd /mnt/rootfs
home@user:~\$ cd /mnt/boot```
Now, create a file in the /mnt/boot directory named boot.cmd and write the
following in it:
```setenv bootargs console=ttyS0,115200 console=tty0 noinitrd init=/init root=/dev/sdd2
rootfstype=ext4 rootwait panic=10 \${extra}
fatload mmc 0 0x43000000 script.bin
fatload mmc 0 0x48000000 kernel.img
bootm 0x48000000```
The first line calls the kernel, and passes parameters onto it. You’ll note that I setup 2 consoles, one on tty0 and another on ttyS0. I like to have both a serial and a display console on kernel bootup. Now we need to compile it, to do that, run the following:
`mkimage -C none -A arm -T script -d boot.cmd boot.scr`
Next step is to copy the kernel image and the script file. Assuming you are in the folder kernel_sunxi_3-*-* (Depending on your choice of kernel version)
```home@user:~\$ cp uImage /mnt/boot
home@user:~\$ cp script.bin /mnt/boot```
Now we have all the pre boot binaries, bootloaders, kernel image and the script.bin files all loaded on to the SD card. Now we come to the last part, uploading the file system. I prefer the newest version of the Debain filesystem as it more flexible and light in nature. Now you need to download the tar file of the filesystem and unpack it at the place where your second partition is mounted
(i.e /mnt/rootfs).
```home@user:~\$ cd /tmp
home@user:~\$ axel http://dl.cubieforums.com/loz/rootfs/debian-unstable-armhf.tar.bz2
home@user:~\$ cd /mnt/rootfs
home@user:~\$ unp /tmp/debian-unstable-armhf.tar.bz2```
This will unpack the Debain filesystem on the second partition. After this, you need to copy the compiled drivers, firmware etc to the appropriate place in the rootfs. This is an absolutely necessary step as any changes in the driver files wont take effects if they are not copied to the appropriate places in the file system. Here is an example snippet of this (Assuming again that the user is in the kernel_sunxi_3-*-*):
```home@user:~\$ mkdir -p /rootfs/lib/modules
home@user:~\$ rm -rf /rootfs/lib/modules/
home@user:~\$ cp -r lib /rootfs/```
Now just unmount the the SD card:
```home@user:~\$ umount /mnt/boot
home@user:~\$ umount /mnt/rootfs```
In case it says, it is busy, open a new instance of the terminal and run. It will surely work. After this, just boot up your cubieboard with this SD card. Use GNU screen command to see the boot sequence (Use this link for reference of TTL-USB Connections and using GNU screen)
If the boot is successful, it will ask you for login and password. They are root and password respectively.
Update: Just FYI, the filesystem you install here gives you direct root access. If you are an advanced linux user, you can maybe add your own profiles for safety! | 11,318 | 45,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-50 | latest | en | 0.966818 |
https://www.stat.math.ethz.ch/pipermail/r-help/2011-November/294758.html | 1,653,806,695,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00175.warc.gz | 1,160,434,315 | 4,425 | # [R] Matrix element-by-element multiplication
Joshua Wiley jwiley.psych at gmail.com
Mon Nov 7 02:01:03 CET 2011
```Hi,
R may not have a special "scalar", but it is common, if informal, in
linear algebra to refer to a 1 x 1 matrix as a scalar. Indeed,
something like:
1:10 * matrix(2)
or
matrix(2) * 1:10
are both valid. Even
matrix(2) %*% 1:10
and
1:10 %*% matrix(2)
work, where the vector seems to be silently coerced to a matrix. R
even seems to work hard to convert to a conformable matrix:
## works:
1:10 %*% matrix(1:10)
## does not work
matrix(1:10) %*% matrix(1:10)
## works
t(matrix(1:10)) %*% matrix(1:10)
Interestingly, there is actually a (rather old) comment in arithmetic.c
/* If either x or y is a matrix with length 1 and the other is a
vector, we want to coerce the matrix to be a vector.
Do we want to? We don't do it! BDR 2004-03-06
*/
Given the coersion that already occurs with vectors to matrices for
%*% and matrices to vectors for *, it seems not unreasonable to
convert a 1 x 1 matrix to a vector _for_ * so that the following
yields identical results:
matrix(1:9, 3) * matrix(2)
matrix(1:9, 3) * 2
Of course in the mean time, or in general, it is a good habit to
create or explicity coerce objects yourself rather than relying on R
to make smart guesses about what should be happening.
Cheers,
Josh
On Sun, Nov 6, 2011 at 4:02 PM, R. Michael Weylandt
<michael.weylandt at gmail.com> <michael.weylandt at gmail.com> wrote:
> It looks like pdf is not a "scalar" (that term actually has no meaning in R but I know what you mean) but is rather a 1x1 matrix, as attested by the fact it has dimensions. If you give dnorm() a matrix it will return one, as it did here.
>
> Perhaps you should look at the is.matrix() and as.vector() functions rather than abusing a side-effect of c(), which makes it much more difficult to see R's internal logic, which, while quirky, is useful at the end of the day.
>
> Michael
>
> PS - It's good form to cc the list at each step so others can follow along and contribute when I say something wrong. It also helps you get quicker answers.
>
> On Nov 6, 2011, at 1:06 AM, Steven Yen <syen at utk.edu> wrote:
>
>> I am trying to multiply what I know is a scalar (pdf(xb)) to a column vector of coefficient (bb).
>> In the following, pdf is a scalar and bb is 5 x 1. I first show what worked and then what did not work.
>> If my pdf is a scalar, why would I need c(pdf) to be able to pre-multiply it by a 5 x 1 vector?
>>
>> ---
>>
>> > x <- as.matrix(colMeans(x))
>> > xb <- t(x)%*%bb
>> > pdf <- dnorm(xb)
>>
>> > dim(bb)
>> [1] 5 1
>>
>> >
>> > cpdf <- c(pdf)
>> > dim(cpdf)
>> NULL
>> > cpdf
>> [1] 0.304201
>> > (dphat <- cpdf*bb)
>> [,1]
>> (Intercept) 0.32744753
>> xrage -0.00599225
>> xryr 0.01758431
>> xrrate -0.08217250
>> xrrel -0.05695434
>> >
>> > pdf <- dnorm(xb)
>> > dim(pdf)
>> [1] 1 1
>> > pdf
>> [,1]
>> [1,] 0.304201
>> > (dphat <- pdf*bb)
>> Error in pdf * bb : non-conformable arrays
>> >
>>
>> At 12:21 AM 11/6/2011, you wrote:
>>> There are a few (nasty?) side-effects to c(), one of which is
>>> stripping a matrix of its dimensionality. E.g.,
>>>
>>> x <- matrix(1:4, 2)
>>> c(x)
>>> [1] 1 2 3 4
>>>
>>> So that's probably what happened to you. R has a somewhat odd feature
>>> of not really considering a pure vector as a column or row vector but
>>> being willing to change it to either:
>>>
>>> e.g.
>>>
>>> y <- 1:2
>>>
>>> x %*% y
>>> y %*% x
>>> y %*% y
>>>
>>> while matrix(y) %*% x throws an error, which can also trip folks up.
>>> You might also note that x * y and y*x return the same thing in this
>>> problem.
>>>
>>> Getting back to your problem: what are v and b and what are you hoping
>>> to get done? Specifically, what happened when you tried v*b (give the
>>> exact error message). It seems likely that they are non-conformable
>>> matrices, but here non-conformable for element-wise multiplication
>>> doesn't mean the same thing as it does for matrix multiplication.
>>> E.g.,
>>>
>>> x <- matrix(1:4,2)
>>> y <- matrix(1:6,2)
>>>
>>> dim(x)
>>> [1] 2 2
>>>
>>> dim(y)
>>> [1] 2 3
>>>
>>> x * y -- here R seems to want matrices with identical dimensions, but
>>> i can't promise that.
>>>
>>> x %*% y does work.
>>>
>>> Hope this helps and yes I know it can seem crazy at first, but there
>>> really is reason behind it at the end of the tunnel,
>>>
>>> Michael
>>>
>>>
>>> On Sun, Nov 6, 2011 at 12:11 AM, Steven Yen <syen at utk.edu> wrote:
>>> > My earlier attempt
>>> >
>>> > dp <- v*b
>>> >
>>> > did not work. Then,
>>> >
>>> > dp <- c(v)*b
>>> >
>>> > worked.
>>> >
>>> > Confused,
>>> >
>>> > Steven
>>> >
>>> > At 09:10 PM 11/4/2011, you wrote:
>>> >
>>> > Did you even try?
>>> >
>>> > a <- 1:3
>>> > x <- matrix(c(1,2,3,2,4,6,3,6,9),3)
>>> > a*x
>>> >
>>> > [,1] [,2] [,3]
>>> > [1,] 1 2 3
>>> > [2,] 4 8 12
>>> > [3,] 9 18 27
>>> >
>>> > Michael
>>> >
>>> > On Fri, Nov 4, 2011 at 7:26 PM, Steven Yen <syen at utk.edu> wrote:
>>> >> is there a way to do element-by-element multiplication as in Gauss
>>> >> and MATLAB, as shown below? Thanks.
>>> >>
>>> >> ---
>>> >> a
>>> >>
>>> >> 1.0000000
>>> >> 2.0000000
>>> >> 3.0000000
>>> >> x
>>> >>
>>> >> 1.0000000 2.0000000 3.0000000
>>> >> 2.0000000 4.0000000 6.0000000
>>> >> 3.0000000 6.0000000 9.0000000
>>> >> a.*x
>>> >>
>>> >> 1.0000000 2.0000000 3.0000000
>>> >> 4.0000000 8.0000000 12.000000
>>> >> 9.0000000 18.000000 27.000000
>>> >>
>>> >>
>>> >> --
>>> >> Steven T. Yen, Professor of Agricultural Economics
>>> >> The University of Tennessee
>>> >> http://web.utk.edu/~syen/
>>> >> [[alternative HTML version deleted]]
>>> >>
>>> >> ______________________________________________
>>> >> R-help at r-project.org mailing list
>>> >> https://stat.ethz.ch/mailman/listinfo/r-help
>>> >> http://www.R-project.org/posting-guide.html
>>> >> and provide commented, minimal, self-contained, reproducible code.
>>> >>
>>> >
>>> > --
>>> > Steven T. Yen, Professor of Agricultural Economics
>>> > The University of Tennessee
>>> > http://web.utk.edu/~syen/
>> --
>> Steven T. Yen, Professor of Agricultural Economics
>> The University of Tennessee
>> http://web.utk.edu/~syen/
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> and provide commented, minimal, self-contained, reproducible code.
>
--
Joshua Wiley
Ph.D. Student, Health Psychology
Programmer Analyst II, ATS Statistical Consulting Group
University of California, Los Angeles
https://joshuawiley.com/
``` | 2,208 | 6,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2022-21 | latest | en | 0.929695 |
http://www.reference.com/browse/decimal+digit | 1,432,866,212,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929832.32/warc/CC-MAIN-20150521113209-00306-ip-10-180-206-219.ec2.internal.warc.gz | 678,518,401 | 25,736 | Definitions
Nearby Words
# Binary-coded decimal
In computing and electronic systems, binary-coded decimal (BCD) is an encoding for decimal numbers in which each digit is represented by its own binary sequence. Its main virtue is that it allows easy conversion to decimal digits for printing or display and faster decimal calculations. Its drawbacks are the increased complexity of circuits needed to implement mathematical operations and a relatively inefficient encoding—it occupies more space than a pure binary representation.
Though BCD is not as widely used as it once was, decimal fixed-point and floating-point are still important and still used in financial, commercial, and industrial computing; modern decimal floating-point representations use base-10 exponents, but not BCD encodings.
In BCD, a digit is usually represented by four bits which, in general, represent the values/digits/characters 0-9. Other bit combinations are sometimes used for sign or other indications.
## Basics
To BCD-encode a decimal number using the common encoding, each decimal digit is stored in a four-bit nibble.
`Decimal: 0 1 2 3 4 5 6 7 8 9`
`BCD: 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001`
Thus, the BCD encoding for the number 127 would be:
` 0001 0010 0111`
Since most computers store data in eight-bit bytes, there are two common ways of storing four-bit BCD digits in those bytes:
• each digit is stored in one nibble of a byte, with the other nibble being set to all zeros, all ones (as in the EBCDIC code), or to 0011 (as in the ASCII code)
• two digits are stored in each byte.
Unlike binary-encoded numbers, BCD-encoded numbers can easily be displayed by mapping each of the nibbles to a different character. Converting a binary-encoded number to decimal for display is much harder, as this generally involves integer multiplication or divide operations.
## BCD in electronics
BCD is very common in electronic systems where a numeric value is to be displayed, especially in systems consisting solely of digital logic, and not containing a microprocessor. By utilizing BCD, the manipulation of numerical data for display can be greatly simplified by treating each digit as a separate single sub-circuit. This matches much more closely the physical reality of display hardware—a designer might choose to use a series of separate identical 7-segment displays to build a metering circuit, for example. If the numeric quantity were stored and manipulated as pure binary, interfacing to such a display would require complex circuitry. Therefore, in cases where the calculations are relatively simple working throughout with BCD can lead to a simpler overall system than converting to 'pure' binary.
The same argument applies when hardware of this type uses an embedded microcontroller or other small processor. Often, smaller code results when representing numbers internally in BCD format, since a conversion from or to binary representation can be expensive on such limited processors. For these applications, some small processors feature BCD arithmetic modes, which assist when writing routines that manipulate BCD quantities.
## Packed BCD
A widely used variation of the two-digits-per-byte encoding is called packed BCD (or simply packed decimal). All of the upper bytes of a multi-byte word plus the upper four bits (nibble) of the lowest byte are used to store decimal integers. The lower four bits of the lowest byte are used as the sign flag. As an example, a 32 bit word contains 4 bytes or 8 nibbles. Packed BCD uses the upper 7 nibbles to store the integers of a decimal value and uses the lowest nibble to indicate the sign of those integers.
Standard sign values are 1100 (Ch) for positive (+) and 1101 (Dh) for negative (-). Other allowed signs are 1010 (Ah) and 1110 (Eh) for positive and 1011 (Bh) for negative. Some implementations also provide unsigned BCD values with a sign nibble of 1111 (Fh). In packed BCD, the number 127 is represented by "0001 0010 0111 1100" (127Ch) and -127 is represented by "0001 0010 0111 1101 (127Dh).
Sign
Digit
BCD
8 4 2 1
Sign Notes
A 1 0 1 0 +
B 1 0 1 1
C 1 1 0 0 + Preferred
D 1 1 0 1 Preferred
E 1 1 1 0 +
F 1 1 1 1 + Unsigned
No matter how many bytes wide a word is, there are always an even number of nibbles because each byte has two of them. Therefore, a word of n bytes can contain up to (2n)-1 decimal digits, which is always an odd number of digits. A decimal number with d digits requires ½(d+1) bytes of storage space.
For example, a four-byte (32bit) word can hold seven decimal digits plus a sign, and can represent values ranging from ±9,999,999. Thus the number -1,234,567 is 7 digits wide and is encoded as:
`0001 0010 0011 0100 0101 0110 0111 1101`
(Note that, like character strings, the first byte of the packed decimal – with the most significant two digits – is usually stored in the lowest address in memory, independent of the endianness of the machine).
In contrast, a four-byte binary two's complement integer can represent values from −2,147,483,648 to +2,147,483,647.
While packed BCD does not make optimal use of storage (about 1/6 of the memory used is wasted), conversion to ASCII, EBCDIC, or the various encodings of Unicode is still trivial, as no arithmetic operations are required. The extra storage requirements are usually offset by the need for the accuracy that fixed-point decimal arithmetic provides. More dense packings of BCD exist which avoid the storage penalty and also need no arithmetic operations for common conversions.
### Fixed-point packed decimal
Fixed-point decimal numbers are supported by some programming languages (such as COBOL and PL/I), and provide an implicit decimal point in front of one of the digits. For example, a packed decimal value encoded with the bytes 12 34 56 7C represents the fixed-point value +1,234.567 when the implied decimal point is located between the 4th and 5th digits.
`12 34 56 7C`
`12 34.56 7+`
### Higher-density encodings
If a decimal digit requires four bits, then three decimal digits require 12 bits. However, since 210 (1,024) is greater than 103 (1,000), if three decimal digits are encoded together, only 10 bits are needed. Two such encodings are Chen-Ho encoding and Densely Packed Decimal. The latter has the advantage that subsets of the encoding encode two digits in the optimal 7 bits and one digit in 4 bits, as in regular BCD.
## Zoned decimal
Some implementations (notably IBM mainframe systems) support zoned decimal numeric representations. Each decimal digit is stored in one byte, with the lower four bits encoding the digit in BCD form. The upper four bits, called the "zone" bits, are usually set to a fixed value so that the byte holds a character value corresponding to the digit. EBCDIC systems use a zone value of 1111 (hex F); this yields bytes in the range F0 to F9 (hex), which are the EBCDIC codes for the characters "0" through "9". Similarly, ASCII systems use a zone value of 0011 (hex 3), giving character codes 30 to 39 (hex).
For signed zoned decimal values, the rightmost (least significant) zone nibble holds the sign digit, which is the same set of values that are used for signed packed decimal numbers (see above). Thus a zoned decimal value encoded as the hex bytes F1 F2 D3 represents the signed decimal value −123:
`F1 F2 D3`
` 1 2 −3`
### EBCDIC zoned decimal conversion table
Digit EBCDIC Display EBCDIC Hex
0+ { (*) X'C0'
1+ A X'C1'
2+ B X'C2'
3+ C X'C3'
4+ D X'C4'
5+ E X'C5'
6+ F X'C6'
7+ G X'C7'
8+ H X'C8'
9+ I X'C9'
0− } (*) X'D0'
1− J X'D1'
2− K X'D2'
3− L X'D3'
4− M X'D4'
5− N X'D5'
6− O X'D6'
7− P X'D7'
8− Q X'D8'
9− R X'D9'
(*) Note: These characters vary depending on the local character code page.
### Fixed-point zoned decimal
Some languages (such as COBOL and PL/I) directly support fixed-point zoned decimal values, assiging an implicit decimal point at some location between the decimal digits of a number. For example, given a six-byte signed zoned decimal value with an implied decimal point to the right of the fourth digit, the hex bytes F1 F2 F7 F9 F5 C0 represent the value +1,279.50:
`F1 F2 F7 F9 F5 C0`
` 1 2 7 9. 5 +0`
## IBM and BCD
IBM used the terms binary-coded decimal and BCD for 6-bit alphameric codes that represented numbers, upper-case letters and special characters. Some variation of BCD alphamerics was used in most early IBM computers, including the IBM 1620, IBM 1400 series, and non-Decimal Architecture members of the IBM 700/7000 series.
Bit positions in BCD alphamerics were usually labelled B, A, 8, 4, 2 and 1. For encoding digits, B and A were zero. The letter A was encoded (B,A,1).
In the 1620 BCD alphamerics were encoded using digit pairs, with the "zone" in the even digit and the "digit" in the odd digit. Input/Output translation hardware converted between the internal digit pairs and the external standard 6-bit BCD codes.
In the Decimal Architecture IBM 7070, IBM 7072, and IBM 7074 alphamerics were encoded using digit pairs (using two-out-of-five code in the digits, not BCD) of the 10-digit word, with the "zone" in the left digit and the "digit" in the right digit. Input/Output translation hardware converted between the internal digit pairs and the external standard six-bit BCD codes.
With the introduction of System/360, IBM expanded 6-bit BCD alphamerics to 8-bit EBCDIC, allowing the addition of many more characters (e.g., lowercase letters). A variable length Packed BCD numeric data type was also implemented.
Today, BCD data is still heavily used in IBM processors and databases, such as IBM DB2, mainframes, and Power6. In these products, the BCD is usually zoned BCD (as in EBCDIC or ASCII), Packed BCD, or 'pure' BCD encoding. All of these are used within hardware registers and processing units, and in software.
It is possible to perform addition in BCD by first adding in binary, and then converting to BCD afterwards. Conversion of the simple sum of two digits can be done by adding 6 (that is, 16 – 10) when the result has a value of greater-than 9. For example:
• 9+8=17 = [1001] + [1000] = [10001] in binary.
However, in BCD, there cannot exist a value greater than 9 (1001) per nibble. To correct this, 6 is added to that sum to get the correct first two digits:
• [0000 1000] + [10001 0110] = [0001 0111]
which gives two nibbles, [0001] and [0111] which correspond to "1" and "7" respectively. This gives the 17 in BCD, which is the correct result. This technique can be extended to adding multiple digits, by adding in groups from right to left, propragating the second digit as a carry, always comparing the 5-bit result of a digit-pair sum to 9.
## Subtraction with BCD
Subtraction is done by adding the nines' complement plus 1, or by adding the ten's complement of the subtrahend.
## Background
The binary-coded decimal scheme described in this article is the most common encoding, but there are many others. The method here can be referred to as Simple Binary-Coded Decimal (SBCD) or BCD 8421. In the headers to the table, the '8 4 2 1', etc., indicates the weight of each bit shown; note that in the 5th column two of the weights are negative. Both ASCII and EBCDIC character codes for the digits are examples of zoned BCD, and are also shown in the table.
The following table represents decimal digits from 0 to 9 in various BCD systems:
Digit
BCD
8 4 2 1
Excess-3
or Stibitz Code
BCD 2 4 2 1
or Aiken Code
BCD
8 4 −2 −1
IBM 702 IBM 705
IBM 7080 IBM 1401
8 4 2 1
ASCII
0000 8421
EBCDIC
0000 8421
0 0000 0011 0000 0000 1010 0011 0000 1111 0000
1 0001 0100 0001 0111 0001 0011 0001 1111 0001
2 0010 0101 0010 0110 0010 0011 0010 1111 0010
3 0011 0110 0011 0101 0011 0011 0011 1111 0011
4 0100 0111 0100 0100 0100 0011 0100 1111 0100
5 0101 1000 1011 1011 0101 0011 0101 1111 0101
6 0110 1001 1100 1010 0110 0011 0110 1111 0110
7 0111 1010 1101 1001 0111 0011 0111 1111 0111
8 1000 1011 1110 1000 1000 0011 1000 1111 1000
9 1001 1100 1111 1111 1001 0011 1001 1111 1001
## Legal history
In 1972, the U.S. Supreme Court overturned a lower court decision which had allowed a patent for converting BCD encoded numbers to binary on a computer (see Gottschalk v Benson). This was an important case in determining the patentability of software and algorithms.
## Comparison with pure binary
• Scaling by a factor of 10 (or a power of 10) is simple; this is useful when a decimal scaling factor is needed to represent a non-integer quantity (e.g., in financial calculations)
• Rounding at a decimal digit boundary is simpler.
• Alignment of two decimal numbers (for example 1.3 + 27.08) is a simple, exact, shift
• Conversion to a character form or for display (e.g., to a text-based format such as XML, or to drive signals for a seven-segment display) is a simple per-digit mapping, and can be done in linear (O(n)) time. Conversion from pure binary involves relatively complex logic that spans digits, and for large numbers no linear-time conversion algorithm is known (see Binary numeral system).
• Some non-integral values, such as 0.2, have a finite place-value representation in decimal but not in binary; consequently a system based on binary place-value representations would introduce a small error representing such a value, which may be compounded by further computation if careful numerical considerations are not made. Note that if computation is not performed on the value this is not an issue, since it suffices to represent it using enough bits that when rounded to the original number of decimal digits the original value is correctly recovered.
• Some operations are more complex to implement. Adders require extra logic to cause them to wrap and generate a carry early. 15–20% more circuitry is needed for BCD add compared to pure binary. Multiplication requires the use of algorithms that are somewhat more complex than shift-mask-add (a binary multiplication, requiring binary shifts and adds or the equivalent, per-digit or group of digits is required)
• Standard BCD requires four bits per digit, roughly 20% more space than a binary encoding. When packed so that three digits are encoded in ten bits, the storage overhead is reduced to about 0.34%, at the expense of an encoding that is unaligned with the 8-bit byte boundaries common on existing hardware, resulting in slower implementations on these systems.
• Practical existing implementations of BCD are typically slower than operations on binary representations, especially on embedded systems, due to limited processor support for native BCD operations.
## Applications
The BIOS in many PCs keeps the date and time in BCD format, probably for historical reasons (it avoided the need for binary to ASCII conversion).
## Representational variations
Various BCD implementations exist that employ other representations for numbers. Programmable calculators manufactured by Texas Instruments, Hewlett-Packard, and others typically employ a floating-point BCD format, typically with two or three digits for the (decimal) exponent. The extra bits of the sign digit may be used to indicate special numeric values, such as infinity, underflow/overflow, and error (a blinking display).
## Alternative encodings
If error in representation and computation is the primary concern, rather than efficiency of conversion to and from display form, a scaled binary representation may be used, which stores a decimal number as a binary-encoded integer and a binary-encoded signed decimal exponent. For example, 0.2 can be represented as 2. This representation allows rapid multiplication and division, but may require multiplication by a power of 10 during addition and subtraction to align the decimals. It is particularly appropriate for applications with a fixed number of decimal places, which do not require adjustment during addition and subtraction and need not store the exponent explicitly.
Chen-Ho encoding provides a boolean transformation for converting groups of three BCD-encoded digits to and from 10-bit values that can be efficiently encoded in hardware with only 2 or 3 gate delays. Densely Packed Decimal is a similar scheme that deals more efficiently and conveniently with the case where the number of digits is not a multiple of 3.
## References
• Arithmetic Operations in Digital Computers, R. K. Richards, 397pp, D. Van Nostrand Co., NY, 1955
• Schmid, Hermann, Decimal computation, ISBN 047176180X, 266pp, Wiley, 1974
• Superoptimizer: A Look at the Smallest Program, Henry Massalin, ACM Sigplan Notices, Vol. 22 #10 (Proceedings of the Second International Conference on Architectural support for Programming Languages and Operating Systems), pp122-126, ACM, also IEEE Computer Society Press #87CH2440-6, October 1987
• VLSI designs for redundant binary-coded decimal addition, Behrooz Shirazi, David Y. Y. Yun, and Chang N. Zhang, IEEE Seventh Annual International Phoenix Conference on Computers and Communications, 1988, pp52-56, IEEE, March 1988
• Fundamentals of Digital Logic by Brown and Vranesic, 2003
• Modified Carry Look Ahead BCD Adder With CMOS and Reversible Logic Implementation, Himanshu Thapliyal and Hamid R. Arabnia, Proceedings of the 2006 International Conference on Computer Design (CDES'06), ISBN 1-60132-009-4, pp64-69, CSREA Press, November 2006
• Reversible Implementation of Densely-Packed-Decimal Converter to and from Binary-Coded-Decimal Format Using in IEEE-754R, A. Kaivani, A. Zaker Alhosseini, S. Gorgin, and M. Fazlali, 9th International Conference on Information Technology (ICIT'06), pp273-276, IEEE, December 2006. | 4,507 | 17,647 | {"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} | 3.5625 | 4 | CC-MAIN-2015-22 | latest | en | 0.876092 |
http://1.bigm-event.com/writing-linear-equations-from-graph-worksheet/ | 1,601,571,768,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402131777.95/warc/CC-MAIN-20201001143636-20201001173636-00058.warc.gz | 1,241,351 | 31,791 | # Writing Linear Equations From Graph Worksheet
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Top | 2,011 | 9,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-40 | latest | en | 0.793839 |
https://scicomp.stackexchange.com/questions/21448/is-it-possible-to-show-global-conservative-properties-fem-as-it-is-done-in-fvm/29990 | 1,627,148,452,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150307.84/warc/CC-MAIN-20210724160723-20210724190723-00485.warc.gz | 499,091,338 | 38,846 | Is it possible to show global conservative properties FEM as it is done in FVM?
I know that in FVM, it is possible to show that a discretisation scheme is conservative by adding the discrete terms over a few control volumes and showing that all terms cancel apart from those relating to the flux in and out of the entire domain.
I'm not sure what happens in FEM? My thoughts are: for instance for a 1D case where there are two neighbouring elements sharing a node, and hence 4 equations exists for a 3-element grid, one adds the relevant terms from the 4 equations to see which terms cancel out and which remain in the end. Am I right?
Thank you.
The way this is usually proven in the finite element context is different, but many finite element schemes satisfy conservation properties. For example, if you think of the Stokes equations, as long as the pressure space contains the piecewise constant functions, then mass is conserved. Similar properties can often be shown for the mixed Laplace equation typically used for porous media flow.
It is typically more complicated to show such properties for first order conservation equations, but even there it is sometimes possible if finite element spaces are appropriately chosen.
However, the ways you would show this results from the weak form of the equations, with particularly chosen test functions. It doesn't quite work as you suggest in your question.
• Thanks again. Can I ask if the same hold for a DGFEM case where the test functions are shape functions located on the roots of a Legendre polynomial? – melody Dec 1 '15 at 15:05
• @melody, it is, but it's because the functions are discontinuous (like in FVMs), and you can do the proof over one element at a time. – Bill Barth Dec 1 '15 at 18:05
• Let me add that even for almost any Stokes discretization with continuous pressure spaces there is some sort of local conservation preserved. Understanding this requires a bit of a different viewpoint, e.g. dual meshes instead of primal mesh elements. We did some work on this recently. – Christian Waluga Dec 1 '15 at 18:29
• @Bill, thank you. just to clarify do you mean that it is possible to prove conservation as in FVM or is it not or do you mean that only local conservation can be proved at an element level? – melody Dec 1 '15 at 23:23
• @Christian, I had a quick looked papers, I will have to read it more carefully. For the time being I am just concerned with a simple 1D structured grid and inviscid laminar flow equations. – melody Dec 1 '15 at 23:23
Why don't you include constant function $\psi(x)=1,x\in\Omega$ in your test function space? Then if you integrate it with conservative form of your differential equation. You can recover the conservation laws.
I think for continuous Galerkin, because the test function space must be zero at boundary, it cannot include constant function. So conservation might be in question. But for DG, the function space includes constant function and the BC is imposed weakly. Conservation is no longer a problem. | 672 | 3,032 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-31 | latest | en | 0.944023 |
https://myquestions.in/2014/03/08/geometric-placement-paper/ | 1,716,275,042,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058385.38/warc/CC-MAIN-20240521060250-20240521090250-00399.warc.gz | 367,334,174 | 14,365 | # Geometric Placement-Paper
1) you should learn pointers(they may ask in i.w)
2)****should learn c++(this is very very imp,based on this only i got job.one more thing is class,inheratenci&polymorphism is sufficient)
3)you should prepare one fem project and tell them that as your own work(this will increase chance to 99%)
written test pattern
——————–
1] c test 10q 20minuits
2]f e m test 19q 20minuits
3]aptitude test 15q 20minuits
—————————–
f e m test
———-
1)who used the term finite element for the first time?
a)
b)
c) clough
2)derive the jacobien |j| for beam element with strainĀ energy?
(ans:promlem is not correct please do not attempt)
3)for an element sigma ni=1,which type of element it is?
a)
b)natural co-ordinate
(ans 100% correct)
4)timoshenko beam element theory to consider — – – -?
a)
b)
c)shear deformation
5)shear locking – – – – ?
(ans is very stiff ‘k’)
6)membrane locking
(ans is arch element)
7)ex(epsiolan x)=du/dx,ey=dv/dy,r(x,y)=?(gama(x,y)=?)
(ans is du/dy+dv/dx )
8)k=integral b(transpose)*d*b for large deformation which matrix will get effected?
(ans is d matrix )100%correct
9)for plane strain f(ex,ey,ez,r(x,y) ) – – –
(ans is ez=0)
10)serendipity element is
(ans 8 noded element)
(the element which is having nodes only on boundary is called serendipity element)
11)if the rotation of element and the displacement about n-a is same then the order of continuity
(ans is c1)
12)frontal theory is applied for
13)mindlins theory is applied for
c) this is the answer(both co&c1problems)
14)x=sigmani*xi,u=sigmani*ui which type of element
(refer book)
15)beam subjected to udl find the moments at the
2 nodes
16)integral b(transpose)*sigma(here sigma means stress)*dvĀ represents?
17)integral et(epsiolan transpose)*sigma*dv
p=strain displacement vector
q=stress-strain deformation
find [k]
(ans [k]=[p]t(p transpose)*[q]*[p] )
18)
19)
remember order may not be correct
aptitude test
1)33 1/3 of 101 + 296 is
(ans 1200) check
2)0.625= ? (ans 27/40)
3)one ship goes along the stream direction 28 km and in opposite
direction 13 km in 5 hrs for each direction.what is the velocity
of stream?
(ans 1.5 kmph)
4)cubic root of 3375=?
(ans 15)
5)2020201-565656=?
(ans 1454545)
6)chairs problem
5 chairs=9 tables,12 tables = 7 stools likethat- – –
(ans is 80rs)
7)one clock ringes 7 o’clock in 7 sec.in how many seconds it will
ring 10 o’clock.
(ans 10.5 sec)
8)one watch is showing 30 past 3 .what is the angle between minutes &
hours hand?
(ans 75 degrees)
9)the average of 4 consecutive even numbers is 27. what is the largest
number?
(ans 30)
10) 25 stations ,24 stations are inbetween- – – – – how many tickets should be required.
(ans 25*24=600)puzzles to puzzle you "s.devi"prob 24
11)one ball was dropped from 8ft height and every time it goes half of the height. how much distance it will travell before coming to
rest. (ans 24 approximately)
12)two trains are travelline at equilateral .train a is travelling in the direction of earths spin.other train b is travelling in
opposite direction of earths spin.which trains wheels will wear first?and why? (ans train b .because of less centrifugal force.)
c questions:
what is the out put for following programms
1)main()
{
char a[2];
*a[0]=7;
*a[1]=5;
printf("%d",&a[1]-a)
ans:
ans may be 1.(illegal initialization)
2)
#include<stdio.h>
main(){
char a[]="hellow";
char *b="hellow";
char c[5]="hellow";
printf("%s %s %s ",a,b,c);
printf(" ",sizeof(a),sizeof(b),sizeof(c));
}
(ans is hellow,hellow,hellow
6,2,5 )
3)
#include<stdio.h>
main()
float value=10.00;
printf("%g %0.2g %0.4g %f",value,value,value,value)
}
(ans is 10,10,10,10.000000)
4)
#include<stdio.h>
void function1;
int i-value=100;
main()
{
i-value=50;
function1;
printf("i-value in the function=",i-value);
printf("i-value after the function=",i-value);
}
printf("i-value at the end of main=",i-value);
functioni()
i-value=25;
this is rough idea of the program
ans are
1)i-value in the function=25;
2)i-value after the function=50;
3)i-value at the end of the main=100;
5)
main()
{
funct(int n);
{
switch(n)
case1:
m=2;
break;
case2:
m=5;
break;
case3:
m=7;
break;
default:
m=0;
}
this is rough idea:
(ans:out put is m=0) | 1,355 | 4,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-22 | latest | en | 0.787978 |
https://rdrr.io/cran/tmvtnorm/man/ptmvnorm.html | 1,529,612,493,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864257.17/warc/CC-MAIN-20180621192119-20180621212119-00405.warc.gz | 698,616,023 | 15,031 | # ptmvnorm: Truncated Multivariate Normal Distribution In tmvtnorm: Truncated Multivariate Normal and Student t Distribution
## Description
Computes the distribution function of the truncated multivariate normal distribution for arbitrary limits and correlation matrices based on the `pmvnorm()` implementation of the algorithms by Genz and Bretz.
## Usage
```1 2 3 4``` ```ptmvnorm(lowerx, upperx, mean=rep(0, length(lowerx)), sigma, lower = rep(-Inf, length = length(mean)), upper = rep( Inf, length = length(mean)), maxpts = 25000, abseps = 0.001, releps = 0) ```
## Arguments
`lowerx` the vector of lower limits of length n. `upperx` the vector of upper limits of length n. `mean` the mean vector of length n. `sigma` the covariance matrix of dimension n. Either `corr` or `sigma` can be specified. If `sigma` is given, the problem is standardized. If neither `corr` nor `sigma` is given, the identity matrix is used for `sigma`. `lower` Vector of lower truncation points,\ default is `rep(-Inf, length = length(mean))`. `upper` Vector of upper truncation points,\ default is `rep( Inf, length = length(mean))`. `maxpts` maximum number of function values as integer. `abseps` absolute error tolerance as double. `releps` relative error tolerance as double.
## Details
The computation of truncated multivariate normal probabilities and densities is done using conditional probabilities from the standard/untruncated multivariate normal distribution. So we refer to the documentation of the `mvtnorm` package and the methodology is described in Genz (1992, 1993) and Genz/Bretz (2009).
For properties of the truncated multivariate normal distribution see for example Johnson/Kotz (1970) and Horrace (2005).
## Value
The evaluated distribution function is returned with attributes
`error` estimated absolute error and `msg` status messages.
## References
Genz, A. (1992). Numerical computation of multivariate normal probabilities. Journal of Computational and Graphical Statistics, 1, 141–150
Genz, A. (1993). Comparison of methods for the computation of multivariate normal probabilities. Computing Science and Statistics, 25, 400–405
Genz, A. and Bretz, F. (2009). Computation of Multivariate Normal and t Probabilities. Lecture Notes in Statistics, Vol. 195, Springer-Verlag, Heidelberg.
Johnson, N./Kotz, S. (1970). Distributions in Statistics: Continuous Multivariate Distributions Wiley & Sons, pp. 70–73
Horrace, W. (2005). Some Results on the Multivariate Truncated Normal Distribution. Journal of Multivariate Analysis, 94, 209–221
## Examples
```1 2 3``` ``` sigma <- matrix(c(5, 0.8, 0.8, 1), 2, 2) Fx <- ptmvnorm(lowerx=c(-1,-1), upperx=c(0.5,0), mean=c(0,0), sigma=sigma, lower=c(-1,-1), upper=c(1,1)) ```
### Example output
```Loading required package: mvtnorm | 728 | 2,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-26 | latest | en | 0.710744 |
http://hondaswap.com/threads/for-all-of-u-who-are-good-at-math.12192/ | 1,495,737,198,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608120.92/warc/CC-MAIN-20170525180025-20170525200025-00348.warc.gz | 175,894,696 | 16,897 | # For All Of U Who Are Good At Math..
Discussion in 'Members' Lounge' started by 22crazy, May 4, 2003.
1. ### 22crazySenior Member
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whats up. i have a question that been pissing me off and i was wondering if any of you could solve it. the question: if you construct equalateral triangles of the sides of a quaderlateral, alternately inward and outward, when you connect the vertex of each triangle, the result is a parallelogram. Prove it (the parallelogram that is formed). ill see if i can make the picture.
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3. ### BigJI'm just about that action Boss.VIP
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Yeah looks to me like some seriously hard proof that must be written to prove that one. I am just in mathanalysis in sophomore year of high school so I am little under that problem for now.
4. ### vpSPOONsolSenior Member
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well simply u just take the second derivative of the intergral of A to B, assuming that the distance from A to C is equal.... u then muliply that by the sides of the parrallogram, which gives u = j/p... im taking calculus for my senior year right now, and i dont have ur answer.
im sure some of the older college grad guys can answer that though...
5. ### BakerSenior Member
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I could probably do it but the resulting headache would not make it worthwhile
6. ### 22crazySenior Member
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well can anyone help me, its been bugging me alot and i need to complete it. thanx for the help though. i no it has to do something with the 60 degrees in the equalateral triangles and also, u have to construct a parallelogram within the quaderlateral: that is done by connecting the midpoints.
7. ### BodyDroppedNikes...PENDEJO....VIP
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caught in a mosh...
<--this is me after reading this topic
8. ### DazedCivic99Senior Member
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is this supposed to be a 3D object?
If it is i sorta see what you are going for but a little lost at the moment
as for the calculas answer vpspoon you just reminded me that i need to study for my AP calculas test, damn you j/k
9. ### 22crazySenior Member
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no its a 2D object...can anyone still help me
10. ### pills_PMDSuper Moderator
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what math class is it for?
there are probably several ways of doing it..
and which part am i suposed to prove.. that the red lines are a parallelogra?M
11. ### pissedoffsolRETIRED
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Calesta, i think this ones for you man
12. ### CRX-YEMSuper ModeratorModeratorVIP
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man I can't remember the last time I did a geometric proof. I'd have to bust out my books, but I've got finals this week so that ain't happening.
13. ### doofnoilSenior Member
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If you just have to prove that a parallelogram would exist, I would look at the relation between the sides of the quad. and the parallel. to see if they correspond ?
14. ### lsvtecGNU/Linux Evangelist
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You need to prove that Angles 1 & 6 are equal, and anges 3 & 4 are equal. This proves that angles 2&5 are equal because the sums of 1, 2, & 3 and 4, 5, & 6 are both 60 degrees. Then all you have to do is prove that each pair of opposing sides are equal length. Then you have a parallelogram.
15. ### dohcvtec_accordWRX Sellout
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I was wondering how long it would take you to get involved Eric.
16. ### lsvtecGNU/Linux Evangelist
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I had to sit and think for a while. It has been forever since I did a Geometric proof. And even then (freshman year of HS) I wasn't very good at them.
You could also simply prove that the opposite inside angles of the parallelogram are equal instead of showing the opposite sides equal.
17. ### dohcvtec_accordWRX Sellout
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Exactly what I was going to say.
*pushes non-existent glasses up on nose, goes for high-five with lsvtec, misses horribly, gives nerdy, snorty laugh*
18. ### 22crazySenior Member
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actually those angles 4 3 and 1 6 are not equal, but yet 5 and 2 are....i was able to prove it using geometer sketch pad, but i dont no how to do it paper and pencil method
19. ### Gen2TegSenior Member
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i didn't even read the question, i just looked at the picture and it made my head hurt
Messages:
5,453 | 1,437 | 4,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2017-22 | longest | en | 0.934564 |
http://inform7.com/extensions/Michael%20Callaghan/Fixed%20Point%20Maths/source_26.html | 1,516,522,721,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890394.46/warc/CC-MAIN-20180121080507-20180121100507-00205.warc.gz | 177,457,243 | 1,673 | # Fixed Point Maths
## version 5 by Michael Callaghan
Section 3 - Division of one number by a fixed point number
To decide which real number is (A - a number) real divided by (B - a real number):
let temp be A as a fixed point number;
decide on temp real divided by B. | 70 | 271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-05 | latest | en | 0.912199 |
http://bethanystruble.com/y3e4o3q/3134ba-cardinality-bijective-proof | 1,623,982,367,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487634616.65/warc/CC-MAIN-20210618013013-20210618043013-00034.warc.gz | 5,951,324 | 17,528 | # cardinality bijective proof
Let S and T be sets, and let be a function. But there is no harm in taking the intuitive approach and just interpreting the cardinality $$|X|$$ of a set X to be a measure of the “size” of X. So, we have a means of determining when two sets have the same or different cardinalities. A set is Kurt Gödel . (f is called an inclusion By the lemma, is a result by contradiction. Example. one-to-one) if implies . Therefore, the interval must be uncountably infinite. c) $(0,\infty)$, $\R$ d) $(0,1)$, $\R$ Ex 4.7.4 Show that $\Q$ is countably infinite. Consider the interval $$(0, \infty)$$ as the positive x-axis of $$\mathbb{R}^2$$. Therefore, it's valid to write Let S and T be sets, and let be a function from S to T. A function is called the inverse of f if. has the same cardinality as the real line. uncountable. First, notice that the open interval has the same cardinality as the real line. Cardinality Problem Set Three checkpoint due in the box up front. On one hand it makes sense that $$|\mathbb{N}| = |\mathbb{Z}|$$ because $$\mathbb{N}$$ and $$\mathbb{Z}$$ are both infinite, so their cardinalities are both “infinity.” On the other hand, $$\mathbb{Z}$$ may seem twice as large as $$\mathbb{N}$$ because $$\mathbb{Z}$$ has all the negative integers as well as the positive ones. S and T The 3rd decimal place of $$f(3)$$ is the 3rd entry on the diagonal. Show that the open interval and the closed interval have the same relation. Example. (unless both sets have a single element). If f: A → B is a surjection then f is a bijection. Thus, even though $$f(1)$$ happens to be the real number 0.4, we write it as 0.40000000...., etc. If A;B are nite sets of the same cardinality then any injection or surjection from A to B must be a bijection. For transitivity, suppose $$|A| = |B|$$ and $$|B| = |C|$$. Definition same_cardinality (X Y: Type) : Prop:= ∃ f: X → Y, bijective f. For example, we can define a set with two elements, two , and prove that it has the same cardinality as bool . I know that some infinite sets --- the even integers, for instance This third article further compounds this knowledge by zoning in on the most important property of any given set: the total number of unique elements it contains. Now I have injective functions and . We now describe Cantor’s argument for why there are no surjections $$f : \mathbb{N} \rightarrow \mathbb{R}$$. We emphasize and reiterate that Definition 14.1 applies to finite as well as infinite sets. Also known as the cardinality, the number of disti n ct elements within a set provides a foundational jump-off point for further, richer analysis of a given set. , and hence g is injective. a combinatorial proof is known. View CS011Cardinality7.12.2020.pdf from CS 011 at University of California, Riverside. n or that S has n elements. Because of this bijection $$f : \mathbb{N} \rightarrow \mathbb{Z}$$, we must conclude from Definition 14.1 that $$|\mathbb{N}| = |\mathbb{Z}|$$. If no such bijection exists, then $$|A| \ne |B|$$. There is a bijection between fa;bgand f1;2ggiven by f(a) = 1, f(b) = 2. To begin we will need a lemma. If there is an injective map f : X !Y, then jXj jYj: If there is an injection from X into Y but no bijection between X and Y, we write jXj< jYj: When A = ;, we set jAj= 0: Let n 2N. Inc., 1966 [ISBN 0-8053-2327]. Proof of cardinality. Prove that has the same cardinality as . Of course, everyday which is not countably infinite is uncountably infinite or ought to behave. I'll use the [2] proved around 1940 that the Continuum Hypothesis was consistent Two sets A and B have the same cardinality, written | A | = | B |, if there exists a bijective function f: A → B. May 2009 57 1. We need a new approach that applies to both finite and infinite sets. If, in trying to show two sets A and C have the same cardinality, we can produce a third set B for which $$|A| = |B|$$ and $$|B| = |C|$$, then transitivity assures us that indeed $$|A| = |C|$$. When a set Ais nite, its cardinality is the number of elements of the set, usually denoted by jAj. (c) If and , then there are cardinality. A useful application of cardinality is the following result. a factor of 2". We say two sets Aand Bare related by cardinality if jAj= jBj. Math 127: In nite Cardinality Mary Radcli e 1 De nitions Recall that when we de ned niteness, we used the notion of bijection to de ne the size of a nite set. countably infinite. The next example uses this idea. endpoints) won't fit in either of the intervals that make up the By similar triangles, we have and therefore, If it is not clear from the figure that $$f : (0, \infty) \rightarrow (0, 1)$$ is bijective, then you can verify it using the techniques from Section 12.2. Further gradations are indicated by + and –; e.g., [3–] is a little easier than [3]. reviewing the some definitions and results about functions. If I multiply by , I'll shrink to , which has a total length of 1. anyone has given a direct bijective proof of (2). Hence, while , and I'll prove that is the Acad. Thus, for the function f illustrated in the above table, we have. Because the bijection $$f : \mathbb{N} \rightarrow \mathbb{Z}$$ matches up $$\mathbb{N}$$ with $$\mathbb{Z}$$, it follows that $$|\mathbb{N}| = |\mathbb{Z}|$$. Here's the proof that g and are inverses: Therefore, g is a bijection, so and have the If , then by definition of T, . Using this idea, we showed that $$|\mathbb{Z}| = |\mathbb{N}| \ne |\mathbb{R}| = |(0, \infty)| = |(0,1)|$$. examples of infinite sets which have the same cardinality. Then S and T have the same Now occupies a total length of , whereas the target interval has length 2. Suppose . through is. Definition. Imagine a light source at point P. Then $$f(x)$$ is the point on the y-axis whose shadow is x. define a bijection by "scaling up by Cardinality Lectures Enrique Trevino~ November 22, 2013 1 De nition of cardinality The cardinality of a set is a measure of the size of a set. characteristic of infinite sets that they have the same This is a contradiction. I'll begin by ... We will show that gis bijective, from which the conclusion follows. and conceived of 5 as the thing common to all such sets. So s is an element which is Let and be their the elements of an infinite set can be listed: In fact, to define listable precisely, you'd end up saying --- are countably infinite. Now suppose that . For in nite sets, this strategy doesn’t quite work. Suppose first that . f takes an element of S to a subset of S, and that subset either Proof. one-to-one correspondence) if it is injective and surjective. Schröder-Bernstein theorem, and have the same cardinality. f is invertible if and only if f is Of course, . It's an (Note that there are many functions you could use to do The proof we just worked through is called a proof by diagonalization and is a powerful proof … B. Definition13.1settlestheissue. constructing a function . 21. experience says that this is impossible. Therefore the definition says $$|A| \ne |B|$$ in these cases. In counting, as it is learned in childhood, the set {1, 2, 3, . be overdoing it a bit.). (In fact, g is bijective, and you could (Hint: you can arrange $\Q^+$ in a sequence; use this to arrange $\Q$ into a sequence.) , but I've just shown that the two sets "have In fact, we could be concrete and define $$|X|$$ to be the equivalence class of all sets whose cardinality is the same as that of X . Also, Example 14.3 shows that $$|(0, \infty)| = |(0, 1)|$$. By transitivity, and have the same cardinality. All In this So define by, First, I have to show that this makes sense --- that is, that f 0.25 to shift to . Cardinality Problem Set Three checkpoint due in the box up front. (Of course, does not imply that . Since , obviously , so g does map into . Forums. To show that f is bijective, I have to show that it has an inverse; the inverse is f−1(x) = arctanx. You can also turn in Problem Set Two using a late period. --- but it's true, and I'll omit the proof. no sets which are "between" and in cardinality; it was first The Continuum Hypothesis states that there are Therefore, f and g are bijections. same cardinality. of 9's, rewrite it as a finite decimal --- so, for instance, becomes 0.135.) Now I know that and have the same . There exists no bijection $$f : \mathbb{N} \rightarrow \mathbb{R}$$. Question: Is ? contains the element or it doesn't. is countably infinite; how big is ? Definition. Definition. So just what kind of mathematical entity is $$|\mathbb{Z}|$$? Consider the set T = { a ∈ A | a ∉ f(a) } Thus, . Here's the proof that f and are inverses: This situation looks a little strange. More importantly, we would like to develop some notion of cardinality for infinite sets aswell. }\] The concept of cardinality can be generalized to infinite sets. . There exists a bijection $$f : \mathbb{N} \rightarrow \mathbb{Z}$$. When two sets don't look alike Every integer appears exactly once on the infinitely long second row. 3. bijection. intervals. Theorem 2. jZj= jNj Note: Even though the rst function you may think of, namely f : N !Z given by f(x) = x, is not a bijection, that doesn’t mean there isn’t some other function that is a bijection. consists of two open intervals. Then. A. Proof. important fact that not all infinite sets have the same cardinality • A function f: R → R is bijective if and only if its graph meets every horizontal and vertical line exactly once. The sets $$A = \{n \in \mathbb{Z} : 0 \le n \le 5\}$$ and $$B = \{n \in \mathbb{Z} : -5 \le n \le 0\}$$ have the same cardinality because there is a bijective function $$f : A \rightarrow B$$ given by the rule $$f(n) = -n$$. So. Define by . Find a formula for the bijection f in Example 14.2 (page 270). To avoid repeating this proof twice, we say “without loss of generality” to say that “we will prove the case when a i ∈ S and a i 6∈ T, and the other case is the same so we skip its proof”. Let X and Y be sets and let be a function. Prove that the interval (0,1) has the same cardinality as R. First, notice that the open interval − π 2, π 2 has the same cardinality as the real line. I'm going to be a little informal in cardinality. Proof. Next, I have to define an injective function . Here's the proof that f and are inverses: . the same cardinality as a set with 42 elements. "the set has the same cardinality as the natural numbers". The number of Example 7.2.4. Problem Set Three checkpoint due in the box up front. The open interval is a subset of the closed constructing a bijection from one to the other. deals with finite objects. Since the second set's intervals don't have takes a and d to subsets which don't contain them. Hence, . standard "swap the x's and y's" procedure works; you get. Both individuals and organizations that work with arXivLabs have embraced and accepted our values of openness, community, excellence, and user data privacy. Then the nth decimal place of b differs from the nth decimal place of $$f(n)$$. The first set is an interval of length 2, which (because of its inverse, namely itself. Although this is a fine strategy if the sets are finite (and not too big! Next, I'll add Suppose . In order to be definite, define b to be the positive number less than 1 whose nth decimal place is 0 if the nth decimal place of f (n) does not equal 0, and whose nth decimal place is 1 if the nth decimal place of f (n) equals 0. Let’s see an example of this in action. (c) Suppose that and are bijections. resolved: Could a finite set be bijective with both and (say)? So if the Theorem. A cardinal number is thought as an equivalence class of sets. I'll construct an inverse for f. The inverse should "undo" we'll take in this example. This means that there is a bijection . This shows that g takes inputs in and produces We have the following properties. By Schröder-Bernstein, . same cardinality. A number, say 5, is an abstraction, not a physical thing. It is injective because the way the table is constructed forces $$f(m) \ne f(n)$$ whenever $$m \ne n$$. Suppose . Example. (b) If , then there is a bijection . For any set | P(A) | > | A |. arXivLabs is a framework that allows collaborators to develop and share new arXiv features directly on our website. When the set is in nite, comparing if two sets have the \same size" is a little di erent. Next, I We describe this function geometrically. Theorem13.1 Thereexistsabijection f :N!Z.Therefore jNj˘jZ. The function $$f$$ that we opened this section with is bijective. 3: The hook of the cell (2; 3). Thus, according to the table, given any $$b \in \mathbb{Z}$$ there is some natural number n with $$f(n) = b$$, so f is surjective. can slide inside by subtracting 0.7, which should give . 9's.). With the bijections f and g, I have , so and have the same bijection. Example. It's easy: just define. map to . As usual, I'll show f is bijective by constructing an inverse . every subset of S --- is paired up with an element of S. For example, In this table, the real numbers $$f(n)$$ are written with all their decimal places trailing off to the right. Here it is: Two sets A and B have the same cardinality, written $$|A| = |B|$$, if there exists a bijective function $$f : A \rightarrow B$$. If the function $$f$$ is a bijection, we also say that $$f$$ is one-to-one and onto and that $$f$$ is a bijective function. We begin with a discussion of what it means for two sets to have the cardinality... Diagonal line whose elements add up cardinality bijective proof us to find one of them in order to able! Occupies a total length of, cardinality bijective proof it follows that is,,... Are too big CC BY-NC-SA 3.0, everyday experience says that this is bijection. Think of f as describing how to overlay a onto B so that they have the cardinality... Formal proof of this in action graph meets every horizontal and vertical line exactly once actually... View CS011Cardinality7.12.2020.pdf from CS 011 at University of California, Riverside to grasp our... } \rightarrow \mathbb { R } \ ) is arbitrary naturals have the cardinality. It has an inverse indication of how there are different kinds of infinity '' be lots of correspondence. Theme throughout this chapter we have a means of deciding whether two sets have. These values and only if f: a → B is an obvious injective! One such element, namely itself we ’ d never be done counting their elements think they the! Of things ( five apples, five oranges, etc. =m-n is surjective using a late period in cases... Fit together perfectly countable set sense -- - are countably infinite, so (! Them in order to conclude \ ( |\mathbb { R } \ compare... If, then there are bijections and go down the diagonal line whose elements add up to an explicit to. Following lemma about subsets of S. for instance -- - that element give possibilities in all,. Might make it bigger function for all subset of S is an element which f to., [ 3– ] is a bijection useful application of cardinality can be generalized to sets! Called the diagonalization argument T has 5 elements, and 1413739 a single of... We have declared that two sets have unequal cardinalities, written \ ( g\ ) is arbitrary are any! Features directly on our website numbers, for they are “ infinity ” is accurate. Other words, having the same cardinality by actually constructing a bijection \ ( |\mathbb Z... Well, because every element of B differ from the nth decimal place of \ ( f ( N \! A to B be generalized to infinite sets take in this chapter we a! Following lemma about subsets of the digits in this particular point is n't lost in a in. Is a bijection, then \ ( f ( cardinality bijective proof ) \ ) is bijective, I need to some! 'Ll define injective functions going from each set into the other sets Aand said... 'S '' procedure works ; you get Save as PDF page ID 10902 ; headers. Of mathematical entity is \ ( f ( N ) \ne B\ ) and \ ( =.: you can also turn in Problem... is called an injection between two finite of..., onto functions ( because it is its own inverse function ), while, I... To finite as well as the positive ones injection between two finite sets of the cardinality. Are natural numbers might make it bigger cardinality \ cardinality bijective proof f: a \rightarrow B\ ) is arbitrary other sets... 'Ve just shown that the function f de ned in equation ( 1 ) 2S n't look alike but think! From a more general perspective, in which variables are allowed to range elements! Details left as an example of this section is the following questions bijections. Even, so \ ( f ( 4 ) \ ) is arbitrary for example, if is. Do it here the idea is to multiply by 0.5, I get, interval... Conclude that there are many functions you could add 1 to each digit from 0 to 7 and change to. For any set | P ( a ) is arbitrary handle that kind mathematical! Of cardinalities is an injective function, and explains why it is reflexive, symmetric and transitive a the. Onto B so that they fit together perfectly theorem is tricky but really quite beautiful accurate, because we know... Not surjective way that it 's an important fact that not all infinite,. Already from basic ( pre ) calculus emphasized for two sets have the same number of elements of the functions... = |C|\ ) 's important that both … a surjection n't lost in list... Interval ( 0, 1 ) |\ ) ; ; N 1g uncountable, and we the! Definition says \ ( f ( m, N ) = -n\ ) from example 14.1 a! Never be done counting their elements evens and the function of two infinite sets a and be..., by the cardinality bijective proof like asking what a number using the digits unequal cardinalities written! Of elements whereas the target interval has the advantage of giving an explicit meaning to |X| bijection between them actually... Knowing if the set is equal to N, meaning f is injective my first --! Describing a bijection \ ( | ( 0, 1 ) prove that the open interval and the of! Than writing out an exact proof as an exercise ) the conclusion.! } \rightarrow \mathbb { R } |\ ) bijective f exists, then the sets have equal cardinality by an... A way that it is a subset that does not belong to the other first suppose y∈ f N. 0.5 to shrink to follows that is uncountably infinite as well as infinite sets able to define what it for. Situation later a copy '' of in, then, so it follows that is either or! Justify -- - which means for more information contact us at info @ libretexts.org or check out status. Actually constructing a bijection ( details left as an exercise Q is countably infinite R... A late period ; N 1g definition 14.1 applies to both finite and, then and! 'S important that both of these intervals are closed intervals number, then confirms! Nor surjective uncountable, and U be sets scaling up by a factor 2! Acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057 and... Is the number of elements in a set S itself Xand Y countable. In action gis bijective, and I 'll omit the proof that f and are inverses is in.. $\Q^+$ in a lot like asking what a number using digits! Or surjection from a to B bijection between them how big is this confirms the theorem that follows gives indirect. 4Th decimal place of \ ( \mathbb { N } \ ) is the number of elements contact us info. Contain it surjection between finite sets of the natural numbers for instance, suppose I. Discussion of what it means for two reasons is also clearly a bijection ( details as... Takes into Hypothesis, Proc continue to develop some notion of cardinality can generalized. Some implementation that actually knows how to handle that kind of situation later claim is a bijection from one the. I get, an interval of length cardinality bijective proof theorem in this particular case \$ in a set is the! 0.5 to shrink first, I 'll used to describe the situation by! Situation looks a little tricky to show that gis bijective, I add! Basic ( pre ) calculus over elements of S is finite if it has inverse. An example, you could add 1 to each digit from 0 to 7 and change it to other. To try to match up the elements of two infinite sets, then the sets have the same cardinality ’. Generalized Continuum Hypothesis, Proc by induction on N = card ( a ) the identity function an. Which the conclusion follows that R and the interval has length 4 National Science support... One-To-One correspondence ) if it is: here is why this works 's '' procedure ;! N'T do it here in such a set X, exactly what is cardinality... Bijective is n't lost in a set is and the closed interval have the same as! To, which should give by is a little tricky, so g does into. M, N ) = -n\ ) from example 14.1 is a diagonal shaded band in the box front! An obvious '' injective function from to: if, then \ ( | 0... Exercise ) is so innate are infinite injective, so f does map.! The element which f takes to the other functions from a to B must be a prime know. Values and only works with partners that adhere to them the generalized Continuum Hypothesis was relative. What it means for two reasons Illinois, Chicago Real-valued functions of a bijection cardinality is bijective take any function... Here 's the proof instance, ca n't be arranged in a set,! 4 cardinality is obviously, so this confirms the theorem in this with... Has the advantage of giving an explicit meaning to |X| from each set into the other is clearly a function... In life we instinctively grouped together certain sets of the bijective functions are also one-to-one! This example shows that \ ( \mathbb { N } \rightarrow \mathbb { N \rightarrow. So m is divisible by 2 and is called the diagonalization argument \ ( \mathbb { N } =. Recall that this is a bijection, so contain them: R → R is uncountable, and are... Numbers 1246120, 1525057, and is called the diagonalization argument element give possibilities in all,. So f does map into ; 1 ; 2 ; 3 ) \ ) is,. | 5,627 | 21,832 | {"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} | 4.28125 | 4 | CC-MAIN-2021-25 | latest | en | 0.924354 |
https://mycoursewriter.com/momentum-energy-and-velocity-questions/ | 1,643,360,479,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305423.58/warc/CC-MAIN-20220128074016-20220128104016-00698.warc.gz | 460,831,346 | 13,122 | Momentum, Energy, and Velocity Questions
1. (a) What is the momentum of a 1.20 x 10^4 kg garbage truck moving at 30.0 m/s? (b) At what speed would an 8.00 kg trash can have the same momentum?
2. One hazard of space travel is debris left by previous missions. There are several thousand masses large enough to detect by radar orbiting the earth, but there are far greater numbers of very small masses, such as flakes of paint. Calculate the force exerted by a 0.100 mg chip of paint that strikes a space shuttle window at a relative speed of 4000 m/s, given the collision lasts 6.00 x 10^(-8) s. Such a collision chipped the window of the ill-fated Challenger in June 1983, causing \$50,000 of damage.
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3. A 0.450 kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board. (a) Calculate the duration of the impact. (b) What was the average force exerted on the nail?
Note: There are (at least) two strategies here. One is to assume constant acceleration over the 1-cm distance and determine the time using the equations for constant acceleration. From that, you can find the average force. However, a more “relevant” strategy, in terms of what we have been learning, is to first use the work-energy principle to find the average force. Then use the impulse-momentum principle to find the duration of impact.
4. It is possible for the velocity of a rocket to be greater than the exhaust velocity of the gases it ejects. When that is the case, the gas velocity and momentum are in the same direction as the rocket’s. How does the rocket still obtain thrust by ejecting the gases?
5. Water from a fire hose is directed horizontally against a wall at a rate of 50.0 kg/s and a speed of 42.0 m/s. Calculate the force exerted on the wall, assuming the water’s horizontal momentum is reduced to zero.
6. Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 150,000 kg and a velocity of 0.300 m/s, and the second having a mass of 110,000 kg and a velocity of -0.120 m/s. What is their final velocity?
7. A 0.240 kg billiard ball moving at 3.00 m/s strikes the bumper and bounces straight back at 2.40 m/s (80% of the original speed.) The collision lasts 0.0150 s. (a) Calculate the average force exerted on the ball by the bumper. (b) How much kinetic energy in joules is lost during the collision? (c) What percent of the original energy is left?
8. Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and was approaching at 8.00 m/s due south. The second car has a mass of 850 kg and was approaching at 17.0 m/s due west. (a) Draw arrows representing the momentum vector of each car, and draw an arrow representing the resultant (vector sum) of these two. Label each arrow with its magnitude in kg-m/s. (Make this a clear diagram, with arrows at least 3 inches long. Make the relative length of the arrows accurate.) (b) Calculate the final velocity of the cars. (Magnitude and direction.) (c) How much kinetic energy is lost in the collision? (This energy goes into deformation of the cars.) The solution has the eight solutions to the momentum, energy and velocity questions. These are standard classical mechanics problems pertaining to work.To continue with the answer check on mycoursewriter.com/
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Essay writing is a common assignment writing for most college students. It takes a lot of time to craft a complete perfect essay. Mycourserwiter offers the best essay writing and proofreading services for most college students. Your paper is assigned to an expert writer who does thorough research on fresh ideas and personalizes your essay paper according to your assignment writing guidelines. Our essays are entirely exceptional and undergo a strict procedure that includes papers free from plagiarism and thorough proofreading. | 1,620 | 7,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-05 | latest | en | 0.938385 |
http://help.4matrix.com/knowledgebase.php?article=136 | 1,550,952,049,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249530087.75/warc/CC-MAIN-20190223183059-20190223205059-00164.warc.gz | 123,886,266 | 5,207 | Notices 20/02/2019
1) No Progress 8 for Y7 - Y9? See https://www.4matrix.com/node/6222
2) KS5 Value Added showing warning triangles? See https://www.4matrix.com/help/ks5_va_2018.pdf
3) Up-to-date information on getting help with 4Matrix: https://www.4matrix.com/help/getting_help.pdf
### How are Key Stage 2 Fine Points calculated?
#### Solution
The following information relates to Y11 2015/16. When using Y11 2016/17 and beyond, the Reading mark is imported and a Fine Points score calculated from this instead.
Example pupil data:
READMRK = 31 (This is her test mark from the English Reading element)
ENGWRITMRK = 21 (This is her test mark from the English Writing element)
ENGLEV = 4 (This is her overall English level)
To work out her Fine Points, we need to use the calculation:
Basic Level + ((actual test mark - bottom of level threshold) / (top of level threshold - bottom of level threshold + 1))
Using the data above, the calculation becomes: 6 x (4 + (52 - 41)/(69-41+1))
BUT, before multiplying by 6, the score must be rounded to 2dp. (4 + (52 - 41)/(69-41+1)) = 4.38 (rounded to 2dp).
Now, multiply this score by 6 to get her Fine Points score = 26.28
Her Maths Fine Points will be:
6 x (5 + (88 -77)/(100 - 77 + 1))
This equals 32.76.
Using these values we can then determine her Fine Test Levels using the boundaries from RaiseOnline:
5A 34-35.99
5B 33 32-33.99
5C 30-31.99
4A 28-29.99
4B 27 26-27.99
4C 24-25.99
3A 22-23.99
3B 21 20-21.99
3C 18-19.99
Her English and Maths Fine Levels will be 4B and 5B respectively.
To calculate her Progress 8 baseline, add the English and Maths Fine Points together, divide by 2 and then by 6, and then round to 1dp.
((26.28 + 32.75) / 2) / 6 = 4.9 rounded to 1dp.
For the purposes of calculating a pupil's Progress 8 baseline, English and Maths Level 6's are always ignored and so it should be the mark/level from the standard KS2 test that is imported.
There are some exceptions to the rules above and users should refer to page 10 in the Official VA Guide for further information on KS2 levels that fall below a 3. An example for a pupil with an overall English level N is provided below:
READMRK = 5 (This is her test mark from the English Reading element)
ENGWRITMRK = 5 (This is her test mark from the English Writing element)
ENGLEV = N (This is her overall English level)
In this scenario, the Fine Points calculation is ignored. Provided that she has a Teacher Assessment, she will receive default points according to the VA guide. If she does not have a Teacher Assessment, her Fine Points are left blank and her English score is excluded from calculating the Progress 8 baseline.
Test Level 2 and TA or 3 or higher:
Prior to update 3.5.12.0, if the pupil had a Teacher Assessment higher than 2, the minimum points were capped at 15. This is because, for example, if the pupil had a TA of a 4, when running this through the calculation above they would only receive 12 points, which is less than if they had a TA of 2 (for which they would receive 15 by default). As of 9th September 2016, the DfE have clarified the exact methodology for these pupils as follows (and this is included as part of update 3.5.12.0);
Pupils with a test level of 2 and a teacher assessment of W, 1 or 2 would have a fine point score of 3, 9 and 15 respectively. For pupils with a teacher assessment higher than 2, the following formula is used:
FINEGRADE = 3.0 - ((MINLEVL3MARK - MAINMARK) / (MAXLEVL3MARK - MINLEV3MARK +1))
Where MAINMARK is the mark for the test, i.e. ENGTOTMRK or MATTOTMRK, and MIN/MAXLEV3MARK are the bottom and top of the mark range for level 3 in the relevant subject. When the mark is less than the minimum mark in the level 2 range, MAINMARK is set to be the minimum mark in the range. Where it is greater than the maximum mark in the level 2 range, MAINMARK is set to be the maximum mark.
The Fine grade is then rounded to 2 decimal places and multiplied by 6 to calculate fine points as usual.
Article ID: 136 | 1,126 | 4,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2019-09 | latest | en | 0.826068 |
https://newbedev.com/can-water-pressure-ever-be-high-enough-to-trap-gas-bubbles-or-keep-them-from-surfacing | 1,669,470,422,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446706291.88/warc/CC-MAIN-20221126112341-20221126142341-00378.warc.gz | 469,346,554 | 7,007 | # Can water pressure ever be high enough to trap gas bubbles or keep them from surfacing?
The highest pressure in the ocean is at the bottom of the Mariana trench, where the pressure is 1 086 atmospheres. Using the online calculator for the properties on nitrogen at 4 °C and 1 000 atmospheres the density comes out as 602 kg/m³, which is still less than water. So a bubble of nitrogen would rise even at the deepest point in the ocean.
Response to comment:
In principle we can continue increasing the pressure and the nitrogen should get denser. However at temperatures above 0 °C and the sorts of pressures we are talking about, nitrogen is a supercritical fluid so it does not obey anything like an ideal gas law. Calculating at what point the density would exceed the density of water is far from easy.
The effect of pressure on water is straightforward. At sea bottom temperatures (about 4 °C) the density of water increases only slowly with pressure to about 1 050 kg/m$$^3$$ at 6 000 atmospheres, at which point the water freezes to form ice V. So the question is whether the density of nitrogen exceeds 1 050 kg/m³ below a pressure of 6 000 bar.
I can’t find any figures for the density of nitrogen at these sorts of pressures and temperatures, though I did find this paper that gives a Mie-Grüneisen type equation relating the density, pressure and temperature. Unfortunately the preview only shows two pages and the rest of it is behind a paywall. However using the figures they give and waving my arms around a bit I find the density of nitrogen rises to 1 050 kg/m³ at around 4 000 atmospheres.
So, it might just be possible to get a nitrogen bubble that is denser than water and will sink instead of floating. But I don’t know whether the equation from the paper I cited is accurate at these sorts of pressures and temperatures, and it’s possible the nitrogen will solidify before the water does (though I’d guess not). | 435 | 1,938 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-49 | latest | en | 0.907563 |
http://www.moomoomathblog.com/2016/10/how-much-do-you-weigh-on-moon.html | 1,721,031,015,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514680.75/warc/CC-MAIN-20240715071424-20240715101424-00042.warc.gz | 49,940,163 | 21,789 | # How much do you weigh on the moon?
You only way 16.5 % of your weight on the moon as you do on the Earth. Because of the smaller mass of the moon, it has a smaller gravitational force. As a result, if you weigh 100 pounds on Earth you would tip the scale at 16.5 pounds on the Moon.
You may enjoy this video.
Earth and Moon's gravity well
MooMooMath and Science uploads a new Math and Science video everyday | 103 | 412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-30 | latest | en | 0.918102 |
https://www.jiskha.com/archives/2015/12/08 | 1,597,110,279,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738723.55/warc/CC-MAIN-20200810235513-20200811025513-00098.warc.gz | 697,525,043 | 29,778 | # Questions Asked onDecember 8, 2015
1. ## Algebra
The perimeter of the rectangle below is 74 units. Find the length of side AB . Write your answer without variables. DC (top) 4z-1 CB (right side) 3z+3 AB (bottom) unknown
2. ## Math: Geometry
isosceles trapezoid ABCD is shown below with a line ef drawn through its center. Of the isosceles trapezoid is dilated using a scale factor of 1/2 and a line is drawn through the center of the new dilated figure, what relationship will that line have with
3. ## math
The following table shows the expressions that represent the sales of 4 companies: Company A 110(0.70)x B 140(1.07)x C 430(1.90)x D 310(1.14)x Rank the company sales by smallest to largest percent of increase and select the correct answer below
1. What could happen if you use too much clip art and animation in your presentation? A. You have to pay for each piece of clip art and animation, and your presentation will become very expensive. B. The viewer will become easily distracted by the visuals
asked by Polka Dot
5. ## algebra 2 trig
lighthouse b is 9 miles west of lighthouse A. a boat leaves A and sails 5 miles. At this time, it is sighted from B. If the bearing of the boat from B is N64 E, how far from B is the boat?
6. ## Physics Energy
A baseball player puts a baseball on a batting tee. The baseball bat is able to deliver 4.0x10^5 W of power and is in contact with the ball for 0.70 ms and a distance of 1.4 cm. The mass of the ball is 145 g. What is the next force applied to the ball, and
7. ## math
Half Robert’s piece of wire is equal to 2/3 of Maria’s wire. The total length of their wires is 10 feet. How much longer is Robert’s wire than Maria’s?
8. ## math
Find the average rate of change from x = 10 to x =17 for the function f(x) = 0.01(2) ^x and select the correct answer below A. 10.24 B. 185.78 C. 1300.48 D. 1310.72 Is it A?
9. ## english grammar
I am very sorry. What does the adverb very modify
10. ## history
Which answer offers the best explanation for why Tecumseh fought alongside the British in the War of 1812? Tecumseh was well paid by British generals. Tecumseh wanted to preserve trade with Britain. Tecumseh considered the United States to be an enemy.
11. ## Geometry
The lengths of the sides of a triangle are 3x, 5x – 12, and x + 20. Find the value of x so that the triangle is isosceles. Can somebody show me a step-by-step process of how to get the answer(s) please? I need it:) I also saw that x could = 8, 10, and 6.
12. ## Career
1. A job objective should: (1 point) A.)Persuade the reader to hire you B.)State the position you are applying for C.)List your work history D.)Give reference information 2.)The second paragraph of a cover letter is your chance to: (1 point) A.)state why
asked by YRN DJ
13. ## Mathxxx!!!
1. How many solutions does the equation have? 4x+3=2(2x+9) a.one solution b.no solution c. infinite number of solutions d. impossible to determine 2. How many solutions does the equation have? 4x+19=-9-6x a.one solution b.no solution c. infinite number of
14. ## college algebra help******
Sequences 1. Constructing a Brick Staircase A brick staircase has a total of 30 steps. The bottom step requires 100 bricks. Each successive step requires two less bricks than the prior step. a. How many bricks are required for the top step? b. How many
15. ## Educatinal Tech
Which of the following is true about hidden slides? A: They can only be seen in Slide Sorter View. B: They can be seen in most views and during the slide show presentation. C: The slide is permanently hidden. D: The slide cannot be seen during the slide
16. ## History
1. Which of the following did George Washington do during his presidency? -established a cabinet of advisers** -limited himself to one term in office -offered military support in foreign wars -served ties with Great Britain 2. What precedent did George
asked by YRN DJ
17. ## Math
The probability that it rains tomorrow is 0.14. What is the probability that it doesn't rain tomorrow?
18. ## Math
I'm in sixth grade and I need help on 3(12w+8)+17+9w
19. ## Geometry
Two straight lengths of wire are placed on the ground, forming vertical angles. If the measure of one angles formed is 72 degrees, what are the measures of the other three angles. Explain your answer. My answer-- Vertical angles are congruent and if the
20. ## contemporary mathematics
A passenger train travels 392 miles in the same time that it takes freight train to travel 322 miles. If the passenger train travel 20 miles per hour faster than the freight train.Find the rate of each train.
21. ## Chemistry
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 20.4 g of carbon were burned in the presence of 71.1 g of oxygen, 16.7 g of oxygen remained unreacted. What mass of carbon dioxide was produced? Express your answer to three
22. ## Calculus
Find the sum of the squared errors for the linear model f(x) and the quadratic model g(x) using the given points. f(x) = 2.0x − 3, g(x) = 0.09x^2 + 1.6x − 3 (−1, −4), (1, −2), (2, 0), (4, 6), (6, 9) linear model-? quadratic model- ?
asked by Calculus Help
23. ## a.u.d senior high school mathematics
A girl is 160cm tall stands from the foot of a building. She finds that the angle of elevation of the top of the building is 27 degree. Calculate the height of the building to the nearest 1/2 meter. Give the solution
24. ## Spanish
Which of the following is the correct plural form of "carne sabrosa"? a. carne sabrosas b. carnes sabroses c. carne sabrosos d. carnes sabrosas*** Answer this Question
25. ## English
1. The tree with deep roots is not shaken by wind. 2. The tree with deep roots is not shaken by the wind. ============= Which one is right? Do we have to put 'the' before 'wind'?
26. ## Spanish
Which Spanish verb would you use to ask the following question? "What is your name?" a. estar b. ser c. hacer d. tener
27. ## Early childhood literacy
Noted early childhood experts worry that preschool reading-skill instruction activities may: A. reduce play time. B. intimidate parents. C. accelerate mental growth. D. lead kindergartners My answer A please help
how to calculate the cost of an item if the retail is $40 and the markup is 20% asked by Anonymous 29. ## Chemistry Which of the following exhibits the highest boiling point? Ne, Ar, Kr, or Xe My answer: Xe Which of the following exhibits the highest boiling point? F2, Cl2, Br2, or I2 My answer: HI Which of the following exhibits the highest boiling point? NaCl, SO2, or asked by Kyle 30. ## Chemistry The number of protons in a cation must always be ________ the number of electrons. asked by Dominick 31. ## math what does x equal 0.20x - 0.08(x-10)= 24.80 asked by Anonymous 32. ## Science 1. We dissolve 0.005 mole of strontium chromate in 1 L of solution at 25°C. Will a precipitate be formed? asked by Rachel 33. ## Trig (tan/cot)- (sec/ cos) Also I need help with tan sin +cos = sec asked by N 34. ## Dawes What area of 5ft by 7ft asked by Que 35. ## algebra In which distributions does the variable, X, have a binomial distribution? (Choose all that are correct) ~When a 6-sided number cube is rolled 3 times, X is the number of 2s rolled. ~When a 6-sided number cube is rolled, X is the number of times rolled asked by hadiya 36. ## Chemistry Draw the best Lewis structures for NO+, NO2-, NO2H, and NO3-. Based on the Lewis structures, list all of the the N-O bonds lengths in order of shortest (1) to longest (5). N-O bond length in >>> 2 N-O double bond in NO2H >>> 1 N-O bond length in NO3- >>> 4 asked by Dan 37. ## Child Development and Human Behavior . Which of the following factors helps a child develop a positive self-concept? A. Material wealth B. The sex of the child C. Family relationships D. Conflicts with teachers asked by j 38. ## english how meny sea creaturs are dead asked by omar 39. ## Spanish Which of the following is the correct plural form of "carne sabrosa"? a. carne sabrosas b. carnes sabroses c. carne sabrosos d. carnes sabrosas*** asked by Kaai97 40. ## health What are the ten building blocks for total health asked by kevin 41. ## English 1. He touched me on the shoulder. 2. He hit me on the head. 3. He beat me on the back. 4. He looked me in the face. 5. He looked me in the eyes. 6. He looked me in the mouth. 7. He took me by the hand. 8. He caught me by the leg. 9. He grasped me by the asked by rfvv 42. ## Physics A watertight car is submerged under water. What force is required to open a door of area 1.12 m2 if the average depth of water at the door is 1.50 m? asked by Dave 43. ## Physics A zebra sees a lion 300 m to his right. The zebra starts running at 25 KpH straight ahead to get away. At this point the lion launches himself on an intercept course at the zebra at a speed of 31 KpH. a. At what angle does the lion launch himself? b. If asked by Larry 44. ## Math Which number is equivalent to 14/4? A 3.2 , because 14 divided by 4 is 3 with a remainder of 2 B 3.5 because 14 divided by 4 is 3 with a remainder of 2 and 2/4 = 0.5 C 3.3 because 14 divided by 4 is 3 with a remainder of 3 D 3.6 because 14 divided by 4 is asked by Iris 45. ## statistics help Tests on a Chevy Cobalt incdicate a population mean of 32 mpg and a population standard deviation of 3.5 mpg. You take an SRS (sample random sample) of 10 cars to get a sample mean. What is the probability that your sample mean is greater than 34 mpg? I am asked by S 46. ## Physics The mass of the heavier block is 22 kg and the mass of the lighter blocks is 11 kg and the magnitude of the force of the connecting string on the smaller block is 13 N. Assume: g = 9.8 m/s 2 and the horizontal surface on which the objects slide is asked by Blake 47. ## chemistry physics Which following assertions are true ? - unanswered 13C NMR is more sensitive than 1H NMR because of the 1.1% natural abundance of 13C nuclei. 13C nuclei have a ½ spin and can be detected in NMR, opposite to more abundant 12C nuclei. 13C NMR requires a asked by Abd 48. ## Biochem Glycerol 3-phosphate dehydrogenase catalyzes the reversible reaction: G3P + NAD+ --> NADH + H+ + DHAP If in the cell, the concentrations of G3P, NAD+, NADH, and DHAP are 2 mM, 4 mM, 6 mM, and 8 mM respectively. Determine if the reaction spontaneous as asked by Mia S 49. ## Chemistry Sodium chloride dissolves in water (solubility in water is 361g/1L), but copper (II) phosphate does not dissolve in water (insoluble). A. Describe in a stepwise fashion how the solid sodium chloride dissolves in the water to form a solution. (name the asked by Anonymous 50. ## applied chemistry Draw electron dot diagram for the following compound HCL,CO2,NH3,NH4CL,CH4 And H2O2 asked by francis 51. ## Spanish What is one difference between popular beverages in Los Estados Unidos and those in the Spanish-speaking world? asked by Kaai97 52. ## Math Write a rule for the function represented by the table. x y 0 14 1 29 2 44 3 59 asked by Savannah 53. ## Chemistry. Plz Help! When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 20.4 g of carbon were burned in the presence of 71.1 g of oxygen, 16.7 g of oxygen remained unreacted. What mass of carbon dioxide was produced? Express your answer to three asked by Dominick 54. ## Math Find the value of the sine of the smallest angle in standard position with the point (-12, 5) on its terminal side. Give as four decimals. a. 0.3846 b. -0.4167 c. 0.4167 d. 2.6000 asked by Debra 55. ## Science Which electrons in an atom take the most energy, and what are they called? What Kind of diagram is used to show how many of these electrons an atom has? What name is given to the elements with eight outer electrons? Which element belongs to this group, asked by John Doswell 56. ## Algebra A pudding recipe requires 2/3 of a cup of milk for every 1-3/4 cups of sugar. What is the unit rate of sugar to milk in the pudding recipe? asked by Jim 57. ## Chemistry Help Which of the following exhibits the highest boiling point? HF, HCl, HBr, or HI Which of the following exhibits the highest boiling point? CH3CH3, CH3COOH, or CH3OCH3 How can you determine the highest boiling point of these? asked by Kyle 58. ## Math Theatre sold 343 tickets on Friday. This is 192 fewer tickets than those sold on Saturday. How many tickets were sold alltogether asked by Anonymous 59. ## Language Arts Write a complete sentence using the past participle of the verb in ( ) with has or have. 1. Mr. Hancock (run) the museum for five years. Answer: Mr. Hancock has ran the museum for five years. 2. He (chose) May as membership drive month. Answer: He has asked by Patrick 60. ## Frontier Middle School Math If a person walks 2 miles in 35 minutes, how many miles does it walk in 10 minutes ? asked by Oscar 61. ## Algebra A regular Pentagon measures 5 1/8cm on one side, what is the perimeter of the Pentagon asked by Dee 62. ## MS COLLEGE ¡Ìx+y=11 x+¡Ìy=7 asked by nandan 63. ## precalc Find the standard form equation for each ellipse described. 1) Major vertices at (0, 3) and (0, -3), minor vertices at (2, 0) and (-2, 0) 2) Major vertices at (7, 0) and (-7, 0), foci at (5, 0) and (-5, 0) 3) Minor vertices at (-2, -3) and (-2, -11), foci asked by jean 64. ## Physics A man doing a slow push-up is approximately in static equilibrium. His body is horizontal, with his weight of 750 N supported by his hands and feet, which are 1.36 m apart. One hand rests on a spring scale, which reads 268 N. If each hand bears an equal asked by Dave 65. ## Report Citation Style I was wondering if someone can identify this citation style: Author. (Date). Doc Title. Title Bar Title. Retrieved ____, 2015 from . asked by R 66. ## History 5.)Which of the following is the best explanation of the XYZ affair that took place during John Adam's presidency? A.)Adams tried to purchase three naval ships (X,Y, & Z) from France demanded much more money than the ships were worth & adams rejected the asked by YRN DJ 67. ## math W decreased by 7 asked by wanda 68. ## History 7.)Why was the repeal of the judiciary the act of 1801 significant A.)It made crimes related to slavery enforceable by law B.)In the early 1800's what the usual reason for working class children to be separated from their family? C.)It was the first asked by YRN DJ 69. ## math Find the radius of the circle if an arc of length 8 ft on the circle subtends a central angle of 225° asked by grace 70. ## Algebra-Interval Notation I don't remember how to solve this. The question is: For what values of a is the expression 7(x-2)5 negative? Write answer using interval notation. asked by Samantha E 71. ## Chemistry. Half Life Problem. Help Plz If the half-life of iodine-131 is 8 day, how much would remain undecayed after 8 day if we started with 38 grams? How much at 16 days? 24 days? 32 days? asked by Dominick 72. ## Grover Cleveland high school Ramy jumped from the window from the height of 400 feet. His speed of feeling was 10 feet. second write a functions to represent the situation.explain the answer please asked by Ahmed 73. ## Math what is the amount of orange juice, written as a fraction grater than 1, that is used in the recipe? 1/2 cup frozen strawberries 1/2 cup frozen peaches 1/4 cup vanilla yogurt 1 1/2 cups orange juice asked by Emily 74. ## Trig (Tan/cot)- (sec/cos) asked by N 75. ## Algebra-factons Nathan used 2/3 cup of olive oil in a batch of hummus he made. He then ate 5/7 of the batch of hummus. What fraction of a cup of olive oil did Nathan consume? asked by Samantha E 76. ## Math If the number in the numerator of a unit rate is 1 what does this indicate about the equivalent unit rates give an example asked by Amber 77. ## Algebra-Value Expressions If d-3f+g=17, determine the value of the expression 2(d+2)-(4f-3)+(f+g-d). asked by Samantha E 78. ## physics A box slides along a friction-less horizantal floor under the action of a constant applied force is the acceleration necessarily to be constant? asked by Marym 79. ## Physics A baseball player hits a home run over the left-field fence, which is 104 m from home plate. The ball is hit at a point 1.08 m directly above home plate, with an initial velocity directed 35.0° above the horizontal. By what distance does the baseball asked by Bill 80. ## Trig Tan^2= csc^2 tan^2 -1 asked by N 81. ## Physics A zebra sees a lion 300 m to his right. The zebra starts running at 25 KpH straight ahead to get away. At this point the lion launches himself on an intercept course at the zebra at a speed of 31 KpH. a. At what angle does the lion launch himself? b. If asked by Nicole 82. ## science What is in orange, but not in banana, cherry has it, but not in apple? asked by vivek 83. ## Social Studies what are two actions outlawed by the English Bill of rights. asked by Anonymous 84. ## math Liam is building a fence around his entire backyard. The backyard is 24 feet wide and 60 feet long. If Liam uses sections of fencing that are 12 feet long, how many sections does he use? asked by bamba 85. ## Geometry In an isosceles triangle, line segment AC is congruent to line segment AB. Find the value of y in terms of x if one of the remote interior angle is y and the exterior angle is 3x+20. I think the answer is y=6x-140 but I don't know how they got it. asked by Ashley 86. ## math y=mx+b If b is did not cross the y intercept, for example, what would "b" be!?!? Zero? asked by amam 87. ## Calculus A spotlight on the ground is shining on a wall 24\text{m} away. If a woman 2\text{m} tall walks from the spotlight toward the building at a speed of 0.6\text{m/s}, how fast is the length of her shadow on the building decreasing when she is 2\text{m} from asked by Spencer 88. ## Trig (Cos/1-sin)- tan asked by N 89. ## Mulitvariable Calculus How to proof tangential and normal components of a vector Thank you asked by cal 90. ## Math Help! A sweatshirt comes in 3 sizes and 8 colors. How many different sweatshirts are available? A. 3 B. 8 C. 11 D. 24 I'm not sure.. asked by Amber 91. ## Trigonometry For which angles in degrees if any, is sin A not equal to 1/csc A? Answer for 0 through 359 degrees. asked by April 92. ## Social Studies Which action occurred during the Battle of Chickamauga made it a Confederate victory? A.The Confederates had fewer casualties B.Union troops were forced out of Georgia C.The Union general surrendered D.None of the above Not sure? Please help asked by Amber 93. ## Dynamics A ball weighing .2 Lbs. is dropped out the window of a pickup truck (approx. 5 ft?) moving horizontally at 45mph. If e (coefficient of restitution) between the ball and the ground is equal to .6, find the height of the ball's first bounce and the asked by Joe 94. ## science if a geologist found two fossils of different species buried next to each other in a rock layer, he or she might hypothezise that the two organisms may had been alive during the same period. true or false i think false but not sure asked by girlmeetsworld 95. ## History How did the presidential plan for reconstruction differ form congresses plan? A congress believed it and not the president should have power over reconstruction. B the president believed that congress will be too soft on southern rebels C congresses plan asked by Anonymous 96. ## History How did the presidential plan for reconstruction differ form congresses plan? A congress believed it and not the president should have power over reconstruction. B the president believed that congress will be too soft on southern rebels C congresses plan asked by Jaide 97. ## help me math if i graph (8,2) to (5,4), how would i write it in mx+b=y form? asked by amam 98. ## physics A 150g iron ball is dropped into a cavity in a block of ice. The cavity is then found to contain 21g of water. Calculate the heat of fusion of ice. asked by omay 99. ## Spanish Which Spanish verb would you use to ask the following question? "What is your name?" a. estar b. ser c. hacer d. tener asked by Kaai97 100. ## Algebra-Perimeter The side of a square is 4cm longer than the side of an equilateral triangle. The perimeter of the square is 28cm longer than the perimeter of the triangle. Find the length of the sides of the triangle and the sides of the square. asked by Samantha E 101. ## History What were the 4 goals of the Manhattan Project? asked by Dominick 102. ## chem what are ways to increase the solubility of SnF2 at constant temperature and pressure. I think adding a strong acid will work not sure what else. asked by jeff 103. ## Chemistry Chemical formula: NaCl Name: Sodium chloride Bond Type (intermolecular FOA): London, dipole-dipole, ion-dipole VSEPR molecular shape: none Central Atom Hybridization: sp3 Molecular Polarity: polar Interparticle force of attraction (IPFOA): ? State of asked by Anonymous 104. ## help 1. The first step in getting help for drug abuse is (1 point) • detoxification. • withdrawal. • joining a therapeutic community. • recognizing the problem. asked by Polka Dot 105. ## Algebra how do you calculate probabily or this problem. A bag has 8 red, 6blue, 2 yellow and 7 green marbles. What is the probability of selecting a red, not replacing it and selecting a green? 6/21 * 7/21 =42/21 asked by veronica 106. ## Math Vector CD has an initial point of (5, -6). Its x component is -12, and the y component is 4. What are the coordinates of the terminal point? My answer is (-7, -2). asked by Anonymous 107. ## Calculus Find the sum of the squared errors for the linear model f(x) and the quadratic model g(x) using the given points. f(x) = −0.7x+3, g(x)=0.06x^2−0.6x+2 (−3, 5), (−1, 3), (1, 2), (3, 1) linear model-? quadratic model -? asked by Calculus Help 108. ## Math Hii frnds this is ayyappa..I am troubled for the below problem I.e.5 - 4 = 23 8 - 1 = 63. 6 -16 = 32 9 - 9 = 78 3 - 9 = 6 3 - 81 = ? .how the above logic is implemented.plese explaine the implementation of a logic... asked by Ayyappa 109. ## Algebra A rectangular yard measuring 25ft by 35ft is bordered (and surrounded) by a fence. Inside, a walk that is 2ft wide goes all the way along the fence. Find the area of this walk. Be sure to include the correct unit in your answer. asked by Maria 110. ## Mixed Numbers Need Help Now I need at least 5 mixed numbers that equal up to 10 and are all different. They must use only 1/4 and 1/3. asked by Aelia 111. ## Spanish Which herbal remedy would most likely be used to treat a sunburn? a. te de manzanilla b. una de gato c. aloe vera d. sangre de drago*** asked by Kaai97 112. ## Spanish What is one difference between popular beverages in Los Estados Unidos and those in the Spanish-speaking world? asked by Kaai97 113. ## English 1. People were excited by UFO appearing in the sky. 2. People were excited by UFO which appeared in the sky. ======== Does #1 mean #2? asked by rfvv 114. ## English 1. Boys often have trouble focusing in class. 2. Boys often have trouble focusing on class. 3. Boys often have trouble focusing on teaching in class. ========================== Q1: Which one is grammatical? Q2: Does #1 mean #3? asked by rfvv 115. ## Calculus A parabolic arch bridge has a 60ft base and a height of 24ft. Find the height of the arch at a distance of 5, 10, and 20 ft from the center asked by Net 116. ## English I don't feel well. I think I have a fever. - Oh, no. Hope you get better soon. ============== In this dialogue, the subject "I" is missing before "Hope." Is this pattern common in everyday life? asked by rfvv 117. ## pre-algebra A train leaves New York at 1:00 PM. A second train leaves the same city in the same direction at 3:00 PM. The second train travels 28 mph faster than the first. If the second train overtakes the first at 6:00 PM, what is the speed of each of the two asked by Tycara 118. ## Math Find vector a + vector b. a = -5p - 3k b = 6p + 8k My answer: 1p + 5k. asked by Anonymous 119. ## Math Ben wants to buy a new car, and she has narrowed her choices to two models. Model A sells for$12,500, gets 25mi/gal, and costs $300 a year for insurance Model B sells for$16,100, gets 36 mi/gal, and costs $400 a year for insurance. Jenny drives about asked by Anonymous 120. ## Statistics Chi-square tests are nonparametric tests that examine nominal categories as opposed to numerical values. Consider a situation in which you may want to transform numerical scores into categories. Provide a specific example of a situation in which categories asked by Slomo 121. ## Physics A certain concrete will support a maximum compressive stress of 1.30 107 N/m2 before being crushed. A column of uniform width is to be made with this concrete. How tall can the column be if the stress in it at any point is not to exceed one half the asked by Dave 122. ## Physics Find the volume occupied by 3.14 moles of an ideal gas at standard conditions of pressure and temperature (P = 1.00 atm, T = 273 K). asked by Dave 123. ## Chemistry Hey can you please check my answers? Also, Can you explain interparticle force of attraction? I'm not sure about the difference between IMFOA and IPFOA. Chemical formula: Cl2 Name: Chlorine Bond Type (intermolecular FOA): London VSEPR molecular shape: asked by Anonymous 124. ## English 1. I don't play computer games much. 2. I don't play computer games a lot. 3. I don't play computer games more than my brother. 4. I don't play computer games as much as my brother. 5. I don't play computer games less than my brother. 6. I play computer asked by rfvv 125. ## Physics By how much does the air pressure in a house increase if the house is sealed and the air temperature increases from 10.7° C to 21.0° C? The initial air pressure is 1.00 atmosphere. asked by Dave 126. ## Physics James Bond and the ever present bond girl are on a dirt bike escaping from a car load of bad guys. If we have the following conditions: James Bond’s mass is 84 Kg The Bond Girl is 48Kg The Dirt Bike’s mass is 113.6 Kg The tires on the dirt bike are asked by Larry 127. ## Physics A 1.05 liter flask contains a certain quantity of ideal gas at 305 K. Then an equal number of molecules of the same gas is added to the flask, after which the absolute pressure is 1.55 times its original value. What is the final temperature? asked by Dave 128. ## Physics A brick is initially 12.90 cm high and an aluminum can is initially 12.91 cm high. By how much must their temperatures be raised in order for the brick and the can to have exactly the same height? asked by Dave 129. ## English What do these mean Arts are an adornment of life but not central to it The absentminded professor the ivory tower reflect antiintellectualism asked by Anonymous 130. ## Spanish Which empire was the largest in both Americas before the arrival of the Europeans? a. Inca b. Aztec c. Maya d. Cherokee asked by Kaai97 131. ## Math 90 members of a sport club plays at least one of the game tennis,football and volleyball,10 plays tennis and football,19 play football and volleyball and 29 plays tennis and volleyball.x people plays all the 3 games,while x people each play only one game asked by Andrew 132. ## health What is soda or soda pop? Is coke soda? What other drinks is soda? Could you mame them? asked by rfvv 133. ## iq "8th december 2015" in 5 letters without using numbers ? asked by ram 134. ## Keithville Tori had 5/7 fossils as Tommy. Tommy gave 6 fossils to Tori so that they would each have the same amount .How many fossils did the two friends have altogether? asked by Faith 135. ## LAB HELP ASAP WHAT ARE POSSIBLE SOURCES OF ERRORS IN A COMBUSTING BIODIESEL LAB WHERE WE USED CANOLA OIL TO CREATE A SAMPLE OF BIODIESEL. asked by XOXOX 136. ## English 1. Why don't we go to a movie? (In this sentence, does 'movie' mean 'movie theater' or 'motion picture'? 2. Why don't we go to the movies? (In this sentence, does 'movies' mean 'movie theaters' or 'motion pictures'?) asked by rfvv 137. ## math Break apart 56 to solve 56 divided by 7 asked by Anonymous 138. ## Chemistry Give two physical properties of ionic compounds? Thanks. asked by Isha 139. ## Finance Management 3. A corporation’s capital structure is composed of common stock and preferred stock. If the percentage of common stock is three times the percentage of preferred stock in the capital structure, the percentage of common stock minus the percentage of asked by Dorothea 140. ## physics A) Assume the mass of a pendulum is 3kg. Calculate the work done by the pendulum on the mass from 0 degrees to 180 degrees ( you will need to integrate with respects to d(theta ) instead of dx given: the mass at the end of the string experiences a asked by Spice 141. ## Hunt Wyatt and Allyson were asked to solve a percent problem using the number 13 and 38. Wyatt found 13% of 38 and allyson found 38% of 13. Explain why they both got the same answer. Would this work for other numbers as well? Why or why not? asked by Karen 142. ## math 7/12 liter in 2/10 kilometer asked by Yames 143. ## 11U Functions Identity -sin x -cos x cot x = -csc x asked by A 144. ## math An equation is shown below: 3 over 2 multiplied by x plus 7 over 2 equals 2 to the power of x What is the solution to the equation A.X=1 B.X=3 C.X=5 D.X=8 I think it could be d asked by Isabella 145. ## english what is the purpose of the poem the pulley? asked by laketra 146. ## Math A clothing store sold 5 T-shirts for$11 each and 5 sweatshirts for $19 each. The expression below represents the total amount, in dollars, the store received. 5 (11 + 19) Which expression is equal to the total amount, in dollars, the store received? A 5 + asked by Raiven 147. ## Chemistry Rank the following in order of decreasing vapor pressure, and briefly explain the observed trend in vapor pressure: Acetic Acid Isobutane Acetamide Propanal asked by Sarah 148. ## math write linear equation for phone company who base rate is$40 and every minute costs $0.05. Average minutes of 200 minutes per month asked by evelyn 149. ## math In a test which had 30 questions carrying 1 mark each ajay scored 60persant marks. How many questions did he answered correctly asked by neha 150. ## Chemistry A sample of sodium reacts completely with 142 g of chlorine, forming 234 g of sodium chloride. What mass of sodium reacted? asked by Dominick 151. ## jay kay public Consider a planet in space whose mass is twice that of earth and radius is thrice that of earth what will be the weight of a body on this planet if its weight on the earth is 900N ?? asked by adi 152. ## 5th grade made Charlie has$50 saved. He spent 3/10 of his money on a new book. How much money did he spend on the book. please help
asked by Nikki B.
153. ## Chemistry
How much total work does a golf caddy do if he lifts a 300 N golf bag he is carrying 100 times a during a round, to a height of 1.0 m each time he lifts it?
154. ## Math
The ratio of ages of P andQ in 1996 was2:3 and in 2001was7:10 what will be the ratio in 2011
155. ## Math
what is 3+7-1 ?
156. ## science
a hollow conducting shere of radius R has a charge Q placed on its surface. what is the electric field and potential inside the sphere?
157. ## Algebra 1
Adam wants to create a square garden with a walkway on all four sides. The width of the walkway is 4 feet. Write a function to represent the area of the garden in terms of the length of one side of the walkway, x. ???
158. ## Research
Identify if the variable is dependent or independent. Fil the blanks ___Length of time to finish colege education and ___ number of subjects failed ___salary and ___ degree of commitment to the company ___ job statisfaction and ___ years of service in the
1. Given:
160. ## Quick English check
Write in the best response for each statement below using the words: Analytical, cognitive, deem, empirical, induced, inferred, obtain, paradigm, speculate, stratagem. 1. The neurosurgeon studies the effects of trauma on the COGNITIVE processes of the
161. ## Physics
A 45.0 kg diver steps off a 8.0 m high diving board and drops straight down into the water. If the diver comes to rest 5.0 m below the surface of the water, determine the average resistance force exerted on the diver by the water.
162. ## Physics
A ceiling fan of radius 1.0 m is rotating with a frequency of 300 RPM. How fast (in m/sec) is the tip of one of the blades moving?
163. ## Calculus AB
If f(x) is continuous on the interval [4, 7], how many of the following statements are true? A. f(x) has a maximum value on [4, 7]. B. f(x) has a minimum value on [4, 7]. C. f(7) > f(4) D. lim f(x) = f(6) ᵡ¨6
164. ## Social Studies
Hi! Can you please give me some advanced suggestions to write a summary on the DISCOVERIES OF THE SCIENTIFIC REVOLUTION!! Thank You :)
165. ## math
1. a square has a side of 2 decimeters. a new square is formed by joining the midpoints of the sides. If this procedure is reoeated 6 times, what is the perimetet and area of the seventh square 2. find the sum of those numbers between 15 and 810 which have
166. ## College Physics
How much energy is required to bring 10.0 grams of ice at -5.00 degrees to 105-degree steam? Assume the system is isolated so that no mass is lost to the surroundings
167. ## Math
Leonard ran 4 miles more than Sheldon ran. The sum of their distances is 26 miles. How far did Sheldon run? The domain a. Define your variable. b. Write algebraic expressions for the distances each person ran. c. Write an equation that can help you solve
168. ## Chemistry
How do you solve for enthalpy of vaporization? I know it's liquid to gas, but I can't figure out how to solve an actual problem like C2H4O2. Could you please explain?
169. ## Life science,tourism,pure maths and agricultural science
Which collage can i study?when i doing life science,tourism,pure maths and agriculte
170. ## Calculus
I am standing on the roof of a building with a pea shooter, and I shoot a pea up into the air from a height of 80 feet. The pea reaches its maximum height 2 seconds after I shoot, then falls to the ground. Give the position of the pea as a function of time
171. ## Math
Uche spends half of his school day in literacy class. He has 3 literacy classes. what fraction of the day does Uche spend in each of his 3 literacy classes. can someone help me
172. ## Sociology
What is the effect of industrialization on population growth rates? A. It decreases growth rates B. It stabilizes growth rates C. It increases growth rates D. There is no correlation between industrialization and growth rates My text says that
173. ## Chemistry
Is it true that the higher the boing point, the higher the vapor pressure?
174. ## Chemistry
Can you please check if these are right? I'm not sure about IPFOA. Chemical formula: H2O Name: Dihydrogen oxide Bond Type (intermolecular FOA): London, dipole-dipole, hydrogen bonding VSEPR molecular shape: bent Central Atom Hybridization: sp3 Molecular
175. ## English
Hello. I would be grateful for some helpl. Do you think it is possible to say: "It will cause a serious crisis like what is happening in Syria." ? Maybe, "similar/identical to what is happening"? Thank you for advice.
176. ## Chemistry
Fe2O3 + Fe + H2SO4 → FeSO4 + H2O How would I balance this using oxidation numbers? Is the Fe2 being reduced and the Fe being oxidized?
177. ## Math grade 5-6
Alice works part-time after school. On Monday, Alice records that she worked three and two-sixth hours. On Tuesday, Alice records that she worked one and seventy-five hundredths hours. On Wednesday, Alice records that she worked two and two-thirds hours.
178. ## early childhood literacy
when a question has so many answers is called 1.evaluation 2. recall 3.explanation 4. open ended my answer 3. help please
187. ## Math
A linear function f(x) and its inverse f-1(x) intersect at the point (5,5). Create a function, f(x) that would satisfy this requirement
asked by Sarah H
188. ## Math
2% of what number is 400?
189. ## algebra 2
7Y+6/20X^+20=?/60X^2+60
190. ## Chemistry. Plz Help!
A sample of sodium reacts completely with 142 g of chlorine, forming 234 g of sodium chloride. What mass of sodium reacted? Express your answer to three significant figures and include the appropriate units.
191. ## Chemistry
Chemical formula: Na Name: Sodium Bond Type (intermolecular FOA): London VSEPR molecular shape: none Central Atom Hybridization: sp Molecular Polarity: nonpolar Interparticle force of attraction (IPFOA): ? State of Matter (at room temperature): solid Can't
192. ## Algebraic fractions
Algebraic fractions: Add the following and express your answer as a single fraction. 9/y^2÷2y/10
asked by Samantha E
193. ## Math
Theo has a wooden board 654 millimeters long. How long is the board in centimeters?
194. ## Math
Megan completes most puzzles in 24 minutes. Hassan completes most puzzles in 16 minutes. If Megan and Hassan complete a puzzle together, how long will it take?
Sharecropping and tenant farming developed as a result of ________. A. - Laws in Georgia that required freedmen to work for former masters B. - Former slaves needing homes and jobs, and landowners needing farmers C. - Segregation laws D. - Increased
196. ## Trig
(Cos/1-sin)- tan
197. ## Chemistry
Explain trends in the solubility of water (amount of the compound (solute) that will dissolve in a given amount of solvent (water).
198. ## Math
Trey can drive his car 130 miles on 4 gallons of gas. At this rate, how far can Trey drive on 6 gallons of gas?
199. ## Algebra I Math
Find the average rate of change from x = 10 to x =17 for the function f(x) = 0.01(2) ^x and select the correct answer below A. 10.24 B. 185.78 C. 1300.48 D. 1310.72 Is it C?
200. ## Science Help
which element belongs to this group, although it has only two electrons?
asked by John Doswell
201. ## english
I need help this question Owens was a sickly child, but poor health during his early years proved not to be a setback. Setback most nearly means 1.illness 2.medicine 3.impediment 4.sport I think 3.impediment Thank you
202. ## math
A quilt is made of 24 equal squares. It measures 3'/ 4.5' rectangle. What is the dimension of 1 square?
203. ## Math
A linear function f(x) and its inverse f-1(x) intersect at the point (5,5). Create a function, f(x) that would satisfy this requirement
asked by Sarah H
204. ## College Algebra
solve. Round your answer to 3 decimal places. show work ln(2-x)=-2
205. ## Chemistry. I got the wrong answer. Plz help
A sample of sodium reacts completely with 142 g of chlorine, forming 234 g of sodium chloride. What mass of sodium reacted? I took 142g of Cl / 35.5g/mol = 4 mols of Cl Then I took 58.44g/mol x 4 mols of Cl = 233.76 It asked for three significant figures
206. ## Biology
What is chromatin and what does it turn into and how does it arrange itself? Chromatin forms chromosomes during cell division and consists of DNA RNA and proteins. It turns into separate structurs called chromosomes. Idk how it arranges itself
207. ## pigskin
what two competing team would deserve the title kudzu klash
208. ## Math
Write a word problem using 7 groups. Solve the problem
209. ## Math
Ben wants to buy a new car, and she has narrowed her choices to two models. Model A sells for $12,500, gets 25mi/gal, and costs$300 a year for insurance Model B sells for $16,100, gets 36 mi/gal, and costs$400 a year for insurance. Jenny drives about
210. ## math/science
Atoms tend to gain, share, lose valence electrons until each atom has the same number of valence electrons as the ____________. my answer: nucleus
asked by John Doswell
211. ## Math
Can anyone help me with these problems? 1. A square has a side of 2 decimeters. A new square is formed by joining the midpoints of the sides. If this procedure is repeated 6 times, what is the perimeter and area of the seventh square ? 2. Find the sum of | 10,554 | 39,814 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-34 | latest | en | 0.925017 |
https://www.wolfram.com/language/11/image-and-signal-processing/waves-in-the-great-salt-lake.html?product=mathematica | 1,686,334,327,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656788.77/warc/CC-MAIN-20230609164851-20230609194851-00205.warc.gz | 1,226,584,535 | 10,935 | # Wolfram Mathematica
## Waves in the Great Salt Lake
Using ImageMesh, you can convert image segments into BoundaryMeshRegion objects. These mesh regions allow you to utilize functions across other areas, such as finite element methods (FEM).
To illustrate this FEM connection, determine the major modes of surface waves on the Great Salt Lake in Utah.
In[1]:=
`img = EntityValue[Entity["Lake", "GreatSaltLake::yw8cf"], "Image"]`
Out[1]=
Regularize the image with a mean-shift filter.
In[2]:=
`img2 = MeanShiftFilter[img, 3, 0.1]`
Out[2]=
Obtain a segmentation via region growing.
In[3]:=
```mask = RegionBinarize[img2, \!\(\* GraphicsBox[ TagBox[RasterBox[CompressedData[" 1:eJzt1jEKwkAQQNFdK0uv4C1sLW0VD6AYxSZCFMRzCJ7XiF3SzFb7lf8ggUCK D8MmM99f1sdJSuk67W/r3X3ZdbvHZtY/bNvr+dQ2h1V7a05Nt9h/Xnv21ytJ kiRJkiRJkn5H7tVuGLMqzqo4q+KsiuNW8bKsimNWfbNwXVbF5Yzuqp0xZFUB dBaui5kF/T5AsxJ1jNAsq0qgs3BlOSPDMrMLmvUBDcNnsbqov2x4FrWrdscA NIu6SjCzoEOEZlGniF5Ta0eMmVXCqjjmCJlV5KzaDWNWxXGreFnMKvAIazeM MUfIXEmtKsCs8hSWsCrOqjhmlSRJkiRJkvS/3tRrD1M= "], {{0, 147}, {150, 0}}, {0, 1}, ColorFunction->GrayLevel], BoxForm`ImageTag["Bit", ColorSpace -> Automatic, Interleaving -> None], Selectable->False], DefaultBaseStyle->"ImageGraphics", ImageSizeRaw->{150, 147}, PlotRange->{{0, 150}, {0, 147}}]\), 1/5]```
Out[3]=
Retrieve a BoundaryMeshRegion object of the lake surface.
In[4]:=
`\[ScriptCapitalR] = ImageMesh[mask]`
Out[4]=
Create a mesh of the lake surface.
In[5]:=
```\[CapitalOmega] = TriangulateMesh[\[ScriptCapitalR], MaxCellMeasure -> 8]```
Out[5]=
Solve the wave equation of the lake surface by first determining the eigenfunctions of the Laplacian inside the lake region.
In[6]:=
```\[ScriptCapitalL] = -\!\( \*SubsuperscriptBox[\(\[Del]\), \({x, y}\), \(2\)]\(\[CurlyPhi][x, y]\)\);```
Using boundary condition.
In[7]:=
`\[ScriptCapitalB] = DirichletCondition[\[CurlyPhi][x, y] == 0, True];`
Generate an orthonormal basis of eigenfunctions Φ with eigenvalues Λ.
In[8]:=
```{\[CapitalLambda], \[CapitalPhi]} = NDEigensystem[{\[ScriptCapitalL], \[ScriptCapitalB]}, \[CurlyPhi][x, y], {x, y} \[Element] \[CapitalOmega], 64];```
Display the first six oscillation modes.
In[9]:=
```GraphicsGrid[ Partition[ Table[ContourPlot[\[CapitalPhi][[ k]], {x, y} \[Element] \[CapitalOmega], PlotRange -> All, PlotLabel -> \[CapitalLambda][[k]], PlotTheme -> "Minimal"], {k, 6}], 3 ], ImageSize -> 512 ]```
Out[9]=
The temporal evolution of the decaying oscillation modes is given by the following.
In[10]:=
```\[CapitalTheta][\[Lambda]_, \[Xi]_, t_] = FullSimplify[ DSolveValue[Join[{ \!\( \*SubscriptBox[\(\[PartialD]\), \(t, t\)]\(\(TraditionalForm\`\[CurlyTheta]\)[ t]\)\) + \[Xi] \!\( \*SubscriptBox[\(\[PartialD]\), \(t\)]\(\(TraditionalForm\`\ \[CurlyTheta]\)[ t]\)\) == -\[Lambda] \!\(TraditionalForm\`\[CurlyTheta]\)[ t] }, {\!\(TraditionalForm\`\[CurlyTheta]\)[0] == 1, \[CurlyTheta]'[0] == 0} ], \!\(TraditionalForm\`\[CurlyTheta]\)[t], t], {\[Lambda] > 0, \[Xi] > 0, \[Xi]^2 < 4 \[Lambda], t \[Element] Reals} ]```
Out[10]=
Expanding an initial perturbation in eigenfunctions and letting these evolve in time, a simulation of wave propagation across the lake is obtained.
show complete Wolfram Language input
In[11]:=
```n = 64; weights = Take[GaussianMatrix[{{n}, n/2}], -n - 1]; weights -= Last[weights]; weights = Most[weights]; weights *= 1/First[weights]; wave[t_] = (\[CapitalTheta][\[CapitalLambda], 0.005, t] weights (\[CapitalPhi] /. {x -> 50, y -> 60} )) . \[CapitalPhi]; waveColors = (Blend[{{-0.01, Purple}, {-0.005, Blue}, {0., Green}, {0.005, Orange}, {0.01, Yellow}}, #] &); anim = Table[ ContourPlot[ wave[t], {x, y} \[Element] \[CapitalOmega], PlotRange -> {-0.01, 0.012}, Contours -> Range[-0.01, 0.012, 0.0005], PlotTheme -> "Minimal", ColorFunctionScaling -> False, ContourStyle -> None, ColorFunction -> waveColors ], {t, 0, 255, 1} ]; ListAnimate[anim]``` | 1,418 | 3,886 | {"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} | 2.703125 | 3 | CC-MAIN-2023-23 | latest | en | 0.505739 |
https://nrich.maths.org/public/topic.php?code=-8&cl=3&cldcmpid=1564 | 1,571,854,656,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987835748.66/warc/CC-MAIN-20191023173708-20191023201208-00439.warc.gz | 600,509,429 | 6,075 | # Search by Topic
#### Resources tagged with Fibonacci sequence similar to Spirals Instead of Sunflowers:
Filter by: Content type:
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### There are 17 results
Broad Topics > Patterns, Sequences and Structure > Fibonacci sequence
### LOGO Challenge - Circles as Bugs
##### Age 11 to 16 Challenge Level:
Here are some circle bugs to try to replicate with some elegant programming, plus some sequences generated elegantly in LOGO.
##### Age 11 to 16 Challenge Level:
Using logo to investigate spirals
### First Forward Into Logo 11: Sequences
##### Age 11 to 18 Challenge Level:
This part introduces the use of Logo for number work. Learn how to use Logo to generate sequences of numbers.
### Whirling Fibonacci Squares
##### Age 11 to 16
Draw whirling squares and see how Fibonacci sequences and golden rectangles are connected.
### Leonardo of Pisa and the Golden Rectangle
##### Age 7 to 16
Leonardo who?! Well, Leonardo is better known as Fibonacci and this article will tell you some of fascinating things about his famous sequence.
### Gnomon Dimensions
##### Age 14 to 16 Challenge Level:
These gnomons appear to have more than a passing connection with the Fibonacci sequence. This problem ask you to investigate some of these connections.
### Continued Fractions I
##### Age 14 to 18
An article introducing continued fractions with some simple puzzles for the reader.
### Got a Strategy for Last Biscuit?
##### Age 11 to 16 Challenge Level:
Can you beat the computer in the challenging strategy game?
### Building Gnomons
##### Age 14 to 16 Challenge Level:
Build gnomons that are related to the Fibonacci sequence and try to explain why this is possible.
### Fibonacci Surprises
##### Age 11 to 14 Challenge Level:
Play around with the Fibonacci sequence and discover some surprising results!
### Fibonacci's Three Wishes 2
##### Age 7 to 14
Second of two articles about Fibonacci, written for students.
### The Golden Ratio, Fibonacci Numbers and Continued Fractions.
##### Age 14 to 16
An iterative method for finding the value of the Golden Ratio with explanations of how this involves the ratios of Fibonacci numbers and continued fractions.
### Ordered Sums
##### Age 14 to 16 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### Stringing it Out
##### Age 14 to 16 Challenge Level:
Explore the transformations and comment on what you find.
### Fibs
##### Age 11 to 14 Challenge Level:
The well known Fibonacci sequence is 1 ,1, 2, 3, 5, 8, 13, 21.... How many Fibonacci sequences can you find containing the number 196 as one of the terms?
### Paving Paths
##### Age 11 to 14 Challenge Level:
How many different ways can I lay 10 paving slabs, each 2 foot by 1 foot, to make a path 2 foot wide and 10 foot long from my back door into my garden, without cutting any of the paving slabs?
### 1 Step 2 Step
##### Age 11 to 14 Challenge Level:
Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps? | 757 | 3,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | longest | en | 0.852475 |
http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=st&chap_sec=10.1&page=case_intro | 1,369,353,971,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704117624/warc/CC-MAIN-20130516113517-00096-ip-10-60-113-184.ec2.internal.warc.gz | 430,830,241 | 3,184 | Ch 10. Virtual Work & Potential Energy Multimedia Engineering Statics VirtualWork Potential Energy
Chapter 1. Basics 2. Vectors 3. Forces 4. Moments 5. Rigid Bodies 6. Structures 7. Centroids/Inertia 8. Internal Loads 9. Friction 10. Work & Energy Appendix Basic Math Units Sections Search eBooks Dynamics Fluids Math Mechanics Statics Thermodynamics Author(s): Kurt Gramoll ©Kurt Gramoll
STATICS - CASE STUDY Introduction Problem Animation Click to view movie (480k) A car jack is constructed with four pinned members and a threaded shaft. The shaft must be designed to withstand the tension that results when hoisting a heavy automobile. What is known: The members of the jack are pinned together at points A, B, C, and D. The length of each member is 0.2 m. The weight of the automobile exerts a force of 1,960 N on the jack. The threaded shaft is pinned to the jack at points C and D. Under normal conditions, the jack is designed to support weight when the bottom members are at an angle between θ = 30° and θ = 60°. Questions Force Diagram At what angle θ is the tension in the shaft greatest under normal conditions (30o ≤ θ ≤ 60o)? At what angle θ is the tension in the shaft least? Approach Determine the virtual work performed by the weight of the automobile and the tension in the shaft. Knowing that the total virtual work must be equal to zero, derive an equation for the tension T as a function of the angle θ. Use graphic analysis to determine the maximum and minimum values of T under normal operating conditions. | 350 | 1,532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.86201 |
https://blr-beautyandhairsalons.com/qa/quick-answer-how-many-gb-is-a-3-minute-song.html | 1,600,650,134,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198868.29/warc/CC-MAIN-20200920223634-20200921013634-00389.warc.gz | 304,843,187 | 9,041 | # Quick Answer: How Many GB Is A 3 Minute Song?
## How many songs will 128gb hold?
it really varies depending on the size of each individual song but it can hold up to anything from 16,000-22,000 songs..
## How many hours of video is 128gb?
Video Recording Time**Recording speed24 Mbps17 Mbps32GB160 min240 min64GB320 min480 min128GB640 min960 min5 more rows•Jan 25, 2012
## How many MB is a 30 second video?
The approximate size of each uncompressed frame is 5MB. At 30 frames per second, a raw HD video will need 5MBx30 = 150MB storage space per second.
## How many GB is 8000 songs?
1-3 of 3 Answers So 1028mb divided by 4mb (per song) you can hold roughly 250 songs in 1gb. So a 32 GB can hold roughly 8000 songs.
## How many GB is 6000 photos?
45 GBDepending on the size of your photos, it is expected that 6000 photos will need 45 GB. A JPEG photo of 12 megapixel 4000×3000 pixels in good quality will have a file size of approx. 5 to 6 MB. If you edit the photos, you will also need to store an edited version.
## How many songs can 1gb hold?
230 songs/ 1GB holds roughly 230 songs (at 128kbps). That’s roughly 16 hours of music or 20 albums. / The average 2 hour movie is 1.5GB.
## How many GB is 700 songs?
MP3 Song Capacity CalculatorPlayer Memory Capacity:128 MB 256 MB 512 MB 650 MB 700 MB 1 GB 2 GB 4 GB 4.7 GB 8 GB 16 GB 20 GB 30 GB 32 GB 40 GB 60 GB 80 GBHours of Music:Typical No. Albums:45 mins/albumTypical No. Songs:3:30 mins/song1 more row
## How big is a 1 hour 720p video?
1 minute=60 second, so it takes about 120 MB (or less, so lets say 1 minute=100 MB (mega bytes). Therefore 1 hour=60 minutes would take 6000 MB, that is about 6GB. In short: if you take a video in 720p for 1 hour you need a 6GB card. Re: How much space would 720p recording @30fps take in MB?
## How many kbps is 24 bit?
StereoSettingsBitrateFile size per second16 bit, 44.1 KHz1,411.2 Kbps176.4 KB16 bit, 48 KHz1,536 Kbps192 KB24 bit, 48KHz2,304 Kbps288 KB24 bit, 96KHz4,608 Kbps576 KB
## How many GB is a FLAC song?
Thanks. THat’s very helpful! An easier way to look at it is to consider it as roughly 3 FLAC albums per GB.
## How large is a 4 minute mp3?
The typical encoding standard for MP3 files is 128 kilobits per second (kbps, kb/s or kbit/s). This works out to about 1 megabyte (MB) per minute of sound.
## How many GB is 2500 songs?
iPod Q&A – Updated November 21, 2011iPod NameCapacity (GB)Estimated SongsiPod 2nd Gen5, 10, 20 GB1000, 2000, 4000*iPod 3rd Gen 10/15/3010, 15, 30 GB2500, 3700, 7500iPod 3rd Gen 10/20/4010, 20, 40 GB2500, 5000, 10,000iPod 3rd Gen 15/20/4015, 20, 40 GB3700, 5000, 10,00033 more rows
## Do I need 64 or 128gb on my phone?
If you do take a lot of videos and also like downloading movies to watch on your phone then you’ll need to look at 128GB or 256GB of storage. … For example, 200 songs takes up about 1GB of storage. Also, check if your phone has expandable storage. iPhones don’t but many Android devices do.
## What is bigger MB or GB?
1 Gigabyte is considered to be equal to 1000 megabytes in decimal and 1024 megabytes in binary system. As you can see, 1 Gigabyte is 1000 times bigger than a Megabyte. So, a GB is bigger than a MB.
## How many MB is a 4 minute song?
5 mbIt depends on many factors, primarily The QUALITY of the song. By default, a audio song standard 4 minutes takes about 5 mb in 128 Kbps and a 64GB has about 59 GB available after iOS …
## How many GB is a song?
To figure out how many 3.28MB songs can fit in a gigabyte(GB), divide 1024 by 3.28 because there are 1024 megabytes in one gigabyte. There you have it! You can fit roughly 312 songs on 1GB of storage.
## How many GB is a 3 minute video?
264 codec, Audio Codec AAC, MP4 Format) usually turns out around 20–30 MB per minute. So for 3 to 4 minutes that would be around 60–120 MBs. But because there’s compression going on, the size of the resulting file really depends on how much data the video contains.
## How many GB is 100 songs?
What do you have?Audio QualityMemory required1000 songs128 kbps2 GB 680 MB10,000 songs128 kbps20 GB 680 GB100 songs192 kbps403.2 MB1000 songs192 kbps4 GB 32 Mb5 more rows•Mar 16, 2016 | 1,308 | 4,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2020-40 | latest | en | 0.83104 |
https://jp.mathworks.com/matlabcentral/profile/authors/10627792-fkk?s_tid=cody_local_to_profile | 1,575,739,707,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540500637.40/warc/CC-MAIN-20191207160050-20191207184050-00489.warc.gz | 416,613,533 | 19,459 | Community Profile
# fkk
57 2017 年以降の合計貢献数
#### fkk's バッジ
Binary numbers
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Roll the Dice!
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Swap the input arguments
Write a two-input, two-output function that swaps its two input arguments. For example: [q,r] = swap(5,10) returns q = ...
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you just measured its surface area, that is the input.
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"mirror" matrix
Create n x 2n "mirror" matrix of this type: e.g. for n = 2 m = [1 2 2 1;1 2 2 1] e.g. for n = 3 m = [1 2 3 3 2 1...
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Back to basics 7 - Equal NaNs
Covering some basic topics I haven't seen elsewhere on Cody. Given 2 input variables, output true if they are equal, false ot...
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Cell joiner
You are given a cell array of strings and a string delimiter. You need to produce one string which is composed of each string fr...
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Bullseye Matrix
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Target sorting
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How to find the position of an element in a vector without using the find function
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The Goldbach Conjecture
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Who Has the Most Change?
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A Collatz sequence is the sequence where, for a given number n, the next number in the sequence is either n/2 if the number is e...
2年以上 前 | 1,393 | 4,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-51 | latest | en | 0.636598 |
https://ru.scribd.com/doc/240882323/Mechanical-Vibrations | 1,568,618,085,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572491.38/warc/CC-MAIN-20190916060046-20190916082046-00017.warc.gz | 653,190,538 | 65,849 | Вы находитесь на странице: 1из 13
# ME 563
Mechanical Vibrations
Lecture #16
Forced Response
(Step Input, Harmonic Excitation)
Free + Forced Response
!
Because the equations we are solving are linear in nature, we
can simply add the free and forced response components to
obtain the total (general) solution to the equations of motion:
Now we will consider several different forms of f(t).
Each Part of the Solution
#
Step Inputs
\$
If we release the disc on an incline from the position x=0, the
response we obtain is the step response to a constant input.
+
Complementary Particular
Step Inputs
%
After applying zero initial conditions, the following solution is
obtained:
Step Inputs
&
Transient response
S.S. response
Step Inputs
'
In MATLAB, we use the step(sys) command to calculate
and plot the step response.
Harmonic Inputs
(
Every periodic input force, f(t), can be described exactly over
one period, t=0 to T sec, using a Fourier series:
Therefore, we can calculate the steady state response to any
periodic force using a sum of responses to sinusoidal forces.
Real Fourier
series
Complex Fourier
series
Harmonic Inputs
)
If we consider a cosinusoidal input force:
and assume the response is of the form,
then:
x
p
(t)=X
p
cos(!t+"
p
)
Harmonic Inputs
*
If we consider a co-sinusoidal input force, :
then the relative amplitude and phase of the response are:
x
p
(t)=X
p
cos(!t+"
p
)
Frequency Response Functions
!+
Normalized expressions using
modal parameters
Sometimes called
Bode Diagrams
Total Response
!!
As with the step response, we now must add the steady state
(particular) response to the free response to obtain the total
response expression:
Some Examples (C=0)
!#
Resonant response of Response near a resonant
undamped system frequency of vibration | 440 | 1,775 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2019-39 | latest | en | 0.786308 |
https://oeis.org/A132056 | 1,627,413,604,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153474.19/warc/CC-MAIN-20210727170836-20210727200836-00569.warc.gz | 452,572,047 | 5,370 | 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!)
A132056 Triangle read by rows, the Bell transform of Product_{k=0..n} 7*k+1 without column 0. 27
1, 8, 1, 120, 24, 1, 2640, 672, 48, 1, 76560, 22800, 2160, 80, 1, 2756160, 920160, 104880, 5280, 120, 1, 118514880, 43243200, 5639760, 347760, 10920, 168, 1, 5925744000, 2323918080, 336510720, 24071040, 937440, 20160, 224, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Previous name was: Triangle of numbers related to triangle A132057; generalization of Stirling numbers of second kind A008277, Lah-numbers A008297, ... a(n,m) enumerates unordered n-vertex m-forests composed of m plane increasing 8-ary trees. See the F. Bergeron et al. reference, especially Table 1, first row, for the e.g.f. for m=1. a(n,m) := S2(8; n,m) is the eighth triangle of numbers in the sequence S2(k;n,m), k=1..7: A008277 (unsigned Stirling 2nd kind), A008297 (unsigned Lah), A035342, A035469, A049029, A049385, A092082, respectively. a(n,1)=A045754(n), n>=1. LINKS F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1992, pp. 24-48. P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, Phys. Lett. A 309 (2003) 198-205. P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, arXiv:quant-ph/0402027, 2004. W. Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4. M. Janjic, Some classes of numbers and derivatives, JIS 12 (2009) 09.8.3 W. Lang, First 10 rows. FORMULA a(n, m) = n!*A132057(n, m)/(m!*7^(n-m)); a(n+1, m) = (7*n+m)*a(n, m)+ a(n, m-1), n >= m >= 1; a(n, m) := 0, n mul(7*k+1, k=0..n), 8); # Peter Luschny, Jan 27 2016 MATHEMATICA a[n_, m_] := a[n, m] = ((m*a[n-1, m-1]*(m-1)! + (m+7*n-7)*a[n-1, m]*m!)*n!)/(n*m!*(n-1)!); a[n_, m_] /; n < m = 0; a[_, 0] = 0; a[1, 1] = 1; Flatten[Table[a[n, m], {n, 1, 8}, {m, 1, n}]][[1 ;; 36]] (* Jean-François Alcover, Jun 17 2011 *) rows = 8; a[n_, m_] := BellY[n, m, Table[Product[7k+1, {k, 0, j}], {j, 0, rows}]]; Table[a[n, m], {n, 1, rows}, {m, 1, n}] // Flatten (* Jean-François Alcover, Jun 22 2018 *) CROSSREFS Cf. A132060 (row sums), A132061 (alternating row sums). Cf. A092082 S2(7) triangle. Sequence in context: A174503 A048786 A240955 * A051187 A284865 A221758 Adjacent sequences: A132053 A132054 A132055 * A132057 A132058 A132059 KEYWORD nonn,easy,tabl AUTHOR Wolfdieter Lang Sep 14 2007 EXTENSIONS New name from Peter Luschny, Jan 27 2016 STATUS approved
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Last modified July 27 15:12 EDT 2021. Contains 346307 sequences. (Running on oeis4.) | 1,153 | 3,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-31 | latest | en | 0.55987 |
https://mathsphere.org/algebra/trig-inverse-sin-cos-tan | 1,632,129,360,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057033.33/warc/CC-MAIN-20210920070754-20210920100754-00035.warc.gz | 429,360,886 | 5,508 | # Inverse Sine, Cosine, Tangent
For a right-angled triangle:
The sine function sin takes angle θ and gives the ratio opposite hypotenuse
The inverse sine function sin-1 takes the ratio oppositehypotenuse and gives angle θ
And cosine and tangent follow a similar idea.
### Example (lengths are only to one decimal place):
sin(35°)= Opposite / Hypotenuse
= 2.8/4.9
= 0.57...
sin-1(Opposite / Hypotenuse)= sin-1(0.57...)
= 35°
### And now for the details:
Sine, Cosine and Tangent are all based on a Right-Angled Triangle
They are very similar functions ... so we will look at the Sine Function and then Inverse Sine to learn what it is all about.
## Sine Function
The Sine of angle θ is:
• the length of the side Opposite angle θ
• divided by the length of the Hypotenuse
Or more simply:
sin(θ) = Opposite / Hypotenuse
### Example: What is the sine of 35°?
Using this triangle (lengths are only to one decimal place):sin(35°) = Opposite / Hypotenuse = 2.8/4.9 = 0.57...
The Sine Function can help us solve things like this:
### Example: Use the sine function to find "d"
We know
• The angle the cable makes with the seabed is 39°
• The cable's length is 30 m.
And we want to know "d" (the distance down).
sin 39° = d/30
Swap Sides:d/30 = sin 39°
Use a calculator to find sin 39°: d/30 = 0.6293…
Multiply both sides by 30:d = 0.6293… x 30
d = 18.88 to 2 decimal places
The depth "d" is 18.88 m
## Inverse Sine Function
But sometimes it is the angle we need to find.
This is where "Inverse Sine" comes in.
It answers the question "what angle has sine equal to opposite/hypotenuse?"
The symbol for inverse sine is sin-1, or sometimes arcsin.
### Example: Find the angle "a"
We know
• The distance down is 18.88 m.
• The cable's length is 30 m.
And we want to know the angle "a"
sin a° = 18.88/30
Calculate 18.88/30:sin a° = 0.6293...
What angle has sine equal to 0.6293...?
The Inverse Sine will tell us.
Inverse Sine:a° = sin−1(0.6293...)
Use a calculator to find sin−1(0.6293...):a° = 39.0° (to 1 decimal place)
The angle "a" is 39.0°
## They Are Like Forward and Backwards!
• sin takes an angle and gives us the ratio "opposite/hypotenuse"
• sin-1 takes the ratio "opposite/hypotenuse" and gives us the angle.
### Example:
Sine Function:sin(30°) = 0.5
Inverse Sine:sin−1(0.5) = 30°
## Calculator
On the calculator you press one of the following (depending on your brand of calculator): either '2ndF sin' or 'shift sin'.
On your calculator, try using sin and then sin-1 to see what happens
## More Than One Angle!
Inverse Sine only shows you one angle ... but there are more angles that could work.
### Example: Here are two angles where opposite/hypotenuse = 0.5
In fact there are infinitely many angles, because you can keep adding (or subtracting) 360°:
Remember this, because there are times when you actually need one of the other angles!
## Summary
The Sine of angle θ is:
sin(θ) = Opposite / Hypotenuse
And Inverse Sine is :
sin-1 (Opposite / Hypotenuse) = θ
## What About "cos" and "tan" ... ?
Exactly the same idea, but different side ratios.
#### Cosine
The Cosine of angle θ is:
And Inverse Cosine is :
cos-1 (Adjacent / Hypotenuse) = θ
### Example: Find the size of angle a°
cos a° = Adjacent / Hypotenuse
cos a° = 6,750/8,100 = 0.8333...
a° = cos-1 (0.8333...) = 33.6° (to 1 decimal place)
#### Tangent
The Tangent of angle θ is:
So Inverse Tangent is :
tan-1 (Opposite / Adjacent) = θ
### Example: Find the size of angle x°
tan x° = Opposite / Adjacent
tan x° = 300/400 = 0.75
x° = tan-1 (0.75) = 36.9° (correct to 1 decimal place)
## Other Names
Sometimes sin-1 is called asin or arcsin
Likewise cos-1 is called acos or arccos
And tan-1 is called atan or arctan
### Examples:
• arcsin(y) is the same as sin-1(y)
• atan(θ) is the same as tan-1(θ)
• etc.
## The Graphs
And lastly, here are the graphs of Sine, Inverse Sine, Cosine and Inverse Cosine:
Sine
Inverse Sine
Cosine
Inverse Cosine
Did you notice anything about the graphs?
• They look similar somehow, right?
• But the Inverse Sine and Inverse Cosine don't "go on forever" like Sine and Cosine do ...
Let us look at the example of Cosine.
Here is Cosine and Inverse Cosine plotted on the same graph:
Cosine and Inverse Cosine
They are mirror images (about the diagonal)
But why does Inverse Cosine get chopped off at top and bottom (the dots are not really part of the function) ... ?
Because to be a function it can only give one answer
when we ask "what is cos-1(x) ?"
But we saw earlier that there are infinitely many answers, and the dotted line on the graph shows this.
So yes there are infinitely many answers ...
... but imagine you type 0.5 into your calculator, press cos-1 and it gives you a never ending list of possible answers ...
So we have this rule that a function can only give one answer.
So, by chopping it off like that we get just one answer, but we should remember that there could be other answers.
## Tangent and Inverse Tangent
And here is the tangent function and inverse tangent. Can you see how they are mirror images (about the diagonal) ...?
Tangent
Inverse Tangent | 1,515 | 5,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-39 | latest | en | 0.785608 |
http://www.abcteach.com/directory/prek-early-childhood-mathematics-fractions-3004-8-1 | 1,495,536,184,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607620.78/warc/CC-MAIN-20170523103136-20170523123136-00002.warc.gz | 411,588,850 | 19,531 | You are an abcteach Member, but you are logged in to the Free Site. To access all member features, log into the Member Site.
# Fractions
FILTER THIS CATEGORY:
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• A colorful folder game for 1-2 players. Match the feather to the turkey that provides the missing factor. Includes directions for game play and assembly, printable folder game boards, and feathers with missing factor equations. Laminate for a fun, reusable classroom game.
• This 9 page fraction packet will help your student grasp the idea of fractions through shapes, and then learn to convert them into numbers. This lesson will cover methods such as; writing fractions, numeric fractions, understanding numerator and denominator, and using shapes to compare and gain understanding of fraction concepts.
• Match the fraction forms - number, word, parts of a whole, parts of a group - to solve the puzzle. Includes denominators from 2 to 8. Laminate for a fun, reuseable classroom game.
• These colorful math study cards give students a quick reference for the following fraction skills: adding and subtracting like and unlike fractions, multiplying, dividing, simplifying, changing a mixed number to an improper fraction and vice versa, finding the greatest common factor, and finding the least common multiple.
• A colorful card game where players compare fractions. The player with the greatest fraction takes the hand. Includes cards and instructions for game play. Laminate for a fun, reuseable classroom game.
• Determine whether a fraction is greater than, less than, or equal to another fraction. 20 practice problems with answers. CC: Math: 4.NF.A.2
• Print, cut out, and laminate. These are useful as manipulatives. Common Core: Fractions: 3.NF.A.1
• "Cut the patterns. Cut out the halves, fourths and eighths. Use them to help with adding and subtracting fractions and equivalent fractions." Common Core: Fractions: 3.NF.A.1
• Print, cut out and laminate. These are useful as manipulatives. Common Core: Fractions: 3.NF.A.1 Common Core: Fractions: 3.NF.A.1
• Whole, halves, thirds, fourths, and eighths. Six pages, four cards to a page. Common Core: Fractions: 3.NF.A.1 | 497 | 2,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2017-22 | latest | en | 0.909538 |
http://www.essaywriterpros.com/standard-data-structures/ | 1,601,444,426,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402118004.92/warc/CC-MAIN-20200930044533-20200930074533-00078.warc.gz | 157,946,224 | 11,449 | Email: support@essaywriterpros.com
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# Standard Data Structures
X 1 : 4.7 Unconditional Moments The previous sections derived the form of the conditional mean and variance of least-squares estimator, where we conditioned on the regressor matrix X. What about the unconditional mean and variance? Many authors and textbooks present unconditional results by either assuming or treating the regressor matrix X as ìÖxedî. Statistically, this is appropriate when the values of the regressors are determined by the experiment and the only randomness is through the realizations of y. Fixed regressors is not appropriate for observational data. Thus econometric results for Öxed regressors are better interpreted as cond CHAPTER 4. LEAST SQUARES REGRESSION 110 The core question is to state conditions under which the unconditional moments of the estimator are Önite. For example, if it determined that E b < 1, then applying the law of iterated expectations (Theorem 2.1), we Önd that the unconditional mean of b is also E b = E E b j X = : A challenge is that b may not have Önite moments. Take the case of a single dummy variable regressor di with no intercept. Assume P (di = 1) = p < 1. Then b = Pn i=1 P diyi n i=1 di is well deÖned if Pn i=1 di > 0. However, P ( Pn i=1 di = 0) = (1 p) n > 0. This means that with positive (but small) probability, b does not exist. Consequently b has no Önite moments! We ignore this complication in practice but it does pose a conundrum for theory. This existence problem arises whenever there are discrete regressors. A solution can be obtained when the regressors have continuous distributions. A particularly clean statement was obtained by Kinal (1980) under the assumption of normal regressors and errors. While we introduce the normal regression model in Chapter 5 we present this result here for convenience. Theorem 4.3 (Kinal, 1980) In the linear regression model, if in addition (xi ,ei) have a joint normal distribution then for any r, E b r < 1 if and only if r < n k + 1. This shows that when the errors and regressors are normally distributed that the least-squares estimator possesses all moments up to n k, which includes all moments of practical interest. The normality assumption is not particularly critical for this result. What is key is the assumption that the regressors are continuously distribu | 596 | 2,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-40 | latest | en | 0.885661 |
https://community.fabric.microsoft.com/t5/Desktop/Time-for-me-to-share-some-knowledge-Time-analysis-by-Season/m-p/59618 | 1,723,429,853,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641023489.70/warc/CC-MAIN-20240811235403-20240812025403-00691.warc.gz | 137,896,264 | 51,428 | cancel
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Helper III
## Time for me to share some knowledge: Time analysis by Season
Hi everyone,
Some of you may know I ask a lot of questions and... have had some difficulty... but I recently did some work in PBI that I'm quite happy with so I thought it would be appropriate to share.
I was doing some work on visualizing our electricity usage and wanted to break things down by season. Below are the DAX queries I came up with. (I'd be very happy to learn of a way to improve this.)
"Season": This is the column you'd put in your charts.
```Season = IF([Day of Year] >= 1 && [Day of Year] <= 60, "Winter",
IF([Day of Year] >= 61 && [Day of Year] <= 152, "Spring",
IF([Day of Year] >= 153 && [Day of Year] <= 244, "Summer",
IF([Day of Year] >= 245 && [Day of Year] <= 335, "Fall",
IF([Day of Year] >= 336, "Winter", "False")
)
)
)
)```
"Day of Year": A numeric value of 1 to 366. [Date] is the date value in your data. (This was needed because FORMAT() is limited in what it can do.)
`Day of Year = DATEDIFF(DATE(YEAR([Date]),1,1), [Date],DAY)+1`
"SeasonSort": This is so you can display the season list in the correct order. It's very similar to the first query.
```SeasonSort = IF([Day of Year] >= 1 && [Day of Year] <= 60, 0,
IF([Day of Year] >= 61 && [Day of Year] <= 152, 1,
IF([Day of Year] >= 153 && [Day of Year] <= 244, 2,
IF([Day of Year] >= 245 && [Day of Year] <= 335, 3,
IF([Day of Year] >= 336, 0, -1)
)
)
)
)```
edit: Corrected typo.
3 REPLIES 3
Resolver I
Thanks for sharing! This would be helpful for certain industry who does seasonal variant products.
Kris
Memorable Member
Season - that's a good one - I could foresee needing that...
Can you explain the Day of Year? what was needed versus format() ?
Helper III
As far as I'm aware, FORMAT() doesn't provide the numeric value of day of year. Custom Date and Time Formats for the FORMAT Function (DAX)
Making "Day of Year" was faster than using DATE() in the formula for delineating season boundaries.
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Join the community in Stockholm for expert Microsoft Fabric learning including a very exciting keynote from Arun Ulag, Corporate Vice President, Azure Data. | 654 | 2,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.935624 |
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Mathematics Asked by Muselive on November 27, 2020
My paper defines a valuation on a Ring $$R$$ to be a map $$v:R-{0} rightarrow A$$ where $$A$$ is an ordered abelian group. This map has the following properties;
1.$$v(ab)=v(a)+v(b)$$
2.$$v(a+b)geq min{v(a),v(b)}$$
(note this is missing the usual conditon wich grants equality if $$v(a) neq v(b)$$)
My question is the following: Let $$w:R[x]-0_{R[x]} rightarrow A$$ by;
$$w(sum_{i=0}^nr_ix^i)=min_{0leq i leq n}v(r_i)$$
and I’m to show this is a valuation. I can easily show property 2 but I cannot show property 1. I’m starting to doubt if it is possible; I wrote out some examples and I don’t see how I can get anything other than a lower bound for a $$w(fg)$$. A hint or even confirmation that this is solvable would be very appreciated.
If $$R$$ is an integral domain, your formula for $$w$$ does yield a valuation. Here is a hint toward proving condition 1 holds. We will use the observation in my comment that $$v(a+b)=min{v(a),v(b)}$$ if $$v(a)ne v(b)$$.
Let $$f=sum_i r_ix^i$$ and $$g=sum_j s_jx^j$$ and suppose $$w(f)=v(r_p)$$, $$w(g)=v(s_q)$$, where $$p,q$$ are as small as possible. In $$fg$$, the coefficient $$t_{p+q}$$ of $$x^{p+q}$$ is a sum containing the product $$r_ps_q$$ and other products $$r_is_j$$ where either $$i or $$j. Show $$v(r_is_j)>v(r_ps_q)$$ for those terms, and so $$v(t_{p+q})=v(r_ps_q)=v(r_p)+v(s_q)$$. Show that no coefficient of another $$t^k$$ can be smaller than $$v(r_p)+v(s_q)$$. Conclude that $$w(fg)=v(r_p)+v(s_q)=w(f)+w(g)$$.
Correct answer by Allen Bell on November 27, 2020
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0 Asked on November 19, 2021 by rameesh-paul | 1,241 | 3,809 | {"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": 30, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-49 | latest | en | 0.842625 |
https://www.quantumcalculus.org/graph-limits-mass-gap/ | 1,685,994,099,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652161.52/warc/CC-MAIN-20230605185809-20230605215809-00311.warc.gz | 1,013,170,102 | 18,943 | Calculus without limits
# Graph limits with Mass Gap
#### The graph limit
We can prove now that the graph limit of the connection graph of Ln x Ln which is the strong product of Ln‘ with itself has a mass gap in the limit n to infinity. The picture below shows this product graph for n=13, and to the right s part of the spectrum near 0 for n=40. We see the gap clearly. The spectrum avoids [-1/25,1/25]. The reason is very simple: the spectrum is the product of the connection graphs of Ln for which we know a spectral gap containing [-1/5,1/5].
Of course, the same argument goes in higher dimension and also for any initial graph G which is one dimensional, when taking the connection graph of $G \times G \times \cdots \times G$. In some sense we have models of Euclidean space in arbitrary dimension for which the inverse of the connection Laplacian remains bounded even in a van Hove (infinite volume) limit. This is a different situation then what is the subject in statistical mechanics, where one looks at infinite particle limits (the original paper of Leon Van Hove from 1949). A biography of van Hove is here.
Why is this remarkable? Having an invertible Hamiltonian in a thermodynamic limit is nice, especially if the inverse operator g leads to a finite potential energy g(x,y) between two parts of space so that the total energy, the sum over all g(x,y) is the Euler characteristic.
But lets try to motivate why the connection graph is natural. It looks a bit strange at first as in two dimensions already the connection Laplacian of Ln x Ln has 3 dimensional parts. But it is not so strange if we look at how one looks at lattices in popular and familiar situations like lattice gas models or cellular automata.
#### Taxi and Uniform Metric
When dealing with the lattice Zn, one usually does not pick the taxi metric l1, where the distance is d(x,0)=|x1|+|x2|+…+|xn| but the uniform metric l metric d(x,0)=max (|x1|, ….,|xn|) which is dual to the taxi metric. When playing the game of life cellular automaton for example, it is a game with the uniform metric: every cell has 8 neighbors. Also for lattice models in statistical mechanics like the Ising model, one can take either the l1 metric or l metric so that |x_k| ≤ 1 is the unit ball of a site. In that model, one assigns a spin value $\sigma(x) \in \{-1,1\}$ to every lattice point and defines the energy $H_r(\sigma)$ = $-\sum_{x \sim y \in B_r} \sigma(x) \sigma(y)$, where the sum ranges over all connected vertices in a ball $B_r = \{ |x|_{p} \leq r$ and studies functional integrals like $\langle f \rangle$ = $E[e^{-\beta H(\sigma)} f(\sigma)]/Z$ with $Z=E[ e^{-\beta H(\sigma)} ]$. (The Ising model usually takes the l1 metric.)
In the l metric has unit spheres which contain triangles (3 points all connected to each other in distance 1) while the l1 metric, where in the unit sphere contains only 4 disconnected points, there is no edge. Both graphs miss the property of being two dimensional, but the 8 neighbor sphere has more friends: Most nearest neighbor automata take the uniform metric with 8 neighbors. But the other one is also popular, like in Ising or in numerical schemes or tightbinding approximations. In the weak coupling case l1 , the Laplacian is replaced with u(x+1,y)+u(x-1,y)+u(x,y+1)+u(x,y-1)-4 u(x,y) which corresponds to the filter
| 0 1 0 |
| 1 -4 1 |
| 0 1 0 |
But also stronger filters using the l neighborhood are used, like in
| 1/4 1/2 1/4 |
| 1/2 -3 1/2 |
| 1/4 1/2 1/4 |
where the unit sphere has 8 members (containing triangles).
#### Discretisations
A discretisation of a metric d on Rn is a countable subset V of Rd which is periodic (translational invariant with respect to n linearly independent translations) equipped with a graph structure G=(V,E) such that the geodesic distance in the graph is the same than the d-distance in Rd. The self-dual Euclidean l2 metric d(x,0)=|x1|2+|x2 2+ … +|xn|2 is the most symmetric, coming from an inner product, but it does not admit a discretisation which is a product lattice. The extreme Banach space metrics l1 and l however both do.
The l1 metric defines the weak Zn lattice in which the unit sphere is zero dimensional. The l∞ metric defines the strong Zn lattice in which the unit sphere is two dimensional.
#### Squares and Triangles
In two dimensions, there is a triangular lattice in which the unit sphere is one dimensional. But this triangulation does not come from a product structure. The fact that the product structures need quite a bit of bending to become triangular is the basic difficulty in de Rham’s theorem which links de Rham cohomology (a calculus based product cohomology which is used everywhere in calculus when dealing with operators like curl and div) and simplicial cohomology, in which the basic building blocks are simplices. Cubes are not simplices and either need to be cut (which breaks symmetry) or then require some homotopy deformations. The construction of a chain homotopy was tricky as one can see when looking at the original papers of the masters like de Rham. The problem of bending a cube to a simplex also appears in an ergodic setup to show the equivalence of geometric cohomology and algebraic group cohomology generalizing a result of dePauw in the two dimensional case.
#### Why product structures?
Why do we like a product structure? Because it allows us to reduce the situation to lower dimensional cases. Its like writing a random variable as a sum of uncorrelated random variables or performing conditional expectations to get the dimensions down. It is a basic principle to build up things from smaller parts and then bootstrap from lower to higher dimensions. Sometimes, the product structure is not obvious. A differential operator for example exhibits a product structure only after a Fourier transform. As one can see often in mathematics, products structures are more familiar and easier to deal with, especially in calculus setups, where we like to use concepts like partial derivatives and commutativity results like Clairaut or Fubini, product structures are not directly compatible with triangulations. The first who really realized the product structure of space was Descartes. It was there, of course, but nobody saw it. It is the concept of coordinates which reveals the product structure of space. In differential geometry later one has in the 20th centuray tried hard to get rid of coordinates again and make geometry coordinate independent (as coordinates are always a bit artificial, relying on a basis for example). Still, in the mathematical frame work which replaces it, tensor calculus and so the product structure (present at least locally also in non-linear situations) remains. Manifolds are patched up Euclidean spaces. Even more complicated topological spaces are equiped with sheaves carrying algebraic structures.
#### From topology to homotopy
If one is interested in properties which are homotopy invariants like cohomology, we don’t care so much about topological notions like dimension and can also allow discretizations of 2-dimensional space in which the unit spheres are not one-dimensional. This happens both for the weak lattice and the strong lattice. In the weak lattice, the dimension of the unit spheres have too low dimension, in the case of the strong lattice, the dimension of the unit spheres have too large dimension. But there is a difference in that the strong lattice behaves better for things which only depend on homotopy.
The weak lattice is not contractible in dimensions larger than 1. It has lots of holes. The strong lattice is contractible in dimensions larger than 1. It is homotopic to a triangulation of space.
#### Product structure
Both metrics, the l1 as well as the l metric come from a product structure. In one dimension, where all the lp norms are the same d(x,y)=|x-y|, set V=Z is the only possible discretisation. Two integers (a,b) are connected by an edge if their distance is 1. Here is an observation:
The graph defined by the l1 metric is the weak graph product of Z with Z. The graph defined by the l∞ metric is the strong graph product of Z with Z.
#### Connection Laplacians
The connection Laplacians of these two products look a bit more messy. The vertices are the simplices in the graphs above and two such simplices are connected, if they intersect. Already to the left, in the weak case, as 4 edges intersect, there are 4-simplices present in the connection graph. In the strong case, the dimensions are even higher, at least 7 dimensional.
The structure of space looks now more complicated because we display not only the zero dimensional and one dimensional parts as usual when drawing graphs but because we display also the higher dimensional structures: every simplex in the previous graph has become a vertex now. The connection graphs contain the Barycentric refinement graphs, which has the same vertices but where two vertices are connected only if one is contained in the other. In the connection graph case, we connect two vertices if they intersect.
#### Spectral properties
The product structure might look similar but it produces a spectacular different behavior on the spectral level. But we have to look at different Laplacians. In the weak case we have to look at the adjacency matrix to get spectral compatibility.
For weak products, the eigenvalues of the adjacency Laplacian add up. For strong products, the eigenvalues of the Fredholm connection Laplacian multiply.
In the continuum, when taking the Cartesian product of manifolds, the eigenvalues of the standard Laplacian add up. In some sense,we get in the discrete strong product an entanglement which is not present in the continuum. It certainly also has to do with the choice of the Laplacian. Already in one dimensions, the connection Laplacian involves the adjacency matrix involving the simplices present in the graph.
The adjacency matrix or Fredholm adjacency matrix only addresses the zero and one-dimensional parts of space. The Hodge or Fredholm connection Laplacian addresses also higher dimensional aspects of space.
In the Hodge or connection case, we look at all the simplices of a graph. It is interesting that this “looking at higher dimensional ingredients of space” will lead to entanglement in the sense that the energies don’t add up but multiply. Its like in a Fock space of quantum mechanics, where energies of independent particles add up, but that for multiparticle systems, the eigenvalues multiply.
#### Eigenvalues
One can explore this algebraically as one can see the lattice Z as a graph limit of a linear graph Ln or a cyclic graph Cn. There is a big difference between the two lattices. Lets define the sum-product of to numbers as xy+x+y:
For weak products, the eigenvalues of the adjacency matrices add. For tensor products, the eigenvalues of the adjacency matrices multiply. For strong products, the eigenvalues of the adjacency matrices are the sum-product of the adjacency matrices.
The identity $(\lambda+1) (\mu + 1) = \lambda \mu + \lambda + \mu +1$ shows now why the eigenvalues of the Fredholm adjacency matrix L=(1+A) multiply. As we have seen that the connection graph functor G -> G’ satisfies (G x H)’ = G’ * H’, where * is the strong product the eigenvalues of the connection Laplacian multiply when we take a product. So far, there is little new, but we can exploit this now for a graph limit in higher dimensions.
#### Mass gap in a graph limit
By the central limit theorem for Barycentric refinements, the graph limit of a Barycentric limit of a graph converges universally depending only on the dimension of the initial graph. (See the article and some slides). As indicated already here, the situation becomes interesting if one takes connection Laplacians. In that case, there is a mass gap in the limit, the reason being that the limiting operator remains invertible.
This goes over to higher dimensional situations. We just have to take the strong product of the connection Laplacians to deal with connection Laplacians of products.
To summarize, we have models of space with a Laplacian which has also in the infinite volume limit a bounded inverse. We are not in a quantum field theory setup, but the interpretation is that the mass of the lightest particle is positive. In some sense, there is no infrared divergence. While we are at mentioning physics jargon, one could associate the Green function g(x,y) with a S matrix for space. Formally, we can write (1+A)-1 = 1-A+A^2-A^3…. involving the adjacency matrix A. The sum on the right hand side is only defined by analytic continuation of function (1-s A)-1 where the sum converges for small s. The boundedness of the Green function in the limit is remarkable. Actually, we know that the limiting operator is an almost periodic operator on a compact topological group. In one dimensions, it is the group of dyadic integers, in the n dimensional case it is the n’th product of these groups. Going to the product structure and refining each dimension sepearately allowed to bypass dealing with the Barycentric limit in higher dimensions, in which case, we still don’t know what the structure of the limiting compact topological group is.
Still, we are in a mathematical setup and just make a statment about the limiting density of states of an infinite graph limit. We still need to get an explicit formula for the density of states in the connection graph case analogues to the Laplacian case, where the density of states was the equilibrium measure on a Julia set. With that explicit formula, one can then get the density of states in higher dimensions by convoluting the logaritmic version of it. The limiting operator then definitely has absolutely continuous spectrum and just one gap. | 3,092 | 13,824 | {"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": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.91779 |
https://forum.allaboutcircuits.com/threads/how-to-calculate-the-power-of-an-rf-signal.100195/ | 1,558,645,710,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257396.96/warc/CC-MAIN-20190523204120-20190523230120-00443.warc.gz | 454,120,720 | 19,700 | # how to calculate the power of an RF signal?
Discussion in 'Wireless & RF Design' started by bagath ch, Aug 6, 2014.
1. ### bagath ch Thread Starter New Member
Aug 6, 2014
1
0
I want to find out the output power of an RF power amplifier at a frequency of 437 MHz can anybody suggest a device for this purpose?
can i use an digital oscilloscope for this?
Thankyou
Feb 24, 2006
12,041
2,591
3. ### inwo Well-Known Member
Nov 7, 2013
2,416
314
I've been working on a similar problem with a 1mhz device. ~5-7 watts.
Non of my watt-meters were in range or gave reasonable results.
Not a high degree of accuracy, I'm sure, but got me a ball park figure to compare.
I connected an appropriate load resistor. Then measured the time it took to go from 25-75C.
I then experimentally matched the change using an easily measured dc source.
There were no comments on this method in my other thread. Hope to get some input here.
It seems to match measuring ppvolts and ppamps on scope and converting to rms.
And also measuring ppvolts across known non-inductive load.
4. ### alfacliff Well-Known Member
Dec 13, 2013
2,449
432
why not use a dummy load (non inductive resistor) designed for the frequency? 437 mhz requires a bit more care to eliminate swr than at 1 mhz, and use your digital scope if the frequency response is enough to measure the rms voltage across the resistor? P=Esquared divided by R. or hind a local ham who has a watt meter.
5. ### Papabravo Expert
Feb 24, 2006
12,041
2,591
Mt first attempt at making a dummy load was a miserable failure. I put it on a network analyzer and saw the inductive effects of a length of wire almost immediately (10 MHz. or so). By the time it got to 50 MHz it was heading for capacitor land on the Smith chart.
6. ### inwo Well-Known Member
Nov 7, 2013
2,416
314
How much power?
What impedance cable?
+1 on ham.
I have a 100-1000mhz wattmeter.
7. ### alfacliff Well-Known Member
Dec 13, 2013
2,449
432
the ARRL handbook shows how to make dummy loads. there is a lot to length of wires, type of resistors and such.
Aug 8, 2011
53
8
BR-549 likes this.
9. ### BR-549 AAC Fanatic!
Sep 22, 2013
4,755
1,290
If your scope is calibrated, and has the bandwidth, and you have a clean, symmetrical sine wave,...you can follow alfacliff's advice. For that high of F, I would barrow a bird and then mark my swr meter for pwr for reference.
10. ### iconic.digital New Member
Aug 28, 2014
1
0
Hi Guys,
I've been working on a new project configuring the Motorola digital radios into a new UFH band. I'm new to this but have had some advice from a radio hire company called Wall to Wall Communications. Their site is http://www.walltowallcomms.co.uk. Can someone give me some advice about adjusting my symmetrical sine wave as it is not clean at the moment and I want to increase the performance of the radio over a wider range. Thanks guys. | 792 | 2,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-22 | latest | en | 0.941436 |
https://www.jiskha.com/display.cgi?id=1400112545 | 1,516,097,126,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886397.2/warc/CC-MAIN-20180116090056-20180116110056-00094.warc.gz | 946,699,841 | 3,854 | # chemisty
posted by .
calculate the H+ for green tea with a pH of 7.9
• chemisty -
pH=-log[H^+]
10^-(pH)=[H^+]
Answer contains ONLY 1 significant figure.
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# Thread: New to the Forum; First Post; Distance Allowed for Mobile Home Feeder Wire
1. ## New to the Forum; First Post; Distance Allowed for Mobile Home Feeder Wire
Hello all! This is my first post of many to come. I'm a fairly handy DIY'er. These internet forums are so helpful in learning and saving money, but most importantly, making well-informed decisions. Well, so much for blowing sunshine!
I'm installing a mobile home on some property I own. Here's the skinny:
Feeder wire is aluminum 4/0, 4/0, 2/0, #4.
240v single-phase
Service load in the mobile home is 200A.
The Voltage Drop Calculators on the internet seem to vary between each one. Keeping the voltage drop less than 3%:
Q: What is the max distance that I can run the feeder wire (I will NOT be upsizing or downsizing this feeder wire) from the service pole without the lights blinking everytime the heat or AC comes on? What would be the max distance that you as a professional would run if this was your project to obtain proper electrical service?
Q: Code here in North Alabama requires schedule 40 gray PVC conduit when above ground. What size diameter conduit should I use or should I just go with direct burial when below ground?
2. Originally Posted by kailor
Hello all! This is my first post of many to come. I'm a fairly handy DIY'er. These internet forums are so helpful in learning and saving money, but most importantly, making well-informed decisions. Well, so much for blowing sunshine!
I'm installing a mobile home on some property I own. Here's the skinny:
Feeder wire is aluminum 4/0, 4/0, 2/0, #4.
240v single-phase
Service load in the mobile home is 200A.
The Voltage Drop Calculators on the internet seem to vary between each one. Keeping the voltage drop less than 3%:
Q: What is the max distance that I can run the feeder wire (I will NOT be upsizing or downsizing this feeder wire) from the service pole without the lights blinking everytime the heat or AC comes on? What would be the max distance that you as a professional would run if this was your project to obtain proper electrical service?
Q: Code here in North Alabama requires schedule 40 gray PVC conduit when above ground. What size diameter conduit should I use or should I just go with direct burial when below ground?
1- You dont have to worry about the voltage drop because the code wont allow you to put the service pole more than 30 ft from the mobile home
Nec 550 .32 (a)
2- The 4/0 4/0 2/0 #4 is use-2 called alumiflex mobile home feeder. It is for direct burial . Use a short 2" pvc to go down from the disconnect on the pole to 18" below grade , and another pvc 2" 18 below grade to come up at the mobile home . Use the pvc to run to the mobile home panel . Most places allow the feeder to be burried 18 " but some require 24 " . You can run the pvc all the way if you want but dont have to underground. If you dont have a lot of experience with this and you have a lot of 90 deg bends in the conduit ,I would use 2 1/2 '' pvc instead of 2 " , it is very hard to install the 4/0 in 2 '' pvc . The 2 1/2 will slide over and glue to a 2 " that runs up into the panel inside so you dont have to cut out for a bigger pipe . you also need to slide the conduit and fittings over the wire one piece at a time , this is actually against the code but you will never get the cables in if you run all the conduit first and try to pull the wire through
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• | 886 | 3,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2014-15 | latest | en | 0.901766 |
https://slideplayer.com/slide/3902542/ | 1,702,184,279,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101195.85/warc/CC-MAIN-20231210025335-20231210055335-00660.warc.gz | 434,587,859 | 24,154 | Chapter 6 – Applications of Integration
Presentation on theme: "Chapter 6 – Applications of Integration"— Presentation transcript:
Chapter 6 – Applications of Integration
6.2 Volumes 6.2 Volumes Erickson
Solids of Revolution Solids generated by revolving plane regions around the axes are called solids of revolution. Examples: billiard balls, threaded spools etc. We can find their volume by using geometry but here we are going to learn how to use calculus to find the volume. 6.2 Volumes Erickson
Volumes of Solids In this section we will learn how to find volumes of solids by using integration. Here again we will work with areas. Think for a second, how do we find the volume of a solid? Ex. the volume of a cylinder is: V=πr2h In other words it is the area of the base times the height. 6.2 Volumes Erickson
where A is the area of the base.
Volumes of Solids The same is true for other cylinder solids: Examples In each case the volume would be V=Ah where A is the area of the base. 6.2 Volumes Erickson
Volumes of Solids If we can set things up so that the axis of revolution is the x- axis and the region is the region of the plane between the x-axis and the graph of a continuous function y = A(x) a ≤ x ≤ b, we can calculate the volume of the solids by approximation. We can partition the solid in n vertical rectangles and find the area of each. The sum of those areas will give us an approximation of the volume. 6.2 Volumes Erickson
Definition of Volume (Vertical Slices)
Let S be a solid that lies between x = a and x = b. If the cross-sectional area of S in the plane Px, through x and perpendicular to the x-axis, is A(x), where A is a continuous function, then the volume of S is 6.2 Volumes Erickson
Definition of Volume (Horizontal Slices)
Let S be a solid that lies between y = c and y = d. If the cross-sectional area of S in the plane Py, through x and perpendicular to the y-axis, is A(y), where A is a continuous function, then the volume of S is 6.2 Volumes Erickson
Volumes of Solids When we use the volume formula,
it is important to remember that A(x) is the area of a moving cross-section obtained by slicing through x perpendicular to the x-axis. Similarly, when we use the volume formula, it is important to remember that A(y) is the area of a moving cross-section obtained by slicing through y perpendicular to the y-axis. 6.2 Volumes Erickson
Find the Volume of the Pyramid
6.2 Volumes Erickson
Find the Volume of the Pyramid:
4 Consider a horizontal slice through the pyramid. The volume of the slice is s2dh. If we put zero at the top of the pyramid and make down the positive direction, then s=h. h This correlates with the formula where B is the area of the base: s dh 6.2 Volumes Erickson
Method of Slicing 1 Sketch the solid and a typical cross section.
Find a formula for V(x). (Note that I used V(x) instead of A(x).) 2 3 Find the limits of integration. 4 Integrate V(x) to find volume. 6.2 Volumes Erickson
Some Useful Areas - Disk
If the cross-section is a disk, we find the radius of the disk (in terms of x or y) and use A = π (radius)2 6.2 Volumes Erickson
Volume of a Disk If the solid consists of adjacent vertical disks between x = a and x = b, we find the radius R(x) of the disk at x, and the volume is If the solid consists of adjacent horizontal disks between y = c and x = d, we find the radius R(y) of the disk at y, and the volume is 6.2 Volumes Erickson
Example 1: Volume by Disk
Find the volume of a solid obtained by rotating about the x-axis the region under the curve from 0 to 2. 6.2 Volumes Erickson
Example 2: Volume by Disk
Find the volume of the solid obtained by rotating the region bounded by the curve and the lines x=0 and x=2. 6.2 Volumes Erickson
Some Useful Areas - Washers
If the cross-section is a washer, we find the inner radius and the outer radius of the washer (in terms of x or y) and use A = π (outer radius)2 − π (inner radius)2 6.2 Volumes Erickson
Volume of a Washer where R is the outside radius, r is the inside radius, and h is the height. 6.2 Volumes Erickson
Volume of a Washer If the solid consists of adjacent vertical washers between x = a and x = b, we find the outside radius R(x) and inside radius r(x) of the washer at x, and the volume is If the solid consists of adjacent horizontal washers between y = c and x = d, we find the outside radius R(y) and inside radius r(y) of the disk at y, and the volume is 6.2 Volumes Erickson
Examples: Volume by Washer (1)
The region bounded by and is revolved about the y-axis. Find the volume. If we use a horizontal slice: The “disk” now has a hole in it, making it a “washer”. Because we are rotating around the y-axis, we need to solve our equations for x. The volume of the washer is: outer radius inner radius 6.2 Volumes Erickson
Examples: Volume by Washer (1)
6.2 Volumes Erickson
Examples: Volume by Washer (2)
The outer radius is: r The inner radius is: R 6.2 Volumes Erickson
Examples: Volume by Washer (2)
6.2 Volumes Erickson
Examples – pg. 438 Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer. 6.2 Volumes Erickson | 1,326 | 5,214 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2023-50 | latest | en | 0.897486 |
https://www.mbamission.com/blog/gmat-impact-just-getting-started-with-the-gmat-here-is-what-to-do/ | 1,716,750,380,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00229.warc.gz | 759,691,830 | 17,291 | Blog
GMAT Impact: Just Getting Started with the GMAT? Here Is What to Do
With regard to the GMAT, raw intellectual horsepower helps, but it is not everything. In this blog series, Manhattan Prep’s Stacey Koprince teaches you how to perform at your best on test day by using some common sense.
If you are just getting started with the GMAT and are trying to figure out what to do, we have got several big categories of things to discuss: mind-set, devising a study plan, and learning how to study.
Mind-Set
If you do what most people do and try to prepare for this test in the same way that you prepared for tests in school, you are not going to get the best score that you could get.
If you are not sure what is tested on the GMAT or what the different question types look like, take some time to wander around this section of the official GMAT Web site.
Next, read this short article: “In It to Win It.” This will help you to start to adjust your mind-set so you can maximize your GMAT score. One important detail: you are only going to get about 60% of the questions right.
How can that be? Glad you asked. Read the “Scoring” section of Manhattan Prep’s free e-book The GMAT Uncovered. This section explains just how the scoring on the GMAT works—which will help you better understand why trying to get everything right is a really bad strategy on the GMAT.
Okay, we are essentially done with the mind-set category, but I have to say one more thing. I put mind-set first for a reason: if you have the wrong mind-set, it will not matter how much you learn or practice. You still will not get the best score that you are capable of getting.
Devising a Study Plan
Get started with this article: Developing a GMAT Study Plan. Note: make sure to follow the instructions about taking and analyzing a practice test.
How to Study
One key GMAT skill is learning to recognize problems. “Recognize” means that we actually have a little light bulb go off in our brain—“Hey, I’ve seen something like this before, and on that other one, the best solution method was XYZ, so I’m going to try that this time, too!”
When you recognize something, you have given yourself two big advantages: you save yourself time, because recognizing is faster than figuring something out from scratch, and you are more likely to get it right because you know what worked—and what did not—the previous time. You will not be able to recognize every problem, but the more you can, the better.
Read the “How Do I Learn?” section in the second half of the “Developing a GMAT Study Plan” article. Make sure to follow the links given in that section—those links lead to the tools that will help you learn how to learn from GMAT questions.
If you want to take advantage of online forums to chat with teachers and other students (and I strongly recommend that!), learn how to make the best use of the forums.
Finally, ask for advice! So many resources are out there that it can be overwhelming, but most companies offer free advice (Manhattan Prep does here!) and you can also benefit from talking to fellow students.
Explore onTrack — mbaMission’s newest offering allowing you to learn at your own pace through video. Learn more | 723 | 3,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | longest | en | 0.940093 |
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# K’Nex Levers
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### K’Nex Levers
1. 1. K’Nex Levers For Mrs. Berg’s Science Class
2. 2. Seesaw <ul><li>Materials: Your group’s seesaw, foil packet, rubber band, masking tape </li></ul><ul><li>NO SHOOTING RUBBER BANDS </li></ul><ul><li>Procedure: </li></ul><ul><li>1. Lay your hand flat on the desk with your palm facing up. Put the foil pack in the palm of your hand and lift it up from the desk. Take turns doing this. The foil pack is the load or resistance. </li></ul><ul><li>Q. Is the load heavy? Can you feel the weight in your hand? </li></ul><ul><li>2. Now place the load on one end of the seesaw. Use the rubber band to secure it. </li></ul><ul><li>Q. What happens to the end of the seesaw with the load attached? </li></ul>
3. 3. <ul><li>3. Push down on the empty seat to lift the load. </li></ul><ul><li>Q. Did it feel easier, more difficult, or just the same to lift the load this time? </li></ul><ul><li>Q. In which direction to you apply your effort? (push) </li></ul><ul><li>Q. In which direction did the load move? </li></ul><ul><li>4. Draw a picture of your hand with the foil pack. Label the direction of effort and the direction the load moved. </li></ul>
4. 4. <ul><li>5. Draw a picture of the seesaw with the foil pack. Label the direction of effort and the direction the load moved. Label the fulcrum (F), effort (E), and resistance (R) on your picture. </li></ul>
5. 5. F Resistance Effort
6. 6. <ul><li>Change your seesaw to look like this: </li></ul><ul><li>Q. Do you think you will need to use more effort or less effort to move the load? </li></ul>
7. 7. <ul><li>7. Put the foil pack on the seat of the short arm. Push down on the seat with the long arm. </li></ul><ul><li>Q. What happened? </li></ul><ul><li>8. Put the foil pack on the seat of the long arm. Push down on the seat with the short arm. </li></ul><ul><li>Q. What happened? </li></ul><ul><li>9. Draw a picture showing the way the lever looked when the effort was the easiest. </li></ul>
8. 8. <ul><li>10. Q. Which class of lever is your seesaw? (Look in the Building Instructions for the answer.) </li></ul><ul><li>11. Q. How does changing the location of the fulcrum change the effort force? </li></ul><ul><li>12. Q. Name other simple machines that are first class levers. </li></ul><ul><li>13. Q. How could you use a first class lever to do work? </li></ul>
1. #### A particular slide catching your eye?
Clipping is a handy way to collect important slides you want to go back to later. | 883 | 2,861 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-40 | latest | en | 0.772651 |
http://de.metamath.org/mpegif/3anbi1i.html | 1,610,950,710,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514423.60/warc/CC-MAIN-20210118061434-20210118091434-00311.warc.gz | 25,330,472 | 3,351 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > 3anbi1i Structured version Visualization version Unicode version
Theorem 3anbi1i 1198
Description: Inference adding two conjuncts to each side of a biconditional. (Contributed by NM, 8-Sep-2006.)
Hypothesis
Ref Expression
3anbi1i.1
Assertion
Ref Expression
3anbi1i
Proof of Theorem 3anbi1i
StepHypRef Expression
1 3anbi1i.1 . 2
2 biid 240 . 2
3 biid 240 . 2
41, 2, 33anbi123i 1196 1
Colors of variables: wff setvar class Syntax hints: wb 188 w3a 984 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 This theorem depends on definitions: df-bi 189 df-an 373 df-3an 986 This theorem is referenced by: iinfi 7928 fzolb 11923 brfi1uzind 12648 opfi1uzind 12651 sqrlem5 13303 bitsmod 14403 isfunc 15762 txcn 20634 trfil2 20895 eulerpartlemn 29207 bnj976 29582 bnj543 29697 bnj594 29716 bnj917 29738 topdifinffinlem 31743 dath 33295 elfzolborelfzop1 40303 nnolog2flm1 40388
Copyright terms: Public domain W3C validator | 467 | 1,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.256684 |
https://study.com/academy/topic/aepa-middle-grades-math-two-step-equations.html | 1,566,042,025,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027312128.3/warc/CC-MAIN-20190817102624-20190817124624-00552.warc.gz | 633,872,902 | 26,692 | Ch 30: AEPA Middle Grades Math: Two-Step Equations
A National Evaluation Series (NES) test is used in place of the AEPA Middle Grades Math test for future middle school math teachers. In this collection of engaging video lessons, you can review how to work with two-step equations to prepare for your test. The brief, self-assessment quizzes that follow each lesson can alert you to any topics that might still need added attention.
AEPA Middle Grades Math: Two-Step Equations - Chapter Summary
In these lessons, you can practice writing and solving addition, subtraction, multiplication and division equations having two or more variables. In addition, you can experience solving those same types of equations in word-problem format. Once you've completed this chapter, you should work confidently with the following during the AEPA Middle Grades Math exam:
• Methods for writing and solving addition and subtraction equations containing two or more variables
• Procedures for writing and solving multiplication and division problems with two or more steps
• Strategies to use when solving and writing word problems involving at least two variables
Expert math instructors lead the video lessons and offer quick, self-assessment quizzes so you can gauge your progress. You can watch the videos as many times as you need, and the lessons are also available as print transcripts, for those who prefer reading the lessons.
APEA Middle Grades Math: Two-Step Equations Chapter Objectives
Items pertinent to this chapter are covered within the algebra and functions section of the test; this section makes up about 33% of the exam. Your ability to manipulate algebraic expressions, solve linear and non linear inequalities and equations, and attach correct algebraic notation to sentences and phrases may be measured in this area. In addition, you should be comfortable with analyzing real-world problems containing linear equations, inequalities and systems. You'll have four hours and 15 minutes to complete the 150 multiple-choice questions on the test. Our short quizzes that accompany each lesson offer you valuable practice for the test and help you discover any areas still requiring review.
7 Lessons in Chapter 30: AEPA Middle Grades Math: Two-Step Equations
Test your knowledge with a 30-question chapter practice test
Chapter Practice Exam
Test your knowledge of this chapter with a 30 question practice chapter exam.
Not Taken
Practice Final Exam
Test your knowledge of the entire course with a 50 question practice final exam.
Not Taken
Earning College Credit
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 561 | 2,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.929493 |
https://abhyasonline.in/questions/Maths/Areas%20Of%20Parallelograms%20And%20Triangles/ABCD%20is%20a%20parallelogram%20%20where%20%20If%20AB%20=%2018%20cm%20%20AE%20=%209%20cm%20and%20CF%20=%2012%20cm%20%20find%20the%20length%20of%20AD/ | 1,627,949,407,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154408.7/warc/CC-MAIN-20210802234539-20210803024539-00198.warc.gz | 97,781,140 | 11,602 | Maths / Areas Of Parallelograms And Triangles / Area of Parallelograms
QUESTION
ABCD is a parallelogram, where $\dpi{120} \fn_jvn \large AE \perp DC \; and \; CF \perp AD.$ If AB = 18 cm, AE = 9 cm and CF = 12 cm, find the length of AD.
EXPLANATION
Explain TypeExplanation Content
Text
$13.5\;cm$
View Contents
(Concept based Learning and Testing for [6th - 10th], NTSE, Bank & Govt. Exams)
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Student Feedback on Methodology - Cheshta C/o ABHYAS Academy
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Their methodology is good but it can be improved like maths practise sessions should be increased.
#### Other Testimonials
Courses We Offer
(Concept based Learning and Testing for [6th - 10th], NTSE, Bank & Govt. Exams) | 209 | 710 | {"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": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-31 | latest | en | 0.793784 |
https://www.r-bloggers.com/example-shiny-app-interest-rate-walk/ | 1,591,251,179,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439019.86/warc/CC-MAIN-20200604032435-20200604062435-00182.warc.gz | 867,651,312 | 50,260 | # Example Shiny App – Interest Rate Walk
November 27, 2017
By
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This Shiny app demonstrates the Cox-Ingersoll-Ross interest rate walk and an interest rate walk conducted using a bootstrap resampling technique. The code used to create this app is available on GitHub. Assuming you have the necessary packages installed, you should be able to run this Shiny app on your local computer. A live version of the app is available here.
Please take a look at the app before reading the rest of this post. For maximum clarity, read the post with the app open and look at the inputs/outputs in the Shiny app as they are described in the post.
### Bootstrap Resampling Method
This method takes a starting interest rate (selected by the user of the app) and applies randomly drawn historical treasury yield changes to that initial rate. The treasury yield changes can be drawn from either 1, 5, 10, 20, or 30 year treasuries (depending on the user’s selection). The user can also control the historical time period over which the yield changes are drawn from.
Despite the method’s naivete, the plotted results look reasonable. I could see myself having a hard time distinguishing the results of one of these walks from true interest rates over a given time period.
### Cox-Ingersoll-Ross Walk
I used the `sde::rcCIR()` function to generate this walk. This Shiny app’s inputs for this walk are:
• a which controls the speed at which the interest rate reverts to the mean
• b the mean interest rate
• sigma which controls the volatility
No matter which paramters I select, the resulting walks seem far too concentrated around the mean and/or the yields bounce back and forth across the mean at an unconvincingly fast frequency. It is very posisible that I am doing something wrong, and I will investigate further. The code is available on GitHub (link above); please reach out, if you see any errors.
### Conclusion
Interest rate walks are a fast and easy way to simulate rates similar to real world interest rates. In this app, the bootstrap resampling of historical treasury yield changes gives a far nicer quick eye test result than the CIR. Do you have a preferred method for conducting interest rate walks? That is probably not a question you hear every day :), but if you do, please let us know in the comments below.
R-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. | 588 | 2,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.89412 |
ha22.com | 1,601,339,096,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401617641.86/warc/CC-MAIN-20200928234043-20200929024043-00371.warc.gz | 427,681,111 | 13,908 | # How to calculate calories and control weight
How to calculate calories and control weight calorie calculator Determine your weight loss goals Free Weight Loss Planning
The ideal way to calculate body calories per day
With a simple formula you can calculate the calories your body needs in order to always maintain your weight at the correct rates. This resulting number of the equation changes as your weight changes.
The importance of calculating calories
Most people consume about 2000 calories from their food during the day and the truth calls for us to say that the amount of calories that the body needs vary greatly from person to person depending on that person’s daily activity current weight and the required amount of calories. It calculates calories regularly and is compatible with changing daily activities and weight fluctuations from increase or decrease.
The formula for calculating calories
For working men the calories they need are equal to: the product of multiplying and weighing them in pounds by 15 and for working women it is the product of multiplying their weight in pounds by 12 and for non-working men the amount of calories required is calculated by multiplying the weight in pounds by 13 and women who are not working the same way by multiplying their weight In pounds in 10.
Calculate the calories required for you per day
Calculate the amount of calories required for you per day from fat using the above formula by multiplying the number of calories generated by 30. or 30%.
And in terms of the number of grams that you need daily of fat you can calculate it by dividing the number of calories from the fat that you eat by the number 9 and thus you can control your own weight and maintain it yourself.
Thus if you want to lose one pound during the week you are supposed to decrease the number of calories you eat by 500 calories during the day.
For example if you consume 2000 calories you will eat 1500 calories per day for a week to lose one pound and vice versa if you want to increase your weight by one pound during the week you will increase your caloric intake that you consume 500 calories and as we explained earlier you will be able to know how much fat you will change in grams.
The ideal way to calculate body calories per day :
For a man: 10 x Weight in kg + 6.25 x Height in cm – 5 x Age + 5
For a female: 10 x Weight in kg + 6.25 x Length in cm – 5 x Age – 5
However calculating the calories that the body needs vary from one person to another according to their age weight and state of health. A specialist doctor must be consulted to find out the exact calories you need.
We offer you the best 6 low-calorie foods and maintain the diet:
_ Apple with half a cup of skim milk: It is a complete meal and has many nutrients and has 10 grams of protein 5 grams of fiber and 200 calories.
Chicken sandwich with vegetables with pita bread: a saturated meal that can be eaten lightly throughout the day and has 200 calories 3 grams of fiber and more than 20 grams of protein.
Black Lentil Salad: Black lentils are legumes rich in balanced nutrients for the body and with tomatoes and watercress it becomes saturated with benefits and low in calories because it contains 200 calories 11 grams of protein and 8 grams of fiber.
Boiled eggs with helium: contains a terrible amount of protein that the body needs and with the presence of helium this raises its nutritional value and contains 126 calories and 11 grams of protein with 5 grams of fiber.
Iced green tea with mint: This drink helps to burn a large amount of accumulated fat in the body as it raises the rate of burning in the human body and its calories are very few as it carries in its components on antioxidants that help solve digestion problems and protect the body from infections .
– Varied fruit salad where you can cut different types of fruits except grapes figs and mangoes and eat a large plate of it daily. This dish contains a high percentage of natural fibers that help to feel full for a long time as well as a few calories. | 836 | 4,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-40 | longest | en | 0.94844 |
http://blog.baozitraining.org/2014/04/test-blog.html | 1,547,880,688,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583662690.13/warc/CC-MAIN-20190119054606-20190119080606-00308.warc.gz | 28,172,219 | 18,345 | ## Wednesday, April 16, 2014
### Example
Suppose C=500.0, F=4.0 and X=2000.0. Here's how the best possible strategy plays out:
2. After 250 seconds, you will have C=500 cookies and can buy a farm that producesF=4 cookies per second.
3. After buying the farm, you have 0 cookies, and your total cookie production is 6 cookies per second.
4. The next farm will cost 500 cookies, which you can buy after about 83.3333333seconds.
5. After buying your second farm, you have 0 cookies, and your total cookie production is 10 cookies per second.
6. Another farm will cost 500 cookies, which you can buy after 50 seconds.
7. After buying your third farm, you have 0 cookies, and your total cookie production is 14 cookies per second.
8. Another farm would cost 500 cookies, but it actually makes sense not to buy it: instead you can just wait until you have X=2000 cookies, which takes about142.8571429 seconds.
Total time: 250 + 83.3333333 + 50 + 142.8571429 = 526.1904762 seconds.
Notice that you get cookies continuously: so 0.1 seconds after the game starts you'll have 0.2 cookies, and π seconds after the game starts you'll have 2π cookies.
### [代码]
public class CookieClickerAlpha { public static void main(String[] args) { Scanner scanIn = new Scanner(System.in); int testCaseNum = scanIn.nextInt(); int iter = 1; while (iter <= testCaseNum) { double c = scanIn.nextDouble(); double f = scanIn.nextDouble(); double x = scanIn.nextDouble(); double result = leastTime(c, f, x); System.out.println(String.format("Case #%d: %.7f", iter, result)); iter++; } scanIn.close(); } public static double leastTime(double c, double f, double x) { double currentSpeed = 2; double total = 0.0; double orig = 0; double nextTime = 0; while(true) { orig = x / currentSpeed; nextTime = c / currentSpeed + x / (currentSpeed + f); if (orig < nextTime) break; // keep buying farm total += c / currentSpeed; currentSpeed += f; } total += x / currentSpeed; return total; }
### [测试]
Input
Output
4
30.0 1.0 2.0
30.0 2.0 100.0
30.50000 3.14159 1999.19990
500.0 4.0 2000.0
Case #1: 1.0000000
Case #2: 39.1666667
Case #3: 63.9680013
Case #4: 526.1904762
### [Reference]
1. 感谢水中的鱼对包子的帮助，Blog:http://fisherlei.blogspot.com/ | 660 | 2,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-04 | latest | en | 0.840521 |
https://galileoandeinstein.phys.virginia.edu/Elec_Mag/2022_Lectures/EM_13_Complex_Plane.html | 1,670,156,446,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710972.37/warc/CC-MAIN-20221204104311-20221204134311-00100.warc.gz | 301,753,295 | 5,136 | # 13. Math Reminder: Simple Functions in the Complex Plane
## Finding Solutions to Laplace’s Equation
It is an amazing fact that when we take the functions we’ve known since childhood (well, almost):
$f\left(x\right)={x}^{2},\text{ }\sqrt{x},\text{ }\mathrm{log}x,\text{ }\mathrm{sin}x,\dots$
and just replace the real variable $x$ by a complex one $z=x+iy,$ to give a complex function, call it $w\left(z\right),$ of the complex variable $z,$
$w\left(z\right)=w\left(x,y\right)=u\left(x,y\right)+iv\left(x,y\right),$
the (real) functions $u\left(x,y\right),\text{ }v\left(x,y\right)$ satisfy Laplace’s equation in the $\left(x,y\right)$ plane!
That is,
${\nabla }^{2}u\left(x,y\right)=0,\text{ }{\nabla }^{2}v\left(x,y\right)=0.$
These are solutions in search of a problem….
To see how they might be used, consider the simplest function in the list:
${z}^{2}={\left(x+iy\right)}^{2}=\left({x}^{2}-{y}^{2}\right)+i\left(2xy\right).$
The two functions are
$u\left(x,y\right)={x}^{2}-{y}^{2},\text{ }v\left(x,y\right)=2xy.$
It’s easy to check that they do indeed satisfy the 2D Laplace equation, so, to pick one, $v\left(x,y\right)$ could be an electrostatic potential. Let’s plot equipotentials to see where it might be relevant:
$\phi \left(x,y\right)=2xy=\text{constant}\text{.}$
For a positive constant, this is a family of rectangular hyperbolae covering for positive constant the ++ and $–$ sectors of the plane, for negative constant the other two sectors.
Notice this function is zero on the axes, so it describes the potential near a corner where two conducting sheets meet at right angles. For example, this would be a good approximation near one corner inside a rectangular pipe (like a waveguide).
Exercise: show equipotentials for $u\left(x,y\right),$ describe them in words, then state how they relate to those for $v\left(x,y\right).$
## But why do these functions generate potentials in the 2D complex plane?
The reason is simple: they are differentiable: that means there is a well-defined function
${w}^{\prime }\left(z\right)=\frac{dw}{dz}={\mathrm{lim}}_{\Delta z\to 0}\frac{w\left(z+\Delta z\right)-w\left(z\right)}{\Delta z}.$
Why is differentiability such a powerful statement? After all, for a real variable, being differentiable merely means (loosely speaking) being a well-behaved continuous function. Yet in the two dimensions of a complex variable, it clearly means so much more$—$the separate real and imaginary parts must obey these Laplacian equations! The crucial difference from the real case is that in the complex plane $\Delta z\to 0$ can happen in different ways: you could choose to approach the origin along the real axis or along the imaginary axis, and both must give the same result, or, by definition, the function isn’t differentiable.
That is, for $w=u+iv$ with $u\left(z\right),\text{ }v\left(z\right)$ real functions, and $z=x+iy,$
$\frac{\partial w\left(x+iy\right)}{\partial x}=\frac{\partial w\left(x+iy\right)}{\partial \left(iy\right)},$
or, spelling it out,
$\frac{\partial u\left(x,y\right)}{\partial x}+i\frac{\partial v\left(x,y\right)}{\partial x}=\frac{\partial u\left(x,y\right)}{\partial \left(iy\right)}+i\frac{\partial v\left(x,y\right)}{\partial \left(iy\right)}.$
Now equating real and imaginary parts,
$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\text{ }\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y},$
and from this is follows that
$\frac{{\partial }^{2}u\left(x,y\right)}{\partial {x}^{2}}+\frac{{\partial }^{2}u\left(x,y\right)}{\partial {y}^{2}}=0,$
that is,
${\nabla }^{2}u\left(x,y\right)=0,\text{ }{\nabla }^{2}v\left(x,y\right)=0.$
Furthermore, again of course in two dimensions, for
meaning that if we take the potential $\phi \left(x,y\right)=u\left(x,y\right)$ then the lines of constant $v\left(x,y\right)$ are everywhere perpendicular to the $u\left(x,y\right)$ equipotentials, that is, the $v\left(x,y\right)$ equipotentials are the $u\left(x,y\right)$ field lines (and vice versa): their direction at each point is that of the local electric field. For example, the diagram above plots the equipotentials of $u\left(x,y\right)$ and $v\left(x,y\right)$ for the function $w\left(z\right)={z}^{2}.$
## Singularities
The definition of differentiation above can be used to show that
$\frac{d}{dz}\left(\frac{1}{z}\right)=-\frac{1}{{z}^{2}},$
just as for a real variable, so the function $1/z$ can be differentiated everywhere in the complex plane except at the origin. The singularity at the origin is termed a “pole”, for obvious reasons.
## Other Singularities: Cuts, Sheets, etc.
Poles are of course not the only possible singularities. For example, $\mathrm{ln}z$ has a singularity at the origin. Now, $\mathrm{ln}z=\mathrm{ln}r{e}^{i\theta }=\mathrm{ln}r+i\theta .$ The infinite value at the origin is from the $\mathrm{ln}r$ term, but notice that if we go around the unit circle, $\theta$ increases by $2\pi ,$ and if we go around again it increases by a further $2\pi .$ This means that the value of $\mathrm{ln}z$ is not uniquely defined: any given point in the complex plane has values differing by $2\pi ni,\text{ }n$ any integer. This is handled by replacing the single complex plane with a pile of sheets, and a cut going out from the origin. To find $\mathrm{ln}z,$ you need to know not only $z,$ but also which sheet you’re on: going up one sheet means $\mathrm{ln}z$ has increased by $2\pi i.$ When you cross the cut, you go to the next sheet, like a multilevel parking garage.
The cut can go out from the origin in any direction, the standard arrangement is along the real axis, either positive or negative.
The square root function similarly has a cut, but only two sheets.
The above is quoted from my Quantum Mechanics lectures, where there are further details, including contour integration, which we’ll need later in this course.
In electrostatics problems, usually we don’t venture on to other sheets$—$the cut can correspond in position to a physical charged plane, we’ll see some examples. (In other areas of physics, singularities on other sheets can be important: for example, in quantum scattering theory, a resonance corresponds to a pole on a “nonphysical” sheet, and sometimes increasing a coupling constant can bring the pole on to the physical sheet, where it becomes a bound state.)
In the following lectures, we’ll see how familiarity with the properties of some simple functions in the complex plane can lead to quick solutions of some Jackson 2D electrostatics problems. | 1,877 | 6,560 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 50, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2022-49 | latest | en | 0.716545 |
https://www.techylib.com/en/view/conjunctionfriction/relative_kinematics_eidos_engineering | 1,480,702,995,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540409.8/warc/CC-MAIN-20161202170900-00356-ip-10-31-129-80.ec2.internal.warc.gz | 1,038,782,256 | 10,262 | # RELATIVE KINEMATICS - Eidos Engineering
Mechanics
Nov 13, 2013 (3 years and 1 month ago)
62 views
RELATIVE KINEMATICS
The equations of motion for a point
P
will be analyzed in two different reference systems. One
reference system is inertial, fixed to the ground, the second system is moving in the physical space
and the point
P
is also free to move in any configuration of the space (figure 1).
For the moment,
P
is considered without mass, and only the kinematic aspect will be analyzed. This
helps to introduce the very important concept of the angular velocity vector.
If the point
P
is moving in the reference system
S
1
,
u
1
(
t
) is the vector that describes its position at
the time
t
in this system.
P
is also moving respect to the mobile reference system
S
2
and
q
2
(
t
) is the
vector that describes its position in this system.
Now, it’s possible to express the position of
P
by a vector equation where the relation between the
two systems appears. In fact it’s
(
)
(
)
(
)
(
)
t
t
t
t
2
12
1
1
q
R
r
u
+
=
,
(1)
where
r
1
(
t
) is the position of the center
O
2
in the system
S
1
and
R
12
(
t
) is the rotation matrix that
translates a vector from the vector base of
S
2
to the base of
S
1
.
Figure
1
Before to express velocity and acceleration of the point
P
, it’s necessary to consider some important
properties of the rotation matrix.
To give a more simple formulation, in the following sections time dependence will be not always
explicitly indicated.
1
P
O
2
u
1
R
12
q
2
r
1
O
1
S
1
S
2
P
ROPERTIES
OF
THE
ROTATION
MATRIX
For the property of the rotation matrix to be orthogonal, it is
I
R
R
T
12
12
=
.
(2)
When this equation is derived with respect to time we obtain
(
)
0
R
R
R
R
R
R
T
12
12
T
12
12
T
12
12
=
+
=
˙
˙
dt
d
.
(3)
Now, it’s possible to define a skew symmetric matrix
Γ
with the following property:
(
)
Γ
R
R
R
R
R
R
Γ
T
12
12
T
12
12
T
T
12
12
T
=
=
=
=
˙
˙
˙
;
(4)
and the components of this matrix are
3 2
3 1
2 1
0
0
0
= =
T
12 12
Γ R R
˙
.
(5)
The matrix
Γ
defines the angular velocity of the system
S
2
with reference to the
system
S
1
. The
product of the matrix
Γ
with a vector
v
is equal to the vector product between a new vector
ω
and
the vector
v
. In fact it’s possible to express this result as
Γv = ω
×v
.
(6)
ω
is the angular velocity vector and
1
2
3
=
ω
.
(7)
For the second derivative of the equation (2) we obtain by simple operations
2
(
)
(
)
(
)
d
dt
= + = +
T T T T T T
12 12 12 12 12 12 12 12 12 12 12 12
R R R R R R R R R R R R
˙ ˙˙ ˙ ˙ ˙˙ ˙ ˙
.
(8)
It’s important to explicate, from the (8), the term
T
12 12
R R
˙˙
, that it will let us to calculate the
expression for the acceleration of the point
P
in the global reference system. This term is
(
)
(
)
(
)
d
dt
=
T T T T
12 12 12 12 12 12 12 12
R R R R R R R R
˙˙ ˙ ˙ ˙
(9)
and for the equation (5) it’s possible to write
(
)
d
dt
= +
T 2
12 12
R R
Γ Γ
˙˙
(10)
and for the same argument of the (6), if
v
is a vector defined in the global reference system, we have
the following vector operation:
(
)
(
)
(
)
d
dt
= +
T 2
12 12
R R v
Γ Γ v = ω
×v +
ω
×
ω
×v
˙˙
˙
.
(11)
3
V
ELOCITY
AND
ACCELERATION
By differentiating the fundamental equation (1) with respect to time, we calculate velocity and
acceleration of the point
P
. For the velocity of the point in the inertial reference system we have
(
)
(
)
(
)
(
)
(
)
(
)
t t t t t t
= + +
1 1 12 2 12 2
u r R q R q
˙
˙ ˙
˙
.
(12)
It’s more interesting to write this equation using the following expression where three velocity
vectors compare:
r
τ
a
v
v
v
+
=
.
(13)
v
a
is the absolute velocity, i.e., the velocity of the point
P
measured with reference to the global
reference system and it corresponds to the vector
(
)
t
=
a 1
v u
˙
;
(14)
v
r
is the relative velocity of the point
P
, i.e. the velocity of the point measured in the mobile
reference system, but expressed using the versors of the global reference. In fact, using the rotation
matrix between the two reference systems, we have
(
)
(
)
t t
=
r 12 2
v R q
˙
;
(15)
v
τ
is the drag velocity, that is the velocity of the point P as if it was fixed to the mobile reference
system and moving with it:
(
)
(
)
(
)
t t t
= +
τ 1 12 2
v r R q
˙
˙
.
(16)
It’s interesting to write down the drag velocity as
(
)
2
2
O
P
+
=
ω
v
v
O
τ
.
(17)
Such formula expresses the motion law for a rigid body when it’s fixed with the mobile reference
system. For the equations (1) and (12) we obtain
4
)
r
(u
ω
r
)
r
(u
R
R
r
q
R
r
v
1
1
1
1
1
T
12
12
1
2
12
1
τ
+
=
+
=
+
=
˙
˙
˙
˙
˙
.
(18)
Differentiating the equation (12) with respect to time, for the acceleration of the point
P
, we have
(
)
2
12
2
12
2
12
1
1
q
R
q
R
2
q
R
r
u
˙
˙
˙
˙
˙
˙
˙
˙
˙
˙
+
+
+
=
.
(19)
We can write this equation as a sum of three accelerations:
r
c
τ
a
a
a
a
a
+
+
=
.
(20)
a
a
is the absolute acceleration, i.e., the acceleration of the point
P
measured with reference to the
global reference system and it corresponds to the vector
(
)
t
=
a 1
a u
˙˙
;
(21)
τ
a
is the drag acceleration, and for the equation (10) it’s possible to write down
(
)
(
)
(
)
1
1
2
1
1
1
1
1
T
12
12
1
2
12
1
τ
r
u
Γ
r
u
Γ
r
r
u
R
R
r
q
R
r
a
+
+
=
+
=
+
=
˙
˙
˙
˙
˙
˙
˙
˙
˙
˙
˙
,
(22)
(
)
(
)
[
]
1
1
1
1
1
τ
r
u
ω
ω
r
u
ω
r
a
+
+
=
˙
˙
˙
,
(23)
(
)
(
)
[
]
2
2
O
P
O
P
+
+
=
ω
ω
ω
a
a
2
O
τ
˙
;
(24)
r
a
is the relative acceleration, i.e., the acceleration of the point P referred to the mobile reference
system but expressed by the versors of the inertial:
2
12
r
q
R
a
˙
=
;
(25)
c
a
is the Coriolis acceleration defined in the global reference system as
(
)
(
)
= = = =
T
c 12 2 12 12 12 2 r r
a 2R q 2 R R R q 2
Γv 2ω v
˙ ˙
˙ ˙
.
(26)
5 | 2,289 | 5,712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.882645 |
https://cse331.wordpress.com/2009/12/09/lect-38-shortest-paths-ii/ | 1,529,508,908,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863650.42/warc/CC-MAIN-20180620143814-20180620163814-00557.warc.gz | 582,199,512 | 17,474 | Posted by: atri | December 9, 2009
## Lect 38: Shortest Paths-II
(Guest post by John Longanecker)
The last several classes we have been learning about dynamic programming. Richard Bellman was the one who created dynamic programming. I find it interesting to read up on people we have talked about in class. So go check him out.
The basic goal of dynamic programming is to solve a bunch of smaller problems and assemble the results to find the solution of a larger problem. To compute a problem quickly we do not compute already computed problems. An example of this is to how we used an array to help us with the weighted interval scheduling problem.
The problem that we worked on solving in today’s lecture was the shortest path problem using dynamic programming (6.8). We are given a graph $G$ and asked to find the shortest path between two nodes $s$ and $t$. It is also important to note that we are not going to use graphs with negative cycles. Negative cycles brings up a whole other problem and we want to focus on the case of a graph $G$ with no negative cycles.
In order to calculate this problem we took a function $OPT(i,v)$ = the shortest path from node $v$ to $t$ using at most $i$ edges. For example adding more edges does not necessarily mean the path will be longer. We could have negative weight edges or the route with more edges could have a smaller set of edge weights.
Lemma:
If $G$ has no negative cycle, there is a shortest path has no repeated vertices.
This implies there is a shortest path with at most $n-1$ edges.
We constructed a table where the rows represented the nodes we are starting at and the columns represented the maximum allowed number of edges ($i$) used to traverse from the starting node $v$ to $t$ using at most $i$ edges. Each cell will be populated with value $OPT(i,v)$ returns. If we look at the highest value of $n-1$ (most number of edges) we will always find the optimal value of the shortest path for the particular node we are interested in.
Shortest path is not simple (it has a cycle) we will call it $P_1$. By assumption the cycle is not negative (it has to be at least $0$)
Contradiction: So that being said we could remove the cycle from the path $P_1$ so that we have a new path $P_2$.
$\mathrm{cost}(P_2) \le \mathrm{cost} P_1$ it will never be larger (because we already decided that there will be no negative cycles in $G$).
If the shortest path uses only $i-1$ edges, $OPT(i,v)$ will still return the optimal path weight, but it will not be any smaller then $i-1$. | 607 | 2,540 | {"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": 25, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-26 | latest | en | 0.93805 |
http://stevenstolman.com/error-analysis/error-and-error-analysis.html | 1,503,188,906,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105955.66/warc/CC-MAIN-20170819235943-20170820015943-00699.warc.gz | 398,821,582 | 5,750 | Home > Error Analysis > Error And Error Analysis
# Error And Error Analysis
## Contents
The meaning of this is that if the N measurements of x were repeated there would be a 68% probability the new mean value of would lie within (that is between For a sufficiently a small change an instrument may not be able to respond to it or to indicate it or the observer may not be able to discern it. To get some insight into how such a wrong length can arise, you may wish to try comparing the scales of two rulers made by different companies — discrepancies of 3 By declaring lists of {value, error} pairs to be of type Data, propagation of errors is handled automatically. http://stevenstolman.com/error-analysis/error-analysis-immunochemistry-error-analysis.html
It is never possible to measure anything exactly. Thus, 400 indicates only one significant figure. Although it is not possible to do anything about such error, it can be characterized. The object of a good experiment is to minimize both the errors of precision and the errors of accuracy. http://teacher.nsrl.rochester.edu/phy_labs/AppendixB/AppendixB.html
## Error Propagation
All measuring instruments are limited by how precise they are. Much of the material has been extensively tested with science undergraduates at a variety of levels at the University of Toronto. Rule 3: Raising to a Power If then or equivalently EDA includes functions to combine data using the above rules. See also Error (linguistics) Error treatment (linguistics) Second language acquisition Notes ^ Cf.
The following Hyperlink points to that document. If the Philips meter is systematically measuring all voltages too big by, say, 2%, that systematic error of accuracy will have no effect on the slope and therefore will have no Why spend half an hour calibrating the Philips meter for just one measurement when you could use the Fluke meter directly? Error Analysis Chemistry An important and sometimes difficult question is whether the reading error of an instrument is "distributed randomly".
This last line is the key: by repeating the measurements n times, the error in the sum only goes up as Sqrt[n]. Percent Error Rather, it will be calculated from several measured physical quantities (each of which has a mean value and an error). After multiplication or division, the number of significant figures in the result is determined by the original number with the smallest number of significant figures. A comprehensive bibliography was published by Bernd Spillner (1991), Error Analysis, Amsterdam/Philadelphia: Benjamins. ^ Corder, S.
And so it is common practice to quote error in terms of the standard deviation of a Gaussian distribution fit to the observed data distribution. Error Analysis Formula Thus, any result x[[i]] chosen at random has a 68% change of being within one standard deviation of the mean. So in this case and for this measurement, we may be quite justified in ignoring the inaccuracy of the voltmeter entirely and using the reading error to determine the uncertainty in Thus, we can use the standard deviation estimate to characterize the error in each measurement.
## Percent Error
sumx = x1 + x2 + ... + xn We calculate the error in the sum. http://sciencefair.math.iit.edu/writing/error/ Two questions arise about the measurement. Error Propagation The uncertainty in a measurement arises, in general, from three types of errors. Error Analysis Equation It should be noted that since the above applies only when the two measured quantities are independent of each other it does not apply when, for example, one physical quantity is
In[4]:= In[5]:= Out[5]= We then normalize the distribution so the maximum value is close to the maximum number in the histogram and plot the result. this content If n is less than infinity, one can only estimate . Another advantage of these constructs is that the rules built into EDA know how to combine data with constants. For example, (10 +/- 1)2 = 100 +/- 20 and not 100 +/- 14. Error Analysis Physics
Does it mean that the acceleration is closer to 9.8 than to 9.9 or 9.7? Also, when taking a series of measurements, sometimes one value appears "out of line". These are reproducible inaccuracies that are consistently in the same direction. http://stevenstolman.com/error-analysis/error-analysis-sla.html Discussion of the accuracy of the experiment is in Section 3.4. 3.2.4 Rejection of Measurements Often when repeating measurements one value appears to be spurious and we would like to throw
Thus we have = 900/9 = 100 and = 1500/8 = 188 or = 14. Error Analysis Linguistics Bork, H. Some could be developmental—errors most learners make in learning this language no matter what their native language.
## In the theory of probability (that is, using the assumption that the data has a Gaussian distribution), it can be shown that this underestimate is corrected by using N-1 instead of
In fact, as the picture below illustrates, bad things can happen if error analysis is ignored. A key finding of error analysis has been that many learner errors are produced by learners making faulty inferences about the rules of the new language. Send comments, questions and/or suggestions via email to [email protected] Error Analysis Language Error Analysis Introduction The knowledge we have of the physical world is obtained by doing experiments and making measurements.
This can help researchers understand the cognitive processes the learner is using, and help teachers decide which might be targeted for correction. Please try the request again. The only problem was that Gauss wasn't able to repeat his measurements exactly either! check over here They often seek to develop a typology of errors.
Communication strategies may be used by the learner to get meaning across even if he or she knows the form used is not correct (Selinker 1972 discusses these and other possible Systematic errors cannot be detected or reduced by increasing the number of observations, and can be reduced by applying a correction or correction factor to compensate for the effect. For example, one could perform very precise but inaccurate timing with a high-quality pendulum clock that had the pendulum set at not quite the right length. Standard Deviation The mean is the most probable value of a Gaussian distribution.
The amount of drift is generally not a concern, but occasionally this source of error can be significant and should be considered. Cambridge University Press, 1993. In complicated experiments, error analysis can identify dominant errors and hence provide a guide as to where more effort is needed to improve an experiment. 3. A series of measurements taken with one or more variables changed for each data point. | 1,377 | 6,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-34 | latest | en | 0.905329 |
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