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https://docsbay.net/inquiry-into-refraction | 1,627,149,822,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150307.84/warc/CC-MAIN-20210724160723-20210724190723-00123.warc.gz | 219,746,427 | 5,407 | # Inquiry Into Refraction
Unit / Properties of Light
Enduring Understanding / 1. Humans use science to organize their understanding of the natural
world.
2. Our universe is composed of various forms of matter and energy that exist, interact, and change in a variety of ways.
6. Patterns and cycles exist in nature
SOL Objectives
Title / Inquiry Lab: Refraction of Light
Lesson Objective / This lab is designed to hit on a QUALATATIVE property of light refraction. Students will be able to:
1. Observe how the path of a light ray changes when it travels from one medium to another
2. Compare the light bending properties of different liquids
Inquiry Level / 2
Materials Required /
• Semicircular transparent dish
• Different liquids (water, baby oil, corn oil, etc)
• Laser ray box
• Diagram sheets (idealy graph paper – 2-3 per group)
• Protractor
• Ruler
Light Properties: Refraction
Name: ______Pd: ______Date: ______
Inquiry Lab: Refraction of Light
Objectives:
• To observe how the path of a light ray changes when it travels from one medium to another
• To compare the light bending properties of different liquids
Materials:
• Semicircular transparent dish
• Different liquids (water, baby oil, corn oil)
• Laser ray box
• Diagram sheets
• Protractor
• Ruler
Procedure:
1. Label your first sheet of paper Activity 1. Place the semicircular dish on the diagram paper.
1. Point the single laser beam at the dish, along the normal (dashed line) to the flat side of the dish. This is an angle of zero degrees. Trace the incoming beam (called the incident ray), label it “1”
1. Trace the beam WITH A RULER that leaves the dish on the curved side (called the refracted ray). Label it “IV 1”
(instructions continue next page)
1. For each trial, increase the angle the incident ray makes with the normal by 10 degrees. Trace the incident ray WITH A RULER and the refracted ray. Label each line (IV 1, IV 2, etc).
1. Measure the angle the refracted rays make with the normal and record in Table 1 (below).
Table 1 – Light passing from air into water
IV Level / Angle in Air
Incident Angle, i
(degrees) / Angle in Water
Refraction angle, r
(degrees)
1 / 0.0
2 / 10.0
3 / 20.0
4 / 30.0
5 / 40.0
Observations:
1. As the angle of the incident ray was increased, what happened to the angle the refracted ray made with the normal?
1. Did the angle of refraction increase at the same rate as the angle of incidence? Explain.
1. What happens to the refraction angle if you put the laser on the other side of the normal line? (10 degrees away, 20 degrees away, etc). Make a conclusion from your observations.
1. Prediction: Will this observed behavior from Activity 1 continue beyond the angle of incidence of 40 degrees? Why or why not?
1. Test your prediction. Gradually increase the angle of incidence. Locate the refracted ray as you keep increasing the angle of incidence. Was your prediction correct? Explain.
1. Label your second sheet of paper Activity 3. Place the semicircular dish on the diagram paper.
1. Point the single laser beam at the dish, along the normal (dashed line) to the curved side of the dish. This is an angle of zero degrees. Trace the incoming beam (called the incident ray), label it “IV 1”
1. Trace the beam that leaves the dish on the flat side (called the refracted ray). Label it “1”
1. For each trial, increase the angle the incident ray makes with the normal by 10 degrees. You may have to adjust the angle of the laser beam so that it hits the flat side of the dish at the point where the normal is drawn. Trace the incident ray and the refracted. Label each line.
1. Measure the angle the refracted rays make with the normal and record in Table 2 (next page).
Table 2 - Light passing from water to air
Trial / Angle in Water
Incident Angle, i
(degrees) / Angle in Air
Refraction angle,
(degrees)
1 / 0.0
2 / 10.0
3 / 20.0
4 / 30.0
5 / 40.0
Observations:
1. As the angle of the incident ray was increased, what happened to the angle the refracted ray made with the normal?
1. Did the angle of refraction increase at the same rate as the angle of incidence? Explain.
1. Prediction: If a different substance is used instead of water in the dish, how will the results of activities 1-4 be different? Explain and be specific.
1. Redo Activity 1 with two other substances and summarize your results below. (You may draw on the same “activity 1” paper if you use two different colored pencils to reflect each substance. Otherwise, get a new piece and label it Activity 4.)
Turn in this lab packet, and staple your ray drawings to the back of the packet. | 1,144 | 4,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | longest | en | 0.831753 |
https://www.crazy-numbers.com/en/32274 | 1,542,360,017,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743007.0/warc/CC-MAIN-20181116091028-20181116113028-00191.warc.gz | 839,113,252 | 4,428 | Discover a lot of information on the number 32274: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 32274
Is 32274 a prime number? No
Is 32274 a perfect number? No
Number of divisors 24
List of dividers 1, 2, 3, 6, 9, 11, 18, 22, 33, 66, 99, 163, 198, 326, 489, 978, 1467, 1793, 2934, 3586, 5379, 10758, 16137, 32274
Sum of divisors 76752
## How to write / spell 32274 in letters?
In letters, the number 32274 is written as: Thirty-two thousand two hundred and seventy-four. And in other languages? how does it spell?
32274 in other languages
Write 32274 in english Thirty-two thousand two hundred and seventy-four
Write 32274 in french Trente-deux mille deux cent soixante-quatorze
Write 32274 in spanish Treinta y dos mil doscientos setenta y cuatro
Write 32274 in portuguese Trinta e dois mil duzentos setenta e quatro
## Decomposition of the number 32274
The number 32274 is composed of:
1 iteration of the number 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3
2 iterations of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
1 iteration of the number 7 : The number 7 (seven) represents faith, teaching. It symbolizes reflection, the spiritual life.... Find out more about the number 7
1 iteration of the number 4 : The number 4 (four) is the symbol of the square. It represents structuring, organization, work and construction.... Find out more about the number 4
Other ways to write 32274
In letter Thirty-two thousand two hundred and seventy-four
In roman numeral
In binary 111111000010010
In octal 77022
In US dollars USD 32,274.00 (\$)
In euros 32 274,00 EUR (€)
Some related numbers
Previous number 32273
Next number 32275
Next prime number 32297
## Mathematical operations
Operations and solutions
32274*2 = 64548 The double of 32274 is 64548
32274*3 = 96822 The triple of 32274 is 96822
32274/2 = 16137 The half of 32274 is 16137.000000
32274/3 = 10758 The third of 32274 is 10758.000000
322742 = 1041611076 The square of 32274 is 1041611076.000000
322743 = 33616955866824 The cube of 32274 is 33616955866824.000000
√32274 = 179.64965905896 The square root of 32274 is 179.649659
log(32274) = 10.382017231501 The natural (Neperian) logarithm of 32274 is 10.382017
log10(32274) = 4.5088527946655 The decimal logarithm (base 10) of 32274 is 4.508853
sin(32274) = -0.40654538486188 The sine of 32274 is -0.406545
cos(32274) = -0.91363058729855 The cosine of 32274 is -0.913631
tan(32274) = 0.44497786141767 The tangent of 32274 is 0.444978 | 869 | 2,765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-47 | longest | en | 0.743975 |
https://acikerisim.aku.edu.tr/xmlui/handle/11630/4629?show=full | 1,624,149,944,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487653461.74/warc/CC-MAIN-20210619233720-20210620023720-00385.warc.gz | 96,081,482 | 8,629 | dc.contributor.author ÜZAR, Neslihan dc.date.accessioned 2017-10-11T10:30:24Z dc.date.available 2017-10-11T10:30:24Z dc.date.issued 2017 dc.identifier.issn 2149-3367 dc.identifier.uri http://fenbildergi.aku.edu.tr/wp-content/uploads/2017/10/021103-415-425.pdf dc.identifier.uri http://hdl.handle.net/11630/4629 dc.description.abstract In this study, the fractional Sine-Gordon (SG) equations (time-fractional, space-fractional and timespace- en_US fractional) are solved using Homotopy Perturbation Method (HPM). The crucial point is the attained remarkable result from these solutions. While the solutions of classical and time-fractional SG equations are kink of type (Although of being the same type, they are different from each other), solution of the space-fractional SG equation is breather of type i.e., different types of soliton solutions are obtained using similar initial conditions for time and space fractional SG equation. Also these results show that some events such as vortex-antivortex couples in a Josephson junction or losses in signal dispersion of fiber optics communication can be modelled by fractional SG equations. In other words, this study may be very important for bringing to light the real behaviour of physical systems which have usually been described by classical SG equation. Because some physical events such as the memory effects of non-Markovian processes, the effects of non-Gaussian distribution, interactions between the systems and environment and some physical losses in the systems which are neglected in classical SG equation can be taken into account with fractional SG equations. dc.description.abstract Bu çalışmada kesirli Sine-Gordon denklemleri (zaman-kesirli, uzay-kesirli ve zaman-uzay-kesirli) en_US homotopy pertürbasyon metodu (HPM) kullanılarak çözülmüştür. Bu çözümlerden dikkat çekici sonuçlar elde edilmiştir. Klasik ve zaman-kesirli SG denklemlerinin çözümü kink tipi iken (aynı tipte olmalarına rağmen, birbirlerinden farklıdır), uzay-kesirli SG denkleminin çözümü breather tiptir; başka bir ifadeyle zaman ve uzay kesirli SG denklemi için benzer başlangıç koşulları kullanıldığında farklı tip soliton çözümleri elde edilmiştir. Ayrıca bu sonuçlar josephson eklemlerindeki vorteks-antivorteks çifleri veya fiber optik iletişimde sinyal dağılımındaki kayıplar gibi bazı olayları kesirli SG denklemleri ile modelleyebileceğini göstermiştir. Diğer bir deyişle, bu çalışma genellikle klasik SG denklemleri ile tanımlanan fiziksel sistemlerin gerçek davranışlarına ışık tutabilir. Çünkü Markovian olmayan süreçlerin bellek etkileri, Gaussian olmayan dağılımların etkileri, sistem ile dış çevre arasındaki etkileşmeler ve klasik SG denkleminin ihmal ettiği sistemler içerisindeki bazı fiziksel kayıplar gibi bazı fiziksel olaylar kesirli SG denklemleri ile hesaba katılabilir. dc.language.iso eng en_US dc.publisher Afyon Kocatepe Üniversitesi, Fen ve Mühendislik Bilimleri Dergisi en_US dc.identifier.doi 10.5578/fmbd.57216 en_US dc.rights info:eu-repo/semantics/openAccess en_US dc.subject Caputo fractional derivative operator en_US dc.subject HPM en_US dc.subject SG equation en_US dc.subject fractional non-linear equation en_US dc.title Approximate Soliton Solutions of Real Order Sine- Gordon Equations en_US dc.title.alternative Reel Mertebeli Sine-Gordon Denklemlerinin Yaklaşık Soliton Çözümleri en_US dc.type article en_US dc.relation.journal Fen ve Mühendislik Bilimleri Dergisi en_US dc.department Istanbul University, Science Faculty, Department of Physics en_US dc.identifier.volume 17 en_US dc.identifier.startpage 415 en_US dc.identifier.endpage 425 en_US dc.identifier.issue 2 en_US dc.relation.publicationcategory Makale - Ulusal Hakemli Dergi - Kurum Yayını en_US
| 928 | 3,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.749519 |
https://webob.info/relationship/female-family-relationship-tree.php | 1,568,667,579,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572934.73/warc/CC-MAIN-20190916200355-20190916222355-00248.warc.gz | 713,082,828 | 9,453 | # Female family relationship tree
### Niece and nephew - Wikipedia
(Lord Dark Helmet and Lone Starr, discussing family relationships in the this fattens the family tree, creating new relationships like sister and niece . ants the males have only half as much genetic information as females. A genogram is composed of a series of symbols representing male, female, marriage, A genogram is a graphic representation of a family tree that displays the the user to analyze family, emotional and social relationships within a group. Find family tree Stock Images in HD and millions of other royalty-free stock photos, illustrations, and vectors in the Shutterstock collection. Thousands of new .
A family is shown by an horizontal line connecting the two. The children are placed below the family line from the oldest to the youngest, left to right. Although this may sound obvious, it will be very important to remember these rules when the situation becomes a bit more complex. The following case is a husband with three spouses.
## Symbols used in Genogram
The husband had three children with the first wife, then divorced. The husband married the second wife, had one child and separated.
The husband currently lives with another woman. A Husband with Multiple Spouses Reversing the scenario where the wife had multiple husbands, we get the following genogram: A Wife with Multiple Husbands The second scenario is the same as the first one, except the female spouse had three husbands.
She had three children with her first husband and divorced.
### The mathematics of your next family reunion | webob.info
The wife married another man, had one child with him and now lives with someone else. Please notice the oldest child is always at the left most position of the family of his biological parents.
In this scenario, the "Oldest Brother" is older than the twins and the half sister, however the half-sister must be placed under the family of her biological parents. Since the second marriage is after the first marriage, it follows that the half sister is younger than the children from the first marriage.
## The mathematics of your next family reunion
The half sister therefore appears to the left, although she is not the oldest child. If you are confused, please read this paragraph again. In summary, here are the four rules to build a genogram: The male parent is always at the left of the family and the female parent is always at the right of the family.
In the case of ambiguity, assume a male-female relationship, rather than male-male or female-female relationship. The oldest child is always at the left his family, the youngest child is always at the right his family. To simplify your genealogy layout, it is acceptable to swap the husband and wife as long as there is only a single family involved.
There is no ambiguity to have the wife at the left position, as long as each spouse have had only one partner. The four rules are there to remove ambiguity. Possible ambiguity if rule 2 is ignored By ignoring rule 2, we could interpret the scenario of a wife with multiple husbands as a man who had two male partners and one female partner.
The figure above has the same arrangement as a wife with multiple husbandsexcept the children have been removed and the families are all colored black. As you can see, it could be possible Max and Carl were together, then Max lived with Joe and finally Max lived with Kathy. By following rule 2the Joe-Kathy relationship has precedence over the Max-Carl relationship.
It follows Carl is Kathy's second husband, and Max the third husband. Wrong Genogram Layout The following figure represent a wrong genogram layout for Kathy with three husbands. Wrong genogram layout for Kathy having three husbands The idea here is to show Kathy had three husbands, however there is a problem with the figure: The problem is that rule 3 has not been followed.
By looking carefully at this genogram, we get a completely different outcome. Joe and Kathy have never been together because it breaks rule 1. Carl was Kathy's first husband rule 1 and rule 3. Carl left Kathy and decided to live with Joe rule 3. In the case of a miscarriage, there is a diagonal cross drawn on top of the triangle to indicate death.
### Genogram Symbols - GenoPro
Abortions have a similar display to miscarriages, only they have an additional horizontal line. A still birth is displayed by the gender symbol; the diagonal cross remains the same size, but the gender symbol is twice as small.
Genogram Symbols for Children's Links and Pregnancy Terminations In the case of multiple births such as twins, triplets, quadruplets, quintuplets, sextuplets, septuplets, octuplets, or more, the child links are joined together. GenoPro uses the term twin to describe any type of multiple birth. With GenoPro, creating twins is as simple as a single click on the toolbar button "New Twins". GenoPro take cares of all the drawing, including joining the lines together.
Identical twins or triplets In the example below, the mother gave birth to fraternal twin brothers, identical twin sisters and triplets, one of whom died at birth. Child links are joined for multiple births such as twins and triplets In addition to this, GenoPro supports medical genograms by using color codes and special drawing in the gender symbol. To learn more, please visit medical genograms. | 1,101 | 5,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | longest | en | 0.974825 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-mathematics-for-calculus-7th-edition/chapter-1-section-1-8-inequalities-1-8-exercises-page-89/33 | 1,539,591,064,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583508988.18/warc/CC-MAIN-20181015080248-20181015101748-00163.warc.gz | 951,781,667 | 15,082 | ## Precalculus: Mathematics for Calculus, 7th Edition
$[4.5, 5)$ Refer to the attached image for the graph.
Subtract 8 to each part to have: $\\-2-8 \lt 8-2x-8 \le -1-8 \\-10 \lt -2x \le -9$ Divide $-2$ to each part. Note that since you are dividing a negative to the inequality, the inequality symbols will flip. Thus, $\\\frac{-10}{-2} \gt x \gt \frac{-9}{-2} \\5 \gt x \ge 4.5 \\4.5 \le x \lt 5$ | 152 | 399 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-43 | longest | en | 0.762093 |
http://www.justanswer.com/homework/92mtk-implement-simplified-bridge-game-betting-user.html | 1,492,978,267,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118743.41/warc/CC-MAIN-20170423031158-00449-ip-10-145-167-34.ec2.internal.warc.gz | 587,138,676 | 19,334 | • 100% Satisfaction Guarantee
Category: Homework
Satisfied Customers: 3040
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# Implement a simplified Bridge game with betting between 1 user
### Customer Question
Implement a simplified Bridge game with betting between 1 user player and the computer 3 AI agents for 2 games. Our Contract Bridge game is a trick-taking game using a standard 52-card deck without any joker.
The players will be positioned: down (the user), left, up, right (the 3 AI players). The player is paired with the up AI player, while the left and right AI players are grouped together.
The computer deals out all the cards equally to all the players, so that each player receives a hand of thirteen cards. The user can only see his own cards and none of the other AI players hands.
The user then looks at his cards and starts playing the game:
- the player selects a card and puts it in the middle face-up;
- each player will play clockwise (left player first, up player next, down player last) one card face-up that must "follow suit" (that is, play a card of the suit lead to the trick) if able. A player that cannot follow suit may play "sluff" (i.e., play a card of any other suit). The player that plays the highest card that matches the suite of the starting card wins 1 point the trick for its side and proceeds to lead to the next trick.
At the end of the game you should print if the user's team won or lost.
Submitted: 1 year ago.
Category: Homework
Customer: replied 1 year ago.
that's how should be look like.
Customer: replied 1 year ago.
Sample run:
Expert: Mr. Gregory White replied 1 year ago.
Hello, my name is ***** ***** this might be time sensitive.Do you still need this answered?
Customer: replied 1 year ago.
yes i still need
Expert: Mr. Gregory White replied 1 year ago.
After going through my resources, I do not have what is necessary to complete at this time.I am opting out and opening up to the other professionals and messaging a couple who might be able to help. Someone should be with you shortly. | 485 | 2,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.946834 |
https://www.customcoils.com/blog/getting-literate-about-flyback-transformers/ | 1,719,312,914,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865972.21/warc/CC-MAIN-20240625104040-20240625134040-00033.warc.gz | 637,475,964 | 13,005 | The electrical industry has come a long way and has seen several advancements in the equipment, components, devices, and technologies used. Transformers are such devices. There are several different types of transformers used today in a variety of applications. Step-down, step-up, and flyback transformers are a few examples of transformers used in the electrical industry. What are Flyback transformers? What are they used for? This post will shed some light upon the design, operation, and applications of flyback transformers. Read on to know more.
What are Flyback Transformers?
Flyback transformers are a type of electrical transformer, which are intended to produce high voltage signals at a comparatively high frequency. These signals are typically sawed tooth-shaped. Most of the time, these transformers are also referred to as line output transformers (LOPT). The main purpose of inventing these transformers was to make it easier to control the movement of electron beams in a cathode ray tube (CRT). Nowadays, these transformers are being used largely in switching mode power supplies (SMPS). The line output transformers can be used in both, the high voltage (above 10 kV) as well as low voltage (3 V) supplies.
How Do Flyback Transformers Operate?
The operation of Flyback transformers involves two or three stages that are repeated. These three stages of operation are as follows:
1. Charging Stage – The line output transformer uses power source to draw current. The current goes on increasing and causes the magnetic field flux to increase. The magnetic field helps to store the energy. The energy stored is given by the following equation:
E = (I x I x L) / 2
Where,
E = Stored Energy
I = Current in Amps
L = Inductance in Henry
The current is calculated by the following differential equation:
V(t) = L x di/dt
1. Discharging Stage – When a switch is opened, it interrupts the current from the source. This causes the magnetic flux to decrease, which results in negative flux change. A voltage is induced in the opposite direction than that induced during the charging stage. The reversed voltage induces current flow. However, this current is not allowed to flow through power source by the open switch. The decrease in the magnetic field is slowed down by the induced current, which results in a lower induced flyback voltage.
1. Idle Stage – As the name suggests, this is the stage in which the input current, as well as the output current of the transformer or inductor is at zero value. This occurs when the energy stored in the transformer is completely discharged.
Where are the Flyback Transformers Used?
There are a number of areas, where flyback transformers are used on a large scale. This is because of the several benefits that they offer.
1. Communication Devices
2. Aeronautical Appliances
3. Cathode Ray Tubes
4. Electrical Converters and Inverters
This was just an insight into flyback transformers that you need to know before actually purchasing one for your applications. For more information on the topic, you can always consult our engineering that have experience and expertise in transformers manufacturing. If you are looking for an experienced, reliable, and well-known manufacturer, then you should consult Custom Coils. The company has a team of skilled staff, who is always ready to help you with all your application requirements related to flyback transformers. | 684 | 3,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-26 | latest | en | 0.952184 |
https://calculationcalculator.com/pole-to-gaj | 1,721,461,334,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515020.57/warc/CC-MAIN-20240720052626-20240720082626-00290.warc.gz | 131,032,792 | 22,578 | # Pole to Gaj Conversion
## 1 Pole is equal to how many Gaj?
### 1210 Gaj
##### Reference This Converter:
Pole and Gaj both are the Land measurement unit. Compare values between unit Pole with other Land measurement units. You can also calculate other Land conversion units that are available on the select box, having on this same page.
Pole to Gaj conversion allows you to convert value between Pole to Gaj easily. Just enter the Pole value into the input box, the system will automatically calculate Gaj value. 1 Pole in Gaj? In mathematical terms, 1 Pole = 1210 Gaj.
To conversion value between Pole to Gaj, just multiply the value by the conversion ratio. One Pole is equal to 1210 Gaj, so use this simple formula to convert -
The value in Gaj is equal to the value of Pole multiplied by 1210.
Gaj = Pole * 1210;
For calculation, here's how to convert 10 Pole to Gaj using the formula above -
10 Pole = (10 * 1210) = 12100 Gaj
Pole Gaj Conversion
0.001 1.21 0.001 Pole = 1.21 Gaj
0.002 2.42 0.002 Pole = 2.42 Gaj
0.003 3.63 0.003 Pole = 3.63 Gaj
0.004 4.84 0.004 Pole = 4.84 Gaj
0.005 6.05 0.005 Pole = 6.05 Gaj
0.006 7.26 0.006 Pole = 7.26 Gaj
0.007 8.47 0.007 Pole = 8.47 Gaj
0.008 9.68 0.008 Pole = 9.68 Gaj
0.009 10.89 0.009 Pole = 10.89 Gaj
0.01 12.1 0.01 Pole = 12.1 Gaj
0.02 24.2 0.02 Pole = 24.2 Gaj
0.03 36.3 0.03 Pole = 36.3 Gaj
0.04 48.4 0.04 Pole = 48.4 Gaj
0.05 60.5 0.05 Pole = 60.5 Gaj
0.06 72.6 0.06 Pole = 72.6 Gaj
0.07 84.7 0.07 Pole = 84.7 Gaj
0.08 96.8 0.08 Pole = 96.8 Gaj
0.09 108.9 0.09 Pole = 108.9 Gaj
0.1 121 0.1 Pole = 121 Gaj
0.2 242 0.2 Pole = 242 Gaj
0.3 363 0.3 Pole = 363 Gaj
0.4 484 0.4 Pole = 484 Gaj
0.5 605 0.5 Pole = 605 Gaj
0.6 726 0.6 Pole = 726 Gaj
0.7 847 0.7 Pole = 847 Gaj
0.8 968 0.8 Pole = 968 Gaj
0.9 1089 0.9 Pole = 1089 Gaj
1 1210 1 Pole = 1210 Gaj
2 2420 2 Pole = 2420 Gaj
3 3630 3 Pole = 3630 Gaj
4 4840 4 Pole = 4840 Gaj
5 6050 5 Pole = 6050 Gaj
6 7260 6 Pole = 7260 Gaj
7 8470 7 Pole = 8470 Gaj
8 9680 8 Pole = 9680 Gaj
9 10890 9 Pole = 10890 Gaj
10 12100 10 Pole = 12100 Gaj
11 13310 11 Pole = 13310 Gaj
12 14520 12 Pole = 14520 Gaj
13 15730 13 Pole = 15730 Gaj
14 16940 14 Pole = 16940 Gaj
15 18150 15 Pole = 18150 Gaj
16 19360 16 Pole = 19360 Gaj
17 20570 17 Pole = 20570 Gaj
18 21780 18 Pole = 21780 Gaj
19 22990 19 Pole = 22990 Gaj
20 24200 20 Pole = 24200 Gaj
21 25410 21 Pole = 25410 Gaj
22 26620 22 Pole = 26620 Gaj
23 27830 23 Pole = 27830 Gaj
24 29040 24 Pole = 29040 Gaj
25 30250 25 Pole = 30250 Gaj
26 31460 26 Pole = 31460 Gaj
27 32670 27 Pole = 32670 Gaj
28 33880 28 Pole = 33880 Gaj
29 35090 29 Pole = 35090 Gaj
30 36300 30 Pole = 36300 Gaj
31 37510 31 Pole = 37510 Gaj
32 38720 32 Pole = 38720 Gaj
33 39930 33 Pole = 39930 Gaj
34 41140 34 Pole = 41140 Gaj
35 42350 35 Pole = 42350 Gaj
36 43560 36 Pole = 43560 Gaj
37 44770 37 Pole = 44770 Gaj
38 45980 38 Pole = 45980 Gaj
39 47190 39 Pole = 47190 Gaj
40 48400 40 Pole = 48400 Gaj
41 49610 41 Pole = 49610 Gaj
42 50820 42 Pole = 50820 Gaj
43 52030 43 Pole = 52030 Gaj
44 53240 44 Pole = 53240 Gaj
45 54450 45 Pole = 54450 Gaj
46 55660 46 Pole = 55660 Gaj
47 56870 47 Pole = 56870 Gaj
48 58080 48 Pole = 58080 Gaj
49 59290 49 Pole = 59290 Gaj
50 60500 50 Pole = 60500 Gaj
51 61710 51 Pole = 61710 Gaj
52 62920 52 Pole = 62920 Gaj
53 64130 53 Pole = 64130 Gaj
54 65340 54 Pole = 65340 Gaj
55 66550 55 Pole = 66550 Gaj
56 67760 56 Pole = 67760 Gaj
57 68970 57 Pole = 68970 Gaj
58 70180 58 Pole = 70180 Gaj
59 71390 59 Pole = 71390 Gaj
60 72600 60 Pole = 72600 Gaj
61 73810 61 Pole = 73810 Gaj
62 75020 62 Pole = 75020 Gaj
63 76230 63 Pole = 76230 Gaj
64 77440 64 Pole = 77440 Gaj
65 78650 65 Pole = 78650 Gaj
66 79860 66 Pole = 79860 Gaj
67 81070 67 Pole = 81070 Gaj
68 82280 68 Pole = 82280 Gaj
69 83490 69 Pole = 83490 Gaj
70 84700 70 Pole = 84700 Gaj
71 85910 71 Pole = 85910 Gaj
72 87120 72 Pole = 87120 Gaj
73 88330 73 Pole = 88330 Gaj
74 89540 74 Pole = 89540 Gaj
75 90750 75 Pole = 90750 Gaj
76 91960 76 Pole = 91960 Gaj
77 93170 77 Pole = 93170 Gaj
78 94380 78 Pole = 94380 Gaj
79 95590 79 Pole = 95590 Gaj
80 96800 80 Pole = 96800 Gaj
81 98010 81 Pole = 98010 Gaj
82 99220 82 Pole = 99220 Gaj
83 100430 83 Pole = 100430 Gaj
84 101640 84 Pole = 101640 Gaj
85 102850 85 Pole = 102850 Gaj
86 104060 86 Pole = 104060 Gaj
87 105270 87 Pole = 105270 Gaj
88 106480 88 Pole = 106480 Gaj
89 107690 89 Pole = 107690 Gaj
90 108900 90 Pole = 108900 Gaj
91 110110 91 Pole = 110110 Gaj
92 111320 92 Pole = 111320 Gaj
93 112530 93 Pole = 112530 Gaj
94 113740 94 Pole = 113740 Gaj
95 114950 95 Pole = 114950 Gaj
96 116160 96 Pole = 116160 Gaj
97 117370 97 Pole = 117370 Gaj
98 118580 98 Pole = 118580 Gaj
99 119790 99 Pole = 119790 Gaj
100 121000 100 Pole = 121000 Gaj
A Pole is a land area measurement unit, also known as a perch or rood. 1 Pole is equal to 0.25 Acre (ac). Pole is a large area measurement unit.
A Gaj / Gaz or yard is a unit of length used in parts of Asia. Historically, it was a regionally variable measurement. It was often used for measuring textiles.
The value in Gaj is equal to the value of Pole multiplied by 1210.
Gaj = Pole * 1210;
1 Pole is equal to 1210 Gaj.
1 Pole = 1210 Gaj.
• 1 pole = gaj
• pole into gaj
• poles to gajs
• convert pole to gaj
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→ | 2,334 | 5,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-30 | latest | en | 0.502811 |
https://www.wallstreetoasis.com/finance-dictionary/what-is-standard-deviation | 1,555,864,107,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578531994.14/warc/CC-MAIN-20190421160020-20190421182020-00140.warc.gz | 875,742,205 | 28,865 | Standard deviation is a mathematical measure which approximates the average deviation of any given data point from the mean of the entire data set. For example, if a data set of 100 has a standard deviation of 5 and a mean of 50, then on average no value will be outside the range of 45-55.
The calculation for standard deviation is simply the square root of the variance.
In finance, standard deviation is used to measure risk. If an asset has a return of 10% with a high standard deviation, then it is less likely that it will return that 10% consistently and is therefore more risky.
Related Terms | 130 | 603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | longest | en | 0.911829 |
https://wiki.zeropage.org/wiki.php/TugOfWar/%EC%9D%B4%EC%8A%B9%ED%95%9C | 1,627,598,439,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153897.89/warc/CC-MAIN-20210729203133-20210729233133-00520.warc.gz | 622,129,224 | 4,247 | # TugOfWar/이승한¶
150/200 빼고는 답이 나오긴 하지만. 분명히 틀린 알고리즘.
급기야는 오류가 보이자 땜빵으로 매꿔버리기도 하고...-_-ㅋ;;
```~cpp
#include <iostream>
using namespace std;
const int MAX= 100;
int maxToZero(int input[]); //가장큰수를 0로 바꾸어주는 함
int returnAver(int input[]);
void main(){
int weight[MAX] = {0};
int outA[MAX] = {0};
int outB[MAX] = {0};
int teamA, teamB;
teamA = teamB = 0;
int aver = 0;
int maxInInput = 0;
int N = 0;
int inputN = 0;
int input = 0;
int A, B;
A = B = 0;
cin>>N;
//23478
for(int n = 0; n< N; n++)
{
cin.get();
cin>> inputN;
for(input =0; input<inputN; input++){
cin>> weight[input];
}
cin.get();
teamA = teamB = A = B = 0;
aver = returnAver(weight);
for(int cycle=0; cycle < MAX; cycle++){
maxInInput = maxToZero( weight);
if( teamA + maxInInput == aver ){
teamA += maxInInput;
}else if( teamB + maxInInput == aver ){
teamB += maxInInput;
}else if( teamA < teamB ){
teamA += maxInInput;
}else{
teamB += maxInInput;;
}
}
outA[n] = teamA;
outB[n] = teamB;
}
for(n=0; n<N; n++)
cout<<outA[n]<<" "<<outB[n]<<endl;
}
int maxToZero(int * weight){
int max = 0;
int maxsIndex = 0;
for(int cycle=0; cycle < MAX; cycle++){
if(max < weight[cycle]){
max = weight[cycle];
maxsIndex = cycle;
}
}
weight[ maxsIndex ] = 0;
return max;
}
int returnAver(int * weight){
int sum =0;
for(int cycle=0; cycle<MAX ; cycle++){
sum += weight[cycle];
}
return ( sum / 2);
}
``` | 525 | 1,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-31 | latest | en | 0.237534 |
https://spiritualmediumrose.com/o2cqu/a5d0fb-consecutive-exterior-angles-converse | 1,620,750,511,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991648.10/warc/CC-MAIN-20210511153555-20210511183555-00385.warc.gz | 548,408,797 | 8,504 | Consecutive Exterior Theorem Pairs Congruent Congruent: Geometry Definition: exterior. Search Toyota lease deals in our database. ANSWER: consecutive interior angles converse 5. Theorem 3-6 Consecutive Angles (converse) Report an … Also the angles 4 and 6 are consecutive interior angles. Notes: j k j k 6 2 D. Use the Alternate Interior Angles Converse Theorem. Consecutive Interior Angles Theorem If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles formed are supplementary. Hi, Since the marked angles are alternate exterior angles, use: B) Alternate Exterior Angles Converse <==ANSWER. Ungraded . 3. (4) ∠1 ≅ ∠5 //(3), the definition of congruent angles (5) AB||CD //converse of the Corresponding Angles Theorem. PLZ HELP DUE IN 20 MIN A. Converse of Alternate Interior Angles Theorem B. Converse of Corresponding Angles Theorem C. Converse of Consecutive Angles Theorem D. Converse of Alternate Exterior Angles Theorem are called Consecutive Interior Angles. Consecutive exterior angles: These angles lie on the outside of the two parallel lines and on the same side of the transversal. If two lines in a plane are cut by a transversal so that corresponding angles are congruent, then the lines are parallel. Directions: Move point E or F and analyze the relationship between the measurements. Give the converse to justify your answer. < 8 ≅ <19 b. See below for explanation. 2. Theorem 3-10 Converse of the Consecutive Interior Angles Theorem If two lines are intersected by a transversal so that the consecutive interior angles are THEOREM 3.5 Alternate Exterior Angles Converse If two lines are cut by a transversal so the alternate exterior angles are congruent, then the lines are parallel. Converse of the Alternate Exterior Angles Theorem If two lines are intersected by a transversal so that the alternate exterior angles are congruent, then the lines are parallel. SURVEY . alternate exterior angles converse 4. m 6 + m 8 = 180 SOLUTION: and are consecutive interior angles of lines and m. Since , by the Converse of Consecutive Interior Angles Theorem. Swapalease.com, the web’s largest leasing marketplace, has great Toyota lease specials to browse. Alternate Interior Angles Theorem states that, when two parallel lines are cut by a transversal, the resulting alternate interior angles are congruent. Consecutive Exterior Angles Converse Use the diagram below for #14. Consider two statements: (A) Two lines that are cut by a transversal are parallel (B) Alternate interior angle formed by these lines are congruent They are equivalent. \$16:(5 20 PROOF Copy and complete the proof of Theorem 3.5. These are expectations for students that every current and prospective Math teacher ought to be familiar with. Play with it below (try dragging the points): The following content standards apply for Geometry and … Transitive Property of Parallel Lines: Example: Is it possible to prove the lines are parallel or not parallel? 120 seconds . Let's represent it in a form "if A … SOLUTION: angle and angle are alternate < 13 ≅ <15 f. Use the Consecutive Interior Angles Converse Theorem. If lines are parallel, then same side exterior angles are supplementary. As we have noticed here that both the above … Las Vegas Mayor Carolyn Goodman, on April 23, called for the city's businesses to reopen while refusing to provide any social distancing guidelines. ∠9 and ∠13 F alternate exterior H consecutive interior G alternate interior J corresponding 5. ∠3 and ∠10 A alternate exterior C consecutive interior B alternate interior 3. <3 and <8 are alternate exterior angles of line l and n. Since <3 ≅ <8, l is || to n by the Converse of the Alternate Exterior Angles Theorem. 215 3 20 180()( ) 2303 20180 550180 5130 26 xx xx x x x +°+ + °= ° +++ = += = = 3. yes; Alternate Exterior Angles Converse (Thm. If so, state the postulate or theorem you would use. :-) To contact Las Vegas please call 702-439-5636 Sunguard custom painting, a family owned and operated business has been beautifying the valley since 1979.
Yes, by Alternate Exterior Angles Converse.
Yes, by Consecutive Interior Angles Converse.
Yes, by Corresponding Angles Converse.
answer explanation . Converse of the Alternate Exterior Angles Theorem If two lines are crossed by a transversal and the alternate exterior angles are congruent, then the lines crossed by the transversal are parallel. Consecutive Interior Angles Converse no 115 80 Practice B LESSON 3.3 For use with pages 161—169 In Exercises 14-18, complete the two-column proof. The converse of this is if the two alternate exterior angles are congruent and when the exterior angle passes through the two lines these two lines will be parallel. Proof: Ex. 13. D corresponding 4. cl1 Alternate Interior Angles Converse clle Alternate Exterior Angles Converse dlle Consecutive Exterior Angles Converse 1le Transitive Property of Paralel Lines d11 Transitive Property of Paralel Lines State the converse used to support your For questions 16-17, find the value of x that makes r Il answer 17 16. A. Read More Find x so that m || n. Identify the postulate or theorem you used. Converse Of The Corresponding Angles Postulate search trends: Gallery. Simply give … alternate exterior angles are congruent. Don’t Get alternate interior consecutive interior same side interior yet, first read this. Using only premium products from start to finish on each job, we strive to provide a wonderful experience with each customer. Angle Relationship Parallel Lines Converse a. alternate interior angles alternate exterior angles consecutive interior angles Theorems Theorem 3.5 Corresponding Angles Converse If two lines are cut by a transversal so the corresponding angles are congruent, then the lines are parallel. C. Use the Consecutive Interior Angles Converse Theorem. If two lines are cut by a transversal so that alternate exterior angles are congruent, then the lines are parallel. Practice B continued LESSON 3.1 For use with pages 146—152 Copy and complete the statement with sometimes, always. or consecutive interior angles. Exterior angle pairs are on the exterior of the figure (above and below the parallel lines). Consecutive Exterior Angles Converse Theorem: If 2 lines are cut by a transversal so the consecutive exterior angles are _____, then the lines are _____. Check oll that opply. c and e are Consecutive Interior Angles. When the two lines are parallel, any pair of Consecutive Interior Angles add to 180 degrees. Given: Prove: Proof: Since the TExES Mathematics 7–12 (235) Exam enables one teach High School Math, it's an extremely profitable exercise to scrutinize the Math content standards applicable for High School Math teachers. B. 3.8) 5. a. The consecutive and exterior angle theorem states that if the transversal passes through the two parallel lines then any two exterior angles are congruent. Identify the special name for each angle pair. How would you show that the lines b and c are parallel? The pair of consecutive exterior angles for the two sections in the above figure can be named as (∠C,∠D) (∠ C, ∠ D) and (∠G,∠H) (∠ G, ∠ H). And similarly, for the exterior pair: (1) m∠1 = m∠7 //given (2) m∠5 = m∠7 //vertical, or opposite angles (3) m∠1 = m∠5 //using (1) and (2) and transitive property of equality, both equal m∠7 Putting together the alternate exterior angles theorem and its converse, we get the biconditional statement: Two lines crossed by a transversal are parallel if and only if … Consecutive exterior angles posted Dec 1, 2012, 4:24 PM by Ray Kim If parallel lines are cut by a transversal line, then consecutive exterior angles are supplementary. Proving Lines are Parallel (Using Converses) Sort, Cut, and Paste Activity Students will practice identifying parallel lines using the Corresponding Angles Converse, Alternate Interior Angles Converse, Alternate Exterior Angles Converse, and Consecutive Interior Angles Converse. j || k; converse of corresponding angles postulate \$16:(5 j || k; alternate interior angles converse \$16:(5 alternate exterior angles converse m 6 + m 8 = 180 \$16:(5 consecutive interior angles converse SHORT RESPONSE Find x so that m || n. Show your work. ∠2 and ∠7 are same side exterior angles (sometimes consecutive exterior angles). Determine which lines, if any, can be proved parallel given the angle relationship. 36, p. 168 THEOREM 3.6 Consecutive Interior Angles Converse If two lines are cut by a transversal so the consecutive interior angles are Theorem 3-5 Alt Ext Angles (converse) If two lines in a plane are cut by a transversal so that a pair of alternate exterior angles is congruent, then the two lines are parallel. Winner of Angie's List super service for 5 consecutive years! 14. Converse of the Alternate Exterior Angles Theorem If two lines are crossed by a transversal and the alternate exterior angles are congruent, then the lines crossed by the transversal are parallel. d and f are Consecutive Interior Angles. or never. Proof: Given: k ∥ l , t is a transversal I hope that helps!! In the figure, the angles 3 and 5 are consecutive interior angles. Given p ║ q … Tags: Topics: Question 15 . Another example of alternate exterior angles is ∠2 and ∠8. < 6 ≅ <7 <6 and <7 are Consecutive Interior angles of lines m and n. Since <6 and <7, m is || to n by the Converse of the consecutive Interior angles. Use the Alternate Exterior Angles Converse Theorem. Las Vegas-based construction company, Forté Specialty Contractors, celebrating its 10-year anniversary, is pivoting from constructing entertainment, restaurant … 3.7) 4. yes; Consecutive Interior Angles Converse (Thm. Use what you notice to complete the definition. ()( )416 7 20 36 3 12 xx x x +°= − ° = = 2. x = 26; Lines s and t are parallel when the marked consecutive interior angles are supplementary. As seen from the above picture, the two consecutive exterior angles are supplementary because the transversal line cuts the parallel lines. Use the diagram below in the following exercise. | 2,366 | 10,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.827329 |
http://www.cut-the-knot.org/Probability/RandomIntervalsWithDominant.shtml | 1,539,635,787,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509690.35/warc/CC-MAIN-20181015184452-20181015205952-00181.warc.gz | 438,148,115 | 10,946 | # Random Intervals with One Dominant
### Solution 1
Our sample space consists of $(2n)!$ permutations of the given point indices. An interval $J_j$ is dominant in $2(2n-2)!$ cases, as the $2n-2$ indices can be arbitrarily permuted between the two extremes. One of the intervals is dominant in $n\cdot 2(2n-2)!=2n(2n-2)!$ cases. Thus the probability of there being a dominant interval is $\displaystyle \frac{2n(2n-2)!}{(2n)!}=\frac{1}{2n-1}.$
### Solution 2
The problem is reduced to finding adjacent $X_{max}$and $X_{min}$(since only the interval $[X_{max}, X_{min}]$ covers all other intervals) in permutations of $X_1,\ldots,X_{2n}$.
Note that, given that $2 j$ is even, to be compatible with the previous formulation of the problem, we consider intervals $[X_1,X_2]$ and$[X_3,X_4]$,but not $[X_2,X_3]$.
• $X_{max}$ followed by $X_{min}$
There are $(2n-2)!$ sequences of each
$n \text{ lines}\left\{ \begin{array}{l} (X_{max},X_{min},X_3,X_4,\ldots,X_{2n})\\ (X_1,X_2,X_{max},X_{min},\ldots,X_{2n})\\ ...\\ (X_1,X_2,\ldots,\ldots,X_{max},X_{min})\\ \end{array} \right.$
for a total of $n (2n-2)!$
• $X_{min}$ followed by $X_{max}$
• By mirroring, same $n (2n-2)!$
• Thus
Numerator= $2 n (2 n-2)!$
Denominator(total permutations)= $(2n)!$
$\displaystyle p=\frac{1}{2 n-1}$
Note: thanks to Michael Wiener for correcting for the count of the number of lines.
### Solution 3
Easier path is that $X_{min}$ is equally likely to be paired with any of the other $2n-1$ values. Only $1$ is $X_{max}.$
### Acknowledgment
The problem has been proposed by N. N. Taleb as a follow-up to an earlier question about overlapping intervals. Solution 2 is by N. N. Taleb; Solution 3 is by Michael Weiner; Solution 4 is by Zhuo Xi. Michael Weiner came up with a variant of the problem. Solution 1 to the Variant problem is by Josh Jordan; Solution 2 is by Michael Weiner.
[an error occurred while processing this directive] | 630 | 1,926 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-43 | longest | en | 0.823837 |
https://brainmass.com/math/real-analysis/define-the-function-and-determine-the-pointwise-limit-245776 | 1,477,367,302,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719877.27/warc/CC-MAIN-20161020183839-00170-ip-10-171-6-4.ec2.internal.warc.gz | 825,943,872 | 17,865 | Share
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# Define the Function and Determine the Pointwise Limit
For each natural number n, define the function g_n: [0,1] --> R by (g_n)(x) = nx(1- x^2)^n.
a. Determine the pointwise limit of the sequence {g_n}. Be sure to prove that your claim for the pointwise limit is correct.
b. Does the sequence {g_n} converge uniformly on {0,1]? Explain.
c. Evaluate and compare the two limits of g_n. You are welcome to use the fundamental theorem of calculus and/or ideas about area to evaluate the integrals.
A Word document is attached which presents the question under proper format.
#### Solution Summary
This solution is presented within an attached Word document and explains the three proofs in a detailed, step by step manner.
\$2.19 | 180 | 759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | latest | en | 0.839772 |
https://math.stackexchange.com/questions/4198/on-the-definition-of-weakly-compact-cardinals | 1,571,687,998,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987787444.85/warc/CC-MAIN-20191021194506-20191021222006-00237.warc.gz | 611,435,129 | 34,129 | # On the definition of weakly compact cardinals
I am reading in Jech's Set Theory the chapter about large cardinals. After discussing measurable cardinals he moves on to weakly-compact cardinals, which have been discussed far earlier in the book.
I went back to the chapter dealing with weakly-compact cardinals and began retracing the definition.
Eventually, it came into this:
We denote $[k]^n = \{X \subseteq \kappa\mid |X| = n\}$. If $\lambda$ is a cardinal, we denote $\kappa \to (\lambda)^2$ when for every partition of $[\kappa]^2$ into $2$ we have $H \subseteq \kappa$ that is of cardinality $\lambda$, and for which $[H]^2$ is strictly in one part.
And we say that $\kappa$ is weakly-compact if it satisfies the property $\kappa \to (\kappa)^2$.
The problem is that I'm a bit lost in all those definitions, and not even sure about the $\kappa \to (\lambda)^2$ notation.
My questions are, if so, can someone help me make some sense into those definitions, and is there an equivalent definition for weakly-compact cardinals which can help me understand their properties better?
• With three downvotes, I finally read the message LOUD and clear. The first sentence was just so badly written it urged people to downvote! I hope those who casted their downvotes will reconsider them, now that the only visible problem has been fixed. – Asaf Karagila May 3 '15 at 21:52
• (In the unlikely case that the recent downvote, well and the other downvotes, were cast solely as revenge or as ways to express anger with whomever posted the question, all that time ago; I'd like to take a moment and applaud you for your diligent service to this website, and thank you personally for taking the time to peruse my profile for a question which is sufficiently old so downvoting on it will serve no practical purpose than to annoy me. Kudos!) – Asaf Karagila May 3 '15 at 21:57
There are many ways to think of these definitions. Here's a way one can understand for example what $[\kappa]^2$ means and what it means to have a homogeneous subset, that I find intuitive.
Assume that you have a complete undirected graph with $\kappa$ many nodes. That means that you have $\kappa$ points and you connect every pair of them with a line. Now assume that we have two colours, red and blue, and that every line between two nodes is coloured either red or blue. A subset of these $\kappa$ nodes is called homogeneous if the lines between each and every of its nodes have the same colour (that's the same as saying that it has a complete subgraph whose lines are of one colour only).
Now we say that $\kappa\to(\lambda)^2$ is true if no matter how we paint those lines using two colours, we can find a homogeneous set of cardinality $\lambda$. That is, for every way we colour the lines we can find $\lambda$ many nodes that every line between them is of the same colour.
In other words, for every function $f:[\kappa]^2\to 2$ (you can think of this as a function that sends every two elements of $\kappa$ to one of the 2 colours) we can find a subset of $\kappa$ (let's call it$H$) that has cardinality $\lambda$ such that for every $x,y,z,w\in H$ we have $f(\{ x,y\} )=f(\{ z,w\} )$.
To generalize this if for every function $f:[\kappa]^n\to\mu$ (again you can see this as a function that sends every $n$ elements of $\kappa$ to one of the $\mu$ colours, or that it partitions the subsets of $\kappa$ of cardinality $n$ into $\mu$ partitions) we can find a set $H$ such that $\left|H\right|=\lambda$ and for every $x_1,\ldots,x_n,y_1,\ldots,y_n\in H$ we have $f(\{ x_1,\ldots,x_n\} )=f(\{ y_1,\ldots,y_n\} )$ then we say that $\kappa\to(\lambda)^n_\mu$ is true.
The arrow notation, even though appears weird at first is used because the property remains true if we replace the cardinal in the left side of the arrow with a larger cardinal or if we replace any cardinal in the right side of the arrow with a smaller cardinal (if the subscript in the left side of the arrow is omitted then it is assumed that it is 2). It should be obvious that the notation only has meaning if $\lambda<\kappa$.
There are many equivalent definitions of weakly compact cardinals. Jech's book mentions most of them, but here are some. If $\kappa$ is a cardinal then it is weakly compact iff:
$\kappa\to(\kappa)^2$ is true
$\kappa\to(\kappa)^n_\lambda$ where $\lambda<\kappa$ is true
$\kappa$ is inaccessible and has the tree property, that is, if you have a tree with height $\kappa$ and each level of the tree has less than $\kappa$ elements then it has a branch of length $\kappa$
$\kappa$ is $\Pi^1_1\textrm{-indescribable}$.
$\kappa$ is inaccessible and the languages $\mathcal{L}_{\kappa,\omega}$ satisfy the weak compactness theorem (hence the name). This means that in a language where we allow $\lambda$ conjunctions and disjunctions (where $\lambda<\kappa$), if we have a set of sentences $\Sigma$ where$\left|\Sigma\right|=\kappa$ and every subset of $\Sigma$ with less than $\kappa$ sentences is satisfiable then $\Sigma$ is satisfiable.
Let me add to Ivan's answer another equivalent characterization of weak compactness, which might appeal to you, as it makes them resemble miniature measurable cardinals.
Namely, if $\kappa$ is a cardinal and $\kappa^{<\kappa}=\kappa$, then $\kappa$ is weakly compact if and only if for every transitive set $M$ of size $\kappa$ with $\kappa\in M$, there is another transitive set $N$ and an elementary embedding $j:M\to N$ having critical point $\kappa$.
This embedding characterization admits myriad forms. For example, one can insist that $M\models ZF^-$ or even $M\models ZFC$, and that $M^{<\kappa}\subset M$, or that every $A\subset \kappa$ can be placed into such an $M$, and so on. One can even insist that $j\in N$, a property known as the Hauser property.
These various embedding formulations of weak compactness allow one to borrow many of the methods and techniques from much larger large cardinals, which are most often described in terms of embeddings, and apply them with weakly compact cardinals. For example, using Easton support forcing iterations, one can control the value of $2^\kappa$ while preserving the weak compactness of $\kappa$. | 1,607 | 6,190 | {"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} | 2.890625 | 3 | CC-MAIN-2019-43 | latest | en | 0.956944 |
https://es.planetcalc.com/480/ | 1,726,507,365,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00742.warc.gz | 219,718,717 | 13,735 | # Seasonal fluctuations. Seasonal indices. Method of simple averages
The study of periodic (seasonal) fluctuations. Calculation of average seasonal indices by the method of simple averages.
Note that this calculator calculates seasonal indices for monthly data. For quarterly data please use Seasonal Indices for Quarterly Data.
This is the continuation of a theme started in the article Analytical performance indicators.
Here we will talk about average seasonal indices - analytical indicators of time series characterizing the seasonal fluctuations
The seasonal fluctuations are annual, constantly repeating changes of the studied phenomena. During the analysis of the annual dynamics, you obtain the quantitative characteristics, reflecting the nature of the indicators' changes by months of the annual cycle.
Seasonal fluctuations are described by seasonal indices, which are calculated as a ratio of the indicator's actual value to some theoretical (predicted) level.
$I_{ij}=\frac{Y_{ij}}{Y_t_{ij}}$
Where i - the number of the seasonal cycle (years), j - the season's ordinal (months).
The obtained values are subject to random deviations. That's why these values are averaged out by years, so you get the average seasonal indices for each period of the annual cycle (months)
$I_{sj}=\frac{\sum_{i=1}^n I_{ij}}{n}$
Depending on the nature of the time series changes, you can calculate the formula with different methods.
I'll review the easiest method - the method of simple averages. The method can be used for the time series with no or negligible downward/upward tendencies. In other words, the observed value fluctuates around a certain constant value.
$I_{sj}=\frac{Y_{sj}}{Y_{s0}}$,
where
$Y_{sj}=\frac{\sum_{i=1}^n Y_{ij}}{n}$, average for each season j (months) for all n periods
$Y_{s0}=\frac{\sum_{i=1, j=1}^{i=n, j=m} Y_{ij}}{nm}$, an average for all periods (n) and seasons (m)
#### Indicators
JanuaryFebruaryMarchAprilMayJuneJulyAugustSeptemberOctoberNovemberDecember
Items per page:
Digits after the decimal point: 1
Seasonal indices
The file is very large. Browser slowdown may occur during loading and creation. | 486 | 2,154 | {"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": 5, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-38 | latest | en | 0.881361 |
https://blog.csdn.net/zwizhy/article/details/90488950 | 1,638,649,307,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363006.60/warc/CC-MAIN-20211204185021-20211204215021-00451.warc.gz | 221,794,198 | 20,620 | # 三维闭合B样条曲线拟合算法Matlab代码
1 篇文章 0 订阅
% ref: 闭合 B 样条曲线控制点的快速求解算法及应用
% http://www.doc88.com/p-5714423317458.html
% https://blog.csdn.net/liumangmao1314/article/details/54588155
=============================
%计算B样条曲线控制点
function px = LU_B1(CPnum, V)
a = 1;
b = 4;
c = 1;
d = 1;
e = 1;
f = zeros(CPnum-1,1);
g = zeros(CPnum-2,1);
h = zeros(CPnum,1);
k = zeros(CPnum-1,1);
% get h[] & f[]
h(1) = b;
for i=1:CPnum-2
f(i) = a/h(i);
h(i+1) = b - f(i)*c;
end
% get g[] & f[n-1]
g(1) = d/h(1);
for i=1:CPnum-3
g(i+1) = -g(i)*c/h(i+1);
end
f(CPnum-1) = ( a-g(CPnum-2)*c )/h(CPnum-1);
% get k[] & h[n]
k(1) = e;
for i=1:CPnum-3
k(i+1) = -f(i)*k(i);
end
k(CPnum-1) = c - f(CPnum-2)*k(CPnum-2);
gksum = 0;
for i=1:CPnum-2
gksum = gksum + g(i)*k(i);
end
h(CPnum) = b - gksum - f(CPnum-1)*c;
% 矩阵求解过程，追的过程
x = zeros(CPnum,1);
x(1) = 6*V(end);
for i=1:CPnum-2
x(i+1) = 6*V(i) - f(i)*x(i);
end
gxsum = 0;
for i=1:CPnum-2
gxsum = gxsum + g(i)*x(i);
end
x(CPnum) = 6*V(CPnum-1) - gxsum - f(CPnum-1)*x(CPnum-1);
% 赶的过程
px = zeros(CPnum+2,1);
px(CPnum) = x(CPnum)/h(CPnum);
px(CPnum-1) = ( x(CPnum-1)-k(CPnum-1)*px(CPnum) )/h(CPnum-1);
for i=CPnum-2:-1:1
px(i) = ( x(i)-c*px(i+1)-k(i)*px(CPnum) )/h(i);
end
px(CPnum+1) = px(1);
px(CPnum+2) = px(2);
end
% 插值计算三次周期性b样条曲线
function p = Cubic_Bsp(u,x)
b(1) = ((1-u)^3)/6;
b(2) = (3*u^3-6*u^2+4)/6;
b(3) = (-3*u^3+3*u^2+3*u+1)/6;
b(4) = (u^3)/6;
p = x*b';
end
%test.m 测试
clear;clc
x = [-1;-1;1;1;0.2];
y = [1;-0.5;-1;1;0.8];
z = zeros(6,1);
z(3) = 1;
CPnum = size(x,1);
%x,y,z坐标分别计算B样条曲线控制点
px = LU_B1(CPnum, x);
py = LU_B1(CPnum, y);
pz = LU_B1(CPnum, z);
%首尾相连，曲线闭合
x(CPnum+1) = x(1);
y(CPnum+1) = y(1);
z(CPnum+1) = z(1);
figure;
plot3(x,y,z,'ro-');
hold on;
for i=1:CPnum%用这个循环
c=num2str(i);
c=[' ',c];
text(x(i),y(i),z(i),c)
%text(x(i),y(i),c)
end
plot3(px,py,pz,'g-');
for i=1:CPnum+1%用这个循环
c=num2str(i);
c=[' ',c];
text(px(i),py(i),pz(i),c)
plot3(px(i),py(i),pz(i),'*');
end
axis equal
px(CPnum+3) = px(3);
py(CPnum+3) = py(3);
pz(CPnum+3) = pz(3);
px(CPnum+4) = px(4);
py(CPnum+4) = py(4);
pz(CPnum+4) = pz(4);
%两点之间对10个点进行插值计算
nP = 10;
delta = 1.0/nP;
for j=1:CPnum
u = 0.0;
for i=1:nP
p = px';
Xi = p(j:j+3);
xx(i+(j-1)*nP) = Cubic_Bsp(u,Xi);
p = py';
Yi = p(j:j+3);
yy(i+(j-1)*nP) = Cubic_Bsp(u,Yi);
p = pz';
Zi = p(j:j+3);
zz(i+(j-1)*nP) = Cubic_Bsp(u,Zi);
u = u + delta;
end
end
xx(end+1) = x(1);
yy(end+1) = y(1);
zz(end+1) = z(1);
plot3(xx,yy,zz,'b--')
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03-28 | 1,421 | 2,536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2021-49 | latest | en | 0.296228 |
https://math.stackexchange.com/questions/1045404/integrating-chebyshev-theta-function | 1,569,078,280,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574532.44/warc/CC-MAIN-20190921145904-20190921171904-00454.warc.gz | 569,669,232 | 30,568 | # Integrating Chebyshev theta function
I'm trying to compute the following integral ($\vartheta(x) = \sum\limits_{p \leq x}\log(p)$)
$$\int\limits_{0}^{\infty}\vartheta(e^x) e^{-(1+s)x} \text{dx}$$
The result is supposed to be $\frac{\sum\log(p)/p^{s+1}}{1+s}$, but I'm having trouble doing it. I tried to substitute $t=e^x, dt/t=dx$ and I got
$$\int\limits_{1}^{\infty} \vartheta(t)t^{-(2+s)} \text{dx}$$
And I'm stuck. I don't know how to evaluate an integral with a sum inside that depends on in
where $[\,\cdot\,]$ is the Iverson bracket, $[x \geqslant \log p]$ is $1$ if $x\geqslant \log p$, and $0$ if $x < \log p$. | 233 | 627 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-39 | latest | en | 0.750395 |
http://openstudy.com/updates/50b8308ce4b0c789d5100d80 | 1,448,917,247,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398463315.2/warc/CC-MAIN-20151124205423-00319-ip-10-71-132-137.ec2.internal.warc.gz | 169,356,010 | 10,118 | ## spndsh 3 years ago e^ln2 ?
1. rsilvabenevides
2
2. spndsh
so e^-ln2 = -2 right?
3. Algebraic!
no...
4. spndsh
1/2?
5. Algebraic!
yes:)
6. spndsh
how?
7. rsilvabenevides
ln(a^b) = b*ln(a) So ln(1/2) = ln(2^(-1)) = - ln (2) | 115 | 238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2015-48 | longest | en | 0.463127 |
https://airwiki.elet.polimi.it/index.php?title=Talk:RunBot:_a_Robogame_Robot&diff=13329&oldid=13327 | 1,638,595,317,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362930.53/warc/CC-MAIN-20211204033320-20211204063320-00369.warc.gz | 154,026,652 | 10,849 | Difference between revisions of "Talk:RunBot: a Robogame Robot"
Thesis should be find in the link.Submitted version of the thesis
Requirements
The robot should have:
• a speed of about 1 m/sec
• omnidirectional movement (Kamro wheels)
• sensors to avoid obstacles (sonars)
• a camera that can be moved up and down (and eventually left and right)
• wireless connection to a computer (Wi-fi)
• Bluetooth connection
• have power enough to move and transmit for at least 2 hours without recharging
• Have the possibility to recharge autonomously
The robot should cost as less as possible
TO DO
• Select HW
** ST ARM HW (ask Martino Migliavacca <martino.migliavacca@gmail.com>, GUMSTIX (http://www.gumstix.org/), ARDUINO (http://www.freeduino.org/), other...)
** Engines (WWW.robot-Italy.com, http://www.jonathan.it/)
** batteries (above, or standard A, AA or AAA type)
** Camera (ST smart cameras, with Ethernet wired link)
** Sonar (already available (see ROBOWII2.0))
** Blue tooth and WI-FI
• Design body (structure and appearance), sensor placement, eventual movements of the camera
• Implement the robot
• test the robot with a simple game (e.g. RoboWII2.0)
Source Code Backup The backup file
Camera
Wheels
Motors
Lenses
Batteries
Mice
The results obtained so far with that setup is in the pdf file below.
The results obtained so far
Camera
The demo software given by ST is used in order to test the colr blob detection.
Detecting 3 blob color
Detecting 4 blob color
Motion Control
The angle between each wheel is 2*pi/3, which 120 degree.
The front wheels are making a pi/6 angle between the ground.
The positions of the wheels in the coordinate system is as follows.
Wheel 1 is (cos(pi/6),sin(pi/6)) = (0.866,0.5)
Wheel 2 is (cos(pi/6),sin(pi/6)) = (0.866,0.5)
Wheel 3 is (cos(3*pi/2),sin(3*pi/2))=(0,-1)
I wrote simulations in matlab, in two ways.
The first one is using the TriskarOdometry directly, and estimates the necessary wheel speeds to reach the desired frontal, lateral and angular speed of the robot.
The second one is instead giving the motor speeds in terms of rotations and estimates the frontal ,lateral ,angular speed of the robot according to motor speeds.
The matlab code for the motion control simulation is consist of the following files:
-motor_speed.m --This file which takes the motor speed as input and calculates the Vf,Vl,w of the robot.
-TriskarOdometry.m --Giving the input Vf,Vl,w calculates the motor speed
-trajectory.m --The function which draws some trajectories based on the following schema
1) DIRECT MOVEMENT
THE HEAD OF THE ROBOT ALWAYS LOOKING IN A FIXED DIRECTION
2) ANGULAR MOVEMENT
THE ROBOT WILL MOVE IN A CIRCULAR TRAJECTORY WITH FIXED HEAD
DIRECTION
3) SPIN & GO
THE ROBOT EITHER GOES ANGULAR BUT THE HEAD POSITION
IS ALSO CHANGES
4) SPIN & GO DIRECT
THE ROBOT EITHER GOES DIRECT BUT THE HEAD POSITION
IS ALSO CHANGES
-direc_motion.m --DIRECT MOVEMENT
-angular_motion.m --ANGULAR MOVEMENT
-spin_go.m -- SPIN & GO
-spin_go_d.m --SPIN & GO DIRECT
-draw_arrow.m --A function that is used to draw arrow.
File with the matlab files listed above.
Processor Comparison
This is the processor comparison table, the cells without entry is left blank
Name Clock Speed Ram Storage Onboard Devices Power Consumption Price OS Source Page
1 ESOM270 PXA270@520MHz 128 MB SDRAM 32 MB FLASH PMIC USB device,Camera interface(QCI),USB host 0.8 W @ Full running Mode,100mW @ Deep Sleep mode 76 € Comes with preinstalled uboot & Linux 2.6.25 or eboot Windows CE 6.0 R2 http://www.e-consystems.com/esom270.asp
2 IGEPv2 BOARD OMAP3530@720 Mhz 512MB RAM 512MB ONENAND-FLASH Ethernet 10/100 Mb BaseT.,Wifi IEEE 802.11b/g + Bluetooth 2.0 (Integrated antenna).,1 x USB OTG,1x USB Host 145 € IGEPv2 running Ubuntu 9.04 http://www.igep-platform.com/index.php?option=com_content&view=article&id=46&Itemid=55
3 Em-x270 Intel's XScale PXA270 CPU, up to 520 MHz 128 Mbyte SDRAM 512 Mbyte Flash Disk WLAN / WiFi 802.11b/g Interface,Bluetooth interface,Slave and host USB ports, including keyboard and mouse support,100 Mbps Ethernet port 0.2 - 2 W 76 \$ ce OR linux http://www.compulab.co.il/x270cm/html/x270-cm-datasheet.htm
4 S3C2440 Core Board II Samsung S3C2440A based on ARM920T, 400MHz 64MB SDRAM 64MB NAND Flash 2-ch USB Host controller / 1-ch USB Device controller (ver 1.1),Camera interface ,(Max. 4096 x 4096 pixels input support. 2048 x 2048 pixel input support for scaling) 110\$ Windows CE4.2/5.0, 6.0 and Linux2.6 http://www.embedinfo.com/en/list.asp?id=64
5 Overo™ Air COM [GUM3503A] OMAP 3503 Application Processor with ARM Cortex-A8 CPU 600 MHz 256MB RAM 256MB Flash 802.11(g) and Bluetooth®,USB OTG signals, USB HS Host 199\$ Linux 2.6.31 or higher OpenEmbedded http://www.gumstix.com/store/catalog/product_info.php?products_id=226
6 MX31 TurboG5 Module Freescale i.MX31 @ 532MHz 128 MB of Mobile DDR 32 MB of Flash Support for 802.11b/g embedded wireless module,One SD/MMC card slot,One USB 2.0 On-The-Go (OTG) port (H/F/L speed),One USB 2.0 host port (H/F/L speeds),One Camera Sensor Interface,One 10/100BASE-T Ethernet port Windows CE 5.0 and 6.0 Linux http://www.eurotech-inc.com/single-board-computer-imx31-com-turbog5.asp
Motor Suggestion
no load speed: 500 rpm stall torque: 29 kg-cm ~= 0.282 Nm
maximum power output = Pmax = 26.190 rad/s * 0.141 Nm = 3.692 W occuring at W= 250 rpm and t=0.141 Nm
no load speed: 470 rpm stall torque : 7 oz ( 0.5 kg-cm ) ~= 0.049 Nm
maximum power output = Pmax = 24.619 rad/s * 0.0245 Nm = 0.603 W occuring at W=235 rpm and t = 0.0245 Nm
calculations made according to the reference paper at the link http://lancet.mit.edu/motors/motors3.html http://lancet.mit.edu/motors/motors4.html
Wheel Specification
2052-3/8X CAT-TRAK
Description
• Standard 2" O.D. - 3/8" Bore Double Row
• Metric 49.2mm O.D. - 9.5mm Plain Bore.
• Steel Bottom = 25 lbs. 11.3kg.
• Plywood Surface = 15 lbs. 6.8kg.
• 200# Test Corrugated Bottom = 10 lbs. 4.5kg.
• Weight = 1.75 oz.
• Synthetic rubber coated polypropylene rollers.
• Price:\$7.48.
• Source: http://store.kornylak.com/ProductDetails.asp?ProductCode=FXA314
2052BX CAT-TRAK
Description
• Standard 2" O.D. - 1/4" I.D. Bushing Double Row
• Metric 49.2mm O.D. - 6.35mm Bushing
• Steel Bottom = 25 lbs. 11.3kg
• Plywood Surface = 15 lbs. 6.8kg
• 200# Test Corrugated Bottom = 10 lbs. 4.5kg
• Weight = 1.75 oz
• Synthetic rubber coated polypropylene rollers.
• Price :\$7.48
• Source: http://store.kornylak.com/ProductDetails.asp?ProductCode=FXA317
2052KX CAT-TRAK
Description
• Standard 2" O.D. - 1/2" Bore Double Row w/ Keyway
• Metric 49.2mm O.D. - 12.7mm Bore w/ Keyway
• Steel Bottom = 25 lbs. 11.3kg
• Plywood Surface = 15 lbs. 6.8kg
• 200# Test Corrugated Bottom = 10 lbs. 4.5kg
• Weight = 1.75 oz
• Synthetic rubber coated polypropylene rollers.
• Price :\$7.48
• Source: http://store.kornylak.com/ProductDetails.asp?ProductCode=FXA315
Related Files
Matlab scripts, that are used to calculate, color; motor contributions; and object position.Helper scripts
The final version of microcontroller code with the robogame developed.Micro-code
Microcontroller code coming with ST, offers demo functionalities, and used mainly for color selection.Micro-code of ST | 2,182 | 7,211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.860581 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-11-additional-topics-11-7-relations-and-functions-problem-set-11-7-page-508/24 | 1,679,585,449,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945168.36/warc/CC-MAIN-20230323132026-20230323162026-00764.warc.gz | 885,969,754 | 12,931 | Elementary Algebra
The domain of the given function is the set of all real numbers except $\frac{-3}{7}$.
When the question asks for the domain of the function, it is asking for what values of x can be legally substituted into the function. (For instance, in the function $\frac{2}{x}$, x $\ne$ 0, for this would make the denominator of the fraction equal to 0.) Look for the number that makes the denominator zero by equating the denominator to 0: $7x+3=0 \\7x = 0-3 \\7x=-3 \\x = \frac{-3}{7}$ Thus, when $x=\frac{-3}{7}$, the denominator is equal to zero. Therefore, the domain of the given function is the set of all real numbers except $\frac{-3}{7}$. | 182 | 657 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-14 | latest | en | 0.801491 |
https://logosconcarne.com/tag/gamma-factor/ | 1,585,561,475,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496901.28/warc/CC-MAIN-20200330085157-20200330115157-00218.warc.gz | 564,338,553 | 22,443 | # Tag Archives: gamma factor
## SR #22: Relative Time
Last time we saw that Em non-paradoxically time-travels over three years into Al’s future by flying 12 light years at half the speed of light for just over two decades. Her journey completed, Em has aged only 20.8 years while Al has aged 24.
That may not seem like much of a gain, but Em was only moving really fast — not really, really fast. If she travels at 99% of light-speed, her round trip shortens to 1.7 years while Al doesn’t wait much longer than it takes light to make the six light-year round trip: 12.12 years! And at 99.9% c, Em’s whole trip takes her only half a year!
Today we break down dime tilation. I mean, time dilation!
## SR #18: Light Cones
Last time I focused on how it was possible for Al to see — even enclose in a tunnel — a train that appears shorter to him due to its motion. It turns out that the train Al sees is a stack of time slices of the train at different moments. As we’ve seen, lots of things look different in a moving frame.
Today I want to say a little about Em’s point of view, run some numbers, and take you through a little math (just one equation, I promise). Then, because it’s Friday (when I try to write about light), I’ll introduce you to light cones.
They’re not actually necessary, but they’re kinda cool. | 329 | 1,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | longest | en | 0.94789 |
http://mathoverflow.net/questions/33730?sort=newest | 1,369,292,379,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702957608/warc/CC-MAIN-20130516111557-00079-ip-10-60-113-184.ec2.internal.warc.gz | 141,390,937 | 12,473 | ## “In the sequel” - outdated mathematical jargon or precise technical term? [closed]
As a non-native speaker of English, I have been perplexed by the phrase "in the sequel" as used in textbooks, lecture notes, and even research articles. None of my linguistically inclined native speaker friends have seen it outside of mathematical literature, and the relevant Oxford English Dictionary definition of "sequel" suggests mathematical usage is non-standard:
The ensuing narrative, discourse, etc.; the following or remaining part of a narrative, etc.; that which follows as a continuation; esp. a literary work that, although complete in itself, forms a continuation of a preceding one.
What I've inferred from context is that it means something along the lines of "for the rest of this book/paper/text", especially as it's usually used to introduce notation and/or convention.
Some examples:
We hope that the relation between linear transformations and matrices is by now sufficiently clear that the reader will not object if in the sequel, when we wish to give examples of linear transformations with various properties, we content ourselves with writing down a matrix.
--Paul Halmos, Finite-Dimensional Vector Spaces, p. 86
[Here, and in the sequel, Card(S) denotes the number of elements in the finite set S.]
--J.P. Serre, Local Fields, p. 64
In the sequel we shall denote by ∅ the empty set and by {pt} a set with one element.
-- Pierre Schapira, Algebra and Topology course notes, p. 8
In the sequel, we will denote by L(C ) the configuration space of any convergent CFG C.
--E. Goles et. al., Sandpile Models and Lattices: a comprehensive survey, Theoretical Computer Science, 2004, Vol 322, Issue 2, p. 398
So my actual question is two-fold.
1. What does the phrase actually mean?
2. When is its use warranted over the use of phrases such as "for the rest of the book/paper/text"?
-
It means exactly what the OED says it means. – Robin Chapman Jul 28 2010 at 23:32
.. This was also on mathunderflow, where Pete Clark gave a satisfactory answer to it. – Harry Gindi Jul 28 2010 at 23:38
I wonder why was this copied from Math.SE? – Mariano Suárez-Alvarez Jul 28 2010 at 23:43
Indeed, it is because of this brilliant answer math.stackexchange.com/questions/907/… that I gained the ability to comment on posts at Math.SE. I agree that it is a little curious that this question appeared here less than a day after it got asked and answered on the other site. – Pete L. Clark Jul 28 2010 at 23:59
I don't understand why people are voting to close this question. – Kevin Lin Jul 29 2010 at 1:11
## closed as off topic by Robin Chapman, Harry Gindi, Mariano Suárez-Alvarez, Akhil Mathew, Andrew StaceyJul 29 2010 at 8:00
I have always understood "in the sequel" to mean "in what follows". I have never checked but I assumed this was from the Latin as in "sequence" and "non sequitur" http://en.wikipedia.org/wiki/Non_sequitur
In the usage you refer to it means what follows in the same article. The sequel to a film is the next film.
I see Franklin has checked the etymology.
-
Related also is the Latin abbreviation et seq., which means "and what follows." Use it like this, "For a summary of this fact, see Lemma 2.11 et seq. in Fakename's article." – Willie Wong Jul 29 2010 at 0:24
Right. This is an abbreviation for "et sequitur"; similarly we have et al. for "et aliter" (and others) and etc. for "et cetera" (and things). – Bruce Westbury Jul 29 2010 at 1:22
Actually, it is 'et sequens'. The verb is 'sequor'. 'Sequitur' is the third person singular form from which 'sequitur' the noun is derived. 'Sequens' is the present active participle. This is not just a quibble on etymology: 'sequitur' the noun has a specific meaning as logically following, whereas 'sequens' the participle just means "following" in general. So as 'et seq.' is usually used to refer to the "narrative which follows", a reader thinking of "et sequitur" runs the risk of finding a non sequitur. =) Okay, this is getting too much to be LatinOverflow... – Willie Wong Jul 29 2010 at 1:48
(Okay, I can't resist: I was taught that 'et al.' expands to 'et alii/aliae/alia' for people, and 'et alibi' for places. I don't recall the word 'aliter', so I looked it up, and it seems to be the adverb meaning 'otherwise'.) – Willie Wong Jul 29 2010 at 1:57
Willie, Thanks. – Bruce Westbury Jul 29 2010 at 7:08
It means "from now on." (Such as, "in the sequel, $K$ will denote a perfect field...") This does, in fact, tend to appear only in older books, or in books in translation. I also was confused when I first saw the term. | 1,203 | 4,620 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.84375 | 3 | CC-MAIN-2013-20 | latest | en | 0.919439 |
https://social.msdn.microsoft.com/Forums/en-US/86441f95-58e6-4473-ba9b-5b88b9c30af8/salary-textbox-have-many-0000-for-a-decimal?forum=aspwebforms | 1,680,418,624,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950383.8/warc/CC-MAIN-20230402043600-20230402073600-00595.warc.gz | 596,614,030 | 12,288 | # salary textbox have many 0000 for a decimal
• ### Question
• User-944424728 posted
<br>
<p>Hello, I can input 50.0000 on my salary text box for hour and 60000.000 for salary become 60000 with no decimal. How can I make it to be 50.00 and 60000.00 or 60,000.00 on the input textbox. I had a pop up validation error but it not change in the textbox.
The code below will work but how can I use the same textbox so the number change by itself?<br>
thank you.</p><br>
<br>
Textbox2.text=Formatcurrency(textbox1.text)<br>
Or
Dim dec as decimal
dec=convert.todecimal(textbox1.text)
Textbox1.text=dec.tostring(“n2”)
<p>--</p><br>
<p></p><br>
<br>
Thursday, June 20, 2019 4:40 PM
### All replies
• User-719153870 posted
Hi aspvbnet,
According to your description, you want to control the data entered into textbox to keep two decimal places. I recommend using JS and regular expressions.
First, make sure that the input is a number, and then remove the excess decimal parts.
```<script type="text/javascript">
function num(obj) {
obj.value = obj.value.replace(/[^\d.]/g, ""); // Clear Characters Other than Numbers and'. '
obj.value = obj.value.replace(/^\./g, ""); // Verify that the first character is a number
obj.value = obj.value.replace(/\.{2,}/g, "."); //Retain only the first . and remove the redundant
obj.value = obj.value.replace(".", "\$#\$").replace(/\./g, "").replace("\$#\$", ".");
obj.value = obj.value.replace(/^(\-)*(\d+)\.(\d\d).*\$/, '\$1\$2.\$3'); // Only two decimal digits can be entered
}
</script>
<div>
Test: 50.0000, 60000.000
<input name="ctl00\$Main_content\$ucIndEmpHistory\$txtSalary" type="text"
id="Text1" onkeyup="num(this)" onpaste="num(this)"
style="width: 100px; text-align: right;" wiid="C1286" runat="server" />
</div>
```
Hereis result of my demo:
Best Regards,
Yang Shen
Friday, June 21, 2019 3:12 AM
• User-1038772411 posted
Hello aspvbnet,
Try with this
`decimalVar.ToString ("0.##"); // returns "0" when decimalVar == 0`
or Use FormatNumber:
``````Dim myStr As String = "38"
MsgBox(FormatNumber(CDbl(myStr), 2))
//display as 38.00
Dim myStr2 As String = "6.4"
MsgBox(FormatNumber(CDbl(myStr2), 2))//display as 6.40``````
or you can use
`Convert.ToDecimal(TextBox1.Text).ToString("0.00");` | 696 | 2,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-14 | latest | en | 0.528186 |
http://slideplayer.com/slide/2355885/ | 1,547,784,619,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659677.17/warc/CC-MAIN-20190118025529-20190118051529-00500.warc.gz | 211,313,243 | 27,242 | # Acid/Base solutions.
## Presentation on theme: "Acid/Base solutions."— Presentation transcript:
Acid/Base solutions
What is an acid? Bronsted-Lowry definition: An acid is a proton donor. So, what’s a base? Bronsted-Lowry definition: A base is a proton acceptor.
They go together…like carrots and peas, Forrest.
If you are going to donate a proton, something must accept it. You can’t really be an acid without a base.
What’s the most common acid?
Water!! H-OH, it has a proton it can donate.
What’s the most common base?
Water! ° ° H - O - H It has extra electrons on the oxygen, so a proton can stick to it.
Water is special… …it is amphoteric: it can act as an acid or a base. It’s not the only compound that can, we’ll see other’s later. It also means that most Bronsted-Lowry acids or bases can dissolve in water.
We like water… Acids and bases like water… So, acids and bases are mostly found as aqueous solutions here. Like all solutions, the concentration is a critical parameter.
All solutions are created equal…
Like any other aqueous solution, a solution of either an acid or base is defined by its concentration. So what’s this thing called pH?
pH is concentration The pH scale is just a logarithmic scale for the Molarity of the protons in the solution. The pH scale is logarithmic (the difference between pH=1 and pH=2 is a factor of 10) pH is concentration
Damn those logs pH = - log [H+] [x] always means “concentration of x” [H+] should be in M. pH is ONLY the concentration of H+.
Example 1 M HCl solution. What’s the pH? Implicitly, you must recognize that: HCl(aq) → H+(aq) + Cl-(aq) Or, HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq) pH = - log [H+] = - log [H3O+] pH = - log (1 M) = 0
Another example 2.4 M H2SO4. What’s the pH? In this case, you must recognize: H2SO4 (aq) → 2H+(aq) + SO42-(aq) So [H+] = 2[H2SO4] [H+] = 2(2.4 M) = 4.8 M
ICE chart I think it is easier to see what’s going on if you follow the concentrations in an “ICE” chart. I = initial C = change (during reaction) E = End (or equilibrium) amounts
H2SO4 (aq) → 2H+(aq) + SO42-(aq) I 2
H2SO4 (aq) → 2H+(aq) + SO42-(aq) I 2.4 M 0 0 C –1x +2x +1x E 0 2x 1x Each column is a little algebra problem: Initial + change = End (-1x) = 0 X = 2.4
H2SO4 (aq) → 2H+(aq) + SO42-(aq) I 2. 4 M 0 0 C –2. 4 +2(2. 4) +2
H2SO4 (aq) → 2H+(aq) + SO42-(aq) I 2.4 M 0 0 C –2.4 +2(2.4) +2.4 E Each column is a little algebra problem: Initial + change = End (-1x) = 0 X = 2.4
The lies you were told in HS…
pH = - log [H+] = - log (4.8) = Now, pH=-0.68 might be surprising to you. A common “Earth Science lie” is that the pH scale goes from 0 to 14. I am here to tell you the truth!
The Truth according to Joe
pH = - log [H+] by definition So, pH is governed by the limits of the concentration of H+ only.
The Truth according to Joe
The highest [H+] is 56 M (because pure water is 56 M!) The smallest pH is, therefore -1.7 The highest pH is a little harder to define. What if you had 1 atom of H+ in a liter of water? Then the concentration would be 1.66x10-24 and the pH=23.78 (although is more common as the upper end of the pH scale)
Bottom Line according to Joe
pH is simply a logarithmic concentration scale.
New example What is the pH of 2.4 M NaOH solution?
New example What is the pH of 2.4 M NaOH solution? What is NaOH? It’s a Base!
New example What is the pH of 2.4 M NaOH solution? Where’s the H+? What does NaOH do in solution?
New example What is the pH of 2.4 M NaOH solution? NaOH(aq) → Na+(aq) + OH-(aq) I still don’t see the H+?
Autoionization of water
Water is a special molecule for many reasons: Polar solvent Unusually high boiling point It expands when it freezes It is amphiprotic
Amphi-who? Amphiprotic means that water can act as either an acid or a base. What happens when you mix an acid and base together? They neutralize each other.
If you mix water with water…
It neutralizes itself!
Autoionization of water
H2O + H2O ↔ H3O+ + OH- What’s with the “↔”? ↔ means that the reaction goes both ways. It is an “equilibrium reaction”
Equilibrium Reactions
Equilibrium reactions have a balance between the reactants and the products. In the case of the autoionization of water: [H3O+][OH-] = 1x10-14 (the equilibrium constant expression – is this familiar?)
What does this mean? It means that in aqueous solutions, you ALWAYS have BOTH H+ and OH- present. In fact, you can define pOH similarly to pH: pOH = - log [OH-] And pOH and pH should be related.
The Relationship How are they related? [OH-][H+] = 1x log([OH-][H+]) = - log(1x10-14) -log([OH-] + (- log[H+]) = - log(1x10-14) pOH + pH = 14 (this is the reason for the Earth Science Lie)
Returning to our base problem
What is the pH of 2.4 M NaOH solution? NaOH(aq) → Na+(aq) + OH-(aq) How would you solve it?
Returning to our base problem
What is the pH of 2.4 M NaOH solution? NaOH(aq) → Na+(aq) + OH-(aq) How would you solve it? 2 ways: Calculate [H+] from [OH-] Calculate pH from pOH
[H+] from [OH-] What is the pH of 2.4 M NaOH solution? NaOH(aq) → Na+(aq) + OH-(aq) [H+] [OH-] = 1x10-14 [H+] [2.4] = 1x10-14 [H+] = 4.17x10-15 pH = -log (4.17x10-15) = 14.38
pH from pOH What is the pH of 2.4 M NaOH solution? NaOH(aq) → Na+(aq) + OH-(aq) pOH = - log (2.4) = pH + pOH = 14 pH + (-0.38) = 14 pH = 14.38
Why is pH important? pH is a measure of the concentration of acid or base. Why do we care? It’s still all about the water. Acidic water can be corrosive to many materials. Life forms on Earth do all of their chemistry in water. This chemistry is optimized in certain pH ranges (usually 6-8).
Measuring pH There are many ways to measure pH:
Titration – most accurate – we do this a LOT Electrodes – pretty accurate & easy (pH meters) Chemical indicators – cheap and “ballpark” (this is the pool test kit)
Little problems What is the pH of an aqueous solution that is 1.25 M HCl? pH = - log [H+] = - log (1.25 M) pH =
What is the pH of an aqueous solution that is 0. 25 M NaOH
What is the pH of an aqueous solution that is 0.25 M NaOH? pOH = - log (0.25 M) = 0.60 pH + pOH = 14 pH = 14 – pOH = = 13.4 OR [H+][OH-]=1x10-14 [H+](0.25) = 1x10-14 [H+] = 4x10-14 pH = - log (4x10-14) = 13.4
Similar presentations | 2,000 | 6,188 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.896655 |
http://www.enotes.com/homework-help/identify-by-name-results-oflimit-limx-gt-0-1-cosx-332210 | 1,477,657,122,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988722459.85/warc/CC-MAIN-20161020183842-00444-ip-10-171-6-4.ec2.internal.warc.gz | 449,581,200 | 10,577 | # Identify by name or # the results oflimit limx->0 1-cosx/x^3+x^2 = limx->0 1-cos^2x/(x^3+x^2)(1+cos x) =limx->0 sin^2x/(x^3+x^2)(1+cos x) Continues from above: =limx->0 (sinx/x)^2...
Identify by name or # the results oflimit
limx->0 1-cosx/x^3+x^2 = limx->0 1-cos^2x/(x^3+x^2)(1+cos x)
=limx->0 sin^2x/(x^3+x^2)(1+cos x)
Continues from above:
=limx->0 (sinx/x)^2 limx->0 1/(x+1)(1+cosx)
=(1)^2 (1/1(1+1))=1/2
sciencesolve | Teacher | (Level 3) Educator Emeritus
Posted on
You may evaluate the limit using the half of angle formula such that:
`1 - cos x = 2 sin^2(x/2)`
You need to substitute `2 sin^2(x/2)` for `1 - cos x` in equation under limit such that:
`lim_(x-gt0) (2 sin^2(x/2))/(x^3+x^2)`
You need to use special limit `lim_(x-gt0) sin x/x = 1` , hence, you may form special limit such that:
`lim_(x-gt0) (2 sin^2(x/2))/(x^3+x^2) = 2lim_(x-gt0) ((sin^2(x/2))(x^2/4))/((x^2/4)(x^3+x^2))`
`lim_(x-gt0) (2 sin^2(x/2))/(x^3+x^2) = (2/4) lim_(x-gt0) (sin^2(x/2))/(x^2/4)*lim_(x-gt0) x^2/((x^3+x^2))`
`lim_(x-gt0) (2 sin^2(x/2))/(x^3+x^2) = (2/4)*1*lim_(x-gt0) x^2/((x^3+x^2))`
Notice that the limit `lim_(x-gt0) x^2/((x^3+x^2)) = 0/0` is indetermination, hence you may use l'Hospital's theorem such that:
`lim_(x-gt0) (x^2)/((x^3+x^2)) = lim_(x-gt0) ((x^2)')/((x^3+x^2)')`
`lim_(x-gt0) ((x^2)')/((x^3+x^2)') = lim_(x-gt0) (2x)/(3x^2+2x) `
`lim_(x-gt0) (2x)/(3x^2+2x) = lim_(x-gt0) ((2x)')/((3x^2+2x)')`
`lim_(x-gt0) ((2x)')/((3x^2+2x)') = lim_(x-gt0) 2/(6x+2)`
You need to substitute 0 for x in equation under limit such that:
`lim_(x-gt0) 2/(6x+2) = 2/(0+2) = 2/2 = 1`
`lim_(x-gt0) (2 sin^2(x/2))/(x^3+x^2) = (2/4)*1*1 = 1/2`
Hence, evaluating the limit to the function yields `lim_(x-gt0) (1- cos x)/(x^3+x^2)` yields `lim_(x-gt0) (1- cos x)/(x^3+x^2)= 1/2.`
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
The limit `lim_(x->0)(1 - cos x)/(x^3 + x^2)` has to be determined.
Substituting x = 0 in `lim_(x->0)(1 - cos x)/(x^3 + x^2)` gives the form `0/0` which is indeterminate. This allows the use of the l'Hopital's Theorem and the numerator and denominator can be substituted with their derivatives.
=> ` lim_(x->0)(sin x)/(3x^2 + 2x)`
Substituting x = 0 gives an indeterminate form `0/0` . Again replace the denominator and numerator by their derivatives
=> ` lim_(x->0)(cos x)/(6x + 2)`
Substituting x = 0 gives `1/2`
The value of `lim_(x->0)(1 - cos x)/(x^3 + x^2)` = `1/2` | 1,136 | 2,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2016-44 | latest | en | 0.578843 |
https://www.mathworks.com/matlabcentral/cody/problems/753-solitaire-cipher/solutions/253873 | 1,496,120,020,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613780.89/warc/CC-MAIN-20170530031818-20170530051818-00273.warc.gz | 1,137,513,519 | 11,787 | Cody
# Problem 753. Solitaire Cipher
Solution 253873
Submitted on 4 Jun 2013 by andrea84
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
deck = [1 4 7 10 13 16 19 22 25 28 3 6 9 12 15 18 21 24 27 2 5 8 11 14 17 20 23 26]; n = 10; out = [8 26 1 4 7 23 8 8 4 24]; assert(isequal(solitaire(deck, n),out))
``` k = 8 k = 8 26 k = 8 26 1 k = 8 26 1 4 k = 8 26 1 4 7 k = 8 26 1 4 7 23 k = 8 26 1 4 7 23 8 k = 8 26 1 4 7 23 8 8 k = 8 26 1 4 7 23 8 8 4 k = 8 26 1 4 7 23 8 8 4 24 ```
2 Pass
%% deck = 1:28; n = 10; out = [3 22 9 4 23 7 25 16 14 14]; assert(isequal(solitaire(deck, n),out))
``` k = 3 k = 3 22 k = 3 22 9 k = 3 22 9 4 k = 3 22 9 4 23 k = 3 22 9 4 23 7 k = 3 22 9 4 23 7 25 k = 3 22 9 4 23 7 25 16 k = 3 22 9 4 23 7 25 16 14 k = 3 22 9 4 23 7 25 16 14 14 ```
3 Pass
%% deck = 1:28; n = 9; out = [3 22 9 4 23 7 25 16 14]; assert(isequal(solitaire(deck, n),out))
``` k = 3 k = 3 22 k = 3 22 9 k = 3 22 9 4 k = 3 22 9 4 23 k = 3 22 9 4 23 7 k = 3 22 9 4 23 7 25 k = 3 22 9 4 23 7 25 16 k = 3 22 9 4 23 7 25 16 14 ``` | 646 | 1,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.454465 |
https://progzoo.net/wiki/Draw_a_many_pointed_star | 1,597,120,291,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738727.76/warc/CC-MAIN-20200811025355-20200811055355-00472.warc.gz | 446,963,707 | 6,771 | # Draw a many pointed star
You can draw a complex star as a polygon. A 14 point star will have 14 points on the circumference of a large circle alternating with 14 points on the circumference of a smaller circle. The parameters are:
• `s1`: radius of the large circle
• `s2`: radius of the small circle
• `n`: number of points
A point on the circumference of a circle, radius `R` at an angle of `a` from the x-axis has coordinates [1]:
```(R*cos(a), R*sin(a))
``` | 122 | 467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-34 | latest | en | 0.815119 |
https://gmatclub.com/forum/what-was-a-certain-company-s-revenue-last-year-85524.html?kudos=1 | 1,487,836,128,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171162.4/warc/CC-MAIN-20170219104611-00515-ip-10-171-10-108.ec2.internal.warc.gz | 705,966,948 | 60,459 | What was a certain company's revenue last year? : GMAT Data Sufficiency (DS)
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# What was a certain company's revenue last year?
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What was a certain company's revenue last year? [#permalink]
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20 Oct 2009, 05:51
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What was a certain company's revenue last year?
(1) Last year gross profit was \$4,100
(2) Last year revenue was 50% greater than expenses
[Reveal] Spoiler:
So, I should probably state my understanding of a P&L statement:
Total Sales/Revenues - COGS = Gross Profit
Gross Profit - Expenses = Net Income
I assumed that you would need to also know COGS, but it appears from the official answer that the GMAT is taking a more simplified approach...?
Can someone point me in the right direction here?
[Reveal] Spoiler: OA
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Re: What was a certain company's revenue last year? [#permalink]
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20 Oct 2009, 07:10
Profit = Revenue - Charge
both 1 & 2 insuff
but...
1+2 =>
P = 4100
R - 2/3R = 4100 => then R = 12300
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Re: What was a certain company's revenue last year? [#permalink]
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20 Oct 2009, 16:34
Thanks, but I'm not so much interested in an explanation to the problem as much as I am finding out more about GMAT terminology for Income Statements.
Does anyone know if GMAT questions such as this will always assume a more simplified approach of just Rev - Exp = Profit?
I guess I was taken by surprise because this is an entrance test to Business School.
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Re: What was a certain company's revenue last year? [#permalink]
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20 Oct 2009, 23:32
Actually I read a book which told that this information was assumed.
actually there are not so many information as this one that a test taker should know
you should buy an official book, you will find all of them
regards,
Alex
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Re: What was a certain company's revenue last year? [#permalink]
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29 Apr 2012, 14:07
I would like to actually answer your concern. GMAT's Gross Profit is not real accounting Gross Profit.
I did a similar question also on GMAT Prep that wanted Gross Profit but in reality, the calculation was Net Profit.
So to clarify, based on the OG and GMAT Prep, forget everything you know in the real business world, for GMAT purposes PROFIT = REVENUE - COST. GROSS PROFIT on the GMAT does not equal Number of ITEMS x Sale price per item. I know this sounds strange because when the question asks you about GROSS Profit, you automatically want to use Accounting terms and then use COGS to subtract and get NET PROFITS. For some strange reason GMAT assumes all PROFIT is the same and it is a net profit type of calculation that you tweak a bit based upon what you think is being asked. I know this isn't a perfect answer but I did come across this post trying to figure out a definitive answer for all scenarios.
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Re: What was a certain company's revenue last year? [#permalink]
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29 Apr 2012, 22:00
Jones105,
Remember one thing here. Even though GMAT is a test of entry in to a Business school, the level of stuff tested in the exam is "High school" level. "Profit = Revenue - Cost" is the funda taught in that level and I guess, GMAC won't expect more than that at least for GMAT. I hope I am clear.
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Re: What was a certain company's revenue last year? [#permalink]
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03 Jan 2015, 09:59
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: What was a certain company's revenue last year? [#permalink]
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20 May 2015, 21:18
This can't be a 600-700 question definitely.
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Re: What was a certain company's revenue last year? [#permalink] 20 May 2015, 21:18
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1. CJM 2011 (vol 63 pp. 1238)
Bump, Daniel; Nakasuji, Maki
Casselman's Basis of Iwahori Vectors and the Bruhat Order W. Casselman defined a basis $f_u$ of Iwahori fixed vectors of a spherical representation $(\pi, V)$ of a split semisimple $p$-adic group $G$ over a nonarchimedean local field $F$ by the condition that it be dual to the intertwining operators, indexed by elements $u$ of the Weyl group $W$. On the other hand, there is a natural basis $\psi_u$, and one seeks to find the transition matrices between the two bases. Thus, let $f_u = \sum_v \tilde{m} (u, v) \psi_v$ and $\psi_u = \sum_v m (u, v) f_v$. Using the Iwahori-Hecke algebra we prove that if a combinatorial condition is satisfied, then $m (u, v) = \prod_{\alpha} \frac{1 - q^{- 1} \mathbf{z}^{\alpha}}{1 -\mathbf{z}^{\alpha}}$, where $\mathbf z$ are the Langlands parameters for the representation and $\alpha$ runs through the set $S (u, v)$ of positive coroots $\alpha \in \hat{\Phi}$ (the dual root system of $G$) such that $u \leqslant v r_{\alpha} < v$ with $r_{\alpha}$ the reflection corresponding to $\alpha$. The condition is conjecturally always satisfied if $G$ is simply-laced and the Kazhdan-Lusztig polynomial $P_{w_0 v, w_0 u} = 1$ with $w_0$ the long Weyl group element. There is a similar formula for $\tilde{m}$ conjecturally satisfied if $P_{u, v} = 1$. This leads to various combinatorial conjectures. Keywords:Iwahori fixed vector, Iwahori Hecke algebra, Bruhat order, intertwining integralsCategories:20C08, 20F55, 22E50
top of page | contact us | privacy | site map | | 543 | 1,753 | {"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-2015-40 | longest | en | 0.714773 |
http://mathhelpforum.com/pre-calculus/34757-integration.html | 1,508,613,307,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824894.98/warc/CC-MAIN-20171021190701-20171021210701-00240.warc.gz | 212,355,401 | 10,342 | 1. ## integration
would you possibly be able to show me step by step how to integrate this with respect to x:
((-x)^2)/(1+x^3)
Thanks
2. Originally Posted by daaavo
would you possibly be able to show me step by step how to integrate this with respect to x:
((-x)^2)/(1+x^3)
Thanks
$\int \frac{(-x)^2}{1+x^3}dx=\int \frac{x^2}{1+x^3}dx$
let $u=1+x^3 \mbox{ then } du=3x^2dx \iff \frac{1}{3}du=x^2dx$
so
$\int \frac{x^2}{1+x^3}dx=\frac{1}{3} \int \frac{1}{u}du =\frac{1}{3}\ln|u|+C =\frac{1}{3}\ln|1+x^3|+C$
3. Originally Posted by daaavo
would you possibly be able to show me step by step how to integrate this with respect to x:
((-x)^2)/(1+x^3)
Thanks
$\int \frac{(-x)^2}{1+x^3} \, dx$
$\int \frac{x^2}{1+x^3} \, dx$
$\frac13 \int \frac{3x^2}{1+x^3} \, dx$
Now make the substitution $u = x^3 \Rightarrow 3x^2 \, dx = du$
$\frac13 \int \frac{1}{1+u} \, du = \frac13\ln |u + 1| + C= \frac13\ln|1+x^3| +C$ | 379 | 920 | {"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": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-43 | longest | en | 0.688315 |
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• martingreen10
# Purposeful Maths Calculation Policy
Calculation Policies are a contentious topic, and we know there are many sides to this debate. We have taken the time to write this short blog to introduce our calculation policy for several reasons:
● Explain the background to why we have introduced one
● Explain how we use it
● To receive feedback (both positive and constructive) so we can improve it!
Background:
We think a great starting point for this is to read Jo Morgan’s blog on “Autonomy Vs Consistency” which can be found here (https://www.resourceaholic.com/2021/05/autonomy-vs-consistency.html). Like all of her work, it gives you something to think about. It goes into great detail and is a fascinating read. What resonated with us is the “Autonomy Spectrum'' she explained, shown below:
I am lucky to lead an experienced department made up of great teachers who are all subject specialists, so our intention is definitely not the red zone. Whilst this may work best in some situations it certainly would not be the best for my department. I believe the sweet spot for my department is somewhere between the third and fifth block from the left. I would like some consistency to help students when they move between teachers (due to split classes, changing sets, or new academic year) without turning my team into robots who would be restricted with zero autonomy.
We have centralised assessments and feedback policies to ensure each student is hitting the same check points and receiving consistent feedback. Our schemes of learning have clear paths to take and what topics should be taught. However, the class teacher can choose which “abstract” method they believe fits the class best.
The four stages: Physical, Pictorial, Semi-Abstract, Abstract
Each “sub unit” in our schemes of learning has these 4 stages mapped out in our calculation policy using a specific example.
We had spent some time looking at the concrete – pictorial – abstract model but found the gap between pictorial and abstract was too wide. Hence, we introduced a “semi-abstract” stage with the aim of making this step more manageable.
Each class starts at the “physical stage” and moves right as they understand the steps. How long a class spends on each stage is up to the class teacher. The aim is for each student to eventually get to the abstract stage so they can solve problems efficiently. However, if students are comfortable at a previous stage and are understanding the work they don’t have to get to the abstract stage. For example, expanding double brackets:
Some students may feel comfortable using the semi abstract method (grid method in this case) but start to make mistakes when moving to the abstract method. Staff don’t have to move students to the abstract stage initially. It is important all students have the same experience though, as when this topic is revisited later in the year (or future years) the child may have a different teacher and we want to be able to go back to familiar methods.
Didn’t you say teachers can choose the methods they teach?
They can, and they do! The “abstract” method is where we would traditionally start teaching and staff can still choose methods that they prefer. The ones in the policy are methods we often use – but are not limited to. The introduction of manipulatives is new to the department and we are doing a lot of work to embed them in our practice. Therefore, the policy is designed to show teachers how best to use them for each topic – as this is where we lack experience. The policy was put together to reduce workload so individual staff members didn’t have to think about how to use the manipulatives. We then centrally plan examples, and spend department time solving problems. It is a working document that the department are trialing this year. At regular intervals we will discuss what is working, what isn’t, and what needs tweaking.
We welcome feedback on it, as we aren’t for one second saying it is perfect! However we think it is a great starting point for discussions!
Phil and Martin | 834 | 4,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-22 | longest | en | 0.954991 |
http://mathhelpforum.com/advanced-algebra/130152-3-rotated-coordinate-systems-print.html | 1,527,089,960,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865679.51/warc/CC-MAIN-20180523141759-20180523161759-00572.warc.gz | 182,734,564 | 3,026 | # 3 rotated coordinate systems.
• Feb 22nd 2010, 11:45 AM
Jazzjohn
3 rotated coordinate systems.
I have 3 sets of X,Y,Z axes which differ by rotations only. Call them A, B, and C.
In A, there is a unit vector (0, m, n). The same unit vector relative to A exists in both B and C as (bx, by, bz) and (cx, cy, cz).
I need to find the angle between the X axis of B and the X axis of C as projected ("flattened") onto A's XY plane.
It's been 35 years since my last linear algebra class, so any clues would be greatly appreciated!(Thinking)
John
• Feb 23rd 2010, 08:29 AM
Jazzjohn
As an afterthought, the problem can be simplified.
A unit vector in X,Y,Z is (0,0.34,-0.94). (Please excuse the rounding error)
An axes rotation occurred which caused the same vector to be represented as (a,b,c).
How much did the axes rotate around the Z axis for a given a,b,c?
• Feb 23rd 2010, 03:00 PM
Jazzjohn
I seem to be answering my own question as things come back to me...
Here's a plan that might work:
The rotation matrix that rotates (0, .34, -.94) to the X axis is
0 , .34 , -.94
-1 , 0 , 0
0 , .94 , .34
Apply this to (a, b, c) to get the rotated X axis, X' (in terms of a,b and c).
Next find the rotation matrix T that rotates the X axis to X'
T will be the product of 3 rotations. From T, solve for the Z rotation matrix which will give the angle rotated around Z. The angle will be in terms of a,b and c.
At this point I don't know how to find T or if there is a better way. | 436 | 1,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-22 | latest | en | 0.938588 |
https://ltwork.net/abox-has-5-beads-of-the-same-size-but-all-are-different--12157662 | 1,674,844,107,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764495001.99/warc/CC-MAIN-20230127164242-20230127194242-00280.warc.gz | 366,387,503 | 12,408 | # Abox has 5 beads of the same size, but all are different colors. tina draws a bead randomly from the
###### Question:
Abox has 5 beads of the same size, but all are different colors. tina draws a bead randomly from the box, notes its color, and then puts the bead back in the box. she repeats this 3 times. what is the probability that tina would pick a red bead on the first draw, then a green bead, and finally a red bead again? 1 over 625 1 over 180 1 over 150 1 over 125
### Chartered colonies were those that were founded .A.for economic purposesB.for religious purposesC.to establish trading postsD.with the permission
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### Table 1:x 0 -4 3 -2 9 3y 5 8 7 8 5 -2function or not function?Table 2:x 2 -7 5 1 -6 3y 5 8 -6 8 5 7function or not function?Which
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Idk what the answer is: Carlos mixed \dfrac4{15}\text{ L} 15 4 Lstart fraction, 4, divided by, 15, end fraction, start text, space, L, end text of chocolate syrup with \dfrac12\text{ L} 2 1 Lstart fraction, 1, divided by, 2, end fraction, start text, space, L, end text of milk....
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### 4/3*w = 36 what is w? how do i check?
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### What is the total number of unique triangles that can be formed with side lengths of 6.5 centimeters,6 centimeters, and 2.5 centimeters?
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Tamara has decided to start saving for spending money for her first year of college. her money is currently in a large suitcase under her bed, modeled by the function s(x) = 450. she is able to babysit to earn extra money and that function would be a(x) = 6(x − 2), where x is measured in hours. ex...
PLS EASY HELP ME ITS SCIENCE PLS ✋✋✋✋ $PLSSSS EASY HELP MEEEEE ITS SCIENCE PLSSSS ✋✋✋✋$... | 1,486 | 5,449 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-06 | latest | en | 0.94939 |
https://bafybeiemxf5abjwjbikoz4mc3a3dla6ual3jsgpdr4cjr3oz3evfyavhwq.ipfs.dweb.link/wiki/Bretschneider's_formula.html | 1,660,798,727,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573163.7/warc/CC-MAIN-20220818033705-20220818063705-00230.warc.gz | 134,512,429 | 5,367 | Bretschneider's formula
In geometry, Bretschneider's formula is the following expression for the area of a general quadrilateral:
Here, a, b, c, d are the sides of the quadrilateral, s is the semiperimeter, and α and γ are two opposite angles.
Bretschneider's formula works on any quadrilateral, whether it is cyclic or not.
The German mathematician Carl Anton Bretschneider discovered the formula in 1842. The formula was also derived in the same year by the German mathematician Karl Georg Christian von Staudt.
Proof
Denote the area of the quadrilateral by K. Then we have
Therefore
The law of cosines implies that
because both sides equal the square of the length of the diagonal BD. This can be rewritten as
Adding this to the above formula for 4K2 yields
Note:
Following the same steps as in Brahmagupta's formula, this can be written as
Introducing the semiperimeter
the above becomes
and Bretschneider's formula follows.
Related formulas
Bretschneider's formula generalizes Brahmagupta's formula for the area of a cyclic quadrilateral, which in turn generalizes Heron's formula for the area of a triangle.
The trigonometric adjustment in Bretschneider's formula for non-cyclicality of the quadrilateral can be rewritten non-trigonometrically in terms of the sides and the diagonals e and f to give[1][2]
Notes
1. J. L. Coolidge, "A historically interesting formula for the area of a quadrilateral", American Mathematical Monthly, 46 (1939) 345–347. (JSTOR)
2. E. W. Hobson: A Treatise on Plane Trigonometry. Cambridge University Press, 1918, pp. 204-205 | 385 | 1,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.871503 |
https://stats.stackexchange.com/questions/486060/do-classical-statistical-methods-assume-that-they-contain-or-have-access-to-the | 1,621,097,828,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990551.51/warc/CC-MAIN-20210515161657-20210515191657-00014.warc.gz | 577,342,075 | 39,624 | # Do classical statistical methods assume that they contain or have access to the true distribution?
I was reading recently about Minimum Description Length (MDL) that does not have to assume that the true model is in the model class. When this feature of MDL is highlighted in the literature, the standard statistical inference is usually criticised at the same time that its methods are designed under the assumption that they contain the true distribution. The pioneering papers of Rissanen and others mention this.
My question is: Which of the classic statistical methods make the assumption that they have the true model in the model class or that the true model has to be considered in the analysis? What happens to the results if the true model is not in the model class (which will be the case most of the time, I believe)? Are the results still valid in such cases?
I can clearly see why MDL does not assume that it has access to the true model, but it is not clear to me why the opposite is true in more traditional statistical models.
• "...the assumption that they have the true model in the model class..."---this is assumed in all of classical (parametric) statistics. E.g. take any foundational result---just to name a few, Neyman-Pearson, existence of most powerful or minimax test, Maximum Likelihood Principle, etc. Some results, however, can be extended to the setting where the true model does not lie in the class of models being considered, i.e. under mis-specification. For example, the maximum likelihood estimator minimizes KL divergence between the class over which likelihood is maximized and the true model. – Michael Sep 4 '20 at 21:49
• MDL drops the whole concept of "true model". – carlo Sep 5 '20 at 1:22
Perhaps an example will assist here. Suppose you have a parametric model for data $$\mathbf{x}$$ that depends on an unknown parameter $$\theta \in \Theta$$. One of the concepts formed in classical statistics is the idea of "consistency" of an estimator. Broadly speaking, for an estimator $$\hat{\theta}$$ of the parameter $$\theta$$, this means that $$\hat{\theta} \rightarrow \theta$$ (in some appropriate probabilistic sense) as $$n \rightarrow \infty$$. In order for the estimator to be considered "consistent", this property must hold for all $$\theta \in \Theta$$. Thus, so long as the assumed model form for the likelihood function is correct, a consistent estimator will tend to estimate well in the long run --- this is guaranteed regardless of the unknown parameter value. (Note here that the theoretical derivation of the consistency property involves looking at the behaviour of an estimator under an assumed true parameter value, and varying the assumed true value over the entire parameter space. However, once a particular estimator is known to be consistent within a particular model, for applied work we do not assume that the true parameter is accessible.) | 617 | 2,917 | {"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": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-21 | latest | en | 0.918293 |
https://www.physicsforums.com/threads/2-d-motion-help.155608/ | 1,579,337,780,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250592394.9/warc/CC-MAIN-20200118081234-20200118105234-00536.warc.gz | 1,041,759,050 | 15,080 | # 2-D motion help
A particle's trajectory is defined as x = (1/2t^3 - 2t^2) m and y = (1/2t^2 - 2t)m where t is in s.
What is the particle's direction of motion, measured from the x axis, at t=0 and t= 3.5, measured in degrees counterclockwise from the x axis.
I started out by plugging in t = 0 and t = 3.5 to the equations given. I then attemped to take arctan(y/x) in both cases to find the direction of motion. my answers for these, 0 degrees and 15.9 degrees respectively, both came back as wrong.
Any clues as to what I am doing wrong/how to do it right?
Related Introductory Physics Homework Help News on Phys.org
Integral
Staff Emeritus | 191 | 649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-05 | longest | en | 0.947543 |
https://creativeakademy.org/2015/11/how-to-make-star-without-crossing-lines-with-mswlogo.html | 1,680,004,321,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00651.warc.gz | 229,474,459 | 24,203 | # How to Make Star without Crossing Lines with MSWLogo
#### Byxakbcreative
Nov 30, 2015
Step 1 : Firs we need to calculate the total angle of pentagon formed inside the star. The formula of total angle of a pentagon is 90x(2n-4). One internal angle is {90x(2n-4)}/n. Thus putting the value of n=5, we get internal angle as 1080 .
Step 2 : Therefore base angle of the highlighted triangle will be:
1800-1080 = 720.
Step 3 : Therefore angle on the joint of the lines at second turn as shown in the diagram will also be 72 as shown in the image.
Step 4 : Therefore the final command for making the star without crossing lines will be :
repeat 5 [ fd 200 rt 144 fd 200 lt 72] ↵
That’s all. If you have any problem in creating any shape with MSWLogo, then feel free to comment below.
READ : Main Commands of Logo Part 2 - MSWLogo | 234 | 832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-14 | longest | en | 0.821812 |
https://tex.stackexchange.com/questions/570834/align-equations-in-cases-around-the-equal-sign | 1,696,067,978,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510671.0/warc/CC-MAIN-20230930082033-20230930112033-00816.warc.gz | 615,804,896 | 36,706 | # Align equations in cases around the equal sign
So I'm trying to do a simple thing with cases:
$\begin{cases} -x+5y&=6\\ x-3y&=-4\\ \end{cases} \\ \begin{cases} -x+5y&=6\\ x&=3y-4\\ \end{cases}$
And of course, LaTeX doesn't do what it should. For some reason, the "x" is far to the left.
How to make it look centered around the equal sign?
• The alignment point in cases is for the “conditions”, not for aligning equals signs. Nov 14, 2020 at 15:48
• .... So how do I write the system of equations like people do?! Nov 14, 2020 at 15:49
• 'LaTeX doesn't do what it should' --- It did what your code told it to do. Also you don't have to dig through a mountain of websites. Try starting with the Not So Short Guide and the Short Math Guide. Nov 14, 2020 at 16:15
• I dug through the first one years ago. My point is that it's not intuitive, unlike regular word processors. Nov 14, 2020 at 16:28
• It's intuitive once you get to grips with it. It's like a programming language: it does what you tell it to. Nov 14, 2020 at 16:31
The cases environment is not meant for systems of equations as the name itself hints.
You want a proper alignment environment, most likely aligned:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align*}
&\left\{
\begin{aligned}
-x+5y&=6\\
x-3y&=-4\\
\end{aligned}
\right.
\\
&\left\{
\begin{aligned}
-x+5y&=6\\
x&=3y-4\\
\end{aligned}
\right.
\end{align*}
\end{document}
• Well, okay, that works, Cases automatically added the curly brackets, so I wouldn't have to reinvent the wheel. One more problem, now the whole thing is in the center of the page, previously it was to the left of the oage like I wanted. Nov 14, 2020 at 16:02
• Also, what about the columns? What if I'd like to insert second system of equations next to it? Nov 14, 2020 at 16:25
For the first ‘cases’, thez systeme package yields a nicer layout (and a simpler code). For the second “cases’, you can use the aligned environment inside cases:
\documentclass[12pt,a4paper]{article}
\usepackage{systeme}
\usepackage{mathtools}
\begin{document}
\noindent\systeme{-x+5y=6, x-3y=-4} \\[1ex] \begin{cases} \begin{aligned} -x+5y & =6,\\ x & =3y-4 \end{aligned} \end{cases}
\end{document}
• Oh, cool, aligned inside cases. That simplifies things. Thanks! Nov 14, 2020 at 16:19
• Don't forget that cases adds unwanted space on the right. Nov 14, 2020 at 16:20 | 746 | 2,393 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-40 | latest | en | 0.883501 |
https://www.scholastic.com/teachers/articles/teaching-content/count-cause/ | 1,537,973,173,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267165261.94/warc/CC-MAIN-20180926140948-20180926161348-00216.warc.gz | 859,993,130 | 20,515 | Chores for Charity
At Gethsemane Lutheran School in Tempe, Arizona, third-grade teacher Marissa Moffitt encourages students to find additional chores to do around the house and asks parents to pay children for their extra work—the money they earn is used to go shopping for local charities. In past years, the class has bought items for a food pantry, a pet shelter, and needy children (via the Salvation Army).
“The children use their math skills as they are shopping to estimate the cost of the item, and then use their mental math to keep a running total of how much they’ve spent,” Moffitt says. “They’re also encouraged to look for bargains as they shop and to remember that store brands are a cheaper option. They do everything—earn the money, shop, check out, and even deliver the items. It’s not only a service project but also a great learning experience.”
Delicious Fractions
“Cooking is excellent at teaching the basics of math through collecting and measuring ingredients for a given batch size, and it introduces fractions in a tasty way,” says Rebecca Klemm, a former teacher and founder of Numbers Alive!, a nonprofit focused on demystifying numbers and making math fun.
When your class holds a bake sale to benefit a local charity, students gain math skills in two ways—through the baking process and then by taking responsibility for money handling during the bake sale itself. You can also have students keep track of bake sale costs, then compare them with the sales totals to calculate net profits for the event.
Cause-in-a-Can
Involve students in helping to choose a cause to support, whether it’s buying books for an impoverished school, gathering funds to support an environmental cause, or helping with medical bills for a sick student at your school. Set a fundraising goal and discuss what, specifically, will be accomplished with the funds. Have kids work together to determine how many quarters (or pennies, or dollars) each student in the class would need to raise to meet the goal. If you involve other classes or the entire school, how much would each class need to raise? Ask students to bring in mason jars or tin cans and decorate them with images about the cause. Let students collect change from the jars in each classroom and count the money on a daily or weekly basis, then post the totals where everyone can see. At the end of the project, celebrate the money raised and the difference it will make.
The Math of Sports
Klemm recommends organizing a track meet, basketball tournament, or other athletic event for students from underprivileged schools or organizations like the Boys and Girls Club. (If that’s too complicated, students could organize their own event and donate proceeds from ticket sales.)
Use the sporting event students choose to teach them the principles of math and science. In a track meet, for example, students must organize the runners into heats by age or ability and use stopwatches to track runners’ times; prior to the meet, they could spend time learning about force, motion, and resistance and how they affect performance in track and field. For basketball, students can learn about arcs and angles. In hosting the event, students will also need to create a comprehensive budget that accounts for all of the event’s costs, such as advertising, invitations, prizes, and refreshments.
Recycling by Numbers
If aluminum cans or glass bottles are returnable for a refund or recyclable for payment in your state, help your class organize a neighborhood recycling day. Students can collect returnable cans or bottles from local households or businesses, and they will learn counting, money handling, and the importance of recycling, along the way. Select a neighborhood, distribute flyers notifying residents or businesses of the project and the pickup day, then send students and chaperones around to pick up returnable cans and bottles from their designated areas. Each child should take note of the total he or she collects so the class can compile data on the amount they’ve collected overall. Students can also study the variation in the totals from different areas. Use the money earned from returned bottles or cans to purchase recycling containers for your school or neighborhood.
Cost-Benefit Analysis
Teach your class about how to start a business, including planning for costs, expenses, and revenues, by asking each student to start his or her own “company.” Each student should determine what type of business he or she will operate—selling homemade bracelets, snacks, or artwork, for example, or providing services such as manicures or temporary tattoos. | 916 | 4,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-39 | longest | en | 0.962252 |
http://www.gurufocus.com/term/e10/CSI/E10/Cutwater%2BSelect%2BIncome%2BFund | 1,419,055,079,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802769419.87/warc/CC-MAIN-20141217075249-00054-ip-10-231-17-201.ec2.internal.warc.gz | 551,359,608 | 24,815 | Switch to:
Cutwater Select Income Fund (:CSI)
E10
\$0.00 (As of . 20)
E10 is a concept invented by Prof. Robert Shiller, who uses E10 for his Shiller P/E calculation. E10 is the average of the inflation adjusted earnings of a company over the past 10 years.
Cutwater Select Income Fund's adjusted earnings per share data for the fiscal year that ended in . 20 was \$. Since most companies do not have as long as 10 years history, here we use 6 years to calculate. Add all the adjusted EPS for the past 6 years together and divide 6 will get our e10, which is \$0.00 for the trailing six years ended in . 20.
As of today, Cutwater Select Income Fund's current stock price is \$19.49. Cutwater Select Income Fund's E10 for the trailing six years ended in . 20 was \$0.00. Cutwater Select Income Fund's Shiller P/E Ratio of today is .
Definition
E10 is a concept invented by Prof. Robert Shiller, who uses E10 for his Shiller P/E calculation. When we calculate the today's Shiller P/E ratio of a stock, we use todays price divided by E10.
What is E10? How do we calculate E10?
E10 is the average of the inflation adjusted earnings of a company over the past 10 years. Lets use an example to explain.
If we want to calculate the E10 of Wal-Mart (WMT) for Dec. 31, 2010, we need to have the inflation data and the earnings from 2001 through 2010.
We adjusted the earnings of 2001 earnings data with the total inflation from 2001 through 2010 to the equivalent earnings in 2010. If the total inflation from 2001 to 2010 is 40%, and Wal-Mart earned \$1 a share in 2001, then the 2001s equivalent earnings in 2010 is \$1.4 a share. If Wal-Mart earns \$1 again in 2002, and the total inflation from 2002 through 2010 is 35%, then the equivalent 2002 earnings in 2010 is \$1.35. So on and so forth, you get the equivalent earnings of past 10 years. Then you add them together and divided the sum by 10 to get E10.
For example, Cutwater Select Income Fund's adjusted earnings per share data for the fiscal year that ended in . 20 was:
Adj_EPS = Earnigns per Share / CPI of . 20 * CPI of Current = / * 235.51 =
Since most companies do not have as long as 10 years history, GuruFocus uses 6 years to do the calculation.
Current CPI = 235.51.
Cutwater Select Income Fund does not have a history long enough to calculate E10. Therefore GuruFocus does not calculate it.
Explanation
If a company grows much fast than inflation, E10 may underestimate the company's earnings power. Shiller P/E Ratio can seem to be too high even the actual P/E is low.
For the Shiller P/E, the earnings of the past 10 years are inflation-adjusted and averaged. The result is used for P/E calculation. Since it looks at the average over the last 10 years, the Shiller P/E is also called PE10.
The Shiller P/E was first used by professor Robert Shiller to measure the valuation of the overall market. The same calculation is applied here to individual companies.
Cutwater Select Income Fund's Shiller P/E Ratio of today is calculated as
Shiller P/E Ratio = Share Price / E10 = 19.49 / =
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Be Aware
Shiller P/E Ratio works better for cyclical companies. It gives you a better idea on the company's real earnings power.
Related Terms
Shiller P/E Ratio
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Cutwater Select Income Fund Annual Data
e10 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
Cutwater Select Income Fund Semi-Annual Data
e10 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 1,019 | 3,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2014-52 | longest | en | 0.937799 |
http://stackoverflow.com/questions/10130701/how-to-find-if-a-point-is-within-a-set-of-intervals | 1,394,857,508,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678695829/warc/CC-MAIN-20140313024455-00039-ip-10-183-142-35.ec2.internal.warc.gz | 139,216,205 | 19,148 | # How to find if a point is within a set of intervals?
I am looking for the fastest way to decide whether or not a point on a line is within a subset of this line. I am given an integer Point, and I also have a "list" of either:
1. Points, represented by an integer ( 3, 10, 1000, etc)
2. Intervals, that I represent by 2 integers ( 2:10 is all integers from 2 to 10 inluded, 50:60, etc)
In this example, if the value of my point is 5, then I return true because it is included in an interval, same for 55. If my point is equal to 1000, I also return true because it matches the list of points.
I am looking for a fast way (quicker than linear) to check for this condition, WITHOUT having to instanciate as many integer as there are possible points (ie, for a 1:1000 interval I don't want to instanciate 1000 integers). Can this be done in a logarithmic time?
Thanks
edit : you can consider that any time taken to pre-process the list of data is equal to 0, because once my initial intervals are processed I need to apply this test to 10k points
-
can the intervals overlap? I don't know for sure if that matters, but it feels like it should. – Almo Apr 12 '12 at 19:52
they could, but I can pre-process my data so that they don't anymore, which is no a problem time-wise because I'm using the same interval sets to then process 10k points – lezebulon Apr 12 '12 at 19:54
Are they ordered? – Freddy Apr 12 '12 at 19:55
->check the edit – lezebulon Apr 12 '12 at 20:01
Do you need to know which intervals a point is in, or merely whether it is in any of them or not? – Karl Bielefeldt Apr 12 '12 at 20:29
Hm, maybe you can use an interval or a segment tree:
-
+1 for the correct answer. This is a well-studied computational geometry problem ("1D stabbing query"). – Nemo Apr 12 '12 at 23:45
If you have the integers ranges sorted and the ranges are non-overlapping, you can perform binary search to find the correct range in logarithmic time.
Are there any constraint on the range? Based on that you can probably come up with hashing function to search in constant time. But this depends on how your constraints are.
-
I think I can assume that the range is between 0 and 10 millions. – lezebulon Apr 12 '12 at 19:59
If some ranges overlap you can sort them and collapse the overlapping ones into a single range. – Mark Ransom Apr 12 '12 at 20:26
First check a hash_map of points. That's the simple check.
Then simply order a map of intervals by the first coordinate and then find lower_bound of the point.
Then check if you are contained in the element returned. If you aren't in that, you aren't in any.
-
There seem to be assumptions in some of the responses that the intervals might overlap. You are in control of the data structure you use to solve this issue - it has no dependency necessary on outside or the initial interval set. So you should not be storing overlapping intervals in general - join them when inserting into the map. Whenever dealing with intervals this is fairly standard. – ex0du5 Apr 12 '12 at 20:02
You could do that in sublinear time GIVEN a tree data structure (I'd recommend a B-tree),if you don't count the time taken to build the tree (most trees take n log n or similar time to build).
If you just have a plain list, then you cannot do better than linear because in the worst case you potentially have to check all points and intervals.
-
You can use a Bloom Filter to test a point and see if it's not in an interval, in linear O(1) time. If it passes that test you must use another method such as a binary search to see if it's definitely part of an interval, in O(log n) time.
-
Is the idea to hash each point in the interval? – Matthias Vallentin Apr 12 '12 at 20:13
@MatthiasVallentin, yes it is. The size of the Bloom Filter depends on the number of points set and the probability of false positives, not on the possible range of inputs. – Mark Ransom Apr 12 '12 at 20:24
Thanks, I understand your idea now. However, there are many choices which Bloom filter parameters to fix initially. Since this data structure is often used in space-constrained environments, a common approach is to assume a fixed size and set cardinality to derive the optimal value of k, the number of hash functions. Could you clarify what you mean by "size"? Once instantiated, the size of the (basic) Bloom filter typically does not change anymore. – Matthias Vallentin Apr 12 '12 at 21:30
@MatthiasVallentin, sorry I was unclear. What I meant to say is that once the number of points and the probability of false positives (and the number of hash functions) has been selected, the size of the filter arrays can be calculated. The point being it is not dependent on the range of inputs. – Mark Ransom Apr 12 '12 at 21:42
After reflexion, I think that the following code should work in logarithmic time, excluding the time needed to build the map:
``````enum pointType {
point,
open,
close
};
std::map<long int, pointType> mapPoints;
mapPoints.insert(std::pair<long int, pointType>(3, point));
//create the 5:10 interval:
mapPoints.insert(std::pair<long int, pointType>(5, open));
mapPoints.insert(std::pair<long int, pointType>(10, close));
int number = 4;
bool inside = false;
std::map<long int, pointType>::iterator it1 = mapPoints.lower_bound(number);
if(it1->first == number || it1->second == close) {
inside = true;
}
``````
I think this should work as long as the map is filled properly with non-overlapping intervals
- | 1,382 | 5,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | latest | en | 0.928943 |
https://cs.stackexchange.com/questions/142528/reducing-subsetsum-to-g-l-u-g-is-a-weighted-graph-that-has-a-spanning-tre/142530 | 1,631,927,451,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056120.36/warc/CC-MAIN-20210918002951-20210918032951-00326.warc.gz | 245,267,413 | 32,612 | # Reducing subsetsum to {<G, l, u> | G is a weighted graph that has a spanning tree with weight between l and u} [duplicate]
How can I reduce Subsetsum (or maybe other np-complete problem) problem to the problem below?
input : a weighted graph $$G$$ and numbers $$l$$ and $$u$$.
output : Does $$G$$ has spanning tree, $$S$$, such that $$l \leq \mathrm{weight}(S) \leq u$$?
Let $$X = \{ x_1, x_2, \dots, x_n \}$$ be the (multi-)set of elements in your subset-sum instance and let $$t$$ be the target value.
Create the undirected graph $$G=(V,E)$$ where $$V=\{a,b\} \cup X$$ and $$E = ( \{a,b\} \times X) \cup \{(a,b)\}$$. The weight of edge $$(a, x_i)$$, for $$x_i \in X$$, is $$x_i$$. The weight of edge $$(b, z)$$ for $$z \in \{a\} \cup X$$ is $$0$$. Finally, pick $$l=u=t$$.
There is some $$Y \subset X$$ such that $$\sum_{x_i \in Y} x_i = t$$ if and only if there is a spanning tree $$T$$ of $$G$$ of total weight between $$l$$ and $$u$$.
Indeed, given $$Y$$, you can select $$T$$ as the tree induced by the edges in $$\{(a,x_i) \mid x_i \in Y\} \cup \{(b,z) \mid z \in V \setminus Y\}$$. Moreover, given a tree $$T=(V,F)$$ of total weight between $$l$$ and $$u$$, you can select $$Y = \{ x_i \mid (a,x_i) \in F \}$$. | 427 | 1,223 | {"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": 31, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2021-39 | latest | en | 0.828257 |
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## #1 2006-04-10 00:12:20
RauLiTo
Full Member
Offline
### in the circle
get A and B which make ( AX + BY ) ² = X (4 Y - 3 X ) + 1 an equation for a circle and then get the centre and the radius .
ImPo\$\$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...
## #2 2006-04-10 02:07:53
George,Y
Super Member
Offline
### Re: in the circle
A²X²+2ABXY+B²Y²=4XY-3X²+1
(A²+3)X²+(2AB-4)XY+B²Y²=1
∴2AB-4=0 AND A²+3=B² AND CENTER IS (0,0)
X²/P²+Y²/P²=1 WHERE 1/P²=A²+3=B²
X²+Y²=P² SO THE RADIUS IS P
A=2/B, 4/B²+3=B²=DEF=X
4/X+3=X
you can go on from here
X'(y-Xβ)=0
## #3 2006-04-10 05:47:13
RauLiTo
Full Member
Offline
### Re: in the circle
excuse me can you finish it please ?
it sometimes seems difficult for me to translate that to arabic u know ... sorry everybody for botering
ImPo\$\$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...
## #4 2006-04-10 11:23:25
George,Y
Super Member
Offline
### Re: in the circle
4/X+3=X
4+3X=X²
(X-4)(X+1)=0
center is O, (0,0)
X'(y-Xβ)=0
## #5 2006-04-10 16:20:18
RauLiTo
Full Member
Offline
### Re: in the circle
thanks man
ImPo\$\$!BLe = NoTH!nG
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1. In a game, players take it in turns to flip two fair coins
If a player gets 1 head they score 1 point.
If a player gets 2 heads they score 3 points
The Object of the game is to get 7 or more points
Work out the probability of a player getting 7 or more points on their third turn
2. (Original post by hbos14)
In a game, players take it in turns to flip two fair coins
If a player gets 1 head they score 1 point.
If a player gets 2 heads they score 3 points
The Object of the game is to get 7 or more points
Work out the probability of a player getting 7 or more points on their third turn
The probability is 50/50 each time so to get 7 points you need 3 heads in a row for that you multiply the probabilities together so 0.5*0.5*0.5 which gives you 12.5%. Make sense ?
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100s of GCSE past papers for all your subjects at your fingertips. | 497 | 1,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | latest | en | 0.904082 |
https://socratic.org/questions/how-do-you-rationalize-the-denominator-and-simplify-6sqrt5-5sqrt3 | 1,624,171,035,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487658814.62/warc/CC-MAIN-20210620054240-20210620084240-00328.warc.gz | 446,531,409 | 5,703 | # How do you rationalize the denominator and simplify (6sqrt5)/(5sqrt3)?
Aug 29, 2016
$\left(\frac{2}{5}\right) \sqrt{15}$.
#### Explanation:
The Exp.$= \frac{6 \sqrt{5}}{5 \sqrt{3}}$
$= \frac{6 \sqrt{5}}{5 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$= \frac{\left(\cancel{3} \times 2\right) \left(\sqrt{5 \times 3}\right)}{\cancel{3} \times 5}$
$= \left(\frac{2}{5}\right) \sqrt{15}$. | 163 | 393 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-25 | latest | en | 0.371313 |
https://www.gradesaver.com/textbooks/math/other-math/thinking-mathematically-6th-edition/chapter-5-number-theory-and-the-real-number-system-5-2-the-integers-order-of-operations-exercise-set-5-2-page-269/73 | 1,534,230,331,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221208676.20/warc/CC-MAIN-20180814062251-20180814082251-00308.warc.gz | 1,183,438,298 | 14,844 | ## Thinking Mathematically (6th Edition)
$\frac{0}{-7}$ = 0 Any number divided by zero is equal to zero. | 32 | 105 | {"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} | 3.015625 | 3 | CC-MAIN-2018-34 | longest | en | 0.884166 |
http://math.stackexchange.com/questions/216973/find-the-value-of-f343-56/216979 | 1,464,627,779,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464051036499.80/warc/CC-MAIN-20160524005036-00123-ip-10-185-217-139.ec2.internal.warc.gz | 177,469,497 | 19,068 | # Find the value of f(343, 56)?
I have got a problem and I am unable to think how to proceed.
$a$ and $b$ are natural numbers. Let $f(a, b)$ be the number of cells that the line joining $(a, b)$ to $(0, 0)$ cuts in the region $0 ≤ x ≤ a$ and $0 ≤ y ≤ b$. For example $f(1, 1)$ is $1$ because the line joining $(1, 1)$ and $(0, 0)$ cuts just one cell. Similarly $f(2, 1)$ is $2$ and $f(3, 2) = 4$.
Find $f(343, 56)$.
I have tried by making the equation of line joining $(0,0)$ and $(343, 56)$.
I got the equation as $8x = 49y$.
Now I tried it by randomly putting the values of $x$ and $y$ which are both less that $343$ and $56$ respectively.
But I am unable to get it.
Is there is any better approach?
-
A start: Note that $343=7\cdot 49$, and $56=7\cdot 8$. First find $f(49,8)$.
More: If $a$ and $b$ are relatively prime, draw an $a\times b$ chessboard, and think of the chessboard squares you travel through as you go from the beginning to the end.
-
Sir , So according to your answer f(4,4) = f(1.1) . But f(4,4) passes through 4 blocks. and f(1,1) pass through 4 blocks. – vikiiii Oct 19 '12 at 16:36
@vikiiii: no, he suggested the pattern will repeat. – Ross Millikan Oct 19 '12 at 16:41
@vikiiii: Your line will go through $(49,8)$, $(98,16)$, $(147,24)$, and so on. Once you figure out what happens up to $(49,8)$, everything will be clear. – André Nicolas Oct 19 '12 at 16:48
@Andreas Thanks for the answer . And also Ross Millikan who cleared my doubt . Now i have got the answer. In my next comment i will tell you how i got it . But i want to ask is it a right way . – vikiiii Oct 19 '12 at 16:48
What i did is take your suggestion Find f(49,8) . So Ross told pattern will repeat 7 times. So instead of finding f(49,8) . I found f(6.12 , 1 ) . So this passes exactly through 7 blocks you can imagine this . So f(343,56) will be 7 * 8 * 7 = 392 which is the answer. – vikiiii Oct 19 '12 at 16:51
Hint The line joining these to points can be parametrized as $l(t)=( 7 \cdot 7 \cdot 7t,8 \cdot 7 t).$ Now the number which that line cut cells is the number that when an ordinate change from non integer to integer. This can be easily understood if you draw the grid. The only thing you should be careful is not to count twice the times when both ordinates turn to integer because it the time when you hit a vertex. So you should count only one time. Denote $N(x)$ the number when the first ordinate goes is integer and $N(y)$ the times when the second ordinate is integer. So what you need is when one ordinate is integer or when the other ordinate is integer. $N(x \cup y)= N(x)+N(y)-N(x \cap y)$.
-
Your answer is beyond my thinking. Anyways thanks for answering. – vikiiii Oct 19 '12 at 16:55
@vikiiii This is very similar to my answer, put more formally. – Mark Bennet Oct 19 '12 at 21:27
Each time your line passes through a vertical or horizontal gridline it enters a new square. Count the horizontals and verticals it crosses.
When it hits a corner it goes through a vertical and a horizontal at the same time so you need to adjust for corners. This is determined by considering the common factors of $a$ and $b$.
The first square is given to you, so add 1.
-
It is then possible to derive a formula for the number of squares - if $hcf(a,b)=h$ then the line crosses $h-1$ corners so we get $(a-1)+(b-1)-(h-1)+1=a+b-h$. There are questions involving a black/white checkerboard grid asking how many black squares you cross - note that black and white alternate except for a parity change at corners. [Note that 392=343+56-7] – Mark Bennet Oct 19 '12 at 18:49
The squares might also be black/red – Mark Bennet Oct 19 '12 at 21:28 | 1,107 | 3,658 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2016-22 | latest | en | 0.886104 |
https://us.metamath.org/mpeuni/sbex.html | 1,713,068,369,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816864.66/warc/CC-MAIN-20240414033458-20240414063458-00654.warc.gz | 555,147,200 | 3,830 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > sbex Structured version Visualization version GIF version
Theorem sbex 2281
Description: Move existential quantifier in and out of substitution. (Contributed by NM, 27-Sep-2003.)
Assertion
Ref Expression
sbex ([𝑧 / 𝑦]∃𝑥𝜑 ↔ ∃𝑥[𝑧 / 𝑦]𝜑)
Distinct variable groups: 𝑥,𝑦 𝑥,𝑧
Allowed substitution hints: 𝜑(𝑥,𝑦,𝑧)
Proof of Theorem sbex
StepHypRef Expression
1 sbn 2280 . . 3 ([𝑧 / 𝑦] ¬ ∀𝑥 ¬ 𝜑 ↔ ¬ [𝑧 / 𝑦]∀𝑥 ¬ 𝜑)
2 sbn 2280 . . . 4 ([𝑧 / 𝑦] ¬ 𝜑 ↔ ¬ [𝑧 / 𝑦]𝜑)
32sbalv 2159 . . 3 ([𝑧 / 𝑦]∀𝑥 ¬ 𝜑 ↔ ∀𝑥 ¬ [𝑧 / 𝑦]𝜑)
41, 3xchbinx 336 . 2 ([𝑧 / 𝑦] ¬ ∀𝑥 ¬ 𝜑 ↔ ¬ ∀𝑥 ¬ [𝑧 / 𝑦]𝜑)
5 df-ex 1774 . . 3 (∃𝑥𝜑 ↔ ¬ ∀𝑥 ¬ 𝜑)
65sbbii 2074 . 2 ([𝑧 / 𝑦]∃𝑥𝜑 ↔ [𝑧 / 𝑦] ¬ ∀𝑥 ¬ 𝜑)
7 df-ex 1774 . 2 (∃𝑥[𝑧 / 𝑦]𝜑 ↔ ¬ ∀𝑥 ¬ [𝑧 / 𝑦]𝜑)
84, 6, 73bitr4i 305 1 ([𝑧 / 𝑦]∃𝑥𝜑 ↔ ∃𝑥[𝑧 / 𝑦]𝜑)
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 ↔ wb 208 ∀wal 1528 ∃wex 1773 [wsb 2062 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1789 ax-4 1803 ax-5 1904 ax-6 1963 ax-7 2008 ax-10 2138 ax-11 2153 ax-12 2169 This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-ex 1774 df-nf 1778 df-sb 2063 This theorem is referenced by: sbmo 2692 sbabel 3013 sbcex2 3832
Copyright terms: Public domain W3C validator | 780 | 1,350 | {"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} | 3.515625 | 4 | CC-MAIN-2024-18 | latest | en | 0.189797 |
https://www.thedonutwhole.com/how-many-ounces-are-in-a-average-shot-glass/ | 1,708,709,061,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474440.42/warc/CC-MAIN-20240223153350-20240223183350-00603.warc.gz | 1,033,000,772 | 25,893 | # How many ounces are in a average shot glass?
The average shot glass holds approximately 1. 5 ounces of liquid. However, this can vary greatly depending on the size and shape of the shot glass. Shot glasses can range in size from 1 ounce to 3 ounces or more.
The most common size for a shot glass is generally around 1. 5 to 2 ounces. If it is a standard bar shot glass, common in many bars, it typically holds 1. 5 ounces.
## Is a shot glass 1 or 2 oz?
A shot glass is usually 2 ounces, though it can range anywhere from 1 to 3 ounces depending on which type of shot glass you are using. The standard size for a shot glass is 2 ounces and is considered the industry standard.
When bartenders are asked to pour a shot, they usually use a 2 ounce shot glass. These standard shot glasses are typically made of glass, though they are also available in plastic. Additionally, plastic shot glasses are often used in places such as restaurants and bars where glass shot glasses are not allowed.
This is because they are less likely to break if accidentally dropped.
## Is 1 oz the same as a shot glass?
No, a 1 oz shot is not the same as a shot glass. A shot glass typically holds between 1. 5 and 2 ounces of liquid, depending on the size and style of the shot glass. So, a 1 oz shot is smaller than an average shot glass.
However, some miniature shot glasses can hold as little as 0. 5 ounces, so a 1 oz shot would be larger than one of these.
## How much is 1 oz in a shot?
One ounce is the standard amount of liquid used to measure a “shot” of an alcoholic beverage. Depending on the exact size of the shot glass or other vessel being used, that could be anywhere from 30 milliliters (1 fluid ounce) up to 45 milliliters (1.
5 fluid ounces) when measured to the top of the vessel. However, the “standard” measure prescribed by most establishments, such as bars, is a single fluid ounce of a pour, which is equivalent to 1 ounce in a shot.
## What is a 2 oz shot glass called?
A 2 oz shot glass is also known as a pony shot. This is a small glass typically used for measuring out spirits to make small drinks or shots. The pony shot is a convenient size and can also be used for other measurements such as ingredients for cocktails.
The name ‘pony’ comes from the fact that it is a smaller size than other shot glasses, like the traditional one and a half ounce shot glass. Pony shots can also be used to measure out other liquids, and they make great little bar accessories.
## How many Oz is a standard shot glass?
A standard shot glass is typically 1. 5 ounces (Oz) of liquid. However, the size of a shotglass can range from 1-2. 5 ounces (Oz) depending on the size and shape of the glass. The standard size of a shot glass is generally accepted to be 1.
5 ounces (Oz). | 665 | 2,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-10 | latest | en | 0.970336 |
https://nz.education.com/lesson-plan/writing-christmas-themed-word-problems/ | 1,603,554,280,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107883636.39/warc/CC-MAIN-20201024135444-20201024165444-00055.warc.gz | 431,085,650 | 31,307 | Guided Lessons
# Writing Christmas-Themed Word Problems
Your students will turn boring number sentences into fun, Christmas-themed word problems! They will get practise with maths and writing as they use their creative juices to write and solve single digit addition word problems.
No standards associated with this content.
Which set of standards are you looking for?
Students will be able to write and solve one digit addition word problems.
(5 minutes)
• On one side of the board write: 5 + 3 = ____. On the other side of the board write: Tiny the elf wrapped 5 presents. Jumbo the elf wrapped 3 presents. How many presents did they wrap all together?
• Ask students to determine what is the same and what is different about the two items written on the board.
• Support students to see that both items contain numbers that need to be added together, but one of the items only contains numbers and symbols while the other uses numbers and words.
• Tell students that “5 + 3 = ____” is a Number sentence, an equation using numbers and common symbols, such as a plus sign and an equals sign.
• Tell students that “Tiny the elf wrapped 5 presents. Jumbo the elf wrapped 3 presents. How many presents did they wrap all together?” is a word problem.
• Explain that a Word problemIs one or more sentences describing a problem that needs to be solved using a mathematical calculation.
• Call on a volunteer to solve both the number sentence and the word problem. Ask him or her to explain their thinking as they solve each problem.
(10 minutes)
• Tell students that they will receive number sentences, and their job is to write Christmas-themed word problems for each number sentence.
• Write “Addition Words” at the top of a piece of chart paper.
• Ask students to brainstorm a list of maths words that they would see in addition word problems. These may include: total, join, both, sum, all together, plus, combined, increase, add, and in all. Encourage students to use these words as they are writing their word problems. Write these ideas on chart paper for student to use during Guided practise/Modeling and Independent Work Time.
• Tell the class that there is a list of steps to follow as they are writing their word problems.
• Write these steps down on a new piece of chart paper for students to use during Guided practise/Modeling and Independent Work Time with the heading “Steps to Write a Word Problem.”
• Tell students that first, they will choose a main character. For example, I selected Tiny the elf.
• Inform your class that second, they will decide what kind of object the character has. The number sentence will tell them how many of this object their character has. For example, “Tiny the elf wrapped five presents.”
• Tell your group that third, they will choose a second character and identify what object he or she has. Remind students that the number sentence will tell them the quantity. For example, “Jumbo the elf wrapped three presents.”
• Finally, students will ask a question using the maths words we brainstormed. For example, “How many presents did they wrap all together?”
• Remind students that they will follow these steps as they are writing and solving their addition word problems.
(10 minutes)
• Project the Christmas Addition #1 worksheet onto the board.
• Tell students that you will be writing Christmas-themed word problems for these addition number sentences as a class.
• Refer students to the first number sentence: 3 + 2 = ____ .
• Call on a volunteer to select a main character (e.g. Frosty the Snowman).
• Call on a volunteer to select what object the character has. Remind students that the number sentence will tell them how many objects the character has (e.g. Frosty the Snowman has three buttons).
• Choose a student to select a second character and identify what object they have (e.g. Rudolph the Reindeer gave Frosty two more buttons).
• Select someone to finish the word problem by asking a question (e.g. How many buttons does Frosty the Snowman have in all?).
• Write this word problem on the board and select a student to solve it.
• Continue this process with 3-5 more number sentences.
(20 minutes)
• Distribute one piece of lined paper to each student.
• Preview and handout the Christmas Addition #2 worksheet for students to complete independently.
• Remind students to use the “Addition Words” and “Steps to Write a Word Problem” posters as a resource as they are working on converting their number sentences into word problems.
Support:
• Provide students with manipulatives (e.g. unifix cubes or counters) to assist them with the addition word problems.
• Allow students to focus on solving the number sentences instead of writing word problems.
Enrichment:
• Encourage students to make their word problems more complex by adding in a third character and a third number to their addition word problems.
• Give students the opportunity to work with solving word problems that contain subtraction and double digit addition with the Christmas Word Problems #1 and Christmas Word Problems #2 worksheets.
(10 minutes)
• Distribute whiteboards to each student.
• Tell students that you will say aloud some Christmas-themed word problems, and they will write a number sentence that matches each word problem. Ask students to hold up their whiteboards to show you their number sentences each time.
• Tell students at least five word problems for them to write number sentences. Remember to speak slowly and clearly. Alternatively, you may also write the word problems on the board instead of just reciting them orally.
(5 minutes)
• Ask students to reflect and think about this lesson by going around in a circle and having students share one comment. It can either be one thing they learned, one thing they liked doing, one thing they did not like, or one thing they found challenging.
• Explain to students that thinking about their learning is very important because it will help them to explore what type of learner they are, how they learn best (where they need extra support, what kind of learning they prefer), and it helps you as a teacher create lessons that are interesting and fun for your students.
• Thank your students for their contributions during this reflection time. | 1,287 | 6,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2020-45 | latest | en | 0.955508 |
https://studylib.net/doc/7674505/lesson--measuring-volume-in-cubic-centimetres | 1,632,394,372,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057421.82/warc/CC-MAIN-20210923104706-20210923134706-00375.warc.gz | 571,436,812 | 11,433 | # Lesson: Measuring Volume in Cubic Centimetres
```Math 5B - Extra Practice
Name: _______________________
Date: _________________
Lesson: Measuring Length
Copy and complete.
1. a) 9 cm = ______ mm b) 40 cm = _____ mm
c) 23 cm = _____ mm
2. a) 70 mm = ____ cm
c) 90 mm = _____ cm
b) 50 mm = _____ cm
3. a) 3000 mm _____ m b) 8000 mm = _____ m c) 5000 mm = _____ m
4. a) 4 m = _____ mm
b) 7 m = _____ mm
c) 1 m = _____ mm
5. Use >, <, or =
a) 7 cm 70 mm
b) 140 mm 11 cm
c) 80 mm 9 cm
d) 24 mm 2.4 cm
6. Which unit would you use to measure each item?
a)
b)
c)
d)
the width of a slice of bread
the thickness of a sandwich
the length of a playground
the length of a staple
1
Lesson: Exploring Rectangles with Equal Perimeters
Use 1-cm grid paper.
7. Andrew has 48 m of fencing to put around his garden.
a) List all the possible lengths and widths of Andrew’s garden.
b) Which dimensions will Andrew choose if he wants the garden with the
greatest possible area? The least possible area?
Lesson: Exploring Rectangles with Equal Areas
Use 1-cm grid paper.
8. Draw a rectangle with each area and perimeter.
a) area 24 cm2 and perimeter 28 cm
b) area 16 cm2 and perimeter 16 cm
Lesson: Measuring Volume in Cubic Centimetres
9. Find the volume of each rectangular prism.
10. Order the prisms in question 9 from greatest to least volume.
11. A box has a volume of 16 cm3 .
The box is 4 cm tall.
a) How many centimetre cubes will fit in one layer of the box?
How do you know?
b) How long and how wide might the box be?
Give as many answers as possible.
Lesson : Constructing Rectangular Prisms with a Given Volume
Use centimetre cubes.
12. Build a rectangular prism with each volume.
Record your work in a table.
3
a) 12 cm
c) 16 cm3
3
Volume
Length
Width
Height
b) 24 cm
d) 11 cm3
13. Build all the possible rectangular prisms with volume 18 cm3.
Record your work in a table.
Length
(cm)
Width
(cm)
Height
(cm)
Volume
(cm3)
18
b) Suppose the number of centimetre cubes were halved. How many
different rectangular prisms could be made? Write their dimensions.
14. Suppose you want to build a rectangular prism with 35 centimetre
cubes. You put 7 cubes in the bottom layer.
a) How many layers of cubes will you need?
b) What are the dimensions of the prism?
Lesson : Measuring Volume in Cubic Metres
Use centimetre cubes.
15. Would you use cubic centimetres or cubic metres to measure the
volume of each item?
a) a donut box
b) your classroom c) a cargo ship
d) a pencil box
e) a tissue box
f) a garage
Lesson: Exploring Capacity: The Litre
16. One litre fills about 4 glasses.
About how many glasses can you fill with each?
a) a 4-L jug of punch
b) a 2-L bottle of soda
c) a 3-L jug of lemonade
d) a 10-L container of water
17. Which containers hold more than one litre?
a) an automobile’s gasoline tank
b) a baby bottle
c) an eyedropper
d) a punch bowl
Lesson: Exploring Capacity: The Millilitre
18. Choose the better estimate.
a) an eyedropper
b) a teacup
c) a bottle of shampoo
d) a water bottle for a gerbil
1 mL or 200 mL
25 mL or 250 mL
75 mL or 750 mL
6 mL or 250 mL
19. Would you use millilitres or litres to measure each container?
a) a teaspoon
b) a drinking glass c) a vinegar jug
d) an aquarium e) a soup bowl
f) a drink box
20. Order from least to greatest capacity.
a) 2 L, 1000 mL, 40 mL, 750 mL
b) 76 mL, 14 mL, 5 L, 17 mL, 17 L
21. Copy and complete.
a) 3 L = _____ mL
b) 7 L = _____ mL
c) 10 L = _____ mL
d) 2000 mL = _____ L
e) 9000 mL = _____ L f) 1000 mL = _____ L
22. Jennifer drank 365 mL of water from her 1-L bottle.
How much water is left in Jennifer’s bottle?
23. Maya poured 880 mL of juice into a 2-L jug.
How many more millilitres will the jug hold?
Lesson: Relating Capacity and Volume
24. Describe how you could find the volume of a basketball in
cubic centimetres.
25. Chantal says that the volume of a rectangular prism is 32 cm3.
John says the volume is 32 mL.
Who is correct? Explain.
``` | 1,223 | 3,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2021-39 | latest | en | 0.768675 |
https://testbook.com/question-answer/what-will-be-the-simple-interest-on-rs-5000-at-8--610449b2d10f85795bab4bc3 | 1,638,618,993,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362969.51/warc/CC-MAIN-20211204094103-20211204124103-00243.warc.gz | 607,446,400 | 30,388 | # What will be the simple interest on Rs. 5000 at 8% per annum for the period from 4th February 2005 to 18th April 2005?
1. Rs. 80
2. Rs. 100
3. Rs. 110
4. Rs. 121
Option 1 : Rs. 80
## Detailed Solution
Given:
Principal = Rs. 5000
Rate = 8% p.a
Time = 4th February, 2005 to 18th April, 2005 = 24 + 31 + 18 = 73 days
Concept used:
February has 28 days in a non-leap year
March has 31 days
1 year = 365 days
Formula used:
SI = (P × R × T)/100
Where, P = Principal, R = Rate, T = Time
Calculation:
SI = (P × R × T)/100
⇒ (5000 × 8 × 73)/(100 × 365)
⇒ 80
∴ The required Simple interest is Rs. 80 | 240 | 610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-49 | latest | en | 0.899266 |
https://ptcouncil.net/is-1-is-greater-than-2/ | 1,659,942,525,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570767.11/warc/CC-MAIN-20220808061828-20220808091828-00671.warc.gz | 419,913,400 | 4,489 | Students will recognize if given fractions are less than, equal to, or better than 1. Students who space successful in ~ this have currently generalized the rule: fractions better than 1 have actually numerators bigger than their denominators; those that are much less than 1 have numerators smaller than their denominators; the rest are same to 1. If students discovered this activity hard, you might want them to walk over the perfect lists and shot to articulate the rule.
You are watching: Is -1 is greater than -2
Part 1
Let’s recognize if each portion is less than 1, equal to 1, or greater than 1. Because that example,
equals 1 therefore it would certainly be placed in the “equal to 1” category. Now let’s try some more.
Examples:
(less 보다 1 & clue is numerator is smaller than denominator)
(greater than 1 & clue is numerator is bigger than denominator)
(equal come 1)
(less 보다 1)
(less 보다 1)
(equal come 1)
(less 보다 1)
(more 보다 1)
(less than 1)
While youngsters are enjoying their building of mastery, feel cost-free to repeat. When kids are eager for more, try component 2.
Extension
Here room some more fractions, combined numbers and decimals come classify.
See more: Acrylic On Oil: Why Exactly Can You Paint Acrylic Over Oil ?
Examples:
1
Eight fifthsTwo thirds50.25 | 314 | 1,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.944273 |
https://teacheradvisor.org/strategies/supportingContent_111 | 1,618,847,063,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038887646.69/warc/CC-MAIN-20210419142428-20210419172428-00421.warc.gz | 638,417,956 | 7,030 | Document
# Unit Fractions and Their Relation to the Whole
EngageNY
## Supporting grades: 3
Description: In Topic A, students divided a given whole into equal parts to create fractional units (halves, thirds, fourths, etc.). Now, they associate one of the fractional units with a number called the unit fraction. An advantage of the term fractional unit is that it distinguishes the nature of the equal parts generated by partitioning a whole from the whole number division students studied in Modules 1 and 3. In Topic B, to avoid confusion, the term fractional unit is mostly replaced by the term equal part. The equal part is represented by the unit fraction. Students recognize that any non-unit fraction is composed of multiple copies of a unit fraction. They use number bonds to represent this. | 166 | 803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-17 | latest | en | 0.923838 |
http://en.wikipedia.org/wiki/Toss_juggling | 1,419,632,975,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447549808.107/warc/CC-MAIN-20141224185909-00072-ip-10-231-17-201.ec2.internal.warc.gz | 39,051,594 | 18,191 | # Toss juggling
Juggling pair performing basic two-person "pass and feed" with clubs
Toss juggling is the form of juggling which is most recognisable as 'juggling'.[1] Toss juggling can be used as: a performing art, a sport, a form of exercise, as meditation, a recreational pursuit or hobby.
In toss juggling, objects — such as balls, bean bags, rings, clubs, etc. — are thrown or tossed into the air and caught. Toss juggling is a form of object manipulation.
## Juggling techniques and patterns
'Toss juggling' has a number of basic principles and patterns. Most more complex patterns are variations of the basic ones and all toss juggling must conform to these basic principles.
### Basic principles
Toss juggling is, according to most sources, the throwing and catching of objects, where there are more objects than there are hands (or sometimes other parts of the body) doing the throwing and catching.[2] Three balls thrown and caught between two hands is toss juggling, as there are more balls than hands. Two balls between two hands would not be considered toss juggling, as the number of balls and hands is equal.
### Basic patterns
There are three basic patterns in toss juggling with many variations, some exceedingly complex, that are based on these basic patterns.
#### Shower
The shower is the most commonly depicted pattern in pictures and illustrations of jugglers (although often wildly exaggerated). The objects juggled follow a circular pattern with one hand doing higher throws and the other passing or doing low throws to the first hand. This pattern can be used to juggle an odd number or even number of objects. In siteswap notation, a three object shower would be called 51.
The cascade pattern is often the first pattern taught to beginner jugglers. The objects juggled follow a horizontal figure of eight pattern with each hand throwing the object to the same height when the previous throw has reached it peak. This pattern is most commonly used to juggle an odd number of objects. In siteswap notation, a three object cascade would be called 3.
#### Fountain
The fountain is where each hand throws and catches objects and they do not transfer to the opposite hand. This pattern is most commonly used to juggle an even number of objects. In siteswap notation, a four object fountain would be called 4.
Main article: Juggling pattern
Besides the cascade (or reverse cascade), other toss patterns include the box or column, and any number of multiplex patterns or contact juggling tricks within the pattern.
There are also amusing stunts such as the Yo-yo and the Robot, which give the illusion that the balls are connected to each other or to the juggler's body. By adding elementary skills to any juggling pattern the pattern can be increased, incrementally, in complexity.
There are even wildly intricate patterns such as the Mills Mess, which is widely attributed to Steve Mills of The Dazzling Mills Family. For a given pattern any number of variants may occur to the juggler, for instance, Mills Mess has at least three well-known variants, produced by adding flourishes such as chops, holds, stalls, and carries. "Mills" colloquially refers to any pattern in juggling where the hands cross back and forth over one another. Any pattern valid in siteswap notation can be done "vanilla", reverse, and in Mills.
### Other patterns
Below is a list of other common three-ball patterns.
## Performance
Toss juggling - street performance in Denmark
Juggling has been used as a performance skill for thousands of years. Toss juggling has been significant in such performances and has continued to be a major part of juggling performances up to the present day. Juggling performances are usually of two types: a performance which shows off the technical skill of the juggler which is often performed to music; and a performance of juggling with comedy. The two types are not mutually exclusive with both technical skill, comedy, music and sometimes other skills being incorporated into an performance or show.
## Sport
As a sport, juggling can be done competitively, with jugglers taking turns to try to best each other's abilities, or juggling together, in teams—called passing. In passing, two or more jugglers will throw objects to each other, performing such feats as may be accomplished 'solo', but juggling more objects. Jason Garfield is a leader/initiator of the concept "Sport Juggling". This concept really took off in 2004 when he put together the World Juggling Federation competition. This is an annual event that has been previously held in Las Vegas, Nevada.
## Passing and feeding
Main article: Passing (juggling)
If three jugglers can each juggle three items, together they may readily juggle nine items. A variation on simple passing, called feeding , is performed by two or more jugglers all tossing to one specific juggler—called the feeder. The feeder is usually the juggler with the better-developed skill. In a 3-person feed, the feeder may pass on every second beat (called "solids" or "2-count") -- alternately passing to the other two jugglers, where every right hand throw is a pass, while they in turn pass back to him on "every others" (every fourth beat (sometimes called "4-count"), where every other right hand is a pass). Meanwhile all the balls not passed to the feeder are 'self-tossed' or thrown to oneself. Of course, a seemingly infinite variety of combinations of number of balls, number of jugglers, feeding directions or rhythms, etc. may be accomplished. Pass juggling helps to develop a better sense of rhythm.
## Exercise and fitness
As exercise, juggling is a highly aerobic activity, increasing the heart rate and respiration. Juggling helps one to develop good hand-eye coordination, physical fitness and balance.
### Reflexes
Juggling helps to develop quick reflexes, and in fact, jugglers develop "higher-order reflexes", reflexes not typically associated with normal human activity. These reflexes are formed through repetition of what are, at first, slowly learned and difficult processes. As the various processes develop into reflexive actions, additional, more difficult or complex skills are 'layered' atop previous, well-developed skills. These new more complex skills eventually become more reflexive, and eventually, various unusual high order reflexes develop. An example is that, while catching a raw chicken's egg may be quite challenging to a novice juggler, a more skilled juggler might be able to easily catch — unbroken — an egg thrown towards him without warning.
## Meditation
Paradoxically, the same processes that are well known for exercising the body, can also be a very relaxing activity. As meditation, juggling a repeating pattern or patterns can take one's mind off the stresses they might encounter in their daily lives. Jugglers have described a phenomenon of near-disembodiment and tranquility which may come over them while juggling.[citation needed] The constant rising and falling of the objects, the regularity of the rhythms, can become almost hypnotic, and the attention of a juggler while tightly focused on the juggling pattern may seem to expand and even to "encompass the universe".
## Recreation
As a recreational pursuit, juggling excels in many ways. Besides the previously mentioned health benefits, any form of juggling is at its best when done socially. The equipment is inexpensive or free — though very costly equipment is also available — and easily portable. Juggling is great for "breaking the ice" at parties and other social gatherings[citation needed]. Jugglers are a friendly sort, usually, and are often very willing to help beginning jugglers with advice. Juggling conventions, clubs and other gatherings where jugglers congregate can be great places to meet and share the art of juggling, for experts novices and even 'non-jugglers' — who often find themselves doing some form of juggling themselves.
In 1990 it was estimated by the International Jugglers' Association that twenty-three percent of Americans can juggle a three-ball cascade.[3] | 1,684 | 8,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | longest | en | 0.95435 |
http://forum.allaboutcircuits.com/threads/power-and-voltage-help-i-blanked.119122/ | 1,485,035,838,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281226.52/warc/CC-MAIN-20170116095121-00413-ip-10-171-10-70.ec2.internal.warc.gz | 104,287,741 | 13,654 | Power and Voltage Help (I blanked)!
Discussion in 'General Science' started by NinaFarac, Jan 1, 2016.
1. NinaFarac Thread Starter New Member
Jan 1, 2016
2
0
Hey everyone, I'm having a 'blanking' moment:
Today, my teacher gave us an example of an electrical source that had enough power to power a LED light, but not enough voltage. We then had to offer a suggestion to change the circuit in order to make enough voltage.
Can someone please explain how you can have enough power, but not enough voltage? I don't see how this fits in the general power-voltage relationship.
2. djsfantasi AAC Fanatic!
Apr 11, 2010
2,905
879
Power is what? In other words, power is the relationship between what two values!
The answer is P=VxI. If the voltage is too low, and the current available is higher than necessary - you have sufficient power but not enough voltage.
Take a typical but mythical LED circuit for illustration. The LED specs say it has a forward voltage of 2V and typical current of 15ma. It's power requirement would be 0.03W
Now consider a power supply of 1V and 30ma. Power is also 0.03W. But there is insufficient voltage to illuminate the LED; it requires 2V.
NinaFarac likes this.
3. NinaFarac Thread Starter New Member
Jan 1, 2016
2
0
Thank you for your reply djsfantasi! I was able to work it out after I posted the thread | 350 | 1,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-04 | latest | en | 0.953856 |
https://www.queryhome.com/puzzle/2524/what-will-be-the-duration-of-time | 1,542,063,211,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741151.56/warc/CC-MAIN-20181112215517-20181113001023-00009.warc.gz | 969,580,391 | 31,398 | # What will be the duration of time ?
88 views
A picnic bus starts at 8 am at constant speed 40 km/hr for 3 hours.The picnic place is at a distance of 120 km.The whole route is a straight path. After some travelling rain starts due to which the driver decreases the speed to 20 km/hr upto the time of raining. When rain fall stops he increases his speed to 1.5 times and reach the destination on time.
What will be the duration of time ?
posted Aug 14, 2014
As per the starting Two statement
A picnic bus starts at 8 am at constant speed 40 km/hr for 3 hours.The picnic place is at a distance of 120 km. what-will-be-the-duration-of-time
If the distance is 120 Km then in first 3 hours the bus should reached at the destination.
As it was running as 40km/hr speed for 3 hour.
duration of time Should be 3 hours.
Similar Puzzles
If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days.
What will be the time taken by 15 men and 20 boys in doing the same type of work?
Rs 20 is the true discount on Rs 260 due after a certain time.
What will be the true discount on the same sum due after half of the former time , the rate of interest being the same ? | 310 | 1,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-47 | longest | en | 0.939388 |
http://stackoverflow.com/questions/9252764/how-to-create-a-custom-mesh-on-three-js | 1,386,350,502,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163052216/warc/CC-MAIN-20131204131732-00013-ip-10-33-133-15.ec2.internal.warc.gz | 170,241,072 | 12,474 | # How to create a custom mesh on THREE.JS?
I've asked this and got the answer:
``````var geom = new THREE.Geometry();
var v1 = new THREE.Vector3(0,0,0);
var v2 = new THREE.Vector3(0,500,0);
var v3 = new THREE.Vector3(0,500,500);
geom.vertices.push(new THREE.Vertex(v1));
geom.vertices.push(new THREE.Vertex(v2));
geom.vertices.push(new THREE.Vertex(v3));
var object = new THREE.Mesh( geom, new THREE.MeshNormalMaterial() );
``````
I expected this to work but it didn't.
-
You've added vertices, but forgot to put those vertices into a face and add that to the geometry:
``````geom.faces.push( new THREE.Face3( 0, 1, 2 ) );
``````
so your snippet becomes:
``````var geom = new THREE.Geometry();
var v1 = new THREE.Vector3(0,0,0);
var v2 = new THREE.Vector3(0,500,0);
var v3 = new THREE.Vector3(0,500,500);
geom.vertices.push(v1);
geom.vertices.push(v2);
geom.vertices.push(v3);
geom.faces.push( new THREE.Face3( 0, 1, 2 ) );
var object = new THREE.Mesh( geom, new THREE.MeshNormalMaterial() );
``````
Since you're using a mesh normals material, you might want to compute normals for the geometry. Also, make sure your object can be visible (is not to big or to close to the camera to be clipped out, is facing the right direction - towards the camera, etc.) Since you're drawing in the YZ plane, to see your triangle, something like this should work:
``````var geom = new THREE.Geometry();
var v1 = new THREE.Vector3(0,0,0);
var v2 = new THREE.Vector3(0,500,0);
var v3 = new THREE.Vector3(0,500,500);
geom.vertices.push(v1);
geom.vertices.push(v2);
geom.vertices.push(v3);
geom.faces.push( new THREE.Face3( 0, 1, 2 ) );
geom.computeFaceNormals();
var object = new THREE.Mesh( geom, new THREE.MeshNormalMaterial() );
object.position.z = -100;//move a bit back - size of 500 is a bit big
object.rotation.y = -Math.PI * .5;//triangle is pointing in depth, rotate it -90 degrees on Y
``````geom.vertices.push(v1); | 544 | 1,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2013-48 | latest | en | 0.743288 |
www.royalcircuits.com | 1,611,352,214,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703531429.49/warc/CC-MAIN-20210122210653-20210123000653-00335.warc.gz | 959,452,649 | 76,790 | # Simulation and Measurement Tutorial
## Introduction
Civil Engineers have a saying: “Any idiot can build a bridge that stands. It takes an engineer to make a bridge that can barely stand.” To achieve their goals, Engineers have finite element and lumped element analysis available to them that allows them to simulate all manner of loading and failure analysis. With that safety information in hand, they can make decisions to decrease cost. The technical aspects are specific to the field of Civil Engineering, but the processes are universal to all disciplines – make it work, make it safe, and then make it inexpensive. Engineering is at its heart, a compromise between competing requirements and risk, with cost and safety usually winning the day.
The same idea can be applied to PCB design. While not every idiot can build a HDI PCB, “It takes an engineer to make an HDI PCB that barely works.” That is because the “barely” can come with significant cost savings. You can absolutely go out and purchase Roger’s core material, use 2 oz. copper, and have a design that works. But can you replace iTera MT40 with 1080 FR4, and the 2 oz. with 1 oz. copper? You will never know until you either simulate your design or take the chance of building your board with less expensive materials. The purpose of this article is to convince your company to invest in the tools you need to do your job so you can purchase less expensive materials from us, and make your overall design less expensive to produce.
The operating frequency of your design isn’t all that important. In your decision to simulate or your selection of test equipment, the critical number you need to pay attention to is the rise-time / fall time of your fastest signal in your circuit.
#### This time-domain graph shows the potential difference on a hypothetical trace during a logic transition from low to high. Rise time is measured as the interval between 20% and 80% of the logic-high voltage or the 10% and 90% interval.
It is possible (although unlikely) to have a leisurely 40 kHz switching frequency with a 20 GHz equivalent rise time by using a fast GaN switch. Frequency is the reciprocal of the period, so 1/50 ps = 20 GHz (20 GHz is an approximation since the choice of 80/20 or 90/10 is somewhat arbitrary and I’m ignoring a few other details.) To reiterate — you might be designing a “low-speed” circuit on paper, but in practice, it could be a very, very high-speed design. And those high-speed harmonics, reflections, etc… might completely disrupt your design.
And to make matters worse, it’s not just the fundamental frequency (determined from rise-time) that you need to worry about. In the frequency domain, square waves are made of a series of lower-amplitude, higher-frequency sine waves. And unfortunately, the first few harmonics have sufficient amplitude to disrupt your design as well.
## Measurement
Is your scope able to accurately measure the rise/fall time of your signal? The rise time is usually defined as either the time it takes for the electric potential to rise from 20% to 80% of the maximum value or from 10% to 90% of the maximum value. But most affordable oscilloscopes lack the ability to capture the high-speed signals that modern switching circuits can generate.
As a very quick rule-of-thumb, your scope and probe bandwidth should be at least 60% of the reciprocal of the shortest rise-time interval in your circuit.
For example, a 300 ps signal would require a scope with a 2 GHz bandwidth, and a 1 GHz scope could capture at most a 600 ps signal. Please keep in mind this equation is a “back-of-an-envelope” estimation. To truly determine your needs, you’ll need to know the measurement accuracy required, the frequency response of your scope, the resistance and capacitance of your scope probes, and a few other things. It’s best to just get a test-equipment representative on the phone and explain your requirements and let them tell you which test instruments they have that can meet your needs.
If you don’t have sufficient sampling points, you won’t be able to capture the actual rise and fall times of your signal.
What does that look like in real life? Let’s take a moment to look at some actual sampled signals from a ~40 ps rise-time source on a Tektronix MDO3104 oscilloscope with a rated combined bandwidth of 1 Gsps. The Rule-Of-Thumb provided earlier says that this scope would, at best, be able to see a 600 ps rise time, a difference from the datasheet’s stated rise time by over an order-of-magnitude.
First, I set the scope for 250 MHz — as would be the maximum bandwidth on a single channel if all four channels were in use. There we see a signal that’s pretty similar to the first image that appeared in this article. It looks remarkably clean, so you might be tempted to think everything worked out as it should, write down the number, and move on. But don’t be bamboozled!
#### This MDO3104 oscilloscope image shows the 40 ps signal recorded with a 250 MHz bandwidth. The curve has a measured rise-time of approximately 725 ps.
Now let’s find out what that signal looks like when the entire 1 GHz bandwidth is used and the sampling increases by a factor of four.
#### This oscilloscope image shows the 40 ps signal recorded with a 250 MHz bandwidth. The curve has a measured rise-time of approximately 260 ps.
That’s a different signal entirely. And the measured rise time is significantly higher than before as well. We’re not seeing the true rise time of the signal — we’re seeing the limitations of the oscilloscope’s sampling rate.
This image isn’t the true representation of the signal either! We’re closer to the truth, but even though we’ve quadrupled the number of sampling points, there still aren’t enough sampled points to accurately capture and reproduce the waveform.
#### As an example, the blue signal is sampled by ~20 evenly spaced points on the left and 4x that number on the right. But as you can see, the rise and fall of the signal are not accurately sampled.
In short — for high speed signals, you need quality test equipment in your lab to truly understand what is happening to the signals in your circuits.
## Simulation
Before you ever power on your test equipment or send your artwork to your fab house, you should have a thorough understanding of what is happening in your design. For that you need to find a simulation software suite that fits your requirements and your budget.
Your simulation needs will again depend on the rise-time of your devices. There are a variety of software options out there that fill numerous needs. The following list is non-exhaustive and inclusion does not mean that Royal Circuits recommends or endorses the software.
### Polar Instruments PCB Stackup, Signal Integrity, and Controlled Impedance Calculators
#### PCB signal integrity, stackup and controlled impedance software by Polar Instruments
This software is relatively well known amongst high-speed design engineers. While it is primarily a 2D field solver, that can usually be enough for many designs.
### Ansys HFSS and SIWave
#### RF and Microwave Simulation software from Ansys.
Ansys has multi-physics capable solvers for electromagnetic, thermal, and mechanical simulation of chip and board level designs. Two popular tools are HFSS (High Frequency Structure Simulator) and SIWave. HFSS provides 3D electromagnetic simulation for antennas, high speed interconnects, filters, etc… SIWave provides power integrity, signal integrity, and EMI analysis. It can determine power noise margins and electromagnetic compliance.
### Simberian
Simberian’s Simbeor Electromagnetic Signal Integrity Software provides full 3D-wave analysis of PCB interconnects. It allows engineers to visualize and calculate cross-talk,
## Summary
You need the right tools to do your job. You can either ask purchasing to make a one-time investment in better equipment, or let them know that they will have to pay for higher-grade dielectrics, repeatedly, every time a new panel or a new design is produced. And every time a board doesn’t work because you can’t accurately measure the signal.
Good luck!
## Never Pay for a Programming Header Again!
Off-board connectors are expensive to purchase and usually expensive to install. If they are necessary for a design, then you have to bite the bullet and purchase them. Fortunately, the cost is amortized over the lifetime of the project. For most electronics projects though, the often-used rectangular programming headers are one-time-use devices — and it’s… View Article
## New Customer Special 2021
There’s Never Been a Better Time to Give Us a Try! Throughout 2020, Royal Circuits reinvested in the company so that our customers can get the best quality PCBs in the fastest possible turns. Thousands of engineers trust us to deliver their PCBs on-time and on-budget. 5 x 5 Rigid Special 5 Boards 5-Day Turn… View Article | 1,937 | 8,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | longest | en | 0.934717 |
https://www.futurestarr.com/blog/mathematics/what-percent-is-15-out-of-18 | 1,656,239,712,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103205617.12/warc/CC-MAIN-20220626101442-20220626131442-00231.warc.gz | 856,820,650 | 22,181 | FutureStarr
What Percent Is 15 Out of 18
# What Percent Is 15 Out of 18
The percentage of X is 15 out of what? The answer is 3 out of 18, which constitutes 14%. This is largely a way for a person to divvy up a set of numbers to answer what percent is X out of what number. The answer is 3 out of 18, which constitutes 14%.
### Simple
This percentage calculator is a tool that lets you do a simple calculation: what percent of X is Y? The tool is pretty straightforward. All you need to do is fill in two fields, and the third one will be calculated for you automatically. This method will allow you to answer the question of how to find a percentage of two numbers. Furthermore, our percentage calculator also allows you to perform calculations in the opposite way, i.e., how to find a percentage of a number. Try entering various values into the different fields and see how quick and easy-to-use this handy tool is. Is only knowing how to get a percentage of a number is not enough for you? If you are looking for more extensive calculations, hit the advanced mode button under the calculator.
In calculating 12% of a number, sales tax, credit cards cash back bonus, interest, discounts, interest per annum, dollars, pounds, coupons,12% off, 12% of price or something, we use the formula above to find the answer. The equation for the calculation is very simple and direct. You can also compute other number values by using the calculator above and enter any value you want to compute.percent dollar to pound = 0 pound (Source: www.percentage-off-calculator.com)
### Number
This is all nice, but we usually do not use percents just by themselves. Mostly, we want to answer how big is one number in relation to another number?. To try to visualize it, imagine that we have something everyone likes, for example, a large packet of cookies (or donuts or chocolates, whatever you prefer 😉 - we will stick to cookies). Let's try to find an answer to the question of what is 40% of 20? It is 40 hundredths of 20, so if we divided 20 cookies into 100 even parts (good luck with that!), 40 of those parts would be 40% of 20 cookies. Let's do the math.
Percentage is one of many ways to express a dimensionless relation of two numbers (the other methods being ratios, described in our ratio calculator, and fractions). Percentages are very popular since they can describe situations that involve large numbers (e.g., estimating chances for winning the lottery), average (e.g., determining final grade of your course) as well as very small ones (like volumetric proportion of NOâ‚‚ in the air, also frequently expressed by PPM - parts per million). (Source: www.omnicalculator.com)
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https://1library.net/document/zpk5p9oy-cartesian-coordinates-functions-trigonometry-functions-trigonometric-exponentials-logarithms.html | 1,670,363,995,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711114.3/warc/CC-MAIN-20221206192947-20221206222947-00768.warc.gz | 103,977,236 | 35,332 | # cartesian coordinates, functions, trigonometry. radical functions, trigonometric functions, exponentials, and logarithms.
## Full text
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### Review of some Prerequisites
• This is a review of an entire semester. We will be moving much more slowly (like 40 times as slowly) through the subject of Math 1220. • If any item does not make sense to you then
your should review the relevant material. These Notes are numbered for easy ref-erence.
### 1.
Prerequisites: know arithmetic, algebra,
ge-ometry, cartesian coordinates, functions, trigonom-etry.
### 2.
Basic functions: polynomials, rational tions, radical functions, trigonometric func-tions, exponentials, and logarithms.
### 3.
Exponential:
f(x) = ax.
a > 0 is the base, x is the exponent.
### 4.
Logarithms are the inverse of the exponential: alogax = x and log
a ax = x.
### 5.
Particularly important bases are e = 2.71828 . . . (natural logarithm and exponential), 2, and 10 (common logarithm).
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### 6.
Some Examples: 23 = 32 = 30 = 3−1 = 912 = 813 = 823 = 8−32 = log10 100 = log2 1 2 =
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–3 –2 –1 0 1 2 3 y –3 –2 –1 1 2 3 x
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### 7.
Some properties of exponentials and logarithms a0 = 1 aras = ar+s ar as = a r−s (ar)s = ars
log(uv) = log u + log v log u v = log u − log v log ax = x log a loga x = logb x logb a = ln x ln a (1)
### 8.
Know the difference between simplifying an expression and solving an equation.
### 9.
Example of simplifying an expression:
z = ln(x2 + 5x + 6) − ln(x + 3) = =
### 10.
To solve an equation means to figure out for which values of its variables it is true. We say that those values satisfy the equation.
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### 12.
However, there is just one principle: Apply the same operation on both sides of the equa-tion until you have the variable by itself.
### 13.
How to pick the operations is the crux of the matter, of course, and depends on the con-text.
### 14.
How not to solve x2
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How to solve
x2
x+ 2 = x − 1
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### 16.
Major procedure: come up with a concept, make it precise, derive its properties, and then use the properties to work with the concept.
### 17.
In Math 1210, we applied this procedure to three major concepts: limits, derivatives, and integrals.
### 18.
Limits. limx−→cf(x) = L means that for all
ǫ > 0 there is a δ > 0 such that 0 < |x−c| < δ implies that |f (x) − c| < ǫ.
### 19.
Most functions we deal with are continuous, i.e.,
lim
x−→cf(x) = f (c) (2)
for all c in the domain of f .
### 20.
Intuitively, continuity means that we can draw the graph without lifting the pencil.
### 21.
If we have an expression that is undefined at a point, then manipulate it to get an expres-sion that has the same value as the original everywhere and that can be evaluated at that point. We can do this because of the Fun-damental Limit Theorem that states facts like that the limit of the sum, difference, uct, or quotient is the sum, difference, prod-uct, or quotient of the limits.
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### 22.
Example: lim h−→0 (x + h)2 − x2 h = limh−→0 x2 + 2hx + h2 − x2 h = lim h−→0 2hx + h2 h = lim h−→0(2x + h) = 2x. (3)
### 23.
The main concept we introduced in Math 1210 was that of a derivative of a function f :
f′(x) = d dxf(x) = lim h−→0 f(x + h) − f (x) h = lim y−→x f(y) − f (x) y − x (4)
### 24.
Geometrically: the slope of the tangent is the limit of the slopes of the secants.
### 25.
The derivative tell us how quickly a function is changing at the point where we evaluate the derivative.
### 26.
Locally the derivative approximates the func-tion.
### 27.
To compute derivatives we apply their prop-erties, i.e., Differentiation Rules
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d dxxr = rxr−1 Power Rule (f + g)′ = f′ + g′ Sum Rule (f − g)′ = f′ − g′ Difference Rule (kf )′ = kf′
Constant Multiple Rule d
dx sin x = cos x Sine Rule
d
dx cos x = − sin x Cosine Rule (uv)′ = u′ v + uv′ Product Rule u v ′ = u ′ v − uv′ v2 Quotient Rule d dxf g(x) = f′ g(x)g′ (x) Chain Rule (5)
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### 28.
Differentiation can be repeated, giving rise to higher derivatives, for example:
f(x) = x6, f′ (x) = 6x5, f′′(x) = 30x4, f′′′ (x) = 120x3, f(4)(x) = 360x2, f(5)(x) = 720x, f(6)(x) = 720, f(7)(x) = 0. (6)
### 29.
Differentiating a polynomial reduces its de-gree by 1.
### 30.
In general, a function f is a polynomial of degree up to n if and only if the (n + 1)-th derivative of f is everywhere zero.
### 31.
Newton’s Method can be used to find a root of a function f , i.e., a solution of the equation f (x) = 0. The basic idea is to con-struct a sequence where each term is the x-intercept to the tangent at the point corre-sponding to the previous term. You want to understand the formula
xk+1 = xk −
f(xk)
f′
(xk)
. (7)
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### 32.
Differentiation can be done implicitly, for example, thinking of y as a function of x, we get
x2+y2 = 1 =⇒ 2x+2yy′
= 0 =⇒ y′ = −x y (8)
### 33.
Implicit Differentiation occurs frequently in Related Rates Problems: Write one or more equations that hold at all time, differ-entiate, obtain equations that involve rates (derivatives), solve for what you want to know.
### 34.
Differentials: The change in a function value is approximately equal to the change in the in-dependent variable, multiplied with the deriva-tive.
∆y ≈ dy = f′
(x)dx = f′
(x)∆x. (9)
### 35.
Source of word problems: the derivative of position is velocity, the derivative of velocity is acceleration.
### 36.
Minima and maxima can occur only at crit-ical points:
- end points of intervals,
- singular points where the derivative does not exist,
- stationary points, where the derivative is zero.
### 37.
If f′
(x) = 0 and f′′
(x) > 0 then f (x) is a local minimum.
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### 38.
If f′
(x) is positive (negative) in some inter-val then f is increasing (decreasing) in that interval.
### 39.
If f′′
(x) is positive (negative) in some interval then f is concave up (down) in that interval.
### 40.
A point of inflection is a point on the graph where the second derivative changes sign.
### 41.
You should be able to draw graphs of func-tions using many sources of information, for example symmetry, singularities, asymptotes, and first and second derivatives. It’s only rarely appropriate simply to compute a large number of points and plot them in a coordi-nate system.
### 42.
Rational Functions have asymptotes, vertical ones where the denominator is zero, the x-axis if the degree of the denominator exceeds that of the numerator, horizontal ones if the degrees of numerator and denominator are the same, and slanted ones if the degree of the numerator exceeds that of the denominator by 1.
### 43.
The Mean Value Theorem for deriva-tives: If f is differentiable in (a, b) and con-tinuous in [a, b] then there is a point c in (a, b) such that
f(b) − f (a) = f′
(c)(b − a). (10)
### 44.
Differential Equations: Equations that in-volve a function and some of its derivatives. Usually the goal is to find the function.
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### 45.
Antiderivatives. F is an antiderivative of f if F′
= f . An integrable function f has infinitely many antiderivatives. Any two an-tiderivatives differ only by a constant.
### 46.
Indefinite Integrals. Z
f(x)dx = F (x) + C (11) where f is the integrand, F′
= f , and C is the integration constant. F is an an-tiderivative of f . The indefinite integral is the set of all antiderivatives. The value of the integration constant may be determined by a side condition.
### 47.
Definite Integrals as limits of Riemann Sums: Z b a f(x)dx = lim n−→∞ n X i=1 f(xi)∆x where ∆x = b − a n and xi = a + i∆x (12)
### 48.
When computing Riemann Sums, some spe-cial sum rules are useful:
n X i=1 1 = n, n X i=1 i = n(n + 1) 2 , n X i=1 i2 = n(n + 1)(2n + 1) 6 . (13)
(14)
### 49.
If f (x) is non-negative everywhere in [a, b] then Rb
a f(x)dx is the area of the region bounded
by the x-axis, the graph of f , and the vertical lines x = a and x = b.
### 50.
We use Riemann sums to recognize as a def-inite integer what we are trying to compute. However, we usually compute definite inte-grals by one version of the Fundamental The-orem of Calculus: Z b a f(x)dx = F (b) − F (a) where F′ = f. (14)
### 51.
We usually compute definite integrals by find-ing an antiderivative. A major topic this semester (in 1220) will be to find ways of doing this, i.e., to find integration techniques.
### 52.
In 1210 we discussed (briefly) only one major technique, integration by substitution.
### 53.
This is a systematic implementation of the inverse process of the chain rule:
Z
f′ g′(x)dx = f g(x) + C
### 54.
This works, by the fundamental theorem of calculus, since d dxf g(x) = f ′ g(x)g′ (x).
(15)
### 55.
Example: I = Z 1 0 10x (x2 + 4)2 dx
(16)
(17)
### 56.
However, some integrals can be computed with-out knowing an antiderivative. In Math 1210, we discussed in particular: Z c −c f(x)dx = 0 if f is odd, (15) Z a+2π a sin2xdx = Z a+2π a cos2 xdx = π, (16) and Z r −r p r2 − x2dx = πr 2 2 . (17)
### 57.
On the final exam for Math 1210 it was a common misconception that the integral (17) was zero, because of the symmetry. This can’t possibly be true since the integrand is never negative.
### 58.
The Mean Value Theorem for Integrals: There is a point c in [a, b] such that
Z b
a
f(x)dx = f (c)(b − a). (18)
### 59.
You can be casual when computing an an-tiderivative, once you have it check it by dif-ferentiation.
### 60.
We used definite integrals to solve the follow-ing problems:
(18)
- Computation of volumes by integrating the area of the cross section, using the methods of slabs, disks, washers, or shells.
- Computation of the length of a plane curve. - Computation of the surface area of a solid of revolution
- Computation of Work.
- Computation of the center of mass.
### 61.
The limits of integration may depend on a variable. We can differentiate with respect to that variable without actually computing an antiderivative: d dx Z U(x) L(x) f(t)dt = d dx F U(x) − F L(x) = f (U (x))U′ (x) − f (L(x))L′ (x) (19) where, as usual, F is any antiderivative of f .
### 62.
A special case of that formula is this version of the Fundamental Theorem of Calculus:
d dx Z x a f(t)dt = f (x). (20)
### 63.
The Fundamental Theorem of Calculus says that differentiation and integration are inverse processes of each other.
### 64.
Following is a list of some words and phrases, listed in alphabetical order, that you should be able to define, use, and understand.
(19)
acceleration, antiderivative, asymptote, base, chain rule, concave down, concave up, constant, continuity, critical points, cubic, decreasing func-tion, definite integral, degree of a polynomial, denominator, dependent variable, derivative, dif-ferential, differential equation, domain, even func-tion, equafunc-tion, exponent, expression, first deriva-tive, function, Fundamental Limit Theorem, Fun-damental Theorem of Calculus, graph (of a func-tion or an equafunc-tion), implicit differentiafunc-tion, in-creasing function, indefinite integral, indepen-dent variable, inflection point, integrand, inte-gration constant, inteinte-gration variable, Leibniz notation, limit, limits of integration (upper and lower), linear, Mean Value Theorem for deriva-tives, Mean Value Theorem for integrals, method of disks, method of shells, method of slabs, method of washers, Newton’s method, numerator, odd functions, points of inflection, polynomial, posi-tion, power rule, power, product rule, quadratic, quartic, quintic, quotient rule, radical, range, ra-tional function, related rates, Riemann Sum, se-cant, second derivative, singular point, solid of revolution, stationary point, sum rules, tangent, velocity, work.
### 65.
Contents of Math 1220 More differenti-ation and integrdifferenti-ation rules (particularly ex-ponentials, logarithms, inverse trig functions, integration by parts, logarithmic differenti-ation), more applications, indeterminate ex-pressions, improper integrals, sequences and series (particularly their convergence, power series, Taylor series).
Updating...
## References
Related subjects : Functions and Logarithms | 3,543 | 12,116 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-49 | latest | en | 0.855312 |
https://madrury.github.io/jekyll/update/statistics/2017/08/04/basis-expansions.html | 1,696,103,094,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510707.90/warc/CC-MAIN-20230930181852-20230930211852-00309.warc.gz | 402,630,935 | 12,521 | Beginners in machine learning often write off regression methods after learning of more exotic algorithms like Boosting, Random Forests, and Support Vector machines; why bother with a linear method when powerful non-parametric methods are readily available?
It’s easy to point out that the linear in linear regression is not meant to convey that the resulting model predictions are linear in the raw feaures, just the estimated parameters! The modeler can capture non-linearities in their regression, they only need to transform the raw predictors!
A common response is that the process of feature engineering is error prone, has no definite goal, and that it can be annoying work to explore data by hand and somehow divine correct transformations of predictors: why bother when other methods do it automatically.
Yet regression offers advantages that more high-tech methods do not. It is simple, and hardly a black box. Simple and common assumptions allow the modeler to perform statistical inference. It has many generalizations that advance both goals of predictive and inferential power simultaneously. And, more mundanely, it has a long history and solid foothold in many industries and sciences. Because of all this, we should seek to use this tool more skillfully.
It would be wise for us to seek a more non-parametric way to capture non-linear effects in regression models. Usually the only truly flexible method beginners learn is polynomial regression. This is a real shame, because as we will demonstrate below, it is about the worst performing method available.
The purpose of this post is to spread awareness of better options for capturing non-linearities in regression models; in particular, we would like to advocate more widespread adoption of linear and cubic splines.
### Software
I’ve taken the opportunity to write a small python module that is useful for using the basis expansions in this post. It conforms to the sklearn transformation interface, so can be used in pipelines and other high level processes in sklearn.
For example, to create a simple regression model using a piecewise linear spline on a single feature, we can use the following pipelining code:
from sklearn.linear_model import LinearRegression
from basis_expansions import LinearSpline
def make_pl_regression(n_knots):
return Pipeline([
('pl', LinearSpline(0, 1, n_knots=n_knots)),
('regression', LinearRegression(fit_intercept=True))
])
Users of sklearn will easily be able to adapt this to more complicated examples. For examples of use, the code used to run all the experiments and create images for this post can be found in this notebook.
### Acknowledgements
The idea for this post is based on my answer to hxd1011s question regarding grouping vs. splines at CrossValidated.
## Basis Expansions in Regression
To capture non-linearities in regression models, we need to transform some or all of the predictors. To avoid having to treat every predictor as a special case needing detailed investigation, we would like some way of applying a very general family of transformations to our predictors. The family should be flexible enough to adapt (when the model is fit) to a wide variety of shapes, but not too flexible as to overfit before adapting itself to the shape of the signal being modeled.
This concept of a family of transformations that can fit together to capture general shapes is called a basis expansion. The word basis here is used in the linear algebraic sense: a linearly independent set of objects. In this case our objects are functions:
and we create new sets of features by applying every function in our basis to a single raw feature:
### Polynomial Expansion
The most commonly, and often only, example taught in introductory modeling courses or textbooks is polynomial regression.
In polynomial regression we choose as our basis a set of polynomial terms of increasing degree1:
This allows us to fit polynomial curves to features:
Unfortunately, polynomial regression has a fair number of issues. The most often observed is a very high variance, especially near the boundaries of the data:
Above we have a fixed data set, and we have fit and plotted polynomial regressions of various degrees. The most striking feature is how badly the higher degree polynomials fit the data near the edges. The variance explodes! This is especially problematic in high dimensional situations where, due to the curse of dimensionality, almost all of the data is near the boundaries.
Another way to look at this is to plot residuals for each data point $x$ over many samples from the same population, as we fit polynomials of various degrees to the sampled data:
Here we see the same pattern from earlier: the instability in our fits starts at the edges of the data, and moves inward as we increase the degree.
Another final way to observe this effect is to estimate the average testing error of polynomial regressions fit repeatedly to the same population as the degree is changed:
The polynomial regression eventually drastically overfits, even to this simple one dimensional data set.
There are other issues with polynomial regression; for example, it is inherently non-local, changing the value of $y$ at one point in the training set can affect the fit of the polynomial at data points very far away, resulting in tight coupling across the space of our data (often referred to as rigidity). You can get a feel for this by playing around with the interactive scatterplot somoothers app hosted on this site.
The methods we will lay out in the rest of this post will go some way to alleviate these issues with polynomial regression, and serve as superior solutions to the same underlying problems.
### Binning Expansion
Probably the first thing that occurs to most modelers when reflecting on other ways to capture non-linear effects in regression is to bin the predictor variable:
In binned regression we simply cut the range of the predictor variable into equally sized intervals (though we could use a more sophisticated rule, like cutting into intervals at percentiles of the marginal distribution of the predictor). Membership in any interval is used to create a set of indicator variables, which are then regressed upon. In the one predictor case, this results in our regression predicting the mean value of $y$ in each bin.
Binning has its obvious conceptual issues. Most prominently, we expect most phenomena we study to vary continuously with inputs. Binned regression does not create continuous functions of the predictor, so in most cases we would expect there to be some unavoidable bias within each bin. In the simple case where the true relationship is monotonic in an interval, we expect to be under predicting the truth on the left hand side of each bin, and overpredicting on the right hand side.
Even so, binning is popular. It is easy to discover, implement, and explain, and often does a good enough job of capturing the non-linear behaviour of the predictor response relationship. There are better options though, we will see later that other methods do a better job of capturing the predictive power in the data with less estimated parameters.
### Piecewise Linear Splines
As a first step towards a general non-parametric continuous basis expansion, we would like to fit a piecewise linear funtion to our data. This turns out the be rather easy to do using translations of the template function $f(x) = max(0, x)$.
Taking linear combinations of these functions, we can create many piecewise linear shapes:
If we set down a fixed set of points where we allow the slope of the linear segment to change, then we can use these as basis functions in our regression:
The result will be to fit a continuous piecewise linear function to our data:
The values $\{k_1, k_2, \ldots, k_r\}$, which are the $x$ coordinates at which the slope of the segments may change, are called knots.
Note that we include the basis function $f_0(x) = x$ which allows the spline to assume some non-zero slope before the first knot. If we had not included this basis function, we would have been forced to use a zero slope in the interval $(\infty, k_1)$.
The estimated parameters for the basis function elements have a simple interpretation, they represent the change in slope when crossing from the left hand to the right hand side of a knot.
Fitting piecewise linear curves instead of polynomials prevents the explosion of variance when estimating a large number of parameters:
We see that as the number of knots increases, the linear spline does begin to overfit, but much more slowly than the polynomial regression with the same number of parameters.
The variance does not tend to accumulate in any one area (as it did near the boundaries of the data in polynomial regression). Instead the pockets of high variance are dependent on the specific data we happen to be fitting to:
### Natural Cubic Splines
Our final example of a basis expansion will be natural cubic splines. Cubic splines are motivated by the philosophy that most phenomena we would like to study vary as a smooth function of their inputs. While the linear splines changed abruptly at the knots, we would like some way to fit curves to our data where the dependence is less violent.
A natural cubic spline is a curve that is continuous, has continuous first derivatives (the tangent line makes no abrupt changes), continuous second derivatives (the tangent lines rate of rotation makes no abrupt changes); and it equal to a cubic polynomial except a points where it is allowed to high higher order discontinuities, these points are the knots. Additionally, we require that the curve be linear beyond the knots, i.e. to the left of the first knot, and right of the final knot.
We will skip writing down the exact form of the basis functions for natural cubic spline, the interested reader can consult The Elements of Statistical Learning or the source code used in this post.
All thought they seem more complex, it is important to realize that the continuity constraints on the shape of the spline are strong. While estimating a piecewise linear spline with $r$ knots uses $r+1$ parameters (ignoring the intercept), estimating a cubic spline takes only $r$ parameters. Indeed, in the above picture, the spline with only one knot is a line; this is because the slopes of the left hand and right hand line must match.
The linear constraints near the edge of the data are intended to prevent the spline for overfitting near the boundaries of the data, as polynomial regression tends to do:
## Experiments
We describe below four experiments comparing the behaviour of the basis expansions above. In each experiment we fit a regression to a dataset using one of the above expansions. The degrees of freedom (number of estimated parameters) was varied, and the performance of the fit was measured on a held out dataset2. This procedure what then repeated many times3 for each expansion and degree of freedom, and the average and variance of the hold out error was computed.
Our four experiments each fit to datasets sampled from a different population:
• A line plus random Gaussian noise.
• A sin curve plus random Gaussian noise.
• A weird continuous curve of mostly random design, plus random Gaussian noise.
• A sin curve with random discontinuities added, plus random Gaussian noise.
### Fitting to a Noisy Line
As a first experiment, let’s fit to a linear truth. This is the simplest possible case for our regression methods, and serves to test how badly they will overfit.
Let’s look at how the training and testing error varies as we increase the number of fit parameters for each of our basis expansions
The polynomial expansion eventually overfits drastically, while the other tree transformations only overfit gradually. None of the models have any bias, so the minimum expected testing error occurs for the minimum number of fit parameters.
The binning, piecewise linear, and piecewise cubic transformations do about equally well in terms of average testing error throughout, and seem to have about the same variation in testing error (i.e. the bands around the testing error curves area ll approximately the same width).
We can get a clearer comparison by plotting only the testing errors all on the same axis:
Between two and fifteen fit parameters, all the methods but polynomial regression perform about equally well, and overfit the training data at the same rate. So in this simple case, if we are only going for test performance, it doesn’t so much matter what we choose.
To examine the variance of the methods, we can plot the width of the variance bands all on the same axis:
In this example the variance of the testing error stays the same for all basis expansions throughout the entire range of complexity, with the obvious exception of polynomial regression, where the variance explodes along with the expectation.
### Fitting to a Noisy Sinusoid
Our next experiment will fit the various basis expansions to a sinusoidal signal:
The training / testing error plots are created in the same way as for the linear signal:
The most noticeable difference in this situation is that at a low number of estimated parameters, all the models are badly biased: while the fits to the linear function started at their lowest testing and training errors, for the sinusoid each basis expansion regression starts with high error, and decreases quickly as more parameters enter the model and the bias decreases.
The rates of overfitting tell the same story as before. The polynomial regression eventually overfits drastically, while the others overfit slowly.
Plotting the hold out error curves all on the same plot highlights another interesting feature:
The binned regression fits to the data much more slowly than polynomial and the two splines. While the splines achieve their lowest average testing error at around five estimated parameters, it takes around eighteen bins for the binned regression to achieve its minumum. Furthermore, the minimum testing error for the polynomial and spline regressions is lower than for the bins, so we have done worse overall, but payed a higher price.
A similar story holds for the testing error variances: the binning transformation has slightly higher variance than the two spline methods.
### Fitting to a Noisy Weird Curve
Our next example is a weird function constructed by hand to have an unusual shape
def weird_signal(x):
return (x*x*x*(x-1)
+ 2*(1/(1 + np.exp(-0.5*(x - 0.5))))
- 3.5*(x > 0.2)*(x < 0.5)*(x - 0.2)*(x - 0.5)
- 0.95)
The graph of this function has exponential, quadratic, and linear qualities in different segments of its domain.
We follow the usual experiment at this point. Here are the training and testing curves for the various basis expansion methods we are investigating:
The story here captures the same general themes as in the previous example:
• The quadratic basis expansion eventually drastically overfits rapidly.
• The other basis expansions overfit slowly.
• For very a very small number of estimated parameters, each of the basis expansions is biased.
• The variance in test error is generally stable as more parameters are added to the model, except for the polynomial regression, which explodes.
Comparing all the test errors on the same plot again highlights similar points to the sin curve
The bias of the binned regression decreases more slowly than the spline methods, and its minimum hold out error is both larger than the splines, and achieved at more estimated parameters. The linear and cubic spline generally achieve a similar testing error for the same number of estimated parameters.
The variances for the non-polynomial models are stable, with a hint that the binned regression has a larger overall variance across a range of model complexities.
### Fitting to a Noisy Broken Sinusoid
For our final example, we try a discontinuous function.
cutpoints = sorted(np.random.uniform(size=6))
def broken_sin_signal(x):
return (np.sin(2*np.pi*x)
- (cutpoints[0] <= x)*(x <= cutpoints[2])
- (cutpoints[1] <= x)*(x <= cutpoints[2])
- 2*(cutpoints[3] <= x)*(x <= cutpoints[4]))
To create this function, we have taken the sin curve from before, and created discontinuities by shifting intervals of the graph up or down. The result is a broken sin curve
In this situation, all of our models are biased, the bins cannot capture the continuous parts correctly, while the others can not capture the discontinuities.
Let’s repeat our experiments and see if the discontinuities change the patterns we have been observing among the different basis expansions:
The major difference in this example is the difficulty all methods have in decreasing the bias of the fit. Since the models we are fitting are all continuous, it is impossible for any of them to be unbiased. The best we can do is trap the discontinuities between two ever closer knots (in the case of splines), or inside a small bin. Indeed, even at 29 estimated parameters, the test error is still decreasing (slowly), indicating the our models can still do a better job decreasing the bias of their fits before becoming overwhelmed with variance.
Otherwise, there are no striking differences to the patterns we observe when comparing the various basis expansions.
Comparing the testing errors of the various methods, this time the binned expansions does noticeably worse than the other methods throughout the entire range of complexities. Otherwise, the two spline models are mostly equivalent in performance.
Finally, the variance of the binned model is also significantly worse in terms of variance than the other methods, while the two spline methodologies are equivalent.
## Conclusions
In this post we have compared four different approaches to capturing general non-linear behaviour in regression.
### Polynomial Regression
The major problem with polynomial regression is its instability once a moderate number of estimated parameters is reached. Polynomial regressions as highly sensitive to the noise in our data, and in every example were seen to eventually overfit violently; this feature was not seen in any of the other methods we surveyed.
The polynomial regression also does not accumulate variance uniformly throughout the support of the data. We saw that polynomial fits become highly unstable near the boundaries of the available data. This can be a serious issue in higher dimensional situations (though we did not explore this), where the curse of dimensionality tells us that the majority of our data will lie close to the boundary.
### Binning
Binning, which conceptually simple, was seen to suffer from a few issues in comparison to the other methods
• With a small number of estimated parameters, the binned regression suffered from a higher bias than its competitors. It was seen across multiple experiments to achieve its minimal hold out error at a larger number of estimated parameters than the other methods, and furthermore, often the minimal error achieved by the binning method was larger than the minimum achieved by the other methods.
• The binned regression’s hold out error often had a higher variance than the other method’s. This means that, even if it performed just as well as another method on average, any individual binned regression is less trustworthy than if using another method.
Additionally, the binned regression method has the disadvantage of producing discontinuous functions, while we expect most processes we encounter in nature or business to vary continuously in their inputs. This is philosophically unappealing, and also accounts for some of the bias seen when comparing the binning regressions to the other basis expansions.
### Linear and Cubic Splines
These methods performed well across the range of our experiments.
• They overfit slowly, especially compared to polynomial regression.
• They generally achieve a lower hold out error than the binned regressions, and achieve their minimum hold out error at a lower number of estimated parameters.
• Their hold out error is generally the lowest out of the methods we surveyed, making each individual regression more trustworthy.
The two spline models are clear winners over the binned and polynomial regressions. It is a shame that they are often not taught in introductory courses in predictive modeling. The author sees no real reason to fit polynomial or binned expansions in any predictive or inferential model, and hopes that these methods will enter the standard curriculum taught to new users of regression methods. At the very least, we hope you will use these methods in your own modeling work.
1. Of course, to avoid numerical issues we follow the usual advice of standardizing a predictor before applying a polynomial expansion.
2. The training and testing data were both sampled from the same population. Each model was trained on 250 data points, and tested on 250 data points.
3. Each experiment (sampling train and test data, fitting the model to train, evaluating performance on the hold out) was performed 500 times. | 4,191 | 21,160 | {"found_math": true, "script_math_tex": 10, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-40 | latest | en | 0.919628 |
https://ubraintv-jp.com/what-does-a-rectangular-pyramid-look-like/ | 1,653,273,145,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662552994.41/warc/CC-MAIN-20220523011006-20220523041006-00425.warc.gz | 650,330,723 | 4,800 | Pyramids space three-dimensional structures having triangle faces and also have one encompassing polygon form at that base. If the base of a pyramid is rectangular, climate it is called a rectangular pyramid. Despite the form of the basic is rectangular, the political parties of all pyramids space in a triangle shape. So, whenever we watch a pyramid, it each other a triangle from every of that is sides. The shape of the pyramid will assist us to determine the surface area and also volume of the pyramid.
You are watching: What does a rectangular pyramid look like
Like various other shapes, we have learned so far in geometry, the pyramid is likewise defined by their properties. The significant properties the the rectangle-shaped pyramid are based upon edges, faces and also vertices.
## Definition
As disputed in the introduction, a rectangular pyramid is a pyramid, which has a rectangular-shaped base. If we look at the bottom view of this pyramid, the looks prefer a rectangle. Hence, the base has actually two parallel sides equal.
A pyramid crowns at a suggest on optimal of the base, i m sorry is known as the apex. A rectangle-shaped pyramid can be a right pyramid or slope pyramid. If it is a right rectangle-shaped pyramid, climate the apex lies appropriate over the facility of the base and if it is an oblique rectangular pyramid, the apex is not directly over the center of the base, yet at some angle from the center.
Apart native the rectangle-shaped pyramid, there are other species of pyramids based upon the shape of your bases, together as:
Triangular PyramidSquare PyramidPentagonal PyramidHexagonal Pyramid
## Faces, Edges and also Vertices
The faces, edges and vertices of any type of pyramid are their key features. Let us comment on these 3 properties the the rectangle-shaped pyramid.
Faces: A rectangle-shaped pyramid has a total of five faces. One face is its rectangle-shaped base and also the remainder four encounters are its four triangular-shaped faces. Every the triangular encounters here space congruent to the opposite triangle face.
Vertex: there is a complete of 5 vertices of a rectangle-shaped pyramid. These vertices are the allude where the edges accomplish or intersect. One of the vertices is in ~ the top over the base, whereby the triangular faces of the pyramid meet. Rest four vertices lie at the edge of the rectangular base.
See more: What Channel Is The Superbowl On On Dish, What Tv Channel Is The Super Bowl On
Edges: A rectangular pyramid has actually eight edges. Each edge is formed by the intersection of two encounters or surfaces. 4 out of eight edges are located on the rectangular base, while the other four edges form slopes above the basic to a optimal point, which is the peak of the pyramid.
## Rectangular Pyramid Formula
The formula for surface area and volume the a rectangular pyramid is given here: | 580 | 2,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2022-21 | latest | en | 0.935897 |
https://tractorfriends.org/answers-on-questions/quick-answer-what-scale-is-my-tractor.html | 1,669,904,925,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710813.48/warc/CC-MAIN-20221201121601-20221201151601-00186.warc.gz | 621,051,577 | 20,756 | # Quick Answer: What Scale Is My Tractor?
## What size is a 1/32 scale tractor?
1:32 scale means that the toy is 1/:32nd as large as the real machine. For example if an 8320 tractor is 16-feet long the reproduction will be 6-inches long. 1:64 scale is the versatile scale.
## What size is 1.16 scale?
Every inch on the replica equals 16 inches on the real tractor. The larger the scale, the smaller the denominator; a 1/16 scale toy is much larger than a 1/87. “ Scale ” doesn’t refer to actual size. For example, a 1/16 scale toy could measure 8, 12, or 14 inches long.
## What size is 1/18 scale in inches?
Most models are approximately 11 inches (280 mm) long by 5 inches (130 mm) wide by 4 inches (100 mm) tall, depending on what vehicle is being represented.
## What size is a 1 50 scale model?
Model scales
Ratio Millimetres per foot
1:56 5.442 mm
1:55 5.644 mm
1: 50 6.096 mm
1:48 6.350 mm
You might be interested: FAQ: How To Change A Tire On A Kubota Tractor?
97
## What are the scale sizes for models?
Models come in a range of scales, the most common being 1:4, 1:8, 1:12, 1:16, 1:18, 1:24, 1:48, and 1:72. Choosing a scale that works for you is the first big step in mastering your model builds.
## How many cm is 1/32 scale?
A 6 ft (183 cm ) tall person is modeled as 2 14 in (57 mm) tall in 1:32 scale. 1:32 was once so common a scale for toy trains, autos, and soldiers that it was known as “standard size ” in the industry (not to be confused with Lionel’s “Standard Gauge”). 1:32 is the scale for Gauge 1 toy and model trains.
## What scale is an 8 inch figure?
Scale
Ratio Inches per foot Size
1:15 0.8″ [20.32 mm] 5″ [127 mm]
1:12 1″ [25.40 mm] 6″ [152.4 mm]
1:10 1.2″ [30.48 mm] 7″ [177.8 mm]
1:9 1.33″ [33.87 mm] 8 ” [203.2 mm]
## What is the difference between 1 24 and 1 25 scale?
For display purposes, 1/24th and 1/25th scale are close enough for horseshoes, as they say. The difference in size is about 4 percent. You won’t be able to swap engine or chassis parts between the two scales, though — one won’t fit with the other.
## How tall is a 1/12 scale figure?
For starters, the Mezco Toyz action figure is 1:12 scale which puts most figures at roughly 6 inches tall. These 1:12 scale figures are highly articulated and able to capture a very wide array of poses and stances.
## What is the biggest scale model size?
From front to back, the scales are: 1/144, 1/72, 1/48, 1/32, and 1/24. The smallest (1/144 scale ) would be a little more than 2½” long; the biggest (1/24 scale ), slightly more than 16″.
You might be interested: Question: Where To Sale Small Tractor In Ky?
## Whats bigger 1/12 or 1/18 scale?
1:12 scale – Very large, highly detailed models; usually about 14 or 15 in (36 or 38 cm) long; mainly targeted at adult collectors. Many diecast motorcycles are also produced in this scale. 1:18 scale – Large, detailed models, usually about 8 to 11 in (18 or 20 cm) long; mostly targeted at adults.
## What size is 1.24 scale?
1:24 means that a unit of measurement, such as one inch or one centimeter, on the model represents 24 units on the actual object. An example would be one inch of length on a model automobile would represent 24 inches on the real vehicle.
## What does a scale of 1 50 mean?
1: 50 is a ratio. it means you’re scaling 1 unit to 50 units. that could be inches ( 1 “= 50 “) or miles ( 1 mile= 50 miles) or anything else, but it’s a direct scale.
## What size is a 1/72 scale model?
1:72 scale is a scale used for scale models, most commonly model aircraft, corresponding to one sixth of an inch representing one foot (or 1 inch to 6 feet). In other words, 72 of a given model placed end to end would represent the length of the real thing.
## What scale are Lledo models?
These models were a big departure from the multiple-livery classic vehicle fare that Lledo normally made – they were entirely new castings and were not subsequently reproduced in any other promotional form. The models were in varied scales from 1:80 to 1:100 (the real cars were very large).
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detector | 1,191 | 4,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.901325 |
https://forum.arduino.cc/t/latching-an-internal-bit/694706 | 1,627,862,357,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154277.15/warc/CC-MAIN-20210801221329-20210802011329-00326.warc.gz | 266,985,221 | 8,381 | # Latching an internal Bit
Team, I hope this is an easy one
I have 2 inputs, and one output
for Argument,
Input A (physical switch) (OFF- (ON)) Switch (Magnetically held)
Input B (internal Driven Bit - input is serial from elsewhere)
Output C (physical output) (Magnet)
When A is High --> Output C is High, this also triggers input B to be sent after X time
When A is High and B is High --> C is also High, BUT
After a period, Input B goes low, and this needs to trigger Output C to go Low (which was triggered originally by Input A which is still high at this stage, sending Output C low will Physically set Input A to LOW also)
(Magnetically held Switch)
In Basic Steps
A Low + B Low = C Low (Normal off state)
A High + B Low = C High (Held) - Holds A High
A High + B High = C High (held) - Hold A high
A High + B Low = C Low (Released)
And
A Low at anytime = C is LOW, regardless of B
A is Held in the ON By C (The Magnet)
A is a Toggle OFF-(ON)
!A = !C
!A & !B = !C
A & !B = C
A & B = C
A & !B = !C (only after the above)
any thoughts Ben
jabmel:
A is Held in the ON By C (The Magnet)
A is a Toggle OFF-(ON)
A & !B = C
A & B = C
A & !B = !C (only after the above)
very confusing - conflicting statements
sees like there is a time element that isn't represented by you logic statements
sounds like maybe A will "arm" a system. B rising makes C HIGH. B going low causes C to go LOW as well as dis-arming the system so that without "reseting(?)" A, B rising has no further effect
Thanks for that. Yes. It’s a bit of a weird one.
It is part of DCS simulator.
The APU (aircraft Auxiliary Power Unit)
The switch that operates it is a 2 position switch.
OFF and Monetary ON. “OFF (ON)”
But. The monetary on is held on by a magnet
So the onboard computer can turn it off as no longer needed.
But the simulator when written failed to provide this output.
So the sequence is.
Switch ON (digital input)
This also triggers the output to hold the magnet
After a time the APU light is lit (internal game bit) this can be used.
Once the APU is no longer required this light goes out and the APU is turned off. I need to send the (Output low) for the magnet but it’s still held high by the same switch.
So the issue is.
The switch can trigger the magnet, but the magnet is needed to turn the same switch off. Once the light turns off.
Or at any time the switch is off. The output is off
Then all Over again if needed.
are you looking at solving for this with logic gates / delay components or with code ?
No time delay.
sounds like maybe A will "arm" a system. B rising makes C HIGH. B going low causes C to go LOW as well as dis-arming the system so that without "reseting(?)" A, B rising has no further effect
i believe my earlier description is accurate.
don't understand the need for a delay.
believe you need a state variable, "arm", that is set when A is HIGH.
so C is set if (arm & B)
the C falling resets arm
how do you capture the C falling?
I don’t set the time delay. That’s in the game software.
The switch is then held by the magnet to hold the switch in the on position.
I need to somehow release this output to turn the switch off.
When the game turns the light off after X time
But the light does not turn on as soon as the switch is turned on.
And the duration the light stays on is completely variable. Can be 1 min to XX mins
It's just a matter of states as gcjr said. Keep an enum to tell you where you are - a little state machine.
When you flip the switch to APU start, I assume you send that info to DCS. At that time set your state variable to APUStarting and use the magnet to hold the switch.
DCS will tell you to light the APU light. State= APURunning.
DCS tells you to turn the light off. State = APUOff, release the magnet.
I don't get the
I need to send the (Output low) for the magnet but it's still held high by the same switch.
. Do you have control over the magnet?
Once the APU is no longer required this light goes
how do you know it's no longer required? user action? time-out? black-magic?
that's what I understand in terms of state machine, is that correct?
J-M-L:
how do you know it's no longer required? user action? time-out? black-magic?
I think the APU is being used to start the main engines. There's user action required in the simulator to command them to spool up and an amount of time for that to complete.
Once they're running, the APU is not needed. So the Arduino doesn't know when the APU is shutdown without notification from the sim. It could probably estimate it from timing when the second engine was started, but maybe ambient temperatures or other factors make that vary.
The diagram looks perfect.
I can only control the switch. (On or off) but the game
Can turn it off to. Hence the magnet.
The switch needs to be held by the magnet until the APU light is out. (Or turned off physically)
The APU light is controlled by the DCS game. I can’t control this. But it is an output from the game (input to Arduino) I can use.
Yes correct. The APU starts the main engines. When they run and are stable the game (and in the real aircraft) turns the APU off. But the writer is the game interface forgot this. So I need to emulate it. It can’t be a timer.
Yeh only bits i have are.
Import Switch on off. (Real life)
APU run light on and off. Internal game bit)
Output the magnet on and off (output I drive)
Does DCS spam you with the state of the APU light or does it just let you know when it changes?
can you fill the blanks in the scenarios:
press the switch ON
monitor APU light, when it goes out, DO WHAT ?
press the switch ON
press the switch OFF, DO WHAT ?
wildbill:
Does DCS spam you with the state of the APU light or does it just let you know when it changes?
wildbill:
Does DCS spam you with the state of the APU light or does it just let you know when it changes?
it gives an output to drive a physical LED, the Physical LED matches the Game indication
J-M-L:
can you fill the blanks in the scenarios:
press the switch ON
monitor APU light, when it goes out, DO WHAT ?
press the switch ON
press the switch OFF, DO WHAT ?
J-M-L:
can you fill the blanks in the scenarios:
press the switch ON
monitor APU light, when it goes out, DO WHAT ?
press the switch ON
press the switch OFF, DO WHAT ?
press the switch ON ( MAG OUTPUT HIGH)
monitor APU light, when it goes out, DO WHAT ? (MAG OUTPUT LOW)
press the switch ON ( MAG OUTPUT HIGH)
press the switch OFF, DO WHAT ? (MAG OUTPUT LOW)
But remember the MAGNET (or output drving the magnet) is holding the switch physically in the ON position, thus the Switch input High, I need to logically break (manipulate) this output when the APU light turns off, whilst the switch is still on to allow it to turn off by releasing the magnet holding the toggle switch it is a spring toggle back to off
so you need some sort of actuator pressing the button physically for you? how do you plan to release the MAGNET?
Can we somehow just record the “on”input as a momentary bit. And use it to toggle an internal bit.
Then it won’t care that it stays on. Until it’s turned off ie the change of state.
J-M-L:
so you need some sort of actuator pressing the button physically for you? how do you plan to release the MAGNET?
The magnet releases when the switch is turned off, this in turn drives the output low until next turn on.
But the hard bit is I need to break this output logically when the APU light (game input) turns off. But I can’t break it as the switch is also holding it on. I need to let if forget the switch is on. Can the first turn in monetary drive an internal toggle but somehow
The switch is turned off 2 ways.
Either physically moving the real switch. Or the aircraft telling it it’s no longer needed.
This can only be done at this stage by trying to use the APU light turning off.
It might be useful to see some code. Do you tell DCS that the switch is pressed when it BECOMES pressed or repeatedly that it IS pressed.
Similarly, does DCS keep telling you the state of the APU light whether it’s changed or not? I suspect it does. | 2,014 | 8,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-31 | latest | en | 0.939765 |
https://math.stackexchange.com/questions/384835/does-f-analytic-imply-f-analytic | 1,726,237,994,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00232.warc.gz | 339,013,724 | 38,496 | # Does $f'$ analytic imply $f$ analytic?
If $f'$ is known to be analytic, does it mean that $f$ is analytic as well?
I've tried to expand $f$ and then to replace the tail of it by the expansion of $f'$, yet the factorials don't add up. I also tried to start with the known-to-converge expansion of $f'$ yet it was unclear how to move to $f$ (I didn't have integration yet).
If the statement isn't true then how does one prove, for example, that $f(x)=-\log\cos(x)$ is analytic in zero by using the fact that its derivative $\tan(x) = \sum_{n=1}^\infty (-1)^{n-1} 2^{2n}(2^{2n}-1) B_{2n}x^{2n-1}/(2n)!$ is analytic in zero?
• Do you mean real analytic? If $f'$ is known to be analytic (i.e. holomorphic), then $f'$ exists, which already implies that $f$ is analytic. Commented May 7, 2013 at 20:17
• @AWalker: yes, I meant it to be real analytic. Thank you for pointing it out about the complex case.
– Leo
Commented May 7, 2013 at 20:24
• If $g(x):=f'(x)=\sum_n a_nx^n$, then find $b_n$ ($n\ge 1$) such that for $G(x):=\sum_n b_nx^n$ we have $G'=g$.
Hint: $(x^n)'=nx^{n-1}$.
• If you know that $h'=0\implies h=$const., then you get $b_0$ as $(f-G)'=g-g=0$
since $f'$ is analytic in $\mathbb{c}$, it has a power series expansion say $f'(z)=\displaystyle\sum_{n=1}^{\infty}a_n z^n$ so $\int f'(z)dz= \int\displaystyle\sum_{n=1}^{\infty}a_n z^n$. This implies that $f(z)=\displaystyle\sum_{n=1}^{\infty}\int a_n z^n =\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{n+1} z^{n+1}= \displaystyle \sum_{n=1}^{\infty}b_n z^n$ which is anlytic. | 565 | 1,535 | {"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} | 3.6875 | 4 | CC-MAIN-2024-38 | latest | en | 0.890061 |
https://www.coursehero.com/file/5830098/368-pdfsam-math-54-differential-equation-solutions-odd/ | 1,513,058,126,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948515165.6/warc/CC-MAIN-20171212041010-20171212061010-00192.warc.gz | 721,078,490 | 81,038 | 368_pdfsam_math 54 differential equation solutions odd
# 368_pdfsam_math 54 differential equation solutions odd -...
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Chapter 6 Diferentiating y p ( x ) yields y 0 p ( x )=2 C 1 xe 2 x 2 C 1 x 2 e 2 x =2 C 1 ( x x 2 ) e 2 x y 0 p ( x )= 4 C 1 ( x x 2 ) e 2 x +2 C 1 (1 2 x ) e 2 x C 1 (2 x 2 4 x +1) e 2 x y 0 p ( x 4 C 1 (2 x 2 4 x e 2 x C 1 (4 x 4) e 2 x =4 C 1 ( 2 x 2 +6 x 3) e 2 x . By substituting these expressions into the nonhomogeneous equation, we obtain y 0 p ( x )+3 y 0 p ( x ) 4 y p ( x )=4 C 1 ( 2 x 2 x 3) e 2 x +6 C 1 (2 x 2 4 x e 2 x 4 C 1 x 2 e 2 x = e 2 x ⇒− 6 C 1 e 2 x = e 2 x . By equating coefficients, we see that C 1 = 1 / 6. Thus, a general solution to the nonhomoge- neous diferential equation is given by y ( x y h ( x )+ y p ( x c 1 e x + c 2 e 2 x + c 3 xe 2 x 1 6 x 2 e 2 x . 9. Solving the auxiliary equation, r 3 3 r 2 +3 r 1=( r 1) 3 = 0, we Fnd that r = 1 is its root o± multiplicity three. There±ore, a general solution to the associated homogeneous equation is
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Ask a homework question - tutors are online | 506 | 1,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-51 | latest | en | 0.721744 |
https://www.effortlessmath.com/math-topics/10-most-common-grade-3-mcas-math-questions/ | 1,624,596,256,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488567696.99/warc/CC-MAIN-20210625023840-20210625053840-00586.warc.gz | 662,318,433 | 15,777 | # 10 Most Common 3rd Grade MCAS Math Questions
Preparing 3rd-grade students for the MCAS Math test? Searching for a preview of the most common mathematics questions on the 3rd Grade MCAS Math test? If so, then you are in the right place.
The mathematics section of 3rd Grade MCAS can be a challenging area for some students, but with enough patience, it can be easy and even enjoyable!
Make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions your students need to practice.
## 10 Sample 3rd Grade MCAS Math Practice Questions
1- Classroom A contains 8 rows of chairs with 4 chairs per row. If classroom B has three times as many chairs, which number sentence can be used to find the number of chairs in classroom B?
A. $$8 × 4 + 3$$
B. $$8 + 4 × 3$$
C. $$8 × 4 × 3$$
D. $$8 + 4 + 3$$
2- A cafeteria menu had spaghetti with meatballs for $8 and bean soup for$6. How much would it cost to buy three plates of spaghetti with meatballs and three bowls of bean soup? __________
3- There are 7 days in a week. There are 24 hours in day. How many hours are in a week?
A. 48
B. 96
C. 168
D. 200
4- Emily described a number using these clues:
Three-digit odd numbers that have a 6 in the hundreds place and a 3 in the tens place
A. 627
B. 637
C. 632
D. 636
5- This clock shows a time after 12:00 PM. What time was it 1 hours and 30 minutes ago?
A. 12:45 PM
B. 1:45 PM
C. 1:15 PM
D. 12:15 PM
6- Olivia has 84 pastilles. She wants to put them in boxes of 4 pastilles. How many boxes does she need?
A. 20
B. 21
C. 24
D. 28
7- A football team is buying new uniforms. Each uniform costs $20. The team wants to buy 11 uniforms. Which equation represents a way to find the total cost of the uniforms? A. $$(20 × 10) + (1 × 11) = 200 + 11$$ B. $$(20 × 10) + (10 × 1) = 200 + 10$$ C. $$(20 × 10) + (20 × 1) = 200 + 20$$ D. $$(11 × 10) + (10 × 20) = 110 + 200$$ 8- There are 92 students from Lexington Elementary school at the library on Tuesday. The other 54 students in the school are practicing in the classroom. Which number sentence shows the total number of students in Riddle Elementary school? A. $$92 + 54$$ B.$$92 -54$$ C. $$92 × 54$$ D. $$92 ÷54$$ 9- Use the picture below to answer the question. Which fraction shows the shaded part of this square? A. $$\frac{92}{100}$$ B. $$\frac{92}{10}$$ C. $$\frac{90}{100}$$ D. $$\frac{8}{100}$$ 10- Which number correctly completes the number sentence $$90 × 25 =$$ ? A. 225 B. 900 C. 1250 D. 2250 ## Best 3rd Grade MCAS Math Prep Resource for 2021 ## Answers: 1- C 3times of 8 rows of chairs with 4 chairs per day is: $$3 × 8 × 4$$ 2- 42 3spaghetti with meatballs cost: $$3 × 8 = 24$$ 3bowls of bean soup cost: $$3 × 6 = 18$$ 3spaghetti with meatballs + 3bowls of bean soup cost: $$24 + 18 = 42$$ 3- C 1 day: 24 hours so 7 days $$= 7 × 24 = 168$$ hours 4- B Three-digit odd numbers that have a 6 in the hundreds place and a 3 in the tens place is 637. 632 and 636 are even numbers. 5- A Subtract hours: $$14– 1 = 13$$ Subtract the minutes: $$15 – 30 = – 15$$ The minutes are less than 0, so: • Add 60 to minutes ( $$-15 +60 =45$$ minutes) • Subtract 1 from hours $$(13 – 1 =12)$$ 6- B Olivia wants to divide 84 pastilles into boxes of 4 pastilles, so $$84 ÷ 4 = 21$$ is the amount of boxes. 7- C Football team should buy 11 uniforms that each uniform cost$20 so they should pay$$(11×20) 220$$.
Therefore, choice C is correct answer:
$$(20 × 10) + (20 × 1) = 20(10+1) =20×11=220$$
8- A
Add 92 and 54 students to know the whole number of students.
9- A
the model for the fraction is divided into 100 equal parts. We shade 92 parts of these 100 parts that it’s equal to $$\frac{92}{100}$$
10- D
$$90 × 25 =2,250$$
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Help save the environment | 1,298 | 4,047 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-25 | longest | en | 0.93494 |
https://wiki.haskell.org/index.php?title=Euler_problems/21_to_30&oldid=19230 | 1,440,999,891,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644065534.46/warc/CC-MAIN-20150827025425-00187-ip-10-171-96-226.ec2.internal.warc.gz | 880,104,370 | 7,694 | # Euler problems/21 to 30
Jump to: navigation, search
## 1 Problem 21
Evaluate the sum of all amicable pairs under 10000.
Solution:
```--http://www.research.att.com/~njas/sequences/A063990
problem_21 = sum [220, 284, 1184, 1210, 2620, 2924, 5020, 5564, 6232, 6368]```
## 2 Problem 22
What is the total of all the name scores in the file of first names?
Solution:
```import Data.List
import Data.Char
problem_22 =
do input <- readFile "names.txt"
let names = sort \$ read\$"["++ input++"]"
let scores = zipWith score names [1..]
print . show . sum \$ scores
where score w i = (i *) . sum . map (\c -> ord c - ord 'A' + 1) \$ w```
## 3 Problem 23
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
Solution:
```--http://www.research.att.com/~njas/sequences/A048242
import Data.Array
n = 28124
abundant n = eulerTotient n - n > n
abunds_array = listArray (1,n) \$ map abundant [1..n]
abunds = filter (abunds_array !) [1..n]
rests x = map (x-) \$ takeWhile (<= x `div` 2) abunds
isSum = any (abunds_array !) . rests
problem_23 = putStrLn . show . foldl1 (+) . filter (not . isSum) \$ [1..n]```
## 4 Problem 24
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
Solution:
```import Data.List
fac 0 = 1
fac n = n * fac (n - 1)
perms [] _= []
perms xs n= x : perms (delete x xs) (mod n m)
where m = fac \$ length xs - 1
y = div n m
x = xs!!y
problem_24 = perms "0123456789" 999999```
## 5 Problem 25
What is the first term in the Fibonacci sequence to contain 1000 digits?
Solution:
```valid ( i, n ) = length ( show n ) == 1000
problem_25 = fst . head . filter valid . zip [ 1 .. ] \$ fibs
where fibs = 1 : 1 : 2 : zipWith (+) fibs ( tail fibs )```
## 6 Problem 26
Find the value of d < 1000 for which 1/d contains the longest recurring cycle.
Solution:
`problem_26 = head [a | a<-[999,997..], and [isPrime a, isPrime \$ a `div` 2]]`
## 7 Problem 27
Find a quadratic formula that produces the maximum number of primes for consecutive values of n.
Solution:
```problem_27 = -(2*a-1)*(a^2-a+41)
where n = 1000
m = head \$ filter (\x->x^2-x+41>n) [1..]
a = m-1```
## 8 Problem 28
What is the sum of both diagonals in a 1001 by 1001 spiral?
Solution:
`problem_28 = sum (map (\n -> 4*(n-2)^2+10*(n-1)) [3,5..1001]) + 1`
## 9 Problem 29
How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?
Solution:
```import Control.Monad
problem_29 = length . group . sort \$ liftM2 (^) [2..100] [2..100]```
## 10 Problem 30
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
Solution:
```--http://www.research.att.com/~njas/sequences/A052464
problem_30 = sum [4150, 4151, 54748, 92727, 93084, 194979]``` | 964 | 2,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2015-35 | longest | en | 0.658913 |
https://answers.search.yahoo.com/search?p=1+to+100+numbers&ei=UTF-8&flt=cat%3APolitics&fr2=rs-bottom%2Cp%3As%2Cv%3Aw%2Cm%3Aat-s&b=21&pz=10&bct=0&xargs=0 | 1,603,962,864,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107903419.77/warc/CC-MAIN-20201029065424-20201029095424-00138.warc.gz | 216,139,675 | 32,995 | # Yahoo Web Search
1. ### How many prime numbers between 1 to 100 ?
25 - I don't know of a mathematical way to figure it out. ref: http://learning.mgccc.cc.ms.us/pk/mathdocs/devmath/primefact.htm 39 = 3 * 13 51 = 3 * 17 57 = 3 * 19 69 = 3 * 23 89 = 3 * 29
7 Answers · Science & Mathematics · 26/06/2007
2. ### what do you get when you add all of the numbers from 1 to 100 ?
You pair them off: 1 +99 = 100 2 + 98 = 100 3 + 97 = 100 ... 48+52 = 100 49 + 51 = 100 That's 49...http://en.wikipedia.org/wiki/Arithmetic_series#Sum_.28arithmetic_series.29 1 + 2 + ... + n = n(n+ 1 ) / 2
7 Answers · Science & Mathematics · 01/08/2007
3. ### Write a program in C++ that prints the numbers from 1 to 100 .?
...;< "Would you like to play again? y or n: "; cin >> answer...; endl; } while (num >= 1 && num <= 100 && answer == 'y'); system("PAUSE...
3 Answers · Computers & Internet · 24/10/2011
4. ### can any body provide program for first 100 (not 1 to 100 )prime numbers in c?
... any number: ");//enter 100 scanf("%d", &num); printf("\...n++) { if(n ==2) p= 1 ; for(div=2; div<n...0; } If you copy and paste remember to edit out white spaces needed...
2 Answers · Education & Reference · 20/02/2013
5. ### do you know the formula total 1 to 100 numbers ?
The sum of any 'n' numbers from 1 to 'n' is n(n + 1 )/2. Thus if n = 100 , the sum will be 100 * 101 / 2 = 5050.
8 Answers · Science & Mathematics · 24/03/2008
6. ### which whole numbers from 1 to 100 would not appear in any of the pairs of numbers to equal 100 ?
...can add to 100 to get to 100 is 0, and 0 is not included in 1 to 100 .
2 Answers · Science & Mathematics · 26/09/2007
7. ### roman numbers from 1 to 100 (15 points) ?
... 20 XX 40 XL 60 LX 80 LXXX 100 C
4 Answers · Education & Reference · 09/12/2012
8. ### i need a simple program from printing odd and even numbers from 1 to 100 ?
...main(String arg[]) { int s=0; System.out.print("Even Numbers are"); for(int i= 1 ;i<= 100 ;i++) { if(i%2==0) { System.out.println(i); } } System.out.println...
2 Answers · Computers & Internet · 18/12/2008
9. ### What are all of the Prime Numbers from 1 to 100 ?
... prime itself. Fine I give up its kinda hard to explain so here. 1 2 3 5 7 11 13 17 19 OH JUST FIGURE IT...
11 Answers · Education & Reference · 20/04/2008
10. ### how many times is the digit 5 written when listing all numbers from 1 to 100 ,000?
...there are a total of 100 *2 = 200 digits. Thus, each digit 0 to 9 comprises 10... in that range of numbers . This pattern ... that for 1 to 100000, you have one more...
4 Answers · Science & Mathematics · 17/04/2007
1. Try asking your question on Yahoo Answers | 890 | 2,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-45 | latest | en | 0.665536 |
https://knowt.com/flashcards/5713bd25-9030-4597-9529-b826ffe6ee7c | 1,717,037,873,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059418.31/warc/CC-MAIN-20240530021529-20240530051529-00245.warc.gz | 276,536,522 | 43,117 | # AP Statistics Unit 2 KNIGHT
Studied by 40 people
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get a hint
hint
population
1 / 37
## Tags and Description
### 38 Terms
1
population
the whole group that we want info about
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2
bias
results of a study consistently over or underestimate the value of a parameter we want to know
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3
convenience sampling
taking a sample consisting of individuals that are easy for the experimenter to reach
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4
Why is convenience sampling biased?
The individuals that are easiest to reach will likely differ from the population at large in some systemic fashion
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5
Voluntary response sample
consists of individuals who choose to respond to a survey
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6
Why is a voluntary response sample biased?
The people who choose to respond most likely have stronger opinions than the average individual, which cannot be generalized to the population at large
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7
Random sampling
requires a chance process
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8
Simple random sample
a sample of size n that is chosen so that every possible group of n individuals in the population have an equal chance of being selected for the sample
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9
sampling frame
a list of individuals from whom the sample is drawn
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10
downsides of simple random sampling (srs)
1. sampling frames are difficult to construct accurately (especially for large populations) 2. selection of subjects takes a long time
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11
stratified random sampling
population is divided into groups of similar individuals (strata) and a simple random sample will then be taken in each stratum and the combined results will be the actual sample (SIMILAR WITHIN, DIFFERENT BETWEEN)
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12
Cluster sample
population is divided into groups of individuals located near each other (clusters) (DIFFERENT WITHIN, SIMILAR BETWEEN) each cluster should be a small-scale population
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13
larger samples....
are more precise and decrease variation
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14
selection bias (undercoverage)
some members of the population cannot be chosen in a sample (ex: survey mailed to homes dont cover homeless ppl)
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15
nonresponse bias
individual who is chosen for the saample cannot be contacted or refuses to participate
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16
wording of question bias
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17
response bias
gender, age, race, ethnicity, or the behavior of the interviewer affects the responses in some systematic way
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18
response variable
a measured outcome of a study
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19
explanatory variable
something that can explain the response variable
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20
observational study
individuals are observed and variables of interest are measured; no treatment is imposed (NO CAUSE AND EFFECT CAN BE DETERMINED)
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21
experiment
deliberately imposes treatment on the subject (CAN BE USED TO DETERMINE CAUSE AND EFFECT)
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22
controlled experiment
some subjects are given a treatment and others given a placebo (used a comparison group)
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23
confounding
2 or more variable are associated in such a way that their effects on the response cannot be distinguished from each other
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24
experimental units
smallest collection of individuals to which treatments are applied
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25
4 principles of good experimental design
comparison, random assignment, control, replication
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26
principle of comparison
a good experimental design will be used to compare 2 or more treatments
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27
principle of random assignment
units are to be assigned to treatments by some chance process (purpose of random assignment is to CREATE APPROX EQUAL GROUPS FOR COMPARISON)
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28
principle of control
keeping outside variables that might affect the response constant (as much as possible). This REDUCES VARIABILITY in the response variable
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29
principle of replication
to use enough experimental units in each group to distinguish the results (from the treatment) from chance differences between the groups
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30
statistically significant
the effect is large enough to rarely occur by chance alone
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31
practical importance
Something can be statistically significant but not practically important (Ex: avg test score goes from 90 to 92% due to a test prep site can be statistically significant but not practically important)
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32
block
a group of experimental units that are known before the experiment begins to be similar in some way that is thought to affect the response (purpose: TO REDUCE VARIABILITY THAT MIGHT ARISE FROM RANDOM ASSIGNMENT)
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33
matched pairs design
pairs of similar experimental units are matced up and one of the units in the pair is randomly assigned the treatment (and other the control) Alternatively, both treatment and control can be assigned to the same unit with the order of reception randomized
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34
random selection
if so, you can make an inference about the population
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35
random assignment to groups
if so, you can assume cause and effect
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36
Types of bias
response bias, nonresponse bias, wording of question bias, selection bias (undercoverage)
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37
factor
explanatory variables in an experiment
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38
level
each possible value a factor can take on
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4.1 Stars(393) | 1,514 | 6,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-22 | latest | en | 0.925796 |
http://slideplayer.com/slide/4180400/ | 1,524,499,689,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946077.4/warc/CC-MAIN-20180423144933-20180423164933-00489.warc.gz | 287,913,778 | 21,131 | # ©UCL 2004 17 TH IEEE/LEOS Conference Puerto Rico, 7-11 th November, 2004 Multimode Laterally Tapered Bent Waveguide Ioannis Papakonstantinou, David R.
## Presentation on theme: "©UCL 2004 17 TH IEEE/LEOS Conference Puerto Rico, 7-11 th November, 2004 Multimode Laterally Tapered Bent Waveguide Ioannis Papakonstantinou, David R."— Presentation transcript:
©UCL 2004 17 TH IEEE/LEOS Conference Puerto Rico, 7-11 th November, 2004 Multimode Laterally Tapered Bent Waveguide Ioannis Papakonstantinou, David R. Selviah and F. Anibal Fernandez Department of Electronic and Electrical Engineering University College London Outline Research Motivation Modelling Approach Results - Discussion
©UCL 2004 2 Research Motivation To minimise cost of connectors between the laser-detector arrays and the backplane waveguides Passive alignment of the optical connectors A large amount of misalignment must be tolerated Tapered waveguide entrances seem ideal In a dense configuration of boards and connectors the waveguides are curved to avoid the neighbouring connector A bent taper conserves space Optical Backplane Optical waveguides Laser – Detector array Connector area
©UCL 2004 3 The Bent Taper In a “bent taper” the lateral dimension, a, tapers linearly with respect to angle, θ to the final width, b x c z y c a b θ r Bend Taper a c b y z x Linear Taper In a “linear taper” the lateral dimension, a, tapers linearly with respect to the – z axis to the final width, b
©UCL 2004 4 ≡ Co-ordinate Transform The transform u = r – R, v = Rθ maps the bent taper to a straight taper The effective index of the structure is tilted in comparison with the usual step index guide The slope of the tilt depends on the radius of curvature For u > u o, n cladding > n core. A bend is always lossy Index in the core is asymmetric resulting to asymmetric modes Solid line: Index of transformed guide Dashed line: Step-index guide
©UCL 2004 5 Simulation Technique (A) Bent Taper (B) Transformed straight taper z z FD – BPM 3D – Mesh of 0.1 μm × 0.1 μm and 1 μm axial step (1,1) Padé Coefficients Full TBC boundary conditions Benefits by using the transform with BPM A.BPM paraxial limitations are altered B.Significant reduction of the simulation area/time θ w R A2A2 A1A1
©UCL 2004 6 Physical Parameters Channel waveguide with initial dimensions a = 50 μm, c = 50 μm Dimension b varies from 25 μm to 2 μm Variable taper ratio (a/b): 2 < a/b < 25 n core = 1.54, n cladding = 1.5107. N.A = 0.3 R > 20 mm to minimize bend losses Material intrinsic losses and scattering losses all ignored Launching field: Gaussian 7 μm 1/e width, TE – polarised, λ = 850 nm VCSEL fundamental mode z y x c c a b θ r Bent Taper
©UCL 2004 7 Lateral Misalignment Input Gaussian field is translated along the x-axis Position 0 is at the centre of the guide Maximum transmittance NOT when the source is centred to the guide Coupling is better towards the outer side of the bend This is due to the asymmetric nature of the modes inside the bend VCSEL Transmittance (dB) Field axial misalignment (μm)
©UCL 2004 8 φ Angular Misalignment Input field is positioned at the maximum position on the x - axis Then it is rotated on the xz - plane As the taper ratio increases losses increase For < 3 dB losses we can tolerate just a few degrees of misalignment in any case Therefore angular misalignment might be more critical than translational VCSEL Transmittance (dB) Field rotational misalignment (degrees)
©UCL 2004 9 Comparison with Linear Tapers FWHM of the lateral and rotational misalignment graphs for bent and linear tapers are compared Linear tapers show higher insertion loss but better lateral misalignment tolerance Bent tapers show better angular misalignment tolerance All FWHM degrade as taper ratios increase Taper ratio (a/b) Lateral offset FWHM (μm) Max. normalized power (dB) Angular rotational FWHM (degrees) Solid lines: Bent taper Dashed lines: Linear taper Solid line: Bent taper Dashed line: Linear taper
©UCL 2004 10 Conclusions Acknowledgements Bent taper simulations using FD-BPM revealed: As taper ratio varies from 1 < a/b < 25 lateral misalignment FWHM x degrades from 50 μm down to 7 μm Proportionally angular misalignment FWHM θ degrades from 10 0 to 2 0 Xyratex Ltd. for financial support and useful discussion Frank Tooley for useful discussion
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https://de.mathworks.com/matlabcentral/cody/problems/2023-is-this-triangle-right-angled/solutions/687399 | 1,579,450,353,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594662.6/warc/CC-MAIN-20200119151736-20200119175736-00498.warc.gz | 414,181,097 | 15,613 | Cody
Problem 2023. Is this triangle right-angled?
Solution 687399
Submitted on 16 Jun 2015 by Cory Johnson
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
%% a = 3; b = 4; c = 5; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))
2 Pass
%% a = 3; b = 4; c = 6; flag_correct = false; assert(isequal(isRightAngled(a,b,c),flag_correct))
3 Pass
%% a = 3; b = 5; c = 4; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))
4 Pass
%% a = 4; b = 3; c = 5; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))
5 Pass
%% a = 4; b = 5; c = 3; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))
6 Pass
%% a = 5; b = 3; c = 4; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))
7 Pass
%% a = 5; b = 4; c = 3; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct))
8 Pass
%% a = 5; b = 12; c = 13; flag_correct = true; assert(isequal(isRightAngled(a,b,c),flag_correct)) | 386 | 1,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.338606 |
https://www.mathhomeworkanswers.org/258527/96-8%C2%B0f-%E2%89%A4-0-4-%E2%89%A4-102-2%C2%B0f?show=258761 | 1,552,989,064,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201953.19/warc/CC-MAIN-20190319093341-20190319115341-00122.warc.gz | 835,686,681 | 13,168 | Write and solve an absolute value inequality that represents the range of possible temperatures for a reading of 99.1°f . Explain what you need to consider and calculate. Than graph the solution set
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## 1 Answer
Assuming that 99.1℉ is a temperature rounded to 1 decimal place then the true temperature T must satisfy T<99.15 and T>99.05, that is, 99.05<T<99.15 (ignoring the traditionally accepted rule for rounding T≥99.05).
To express this as an absolute value inequality we subtract 99.1 from each term in the multiple inequality:
-0.05<T-99.1<0.05. This is now in the form ready to be converted to an absolute value inequality:
|T-99.1|<0.05. The range of values of T making this true is shown graphically below.
Continued in comment...
by Top Rated User (613k points)
The blue section of the line represents the inequality, excluding the orange points and line. What we’ve done here is to consider what’s meant by “absolute value”: the distance between zero and two points equidistant from zero. Then we adjust the zero point—T-99.1=0 so that we have the two points 99.05 and 99.15 equidistant from this zero point. This gives us the blue range of values for T that produce a reading of 99.1℉ accurate to one decimal place.
If we now consider the temperature range 96.8 and 102.2 we take the average=(96.8+102.2)/2=99.5℉, and we subtract this from each of the other temperatures and we get -2.7<T-99.5<2.7, which becomes |T-99.5|<2.7. The number line graph can be altered so that the zero point is 99.5 and the end points are 96.8 and 102.2.
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Star Connection vs Delta Connection - Difference between Star and Delta Connection
As we know that Star & Delta Connection is only possible in 3 phase system, so in our domestic system it is not possible, because generally all house hold electrical equipment are designed with single phase supply. In a three phase circuit, there are two types of connections. One is known as Star Connection, and the other one is Delta Connection.
omparison between Star and Delta Connections
We mostly use the terms star and delta in electrical systems while discussing three phase AC circuits and electric motors. Below is a given table which compares both Star and delta connections which shows the exact difference between Star (Y) and Delta (Δ) Connections.
Star Connection:
Star connection “Y” is obtained by joining together similar ends of coils either “starting” or finishing. The other ends are joined to the line wires. The common point is called the Neutral or Star Point. This three-phase, 4-wires system is used in power distribution, transformers and small scale domestic and residential applications.
Delta Connection
Delta or Mesh Connection “Δ” is obtained by connecting the starting end of the first coil to the finishing end of the second coil and so on (for all three coils) which forms like a closed loop or mesh circuit. This three-phase, 3-wires system is used in power transmission, transformers, and large scale industrial and commercial applications.
In both systems, the Voltage between two phases (Line to Line) is known as Line Voltage, while the voltage between Phase to Neutral (Line to Neutral) is known as Phase voltage. The voltage between any line (or phase) and Neutral is single phase while the voltage between all three lines (or phases) is known as three phase voltage. Keep in mind that the power in both systems are always equal and the same as it is the game of different levels of voltages and currents only used in different systems based on the requirement.
Star Connection vs Delta Connection - Difference between Star and Delta Connection
As we know that Star & Delta Connection is only possible in 3 phase system, so in our domestic system it is not possible, because generally all house hold electrical equipment are designed with single phase supply. In a three phase circuit, there are two types of connections. One is known as Star Connection, and the other one is Delta Connection.
omparison between Star and Delta Connections
We mostly use the terms star and delta in electrical systems while discussing three phase AC circuits and electric motors. Below is a given table which compares both Star and delta connections which shows the exact difference between Star (Y) and Delta (Δ) Connections.
Star Connection:
Star connection “Y” is obtained by joining together similar ends of coils either “starting” or finishing. The other ends are joined to the line wires. The common point is called the Neutral or Star Point. This three-phase, 4-wires system is used in power distribution, transformers and small scale domestic and residential applications.
Delta Connection
Delta or Mesh Connection “Δ” is obtained by connecting the starting end of the first coil to the finishing end of the second coil and so on (for all three coils) which forms like a closed loop or mesh circuit. This three-phase, 3-wires system is used in power transmission, transformers, and large scale industrial and commercial applications.
In both systems, the Voltage between two phases (Line to Line) is known as Line Voltage, while the voltage between Phase to Neutral (Line to Neutral) is known as Phase voltage. The voltage between any line (or phase) and Neutral is single phase while the voltage between all three lines (or phases) is known as three phase voltage. Keep in mind that the power in both systems are always equal and the same as it is the game of different levels of voltages and currents only used in different systems based on the requirement. | 794 | 4,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.967521 |
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# sherita and rachel007 new workout numbers for OCTOBER
Topic closed. 87 replies. Last post 4 years ago by meriwetherman.
Page 3 of 6
Vtracs is My Game!
Georgia
United States
Member #3617
February 6, 2004
7421 Posts
Offline
Posted: October 13, 2012, 1:41 pm - IP Logged
Wow!! that just confirm we are going in the right direction.Just know I'm playing the (9) numbers all this month.
in addition im liking the 2389 cash 4 which by the way is my address.I noticed how close that number has been all this month. So basically right now I'M TOTALLY focus on my 9 numbers and 2389.If I see anything else interesting I'll let you know.
Oh by the way! you know when we talked about the digit 1 and how hot it is this month and that due to digit 1 being hot I;m seeing the number 11 multiplying with other numbers such as
for example 11x1 =11 number that fell 1711
11x2 =22 number that fell 1122
11x3 =33 number that fell 5633
11x6 =66 number that fell 4686
11x8 =88 number thats fell 8892
so that leaves the 11x4=44
11x5=55
11x7=77
11x9=99
These pairs have not appeared yet all I do see is odd numbers showing up with the 11 it makes since 11 is a odd number now if I can only figure out the rest hopefully as the month progress we will be able to see more interesting movements.
Good Luck!
Ga Mid 1619 your 11 pair with odd digit 9..119x. I would have been so sweet if she had let 1919 come instead of 1619!
Congrats To All Winners and Posters!
We are all in it to win! My Pet numbers 103,724,152,397,189,118,205.
Ga Lottery Player!
Hot-Due-Cold. Short Sums/Last Digit Sum. Pairs. Vtracs. Vtrac Pairs. SUM OF VTRACS CHART.
Don't forget to FLIP 6/9 in all WORKOUTS! 66=99=69. Vtracs Code sheets available....
GEORGIA PEACH
United States
Member #70173
January 25, 2009
4009 Posts
Offline
Posted: October 13, 2012, 6:42 pm - IP Logged
As you can see the 11 pair came back with the 96 how interesting.She flipped the 9 to a 6 that is one sneaky woman lol.
Vtracs is My Game!
Georgia
United States
Member #3617
February 6, 2004
7421 Posts
Offline
Posted: October 13, 2012, 10:45 pm - IP Logged
Rachel, these Triplets are now showing....
441x 451x 491x
443x 453x 493x
446x 456x 496x
447x 457x 497x
448x 458x 498x
Congrats To All Winners and Posters!
We are all in it to win! My Pet numbers 103,724,152,397,189,118,205.
Ga Lottery Player!
Hot-Due-Cold. Short Sums/Last Digit Sum. Pairs. Vtracs. Vtrac Pairs. SUM OF VTRACS CHART.
Don't forget to FLIP 6/9 in all WORKOUTS! 66=99=69. Vtracs Code sheets available....
SWATS Shawty
United States
Member #123454
February 21, 2012
22176 Posts
Offline
Posted: October 13, 2012, 11:17 pm - IP Logged
Rachel, these Triplets are now showing....
441x 451x 491x
443x 453x 493x
446x 456x 496x
447x 457x 497x
448x 458x 498x
I played 3446 and 3449. Hope she dont come with 3443!
Vtracs is My Game!
Georgia
United States
Member #3617
February 6, 2004
7421 Posts
Offline
Posted: October 13, 2012, 11:53 pm - IP Logged
I played 3446 and 3449. Hope she dont come with 3443!
You're safe.
Ga Eve
0391
Congrats To All Winners and Posters!
We are all in it to win! My Pet numbers 103,724,152,397,189,118,205.
Ga Lottery Player!
Hot-Due-Cold. Short Sums/Last Digit Sum. Pairs. Vtracs. Vtrac Pairs. SUM OF VTRACS CHART.
Don't forget to FLIP 6/9 in all WORKOUTS! 66=99=69. Vtracs Code sheets available....
SWATS Shawty
United States
Member #123454
February 21, 2012
22176 Posts
Offline
Posted: October 13, 2012, 11:57 pm - IP Logged
You're safe.
Ga Eve
0391
Well sorta, I played 3209!
Vtracs is My Game!
Georgia
United States
Member #3617
February 6, 2004
7421 Posts
Offline
Posted: October 14, 2012, 12:00 am - IP Logged
Well sorta, I played 3209!
Ouch!
Congrats To All Winners and Posters!
We are all in it to win! My Pet numbers 103,724,152,397,189,118,205.
Ga Lottery Player!
Hot-Due-Cold. Short Sums/Last Digit Sum. Pairs. Vtracs. Vtrac Pairs. SUM OF VTRACS CHART.
Don't forget to FLIP 6/9 in all WORKOUTS! 66=99=69. Vtracs Code sheets available....
United States
Member #108847
April 1, 2011
1619 Posts
Offline
Posted: October 14, 2012, 5:21 pm - IP Logged
Rachel, these Triplets are now showing....
441x 451x 491x
443x 453x 493x
446x 456x 496x
447x 457x 497x
448x 458x 498x
4401 NY mid. Good call,I like those Oct. #s Good Luck and thanks!
GEORGIA PEACH
United States
Member #70173
January 25, 2009
4009 Posts
Offline
Posted: October 14, 2012, 6:15 pm - IP Logged
On behalf of Sherita and I Thank you. At this time i have no additional numbers to add. I;m still waiting on the lucky 9 and of course newly added 9238 or 2389.
Good Luck!
Vtracs is My Game!
Georgia
United States
Member #3617
February 6, 2004
7421 Posts
Offline
Posted: October 14, 2012, 6:47 pm - IP Logged
Rachel, these Triplets are now showing....
441x 451x 491x
443x 453x 493x
446x 456x 496x
447x 457x 497x
448x 458x 498x
Yes, these triplets are Falling everywhere except Ga...
10/14/2012 Midday...
Conn-9214
Illinois/Iowa-2947
NY-4401
Oregon-5846
Congrats To All Winners and Posters!
We are all in it to win! My Pet numbers 103,724,152,397,189,118,205.
Ga Lottery Player!
Hot-Due-Cold. Short Sums/Last Digit Sum. Pairs. Vtracs. Vtrac Pairs. SUM OF VTRACS CHART.
Don't forget to FLIP 6/9 in all WORKOUTS! 66=99=69. Vtracs Code sheets available....
Vtracs is My Game!
Georgia
United States
Member #3617
February 6, 2004
7421 Posts
Offline
Posted: October 14, 2012, 7:38 pm - IP Logged
Rachel, these Triplets are now showing....
441x 451x 491x
443x 453x 493x
446x 456x 496x
447x 457x 497x
448x 458x 498x
I was trying to figure out where I saw that 451 at and just realized Mid and Eve Cash 3 hits where in this!
Congrats To All Winners and Posters!
We are all in it to win! My Pet numbers 103,724,152,397,189,118,205.
Ga Lottery Player!
Hot-Due-Cold. Short Sums/Last Digit Sum. Pairs. Vtracs. Vtrac Pairs. SUM OF VTRACS CHART.
Don't forget to FLIP 6/9 in all WORKOUTS! 66=99=69. Vtracs Code sheets available....
Vtracs is My Game!
Georgia
United States
Member #3617
February 6, 2004
7421 Posts
Offline
Posted: October 14, 2012, 9:29 pm - IP Logged
Rachel, these Triplets are now showing....
441x 451x 491x
443x 453x 493x
446x 456x 496x
447x 457x 497x
448x 458x 498x
MARYLAND EVE-4908
SOUTH CAROLINA-4789
Congrats To All Winners and Posters!
We are all in it to win! My Pet numbers 103,724,152,397,189,118,205.
Ga Lottery Player!
Hot-Due-Cold. Short Sums/Last Digit Sum. Pairs. Vtracs. Vtrac Pairs. SUM OF VTRACS CHART.
Don't forget to FLIP 6/9 in all WORKOUTS! 66=99=69. Vtracs Code sheets available....
Vtracs is My Game!
Georgia
United States
Member #3617
February 6, 2004
7421 Posts
Offline
Posted: October 14, 2012, 10:31 pm - IP Logged
UNHIT GA CASH 4 PAIRS OCTOBER- 18,44,45,49,55
Congrats To All Winners and Posters!
We are all in it to win! My Pet numbers 103,724,152,397,189,118,205.
Ga Lottery Player!
Hot-Due-Cold. Short Sums/Last Digit Sum. Pairs. Vtracs. Vtrac Pairs. SUM OF VTRACS CHART.
Don't forget to FLIP 6/9 in all WORKOUTS! 66=99=69. Vtracs Code sheets available....
Vtracs is My Game!
Georgia
United States
Member #3617
February 6, 2004
7421 Posts
Offline
Posted: October 14, 2012, 11:35 pm - IP Logged
GA EVE
8643
Congrats To All Winners and Posters!
We are all in it to win! My Pet numbers 103,724,152,397,189,118,205.
Ga Lottery Player!
Hot-Due-Cold. Short Sums/Last Digit Sum. Pairs. Vtracs. Vtrac Pairs. SUM OF VTRACS CHART.
Don't forget to FLIP 6/9 in all WORKOUTS! 66=99=69. Vtracs Code sheets available....
GEORGIA PEACH
United States
Member #70173
January 25, 2009
4009 Posts
Offline
Posted: October 15, 2012, 5:55 pm - IP Logged
There is a method we are trying out with my pairs method and Sherita VTRAC method we are ready to take GA LOTTERY to a whole new level. We actually started yesterday and came up with the 9 numbers listed below.We wanted to share our numbers with you.
Good Luck to us!
1076
1872
1070
1876
1676
1670
1072
1672
1870
this is freaking crazy lol
CASH 4 NUMBER 7671
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Assessment title: Perception Lab Report
1 Background
Müller-Lyer illusion laboratory: The role of local and global processing in the Müller-Lyer illusion
In the Müller-Lyer illusion the right vertical line (see the figure above) appears to be longer than the left vertical line, even though they are both exactly the same length. The presence of the fins at either end makes the lines appear to be different in length, with the fins-in arrangement causing the left line to look shorter and the fins-out arrangement causing the right line to look longer. Müller-Lyer coined the term “confluxion” to describe this illusion (“Müller-Lyer,” n.d.). The exact nature of this effect has been studied extensively without consensus (c.f., Dewar, 1967; Presey & Martin, 1990; Restle & Decker, 1977) about which perceptual principles account for the illusion (“Müller-Lyer,” n.d.).
Gregory (1966) explains the illusion on the basis of a mechanism called misapplied size-constancy scaling. Size constancy normally helps us maintain a stable perception of objects by taking distance into account (remember the size-distance scaling equation covered in your readings this week). Gregory proposes that it is this mechanism, normally useful in a three-dimensional world, which creates illusions when it is applied to two-dimensional surfaces, like the figure on this page. Gregory suggests that the fins on the right line make this line look like part of an inside corner, thus appearing further away. According to the size-distance scaling equation (S= R x D) as the distance (D) is larger, the size product (S) is longer. The fins on the left line create the impression of an outside corner and therefore appear closer.
Do you really think the Müller-Lyer looks like the corners of a wall? How about this “dumbbell” version of the illusion:
There is no obvious perspective or depth information available here, yet the illusion is still there!
Day (1989) proposed the conflicting cues theory, which states that our perception of the line length in this figure depends on two cues: (1) the actual length of the lines; a ‘local’ feature and (2) the overall length of the figure; a ‘global’ or ‘whole’ representation. According to Day, these two conflicting cues are integrated to form a compromise of the perception of length. As the overall length of the left part of the figure is shorter than the overall length of the right, the line appears slightly longer. Do you think the visual system can be fooled that easily?
The Müller-Lyer illusion has been tested, challenged and tested again many times. There is still no consensus as to why it occurs. You might be wondering what you can contribute as a university student to this decades-old conundrum, what new can possibly be found?
There has only ever been one direct test of Day’s proposal (see Mundy, 2014), and that produced some tantalising data: What if we could encourage our visual system to process more globally than it does normally? To always see the ‘whole’ instead of its parts? Assuming Day is correct, the illusion should be strengthened by greater attention to the whole figure – biasing the compromise we make when processing length. On the other hand, what if we were to encourage our visual system to process more locally? In that case, the illusion should be weakened by focussing perception to just single features, that is the line itself – ignoring the ‘whole’ and making our perception of length more accurate. Mundy (2014) found a particularly effective way to influence and modulate global and local processing through the use of Navon (1977) stimuli. The beauty of such stimuli relates to the fact that participants can respond to the large letter shape of such stimuli (global processing) or the smaller letters that make up the shape (local processing) whilst keeping the physical stimuli identical:
Processed globally, the figure reads E.S. Processed locally, it reads A.H. If a series of such stimuli are presented in succession and a participant is instructed to focus on one form of response (either global or local) then a bias towards that type of processing is generated for a short time after the presentation (for example, see Macrae & Lewis, 2002). If the strength of the Müller-Lyer illusion was then measured following the induction of such a bias, we might see a change, according to Day’s hypothesis. Mundy (2014) used a relatively short exposure to Navon stimuli and found that the Muller-Lyer illusion was significantly stronger for participants with a global bias, and significantly weaker for those with a local bias, compared with the control condition, thus supporting Day’s hypothesis. However, further studies are required to examine the merit of of Day’s hypothesis for the Müller-Lyer illusion by biasing participants toward global or local visual processing.
In order to replicate and extend the findings of Mundy (2014), we have selected an alternative method of generating global and local bias – mood induction. It has been shown by a number of researchers (e.g., Gasper & Clore, 2002) that mood has a significant bearing on perception. Sad people tend to favour processing of local features, whereas happy people tend to prefer processing whole (global) stimuli. Moreover, it has also been shown that listening to music can alter your mood. So, if we listen to some happy music, do we become more susceptible to the Muller-Lyer illusion, since our perceptual system is now biased more globally? On the other hand, if we listen to sad music, do we see the illusion more weakly, due to enhanced local processing? You should do a literature search to find more examples of this mood-perception phenomenon – there are a number within the face processing literature (e.g., Hills & Lewis, 2011) and further afield.
2 Aims
The aims of this experiment were to carry out a novel and innovative study to examine Day’s hypothesis that the Müller-Lyer illusion is created by ‘conflicting cues’ created by global and local features, replicating and strengthening the original findings of Mundy (2014).
Global processing should enhance the illusion by causing the viewer to judge length based on the whole figure.
A bias toward more local processing should weaken it as the viewer is more likely to focus on the line itself instead of the fins.
3 Procedure
CLASS EXPERIMENT
The experimental procedure is very simple and is, for the most part, modelled on that of the early experiments in which the illusion was quantified, such as that reported by Dewar (1967). You will note that the original experiment was far less high-tech than the current computerised study, but the principles are exactly the same. However, using music stimuli to manipulate processing level before and during the Müller-Lyer test has never been attempted.
Apparatus
A personal computer running a custom designed Java applet from the Online Psychology Laboratory, authored by Mark Tew and Ken McGraw, University of Mississippi. The experiment can be accessed from the following URL: http://opl.apa.org/Experiments/Start.aspx?EID=12 Your class ID code will be given to you by your instructor.
Method
Stage 1
The class will split into three groups or ‘levels of processing’:
Global processing condition (Happy Mood Induction) –
Local processing condition (Sad Mood Induction) –
Control condition
Participants in groups 1 and 2 will listen to pre determined music 1 minute before and during the entire experiment. The Global participants will listen to Mozart Jubilate and Exultate to induce a happy mood; the Local participants will listen to Mozart Requiem to induce a sad mood (music was chosen from examples given in Hills, Werno and Lewis (2011)). The Control participants will not listen to music before and during the experiment.
Stage 2
All participants will now access the experiment URL and enter their class ID code according to which condition they are in.
The study contained in this URL is a variation of the original Muller-Lyer illusion, one which enables investigators to study the effect of changes in fin angle on the apparent length of lines. Participants in the study are presented with two lines, as in the standard Muller-Lyer illusion presentation, but one of the lines has fins and one does not. Moreover, the two lines are initially different in length. The participant’s task is to adjust the plain line (without fins) to make the lengths the same. The adjustment is made using a slider (arrow in the figure below) that can be dragged using the computer mouse. As the slider is moved up and down the scale, the adjustable line changes in length.
Design
Within the Müller-Lyer illusion experiment the independent variable is fin angle, which varies from 15 degrees to 165 degrees in 15 degree steps (i.e., 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165). Thus, there are 11 levels to the independent variable. Angles from 15 to 75 degrees are used to create “fins in” stimuli. Angles from 105 to 165 degrees are used to create “fins out” stimuli.
Two sets of 11 trials are conducted. In each set, the angle used on any one trial is chosen at random from the set of 11, but once the angle has been used, it does not reappear until the next set of trials. The total number of trials, therefore, is 22 with each angle being used twice.
The dependent variable is the difference in length between the two lines. Lengths are measured in pixels. The red line with fins attached–call it the “illusory” stimulus–has a random length of between 100 and 150 pixels. The adjustable line is randomly set to either 90 or 160 pixels at the start of each trial. Thus sometimes the adjustable line must be made longer and sometimes shorter in order to create a perceptual match with the illusory line. To compute the difference, the length of the illusory stimulus is subtracted from the length of the adjustable line following adjustment. A positive difference indicates that the illusory stimulus appears longer than it in fact is; a negative difference indicates that the illusory stimulus appears shorter than it in fact is. Because non-zero differences represent errors in judging the length of lines, the dependent variable in this part of the experiment is a measure of adjustment error.
The overall independent variable is level of processing (Global, Local, or Control), which is modulated by mood induction via music exposure.
The overall dependent variable is the slope of the regression line produced by plotting adjustment errors across fin angle (i.e., the strength of the illusion, see below).
4 Data analysis
Examination of data
Adjustment errors can be graphed to judge the effect of fin angle on adjustment error.The figure below is a plot of data points that represent how the average adjustment error changed as a function of fin angle for one small set of participants. Adding a regression line (otherwise known as a trend line or line of best fit) to the plot shows the general linear trend – a good measure of the strength of the illusion.
You should produce a plot containing each of the three conditions (i.e., one per group: Global, Local and Control). You can do this all on one graph. A steeper regression line compared with Control would provide confirmation that the visual illusion has become more severe, producing greater adjustment errors. A more shallow regression line compared with Control would indicate that the illusion has weakened, creating fewer errors. You may also wish to calculate the slope/gradient of the regression line for each individual participant (tip: use the SLOPE function in Excel) and use ANOVA to determine if any statistical differences exist between level of processing conditions. No difference would imply that perhaps Day’s hypothesis is incorrect and that other explanations of the illusion must be sought.
Data are downloadable in three formats (XML, Excel spreadsheet format, and comma delimited for statistical software packages like SPSS), from, the following URL:
Remember to download the data from all three IV level of processing conditions (there will be three different class ID codes for Monash PSY3051 – please only use data from 2016).
The five columns provide classification data (participant ID number, gender, the class ID number, age, and date of participation)
Experimental data are recorded in columns 7- 17, reflecting the 11 different angles or levels of the IV presented in the experiment. Each value recorded in the respective column is a mean that represents a bias value that was derived by averaging the four trials.
5 Write up & References
Laboratory report
Write up your findings in the form of a laboratory report/journal article, with a length of 2000 words. The submission dates for on-campus, off-campus students, can be found in the PSY3051 Unit Guide. Further information on marking is available in the next chapter of this lab manual.
Day, R.H (1989). Natural and artificial cues perceptual compromise and the basis of veridical and illusory perception. In D. Vickers & P.L. Smith (Eds.), Human information processing: Mechanisms and models (pp. 107-129). London:
Dewar, R.E. (1967). Stimulus determinants of the magnitude of the Muller-Lyer illusion. Perceptual and Motor Skills, 24, 708-710.
Gasper, K., & Clore, G. L. (2002). Attending to the big picture: Mood and global versus local processing of visual information. Psychological Science, 3(1), 34-40.
Goldstein, E.B. (2014). Sensation and perception (9th ed.). Sydney: Wadsworth, Cengage Learning.
Gregory, R.L. (1966). Eye and brain. New York: McGraw-Hill.
Hills, P. J. & Lewis, M. B. (2011). Sad people avoid the eyes or happy people focus on the eyes? Mood induction affects facial feature discrimination. British Journal of Psychology, 102(2), 260-274.
Hills, P.J., Werno, M.A., & Lewis, M.B. (2011). Sad people are more accurate at face recognition than happy people. Consciousness and Cognition, 20, 1502-1517.
Macrae, C.N. & Lewis, H.L. (2002). Do I know you?: Processing orientation and face recognition. Psychological Science, 13, 194-196.
Mundy, M.E. (2014). Testing day: The effects of processing bias induced by Navon stimuli on the strength of the Muller-Lyer illusion. Advances in Cognitive Psychology, 10, 9-14.
Navon, D. (1977). Forest before the trees: The precedence of global features in visual perception. Cognitive Psychology, 9, 353-383.
Pressey, A. & Martin, N.S. (1990). The effects of varying fins in Muller-Lyer and holding illusions. Psychological Research, 52, 46-53.
Restle, F. & Decker, J. (1977). Size of the Muller-Lyer illusions a function of its dimensions: Theory and data. Perception and Psychophysics, 21, 489-503.
6 Marking Guidelines
Word limit: 2000 words (not to exceed 1850 words, excluding the abstract, figures or tables of results, the reference list and appendix). The abstract has a separate word limit of 150 words. Word limits include in text citations and subheadings. There is no 10% leeway on the abstract or main report.
Value: 20%
Estimated return date: 4 weeks from submission
Criteria for Marking:
Reports should be structured with a title page, abstract, introduction, method, results, discussion and references as per the specifications in Findlay (2012). Reports should each typed double spaced on A4. Reports will be marked out of 10, in 0.5 intervals:
8-10 An outstanding report showing extensive knowledge and understanding and an exceptional ability to employ analysis, synthesis, and evaluation in the development and justification of hypotheses. An exemplary method showing complete command of issues of design, procedure, stimulus materials, and sampling. Accurate and clearly presented statistical analysis. Outstanding interpretation and evaluation of the results coupled with relation to the literature and awareness of problems and limitations. Writing is fluent, clear, and grammatical.
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or | 4,163 | 19,156 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.884739 |
https://www.hpmuseum.org/forum/thread-3771.html | 1,611,423,856,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538226.66/warc/CC-MAIN-20210123160717-20210123190717-00062.warc.gz | 827,712,715 | 6,486 | (50g) Cordic Algorithm [TAN(X)]
05-03-2015, 04:00 PM
Post: #1
peacecalc Member Posts: 176 Joined: Dec 2013
(50g) Cordic Algorithm [TAN(X)]
Hello 50g fans,
I never heard about Jaques Laporte. Some days ago I read in this forum about his passing away. A lot of forum member mourn his lost and appreciate his work for thiat community. So I got curious what his work was and the connection to the hp calculators.
After reading his old websites, I learned from his "cordic for dummies", how the cordic algorithm works. So I've had the idea that this would be a nice to have a little program for the hp 50g. The program is "brutal force" there is no elegance (with complex numbers or matrices calculation or anything else optimized).
The original algorithm from Volder works with a binary shift right. So he didn't need
the multiplication like my program.
Tip: compare the values from hp 50g with the values of the program.
Code:
%%HP: T(3)A(R)F(,); @@Program name: CTAN \<< { } { 1, ,1 ,01 ,001 ,0001 } \-> X @@ X: Argument for tangens WLST @@ WLST: List how often subtracted DLST @@ DLST: List with negative powers of ten @@ simulating right shift \<< 0, DUPDUP 1, X \-> X2 Y2 @@ X2, Y2: new values CNT @@ CNT: value how often subtracted X1 Y1 @@ X1, Y1: old values (X1 is now 1, and @@ Y1 is the argument from TAN) \<< -105, SF 1, 5, FOR J CORP J GET @@ CORP: is a list with the values @@ {ATAN(1) ATAN(0,1) ATAN(0,01) ATAN(0,001) ATAN(0,0001)} \-> TB @@ TB: is the actual value f. e. ATAN(0,01) \<< WHILE Y1 TB \>= @@ As long Y1 is greater then TB REPEAT Y1 TB - 'Y1' STO @@ Y1 is subtracted by TB 'CNT' INCR DROP @@ CNT counts how often this takes place END WLST CNT + 'WLST' STO @@ IF Y1 is smaller then TB the the while loop is left. @@ The value of CNT is stored in WLST. 0, 'CNT' STO @@ and CNT is initialised with zero \>> NEXT 1, 5, FOR J 1, WLST J GET DUP @@ 1, is the start value for the START-loop @@ In WLST the end value is found and is doubled on stack IF 0, == @@ IF the end-value is zero, THEN DROP DROP @@ the remaining values are cleared ELSE DLST J GET @@ IF NOT the actual negativ power of ten is stored in TB \-> TB \<< START X1 Y1 TB * - 'X2' STO @@ the heart of the cordic-algorithm, the Y1 X1 TB * + 'Y2' STO @@ 'transformation of the coordinates' Y2 'Y1' STO @@ old values initialised with new values X2 'X1' STO NEXT \>> END NEXT Y2 X2 / @@ calculate tan of y2 and x2 -105, CF \>> \>> \>>
05-03-2015, 04:19 PM
Post: #2
Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: (50g) Cordic Algorithm [TAN(X)]
(05-03-2015 04:00 PM)peacecalc Wrote: The program is "brutal force" there is no elegance (with complex numbers or matrices calculation or anything else optimized).
Cf. Exploring the CORDIC algorithm with the WP-34S
Related but uses complex multiplication.
Cheers
Thomas
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User(s) browsing this thread: 1 Guest(s) | 1,376 | 4,225 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2021-04 | latest | en | 0.585425 |
https://interviewmania.com/aptitude/speed-time-and-distance/1/90 | 1,631,940,931,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056297.61/warc/CC-MAIN-20210918032926-20210918062926-00541.warc.gz | 393,084,080 | 9,297 | ## Speed, Time and Distance
#### Speed, Time and Distance
1. An athlete runs 200 metres race in 24 seconds. His speed (in km/hr) is :
1. 20
2. 24
3. 28.5
4. 30
1. Speed = Distance Time
= 200 m/s 24
200 m/s = 10 × 18 = 30 km/hr. 24 24 5
∵ x m/s = 18 x km/hr 5
##### Correct Option: D
Speed = Distance Time
= 200 m/s 24
200 m/s = 10 × 18 = 30 km/hr. 24 24 5
∵ x m/s = 18 x km/hr 5
1. A man crosses a road 250 metres wide in 75 seconds. His speed in km/hr is :
1. 10
2. 12
3. 12.5
4. 15
1. Speed = Distance = 250 75 3
= 10 m/sec = 10 × 18 km/hr. 3 3 5
∵ 1 m/s = 18 km/hr 5
= 2 × 6 km/hr. = 12 km/hr.
##### Correct Option: B
Speed = Distance = 250 75 3
= 10 m/sec = 10 × 18 km/hr. 3 3 5
∵ 1 m/s = 18 km/hr 5
= 2 × 6 km/hr. = 12 km/hr.
1. A man walking at the rate of 5 km/hr. crosses a bridge in 15 minutes. The length of the bridge (in metres) is :
1. 600
2. 750
3. 1000
4. 1250
1. Speed of the man = 5km/hr
= 5 × 1000 m / min = 250 m / min 60 3
Time taken to cross the bridge
= 15 minutes
Length of the bridge
= speed × time
= 250 × 15m = 1250m 3
##### Correct Option: D
Speed of the man = 5km/hr
= 5 × 1000 m / min = 250 m / min 60 3
Time taken to cross the bridge
= 15 minutes
Length of the bridge
= speed × time
= 250 × 15m = 1250m 3
1. An aeroplane covers a certain distance at a speed of 240 km hour in 5 hours. To cover the
same distance in 1 2 hours, it must 3
travel at a speed of :
1. 300 km./hr.
2. 360 km./hr.
3. 600 km./hr.
4. 720 km./hr.
1. Let the required speed is x km/hr
Then, 240 × 5 = 5 × x 3
∴ x = 720 km/hr.
##### Correct Option: D
Let the required speed is x km/hr
Then, 240 × 5 = 5 × x 3
∴ x = 720 km/hr.
1. A train is travelling at the rate of 45km/hr. How many seconds it
will take to cover a distance of 4 km ? 5
1. 36 sec.
2. 64 sec.
3. 90 sec.
4. 120 sec.
1. Time taken = Distance time
= (4/5) hour 45
= 4 × 60 × 60 sec. 5 × 45
= 64 seconds
##### Correct Option: B
Time taken = Distance time
= (4/5) hour 45
= 4 × 60 × 60 sec. 5 × 45
= 64 seconds | 845 | 2,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-39 | latest | en | 0.797545 |
http://spmaddmaths.blog.onlinetuition.com.my/2013/02/inverse-function-example-7.html | 1,585,662,814,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500482.27/warc/CC-MAIN-20200331115844-20200331145844-00021.warc.gz | 166,842,711 | 10,213 | # 1.5.7 Inverse Function Example 7
Example 7 (Comparison Method)
Given that $f:x\to \frac{2h}{x-3k}$ , x≠3k , where h and k are constants and ${f}^{-1}:x\to \frac{14+24x}{x}$ , x≠0, find the value of h and of k.
Solution: | 88 | 224 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.555974 |
https://www.clutchprep.com/chemistry/practice-problems/143466/multiplication-and-divisionwhen-multiplying-or-dividing-the-final-answer-is-roun-3 | 1,603,276,598,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107876307.21/warc/CC-MAIN-20201021093214-20201021123214-00276.warc.gz | 682,756,528 | 29,588 | # Problem: Part D. Compute 4.659 × 104 − 2.14 × 104. Round the answer appropriately.Express your answer numerically using the proper number of significant figures.Multiplication and divisionWhen multiplying or dividing, the final answer is rounded to the same number of significant figures as the measured number with the fewest significant figures.Addition and subtractionWhen adding or subtracting, the final answer is rounded to the same number of digits to the right of the decimal point as the measured number with the fewest digits to the right of the decimal point. This item does not cover measured values without a decimal point, which may involve interpreting values with an ambiguous number of significant figures.
###### Problem Details
Part D. Compute 4.659 × 104 − 2.14 × 104. Round the answer appropriately.
Multiplication and division
When multiplying or dividing, the final answer is rounded to the same number of significant figures as the measured number with the fewest significant figures. | 207 | 1,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.865987 |
https://etc.usf.edu/clipart/62300/62391/62391_cube_add1.htm | 1,632,019,684,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056656.6/warc/CC-MAIN-20210919005057-20210919035057-00128.warc.gz | 317,024,451 | 10,390 | A cube (A) has sides of 20 inches in length each, making its solid contents equal 8000 cubic inches. Being added are 3 equal portions 20x20x5, equaling 2000 cubic inches. The sum of these are 6000. You can find the second portion of the problem here.
### Galleries
Miscellaneous Solid Forms
### Source
Moore, John H. New Commercial Arithmetic (New York, NY: American Book Company, 1907)
TIFF (full resolution)
2400×1698, 1.7 MiB
Large GIF
1024×724, 235.0 KiB
Medium GIF
640×452, 118.8 KiB
Small GIF
320×226, 37.8 KiB | 165 | 528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-39 | longest | en | 0.796429 |
http://www.cut-the-knot.org/Curriculum/Geometry/FourSimsons.shtml | 1,500,668,933,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423809.62/warc/CC-MAIN-20170721202430-20170721222430-00519.warc.gz | 418,938,251 | 13,240 | ## Simsons and 9-Point Circles in Cyclic QuadrilateralWhat is this about? A Mathematical Droodle
### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.
What if applet does not run?
(When the simsons are shown, you may want to verify their definition. Click on them in turn, or click away from any to remove additional information.)
Explanation
### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.
What if applet does not run?
Any quadrilateral ABCD defines four triangles: BCD, CDA, DAB, ABC. In each of the triangles one may consider associate remarkable points and lines. The above applet is specifically concerned with their 9-point circles and also the Simson lines of each of the points A, B, C, D with respect to the triangle formed by the other three. The applet purports to illustrate the fact that, provided the quadrilateral ABCD is cyclic, the eight aforementioned elements all concur in a point.
First let's prove the concurrence of the four 9-point circles. As we know, the quadrilateral NANBNCND formed by the 9-point centers of the four triangles BCD, CDA, DAB, ABC is similar to ABCD with the factor of 1/2. So that the distance from the circumcenter of NANBNCND to its vertices is R/2, where R is the circumradius of ABCD. We also know that in any triangle the radius of its 9-point circle equals half that of its circumradius. It therefore follows that the circles of radii R/2 centered at the points NA, NB, NC, and ND meet at the circumcenter of the quadrilateral NANBNCND. In other words, the 9-point circles of the triangles BCD, CDA, DAB, ABC meet at the circumcenter of NANBNCND.
Now let's tackle the simsons. As we showed elsewhere, the circumcenter of NANBNCND coincides with H, the center of homothety that maps ABCD onto HAHBHCHD, the quadrilateral formed by the orthocenters of the triangles BCD, CDA, DAB, ABC. The two quadrilaterals ABCD and HAHBHCHD are equal so that H is the midpoint of each of the segments AHA, BHB, CHC, and DHD. By a well known property of triangles [Honsberger, p. 43], the simson of A with respect to ΔBCD passes through the midpoint of AHA, i.e., H. This of course applies to the other three simsons.
Note, that the result admits a natural generalization: if the quadrilateral is not cyclic, the simsons convert into circles, and the eight circles involved are still concurrent in a point.
### References
1. R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, 1995.
### Simson Line - the simson
• Simson Line: Introduction
• Simson Line
• Three Concurrent Circles
• 9-point Circle as a locus of concurrency
• Miquel's Point
• Circumcircle of Three Parabola Tangents
• Angle Bisector in Parallelogram
• Reflections of a Point on the Circumcircle
• Simsons of Diametrically Opposite Points
• Simson Line From Isogonal Perspective
• Pentagon in a Semicircle
• Simson Line in Disguise
• Two Simsons in a Triangle
• Carnot's Theorem
• A Generalization of Simson Line | 859 | 3,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-30 | latest | en | 0.920117 |
http://de.metamath.org/mpeuni/resdmres.html | 1,721,657,215,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00721.warc.gz | 10,258,088 | 4,598 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > resdmres Structured version Visualization version GIF version
Theorem resdmres 5543
Description: Restriction to the domain of a restriction. (Contributed by NM, 8-Apr-2007.)
Assertion
Ref Expression
resdmres (𝐴 ↾ dom (𝐴𝐵)) = (𝐴𝐵)
Proof of Theorem resdmres
StepHypRef Expression
1 in12 3786 . . . 4 (𝐴 ∩ ((𝐵 × V) ∩ (dom 𝐴 × V))) = ((𝐵 × V) ∩ (𝐴 ∩ (dom 𝐴 × V)))
2 df-res 5050 . . . . . 6 (𝐴 ↾ dom 𝐴) = (𝐴 ∩ (dom 𝐴 × V))
3 resdm2 5542 . . . . . 6 (𝐴 ↾ dom 𝐴) = 𝐴
42, 3eqtr3i 2634 . . . . 5 (𝐴 ∩ (dom 𝐴 × V)) = 𝐴
54ineq2i 3773 . . . 4 ((𝐵 × V) ∩ (𝐴 ∩ (dom 𝐴 × V))) = ((𝐵 × V) ∩ 𝐴)
6 incom 3767 . . . 4 ((𝐵 × V) ∩ 𝐴) = (𝐴 ∩ (𝐵 × V))
71, 5, 63eqtri 2636 . . 3 (𝐴 ∩ ((𝐵 × V) ∩ (dom 𝐴 × V))) = (𝐴 ∩ (𝐵 × V))
8 df-res 5050 . . . 4 (𝐴 ↾ dom (𝐴𝐵)) = (𝐴 ∩ (dom (𝐴𝐵) × V))
9 dmres 5339 . . . . . . 7 dom (𝐴𝐵) = (𝐵 ∩ dom 𝐴)
109xpeq1i 5059 . . . . . 6 (dom (𝐴𝐵) × V) = ((𝐵 ∩ dom 𝐴) × V)
11 xpindir 5178 . . . . . 6 ((𝐵 ∩ dom 𝐴) × V) = ((𝐵 × V) ∩ (dom 𝐴 × V))
1210, 11eqtri 2632 . . . . 5 (dom (𝐴𝐵) × V) = ((𝐵 × V) ∩ (dom 𝐴 × V))
1312ineq2i 3773 . . . 4 (𝐴 ∩ (dom (𝐴𝐵) × V)) = (𝐴 ∩ ((𝐵 × V) ∩ (dom 𝐴 × V)))
148, 13eqtri 2632 . . 3 (𝐴 ↾ dom (𝐴𝐵)) = (𝐴 ∩ ((𝐵 × V) ∩ (dom 𝐴 × V)))
15 df-res 5050 . . 3 (𝐴𝐵) = (𝐴 ∩ (𝐵 × V))
167, 14, 153eqtr4i 2642 . 2 (𝐴 ↾ dom (𝐴𝐵)) = (𝐴𝐵)
17 rescnvcnv 5515 . 2 (𝐴𝐵) = (𝐴𝐵)
1816, 17eqtri 2632 1 (𝐴 ↾ dom (𝐴𝐵)) = (𝐴𝐵)
Colors of variables: wff setvar class Syntax hints: = wceq 1475 Vcvv 3173 ∩ cin 3539 × cxp 5036 ◡ccnv 5037 dom cdm 5038 ↾ cres 5040 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-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-sep 4709 ax-nul 4717 ax-pr 4833 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 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-ral 2901 df-rex 2902 df-rab 2905 df-v 3175 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-nul 3875 df-if 4037 df-sn 4126 df-pr 4128 df-op 4132 df-br 4584 df-opab 4644 df-xp 5044 df-rel 5045 df-cnv 5046 df-dm 5048 df-rn 5049 df-res 5050 This theorem is referenced by: imadmres 5544 lindfres 19981 imacmp 21010 metreslem 21977 volres 23103 uhgrares 25837 umgrares 25853 usgrares 25898 resresdm 40319
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https://www.beatthegmat.com/what-is-the-length-of-ad-t299613.html | 1,563,951,407,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195531106.93/warc/CC-MAIN-20190724061728-20190724083728-00421.warc.gz | 631,273,970 | 38,381 | • NEW! FREE Beat The GMAT Quizzes
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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## What is the length of AD? tagged by: swerve ##### This topic has 1 expert reply and 0 member replies ### Top Member ## What is the length of AD? What is the length of AD? $$A.\ \ 2\sqrt{10}$$ $$B.\ \ 4\sqrt{10}$$ $$C.\ \ 2\sqrt{5}$$ $$D.\ \ 4\sqrt{5}$$ $$E.\ \ \sqrt{10}$$ The OA is A. Please, can any expert explain this PS question for me? I tried to solve it but I'm not sure and I can't get the correct answer. I need your help. Thanks. ### GMAT/MBA Expert GMAT Instructor Joined 04 Oct 2017 Posted: 551 messages Followed by: 11 members Upvotes: 180 Top Reply Quote: What is the length of AD? $$A.\ \ 2\sqrt{10}$$ $$B.\ \ 4\sqrt{10}$$ $$C.\ \ 2\sqrt{5}$$ $$D.\ \ 4\sqrt{5}$$ $$E.\ \ \sqrt{10}$$ The OA is A. Please, can any expert explain this PS question for me? I tried to solve it but I'm not sure and I can't get the correct answer. I need your help. Thanks. Hi swerve, Let's take a look at your question. Draw a horizontal line parallel to CB, from point D that intersects the line Ab at point E, such that AED is a right triangle. Since, $$DE=CB$$ Therefore, $$DE=6$$ Now, let's find AE. $$AE=AB-EB$$ $$AE=AB-DC$$ $$AE=12-10$$ $$AE=2$$ Now, we can find AD using Pythagoras Theorem, $$AD=\sqrt{\left(AE\right)^2+\left(DE\right)^2}$$ $$AD=\sqrt{\left(2\right)^2+\left(6\right)^2}$$ $$AD=\sqrt{4+36}$$ $$AD=\sqrt{40}$$ $$AD=2\sqrt{10}$$ Therefore, option A is correct. Hope it helps. I am available if you'd like any follow up. _________________ GMAT Prep From The Economist We offer 70+ point score improvement money back guarantee. Our average student improves 98 points. Free 7-Day Test Prep with Economist GMAT Tutor - Receive free access to the top-rated GMAT prep course including a 1-on-1 strategy session, 2 full-length tests, and 5 ask-a-tutor messages. Get started now. • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200
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https://vroomlab.wordpress.com/2017/06/17/herons-formula-saves-the-day-again/ | 1,656,230,517,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037649.11/warc/CC-MAIN-20220626071255-20220626101255-00290.warc.gz | 669,842,435 | 28,984 | # a x + b y + c = 0 : Why It Applies to All Straight Lines
In the traditional teaching of Analytical Geometry, the governing equation for a straight line has the following five forms, along with limitations for the first four:
[1] Point-Slope form: $y - y_1 = k (x-x_1)$ where $(x_1, y_1)$ is a point on the line, and $k$ is the slope. The limitation for this form is that it can not represent line perpendicular to the x-axis since it has no slope.
[2] Slope-Intercept form: $y = k x + b$ where $k$ is the slope, $b$ is the intersect the line made on y-axis. Its limitation is that it can not represent line perpendicular to the x-axis.
[3] Two-Point form: $\frac{y-y1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$ where $(x_1, y_1), (x_2, y_2)$ are two points on the line. However, this form can represent neither line perpendicular nor parallel to x-axis due to the fact when $x_1 = x_2$ or $y_1 = y_2$, the form breaks down from dividing by zero.
[4] Point-Intercept form: $\frac{x}{a} + \frac{y}{b} = 1$ where $a, b$ are the intersects the line made on x-axis and y-axis respectively, and $a\neq 0, b\neq0$. Again, this form can represent neither line perpenticular nor parallel to the x-axis. It does not work for any line that passes the point of origin either.
[5] General form: $a x +b y +c = 0 (a^2+b^2 \neq 0)$, this form can represent all lines.
Here I am presenting a proof to show [5] is indeed capable of representing all straight lines.
In a rectangular coordinate system, given two distinct points $(x_1, y_1), (x_2, y_2)$, and any point $(x, y)$ on the line connecting $(x_1, y_1)$ and $(x_2, y_2)$, the area of triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x, y)$ must be zero!
Recall a theorem proved in my blog “Had Heron Known Analytic Geometry“, it means for such $(x_1, y_1), (x_2, y_2)$ and $(x, y)$,
$\left|\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{array}\right|= 0$.
Therefore, we can define the line connecting two distinct points as a set of $(x, y)$ such that the area of the triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x, y)$ is zero, mathematically written as
$A \triangleq \{ (x, y) | \left|\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{array}\right|= 0, (x_1-x_2)^2+(y_1-y_2)^2 \neq 0\}$.
Since $\forall (x, y) \in A$,
$\left|\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{array}\right|= x_1 y_2-x y_2-x_2 y_1+x y_1+x_2 y-x_1y =$
$(y-y_1)(x_1-x_2)-(x-x_1)(y_1-y_2)=0\quad\quad\quad\quad(1)$
is an algebraic representation of the line connecting two distinct points $(x_1, y_1)$ and $(x_2, y_2)$.
When $x_1=x_2$, (1) becomes
$(x-x_1)(y_1-y_2)=0$,
and when $x_1 = x_2, y_1-y_2 \neq 0$, we have
$x = x_1$,
a line perpendicular to the horizontal axis.
When $y_1=y_2$, (1) becomes
$y = y_1$,
a line parallel to the horizontal axis.
Evaluate (1) with $x_2=0, y_2=0$ yields:
$(y-y_1) x_1 -(x-x_1)y_1=0$.
Collecting terms in (1), and letting
$a=y_1-y_2$,
$b=x_2-x_1$,
$c=x_1y_2-x_2y_1$,
(1) can be expressed as
$ax + by + c = 0$.
In fact, we can prove the following theorem:
$B \triangleq \{ (x, y) | \exists a, b, a^2+b^2 \neq 0, a x +b y+c=0\} \implies A=B$.
To prove $A=B$, we need to show
$\forall (x, y) \in A \implies (x, y) \in B\quad\quad\quad\quad(2)$
$\forall (x, y) \in B \implies (x, y) \in A\quad\quad\quad\quad(3)$
We have already shown (2) by setting the values of $a, b$ and $c$ earlier.
We will prove (3) now:
$\forall (x_1, y_1), (x_2, y_2)$ and $(x, y) \in B$, we have
$\begin{cases}a x_1 + b y_1 +c =0 \\ a x_2 + b y_2 +c =0 \\ a x + b y+c =0\end{cases}$.
Written in matrix form,
$\left(\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{array}\right)$ $\left(\begin{array}{rrr} a \\ b \\ c \end{array}\right)= 0$.
If
$\left|\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{array}\right| \neq 0$,
then by Cramer’s rule,
$\left(\begin{array}{rrr} a \\ b \\ c \end{array}\right)$ is a column vector of zeros,
i.e.,
$a=b=c=0$
$a,b$ are not all zero.
Hence,
$\left|\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{array}\right| = 0$
which implies:
$(x, y) \in A$.
The consequence of $A=B$ is that every point $(x, y)$ on a line connecting two distinct points satisfies equation $a x + b y + c =0$ for some $a, b (a^2+b^2\neq 0)$.
Stated differently,
$a x + b y +c = 0$ where $a, b$ are not all zero is the governing equation of any straight line. | 1,730 | 4,484 | {"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": 67, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2022-27 | latest | en | 0.82175 |
https://www.lidolearning.com/questions/m_bb_rdsharma10_ch12_ex12_p1_q5/a-kite-is-flying-at-a-height-o/ | 1,680,086,557,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948965.80/warc/CC-MAIN-20230329085436-20230329115436-00534.warc.gz | 927,980,625 | 13,019 | # RD Sharma Solutions Class 10 Mathematics Solutions for Some Applications Of Trigonometry Exercise 12.1 in Chapter 12 - Some Applications Of Trigonometry
Question 6 Some Applications Of Trigonometry Exercise 12.1
A kite is flying at a height of 75 meters from the ground level, attached to a string inclined at 60° to the horizontal.
Find the length of the string to the nearest meter.
Solution:
A common notion in trigonometry, specifically, is the angle of elevation, which has to do with height and distance. It is described as an angle formed by the horizontal plane and an oblique line between the observer's eye and a target above it.
Given,
Height of kite flying from the ground level = 75 m = AB
Angle of inclination of the string with the ground (θ) = 60°
Let the length of the string be L = AC
So, from the figure formed we have △ABC as a right triangle.
Hence,
\begin{aligned} &\sin \theta=\frac{\text { opposite side }}{\text { hypotenuse }} \\ &\sin 60^{\circ}=\frac{\mathrm{AB}}{\mathrm{AC}} \end{aligned}
\begin{aligned} &\frac{\sqrt{3}}{2}=\frac{75}{L} \\ &L=50 \sqrt{3} \mathrm{~m} \end{aligned}
\text { Length of string } \mathrm{L}=50 \sqrt{3} \mathrm{~m}
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Questions about how Mathematica evaluates expressions, tracing evaluation (Trace*), handling expressions in unevaluated form (Hold*, Unevaluated), nonstandard evaluation, etc.
You can watch the evaluations of a function using On: SetAttributes[zot, HoldFirst] zot[x_, y_] := {x, y} On[zot] In[11]:= zot[1+1, 2+2] During evaluation of In[11]:= zot::trace:zot[1+1,2+2 … ] --> zot[1+1,4]. During evaluation of In[11]:=zot::trace: zot[1+1,4] --> {1+1,4}. Out[11]= {2,4} Here we can see the original unevaluated arguments to zot and also the partially evaluated arguments …
answered Feb 28 '14 by WReach
is given the HoldAll attribute to delay the evaluation of the arguments until after we have had a chance to process them. It wraps the supplied sequence of symbols in Hold and then uses Replace to …
answered Apr 5 '14 by WReach
: Iteration limit of 4096 exceeded. >> *) To guard against this possibility, we make sure that the last definition only takes effect if the evaluation of its argument actually resulted in a change … themselves HoldAll because Block will temporarily disable that attribute as well, leading to evaluation leaks. It would also be unsuitable if the application will not tolerate the partial evaluation of expressions such as we see in the last example above. …
answered Jan 31 '15 by WReach
h[e1,e2,...]. The evaluation sequence is described in more detail in Chapter 7 of Power Programming with Mathematica by David B. Wagner. It is freely available to StackExchangers via (16485). When … they self-evaluate. Such symbols are said to be "inert". In the present case, f is inert. As a result, when evaluation gets down to the eleventh step the down-value of f is used to produce the …
answered Oct 11 '16 by WReach
We can define a new "variable container" that can be used to assign the same value to multiple variables: ClearAll[vars] SetAttributes[vars, HoldAll] vars /: s:(_vars = _) := CompoundExpression @@ Th …
answered Jan 21 '12 by WReach
Evaluation stops when there is no definition in place whose pattern matches the expression being evaluated. Conversely, evaluation will continue as long as there is a matching definition. Thus, if … I have this definition: zot[x_] := zot[x] and I evaluate zot[1], the evaluation will never terminate even though the expression never changes. (Well, in principle it will never terminate but …
answered Mar 17 '11 by WReach
prints the tracing information as execution proceeds instead of returning that information in the final evaluation result. Thus, TracePrint meets the stated requirement. However... The output of … the draw, it will crash. Also, the front-end is slow to respond to Abort Evaluation requests when large quantities of output are present. TraceScan TraceScan is an alternative to TracePrint that …
answered Jan 17 '15 by WReach
We can use some of Mathematica's built-in tracing facilities to help us answer this question. Let's start by ensuring that the symbols we are about to use carry no extraneous definitions: ClearAll[f …
answered Jan 28 '15 by WReach
. Responding to the Updated Question If we wish to update the original variables, we need to prevent the evaluation of the individual variables. For example: Scan[Function[Null, PrependTo[#, {"A", "B … programs the chance for so-called "evaluation leaks" to spring up grows quickly. All of the built-in destructive operators (like PrependTo, Set, SetDelayed etc) require held arguments. In other …
answered Mar 8 '17 by WReach
I agree with @Ymareth that the simplest thing would be to have the calling notebook take explicit measures to communicate the context to the target notebook. However, if for some reason it is undesir …
answered Dec 1 '14 by WReach
Short Version The behaviour we see is by design. It is due to the fact that we are using the special head Evaluate to force early evaluation of the second argument of CompoundExpression -- an … wrap the second expression within Evaluate, it is evaluated before the compound expression itself is evaluated: Print["one"]; Evaluate[Print["two"]] (* two one *) After that early evaluation and … | 964 | 4,170 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-35 | latest | en | 0.854058 |
https://an-com.org/23-what-is-inequality-notation-pics/ | 1,627,169,661,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151531.67/warc/CC-MAIN-20210724223025-20210725013025-00299.warc.gz | 116,434,372 | 14,635 | 23+ What Is Inequality Notation
Pics
. It's a notation for homogeneous polynomials in three variables that makes working with them easy. The inequality —4 < x < 6 represents the interval notation [—4, 6).
The colon means such that. The real number line, interval notation and set notation. Means that a is not smaller than b, that is, it is either bigger, or equal to b).
### People also asked, what is an inequality notation?
Inequalities, the number line and interval notation. Converting an inequality to interval notation. The last notation is more of an illustration. Smaller than the other (.
(Visited 1 times, 1 visits today) | 148 | 636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.924279 |
http://www.popflock.com/learn?s=Wave_propagation | 1,604,031,985,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107907213.64/warc/CC-MAIN-20201030033658-20201030063658-00287.warc.gz | 158,522,024 | 16,866 | Wave Propagation
Get Wave Propagation essential facts below. View Videos or join the Wave Propagation discussion. Add Wave Propagation to your PopFlock.com topic list for future reference or share this resource on social media.
Wave Propagation
Wave propagation is any of the ways in which waves travel.
With respect to the direction of the oscillation relative to the propagation direction, we can distinguish between longitudinal wave and transverse waves.
For electromagnetic waves, propagation may occur in a vacuum as well as in a material medium. Other wave types cannot propagate through a vacuum and need a transmission medium to exist[].
## Reflection of plane waves in a half-space
The propagation and reflection of plane waves-- e.g. Pressure waves (P-wave) or Shear waves (SH or SV-waves) are phenomena that were first characterized within the field of classical seismology, and are now considered fundamental concepts in modern seismic tomography. The analytical solution to this problem exists and is well known. The frequency domain solution can be obtained by first finding the Helmholtz decomposition of the displacement field, which is then substituted into the wave equation. From here, the plane wave eigenmodes can be calculated.
### SV wave propagation
The propagation of SV-wave in a homogeneous half-space (The horizontal displacement field)
The propagation of SV-wave in a homogeneous half-space (The vertical displacement field)
The analytical solution of SV-wave in a half-space indicates that the plane SV wave reflects back to the domain as a P and SV waves, leaving out special cases. The angle of the reflected SV wave is identical to the incidence wave, while the angle of the reflected P wave is greater than the SV wave. For the same wave frequency, the SV wavelength is smaller than the P wavelength. This fact has been depicted in this animated picture. [1]
### P wave propagation
Similar to the SV wave, the P incidence, in general, reflects as the P and SV wave. There are some special cases where the regime is different.
## Wave velocity
Seismic wave propagation in 2D modelled using FDTD method in the presence of a landmine
Wave velocity is a general concept, of various kinds of wave velocities, for a wave's phase and speed concerning energy (and information) propagation. The phase velocity is given as:
${\displaystyle v_{p}={\frac {\omega }{k}},}$
where:
The phase speed gives you the speed at which a point of constant phase of the wave will travel for a discrete frequency. The angular frequency ? cannot be chosen independently from the wavenumber k, but both are related through the dispersion relationship:
${\displaystyle \omega =\Omega (k).\,}$
In the special case , with c a constant, the waves are called non-dispersive, since all frequencies travel at the same phase speed c. For instance electromagnetic waves in vacuum are non-dispersive. In case of other forms of the dispersion relation, we have dispersive waves. The dispersion relationship depends on the medium through which the waves propagate and on the type of waves (for instance electromagnetic, sound or water waves).
The speed at which a resultant wave packet from a narrow range of frequencies will travel is called the group velocity and is determined from the gradient of the dispersion relation:
${\displaystyle v_{g}={\frac {\partial \omega }{\partial k}}}$
In almost all cases, a wave is mainly a movement of energy through a medium. Most often, the group velocity is the velocity at which the energy moves through this medium.
## References
1. ^ The animations are taken from Poursartip, Babak (2015). "Topographic amplification of seismic waves". UT Austin. | 762 | 3,709 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.901444 |
https://au.mathworks.com/matlabcentral/cody/problems/42643-matlab-basic-rounding-iii/solutions/2733439 | 1,606,543,435,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195069.35/warc/CC-MAIN-20201128040731-20201128070731-00474.warc.gz | 193,971,604 | 17,069 | Cody
# Problem 42643. MATLAB Basic: rounding III
Solution 2733439
Submitted on 23 Jul 2020 by David K
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = -8.8; y_correct = -9; assert(isequal(round_x(x),y_correct))
2 Pass
x = -8.4; y_correct = -9; assert(isequal(round_x(x),y_correct))
3 Pass
x = 8.8; y_correct = 8; assert(isequal(round_x(x),y_correct))
4 Pass
x = 8.4; y_correct = 8; assert(isequal(round_x(x),y_correct))
5 Pass
x = 8.49; y_correct = 8; assert(isequal(round_x(x),y_correct))
6 Pass
x = 128.52; y_correct = 128; assert(isequal(round_x(x),y_correct))
7 Pass
x = pi; y_correct = 3; assert(isequal(round_x(x),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 284 | 903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-50 | latest | en | 0.665989 |
https://essaymount.com/2021/05/07/write-a-sinusoidal-function-that-models-the-number-of-daylight-hours-for-the-year-in-this-city/ | 1,725,867,130,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00158.warc.gz | 215,939,236 | 12,107 | # Write a sinusoidal function that models the number of daylight hours for the year in this city.
REGRESSION CURVE
Do a web search to find the sunrise and sunset times for the last year in the city in (or near) which you live. Print out the information you find.
Starting with January 1, calculate the number of daylight hours every 30 days. (You should have 13 items of data.) Convert the time from hours and minutes to the nearest tenth of an hour. For example, if the sunrise time is 0708, write the decimal 7.1, by dividing 8 by 60.
Graph this set of data on graph paper. The REGRESSION CURVE will represent the day of the year, and the vertical axis will represent the number of daylight hours. Be sure to title your graph and label the axis.
Draw a regression curve, a curve that best represents your set of data. (This is similar to drawing a line of best fit.) | 202 | 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} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.895312 |
https://puzzles.mit.edu/2013/coinheist.com/indiana/phone_tag/answer/index.html | 1,675,937,519,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501555.34/warc/CC-MAIN-20230209081052-20230209111052-00550.warc.gz | 490,444,726 | 2,008 | Phone Tag (Solution)
by David Roe, Dimah Eroshkin, John Stumpo
The first thing to note is that the last names of the salesmen are names of American cities. Moreover, the extension listed in the voice mail message is the area code of the corresponding city.
Each directory gives an unordered list of other salesmen and an unordered list of area codes. Note that dialing the extension while in the directory does not take you to the person that's supposed to correspond to that mailbox. Instead, these directories reveal two parallel graphs on the same set of vertices. One graph corresponds to the names of cities, the other to their area codes.
Wagner and Willy L are clues to Arthur Miller's Death of a Salesman, which suggests looking at the traveling salesman problem on these two graphs. Solving the traveling salesman problem with the graph defined by the city names in the directory yields the path suggested in the voice mail messages. Solving it with the other graph (area codes), yields another path. Dialing through the phone tree along this path reaches a live person, who gives you the answer.
Secret graph solution:
```McAllen, TX
Lincoln, NE
Hampton, VA
Montgomery, AL
Flint, MI
Stamford, CT
Norman, OK
Yonkers, NY
Memphis, TN
Pomona, CA
Irvine, CA
Corona, CA
Eugene, OR
Stockton, CA
Hayward, CA
Chandler, AZ
Fargo, ND
Billings, MT
Buffalo, NY
McAllen TX
```
Total distance: 25872.5 km
Following the path either forwards or backwards leads to the answer, RIOT on the step before last. | 369 | 1,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-06 | latest | en | 0.895969 |
https://www.jiskha.com/display.cgi?id=1341365669 | 1,503,019,626,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104204.40/warc/CC-MAIN-20170818005345-20170818025345-00285.warc.gz | 913,239,478 | 3,372 | # algebra
posted by .
Solve using the elimination method. Show your work. If the system has not solution or an infinite number of solutions, state this.
-2x-8y=-42
-4x+6y=-18
• algebra -
Multiply top equation by 2 and then subtract equation #2 from #1.
## Similar Questions
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Solve using the elimination method. Show your work. If the system has no solution or an infinite number of solutions, state this. 1/5x -1/4 y = 3 2/5x + 1/2y = 2
2. ### algebra
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3. ### algebra
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6. ### algebra
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Solve using the elimination method. Show your work. If the system has no solution or an infinite number of solutions,state this. 3x+7y=77 -2x+7y=42
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9. ### Algebra
Solve using the elimination method.Show work.If the system has no solution or an infinite number of solutions state this. 7x+8y=51,7x+10y=55
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More Similar Questions | 592 | 2,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.80817 |
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Database & Sequential File Processing
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I need to know the best way to answer these questions. I've read the chapters and am clueless to computers. Please provide examples. The book looks like jargon. Thanks.
1. Sequential File Processing:
Give a brief definition of sequential file processing.
Develop an example to illustrate your definition. (The example does not have to be as elaborate. Just a record with a simple explanation)
Describe where sequential file processing might be used.
2. Database Processing:
Give a brief definition of database processing.
Develop an example to illustrate you definition. (The example does not have to be as elaborate as in #4 above. Just a few tables with a simple explanation)
Describe where database processing might be used.
3. In "5. Data Representation" above several different numbering systems are illustrated.
Please develop your own numbering system using a base number less than 30 and other than 2, 8, 10 and 16. Illustrate your numbering system by completing the "My System" column in #4 above. Make sure to share your base number with us. (see below)
Decimal Binary Octal Hexadecimal My System
Base Ten Base Two Base Eight Base Sixteen Base ?
0 0 0 0
1 1 1 1
2 10 2 2
3 11 3 3
4 100 4 4
5 101 5 5
6 110 6 6
7 111 7 7
8 1000 10 8
9 1001 11 9
10 1010 12 A
11 1011 13 B
12 1100 14 C
13 1101 15 D
14 1110 16 E
15 1111 17 F
16 10000 20 10
17 10001 21 11
18 10010 22 12
19 10011 23 13
20 10100 24 14
21 10101 25 15
22 10110 26 16
23 10111 27 17
24 11000 30 18
25 11001 31 19
26 11010 32 1A
27 11011 33 1B
28 11100 34 1C
29 11101 35 1D
30 11110 36 1E
31 11111 37 1F
32 100000 40 20
33 100001 41 21 | 589 | 1,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.790258 |
https://oeis.org/A030231 | 1,624,091,411,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487647232.60/warc/CC-MAIN-20210619081502-20210619111502-00411.warc.gz | 406,284,493 | 4,760 | 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!)
A030231 Number of distinct primes dividing n is even. 18
1, 6, 10, 12, 14, 15, 18, 20, 21, 22, 24, 26, 28, 33, 34, 35, 36, 38, 39, 40, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58, 62, 63, 65, 68, 69, 72, 74, 75, 76, 77, 80, 82, 85, 86, 87, 88, 91, 92, 93, 94, 95, 96, 98, 99, 100, 104, 106, 108, 111, 112, 115, 116, 117, 118, 119 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Gcd(A008472(a(n)), A007947(a(n)))=1; see A014963. - Labos Elemer, Mar 26 2003 Superset of A007774. - R. J. Mathar, Oct 23 2008 A076479(a(n)) = +1. - Reinhard Zumkeller, Jun 01 2013 LINKS T. D. Noe, Table of n, a(n) for n = 1..1000 H. Helfgott and A. Ubis, Primos, paridad y análisis, arXiv:1812.08707 [math.NT], Dec. 2018. FORMULA From Benoit Cloitre, Dec 08 2002: (Start) n such that Sum_{d|n} mu(d)*tau(d) = (-1)^omega(n) = +1 where mu(d)=A008683(d), tau(d)=A000005(d) and omega(d)=A001221(d). n such that A023900(n)>0. (End) MATHEMATICA Select[Range[200], EvenQ[PrimeNu[#]]&] (* Harvey P. Dale, Jun 22 2011 *) PROG (PARI) j=[]; for(n=1, 200, x=omega(n); if(Mod(x, 2)==0, j=concat(j, n))); j (PARI) is(n)=omega(n)%2==0 \\ Charles R Greathouse IV, Sep 14 2015 (Haskell) a030231 n = a030231_list !! (n-1) a030231_list = filter (even . a001221) [1..] -- Reinhard Zumkeller, Mar 26 2013 CROSSREFS Cf. A028260, A030230, A123066. Cf. A008472, A007947, A014963. Cf. A007774, A076479. Cf. A008683, A000005, A001221, A023900. Sequence in context: A324455 A327476 A007774 * A267114 A275665 A056760 Adjacent sequences: A030228 A030229 A030230 * A030232 A030233 A030234 KEYWORD nonn,easy,nice AUTHOR EXTENSIONS Corrected by Dan Pritikin (pritikd(AT)muohio.edu), May 29 2002 STATUS approved
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Last modified June 19 04:01 EDT 2021. Contains 345125 sequences. (Running on oeis4.) | 882 | 2,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.512224 |
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## Quiz of the Month (June 2000)
### Hector C. Parr
***
#### THIS MONTH'S QUIZ
``` By writing the number "4" four times, with ordinary
mathematical signs, you can obtain the number 9 as follows:
4/.4 - 4/4 = 10 - 1 = 9
Now try to obtain in the same way, using exactly four "4's"
each time:
1) 2
2) 8
3) 114
```
***
(c) Hector C. Parr (2001)
Back to Quiz Archive | 234 | 704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-17 | latest | en | 0.688898 |
https://cboard.cprogramming.com/c-programming/144226-how-reprint-food-within-borders-snake.html | 1,490,417,748,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188785.81/warc/CC-MAIN-20170322212948-00593-ip-10-233-31-227.ec2.internal.warc.gz | 785,871,870 | 12,343 | # Thread: how to reprint food WITHIN the borders in snake..
1. ## how to reprint food WITHIN the borders in snake..
If you visit this forum daily, you would notice that I have been posting topics about the snake game recently. I've been able to move the snake already and now, my problem is to reprint the food WITHIN the borders after it has been eaten because when it prints after it has been eaten, it sometimes prints outside the border. Need help here guys. Thank you. Here's my code btw.
Code:
```while (snake == 1)
{
delay(40000000); // Delay of the loop.
gotoxy (snakeX[0], snakeY[0]);
printf("*");
gotoxy (snakeX[nLength], snakeY[nLength]); // This erases the "trail" the snake leaves behind.
printf(" ");
for (i = nLength; i >= 1; i--) // Passing of values on the array. This means that the value of nSnakeX/Y[6] will be equivalent to the value of nSnakeX/Y[5] and so on. So nSnakeX/Y[1] would be equal to nSnake[0] and that makes the body of the snake.
{
snakeX[i] = snakeX[i - 1];
snakeY[i] = snakeY[i - 1];
}
if ( kbhit() )
{
nDir = getch();
if (nDir == 0)
nDir = getch(); // Detects arrow key input.
}
if ((nDir == UP && nDir2 != DOWN) || (nDir == DOWN && nDir2 != UP) || (nDir == LEFT && nDir2 != RIGHT) || (nDir == RIGHT && nDir2 != LEFT))
nDir2 = nDir; // This is used in order for the snake not to move in reverse. Example, if the snake is moving up, it will NOT move down because of the if statement.
switch (nDir2)
{
case UP: snakeY[0]--; break;
case DOWN: snakeY[0]++; break;
case LEFT: snakeX[0]--; break;
case RIGHT: snakeX[0]++; break;
}
if (snakeX[0] == food_x && snakeY[0] == food_y || snakeX[0] == food_y || snakeY[0] == food_x) // Collision detector with the food.
{
srand(time(NULL));
food_x = rand() % LENGTH + 2;
food_y = rand() % WIDTH + 2;
gotoxy (food_x, food_y);
printf("A");
}
if (nScore % 2 == 0 && nScore != 0) // Snake gets longer if the updated/current score is divisible by two.
nLength++;
if (snakeY[0] == 0 || snakeX[0] == 0 || snakeY[0] == LENGTH * 2 - 2 || snakeX[0] == WIDTH - 1) // Collision detector with the walls.
{
snake = 0;
gotoxy(33, 10);
printf("Game Over!");
}
}
}```
2. > srand(time(NULL));
Do this ONCE at the start of main.
Calling srand() more than once - and especially if in a loop can make things decidedly non-random.
> food_x = rand() % LENGTH + 2;
> food_y = rand() % WIDTH + 2;
How does 2..LENGTH+1 seem as a valid range to you?
Perhaps some more constants which describe the actual dimensions of the field would help you make better calculations.
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https://www.pagalguy.com/discussions/permutations-and-combinations-28334788 | 1,606,326,345,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141183514.25/warc/CC-MAIN-20201125154647-20201125184647-00091.warc.gz | 794,552,767 | 21,946 | # Permutations and Combinations
1 Post · 6 Users
There are 5 teachers, 4 doctors, 3 players who are involved in various activities like carrom, billiards and playing cards. Find 1\. In how many different ways 3 persons are to be selected to play billiards but there is no doctor?2\. In ho...
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There are 5 teachers, 4 doctors, 3 players who are involved in various activities like carrom, billiards and playing cards. Find
1. In how many different ways 3 persons are to be selected to play billiards but there is no doctor?
2. In how many different ways 5 persons are to be selected to play cards in which maximum number of players?
3. In how many different ways 2 persons are to be selected to play carrom and both are from same post?
(Plz explain in detail)
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