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George Latimer, the father of inventor Lewis Howard Latimer, was the first fugitive from the Maafa (slavery) whose arrest, imprisonment, trial, and emancipation, as well as the numerous public meetings held all over Massachusetts on his behalf, made him a cause célèbre. His supporters, including, Frederick Douglass, Charles Lenox Remond, and other abolitionists fought round the clock for his freedom, and won.
George W. Latimer was born in Norfolk, Virginia on July 4, 1819. His father, Mitchell Latimer, was a whyte man and his mother, Margaret Olmsted, an enslaved woman in bondage to his uncle Edward A. Latimer. In the early part of his life he was enslaved by a man named Edward Mallery, for whom he worked as a domestic servant until the age of sixteen. After that time, his labor was hired out and he primarily worked driving a dray and as a shopkeeper. On two separate occasions he spent time in prison as a result of the debts of his enslaver. He was eventually sold to James B. Gray, a shop owner whose store Latimer manned. He abused Latimer and it is thought that this abuse precipitated Latimer’s flight to Boston.
On October 4, 1842, Latimer and his wife, Rebecca, who was pregnant at the time, ran away. The pair hid beneath the deck of a northbound ship that took them to Baltimore. From there they traveled to Philadelphia, with Rebecca posing as a servant to her lighter-skinned husband. At last, they made their way to Boston, arriving on either October 7 or 8th. Gray offered a reward of $25 if Latimer was captured in Virginia and $50 plus expenses if he was captured outside Virginia.
On the day the couple arrived in Boston, Latimer was recognized by a former employee of Gray. On October 20, Latimer was arrested. The initial charge was larceny. Latimer was brought before Justice Joseph Story, who ordered that he be held.
Latimer’s arrest resulted in an uproar so great that “Boston was, without a doubt, the most potentially violent city in America.” When news of his arrest spread through the Black community, the following day, nearly 300 Black males assembled around the court house “to prevent the [fugitive] from being moved out of the city until word was pledged that Mr. Gray would take no steps not authorized by law.” Abolotionists rallied to his defense, attempting to have him released by a writ of habeas corpus. When this proved unscussessful the abolitionists held a mass meeting at Faneuil Hall, where attendees not only vowed resistance to slave-catching but also voted for disunion. Additional meetings were held throughout the state, called “Latimer Meetings.” The Latimer Meetings were addressed by abolitionists such as Frederick Douglass and Charles Lenox Remond; and the money raised from these meetings was eventually used to purchase Latimer’s freedom.
To coordinate the abolitionist protest a Latimer Committee was appointed, made up of Dr. Henry Bowditch, William Francis Channing, and Frederick Cabot. The trio founded a newspaper called the Latimer Journal and North Star, which first appeared in Boston on November 11, 1842. Its purpose was “to meet the urgency of the first enslavement in Boston” and to rescue a fugitive from the custody in which he was detained. The Latimer Journal reported that the social unrest related to Latimer’s imprisonment was such that “fire and bloodshed threatened in every direction.” Critics claimed that the Latimer Journal “greatly excited and alarmed the credulous, vexed the irritable, inflamed the passionate, and exasperated those whose sympathies ran beyond their judgments.” Six issues appeared subsequently from November 11, 1842 to May 16, 1843, with a circulation of 20,000.
Another committee operation was the promotion of two petitions, one to the state legislature and another to the national legislature. The “Great Massachusetts Petition” requested a law banning the involvement of state officials or public property in the detention or arrest of suspected fugitives. The “Great Petition to Congress” demanded that laws or amendments be passed severing any connection between Massachusetts and the Maafa.
Gray, surprised by the depth of feeling which the case aroused throughout New England and fearful of the counter-charges launched by Latimer’s legal advisers, decided to sell him for $400. The abolitionist highly pleased at this turn of events, did not abandon their petition drive. By mid-February 1843, the two petitions were delivered to their respective destinations. The petition delivered to the State Assembly contained 64,526 signatures and weighed 150 pounds by the time it was delivered on February 17, 1843. Five weeks later the legislature passed a measure dubbed the Latimer Law, the 1843 Personal Liberty Act. The act prevented Massachusetts officials from assisting in the detention of suspected fugitives and banned the use of state facilities to detain such suspects.
After his freedom was purchased, George Latimer remained involved in the abolitionist cause, attending anti-slavery conventions and helping to gather signatures for the two petitions that were started while he was imprisoned.
There is not a great deal of information available about Latimer’s life as a free man. He continued to be involved in, and connected to, the abolitionist movement. In 1851 he was involved in the rescue of an escaped enslaved man, Shadrach Minkins, when he was paid to keep Minkins’s owner under surveillance.
Latimer’s primary occupation was as a paperhanger and he worked in this capacity for forty-five years in Lynn, Massachusetts. The first of the Latimers’ four children was born shortly after his freedom was purchased. The youngest, Lewis Howard Latimer, who was born in 1848, went on to become a famous inventor.
Black Abolitionists by Benjamin Quarles<|endoftext|>
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Transmission electron microscope image of negative-stained, Fortaleza-strain Zika virus (red), isolated from a microcephaly case in Brazil. Credit: NIAID.
Researchers have found that Zika virus (ZIKV), the mosquito-borne virus which is associated with ZIKV congenital syndrome, may have been present in many affected regions months before it was first detected.
What is Zika?
ZIKV is a flavivirus which is transmitted primarily through the Aedes mosquito. Infection with ZIKV causes similar symptoms to the tropical disease Dengue fever such as rash, skeletomuscular pain and fever. These symptoms are usually mild and last 2 – 7 days. However, ZIKA has more serious implications if contracted in pregnancy, as it can cause congenital brain abnormalities, such as microcephaly . The unprecedented outbreak of ZIKV in the world’s tropical belt has led to a rapid escalation of research into the pathogenesis and transmission of ZIKV.
Why do we need genetic information about Zika?
A consortium of researchers funded in part by the National Institute of Allergy and Infectious Diseases (NIAID), part of the National Institutes of Health (NIH), has published a new study in the journal Nature that investigated ZIKV phylogenetic tree (scroll down for a description) and determined if there are any mutations in the viral genetic code. This information is essential to fully understand the trajectory of the outbreak and if there are any new characteristics of the virus which may make it more pathogenic. Additionally, this also helps determine whether current diagnostic tests are still effective at detecting ZIKV in at risk populations .
Despite the fact that ZIKA was discovered 70 years ago, fewer than 100 ZIKV genomes have been sequenced. This is predominantly due to improper processing of clinical samples and low amounts of virus present in the plasma of infected individuals (significantly lower than other viruses such as Ebola).
Genetic sequencing predicts outbreak origins and mutations
Using human and mosquito samples, the investigators of this study used multiple strategies to create a genetic dataset of 110 ZIKV genomes. Together with 64 published genomes from the scientific literature, they generated a phylogenetic tree and analysed the patterns and timings of when the virus was introduced into geographical regions within the Americas.
The analysis indicates that ZIKV originated in Brazil and rapidly spread to neighboring countries. The authors commented that the speed of transmission early in the outbreak, was probably due to the naivety of the South American population to the virus; meaning their immune systems had not become resistant to ZIKV. The virus diversified into 4 genetically distinct clades or branches of the phylogentic tree (a common ancestor and all its descendants), which spread to Puerto Rico, Honduras, Colombia and the fourth clade included certain Caribbean islands and the continental United States .
The researchers also estimate that ZIKV arrived and circulated in certain areas much earlier than the first confirmed case; approximately 4.5 months earlier in Puerto Rico and 9 months earlier in the Caribbean-US areas .
Just like humans, small mutations in the viral genetic code can occur over time. Viral mutations can facilitate drug resistance or help the virus to escape the host immune system. 1,030 mutations were detected in the dataset across the whole genome. While the researchers commented that further investigation would be needed to fully understand the functional implications of these mutations, information about ZIKVs genetic variability will provide vital insight into its biology and pathogenesis. Moreover, this knowledge will also no doubt be a valuable resource for the entire ZIKV research community.
In addition, these mutations can render diagnostic tests obsolete. Fortunately, the mutations detected in the genomes are not predicted to affect the performance of the current diagnostic tests. However, the authors warned, that as viral mutation is an ongoing process, continuous monitoring is recommended to ensure there is suitable health surveillance in vulnerable populations .
This work is the first to provide significant insight into the evolution and genetic changes in the ZIKV throughout the recent outbreak. This study highlights the importance of appropriate clinical sample handling and provides a strategy to obtain genetic information from viruses with low plasma levels. Details about genomic evolution will help to inform authorities and healthcare workers about best practice for disease monitoring and diagnosis. Furthermore, understanding the intricacies of viral spread can support public policy decisions and public health strategies.
Terms and definitions:
What is a phylogenetic tree?
A phylogenetic tree is a diagram that describes the diversification of a group of organisms through time. The main branch represents a common ancestral lineage and the branches are the descendants of that lineage. Following an event which causes the common ancestor to split or change (speciation), a daughter lineage is created .
Eric Gaba, NASA Astrobiology Institute
What is a lineage?
A succession of species, where each member has evolved from its predecessor .
HC Metsky et al (2017). Genome sequencing reveals Zika virus diversity and spread in the Americas. Nature. 0028-0836. https://www.nature.com/nature/journal/vaop/ncurrent/full/nature22402.html
Larson, Allan, et al (2014). Concepts in Character Macroevolution: Adaptation, Homology, and Evolvability.The Princeton Guide to Evolution, edited by Jonathan B. Losos et al., Princeton University Press, pp. 89–99. http://www.jstor.org/stable/j.ctt4cgc5m
Baum, D. (2008). Reading a Phylogenetic Tree: The Meaning of Monophyletic Groups. Nature Education 1(1):190. https://www.nature.com/scitable/topicpage/reading-a-phylogenetic-tree-the-meaning-of-41956<|endoftext|>
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The focus of Lesson 1 is Newton's first law of motion - sometimes referred to as the law of inertia. An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
## What are Newton's first 3 laws?
The Newton's three laws of motion are Law of Inertia, Law of Mass and Acceleration, and the Third Law of Motion. A body at rest persists in its state of rest, and a body in motion remains in constant motion along a straight line unless acted upon by an external force.
## What does Newton's 2nd law state?
Newton's second law is a quantitative description of the changes that a force can produce on the motion of a body. It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.
## What Newton's laws explain?
Answer: Newton's first law of motion explains how inertia affects moving and nonmoving objects. Newton's first law states that an object will remain at rest or move at a constant speed in a straight line unless it is acted on by an unbalanced force.
## What is the state of motion?
The state of motion of an object is defined by its velocity - the speed with a direction. Thus, inertia could be redefined as follows: Inertia: tendency of an object to resist changes in its velocity. An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity.
## Why do we call Newton's first law as law of inertia explain?
It is called the law of inertia because it tells that every material body has a property by virtue of which it resists the change in its state of rest or in its state of motion.
## How do you demonstrate Newton's first law?
Newton's first law states than an object at rest, assuming no force acts on it, will try to stay at rest. Demonstrate this by placing an index card on top of the plastic cup. After placing the penny on top of the index card, quickly remove the index card horizontally.<|endoftext|>
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Because they can, or possibly because they have to.
Let me explain:
First of all, isotopes are simply atomic nuclei with the same number of protons and different numbers of neutrons. They may range anywhere from stable isotopes to those that last less than 10^-14 seconds. With today’s knowledge of nuclear science, it is not very difficult to add a neutron to a nucleus and produce an isotope, even if its lifetime is fleeting.
Perhaps you were referring more to isotopes in nature. In that case, the answer is more difficult but still centers around the concept of “because they can/have to”. ‘Heavy’ nuclei are formed within stars via two main processes, the r process and the s process. The difference between these processes is not important; however they both involve the absorption of neutrons by nuclei followed by beta decay (the process by which a neutron in the nucleus is converted into a proton or vice verse). In large stars, there are enough neutrons that a nucleus can absorb many neutrons before it has time to decay. Depending on the number of neutrons absorbed, the isotope of the final product can vary.
Unfortunately, if you are looking for a more in depth answer, then we must begin delve into nuclear physics (always a fun topic
). The stability of a nucleus is governed largely by the binding energy which incorporates the volume, surface, coulomb, symmetry and pairing energy of the nucleons. Each of these terms is dependent upon the mass and charge of the nucleus. Without going into the physics involved, since five different terms determine the stability of the nucleus, it is possible to come to several answers for each mass or charge that will result in ‘stable’ nuclei. Ergo, Isotopes.<|endoftext|>
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Math Games to Make Math Fun!
The best ideas from Japan...
# Games to Math FUN!
Main Menu - About the site - Contact
# Body Building Game!
## Target Grade:1-6 Target Maths: Any questions or sums Preparation: an inflatable dice
This is the perfect game to be used in a variety of ways for reviewing different topics.
Make sure the kids remember the words for parts of the body (head, eye, ear, mouth, nose, arm and leg) and then...
1. Split the class into 6 teams. If they are in columns of desks then saying the left had column is team 1, the next team 2 etc. is usually the best.
2.On the board draw 6 limbless, featureless faces and bodies.
3. Tell the front person in each group to stand up.
4. You then ask the kids who are stood up a question e.g. "What's 3 x 6?" or "What's this shape?" or even things like "How many legs does a spider have?" The quickest one to answer is the winner.
5. The winner rolls the inflatable dice.
6. If they get a "1" they can draw an eye on their team's face. If they have a "2" they can draw an ear. Similarly,"3" is a nose, "4" is a mouth, "5" is an arm and "6" is a leg.
Here we threw a "5" so can draw an arm!
7. But each team can only have one nose, one mouth, and two each of the ears, eyes, arms and legs. For example if your team has already 2 arms, and you roll another number "5" then that go is a "pass" because you can't add another arm! This really helps the weaker teams catch up (and sometimes win!!!)
8. The next person in each group stands up and repeat from step 4
9. The winning team is the first team to get 2 eyes, 2 ears, 1 nose, 1 mouth, 2 arms and 2 legs. If you run out of time then the team with the most features is the winner!
The eyes, ears, etc. are drawn on one at a time.
In Primary School the kids can get very good and it can often be difficult to tell who has answered the question the fastest! I'm still working on this problem, trying making the questions more difficult helps a bit.
Different numbers of students on each team can be good as you have different opponents each time.
Math Games to Make Math Fun!
If you like this idea, please tell your friends!<|endoftext|>
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Dividing Fractions Word Problem
• 0:00 - 0:01
• 0:01 - 0:06
A baby's T-shirt requires 4/5
yards of fabric, or 4/5 of a
• 0:06 - 0:07
yard of fabric.
• 0:07 - 0:11
How many T-shirts can be
• 0:11 - 0:13
So what we want to do is we
essentially want to say how
• 0:13 - 0:18
many groups of 4/5 of a yard
can we make with 48 yards?
• 0:18 - 0:25
So you literally view this as
we want to take 48 yards and
• 0:25 - 0:30
divide it into groups of 4/5
of a yard, and say how many
• 0:30 - 0:31
groups are there?
• 0:31 - 0:32
Because each of those
groups can make
• 0:32 - 0:34
one to baby's T-shirt.
• 0:34 - 0:37
If you give me 4/5 of a yard,
one baby's T-shirt, so the
• 0:37 - 0:41
number of groups of 4/5 is the
number of babies' T-shirts.
• 0:41 - 0:43
Now, when we divide by a
fraction, we just have to
• 0:43 - 0:46
remember that that is the same
thing, that is completely
• 0:46 - 0:51
equivalent to multiplying
times the
• 0:51 - 0:52
reciprocal of the fraction.
• 0:52 - 0:57
So if we have 4/5 here,
that'll be 5/4, the
• 0:57 - 0:58
reciprocal.
• 0:58 - 1:00
Now, you still might say, hey, I
have a whole number here and
• 1:00 - 1:02
a fraction, and you just have
to remember any whole number
• 1:02 - 1:04
can be written as a fraction.
• 1:04 - 1:11
This is the same thing
as 48/1 times 5/4.
• 1:11 - 1:14
Now, we could just multiply it
out at this point and figure
• 1:14 - 1:19
out what 48 times 5 is and
that'll be over 4, but that'll
• 1:19 - 1:22
get big numbers and it'll be
hard to kind of divide and all
• 1:22 - 1:24
that, but we could divide at
this stage right here.
• 1:24 - 1:26
We could divide the numerator
and the denominator by 4.
• 1:26 - 1:30
Or we could say, look, this is
going to be equal to 48 times
• 1:30 - 1:34
5, whatever that is, over 4.
• 1:34 - 1:36
Now, let's divide the
numerator by 4.
• 1:36 - 1:40
Well, we could divide 48 by 4,
and we will get to 12, and
• 1:40 - 1:42
whatever we did to the
numerator, we have to do to
• 1:42 - 1:46
the denominator, so if we
divide 4 by 4, we get 1.
• 1:46 - 1:51
So then we're left with 12 times
5, which is equal to 60.
• 1:51 - 1:57
12 times 5, which is equal to
60/1, which is the exact same
• 1:57 - 1:59
thing as 60.
• 1:59 - 2:03
So you can actually make 60
children's or babies' T-shirts
• 2:03 - 2:07
from 48 yards if each of
them use 4/5 of a yard.
• 2:07 - 2:08
Title:
Dividing Fractions Word Problem
Description:
U02_L2_T3_we5 Dividing Fractions Word Problem
more » « less
Video Language:
English
Duration:
02:08
brettle edited English subtitles for Dividing Fractions Word Problem brettle edited English subtitles for Dividing Fractions Word Problem brettle edited English subtitles for Dividing Fractions Word Problem
Incomplete
• brettle<|endoftext|>
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How do I teach my 12-year-old grandchild about the Holocaust?
The Holocaust is an important topic not only in Jewish history, but in the history of humankind. The topic is disturbing, and it is appropriate to feel uncomfortable and upset by the stories, the facts, and especially the images. I applaud your thoughtful approach to how – not if – your tween grandchild should be taught about the Holocaust. Here's one teacher's story about why it's so important.
There are many ways to lead into the subject at younger ages, depending on the child’s temperament and emotional development, how much the child already knows, and what is prompting the question. There is general agreement that parents and educators should be cautious about how much explicit detail children are exposed to before middle school, but that isn’t to say you should avoid the topic altogether.
In the field of Jewish education, the general guideline is to teach topics “up to the gates” until fifth grade. Yad Vashem - The World Holocaust Remembrance Center in Israel requires visitors to its main exhibits to be at least 10 years old. Of course, parents know their own kids best – and as with all difficult subjects, it is best to be sure that you and the child’s parents are on the same page about what is appropriate.
Reading and discussing books together can be a doorway into topics related to Holocaust history. In some public schools, students read Lois Lowry’s Number the Stars in 4th grade and The Diary of Anne Frank in 5th or 6th grade. There are developmentally appropriate books for even younger children that explore the importance of memory, of human dignity, of standing up for others, as well as picture books set in that historical time period in Europe. For additional book suggestions, we recommend searching the titles curated by PJ Library® and PJ Our Way®.
I encourage you also to reach out to the rabbi and educator at your synagogue (and if you don’t yet belong to a synagogue, you can find one through our directory). This is an area in which they are likely to have some expertise, and they should be able to provide some helpful resources.<|endoftext|>
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How were the time zones decided?
Before time zones were set up, there was a great deal of confusion, especially when people had to use railroad timetables. To end this confusion the United States in 1883 began using a system of standard time zones.
In 1884 an international conference was held in Washington, D.C., to set up a system to fit the whole world. The earth was divided into 24 zones, each covering 15 degrees of longitude. This is a natural division, for the earth rotates at the rate of 15 degrees each hour.
Within each zone, the time is the same, and the difference between one zone and the next is exactly one hour. Greenwich (London), England, was selected as the starting point. Thus, when it is noon in Greenwich, the time in the next zone eastward is 1 p.m. The time in the next zone westward is 11 a.m. In New York, five zones west of Greenwich, the time is 7 a.m.
The United States is divided into four zones based on the 75th, 90th, 105th, and 120th meridians. The times in these zones are called Eastern, Central, Mountain, and Pacific Standard Time.
On the opposite side of the world from Greenwich is another dividing line, the International Date Line. This line is approximately the 180th meridian. When it is noon at Greenwich it is midnight at the International Date Line. Crossing the line, a person gains or loses a day, depending on whether he is moving east or west.<|endoftext|>
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What is an Acceptable Use Policy?
An acceptable use policy (AUP) “outlines, in writing, how a school or district expects its community members to behave with technology.” (“1-1 Essentials”) It is common for schools and districts to outline both acceptable and unacceptable use, being careful not to create a negative tone by only listing the dont’s. Encounters with inappropriate material, digital citizenship, network fraud and protecting personal information are just a few of the many topics that should be included in a comprehensive AUP.
Creating the AUP is only the jumping off point. Using the policy to teach students appropriate and responsible online behavior is the next crucial step. Depending on the age and maturity level of students, this conversation will look and sound different. In the 12 technology classes I teach, we learn, review and sign the AUP before we access the internet or any of the school’s technology tools. The same AUP is used for students K-6, but the explanation and discussion of the AUP is simplified depending on the age of the students. At the end of our discussions and lessons, all students sign the AUP.
After creating and teaching the AUP, the next part of successful implementation is follow through. The folks at Common Sense Media remind us that “an acceptable use policy is only as strong as your commitment to enforce it.” (“1-1 Essentials”) Therefore, someone, or a group of people, should be appointed to enforce the consequences set forth in the policy. Let students know they are being monitored and then apply consequences as necessary.
As the use of the internet and technology tools are implemented in education, we are faced with a difficult challenge as dictated by our friends at Common Sense Media. “Today’s educators have the tough job of maintaining the delicate balance of protecting students while providing access to the digital world. Educators need to be comprehensive yet not limiting when creating a stimulating but safe learning environment. An AUP is a first step in framing these opportunities.” (“1-1 Essentials”)
What Should be Included in an AUP?
“The Acceptable Use Policy for Internet use is one of the most important documents a school will produce.” (“Getting Started”) With this in mind, here are some common elements of Acceptable Use Policies:
1-to-1 Essentials - Acceptable Use Policies. (n.d.). Retrieved February 23, 2017, from https://www.commonsensemedia.org/educators/1to1/aups
Why Have a Technology Policy in Your School or Library? | Librarians. (n.d.). Retrieved February 24, 2017, from http://www.scholastic.com/librarians/tech/techpolicy.htm
Education World: Getting Started on the Internet: Acceptable Use Policies. (n.d.). Retrieved February 24, 2017, from http://www.educationworld.com/a_curr/curr093.shtml
Links to Acceptable Use Policies
1. Compass Public Charter School
2. West Ada
3. North Star Charter School
4. Liberty Charter School<|endoftext|>
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Following the end of his formal education, Marx entered the world of journalism and began to write articles for various newspapers. After contributing to the Rheinische Zeitung, a liberal paper based in Cologne, Germany, Marx became editor-in-chief of the paper and utilized the platform to project his revolutionary and socialist-centered ideas. Shortly thereafter, Marx first met Friedrich Engels, an event of great importance and one that marked the beginning of an important friendship. The Prussian government banned the Rheinische Zeitung, and Marx was forced to resign as editor-in-chief, ending his affiliation with the newspaper.
Key Words: Journalism, Rheinische Zeitung, Friedrich Engels, Prussian government
To explore the five other eras of Marx’s life, please click on one of the following links:Childhood and Education, Exile in Paris and Brussels, 1848 Revolutionary Events and Writings, Exile in London: Before Das Kapital, and Exile in London: After Das Kapital.
*To see a visual tour of Marx’s movements throughout this era and his entire lifetime, please click on the following link to download the file and open it up in a Google Earth program: Karl Marx Tour<|endoftext|>
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Before we can understand a prepreg, we need to know what a PCB is. A Printed Circuit Board (PCB) is what makes all of the electronics you use today operate. It is a series of electrical components interconnected onto a small device. Without the PCB, our electronics wouldn’t be able to function as intended.
A PCB is mostly made up of multiple layers of the substrate, copper layering, small holes drilled into the device, solder mask, a coating (quite often in tin-lead), multiple components, before being fused together. Although, when put this way, developing a PCB doesn’t seem like it involves too much. However, the actual creation process is extensive, as you must be precise with every calculation to avoid malfunction.
What is a Prepreg PCB?
A prepreg (pre-impregnated) is one of the main materials used in multilayer boards and is what holds the cores together. It is composed of fiberglass impregnated with resin (an epoxy-based material). The layers get pressed together at a temperature to create the required board thickness.
The prepreg and core quite often get mixed up. However, they are two separate components of the PCB. The core of the PCB is the FR4 layers of copper traces and glass-reinforced epoxy laminate sheets. Once heated, the prepreg holds the core of the PCB and the layers together.
Application in the PCB Manufacturing Process
The prepreg in a multi-layered board is what holds the layers together. Once you stack all the layers of the board, you fuse them together by exposing the board to high temperatures. The thickness of the prepreg will be in accordance with the thickness of the overall board.
The prepreg should have specific characteristics to it when used on a multi-layered board. The cloth surface should look and feel smooth and be free of any oil, stains, defects, foreign matter, cracks, or excessive resin powder.
If you have a complex PCB, you’ll likely have a more complex prepreg. You’ll need to use different types of the prepreg to achieve the required thickness of both the prepreg and the board itself.
How it is Used
Prior to use, the prepreg will be dry but not hard. This allows the prepreg to flow through the PCB when heated. The fiberglass gets impregnated with the resin before being put onto the layer. Once aligned, the prepreg is heated and put under pressure to soften the material. Once done, as the PCB cools, the prepreg begins to solidify and hold everything together.
Why is it Important?
The prepreg plays an essential role in the development process by the PCB manufacturer. Without the prepreg, there would be nothing there to hold the layers together.
What Role Does it Play?
The role of the prepreg is to fuse etched cores together or to glue a copper foil to an etched core. You will find a layer of prepreg between each layer of the PCB.
The prepreg is not just an important part in the PCB manufacturing process, it is a necessity if the board has multiple layers. Without the prepreg, there would not be anything holding the different layers together.
The prepreg and core are two different parts of the PCB. For starters, the core is an FR4 material with copper traces. Whereas, the prepreg is fiberglass impregnated with resin. It is the prepreg that holds the core together on the PCB.<|endoftext|>
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# If X and Y are random variables and c is any constant, show that E(cX)=cE(X).
If X and Y are random variables and c is any constant, show that $E\left(cX\right)=cE\left(X\right)$.
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Approach:
Let X represent any random variable, the probability distribution pattern is as follows,
$E\left(X\right)={x}_{1}{p}_{1}+{x}_{2}{p}_{2}+{x}_{3}{p}_{3}+.....+{x}_{n}{p}_{n}-$
Here, ${x}_{1},{x}_{2},{x}_{3},....{x}_{n}$ are all possible favorable outcomes and ${p}_{1},{p}_{2},{p}_{3},.....{P}_{n}$ are their respective probabilities.
Calculation:
Consider the expected value of random variable X as,
$E\left(X\right)={x}_{1}{p}_{i}+{x}_{2}{p}_{2}+{x}_{3}{p}_{3}+{x}_{n}{p}_{n}$
The associated probabilities will not change of the variable X is changed to eX, where, c is any constant value.
From part (a),
$E\left(c\right)=c$.
Therefore,
$E\left(cX\right)=cE\left(X\right)$.
Therefore, for any number c, $E\left(cX\right)=cE\left(X\right)$.
Conclusion:
Hence, the relation $E\left(cX\right)=cE\left(X\right)$ is proved if X and Y are random variables and c is any constant.<|endoftext|>
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Question
# Solving the following equations will require you to use the quadratic formula. Solve each equation for u between 0^(circ) and 360^(circ), and round your answers to the nearest tenth of a degree. cos2 u + sin u = 0
Equations and inequalities
Solving the following equations will require you to use the quadratic formula. Solve each equation for u between $$\displaystyle{0}^{{\circ}}{\quad\text{and}\quad}{360}^{{\circ}}$$, and round your answers to the nearest tenth of a degree.
$$\displaystyle{\cos{{2}}}{u}+{\sin{{u}}}={0}$$
2021-02-13
We have,
$$\displaystyle{{\cos}^{{{2}}}{u}}+{\sin{{u}}}={0}$$
By using the identity, $$\displaystyle{{\cos}^{{{2}}}{u}}={1}−{\sin{{2}}}{u}$$ and solving further,
$$\displaystyle{{\cos}^{{{2}}}{u}}+{\sin{{u}}}={0}$$
$$\displaystyle{1}−{{\sin}^{{{2}}}{u}}+{\sin{=}}{0}$$
$$\displaystyle{{\sin}^{{{2}}}{u}}−{\sin{{u}}}−{1}={0}$$
The above equation is quadratic equation in variable sinu.
Using the quadratic equation formula and solving it further, we get our result as
$$\displaystyle{\sin{{u}}}={\frac{{-{\left(-{1}\right)}\pm\sqrt{{{\left(-{1}\right)}^{{{2}}}-{4}{\left({1}\right)}{\left(-{1}\right)}}}}}{{{2}{\left({1}\right)}}}}$$
$$\displaystyle{\sin{{u}}}={\frac{{{1}+\sqrt{{{5}}}}}{{{2}}}},{\frac{{{1}-\sqrt{{{5}}}}}{{{2}}}}$$
Now,
$$\displaystyle{\sin{{u}}}={\frac{{{1}+\sqrt{{{5}}}}}{{{2}}}}$$
No solution occurs because range for sine function lies between -1 to 1
For,
$$\displaystyle{\sin{{u}}}={\frac{{{1}-\sqrt{{{5}}}}}{{{2}}}}$$
$$\displaystyle{\sin{{u}}}={0.618}$$
$$\displaystyle{u}={{\sin}^{{-{1}}}{\left(-{0.618}\right)}}$$
$$\displaystyle{u}=-{{\sin}^{{-{1}}}{\left({0.618}\right)}}$$
$$\displaystyle{u}={38.17}^{{\circ}}$$
$$\displaystyle{u}={360}^{{\circ}}-{38.17}^{{\circ}}$$
$$\displaystyle{u}={321.82}^{{\circ}}$$
Hence,value of u will be $$\displaystyle{321.82}^{{\circ}}$$<|endoftext|>
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On July 10, 1890, the Territory of Wyoming entered the American republic as the 44th state of the union. It’s nickname, the “Equality State” can seem somewhat of a misnomer given Wyoming’s conservative politics and high GDP (PPP) per capita (it ranks seventh in the union), but a closer look at this state’s relatively unknown history highlights how and why Wyoming came to be known as the Equality State.
First, though, let’s get some facts out of the way. Wyoming is the least populous state in the union, with only half a million people residing there. The largest city (and capital), Cheyenne, is home to about 64,000 souls, and the city of Laramie is its cultural capital. The University of Wyoming calls Laramie home.
Wyoming is considered to be the home state of the Cheney political dynasty, too, and Wyoming’s sole representative in the U.S. House is Liz Cheney, daughter of the much-loathed Dick Cheney, former vice president of the United States. Wyoming turned decidedly Republican in the mid-1960s with the advent of a new mining boom in the country. Somewhat ironically, Wyoming gets more federal money than any other state in the union save for Alaska (which is also dominated by the GOP, but not nearly as completely as Wyoming).
In 1889, Wyoming’s voters approved the first constitution in the world to grant women the right to vote. When the territory became a state one year later, suffrage was carried over and Wyoming became the first state in the union to grant voting rights to women. (Hence the nickname Equality State.)
Other states in the Rockies quickly followed Wyoming’s lead, with Idaho (1896), Utah (1896), and Colorado (1893) all granting full voting rights to women before the turn of the century. The fifth state to grant women’s suffrage, Washington, did not do so until 1910, well over a decade after Utah and Idaho granted suffrage.
So why did these “mountain west” states grant suffrage so much earlier than the rest of the country (and, indeed, the rest of the world)? There are two answers I’d to put forth here. One is relatively straightforward: the mountain west states were less populated than most states throughout the country, and this meant that these states needed all the help they could get when it came to churning out votes for political goodies from Washington.
A second, more controversial, answer is that these states - with their individualistic mores and relatively conservative politics - were simply more likely to grant every adult full voting rights due to their ideological and economic underpinnings. There’s an open secret in this country that Republican women are numerous, well-educated, and civically engaged, rather than shuttered up in the home and forced to cook and clean for their men.
Could the egalitarian impulse of the libertarian-ish mountain west simply be a different understanding of the notion of equality? It’s a question worth asking, especially now, given the political divisions that confront the republic.<|endoftext|>
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# Write a quadratic equation with one solution
Separable Equations — In this section we solve separable first order differential equations, i. Solving the Heat Equation — In this section we go through the complete separation of variables process, including solving the two ordinary differential equations the process generates.
Anyway, hopefully you, found that fun. So let's just solve this the way we solve everything. Students will connect functions and their associated solutions in both mathematical and real-world situations.
Undetermined Coefficients — In this section we introduce the method of undetermined coefficients to find particular solutions to nonhomogeneous differential equation. So, I'm assuming that you have tried that, so let's work through each of them, step one by one.
So this is the interval notation for this compound inequality right there. The student applies the mathematical process standards and algebraic methods to rewrite in equivalent forms and perform operations on polynomial expressions. For cwe need to see when the graph goes back down to 0; this is when there are no rabbits left on the island.
The student applies mathematical processes to analyze data, select appropriate models, write corresponding functions, and make predictions. So, once again, we're going to do five times our x, which is going to be six, actually let me just write it out, minus three needs to be equal to four times our x, plus three, and in this case our x is six, so it's going to be five times six minus three needs to be equal to four times six plus three.
Reduction of Order — In this section we will discuss reduction of order, the process used to derive the solution to the repeated roots case for homogeneous linear second order differential equations, in greater detail.
Using these notes as a substitute for class is liable to get you in trouble. Then we would have a negative 1 right there, maybe a negative 2. The student applies the mathematical process standards when using graphs of quadratic functions and their related transformations to represent in multiple ways and determine, with and without technology, the solutions to equations.
So if you subtract 2 from both sides of the equation, the left-hand side becomes negative 5x. I'm hoping that these three examples will help you as you solve real world problems in Algebra.
So that is our number line. Remember y and f x represent the same quantity.
Due to the nature of the mathematics on this site it is best views in landscape mode. Then we take the square root of each side, remember that we need to include the plus and minus of the right hand side, since by definition, the square root is just the positive.
Quadratic Equations make nice curves, like this one: Name The name Quadratic comes from "quad" meaning square, because the variable gets squared (like x 2). One property of this form is that it yields one valid root when a = 0, while the other root contains division by zero, because when a = 0, the quadratic equation becomes a linear equation, which has one root.
QUADRATIC EQUATIONS. A quadratic equation is always written in the form of. 2. ax +bx +c =0 where. a ≠0. The form. ax. 2 +bx +c =0 is called the.
standard form.
of a quadratic equation. Examples: x2 −5x +6 =0 This is a quadratic equation written in standard form. x2 +4x =−4 This is a quadratic equation that is not written in standard form but. For a quadratic equation ax 2 +bx+c = 0 (where a, b and c are coefficients), it's roots is given by following the formula.
The term b ac is known as the discriminant of a quadratic equation. The discriminant tells the nature of the roots. If discriminant is greater than 0, the roots are real and different. In this section we will discuss how to solve Euler’s differential equation, ax^2y'' + bxy' +cy = 0.
Note that while this does not involve a series solution it is included in the series solution chapter because it illustrates how to get a solution to at least one type of differential equation at a singular point. The solution(s) to a quadratic equation can be calculated using the Quadratic Formula: The "±" means we need to do a plus AND a minus, so there are normally TWO solutions!
The blue part (b 2 - 4ac) is called the "discriminant", because it can "discriminate" between the possible types of answer.
Write a quadratic equation with one solution
Rated 3/5 based on 46 review
Create a System of Equations, Given 1 Equation and the Solution | Open Middle™<|endoftext|>
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Knights and castles
(English learner's edition)
Part of the English Learners series
This title takes beginner readers on a trip back to medieval times to learn what it was like to be a knight and live in a castle, finding out how they fought and what they did for fun.
It includes short biographies of famous knights and descriptions of castles that can still be seen today.
This title is part of the Usborne English Language Learners programme, featuring books from the Usborne Reading Programme, with audio CDs in British and American English, downloadable worksheets and teacher's notes.
4KHA For National Curriculum Early Years, 4KHF For National Curriculum Key Stage 1, 5AD Interest age: from c 4 years, YFJ Traditional stories (Children's / Teenage), YQCR Educational: English language: readers & reading schemes<|endoftext|>
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Volume of irregular shapes
You can calculate the volume of irregular shapes easily once you know how to get the volume of a single three-dimensional shape.
The house below is made of two 3-d shapes: A triangular prism and a rectangular prism.
The roof of the house is the one shaped like a triangular prism.
Volume of house = volume of triangular prism + volume of rectangular prism.
volume of triangular prism = area of triangle × length of house.
volume of triangular prism =
base of triangle × height of triangle / 2
× length of house
length of house = 50 feet, base of triangle = 15 feet, and height of triangle = 10 feet
volume of triangular prism =
15 × 10 / 2
× 50
volume of triangular prism = 75 × 50 = 3750 feet3
volume of rectangular prism = 15 × 15 × 50 = 11250 feet3
Volume of house = 3750 + 11250 = 15000 feet3
A couple more great examples showing how to calculate the volume of irregular shapes
Ice cream cone:
When you order ice cream, you may have never realized that it could be a combination of half a sphere and a cone
Of course, we have to assume that the shape made by the ice cream scoop is half a sphere.
The shape of my ice cream shown below is half a sphere and then we put it on a cone. What is the volume?
It may be easier to see how to compute the volume if we remove the ice cream
Pretend that the radius of the cone or r is 1.5 inches and the height of the cone or h is 6.5 inches here is how to compute the volume
Volume = volume of half the sphere + volume of the cone
Volume of a sphere =
4 × π × r3 / 3
Volume of a sphere =
42.39 / 3
Volume = 14.13 inches3
Since it is half a sphere, we get 7.065 inches3
Volume of a cone =
π × r2 × h / 3
Volume of a cone =
3.14 × 1.52 × 6.5 / 3
Volume of a cone =
45.9225 / 3
Volume of cone = 15.3075 inches3
Volume of the whole thing is 7.065 + 15.3075 = 22.3725 inches3
A box containing 4 cream cones:
A box has 4 identical ice cream cones arranged inside of the box as shown below. The radius of the base of the cone is 1 inch and the height is 6 inches
What is the volume of the space left inside the box ?
Space left inside the box = volume of box - volume of 4 cones
Volume of box = 8 × 8 × 8 = 512 inches3
volume of a cone =
π × r2 × h / 3
volume of a cone =
3.14 × 12 × 6 / 3
volume of a cone = 6.28. Therefore, volume of 4 cones = 4 × 6.28 = 25.12
Volume left inside = 512 - 25.12 = 486.88 inches3
Hopefully you know now how to find the volume of irregular shapes!
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
Recommended<|endoftext|>
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## Archive for June 17th, 2009
### The Fourth Bunch of Ways to Multiply
June 17, 2009
Just three more ways today, although with all the ones that have been suggested I think we’ll get to the 25!
(15) Multiplication with Log Tables. People had already been using trig tables to multiply, but when logarithms were discovered they became THE way to multiply numbers. The idea behind log tables is that logarithms turn multiplication (nasty) into addition (fun!) without having to derive a bunch off trig formulas. In particular, we’re going to use the fact that log(x·y)=log(x)+log(y).
Here’s how we can find 875×978 with logarithm tables. We’ll start by writing the two numbers in Scientific Notation: 875 would become 8.75×102 and 978 would become 9.78×102. We’re going to multiply 8.75 and 9.78, and then adjust that product by the appropriate power of 10 (in this case, 104).
The next step for the multiplication is to look in your Table of Common [Base 10] Logarithms. If you don’t have a copy handy, you can look here. It turns out that log(8.75) is 0.9420081 [that’s our log(x)] and log(9.78) is 0.9903389 [that’s our log(y)].
Now we’ll add those together to get 1.9323470. This must be the log of our product! So we work backwards with the table, looking for the number whose log is 1.932347. Unfortunately, the table only gives results between 0 and 1, so initially it seems like we’re stymied, but we can be sneaky and subtract 1. We’ll come back to that in a moment.
So now we’re looking for a number whose log is 0.932347.
Looking back at our table, we see that log(8.55)=0.9319661 and log(8.56)=0.9324738, so the number we’re looking for must be between 8.55 and 8.56. We can pick 8.56, which is the closer number, or do a little interpolation. If we round, we’re looking for 0.9323 and instead we got 0.9320 and 0.9325. The number we wanted was about 60% of the way from the smaller to the larger, and so instead of choosing 8.55 or 8.56 we could go about 60% of the distance between them, and guess that the product was 8.556. Let’s do that. [We could be even more accurate if we used a calculator to figure the exact percentage, but that seems to defeat the purpose of using the log table to multiply.]
So now we have a product of 8.556, but clearly that’s not exactly right. We first have to account for the fact that we subtracted 1. Notice that 1+log(blah) is the same as log(10)+log(blah) [because we’re using the common logarithm], and THAT is the same as log(10·blah). Here log(blah) is 0.9323470, and we just saw that blah was approximately 8.556, so the product that we want — the product of 8.75 and 9.78 — is approximately 85.56.
We’re almost there! Remember how originally we wanted the product of 875 and 978 but we first wrote those in Scientific Notation? We need to adjust our answer of 85.56 by 104, leading us to the conclusion that 875×978 is approximately (drum roll please) 855600. The correct answer is 855750, so we’re certainly in the right ballpark and more accurate than we were with the trig tables, but, well, it’s a wonder to me that this method was so popular for so long when so many other ways are even more accurate. Maybe I’m missing something.
(I’ve multiplied with log tables before, but refreshed my memory with this site on the Obsolete Skills Wiki, which also explains how to Get off the couch to Change Channels on the TV set and how to Make Change in [old] Shilling and Pence.)
(16) Slide Rule. This takes the ideas of the log table, but bypasses the actual looking up. Instead, the numbers are scaled on the slide rule in such a way that you don’t have to do much at all. William Oughtred is credited in many places (online) with the slide rule, and the date 1622 shows up, so this happened just a few years after the invention of logarithms. Pretty quick thinking!
Let’s do a simple example first: 2.5×3. You can use an actual slide rule, or use the java version here. The slide rule looks complicated because it can do a lot of things, but we’re mostly going to be looking at the bottom of the slidey part in the middle (C) and the fixed part at the bottom (D).
To multiply 2.5 by 3, you start with2.5 and align the 1 on C with the 2.5 on D. Then look for 3 on C, and right below it will be the product (in this case, 7.5):
See how once you’ve aligned the 1 with the first factor 2.5, you can move the slider doohickey to the 3 on C so that you can see what’s right below it?
Now let’s look at a more complicated example: back to 875×978. As with log tables, we’ll start by writing the numbers in scientific notation (8.75×102 and 9.78×102) and then we’ll multiply 8.75 and 9.78 and adjust our final answer.
Normally we’d align the 1 of C with 8.75 beneath it on D, and then start looking to the right. But 8.75 is so large, we quickly run out of room:
So we’re sneaky, and instead of aligning the 1 (on C) with the 8.75, we align 10 (on C) with 8.75. We’ll have to adjust by multiplying by an extra power of 10 at the end, kind of like we did with the log tables before (and, really, it’s not a coincidence that this happened in both calculations).
So we’ve aligned the 10 on C with the 8.75 on the bottom. Then we move the slider over to 9.78 on C, and look below to see the product! The product looks like, ummm, it seems to be a bit over 8.55 but not yet at 8.56, so we’ll say 8.555.
As our final step, we need to multiply by 10 because we aligned with the 10 instead of with the 1 (as discussed above), and also by 104 because we’d had to write the numbers in scientific notation. This means our final answer is 8.555×105, or 855500. As we’ve seen before, it’s not exact but it is pretty close and it’s a lot faster than looking up in tables!
(17) The Gunter scale. This is the precursor to the slide rule, invented by Edmund Gunter, and it didn’t slide at all but it was BIG: two feet long was standard. It was a big ole piece of wood with a logarithm scale on it, and if you wanted to measure a product like 2.5×3 you’d measure the physical distances to 2.5 and to 3, add them together, and see where you ended up on the scale. Essentially it was a Slide Rule that didn’t slide: you had to do that part by hand.
(There’s a picture here and more information about Gunter here, including the fact that he coined the terms for Cosine and Cotangent, though Cosecant is a bit older. I hadn’t known about this method until Pat Ballew brought it up in a recent comment: thanks Pat!)<|endoftext|>
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# How do you simplify (m^5)^2/m^-8?
Jul 8, 2016
${m}^{18}$
#### Explanation:
To simplify ${\left({m}^{5}\right)}^{2} / {m}^{-} 8$
Begin by multiplying the two exponents on the numerator
${m}^{10} / {m}^{-} 8$
In order to eliminate the negative exponent in the denominator move the term to the numerator
${m}^{10} {m}^{8}$
Now multiply the terms by adding the exponents
${m}^{10 + 8}$
${m}^{18}$
Jul 8, 2016
${m}^{18}$
#### Explanation:
Using the $\textcolor{b l u e}{\text{laws of exponents}}$
$\textcolor{\mathmr{and} a n \ge}{\text{Reminders}}$
•(a^m)^n=a^(mn)........ (1)
•a^m/a^n=a^(m-n)........ (2)
Using (1) : ${\left({m}^{5}\right)}^{2} = {m}^{5 \times 2} = {m}^{10}$
Using (2): m^(10)/m^(-8)=m^(10-(-8)=m^(10+8)=m^(18)<|endoftext|>
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# Divisibility Rules For 13
### Divisibility Rule
A non zero integer m divides an integer n provided that there is an integer q such that n = mq. We say that m is a divisor of n and m is a factor of n and use the notation m|n.
Divisibility rules are basically to solve problems related to integer division in a very easy way. Divisibility rule has come to check whether the dividend integer can be completely divided by any other divisor integer or not.
In order to check the divisibility of a large number by interest will take about time. That’s why to counter time divisibility rules were introduced. So, in this article, we are going to discuss divisibility rules for 13.
If adding four times the last digit to the number formed by remaining digits is divisible by 13, then the number is divisible by 13.
Apart from 13, there are divisibility rules for 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, and so on.
For Example:
Divisibility by 4 rule, 48 in a number which is completely divided by 4 as the sum of the last two digits of the number is divided by 4. Let’s find another number 47, which is not divisible by 4 as the sum of the last two digits of the number is not completely divided by 4. Using this simple rule, we can find if any number is divisible by 4 or not.
Now, let’s discuss the divisibility rule for 13, using definitions and examples.
Different Divisibility Rules For 13
We have to read 4 different types of divisibility rules for 13. Let us explain to you with examples one by one.
Divisibility Rule 1:
For a given number, form alternating sums of blocks of three numbers from the right and moving towards the left. Suppose (n1, n2, n3, n4, n5, n6…..) is a number N, then if the number formed by the alternative sum of blocks of 3-3 digits from right to left (n1, n2, n3, – n4, n5, n6, + …. ) is divisible by 13, then the number N is additionally divisible by 13.
Example: Let a number is 2,453,674. Find out whether it is divisible by 13 or not.
Solution: By applying Rule 1,
674 – 453 + 2 = 223 is not divisible by 13
Therefore, 2,453,674 also is not divisible by 13
Divisibility Rule 2:
If a number N is given, then multiply the last digit of N with 4 and add it to the rest truncate of the number. If the result is divisible by 13, then the number N is additionally divisible by 13.
Example: Let a number is 780. Find whether it is divisible by 13.
Solution: By applying Rule 2,
780: 78 + 0 x 4 = 78 and, number 78 is divisible by 13 and gives divisor as 6.
Therefore, 780 is also divisible by 13.
Divisibility Rule 3:
For a number N, to check whether it is divisible by 13 or not, subtract the last 2 digits of the number N from the 4 times multiple of the rest of the number.
Example: Let a number is 728. Check whether it is divisible by 13 or not.
Solution: By implementing the divisibility rule of 13, we get,
2197: 21 x 4 – 97 = 97 – 84 = 13, and number 13 is divisible by 13, giving the result as 0.
Divisibility Rule 4:
Multiply the last digit by 9 of a number N and subtract it from the rest of the number. If the result is divisible by 13, then the numeral N is also divisible by 13.
Example: If a number is 858 then find out whether it is divisible by 13 or not.
Solution: By applying rule 4,
936: 93 – 6 x 9 = 39, and 39 is divisible by 13
Therefore, 936 is divisible by 13.
### Questions:
Question 1.
(a). Is 298 Divisible by 13?
Ans. Four times of the last digit = 4 x 8 = 32
Remaining left 29
Addition = 29 + 32 = 61
Since 61 is not divisible by 13
298 is not divisible by 13.
(b). Is 247 Divisible by 13?
Ans. Four times of the last digit = 4 x 7 = 28
Remaining left 24
Addition = 24 + 28 = 52
Since 52 is divisible by 13
247 is divisible by 13.
(c). Is 317 Divisible by 13?
Ans. Four times of the last digit = 4 x 7 = 28
Remaining left 31
Now,
Addition = 28 + 31 = 59
Since 59 is not divisible by 13
317 is not divisible by 13.
(d). Is 50661 Divisible by 13?
Ans. Four times of the last digit = 4 x 1 = 4
Remaining left 5066
Now,
Addition = 5066 + 4 = 5070
Again, Four times of the last digit = 4 x 0 = 0
Now,
Addition = 507 + 0 = 507
Again,
Four times of the last digit = 4 x 7 = 28
Now,
Addition = 50 + 28 = 78
And,
78 is divisible by 13 at 13 x 6.
FAQs (Frequently Asked Questions)
Question 1. What is the Divisibility Rule for 12?
Ans. In order to find whether the number is divisible by 12 or not, you must have to find if the number is divisible by both 3 and 4. The number as we all know 12 is the product of numbers 3 and 4.
Question 2. What is the Rule of Divisibility by 23?
Ans. The divisibility rule for 23 is used to determine if a number is divisible by 23. It can also be known as the divisibility test for 23. To determine if a number is divisible by 23, take the digit at one place and multiply it by 7. Then add the result with the rest of the digits. Then, divide the sum by 23; if the sum is completely divisible by 23 then the number given is divisible by 23.<|endoftext|>
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Pedagogic Principles In The Primary Phase Teaching
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The main aim of pedagogy principle is to spread education so that suitable education can be provided to the respective concerned people. The present research study has been emphasizing on importance of pedagogy principle that promotes primary education standards which makes the teaching procedure more appropriate. Thus, the study is also emphasizing on several principles of pedagogy which are crucial to consider at the time of teaching the children. While teaching, it is crucial for the teachers to ensure that rights and obligations of children are being included in the teaching process (McDonald, Kazemi and Kavanagh, 2013). Several case studies have been accessed for the present research study for the purpose of reaching towards the defined aim and objectives.
Description has also been included regarding different research methodologies which have helped researcher to meet the aim and objective. Further, discussion has also been included regarding efforts of teachers while dealing with children’s education. The main aim of developing the present research study is to showcase the educational issues which teachers usually experience in school context. The value of teaching can only be enhanced at the time when teachers follow all the principles of pedagogy which makes things effectual and suitable (Kong and Song, 2013). Apart from this researcher has also emphasized on pedagogy principles which are most often applied in the teaching procedures.
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The basic principles of pedagogy are motivation, exposition, direction of activity, criticism and inviting imitation. However, these principles may overlap each other; but these principles usually assist teachers in appropriate manner. According to Laurillard, Charlton and Whittlestone, (2013), it is analysed that teachers are using ‘direction of activity’ principle wherein they make children through doing the things. This is the most important principle that is being considered by the teachers at the time of teaching students (Laurillard, Charlton and Whittlestone, 2013). However, at the same time, as per the school context, it is also ascertained that there are various teachers who merely focus on completing the part of teaching through explaining the things in accurate manner. This does not depict any sort of impact on children’s education; hence this impedes the value of education.
Adams (2016), also says that teachers usually consider ‘command and control’ aspects so as to include every part of the study. This also aims to include disciplinary aspects into account which helps the students to grow and prosper more. At the same time, it is also ascertained that learning activity delivery is yet another principle that is being designed for the purpose of managing teaching activities (Fontana and Setterfield, 2016). This basically focuses on facilitation in which delivery of education holds different importance. However, it is also been analysed that in the digital world, delivery can be largely automated in which each and every prospect of study is managed automatically. Apart from this, it can also be said that teachers use this principle because in practice, a good instructional process will represent a blending of several types of activities (Kreijns, Vermeulen and Acker, 2013).
According to Fontana and Setterfield, (2016) pedagogy principles supports for rendering superior quality education to children. Teachers teaching in school should make positive efforts for providing quality education for students. With this, there are five basic principles of pedagogy and they should be included while teaching in the classroom. Motivation, Exposition, Direction of activity, Criticism and Inviting limitation are major principles that are included among the principles of pedagogy. One of the most important principle is motivation and it is vital that children studying in the schools should be encouraged for grasping new knowledge that is being provided to them (Littlewood, 2014).
It depends on the personality of the teacher and capability of teacher for developing healthy relationship with the students. Van Manen, (2016) has mentioned that teachers should develop interest among students so that they can be directed towards acquiring new knowledge and experience. Better and positive outcomes from the educational programs could be achieved and encouraging children will support for enhancing existing experience of children. Moreover, by motivating children they can be made aware about great possibilities that are available for them in outside world.
Canagarajah, (2013) has defined that Exposition is second motivational principle that needs to be considered while teaching children in the classroom. With this, exposition is used for casting the students in a passive role by making use of this principle key points, facts and principles can be provided to the students. It is critical that ability for public performance should be there among teachers and they should also posses good subject knowledge so that they can deliver effective learning in the classroom (Canagarajah, 2013). With this, direction of activity is also one of the major principle that can be included while rendering knowledge to pupils. However, the design of activities that delivers specific objectives and for engaging children so that required objectives of classroom can be accomplished.
The principle of pedagogy can be applied in many contexts and fields such as for academic and professional teachings. Thus, in such respect, it can be said that according to the pedagogy principle, teachers are required to plan the entire curriculum so that suitable education can be delivered. There must be proper assessment of educational policies and procedures so that new things can be added to the teaching process. Focus should also be laid on academic writing so that capability of children can be enhanced.
Further, in this context it is also crucial for teachers to focus on pass, distinction and merit criteria so that student's capability and skills can be amended. According to pedagogy principle, teachers must also focus on demonstrating accurate criteria in all studies as this will have positive impact on student's capabilities.
Teachers must also focus on increasing the confidence level of students so that they can get maximum number in their studies. At the same time, pedagogy principle states that teachers are required to emphasize on student's writing and conceptual capabilities as this will help them to develop more in their career prospects. Similarly, in this way it can also be said that teachers are requisite to develop the critical thinking of students so that they can develop more. This is also essential for the purpose of enhancing student's core competencies.
Teachers managing education services in primary schools usually focus on diverse criteria so that children can get knowledge appropriately (Kumaravadivelu, 2016). At the same time, it is also analysed that teaching at primary phase typically requires new insights so that child management aspects can be managed. Along with the same, it can also be said that teachers usually experience various challenges and constraints at the time of dealing with children studying on primary level.
Research design can be described as a generalised plan that aids the researcher to get answer about the research question. Thus, according to the present study researcher has been emphasizing on exploratory technique under which case study method is being utilized in the research work. This research design does not aim to provide the final and conclusive answers to the research questions (Koh and Chai, 2014). Thus, accordingly it can be said that researcher has emphasized on effectiveness of pedagogy principle which helps teachers to strengthen the teaching process. In the present research study, researcher has made use of case study method in which several cases are being emphasized that focuses on principles of pedagogy.
Case study method is being used in the research report so that teachers can ensure that appropriate and accurate information is being given to the students in every respect. Moreover, it can also be said that researcher has also used the technique of observation so that the effectiveness of pedagogy principle in teaching can be found out (Van Manen, 2016). There are various factors that could affect the study procedure; hence to shed light on all such factors researcher focused on case study method. The key concept of teaching is also described in detailed manner along with its usefulness for students and their competency level. Further, with the help of observation, it is ascertained that pedagogy principles are being applied in teaching sector so that it can create positive impact on the teaching proceedings (Littlewood, 2014).
The most important section of the research study is research methodology in which researcher uses different tools and methods. This is typically done for the purpose of meeting the defined aim of the study. Therefore, looking towards the data collection aspects researcher has made use of both primary and secondary sources. This is essential in terms of collecting valid and accurate data about the study (Cunliffe, 2016). Data has been collected from several primary school teachers who are engaged in delivering suitable education to the children. Along with the same, secondary sources are also being used in the research study wherein focus has been laid on past studies regarding effectiveness of pedagogy principles in managing teaching procedure.
Several journals and articles are also being accessed for the purpose of ascertaining the validity and credibility of the research procedure (Carter and McCarthy, 2014). However, for the same aspect researcher has made inclusion and exclusion criteria so that on such basis, data can be collected easily and appropriately. The studies which are completed after 2010 are being included in the present research work; however studies made after 2010 are not included in the research process. Only English language is being used in the research study so that appropriateness and effectiveness of content can be enhanced. Defining the exclusion criteria, it can be said that researcher has not included any such article which is not made on the current topic.
Afterwards, researcher has also used different Qualitative method for the purpose of describing the findings of the research study in appropriate manner. The method has been included because that does not require application of statistical methods which are based on numbers, tables and charts. Similarly, in this context it can also be said that researcher has also made an implementation plan in which all the activities are properly scheduled (Carter and McCarthy, 2014). This is also essential to consider so that time scale can be followed to manage the research proceedings. Apart from this, it is also vital for the researcher to focus on such things because that aids in minimizing the errors and at the same time, it leads to bring more efficiency to the research study. Since, statistical tools are not being used in the research study; therefore researcher has not applied any such tool.
Apart from this, it can also be said that to specify the findings, proper evaluation is being carried out so that valid and feasible conclusion can be stated. This is also imperative in terms of enhancing success facets of the research method (Van Manen, 2016). On the basis of secondary analysis, researcher has discussed each and every element of the research work. This is also effective for showcasing the creditability and suitability of the research procedure. Moreover, in this context it can also be said that use of such methods helped the researcher to develop valid and feasible conclusion. The study seems to be highly important for further readers as that focuses on education processes and teaching effectiveness.
On the basis of research findings, it is identified that teachers usually experience various issues while delivering primary education to the children. This usually happens because the comprehensive ability of students are less in initial stages (James and Pollard, 2014). According to research findings, it can be said that teachers focus chiefly on developing their skills and abilities so that appropriate knowledge and education can be delivered to the respective children. Moreover, in this respect it can also be articulated that doing and experimenting things aids the teachers to enhance the learning and knowledge aspects of the students. At the same time, it is also vital for teachers to comprehend the knowledge level of children and on that basis only, teachers must adopt necessary principles so that the level of understanding can be enhanced (Adams, 2016).
At the same time, it is also analysed that teachers need to comprehend the skills and abilities so that on such basis value of education can be encouraged. Apart from this, it is also analysed that teachers can overcome the challenges through adopting numerous methods and this can also aid in meeting the requirements of teaching procedure (Wette, 2014). This is also analysed that teachers need to be more capable in terms of understanding diverse aspects of learning and education. According to school context, teachers are required to focus on primary education tools so that they can uphold the value of education in successful manner.
Pedagogy being an art of teaching with five leading principles that greatly assists the teachers to handle the basics of primary education (Richards and Rodgers, 2014). It is on the basis of the findings acquired form the above conducted research, teachers at their initial stage of joining are mostly concerned about their job responsibilities and find it hard to handle the small kids who are the students of primary classes. Being unable to manage them in an effective manner, they often lost their control and intends to leave this occupation by undertaking it as a failure of their own (Kong and Song, 2013).
However, the current research work has greatly assisted in acknowledging the related principles of pedagogy where it is mainly composed of some elementary principles that largely supports the teachers to effectively deal with such concerning situation (Danielewicz, 2014). A foremost principle of motivation is apparent to assist the teachers to conduct instructive sessions to educate children. In addition to which, a vital presence of this factor will also aid the teachers to effectively engage the students in the educational session which is specially arranged for them. Another principle of exposition is evident to reflect as another effective approach by which the students can be taught in a profitable manner (Laurillard, Charlton and Whittlestone, 2013). This is basically in context to its less expensive requirements where the students are usually allotted a passive role to perform.
From the research findings, it has been evaluated that teaching students and specially children is quite a tough job. There has to be a follow up which needs to be considered by the teachers or every prospective individual that is making other people learn (Kong and Song, 2013). The pedagogic principles of teaching should be applied in the education so that students or children are able to learn in a more effective manner (Fontana and Setterfield, 2016). Often it happens that teacher is quite efficient in handling the students but there is no as such improvement in the performance. In these situations when principles are integrated with the teaching attributes then students are more comfortable. The use of teaching ethics and pedagogic principles helps in better learning and effective delivery of the information that is to be taught to the students (McDonald, Kazemi and Kavanagh, 2013).
Concluding the entire research process, it can be said that it is essential for teachers to consider pedagogy principles so that every aspect of education can be managed in effectual manner. Moreover, in this context it is also identified that there must be suitable teaching processes for encouraging the value of primary education. At primary phase, children are unable to comprehend anything; therefore it is essential for the teachers to adopt necessary practice so that to make the process of study more effectual and comprehensive. Apart from this, it is also evident that teachers must emphasize majorly on teaching skills and processes which gives a positive impact on children’s capability and learning skills. However, at initial stage, it is vital for teachers to ensure that every aspect of teaching is being included for meeting the appropriate standard of teaching.
Type: Assignment I
Published: 10 Apr ,2019 I
1. Discussing the research questions used in the study PICO process is the technique which is usually used in evidence-based practice for answering the clinical or health-related questions. In this, PICO stands for patient, intervention, comparison, and outcomes. According to the PICO framework...ReadMore
Type: Case Study I
Published: 29 Mar ,2019 I
Introduction It has been reviewed from the study that students of the age group of 3 years are facing issues related to reading comprehension. Teachers need to plan proper literacy framework for students because implementing an effective literacy program helps students to pay more attention on...ReadMoreView or Download
Type: Minor Case Study I
Published: 28 Mar ,2019 I
Introduction to Professional Practice Professional practice is referred to be a fundamental concern for a scholar who is intending to adopt the practical environment of learning as a way of extending both personal and professional skills and abilities. This will in turn not only result in the...ReadMoreView or Download
Published: 13 Mar ,2019 I
CREATIVE WRITING OF THE IMAGE Essay - This specific place has been used for the purpose of treating a boy (Philip) who is experiencing Alzheimer’s disease which is a type of dementia that changes memory, thinking and behaviour of a person. Thus, firstly it is essential to state about...ReadMoreView or Download
Type: Case Study I
Published: 16 Feb ,2019 I
Introduction In a world where all things has gone online or we could say that it is a era where internet has acquired almost all our work. We all share some sort of information on it which is termed as data so where does this data get stored. Cloud Computing is a term which is a internet based...ReadMore
If the samples impressed you and you want to place an order with us just submit your queries with us and place your order.Chat With Expert
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What determines sex? At first glance, it appears to be a simple matter of X and Y chromosomes (at least in mammals). But the physical and behavioral differences between the sexes involves a complicated interplay between genetics and the environment that can be constantly reassessed through an animal's life. Some startling findings published yesterday in Nature reveals that, in mice at least, some of the most basic aspects of sexual behavior are imposed by the environment.
The study built on previous work regarding olfaction in mice. Most of the odorant receptors in mice are expressed for the detection of smells in their surroundings. But mice also have a dedicated structure, the vomeronasal organ, that expresses receptors for detecting pheromones, the molecules many species use to help send signals among individuals.
Previous studies had shown that pheromone reception in the vomeronasal organ was essential for mice to recognize where to direct some of their sexual behavior. Male mice that lack a key ion channel (Trpc2) that's used in nerve cells to trigger pheromone signaling appeared incapable of realizing they were facing another male: they'd attempt to mate with any other mouse placed in the cage with them, regardless of whether the mouse was male or female.
The new results look at females lacking Trpc2, and the data reveals that the situation is probably more complex than the earlier results has suggested. These female mice also attempted to mate with anything placed in the cage with them but, in an unexpected twist, tried to mate as males. It was not (or at least, not only) that they couldn't identify the new mouse's sex—without the pheromone input, they apparently defaulted to male behavior. Those of you who are interested in mouse porn can check out the paper's supplemental data, which has videos of their behavior, with the genotype of the mouse conveniently color-coded via their fur color.
The behavioral changes were extensive, including presexual courtship such as the ultrasonic songs normally made by male mice. Male mice could still mate with the mutant females, and those females successfully raised their young. They appeared somewhat neglectful, but it's thought that female mice normally bond with their young in part via pheromones, so this may have nothing to do with sexual identity.
Since the Trpc2 gene was missing throughout the mouse's development, it wasn't clear whether the changes in mating behavior were the product of the mutant vomeronasal organ imposing behavior on the brain or the inability to form proper connections between the two during earlier development. To discriminate between these two possibilities, the researchers let normal mice develop into adults, and then surgically removed the vomeronasal organ. These mice also attempted to mate with anything present as a male.
Apparently, the mouse's brain has no hard-wired sexual behavior. It simply does what the pheromones tell it to do and, in their absence, defaults to a male mating approach. In anticipation of the obvious question, humans do appear to have a limited ability to respond to pheromones, but it remains unclear whether they have a vomeronasal equivalent at all. Still, these results are a healthy reminder that sexual behavior may not be the "genes vs. choice" issue that it's sometimes made out to be. It's clear that, for the mice that lack a vomeronasal organ, sexual attraction isn't a matter of the genes, but it sure isn't a choice, either.<|endoftext|>
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The month of September is Disability Awareness Month.
The Rights of People with Disabilities Unit in the DWCPD has just completed a global brainstorming exercise for UNICEF to develop a child-friendly definition of disability (targeting 10 year olds). This exercise goes to the heart of the new conceptualisation of disability as a social construct. This is how you could explain to your children what disability is!
All children are different, but this is what makes children so interesting. Some are tall, some are small, some are black, some are white, some have disabilities, and children speak all different kind of languages.
Children with disabilities are children like all other children, but they do things a bit differently.
A child who cannot see, for example read by using her fingers to read dotted letters called braille.
A child who cannot hear or speak, uses his hands to sign the words and sentences.
Some children move around with crutches, special shoes or wheelchairs, as it is difficult for them to walk and run. Some children take a bit longer to understand, but if we give them time and use simple language, they can learn and do things with other children.
If we remember that each one of us is different, we all will find different ways of playing together
So it is important for us to find playgrounds where wheelchairs can move around easily.
If we all learn to use our hands when we speak, we can all understand each other.
And if we explain to children who cannot see or understand what is happening around them and what the rules of our games are, they can also play with us.
Childline South Africa offers a counselling for Deaf and Mute Children who cannot call us on our Crisisline. Visit our page on our Online Counselling Service now for more information!!!<|endoftext|>
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Chapter 13 Class 8 Introduction to Graphs
Concept wise
### Transcript
Example 2 The given graph (Fig 15.8) describes the distances of a car from a city P at different times when it is travelling from city P to city Q, which are 350km apart. Study the graph and answer the following: (i) What information is given on the two axes? x-axis shows the Time (in hour) y-axis shows the Distance of car from City P (in km) (ii) From where and when did the car begin its journey ? He started his journey from point P at 8 am (iii) How far did the car go in the first hour? Distance travelled from 8 am to 9 am = 50 − 0 = 50 km In 1st hour, the car went 50 km. (ii) From where and when did the car begin its journey ? He started his journey from point P at 8 am (iii) How far did the car go in the first hour? Distance travelled from 8 am to 9 am = 50 − 0 = 50 km In 1st hour, the car went 50 km. (ii) From where and when did the car begin its journey ? He started his journey from point P at 8 am(iii) How far did the car go in the first hour? Distance travelled from 8 am to 9 am = 50 − 0 = 50 km In 1st hour, the car went 50 km. (iv) How far did the car go during (i) the 2nd hour? (ii) the 3rd hour? Distance travelled from 9 am to 10 am Distance travelled from 10 am to 11 am = 150 − 50 = 100 km = 200 − 150 = 50 km In 2nd hour, the car went 100 km, and in 3rd hour, car went 50 km. (v) Was the speed same during the first three hours? How do you know it? We see that In 1st hour, car travels 50km. In 2nd hour, it travels 100 km In 3rd hour, it travels 50 km Since it travels different distances in 1 hour, Its speed is not same in the first three hours (vi) Did the car stop for some duration at any place? Justify your answer.Between 11 am and 12 noon, there is a horizontal line in the graph Thus, the distance from point P is the same. ∴ Car stopped from 11 am to 12 noon (vii) When did the car reach city Q?City Q is 350 km from City P. Car travelled 350 km at 2 pm<|endoftext|>
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A Chinese naval paddleboat of the twelfth century, propelled by muscle power rather than steam. This kind of vessel was for use on the inland waterways and lakes of China. The attention of the Chinese authorities was generally directed towards maintaining the vast internal economic and military networks. The concept of the paddleboat was known in the Roman Empire, but it was impractical for general use in the Mediterranean.
During the Song Dynasty there was also great amount of attention given to the building of efficient automotive vessels known as paddle wheel craft. The latter had been known in China perhaps since the 5th century, and certainly by the Tang Dynasty in 784 with the successful paddle wheel warship design of Li Gao. In 1134 the Deputy Transport Commissioner of Zhejiang, Wu Ge, had paddle wheel warships constructed with a total of nine wheels and others with thirteen wheels. However, there were paddle wheel ships in the Song that were so large that 12 wheels were featured on each side of the vessel. In 1135 the famous general Yue Fei (1103–1142) ambushed a force of rebels under Yang Yao, entangling their paddle wheel craft by filling a lake with floating weeds and rotting logs, thus allowing them to board their ships and gain a strategic victory. In 1161, gunpowder bombs and paddle wheel crafts were used effectively by the Song Chinese in the Battle of Tangdao and the Battle of Caishi against the Jurchen Jin Dynasty, who made an unsuccessful invasion of the Southern Song along the Yangtze River. In 1183 the Nanjing naval commander Chen Tang was given a reward for constructing ninety paddle wheel craft and other warships. In 1176, Emperor Xiaozong of Song (r. 1162–1189) issued an imperial order to the Nanjing official Guo Gang (who desired to convert damaged paddle wheel craft into junk ships and galleys) not to limit the number of paddle wheel craft in the navy's dockyards, since he had high esteem for the fast assault craft that won the Chinese victory at Caishi. However, paddle wheel craft found other uses besides effective assaults in warfare. The Arab or Persian Commissioner of Merchant Shipping for Quanzhou, the Muslim Pu Shougeng (who served from 1250–1275) noted that paddle wheel ships were also used by the Chinese as tugboats for towing.<|endoftext|>
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This month’s Outside the Box takes a look at seaweed. “What’s too look at?” you might ask, some people see food, others see something kind of slimy and a few see a new source of biofuel. To read more about this interesting application of seaweed click here.Seaweed has been used as a staple in human diets in Japan, Korea and China since prehistoric times. There are some 21 species used daily in Japan, and seaweed accounts for close to 10% of Japanese diets. Sometimes called Kelp, seaweed grows rapidly in underwater forests on shallow ocean bottoms. Kelp has high levels of iron, calcium, magnesium, potassium and several trace minerals, and is therefore considered a multi-vitamin. Alongside its nutritional value, what else could seaweeds be useful for? Some think that it could be a viable source of renewable energy.
One of the biggest criticisms aimed at biofuels is that are derived from crops and thus take valuable land away from food production. Bio Architecture Lab (BAL) claims to have developed a technology that makes seaweed a cost-effective source of energy. The process would extract all the major sugars in seaweed and then convert them into renewable fuels and chemicals. BAL says that it will take less than 3% of the global coastal waters to produce enough seaweed to replace 60 billion gallons of fossil fuels annually – to put this in perspective, the United States used 171 billion gallons of fossil fuels in 2005 just for our transportation needs.
Researchers from Israel believe that producing biofuel from seaweed-based sources could solve already existing problems in the marine environment. Many coastal regions suffer from eutrophication due to pollution caused by human waste, agriculture and fish farming. Eutrophication leads to elevated levels of nutrients and harmful algae which can be very damaging to coral reefs. By employing multiple varieties of seaweeds which capture and bio-accumulate pollutants, eutrophication could be reduced in shallow water ecosystems.
One of seaweed’s most abundant sugars is alginate – a polymer that can’t be converted to ethanol. However, scientists at BAL believe that when combined with diced fresh seaweed, a fermentation process would take place that allows for the necessary transformation to take place. Thierry Chopin, at the University of New Brunswick, thinks that this process may not be as cost-efficient as others believe. He is also concerned that you would not be able to harvest year-round.
BAL is quick to respond, saying that they don’t expect it to solve the world’s oil problems, only to make a considerable contribution. Their purpose is to reduce our carbon footprint and for the technology to work in tandem with other alternative energy solutions such as wind and solar. Dutch company Ecofys is currently engaged in a project to extract seaweed biofuels from their coastline by late June 2012. They believe that seaweed and wind turbines are a good combination. When the wind farms are closed for shipping and commercial fishing operations, the seaweed farms can provide proteins for fish and livestock feed and can be used for the creation of energy.
While this is an energy source in the making, it is a promising sign for the future of renewable energy sources. As traditional and alternative energy firms continue to establish new avenues of power generation, it will allow us to reduce our dependence on oil and gas as well as to learn how to use these resources more efficiently. There is more than 200,000 miles of coastline in the world, a distance roughly equal to that from the Earth to the Moon. With the potential area for aquatic agriculture fields throughout the world, we can have high hopes that this technology will be mastered and provide for a great deal of our future energy needs.<|endoftext|>
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Sixty-five years ago, the Supreme Court declared that segregated public schools were “inherently unequal” and unconstitutional, smashing a 1896 ruling that permitted “whites-only” and “Negroes-only” schools. The historic Brown vs. Board of Education decision ordered that public schools must be integrated, launching a decades-long struggle to end systemic inequality in American life.
After all these years, a new report says that while Brown vs. Board may have led to desegregation in other parts of American society, it has been unsuccessful in its stated mission: to integrate public schools.
Now, the promise of the ruling is “at grave risk,” according to the report titled “Harming our Common Future: America’s Segregated Schools 65 Years after Brown.”
The report was issued by the Civil Rights Project at UCLA and the Center for Education and Civil Rights at Pennsylvania State University, with input from researchers at Loyola Marymount University and North Carolina State University.
It says that while intense levels of segregation markedly decreased for black students after the 1954 court ruling, they have been rising again since Supreme Court decisions in the 1990s led to the end of hundreds of desegregation orders and plans across the country. It says:
The growth of racial and economic segregation that began then has now continued unchecked for nearly three decades, placing the promise of Brown at grave risk.
These trends matter for students, and for communities whose futures are determined by how the public schools prepare their students for a diverse future. Research shows that segregation has strong, negative relationships with the achievement, college success, long-term employment and income of students of color.
The report measures segregation based on how much exposure groups of students have to each other, regardless of the demographics of the districts in which they live. It says:
- Public school enrollment stands at nearly 50 million. White students are less than half of the student population: 48.4 percent in 2016. Latinos were 26.3 percent of the student population; blacks, 15.2 percent; Asians, 5.5 percent; multiracial, 3.6 percent; and American Indian, 1 percent.
- Despite the increase in diversity, segregation has intensified and expanded. Over the last three decades, black students have been increasingly segregated in intensely segregated schools (which are defined as being 90 to 100 percent nonwhite). By 2016, 40 percent of all black students were in schools with 90 percent or more students of color. New York, California, Illinois and Maryland are the four states in which a majority of black students attend intensely segregated schools.
- New York remains the most segregated state for African American students, with 65 percent of African American students in intensely segregated schools. California is the most segregated for Latinos, with 58 percent of those students attending intensely segregated schools. The typical Latino student attends a school in which only 15 percent of students are white.
- White students continue to attend schools in which nearly seven out of 10 classmates are also white, a much higher percentage than their overall share of the enrollment.
- In 2016, Latino students on average attended a school in which 55 percent of the students were Latino. The states in which they were most likely to attend intensely segregated schools were California, New Jersey, New York and Texas.
- In suburban schools in the country’s largest metropolitan areas, 47 percent of students were white in 2016, a 10 percentage point decline in a decade. About one-seventh of suburban students were black, and more than one-fourth were Latino. Considerable segregation exists within the suburbs, where African American and Latino students typically attended schools that were about three-fourths nonwhite. White students in these same large suburbs attended schools where, on average, two-thirds of the enrolled students were white.
- In rural schools, the typical white student went to a campus on which 80 percent of students were white, while the typical black or Latino student went to a rural school with 57 percent nonwhite enrollment.
- Charter schools, which are publicly funded but privately operated, are even more segregated than schools in the traditional public systems. Typically, they have no integration policies. School choice plans without policies to ensure equity often end up with the most-connected and best-educated families getting the best choices, actually increasing inequality.
The report makes recommendations for policymakers, including:
- Creating a better way to measure economic segregation and its connection to racial segregation in schools.
- Creating and supporting diverse educational institutions inside polarized communities.
- Training for school faculties, leaders and staffs to respond to growing racial, economic and linguistic diversity.
Richard Rothstein, a longtime research associate at the Economic Policy Institute who has written extensively on segregation in American society, has said housing policies are actually the most important school policies. On the 60th anniversary of Brown v. Board, he wrote:
Schools remain segregated today because neighborhoods in which they are located are segregated. Raising achievement of low-income black children requires residential integration, from which school integration can follow. Education policy is housing policy.
Federal requirements that communities must pursue residential integration have been unenforced, and federal programs to subsidize movement of low-income families to middle-class communities have been weak and ineffective.
Correcting these policy shortcomings is essential if the promise of Brown is to be fulfilled.
Concern about the lack of progress with school desegregation was sounded in Congress last month by Rep. Robert C. “Bobby” Scott (D-Va.), chairman of the House Committee on Education and Labor, who said: “After four decades without federal support for desegregation, we are right back where we started.”<|endoftext|>
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Read the problem situations below. Be sure to include your thinking!a) Ethan and his three brothers plan to mow their yard, which is 1/3 of an acre. How much of an acre will each of the 4 boys mow? Explain your thinking.
b) Candy has 4 boards to use for a project. The plan calls for boards 1/3 the size of the existing boards. How many of the one-third sized pieces will Candy have? Explain your thinking.
Dividing Unit Fractions by a Whole Number
Click on the following links for interactive games.
Click on the following links for more information.
5.3 Number and operations. The student applies mathematical process standards to develop and use strategies and methods for positive rational number computations in order to solve problems with efficiency and accuracy. The student is expected to:<|endoftext|>
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# How do you solve the problem of a rocket with variable mass?
## Homework Statement
Rockets are propelled by the momentum of the exhaust gases expelled from the tail. Since these gases arise from the reaction of the fuels carried in the rocket, the mass of the rocket is not constant, but decreases as the fuel is expended. Show that for a Rocket starting initially from rest, and taking the velocity of the exhaust gases relative to the rocket, ##v'## = 2.1 m/s and a rate of mass loss per second L = 1/60 of the initial mass, to reach the escape velocity of the Earth (##v_e## = 11.2 km/s), the ratio of the weight of the fuel to the weight of the rocket must be almost 300!
## Homework Equations
I know that ##F=m(\frac {dp} {dt})=ma=m\frac {dv} {dt}##
## The Attempt at a Solution
First, I work in the frame of the rocket; so we have two forces, the weight ##W=mg## and the force of the exhaust exiting the rocket ##\frac {dm} {dt}v'##. Both forces are pointing downward, thus I set up the following equation: $$ma=m\frac {dv} {dt}=-v'(\frac {dm} {dt})-mg$$ Now I rewrite ##m\frac {dv} {dt}## as ##m\frac {dv} {dm}\frac {dm} {dt}##; now I have this expression:$$m\frac {dv} {dm}\frac {dm} {dt}=-v'(\frac {dm} {dt})-mg$$As ##\frac {dm} {dt}=L=-\frac {m_0} {60}## and dividing both sides by ##m## we have:$$\frac {dv} {dm}(-\frac {m_0} {60})=-\frac {v'} {m}(-\frac {m_0} {60})-g$$Dividing both sides by ##L## yields:$$\frac {dv} {dm}=-\frac {v'} {m}+\frac {60g} {m_0}$$Now we're ready to write our differential equation as follows:$$dv=-\frac {v'} {m}~dm+\frac {60g} {m_0}~dm$$We can now integrate, this knowing that we start from ##v_0=0## and from a certain initial mass given by ##m_0=m_{rocket}+m_{fuel}## and a final mass ##m_f=m_{rocket}##$$\int_0^v dv=-v'\int_{m_0}^{m_f} \frac {1} {m} \ dm+\frac {60g} {m_0}\int_{m_0}^{m_f} dm$$After integrating and evaluating I get the following expression:$$v=-v'ln(\frac {m_f} {m_0})+\frac {60g} {m_0}(m_f-m_0)$$This can be rewritten as:$$v=-v'ln(\frac {m_{rocket}} {m_{rocket}+m_{fuel}})-\frac {60g} {m_{rocket}+m_{fuel}}(m_{fuel})$$Then, if ##m_{fuel}>>m_{rocket}## we can ignore the term ##m_{rocket}## in ##\frac {1} {m_{rocket}+m_{fuel}}##, and the expression can be rewritten as:$$v=-v'ln(\frac {m_{rocket}} {m_{fuel}})-60g=v'ln(\frac {m_{fuel}} {m_{rocket}})-60g$$Solving for the desired ratio I get:$$\frac {m_{fuel}} {m_{rocket}}=e^{\frac {v+60g} {v'}}$$However when plugging in the equation the values ##v'=2.1 m/s##, ##v=11.2 km/s##, and ##g## the value is much, much bigger than 300.
What did I did wrong? Did I chooses the incorrect frame? To my understanding my equation ##ma=m\frac {dv} {dt}=-v'(\frac {dm} {dt})-mg## should be true for this frame, thus ##v'## indeed is 2.1 m/s
Thank you for your time! As always I appreciate detailed answers, thanks again!
Related Calculus and Beyond Homework Help News on Phys.org
andrewkirk
Homework Helper
Gold Member
As soon as I read the problem, the figure of 2.1 m/s for the gas velocity struck me as ridiculously low. Rockets accelerate really fast, so the velocity of the gases they emit must be several times faster than that. I suspect the figure is supposed to be 2.1 km/s.
A quick check with wiki seems to corroborate this. This article says the exhaust gases travel at hypersonic velocities, which apparently means velocities of Mach 5 and above. 2.1 km/s is about Mach 7.
2.1m/s is a slow walk, which is not going to send a big cylinder of metal anywhere!
EDIT: I forgot to put the link to the article on rocket engines. I've put it in now.
Last edited:
As soon as I read the problem, the figure of 2.1 m/s for the gas velocity struck me as ridiculously low. Rockets accelerate really fast, so the velocity of the gases they emit must be several times faster than that. I suspect the figure is supposed to be 2.1 km/s.
A quick check with wiki seems to corroborate this. This article says the exhaust gases travel at hypersonic velocities, which apparently means velocities of Mach 5 and above. 2.1 km/s is about Mach 7.
2.1m/s is a slow walk, which is not going to send a big cylinder of metal anywhere!
Do you think that, even when considering that ##v'## is relative to the rocket (i.e., measured by someone in the rocket, not an observer on ground), 2.1 m/s would still, anyway, be too small?
andrewkirk
Homework Helper
Gold Member
Do you think that, even when considering that v' is relative to the rocket
Yes, and that co-moving frame of the rocket is the natural one to use. Note that that is the same as the Launchpad frame when the countdown hits zero.
2.1 m/s is a waft, not a rocket blast. It would be overstating it to call it a gentle breeze.
Ray Vickson<|endoftext|>
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# Finite Sets
## Definition of Finite Sets
The meaning of the word ‘Finite’ is ‘having a definite limit or fixed size’. Therefore, the Finite Sets may be defined as the sets that have a definite number of elements or a countable number of elements in them.
In other words, a set $A$ is said to be a Finite Set if it has a definite or countable number of elements present in it. Therefore, Finite Sets have a finite number of elements. Also, the Cardinality of a Finite Set is Finite.
## Finite Set Examples
The above figure shows the $3$ sets $A, B$ and $C$. The elements of the three sets are,
• $A = \{1, 2, 3, 4, 5\}$
• $B = \{b, c, d, f\}$
• $C = \{1, 2, 3, ......., 30\}$
All the sets $A, B$ and $C$ in the above examples have a finite number of elements. The set $A$ has $5$ elements, the set $B$ has $4$ elements and the set $C$ has $30$ elements. Therefore, the above sets are Finite Sets.
## Practical Examples of Finite Sets
• Set of all the months in a year.
We are well aware that there are 12 months in a year i.e., January, February, March, April, May, June, July, August, September, October, November and December. Therefore, the set of all the months in a year has a definite number of elements i.e., 12 and hence it is a Finite Set.
• Set of all the days in a week.
Similarly, there are seven days in a week viz., Sunday, Monday, Tuesday, Wednessday, Thursday, Friday and Saturday. Therefore, the set of all the days in a week has a definite number of elements that is 7 and hence it is a Finite Set.
• Set of all the planets in our solar system.
The set of all the planets in our Solar System is also Finite Set because the solar system has a Finite number (8) of planets viz., Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune.
• Set of all the elements in the periodic table.
The set of all the elements in the Periodic Table is Finite Set because it has 118 number of elements.
• Set of all the books for class 10.
## Properties of a Finite Set
The properties of a Finite sets are as follows:
• The Subset of a Finite Set is always a Finite Set.
• The Superset of a Finite Set may be finite or infinite.
• The Union of two Finite Sets is always a finite set.
### What is a Finite Set?
A set A is said to be finite if it has a countable number of elements present in the list or finite sets have a finite number of elements.
### What is the difference between Finite and Infinite Set?
The main difference between the Finite and Infinite set is that the Finite sets have a definite or countable number of elements, however, the Infinite Sets have an indefinite or uncountable Number of elements.
Also, the Cardinality of a Finite set is definite. However, the Cardinality of an Infinite Set is not defined.
### Is 0 finite or infinite?
0 is neither finite nor infinite.
### What is finite set examples?
The example of some finite sets are:
1. Set of all the natural numbers which are less than 10.
2. Set of all the whole numbers which are less than 10.
3. Set of all the integers which are greater than -5 and less than 5.
4. Set of all the prime numbers less than 50.
### Which of the following is a finite set? 1. Set of all the natural numbers 2. Set of all the positive integers
None of the above sets is a finite set because, both the sets have infinite number of elements. The set of all the natural numbers starts with 1 and ends in infinity. Similarly, the set of all the positive integers starts with 1 and ends in infinity.
## Quiz on Finite Sets
1.
Which of the following set is a finite set?
2.
Finite sets have..
3.
The set of all the months in a year is a..
4.
The set of all the planets in our solar system is an...<|endoftext|>
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# Tag Archives: Recursive Formula
## Recursive Discrete Aggregate Loss
You have 2 six-sided dice. You roll one dice to determine the number of times you will roll the second dice. The sum of the results of each roll of the second dice is the amount of aggregate loss. Since the frequency and severity are discrete, for any aggregate loss amount, the number of combinations of rolls to produce such an amount is clearly countable and finite. For example, an aggregate loss amount of 3 can be arrived at by rolling a 1 on the first dice, then rolling a 3; or rolling a 2, then rolling the combinations (1,2),(2,1); or rolling a 3 and then rolling (1,1,1) on the second dice. The probability of experiencing an aggregate loss of 3 is:
$\begin{array}{rll} \Pr(S=3) \displaystyle &=& \frac{1}{6^2} + \frac{2}{6^3} + \frac{1}{6^4} \\ \\ \displaystyle &=& \frac{49}{6^4} \end{array}$
This method of calculating the probability is called the convolution method. Now imagine the frequency and severity distributions are discrete but infinite. To calculate $\Pr(S=10)$ would require calculating the probability for many possible combinations. If the discrete functions are from the (a,b,0) class, there is a recursive formula that can calculate this. It is given by:
$g_k = \displaystyle \frac{1}{1-af_0}\sum_{j=1}^k \left(a+\frac{bj}{k}\right)f_jg_{k-j}$
where $k$ is an integer, $g_k = \Pr(S=n)=f_S(n)$, $f_n = \Pr(X=n)$, and $p_n = \Pr(N=n)$. This is called the recursive method. To start the recursion, you need to find $g_0$. You can then find any $g_k$. If a problem asks for $F_S(3)$, this is equal to $g_0+g_1+g_2+g_3$. You iterate through the recursion to find each $g_k$ then add them together.<|endoftext|>
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What’s in this article
- What is Gaucher disease?
- When do you require liver surgery?
- Is the disease contagious?
- Treatment and prevention
What is Gaucher disease?
People who suffer from this disease have a lack of glucocerebrosidase enzyme (This enzyme is active in lysosomes, which are structures inside cells that act as recycling centers).
Without the glucocerebrosidase enzyme, excessive amounts of fat accumulate in the liver, lungs, bones, spleen, and, sometimes in the brain. Excess fat in these organs stops them from functioning properly.
There are three different types of Gaucher disease:
- Type I – this is the most common form of Gaucher disease and is not as serious as the other types. According to the Liver Directory type I accounts for around 90% of all cases. It causes liver and spleen enlargement, pain in the bones, broken bones, and kidney or lung problems. Type I can occur at any age to anyone.
- Type II – This often occurs in infants and is a more serious form of the disease which affects the brain. Type II Gaucher disease can cause serious brain damage or even death.
- Type III – This type of Gaucher disease usually appears in childhood and can cause liver, spleen, and gradual brain enlargement.
What causes Gaucher disease?
Gaucher disease is an inherited disease in an autosomal recessive manner (where two copies of an abnormal gene must be present in order for the disease or trait to develop). Normally, the body has two copies of the gene that provide instructions for making the enzyme, glucocerebrosidase. For most people, both genes work properly. When one of the two genes is not functioning properly, the person is a carrier of Gaucher disease.
Signs & symptoms
According to the Mayo Clinic, signs and symptoms of Gaucher disease vary widely. Siblings with the disease can have different levels of severity. Some people who have Gaucher disease have only mild or no symptoms at all, if they do they could experience the following:
- Low platelet count that can lead to easy bruising
- Enlarged spleen and liver (hepatosplenomegaly)
- Easy bleeding that is difficult to stop
- Bone pain and the chance of easily broken bones
- Delayed puberty
Treatment and Prevention
Patients suffering from type I of the disease may only suffer from mild symptoms, therefore will not need any treatment. Unfortunately for those suffering from type II, there is not yet any treatment available. Sufferers of type I and III may be offered Enzyme replacement therapy. Enzyme replacement therapy can often help to reduce liver and spleen enlargement, blood abnormalities and help to improve bone density. It works by having an IV that feeds an artificial enzyme replacement into the body. This is done in two-week intervals and is most effective for type I Gaucher disease.
There are also medications available to treat Gaucher disease, however, there are many side effects that come with this treatment so it is usually not preferred by patients.
People who suffer from severe types of this disease my need to undergo a bone marrow transplant which may help to reverse many of the symptoms, however, this is a high-risk procedure.
Find out more about the basics of The Basics of Liver Disease & Liver Functions with Dr. Tarek Hassanein, M.D.<|endoftext|>
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# Cosc 2150: Computer Organization Chapter 3: Boolean Algebra and Digital Logic.
## Presentation on theme: "Cosc 2150: Computer Organization Chapter 3: Boolean Algebra and Digital Logic."— Presentation transcript:
Cosc 2150: Computer Organization Chapter 3: Boolean Algebra and Digital Logic
Objectives Understand the relationship between Boolean logic and digital computer circuits. Learn how to design simple logic circuits. Understand how digital circuits work together to form complex computer systems.
Introduction In the latter part of the nineteenth century, George Boole incensed philosophers and mathematicians alike when he suggested that logical thought could be represented through mathematical equations. —How dare anyone suggest that human thought could be encapsulated and manipulated like an algebraic formula? Computers, as we know them today, are implementations of Boole’s Laws of Thought. —John Atanasoff and Claude Shannon were among the first to see this connection.
Introduction In the middle of the twentieth century, computers were commonly known as “thinking machines” and “electronic brains.” —Many people were fearful of them. Nowadays, we rarely ponder the relationship between electronic digital computers and human logic. Computers are accepted as part of our lives. —Many people, however, are still fearful of them. In this chapter, you will learn the simplicity that constitutes the essence of the machine.
Boolean Algebra Boolean algebra is a mathematical system for the manipulation of variables that can have one of two values. —In formal logic, these values are “true” and “false.” —In digital systems, these values are “on” and “off,” 1 and 0, or “high” and “low.” Boolean expressions are created by performing operations on Boolean variables. —Common Boolean operators include AND, OR, and NOT.
A Boolean operator can be completely described using a truth table. The truth table for the Boolean operators AND and OR are shown at the right. The AND operator is also known as a Boolean product. The OR operator is the Boolean sum. Boolean Algebra
The truth table for the Boolean NOT operator is shown at the right. The NOT operation is most often designated by an overbar. It is sometimes indicated by a prime mark ( ‘ ) or an “elbow” ( ). Boolean Algebra
A Boolean function has: —At least one Boolean variable, —At least one Boolean operator, and —At least one input from the set {0,1}. It produces an output that is also a member of the set {0,1}. Now you know why the binary numbering system is so handy in digital systems.
The truth table for the Boolean function: is shown at the right. To make evaluation of the Boolean function easier, the truth table contains extra (shaded) columns to hold evaluations of subparts of the function. Boolean Algebra
As with common arithmetic, Boolean operations have rules of precedence. The NOT operator has highest priority, followed by AND and then OR. This is how we chose the (shaded) function subparts in our table. Boolean Algebra
Digital computers contain circuits that implement Boolean functions. The simpler that we can make a Boolean function, the smaller the circuit that will result. —Simpler circuits are cheaper to build, consume less power, and run faster than complex circuits. For the purposes of this course, we will deal only with And, Or, and Not gates —While there are more interesting gates, such as XOR, NOR, and NAND gates.
The three simplest gates are the AND, OR, and NOT gates. They correspond directly to their respective Boolean operations, as you can see by their truth tables. Logic Gates
There is one other graphical convention The NOT gate is sometimes reduced to a small circle attached to some other gate, so
Combinational Circuit A Combinational Circuit has no memory. Its outputs depend only on its inputs —not on a value stored inside the circuit —Each input (and each output) is either 1 or 0. To specify its behavior, we use a truth table Example —Imagine a circuit with 3 inputs and 2 outputs. —The outputs form a binary number = number of incoming 1's
Combinational Circuit Example
Combinational logic circuits give us many useful devices. One of the simplest is the half adder, which finds the sum of two bits. We can gain some insight as to the construction of a half adder by looking at its truth table, shown at the right. Half Adder
We can change our half adder into to a full adder by including gates for processing the carry bit. The truth table for a full adder is shown at the right. Full Adder
Multiplexor A Multiplexor that allows us to choose one of two incoming bits. —S selects either A or B and sends its value on to C. —Left: A two input multiplexor —Right: implementation with gates
Multiplexor Exercise If A,B,S are 0,1,0, —what are the AND gate outputs? —the OR gate output? Now change S to 1 and update the outputs accordingly.
This is what a 4-to-1 multiplexer looks like on the inside. 4-to-1 multiplexer
Building a circuit from a truth table Example: A circuit with 3 inputs (A,B,C) and 3 outputs (D,E,F). Output D is turned on —when at least one input is on. Output E is turned on —when exactly two inputs are on. Output F is turned on —when all three inputs are on.
Building a circuit from a truth table (2) Truth Table —D=1, when one input is on —E=1, when only two inputs are on —F=1, when all three inputs are on
Building a circuit from a truth table (3) To make a circuit that will compute these 3 outputs correctly, we start by creating an AND for each line of the truth table: This AND gate is turned on only when A is 0 (note the inverter), b is 1 and c is 1 —Which corresponds to line 4 of the truth table
Building a circuit from a truth table (4) Now we need an OR gate for each output Now we connect some of the input AND's to some of the output OR's.
Building a circuit from a truth table (5) Before continuing, let finish the diagram for Outputs D (at least one input set to 1) and F (all 3 set to 1).
Building a circuit from a truth table (6) Notice that one of the AND gates (line one of the truth table) never got used. Also F's OR has only one input. We can eliminate these two gates to give
Building a circuit from a truth table (7) The final circuit:
An Adder for 2 bits A single column could be done by a circuit like this
An Adder for 2 bits (2) Here is its truth table EXERCISE: fill in the missing data in the table below
An Adder for 2 bits (3) Now Let’s complete the circuit
Creating a simple ALU We start with a circuit that's similar to our adder. —3 data inputs (a,b,CarryIn) —2 data outputs (Results, CarryOut) —The box labeled + is the full adder we did above. We also have 3 Control Signals: Ainvert, Binvert, Operation —We ignoring the less
Creating a simple ALU (2) Operation is a MUX —Allowing us to choose one of four inputs Exercise: —Ainvert and Binvert are 0, What happens —If Operation is 0? —If operation is 1? —If operation is 2?
Simple 32 bit ALU
Bitwise AND Note, carryIn and carry out don’t contribute anything
Bitwise OR We can also use Bitwise OR Note: carryIn and carry out don’t contribute anything
Lastly It can also do subtraction —a31a30...a0 - b31b30....b0 —EXERCISE: How? —Recall subtraction is done with 2's complement –2's complement, flip all the bits and add 1 So that's an ALU that can do Add, Subtract, AND, OR
Q A &
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In-toeing, sometimes called pigeon-toed, is a common condition in children where the feet point inwards when walking and standing.
There are three conditions that create in-toeing: metatarsus adductus, tibial torsion and femoral anteversion.
This condition is an inward curve of the outer border of the foot. It is detected in childhood, usually noticed during infancy and is typically caused by the position of the baby in the uterus. The Metatarsus is considered: mild, moderate, or severe.
In mild feet, the foot can be overcorrected passively. These feet usually resolve on their own by age 2.
In moderate feet, the foot can be passively corrected so that the lateral border is straight, but it cannot be over-corrected. These feet also usually improve on their own in the majority of cases, and occasionally may require special shoes to help the foot to permanently straighten. Both mild and moderate feet also respond well to stretching to help get the foot straighter. This is usually shown to the parents by the doctor and can be done at home at bath time.
Severe feet cannot be stretched so that the outside border of the foot is straight. These feet may be treated with a series of casts to help stretch the feet, followed by special shoes to maintain the correction. Rarely, surgery may be necessary if casting fails or if the deformity recurs during growth and the child has functional problems as a result.
Internal tibial torsion is an inward twist of the tibia or shinbone. This is usually noticed when the child begins to walk. Inward twisting is normal in many babies and often corrects by age 1. However, the inward twist is slower to correct in some children and these are the ones that usually need to see a doctor. In about 90% of patients the inward twist slowly corrects by about ages 4-6.
A child with tibial torsion will frequently trip and fall when initially walking. As the child grows and muscles develop, he or she is better able to cope with the in-toeing until it ultimately resolves in most cases. Studies have shown that bracing does not speed up the correction of inward tibial torsion, so most doctors do not prescribe any treatment other than observation. In about 10% of patients the tibial torsion does not correct, but most children function perfectly well and there is no evidence that tibial torsion causes arthritis or functional problems in the long run.
In the rare case that the torsion does not resolve by age 6 to 8 and the child does have a functional problem as a result of the torsion, the treatment is to cut the bone and rotate it outward so the feet point straight. Very few normal children without neuromuscular problems need this surgery and careful discussion with the doctor is necessary before deciding on surgery as the best treatment option.
Femoral anteversion is an excessive inward twist of the upper thigh bone at the hip region. This is usually noticed between ages 2 to 4. All children are born with some inward twist of the thigh bone and as they grow and their ligaments around the hip tighten, the anteversion resolves during the first few years of life. In some children, these ligaments never completely tighten up and when the child starts to walk they can become looser, causing the hips to rotate further inward, causing the in-toeing to be noticed between ages 2 and 4.
Most cases of femoral anteversion resolve spontaneously by the time the child is between 6 and 8 years old. Once again, it has been shown that special shoewear and braces do not improve on the natural resolution of the deformity, and may actually cause problems such as discomfort and poor self esteem. A few children will not resolve their anteversion but most function fine without any problems. In the rare case that the anteversion does not resolve by age 6 to 8 and the child does have a functional problem as a result of the torsion, the treatment is to cut the bone and rotate it outward so the feet point straight.
Out-toeing is much less common than in-toeing and often detected by age 2. Most children are born with external rotation contractures of the hips and this resolves shortly after walking begins. In those children where this does not happen quickly, out-toeing is the result when they first start walking. This will almost always resolve within a year from the onset of walking.
Out-toeing may also be caused by outward twisting of the tibia or femur bone, and is more common seen in children with neuromuscular abnormalities. As with in-toeing, bracing and shoewear are not helpful in resolving the deformity. In extreme cases, surgery may be necessary, but very infrequently.
To make an appointment with one of our orthopedic specialists or to learn more about our services, centers and treatment options, please call 410-448-6400.<|endoftext|>
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# How to Calculate the Roof Area Using the Building Square Footage & the Pitch of the Roof
So you're thinking of reroofing the house yourself and want an estimate for how much roofing material you'll need, but you don't want to go climbing around with a tape measure and scare your spouse half to death. You're in luck, because math can rescue you from pulling out that ladder and tape measure.
## Straight to the answer
#### 1
Divide your roof pitch by 12. Roof pitch is given as the number of inches in height change over the distance of one foot. Dividing by 12 gives the ratio of inches in rise per distance. For example, a standard roof pitch is likely to be about 4. So for a pitch angle of 4, divide by 12 and get 1/3.
#### 2
Square the result of Step 1. So keeping with our roof pitch of 4 this gives us 1/9.
#### 3
Add 1, then take the square root. Adding 1 gives us 1 + 1/9 = 10/9, then taking the square root gives us about 1.0541.
#### 4
Multiply by the square footage of a single floor of your house. So continuing our example, if you've got a two story house with a total square footage of 2,700 square feet, with each floor having 1,350 square feet, multiply 1,350 by 1.0541 to get 1,423 square feet. So you'd need 1,423 square feet of roofing.
#### Things You Will Need
• Square footage of your house
• Pitch of your roof
• Calculator
• Paper and Pen
#### Tip
• This calculation will give a solid estimate for a single slope roof and a simple gable roof with a single pitch. For more complex roof lines with multiple pitch angles, it's a little more complicated but the same steps can be followed; just break down the roof into separate sections with a different square footage for each different pitch angle.
#### Warnings
• Use only the square footage directly covered by the roof. On a multistory house, use only the square footage of a single floor.
• Minor differences in calculations occur based on chimneys and other variations.
#### About the Author
Dr. Butler received his PhD in General Physics in 2004. His dissertation topic included gravitational cosmological background radiation detection, digital signal analysis, feedback control, thermal noise, statistical mechanics, and LASER optics. Additional research topics include charge, parity, time inversion investigation, spectroscopy, optical state transitions, and non-linear effects in optical cavities. After his PhD Dr. Butler has worked for 8+ years in industry gaining experience in areas of silicon manufacturing, CPU architecture, design, validation and debug, process management, project management, and computer programming. Time permitting Dr. Butler also enjoys volunteering as a science fair judge for the local high school and junior high school age competitions.
#### Photo Credits
• Jupiterimages/Photos.com/Getty Images<|endoftext|>
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## Note on Sphere and Hemisphere
• Note
• Things to remember
• Videos
• Exercise
• Quiz
We are so much familiar with spherical objects. The sphere is also a solid object whose each point in the outer surface is equidistance from the fixed point inside it. Such a fixed point is called the centre of the sphere. The constant distance is called the radius of the sphere. A solid object such as a globe, volleyball, toy ball, table tennis ball, marble etc is the example of a sphere.
The figure as shown alongside is a sphere. The fixed point 'O' inside it is the centre which is equidistance from each point P on the surface of the sphere. So, OP = r is the radius of the sphere.
If we cut a sphere through its diameter, there are two half spheres called the hemisphere and the cross section is called the great circle. The radius of the sphere is same as the radius of the great circle. The centre of the sphere and its great circle. The centre of the sphere and its great circle is same.
### Surface area of sphere
The surface area of a sphere is the area of its outer part, which is a smooth curved surface.
The surface area of a sphere is given by SA = 4πr2 where r is the radius of the sphere. The total surface area of hemisphere = 2πr2+ πr2 = 3πr2 square unit.
Note Curved surface area of hemisphere = 2$$\pi$$r2square units. The surface area of a sphere (SA) = $$\pi$$d2, if the diameter is given.
### Volume of sphere
The volume of a sphere means the space that it occupies. We can measure the volume of sphere experimentally. Fill up the measuring cylinder with the water level in the cylinder. The difference of two levels is the volume of the sphere.
#### Alternatively
Choose a sphere of given diameter (d) =10 cm (say).
Immerse of the whole sphere into the water in the measuring cylinder, the water level is raised by 523.33 ml. Therefore its volume is 523.33 cm3. From this experiment,
The diameter (d) = 10cm
The volume of sphere (V) = 523.33 cm3
Let us make the ratio
\begin{align*} 6V:d^3 &= \frac {6V} {d^3} \\ &= \frac {6 \times 523.33} {10^3} \\ &= \frac {3140} {1000} \\ &= 3.14 \end{align*}
$$(\because \pi = \frac {22} {7} = 3.14)$$
\begin{align*} \therefore \frac {6V} {d^3} &= \pi \\ or, V &= \frac {\pi (2r)^3} {6} \\ &= \frac {4 \pi r^3} {3} \\ \end{align*}
$$\therefore \text {Volume of a sphere} (V) = \frac {4 \pi r^3} {3} = \frac {\pi d^3} {6} \text {cubic units.}$$
1. Curved surface area of hemisphere = 2$$\pi$$r2square units.
2. The surface area of a sphere (SA) =d2, if the diameter is given.
3. Volume of a sphere(v) = $$\frac{4\ (\pi) r^3}{3}$$ = $$\frac{\ (\pi)d^3}{6}$$ cubic units.
.
### Very Short Questions
Here,
r = 15 cm
\begin{align*} \text{Total Surface Area of hemisphere} &= 3{\pi}r^2\\ &= 3 \times \frac {22}7 \times 15 \times 15\\ &= \frac {14850}{7}\\ &= 2121.43 cm^2_{Ans}\\ \end{align*}
Surface Area of the sphere (A) = 616 cm2
radius of the sphere (r) = ?
By formula,
A = 4$$\pi$$r2
or, r2 = $$\frac A{4\pi}$$
or, r2 = $$\frac {616 \times 7}{4 \times 22}$$
or, r2 = 49
∴ r = 7cmAns
Suppose,
Then,
Surface Area of Sphere = 4$$\pi$$r2
or, $$\pi$$ = 4$$\pi$$r2
or, r2 = $$\frac {\pi}{4\pi}$$
or, r2 = $$\frac 14$$
or, r = $$\sqrt {\frac 14}$$
∴ r = $$\frac 12$$cmAns
Here,
r = $$\frac {42 cm}{2}$$ = 21 cm
\begin{align*} \text{Volume of the hemisphere (V)} &=\frac 23{\pi}r^3\\ &= \frac 23 \times \frac {22}7 \times (21)^3\\ &= 44 \times (21)^2\\ &= 19404 cm^3_{Ans}\\ \end{align*}
Here,
Volume of marble (V) = $$\frac {\pi}6cm^3$$
or, $$\frac 43$$ $$\pi$$r3 = $$\frac {\pi}6cm^3$$
or, r3 = $$\frac {\pi}6$$ $$\times$$ $$\frac 3{4\pi}cm^3$$
or, r3 = $$\frac 1{2^3}cm^3$$
∴ r = $$\frac 12cm$$
Hence, the diameter of the marble = 2 $$\times$$ r = 2 $$\times$$ $$\frac 12$$ cm = 1 cmAns
Let r be the radius of the sphere.
Volume = $$\frac 43$$$$\pi$$r3
or, 38808 = $$\frac 43$$ $$\times$$ $$\frac {22}7$$ $$\times$$ r3
or, r3 = 38808 $$\times$$ $$\frac {21}{88}$$
or, r3 = 441 $$\times$$ 21
or, r3 = (21)3
∴ r = 21Ans
Here,
Volume of sphere = $$\frac {3773}{21}cm^3$$
or, $$\frac 43$$$$\pi$$r3 = $$\frac {3773}{21}cm^3$$
or, $$\frac 43$$ $$\times$$ $$\frac {22}7$$ $$\times$$ r3 = $$\frac {3773}{21}cm^3$$
or, $$\frac {88}{21}r^3$$ = $$\frac {3773}{21}cm^3$$
or, r3 = $$\frac {3773}{21}$$ $$\times$$ $$\frac {21}{88}cm^3$$
or, r3 = $$\frac {343}{8}cm^3$$
or, r3 = ($$\frac 73$$cm)3
∴ r = $$\frac 72$$cm
Hence, Circumference of sphere = 2$$\pi$$r = 2 $$\times$$ $$\frac {22}7$$ $$\times$$ $$\frac 72$$ cm = 22 cmAns
Here,
Volume of spherical solid = $$\frac 43$$$${\pi}cm^3$$
or, $$\frac 43$$$${\pi}r^3$$ = $$\frac 43$$$${\pi}cm^3$$
or, r3 = $$\frac {\frac {4\pi}{3}}{\frac {4\pi}{3}}cm^3$$
or, r3 = 1 cm3
∴ r = 1 cm
Hence, the radius of spherical solid is 1 cm.Ans
Here,
Volume of sphere (V) = 36$${\pi}cm^3$$
Diameter of a sphere (d) = ?
We know that,
V = $$\frac 16$$$${\pi}d^3$$
or, d = $$\sqrt [3]\frac{6V}{d}$$
or, d = $$\sqrt [3]\frac {6 \times 36\pi}{\pi}$$
or, d = $$\sqrt [3]{6 \times 6 \times 6}$$
∴ d = 6cm
\begin{align*} \text{The Surface Area (A)} &= {\pi}d^2\\ &= \frac {22}7 \times (6cm)^2\\ &= \frac {22}7 \times 36 cm^2\\ &= 113.14cm^2_{Ans}\\ \end{align*}
Here,
Volume (V) = $$\frac {1372\pi}{3}cm^3$$
Total Surface Area (S) = ?
By Formula,
V = $$\frac 43$$$${\pi}r^3$$
or, $$\frac {1372\pi}{3}$$ =$$\frac 43$$$${\pi}r^3$$
or, r3 =$$\frac {1372\pi}{3}$$ $$\times$$ $$\frac 3{4\pi}$$
or, r3 = 343
or, r3 = 73
∴ r = 7
We know that,
\begin{align*} S &= 4{\pi}r^2\\ &= 4 \times \frac {22}7 \times 7^2 cm^2\\ &= 616 cm^2_{Ans}\\ \end{align*}
Here,
Surface Area of sphere = 616 cm2
or, 4$${\pi}r^2$$ = 616 cm2
or, $${\pi}r^2$$ = $$\frac {616}4 cm^2$$
∴$${\pi}r^2$$ = 154 cm2
We know that,
\begin{align*} \text{Total Surface Area of a hemisphere} &= 3{\pi}r^2\\ &= 3 \times 154 cm^2\\ &= 462 cm^2_{Ans}\\ \end{align*}
Here,
Total Surface Area of hemisphere (S) = 243 $${\pi}cm^2$$
Volume (V) = ?
We know that,
S = 3$${\pi}r^2$$
or, 243$$\pi$$ = 3$${\pi}r^2$$
or, r2 = $$\frac {243\pi}{3\pi}$$
or, r2 = 81
∴ r = 9 cm
Now,
\begin{align*} V &= \frac 23 {\pi}r^3\\ &= \frac 23 \times \frac {22}7 \times 9^3\\ &= 1527.43 cm^3_{Ans}\\ \end{align*}
Here,
4$${\pi}r^2$$ = $$\frac 1{4\pi}$$
or, r2 = $$\frac 1{(4\pi)^2}$$
∴r = $$\frac 1{4\pi}$$
Now,
\begin{align*} \text{Volume (V)} &= \frac 43 {\pi}r^3\\ &= \frac {4\pi}3 (\frac 1{4\pi})^3\\ &= \frac 13 \times \frac 1{16\pi^2}\\ &= \frac 1{44\pi^2}cm^3_{Ans}\\ \end{align*}
Here,
r1 = $$\frac {6cm}2$$ = 3cm
r2 = $$\frac {8cm}2$$ = 4cm
r3 = $$\frac {10cm}{2}$$ = 5cm
Suppose,
radius of new sphere = R
Now,
Volume of new sphere = sum of the volume of first, second and third spheres
or, $$\frac 43$$$${\pi}R^3$$ = $$\frac 43$$$${\pi}r_1^3$$ +$$\frac 43$$$${\pi}r_2^3$$ +$$\frac 43$$$${\pi}r_3^3$$
or, R3 = r13 + r23 + r33
or, R3 = (33 + 43 + 53)
or, R3 = 27 + 64 + 125
or, R3 = 216
or, R = $$\sqrt [3]{216}$$
∴ R = 6 cm
Hence, the diameter of new sphere (d) = 2R = 2 $$\times$$ 6 cm = 12 cmAns
\begin{align*} \text{Volume of small ball} &= \frac 43 {\pi}r^3\\ &= \frac 43 {\pi}1^3\\ &= \frac 43 {\pi}cm^3\\ \end{align*}
\begin{align*} \text{Radius of big ball} &= \frac {8cm}2\\ &= 4 cm\\ \end{align*}
\begin{align*} \text{Volume of big ball} &= \frac 43 {\pi}r^3\\ &= \frac 43 {\pi}4^3\\ &= \frac 43 {\pi}64cm^3\\ \end{align*}
Now,
\begin{align*} \text{Number of small balls that can be made from the big ball} &= \frac {\frac 43 {\pi} \times 64}{\frac 43 {\pi}}\\ &= 64_{Ans}\\ \end{align*}
Let x be the radius of the first sphere.
Then,
Radius of second sphere is $$\frac r4$$
Volume of first sphere (V1) = $$\frac 43$$ $${\pi}r^3$$
Volume of second sphere (V2) = $$\frac 43$$ $$\pi$$ ($$\frac r4$$)3 = $$\frac {4\pi}{3}$$$$\times$$ $$\frac {r^3}{64}$$
Now,
$$\frac {V_2}{V_1}$$ = $$\frac {\frac {4\pi}{3} \times \frac {r^3}{64}}{\frac 43 {\pi}r^3}$$ = $$\frac 1{64}$$
Hence, V1 :V2 = 1 : 64Ans
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• ### The diameter of a hemisphere is 84 cm, find its total surface area and volume.
15532 cm2,155243 cm3
13904 cm2, 155238 cm3
16635 cm2, 155132 cm3
16632 cm2,155232 cm3
4158 cm2
4159 cm2
4055 cm2
4170 cm2
1849 cm2
1848 cm2
1845cm2
1846cm2
• ## You scored /14
Forum Time Replies Report
##### Pukar poudel
If S snd V be the Surface area and Volume respectively.Prove that :36S^3=V^2
##### Hemisphere
Find the radius of the hemisphere having total surface area 27πsq. cm.\
##### sushila
total surface area of a sphere<|endoftext|>
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# Lesson Plan in Mathematics Iv Bar Graph
Topics: Educational years, Touring car racing, Chart Pages: 12 (1880 words) Published: March 12, 2013
LESSON PLAN IN MATHEMATICS IV
I. OBJECTIVES
Cognitive:Identify the bar graph.
Psychomotor:Read and interpret data presented in a bar graph using the following parts:
* Title
* Number Scale
* Labels
Affective:Appreciate the importance of bar graph
Show helpfulness towards other member of the group.
II. SUBJECT MATTER
Reference:BEC-PELC V.A 1.1 & 1.2
Teacher’s Workbook
Textbook in Math IV
Materials:Pieces of cartolina, cut outs, picture, and graph
Value:Helping one another and sharing’s one’s ideas
III. LEARNING EXPERIENCE
Teacher’s Activity| Pupil’s Activity|
A. Preparatory Activity 1. DrillAnalyse the bar graph.Primary School Enrollment 600
Number of Pupils
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Complete each sentence. 1. The title of the graph is ________. 2. The number scale is __________. 3. The label on the horizontal axis is ____. 4. The label on the vertical axis is ______. 2. ReviewLast week we discuss about the “Parts of a graph” right?Now, who can give me the parts of a bar graph?Now, I have here an example of a bar graph and you are going to study it and answer the questions followed.BOOKS SOLD IN MAY Number of Books Sold
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Filipino
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Question: 1. What is the title of the graph? 2. What is the number scale? 3. What information is on the horizontal axis? 4. What is the label on the vertical axis? 3. Motivation * Look on the board * Analyse the picture separate and count the number of different animals. * The number of pets there is sold from the pet shop. Daily sold Pet in Jasmine Pet Shop Number of Pets Sold
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Fish
Bird
Dog
Cat
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* How many cats were sold? * How many birds were sold? * How many fish were sold? * How many dogs were sold? B. Developmental ActivitiesClass how do you call to the different data that I present to you?Very good those are bar graph.BAR GRAPH – uses bars of different of different height or lengths to show and compare information.There are two kinds of bar graph * Horizontal bar graph * Vertical bar graphThe information in a bar graph can be read and interpreted using the title, number scale, and labels.It’s that clear class? Do you have any question?Example:Favorite Cartoon Characters of Grade 4 Pupils Number of Pupils
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Twetty
Naruto
Hello
Kitty
Dora
Sponge
Bob
Cartoon Character
This is an example of a vertical bar graph. How many pupils prefer Sponge Bob? To answer the question, look for the bar labelled “sponge bob”. This is on the horizontal axis. From here, find the number on the scale where the bar ends. The bar for cartoons ends at 25. So, 25 pupils prefer sponge bob.It’s that clear? 1. Who prefer most number of the cartoon character? 2. How many pupils prefer to Dora? 3. Who prefer least number of cartoon characters? 4. How many more cartoon character did prefer hello kitty than Dora?2. ActivityGroup 1 & 3Read and interpret the bar graph and answer the questions that follow.Native Bag Production 300
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1. What is the title of the graph? 2. How many bags were produced in the month of April? 3. Which month gave biggest number of native bag? 4. How many native bags were produced by month of January and February? 5. Which month gave least number of native bags?Group 2 & 4Read and interpret the bar graph and answer the questions that follow.Number of Pupils Number of Pupils in Grade IV
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Domestication (from the Latin domesticus: "of the home") is the cultivating or taming of a population of organisms in order to accentuate traits that are desirable to the cultivator or tamer. The desired traits may include a particular physical appearance, behavioral characteristic, individual size, litter size, hair/fur quality or color, growth rate, fecundity, lifespan, ability to use marginal grazing resources, production of certain by-products, and many others. Domesticated organisms may become dependent on humans or human activities, since they sometimes lose their ability to survive in the wild.
Domestication differs from taming in that it may refer not simply to a change in organisms' behaviors or environmental socialization, but also potentially even in their phenotypical expressions and genotypes. The word domestication also is more commonly used to mean a change within whole populations, while taming is more commonly used to mean a change within individuals. Furthermore, taming typically applies only to animals and their becoming habituated to human presence, while domestication is a broader term and can include plants, fungi, and other types of organisms.
Plants domesticated primarily for aesthetic enjoyment in and around the home are usually called house plants or ornamentals, while those domesticated for large-scale food production are generally called crops. A distinction can be made between those domesticated plants that have been deliberately altered or selected for special desirable characteristics (see cultigen) and those plants that are used for human benefit, but are essentially no different from the wild populations of the species. Animals domesticated for home companionship are usually called pets, while those domesticated for food or work are called livestock or farm animals. Domestication is the process of changing plants or animals to make them more useful to humans. Domestication of plants led to the development of agriculture. Domestication led to many things such as the use of fibers to make clothes, making food, and the need to stay in one place.
- 1 Background
- 2 Degrees
- 3 Tame or domesticated
- 4 Negative aspects
- 5 Dates and places
- 6 Genetic pollution
- 7 Notes and references
- 8 Bibliography
- 9 Further reading
- 10 See also
- 11 External links
This section possibly contains original research. (December 2014)
Charles Darwin was the first to describe the connection between domestication, selection and evolution. Darwin described how the process of domestication can involve both unconscious and methodical elements. Routine human interactions with animals and plants create selection pressures that cause adaptation to human presence, use or cultivation. Deliberate selective breeding has also been used to create desired changes, often after initial domestication. These two forces, unconscious natural selection and methodical selective breeding, may have both played roles in the processes of domestication throughout history. Both have been described from human perspective as processes of artificial selection.
The domestication of wheat provides an example. Wild wheat falls to the ground to reseed itself when ripe, but domesticated wheat stays on the stem for easier harvesting. There is evidence that this change was possible because of a random mutation that happened in the wild populations at the beginning of wheat's cultivation. Wheat with this mutation was harvested more frequently and became the seed for the next crop. Therefore, without realizing, early farmers selected for this mutation, which may otherwise have died out. The result is domesticated wheat, which relies on farmers for its own reproduction and dissemination.
The domestication of dogs provides another example. It is speculated that tamer than average wolves, less wary of humans, selected themselves as dogs over many generations. These wolves were able to thrive by following humans to scavenge for food near camp fires and garbage dumps, which gave them an advantage over more shy individuals. Eventually a symbiotic relationship developed between people and these proto-dogs. The dogs fed on human food scraps, and humans found that dogs could warn them of approaching dangers, help with hunting, act as pets, provide warmth, or supplement their food supply. As this relationship progressed, humans eventually began to keep these self-tamed wolves and breed from them the types of dogs that we have today.
In recent times, selective breeding may best explain how continuing processes of domestication often work. Some of the best-known evidence of the power of selective breeding comes from the Farm-Fox Experiment by Russian scientist, Dmitri K. Belyaev, in the 1950s. His team spent many years breeding the domesticated silver fox (Vulpes vulpes) and selecting only those individuals that showed the least fear of humans. Eventually, Belyaev's team selected only those that showed the most positive response to humans. He ended up with a population of grey-coloured foxes whose behavior and appearance was significantly changed. They no longer showed any fear of humans and often wagged their tails and licked their human caretakers to show affection. These foxes had floppy ears, smaller skulls, rolled tails and other traits commonly found in dogs.
Despite the success of this experiment, it appears that selective breeding cannot always achieve domestication. Attempts to domesticate many kinds of wild animals have been unsuccessful. The zebra is one example. Despite the fact that four species of zebra can interbreed with and are part of the same genus as the horse and the donkey, attempts at domestication have failed. Factors such as temperament, social structure and ability to breed in captivity play a role in determining whether a species can be successfully domesticated. In human history to date, only a few species of large animal have been domesticated. In approximate order of their earliest domestication these are: dog, sheep, goat, pig, ox, yak, reindeer, water buffalo, horse, donkey, llama, alpaca, Bactrian camel and Arabian camel.
- Flexible diet – Creatures that are willing to consume a wide variety of food sources and can live off less cumulative food from the food pyramid (such as corn or wheat), particularly food that is not utilized by humans (such as grass and forage) are less expensive to keep in captivity. Carnivores by definition feed primarily or only on flesh, which requires the expenditure of many animals, though they may exploit sources of meat not utilized by humans, such as scraps and vermin.
- Reasonably fast growth rate – Fast maturity rate compared to the human life span allows breeding intervention and makes the animal useful within an acceptable duration of caretaking. Some large animals require many years before they reach a useful size.
- Ability to be bred in captivity – Creatures that are reluctant to breed when kept in captivity do not produce useful offspring, and instead are limited to capture in their wild state. Creatures such as the panda, antelope and giant forest hog are territorial when breeding and cannot be maintained in crowded enclosures in captivity.
- Pleasant disposition – Large creatures that are aggressive toward humans are dangerous to keep in captivity. The African buffalo has an unpredictable nature and is highly dangerous to humans; similarly, although the American bison is raised in enclosed ranges in the Western United States, it is much too dangerous to be regarded as truly domesticated. Although similar to the domesticated pig in many ways, Africa's warthog and bushpig are also dangerous in captivity.
- Temperament which makes it unlikely to panic – A creature with a nervous disposition is difficult to keep in captivity as it may attempt to flee whenever startled. The gazelle is very flighty and it has a powerful leap that allows it to escape an enclosed pen. Some animals, such as the domestic sheep, still have a strong tendency to panic when their flight zone is encroached upon. However, most sheep also show a flocking instinct, whereby they stay close together when pressed. Livestock with such an instinct may be herded by people and dogs.
- Modifiable social hierarchy – Social creatures whose herds occupy overlapping ranges and recognize a hierarchy of dominance can be raised to recognize a human as the pack leader:
- tapirs and rhinoceroses are solitary and do not tolerate being penned with each other
- antelope and deer except for reindeer are territorial when breeding and live in herds only for the rest of the year
- bighorn sheep and peccaries have nonhierarchical herd structures and do not follow any definite leader: instead males fight continuously with each other for mating opportunities
- musk ox herds (although having a defined leader) maintain mutually exclusive territories and two herds will fight if kept together.
However, this list is of limited use because it fails to take into account the profound changes that domestication has on a species. While it is true that some animals retain their wild instincts even if born in captivity, e.g. laying hens, pigs and laboratory mice, some factors must be taken into consideration.
Number (5) may not be a prerequisite for domestication, but rather a natural consequence of a species' having been domesticated. In other words, wild animals are naturally timid and flighty because they are constantly faced by predators; domestic animals do not need such a nervous disposition, as they are protected by their human owners. The same holds true for number (4) – aggressive temperament is an adaptation to the danger from predators. A Cape buffalo can kill even an attacking lion, but most modern large domestic animals were descendants of aggressive ancestors. The wild boar, ancestor of the domestic pig, is certainly renowned for its ferocity; other examples include the aurochs (ancestor of modern cattle), horse, Bactrian camels and yaks, all of which are no less dangerous than their undomesticated wild relatives such as zebras and buffalos. Others have argued that the difference lies in the ease with which breeding can improve the disposition of wild animals, a view supported by the failure to domesticate the kiang and onager. On the other hand for thousands of years humans have managed to tame dangerous species like bears and cheetahs whose failed domestications had little to do with their aggressiveness.
Number (6), while it does apply to most domesticated species, also has exceptions, most notably in the domestic cat and ferret, which are both descended from strictly solitary wild ancestors but which tolerate and even seek out social interaction in their domestic forms. Feral domestic cats, for example, naturally form colonies around concentrated food sources and will even share prey and rear kittens communally, while wildcats remain solitary even in the presence of such food sources. Zoologist Marston Bates devoted a chapter on domestication in his 1960 book The Forest and the Sea, in which he talks a great deal about how domestication alters a species: Dispersal mechanisms tend to disappear for the reason stated above, and also because people provide transportation for them. Chickens have practically lost their ability to fly. Similarly, domestic animals cease to have a definite mating season, and so the need to be territorial when mating loses its value; and if some of the males in a herd are castrated, the problem is reduced even further. What he says suggests that the process of domestication can itself make a creature domesticable. Besides, the first steps towards agriculture may have involved hunters keeping young animals, who are always more impressionable than the adults, after killing their mothers.
Another strong factor deciding whether a species will be considered for domestication is quite simply the availability of more suitable (or even better already domesticated) alternatives. For example a community that had been introduced to domestication by neighboring peoples will generally find it much more practical, economical and time saving to import already domesticated species than experiment with wild animals (even if they are of the same species). Generally speaking, the species of animals originally domesticated by early humans in the interconnected landmasses of Eurasia and Africa were far superior, both in working capacity and in food production, than the species found in the other continents, namely the Americas and Oceania.
The earliest human attempts at plant domestication occurred in South-Western Asia. There is early evidence for conscious cultivation and trait selection of plants by pre-Neolithic groups in Syria: grains of rye with domestic traits have been recovered from Epi-Palaeolithic (c. 11,050 BCE) contexts at Abu Hureyra in Syria, but this appears to be a localised phenomenon resulting from cultivation of stands of wild rye, rather than a definitive step towards domestication.
By 10,000 BCE the bottle gourd (Lagenaria siceraria) plant, used as a container before the advent of ceramic technology, appears to have been domesticated. The domesticated bottle gourd reached the Americas from Asia by 8000 BCE, most likely due to the migration of peoples from Asia to America.
Cereal crops were first domesticated around 9000 BCE in the Fertile Crescent in the Middle East. The first domesticated crops were generally annuals with large seeds or fruits. These included pulses such as peas and grains such as wheat.
The Middle East was especially suited to these species; the dry-summer climate was conducive to the evolution of large-seeded annual plants, and the variety of elevations led to a great variety of species. As domestication took place humans began to move from a hunter-gatherer society to a settled agricultural society. This change would eventually lead, some 4000 to 5000 years later, to the first city states and eventually the rise of civilization itself.
Continued domestication was gradual, a process of trial and error that occurred intermittently. Over time perennials and small trees began to be domesticated including apples and olives. Some plants were not domesticated until recently such as the macadamia nut and the pecan.
In other parts of the world very different species were domesticated. In the Americas squash, maize, beans, and perhaps manioc (also known as cassava) formed the core of the diet. In East Asia millet, rice, and soy were the most important crops. Some areas of the world such as Southern Africa, Australia, California and southern South America never saw local species domesticated.
Domesticated plants often differ from their wild relatives in which they
- spread to a more diverse environment and have a wider geographic range
- may have a different ecological preference
- may flower and fruit simultaneously
- may lack shattering or scattering of seeds and may have lost dispersal mechanism completely
- may have larger fruits and seeds, and so lower efficiency of dispersal
- may have been converted from a perennial to annual
- may have lost seed dormancy
- may have lost photoperiodic controls
- may lack normal pollinating organs
- may have a different breeding system
- may lack defensive adaptation such as hairs, spines and thorns
- may lack protective coverings and sturdiness
- may have better palatability and chemical composition, rendering them more likely to be eaten by animals
- may be more susceptible to diseases and pests
- may develop seedless parthenocarpic fruits
- may have undergone selection for double flowers, which may involve conversion of stamens into petals
- may have become sexually sterile and vegetatively reproduced
The boundaries between surviving wild populations and domestic clades can be vague. A classification system that can help solve this confusion surrounding animal populations might be set up on a spectrum of increasing domestication:
- Wild: These populations experience their full life cycles without deliberate human intervention.
- Raised in captivity/captured from wild (in zoos, botanical gardens, or for human gain): These populations are nurtured by humans but (except in zoos) not normally bred under human control. They remain as a group essentially indistinguishable in appearance or behaviour from their wild counterparts. Examples include Asian elephants, animals such as sloth bears and cobras used by showmen in India, and animals such as Asian black bears (farmed for their bile), and zoo animals, kept in captivity as examples of their species. (It should be noted that zoos and botanical gardens sometimes exhibit domesticated or feral animals and plants such as camels, mustangs, and some orchids)
- Raised commercially (captive or semidomesticated): These populations are ranched or farmed in large numbers for food, commodities, or the pet trade, commonly breed in captivity, but as a group are not substantially altered in appearance or behavior from their wild cousins. Examples include the ostrich, various deer, alligator, cricket, honeybees, pearl oyster, raptors used in falconry and ball python. (These species are sometimes referred to as partially domesticated.)
- Domesticated: These populations are bred and raised under human control for many generations and are substantially altered as a group in appearance or behaviour. Examples include sweet potato, garlic, pigs, ferrets, turkeys, canaries, domestic pigeons, budgerigars, goldfish, koi carp, silkworms, dogs, cats, sheep, cattle, chickens, llamas, guinea pigs, laboratory mice, horses, goats and (silver) foxes.
This classification system does not account for several complicating factors: genetically modified organisms, feral populations, and hybridization. Many species that are farmed or ranched are now being genetically modified. This creates a unique category because it alters the organisms as a group but in ways unlike traditional domestication. Feral organisms are members of a population that was once raised under human control, but is now living and multiplying outside of human control. Examples include mustangs. Hybrids can be wild, domesticated, or both: a liger is a hybrid of two wild animals, a mule is a hybrid of two domesticated animals, and a beefalo is a cross between a wild and a domestic animal.
Tame or domesticated
A great difference exists between a tame animal and a domesticated animal. The term "domesticated" refers to an entire species or variety while the term "tame" can refer to just one individual within a species or variety. Humans have tamed many thousands of animals that have never been truly domesticated. These include the elephant, giraffes, and bears and cats. There is debate over whether some species have been domesticated or just tamed. Some state that the elephant has been domesticated, while others argue that the cat has never been domesticated. Dividing lines include whether a specimen born to wild parents would differ in appearance or behavior from one born to domesticated parents. For instance a dog is certainly domesticated because even a wolf (which genetically shares a common ancestor with all dogs) raised from a pup would be very different from a dog, in both appearance and behaviour. Similar problems of definition arise when domesticated cats go feral.
Many other languages use the same word for both concepts.
Selection of animals for visible "desirable" traits may make them unfit in other, unseen, ways. The consequences for the captive and domesticated animals were reduction in size, piebald color, shorter faces with smaller and fewer teeth, diminished horns, weak muscle ridges, and less genetic variability. Poor joint definition, late fusion of the limb bone epiphyses with the diaphyses, hair changes, greater fat accumulation, smaller brains, simplified behavior patterns, extended immaturity, and more pathology are a few of the defects of domestic animals. All of these changes have been documented in direct observations of the rat in the 19th century, by archaeological evidence, and confirmed by animal breeders in the 20th century. A 2014 commentary published in Genetics proposed that many of these features may arise due to mild neural crest deficits that also cause tameness; hence, selectively breeding tame animals also selects for these negative traits.
One side effect of domestication has been zoonotic diseases. For example, cattle have given humanity various viral poxes, measles, and tuberculosis; pigs and ducks have given influenza; and horses have given the rhinoviruses. Humans share over sixty diseases with dogs. Many parasites also have their origins in domestic animals. The advent of domestication resulted in denser human populations which provided ripe conditions for pathogens to reproduce, mutate, spread, and eventually find a new host in humans.
Other negative aspects of domestication have been explored. For example, Paul Shepherd writes "Man substitutes controlled breeding for natural selection; animals are selected for special traits like milk production of passivity, at the expense of overall fitness and naturewide relationships...Though domestication broadens the diversity of forms – that is, increases visible polymorphism – it undermines the crisp demarcations that separate wild species and cripples our recognition of the species as a group. Knowing only domestic animals dulls our understanding of the way in which unity and discontinuity occur as patterns in nature, and substitutes an attention to individuals and breeds. The wide variety of size, color, shape, and form of domestic horses, for example, blurs the distinction among different species of Equus that once were constant and meaningful."
Going further, some anarcho-primitivist authors describe domestication as the process by which previously nomadic human populations shifted towards a sedentary or settled existence through agriculture and animal husbandry. They claim that this kind of domestication demands a totalitarian relationship with both the land and the plants and animals being domesticated. They say that whereas, in a state of wildness, all life shares and competes for resources, domestication destroys this balance. Domesticated landscape (e.g. pastoral lands/agricultural fields and, to a lesser degree, horticulture and gardening) ends the open sharing of resources; where "this was everyone's," it is now "mine." Anarcho-primitivists state that this notion of ownership laid the foundation for social hierarchy as property and power emerged. It also involved the destruction, enslavement, or assimilation of other groups of early people who did not make such a transition.
To primitivists, domestication enslaves both the domesticated species as well as the domesticators. Advances in the fields of psychology, anthropology, and sociology allows humans to quantify and objectify themselves, until they too become commodities.
Dates and places
Since the process of domestication inherently takes many generations over a long period of time, and the spread of breed and husbandry techniques is also slow, it is not meaningful to give a single "date of domestication". However, it is believed that the first attempt at domestication of both animals and plants were made in the Old World by peoples of the Mesolithic Period. The tribes that took part in hunting and gathering wild edible plants, started to make attempts to domesticate dogs, goats, and possibly sheep, which was as early as 9000 BC. However, it was not until the Neolithic Period that primitive agriculture appeared as a form of social activity, and domestication was well under way. The great majority of domesticated animals and plants that still serve humans were selected and developed during the Neolithic Period, a few other examples appeared later. The rabbit for example, was not domesticated until the Middle Ages, while the sugar beet came under cultivation as a sugar-yielding agricultural plant in the 19th century. As recently as the 20th century, mint became an object of agricultural production, and animal breeding programs to produce high-quality fur were started in the same time period.
The methods available to estimate domestication dates introduce further uncertainty, especially when domestication has occurred in the distant past. So the dates given here should be treated with caution; in some cases evidence is scanty and future discoveries may alter the dating significantly.
Dates and places of domestication are mainly estimated by archaeological methods, more precisely archaeozoology. These methods consist of excavating or studying the results of excavation in human prehistorical occupation sites. Animal remains are dated with archaeological methods, the species they belong to is determined, the age at death is also estimated, and if possible the form they had, that is to say a possible domestic form. Various other clues are taken advantage of, such as slaughter or cutting marks. The aim is to determine if they are game or raised animal, and more globally the nature of their relationship with humans. For example the skeleton of a cat found buried close to humans is a clue that it may have been a pet cat. The age structure of animal remains can also be a clue of husbandry, in which animals were killed at the optimal age.
New technologies and especially mitochondrial DNA, which are simple DNA found in the mitochondria that determine its function in the cell provide an alternative angle of investigation, and make it possible to reestimate the dates of domestication based on research into the genealogical tree of modern domestic animals.
It is admitted for several species that domestication occurred in several places distinctly. For example, research on mitochondrial DNA of the modern cattle Bos taurus supports the archaeological assertions of separate domestication events in Asia and Africa. This research also shows that Bos taurus and Bos indicus haplotypes are all descendants of the extinct wild ox Bos primigenius. However, this does not rule out later crossing inside a species; therefore it appears useless to look for a separate wild ancestor for each domestic breed.
The dog was the first domesticated animal dating 18,000-32,000 years ago, which supports the hypothesis that dog domestication preceded the emergence of agriculture and occurred in the context of European hunter-gatherer cultures. This preceded the domestication of other species by several millennia. In the Neolithic a number of important species such as goats, sheep, pigs and cattle were domesticated, as part of the spread of farming which characterises this period. The goat, sheep and pig in particular were domesticated independently in the Levant and Asia.
The earliest secure evidence of horse domestication, bit wear on horse molars at Dereivka in Ukraine, dates to around 4000 BC. The unequivocal date of domestication and use as a means of transport is at the Sintashta chariot burials in the southern Urals, c. 2000 BC. Local equivalents and smaller species were domesticated from the 26th century BC.
The availability of both domesticated vegetable and animal species increased suddenly following the voyages of Christopher Columbus and the contact between the Eastern and Western Hemispheres. This is part of what is referred to as the Columbian Exchange.
Approximate dates and locations of original domestication
|Dog (Canis lupus familiaris)||prior to 33000 BCE||Western Europe|
|Sheep (Ovis orientalis aries)||between 11000 BCE and 9000 BCE||Southwest Asia|
|Pig (Sus scrofa domestica)||9000 BCE[unreliable source?]||Near East, China, Germany|
|Goat (Capra aegagrus hircus)||8000 BCE[unreliable source?]||Iran|
|Cow (Bos primigenius taurus)||8000 BCE[dead link]||India, Middle East, and North Africa|
|Cat (Felis catus)||7500 BCE||Cyprus and Near East|
|Chicken (Gallus gallus domesticus)||6000 BCE[dead link]||India and Southeast Asia|
|Guinea pig (Cavia porcellus)||5000 BCE||Peru|
|Donkey (Equus africanus asinus)||5000 BCE||Egypt|
|Domesticated duck (Anas platyrhynchos domesticus)||4000 BCE||China|
|Water buffalo (Bubalus bubalis)||4000 BCE||India, China|
|Horse (Equus ferus caballus)||4000 BCE[dead link]||Eurasian Steppes|
|Dromedary (Camelus dromedarius)||4000 BCE||Arabia|
|Llama (Lama glama)||6000 BCE||Peru|
|Silkworm (Bombyx mori)||3000 BCE||China|
|Reindeer (Rangifer tarandus)||3000 BCE||Russia|
|Rock pigeon (Columba livia)||3000 BCE||Mediterranean Basin|
|Goose (Anser anser domesticus)||3000 BCE||Egypt|
|Bactrian camel (Camelus bactrianus)||2500 BCE||Central Asia|
|Yak (Bos grunniens)||2500 BCE||Tibet|
|Banteng (Bos javanicus)||Unknown||Southeast Asia|
|Gayal (Bos gaurus frontalis)||Unknown||Southeast Asia|
|Alpaca (Vicugna pacos)||1500 BCE||Peru|
|Ferret (Mustela putorius furo)||1500 BCE||Europe|
|Muscovy duck (Cairina momelanotus)||Unknown||South America|
|Common carp (Cyprinus carpio)||Unknown||East Asia|
|Domesticated turkey (Meleagris gallopavo)||500 BCE||Mexico|
|Goldfish (Carassius auratus auratus)||Unknown||China|
|European rabbit (Oryctolagus cuniculus)||CE 600||Europe|
Second circle[clarification needed]
|Zebu (Bos primigenius indicus)||8000 BCE||India|
|Honey bee||4000 BCE||Multiple places|
|Asian elephant (Elephas maximus) (endangered)||2000 BCE||Indus Valley civilization|
|Fallow deer (Dama dama)||1000 BCE||Mediterranean Basin|
|Indian peafowl (Pavo cristatus)||500 BCE||India|
|Barbary dove (Streptopelia risoria)||500 BCE||North Africa|
|Japanese quail (Coturnix japonica)||1100–1900||Japan|
|Mandarin duck (Aix galericulata)||Unknown||China|
|Mute swan (Cygnus olor)||1000–1500||Europe|
|Canary (Serinus canaria domestica)||1600||Canary Islands, Europe|
This section possibly contains original research. (September 2007)
|Fancy rat (Rattus norvegicus)||1800s||UK|
|Fox (Vulpes vulpes)||1800s||Europe|
|European mink (Mustela lutreola)||1800s||Europe|
|Budgerigar (Melopsittacus undulatus)||1850s||Australia|
|Cockatiel (Nymphicus hollandicus)||1870s||Australia|
|Zebra finch (Taeniopygia guttata)||1900s||Australia|
|Hamster (Mesocricetus auratus)||1930s||United States|
|Silver fox||1950s||Soviet Union|
|Muskox (Ovibos moschatus)||1960s||United States|
|Corn snake (Pantherophis guttatus guttatus)||1960s||United States|
|Ball python (Python regius)||1960s||Africa|
|Madagascar hissing cockroach (Gromphadorhina portentosa)||1960s||Madagascar|
|Red deer (Cervus elaphus)||1970s||New Zealand|
|Hedgehog (Atelerix albiventris)||1980s||United States|
|Sugar glider (Petaurus breviceps)||1980s||Australia|
|Skunk (Mephitis mephitis)||1980s||United States|
|Capybara (Hydrochoerus hydrochaeris)||1990s||United States|
Researchers at the Max Planck institute in Germany are attempting to find a genetic basis for the processes of taming and domestication. They have obtained two strains of grey rats which were bred by Dmitry Konstantinovich Belyaev at the Institute of Cytology and Genetics in Novosibirsk, Russia, research which was later continued by Irina Plyusnina. One strain had been selected for aggressiveness while the other had been selected for tameness, mimicking the process by which neolithic farmers are thought to have first domesticated animals. A similar experiment studying silver foxes has been ongoing at the same institute since 1959. Richard Wrangham of Harvard suggests that similar genes could be involved in human self-domestication.
Some species are said to have been domesticated, but are not any more, either because they have totally disappeared, or since their domestic form no longer exists. Examples include the jaguarundi, the kakapo, the ring-tailed cat, caracal and Bos aegyptiacus.
Hybrid domestic animals
- Alpaca: DNA evidence shows that alpacas are a llama/vicuña hybrid
- Bengal cat
- Cama (animal)
- Domesticated hedgehog: A cross between the Algerian hedgehog and the four-toed hedgehog.
- Sheep-goat hybrid
- Iron Age pig
- Savannah (cat)
Animals of domestic origin and feral ones sometimes can produce fertile hybrids with native, wild animals which leads to genetic pollution in the naturally evolved wild gene pools, many times threatening rare species with extinction. Cases include the mallard duck, wildcat, wild boar, the rock dove or pigeon, the red junglefowl (Gallus gallus) (ancestor of all chickens), carp, and more recently salmon. Another example is the dingo, itself an early feral dog, which hybridizes with dogs of European origin. On the other hand, genetic pollution seems not to be noticed for rabbits. There is much debate over the degree to which feral hybridization compromises the purity of a wild species. In the case of the mallard, for example, some claim there are no populations which are completely free of any domestic ancestor.
Notes and references
- "Domesticate". Oxford Dictionaries. Oxford University Press. 2014.
- ^ a b c d Diamond, Jared (1999). Guns, Germs, and Steel. New York: Norton Press. ISBN 0-393-31755-2.
- "Domestication." Dictionary.com. Based on the Random House Dictionary (Random House, Inc. 2013). http://dictionary.reference.com/browse/domesticate
- Darwin, Charles (1868). The Variation of Animals and Plants under Domestication. London: John Murray. OCLC 156100686.
- Zohary, D. & Hopf, M. (2000). Domestication of Plants in the Old World Oxford: Oxford Univ. Press.[page needed]
- ^ a b Lyudmila N. Trut (1999). "Early Canid Domestication: The Farm-Fox Experiment" (PDF). American Scientist. Sigma Xi, The Scientific Research Society. 87 (March–April): 160–169. Bibcode:1999AmSci..87.....T. doi:10.1511/1999.20.813. Retrieved June 25, 2011.
- Clutton-Brock, J. (1981) Domesticated Animals from Early Times. Austin: Univ. Texas Press.[page needed]
- Ning L., Jinge G. and Aireti. 1997. "Yak in Xinjiang", in Miller D.G., Craig S.R. and Rana G.M. (eds), Proceedings of a workshop on conservation and management of yak genetic diversity held at ICIMOD, Kathmandu, Nepal, October 29–31, 1996. ICIMOD (International Centre for Integrated Mountain Development), Kathmandu, Nepal. pp. 115–122.
- Cronin, M.A.; Renecker, L; Pierson, B.J. and Patton, J.C.; "Genetic variation in domestic reindeer and wild caribou in Alaska"; Animal Genetics, volume 26, Issue 6 (December 1995), pp. 427–434
- Diamond, Jared; Guns, Germs, and Steel: The Fates of Human Societies; p. 147. ISBN 0-393-31755-2
- Diamond, Jared (1998). Guns, Germs, and Steel. Vintage. pp. 169–174. ISBN 978-0-09-930278-0.
- McBride, G., Parer, I.P. and Foenander, F., (1969). The social organization and behaviour of the feral domestic fowl. Animal Behaviour Monographs, 2:125–181
- Stolba, A. and Wood-Gush, D.G.M., (1989). The behaviour of pigs in a semi-natural environment. Animal Production, 48: 419-425
- Sherwin, C.M. (2002). "Comfortable Quarters for Mice in Research Institutions". Animal Welfare Institute. Retrieved November 6, 2013.
- "The Domestication of the Cat". Messybeast.com. October 5, 2009. Retrieved November 5, 2013.
- Hillman G, Hedges R, Moore A, Colledge S, Pettitt P; Hedges; Moore; Colledge; Pettitt (2001). "New evidence of Lateglacial cereal cultivation at Abu Hureyra on the Euphrates". Holocene. 11 (4): 383–393. doi:10.1191/095968301678302823.
- Erickson DL, Smith BD, Clarke AC, Sandweiss DH, Tuross N; Smith; Clarke; Sandweiss; Tuross (December 2005). "An Asian origin for a 10,000-year-old domesticated plant in the Americas". Proc. Natl. Acad. Sci. U.S.A. 102 (51): 18315–20. Bibcode:2005PNAS..10218315E. doi:10.1073/pnas.0509279102. PMC . PMID 16352716.
- Zeven, A. C.; de Wit, J. M. (1982). Dictionary of Cultivated Plants and Their Regions of Diversity, Excluding Most Ornamentals, Forest Trees and Lower Plants. Wageningen, Netherlands: Centre for Agricultural Publishing and Documentation.
- Virányi Z, Gácsi M, Kubinyi E, Topál J, Belényi B, Ujfalussy D, Miklósi Á; Gácsi; Kubinyi; Topál; Belényi; Ujfalussy; Miklósi (2008). "Comprehension of human pointing gestures in young human-reared wolves (Canis lupus) and dogs (Canis familiaris)". Animal Cognition. 11 (3): 373–387. doi:10.1007/s10071-007-0127-y. PMID 18183437.
- Berry, R.J. (1969). "The Genetical Implications of Domestication in Animals". In Ucko, Peter J., Dimbleby, G.W. The Domestication and Exploitation of Plants and Animals. Chicago: Aldine. pp. 207–217.
- Wilkins, Adam S.; Wrangham, Richard W.; Fitch, W. Tecumseh (July 2014). "The 'Domestication Syndrome' in Mammals: A Unified Explanation Based on Neural Crest Cell Behavior and Genetics". Genetics. 197 (3): 795. doi:10.1534/genetics.114.165423.
- Shepherd, Paul (1973). "Ten Thousand Years of Crisis". The Tender Carnivore. Athens, GA: University of Georgia Press. pp. 10–11.
- Boyden, Stephen Vickers (1992). "Biohistory: The interplay between human society and the biosphere, past and present". Man and the Biosphere Series. Pari: UNESCO. 8 (supplement 173): 665. Bibcode:1992EnST...26..665 Check
|bibcode=length (help). doi:10.1021/es00028a604.
- Moore, John. "A Primitivist Primer: What is anarcho-primitivism?".[self-published source]
- [dead link]
- Troy CS, MacHugh DE, Bailey JF; et al. (April 2001). "Genetic evidence for Near-Eastern origins of European cattle". Nature. 410 (6832): 1088–91. doi:10.1038/35074088. PMID 11323670.
- Wendorf F., Schild R.; Schild (1998). "Nabta Playa and its role in ortheastern African prehistory". J. Anthropol. Archaeol. 17 (2): 97–123. doi:10.1006/jaar.1998.0319.
- ^ a b O. Thalmann, B. Shapiro, P. Cui, V. J. Schuenemann, S. K. Sawyer, D. L. Greenfield, M. B. Germonpré, M. V. Sablin, F. López-Giráldez, X. Domingo-Roura, H. Napierala, H-P. Uerpmann, D. M. Loponte, A. A. Acosta, L. Giemsch, R. W. Schmitz, B. Worthington, J. E. Buikstra, A. Druzhkova, A. S. Graphodatsky, N. D. Ovodov, N. Wahlberg, A. H. Freedman, R. M. Schweizer, K.-P. Koepfli, J. A. Leonard, M. Meyer, J. Krause, S. Pääbo, R. E. Green, R. K. Wayne - Complete Mitochondrial Genomes of Ancient Canids Suggest a European Origin of Domestic Dogs - Science 15 November 2013: Vol. 342 no. 6160 pp. 871-874 DOI: 10.1126/science.1243650 Full Text available here
- ^ a b "Oldest Known Pet Cat? 9500-Year-Old Burial Found on Cyprus". National Geographic News. April 8, 2004. Retrieved March 6, 2007.
- ^ a b Muir, Hazel (April 8, 2004). "Ancient remains could be oldest pet cat". New Scientist. Retrieved November 23, 2007.
- ^ a b Walton, Marsha (April 9, 2004). "Ancient burial looks like human and pet cat". CNN. Retrieved November 23, 2007.
- Druzhkova AS, Thalmann O, Trifonov VA, Leonard JA, Vorobieva NV, et al. (2013) Ancient DNA Analysis Affirms the Canid from Altai as a Primitive Dog. PLoS ONE 8(3): e57754. doi:10.1371/journal.pone.0057754
- MSNBCE : World's first dog lived 31,700 years ago, ate big
- Krebs, Robert E. & Carolyn A. (2003). Groundbreaking Scientific Experiments, Inventions & Discoveries of the Ancient World. Westport, CT: Greenwood Press. ISBN 0-313-31342-3.
- Simmons, Paula; Carol Ekarius (2001). Storey's Guide to Raising Sheep. North Adams, MA: Storey Publishing LLC. ISBN 978-1-58017-262-2.
- Giuffra E, Kijas JM, Amarger V, Carlborg O, Jeon JT, Andersson L; Kijas; Amarger; Carlborg; Jeon; Andersson (April 2000). "The origin of the domestic pig: independent domestication and subsequent introgression". Genetics. 154 (4): 1785–91. PMC . PMID 10747069.
- G. Larson, K. Dobney, U. Albarella, M. Fang, E. Matisso-Smith, J. Robins, S. Lowden, H. Finlayson, T. Brand, E. Willerslev, P. Rowley-Conwy, L. Andersson, A. Cooper; Dobney; Albarella; Fang; Matisoo-Smith; Robins; Lowden; Finlayson; Brand; Willerslev; Rowley-Conwy; Andersson; Cooper (March 2005). "Worldwide Phylogeography of Wild Boar Reveals Multiple Centers of Pig Domestication" (PDF). Science. 307 (5715): 1618–21. Bibcode:2005Sci...307.1618L. doi:10.1126/science.1106927. PMID 15761152.
- Melinda A. Zeder, Goat busters track domestication (Physiologic changes and evolution of goats into a domesticated animal), April 2000, (in English) (summarizing research done in Ganj Dareh).
- Source : Laboratoire de Préhistoire et Protohistoire de l'Ouest de la France , (in French).
- , domestication of the cat on Cyprus, National Geographic.
- West B, Zhou B-X.; Zhou (1989). "Did chickens go north? New evidence for domestication" (PDF). World's Poultry Science Journal. 45 (3): 205–218. doi:10.1079/WPS19890012.
- History of the Guinea Pig (Cavia porcellus) in South America, a summary of the current state of knowledge
- Beja-Pereira A, England PR, Ferrand N; et al. (June 2004). "African origins of the domestic donkey". Science. 304 (5678): 1781. doi:10.1126/science.1096008. PMID 15205528. [New Scientist Donkey domestication began in Africa Lay summary] Check
- Roger Blench, (PDF, 235 KB) (in English).
- The Domestication of the Horse; see also Domestication of the horse
- Domestication of Reindeer
- Geese: the underestimated species
- ^ a b Nicholas Wade (July 25, 2006). "Nice Rats, Nasty Rats: Maybe It's All in the Genes". NY times.
- .Sometimes it is because these animals don't breed well in captivity
- Charles Darwin, The Variation of Animals and Plants under Domestication, 1868.
- Jared Diamond, Guns, germs and steel. A short history of everybody for the last 13,000 years, 1997.
- Laura Hobgood-Oster, A Dog's History of the World: Canines and the Domestication of Humans, 2014
- Hope Ryden, Out of the Wild: The Story of Domesticated Animals Hardcover, 1995
- Halcrow, S. E., Harris, N. J., Tayles, N., Ikehara-Quebral, R. and Pietrusewsky, M. (2013), From the mouths of babes: Dental caries in infants and children and the intensification of agriculture in mainland Southeast Asia. Am. J. Phys. Anthropol., 150: 409–420. doi: 10.1002/ajpa.22215
- Hayden, B. (2003). Were luxury foods the first domesticates? Ethnoarchaeological perspectives from Southeast Asia. World Archaeology, 34(3), 458-469.
- Animal husbandry
- Columbian Exchange
- Domesticated silver fox
- Domestication of the horse
- Domestication theory
- Experimental evolution
- Genetic engineering
- Genetic erosion
- Genetic pollution
- Genomics of domestication
- History of plant breeding
- Lion taming
- List of domesticated animals
- List of domesticated fungi and microorganisms
- List of domesticated plants
- Marker assisted selection
- Selective breeding
- Timeline of agriculture and food technology
|Look up domestication or taming in Wiktionary, the free dictionary.|
- Crop Wild Relative Inventory and Gap Analysis: reliable information source on where and what to conserve ex-situ, for crop genepools of global importance
- Discussion of animal domestication
- Guns, Germs and Steel by Jared Diamond (ISBN 0-393-03891-2)
- News story about an early domesticated cat find
- Belyaev experiment with the domestic fox
- Use of Domestic Animals in Zoo Education
- The Initial Domestication of Cucurbita pepo in the Americas 10,000 Years Ago
- Cattle domestication diagram
- Major topic "domestication": free full-text articles (more than 100 plus reviews) in National Library of Medicine
- Why don't we ride zebras? an online children's film about animal domestication
- Isidro A. T. Savillo and Villaluz, Elizabeth A. 2013 this introduces a proposed Domesticity Scale for Wild Birds<|endoftext|>
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WASHINGTON, May 21 (Xinhua) -- A research from the University of Oxford suggested a new theory that for most animals the Cambrian Explosion around 540 to 550 million years ago was a gradual process, but happened no earlier than 550 million years ago.
According to a paper published Monday in the Proceedings of the National Academy of Sciences, a team at Oxford University Museum of Natural History and the University of Lausanne carried out a comprehensive analysis of early fossil euarthropods from every different possible type of fossil preservation.
The Cambrian Explosion produced the largest and most diverse grouping of animals Earth has ever seen, namely the euarthropods.
Euarthropoda contains the insects, crustaceans, spiders, trilobites, and a huge diversity of other animal forms alive and extinct. They comprise over 80 percent of all animal species on the planet and are key components of all of Earth's ecosystems.
The new analysis found, taken together, the total fossil record shows a gradual radiation of euarthropods during the early Cambrian, 540 to 500 million years ago.
It presented a challenge to the two major competing hypotheses about early animal evolution.
The first claimed the nearly instantaneous appearance of euarthropods 540 million years ago because of highly elevated rates of evolution.
The other hypothesis suggested a slow, gradual evolution of euarthropods starting 650 to 600 million years ago.
Instead, the new research adopted a middle-ground between these two hypotheses, with the origin of euarthropods no earlier than 550 million years ago, but with the subsequent diversification taking place over the next 40 million years.
"Each of the major types of fossil evidence has its limitation and they are incomplete in different ways, but when taken together they are mutually illuminating and allow a coherent picture to emerge of the origin and radiation of the euarthropods during the lower to middle Cambrian," said Allison Daley, who carried out the work at Oxford University Museum of Natural History and at the University of Lausanne.
"This indicates that the Cambrian Explosion, rather than being a sudden event, unfolded gradually over the 40 million years of the lower to middle Cambrian," Daley said.
It has been argued that the absence of euarthropods from the Precambrian Period, earlier than around 540 million years ago, is the result of a lack of fossil preservation. But the new fossil study said that this wasn't the case.
"The authors make a very compelling case that the late Precambrian and Cambrian are in fact very similar in terms of how fossils preserve," said Greg Edgecombe from the Natural History Museum, who was not involved in the study. "There is really just one plausible explanation: arthropods hadn't yet evolved."<|endoftext|>
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What Is A Sutra?
This definition of “Sutra” is provided by the UW-Madison Buddhism Study Group:
“The word sutra is the most common term for a Buddhist scripture, so that Buddhists refer to the sutras just as Christians might speak of the Bible.
But although it tends to be used so generally, sutra has a specific meaning.
It comes from a word meaning ‘a thread: so it suggests a number of topics strung together on a common thread of discourse.
The form of a sutra is almost always the same.
First you get a description of where the discourse was given, what was going on, and who was present. That is followed by the main body of the text, which usually consists of a teaching of the Dharma, the real truth, by the Buddha himself. The sutra then ends with an account of the effect of the Buddha’s teaching on the people listening.
It is important to understand that whatever is said in the body of a sutra is not just issuing from the ordinary level of consciousness.
It isn’t something that has been worked out intellectually.
It isn’t a proof or an explanation of something in the mundane sense.
It is a truth, a message, even a revelation, issuing from the depths of the Enlightened consciousness, the depths of the Buddha nature.
This is the essential content of any Buddhist scripture, and this is its purpose: to communicate the nature of Enlightenment and show the way leading to its realization.”
A Brief History of The Diamond Sutra
The World’s Earliest Dated Printed Book
Diamond Sutra Scroll
Diamond Sutra. Cave 17, Dunhuang, ink on paper
British Library Or.8210/ P.2
Copyright © The British Library Board
From “Landmarks in Printing: Diamond Sutra”:
“Hidden for centuries in a sealed-up cave in north-west China, this copy of the ‘Diamond Sutra’ is the world’s earliest complete survival of a dated printed book. It was made in AD 868. Seven strips of yellow-stained paper were printed from carved wooden blocks and pasted together to form a scroll over 5m long. Though written in Chinese, the text is one of the most important sacred works of the Buddhist faith, which was founded in India. Although not the earliest example of a printed book, it is the oldest we have bearing a date. By the time it was made, block-printing had been practised in the Far East for more than a century. The quality of the illustration at the opening of this ‘Diamond Sutra’ shows the carver of the printing blocks to have been a man of considerable experience and skill.
This scroll was found in 1907 by the archaeologist Sir Marc Aurel Stein in a walled-up cave at the ‘Caves of the Thousand Buddhas’, near Dunhuang, in North-West China. It was one of a small number of printed items among many thousands of manuscripts, comprising a library which must have been sealed up in about AD 1000. Although not the earliest example of blockprinting, it is the earliest which bears an actual date.
The colophon, at the inner end, reads: ‘Reverently [caused to be] made for universal free distribution by Wang Jie on behalf of his two parents on the 13th of the 4th moon of the 9th year of Xiantong [i.e. 11th May, AD 868]’. ”
Diamond Sutra Scroll Photo
Here is information from the web page:
The Oldest Printed Text in the World – The Diamond Sutra.
According to National Library of Peking in 1961, the Diamond Sutra is described as: “The Diamond Sutra, printed in the year 868….is the world’s earliest printed book, made of seven strips of paper joined together with an illustration on the first sheet which is cut with great skill.” The writer adds: “This famous scroll was stolen over fifty years ago by the Englishman Ssu-t’an-yin [Stein] which causes people to gnash their teeth in bitter hatred.” It is currently on display in the British Museum. The scroll, some sixteen feet long, 17 an half feet long and 10 and half inches wide, bears the following inscription: ” reverently made for universal free distribution by Wang Jie on behalf of his parents on the fifteenth of the fourth moon of the ninth year of Xian Long (May 11, 868)”
Methodology: How This New Translation Was Done
This new translation was created by taking 15 different previous translations of the Diamond Sutra and carefully reviewing them line by line.
Each chapter was reconstructed line by line, word by word, by comparing each of these different translations.
This new translation kept every element that was common through each of the other 15 translations.
In some cases some translations had more text, and others less. Where the text seemed to be “embellishment” or repetition that I did not believe was necessary to the message I left it out.
If there was any doubt on my part I tended to leave words or passages in rather than remove them if they seemed to work with the tone or tenor of the passage.
I left out most of the long names or names of locations, such as Anathapindika, which was the location in the Jeta Grove where the Buddha spoke.
I also left out words like Tathagata, the Arhat, Bodhisattvas, Bhagavat, Mahasattvas, Bhikshus, Nirvana, Anuttarasamyaksambodhi, and annutara-samyak-sambodhicitta. In every case I came up with a simple word which kept the spirit of the Sutra while making it easier to read.
For forms of address I chose “Most Honored One” or “Buddha” to refer to the Buddha.
I chose this over other forms in other translations such as these: “World-Honored One”, “O Lord”, “O Well-Gone”, “Tathagata”, “Arhat”, “The Fully Enlightened One”, “The Lord”, “O Sugata”, and “Thus Come One”.
There is a balance between a very “formal” style in some translations, and overly “familiar” styles present in some others. This translation finds a middle ground, maintaining and respecting the seriousness of the setting and message, while avoiding cumbersome phrases or attempting to “dumb down” the style presented in the bulk of the other translations.
For example, one translation had the phrase “So listen up, Subhuti”, another translation had “Buddha replied: Listen carefully”, a third said “Therefore, O Subhuti, listen and take it to heart, well and rightly”, and finally another said “Please listen with all of your attention and the Tathagata will respond to your question”.
The final translation that I came up with in this instance is this: “Listen carefully with your full attention, and I will speak to your question.”
After reviewing each chapter line by line, word by word, I worked the final translation over one more time to make any minor adjustments that would make the text flow more like our modern language.
The resulting translation presented here is one that is true to every line and every word of the original Sutra, as passed on through these 15 earlier translations.
Finally, you will find below a list of the other translations I used in preparing this new translation.
I hope you enjoy this site and come back often to read this wonderful Buddhist text, and feel free to show your support for this work presented here by donating to the hosting, registration, and upkeep of this website. Just click here to support this site.
Other Translations Used In Creating This New Translation
The Vajracchedika Prajnaparamita Sutra – A Translation of the Diamond Sutra.
Translation on Buddhism Today
The Oldest Printed Text in the World – The Diamond Sutra
Note: the sites below are no longer available. The translation used on this website was completed in 2003, and all of these sites were active at the time.
www.Sinc.sunysb.edu – (Link No Longer Available)
www.io.com – (Link No Longer Available)
www.buddhistinformation.com – (Link No Longer Available)
www.buddhistinformation.com 2 – (Link No Longer Available)
www.buddhismtoday.com – (Link No Longer Available)
www.buddhismtoday.com 2 – (Link No Longer Available)
www.community.palouse.net – (Link No Longer Available)
home.flash.net/~cameron/texts/diamond.html – (Link No Longer Available)v
www.buddhistinformation.com 2 – (Link No Longer Available)
www.buddhistinformation.com 3 – (Link No Longer Available)
www.gruntose.com – (Link No Longer Available)
www.terebess.hu – (Link No Longer Available)
There are also three other web sites that had translations that are no longer available.
* To begin reading the Diamond Sutra just Click Here for Chapter 1.<|endoftext|>
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Plant-eating dinosaurs (Barosaurus and Dryosaurus) and a meat-eating dinosaur (Allosaurus) of Jurassic North America. Dinosaurs can be divided into plant-eaters or meat-eaters, although some may have been omnivores: they ate both animals and plants. The shape of a dinosaur’s jaws or teeth is often the first clue that reveals to which group it belonged and how it obtained its food. We know from the fossil record what food was available at a certain time. In some fossils, the stomach contents have also been preserved, providing vital clues about diet. We can also work out how dinosaurs fed simply by studying modern animals’ eating habits.
The sauropods’ sheer size meant they would have needed immense amounts of food in the form of plant matter to nourish them. Yet their teeth were small, few in number and incapable of chewing. How could they digest their colossal meals?
The presence of pebbles in some fossil remains provides the answer. Known as gastroliths, these polished, rounded pebbles were a vital part of the digestive system.
From studying how birds—the dinosaurs’ modern relatives—eat, we know that they swallow grit or small stones to fill their gizzard, a part of their stomach. Muscular movements inside the gizzard cause the stones to grind the food into a paste, which can then be digested in the intestines. Sauropods such as Apatosaurus could digest their vast daily intake of unchewed vegetation with the help of gastroliths in the same way.A large herd of Apatosaurus lumber through a Jurassic forest, stripping enormous amounts of leaves from the trees as they go.
Palaeontologists can discover a lot about a dinosaur's feeding habits from studying dinosaurs' jaws and teeth as in, for example,...Read More >>Palaeontologists can discover a lot about a dinosaur's feeding habits from studying dinosaurs' jaws and teeth as in, for example, Saurolophus, whose fossil skull is pictured here.
One unusual group of dinosaurs, the heterodontosaurs, had teeth for eating both plants and flesh in their jaws. Their cheek teeth were good for grinding tough plant material, while the pointed fangs at the front of their jaws, along with the sharp, curved claws on their arms, suggest they were predators, too.
Find the answer<|endoftext|>
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Our engaging, dynamic exhibits and programs are aligned with state and national standards so you can connect your field trip with your classroom curriculum.
Download the Guide to Standards-Aligned Exhibits & Programs.
About the Document
Drafted by George Mason, this declaration of rights later became a model for other state constitutions and the Bill of Rights.
Adopted June 12, 1776
A DECLARATION OF RIGHTS made by the representatives of the good people of Virginia, assembled in full and free convention which rights do pertain to them and their posterity, as the basis and foundation of government .
Section 1. That all men are by nature equally free and independent and have certain inherent rights, of which, when they enter into a state of society, they cannot, by any compact, deprive or divest their posterity; namely, the enjoyment of life and liberty, with the means of acquiring and possessing property, and pursuing and obtaining happiness and safety.
Section 2. That all power is vested in, and consequently derived from, the people; that magistrates are their trustees and servants and at all times amenable to them.
Section 3. That government is, or ought to be, instituted for the common benefit, protection, and security of the people, nation, or community; of all the various modes and forms of government, that is best which is capable of producing the greatest degree of happiness and safety and is most effectually secured against the danger of maladministration. And that, when any government shall be found inadequate or contrary to these purposes, a majority of the community has an indubitable, inalienable, and indefeasible right to reform, alter, or abolish it, in such manner as shall be judged most conducive to the public weal.
Section 4. That no man, or set of men, is entitled to exclusive or separate emoluments or privileges from the community, but in consideration of public services; which, nor being descendible, neither ought the offices of magistrate, legislator, or judge to be hereditary.
Section 5. That the legislative and executive powers of the state should be separate and distinct from the judiciary; and that the members of the two first may be restrained from oppression, by feeling and participating the burdens of the people, they should, at fixed periods, be reduced to a private station, return into that body from which they were originally taken, and the vacancies be supplied by frequent, certain, and regular elections, in which all, or any part, of the former members, to be again eligible, or ineligible, as the laws shall direct.
Section 6. That elections of members to serve as representatives of the people, in assembly ought to be free; and that all men, having sufficient evidence of permanent common interest with, and attachment to, the community, have the right of suffrage and cannot be taxed or deprived of their property for public uses without their own consent or that of their representatives so elected, nor bound by any law to which they have not, in like manner, assembled for the public good.
Section 7. That all power of suspending laws, or the execution of laws, by any authority, without consent of the representatives of the people, is injurious to their rights and ought not to be exercised.
Section 8. That in all capital or criminal prosecutions a man has a right to demand the cause and nature of his accusation, to be confronted with the accusers and witnesses, to call for evidence in his favor, and to a speedy trial by an impartial jury of twelve men of his vicinage, without whose unanimous consent he cannot be found guilty; nor can he be compelled to give evidence against himself; that no man be deprived of his liberty except by the law of the land or the judgment of his peers.
Section 9. That excessive bail ought not to be required, nor excessive fines imposed, nor cruel and unusual punishments inflicted.
Section 10. That general warrants, whereby an officer or messenger may be commanded to search suspected places without evidence of a fact committed, or to seize any person or persons not named, or whose offense is not particularly described and supported by evidence, are grievous and oppressive and ought not to be granted.
Section 11. That in controversies respecting property, and in suits between man and man, the ancient trial by jury is preferable to any other and ought to be held sacred.
Section 12. That the freedom of the press is one of the great bulwarks of liberty, and can never be restrained but by despotic governments.
Section 13. That a well-regulated militia, composed of the body of the people, trained to arms, is the proper, natural, and safe defense of a free state; that standing armies, in time of peace, should be avoided as dangerous to liberty; and that in all cases the military should be under strict subordination to, and governed by, the civil power.
Section 14. That the people have a right to uniform government; and, therefore, that no government separate from or independent of the government of Virginia ought to be erected or established within the limits thereof.
Section 15. That no free government, or the blessings of liberty, can be preserved to any people but by a firm adherence to justice, moderation, temperance, frugality, and virtue and by frequent recurrence to fundamental principles.
Section 16. That religion, or the duty which we owe to our Creator, and the manner of discharging it, can be directed only by reason and conviction, not by force or violence; and therefore all men are equally entitled to the free exercise of religion, according to the dictates of conscience; and that it is the mutual duty of all to practise Christian forbearance, love, and charity toward each other.<|endoftext|>
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This map uses Lambert azimuthal equal-area projection. But what is a Lambert, or an azimuthal and what is projection? Well, lets start about what a projection actually is...
Map projection. Map projection is a way to represent the globe which is an inperfect sphere, onto a flat survace. (For example a large piece of paper or the screen you are currently watching ;-)) And why? Well in order to get a useable map. And that last part, well that is the hard part. You have to deal with the practival use of maps (navigation, geograpgical shapes, distance between objects, etc) as well as very complicated mathematical problems about how to shape the world. As you must surely understand, there is not one answer to this problem. Any 2D projection of the world will always distort the reality in some smaller or larger way. Because every problem has it's own solution, there are numerous amounts of projections. Each was developed to minimize the disortion for that specific use. Some are very famous, some you will probably never have heard from. Maybe click around the globe on mapclicker and find the obscure one you where looking for.
Johann Heinrich Lambert was born in Mulhouse in France and was a German/Swiss scientist. He did more than creating maps, he was also a mathematician, an astronomer, physicist and philosopher. You could say he was quite the cool guy, mostly because he was the first that proofed that pi (3,14∞) is an irrational number. That means you cannot write pi as a fraction. Lambert is seen as the founder of modern cartography with his theory of map projections. The most famous and widely used is LAEA (used for Oceania). Other widely used projections from Lambert are LCC (used for Canada) and Transeverse Mercator (used for Finland).<|endoftext|>
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The obvious place to divide the Northern and Southern Hemispheres was the equator. But the division of the Eastern and Western hemispheres was the source of much political turmoil. Greenwich (Great Britain) won, placing for example The Netherlands in the Eastern and Ireland in the Western Hemisphere.
It takes the earth 24 hours for a full rotation of 360°. Thus, every hour we rotate 15° longitude, see figure 2.
When it is 12:00 UTC (international standard time) - anywhere in the world - it is 12:00 Local Time in Greenwich and 24:00 Local Time at the other side of the planet: 180° E or 180° W: the date line. Crossing this special meridian changes not only the hour but also the date.
The North Pole has a latitude of 90° N and the South Pole 90° S. The meridians cover twice this angle up to 180° W or E.
Meridians converge at the poles, whereas parallels run parallel to each other and never meet. All meridians and the equator - the biggest parallel - form great circles, and the remaining parallels form so-called small circles. A great circle divides the earth in two exact halves.
In figure 3 the position of Boston in the United States is shown using latitude and longitude in degrees, minutes and seconds:
42° 21' 30" N , 71° 03' 37" W
Most sailors will actually notate seconds in metric fractions of minutes:
42° 21,5' N , 71° 03,6' W or
42° 21.5' N , 71° 03.6' W, see the notation style guide
On small scaled charts we want to be accurate within one minute or one nautical mile. On larger scaled charts the accuracy is more likely to be within a tenth of a mile (a cable).
If the earth were a perfect sphere with a circumference of roughly 40000 kilometres all great circles - meridians plus the equator - would have the same length and could be used as a distance unit when divided
into 360 degrees, or 360° x 60' = 21600' minutes.
In 1929, the international community agreed on the definition of 1 international nautical mile as 1852 metres, which is roughly the average length of one minute of latitude i.e. one minute of arc along a line of longitude (a meridian).
Or to put it shortly: 1 nm = 1'
We are now able to describe any position in latitudes and longitudes. Moreover, we can state the distance between two of those positions using nautical miles or minutes. All we need now is a proper way to define speed. For that, sailors use knots, the number of nautical miles an hour.
The portolans contained maps of coastlines, locations of harbours, river mouths, and man-made features visible from the sea. They were a compilation of centuries of seafarer observations. As sailors' skills improved and the use of the compass was more widespread, portolans improved in accuracy.
Also Columbus used these portolans on his journeys. Portuguese chart makers added the meridian line, a point useful for latitude sailing as well as for navigating solely by compass. A geographic feature could now be located through the use of its distance in degrees of latitude from a ship's point of departure. Note that the use of latitude and longitude was understood since the time of Ptolemy, the second century CE.
During the fifteenth century Portugal led the European world in sea exploration. The golden age of discovery for Portugal lasted almost a century until the Dutch eventually seized their trade routes from them.
As we move to the next chapter of this course we enter the sixteenth century when the Mercator chart was invented.
Use the logo to navigate through this course,
...or go to the next chapter,
or download the complete navigation course as PDF - including practice materials, exercises and answers.<|endoftext|>
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Rick Thoman, ACCAP
Sparsely populated, the Arctic has few weather stations and lacks a long record of climate observations, but even these limited data show that the region is highly sensitive to the effects of greenhouse gases. Temperatures in the Arctic are warming faster than any other region on Earth, especially in winter. The resulting dramatic losses of sea ice and thawing permafrost, glaciers, and ice sheets feed back to the global climate and amplify the effects of warming. At the same time, the region’s climate varies widely from year to year and from place to place, making it challenging to predict the future and critical to make the most of the information we have.<|endoftext|>
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Bouncing Ball
Problem
A ball is projected from the bottom left corner of unit square $ABCD$ into its interior. We shall assume that the speed of the ball remains constant and it will continue bouncing off the edges until it arrives at a corner. For example, if the ball strikes $\frac{2}{3}$ of the way from $D$ to $C$ it will terminate at $D$.
Where must the ball strike on $DC$ to finish at $A$?
Solution
We shall solve this problem in two ways.
Method 1
By allowing the square to be reflected we can consider the reflection to be a continuous straight path. In the example below, the reflected path on the left diagram can be represented by the continuous path in the diagram on the right.
Extending this across the entire plane we can place the original unit square $ABCD$ on co-ordinate axes with $A$ at the origin.
It can be seen that for the ball to finish at $A$ it must terminate at one of the image points of $A$, represented by $A'$. As we are dealing with a unit square these points would be represented by the co-ordinate $(x,y)$ where both $x$ and $y$ are even. Let $x = 2a$ and $y = 2b$.
However, if a line passes through the point $(2a,2b)$ then it must first have passed through the point $(a,b)$, which means that it is impossible for the ball to return to its corner of origin before first arriving at one of the other corners.
Method 2
First we shall consider rectangle $ABCD$. Suppose the ball is projected from $A$ towards $E$, a point on $DC$, such that the length $DE$ splits $DC$ into equal sections. In the example below, $DE = \frac{1}{4}DC$.
It can be seen that the ball will terminate at $B$ if the number of sections is even and it will terminate at $C$ if the number of sections is odd.
We shall now approach the problem in a similar manner to the previous method. We shall allow the ball to bounce off the top and bottom edges: DC and $AB$, but instead of allowing the ball to bounce off the vertical edges: $DA$ and $CB$, we shall repeatedly reflect the square in a horizontal direction to consider the path of the ball to be a continuous path.
For example, in the unit square $ABCD$ let $DE = \frac{2}{3}$. As $3 \times \frac{2}{3} = 2$, it will be necessary to have two complete squares for the ball to reach a corner; in this case the ball would terminate at $D$.
In general, if $DE = \frac{p}{q}$, where $GCD(p,q) = 1$, then we shall have $q \times \frac{p}{q} = p$ complete squares and we shall generate $q$ sections.
We have already established that it is necessary for there to be an even number of sections in a rectangle for the ball to terminate in the bottom corner, so $q$ must be even. However, it can be seen that as the square is repeatedly reflected the bottom right corner alternates between the images of $A$ and $B$ Hence there must be an even number of squares for the image of $A$ to be in the bottom right corner.
This would require both $p$ and $q$ to be even which is a contradiction. Hence it is impossible for the ball to terminate at $A$.
Given that $DE = \frac{p}{q}$ where $GCD(p,q) = 1$, can you deduce which corner the ball will terminate in?
What would happen if $DE$ were irrational?
Problem ID: 363 (28 Oct 2009) Difficulty: 3 Star
Only Show Problem<|endoftext|>
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# Standard Algebraic Identities | Class 8 Maths
• Last Updated : 28 Jan, 2022
Algebraic Identities are algebraic equation which is always true for every value of the variable in them. The algebraic equations that are valid for all values of variables in them are called algebraic identities. It is used for the factorization of polynomials. In this way, algebraic identities are used in the computation of algebraic expressions and solving different polynomials. They contain variable and constant on both the side of polynomial i.e. LHS and RHS. In algebraic identity, LHS must be equal to RHS.
### What is Identity?
Consider the equality (x + 1) (x +2) = x2 + 3x + 2. One can evaluate both sides of this equality for some value of a, say x = 5. For x = 5,
• LHS = (x + 1) (x + 2) = (5 + 1) (5 + 2) = 6 × 7 = 42
• RHS = x2 + 3x + 2 = 52 + 3 × 5 + 2 = 25 + 15 + 2 = 42
Thus, the values of the two sides of the equality are equal for a = 5. One can find that for any value of x, LHS = RHS. Such equality, true for every value of the variable in it, is called an identity. Thus, (x + 1) (x + 2) = x2 + 3x + 2 is an identity.
### Standard Identities
All standard Algebraic Identities are derived from the Binomial Theorem. There are number of algebraic identities but few are standard that are listed below.
• (a + b)2 = a2 + 2ab + b2
• (a – b)2 = a2 + b2 – 2ab
• (a + b)(a – b) = a2 – b2
• (a + b)3=a3 + b3 + 3ab(a + b)
• (a – b)3=a3 – b3 – 3ab(a – b)
• (a + b + c)2=a2 + b2 + c2 + 2ab + 2bc + 2ca
### Methods to Solve Identities
• We can verify algebraic identities by substitution method, in which we can put values in variable place and try to make both sides equal. i.e LHS = RHS.
Example:
(a – 2) (a + 2) = a2 – 22
Now we will start putting value in place of a.
starting with a = 1, (-1) x (3) = -3
then we will put a = 2, 0 x 4 = 0
Here we got a = 1 and a = 2 as the value which satisfy the given question.
• Another method is by manipulating identities which are commonly used:
i. (a + b)2 = a2 + b2 + 2ab
ii. (a – b)2 = a2+ b2 – 2ab
iii. (a + b)(a – b) =a2 – b2
iv. (x + a)(x + b) = x2 + (a + b)x + ab
Proof:
i. (a + b)2 = (a + b) (a + b)
= (a + b) (a) + (a + b) (b)
= a2 + ab + ab + b2
= a2 + 2ab + b2
Hence, LHS = RHS.
ii. (a – b)2 = (a – b) (a – b)
= (a – b) (a) + (a – b) (b)
= a2 – ab – ba + b2
= a2 – 2ab + b2
Hence, LHS = RHS.
iii. (a + b) (a – b) = a (a – b) + b (a – b)
= a2 – ab + ab – b2
= a2 – b2
Hence, LHS = RHS.
## Applying Identities
Example 1: Solve (2x + 3) (2x – 3) using algebraic identities?
Solution:
By the algebraic identity (a + b)(a – b) = a2 – b2
We can re-write the given expression as
(2x + 3) (2x – 3) = (2x)2 – (3)2 = 4x2– 9
Example 2: Solve (3x + 5)2 using algebraic identities?
Solution:
By algebraic identity
(a + b)2 = a2 + b2 + 2ab
We can re-write the given expression as;
(3x + 5)2 = (3x)2 + 2(3x)5 + 52
(3x + 5)2 = 9x 2 + 30x + 25
Example 3: Find the product of (x + 1)(x + 1) using standard algebraic identities?
Solution:
(x + 1)(x + 1) can be written as (x + 1)2. Thus, it is of the standard form I where a = x and b = 1.
We have,
(x + 1)2 = (x)2 + 2(x)(1) + (1)2 = x 2 + 2x + 1
Example 4: Expand (3x – 4y)3 using standard algebraic identities?
Solution:
(3x – 4y)3 is of the standard form VII where a = 3x and b = 4y.
We have,
(3x – 4y)3 = (3x)3 – (4y)3 – 3(3x)(4y)(3x – 4y) = 27x 3 – 64y 3 – 108x2y + 144xy 2
My Personal Notes arrow_drop_up<|endoftext|>
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# Section 4.3 & 4.4: Proving s are Congruent - PowerPoint PPT Presentation
Section 4.3 & 4.4: Proving s are Congruent. Goals. Identify figures and corresponding parts Prove that 2 are . Anchors. Identify and/or use properties of congruent and similar polygons Identify and/or use properties of triangles. M. Q. N. R. P. S.
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### Section 4.3 & 4.4: Proving s are Congruent
Goals
• Identify figures and corresponding parts
• Prove that 2 are
Anchors
• Identify and/or use properties of congruent and similar polygons
• Identify and/or use properties of triangles
Q
N
R
P
S
Side-Side-Side (SSS) Postulate
• If 3 sides of 1 are to 3 sides of a 2nd , then the 2 ’s are .
If Side MN QR
Side NP RS and
Side PM SQ
Then MNP QRS
Then we can say: M Q, N R , and P S
Reasons
Given: W is the midpoint of QS PQ TS and PW TWProve: PQW TSW
• W is the mdpt of QS,
• PQ TS and PW TW
• Given
2) QW SW
2) Def. of midpoint
3) PQW TSW
3) SSS
Reasons
Given: D is the midpoint of ACABC is isosceles ABC is the vertex angleProve: ABD CBD
• D is the mdpt of AC,
• ABC is isosceles
• Given
2) Def. of midpoint
3) AB BC
3) Property of Isosceles
4) BD BD
4) Reflexive
5) ABD CBD
5) SSS
X
)
P
W
S
Y
)
Side-Angle-Side (SAS) Postulate
• If 2 sides and the included of 1 are to 2 sides and the included of a 2nd , then the 2 s are .
If Side PQ WX
Angle Q X
Side QS XY
Then PQS WXY
Then we can say: PS WY, P W , and S Y
Reasons
Given: QRS is isosceles RT bisects QRS QRS is the vertex angle Prove: QRT SRT
)
• QRS is isosceles
• RT bisects QRS
• Given
2) QRT SRT
2) bisector
3) QR RS
3) Property of Isosceles
4) RT RT
4) Reflexive
5) QRT SRT
5) SAS
Reasons
Given: BD and AE bisect each otherProve: ABC EDC
)
)
• BD and AE bisect
• each other
• Given
2) BC CD, AC CE
2) Segment bisectors
3) BCA ECD
3) Vertical angles
4) ABC EDC
4) SAS
Q
M
)
R
N
S
P
)
)
Angle-Side-Angle (ASA) Postulate
• If 2 ’s and the included side of 1 are to 2 ’s and the included side of a 2nd, then the 2 are
If Angle N R
Side MN QR
Angle M Q
Then MNP QRS
Then we can say: MP QS, NP RS , and P S
)
Statements
Reasons
Given: B N RW bisects BNProve: BRO NWO
)
)
• B N
• RW bisects BN
• Given
2) BOR WON
2) Vertical Angles
3) BO ON
3) Segment bisector
4) BRO NWO
4) ASA
1
3
4
2
Statements
Reasons
Given: 1 2 CD bisects BCEProve: BCD ECD
)
)
• 1 2
• CD bisects BCE
• Given
2) 3 4
• Supplements of congruent s are congruent
3) BCD ECD
3) Angle bisector
4) BCE is isosceles
4) Property of isosceles
5) BC CE
5) Property of isosceles
6) BCD ECD
6) ASA
Q
(
(
W
P
Y
S
(
(
Angle-Angle-Side (AAS) Theorem
• If 2 ’s and a non-included sideof 1 are to 2 ‘s and a non-included side of a 2nd , then the 2 ’s are .
If Angle P W
Angle S Y
Side QP WX
Then PQS WXY
Then we can say: QS XY, PS WY , and Q X
)
Statements
Reasons
Given: AD ║ EC , B is the mdpt of CDProve: ABD EBC
)
)
1) AD ║ EC , B is the mdpt of CD
• Given
2) A E
2) Alternate Interior s
3) ABD CBE
3) Vertical Angles
4) BD BC
4) Midpoint
5) ABD EBC
5) AAS
)
Statements
Reasons
Given: AD ║ EC , B is the mdpt of CDProve: ABD EBC
)
)
1) AD ║ EC , B is the mdpt of CD
• Given
2) A E, D C
2) Alternate Interior s
3) BD BC
3) Midpoint
4) ABD EBC
4) AAS
40
40
50
50
Why Angle-Angle-Angle (AAA)Doesn’t Work
The angles are , but the sides are proportional.
(
A
D
F
(
C
B
Why Side-Side-Angle (SSA)Doesn’t Work
Two different triangles can be formed if you use two sides and a non-included angle.
Theorem 4.8: Hypotenuse-Leg (HL) Theorem
• If the hypotenuse and a leg of a right are to a hypotenuse and a leg of a 2nd right , then the 2 ’s are
D
A
If BC EF and AC DF,
then ABC DEF
Special case of SSA
B
C
E
F
Then we can say: AB DE, A D , and C F
Reasons
Given: RS QT QRT is isosceles QRT is the vertex angleProve: QRS TRS
1) RS QT, QRT is isosceles
• Given
2) QSR 90,
TSR 90
2) Definition of perpendicular
3) QSR TSR
3) Substitution
4) QR RT
4) Property of isosceles
5) RS RS
5) Reflexive
6) QRS TRS
6) HL<|endoftext|>
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# Lesson 1Go the DistanceDevelop Understanding
## Learning Focus
Find the distance between two points in the coordinate plane.
Find the perimeter of a geometric figure in the coordinate plane.
How can I find the distance between two points if they’re not on vertical or horizontal lines?
## Open Up the Math: Launch, Explore, Discuss
The performances of the Crawford High School drill team are very popular during half-time at the school’s football and basketball games. When the Crawford High School drill team choreographs the dance moves that they will do on the football field, they lay out their positions on a grid like the one shown:
In one of their dances, they plan to make patterns holding long, wide ribbons that will span from one dancer in the middle to six other dancers. On the grid, their pattern looks like this:
The question the dancers have is how long to make the ribbons. Gabriela () is standing in the center, and some dancers think that the ribbon from Gabriela () to Courtney () will be shorter than the one from Gabriela () to Brittney ().
### 1.
How long does each ribbon need to be?
### 2.
Explain how you found the length of each ribbon.
When they have finished with the ribbons in this position, they are considering using them to form a new pattern like this:
### 3.
Will the ribbons they used in the previous pattern be long enough to go between Brittney () and Courtney () in the new pattern? Explain your answer.
Gabriela notices that the calculations she is making for the length of the ribbons remind her of math class. She says to the group, “Hey, I wonder if there is a process that we could use like what we have been doing to find the distance between any two points on the grid.” She decides to think about it like this:
“I’m going to start with two points and draw the line between them that represents the distance that I’m looking for. Since these two points could be anywhere, I named them and . Hmmmmm. . . when I figured the length of the ribbons, what did I do next?”
### 4.
Think back on the process you used to find the length of the ribbon and write down your steps here, in terms of and .
### 5.
Use the process you came up with in problem 4 to find the distance between two points located far enough away from each other that using your formula from problem 4 is more efficient than graphing and counting. For example, find the distance between and .
### 6.
Use your process to find the perimeter of the hexagon pattern shown in problem 3.
Find as many points as you can that are units away from the point . Plot each of the points along with . What do you notice about the graph of the points?
## Takeaways
Steps in words
Steps in symbols
The Distance Formula:
Finding the perimeter of a geometric figure on the coordinate plane:
## Lesson Summary
In this lesson, we learned to find the distance between two points. We used the Pythagorean theorem to develop a formula that could be used whenever we need to find the length of a segment between two points. The formula can be applied to find the length of the sides of a geometric figure in the coordinate plane to calculate the perimeter.
## Retrieval
### 1.
Point is graphed in the coordinate plane.
1. Rotate counterclockwise about the origin, and label the new coordinates. The new point is . Compare the coordinates of with the coordinates of .
2. Rotate point clockwise about the origin. Label the new coordinates. The new point is . Compare the coordinates of with the coordinates of .
### 2.
Fill in the missing coordinates. Then find and .<|endoftext|>
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The mosquito Aedes aegypti transmits viruses that cause diseases such as dengue, chikungunya and zika. Measures are taken to control the mosquito since these infectious diseases represent a significant health problem. This is the case on the island of Saba, a Dutch Caribbean island. In order to fight these diseases a British company has genetically modified the mosquito in such a way that it can suppress local mosquito populations. The modification causes the mosquitoes' offspring to die prematurely. The potential release of these mosquitoes on Saba is considered to result in negligible risks for human health and the environment.
This is the outcome of a technical evaluation of the potential release of these genetically modified mosquitoes. RIVMRijksinstituut voor Volksgezondheid en Milieu's GMO (Genetically Modified Organisms) Office was commissioned by the Executive Council of Saba to perform this evaluation.
Among others, this evaluation looked into effects on the food chain to determine whether an important food source would disappear if the local mosquito population were to be eliminated. It was also considered whether it is unhealthy if people accidentally swallow a genetically modified mosquito. Another element of the evaluation was whether the genetic modification would increase the efficiency of the mosquito to spread diseases.
An evaluation of the efficacy of application of the genetically modified mosquitoes was not part of this technical evaluation. The same applies to socio-economic effects or the desirability of using these mosquitoes.
Technical evaluation of a potential release of OX513A Aedes aegypti mosquitos on the island of Saba PDF | 1.5 MB<|endoftext|>
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# SOLUTION: Gina is going to a party and was asked to bring 3lbs of mixed nuts that includes cashews and peanuts. If the cashews cost \$6.00 per pound and the peanuts cost \$1.50 per pound, how
Algebra -> Algebra -> Customizable Word Problem Solvers -> Mixtures -> SOLUTION: Gina is going to a party and was asked to bring 3lbs of mixed nuts that includes cashews and peanuts. If the cashews cost \$6.00 per pound and the peanuts cost \$1.50 per pound, how Log On
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Word Problems: Mixtures Solvers Lessons Answers archive Quiz In Depth
Question 85592: Gina is going to a party and was asked to bring 3lbs of mixed nuts that includes cashews and peanuts. If the cashews cost \$6.00 per pound and the peanuts cost \$1.50 per pound, how many pounds of each did she buy if her total cost was \$9.00? I know the answer from doing it in my head, I just need help setting up the equation properly on paper. Thanks for your input / time!Answer by ptaylor(2048) (Show Source): You can put this solution on YOUR website! Let x=number of lbs cashews Then 3-x=number of lbs of peanuts We know that \$6.00x=cost of the cashews We also know that \$1.50(3-x)=cost of peanuts Now we are told that her total cost was \$9.00, so our equation to solve is: \$6.00x+\$1.50(3-x)=\$9.00 get rid of parens \$6.00x+\$4.50-\$1.50x=\$9.00 subtract \$4.50 from both sides \$6.00x+\$4.40-\$4.50-\$1.50x=\$9.00-\$4.50 collect like terms \$4.50x=\$4.50 divide both sides by \$4.50 x=1 lb---------------------------------number of lbs of cashews 3-x=3-1=2 lbs-----------------------number of lbs of peanuts CK \$6.00(1)+\$1.50(2)=\$9.00 \$6.00+\$3.00=\$9.00 \$9.00=\$9.00 Hope this helps---ptaylor<|endoftext|>
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# FAQ: How To Square Root Without A Calculator?
## How do you manually calculate square roots?
Take the number you wish to find the square root of, and group the digits in pairs starting from the right end. For example, if you want to calculate the square root of 8254129, write it as 8 25 41 29. Then, put a bar over it as when doing long division.
## How do you find the square root of 5 without a calculator?
Calculate 5 x 45, write that below 245, subtract, bring down the next pair of digits (in this case the decimal digits 00). 504 x 4 = 2016, so 3 works. Calculate 3 x 503, write that below 2000, subtract, bring down the next digits. 5069 x 9 = 45621, which is less than 49100, so 9 works.
## What is the formula of square root?
The two square numbers in between which;3′ lies are 1 and 4. So, the square root of 3 lies between 1 and 2. Find the average of these two numbers to get the square root of 3. Square root of 3 = (1.5 + 2)/2 = 1.75 which is approximately equal to square root of 3. Square Root Formulas with Examples.
2 576
2 18
3 9
3 3
1
## What is the shortcut key for square root?
The Alt code for the symbol for the Square root is Alt +251 or 221A, then Alt+X. Follow these three simple steps to attach the symbol using the Alt code: – Position the pointer in the place where you want the square root symbol inserted. – Press and hold down the Alt key and type 251 from the numeric keypad.
You might be interested: Question: How To Insert Square Root?
## What are all the real square roots of 100?
List of Perfect Squares
NUMBER SQUARE SQUARE ROOT
97 9,409 9.849
98 9,604 9.899
99 9,801 9.950
100 10,000 10.000
## How do you find the square root of 15?
The square root of 15 can be expressed as √ 15 or 15 ½. It is an irrational number. We can find the square root of 15 using the long division method. The square root of 15 by long division method = 3.872 (to three decimal places)
## How do you find the square root of 85?
The square root of 85 is a non-terminating decimal, i.e., √ 85 = 9.219544.
## What are the roots of 4?
Square Root From 1 to 50
Number Square Root Value
3 1.732
4 2
5 2.236
6 2.449<|endoftext|>
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Conic Section
# Normal to a Parabola
## Equation of the Normal at the Point $(x_1,y_1)$ of the Parabola $y^2=4ax$
Let the equation of a parabola be $y^2=4ax$ Take any point $P(x_1,y_1)$ on the parabola. Then, the equation of the tangent at point $P$ is given by $yy_1=2a(x+x_1)$
$\text{Slope of the tangent}=\frac{2a}{y_1}$ $\therefore\text{Slope of the normal}(m)=-\frac{y_1}{2a}$ Hence the equation of the normal is $y-y_1=-\frac{y_1}{2a}(x-x_1)\text{ __(1)}$
## Equation of the Normal in Slope Form (m-form)
We have, the equation of normal to the parabola $y^2=4ax$ at the point $(x_1,y_1)$, $y-y_1=-\frac{y_1}{2a}(x-x_1)$ $\text{Slope}\:(m)=-\frac{y_1}{2a}$ $\therefore y_1=-2am$ Since the point $(x_1,y_1)$ passes through the parabola, we have $y_1^2=4ax_1$ $\therefore x_1=\frac{y_1^2}{4a}=\frac{(-2am)^2}{4a}=am^2$ Thus the equation of the normal to the parabola in slope form is $y-(-2am)=m(x-am^2)$ $\therefore y=mx-2am-am^3$ This form of the normal is also known as the m-form.
## Equation of the Normal to the Parabola in Parametric Form
Let $P(at_1^2,2at_1)$ be a point on a parabola $y^2=4ax$. Then, the equation of the tangent in parametric form is given by $x-yt_1+at_1^2=0$ $\text{Slope of tangent}=\frac{1}{t_1}$ $\therefore\text{Slope of normal}=-t_1$ Hence, the equation of the normal is $y-2at_1=-t_1(x-at_1^2)$ $y=-t_1x+2at_1+at_1^3$ $\therefore y+t_1x=2at_1+at_1^3$
### Obtain the equation of the normal to the parabola $y^2=8x$ at $(2,-4)$.
Equation of the parabola is $y^2=8x$ $\therefore 4a=8\Rightarrow a=2$ Given point is $(x_1,y_1)$$=(2,-4)$. Then the equation of the normal is $y-y_1=-\frac{y_1}{2a}(x-x_1)$ $y+4=\frac{4}{4}(x-2)$ $\therefore x-y-6=0$
### Obtain the equation of the normal to the parabola $y^2=3x$ parallel to the line $y=2x+1$.
Equation of the parabola is $y^2=3x$ $\therefore 4a=3\Rightarrow a=\frac{3}{4}$ Given line is $y=2x+1$ $\text{Its Slope}=2$ Since the required normal line is parallel to the given line, $\text{Slope of normal }(m)=2$ Hence, the equation of the normal is $y=mx-2am-am^3$ $y=2x-2×\frac{3}{4}×2-\frac{3}{4}×2^3$ $\therefore y=2x-9$
### Find the condition that the line $lx+my+n=0$ may be normal to the parabola $y^2=4ax$.
The equation of the normal to the parabola $y^2=4ax$ is $y=m’x-2am’-am’^3$ $y-m’x+(2am’+am’^3)=0\text{ __(1)}$ Given line is $lx+my+n=0\text{ __(2)}$ The equations $\text{(1)}$ and $\text{(2)}$ will represent the same line if $\frac{1}{m}=-\frac{m’}{l}=\frac{2am’+am’^3}{n}$ $\therefore m’=-\frac{l}{m}$ $\text{and, }-\frac{n}{l}=2a+am’^2$ $-\frac{n}{l}=a\left(2+\frac{l^2}{m^2}\right)$ $al(2m^2+l^2)=-m^2n$ $al(2m^2+l^2)+m^2n=0$ This is the required condition.<|endoftext|>
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Project-based learning sounds great: but does it work? Is there a way to find out what effects project-based learning (PBL) has on students? Admittedly, assessing PBL’s benefits can be tricky since it can be applied in a number of different ways and is often not the only form of instruction students receive. Luckily, as PBL’s popularity grows, there is also a growing body of research. Studies measuring the effects of PBL show significant benefits to students’ academic performance.
Students who participate in PBL exhibit better retention and depth of understanding than their traditional schooling counterparts.
Middle-school students who participated in multimedia-based PBL showed more significant “mastery of content” than the students in classes with traditional instruction and were better able to “create complex products that exhibited their skill”. (Penuel & Means, 2000) Another study looked at high school economics instruction and found that “students whose teachers used the problem-based curriculum in their classrooms scored significantly higher” on assessments. (Finkelstein et al., 2010)
PBL has a positive effect on student attitudes toward learning.
A 2000 research review determined that PBL instruction was just as effective, or more effective, than other types of instruction in achieving “gains in general academic achievement and for developing lower-level cognitive skills in traditional subject matter areas.” There was also evidence that PBL may help students become more capable problem-solvers. (Thomas, 2000)
The PBL model is effective at reaching students of various backgrounds and achievement levels.
Results of a 3-year-study published in 2002 found that students taught using a project-based math curriculum outperformed their counterparts receiving traditional education methods. The two schools monitored were both in low income areas and students were predominately from working class families. (Boaler, 2002)
Two groups of 7th and 8th graders in Detroit Public Schools participated in PBL science programs and, as compared to other students in the same schools, showed increased competency in science and higher scores on a high-stakes science test. The study also concluded that participating in PBL helped urban African-American boys close a gender achievement gap. (Geier, 2008)
When it comes to testing, PBL students score the same or better than students receiving traditional instruction.
A project-based approach to teaching an AP course at three high schools resulted in students achieving higher scores on the AP test than those students at the same schools who received traditional AP instruction. The PBL students also understood the course content more comprehensively than their counterparts. (Parker et al., 2011)In 2012, the West Virginia Department of Education published findings on PBL and student achievement. Students whose teachers followed a PBL curriculum performed as well on the state’s summative assessment as those students who didn’t have PBL instruction. (Hixson, Ravitz, & Whisman, 2012)<|endoftext|>
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Eye allergies, called allergic conjunctivitis, are a common condition that occurs when the eyes react to something that irritates them (called an allergen). The eyes produce a substance called histamine to fight off the allergen. As a result, the eyelids and conjunctiva – the thin, filmy membrane that covers the inside of your eyelids and the white part of your eye (sclera) – become red, swollen and itchy, with tearing and burning. Unlike bacterial or viral conjunctivitis, allergic conjunctivitis is not spread from person to person.
The most common eye allergy symptoms include
Red, swollen or itchy eyes
Burning or tearing of the eyes
Sensitivity to light
People who suffer from eye allergies usually (though not always) have nasal allergies as well, with an itchy, stuffy nose and sneezing. It is usually a temporary (acute) condition associated with seasonal allergies. However, in other cases, eye allergies can develop from exposure to other environmental triggers, such as pet dander, dust, smoke, perfumes, or even foods. If the exposure is ongoing, the allergies can be more severe, with significant burning and itching and even sensitivity to light.
Many eye allergies are caused by the body's response to allergens in the air – both indoors and out – such as dust, pet dander, mold, or smoke. Some of the most common airborne allergens include pollen from grass, trees and ragweed, contributing to seasonal allergies.
Some people can inherit eye allergies from their parents. You're more likely to have allergies if both of your parents have them than if only one does.
To provide proper treatment, your ophthalmologist will check to see whether your symptoms are related to an eye infection or allergic conjunctivitis. He or she can usually diagnose allergic conjunctivitis easily by examining your eyes and discussing your medical history – including your history and your family's history of allergies.
The key to treating eye allergies is to avoid or limit contact with the substance causing the problem. But you have to know what to avoid. If necessary, an allergist can perform a skin or blood test to help identify the specific allergen(s). Those most common are
There are various forms of treatment for eye allergies including eyedrops and medicines
Decongestants (with or without antihistamines)
Your doctor can help determine which treatments are best for you.<|endoftext|>
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What Space Station Droplet Tests Could Mean for Automotive
A team of international researchers have revealed a new type of cool-burning flames after experimentation aboard the International Space Station (ISS) allowed the measurement of droplet and flame histories. The microgravity aboard the ISS meant that an unprecedented range of conditions could be simulated to produce long-duration flames of various types using fuel droplets. The discovery could lead to far cleaner and more efficient combustion engines on earth.
The paper, an open access publication in the journal Microgravity Science and Technology, details how the chemistry of the flames' burn operates. This sheds light into how improvements in internal combustion engines could be made to burn at cooler temperatures, thus emitting fewer pollutants at current or better fuel economy numbers.
Spherically symmetrical combustion of liquid fuel droplets has been a common avenue of research as one of its expected advantages is that only one "side" (spacial dimension) of the fuel enters the combustion process, enabling time-independent conservation equations to apply to the fuel's burn. The problem is, normal gravity destroys this spherical symmetry, while microgravity allows for more control and maintenance over the spherical droplets.
What was observed on the ISS was that under highly controlled conditions, droplets of methanol or heptane (representatives of alcohol and aklane fuels respectively) resulted in an initial hot flame reaction, as expected, but it then devolved into a "cool flame" burn. This was repeated over a variety of ambient conditions (atmospheric levels, mixtures, etc). One of those environments was in air of a mixture similar to what is commonly found on earth, diluted with nitrogen, carbon dioxide, and helium. Products of the combustion process were burned off during the cold flame process, leaving far fewer emissions than were found in a purely hot burn under the same circumstances.
The theory is that buoyancy of the gases created in the initial hot burn causes them to fall before they, in turn, completely burn themselves in the cold flames that come after the heat.
These findings, however, indicate that it is possible to have a cool-flame burn from common combustion fuels. NASA plans to conduct further experiments on the ISS to keep narrowing the possibilities.
What this implies for automotive, however, could be amazing. If a way to form droplets that hold gases in a more controlled way during burn is found, then combustion engines could not only run cooler (outputting less heat), but also more efficiently and with fewer emissions. A cooler-running engine is easier to control and requires less engineering for the cooling process, thus reducing the size of the engine and its peripheral parts. Everything from the radiator to the alternator and water pump could be reduced in size, thus reducing weight and power leach from the engine itself. Lower emissions would mean fewer emission controls would be needed on the car, thus eliminating more weight and possible engine choking additions like exhaust recirculation.
This would radically change combustion engines - no matter the fuel source - in many ways, all for the better.
For those interested, the paper can be read here.<|endoftext|>
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### DOTS Division
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Common Divisor
Find the largest integer which divides every member of the following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.
# Hypotenuse Lattice Points
##### Stage: 4 Challenge Level:
The triangle OMN has vertices M(0,m) and N(n,0) with whole number co-ordinates and we have to find how many lattice points there are on the hypotenuse MN.
The hint here was "First test what happens when m and n are co-prime (have no common factor)", so this tells you that somehow the problem involves common factors.
Absolutely the best way to learn is to make mistakes (generally if you make none then the maths has not got enough challenge in it for you).
A solution was sent in with a graph drawn for m=27 and n=20, and the line looked as if it went through the point (3, 23) but it actually goes through (3, 22.95). You have to check things like this. One way is to draw the line yourself, decide what you think about it, then check using a graphic calculator and using the trace function which will give you readings of the co-ordinates of the points on the line. ... When you go 20 squares in the x-direction you have to go DOWN 27 squares in the y direction. You can't divide this big step into smaller steps going across and down by whole number changes. [If you use other numbers which have a common factor, instead of 20 and 27, then there will be smaller steps between points on the line which have whole number co-ordinates].
Think about the line extended on and on in the example above. It has equation:
27 x + 20 y = 540
and we are looking for whole number solutions to this equation (a Diophantine equation).
We know that x = 0, y = 27 is a solution and if there is another solution with a bigger x then it will go with a smaller y .
Looking at this equation we see that 20 divides exactly into 540 - 20y so x has to be a multiple of 20. Similarly y has to be a multiple of 27. Solutions are given by points on the line with integer co-ordinates and they are:
. . , (-20, 54), (0,27), (20, 0), (40, -27), . . .
When you go 20 squares in the x -direction you have to go DOWN 27 squares in the y direction. There are no solutions in the first quadrant apart from the points on the axes.
The following is a short general proof of this result:
Theorem
If m and n are co-prime then there are no lattice points on the line between M(0, m) and N(n, 0) in the first quadrant.
Proof
The line MN has equation mx + ny = mn
Suppose (p,q) is the lattice point on MN and we have
mp + nq = mn (1)
We see from (1) that n must divide nq so that, if m,n are co-prime, then m must divide q so q = 0 or q = m
Similarly n divides mp so p = 0 or p = n. The only two lattice points are the points M and N.
In the example given with m = 27 and n = 20. The only 2 lattice points are the points M and N.
In the example given with m = 27 and n = 20 the general points with whole number co-ordinates are given by:
x = 20t, y = 27 - 27t (where t is any integer).
Now you will need to do some work to check that when the greatest common divisor of m and n is the number 2, so that m = 2r and n = 2s where r and s are co-prime, this introduces one lattice point between M and N.
Next, if the greatest common divisor is 3, there will be two lattice points between M and N, and so on.
We use the notation gcd(m,n) for the greatest common divisor of m and n, so for example gcd(30, 70) = 10.
The general answer is that the number of lattice points (including M and N on the axes) is 1 + gcd(m,n).<|endoftext|>
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# What is the perimeter of triangle a
what is the perimeter of triangle a
## What is the perimeter of triangle a
To determine the perimeter of a triangle, you need to know the lengths of all three sides of the triangle. The perimeter is the sum of these lengths.
### Solution By Steps:
1. Identify the Side Lengths:
• Let’s denote the sides of the triangle as a, b, and c.
• Typically, you need to know these values. If they are given, you can proceed to calculate the perimeter.
For instance, if the sides of the triangle are given by a = 3 units, b = 4 units, and c = 5 units, you can use these values directly.
2. Calculate the Perimeter:
• The formula for the perimeter P of a triangle is:
P = a + b + c
3. Perform the Calculation:
• Substitute the lengths of the sides into the formula.
P = 3 + 4 + 5 = 12 \text{ units}
### Example:
Let’s consider a specific example where the side lengths of triangle a are 6 units, 7 units, and 10 units.
1. Identify the Side Lengths:
• a = 6 units
• b = 7 units
• c = 10 units
2. Calculate the Perimeter:
• Use the formula for perimeter:
P = a + b + c
3. Substitute the Values:
• Substitute the given lengths:
P = 6 + 7 + 10 = 23 \text{ units}<|endoftext|>
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# prob_hyp - B Weaver(31-Oct-2005 Probability Hypothesis...
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B. Weaver (31-Oct-2005) Probability & Hypothesis Testing 1 Probability and Hypothesis Testing 1.1 PROBABILITY AND INFERENCE The area of descriptive statistics is concerned with meaningful and efficient ways of presenting data. When it comes to inferential statistics , though, our goal is to make some statement about a characteristic of a population based on what we know about a sample drawn from that population. Generally speaking, there are two kinds of statements one can make. One type concerns parameter estimation , and the other hypothesis testing . Parameter Estimation In parameter estimation, one is interested in determining the magnitude of some population characteristic. Consider, for example an economist who wishes to estimate the average monthly amount of money spent on food by unmarried college students. Rather than testing all college students, he/she can test a sample of college students, and then apply the techniques of inferential statistics to estimate the population parameter. The conclusion of such a study would be something like: The probability is 0.95 that the population mean falls within the interval of £130-£150. Hypothesis Testing In the hypothesis testing situation, an experimenter wishes to test the hypothesis that some treatment has the effect of changing a population parameter. For example, an educational psychologist believes that a new method of teaching mathematics is superior to the usual way of teaching. The hypothesis to be tested is that all students will perform better (i.e., receive higher grades) if the new method is employed. Again, the experimenter does not test everyone in the population. Rather, he/she draws a sample from the population. Half of the subjects are taught with the Old method, and half with the New method. Finally, the experimenter compares the mean test results of the two groups. It is not enough, however, to simply state that the mean is higher for New than Old (assuming that to be the case). After carrying out the appropriate inference test, the experimenter would hope to conclude with a statement like: The probability that the New-Old mean difference is due to chance (rather than to the different teaching methods) is less than 0.01. Note that in both parameter estimation and hypothesis testing, the conclusions that are drawn have to do with probabilities . Therefore, in order to really understand parameter estimation and hypothesis testing, one has to know a little bit about basic probability. 1.2 RANDOM SAMPLING Random sampling is important because it allows us to apply the laws of probability to sample data, and to draw inferences about the corresponding populations.
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B. Weaver (31-Oct-2005) Probability & Hypothesis Testing 2 Sampling With Replacement A sample is random if each member of the population is equally likely to be selected each time a selection is made . When N is small, the distinction between with and without replacement is very important. If one samples with replacement, the probability of a particular element being selected is constant from trial to trial (e.g., 1/10 if
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## This note was uploaded on 11/09/2009 for the course FINANCE 330 taught by Professor Seri during the Spring '09 term at Birzeit University.
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Ask a homework question - tutors are online<|endoftext|>
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Class 7 Maths NCERT Solutions for Chapter – 5 Lines and Angles EX – 5.1
Lines and Angles
Question 1.
Find the complement of each of the following angles :
Solution:
Since, the sum of the measures of an angle and its complement is 90°, therefore,
1. The complement of an angle of measure 20° is the angle of (90°-20°), f.e., 70°.
2. The complement of an angle of measure 63° is the angle of (90°-63°), i.e., 27°.
3. The complement of an angle of measure 57° is the angle of (90°-57°), i.e., 33°.
Question 2.
Find the supplement of each of the following angles :
Solution:
Since, the sum of the measures of an angle and its supplement is 180°, therefore,
1. The supplement of an angle of measvjre 105° is the angle of (180°-105°), i.e., 75°.
2. The supplement of an angle of measure 87° is the angle of (180°-87°), i.e., 93°.
3. The supplement of an angle of measure 154° is the angle of (180°-154°), i.e., 26°.
Question 3.
Identify which of the following pairs of angles are complementary and which are supplementary :
1. 65°, 115°
2. 63°, 27°
3. 112°, 68°
4. 130°, 50°
5. 45°, 45°
6. 80°, 10°
Solution:
1. Since, 65°+ 115° = 180°
So, this pair of angles is supplementary.
2. Since, 63°+ 27° = 90°
So, this pair of angles is complementary.
3. Since, 112° + 68° = 1800
So, this pair of angles is supplementary.
4. Since, 130°+50° = 180°
So, this pair of angles is supplementary.
5. Since, 45°+ 45° = 90°
So, this pair of angles is complementary.
6. Since, 80°+ 10° = 90°
So, this pair of angles is complementary.
Question 4.
Find the angle which is equal to its complement.
Solution:
Let the measure of the angle be x°. Then, the measure of its complement is given to be x°.
Since, the sum of the measures of an angle and its complement is 90°, therefore,
x° + x° = 90°
⇒ 2x° = 90°
⇒ x° = 45°
Thus, the required angle is 45°.
Question 5.
Find the angle which is equal to its supplement.
Solution:
Let the measure of the angle be x°. Then,
measure of its supplement = x°
Since, the sum of the measures of an angle and its supplement is 180°, therefore,
x° + x° = 180°
⇒ 2x° =180°
⇒ x° = 90°
Hence, the required angle is 90°.
Question 6.
In the given figure, ∠1 and ∠2 are supplementary angles. If ∠1 is decreased, what changes should take place in ∠2 so that both the angles still remain supplementary?
Solution:
∠2 will increase with the same measure as the decrease in ∠1.
Question 7.
Can two angles be supplementary if both of them are :
1. acute?
2. obtuse?
3. right?
Solution:
1. No
2. No
3. Yes
Question 8.
An angle is greater than 45°. Is its complementary angle greater than 45° or equal to 45° or less than 45°?
Solution:
Since, the sum of the measure of ah angle and its complement is 90°.
∴ The complement of an angle of measures 45° + x°, where x > 0 is the angle of [90° – (45° + x°)] = 90° – 45° – x°= 45° – x°.
Clearly, 45° + x° > 45° – x°
Hence, the complement of an angle > 45° is less than 45°.
Question 9.
1. Is ∠1 adjacent to ∠2 ?
2. Is ∠AOC adjacent to ∠AOE?
3. Do ∠COE and ∠EOD form a linear pair?
4. Are ∠BOD and ∠DOA supplementary?
5. Is ∠1 vertically opposite to Z4?
6. What is the vertically opposite angle of ∠5
Solution:
1. Yes
2. No
3. Yes
4. Yes
5. Yes
6. ∠2 + ∠3 = ∠COB
Question 10.
Indicate which pairs of angles are :
1. Vertically opposite angles.
2. Linear pairs.
Solution:
1. Pair of vertically opposite angles are ∠1, ∠4; ∠5, ∠2 + ∠3.
2. Pair of linear angles are ∠1, ∠5; ∠4, ∠5.
Question 11.
Solution:
∠1 is not adjacent to ∠2 because they have no common vertex.
Question 12.
Find the values of the angles x, y and z in each of the following?
Solution:
Question 13.
Fill in the blanks :
(i) If two angles are complementary, then the sum of their measures is __________
(ii) If two angles are supplementary, then the sum of their measures is __________
(iii) Two angles forming a linear pair are __________
(iv) If two adjacent angles are supplementary, they form a __________
(v) If two lines intersect at a point, then the vertically opposite angles are always __________
(vi) If two lines intersect at a point, and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are __________
Solution:
(i) 90°
(ii) 180°
(iii) supplementary
(iv) linear pair
(v) equal
(vi) obtuse angles
Question 14.
In the adjoining figure, name the following pairs of angles :
1. Obtuse vertically opposite angles.
3. Equal supplementary angles.
4. Unequal supplementary angles.
5. Adjacent angles that do not form a linear pair.
Solution:
1. Obtuse vertically opposite angles are ∠AOD and ∠BOC.
2. Adjacent complementary angles are ∠BOA and ∠AOE.
3. Equal supplementary angles are ∠BOE and ∠EOD.
4. Unequal supplementary angles are ∠BOA and ∠AOD, ∠BOC and ∠COD, ∠EOA and ∠EOC.
5. Adjacent angles that do not form a linear pair are ∠AOB and ∠AOE, ∠AOE and ∠EOD; ∠EOD and ∠COD.<|endoftext|>
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# Average Velocity Formula (displacement over time)
Average Velocity Formula (displacement over time)
The velocity of an object is the rate at which it moves from one position to another. The average velocity is the difference between the starting and ending positions, divided by the difference between the starting and ending times. Velocity has a magnitude (a value) and a direction. The unit for velocity is meters per second (m/s).
vavg = average velocity (m/s)
x1 = the start position of an object (m)
x2 = the end position of an object (m)
t1 = the start time of the motion (s)
t2 = the end time of the motion(s)
Average Velocity Formula (displacement over time) Questions:
1) On a driving vacation, a boy sees a road sign that says Toronto is 220 km away. Exactly one hour later, he sees a sign that says Toronto is 100 km away. What is the velocity of the vehicle the boy is in, expressed in meters per second?
Answer: The choice of direction is important when solving velocity problems. In this case, the positions are expressed as distances away from a location. If the location, Toronto, is chosen as x = 0, and the distance values are chosen to be positive, then the vehicle is moving in the negative x direction. Using this definition for direction, the resulting value of velocity will be negative. The start position x1 = 220 km, and the end position x2 = 100 km. The travel time is given as a difference, so the start time is chosen to be t1 = 0 hours. The end time is t2 = 1.0 hours. The question asks for the velocity to be stated in meters per second, so these positions and times must be converted. There are 60 minutes in an hour, and 60 seconds in a minute, and so 1 hour is:
1 hour = 60 x 60 s
1 hour = 3600 s
There are 1000 meters in 1 kilometer, and so 1 kilometer is:
1 km = 1000 m
Using these conversions, the positions and times are:
∴x1 = 220 000 m
∴x2 = 100 000 m
∴t1 = 0 s
∴t2 = 3600 s
The average velocity can be found using the formula:
The average velocity is -33.33 m/s, with the direction defined as above. This means that the average velocity can also be stated as 33.33 m/s, toward Toronto.
2) A homing pigeon was released from its cage at 10:00 am. At that time, the cage was 108.0 km from the pigeon's home. If the average velocity of the pigeon is 20.0 m/s, at what time should the pigeon arrive back at its home?
Answer: The problem is easiest to solve if the position of the cage is assigned to be x1 = 0.0 m. The pigeon's home is at position x2, which is:
∴x2 = 108 000 m
The arrival time can be found by rearranging the average velocity formula:
t2 = (10:00 am) + (5400 s)
t2 = (10:00 am) + 1.5 hours
∴t2 = 11:30 am
The pigeon should arrive at home at 11:30 am.<|endoftext|>
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An introduction to macular degeneration
The retina is vital for vision. It sits at the back of the eye and contains over a hundred million light-sensitive cells: rods, which are sensitive to light, and cones which are sensitive to colour. These cells turn light into nerve impulses which are sent along the optic nerve to the brain where a picture of what we are seeing is created.
The macula is the part of the retina which is responsible for central vision, which is what we see when we focus on an object, as opposed to peripheral vision, which allows us to see movements around us (this is also often referred to as ‘seeing out of the corner of your eye’). It is only about 5mm in diameter but contains millions of light and colour-sensitive cells that help create a picture of what is happening in the centre of your field of vision.
Macular degeneration, therefore, is the term use when the macula becomes damaged. Alternatively, the phrase ‘age-related macular degeneration’ (AMD) is also used because of the critical role that age plays in the development of this condition.
What are the symptoms of macular degeneration?
The symptoms of macular degeneration tend to develop gradually so are often not noticeable for a long time. These symptoms include:
- Blurred vision– this blurring typically only affects the very centre of your field of vision
- Blurred vision that becomes clearer in brighter light
- A black or grey blind spot in the centre of vision
- Straight lines appearing crooked or wavy
- Contrast sensitivity – you may become unable to notice slight differences in the environment around you, for example in different shades of colour, or different textures.
AMD does not cause any pain in the eye, so if you experience this symptom it is advisable to consult your doctor or optician as this is likely to be a sign of a different eye problem.
Types of macular degeneration
There are two types of macular degeneration – dry and wet.
Dry AMD is the most common form of the condition, with around 80-90% of macular degeneration patients being affected by this type. In dry form some of the light-sensitive cells in the macula begin to break down, which affects vision. This condition is indicated by yellow deposits on the macula called drusen. This type of AMD usually takes much longer to develop, so symptoms are not always immediately apparent.
Wet macular degeneration is when there is a growth of abnormal blood vessels underneath the macula. These leak blood into the retina, causing blind spots in the centre of vision as well as distortion of vision – in particular straight lines often appear wavy. This bleeding can eventually cause scarring which leads to permanent loss of vision. Wet macular degeneration can develop very rapidly.
What are the causes and risk factors of macular degeneration?
The exact cause of both of these types of macular degeneration is not yet fully understood, but it is thought that both hereditary and environmental factors play a significant role.
Researchers have identified a number of risk factors that make you more susceptible to developing age-related macular degeneration. These include:
- Age, obviously. The chances of developing macular degeneration increases with age, and it is the leading cause of severe vision loss in people over 60
- Gender – women are more likely to develop this disease than men
- History of macular degeneration in the family
- High blood pressure
- High cholesterol
- Being light skinned or having light-coloured eyes
- Over exposure to sunlight and UV rays.
Natural and herbal treatments
There isn’t a huge amount you can do to treat macular degeneration yourself, but there are steps you can take to slow its progression or reduce the risk of developing it in the first place.
Firstly, to reduce your chances of developing this condition, try to reduce the number of risk factor groups that you fall into. That means no smoking, a healthy, low cholesterol diet and wearing sunglasses to protect your eyes from harmful UV rays.
Increasing the amount of essential eye nutrients like vitamin C, beta-carotene and antioxidants can also support your general eye health. Increasing the amounts of leafy greens such as kale and spinach, as well as brightly coloured vegetables such as bell peppers, tomatoes and carrots in your diet is an easy way to get more of these essential nutrients. For a more comprehensive guide, read our article on the good foods to eat for eye health.
There is significant research to suggest that the development of macular degeneration can be slowed, and the damage it causes reduced, by your intake of lutein and zeaxanthin.1
One great source of lutein and zeaxanthin is our own Vision Complex. In addition to these nutrients, it contains other essential eye nutrients such as zinc and beta-carotene. These nutrients are naturally sourced from Blackcurrants and Marigold, making them easier to absorb than synthetic versions.
There currently isn’t a cure for dry macular degeneration. Many doctors recommend natural treatments to prevent or slow the progression of the condition, such as controlling diet. Your doctor may also be able to prescribe specific supplements that contain useful nutrients.
There are some treatments for wet macular degeneration. These include laser surgery to seal the leaking blood vessels. However, this method can also destroy healthy tissue and can damage vision, so it is important to weigh up the risks and the benefits.
Wet macular degeneration can also be treated with injections of anti-VEGF drugs: VEGF stands for ‘vascular endothelial growth factor’. These are injected directly into the eye to slow the growth of abnormal blood vessels. For effective results repeated injections are required.
1 BBC 'Trust Me, I'm a Doctor' series - Can I improve my eyesight?
See also our blog post discussing the BBC's findings.<|endoftext|>
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# Understanding BODMAS, BIDMAS or BEDMAS (order of operations).
Updated on April 16, 2013
## BODMAS Video
BODMAS, BIDMAS and BEDMAS all help you to carry out operations in the correct order. For younger students it is a good idea to start off with BODMAS:
Brackets
Orders of
Division
Multiplication
Subtraction
Once a student has studied powers (which are also called indices or exponents) then use BIDMAS:
Brackets
Indices
Division
Multiplication
Subtraction
or use BEDMAS:
Brackets
Exponents
Division
Multiplication
Subtraction
Powers are the little numbers that sit on the top right of a number (squares, cubes...)
Basically, if you are given a calculation you always work out what’s inside the brackets first, then do powers, then do division, then do multiplication, then do addition and finally do subtraction.
Let’s take a look at some examples that involve using BODMAS, BIDMAS or BEDMAS.
Example 1
Work out:
7 × (6 – 2)
First do what’s inside the brackets, as brackets is above multiplication in the words BODMAS, BIDMAS or BEDMAS. So:
7 × 4
=28
Example 2
Work out:
9 + 6 × 3
First do the multiplication, as multiplication before addition in the words BODMAS, BIDMAS or BEDMAS.
So:
9 + 18
= 27
Example 3
Work out:
90 – 30 ÷ 6
First do the division, as division comes first in the words BODMAS, BIDMAS or BEDMAS.
So:
90 – 5
= 85.
Example 4
Work out:
4 × 3²
First calculate 3², as powers come before multiplication in the word BODMAS, BIDMAS or BEDMAS.
So:
4 × 9
= 36.
Example 5
Put a bracket in the correct place to make this calculation correct:
11 – 7 × 2 + 6 = 14
The bracket need to do go around the 11-7 as the subtraction needs to be carried out first, to make the calculation correct:
(11 – 7) × 2 + 6
Now let’s check to see if this gives 14.
First do the brackets, as brackets come first in the words BODMAS, BIDMAS or BEDMAS.
This gives 4 × 2 + 6.
Next do multiplication as this comes next in the words BODMAS, BIDMAS or BEDMAS.
This gives 8 + 6, which gives an answer of 14.
59
132
10
31
24<|endoftext|>
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### Direct Variation
```
Equation of Direct Variation
When you see: “y varies direcly as x”
write
Write
the equation of direct variation
that has a coefficient of variation
equal to -3
Replace
k with -3
Assume
y varies directly as x. If the
coefficient of variation is ½, then what
is the value of y when x = -6?
Replace
k with ½
Replace x with -6
Multiply ½ times -6 to get answer
Assume y varies directly as x. If the
coefficient of variation is 4, then what is
the value of x when y = 6?
Replace k with 4 and y with 6
Divide 4 on both sides
Reduce the fraction
Suppose y varies directly as x. If y = 3
when x = 15, then find x when y = 5.
2. Suppose y varies directly as x. Find x
when y = 10 if y = -7 when x = -14.
3. Suppose y varies directly as x. If x = 15
when y = 12, find x when y = 21.
4. Suppose y varies directly as x. If x = 24
when y = 8, then what is the coefficient
of variation?
1.
The amount of a person’s paycheck
varies directly with the number of hours
worked. For 13 hours of work, the
paycheck is \$73.45. Find the pay for 25
hours of work.
“varies directly” means:
Remember, “y” always varies directly as
“x”
The amount of a person’s paycheck
varies directly with the number of hours
worked. For 13 hours of work, the
paycheck is \$73.45. Find the pay for 25
hours of work.
If “y” varies directly as “x” than what
comes before varies directly is y and
what comes after it is x
The amount of a person’s paycheck
varies directly with the number of hours
worked. For 13 hours of work, the
paycheck is \$73.45. Find the pay for 25
hours of work.
Replace paycheck with \$73.45 and
hours with 13
Divide both sides by 13
Plug k back into the equation
The amount of a person’s paycheck
varies directly with the number of hours
worked. For 13 hours of work, the
paycheck is \$73.45. Find the pay for 25
hours of work.
The question asks for how much you’d
make in 25 hours, so replace x with 25
The amount of a person’s paycheck
varies directly with the number of hours
worked. For 13 hours of work, the
paycheck is \$73.45. Find the pay for 25
hours of work.
K = 5.65 which is how much this person
makes per hour.
For 25 hours, this person makes \$141.25
The number of calories in a container of
milk is directly proportional to the
amount of milk in the container. If there
are 160 calories in an 8-ounce glass of
milk, find the number of calories in a 15ounce glass of milk.
Directly proportional means direct
variation
The number of calories in a container of
milk is directly proportional to the
amount of milk in the container. If there
are 160 calories in an 8-ounce glass of
milk, find the number of calories in a 15ounce glass of milk.
Y varies directly as x, so calories is y and
amount of milk is x
The number of calories in a container of
milk is directly proportional to the
amount of milk in the container. If there
are 160 calories in an 8-ounce glass of
milk, find the number of calories in a 15ounce glass of milk.
Calories is 160 and the amount of milk is
8
Divide both sides by 8
Plug k back into the equation
The number of calories in a container of
milk is directly proportional to the
amount of milk in the container. If there
are 160 calories in an 8-ounce glass of
milk, find the number of calories in a 15ounce glass of milk.
K means 20 calories per ounce
Plug in 15 for x to solve the problem
Multiply to solve
The number of calories in a container of
milk is directly proportional to the
amount of milk in the container. If there
are 160 calories in an 8-ounce glass of
milk, find the number of calories in a 15ounce glass of milk.
There are 300 calories in a 15-ounce
glass of milk
Laura typed 176 words correctly in 4
minutes. Assuming a direct variation,
how many words can she type in 30
minutes?
A refund you get varies directly as the
number of cans you recycle. If you get a
\$3.75 refund for 75 cans, how much
should you get for 500 cans?
The number of miles driven varies directly
with the number of gallons of gas used. Erin
drove 297 miles on 9 gallons of gas. How far
would she be able to drive on 14 gallons of
gas?
The kinetic energy of a car varies directly
with the square of the velocity of the car. A
car with a velocity of 9 meters per second
has 33,100 joules of kinetic energy. About
how much kinetic energy does the same
car have when traveling at 12 meters per
second?
The height of a kite from the ground
varies directly with the cube of the wind
speed. A kite flies 8 feet high when the
wind speed is 10 miles per hour. What is
the height of a kite when the wind speed
is 25 miles per hour?
```<|endoftext|>
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## Linear Methods of Applied Mathematics Evans M. Harrell II and James V. Herod*
version of 28 January 2000
(Some remarks for the instructor).
You may wish to review the notion of a linear operator by referring to the Mathematica notebook for Chapter I.
## XIII. Geometry and integral operators
The following four problems illustrate, in a simple way, the primary concerns of the next several chapters. The first is a problem about matrices and vectors, and it will be our guide to solving integral equations and differential equations.
Then
if and only if u = B v.
The equivalence of these two matrix-vector equations is easy to establish, and you have no doubt learned long ago how to construct the inverse matrix B such that statement (b) is equivalent to statement (a).
Model Problem XIII.2. Find the inverse of an integral operator. Example: Let u and v be functions of x in the interval [0,1], and let K(x,t) := 1 + x t. The function u is a solution to
if and only if
If one supposes u is given by the formula (b), then the integral calculus will show that u satisfies (a). On the other hand, the task of deriving a formula for u from the relationship in (a) involves unfamiliar techniques, which we shall discuss in this course.
Model Problem XIII.3. Find the integral solution operator for an ordinary differential equation. Example: Let f and g be continuous functions of x in the interval [0,1], and
The function g is a solution for
(a) g''= -f and g(0) = g(1) = 0
if and only if
Once again, the task of deriving this connection is probably somewhat mysterious, but the result can be checked with the calculus:
VERIFICATION OF MODEL PROBLEM 3.
(a)=>(b) Suppose that f is continuous on [0,1] and g'' = -f with g(0) = g(1) = 0. Suppose also that K is as given by sample problem (3). Then
Using integration by parts this last line can be rewritten as
= -(1-x)[x g'(x) -(g(x)-g(0)}
-x[-(1-x) g'(x) + (g(1) - g(x))]
= (1-x) g(x) + x g(x) = g(x).
To get the last line we used the assumption that g(1) = g(0) = 0.
(b)=>(a) Again, suppose that f is continuous and, now, suppose that
As you can see, it is not hard to show that these two statements are equivalent. In the next few chapters you will learn how, given statement (a), you can construct K such that statement (b) is equivalent to statement (a). Perhaps you can do this already.
uxx + uyy = 0 for y > 0 and all x,
with boundary data that u(x,0) = sin(x), and the "condition at infinity" that u(x,y) remains finite as y -> .
The solution, as can be verified by elementary calculus, is u(x,y) = e-y sin(x). We shall later learn how to obtain such a solution by means of integral transforms of the boundary data.
One of the unifying ideas of this course is that each of these model problems can be written in the form Lu=v for a linear operator L. It is a worthwhile exercise to reformulate each of these model problems in this form.
Most often, we shall take the interval on which our functions are defined to be [0,1]. Of course, we do not work in the class of all functions on [0,1]; rather, in the spirit of Chapters I-II, we ask that the linear space should consist of functions f for which
Then we have the inner product space we called L2( [0,1] ) in Chapter II. We recall that the inner product of two functions is given by
and the norm of f is defined as the square-root of the inner product of a function with itself:
(Compare with the norm in Rn.)
It does not seem appropriate to study in detail the nature of L2[0,1] at this time. Rather, suffice it to say that the space is large enough to contain all continuous functions - even functions which are continuous except at a finite number of places. The interested student can learn more about L2[0,1] by looking in standard books on real analysis or Hilbert space.
Suppose { fp } is a sequence of functions in L2( [0,1]). It is valuable to consider the possible meanings for the statement that limp fp(x) = g(x). There are three useful interpretations for our purposes:
• The sequence {fp} converges pointwise to g at each x in [0,1] provided that for each x in [0,1], limp fp(x) = g(x).
Sometimes we modify this to convergence pointwise a.e. when the set on which the convergence fails is a null set.
• The sequence converges to g uniformly on [0,1] provided that
limp supx |fp(x) - g(x)| = 0.
• The sequence converges to g in norm if
limp || fp - g || = 0.
• Convergence in norm is the same as r.m.s. convergence, as described in chapter II. Uniform convergence on a finite interval implies both of the other two notions of convergence, but examples show that other implications among these notions are not generally valid. (Compare with the notions of convergence for sequences of vectors in Rn.)
In this section we study one of the most common types of integral equation. As an example, given a function called the kernel
K: [0,1]x[0,1] -> R
and a function f: [0,1] -> R, we seek a function y such that for each x in [0,1],
Such equations are called Fredholm equations of the second kind. An equation of the form
is a Fredholm equation of the first kind.
The requirements in this section on K and f will be that
These requirements are met if K and f are continuous.
For simplicity, we denote by K the linear function given by
Note that K has a domain large enough to contain all functions y which are continuous on [0,1]. Also, if y is continuous then K(y) is a function and its value at x is denoted K(y)(x). In spoken conversation, it is not so easy to distinguish the number valued function K and the function valued K. The bold character will be used in these notes to denoted the latter.
It is well to note the resemblance of this function K to the multiplication of a vector u by a matrix A:
This formula has the same form as that for K given above.
It is a historical accident that differential equations were understood before integral equations. Often an integral equation can be converted into a differential equation or vice versa, so many of the laws of nature which we think of as differential equations might just as well have been developed as integral equations initially. In some instances it is easier to differentiate than to integrate, but at other times integral operators are more tractable.
In this course integral operators will be called upon to solve differential equations, and this is one of their main uses. They have many other uses as well, most notably in the theory of filtering and signal processing. In most of these applications the integral and differential operators are linear transformations. The analogy between linear transformations and matrices is deep and useful.
In order to understand K*, one must consider < K(f), g > and seek K* such that < Kf, g > = < f, K*g >.
An examination of these last equations leads one to guess that K* is given by
or, keeping t as the variable of integration,
Those last equations verified that
< K(f), g > = < f, K*(g) >.
Care has to be taken to watch whether the "variable of integration" is t or x in the integrals involved.
In summary, if K is the kernel associated with the linear operator K, then the kernel associated with K* is given by K*(x,y) [[equivalence]] K(y,x). It is of value to compare how to get K* from K with the process of how to get A* from A:
A*p,q = Aq,p.
Consistent with the rather standard notation we have adopted above, it is clear that a briefer representation of the equation
is the concise equation y = K(y) + f, or (1 - K ) y = f.
Example XIII.5: Suppose that
To get K*, let's use other letters for the argument of K* and K to avoid confusion. Suppose that 0 < u < v < 1. Then, K*(u,v) = K(v,u) = 0. In a similar manner, K*(u,v) = (u-v)2 if 0 < v < u < 1. Note that K* is not K.
The discussion of this example has been algebraic to this point. Consider this geometric notion that is suggested by the alternate name for "self-adjoint", namely, some call K "symmetric" if K(x,t) = K(t,x). The geometric name suggests a picture and the picture is the graph of K. The K of this example is not symmetric in x and t. Its graph is not symmetric about the line x = t. The function K is different from the function K*.
THE FREDHOLM ALTERNATIVE THEOREMS
A first understanding of the problem of solving an integral equation
y = Ky + f
can be gotten by referring to the Fredholm Alternative Theorems in this context.
(Review the alternative theorem for matrices.)
I. Exactly one of the following holds:
(a)(First Alternative) if f is in L2{0,1}, then
has one and only one solution.
(b)(Second Alternative)
has a nontrivial solution.
II. (a) If the first alternative holds for the equation
then it also holds for the equation
(b) In either alternative, the equation
have the same number of linearly independent solutions.
III. Suppose the second alternative holds. Then
has a solution if and only if
for each solution z of the adjoint equation
Comparing this context for the Fredholm Alternative Theorems with an understanding of matrix examples seems irresistible. Since these ideas will re-occur in each section, the student should pause to make these comparisons.
Example XIII.6: Suppose that E is the linear space of continuous functions on the interval [-1,1]. with
and that
The equation y = K(y) has a non-trivial solution: the constant function 1. To see this, one computes
One implication of these computations is that the problem y = Ky + f is a second alternative problem. It may be verified that y(x) = 1 is also a nontrivial solution for y = K*y. It follows from the third of the Fredholm alternative theorems that a necessary condition for y = Ky + f to have a solution is that
Note that one such f is f(x) = x + x3.
Exercises.
XIII.1. Reformulate each of Model Problems XIII.2-XIII.4 in the form Lu = v. I.e., carefully identify the to which u and v belong, as well as how the operator L acts. For Model Problem XIII.4 make v a nonzero quantity related to the boundary condition sin(x) by choosing the vector spaces for the linear operator intelligently.
XIII.2. Suppose K(x,t) =1 + 2 x t2 on [0,1]x[0,1] and y(x) = 3 - x. Compute K(y) and K*(y). Ans: (5+3x)/2, (15+14x2)/6
XIII.3. Let
Find B such that, if v is in R2, then these are equivalent:
(a) u is a vector and Au = v.
(b) v is a vector and u = Bv.
XIII.4. Let K be as in Model Problem XIII.2. Show that if u(x) = 3x2 - (25 + 12x )/6 then u solves the equation
XIII.5. Let
Suppose that f is continuous on [0,1]. Show these are equivalent:
XIII.6. Let u(r, = r sin( ). Show that
with u(1,) = sin().
XIII.7.. Suppose K(x,t) = x t if 0 < x < t < 1, and = x t2 if 0 < t < x < 1)). For y(x) = 3 - x, compute K(y) and K*(y).
Ans: K[y](x) = - x5/4 + 4x4/3 - 3x3/2 + 7x/6.
XIII.8.
(1) Suppose that E is the linear space of continuous functions on [0,1] with
and that
(2) Show that y = Ky has non-trivial solution the constant function 1.
(3) Show that y = K*y has non-trivial solution the function + 2 cos( x).
(4) What conditions must hold on f in order that
y = Ky + f
should have a solution?
Find examples of sequences of functions which converge
• pointwise but not uniformly
• pointwise but not in norm
• in norm but not pointwise
• in norm but not uniformly.<|endoftext|>
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Problems & Puzzles: Puzzles
Puzzle 93.- Numbers such that the decimal and hexadecimal representation are the reverse of each other
The 12/01/2000 Jim Howell wrote:
"Some time ago, I noticed that there are a few numbers where the decimal and hexadecimal representations are the reverse of each other:
53 (decimal) = 35 (hex)
371 (decimal) = 173 (hex)
5141 (decimal) = 1415 (hex)
99481 (decimal) = 18499 (hex)
I believe these are the only such numbers (of more than one digit). [Now here is where the primes come in...]
53 = prime
371 = 7 * 53
5141 = 53 * 97
99481 = 53 * 1877
All four of these numbers are divisible by 53 (!)"
(1) are there any other numbers (of more than one digit) where the decimal and hexadecimal representations are reversals?
(2) is there any particular reason for all of the above numbers to be
divisible by 53?
(3) are there other pairs of bases for which something similar happens?
Solution
(1) Jud McCranie wrote "that is all of the solutions. If the number is >=1000000 then it has more decimal digits than hexadecimal digits, so they can't be reversals". But Enoch Haga sent the following additional solution: "8520280 (decimal) = 0820258 (hex)". Maybe Jud was not taking zero leading numbers as valid solutions...
(2) Enoch Haga wrote the following argument: "The reason that all of the decimal numbers above are divisible by 53 is that a sum of powers in one base must equal the sum of powers in another base. Example: 5*10^1=50 and 3*10^0=3, sum 53, and in hex: 3*16^1=48 and 5*16^0=5, sum 53. This is the necessary and sufficient condition for reversibility. The first such sum found will not always be prime, but if prime, will divide all subsequent numbers satisfying the same condition because all of the subsequent numbers are multiples of the first prime"...
(3) Other solutions of the same type are:
13 (10) = 31 (4) (J & E)
23 (10) =32 (7) (J)
46=2*23
2116 = 2^2*23^2
15226 = 2*23*331
1527465 decimal = 5647251 octal. (J & E)
445 = 5*89 (10) = 544 (9) (J)
313725 = 3*5^2*47*89
315231 (10) = 132513 (12) (J)
43 (13) =34 (13) (J)
86 = 2*43
774 = 2*3^2*32
834 (10) = 438 (14) (J)
21 (10) = 12 (19) (J & E)
42
63
84
441
882
1540 (E)
3290 (E)
7721 = 7*1103 - not a multiple of 21
9471
etcetera
Records | Conjectures | Problems | Puzzles<|endoftext|>
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Jump To Section
What Is the Format of a Ballad?
A ballad with lyrics traditionally follows a pattern of rhymed quatrains. This means that for every four-line grouping, either the first and third line will rhyme or the second and fourth lines will rhyme.
The more common of these rhyme schemes is the latter, where the second and fourth lines rhyme with one another. We call this an ABCB quatrain, where the “B” lines rhyme with each other, as they do in the following quatrain:
Upon a horse a knight did ride
Well armed with shield and lance
But when a dragon did appear
He cried and wet his pants
The final word of the second line (“lance”) rhymes with the final word of the fourth line (“pants”). As such we may deem those to each be “B” lines in ABCB analysis. Meanwhile, the first and the third lines do not rhyme; in fact, to ensure proper ABCB form, they must not rhyme.
Another thing to note about the quatrain above is the consistent meter. All the lines are iambic, which means that every even-numbered syllable is accented, as such:
Upon a horse a knight did ride
In addition to consistent iambic form—known as iambic tetrameter in poetic ballads with four lines—each line maintains a fixed set of syllables. The first and third lines each contain eight syllables, while the second and fourth lines each contain six syllables.
Other Examples of Ballad Form
The ABCB form is not the only way to write a verse of a ballad. In fact, even classic ballads took liberties with the ballad format. Consider “La Belle Dame sans Merci,” written by John Keats in 1819. The poem follows the ABCB format, but it takes liberties with the metric pattern of each line. One quatrain reads:
I saw their starved lips in the gloam
With horrid warning gapèd wide
And I awoke and found me here
On the cold hill’s side
The meter is less strict than the one seen in our prior example, but the verse is still unmistakably in the ABCB format.
A Step-by-Step Guide to Writing a Ballad
Although the word “ballad” no longer refers exclusively to story songs, beginning with a story is a great way to compose your first ballad. Here is a step-by-step guide.
1. Choose Your Topic
A ballad can be inspired by a story in the songwriter’s own life, a fictional scenario with fictional characters, or a real event from history or contemporary events. Nobel Prize-winning songwriter Bob Dylan is a noted master of all three:
- Some of Dylan’s most famous compositions are ballads “ripped from the headlines,” whether recent or past. In 1963’s “The Lonesome Death of Hattie Carroll,” Dylan took his listeners through a horrifying event that had occurred only months prior. In 1975’s “Hurricane,” he recounts the trial of boxer Rubin “Hurricane” Carter who, at that point, had been in jail for nine years.
- Other Dylan ballads delve into history. “Tempest,” for instance, is a very loose account of the Titanic tragedy with humor and oddities thrown in. “Highway 61 Revisited” gives similar treatment to the biblical story of Abraham and Isaac.
- Other Dylan ballads concern fictional characters, like “Desolation Row” or “Lily, Rosemary, and the Jack of Hearts.”
- Others are tales of Dylan himself, whether epic ramblers (“Tangled Up in Blue”) plaintive remembrances (“Sara”), humorous fiction (“Bob Dylan’s 115th Dream”) or just downright mysterious (“Highlands”).
2. Choose Your Tone
As Bob Dylan exemplifies, ballads can present a variety of tones, whether purposeful, playful, plaintive, or mysterious. Many of the best ballads will offer multiple tones, sometimes within the same verse.
A strong example of a ballad with a contrasting tone is Samuel Taylor Coleridge’s “The Rime of the Ancient Mariner.” Consider the following pair of quatrains:
And now there came both mist and snow
And it grew wondrous cold
And ice, mast-high, came floating by
As green as emerald
And through the drifts the snowy clifts
Did send a dismal sheen
Nor shapes of men nor beasts we ken—
The ice was all between
The first quatrain describes a sense of wonder and awe. Entities that might portend doom—namely cold and ice—are described using words like “wondrous” and “emerald.” Yet in the subsequent verse, that awe gives way to a sense of foreboding, and words like “dismal” creep in. Suddenly we get a sense of isolation and a fear of what may be in store for the poem’s namesake mariner.
3. Use Rhyme and Meter as Useful Tools
Sometimes it’s easiest to be creative when there are rules to guide you. Remember that most ballads consist of quatrains where either the first and third lines rhyme, or the second and fourth lines rhyme. Don’t regard this as a limitation. Look at it as a structural aid to propel you forward. Perhaps you don’t want your ballad to be as rigidly structured as the earlier example about the pants-wetting knight; then again, perhaps that level of rhythmic precision is helpful. It’s truly up to you.
4. Let the Story Guide You
Writing a full song or poem may be intimidating, but an evolving storyline can easily propel you forward. Case in point: Coleridge’s “Rime of the Ancient Mariner” is 143 verses long. (And Iron Maiden’s adaptation of it as a heavy metal song is thirteen minutes, forty-five seconds long.) Meanwhile, Bob Dylan’s ballad “Highlands” is sixteen minutes, thirty-one seconds long. If you have a good story to tell in your ballad, you should have no difficulty writing verse after verse.
Many musical ballads tell their stories in the verses while continually returning to a repeated chorus, or even just a single repeated line (such as the title phrase in Dylan’s “Tangled Up in Blue”). John Prine’s folk ballad “Lake Marie” contains long spoken verses broken up by anthemic sung choruses that are the same each time. Other ballads, like Iron Maiden’s “Rime of the Ancient Mariner” return to musical motifs but without a repeated lyrical phrase.
The practice of storytelling interspersed with repeated themes or lyrics is called “incremental reception.” One such example is the poem “Lord Randall” by Sir Patrick Spens. Note the repeated phrases in this stanza:
Oh where ha’e ye been, Lord Randall my son?
O where ha’e ye been, my handsome young man?
“I ha’e been to the wild wood: mother, make my bed soon
For I’m weary wi’ hunting, and fair wald lie down”
“Where gat ye your dinner, Lord Randall my son?
Where gat ye your dinner, my handsome young man?”
The story advances, but the repeated phrases give it structure. None other than Bob Dylan himself would emulate this technique in tunes like “A Hard Rain’s a-Gonna Fall.”
Examples of Ballads in Music
Ballads are found in all forms of popular music. These include:
- Folk. Ballads are a key part of the folk tradition. Bob Dylan’s “The Lonesome Death of Hattie Carroll” is one such example. For a lighter folk ballad, seek out “Puff the Magic Dragon” by Peter Paul and Mary.
- Country. Country music has always been a storytelling genre. For a mainstream country ballad, consider “God Bless the Broken Road” by Rascal Flatts. For a more alternative country ballad, you can’t go wrong with “Christian Lady Talkin’ on a Bus” by the eccentric Blaze Foley.
- Rock. The word “ballad” is a bit looser in rock. Some songs with “ballad” in the title really do tell stories, like The Beatles’ “The Ballad of John and Yoko.” Yet ironically the great balladeer Bob Dylan wrote a song called “Ballad of a Thin Man” that’s less of a story and more of a snarling character critique. Plenty of other rock songs tell stories like Led Zeppelin’s Tolkien-inspired “Ramble On.”
- Jazz. In jazz, the word “ballad” typically refers to slow, melodic tunes. The story element is optional, particularly because so many jazz ensembles don’t even feature vocalists. “Misty,” “Darn That Dream,” and “Body and Soul” are examples of classic jazz ballads.<|endoftext|>
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Table of Contents
A machine which can access Internet resources has an IP address, whether that IP address is a public address or a private address hidden behind an SNAT router . With the increasingly common use of linux machines as servers, desktops, and embedded devices and with changing network topologies and re-addressing, the need to be able to determine the current IP address of a machine and modify that address has consequently become a common need.
I assume in this chapter that the reader has some familiarity with CIDR addressing and netmasks. If any of these concepts are unfamiliar, or the reader would like to brush up, I suggest a visit to some of the links which can be found in Section I.1.3, “General IP Networking Resources”.
We'll begin our tour of the utilities for observing, changing, removing, and adding IP addresses to network devices with ifconfig, the traditional utility for IP management. We will also examine the newer and more flexible ip address, a key part of the iproute2 package.
I'm sure somebody will be glad to nitpick here and tell me that s/he has a machine connected to the Internet which uses SNA, DecNET, IPX, or NetBEUI to connect to another host which actually does speak IP, thus proving that not every host which has access to the Internet is actually directly speaking IP. Another example is doubtless, wireless devices, such as telephones. Here, I'll concern myself with the majority case.<|endoftext|>
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# Pow: the Broken Eggs
Topics: Roman numerals, Number, 175 Pages: 4 (1166 words) Published: April 9, 2013
Group 7 Hearts 09-17-10 Period 5
The Broken Eggs
POW #1
1. Problem Statement:
A farmer has some eggs in a cart, and is going to market them. She accidently breaks every egg. She doesn’t remember how many she had, but she remembers some things. She knows that when she put them in groups of 2, 3, 4, 5 and 6, there was one egg left over. When she put them in groups of 13 no eggs were left over. You need to find out how many eggs there are in total.
2. Process:
I first thought about a common multiple of 2, 3, 4, 5, and 6. The least common multiple of those numbers is 60. I know the amount of eggs the farmer has must be a multiple of 60 plus 1. It has to be a multiple of 60 plus 1, because any multiple of 60 plus 1 divided by 2, 3, 4, 5, and 6 will result with a remainder of 1. Then I wrote some multiples of 60 and added 1.
60| +| 1| =| 61|
120| +| 1| =| 121|
180| +| 1| =| 181|
240| +| 1| =| 241|
300| +| 1| =| 301|
360| +| 1| =| 361|
420| +| 1| =| 421|
480| +| 1| =| 481|
540| +| 1| =| 541|
600| +| 1| =| 601|
660| +| 1| =| 661|
Those numbers are possible solutions, but now we have to see if they are divisible by 13. I noticed they all end in 1; therefore the answer must end in 1. I then highlighted the multiples of 13 that end with a 1. 13| 26| 39| 52| 65| 78| 91| 104| 117|
130| 143| 156| 169| 182| 195| 208| 221| 234|
247| 260| 273| 286| 299| 312| 325| 338| 351|
364| 377| 390| 403| 416| 429| 442| 455| 468|
481| 494| 507| 520| 533| 564| 559| 572| 585|
598| 611| 624| 637| 650| 663| 676| 689| 702|
| | | | | | | | |
I checked to...<|endoftext|>
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# Thread: sums and products of roots
1. ## sums and products of roots
i dont understand this question:
in the quadratic equation x²+mx+2=0, the roots are consecutive. find the values of m.
i dont understand the part of the consecutive roots
2. The roots are $\displaystyle a$ and $\displaystyle a+1, \ a\in\mathbb{Z}$
Then $\displaystyle \left\{\begin{array}{ll}a+a+1=-m\\a(a+1)=2\end{array}\right.$
Solve the second equation, find a, then replace a in the first equation and find m.
3. thanks for your help but i havent leant that so could you explain to me how that works?
4. Originally Posted by pogiphilip
thanks for your help but i havent leant that so could you explain to me how that works?
$\displaystyle (-m)^2-4\cdot 2=(a+(a+1))^2-4a(a+1)=(a-(a+1))^2=1$
5. i still dont understand what that is ynj
6. Originally Posted by pogiphilip
i still dont understand what that is ynj
Do you know Vieta's Theorem?
if a quadratic equation $\displaystyle ax^2+bx+c=0$ have two roots $\displaystyle \alpha,\beta$, then $\displaystyle a(x-\alpha)(x-\beta)=ax^2+bx+c\Leftrightarrow ax^2-a(\alpha+\beta)x+a\alpha\beta=0$
Thus $\displaystyle \alpha+\beta=-\frac{b}{a},\alpha\beta=\frac{c}{a}$to keep all the coefficient identical.
So $\displaystyle a+(a+1)=-m,a(a+1)=2$,$\displaystyle (-m)^2-4\cdot 2=(a+(a+1))^2-4a(a+1)=(a-(a+1))^2=1$, Then you can solve m!
7. but also the question says to use roots that are consecutive?, and i don't know vieta's theorem.
8. Originally Posted by pogiphilip
but also the question says to use roots that are consecutive?, and i don't know vieta's theorem.
That means two roots are in the form a,a+1. Vieta's Theorem has been proved in my previous post.
9. thank you for your help and i think i figured it out. thanks ynj and red_dog
10. also what are the roots if the two roots are consecutive?
11. Originally Posted by pogiphilip
also what are the roots if the two roots are consecutive?
you may use formula to solve it .......<|endoftext|>
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# A line segment goes from (1 ,2 ) to (4 ,1 ). The line segment is reflected across x=-1, reflected across y=3, and then dilated about (2 ,2 ) by a factor of 3. How far are the new endpoints from the origin?
##### 1 Answer
Apr 12, 2018
color(crimson)( bar(OA") ~~ 14.87
color(crimson)( bar(OB") ~~ 16.12
#### Explanation:
Given points $A \left(1 , 2\right) , B \left(4 , 1\right)$
Reflected across $x = - 1$, $y = 3$
color(blue)("Reflection Rules :"
color(blue)("reflect over x-axis. (x,-y)"
color(blue)("reflect over y-axis. (-x,y)"
color(blue)("reflect over line y=x. (y,x)"
color(blue)("reflect over line y= -x. (-y,-x)"
color(blue)("reflect thru origin. (-x,-y)"
color(brown)("reflect thru a different point. ex: (5,-1) h=5 k= -1. (2h-x, 2k-y)"
color(blue)("reflect over a line. ex: x=6. (2h-x, y)"
color(blue)("reflect over a line. ex: y= -3. (x, 2k-y)"
A (x,y) -> A’(x, y) = (1,2) -> ((2x’- 1), (2y’ - 2))
A’(x,y) => ((-2 - 1), (6 - 2)) => (-3, 4)
B (x,y) - > B’ (x,y) = (4 , 1) -> ((2x’ - 4), (2y’ - 1)
B’(x,y) => ((-2 - 4), (6 - 1) => (-6, 5)
New coordinates after reflection are A’(-3, 4), B’(-6, 5)
Now we we have to find A”, B” after rotation about point C (2,2) with a dilation factor of 3.
A”(x, y) -> 3 * A’(x,y) - C (x,y)
A”((x),(y)) = 3 * ((-3), (4)) - ((2), (2)) = ((-11),(10))
Similarly, B”((x),(y)) = 2 * ((-6), (5)) - ((2), (2)) = ((-14),(8))
New endpoints are A”(-11, 10), B”(-14, 8)
Distance of new points from origin
vec(OA”) = sqrt(-11^2 + 10^2) ~~ 14.87
vec(OB”) = sqrt(-14^2 + 8^2) ~~ 16.12<|endoftext|>
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# Multiplying 2 digit number by 1 digit number
In this section you will learn multiplying 2 digit number by 1 digit number.
In the earlier section , you have learnt about the multiplication of numbers on the number line as well as without number line.
There are different methods to do the multiplication of two digit number by one digit.
Method 1 :
Example : A flower has five petals. A bunch of flowers has 13 flowers. How many petals are there in the bunch?
Solution : Total flowers = 13
Number of petals in each flower = 5
So total number of petals = 13 X 5
Write 13 as 10 + 3
10 X 5 = 50
3 x 5 = 15
Total petals = 50 + 15 = 65.
Method 2 :
23 x 3 :
This method is known as vertical method.
1) Write the bigger number.
2) Give the multiplication sign (x).
3) Write the 2nd number.
4) Multiply the 2nd number by the digit in ones place and write the answer.
5) Multiply the tens number by the 2nd number and write the answer using tens.
23 ----> X 3 -----> ________ ∴ 23 x 3 = 69 2 tens + 3 ones X 3 ______________ 6 tesn + 9 ones 60 + 9 = 69
3) 36 x 0 36 ----> X 0 -----> ________ ∴ 36 x 0 = 0 When you multiply any number by zero the answer will be 0(zero) only. 3 tens + 6 ones X 0 ______________ 0 tens + 0 ones 0 + 0 = 0
## Examples on multiplying 2 digit number by 1 digit number
1) A design has 3 flowers in it. A piece of cloth has 12 such designs. How many flowers will be on the cloth? Solution : Number of flowers in one design = 3
Total number of designs = 12
∴ Total number of flowers = 12 x 3
12
X 3
_________
3 ← 6
3 tens + 6 ones
= 36
Hence, 12 x 3 = 36.
1) Write the bigger number '12'.
2) Give the multiplication sign (x).
3) Write the 2nd number '3'.
4) Multiply the 2nd number(3) by the digit in ones place(2) and write the answer as 6 ones .
5) Multiply the tens number(1) by the 2nd number(3) and write the answer as 3 tens.
6) 3 tens + 6 ones = 30 + 6 = 36.
∴ Total number of flowers= 36.<|endoftext|>
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In other words, the angle of rotation the radius need to move in order to produce the given arc length. Read on to learn the definition of a central angle and how to use the central angle formula. What would the central angle be for a slice of pizza if the crust length (L) was equal to the radius (r)? Hence, as the proportion between angle and arc length is constant, we can say that: L / θ = C / 2π. To calculate the radius. The relationship between the radius, the arc length and the central angle (when measured in radians) is: a = rθ. three you want to use to determine the arc, and which one to allow to be the rounded-off resultant. To use absolute length you have to set the N input to false.. So multiply both sides by 18 pi. The arc length of a sector is 66 cm and the central angle is 3 0°. Watch an example showing how to find the radius when given the arc length and the central angle measure in radians. How to use the calculator Enter the radius and central angle in DEGREES, RADIANS … The figure explains the various parts we have discussed: Given an angle and the diameter of a circle, we can calculate the length of the arc using the formula: ArcLength = ( 2 * pi * radius ) * ( angle / 360 ) Where pi = 22/7, diameter = 2 * radius, angle is in degree. Therefore the arc length will be half of 8: 4cm. The angle t is a fraction of the central angle of the circle which is … That's the degree measure of 1 radian. Use the formula for S = r Θ and calculate the intercepted arc: 6Π. If we are only given the diameter and not the radius we can enter that instead, though the radius is always half the diameter so it’s not too difficult to calculate. The angle: For more universal calculator regarding circular segment in general, check out the Circular segment calculator. The picture below illustrates the relationship between the radius, and the central angle in radians. Plugging our radius of 3 into the formula, we get C = 6π meters or approximately 18.8495559 m. Now we multiply that by (or its decimal equivalent 0.2) to find our arc length, which is 3.769911 meters. Example 4. Enter central angle =63.8 then click "CALCULATE" and your answer is Arc Length = 4.0087. In a right triangle the sum of the two acute angles is equal to 90°. Where C is the central angle in radians; L is the arc length; r is the radius; Central Angle Definition. The arc segment length is always radius x angle. Find the length of arc whose radius is 42 cm and central angle is 60° Solution : Length of arc = (θ/360) x 2 π r. Here central angle (θ) = 60° and radius (r) = 42 cm = (60°/360) ⋅ 2 ⋅ (22/7) ⋅ 42 = (1/6) ⋅ 2 ⋅ 22 ⋅ 6 = 2 ⋅ 22 = 44 cm. We see that an angle of one radian spans an arc whose length is the radius of the circle. Two acute angles are complementary to each other if their sum is equal to 90°. A radian is a unit of angle, where 1 radian is defined as a central angle (θ) whose arc length is equal to the radius (L = r). Example 4 : Find the radius, central angle and perimeter of a sector whose length of arc and area are 4.4 m and 9.24 m 2 respectively Notice that this question is asking you to find the length of an arc, so you will have to use the Arc Length Formula to solve it! Here you need to calculate the angle, then again use the formula. One radian is the angle where the arc length equals the radius. This math worksheet was created on 2017-03-10 and has been viewed 8 times this week and 37 times this month. The central angle is a quarter of a circle: 360° / 4 = 90°. We get a is equal to-- this is 35 times 18 over 36 pi. The Earth is approximately 149.6 million km away from the Sun. For a given central angle, the ratio of arc to radius is the same. Okay, so the example in my book says that the angle measure *radius = arc length, when the angle measure is in radians. Another example is if the arc length is 2 and the radius is 2, the central angle becomes 1 radian. This allows us to lay out the arc using a large compass. Finding Arc Length of a Circle - Duration: 9:31. Wayne, I would do it in 2 steps. What about if the angle measure is … So all you need is the distance between the end points of your arc and the radius of the circle to compute the angle, $\theta = \arccos (1- {{d^2}\over {2r^2}})$ Lastly, the length is calculated - In circle O, the radius is 8 inches and minor arc is intercepted by a central angle of 110 degrees. Make sure you don’t mix up arc length with the measure of an arc which is the degree size of its central angle. The central angle calculator is here to help; the only variables you need are the arc length and the radius. You can define the chord length with Start, Center, Length, but in this case we want to define the arc length, not the chord. Simplify the problem by assuming the Earth's orbit is circular (. arc length = angle measure x r. when angle measured in deg. Arc Length = θr. Arc Length Formula - Example 1 Discuss the formula for arc length and use it in a couple of examples. And 2π radians = 360°. Now, if you are still hungry, take a look at the sector area calculator to calculate the area of each pizza slice! A radian is the angle subtended by an arc of length equal to the radius of the circle. Since the problem defines L = r, and we know that 1 radian is defined as the central angle when L = r, we can see that the central angle is 1 radian. What is the value of the arc length S in the circle pictured below? Before you can use the Arc Length Formula, you will have to find the value of θ (the central angle that intercepts arc KL) and the length of the radius of circle P.. You know that θ = 120 since it is given that angle KPL equals 120 degrees. Let the arc subtend angle θ at the center Then, Angle at center = Length of Arc/ Radius of circle θ = l/r Note: Here angle is in radians. The simplicity of the central angle formula originates from the definition of a radian. Circular segment. This step gives you The angle: For more universal calculator regarding circular segment in general, check out the Circular segment calculator. Just click and drag the points. Let's approach this problem step-by-step: You can try the final calculation yourself by rearranging the formula as: Then convert the central angle into radians: 90° = 1.57 rad, and solve the equation: When we assume that for a perfectly circular orbit, the Earth travels approximately 234.9 million km each season! An angle of 1 radian refers to a central angle whose subtending arc is equal in length to the radius. 5:14. We could also use the central angle formula as follows: In a complete circular pizza, we know that the central angles of all the slices will add up to 2π radians = 360°. I would be inclined most of the time to let that be the length, since if the arc meets tangent straight lines at both ends, their angles will determine the included angle of the arc, and the radius … Note that our units will always be a length. Arc length = [radius • central angle (radians)] Arc length = circumference • [central angle (degrees) ÷ 360] Proof of the trigonometric ratios of complementary allied angles. You can also use the arc length calculator to find the central angle or the radius of the circle. \] Intuitively, it is obvious that shrinking or magnifying a circle preserves the measure of a central angle even as the radius changes. Because maths can make people hungry, we might better understand the central angle in terms of pizza. The radius: The angle: Finding the arc length by the radius and the height of the circular segment. The formula is $$S = r \theta$$ where s represents the arc length, $$S = r \theta$$ represents the central angle in radians and r is the length of the radius. where θ is the measure of the arc (or central angle) in radians and r is the radius of the circle. When the radius is 1, as in a unit circle, then the arc length is equal to the radius. An arc length R equal to the radius R corresponds to an angle of 1 radian. View solution A sector is cut from a circle of radius 4 2 c m . Use the formula for S = r Θ and calculate the solution. Check out 40 similar 2d geometry calculators . More generally, the magnitude in radians of a subtended angle is equal to the ratio of the arc length to the radius of the circle; that is, θ = s/r, where θ is the subtended angle in radians, s is arc length, and r is radius. Recall that 2πR is the circumference of the whole circle, so the formula simply reduces this by the ratio of the arc angle to a full angle (360). Figure 1. formulas for arc Length, chord and area of a sector In the above formulas t is in radians. What is the radius? You probably know that the circumference of a circle is 2πr. Finding the radius, given the sagitta and chord If you know the sagitta length and arc width (length of the chord) you can find the radius from the formula: where: Length of arc when central angle and radius are given can be defined as the line segment joining any two points on the circumference of the circle provided the value of radius length and central angle for calculation is calculated using Arc Length=(pi*Radius*Central Angle)/180.To calculate Length of arc when central angle and radius are given, you need Radius (r) and Central Angle (θ). Question from pavidthra, a student: Length or arc 11 and angle of subtended 45.need to find a radius A central angle is an angle contained between a radius and an arc length. So the central angle for this sector measures (5/3)π . Central angle in radians* We arrive at the same answer if we think this problem in terms of the pizza crust: we know that the circumference of a circle is 2πr. Welcome to The Calculating Arc Length or Angle from Radius or Diameter (A) Math Worksheet from the Measurement Worksheets Page at Math-Drills.com. So I can plug the radius and the arc length into the arc-length formula, and solve for the measure of the subtended angle. When constructing them, we frequently know the width and height of the arc and need to know the radius. L - arc length h- height c- chord R- radius a- angle. You can find the central angle of a circle using the formula: where θ is the central angle in radians, L is the arc length and r is the radius. One radian is defined as the angle subtended from the center of a circle which intercepts an arc equal in length to the radius of the circle. Use the central angle calculator to find arc length. You can try the final calculation yourself by rearranging the formula as: L = θ * r Try using the central angle calculator in reverse to help solve this problem. Before you can use the Arc Length Formula, you will have to find the value of θ (the central angle that intercepts arc KL) and the length of the radius of circle P.. You know that θ = 120 since it is given that angle KPL equals 120 degrees. The picture below illustrates the relationship between the radius, and the central angle in radians. Notice that this question is asking you to find the length of an arc, so you will have to use the Arc Length Formula to solve it! 1. Do note that the default behaviour is for the length to be ‘normalised’, that is 0.0 = start, 0.5 = middle and 1.0 = end of curve. Where does the central angle formula come from? An arc is a segment of a circle around the circumference. If the Earth travels about one quarter of its orbit each season, how many km does the Earth travel each season (e.g., from spring to summer)? To illustrate, if the arc length is 5.9 and the radius is 3.5329, then the central angle becomes 1.67 radians. Calculate the arc length according to the formula above: L = r * Θ = 15 * π/4 = 11.78 cm . Let’s take some examples If radius of circle is 5 cm, and length of arc is 12 cm. Real World Math Horror Stories from Real encounters. Here central angle (θ) = 60° and radius (r) = 42 cm = (60°/360) ⋅ 2 ⋅ (22/7) ⋅ 42 = (1/6) ⋅ 2 ⋅ 22 ⋅ 6 = 2 ⋅ 22 = 44 cm. The angle around a circle can go from 0 to 2 pi radians. How to Find the Sector Area If the arc has a central angle of π/4, and the arc length is the radius multiplied by the central angle, then (3) (π/4) = 3π/4 = 2.36 m. We know that for the angle equal to 360 degrees (2π), the arc length is equal to circumference. The idea of measuring angles by the length of the arc was already in use by other mathematicians. Arc length. For example, al-Kashi (c. 1400) used so-called diameter parts as units, where one diameter part was 1 / 60 radian. If you know radius and angle you may use the following formulas to calculate remaining segment parameters: This is true for a circle of any size, as illustrated at right: an arclength equal to one radius determines a central angle of one radian, or about $$57.3\degree\text{. Well, same exact logic-- the ratio between our arc length, a, and the circumference of the entire circle, 18 pi, should be the same as the ratio between our central angle that the arc subtends, so 350, over the total number of degrees in a circle, over 360. How to Find the Arc Length Given the Radius and an Angle - Duration: 5:14. The radius is 10, which is r. Plug the known values into the formula. That is not a coincidence. What is the arc length formula? Time for an example. You can imagine the central angle being at the tip of a pizza slice in a large circular pizza. arc length = angle measure x 2x pi x r/ 360. relation betn radian and deg 3. Here you need to calculate the angle, then again use the formula. Arc length formula. Find angle … Both can be calculated using the angle at the centre and the diameter or radius. MrHelpfulNotHurtful 31,487 views. where: C = central angle of the arc (degree) R = is the radius of the circle π = is Pi, which is approximately 3.142 360° = Full angle. To find the length of an arc with an angle measurement of 40 degrees if the circle has a radius of 10, use the following steps: Assign variable names to the values in the problem. ( "Subtended" means produced by joining two lines from the end of the arc to the centre). Have you ever wondered how to find the central angle of a circle? A chord separates the circumference of a circle into two sections - the major arc and the minor arc. Find the length of the arc in terms of π that subtends an angle of 3 0 ∘ at the centre of a circle of radius 4 cm. The angle measurement here is 40 degrees, which is theta. Calculate the area of a sector: A = r² * Θ / 2 = 15² * π/4 / 2 = 88.36 cm² . 5 is half of 10. I am in need of a lisp routine to label arc length, arc radius and total arc angle. Question from pavidthra, a student: Length or arc 11 and angle of subtended 45.need to find a radius Answer. The angle: Finding the arc length by the radius and the height of the circular segment. When the radius is 1, as in a unit circle, then the arc length is equal to the radius. Strictly speaking, there are actually 2 formulas to determine the arc length: one that uses degrees and one that uses radians. 2) A circle has an arc length of 5.9 and a central angle of 1.67 radians. If you want to convert radians to degrees, remember that 1 radian equals 180 degrees divided by π, or 57.2958 degrees. They also used sexagesimal subunits of the diameter part. Area: [1] Arc length: Chord length: Segment height: Calculate the measure of the arc length S in the circle pictured below? Yet it remains to be proved that if an arc is equal to the radius in one circle, it will subtend the same central angle as an arc equal to the radius in another circle. (a) In an angle of 1 radian, the arc length equals the radius (b) An angle of 2 radians has an arc length (c) A full revolution is or about 6.28 radians. The central angle between the first radius and the repositioned radius will be a few degrees less than 60°, approximately 57.3°. Arc Length Formula: The length of an arc along a circle is evaluated using the angle and radius of the circular arc or circle, as represented by the formula below. To calculate the radius. By transposing the above formula, you solve for the radius, central angle, or arc length if you know any two of them. Learn how tosolve problems with arc lengths. when angle measured in radian. 2. So if the circumference of a circle is 2πR = 2π times R, the angle for a full circle will be 2π times one radian = 2π. Arc length is a fraction of circumference. The radius is 10, which is r. It may be printed, downloaded or saved and used in your classroom, home school, or other educational … Once I've got that, I can plug-n-chug to find the sector area. The radius is the distance from the Earth and the Sun: 149.6 million km. The formula is S = r \theta where s represents the arc length, S = r \theta represents the central angle in radians and r is the length of the radius. Then, knowing the radius and half the chord length, proceed as in method 1 above. Area of a sector is a fractions of the area of a circle. When constructing them, we frequently know the width and height of the arc and need to know the radius. For example, an arc measure of 60º is one-sixth of the circle (360º), so the length of that arc will be one-sixth of the circumference of the circle. Central angle when radius and length for minor arc are given is an angle whose apex (vertex) is the center O of a circle and whose legs (sides) are radii intersecting the circle in two distinct points A and B provided the values for radius and length for the minor arc is given and is represented as θ=L/r or Central Angle=Length of Minor Arc/Radius. Remember that the circumference of the whole circle is 2πR, so the Arc Length Formula above simply reduces this by dividing the arc angle to a full angle (360). If you know radius and angle you may use the following formulas to calculate remaining segment parameters: Circular segment formulas. Bonus challenge - How far does the Earth travel in each season? This allows us to lay out the arc using a large compass. Central Angle … Let us consider a circle with radius rArc is a portion of the circle.Let the arc subtend angle θ at the centerThen,Angle at center = Length of Arc/ Radius of circleθ = l/rNote: Here angle is in radians.Let’s take some examplesIf radius of circle is 5 cm, and length of arc is 12 cm. The angle measurement here is 40 degrees, which is theta. An arc’s length means the same commonsense thing length always means — you know, like the length of a piece of string (with an arc, of course, it’d be a curved piece of string). Typically, the interior angle of a circle is measured in degrees, but sometimes angles are measured in radians (rad). The arc segment length is always radius x angle. Use the formula for S = r Θ and calculate the intercepted arc: 4Π. Circular segment - is an area of a circle which is "cut off" from the rest of the circle by a secant (chord).. On the picture: L - arc length h- height c- chord R- radius a- angle. To use the arc length calculator, simply enter the central angle and the radius into the top two boxes. Thank you. Find the length of minor arc to the nearest integer. A practical way to determine the length of an arc in a circle is to plot two lines from the arc's endpoints to the center of the circle, measure the angle where the two lines meet the center, then solve for L by cross-multiplying the statement: measure of angle in degrees/360° = L /circumference. attached dwg. }$$ The interative demonstration below illustrates the relationship between the central angle of a circle, measured in radians, and the length of the intercepted arc. It also separates the area into two segments - the major segment and the minor segment. It could of course be done by calculating the angle mathematically, but an easier way to do it is to draw the arc with the required radius and centre point, then modify it using the lengthen tool on the Modify tab drop down. the Evaluate Length component can be used to find the point at (and the parameter along) the curve at a certain length from the starting point. Click the "Arc Length" button, input radius 3.6 then click the "DEGREES" button. So, the radius of sector is 30 cm. The angle around a circle can go from 0 to 2 pi radians. Worksheet to calculate arc length and area of sector (radians). To find the length of an arc with an angle measurement of 40 degrees if the circle has a radius of 10, use the following steps: Assign variable names to the values in the problem. Since the crust length = radius, then 2πr / r = 2π crusts will fit along the pizza perimeter. You will learn how to find the arc length of a sector, the angle of a sector or the radius of a circle. In this calculator you may enter the angle in degrees, or radians or both. A central angle is an angle with a vertex at the centre of a circle, whose arms extend to the circumference. To elaborate on this idea, consider two circles, one with radius 2 and the other with radius 3. An arc measure is an angle the arc makes at the center of a circle, whereas the arc length is the span along the arc. The circle angle calculator in terms of pizza. There is a formula that relates the arc length of a circle of radius, r, to the central angle, $$\theta$$ in radians. Since each slice has a central angle of 1 radian, we will need 2π / 1 = 2π slices, or 6.28 slices to fill up a complete circle. Interactive simulation the most controversial math riddle ever! Solved: Hello everybody in the forum. If you move the 2nd radius slightly clockwise the right amount, you'll get an arc length that is exactly 1. Step 1: Find the measure of the angle t in the diagram.. The formula is S = r θ where s represents the arc length, S = r θ represents the central angle in radians and r is the length of the radius. The length a of the arc is a fraction of the length of the circumference which is 2 π r.In fact the fraction is .. How many pizza slices with a central angle of 1 radian could you cut from a circular pizza? \theta ~=~ \frac{\text{arc length}}{\text{radius}} ~~. That is often cited as the definition of radian measure. This angle measure can be in radians or degrees, and we can easily convert between each with the formula π radians = 180° π r a d i … The length of an arc depends on the radius of a circle and the central angle θ. In a unit circle, the measure of an arc length is numerically equal to the measurement in radians of the angle that the arc length subtends. Definition. Now, in a circle, the length of an arc is a portion of the circumference. 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DISCOVER
# How to calculate the fluid flow through a hole in a pipe
A common problem with pipes is corrosion. Over time, the corrosion in the pipe can make a hole that causes a leak. Calculating the fluid flow through a hole can be difficult, due to many variables like fluid flow velocity, pressure in the pipe and density of the fluid, just to name a few, but don’t get discouraged. You can find the answer you need by following a simple series of steps.
Obtain measurements: diameter (d) of the hole in the pipe and height (h) of the surface of the fluid above the hole. Make sure all measurements are in the same standard unit. Know that 1 inch = 0.0254 meters, so if you use inches, convert your measurements.
Calculate the cross-sectional area of the hole (A). Divide the diameter of the hole in half to get the radius. Use the formula A = 3.14 (number Pi) * r^2 (radius to the second power). The result will be in square units.
Use the Bernoulli equation to find the fluid velocity (V) if it is not already provided. If the fluid pressure in a pipe is constant (the flow is steady), the fluid leaves through the hole in the pipe at a velocity of V = square root (2_g_h), where g is acceleration due to gravity (g = 9.8m/s^2).
Multiply the cross-sectional area of the hole by the fluid velocity to find the volume of fluid (Q): Q = A * V. This will be the volume of the fluid that leaves the hole in cubic meters per second.
Let's look at an example with numbers. Calculate fluid flow through the hole in the pipe with constant pressure if the water leaves the hole with a velocity of 1.7m/s and the diameter of the hole is d = 1 inch = 1 * 0.0254 = 0.0254 meters.
First, find the cross-sectional area of the hole: A = 3.14 * (0.0254/2)^2 = 0.00051m^2. Since the pressure is constant and the velocity of the water going through the hole is 1.7m/s, use the formula from Step 4 to find the volume of water that leaves the hole: Q = 0.00051m^2 * 1.7m/s = 0.000867m^3/s.
Since 1 cubic meter = 61,024 cubic inches, Q = 0.000867m^3/s * 61,024 = 52.9 inch^3/s. Thus, 52.9 cubic inches of water leaves the hole in the pipe per second.
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# Prove $\log_{\frac{1}{4}} \frac{8}{7}> \log_{\frac{1}{5}} \frac{5}{4}$
Prove $$\log_{\frac{1}{4}} \frac{8}{7}> \log_{\frac{1}{5}} \frac{5}{4}$$
How to prove without a computer?
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Here is another argument using power-series-based computations of $\log$, but slightly different in its details to Ivan's.
$$\log_{1/4} 8/7 = -\log_4 8/7 = \log_4 7/8 = \log_4 (1 - 1/8),$$ while $$\log_{1/5} 5/4 = -\log_5 5/4 = \log_5 4/5 = \log_5 (1 - 1/5).$$
So the problem reduces to showing that $$\log_4(1-1/8) >\log_5 (1- 1/5).$$
This is slightly delicate, because both numbers of which we are taking the $\log$ are close to $1$, so both sides are close to $0$, and we have to estimate how close.
Recall that $$\ln (1- x) = - x - x^2/2 - x^3/3 - \cdots$$ when $|x| < 1$, and that $\log_a (1-x) = (1/\ln a)\ln (1-x)$. So we have to prove $$\ln 4 \, (1/5 + 1/50 + \cdots ) > \ln 5 \,(1/8 + 1/128 + \cdots).$$ Now clearly $$(1/5 + 1/50 + \cdots ) > 8/5 (1/8 + 1/128 + \cdots),$$ so it suffices to show that $$(8/5)\ln 4 > \ln 5,$$ i.e. that $$\ln 4^8 > \ln 5^5,$$ i.e. that $$65536 > 3125,$$ which is true.
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$\newcommand{\Log}{\operatorname{Log}}$I will use approximations by power series. It is easy to show that the inequality is equivalent to:$\Log(3.5) \Log(5) > \Log(4)^2$. On the other hand consider the power series expansion for $\Log(x)$ around $3.5$.It is: $$\Log(3.5)+\frac{2}{7}(x-3.5)-\frac{2}{49}(x-3.5)^2+O(x-3.5)^3$$ For $x=5$ we get: $\Log(5)>33/98+\Log[3.5]$
On the other hand for $x=4$ the linear term gives: $\Log(4)<1/7+\Log(3.5)$. Now we aim to prove the following:$$\Log(3.5) \Log(5) > \Log(3.5)*(33/98 + \Log(3.5)) > (1/7 + \Log(3.5))^2 > \Log(4)^2$$
The first and last are proved already by the approximations. We need the middle. It is equivalent to the positivity of:$$\Log(3.5)*(33/98 + \Log(3.5)) - (1/7 + \Log(3.5))^2 =-(1/49) + 5/98 \Log(3.5) =(1/49)(5/2\Log(3.5)-1)$$
The last is positive because $\Log(3.5)>1>2/5$.
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I would have thought you needed $$\frac{\log(5) - \log(4)}{\log(1) - \log(5)}<\frac{\log(8) - \log(7)}{\log(1) - \log(4)}$$ – Henry Sep 20 '12 at 9:00
This is exactly what I prove. Reversing up the first inequality by multiplying the denominators by -1. – ivan Sep 20 '12 at 9:01
My denominators are not the same as your final line – Henry Sep 20 '12 at 9:02
yes, you are right... this proof is wrong – ivan Sep 20 '12 at 9:07
I have completely redone the proof, so the comments above are obsolete. – ivan Sep 20 '12 at 11:42
My second attempt, hopefully not flawed this time. (Although I think that there must be a simpler way to do this.)
I will use $\ln$ to denote natural logarithm. (Some people use $\log$)
The inequality $$\log_{\frac{1}{4}} \frac{8}{7}> \log_{\frac{1}{5}} \frac{5}{4}$$ can be rewritten as $$\frac{\ln\frac87}{\ln\frac14} > \frac{\ln\frac54}{\ln\frac15}.$$ This is equivalent to the inequality $$\ln\frac87 \ln \frac15 > \ln \frac54 \ln\frac14$$ (we have multiplied both sides of the inequality by $\ln\frac14$ and $\ln\frac15$. So we have multiplied the inequality by a negative number twice - as a final result the inequality signs remains unchanged.)
This leads to equivalent inequalities
$$-\ln 5 \ln\frac87 > -\ln 4 \ln \frac54\\ \ln 4 \ln \frac54 > \ln 5 \ln\frac87. \tag{*}$$
So the above inequality $(*)$ is what we want show.
We know that $\ln(1+x)<x$ and $\ln(1+x)>\frac{x}{1+x}$ for $x>0$. (Both these inequalities can be verified simply by taking derivative of functions $\ln(1+x)-x$ and $\ln(1+x)-\frac{x}{1+x}$ and showing that the derivative is negative/positive for $x>0$. See WolframAlpha: 1, 2. Of course, there can be other ways how to prove this. Both these properties are mentioned at Wikipedia.)
Now if we use one of these inequalities for $x=\frac17$, we get $\ln\frac87 < \frac17$, which gives $$\ln\frac87\ln 5 < \frac17\ln5.$$
For $x=\frac14$ we have $\ln\frac54 > \frac{1/4}{1+1/4} = \frac{1/4}{5/4} = \frac15$ and thus $$\ln \frac54 \ln 4 > \frac15 \ln 4.$$
Now we may notice that the following inequalities are equivalent to each other: $$\frac15 \ln 4 > \frac17\ln5 \tag{**}\\ 7\ln 4 > 5\ln 5\\ \ln(4^7) > \ln{5^5}\\ 4^7 > 5^5\\$$ The last inequality is true, which can be seen by evaluating $4^7=16384$ and $5^5=3125$. (And this can be shown by hand, too, since $4^7=2^{14}=16\cdot2^{10}=16\cdot1024 > 16000$ and $5^5=125\cdot 25 < 128 \cdot 25 = 32 \cdot (4\cdot 25) = 3200$.) This proves $(**)$.
So we have shown the following chain of inequalities $$\ln \frac54 \ln 4> \frac15 \ln 4 > \frac17\ln5 > \ln\frac87\ln 5.$$ If we compare the leftmost and the rightmost expression, this is precisely $(*)$.
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Note that
$\log_{\frac{1}{4}}\frac{8}{7}=-\log_4\frac{8}{7}=\log_47-\log_48=\log_4 7-\frac{3}{2}$,
$\log_{\frac{1}{5}}\frac{5}{4}=-\log_5\frac{5}{4}=\log_54-\log_55=\log_54-1$.
Hence we need to prove $$\log_47-\frac{3}{2}>\log_54-1.\ (*)$$ Now, $$(*)\Leftrightarrow2\log_47-3>2\log_54-2\Leftrightarrow \log_27-3>\log_516-2\Leftrightarrow\log_27>\log_516+1\Leftrightarrow$$ $$\Leftrightarrow\log_27>\log_516+\log_55\Leftrightarrow\log_27>\log_580.$$
I claim that $7^4>2^{11}$ and that $5^{11}>80^4$. Then $$4\log_27=\log_27^4>\log_22^{11}=\log_55^{11}>\log_580^4=4\log_580,$$ proving the desired result.
Since you explicitly ask not to use a computer let me show the two claimed inequalities by hand: $$7^4-2^{11}=(2^3-1)^4-2^{11}=2^{12}-4\cdot 2^9+6\cdot 2^6-4\cdot 2^3+1-2^{11}=$$ $$=2^{12}-2^{11}+3\cdot 2^7-2^5+1-2^{11}=3\cdot 2^7-2^5+1=3\cdot 128-32+1=353>0.$$
Finally, since $80^4=5^4\cdot 2^{16}$ we only need to show that $5^7>2^{16}$. But $$5^7-2^{16}=(5^3)^2\cdot 5-(2^7)^2\cdot 4=125^2\cdot 5-128^2\cdot 4=125^2\cdot 5-(125+3)^2\cdot 4=$$ $$=125^2\cdot 5-125^2\cdot 4-24\cdot 125-36=125^2-24\cdot125-36=125\cdot(125-24)-36=$$ $$=125\cdot 101-36=12625-36=12589>0.$$
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# A bacteria population is 3000 at time t = 0 and its rate of growth is 1000-6^t bacteria per hour...
## Question:
A bacteria population is 3000 at time t = 0 and its rate of growth is {eq}1000-6^t {/eq} bacteria per hour after t hours. What is the population after one hour? (Round your answer to the nearest whole number.)
## Exponential Growth Function
This problem presents us with the rate of exponential growth. We need to find a function that tells us the growth. Once we have the growth function we will use it to find the population after the given time period.
• I wonder if the rate of growth function is properly written but I shall consider the one mentioned in the question.
Growth rate of the population is given by:
{eq}\displaystyle \frac{dp}{dt}=1000-6^t {/eq}
It can be written as,
{eq}\displaystyle dp=(1000-6^t)dt {/eq}
Integrating both the sides, we get,
{eq}\displaystyle \int_{p(0)}^{p(t)}dp=\int_{0}^{t}(1000-6^t)dt {/eq}
{eq}\displaystyle p(t)-p(0)=1000t-\frac{6^t}{\ln 6}+\frac{1}{\ln 6} {/eq}
{eq}\displaystyle p(t)=p(0)+1000t-\frac{6^t}{\ln 6}+\frac{1}{\ln 6} {/eq}
Plugging in the value of initial population, we get,
{eq}\displaystyle p(t)=3000+1000t-\frac{6^t}{\ln 6}+\frac{1}{\ln 6} {/eq}
Therefore, the population after {eq}1 {/eq} hour is:
{eq}\displaystyle p(1)=3000+1000(1)-\frac{6^{1}}{\ln 6}+\frac{1}{\ln 6} {/eq}
{eq}\displaystyle p(1)=3000+1000-\frac{6}{\ln 6}+\frac{1}{\ln 6} {/eq}
{eq}\displaystyle p(1)=4000-\frac{6}{1.79}+\frac{1}{1.79} {/eq}
{eq}\displaystyle \boxed{\displaystyle p(1)=3997.208\approx 3997} {/eq}
So, the population of bacteria after 1 hour is 3997.<|endoftext|>
| 4.46875 |
23,341 |
# The birthday problem
Consider this random experiment. You ask people (one at a time) of their birthdays (month and day only). The process continues until there is a repeat in the series of birthdays, in other words, until two people you’ve surveyed share the same birthday. How many people you have to ask before finding a repeat? What is the probability that you will have to ask $k$ people before finding a repeat? What is the median number of people you have to ask? In this post, we discuss this random variable and how this random variable relates to the birthday problem.
In the problem at hand, we ignore leap year and assume that each of the 365 days in the year is equally likely to be the birthday for a randomly chosen person. The birthday problem is typically the following question. How many people do we need to choose in order to have a 50% or better chance of having a common birthday among the selected individuals?
The random experiment stated at the beginning can be restated as follows. Suppose that balls are randomly thrown (one at a time) into $n$ cells (e.g. boxes). The random process continues until a ball is thrown into a cell that already has one ball (i.e. until a repeat occurs). Let $X_n$ be the number of balls that are required to obtain a repeat. Some of the problems we discuss include the mean (the average number of balls that are to be thrown to get a repeat) and the probability function. We will also show how this random variable is linked to the birthday problem when $n=365$.
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The Birthday Problem
First, we start with the birthday problem. The key is to derive the probability that in a group of $k$ randomly selected people, there are at least two who share the same birthday. It is easier to do the complement – the probability of having different birthdays among the group of $k$ people. We call this probability $p_k$.
\displaystyle \begin{aligned} p_k&=\frac{365}{365} \ \frac{364}{365} \ \frac{363}{365} \cdots \frac{365-(k-1)}{365} \\&=\frac{364}{365} \ \frac{363}{365} \cdots \frac{365-(k-1)}{365} \\&=\biggl[1-\frac{1}{365} \biggr] \ \biggl[1-\frac{2}{365} \biggr] \cdots \biggl[1-\frac{k-1}{365} \biggr] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \end{aligned}
where $k=2,3,4,\cdots,365$. The reasoning for the first line is that there are 365 choices to be picked in the first selection. Each subsequent random selection has to avoid the previous birthday, thus 364 choices for the second person and only $365-(k-1)$ choices for the $k$th person.
To answer the birthday problem, just plug in values of $k$ to compute $p_k$ and $1-p_k$ until reaching the smallest $k$ such that $p_k<0.5$ and $1-p_k>0.5$. The calculation should be done using software (Excel for example). The smallest $k$ is 23 since
$p_{23}= 0.492702766$
$1-p_{23}= 0.507297234$
In a random group of 23 people, there is a less than 50% chance of having distinct birthdays, and thus a more than 50% chance of having at least one identical birthday. This may be a surprising result. Without the benefit of formula (1), some people may think that it will take a larger sample to obtain a repeat.
The benefit of (1) extends beyond the birthday problem. Let’s consider the case for $n$, i.e. randomly pick numbers from the set $\left\{1,2,3,\cdots,n-1,n \right\}$ with replacement until a number is chosen twice (until a repeat occurs). Similarly, let $p_{n,k}$ be the probability that in $k$ draws all chosen numbers are distinct. The probability $p_{n,k}$ is obtained by replacing 365 with $n$.
\displaystyle \begin{aligned} p_{n,k}&=\frac{n}{n} \ \frac{n-1}{n} \ \frac{n-2}{n} \cdots \frac{n-(k-1)}{n} \\&=\frac{n-1}{n} \ \frac{n-2}{n} \cdots \frac{n-(k-1)}{n} \\&=\biggl[1-\frac{1}{n} \biggr] \ \biggl[1-\frac{2}{n} \biggr] \cdots \biggl[1-\frac{k-1}{n} \biggr] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \end{aligned}
Formula (2) will be useful in the next section.
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The Random Variable
We now look into the random variable discussed at the beginning, either the one for picking people at random until a repeated birthday or throwing balls into cells until one cell has two balls. To illustrate the idea, let’s look at an example.
Example 1
Roll a fair die until obtaining a repeated face value. Let $X_6$ be the number of rolls to obtain the repeated value. Find the probability function $P[X_6=k]$ where $k=2,3,4,5,6,7$.
Note that $P[X_6=k]$ is the probability that it will take $k$ rolls to get a repeated die value.
$\displaystyle P(X_6=2)=\frac{6}{6} \times \frac{1}{6}=\frac{1}{6}$
$\displaystyle P(X_6=3)=\frac{6}{6} \times \frac{5}{6} \times \frac{2}{6}=\frac{10}{6^2}$
$\displaystyle P(X_6=4)=\frac{6}{6} \times \frac{5}{6} \times \frac{4}{6} \times \frac{3}{6}=\frac{60}{6^3}$
$\displaystyle P(X_6=5)=\frac{6}{6} \times \frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{4}{6}=\frac{240}{6^4}$
$\displaystyle P(X_6=6)=\frac{6}{6} \times \frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{2}{6} \times \frac{5}{6}=\frac{600}{6^5}$
$\displaystyle P(X_6=7)=\frac{6}{6} \times \frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{2}{6} \times \frac{1}{6} \times \frac{6}{6}=\frac{720}{6^6}$
To get a repeat in 2 rolls, there are 6 choices for the first roll and the second has only one choice – the value of the first roll. To get a repeat in 3 rolls, there are 6 choices for the first roll, 5 choices for the second roll and the third roll must be out of the 2 previous two distinct values. The idea is that the first $k-1$ rolls are distinct and the last roll must be one of the previous values. $\square$
The reasoning process leads nicely to the general case. In the general case, let’s consider the occupancy interpretation. In throwing balls into $n$ cells, let $X_n$ be defined as above, i.e. the number of balls that are required to obtain a repeat. The following gives the probability $P[X_n=k]$.
\displaystyle \begin{aligned} P[X_n=k]&=\frac{n}{n} \times \frac{n-1}{n} \times \cdots \times \frac{n-(k-2)}{n} \times \frac{k-1}{n} \\&=\frac{n-1}{n} \times \cdots \times \frac{n-(k-2)}{n} \times \frac{k-1}{n} \\&=\frac{(n-1) \times (n-2) \times \cdots \times (n-(k-2)) \times (k-1)}{n^{k-1}} \\&=\biggl[1-\frac{1}{n} \biggr] \times \biggl[1-\frac{2}{n} \biggr] \times \cdots \times \biggl[1-\frac{k-2}{n}\biggr] \times \frac{k-1}{n} \ \ \ \ \ \ \ (3) \end{aligned}
where $k=2,3,\cdots,n+1$.
The reasoning is similar to Example 1. To get a repeat in throwing $k$ balls, the first $k-1$ balls must be go into different cells while the last ball would go into one of the $k-1$ occupied cells. For the first $k-1$ balls to go into different cells, there are $n (n-1) \cdots (n-(k-2))$ ways. There are $k-1$ cells for the last ball to land. Thus the product of these two quantities is in the numerator of (3).
Once the probability function (3) is obtained, the mean $E[X_n]$ can be derived accordingly. For the case of $n=365$, $E[X_{365}]=24.62$ (calculated by programming the probability function in Excel). On average it will be required to sample about 25 people to obtain a repeated birthday.
Another interesting quantity is $P[X_n>k]$. This is the probability that it will take throwing more than $k$ balls to get a repeat. Mathematically this can be obtained by first calculating $P[X_n \le k]$ by summing the individual probabilities via (3). This is a workable approach using software. There is another way that is more informative. For the event $X_n>k$ to happen, the first $k$ throws must be in different cells (no repeat). The event $X_n>k$ is identical to the event that there is no repeat in the first $k$ throws of balls. This is how the random variable $X_n$ is linked to the birthday problem since the probability $P[X_n>k]$ should agree with the probability $p_{n,k}$ in (2).
$\displaystyle P[X_n>k]=\biggl[1-\frac{1}{n} \biggr] \ \biggl[1-\frac{2}{n} \biggr] \cdots \biggl[1-\frac{k-1}{n} \biggr] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$
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Back to The Birthday Problem
Consider the case for $X_{365}$. What is the median of $X_{365}$? That would be the median number of people surveyed to obtain a pair of identical birthday. The median of $X_{365}$ would be the least $k$ such that $P[X_{365} \le k]$ is at least 0.5. Note that $P[X_{365}>k]$ is identical to $p_k$ in (1). The above calculation shows that $p_{23}=0.4927$ and $1-p_{23}=0.5073$. Thus the median of $X_{365}$ is 23. Thus when performing the random sampling of surveying birthday, about half of the times you can expect to survey 23 or fewer than 23 people.
The birthday problem is equivalently about finding the median of the random variable $X_{365}$. A little more broadly, the birthday problem is connected to the percentiles of the variable $X_{n}$. In contrast, the mean of $X_{365}$ is $E[X_{365}]=24.62$. The following lists several percentiles for the random variable $X_{365}$.
$\begin{array}{ccccccc} k & \text{ } & P[X_{365}>k] & \text{ } & P[X_{365} \le k] & \text{ } & \text{Percentile} \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ 14 & \text{ } & 0.77690 & & 0.22310 & \\ 15 & \text{ } & 0.74710 & & 0.25290 & \text{ } & \text{25th Percentile} \\ 16 & \text{ } & 0.71640 & & 0.28360 & \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ 22 & \text{ } & 0.52430 & & 0.47570 & \\ 23 & \text{ } & 0.49270 & & 0.50730 & \text{ } & \text{50th Percentile} \\ 24 & \text{ } & 0.46166 & & 0.53834 & \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ 31 & \text{ } & 0.26955 & & 0.73045 & \\ 32 & \text{ } & 0.24665 & & 0.75335 & \text{ } & \text{75th Percentile} \\ 33 & \text{ } & 0.22503 & & 0.77497 & \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ 40 & \text{ } & 0.10877 & & 0.89123 & \\ 41 & \text{ } & 0.09685 & & 0.90315 & \text{ } & \text{90th Percentile} \\ 42 & \text{ } & 0.08597 & & 0.91403 & \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ 46 & \text{ } & 0.05175 & & 0.94825 & \\ 47 & \text{ } & 0.04523 & & 0.95477 & \text{ } & \text{95th Percentile} \\ 48 & \text{ } & 0.03940 & & 0.96060 & \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ 56 & \text{ } & 0.01167 & & 0.98833 & \\ 57 & \text{ } & 0.00988 & & 0.99012 & \text{ } & \text{99th Percentile} \\ 58 & \text{ } & 0.00834 & & 0.99166 & \\ \end{array}$
It is clear that in a group of 366 people, it is certain that there will be at least one repeated birthday (again ignoring leap year). This is due to the pigeon hole principle. As the percentiles in the above table shows, you do not need to survey anywhere close to 366 to get a repeat. The median is 23 as discussed. The 75th percentile of $X_{365}$ is 32.
The preceding calculation shows that you do not need a large group to have a repeated birthday. About 50% of the times, you will survey 23 or fewer people, about 75% of the time, 32 or fewer people, About 99% of the time, you will survey 57 or fewer people, much fewer than 365 or 366. So with around 50 in a random group, there is a near certainty of finding a shared birthday. In a random group of 100 people, there should be an almost absolute certainty that there is a shared birthday.
For a further demonstration, we simulated the random variable $X_{365}$ 10,000 times. The range of the simulated values is 2 to 78. Thus the odds for 100 people to survey is smaller than 1 in 10,000. To get a simulated value of 100, we will have to simulate more than 10,000 values of $X_{365}$. The median of the 10,000 simulated results is 23. The following table summarizes the results.
$\begin{array}{rrrrrr} \text{Interval} & \text{ } & \text{Count} & \text{ } & \text{Proportion} & \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \\ \text{2 to 9} & \text{ } &953 & & 0.09530 & \\ \text{10 to 19} & \text{ } & 2870 & & 0.28700 & \\ \text{20 to 29} & \text{ } & 3055 & & 0.30550 & \\ \text{30 to 39} & \text{ } & 1922 & & 0.19220 & \\ \text{40 to 49} & \text{ } & 886 & & 0.08860 & \\ \text{50 to 59} & \text{ } & 253 & & 0.02530 & \\ \text{60 to 69} & \text{ } & 48 & & 0.00480 & \\ \text{70 to 78} & \text{ } & 13 & & 0.00130 & \\ \end{array}$
Not shown in the table is that 33 of the simulated results are the value of 2. Thus it is possible to ask two people and they both have the same birthday. But the odds of that happening is 33 out of 10,000 according to this particular set of simulations (probability 0.0033). The theoretical probability of 2 is 1/365 = 0.002739726. There are 2 instances of 78 in the 10,000 simulated values. Thus the odds are 2 out of 10,000 with probability 0.0002. The theoretical probability is 0.000037 using (3).
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$\copyright \ 2017 \text{ by Dan Ma}$
# Every character is known but password is hard to crack
In this post, we discusses an example in which you are given a password (every character of it) and yet it is still very hard (or even impossible) to crack. Anyone who understands this example has a solid understanding of the binomial distribution. Here’s the example:
Your friend John tells you that the password to his online bank account has 26 characters. The first character is the first letter in the English alphabet, the second character is the second letter in the English alphabet, the third character is the third letter in the English alphabet and so on.
Now that your friend John has given you the key to his account, does that mean you can log onto his account to find out how much money he has, or to make financial transaction on his behalf or to enrich yourself?
If this example sounds too good to be true, what is the catch?
Even though every character in the 26-character password is known, it is indeed a very strong password. How could this be? You may want to stop here and think about.
Indeed, if every character in John’s password is lower case or if every character is upper case, then his bank account is toast. But John’s password can be made very strong and very hard to crack if the password is case sensitive. The password given by John is not just one password, but is a large collection of passwords. In fact, there are over 67 millions possible passwords (67,108,864 to be exact). The following are two of the most obvious ones.
a b c d e f g h i j k l m n o p q r s t u v w x y z (all lower case)
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z (all upper case)
The following is another possible password. If this is the one John uses, it will be difficult to crack.
a b C d e f G h i j k l M N o p Q R s T U v w X Y z (10 upper case letters)
Here’s another possibility.
A B c D E f G H I J k l M N o P q R S t u v w X y z (14 upper case letters)
Each character in the password has two possibilities – lower case or upper case. Across all 26 characters, there are $2^{26}$ possibilities. This number is 67,108,864. So 2 raised to 26 is a little over 67 millions. So the password given by John is not just one password, but is a generic one with over 67 million possibilities. There is a one in 67 million chance in correctly guessing the correct password if John chooses the upper case letters randomly. This is much better odds than winning the Powerball lottery, one in 292,201,338, which one in 292 million. But it is still an undeniably strong password.
So John tells you the password, but has in fact not given up much secret. This is the case if he makes the case sensitivity truly random. Of course, once he sets his password, unless he has a great memory, he will need to write down the positions that are upper case. Otherwise, he will not be able to log back on. But that is a totally separate story.
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The binomial angle
A good way to represent the passwords is to view each one as a 26-character string of U and L (U stands for upper case and L stands for lower case). Then the above two passwords (the one with 10 upper case letters and the one with 14 upper case letters) are represented as follows:
L L U L L L U L L L L L U U L L U U L U U L L U U L (10 U’s)
U U L U U L U U U U L L U U L U L U U L L L L U L L (14 U’s)
As discussed, there are 67,108,864 many such strings. We can also think of each such string as the record of a random experiment consisting of tossing a coin 26 times. We record a U when we get a head and record an L when the coin toss results in a tail. In fact, this is the kind of records that John would need to keep when he sets the password. Such a string would tell him which characters in the password are in upper case and which characters are in lower case. On the other hand, hacking the password would be essentially trying to pinpoint one such string out of 67,108,864 many possibilities.
We know 67,108,864 is a large number. Let’s further demonstrate the challenge. In each string, the number of U’s can range from 0 to 26. How many of the strings have 0 U’s? Precisely one (all letters are L). This would be what most people would try first (all the letters are lower case). How many of the strings have exactly 1 U? Thinking about it carefully, we should arrive at the conclusion that there are 26 possibilities (the single U can be in any of the 26 positions). So if the would be hacker knows there there is only one upper case letter, then it would be easy to break the password. How many of the strings have exactly 2 U’s? If you try to write out all the possible cases, it may take some effort. There are 325 possibilities. So just having two upper case letters in the password seems to make it something that is approaching a strong password. But the problem is that the two U’s may be guessed easily. John may put the upper case letters on his initials (if his name is John Smith, he may make J and S upper case), or in other obvious letters.
How many of the strings have exactly 3 U’s? This will be really hard to write out by hand. There are 2,600 many possibilities. Why stop at having just 3 upper case letters? It is clear that the more upper case letters used in the password, the stronger it is and the harder it is to crack.
How do we know the precise number of possibilities for a given $k$, the number of U’s? The idea is that of choosing $k$ number of letters out of 26 letters.
The number of ways of choosing $k$ objects from a total of $n$ objects is denoted by the notation $\binom{n}{k}$. Sometimes the notations $C(n,k)$, $_nC_k$ and $C_{n,k}$ are used. Regardless of the notation, the calculation is
$\displaystyle \binom{n}{k}=\frac{n!}{k! (n-k)!}$
The notation $n!$ is the product of all the positive integers up to and including $n$ (this is called $n$ factorial). Thus $1!=1$, $2!=2$, $3!=6$, $4!=24$. To make the formula work correctly, we make $0!=1$.
The following gives the first several calculations in the 26-character password example.
$\displaystyle \binom{26}{2}=\frac{26!}{2! \ 24!}=\frac{26 \cdot 25}{2}=325$
$\displaystyle \binom{26}{3}=\frac{26!}{3! \ 23!}=\frac{26 \cdot 25 \cdot 24}{6}=2600$
$\displaystyle \binom{26}{4}=\frac{26!}{4! \ 22!}=\frac{26 \cdot 25 \cdot 24 \cdot 23}{24}=14950$
If the desire is to see the patterns, the remaining calculations can be done by using software (or at least a hand held calculator). The following table shows the results.
$\displaystyle \begin{array}{rrr} k &\text{ } & \displaystyle \binom{26}{k} \\ \text{ } & \text{ } & \text{ } \\ 0 &\text{ } & 1 \\ 1 &\text{ } & 26 \\ 2 &\text{ } & 325 \\ 3 &\text{ } & 2,600 \\ 4 &\text{ } & 14,950 \\ 5 &\text{ } & 65,780 \\ 6 &\text{ } & 230,230 \\ 7 &\text{ } & 657,800 \\ 8 &\text{ } & 1,562,275 \\ 9 &\text{ } & 3,124,550 \\ 10 &\text{ } & 5,311,735 \\ 11 &\text{ } & 7,726,160 \\ 12 &\text{ } & 9,657,700 \\ 13 &\text{ } & 10,400,600 \\ 14 &\text{ } & 9,657,700 \\ 15 &\text{ } & 7,726,160 \\ 16 &\text{ } & 5,311,735 \\ 17 &\text{ } & 3,124,550 \\ 18 &\text{ } & 1,562,275 \\ 19 &\text{ } & 657,800 \\ 20 &\text{ } & 230,230 \\ 21 &\text{ } & 65,780 \\ 22 &\text{ } & 14,950 \\ 23 &\text{ } & 2,600 \\ 24 &\text{ } & 325 \\ 25 &\text{ } & 26 \\ 26 &\text{ } & 1 \\ \end{array}$
The pattern is symmetrical. Having too few U’s or too many U’s produces weak passwords that may be easy to guess. Having 6 or 7 U’s seems to give strong passwords. Having half of the letters upper case (13 U’s) is the optimal, with the most possibilities (over 10 millions). Even if you are given partial information such as “half of the letters are in upper case”, you are still left with over 10 million possibilities to work with!
Dividing each of the above counts by 67,108,864 gives the relative weight (probability) of each case of having exactly $k$ U’s.
$\displaystyle \begin{array}{rrr} k &\text{ } & P[X=k] \\ \text{ } & \text{ } & \text{ } \\ 0 &\text{ } & 0.00000001 \\ 1 &\text{ } & 0.00000039 \\ 2 &\text{ } & 0.00000484 \\ 3 &\text{ } & 0.00003874 \\ 4 &\text{ } & 0.00022277 \\ 5 &\text{ } & 0.00098020 \\ 6 &\text{ } & 0.00343069 \\ 7 &\text{ } & 0.00980198 \\ 8 &\text{ } & 0.02327971 \\ 9 &\text{ } & 0.04655942 \\ 10 &\text{ } & 0.07915102 \\ 11 &\text{ } & 0.11512876 \\ 12 &\text{ } & 0.14391094 \\ 13 &\text{ } & 0.15498102 \\ 14 &\text{ } & 0.14391094 \\ 15 &\text{ } & 0.11512876 \\ 16 &\text{ } & 0.07915102 \\ 17 &\text{ } & 0.04655942 \\ 18 &\text{ } & 0.02327971 \\ 19 &\text{ } & 0.00980198 \\ 20 &\text{ } & 0.00343069 \\ 21 &\text{ } & 0.00098020 \\ 22 &\text{ } & 0.00022277 \\ 23 &\text{ } & 0.00003874 \\ 24 &\text{ } & 0.00000484 \\ 25 &\text{ } & 0.00000039 \\ 26 &\text{ } & 0.00000001 \\ \end{array}$
The cases of $X=k$ where $k=11,12,13,14,15$ add up to 67.3% of the 67,108,864 possibilities.
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The binomial distribution
Another way to look at it is that in setting the password, John is performing a sequence of 26 independent Bernoulli trials. Here, each trial has two outcomes, a or A, b or B, c or C and so on. For example, the lower case or upper case can be determined by a coin toss. Let $X$ be the number of upper case letters in the 26-character password. Then the random variable $X$ has a binomial distribution with $n=26$ (26 Bernoulli trials) and the probability of success $p=0.5$ in each trial, which is the probability that a character is upper case, assuming that he determines the upper/lower case by a coin toss. The following is the probability function:
$\displaystyle P(X=x)=\binom{26}{x} \biggl[\frac{1}{2}\biggr]^x \biggl[\frac{1}{2}\biggr]^{26-x}=\binom{26}{x} \biggl[\frac{1}{2}\biggr]^{26}$
where $x=0,1,2,\cdots,25,26$. The quantity $\displaystyle P(X=x)$ is the probability that the number of upper case letters is $x$. Here, $\binom{26}{x}$ is the number of ways to choose $x$ letters out of 26 letters and is computed by the formula indicated earlier.
Since the upper/lower case is determined randomly, another way to state the probability function of the random variable $X$ is:
$\displaystyle P(X=x)=\displaystyle \frac{\binom{26}{x}}{2^{26}}=\frac{\binom{26}{x}}{67108864} \ \ \ \ \ \ \ \ \ x=0,1,2,3,\cdots,24,25,26$
The expected value of this random variable $X$ is 13. This is the average number of upper case letters if the case is determined randomly. This obviously produces the most optimally strong password. If John determines the case not at random, the security may not be as strong or the would be hacker may be able to guess.
Stepping away from the 26-character password example, here’s the probability function of a binomial distribution in general.
$\displaystyle P(X=x)=\binom{n}{x} \ p^x \ (1-p)^{n-x} \ \ \ \ \ \ \ \ x=0,1,2,3,\cdots,n-1,n$
This model describes the random experiment of running $n$ independent trials, where each trial has two outcomes (the technical term is Bernoulli trial). In each trial, the probability of one outcome (called success) is $p$ and the probability of the other outcome (called failure) is $1-p$. The random variable $X$ counts the number of successes whenever such an experiment is performed. The probability $P(X=x)$ gives the likelihood of achieving $x$ successes.
As an example, if John has a bias toward lower case letter, then the probability of success (upper case) may be $p=0.4$ (assuming that the random variable $X$ still counts the number of upper case letters). Then the average number of upper case letters in a randomly determined password is 26 x 0.4 = 10.4.
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Interesting binomial distribution problems
The problem of points and the dice problem are two famous probability problems that are in the history book as a result of a gambler seeking help from Pascal and Fermat.
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$\copyright \ 2016 \text{ by Dan Ma}$
# Calculating order statistics using multinomial probabilities
Consider a random sample $X_1,X_2,\cdots,X_n$ drawn from a continuous distribution. Rank the sample items in increasing order, resulting in a ranked sample $Y_1 where $Y_1$ is the smallest sample item, $Y_2$ is the second smallest sample item and so on. The items in the ranked sample are called the order statistics. Recently the author of this blog was calculating a conditional probability such as $P(Y_2>4 \ | \ Y_2>3)$. One way to do this is to calculate the distribution function $P(Y_2 \le t)$. What about the probability $P(Y_5>4 \ | \ Y_2>3)$? Since this one involves two order statistics, the author of this blog initially thought that calculating $P(Y_5>4 \ | \ Y_2>3)$ would require knowing the joint probability distribution of the order statistics $Y_1,Y_2,\cdots ,Y_n$. It turns out that a joint distribution may not be needed. Instead, we can calculate a conditional probability such as $P(Y_5>4 \ | \ Y_2>3)$ using multinomial probabilities. In this post, we demonstrate how this is done using examples. Practice problems are found in here.
The calculation described here can be lengthy and tedious if the sample size is large. To make the calculation more manageable, the examples here have relatively small sample size. To keep the multinomial probabilities easier to calculate, the random samples are drawn from a uniform distribution. The calculation for larger sample sizes from other distributions is better done using a technology solution. In any case, the calculation described here is a great way to practice working with order statistics and multinomial probabilities.
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The multinomial angle
In this post, the order statistics $Y_1 are resulted from ranking the random sample $X_1,X_2,\cdots,X_n$, which is drawn from a continuous distribution with distribution function $F(x)=P(X \le x)$. For the $j$th order statistic, the calculation often begins with its distribution function $P(Y_j \le t)$.
Here’s the thought process for calculating $P(Y_j \le t)$. In drawing the random sample $X_1,X_2,\cdots,X_n$, make a note of the items $\le t$ and the items $>t$. For the event $Y_j \le t$ to happen, there must be at least $j$ many sample items $X_i$ that are $\le t$. For the event $Y_j > t$ to happen, there can be only at most $j-1$ many sample items $X_i$ $\le t$. So to calculate $P(Y_j \le t)$, simply find out the probability of having $j$ or more sample items $\le t$. To calculate $P(Y_j > t)$, find the probability of having at most $j-1$ sample items $\le t$.
$\displaystyle P(Y_j \le t)=\sum \limits_{k=j}^n \binom{n}{k} \ \biggl[ F(t) \biggr]^k \ \biggl[1-F(x) \biggr]^{n-k} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
$\displaystyle P(Y_j > t)=\sum \limits_{k=0}^{j-1} \binom{n}{k} \ \biggl[ F(t) \biggr]^k \ \biggl[1-F(x) \biggr]^{n-k} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$
Both (1) and (2) involve binomial probabilities and are discussed in this previous post. The probability of success is $F(t)=P(X \le t)$ since we are interested in how many sample items that are $\le t$. Both the calculations (1) and (2) are based on counting the number of sample items in the two intervals $\le t$ and $>t$. It turns out that when the probability that is desired involves more than one $Y_j$, we can also count the number of sample items that fall into some appropriate intervals and apply some appropriate multinomial probabilities. Let’s use an example to illustrate the idea.
Example 1
Draw a random sample $X_1,X_2,\cdots,X_{10}$ from the uniform distribution $U(0,4)$. The resulting order statistics are $Y_1. Find the following probabilities:
• $P(Y_4<2
• $P(Y_4<2
For both probabilities, the range of the distribution is broken up into 3 intervals, (0, 2), (2, 3) and (3, 4). Each sample item has probabilities $\frac{2}{4}$, $\frac{1}{4}$, $\frac{1}{4}$ of falling into these intervals, respectively. Multinomial probabilities are calculated on these 3 intervals. It is a matter of counting the numbers of sample items falling into each interval.
The first probability involves the event that there are 4 sample items in the interval (0, 2), 2 sample items in the interval (2, 3) and 4 sample items in the interval (3, 4). Thus the first probability is the following multinomial probability:
\displaystyle \begin{aligned} P(Y_4<2
For the second probability, $Y_5$ does not have to be greater than 2. Thus there could be 5 sample items less than 2. So we need to add one more case to the above probability (5 sample items to the first interval, 1 sample item to the second interval and 4 sample items to the third interval).
\displaystyle \begin{aligned} P(Y_4<2
Example 2
Draw a random sample $X_1,X_2,\cdots,X_6$ from the uniform distribution $U(0,4)$. The resulting order statistics are $Y_1. Find the probability $P(1.
In this problem the range of the distribution is broken up into 3 intervals (0, 1), (1, 3) and (3, 4). Each sample item has probabilities $\frac{1}{4}$, $\frac{2}{4}$, $\frac{1}{4}$ of falling into these intervals, respectively. Multinomial probabilities are calculated on these 3 intervals. It is a matter of counting the numbers of sample items falling into each interval. The counting is a little bit more involved here than in the previous example.
The example is to find the probability that both the second order statistic $Y_2$ and the fourth order statistic $Y_4$ fall into the interval $(1,3)$. To solve this, determine how many sample items that fall into the interval $(0,1)$, $(1,3)$ and $(3,4)$. The following points detail the counting.
• For the event $1 to happen, there can be at most 1 sample item in the interval $(0,1)$.
• For the event $Y_4<3$ to happen, there must be at least 4 sample items in the interval $(0,3)$. Thus if the interval $(0,1)$ has 1 sample item, the interval $(1,3)$ has at least 3 sample items. If the interval $(0,1)$ has no sample item, the interval $(1,3)$ has at least 4 sample items.
The following lists out all the cases that satisfy the above two bullet points. The notation $[a, b, c]$ means that $a$ sample items fall into $(0,1)$, $b$ sample items fall into the interval $(1,3)$ and $c$ sample items fall into the interval $(3,4)$. So $a+b+c=6$. Since the sample items are drawn from $U(0,4)$, the probabilities of a sample item falling into intervals $(0,1)$, $(1,3)$ and $(3,4)$ are $\frac{1}{4}$, $\frac{2}{4}$ and $\frac{1}{4}$, respectively.
[0, 4, 2]
[0, 5, 1]
[0, 6, 0]
[1, 3, 2]
[1, 4, 1]
[1, 5, 0]
\displaystyle \begin{aligned} \frac{6!}{a! \ b! \ c!} \ \biggl[\frac{1}{4} \biggr]^a \ \biggl[\frac{2}{4} \biggr]^b \ \biggl[\frac{1}{4} \biggr]^c&=\frac{6!}{0! \ 4! \ 2!} \ \biggl[\frac{1}{4} \biggr]^0 \ \biggl[\frac{2}{4} \biggr]^4 \ \biggl[\frac{1}{4} \biggr]^2=\frac{240}{4096} \\&\text{ } \\&=\frac{6!}{0! \ 5! \ 1!} \ \biggl[\frac{1}{4} \biggr]^0 \ \biggl[\frac{2}{4} \biggr]^5 \ \biggl[\frac{1}{4} \biggr]^1=\frac{192}{4096} \\&\text{ } \\&=\frac{6!}{0! \ 6! \ 0!} \ \biggl[\frac{1}{4} \biggr]^0 \ \biggl[\frac{2}{4} \biggr]^6 \ \biggl[\frac{1}{4} \biggr]^0=\frac{64}{4096} \\&\text{ } \\&=\frac{6!}{1! \ 3! \ 2!} \ \biggl[\frac{1}{4} \biggr]^1 \ \biggl[\frac{2}{4} \biggr]^3 \ \biggl[\frac{1}{4} \biggr]^2=\frac{480}{4096} \\&\text{ } \\&=\frac{6!}{1! \ 4! \ 1!} \ \biggl[\frac{1}{4} \biggr]^1 \ \biggl[\frac{2}{4} \biggr]^4 \ \biggl[\frac{1}{4} \biggr]^1=\frac{480}{4096} \\&\text{ } \\&=\frac{6!}{1! \ 5! \ 0!} \ \biggl[\frac{1}{4} \biggr]^1 \ \biggl[\frac{2}{4} \biggr]^5 \ \biggl[\frac{1}{4} \biggr]^0=\frac{192}{4096} \\&\text{ } \\&=\text{sum of probabilities }=\frac{1648}{4096}=0.4023\end{aligned}
So in randomly drawing 6 items from the uniform distribution $U(0,4)$, there is a 40% chance that the second order statistic and the fourth order statistic are between 1 and 3.
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More examples
The method described by the above examples is this. When looking at the event described by the probability problem, the entire range of distribution is broken up into several intervals. Imagine the sample items $X_i$ are randomly being thrown into these interval (i.e. we are sampling from a uniform distribution). Then multinomial probabilities are calculated to account for all the different ways sample items can land into these intervals. The following examples further illustrate this idea.
Example 3
Draw a random sample $X_1,X_2,\cdots,X_7$ from the uniform distribution $U(0,5)$. The resulting order statistics are $Y_1. Find the following probabilities:
• $P(1
• $P(3
The range is broken up into the intervals (0, 1), (1, 3), (3, 4) and (4, 5). The sample items fall into these intervals with probabilities $\frac{1}{5}$, $\frac{2}{5}$, $\frac{1}{5}$ and $\frac{1}{5}$. The following details the counting for the event $1:
• There are no sample items in (0, 1) since $1.
• Based on $Y_1<3, there are at least one sample item and at most 3 sample items in (0, 3). Thus in the interval (1, 3), there are at least one sample item and at most 3 sample items since there are none in (0, 1).
• Based on $Y_4<4$, there are at least 4 sample items in the interval (0, 4). Thus the count in (3, 4) combines with the count in (1, 3) must be at least 4.
• The interval (4, 5) simply receives the left over sample items not in the previous intervals.
The notation $[a, b, c, d]$ lists out the counts in the 4 intervals. The following lists out all the cases described by the above 5 bullet points along with the corresponding multinomial probabilities, with two of the probabilities set up.
$\displaystyle [0, 1, 3, 3] \ \ \ \ \ \ \frac{280}{78125}=\frac{7!}{0! \ 1! \ 3! \ 3!} \ \biggl[\frac{1}{5} \biggr]^0 \ \biggl[\frac{2}{5} \biggr]^1 \ \biggl[\frac{1}{5} \biggr]^3 \ \biggl[\frac{1}{5} \biggr]^3$
$\displaystyle [0, 1, 4, 2] \ \ \ \ \ \ \frac{210}{78125}$
$\displaystyle [0, 1, 5, 1] \ \ \ \ \ \ \frac{84}{78125}$
$\displaystyle [0, 1, 6, 0] \ \ \ \ \ \ \frac{14}{78125}$
$\displaystyle [0, 2, 2, 3] \ \ \ \ \ \ \frac{840}{78125}$
$\displaystyle [0, 2, 3, 2] \ \ \ \ \ \ \frac{840}{78125}$
$\displaystyle [0, 2, 4, 1] \ \ \ \ \ \ \frac{420}{78125}$
$\displaystyle [0, 2, 5, 0] \ \ \ \ \ \ \frac{84}{78125}$
$\displaystyle [0, 3, 1, 3] \ \ \ \ \ \ \frac{1120}{78125}=\frac{7!}{0! \ 3! \ 1! \ 3!} \ \biggl[\frac{1}{5} \biggr]^0 \ \biggl[\frac{2}{5} \biggr]^3 \ \biggl[\frac{1}{5} \biggr]^1 \ \biggl[\frac{1}{5} \biggr]^3$
$\displaystyle [0, 3, 2, 2] \ \ \ \ \ \ \frac{1680}{78125}$
$\displaystyle [0, 3, 3, 1] \ \ \ \ \ \ \frac{1120}{78125}$
$\displaystyle [0, 3, 4, 0] \ \ \ \ \ \ \frac{280}{78125}$
Summing all the probabilities, $\displaystyle P(1. Out of the 78125 many different ways the 7 sample items can land into these 4 intervals, 6972 of them would satisfy the event $1.
++++++++++++++++++++++++++++++++++
We now calculate the second probability in Example 3.
$\displaystyle P(3
First calculate $P(1. The probability $P(Y_1 is the probability of having at least 1 sample item less than $t$, which is the complement of the probability of all sample items greater than $t$.
\displaystyle \begin{aligned} P(1
The event $1 can occur in 16256 ways. Out of these many ways, 6972 of these satisfy the event $1. Thus we have:
$\displaystyle P(3
Example 4
Draw a random sample $X_1,X_2,X_3,X_4,X_5$ from the uniform distribution $U(0,5)$. The resulting order statistics are $Y_1. Consider the conditional random variable $Y_4 \ | \ Y_2 >3$. For this conditional distribution, find the following:
• $P( Y_4 \le t \ | \ Y_2 >3)$
• $f_{Y_4}(t \ | \ Y_2 >3)$
• $E(Y_4 \ | \ Y_2 >3)$
where $3. Note that $f_{Y_4}(t | \ Y_2 >3)$ is the density function of $Y_4 \ | \ Y_2 >3$.
Note that
$\displaystyle P( Y_4 \le t \ | \ Y_2 >3)=\frac{P(33)}$
In finding $P(3, the range (0, 5) is broken up into 3 intervals (0, 3), (3, t) and (t, 5). The sample items fall into these intervals with probabilities $\frac{3}{5}$, $\frac{t-3}{5}$ and $\frac{5-t}{5}$.
Since $Y_2 >3$, there is at most 1 sample item in the interval (0, 3). Since $Y_4 \le t$, there are at least 4 sample items in the interval (0, t). So the count in the interval (3, t) and the count in (0, 3) should add up to 4 or more items. The following shows all the cases for the event $3 along with the corresponding multinomial probabilities.
$\displaystyle [0, 4, 1] \ \ \ \ \ \ \frac{5!}{0! \ 4! \ 1!} \ \biggl[\frac{3}{5} \biggr]^0 \ \biggl[\frac{t-3}{5} \biggr]^4 \ \biggl[\frac{5-t}{5} \biggr]^1$
$\displaystyle [0, 5, 0] \ \ \ \ \ \ \frac{5!}{0! \ 5! \ 0!} \ \biggl[\frac{3}{5} \biggr]^0 \ \biggl[\frac{t-3}{5} \biggr]^5 \ \biggl[\frac{5-t}{5} \biggr]^0$
$\displaystyle [1, 3, 1] \ \ \ \ \ \ \frac{5!}{1! \ 3! \ 1!} \ \biggl[\frac{3}{5} \biggr]^1 \ \biggl[\frac{t-3}{5} \biggr]^3 \ \biggl[\frac{5-t}{5} \biggr]^1$
$\displaystyle [1, 4, 0] \ \ \ \ \ \ \frac{5!}{1! \ 4! \ 0!} \ \biggl[\frac{3}{5} \biggr]^1 \ \biggl[\frac{t-3}{5} \biggr]^4 \ \biggl[\frac{5-t}{5} \biggr]^0$
After carrying the algebra and simplifying, we have the following:
$\displaystyle P(3
For the event $Y_2 >3$ to happen, there is at most 1 sample item less than 3. So we have:
$\displaystyle P(Y_2 >3)=\binom{5}{0} \ \biggl[\frac{3}{5} \biggr]^0 \ \biggl[\frac{2}{5} \biggr]^5 +\binom{5}{1} \ \biggl[\frac{3}{5} \biggr]^1 \ \biggl[\frac{2}{5} \biggr]^4=\frac{272}{3125}$
$\displaystyle P( Y_4 \le t \ | \ Y_2 >3)=\frac{-4t^5+25t^4+180t^3-1890t^2+5400t-5103}{272}$
Then the conditional density is obtained by differentiating $P( Y_4 \le t \ | \ Y_2 >3)$.
$\displaystyle f_{Y_4}(t \ | \ Y_2 >3)=\frac{-20t^4+100t^3+540t^2-3750t+5400}{272}$
The following gives the conditional mean $E(Y_4 \ | \ Y_2 >3)$.
\displaystyle \begin{aligned} E(Y_4 \ | \ Y_2 >3)&=\frac{1}{272} \ \int_3^5 t(-20t^4+100t^3+540t^2-3750t+5400) \ dt \\&=\frac{215}{51}=4.216 \end{aligned}
To contrast, the following gives the information on the unconditional distribution of $Y_4$.
$\displaystyle f_{Y_4}(t)=\frac{5!}{3! \ 1! \ 1!} \ \biggl[\frac{t}{5} \biggr]^3 \ \biggl[\frac{1}{5} \biggr] \ \biggl[ \frac{5-t}{5} \biggr]^1=\frac{20}{3125} \ (5t^3-t^4)$
$\displaystyle E(Y_4)=\frac{20}{3125} \ \int_0^5 t(5t^3-t^4) \ dt=\frac{10}{3}=3.33$
The unconditional mean of $Y_4$ is about 3.33. With the additional information that $Y_2 >3$, the average of $Y_4$ is now 4.2. So a higher value of $Y_2$ pulls up the mean of $Y_4$.
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Practice problems
Practice problems to reinforce the calculation are found in the problem blog, a companion blog to this blog.
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$\copyright \ \text{2015 by Dan Ma}$
# Tis the Season for Gift Exchange
Suppose that there are 10 people in a holiday party with each person bearing a gift. The gifts are put in a pile and each person randomly selects one gift from the pile. In order not to bias the random selection of gifts, suppose that the partygoers select gifts by picking slips of papers with numbers identifying the gifts. What is the probability that there is at least one partygoer who ends up selecting his or her own gift? In this example, selecting one’s gift is called a match. What is the probability that there is at least one match if there are more people in the party, say 50 or 100 people? What is the behavior of this probability if the size of the party increases without bound?
The above example is a description of a classic problem in probability called the matching problem. There are many colorful descriptions of the problem. One such description is that there are $n$ married couples in a ballroom dancing class. The instructor pairs the ladies with the gentlemen at random. A match occurs if a married couple is paired as dancing partners.
Whatever the description, the matching problem involves two lists of items that are paired according to a particular order (the original order). When the items in the first list are paired with the items in the second list according to another random ordering, a match occurs if an item in the first list and an item in the second list are paired both in the original order and in the new order. The matching problem discussed here is: what is the probability that there is at least one match? Seen in this light, both examples described above are equivalent mathematically.
We now continue with the discussion of the random gift selection example. Suppose that there are $n$ people in the party. Let $E_i$ be the event that the $i^{th}$ person selects his or her own gift. The event $E=E_1 \cup E_2 \cup \cdots \cup E_n$ is the event that at least one person selects his or her own gift (i.e. there is at least one match). The probability $P(E)=P(E_1 \cup E_2 \cup \cdots \cup E_n)$ is the solution of the matching problem as described in the beginning. The following is the probability $P(E)=P(E_1 \cup E_2 \cup \cdots \cup E_n)$.
$\displaystyle (1) \ \ \ \ \ \ P[E_1 \cup E_2 \cup \cdots \cup E_n]=1-\frac{1}{2!}+\frac{1}{3!}-\cdots+(-1)^{n+1} \frac{1}{n!}$
The answer in $(1)$ is obtained by using an idea called the inclusion-exclusion principle. We will get to that in just a minute. First let’s look at the results of $(1)$ for a few iterations.
$\displaystyle (2) \ \ \ \ \ \ \begin{bmatrix} \text{n}&\text{ }& P[E_1 \cup E_2 \cup \cdots \cup E_n] \\\text{ }&\text{ }&\text{ } \\ 3&\text{ }&0.666667 \\ 4&\text{ }&0.625000 \\ 5&\text{ }&0.633333 \\ 6&\text{ }&0.631944 \\ 7&\text{ }&0.632143 \\ 8&\text{ }&0.632118 \\ 9&\text{ }&0.632121 \\ 10&\text{ }&0.632121 \\ 11&\text{ }&0.632121 \\ 12&\text{ }&0.632121 \end{bmatrix}$
In a party with random gift exchange, it appears that regardless of the size of the party, there is a very good chance that someone will end up picking his or her own gift! A match will happen about 63% of the time.
It turns out that the answers in $(1)$ are related to the Taylor’s series expansion of $e^{-1}$. We show that the probability in $(1)$ will always converge to $1-e^{-1}=0.632121$. Note that the Taylor’s series expansion of $e^{-1}$ is:
$\displaystyle (3) \ \ \ \ \ \ e^{-1}=\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots+(-1)^n \frac{1}{n!}+\cdots$
Consequently, the Taylor’s series expansion of $1-e^{-1}$ is:
\displaystyle \begin{aligned} (4) \ \ \ \ \ \ 1-e^{-1}&=1-\biggl(1-1+\frac{1}{2!}-\frac{1}{3!}+\cdots+(-1)^n \frac{1}{n!}+\cdots \biggr) \\&=1-\frac{1}{2!}+\frac{1}{3!}-\cdots+(-1)^{n+1} \frac{1}{n!}+\cdots \end{aligned}
Note that the sum of the first $n$ terms of the series in $(4)$ is the probability in $(1)$. As the number of partygoers $n$ increases, the probability $P[E_1 \cup E_2 \cup \cdots \cup E_n]$ will be closer and closer to $1-e^{-1}=0.6321$. We have:
$\displaystyle (5) \ \ \ \ \ \ \lim \limits_{n \rightarrow \infty} P[E_1 \cup E_2 \cup \cdots \cup E_n]=1 - e^{-1}$
$\displaystyle (6) \ \ \ \ \ \ \lim \limits_{n \rightarrow \infty} P[(E_1 \cup E_2 \cup \cdots \cup E_n)^c]=e^{-1}$
The equation $(5)$ says that it does not matter how many people are in the random gift exchange, the answer to the matching problem is always $1-e^{-1}$ in the limit. The equation $(6)$ says that the probability of having no matches approaches $e^{-1}$. There is a better than even chance ($0.63$ to be more precise) that there is at least one match. So in a random gift exchange as described at the beginning, it is much easier to see a match than not to see one.
The Inclusion-Exclusion Principle
We now want to give some indication why $(1)$ provides the answers. The inclusion-exclusion principle is a formula for finding the probability of the union of events. For the union of two events and the union of three events, we have:
$\displaystyle (7) \ \ \ \ \ \ P[E_1 \cup E_2]=P[E_1]+P[E_1]-P[E_1 \cap E_2]$
\displaystyle \begin{aligned} (8) \ \ \ \ \ \ P[E_1 \cup E_2 \cup E_3]&=P[E_1]+P[E_1]+P[E_3] \\& \ \ \ \ -P[E_1 \cap E_2]-P[E_1 \cap E_3]-P[E_2 \cap E_3] \\& \ \ \ \ +P[E_1 \cap E_2 \cap E_3] \end{aligned}
To find the probability of the union of $n$ events, we first add up the probabilities of the individual events $E_i$. Because the resulting sum overshoots, we need to subtract the probabilities of the intersections $E_i \cap E_j$. Because the subtractions remove too much, we need to add back the probabilities of the intersections of three individual events $E_i \cap E_j \cap E_k$. The process of inclusion and exclusion continues until we reach the step of adding/removing of the intersection $E_1 \cap \cdots \cap E_n$. The following is the statement of the inclusion-exclusion principle.
$\displaystyle (9) \ \ \ \ \ \ P[E_1 \cup E_2 \cdots \cup E_n]=S_1-S_2+S_3- \ \ \cdots \ \ +(-1)^{n+1}S_n$
In $(9)$, $S_1$ is the sum of all probabilities $P[E_i]$, $S_2$ is the sum of all possible probabilities of the intersection of two events $E_i$ and $E_j$ and $S_3$ is the sum of all possible probabilities of the intersection of three events and so on.
We now apply the inclusion-exclusion principle to derive equation $(1)$. The event $E_i$ is the event that the $i^{th}$ person gets his or her own gift while the other $n-1$ people are free to select gifts. The probability of this event is $\displaystyle P[E_i]=\frac{(n-1)!}{n!}$. There are $\displaystyle \binom{n}{1}=\frac{n!}{1! (n-1)!}$ many ways of fixing 1 gift. So $\displaystyle S_1=\binom{n}{1} \times \frac{(n-1)!}{n!}=1$.
Now consider the intersection of two events. The event $E_i \cap E_j$ is the event that the $i^{th}$ person and the $j^{th}$ person get their own gifts while the other $(n-2)$ people are free to select gifts. The probability of this event is $\displaystyle P[E_i \cap E_j]=\frac{(n-2)!}{n!}$. There are $\displaystyle \binom{n}{2}=\frac{n!}{2! (n-2)!}$ many ways of fixing 2 gifts. So $\displaystyle S_2=\binom{n}{2} \times \frac{(n-2)!}{n!}=\frac{1}{2!}$.
By the same reasoning, we derive that $\displaystyle S_3=\frac{1}{3!}$ and $\displaystyle S_4=\frac{1}{4!}$ and so on. Then plugging $\displaystyle S_i=\frac{1}{i!}$ into $(9)$, we obtain the answers in $(1)$.
The matching problem had been discussed previously in the following posts.
The matching problem
# More about the matching problem
We started the discussion on the matching problem in a previous post (The matching problem). We would like to add a little bit more information. The matching problem has many colorful descriptions. Here are some examples:
• In a dance class with $n$ married couples, suppose that the dance instructor assigns partners randomly. We have a match if a married couple is paired as dancing partners.
• There are $n$ letters with $n$ matching envelops. Suppose that the letters are randomly stuffed into the envelops. We have a match if a letter is placed into the correct envelop.
In this post we use the generic example of a deck of shuffled cards. Suppose a deck of $n$ cards is numbered $1$ through $n$. After the deck is thoroughly shuffled, if the card numbered $i$ is the $i^{th}$ card in the shuffled deck, we say that it is a match. It is clear that the above two examples and the shuffled deck example are equivalent mathematically.
How many matches will there be in a shuffled deck? What is the probability that there will be at least one match? What is the probability that there will be exactly $k$ matches where $k=0,1, \cdots, n$? On average, how many matches will there be? The number of matches is a random variable. We comment on this random variable resulting from the matching problem. We also derive the probability density function and its moments. We also comment the relation this random variable with the Poisson distribution.
Consider the set $S=\lbrace{1,2, \cdots, n}\rbrace$. The matching problem corresponds to the random selection of all the points in $S$ without replacement. The random selection of points of the entire set $S$ results in ordered samples $(Y_1,Y_2, \cdots, Y_n)$ where each $Y_i \in S$ and $Y_i \neq Y_j$ for $i \neq j$. These ordered samples are precisely the permutations of the set $S$ and there are $n!$ many such permutations.
A match occurs when some element $i \in S$ is selected in the $i^{th}$ pick. In the card example, the $i^{th}$ card is a match when the card that is numbered $i$ is in the $i^{th}$ position in the shuffled deck ($Y_i=i$). The matching problem is a counting problem. We demonstrate how to count the $n!$ permutations that result in no matches, at least one match and exactly $k$ matches using the inclusion-exclusion principle.
For each $i \in S$, we define the following indicator variable $I_i$:
$\displaystyle I_i=\left\{\begin{matrix}1&\thinspace Y_i=i \ \ \ \text{(} i^{th} \text{card is a match)}\\{0}&\thinspace Y_i \neq i \ \ \ \text{(} i^{th} \text{card is not a match)}\end{matrix}\right.$
Furthermore, let $X_n=I_1+I_2+ \cdots +I_n$. The random variable $X_n$ is the number of matches when a deck of $n$ cards (numbered $1$ through $n$) are shuffled. We derive the probability function of $X_n$ as well as its first and second moments.
The Inclusion-Exclusion Principle
For any events $A_1,A_2, \cdots, A_n$, the following is the cardinality of the event $A_1 \cup A_2 \cup \cdots \cup A_n$:
$\displaystyle \lvert A_1 \cup A_2 \cup \cdots \cup A_n \lvert=\sum \limits_{m=0}^n (-1)^{m+1} S_m$ where the counts $S_m$ are defined by the following:
$\displaystyle S_1=\sum \limits_{i=1}^n \lvert A_i \lvert$
$\displaystyle S_2=\sum \limits_{i
$\displaystyle S_m=\sum \limits_{i(1)< \cdots
Remark
In the inclusion-exclusion formula, we sum the cardinalities of the sets, subtract the cardinalities of intersections of pairs, add the cardinalities of the intersection of triples and so on. When $n=2$, we have the familiar formula:
$\displaystyle \lvert A_1 \cup A_2 \lvert=\lvert A_1 \lvert +\lvert A_2 \lvert-\lvert A_1 \cap A_2\lvert$
The Probability Density Function
For each $i$ in $S=\lbrace{1,2, \cdots, n}\rbrace$, let $A_i$ be the event that the $i^{th}$ card is a match. The event $A_1 \cup \cdots \cup A_n$ is the event that there is at least one match in the deck of $n$ shuffled cards. We use the inclusion-exclusion principle to derive the count for this event. The union and its complement “no matches in the cards” will become the basis for deriving the density function of $X_n$.
For the event $A_i$, the $i^{th}$ card is fixed and the other $n-1$ cards are free to permute. Thus $\vert A_i \vert=(n-1)!$. As a result, $\displaystyle S_1=\binom{n}{1} (n-1)!=n!$.
For the event $A_i \cap A_j$, the $i^{th}$ card and the $j^{th}$ card are fixed and the other $n-2$ cards are free to permute. Thus we have $\vert A_i \cap A_j \vert=(n-2)!$ and $\displaystyle S_2=\binom{n}{2} (n-2)!=\frac{n!}{2!}$
In the general case, we have $\displaystyle S_i=\binom{n}{i} (n-i)!=\frac{n!}{i!}$
Applying the inclusion-exclusion principle, we have:
$\displaystyle \lvert A_1 \cup A_2 \cup \cdots \cup A_n \lvert=\sum \limits_{i=1}^n (-1)^{i+1} S_i$
\displaystyle \begin{aligned}\lvert (A_1 \cup A_2 \cup \cdots \cup A_n)^c \lvert&=n!-\sum \limits_{i=1}^n (-1)^{i+1} S_i\\&=n!-\sum \limits_{i=1}^n (-1)^{i+1} \frac{n!}{i!}\\&=n!+\sum \limits_{i=1}^n (-1)^{i} \frac{n!}{i!}\\&=\sum \limits_{i=0}^n (-1)^{i} \frac{n!}{i!} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\end{aligned}
Out of the $n!$ permutations, $\displaystyle \lvert (A_1 \cup A_2 \cup \cdots \cup A_n)^c \lvert$ of them have no matches. Thus the probability of having no matches in a shuffled deck of $n$ cards is:
\displaystyle \begin{aligned}P(\text{no matches in n cards})&=\displaystyle \biggl(\sum \limits_{i=0}^n (-1)^{i} \frac{n!}{i!}\biggr) \frac{1}{n!}\\&=\sum \limits_{i=0}^n \frac{(-1)^{i}}{i!}\\&=P(X_n=0)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\end{aligned}
To derive $P(X_n=1)$, we consider the event that $1$ of the cards is a match and the other $n-1$ cards are not matches. Using $(1)$ to obtain the count for the event that $n-1$ cards are not matches, we have:
\displaystyle \begin{aligned}P(X_n=1)&=\displaystyle \binom{n}{1}\displaystyle \biggl(\sum \limits_{i=0}^{n-1} (-1)^{i} \frac{(n-1)!}{i!}\biggr) \frac{1}{n!}\\&=\sum \limits_{i=0}^{n-1} \frac{(-1)^{i}}{i!}\end{aligned}
For the general case, we have:
\displaystyle \begin{aligned}P(X_n=k)&=\displaystyle \binom{n}{k}\displaystyle \biggl(\sum \limits_{i=0}^{n-k} (-1)^{i} \frac{(n-k)!}{i!}\biggr) \frac{1}{n!}\\&=\frac{1}{k!}\sum \limits_{i=0}^{n-k} \frac{(-1)^{i}}{i!}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)\end{aligned}
We point out that $P(X_n=n-1)=0$. It is impossible to have exactly $n-1$ matches while the remaining card is not a match. This can also be verified by looking at the probability density function. On the other hand, $\displaystyle P(X_n=n)=\frac{1}{n!}$. There is only one permutation out of $n!$ many where all the cards are matches.
Poisson Distribution
We compare the probability density functions $P(X_{10}=k)$, $P(X_{20}=k)$ $P(X_{150}=k)$ and the Poisson desnity function with parameter $\lambda=1$. The following matrix shows the results (rounded to eight decimal places).
$\displaystyle \begin{pmatrix} \text{k}&P(X_{10}=k)&P(X_{20}=k)&P(X_{150}=k)&\displaystyle \frac{e^{-1}}{k!} \\{\text{ }}&\text{ }&\text{ } \\{0}&0.36787946&0.36787944&0.36787944&0.36787944 \\{1}&0.36787919&0.36787944&0.36787944&0.36787944 \\{2}&0.18394097&0.18393972&0.18393972&0.18393972 \\{3}&0.06130952&0.06131324&0.06131324&0.06131324 \\{4}&0.01533565&0.01532831&0.01532831&0.01532831 \\{5}&0.00305556&0.00306566&0.00306566&0.00306566 \\{6}&0.00052083&0.00051094&0.00051094&0.00051094 \\{7}&0.00006614&0.00007299&0.00007299&0.00007299 \\{8}&0.00001240&0.00000912&0.00000912&0.00000912 \\{9}&0&0.00000101&0.00000101&0.00000101 \\{10}&0.00000028&0.00000010&0.00000010&0.00000010 \\{\text{ }}&\text{ }&\text{ } \\{\text{Total}}&1.000000&0.99999997&0.99999997&0.99999997\end{pmatrix}$
There is remarkable agreement among the density functions arising from the match problems with the Poisson distribution. The density function $P(X_n=k)$ converges to the Poisson distribution with parameter $\lambda=1$ and the convergence occurs rapidly.
The probability of having exactly $k$ matches in $(3)$ above converges to $\displaystyle \frac{e^{-1}}{k!}$, which is the Poisson distribution with parameter $\lambda=1$.
The probability of no matches among the $n$ cards in $(2)$ above is the Taylor expansion of $e^{-1}$. Thus the probability of no matches in a shuffled deck of $n$ cards when $n$ is large is about $0.3678794412$ and the probability of at least one match is approximately $0.6321205588$. As the above example shows, the convergence occurs fairly rapidly.
The Moments
Since $P(X_n=k)$ converges to the Poisson distribution with parameter $\lambda=1$, we might guess that the mean and variance of $X_n$ approach $\lambda=1$. We now show that $E(X_n)=1$ and $Var(X_n)=1$ regardless of $n$.
Recall that $X_n=I_1+I_2+ \cdots +I_n$ where for each $j=1,2, \cdots, n$, $I_j$ is an indicator variable:
$\displaystyle I_j=\left\{\begin{matrix}1&\thinspace Y_j=j \ \ \ \text{(} j^{th} \text{card is a match)}\\{0}&\thinspace Y_j \neq j \ \ \ \text{(} j^{th} \text{card is not a match)}\end{matrix}\right.$
We claim that for all $j=1,2, \cdots, n$, $\displaystyle P(I_j=1)=\frac{1}{n}$ and for all $i, $\displaystyle P(I_i=1,I_j=1)=\frac{1}{n(n-1)}$. To see this, the event $(I_j=1)$ is the event $A_j$ defined in the derivation of the probability density function of $X_n$. We have $\lvert A_j \vert=(n-1)!$. Thus the probability that the $j^{th}$ card is a match is:
$\displaystyle P(I_j=1)=\frac{(n-1)!}{n!}=\frac{1}{n}$
The event $(I_i=1, I_j=1)$ is the event $A_i \cap A_j$ defined in the derivation of the probability density function of $X_n$. We have $\lvert A_i \cap A_j \vert=(n-2)!$. Thus the probability that both the $i^{th}$ card and the $j^{th}$ card are matches is:
$\displaystyle P(I_i=1,I_j=1)=\frac{(n-2)!}{n!}=\frac{1}{n(n-1)}$
It follows that for all $j=1,2, \cdots, n$, $I_j$ has the Bernoulli distribution with probability of success $\displaystyle P(I_j=1)=\frac{1}{n}$.
Knowing that $I_j$ is Bernoulli, we know its mean and variance. For $j=1,2, \cdots, n$, we have:
$\displaystyle E(I_j)=\frac{1}{n}$
$\displaystyle Var(I_j)=\frac{1}{n} \biggl(1-\frac{1}{n}\biggr)=\frac{n-1}{n^2}$
It is now easy to see that $E(X_n)=1$. To find $Var(X_n)$, we need to find the covariance $Cov(I_i,I_j)$. Note that the indicator variables $I_j$ are not independent. In evaluating $Cov(I_i,I_j)$, we need to consider four possibilities: $(I_i=0,I_j=0)$, $(I_i=0,I_j=1)$, $(I_i=1,I_j=0)$, $(I_i=1,I_j=1)$. The only relevant case is the last one. Thus we have:
\displaystyle \begin{aligned}Cov(I_i,I_j)&=E(I_i I_j)-E(I_i) E(I_j)\\&=\biggl(\sum \limits_{x,y=0,1} x \ y P(I_i=x,I_j=y)\biggr)-E(I_i) E(I_j)\\&=P(I_i=1,I_j=1)-E(I_i) E(I_j)\\&=\frac{1}{n(n-1)}-\frac{1}{n^2}\\&=\frac{1}{n^2 (n-1)} \end{aligned}
The following derives the variance:
\displaystyle \begin{aligned}Var(X_n)&=\sum \limits_{i=1}^n Var(I_i)+2 \sum \limits_{i \ne j} Cov(I_i,I_j)\\&=n \frac{n-1}{n^2}+2 \binom{n}{2} \frac{1}{n^2 (n-1)}\\&=1 \end{aligned}
Reference
1. Feller, W., An Introduction to Probability Theory and its Applications, Vol. I, 3rd ed., John Wiley & Sons, Inc., New York, 1968
# The game of poker dice and the multinomial theorem
This post presents an application of the multinomial theorem and the multinomial coefficients to the game of poker dice. See the previous post The multinomial theorem for a discussion of the multinomial theorem and multinomial coefficients.
The game of poker dice is different from the standard poker game. Instead of drawing five cards from a deck of $52$ cards, five fair dice are rolled. The resulting five scores from the dice form a poker dice hand. The possible hands are ranked as follows:
• Five of a kind: one score occurs five times (e.g. $2,2,2,2,2$).
• Four of a kind: two distinct scores appear, one score occurs four times and the other score occurs one time (e.g. $2,2,2,2,5$).
• Full house: two distinct scores appear, one score occurs three times and the other score occurs two times (e.g. $3,3,3,1,1$).
• Three of a kind: three distinct scores appear, one score occurs three times, the other two scores occur one time each (e.g. $2,2,2,3,6$).
• Two pair: three distinct scores appear, two of the scores occur two times each and the other score occurs once (e.g. $4,4,1,1,5$).
• One pair: four distinct scores appear, one score occurs two times and the other three scores occur one time each (e.g. $5,5,1,2,4$).
• High card: five distinct scores occur (e.g. $2,3,5,4,1$).
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In rolling $5$ dice, there are $6^5=7776$ ordered outcomes. For example, assuming that the five dice are rolled one at a time, $2,3,2,6,2$ indicates outcome that the first die results in a two and the second die results in a three and so on (this is a three of a kind). To find the probability of a three of a kind, we simply divide the number of ways such hands can occur by $7776$. We use the multinomial coefficients to obtain the number of outcomes for each type of poker dice hands.
Rolling $k$ dice (or rolling a die $k$ times) can also be regarded as the occupancy problem of assigning $k$ balls into $6$ cells. As will be shown below, the problem of computing the probabilities of poker dice hands is seen through the lens of the occupancy problem of randonly placing $5$ balls into $6$ cells. For example, five of a kind is equivalent to all five balls being placed in different cells. Three of a kind is equivalent to three of the balls being in one cell, and the other two balls in two other different cells. For discussion on the occupancy problems in this blog, see the following posts:
In placing $5$ balls into $6$ cells, we use $6$-tuples to indicate how many balls are in each of the $6$ cells. For example, $(0,3,1,0,0,1)$ denotes a three of a kind hand of $2,2,2,3,6$ (the score of two appearing three times, the score of three appearing one time and the score of six appearing one time). Note that the $6$ coordinates represent the six scores of a die (six cells) and the sum of the coordinates is $5$. The $6$-tuple of $(3,1,1,0,0,0)$ is also a three of a kind hand, representing the outcome that the score of one appearing three times, the score of two appearing one time and the score of three appearing one time. We use the multinomial coefficients to determine how many of the $7776$ ordered outcomes correspond to a $6$-tuple such as $(3,1,1,0,0,0)$. With respect to the occupancy problem, such $6$-tuples are called occupancy number sets.
The Multinomial Theorem
For any positive integer $n$ and any positive integer $k$, we have the following polynomial expansion:
$\displaystyle \biggl(x_1+x_2+ \cdots +x_n\biggr)^k=\sum \limits_{a_1+ \cdots +a_n=k} \frac{k!}{a_1! \ a_2! \cdots \ a_n!} \ \ x_1^{a_1} x_2^{a_2} \cdots x_n^{a_n}$
Remark
In addition to being a formula for polynomial expansion, there are two important interpretations of the multinomial theorem and multinomial coefficients. One is that of determining the number of ordered strings that can be formed using a set of alphabets. For example, with one $m$, four $i's$, four $s's$ and two $p's$, there are $\displaystyle \frac{11!}{1! \ 4! \ 4! \ 2!}=\text{34,650}$ possible $11$-letter strings that can be formed, of which $mississippi$ is one specific example.
Another interpretation is that of partitioning a set of distinct objects into several groupings where objects in each grouping are indistinguishable. For example, in a group of $11$ candidates, how many ways can we form four committees such that the Committee 1 has only one member, Committee 2 has four members, Committee 3 has four members and Committee 4 has two members (assuming that each person can only serve in one committee)? The answer, as in above example, is $\text{35,650}$.
Example 1
In a random poker dice hand, what is the probability of obtaining a $4$ two times, a $3$ one time, a $5$ one time and a $6$ one time? Note that this is a specific example of the poker dice hand of one pair.
We consider the $6$-tuple of $(0,0,1,2,1,1)$. We are trying to partition $5$ scores into four subgroups, one group having two identical scores of $4$, one group with a score of $3$, one group with a score of $5$ and one group with a score of $6$. Thus consider the following multinomial coefficient:
$\displaystyle \frac{5!}{1! \ 2! \ 1! \ 1!}=60$
So out of $\text{7,776}$ possible hands, $60$ of them satisfiy the condition that a $4$ appearing two times, a $3$ appearing one time, a $5$ appearing one time and a $6$ appearing one time. The probability is:
$\displaystyle \frac{60}{7776}=0.0077160494$
Example 2
What is the probability that one score appears two times, three other scores appear one time each in a random poker dice hand?
Here, we need to count all the possible poker dice hands of one pair. Both $(0,0,1,2,1,1)$ and $(2,1,1,1,0,0)$ are examples of one pair. In essense, we need to count all the occupancy number sets such that among the $6$ coordinates (cells), one of the cells is a $2$ and three of the cells are $1$. To this end, we apply the multinomial theorem twice, one time on the five rolls of dice and one time one the $6$ cells.
Consider the occupancy number set $(2,1,1,1,0,0)$. Note that the multinomial coefficient is $60$ as in Example $1$ (the first application of the multinomial thoerem). Now look at the $6$ coordinates of the occupancy number set $(2,1,1,1,0,0)$. We wish to partition these $6$ coordinates into three groupings, one with one $2$, one with three $1's$ and one with two $0's$. The following is the multinomial coefficient (the second application of the multinomial theorem):
$\displaystyle \frac{6!}{1! \ 3! \ 2!}=60$
Thus the number of possible poker dice hands of one pair is: $60 \times 60=\text{3,600}$ and for a random poker dice hand, the probability that it is a one pair is:
$\displaystyle \frac{3600}{7776}=0.462962963$
Remark
Example $2$ provides the algorithm for computing the remaining poker dice hand probabilities. The key is to apply the multinomial coefficients twice, one time on a representative occupancy number set, the second time on the six cells (the six faces of a die in this case). Then the number of poker dice hands in question is the product of the two multinomial coefficients.
Example 3
What is the probability that a random poker dice hand is three of a kind?
Consider the occupancy number set of $(3,1,1,0,0,0)$. The associated multinomial coefficient for the five rolls of dice is:
$\displaystyle \frac{5!}{3! \ 1! \ 1!}=20$
Now partition the six cells into three groupings (one $3$, two $1's$, three $0's$):
$\displaystyle \frac{6!}{1! \ 2! \ 3!}=60$
Thus the probability that a random poker hand is three of a kind is:
$\displaystyle \frac{20 \times 60}{7776}=0.1543209877$
Summary
The following are the probabilities of poker dice hands.
$\displaystyle P(\text{five of a kind})=\frac{6}{7776}$
$\displaystyle P(\text{four of a kind})=\frac{150}{7776}$
$\displaystyle P(\text{full house})=\frac{300}{7776}$
$\displaystyle P(\text{three of a kind})=\frac{1200}{7776}$
$\displaystyle P(\text{two pair})=\frac{1800}{7776}$
$\displaystyle P(\text{one pair})=\frac{3600}{7776}$
$\displaystyle P(\text{high card})=\frac{720}{7776}$
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$\copyright \ \text{2010 - 2015 by Dan Ma}$ (Revised March 28, 2015)
# The multinomial theorem
The multinomial theorem is a statement about expanding a polynomial when it is raised to an arbitrary power. Rather than just stating the theorem and showing examples, we motivate the theorem by a concrete example of finite random sampling. This example demonstrates that the notion of finite sampling provides another interpretation to the multinomial thoerem. As a result, the multinomial theorem and the multinomial coefficients are useful in combinatorial techniques in finite random sampling models. The multinomial coefficients are also useful in partitioning a set of objects into several subgroups where each subgroup is made up of indistinguishable objects. See the post The game of poker dice and the multinomial theorem for an example of applications of these ideas.
________________________________________________________________________
An example
Suppose we have a finite set $S$ with $n$ elements where $n$ is a positive integer. Suppose we select $k$ elements from the population $S$ with replacement. We consider ordered samples of size $k$ and unordered samples of size $k$. To make this notion more concrete, consider the population consisting of $4$ letters $\lbrace{m,i,s,p}\rbrace$. Suppose we sample $11$ times with replacement. The following are two specific ordered samples of size $11$:
$mississippi \ \ \ \ \ \ \ \ \ \ \ \ mississiipp \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
Thus ordered samples can be represented as strings of characters drawn from the population presented in the order in which the objects are drawn. In this example, there are $4^{11}=4194304$ many ordered samples. Each character drawn has $4$ choices and the drawing is done $11$ times. In general, if the population has size $n$, then there are $n^k$ many ordered samples of size $k$.
We now look at unordered samples obtained from sampling $11$ times from the population $\lbrace{m,i,s,p}\rbrace$. In unordered samples, the order in which the letters are drawn is no longer important (or recorded). We only care about the number of instances each letter is selected. Thus each of the ordered samples shown above yields the same unordered sample: $1 \ m$, $4 \ i's$, $4 \ s's$ and $2 \ p's$. We represent the unordered sample in two ways:
$(1,4,4,2)$
$* \ | \ * \ * \ * \ * \ | \ * \ * \ * \ * \ | \ * \ *$
The first representation is a $4$-tuple where the coordinates are the number of $m's$, the number of $i's$, the number of $s's$ and the number of $p's$ in the $11$ selections. Thus, the sum of the coordinates must be $11$ (the total number of selections). The second representation of the unordered sample is made up of stars and bars. The three bars create $4$ spaces and the stars in each space represent the number of characters drawn. We would like to point out that the “unordered” in the unordered samples refers to the fact that the order in which the objects are drawn is immaterial. However, both the $4$-tuple notation and the stars and bars diagrams are ordered according to the $4$ letters in the population (showing how many instances each object appears in the samples).
The star and bar diagram has a combinatorial advantage. For example, how many unordered samples are there when you draw $11$ letters with replacement from the population $\lbrace{m,i,s,p}\rbrace$? In any unordered sample, there are $3$ bars and $11$ stars. The order of the stars and bars can be in any arbitrary order. Thus there are $\displaystyle \binom{11+3}{11}=364$ many unordered samples. In general, if the population size is $n$, then there are $\displaystyle \binom{k+n-1}{k}=\binom{k+n-1}{n-1}$ many unordered samples, either represented as $k$-tuples or stars and bars diagrams with $n-1$ bars and $k$ stars.
One more note about the number $364$ calculated above. This is also the total number of non-negative integer solutions to the equation $m+i+s+p=11$. Thinking of an unordered sample as a $4$-tuple, the sum of the $4$ coordinates must be $11$. This observation is important to understanding the multinomial theorem.
There is one more count associated with unordered samples. How many unordered samples are there when you draw $11$ letters with replacement from the population $\lbrace{m,i,s,p}\rbrace$ such that each letter is selected at least once? The answer is $120$. Suppose that each letter is already selected once. Then we need to sample $7$ more times out of these $4$ letters. According to the above paragraph, the total count is $\displaystyle \binom{7+3}{3}=120$. To generalize, if the population size is $n$, there are $\displaystyle \binom{k-n+n-1}{n-1}=\binom{k-1}{n-1}$ many unordered samples in which all objects in the population are represented in each sample.
We now tie unordered samples back to ordered samples. How many ordered samples are equivalent to the unordered sample $(1,4,4,2)$? Both ordered samples in $(1)$ are equivalent to $(1,4,4,2)$ (i.e. each letter is drawn the same number of times). In other words, how many $11$-character strings can be formed using $1 \ m$, $4 \ i's$, $4 \ s's$ and $2 \ p's$? The answer is:
\displaystyle \begin{aligned}\binom{11}{1} \binom{10}{4} \binom{6}{4} \binom{2}{2}&=\displaystyle \frac{11!}{1! \ 4! \ 4! \ 2!}\\&=34650\end{aligned}
The reasoning for the above calculation is that out of $11$ positions in the strings, we choose $1$ position for the $m$, choose $4$ positions for the $i$ in the remaining $10$ positions and so on.
________________________________________________________________________
Summary of the example
In sampling $11$ times with replacement from the population $\lbrace{m,i,s,p}\rbrace$, we summarize our observations about the example.
• The number of ordered samples is $4^{11}=4194304$.
• The number of unordered samples is $\displaystyle \binom{11+3}{11}=364$. Furthermore, the number of non-negative integer solutions to the equation $m+i+s+p=11$ is $364$.
• The number of unordered samples where each letter is selected is $\displaystyle \binom{7+3}{3}=120$. Furthermore, the number of positive integer solutions to the equation $m+i+s+p=11$ is $120$.
• For any unordered sample $(a,b,c,d)$ where $a+b+c+d=11$, the total number of ordered samples equivalent to this unordered sample is $\displaystyle \frac{11!}{a! \ b! \ c! \ d!}$. As we shall see, these are called multinomial coefficients.
Note the interplay between the ordered samples and unordered samples. We start out with a large number of ordered samples ($4^{11}$ many). We then collapse these ordered samples to just $364$ unordered samples. We see that each unordered sample corresponds to certain number of ordered samples according to the multinomial coefficients. Thus we have the following sum where $a,b,c,d$ are non-negative integers:
$\displaystyle \sum \limits_{a+b+c+d=11} \frac{11!}{a! \ b! \ c! \ d!}=4^{11}$
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The Multinomial Theorem
With the preceding discussion, we now state the multinomial theorem.
The Multinomial Theorem
For any positive integer $n$ and any positive integer $k$, we have the following polynomial expansion:
$\displaystyle \biggl(x_1+x_2+ \cdots +x_n\biggr)^k=\sum \limits_{a_1+ \cdots +a_n=k} \frac{k!}{a_1! \ a_2! \cdots \ a_n!} \ x_1^{a_1} x_2^{a_2} \cdots x_n^{a_n}$
Remark
The same observations we make about the example apply. For example, the number of terms in the polynomial expansion is $\displaystyle \binom{k+n-1}{k}$, which is the number of non-negative integer solutions to the equation $a_1+ \cdots +a_n=k$. Each term $x_1^{a_1} x_2^{a_2} \cdots x_n^{a_n}$ in the polynomial expansion can be considered as an unordered sample in the finite sampling with replacement. Then the coefficient of each term (called multinomial coefficient) is the number of associated ordered samples. As a result, the multinomial coefficients sum to $n^k$.
We conclude with two interpretations of the multinomial coefficient.
$\displaystyle \frac{k!}{a_1! \ a_2! \cdots \ a_n!} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$
If we have $n$ many distinct symbols (say $x_1,x_2, \cdots, x_n$) and we have $a_1$ many $x_1$, $a_2$ many $x_2$ and so on, then the multinomial coefficient in $(2)$ is the number of $k$-length character strings that can be formed using the available symbols.
Another interpretation is that of partitioning a set of $k$ objects into $n$ subgroups where the objects in each subgroup are indistinguishable.
Both interpretations are one and the same, just looking at the same result in a different angle. For example, all three of the following yield the same answer: $\text{34,650}$. We have $11$ letters (one $m$, four $i's$, four $s's$ and two $p's$), how many character strings can be formed with these letters?
On the other hand, we have $11$ identical candies randomly distributed to Marcus, Issac, Samantha and Paul. How many ordered samples will result if Marcus receives one candy, Issac receives four candies, Samantha receives four candies and Paul receives two candies? Here, we are trying to partition a set of $11$ objects into four subgroups where one group has one element, two of the groups have four elements each and another group has two elements.
If we have $11$ candidates for forming four distinct committees where one committee has one member, two of the committees have four members each and another one has two members. In how many ways can this be done?
Reference
1. Feller, W., An Introduction to Probability Theory and its Applications, Vol. I, 3rd ed., John Wiley & Sons, Inc., New York, 1968
________________________________________________________________________
Revised March 28, 2015
$\copyright \ \text{2010 - 2015 by Dan Ma}$<|endoftext|>
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# Complex Numbers Concept of Rotation for JEE
Last updated date: 11th Sep 2024
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## What is Complex Number?
Being an extension of the real number system, complex numbers have numerous interesting properties that make mathematical calculations even easier. These useful properties model two-dimensional space and trigonometry. A complex number is a point on the complex or argand plane with the horizontal component as the real part and the vertical component as the imaginary part. Take for example z= 2 + 6i. Here, 2 is the horizontal component and 6 is the vertical component.
Besides the several useful properties of complex numbers, one might wonder what is the rotation of complex numbers. Rotation of complex numbers is an easier and streamlined way to correlate the complex numbers and the angle that they make.
## What is Rotation?
The relationship between the complex numbers and the angle that they make is referred to as the rotation of a complex number. A complex rotation can be denoted as $|z| {e}^{i \theta}$ where θ is a real number in radians. ∅ denotes the rotation of the complex number in an anticlockwise or counterclockwise direction and -θ for the clockwise direction.
Let us understand this concept thoroughly.
Concept of Rotation
## Rotation of Complex Numbers
Take another example, in the figure below, there are two complex numbers z and z’. We want to rotate oz’ along with oz. It can be achieved when we rotate the complex number by the argument of $\dfrac{Z}{Z^{\prime}}$.
Rotation of a Complex Number
Further,
$\dfrac{z}{z^{\prime}}=\dfrac{\left|z_{r}\right| e^{i \theta}}{\left|z^{\prime}\right| e^{i θ^{\prime}}} \\ \dfrac{z}{z^{\prime}}=\dfrac{|z|}{\left|z^{\prime}\right|} e^{i \alpha x}$
Or
$\dfrac{z}{z^{\prime}}=\dfrac{|z|}{\left|z^{\prime}\right|}(\cos (\alpha)+i \sin (\alpha))$
The above concept can also be looked at this way.
$z=|z| e^{i \theta}$-----(1)
And
$z^{\prime}=\left|z^{\prime}\right| e^{i \theta^{\prime}}$------(2)
By dividing (1) and (2), we get
$\dfrac{z}{z^{\prime}}=\frac{|z|}{\left|z^{\prime}\right|} e^{i\left(\theta-\theta^{\prime}\right)} \\ \dfrac{z}{z^{\prime}}=\dfrac{|z|}{\left|z^{\prime}\right|} e^{i \alpha}$
It is the same result that we got above. Thus, from the above results, we conclude that to rotate the complex number, we can use $\dfrac{z}{z^{\prime}}=\dfrac{|z|}{\left|z^{\prime}\right|} e^{i \alpha}$ result directly.
## Solved Examples for Rotation of Complex Numbers
Question 1: With its origin at the centre, consider an n-sided polygon. If z1 is the complex number representing a vertex B1 of the polygon, then find the complex number associated with the vertex adjacent to B1, i.e, m B2 and Bn.
N-Sided-Polygon
Solution: From the figure above, we know that the vertex adjacent to B1 is either B2 or Bn.
Now we know from above that
$\dfrac{z}{z^{\prime}}=\dfrac{|z|}{\left|z^{\prime}\right|} e^{i \alpha}$
Using the same relationship,
$\dfrac{z_{2}-0}{z_{1}-0}=\dfrac{\left|z_{2}\right|}{\left|z_{1}\right|} e^{\dfrac{i n}{\pi}} \\ z_{2}=z_{1} e^{\dfrac{i n}{\pi}} \\ \dfrac{z_{n}-0}{z_{1}-0}=\dfrac{\left|z_{n}\right|}{\left|z_{1}\right|} e^{\dfrac{-i n}{\pi}} \\ \\ n=z_{1} e^{\dfrac{-i n}{\pi}}$
Question 2: Complex numbers z1, z2, and z3 are the vertices P, Q, and R, respectively, of an isosceles right-angled triangle with a right angle at R . Show that (z1 – z2)2 = 2(z1 – z3)(z3 – z2).
Rotation of a Complex Number
Solution: In the isosceles triangle PQR, PQ = QR and PQ are at 90 degrees to QR which means that QR has to be rotated to occupy the position PQ.
$\dfrac{z_{2}-z_{3}}{z_{1}-z_{3}}=e^{\dfrac{i \pi}{2}}=i \\ z_{2}-z_{3}=i\left(z_{1}-z_{3}\right)$
On Squaring and solving, we get the result
z22 + z32 – 2 z2 z3 = -(z12 + z32– 2 z1 z3)
(z1 – z2)2 = 2(z1 – z3)(z3 – z2).
Question 3: Rotate the complex number (5+6i) using the multiplicative property of complex numbers.
Solution: Multiply, the above complex number (5+6i) with i. It will then be
$(5+6 i) \times i=5 i+(6 i) \times i \\ (5+6 i) i=5 i+6 \times(-1) \\ (5+6 i) i=-6+5 i$
This gives us the result (5+6i).i = -6+5i.
## Rotation Number
Multiplication of a complex number with i. The rotation property of a complex number can also be correlated with the multiplication property of the complex number. It means that when we multiply a complex number with iota or i of magnitude 1, we find that it leads to a pure rotation of the complex number. Now, this rotation could also be extended to 180, 270 degrees, and so on.
Look at the below diagram. Here, the complex number (4+3i) is multiplied by i to give it a rotation of 90 degrees in the complex plane.
Complex Number Rotation
## Rotation Theorem in Complex Numbers
The rotation theorem of the complex theorem is as follows:
• If P(z1) and Q(z2) are two complex numbers such that |z1| = |z2|, then $z^{2}=z_{1}^{\text {ei } \theta}$, where $\theta$ is the argument between z1 and z2.
• If P(z1), Q(z2), and R(z3) are three complex numbers and θ is the angle between z1, z2, and z3, then
$\dfrac{z_{3-} z_{2}}{z_{1}-z_{2}}=\left|\dfrac{z_{3}-z_{2}}{z_{1}-z_{2}}\right| e^{i \theta}$
• If P(z1), Q(z2), R(z3), and S(z4) are three complex numbers and θ is the angle between the z1, z2, z3, and z4, then
$\dfrac{z_{3}-z_{4}}{z_{1}-z_{2}}=\left|\dfrac{z_{3}-z_{4}}{z_{1}-z_{2}}\right| e^{i \theta}$
$\dfrac{z_{2}-z_{1}}{z_{4}-z_{3}}$
Other things to note are:
• amp (z) = $\theta$ is a line from the origin inclined at an angle from the x-axis.
• Multiplication by $e^{-I\alpha}$ to z rotates the vector in clockwise direction by an angle $\alpha$.
• If z1, z2, z3, and z4 are the vectors defining the points A, B, C, and D, respectively, in the Argand plane, then
• AB is inclined to CD at an angle= arg $\dfrac{z_{2}-z_{1}}{z_{4}-z_{3}}$.
• If CD is inclined at 90o to AB, then arg $=\dfrac{z_{2}-z_{1}}{z_{4}-z_{3}}=\pm \dfrac{\pi}{2}$
## Conclusion
The article traverses through the concepts of rotation of complex numbers. Rotation of complex numbers with reference to multiplication property and the different theorems are listed. The theorems provide an understanding of how the complex numbers rotate on the argand plane, with some particular angle. They also illustrate the rotation of different vectors on the argand plane with respect to one another. It is also seen that on multiplying a complex number with i, it leads to its rotation by 90 degrees. There is a pure rotation as the multiplication factor is i, whose magnitude is 1.
Competitive Exams after 12th Science
## FAQs on Complex Numbers Concept of Rotation for JEE
1. Is the rotation of complex numbers related to complex multiplication?
Yes, the rotation of a complex number can directly be correlated with the multiplication property. In such a case, the rotation property can be denoted as (x+iy)i.
Multiplying a complex number by i represents that the complex number is rotated by 90 degrees in the anticlockwise or counterclockwise direction. Thus, rotation in a complex plane can be visualised as stretching and rotating. The reason this multiplication results in a pure rotation and no stretching is because we are multiplying the complex number by one. This rotation property of the complex number can be extended to common rotations like 0, 90, 180, and 270 degrees.
2. What is a unimodular complex number?
A unimodular complex number is a complex number with a unit magnitude.
Suppose z is a complex number. Thus, the unimodular complex number is represented by |z|=1. Since |z|=1, it means z lies on a circle of radius 1 and with a centre (0,0).
Take for example, z1 and z2 are complex numbers such that $\dfrac{z_{1}-z_{2}}{2-z_{1} \overline{z_{2}}}$ is unimodular and z2 is not unimodular.
To solve the above problem, we take $\left|\dfrac{z_{1}-z_{2}}{4-z_{1} \overline{z_{2}}}\right|=1$. On solving, we get the value of z1 as a circle of radius 2.<|endoftext|>
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The bow is the forward part of the hull of a ship or boat, the point that is usually most forward when the vessel is underway. Both of the adjectives fore and forward mean towards the bow. The other end of the boat is the stern.
- Etymology 1
- Function 2
- Parts 3
- Types 4
- See also 5
- References 6
- Further reading 7
The bow is designed to reduce the resistance of the hull cutting through water and should be tall enough to prevent water from easily washing over the top of it. On slower ships like tankers, a fuller bow shape is used to maximise the volume of the ship for a given length.
A "wet bow" results from seawater washing over the top of the hull. A raked stem can help to reduce the wetness of the bow. Aside from making the deck slippery, water can corrode the metal of the ship. If the temperature is low enough this water can also freeze on the deck, rails, turrets, and other exposed surfaces, increasing the topside weight.
The forward part of the bow, usually on the ship's centreline, is called the stem. Traditionally, the stem was an upright timber or metal bar into which side planks or plates were joined.
Several types of bows exist. These include:
- Boat building
- Bow (rowing)
- Naval architecture
- Stem (ship)
- Sleight, Steve; The New Complete Sailing Manual, Dorling Kindersley Co., (2005) ISBN 0-7566-0944-5
- Steward, Robert; Boatbuilding Manual, 3rd ed. International Marine Publishing Company. Camden, Maine (1987), p2-3. ISBN 0-87742-236-2<|endoftext|>
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Predicting is one of the most ambitious goals of science. It goes beyond describing and explaining, and it attempts to “tell the future”. The prediction process has the following basic steps:
- We have an estimate of the present conditions of a system, for instance, the atmosphere.
- We have a model –i.e. a set of mathematical rules coming from physical principles- which we evolve forward (or integrate) in time.
- We get an estimate of the future state of our system at any given time.
When computing a prediction, it is very important to provide a measure of the quality of this prediction. Intuition tells us that we are more certain, for example, in predicting the temperature in our neighborhood for tomorrow, than in predicting the temperature in the same place a year from now. Where does this certainty/uncertainty come from? Let us explore this next.
For the sake of this discussion, consider that the model mentioned in step (2) is perfect. That is, let us assert that have completely captured in our equation all the processes we are interested in, and that we can solve these equations perfectly with a computer code (this is not true in reality, but we will leave that for another blog entry). In this case the quality of a prediction is determined by the error of our estimate mentioned in step (1) –i.e. the error in our initial conditions– and the error growth in time.
As it turns out, errors grow differently in different dynamical systems. In some systems, making a tiny mistake is irrelevant for a future prediction, while in other systems a tiny initial error can ruin a forecast after a certain lead-time. Let’s take a quick view at different families of dynamical systems with the help of Figure 1. The figure has four panels; for each panel the x-axis corresponds to time, while the y-axis corresponds to the value of a physical variable (it can be wind speed, temperature, etc). Let us run a trajectory starte from a given initial condition; we label this reference trajectory (shown in black in the figure). Also, let us evolve trajectories initialised from ‘nearby’ initial conditions – i.e. initial conditions with errors; we label these trajectories as perturbed trajectories. In the figure, red lines indicate that initial perturbed values are larger than the initial true value, while blue lines indicate that the initial perturbed values are smaller than the initial true value. The behaviour in error growth is different in each case:
- a) In this example, the perturbed trajectories tend towards the reference trajectory. This is a typical dissipative system. Regardless of the initial conditions, the system evolves towards a fixed point, and any initial error disappears. Think of a pendulum with friction: it does not matter at what height you drop it, it will use its gravitational potential energy to swing for a while, but it will eventually stop.
- b) In this example, the errors of the perturbed trajectories grow as time increases, and they do not stop growing, instead, the perturbed trajectories tend towards plus and minus infinity. This system –in which errors grow without limit– is not feasible in reality, since it would require infinite energy. However, if we want to make predictions in a finite-time frame, the accuracy in the initial conditions is crucial, and we will see the quality of the forecast decrease with time.
- c) In this example, the initial error of the perturbed trajectories is preserved as time evolves; it neither grows nor decreases with time. This is typical for periodic systems, such as those found in celestial mechanics, or physical processes related from them, like the tides. If we are wrong in our position of the moon tonight, and do a forecast for the next days, the error will stay constant as time progresses. There is another type of systems, called quasi-periodic, which have similar characteristics, but I will not discuss them further.
- d) The last kind of systems is perhaps the most interesting to us; we are talking about chaotic systems. The atmosphere is a typical forced-dissipative system that presents chaotic behaviour. In this case, errors initially grow slowly, then the error growth turns faster, and eventually the perturbed trajectories do not resemble the reference trajectory at all, and in fact they do not resemble each other. The accuracy of the initial conditions is crucial for a good forecast, and the quality of a forecast decreases with time. In fact, even the tiniest initial errors will ruin a forecast after a given lead-time. What is different with respect to panel (b)? Errors do not grow forever and without limit, instead they saturate. After a long time, the trajectories – both the reference and the perturbed ones – evolve and live within a permissible range of values (without going to plus or minus infinity). This set of values is know as attractor (or climatology).
Figure 1. Error growth for different families of dynamical systems.
Let us discuss chaotic systems a little further using our example in panel (d). A forecast for time t=0.5 is more reliable than that for time t=1, and after approximately t=1.5 we have lost our capacity to predict. Something similar happens in the atmosphere. For large scale features, this limit of predictability is about 2 weeks. Operational centres release forecasts for up to 5 or 7 days in advance, and they equip these forecasts with some probabilistic measure (representing, in simple terms, how different are trajectories initiated from similar initial conditions). Unfortunately, some commercial forecast providers give no information on the accuracy of their forecasts at all. Furthermore, they are known for (irresponsibly) releasing ‘valid’ determinisitic forecasts for up to 45 days in advance (do not confuse this with the proper seasonal outlooks generated by meteorological agencies). As expected, these forecasts change considerably when updated every day, and these changes continue until the lead-time is within the predictability window. Such 45 day ‘forecasts’ are not prediction, they can be considered quasi-random draws from the climatology of different regions. At the end these forecasts have no value and they end up stating the obvious: July will be relatively warm and December will be relatively cold.<|endoftext|>
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Linear Functions:
You are already familiar with the concept of "average rate of change".
When working with straight lines (linear functions) you saw the "average rate of change" to be:
The word "slope" may also be referred to as "gradient", "incline" or "pitch", and be expressed as: A special circumstance exists when working with straight lines (linear functions), in that the "average rate of change" (the slope) is constant. No matter where you check the slope on a straight line, you will get the same answer. Non-Linear Functions: When working with non-linear functions, the "average rate of change" is not constant. The process of computing the "average rate of change", however, remains the same as was used with straight lines: two points are chosen, and is computed. FYI: You will learn in later courses that the "average rate of change" in non-linear functions is actually the slope of the secant line passing through the two chosen points. A secant line cuts a graph in two points.
When you find the "average rate of change" you are finding the rate at which (how fast) the function's y-values (output) are changing as compared to the function's x-values (input).
When working with functions (of all types), the "average rate of change" is expressed using function notation.
Average Rate of Change For the function y = f (x) between x = a and x = b, the
While this new formula may look strange, it is really just a re-write of .
Remember that y = f (x).
So, when working with points (x1, y1) and (x2, y2), we can also write them as
the points .
Then our slope formula can be expressed as .
If we rename x1 to be a, and x2 to be b, we will have the new formula.
The points are , and the
.
Finding average rate of change from a table.
Function f (x) is shown in the table at the right.
Find the average rate of change over the interval 1 < x < 3.
Solution:
If the interval is 1 < x < 3, then you are examining the points (1,4) and (3,16). From the first point, let a = 1, and f (a) = 4. From the second point, let b = 3 and f (b) = 16.
Substitute into the formula:
x f (x) 0 1 1 4 2 9 3 16
The average rate of change is 6 over 1, or just 6.
The y-values change 6 units every time the x-values change 1 unit, on this interval.
Finding average rate of change from a graph.
Function g (x) is shown in the graph at the right. Find the average rate of change over the interval 1 < x < 4. Solution: If the interval is 1 < x < 4, then you are examining the points (1,1) and (4,2), as seen on the graph. From the first point, let a = 1, and g (a) = 1. From the second point, let b = 4 and g (b) = 2. Substitute into the formula: The average rate of change is 1 over 3, or just 1/3. The y-values change 1 unit every time the x-values change 3 units, on this interval.
Finding average rate of change from a word problem.
A ball thrown in the air has a height of h(t) = - 16t² + 50t + 3 feet after t seconds. a) What are the units of measurement for the average rate of change of h? b) Find the average rate of change of h between t = 0 and t = 2? Solution: a) In the formula, , the numerator (top) is measured in feet and the denominator (bottom) is measured in seconds. This ratio is measured in feet per second, which will be the velocity of the ball. b) Start by finding h(t) when t = 0 and t = 2, by plugging the t values into h(t). h(2) = -16(2)² + 50(2) + 3 = 39 h(0) = -16(0)² + 50(0) + 3 = 3 Now, use the average rate of change formula:<|endoftext|>
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By Brian De La Franier, Member-at-large for the GCI
- Inherently Safer Chemistry for Accident Prevention: Substances and the form of a substance used in a chemical process should be chosen to minimize the potential for chemical accidents, including releases, explosions, and fires.
In the 12th and final video of the GCI series on the 12 principles of green chemistry, Gabby and Qusai investigate the 12th principle on inherently safer chemistry and note several common issues found in many labs.
The 12th principle is frequently called the safety principle and is often overlooked when considering green chemistry principles. However, the broad nature of the 12th principle means it both incorporates many of the other principles and is almost impossible to achieve without considering all 12 of them, given that the overall goal of green chemistry is to reduce hazards and pollution.
In the Video #11 blog post, Alex wrote about how driving a car with no windows and mirrors would lead to accidents as there would be no real-time way to analyze your surroundings. The 12th principle is akin to having that car inspected before driving it. It is insuring that all aspects of the car, from the engine, to the brakes, to the steering are all in working order so that the car is less likely to get into an accident. With this principle we consider the ingredients of a reaction (the parts of the car), and make sure that they don’t pose excessive hazards.
An example mentioned in the Video #12 of a hazardous chemical that can be replaced in synthesis is methyl isocyanate, a molecule used in the synthesis of the insecticide carbaryl. In 1984, this toxic compound was released into the air from a pesticide plant in Bhopal, India, immediately killing 3,800 people, and causing premature death in thousands more.1 This disaster could have been avoided had the plant instead used methylamine to carry out the reaction.2Although this principle is specifically about the avoidance of using or producing hazardous compounds, the idea of avoiding hazards can be extended to other areas of the lab. Storing chemicals that are reactive together, such as oxidizers and flammable materials, leads to a risk of release and reaction. If these compounds leak from their containers and react, they will create a large fire. This is a hazard that could be easily avoided by storing these chemical types separately.
Another hazard in the lab is liquid spills. Anything that has been spilled should be immediately cleaned up to prevent people from slipping on it or receiving chemical burns from an unknown substance. If someone comes across an acid spill, but does not know what it is they could easily be burned in attempting to clean it up. Returning to the car analogy, leaving an unknown spill would be like giving someone a damaged car to drive without telling them. The unfortunate driver could be injured as a result of faulty brakes, just as another lab member could be injured by your spill in the lab.
As with our car, the lab should be kept safe and in good repair. If there are damaged parts in a car you should always repair them before driving it, just as if there are hazardous chemicals or situations in our lab they should be replaced before performing reactions.
- Broughton, E. (2005). The Bhopal disaster and its aftermath: a review. Environmental Health, 4(1), 6.
- Thomas A. Unger (1996). Pesticide Synthesis Handbook (Google Books excerpt). William Andrew. pp. 67–68.<|endoftext|>
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# Properties of Numbers - PowerPoint PPT Presentation
Properties of Numbers
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Properties of Numbers
## Properties of Numbers
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Properties of Numbers Flipbook
2. Addition and Multiplication Properties of Numbers Property of Zero Commutative Property Associative Property Identity Property Distributive Property
3. Property of Zero
4. Commutative Property Addition Multiplication *Think of a commute to work – from home to work and work to home - order switched, same distance (same answer)
5. Associative Property Addition Multiplication *Think of associate – at a party it doesn’t matter which group of friends you talk to first. In the end you will talk to everyone. - Talk to everyone (same final answer)
6. Identity Property Addition Multiplication *Think of your identity – No matter what you change about yourself, you are still you. • Adding 0 to 5 won’t change 5’s identity. • Multiplying 1 to 7 won’t change 7’s identity.
7. Distributive Property Multiplication distributed over addition. For any numbers a, b, and c, a(b + c) = ab + ac Example: 8(6 + 7) = 8 · 6 + 8 · 7 OR Multiplication distributed over subtraction. For any numbers a, b, and c, a(b – c) = ab – ac Example: 6(5 – 2) = 6 · 5 – 6 · 2<|endoftext|>
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# Skip counting patterns by 6 and 7 worksheets
## Skip counting patterns by 6 and 7 worksheet
Skip counting is a fundamental mathematical concept that plays a crucial role in developing numerical fluency and understanding patterns within numbers. It involves counting numbers by regularly skipping a fixed value between each number in the sequence. This technique not only helps students grasp the concept of multiplication but also lays the foundation for more advanced mathematical skills. In this discussion, we will explore skip counting patterns, their importance in mathematics education, and practical applications.Skip counting patterns by 6 and 7 worksheets
Skip counting patterns are a systematic way of counting numbers, usually starting from a given number and adding a fixed value to that number to reach the next number in the sequence. For example, when counting by twos, you start from a base number (often 0 or 1) and then add 2 to it to get the next number, and so on. The resulting sequence would be 0, 2, 4, 6, 8, 10, and so forth. Similarly, skip counting can be done for other numbers, such as threes, fours, fives, or any other fixed increment.Skip counting patterns by 6 and 7 worksheets
The importance of skip counting patterns in mathematics education cannot be overstated. Here are some key reasons why skip counting is a valuable concept:
Skip counting patterns by 6 and 7 worksheets
### Foundation for Multiplication Skip counting patterns
Skip counting is a precursor to multiplication. Understanding skip counting patterns helps students grasp the idea of repeated addition, which is the fundamental concept behind multiplication. For instance, counting by fives (5, 10, 15, 20, etc.) is directly related to the concept of 5 times tables (5 x 1, 5 x 2, 5 x 3, 5 x 4, etc.).
1. Enhances Number Sense: Skip counting enhances a student’s number sense by helping them recognize relationships between numbers. It encourages them to observe and understand how numbers change and how patterns emerge. This skill is crucial for more advanced math topics.
2. Counting Efficiency: Skip counting allows for quicker and more efficient counting when dealing with larger quantities. For example, instead of counting individual items one by one, you can use skip counting to count by fives or tens, making the process much faster and more accurate.
3. Preparation for Division: Skip counting can also serve as a foundation for division. Understanding skip counting patterns can help students visualize the idea of dividing a set of objects into equal groups.
4. Real-World Applications: Skip counting has practical applications in everyday life. For instance, when counting money, measuring distances, or organizing items into groups, skip counting can simplify the process and make it more manageable.
5. Pattern Recognition: Skip counting helps develop pattern recognition skills, which are essential in mathematics. Recognizing patterns not only aids in skip counting itself but also in solving more complex mathematical problems.
6. Mental Math Skills: By practicing skip counting, students improve their mental math skills. They become more proficient in performing calculations in their heads, which is a valuable skill in various mathematical contexts and in daily life.
7. Mathematical Fluency: Skip counting contributes to mathematical fluency, ensuring that students can confidently and accurately perform basic arithmetic operations. This fluency serves as a strong foundation for tackling more advanced mathematical concepts.
## Skip counting patterns by 6 and 7
Skip counting patterns by 6 and 7 worksheets
To illustrate the practicality and versatility of skip counting patterns, consider some specific examples:
• ### Time Telling Skip counting patterns
• Learning to count by fives assists children in reading analog clocks. Each number on the clock represents five-minute intervals, and skip counting by fives helps them understand how to read time accurately.
• ### Money Handling Skip counting patterns
• Skip counting is particularly useful when counting coins and bills. For example, counting by twenties is essential when dealing with dollar bills, and counting by fives is vital when dealing with nickels.
• ### Measuring Length Skip counting patterns
• Skip counting is used when measuring distances. For instance, a tape measure typically has markings at intervals of one inch, but you can use skip counting to measure longer distances quickly.
• ### Organizing Data Skip counting patterns
• Skip counting can be applied to organize data into tables and charts. It simplifies the process of grouping data into categories or intervals.
In summary, skip counting patterns are a foundational concept in mathematics that lays the groundwork for more advanced mathematical skills. They promote number sense, help with multiplication and division, enhance mental math abilities, and find applications in various real-world scenarios. By understanding and mastering skip counting, students develop a solid mathematical foundation of Skip counting patterns that will serve them well throughout their mathematical journey and in their daily lives.
Skip counting patterns by 6 and 7 worksheets
Develop number sense and mathematical skills with interactive worksheets on skip counting patterns by 6 and 7. Practice identifying and continuing skip counting sequences in increments of 6 and 7 through engaging exercises and printable worksheets. Strengthen mathematical fluency and develop a solid foundation in skip counting patterns. Explore a variety of educational resources to make learning enjoyable and effective.
Skip counting patterns by 6 and 7 worksheets<|endoftext|>
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Error detection and correction
Error detection and correction is about methods to make sure that information or data is not corrupted, and still makes sense. The techniques help reliable delivery of digital data over unreliable communication channels.
Detecting an error[change | change source]
There are different ways to make sure an error can be detected. This is done by adding more data or information to the data transmitted. Adding more data than needed is called redundancy.
- Repetition codes: The data is broken up into blocks. Each block is then sent a number of times.
- Parity bits can be used. The parity bit is 1 for an odd number of ones, and 0 for an even number (or vice versa)
- Cyclic redundancy checks
- Hamming codes
- Hash function
- Longitudinal redundancy check
- Transverse redundancy check
- Polarity schemes
Correcting an error[change | change source]
There are two main ways to correct an error:
- Automatic repeat request (ARQ) - The receiver detects an error and automatically requests a repetition
- Forward error correction
Other websites[change | change source]
- Error control] --Citizendium<|endoftext|>
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In this animated short from the Walt Disney Company, Uncle Scrooge discusses the history or money and the importance of money in the overall economy. There are A LOT of great teaching opportunities in this clip and would make a great summary of a money supply lesson or a required video to be watched before the lesson.
Opening to 7:15
History of Money
Huey, Dewie, and Louie visit Scrooge McDuck and request that he help them save the money they had earned. Scrooge goes through the history of money and discusses the role of salt as the original salary that Roman soldiers received. He then goes on to describe money from other societies and why money was important following original barter economies. The characters even discuss the role of money as a medium of exchange!
7:15 to 9:59
After learning of the importance of money in the economy, the brothers question why central banks don’t just print more money if everyone wants it. Uncle Scrooge discusses the role of fiat money and why it’s important for the money to be backed by something or someone who can promise to pay the notes that are printed.
10:00 to 13:20
Financial Planning and Taxes
Uncle Scrooge teaches the brothers about the importance of budgeting. People need to make sure that they allocate a portion of their income toward rent, food, and other necessities. He also teaches them about the role of taxes and how important it is for governments to have a budget and make sure that they collect taxes to pay debt.
13:20 to End
Velocity of Money & Investment
The boys are curious why Scrooge keeps so much money in his vault if he tells them that it’s important to put money “to work.” He teaches them that the money in his vault is just his petty cash and then goes on to discuss the importance of money circulating through the economy. The ending portion discusses the role of corporations issuing stocks and shareholders collecting dividends. At the end, he signs the boys up to manage their funds, but charges them a fee. The boys aren’t happy, but he laments that “nothing is ever free.”<|endoftext|>
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Throughout the history of computers, one aspect that has restricted its growth is permanent storage. From floppy disks to refrigerator-sized hard drives, many attempts have been made to find the best possible way to store information for future generations.
Researchers from the University of Michigan and New York University have discovered a new method of storing data in microscopic particles suspended in a solution.
Just one tablespoon of liquid could replace your bulky external hard drive someday
According to this new wet information storage technique, nanoparticles suspended in liquid could be used to encode the same 1s and 0s stored on conventional hard drives. The simulations have shown that a 3% concentration of 12-nanoparticle clusters can store up to 1TB of data.
The researchers used an analogy of a Rubik’s cube to illustrate how the storage works
Like different colors of the cube, nanoparticles are attached to a central sphere that could be twisted and turned in almost eight million unique states, which is the equivalent to 2.86 bytes of data.<|endoftext|>
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European Americans, especially white Anglo-Saxon Protestants, were given special privileges in aspects of citizenship, land acquisition, immigration , education, and criminal procedure as from the midth century to the s. Racially and ethnically structured institutions that have manifested racism are:.
In the midth century, formal structures that propelled racial discrimination were primarily abolished and deemed as socially unacceptable as expressed in this racism essay and other publications. Socioeconomic inequality is the primary manifestation of modern day racism as stratification prevails in education, employment, lending, housing, and government. In the modern days, the narrative continues in the criminal justice system as demonstrated through racial disparities and racial profiling in sentencing and executions.
The conflict between this group and the justice system shows high levels of discrimination, oppression, and injustice of the minority, especially the African American men. Several events have changed race relations between the two, especially law enforcement. The first census in colonial Virginia in showed that the population comprised of Europeans, four Native Americans, and thirty-two Africans. Records from this period showed there were free African Americans and some indentured servants who could secure freedom.
Eastern Shore records of Virginia Northampton County between and showed that there were free African Americans. Historical essay on racism sheds light on the records that show how the oppression was institutionalized and used against African Americans in the United States since the first slave voyage arrived in Point Comfort, Virginia. Oppression and internal colonialism arose in as a means of maintaining domination over the people of African descent, by institutionalizing slavery through the legislative actions of the Virginia House of Burgesses.
After the Civil War and Reconstruction, violence broke out across the country and more than three thousand African Americans were killed between and This number accounts for the official records, and the actual figure of the causalities is impossible to establish and in some cases classified as justifiable homicides conducted by local law enforcement officials. The tipping point for the oppression was in marked by the violent murder of a year-old boy Emmett Till for talking to a white woman in a Mississippi store in the same year, starting the Civil Rights Movement.
Several thousand years before the colonial era, there had been inhabitants in the Northern American continent. They have influenced the history of America and racial relations as well. They lost their land through forced displacement, wars and the imposition of treaties often resulted in hardships. The English had enslaved approximately Choctaws in the early 18th century.
After the United States was established, the idea of removing Indians gained popularity. Those who remained faced official racism. In an essay about racism, the described situation portrayed desperation and hopelessness as they were forced to go through the turmoil of having their livelihoods shattered and personally abused.
Cobb depicted Choctaws as lacking nobility and virtue, and in some regard, he found native Africans more admirable and exciting, in every way. He deemed the Choctaw and Chickasaw, the native tribes he took an interest in, beneath contempt, even worse off than African slaves. There have been as well several debates pertaining to the issues of their sovereignty, upholding some treaty provisions, and their civil rights under US law. Asians were declared ineligible for citizenship by the Naturalization Act of , which limited citizenship to whites only.
Asian Americans have faced racism since the first group of Chinese immigrated to America to fill the labor shortages gap in rail and mining industries during the 19th century rapid industrialization. The First Transcontinental Railroad led to the massive immigration of Chinese laborers to fill the labor gap for the enormous project. The Chinese were majorly despised because they took lesser pay for jobs prescribed to the white.
The Yellow Peril, a phrase that described the possible doom of Western Civilization due to Chinese immigration, was also gaining popularity. It was the first passed law that excluded a significant group from the nation based on ethnicity and class. Perceived as a model minority in the modern time, they are imagined to be educated and successful.
As a result, they are stereotyped as hard-working and intelligent but inept socially. Expecting this treatment form the whites and other minorities, they have faced unreasonable expectations in workplaces as a result of this stereotype. Latino Americans categorized as Hispanic comprise individuals from a variety of racial and ethnic backgrounds.
In the post-Mexican—American War era in to , Mexicans living in the current Southwestern region were subjected to intense discrimination. Nearly Mexicans got lynched from that era till , and there might be more unreported cases. This statistic was second to the African Americans. The Mexican Repatriation program sponsored by the US government during the Great Depression was intended for voluntary return to Mexico, however, it turned into a forcible deportation agenda.
If you make up your mind to write your paper on bias in history, so probably the topic of racism versus slavery will never be settled completely, and you can express your thoughts on it.
You can also consider why prejudice is more experienced in certain sports and who are more bent to it: One more interesting topic is why minorities have less access to healthcare and how it affects their physical and mental well-being. You can also deliberate on a matter of ending racism, what has already been done, and what efforts still have to be made. The main goal of a persuasive essay is to convince a reader that your point of view is right.
When picking a topic for persuasive essay on racism, you will have to look for the most controversial ones. There should be two sides at least, so you will be able to choose the one you believe in or the one you have enough evidence to support. For instance, you can write on negative effects of stereotyping. Blacks are usually considered to be criminals, drug dealers living in ghettos. As a result, they are arrested twice often than white ones.
The treatment of African Americans as being unintelligent and lazy causes huge problems when they are willing to enroll at a college or apply for a decent job. The opinion that almost all Arabs are terrorists makes police check Arab-looking people in airports much more attentively than people of any other race. There are many other stereotypes that can support your point of view on this topic.
If these pieces of advice did not bring you somewhat closer to writing an excellent paper, get some persuasive essay help from us. The problem of discrimination gives a wide range of topics for writing argumentative essays. Generally, you will have to give explanation and convince the readers why racial bias is a bad thing. Thus, to write an argumentative essay, you will have to conduct a good research, it can be some statistical or historical facts that can support your point of view on the issue.
One of the most popular questions on discrimination are whether it still exists in our society and how to overcome it. Some people claim that there no discrimination anymore, but why then we encounter it so often? It is also a good idea to consider the Christian approach to the problem.
The skin shade that what is on the outside has nothing to do with the inferior of a person. However, not everyone practices the holy writing. Thus, in your essay you may encounter ways how to overcome the prejudice. The first one is the necessity not to ignore the situation when it happens, but to help a victim or disrupt a conversation if you hear someone uses racial slurs.
Then try to participate in anti-racial community events not violent protests, but affirmative actions and racist speech bans. One more step is to vote for those the candidates who support ending racism policy. Of course, there are many other ways, which you can mention in your essays.
As it was already mentioned, media plays a crucial role in portraying racial stereotypes. To eradicate racism from a society, actions should be taken to eradicate it from media first.
You may consider racism, for instance, on television, in network news, in commercials or in prime time ads. According to the surveys, despite Latinos constitute one-tenth of U. Moreover, their portrayal is hardly positive, which causes negative attitude of people, especially if these are children or people who rarely encounter Latinos towards them.
The same thing is with the most of network news and prime time programs. The overall majority of news anchors, actors, musicians is whites. The minorities appear seldom on the screen.
Usually, they are involved in natural disasters, foreign affairs, and crime topics. All in all, the news coverage is designated most of the times for white people and their interests. Anti-racism is a rather broad subject, so you will have to concentrate on the aspects that are more interesting for you. For example, you may ground your essay on comparing different types of anti-racism or vice versa focus on the one you want to cover.
Among different types of anti-racism, you may consider everyday anti-racism, multicultural anti-racism, psychological anti-racism, radical anti-racism, and anti-nazi anti-racism. All of them have their own particularities, like the psychological one deals with individual challenging of prejudice, the radical claims that socio-economic powers and privileges mainly promote racial bias. One more good topic for writing an essay on is the Civil Rights movement in the U.
Or you may consider what is the non-violent resistance and does it really help in suppressing racism. When writing any essay, including an essay on racial bias, one should use various hooks. What is a hook? A hook is a sentence or a fact that can attract attention of a reader.
It is always a good way to start your essay with a real-life story or situation that encouraged you to write on this particular topic. One more useful hook is to somehow connect your essay with a popular culture. And, of course, your essay will grab attention if you can suggest any ideas how to solve the problem of racism. Keep in mind that the more sophisticated the methods you invent are, the better your essay will look like.
Australia is known as ethnically and culturally diverse country, which seemed to have no problems with discrimination and inequality. In practice, over 6 million people from the whole world have come to Australia since It had a huge impact on all spheres of Australian society.
And, of course, it had a negative impact on Aboriginal people. One more group of people who face racist harassment is non-English speaking ones. People who have bad skills of language are less likely to get a good employment. Even if people who speak English fluently and comprehend it well, they can be discriminated for having accent.
Currently, among different ethnos and cultures, Arabs, Muslims and African Australians are those who are highly subjected to racism. Thus, it can be concluded that cultural diversity of a country does not guarantee that it is free of prejudice and racial slurs. When writing an essay on racial intolerance in Pakistan, it is essential to mention the long-lasting animosity between Pakistan and India, since these two countries have been involved in 3 wars for the Kashmir region and one indirect war for the Bangladesh independency.
They caused racial intolerance between these South Asian countries. One more thing that should be taken into account is that unlike India, where communalism prevails, Pakistan has many regions which are diverse due to their ethnicity, language, and religion. All of these regions aim for autonomy and are set against the dominant power. The ruling class of the society is Punjabi, which takes the governmental positions. The interesting fact about Pakistan inequality is that linguistic discrimination is even more severe than the regional discrimination.
Canada has a reputation of an extremely loyal society where every ethos and race is welcomed. Nevertheless, this multicultural country, which seemed to be absolutely free of xenophobia, still has huge discrimination issues. The irony of it is that Canadian government and average Canadians are hospitable to foreigners and migrants but immensely hostile to the First Nations people.
During a long period of time, the government has been taking steps to control all spheres of life of Aboriginal people: First Nations females particularly undergo the most severe persecutions and assaults, which results in hundreds of homicides each year. Different physical appearance is usually the main cause of prejudice and in case of Canada, it is the reason for long-lasting racism. Since the very beginning and up to now, the United States of America has faced the problem of color discrimination.
The first people who have experienced prejudice of European colonizers were Indians, also known as Native Americans. The occupation of their territories led to segregation, separate reservations, wars, and slavery. The next wave of racism in the USA started with the introduction of African slavery. Despite the official prohibition of racial bias and eager activity of Martin Luther King Jr in the middle of the previous century, this problem is still vexed. Even the election of the African American president has not exterminated the discrimination of black people.
The reason why the problem still exists in modern society lies in wrong upbringing of children with imposing misleading stereotypes on them. To reduce or end racism, new curriculum of developing respect and understanding towards ethnically diverse people has been introduced into educational programs of all levels. Though over 2 decades have passed since apartheid was officially abolished, the problem has not disappeared. The displays of racial intolerance come both from whites and from blacks.
However, racism is the manifestation of hatred against people of different ethnic origins. For example, over the last decade, more than 3 thousand white farmers have been murdered in South Africa. The cruelty of these murders is shocking. The victims, including females and children, are tortured before being killed. On the other hand, whites also foment hatred by occasional attacks on blacks.
They, for instance, try to restrict them from visiting restaurants and landlords, which cause a new wave of discrimination and vicious insults. Since apartheid has had a long history in South Africa, it probably will take long to eradicate it. Though New Zealand is regarded as a multicultural country the main ethnic groups are European, Maori, Pacific people and Asian , it is regarded as constitutionally bicultural society, thus the main conflict lies between Maori, indigenous inhabitants, and Pakeha people, European descendants.
In comparison to Pakeha, which is considered to be the foremost class of the society, Maori experience lack of medical insurance, lack of learning opportunities and low access to high-paid jobs. Also, there is such a problem as inequality in general income between Pakeha and Maori, and lower life expectancy of the latter.
Another topic for writing an essay on is whether the current population of the country can be considered as New Zealanders. Many claim that it can destroy Maori cultural identity and lead to the deprivation of their rights. Though Lebanon is known as a liberal country due to the vast freedom enjoyed by its female population, and the country attracts visitors with its archeological remnants, good weather and tasty food, the level of racism towards foreign workers grows rapidly.
Most workers come from Nepal, Philippines, Ethiopia and Sri Lanka, usually these are females working as maids and nannies for wealthy Lebanese families. They, as well as dark-skinned men, experience vicious racism and are treated badly by their employers. For example, they are barred from visiting beaches and swimming pools, both private and public ones, and from Lebanese clubs.
The most blatant racism is experienced by Palestinians, who are not allowed to work at almost every job possible. Ethnically-based discrimination in Lebanon is practiced since Lebanese consider migrants only as a cheap labor. The history of Puerto Rico started in early 16 th century when European colonizers have settled on its territory. For a long period of time, particularly for more than 4 hundred years, Puerto Rico has been a possession of Spain and it contributed a lot to the emergency of the so-called mix-race.
The current population of Puerto Rico is primarily the descendants of Native Tainos, Spanish colonizers, and Africans who were brought here as slaves.
At the same time, the U. The most rampant racism has been experienced by Africans, Native Americans, and their posterities.
For instance, blacks have never been elected as governors or senators despite the fact that current president of the U. They are also less likely to get well-paid professions and consequently jobs, which are reserved for white citizens. It becomes obvious that though Puerto Ricans claim they have no prejudice in their country, these examples evidence that it still takes place in the society.
Russian federation is the largest country in the world, which consists of numerous autonomous federations and numerous ethnic groups, but at the same time, it is well known for its racist acts. For example, based on the data of the Sova Centre, in , more than 20 people were murdered and over were injured as a result of racist and neo-Nazi attacks in Moscow.
Among the victims were primarily Central Asia and Caucasus republics natives, dark-skinned and Chinese. Thus, blacks and other ethnically diverse minorities are often warned from coming to study or even to visit Russia.
There are also nationalist organizations in the country, which include thousands of members, who regularly train in gyms or forests. One more thing that reaffirms negative concern to foreigners is the refusal to shelter Syrian refugees and accusing Western European countries of accepting people from those parts of the world. Sport is an important part of our society. Millions of people have their favorite teams and players and watch games all the time.
That is why, whatever happens on the playing field, especially if it is an incident that displays discrimination towards colored players, cannot be left without proper attention. The issue of racism in sports is an old but still hotly disputed problem. They claim that racism has already been conquered, but in reality, colored players are challenged greatly. Though a lot of black sportsmen, known as African Americans, be they individual athletes or team players, have already proved that they can succeed at any sport, they are still discriminated to the present day.
The history gives us many examples of African American black legends, starting from Jesse Owens who has won 4 gold medals during the Berlin Olympics in Unfortunately, the display of racial intolerance still remains a tendency in some sports.
If you write about racism in hockey, we advise you to mention that the majority of hockey players has always been white. Though the hockey leagues, like NHL league, try to change the situation by increasing the number of ethnically diverse players, the issue of racism in the leagues gets more serious as well.
One of the spectators threw a banana at Simmonds while the player was making a shootout attempt. This will also have a positive effect on the younger generation that still doubts whether they should go in for hockey or not.
Football is one the worldwide popular sports, and at the same time, it is considered to be the favorite sports of roughnecks and cads. As a result, the incidents of racist behavior happen on the field here and there.
In most cases, such misconduct is shown by football fans, who often show their racist attitude. According to the surveys, the most ignorant football fans are those from Spain, Italy, Holland, and from Eastern-European countries. They display their intolerance towards ethnically diverse players in different ways: One of the most outrageous displays of racism happened in December , when the Liverpool striker, Luis Suarez, abused Patrice Evra of Manchester United during the game.
For such unacceptable behavior, Suarez received eight-match suspension and a fine, which is just one quarter of his monthly salary. Most experts believe that only more severe punishments for such acts can help to avoid intolerance on the soccer fields. It seems that soccer and racism are closely interrelated.
For example, Euro hosted by Poland and Ukraine, was overflowed with rampant racism. More than one-quarter of all federations of countries participating in the event, primarily Russian, Spanish and Croatian soccer federations, were fined for the uncontrolled racist behavior of their fans during the games. The German federation particularly was fined since its fans exhibited neo-Nazi symbols when their team won a game with Denmark. Even if the most loyal and tolerant countries, like England and Germany, face the problem of racism in sports, then prompt steps should be taken in order to change such a plaguy situation in our community.
Essay on racism Racism is a long-lasting problem that bothers millions of people all over the world. Essay on racism and discrimination When writing an essay on racism and discrimination, first of all, it is necessary to differentiate these two terms and give definitions to both. Racism in the workplace essay Discrimination in the workplace is a current issue since it takes place here and there.
Anti-racism essay Anti-racism is a set of beliefs, policies, and movements that emerged as a response to racism in order to create an egalitarian society where all people could be equal in their rights. Media and racism essay When writing an essay on racial discrimination in media, one may start with a general statement saying that in the 21 st century, people have a regular access to different media sources: Racism in advertising essay Advertisement is something we can hardly imagine our lives without.
Causes and effects of racism essay Racism has always been a delicate subject as it refers to the skin color. Argumentative essays on racism An argumentative essay differs from other types as you cannot rewrite what has already been said hundreds of times, but you will have to do a good research and start writing as soon as all additional information has been accumulated. Short essay on racism Teachers and professors often assign short essays to their students.
Structure Thesis statement about racism Thesis statement is an argumentative sentence that reveals the main idea of your paper. Introduction for racism essay To write a stellar essay, you will have to pay great attention to the introduction. Conclusion for racism essay Conclusion is an important part of an essay, thus a student should pay attention to it.
Racism in America essay outline Before starting to write an essay, you need to make an outline. Good topics for Racism essays Essay about racism in schools Rampant racism is experienced in school in all forms: Racism topics for research paper Research paper is a rather complicated assignment. Topics about racism for essays Discrimination gives a wide range of topics to be discussed. Persuasive essay topics on racism The main goal of a persuasive essay is to convince a reader that your point of view is right.
Argumentative essay topics about racism The problem of discrimination gives a wide range of topics for writing argumentative essays. No racism essay topics One of the most popular questions on discrimination are whether it still exists in our society and how to overcome it. Media and racism essay titles As it was already mentioned, media plays a crucial role in portraying racial stereotypes. Racism essay hooks When writing any essay, including an essay on racial bias, one should use various hooks.
Essays of Racism in Different countries Essay on racism in Australia Australia is known as ethnically and culturally diverse country, which seemed to have no problems with discrimination and inequality. Essay on racism in Pakistan When writing an essay on racial intolerance in Pakistan, it is essential to mention the long-lasting animosity between Pakistan and India, since these two countries have been involved in 3 wars for the Kashmir region and one indirect war for the Bangladesh independency.
Aboriginal racism in Canada essay Canada has a reputation of an extremely loyal society where every ethos and race is welcomed. Racism in the United States essay Since the very beginning and up to now, the United States of America has faced the problem of color discrimination. Racism in Lebanon essay Though Lebanon is known as a liberal country due to the vast freedom enjoyed by its female population, and the country attracts visitors with its archeological remnants, good weather and tasty food, the level of racism towards foreign workers grows rapidly.
Racism in Puerto Rico essay The history of Puerto Rico started in early 16 th century when European colonizers have settled on its territory.
Racism has occurred for centuries, but there is still a chance to end it. I believe racism should end because if it continues, it can split the unity of races. It may .
"Conclusions On Racism" Essays and Research Papers Conclusions On Racism OUTLINE ON RACISM • Racism is the belief that a particular race is superior to another.
Essay Racism in America Today - “Racism can be defined as beliefs, attitudes, actions or behaviors that are based on phenotypic characteristics or ethnic affiliation”(Patcher, ). Racism is a struggle that, to this day, has continued to be a major issue. RACISM ESSAY Racism is one of the world's major issues today. Many people are not aware of how much racism still exists in our schools workforces, and anywhere else where social lives are occurring. It is obvious that racism is bad as it was many decades ago but it sure has not gone away.
Conclusion. As brought out in this racism in America essay, racism has been at the center stage throughout the history and development of the United States. Efforts put in the hope of eradicating racism are futile as there are interest groups that seem to counter these efforts or install new mechanisms to drive the racism agenda for particular gains. Racism in Essays. RACISM IN ESSAYS Is an author’s main purpose of writing only to entertain his readers? Authors sometimes use their literature to demonstrate their opinions about a certain issue.<|endoftext|>
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The story of the ill-fated Franklin Expedition remains one of the most mysterious in the history of maritime exploration. Two British Royal Navy ships, HMS Erebus and HMS Terror, left England in 1845 for an exploratory sailing expedition through the Canadian arctic. Soon after reaching the icy waters of the Northwest Passage in 1845, the ships became icebound. No word was heard nor trace of the ships found until later expeditions discovered evidence enabling them to piece together the grisly end to the entire 129-man crew.
Their ships would remain undiscovered for the next 170 years. The Erebus was discovered in 2014 by Canadian parks workers who were able to recover the ship’s bell, but the fate of the Erebus’ sister ship remained unknown. Now, the story of the Franklin Expedition might get a little more complete thanks to the discovery of the long-lost HMS Terror.
According to Radio Canada International (RCI), the Arctic Research Foundation and Canadian Royal Navy discovered HMS Terror on September 3, 2016 after being led to the wreck by an Inuk crew member familiar with the remote arctic landscape. The ship was found in Terror Bay (coincidentally) near the remote King William Island in Canada’s far north. Researchers still cannot explain why Terror was found 100 kilometers (60 miles) north of where HMS Erebus was discovered.
Ryan Harris, underwater archeologist with Parks Canada, told RCI that the ship’s location adds to the mystery of the doomed Franklin Expedition:
Obviously the first question is how did get through Alexander Strait? Was it indeed remanned at some point as what appear to be the case with Erebus?
Based on notes and anecdotes from local Inuit tribes, explorers and historians have been able to piece together the last months of the stranded crew’s fate. After exhausting their ships of supplies, the men tried to establish camps on nearby islands in September 1846, but began dying of starvation and disease. Reports and eyewitness accounts given by local Inuit communities revealed that the crew might have resorted to cannibalism in their desperate state. According to accounts Inuits gave a British search party in 1848, butchered corpses and cookware containing human remains were found on islands not far from the shipwrecks. One explorer, John Rae, gathered these Inuit accounts in a letter sent to his patron lords in England:
From the mutilated state of many of the corpses and the contents of the kettles, it is evident that our wretched countrymen had been driven to the last resource – cannibalism – as a means of prolonging existence.
How widespread this cannibalism might have been or at what point it began remains unknown. A desperate attempt to walk towards civilization took the lives of the remaining men in April 1848.<|endoftext|>
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# How To Analyze More Complex Series Parallel Circuits
Updated on October 5, 2014
## Theory
In the last Hub, we talked about how to figure out the equivalent resistance of resistors in parallel, and in the Hub before that, we learned how to deal with resistors in series. In this Hub, we'll combine those techniques to deal with circuits that have both series and parallel combinations of resistors, whether they be series of parallel resistors, parallel combinations of series of resistors, or multiple combinations of the two. This can be done, as we'll see, through repeated application of Ohm's Law.
One method of working your way through the finding of the voltage across and current through each resistor in a complex circuit is to work your way "out" from the innermost series or parallel parts of the circuit, substituting equivalent resistances as you go, until you have simplified the circuit into a single resistor. Now using Ohm's Law, you can find the total current draw IT of the circuit. Then, use IT and the circuit as it looked with the equivalent resistances right before you simplified it to just a single resistor, and so on until you've worked the circuit in full. It helps to reduce the amount of recalculation you have to do by using a spreadsheet program. Having a spreadsheet is great because you can change any resistor's value and all the currents and voltages will be automatically recalculated. I'll show an example of using this technique to solve a circuit now.
## An Example Circuit
Shown below is an example of a circuit with series and parallel elements. R1 and R2 form a series, as do R3 and R4. These two series are in parallel with each other, and this parallel subcircuit is in series with resistors R5 and R6.
If you've got a spreadsheet program on your computer, I encourage you to follow along. Learning by doing is usually better than just passive reading!
This circuit has six resistors and one voltage source. Let's put labels for the resistors R1-R6 in cells A1-A6 of the spreadsheet, leave cells B1-B6 for those values, put a label for the voltage source V in C1, and leave D1 for the actual voltage value. You can put in some random numbers for the resistance and voltage values (you can always change them later) so that when we set other cells for equivalent resistances and currents and voltages with formulas dependent upon these cells, we see some numbers instead of just ### or whatever your spreadsheet program shows when it can't resolve a cell's value. Shown below is my spreadsheet so far.
Now let's analyze the schematic. Resistors R1 and R2 are in series, so a single resistor called, say, Ra could be substituted where those two resistors are and it would have a resistance value equal to R1 + R2. So let's put at cell A7 the label RA and in cell B7 let's type in the equation '=B1+B2' (without the quotes, of course).
Similarly, looking at the schematic, we could simplify R3 and R4 to a single resistor called Rb with value R3 + R4. We put RB at cell A8 and '=B3+B4' at cell B8. Now the spreadsheet looks something like the image below.
Now we can picture the two single resistors Ra and Rb in parallel with each other. What value, say we labeled it Rc, would a single resistor representing the two resistors in parallel have? If you remember from the last Hub, the formula for the equivalent resistance of two resistors in parallel is the product over the sum. So we could let cell A9 have the label RC and let cell B9 have the formula '=(B7*B8) / (B7 + B8)'. Now the spreadsheet looks like this:
Now Rc is just in series with resistors R5 and R6, so we can create a cell labeled RD at A10 and its value '=B9+B5+B6' at B10, which would look like this:
## Work Back Down For The Currents
Now that we've found the total resistance of the circuit, we can work backwards to find the currents in each branch. First, let's find out the total current IT drawn from the battery. Using Ohm's Law, we divide voltage V by resistance Rd. Since this is the current flowing through hypothetical equivalent resistor Rd, we can label the current Id, place the label ID in cell C10, and place the value '=D1/B10' in cell D10, giving the following:
To find out how much current is flowing through the branch of Ra and how much is flowing through the branch of Rb, we can use a principle known as Kirchhoff's Current Law (KCL), discovered by German physicist Gustav Kirchhoff, which effectively states that the net current through a node is zero. In other words, what goes in must come out, and what comes out, must have come in. If X amps go out of the node through branch D, then the sum of currents coming into the node through branches A and B must be X.
How is Id divvied up into Ia and Ib? With a two-branch setup like we have, if we want to know how much current is flowing through branch A, i.e., Ia, we divide Rb by the sum of Ra and Rb and multiply the quotient by the total current Id. To find the current Ib flowing through branch B, we perform the same operation but with Ra in the numerator where Rb was used before.
So for IA we give the value '=(B8/(B7+B8))*D10' and for IB we give the value '=(B7/(B7+B8))*D10' as shown below:
## Voltage = Current * Resistance
Now that we have the branch currents for all the real branches in the circuit and the resistance of each resistor, we can use Ohm's Law in voltage solution mode (V = I * R) to find the voltage across each resistor.
R1 is on branch a with current Ia, so V1 = R1 * IA. V2 = R1 * IA. V3 = R3 * IB. V4 = R4 * IB. R5 and R6 are on the branch with current Id, so V5 = R5 * ID and V6 = R6 * ID. This is shown on the spreadsheet below:
Notice that the net voltages in any closed loop through the circuit add up to zero. For instance, starting at the positive terminal of the battery with voltage V and dropping the voltages of V1 and V2 and V5 and V6, or in other words, we have V - V1 - V2 - V5 - V6 = 0. If we loop using the other branch, we get V - V2 - V3 - V5 - V6 = 0. This is known as Kirchhoff's Voltage Law. I encourage you to learn more about that! In my next Hub, we'll look at interesting circuits that can't be broken down using only series and parallel techniques. Thanks for reading!
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# 1 4 Practice Solving Equations Algebra 2
## Inverse Relationships With Two Blanks
Algebra 2: 1.4 Solving Equations
## Linear Equation In Three Variables
A linear equation with three variables, where a, b, c, and d are real numbers and a, b, and c are not all 0, is of the form
Every solution to the equation is an ordered triple, that makes the equation true.
All the points that are solutions to one equation form a plane in three-dimensional space. And, by finding what the planes have in common, weâll find the solution to the system.
When we solve a system of three linear equations represented by a graph of three planes in space, there are three possible cases.
To solve a system of three linear equations, we want to find the values of the variables that are solutions to all three equations. In other words, we are looking for the ordered triple that makes all three equations true. These are called the solutions of the system of three linear equations with three variables.
## Inverse Relationships With One Blank
Inverse relationships worksheets cover a pre-algebra skill meant to help students understand the relationship between multiplication and division and the relationship between addition and subtraction.
## Solutions Of A System Of Linear Equations With Three Variables
Solutions of a system of equations are the values of the variables that make all the equations true. A solution is represented by an ordered triple .
To determine if an ordered triple is a solution to a system of three equations, we substitute the values of the variables into each equation. If the ordered triple makes all three equations true, it is a solution to the system.
## Factoring Expressions That Do Not Include A Squared Variable
Recommended Reading: What Is The Molecular Geometry Of Ccl4
## Determine Whether An Ordered Triple Is A Solution Of A System Of Three Linear Equations With Three Variables
In this section, we will extend our work of solving a system of linear equations. So far we have worked with systems of equations with two equations and two variables. Now we will work with systems of three equations with three variables. But first let’s review what we already know about solving equations and systems involving up to two variables.
We learned earlier that the graph of a linear equation, a is a line. Each point on the line, an ordered pair , is a solution to the equation. For a system of two equations with two variables, we graph two lines. Then we can see that all the points that are solutions to each equation form a line. And, by finding what the lines have in common, weâll find the solution to the system.
Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions
We know when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.
Similarly, for a linear equation with three variables a every solution to the equation is an ordered triple, ( , that makes the equation true.
## Multiplying Factors Of Quadratic Expressions
College Algebra 1.4 Quadratic Equations, Part 2
Read Also: Paris Jackson’s Biological Father
## Solve A System Of Linear Equations With Three Variables
To solve a system of linear equations with three variables, we basically use the same techniques we used with systems that had two variables. We start with two pairs of equations and in each pair we eliminate the same variable. This will then give us a system of equations with only two variables and then we know how to solve that system!
Next, we use the values of the two variables we just found to go back to the original equation and find the third variable. We write our answer as an ordered triple and then check our results.
Read Also: Structural Formula For Ccl4
## Factoring Expressions That Always Include A Squared Variable
Also Check: Geometry Segment Addition Postulate Worksheet
## Missing Numbers Worksheets With Blanks As Unknowns
In these worksheets, the unknown is limited to the question side of the equation which could be on the left or the right of equal sign.
## How To
#### Solve a system of linear equations with three variables.
• Step 1. Write the equations in standard form
• If any coefficients are fractions, clear them.
• Step 2. Eliminate the same variable from two equations.
• Work with a pair of equations to eliminate the chosen variable.
• Multiply one or both equations so that the coefficients of that variable are opposites.
• Add the equations resulting from Step 2 to eliminate one variable
• Step 3. Repeat Step 2 using two other equations and eliminate the same variable as in Step 2.
• Step 4. The two new equations form a system of two equations with two variables. Solve this system.
• Step 5. Use the values of the two variables found in Step 4 to find the third variable.
• Step 6. Write the solution as an ordered triple.
• Step 7. Check that the ordered triple is a solution to all three original equations.
• Also Check: Eoc Algebra 1 Practice Test With Answers 2015<|endoftext|>
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Contributor: Erika Wargo. Lesson ID: 12538
Can you classify a quadrilateral? Can you define a quadrilateral? Can you find a quadrilateral? You'll be able to easily answer all these because they are common shapes! Don't be a square; read on!
categories
## Plane Geometry (2D)
subject
Math
learning style
Visual
personality style
Otter, Golden Retriever
Intermediate (3-5)
Lesson Type
Skill Sharpener
## Lesson Plan - Get It!
Audio:
• Do you see quadrilaterals around you every day?
• How many do you think are in the picture above?
• What makes a shape a quadrilateral?
Polygons with four sides are called quadrilaterals.
Parallelograms, rhombuses, trapezoids, squares, and rectangles are quadrilaterals. On a piece of paper, draw an example of these shapes and describe the properties of each shape out loud.
Quad means four and lateral means side. Quadrilaterals have these special properties:
• four sides and four corners
• 2-dimensional (flat shape)
• straight sides
• closed shape
Quadrilaterals can also have parallel sides — sides that stay the same distance apart and never cross each other. The interior, or inside, angles add up to 360 degrees.
As you watch Math Antics - Quadrilaterals, from mathantics, write down one special property about each quadrilateral:
• square
• rectangle
• rhombus
• parallelogram
• trapezoid
Next, review this quadrilateral chart so you can become more familiar with each shape:
Example 1
Name the quadrilateral and explain your reasoning:
This shape has four sides and four right angles, which are 90 degrees each. The opposite sides are parallel and of equal length, so this quadrilateral is a rectangle.
The tick marks on the sides of the rectangle help you see that the opposite sides are equal. The sides with one tick mark mean that those two sides are equal length and the same is true for the opposite sides with two tick marks, even though the actual measurements are not given.
Example 2
The following shapes are not quadrilaterals. Explain why they are not quadrilaterals:
Quadrilaterals have only four sides and have straight sides. The first figure has more than four sides and is three-dimensional. The second figure has more than four sides. The third figure does not have straight sides.
• Can a quadrilateral have 5 sides? Why or why not?
• What special features are used to classify a quadrilateral?
In the Got It? section, you will play games and complete interactive practice to classify quadrilaterals.
Interactive Video<|endoftext|>
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19320 in words
19320 in words is written as Nineteen Thousand Three Hundred and Twenty. In 19320, 1 has a place value of ten thousand, 9 is in the place value of thousand, 3 is in the place value of hundred and 2 is in the place value of ten. The article on Place Value gives more information. The number 19320 is used in expressions that relate to money, distance, social media views, and many more. For example, “My car has covered Nineteen Thousand Three Hundred and Twenty kilometers and is due for service.”
19320 in words Nineteen Thousand Three Hundred and Twenty Nineteen Thousand Three Hundred and Twenty in Numbers 19320
How to Write 19320 in Words?
We can convert 19320 to words using a place value chart. The number 19320 has 5 digits, so let’s make a chart that shows the place value up to 5 digits.
Ten thousand Thousands Hundreds Tens Ones 1 9 3 2 0
Thus, we can write the expanded form as:
1 × Ten thousand + 9 × Thousand + 3 × Hundred + 2 × Ten + 0 × One
= 1 × 10000 + 9 × 1000 + 3 × 100 + 2 × 10 + 0 × 1
= 19320.
= Nineteen Thousand Three Hundred and Twenty.
19320 is the natural number that is succeeded by 19319 and preceded by 19321.
19320 in words – Nineteen Thousand Three Hundred and Twenty.
Is 19320 an odd number? – No.
Is 19320 an even number? – Yes.
Is 19320 a perfect square number? – No.
Is 19320 a perfect cube number? – No.
Is 19320 a prime number? – No.
Is 19320 a composite number? – Yes.
Solved Example
1. Write the number 19320 in expanded form
Solution: 1 x 10000 + 9 x 1000 + 3 x 100 + 2 x 10 + 0 x 1
We can write 19320 = 10000 + 9000 + 300 + 20 + 0
= 1 x 10000 + 9 x 1000 + 3 x 100 + 2 x 10 + 0 x 1.
Frequently Asked Questions on 19320 in words
Q1
How to write the number 19320 in words?
19320 in words is written as Nineteen Thousand Three Hundred and Twenty.
Q2
Is 19320 a prime number?
No. 19320 is not a prime number.
Q3
Is 19320 divisible by 10?
Yes. 19320 is divisible by 10.<|endoftext|>
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Beowulf (//; Old English: [ˈbeːo̯wulf]) is an Old English epic poem consisting of 3,182 alliterative lines. It is one of the most important works of Old English literature. The date of composition is a matter of contention among scholars; the only certain dating pertains to the manuscript, which was produced between 975 and 1025. The author was an anonymous Anglo-Saxon poet, referred to by scholars as the "Beowulf poet".
First page of Beowulf in Cotton Vitellius A. xv
|Language||West Saxon dialect of Old English|
|Date||c. 700–1000 AD (date of story), c. 975–1010 AD (date of manuscript)|
|State of existence||Manuscript suffered damage from fire in 1731|
|Manuscript(s)||Cotton Vitellius A. xv|
|First printed edition||Thorkelin (1815)|
|Genre||Epic heroic writing|
|Verse form||Alliterative verse|
|Length||c. 3182 lines|
|Subject||The battles of Beowulf, the Geatish hero, in youth and old age|
|Personages||Beowulf, Hygelac, Hrothgar, Wealhþeow, Hrothulf, Æschere, Unferth, Grendel, Grendel's mother, Wiglaf, Hildeburh.|
The story is set in Scandinavia. Beowulf, a hero of the Geats, comes to the aid of Hrothgar, the king of the Danes, whose mead hall in Heorot has been under attack by a monster known as Grendel. After Beowulf slays him, Grendel's mother attacks the hall and is then also defeated. Victorious, Beowulf goes home to Geatland (Götaland in modern Sweden) and later becomes king of the Geats. After a period of fifty years has passed, Beowulf defeats a dragon, but is mortally wounded in the battle. After his death, his attendants cremate his body and erect a tower on a headland in his memory.
The full story survives in the manuscript known as the Nowell Codex. It has no title in the original manuscript, but has become known by the name of the story's protagonist. In 1731, the manuscript was badly damaged by a fire that swept through Ashburnham House in London that had a collection of medieval manuscripts assembled by Sir Robert Bruce Cotton. The Nowell Codex is currently housed in the British Library.
The events in the poem take place over most of the sixth century, after the Anglo-Saxons had started migrating to England and before the beginning of the seventh century, a time when the Anglo-Saxons were either newly arrived or were still in close contact with their Germanic kinsmen in Northern Germany and southern Scandinavia. The stories in the poem may have been brought to England by people of Geatish origins. Some suggest that Beowulf was first composed in the 7th century at Rendlesham in East Anglia, as the Sutton Hoo ship-burial shows close connections with Scandinavia, and the East Anglian royal dynasty, the Wuffingas, may have been descendants of the Geatish Wulfings. Others have associated this poem with the court of King Alfred the Great or with the court of King Cnut the Great.
The poem blends fictional, legendary and historic elements. Although Beowulf himself is not mentioned in any other Anglo-Saxon manuscript, scholars generally agree that many of the other figures referred to in Beowulf also appear in Scandinavian sources. (Specific works are designated in the following section). This concerns not only individuals (e.g., Healfdene, Hroðgar, Halga, Hroðulf, Eadgils and Ohthere), but also clans (e.g., Scyldings, Scylfings and Wulfings) and certain events (e.g., the Battle on the Ice of Lake Vänern). The raid by King Hygelac into Frisia is mentioned by Gregory of Tours in his History of the Franks and can be dated to around 521.
In Denmark, recent archaeological excavations at Lejre, where Scandinavian tradition located the seat of the Scyldings, i.e., Heorot, have revealed that a hall was built in the mid-6th century, exactly the time period of Beowulf. Three halls, each about 50 metres (160 ft) long, were found during the excavation.
The majority view appears to be that people such as King Hroðgar and the Scyldings in Beowulf are based on historical people from 6th-century Scandinavia. Like the Finnesburg Fragment and several shorter surviving poems, Beowulf has consequently been used as a source of information about Scandinavian figures such as Eadgils and Hygelac, and about continental Germanic figures such as Offa, king of the continental Angles.
19th-century archaeological evidence may confirm elements of the Beowulf story. Eadgils was buried at Uppsala according to Snorri Sturluson. When the western mound (to the left in the photo) was excavated in 1874, the finds showed that a powerful man was buried in a large barrow, c. 575, on a bear skin with two dogs and rich grave offerings. The eastern mound was excavated in 1854, and contained the remains of a woman, or a woman and a young man. The middle barrow has not been excavated.
The protagonist Beowulf, a hero of the Geats, comes to the aid of Hrothgar, king of the Danes, whose great hall, Heorot, is plagued by the monster Grendel. Beowulf kills Grendel with his bare hands and Grendel's mother with a giant's sword that he found in her lair.
Later in his life, Beowulf becomes king of the Geats, and finds his realm terrorized by a dragon, some of whose treasure had been stolen from his hoard in a burial mound. He attacks the dragon with the help of his thegns or servants, but they do not succeed. Beowulf decides to follow the dragon to its lair at Earnanæs, but only his young Swedish relative Wiglaf, whose name means "remnant of valour",[a] dares to join him. Beowulf finally slays the dragon, but is mortally wounded in the struggle. He is cremated and a burial mound by the sea is erected in his honour.
Beowulf is considered an epic poem in that the main character is a hero who travels great distances to prove his strength at impossible odds against supernatural demons and beasts. The poem also begins in medias res or simply, "in the middle of things," which is a characteristic of the epics of antiquity. Although the poem begins with Beowulf's arrival, Grendel's attacks have been an ongoing event. An elaborate history of characters and their lineages is spoken of, as well as their interactions with each other, debts owed and repaid, and deeds of valour. The warriors form a kind of brotherhood linked by loyalty to their lord. What is unique about "Beowulf" is that the poem actually begins and ends with a funeral. At the beginning of the poem, the king, hero, Shield Shiefson dies (26–45) and there is a huge funeral for him. At the end of the poem when Beowulf dies, there is also a massive funeral for Beowulf (3140–3170).
First battle: GrendelEdit
Beowulf begins with the story of Hrothgar, who constructed the great hall Heorot for himself and his warriors. In it, he, his wife Wealhtheow, and his warriors spend their time singing and celebrating. Grendel, a troll-like monster said to be descended from the biblical Cain, is pained by the sounds of joy. Grendel attacks the hall and kills and devours many of Hrothgar's warriors while they sleep. Hrothgar and his people, helpless against Grendel, abandon Heorot.
Beowulf, a young warrior from Geatland, hears of Hrothgar's troubles and with his king's permission leaves his homeland to assist Hrothgar.
Beowulf and his men spend the night in Heorot. Beowulf refuses to use any weapon because he holds himself to be the equal of Grendel. When Grendel enters the hall, Beowulf, who has been feigning sleep, leaps up to clench Grendel's hand. Grendel and Beowulf battle each other violently. Beowulf's retainers draw their swords and rush to his aid, but their blades cannot pierce Grendel's skin. Finally, Beowulf tears Grendel's arm from his body at the shoulder and Grendel runs to his home in the marshes where he dies. Beowulf displays "the whole of Grendel's shoulder and arm, his awesome grasp" for all to see at Heorot. This display would fuel Grendel's mother's anger in revenge.
Second battle: Grendel's motherEdit
The next night, after celebrating Grendel's defeat, Hrothgar and his men sleep in Heorot. Grendel's mother, angry that her son has been killed, sets out to get revenge. "Beowulf was elsewhere. Earlier, after the award of treasure, The Geat had been given another lodging"; his assistance would be absent in this battle. Grendel's mother violently kills Æschere, who is Hrothgar's most loyal fighter, and escapes.
Hrothgar, Beowulf, and their men track Grendel's mother to her lair under a lake. Unferð, a warrior who had earlier challenged him, presents Beowulf with his sword Hrunting. After stipulating a number of conditions to Hrothgar in case of his death (including the taking in of his kinsmen and the inheritance by Unferth of Beowulf's estate), Beowulf jumps into the lake, and while harassed by water monsters gets to the bottom, where he finds a cavern. Grendel's mother pulls him in, and she and Beowulf engage in fierce combat.
At first, Grendel's mother appears to prevail, and Hrunting proves incapable of hurting the woman; she throws Beowulf to the ground and, sitting astride him, tries to kill him with a short sword, but Beowulf is saved by his armour. Beowulf spots another sword, hanging on the wall and apparently made for giants, and cuts her head off with it. Travelling further into Grendel's mother's lair, Beowulf discovers Grendel's corpse and severs his head with the sword, whose blade melts because of the "hot blood". Only the hilt remains. Beowulf swims back up to the rim of the pond where his men wait. Carrying the hilt of the sword and Grendel's head, he presents them to Hrothgar upon his return to Heorot. Hrothgar gives Beowulf many gifts, including the sword Nægling, his family's heirloom. The events prompt a long reflection by the king, sometimes referred to as "Hrothgar's sermon", in which he urges Beowulf to be wary of pride and to reward his thegns.
Third battle: The dragonEdit
Beowulf returns home and eventually becomes king of his own people. One day, fifty years after Beowulf's battle with Grendel's mother, a slave steals a golden cup from the lair of a dragon at Earnanæs. When the dragon sees that the cup has been stolen, it leaves its cave in a rage, burning everything in sight. Beowulf and his warriors come to fight the dragon, but Beowulf tells his men that he will fight the dragon alone and that they should wait on the barrow. Beowulf descends to do battle with the dragon, but finds himself outmatched. His men, upon seeing this and fearing for their lives, retreat into the woods. One of his men, Wiglaf, however, in great distress at Beowulf's plight, comes to his aid. The two slay the dragon, but Beowulf is mortally wounded. After Beowulf dies, Wiglaf remains by his side, grief-stricken. When the rest of the men finally return, Wiglaf bitterly admonishes them, blaming their cowardice for Beowulf's death. Afterward, Beowulf is ritually burned on a great pyre in Geatland while his people wail and mourn him, fearing that without him, the Geats are defenceless against attacks from surrounding tribes. Afterwards, a barrow, visible from the sea, is built in his memory (Beowulf lines 2712–3182).
Authorship and dateEdit
The dating of Beowulf has attracted considerable scholarly attention and opinion differs as to whether it was first written in the 8th century or whether the composition of the poem was nearly contemporary with its eleventh century manuscript and whether a proto-version of the poem (possibly a version of the Bear's Son Tale) was orally transmitted before being transcribed in its present form. Albert Lord felt strongly that the manuscript represents the transcription of a performance, though likely taken at more than one sitting. J. R. R. Tolkien believed that the poem retains too genuine a memory of Anglo-Saxon paganism to have been composed more than a few generations after the completion of the Christianisation of England around AD 700, and Tolkien's conviction that the poem dates to the 8th century has been defended by Tom Shippey, Leonard Neidorf, Rafael J. Pascual, and R.D. Fulk, among others. An analysis of several Old English poems by a team including Neidorf suggests that Beowulf is the work of a single author.
The claim to an early 11th-century date depends in part on scholars who argue that, rather than the transcription of a tale from the oral tradition by an earlier literate monk, Beowulf reflects an original interpretation of an earlier version of the story by the manuscript's two scribes. On the other hand, some scholars argue that linguistic, palaeographical, metrical, and onomastic considerations align to support a date of composition in the first half of the eighth century; in particular, the poem's regular observation of etymological length distinctions (Max Kaluza's law) has been thought to demonstrate a date of composition in the first half of the eighth century. However, scholars disagree about whether the metrical phenomena described by Kaluza's Law prove an early date of composition or are evidence of a longer prehistory of the Beowulf meter; B.R. Hutcheson, for instance, does not believe Kaluza's Law can be used to date the poem, while claiming that "the weight of all the evidence Fulk presents in his book[b] tells strongly in favour of an eighth-century date."
The poem is known only from a single manuscript, which is estimated to date from close to AD 1000, in which it appears with other works. The Beowulf manuscript is known as the Nowell Codex, gaining its name from 16th-century scholar Laurence Nowell. The official designation is "British Library, Cotton Vitellius A.XV" because it was one of Sir Robert Bruce Cotton's holdings in the Cotton library in the middle of the 17th century. Many private antiquarians and book collectors, such as Sir Robert Cotton, used their own library classification systems. "Cotton Vitellius A.XV" translates as: the 15th book from the left on shelf A (the top shelf) of the bookcase with the bust of Roman Emperor Vitellius standing on top of it, in Cotton's collection. Kevin Kiernan argues that Nowell most likely acquired it through William Cecil, 1st Baron Burghley, in 1563, when Nowell entered Cecil's household as a tutor to his ward, Edward de Vere, 17th Earl of Oxford.
The earliest extant reference to the first foliation of the Nowell Codex was made sometime between 1628 and 1650 by Franciscus Junius (the younger).:91 The ownership of the codex before Nowell remains a mystery.:120
The Reverend Thomas Smith (1638–1710) and Humfrey Wanley (1672–1726) both catalogued the Cotton library (in which the Nowell Codex was held). Smith's catalogue appeared in 1696, and Wanley's in 1705. The Beowulf manuscript itself is identified by name for the first time in an exchange of letters in 1700 between George Hickes, Wanley's assistant, and Wanley. In the letter to Wanley, Hickes responds to an apparent charge against Smith, made by Wanley, that Smith had failed to mention the Beowulf script when cataloguing Cotton MS. Vitellius A. XV. Hickes replies to Wanley "I can find nothing yet of Beowulph." Kiernan theorised that Smith failed to mention the Beowulf manuscript because of his reliance on previous catalogues or because either he had no idea how to describe it or because it was temporarily out of the codex.
It suffered damage in the Cotton Library fire at Ashburnham House in 1731. Since then, parts of the manuscript have crumbled along with many of the letters. Rebinding efforts, though saving the manuscript from much degeneration, have nonetheless covered up other letters of the poem, causing further loss. Kevin Kiernan, in preparing his electronic edition of the manuscript, used fibre-optic backlighting and ultraviolet lighting to reveal letters in the manuscript lost from binding, erasure, or ink blotting.
The Beowulf manuscript was transcribed from an original by two scribes, one of whom wrote the prose at the beginning of the manuscript and the first 1939 lines before breaking off in mid sentence. The first scribe made a point of carefully regularizing the spelling of the original document by using the common West Saxon language and by avoiding any archaic or dialectical features. The second scribe, who wrote the remainder, with a difference in handwriting noticeable after line 1939, seems to have written more vigorously and with less interest. As a result, the second scribe's script retains more archaic dialectic features which allow modern scholars to ascribe the poem a cultural context. While both scribes appear to proofread their work, there are nevertheless many errors. The second scribe was ultimately the more conservative copyist of the two as he did not modify the spelling of the text as he wrote but rather copied what he saw in front of him. In the way that it is currently bound, the Beowulf manuscript is followed by the Old English poem Judith. Judith was written by the same scribe that completed Beowulf as evidenced through similar writing style. Worm-holes found in the last leaves of the Beowulf manuscript that aren't present in the Judith manuscript suggest that at one point Beowulf ended the volume. The rubbed appearance of some leaves also suggest that the manuscript stood on a shelf unbound, as is known to have been the case with other Old English manuscripts. From knowledge of books held in the library at Malmesbury Abbey and available as source works, and from the identification of certain words particular to the local dialect found in the text, the transcription may have taken place there.
Debate over oral traditionEdit
The question of whether Beowulf was passed down through oral tradition prior to its present manuscript form has been the subject of much debate, and involves more than simply the issue of its composition. Rather, given the implications of the theory of oral-formulaic composition and oral tradition, the question concerns how the poem is to be understood, and what sorts of interpretations are legitimate.
Scholarly discussion about Beowulf in the context of the oral tradition was extremely active throughout the 1960s and 1970s. The debate might be framed starkly as follows: on the one hand, we can hypothesise a poem put together from various tales concerning the hero (the Grendel episode, the Grendel's mother story, and the fire drake narrative). These fragments would have been told for many years in tradition, and learned by apprenticeship from one generation of illiterate poets to the next. The poem is composed orally and extemporaneously, and the archive of tradition on which it draws is oral, pagan, Germanic, heroic, and tribal. On the other hand, one might posit a poem which is composed by a literate scribe, who acquired literacy by way of learning Latin (and absorbing Latinate culture and ways of thinking), probably a monk and therefore profoundly Christian in outlook. On this view, the pagan references would be a sort of decorative archaising. There is a third view that sees merit in both arguments above and attempts to bridge them, and so cannot be articulated as starkly as they can; it sees more than one Christianity and more than one attitude towards paganism at work in the poem; it sees the poem as initially the product of a literate Christian author with one foot in the pagan world and one in the Christian, himself perhaps a convert (or one whose forebears had been pagan), a poet who was conversant in both oral and literary composition and was capable of a masterful "repurposing" of poetry from the oral tradition.
However, scholars such as D.K. Crowne have proposed the idea that the poem was passed down from reciter to reciter under the theory of oral-formulaic composition, which hypothesises that epic poems were (at least to some extent) improvised by whoever was reciting them, and only much later written down. In his landmark work, The Singer of Tales, Albert Lord refers to the work of Francis Peabody Magoun and others, saying "the documentation is complete, thorough, and accurate. This exhaustive analysis is in itself sufficient to prove that Beowulf was composed orally."
Examination of Beowulf and other Old English literature for evidence of oral-formulaic composition has met with mixed response. While "themes" (inherited narrative subunits for representing familiar classes of event, such as the "arming the hero", or the particularly well-studied "hero on the beach" theme) do exist across Anglo-Saxon and other Germanic works, some scholars conclude that Anglo-Saxon poetry is a mix of oral-formulaic and literate patterns, arguing that the poems both were composed on a word-by-word basis and followed larger formulae and patterns.
Larry Benson argued that the interpretation of Beowulf as an entirely formulaic work diminishes the ability of the reader to analyse the poem in a unified manner, and with due attention to the poet's creativity. Instead, he proposed that other pieces of Germanic literature contain "kernels of tradition" from which Beowulf borrows and expands upon. A few years later, Ann Watts argued against the imperfect application of one theory to two different traditions: traditional, Homeric, oral-formulaic poetry and Anglo-Saxon poetry. Thomas Gardner agreed with Watts, arguing that the Beowulf text is of too varied a nature to be completely constructed from set formulae and themes.
John Miles Foley wrote, referring to the Beowulf debate, that while comparative work was both necessary and valid, it must be conducted with a view to the particularities of a given tradition; Foley argued with a view to developments of oral traditional theory that do not assume, or depend upon, ultimately unverifiable assumptions about composition, and instead delineate a more fluid continuum of traditionality and textuality.
Finally, in the view of Ursula Schaefer, the question of whether the poem was "oral" or "literate" becomes something of a red herring. In this model, the poem is created, and is interpretable, within both noetic horizons. Schaefer's concept of "vocality" offers neither a compromise nor a synthesis of the views which see the poem as on the one hand Germanic, pagan, and oral and on the other Latin-derived, Christian, and literate, but, as stated by Monika Otter: "... a 'tertium quid', a modality that participates in both oral and literate culture yet also has a logic and aesthetic of its own."
Transcriptions and translationsEdit
Icelandic scholar Grímur Jónsson Thorkelin made the first transcriptions of the manuscript in 1786 and published the results in 1815, working as part of a Danish government historical research commission. He made one himself, and had another done by a professional copyist who knew no Anglo-Saxon. Since that time, however, the manuscript has crumbled further, making these transcripts a prized witness to the text. While the recovery of at least 2000 letters can be attributed to them, their accuracy has been called into question,[c] and the extent to which the manuscript was actually more readable in Thorkelin's time is uncertain.
Translations and adaptationsEdit
A great number of translations and adaptations are available, in poetry and prose. Andy Orchard, in A Critical Companion to Beowulf, lists 33 "representative" translations in his bibliography, while the Arizona Center for Medieval and Renaissance Studies published Marijane Osborn's annotated list of over 300 translations and adaptations in 2003. Beowulf has been translated into at least 23 other languages.
In 1805, the historian Sharon Turner translated selected verses into modern English. This was followed in 1814 by John Josias Conybeare who published an edition "in English paraphrase and Latin verse translation." In 1815, Grímur Jónsson Thorkelin published the first complete edition in Latin. N. F. S. Grundtvig reviewed this edition in 1815 and created the first complete verse translation in Danish in 1820. In 1837, John Mitchell Kemble created an important literal translation in English. In 1895, William Morris & A. J. Wyatt published the ninth English translation. Many retellings of Beowulf for children began appearing in the 20th century.
First published in 1928, Frederick Klaeber's Beowulf and The Fight at Finnsburg (which included the poem in Old English, an extensive glossary of Old English terms, and general background information) became the "central source used by graduate students for the study of the poem and by scholars and teachers as the basis of their translations."
Following research in the King's College London Archives, Carl Kears proposed that John Porter's translation, published in 1975 by Bill Griffiths' Pirate Press, was the first complete verse translation of the poem entirely accompanied by facing-page Old English.
Translating Beowulf is one of the subjects of the 2012 publication Beowulf at Kalamazoo, containing a section with 10 essays on translation, and a section with 22 reviews of Heaney's translation (some of which compare Heaney's work with that of Anglo-Saxon scholar Roy Liuzza).
J. R. R. Tolkien's long-awaited translation (edited by his son, Christopher) was published in 2014 as Beowulf: A Translation and Commentary. This also includes Tolkien's own retelling of the story of Beowulf in his tale, Sellic Spell.
Sources and analoguesEdit
Neither identified sources nor analogues for Beowulf can be definitively proven, but many conjectures have been made. These are important in helping historians understand the Beowulf manuscript, as possible source-texts or influences would suggest time-frames of composition, geographic boundaries within which it could be composed, or range (both spatial and temporal) of influence (i.e. when it was "popular" and where its "popularity" took it).
19th century studies proposed that Beowulf was translated from a lost original Scandinavian work, but this idea was quickly abandoned. But Scandinavian works have continued to be studied as a possible source. Proponents included Gregor Sarrazin writing in 1886 that an Old Norse original version of Beowulf must have existed, but that view was later debunked by Carl Wilhelm von Sydow (1914) who pointed out that Beowulf is fundamentally Christian and written at a time when any Norse tale would have most likely been pagan.
Grettis saga is a story about Grettir Ásmundarson, a great-grandson of an Icelandic settler, and so cannot be as old as Beowulf. Axel Olrik (1903) claimed that on the contrary, this saga was a reworking of Beowulf, and others followed suit.
However, Friedrich Panzer (1910) wrote a thesis in which both Beowulf and Grettis saga drew from a common folkloric source, and this encouraged even a detractor such as W. W. Lawrence to reposition his view, and entertain the possibility that certain elements in the saga (such as the waterfall in place of the mere) retained an older form.
The viability of this connection has enjoyed enduring support, and was characterized as one of the few Scandinavian analogues to receive a general consensus of potential connection by Theodore M. Andersson (1998). But that same year, Magnús Fjalldal published a volume challenging the perception that there is a close parallel, and arguing that tangential similarities were being overemphasized as analogies.
Hrolf kraki and Bodvar BjarkiEdit
Another candidate for an analogue or possible source is the story of Hrolf kraki and his servant, the legendary bear-shapeshifter Bodvar Bjarki. The story survives in Old Norse Hrólfs saga kraka and Saxo's Gesta Danorum. Hrolf kraki, one of the Skjöldungs, even appears as "Hrothulf" in the Anglo-Saxon epic. Hence a story about him and his followers may have developed as early as the 6th century.
International folktale sourcesEdit
Bear's Son TaleEdit
Friedrich Panzer (1910) wrote a thesis that the first part of Beowulf (the Grendel Story) incorporated preexisting folktale material, and that the folktale in question was of the Bear's Son Tale (Bärensohnmärchen) type, which has surviving examples all over the world.
This tale type was later catalogued as international folktale type 301, now formally entitled "The Three Stolen Princesses" type in Hans Uther's catalogue, although the "Bear's Son" is still used in Beowulf criticism, if not so much in folkloristic circles.
However, although this folkloristic approach was seen as a step in the right direction, "The Bear's Son" tale has later been regarded by many as not a close enough parallel to be a viable choice. Later, Peter Jørgensen, looking for a more concise frame of reference, coined a "two-troll tradition" that covers both Beowulf and Grettis saga: "a Norse 'ecotype' in which a hero enters a cave and kills two giants, usually of different sexes"; which has emerged as a more attractive folk tale parallel, according to a 1998 assessment by Andersson.
Similarity of the epic to the Irish folktale "The Hand and the Child" had already been noted by Albert S. Cook (1899), and others even earlier,[e][f] Swedish folklorist Carl Wilhelm von Sydow (1914) then made a strong argument for the case of parallelism in "The Hand and the Child", because the folktale type demonstrated a "monstrous arm" motif that corresponded with Beowulf wrenching off Grendel's arm. For no such correspondence could be perceived in the Bear's Son Tale or Grettis saga.[g]
James Carney and Martin Puhvel also agree with this "Hand and the Child" contextualisation.[h] Puhvel supported the "Hand and the Child" theory through such motifs as (in Andersson's words) "the more powerful giant mother, the mysterious light in the cave, the melting of the sword in blood, the phenomenon of battle rage, swimming prowess, combat with water monsters, underwater adventures, and the bear-hug style of wrestling."
Attempts to find classical or Late Latin influence or analogue in Beowulf are almost exclusively linked with Homer's Odyssey or Virgil's Aeneid. In 1926, Albert S. Cook suggested a Homeric connection due to equivalent formulas, metonymies, and analogous voyages. In 1930, James A. Work also supported the Homeric influence, stating that encounter between Beowulf and Unferth was parallel to the encounter between Odysseus and Euryalus in Books 7–8 of the Odyssey, even to the point of both characters giving the hero the same gift of a sword upon being proven wrong in their initial assessment of the hero's prowess. This theory of Homer's influence on Beowulf remained very prevalent in the 1920s, but started to die out in the following decade when a handful of critics stated that the two works were merely "comparative literature", although Greek was known in late 7th century England: Bede states that Theodore of Tarsus, a Greek, was appointed Archbishop of Canterbury in 668, and he taught Greek. Several English scholars and churchmen are described by Bede as being fluent in Greek due to being taught by him; Bede claims to be fluent in Greek himself.
Frederick Klaeber, among others, argued for a connection between Beowulf and Virgil near the start of the 20th century, claiming that the very act of writing a secular epic in a Germanic world represents Virgilian influence. Virgil was seen as the pinnacle of Latin literature, and Latin was the dominant literary language of England at the time, therefore making Virgilian influence highly likely. Similarly, in 1971, Alistair Campbell stated that the apologue technique used in Beowulf is so rare in epic poetry aside from Virgil that the poet who composed Beowulf could not have written the poem in such a manner without first coming across Virgil's writings.
It cannot be denied that Biblical parallels occur in the text, whether seen as a pagan work with "Christian colouring" added by scribes or as a "Christian historical novel, with selected bits of paganism deliberately laid on as 'local colour'," as Margaret E. Goldsmith did in "The Christian Theme of Beowulf". Beowulf channels the Book of Genesis, the Book of Exodus, and the Book of Daniel in its inclusion of references to the Genesis creation narrative, the story of Cain and Abel, Noah and the flood, the Devil, Hell, and the Last Judgment.
There is a wide array of linguistic forms in the Beowulf manuscript. It is this fact that leads some scholars to believe that Beowulf has endured a long and complicated transmission through all the main dialect areas. The poem retains a complicated mix of the following dialectical forms: Mercian, Northumbrian, Early West Saxon, Kentish and Late West Saxon.:20–21 There are in Beowulf more than 3100 distinct words, and almost 1300 occur exclusively, or almost exclusively, in this poem and in the other poetical texts. Considerably more than one-third of the total vocabulary is alien from ordinary prose use. There are, in round numbers, three hundred and sixty uncompounded verbs in Beowulf, and forty of them are poetical words in the sense that they are unrecorded or rare in the existing prose writings. One hundred and fifty more occur with the prefix ge- (reckoning a few found only in the past-participle), but of these one hundred occur also as simple verbs, and the prefix is employed to render a shade of meaning which was perfectly known and thoroughly familiar except in the latest Anglo-Saxon period. The nouns number sixteen hundred. Seven hundred of them, including those formed with prefixes, of which fifty (or considerably more than half) have ge-, are simple nouns, at the highest reckoning not more than one-quarter is absent in prose. That this is due in some degree to accident is clear from the character of the words, and from the fact that several reappear and are common after the Norman Conquest.
Form and metreEdit
An Old English poem such as Beowulf is very different from modern poetry. Anglo-Saxon poets typically used alliterative verse, a form of verse in which the first half of the line (the a-verse) is linked to the second half (the b-verse) through similarity in initial sound. In addition, the two halves are divided by a caesura: "Oft Scyld Scefing \\ sceaþena þreatum" (l. 4). This verse form maps stressed and unstressed syllables onto abstract entities known as metrical positions. There is no fixed number of beats per line: the first one cited has three (Oft SCYLD SCEFING, with ictus on the suffix -ING) whereas the second has two (SCEAþena ÞREATum).
The poet has a choice of epithets or formulae to use in order to fulfil the alliteration. When speaking or reading Old English poetry, it is important to remember for alliterative purposes that many of the letters are not pronounced in the same way as in modern English. The letter ⟨h⟩, for example, is always pronounced (Hroðgar: [ˈhroðgar]), and the digraph ⟨cg⟩ is pronounced [dʒ], as in the word edge. Both ⟨f⟩ and ⟨s⟩ vary in pronunciation depending on their phonetic environment. Between vowels or voiced consonants, they are voiced, sounding like modern ⟨v⟩ and ⟨z⟩, respectively. Otherwise they are unvoiced, like modern ⟨f⟩ in fat and ⟨s⟩ in sat. Some letters which are no longer found in modern English, such as thorn, ⟨þ⟩, and eth, ⟨ð⟩ – representing both pronunciations of modern English ⟨th⟩, as // in thing and // this – are used extensively both in the original manuscript and in modern English editions. The voicing of these characters echoes that of ⟨f⟩ and ⟨s⟩. Both are voiced (as in this) between other voiced sounds: oðer, laþleas, suþern. Otherwise they are unvoiced (as in thing): þunor, suð, soþfæst.
Kennings are also a significant technique in Beowulf. They are evocative poetic descriptions of everyday things, often created to fill the alliterative requirements of the metre. For example, a poet might call the sea the "swan-road" or the "whale-road"; a king might be called a "ring-giver." There are many kennings in Beowulf, and the device is typical of much of classic poetry in Old English, which is heavily formulaic. The poem also makes extensive use of elided metaphors.
Interpretation and criticismEdit
The history of modern Beowulf criticism is often said to begin with J. R. R. Tolkien, author and Merton professor of Anglo-Saxon at University of Oxford, who in his 1936 lecture to the British Academy criticised his contemporaries' excessive interest in its historical implications.[citation not found] He noted in Beowulf: The Monsters and the Critics that as a result the poem's literary value had been largely overlooked and argued that the poem "is in fact so interesting as poetry, in places poetry so powerful, that this quite overshadows the historical content..."
Paganism and ChristianityEdit
In historical terms, the poem's characters would have been Norse pagans (the historical events of the poem took place before the Christianisation of Scandinavia), yet the poem was recorded by Christian Anglo-Saxons who had mostly converted from their native Anglo-Saxon paganism around the 7th century – both Anglo-Saxon paganism and Norse paganism share a common origin as both are forms of Germanic paganism. Beowulf thus depicts a Germanic warrior society, in which the relationship between the lord of the region and those who served under him was of paramount importance.
In terms of the relationship between characters in Beowulf to God, one might recall the substantial amount of paganism that is present throughout the work. Literary critics such as Fred C. Robinson argue that the Beowulf poet arguably tries to send a message to readers during the Anglo-Saxon time period regarding the state of Christianity in their own time. Robinson argues that the intensified religious aspects of the Anglo-Saxon period inherently shape the way in which the Poet alludes to paganism as presented in Beowulf. The Poet arguably calls on Anglo-Saxon readers to recognize the imperfect aspects of their supposed Christian lifestyles. In other words, the Poet is referencing their "Anglo-Saxon Heathenism." In terms of the characters of the epic itself, Robinson argues that readers are "impressed" by the courageous acts of Beowulf and the speeches of Hrothgar (181). But one is ultimately left to feel sorry for both men as they are fully detached from supposed "Christian truth" (181). The relationship between the characters of Beowulf, and the overall message of the Poet, regarding their relationship with God is largely debated among readers and literary critics alike.
At the same time, Richard North argues that the Beowulf poet interpreted "Danish myths in Christian form" (as the poem would have served as a form of entertainment for a Christian audience), and states: "As yet we are no closer to finding out why the first audience of Beowulf liked to hear stories about people routinely classified as damned. This question is pressing, given... that Anglo-Saxons saw the Danes as 'heathens' rather than as foreigners." Grendel's mother and Grendel are described as descendants of Cain, a fact which some scholars link to the Cain tradition.
Other scholars disagree, however, as to the meaning and nature of the poem: is it a Christian work set in a Germanic pagan context? The question suggests that the conversion from the Germanic pagan beliefs to Christian ones was a prolonged and gradual process over several centuries, and it remains unclear the ultimate nature of the poem's message in respect to religious belief at the time it was written. Robert F. Yeager notes the facts that form the basis for these questions:
That the scribes of Cotton Vitellius A.XV were Christian beyond doubt, and it is equally sure that Beowulf was composed in a Christianised England since conversion took place in the sixth and seventh centuries. The only Biblical references in Beowulf are to the Old Testament, and Christ is never mentioned. The poem is set in pagan times, and none of the characters is demonstrably Christian. In fact, when we are told what anyone in the poem believes, we learn that they are pagans. Beowulf's own beliefs are not expressed explicitly. He offers eloquent prayers to a higher power, addressing himself to the "Father Almighty" or the "Wielder of All." Were those the prayers of a pagan who used phrases the Christians subsequently appropriated? Or, did the poem's author intend to see Beowulf as a Christian Ur-hero, symbolically refulgent with Christian virtues?
The location of the composition of the poem is also intensely disputed. In 1914, F.W. Moorman, the first professor of English Language at University of Leeds, claimed that Beowulf was composed in Yorkshire, but E. Talbot Donaldson claims that it was probably composed more than twelve hundred years ago, during the first half of the eighth century, and that the writer was a native of what was then called West Mercia, located in the Western Midlands of England. However, the late tenth-century manuscript "which alone preserves the poem" originated in the kingdom of the West Saxons – as it is more commonly known. Donaldson wrote that "the poet who put the materials into their present form was a Christian and ... poem reflects a Christian tradition".
Politics and warfareEdit
Stanley B. Greenfield has suggested that references to the human body throughout Beowulf emphasise the relative position of thanes to their lord. He argues that the term "shoulder-companion" could refer to both a physical arm as well as a thane (Aeschere) who was very valuable to his lord (Hrothgar). With Aeschere's death, Hrothgar turns to Beowulf as his new "arm." Also, Greenfield argues the foot is used for the opposite effect, only appearing four times in the poem. It is used in conjunction with Unferð (a man described by Beowulf as weak, traitorous, and cowardly). Greenfield notes that Unferð is described as "at the king's feet" (line 499). Unferð is also a member of the foot troops, who, throughout the story, do nothing and "generally serve as backdrops for more heroic action."
Daniel Podgorski has argued that the work is best understood as an examination of inter-generational vengeance-based conflict, or feuding. In this context, the poem operates as an indictment of feuding conflicts as a function of its conspicuous, circuitous, and lengthy depiction of the Geatish-Swedish wars—coming into contrast with the poem's depiction of the protagonist Beowulf as being disassociated from the ongoing feuds in every way.
- "wíg" means "fight, battle, war, conflict" and "láf" means "remnant, left-over"
- That is, R.D. Fulk's 1992 A History of Old English Meter.
- For instance, by Chauncey Brewster Tinker in The Translations of Beowulf, a comprehensive survey of 19th-century translations and editions of Beowulf.
- Ecclesiastical or biblical influences are only seen as adding "Christian color", in Andersson's survey. Old English sources hinges on the hypothesis that Genesis A predates Beowulf.
- Ludwig Laistner (1889), II, p. 25; Stopford Brooke, I, p. 120; Albert S. Cook (1899) pp. 154–156.
- In the interim, Max Deutschbein (1909) is credited by Andersson to be the first person to present the Irish argument in academic form. He suggested the Irish Feast of Bricriu (which is not a folktale) as a source for Beowulf—a theory that was soon denied by Oscar Olson.
- von Sydow was also anticipated by Heinz Dehmer in the 1920s as well besides the writers from the 19th century in pointing out "The Hand and the Child" as a parallel.
- Carney also sees the Táin Bó Fráech story (where a half-fairy hero fights a dragon in the "Black Pool (Dubh linn)"), but this has not received much support for forty years, as of Andersson's writing.
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# INMO 2023 - Problems, Solutions and Discussion
Let's Solve INMO 2023
It is a difficult question paper
## Solutions of INMO 2023
Problem 1
Let $S$ be a finite set of positive integers. Assume that there are precisely $2023$ ordered pairs $(x, y)$ in $S \times S$ so that the product $xy$ is a perfect square. Prove that one can find at least four distinct elements in $S$ so that none of their pairwise products is a perfect square.
Note: As an example, if $S=\{1,2,4\},$ there are exactly five such ordered pairs: $(1,1),(1,4),(2,2),(4,1),$ and $(4,4)$.
Solution :
For each $x\in S$, define $S_x = \{y \mid y\in S\;$&$\; xy\text{ is a perfect square}\}$. Now, let us prove the following properties of set $T$ which is the collection of sets $S_x$, for every $x\in S$.
• $T$ covers the entire set $S$
• This is because, $x\in S_x$ and since we consider the set $S_x$ for each $x\in S$, we have covered the entire set $S$.
• If $z \in S_x,$ then $S_z = S_x$
• Because, $xz$ is a perfect square and if $w\in S_x$, then $wx$ is a perfect square, so $xz\times wx = wz\times x^2$ is a perfect square, hence $wz$ is a perfect square $\Rightarrow w\in S_z \Rightarrow S_x \subset S_z$. Similarly we get, $S_z \subset S_x$. So, $S_z = S_x$.
• If $z \notin S_x,$ then $S_z$ and $S_x$ are disjoint sets
• This is because, $xz$ is not a perfect square, but if $S_x$ and $S_z$ share a common element, say $w$, then both $xw$ and $zw$ are perfect squares and so $xw\times zw = xz\times w^2$ is a perfect square and hence, $xz$ must be a perfect square, which would be a contradiction.
So, the collection of all sets $S_x$ will be either the same (or) disjoint and covering the entire set $S$ which forms a partition of $S$. Now according to the question given, we need to prove that there is always greater than 3 partitions in $S$, so that one can choose exactly one number from some 4 partitions of $S$, say $n_1, n_2, n_3, n_4$, and obviously $S_{n_1}, S_{n_2}, S_{n_3}, S_{n_4}$ will be pairwise disjoint and hence their pairwise product will not be a perfect square.
Suppose there are total $n$ partitions of $S$, then multiplying two numbers of $S$ form a perfect square if and only if both of these numbers are from the same partition in $S$ which follows directly from the 3 properties of $T$. Let $x_1, x_2, \ldots, x_n$ represent the number of numbers in each of the $n$ partitions of $S$ as shown above, then the number of ordered pairs from each of the partition will be $x_i\times x_i = x_i^2$. So,
$\text{Total number of ordered pairs} = x_1^2 + x_2^2 + \cdots + x_n^2 = 2023$
Case 1:
If $n=1$, then $x_1^2 = 2023$, but we know that a perfect square cannot end with $3$, so this case is not possible.
Case 2:
If $n=2$, then $x_1^2 + x_2^2 = 2023$. We know that any perfect square is congruent to $0$ (or) $1\pmod{4}$. So, the LHS, $x_1^2+x_2^2 \equiv 0\ (or)\ 1\ (or)\ 2\pmod{4}$, but the RHS, $2023\equiv 3\pmod{4}$, which is not possible.
Case 3:
If $n=3$, then $x_1^2 + x_2^2 + x_3^2 = 2023$. Going$\pmod{4}$ we see that each of $x_1^2,\ x_2^2,\ x_3^2 \equiv 1\pmod{4}$ since $2023\equiv 3\pmod{4}$ and hence they are odd numbers. Any square of an odd number is congruent to $1\pmod{8}$, since $(2k+1)^2 = 8\frac{k(k+1)}{2} + 1 = 8k'+1$, so the LHS, $x_1^2 + x_2^2 + x_3^2 \equiv 3\pmod{8}$, but the RHS, $2023\equiv7\pmod{8}$, which is a contradiction. Hence, this case is also not possible.
So, we conclude that for any such set $S,\ n\geq 4$ and hence one can select $4$ such numbers. Actually, there is a partition of $S$ possible with $n=4$, that is, $(9, 9, 30, 31)$ because $9^2+9^2+30^2+31^2 = 2023$.
Remark: In set theory, the partitions $S_x$ is called equivalence classes of $S$ since multiplying two numbers to get a perfect square is an equivalence relation defined on $S$, which in turn results in a partition of $S$. For further reading, refer to :
• https://mathworld.wolfram.com/EquivalenceClass.html
• https://mathworld.wolfram.com/EquivalenceRelation.html
Problem 2
Suppose $a_0, \ldots, a_{100}$ are positive reals. Consider the following polynomial for each $k$ in $\{0,1, \ldots, 100\}$ :
$a_{100+k} x^{100}+100 a_{99+k} x^{99}+a_{98+k} x^{98}+a_{97+k} x^{97}+\cdots+a_{2+k} x^2+a_{1+k} x+a_k,$where indices are taken modulo $101$, i.e., $a_{100+i}=a_{i-1}$ for any $i$ in $\{1,2, \ldots, 100\}$. Show that it is impossible that each of these $101$ polynomials has all its roots real.
Solution :
Suppose say that all the roots are real for all the polynomials and $\alpha_{ij}$ represent the $i^{th}$ root of the $j^{th}$ polynomial. Then for each $j \in \{1, 2, \ldots, 101\}$,
$\Rightarrow \frac{1}{\alpha_{1j}^2} + \frac{1}{\alpha_{2j}^2} + \cdots + \frac{1}{\alpha_{100j}^2} \geq 0$
$\Rightarrow \left(\frac{1}{\alpha_{1j}} + \frac{1}{\alpha_{2j}} + \cdots + \frac{1}{\alpha_{100j}}\right)^2 - \sum_{0<m<n}^{n\leq 100} \frac{2}{\alpha_{mj}\alpha_{nj}}\geq 0$
$\Rightarrow \left(\frac{\sum_{(i_1,i_2,\ldots,i_{99})\in \text{99 tuples of 1 to 100}}\alpha_{i_1j}\alpha_{i_2j}\cdots\alpha_{i_{99}j}}{\alpha_{1j}\alpha_{2j}\cdots\alpha_{100j}}\right)^2 - \frac{2\sum_{(i_1,i_2,\ldots,i_{98})\in \text{98 tuples of 1 to 100}}\alpha_{i_1j}\alpha_{i_2j}\cdots\alpha_{i_{98}j}}{\alpha_{1j}\alpha_{2j}\cdots\alpha_{100j}} \geq 0$
So, by the Vieta's formula we get,
$\Rightarrow \frac{\left(\frac{-a_{1+k}}{a_{100+k}}\right)^2}{\left(\frac{a_k}{a_{100+k}}\right)^2} - 2\frac{\left(\frac{a_{2+k}}{a_{100+k}}\right)}{\left(\frac{a_k}{a_{100+k}}\right)} \geq 0$
$\Rightarrow \frac{a_{k+1}}{a_k} \geq 2\left(\frac{a_{k+2}}{a_{k+1}}\right),\text{ for each }k\in\{0,1,2,\ldots,100\} \longrightarrow(1)$
Define $b_i \in \mathbb{R_+}$ as, $b_i = \frac{a_i}{a_{i-1}}$, for each $i\in \{1,\ldots,102\}$, so obviously $b_1=b_{102}$, then from the equation $(1)$ we have,
$b_i \leq \frac{b_{i-1}}{2},\text{ for each }i\in\{2,\ldots,102\},\text{ so we have,}$
$b_{102} \leq \frac{b_{101}}{2} \leq \frac{b_{100}}{2^2} \leq \frac{b_{99}}{2^3} \leq \cdots \leq \frac{b_{2}}{2^{100}} \leq \frac{b_{1}}{2^{101}} = \frac{b_{102}}{2^{101}}$
$\Rightarrow b_{102}\leq \frac{b_{102}}{2^{101}}\Rightarrow b_{102} \leq 0,$
which is not possible as $b_i$'s $\in \mathbb{R_+}$. Hence, we arrive at a contradiction and so we conclude that all the roots of all the polynomials cannot be real.
Problem 3
Let $\mathbb{N}$ denote the set of all positive integers. Find all real numbers $c$ for which there exists a function $f: \mathbb{N} \rightarrow \mathbb{N}$ satisfying:
(a) for any $x, a \in \mathbb{N}$, the quantity $\frac{f(x+a)-f(x)}{a}$ is an integer if and only if $a=1$;
(b) for all $x \in \mathbb{N}$, we have $\mid f(x)-cx\mid <2023$.
Solution :
The only possible values of $c = k+\frac{1}{2},\ \forall k\in \mathbb{N_0}$. Here, $\mathbb{N_0}$ means the set of whole numbers. The proof is illustrated below,
Claim 1: For any integers $k \geq 1$ and $x,\ f\left(x+2^k\right)-f(x)$ is divisible by $2^{k-1}$ but not $2^k$.
Proof. We prove this via induction on $k$.
For $k=1$, the claim is trivial as $2^0$ divides any integer. Now assume the statement is true for some $k$, and hence $f\left(x+2^k\right)-f(x)=2^{k-1} y_1$ and $f\left((x+2^k)+2^k\right)-$ $f\left(x+2^k\right)=2^{k-1} y_2$ for some odd integers $y_1, y_2$. Adding these equations, we see that,
$f\left(x+2^{k+1}\right)-f(x)=2^{k-1}\left(y_1+y_2\right)$which is divisible by $2^k$ because $y_1+y_2$ is even. The fact that this is not divisible by $2^{k+1}$ follows from the condition (a) on $f$.
Now using the Claim 1, we see that for any $k \geq 1,\ f\left(1+2^k\right)=f(1)+2^{k-1}\left(2 y_k+1\right)$ for some integer $y_k$, so we get,
$f\left(1+2^k\right)-c\left(1+2^k\right)=f(1)-c+2^k\left(y_k+\frac{1}{2}-c\right),\ \forall k\in \mathbb{N}$
The quantity in the LHS is bounded above by 2023 and hence the RHS is also bounded above by 2023 for all values of $k$, but $f(1)-c$ is a constant with varying value of $k$, thus $2^k\left(y_k+\frac{1}{2}-c\right)$ is bounded above. But if this quantity is never zero, then we would have,
$2^k\left|y_k+\frac{1}{2}-c\right|=2^{k-1}\left|2 y_k+1-2 c\right| \geq 2^{k-1},\ \forall k\in\mathbb{N}$which contradicts the boundedness of $\mid f(x)-cx\mid$. Thus we must have $y_k+\frac{1}{2}-c=0$, for some $k \in \mathbb{N} \Rightarrow \{c\}=\frac{1}{2}$, where $\{x\}$ represents the fractional part of $x$.
Claim 2: If $\{c\}=\frac{1}{2}$ and $c>0$, then there exist function $f(x) = \lfloor cx\rfloor + 1$ that satisfies both conditions given in the question and if $c<0$, then there exist no function satisfying the given conditions
Proof. Since fractional part of $c>0$ is $\frac{1}{2}$, we assume that $c = k+\frac{1}{2},\ \forall k\in \mathbb{N_0}$. For any $x, a\in \mathbb{N}$,$\frac{f(x+a)-f(x)}{a} = \frac{\lfloor c(x+a)\rfloor - \lfloor cx\rfloor}{a} = \frac{1}{a}\left(\left\lfloor kx+ka+\frac{x}{2}+\frac{a}{2}\right\rfloor - \left\lfloor kx+\frac{x}{2}\right\rfloor\right)= k + \frac{1}{a}\left(\left\lfloor \frac{x}{2}+\frac{a}{2}\right\rfloor - \left\lfloor \frac{x}{2}\right\rfloor\right)$
In the above steps, we were able to take $kx$ and $ka$ out of the greatest integer function because they are integers. In general, $\lfloor x+n\rfloor = \lfloor x\rfloor+n$, where $n\in\mathbb{Z}$. So we have,$\frac{f(x+a)-f(x)}{a} =k + \frac{1}{a}\left(\left\lfloor \frac{x}{2}+\frac{a}{2}\right\rfloor - \left\lfloor \frac{x}{2}\right\rfloor\right)$Observe that, it is an integer for $a=1$. But for $a \geq 2$, we have,$\left\lfloor\frac{x+a}{2}\right\rfloor-\left\lfloor\frac{x}{2}\right\rfloor \geq\left\lfloor\frac{x+2}{2}\right\rfloor-\left\lfloor\frac{x}{2}\right\rfloor=1$If $a=2r$, then$\left\lfloor\frac{x+a}{2}\right\rfloor-\left\lfloor\frac{x}{2}\right\rfloor=\left\lfloor\frac{x}{2}+r\right\rfloor-\left\lfloor\frac{x}{2}\right\rfloor=r<2r=a,$and if $a=2r+1$ for $r \geq 1$, then$\left\lfloor\frac{x+a}{2}\right\rfloor-\left\lfloor\frac{x}{2}\right\rfloor \leq\left\lfloor\frac{x+2r+2}{2}\right\rfloor-\left\lfloor\frac{x}{2}\right\rfloor=r+1<2r+1=a$
So in either case, the quantity $\left\lfloor\frac{x+a}{2}\right\rfloor-\left\lfloor\frac{x}{2}\right\rfloor$ is strictly between 0 and $a$, and thus cannot be divisible by $a$. Thus condition (a) holds.
The condition (b) is trivially true because, $\mid f(x)-cx\mid = \lfloor cx\rfloor -cx + 1 = 1-\{cx\}\leq 1 < 2023$
Now coming to the case in which $c<0$, we see that $\mid f(x)-cx\mid$ can go arbitrarily large for a given negative value of $c$ because, given $c<0$, one can definitely choose a natural number $x>\frac{2023}{\mid c\mid}$, so that, $\mid f(x)+(-cx)\mid > 2023$, since $f(x)>0$ and so the second condition would not be satisfied for any $f(x)$.
So, we conclude from result of the Claim 1 that $\{c\} = \frac{1}{2}$ and from the Claim 2 that for all non-negative values of $c$, such that $\{c\}=\frac{1}{2}$, there exist a function $f(x)=\lfloor cx\rfloor + 1$ satisfying both the given conditions and there exist no function for negative values of $c$ that satisfies the given conditions, and hence proving the desired values of $c$.
Problem 4
Let $k \geq 1$ and $N>1$ be two integers. On a circle are placed $2N+1$ coins all showing heads. Calvin and Hobbes play the following game. Calvin starts and on his move can turn any coin from heads to tails. Hobbes on his move can turn at most one coin that is next to the coin that Calvin turned just now from tails to heads. Calvin wins if at any moment there are $k$ coins showing tails after Hobbes has made his move. Determine all values of $k$ for which Calvin wins the game.
Solution :
We claim that Calvin wins if $k \in \{1, 2, 3, \ldots, N+1\}$. First, let us prove the case in which Calvin wins the game and then we will prove the case in which he may not win.
Step 1:
Claim: In the start of the game, Calvin can form two adjacent tails after Hobbes has made his move.
Proof. Calvin starts the game by turning a coin, say $C_0$, from heads to tails. Now, Hobbes cannot make his move as there are no adjacent tails to it. Then, Calvin turns the next to next coin of $C_0$, say $C_1$, from heads to tails. Again, Hobbes cannot make his move because $C_0$ and $C_1$ cannot be adjacent as $N\geq 2$. Then, Calvin turns the coin between $C_0$ and $C_1$, say $C_2$, from heads to tails. Now, Hobbes have an option to turn $C_0$ (or) $C_1$ back to heads. But in any case, we get two adjacent tails after Hobbes has made his move, thus proving the claim.
By the above claim, assume that we have only two adjacent tails and Calvin is supposed to move now. Let the two adjacent tails be $T_1$ and $T_{N+1}$. Calvin turns the next to next coin of $T_1$ that is not adjacent to $T_{N+1}$, say $T_2$, from heads to tails. Now, Hobbes cannot make a move as there are no adjacent tails to $T_2$. Again, Calvin turns the next to next coin of $T_2$ from heads to tails until he reaches the coin $T_{N+1}$. Note that any move of Calvin after forming the adjacent tails $T_1$ and $T_{N+1}$ will not form anymore adjacent tails as the total number of coins is odd and consequently Hobbes cannot make a move. There are $2N-1$ coins except $T_1$ and $T_{N+1}$ and so there are $N-1$ more tails formed by this alternating tails and heads configuration due to Calvin's move. So, in total there are $N+1$ tails after the Hobbes move. Since the number of tails increases by one in the procedure after each of the Hobbes move (which was not possible anyways), for each $k \in \{1, 2, 3, \ldots, N+1\}$, Calvin wins the game by following the above procedure (obviously he cannot win for $k\geq 2$ if he turn the coins randomly). An example of the procedure is illustrated below for $N=3$ attaining $4=N+1$ tails,
Remark: The procedure has total $N+3$ moves and irrespective of value of $N$, Hobbes can make only one move which forms the first adjacent tails after his move.
Step 2:
Pair up the adjacent coins to form $N$ blocks of coin pairs, say $B_1, B_2, \ldots, B_N$, and a single coin, say $C_1$, as shown in the figure below for $N=3$,
Claim: There is a playing strategy for Hobbes that will not allow both coins of any block to show tails after each move of Hobbes.
Proof. Let us prove this claim by induction on the number of moves by Hobbes. Obviously, it is true for the first move of Hobbes, since by the time Calvin has turned only one coin to tails. Let us assume that the claim is true for "$(n-1)^{th}$" move of Hobbes, i.e., there is no block with both coins showing tails in it after $(n-1)^{th}$ move of Hobbes. Now after Calvin's $n^{th}$ move, there is at most one block with both coins showing tails as he can turn only one coin and if there is such a block after Calvin's move, say $B_i$, then the latest turned coin, say $C_0$, must be in the block $B_i$ and hence, Hobbes turns the adjacent coin of $C_0$ that is in the block $B_i$ from tails to heads which makes sure that there is no block with both coins showing tails after his $n^{th}$ move. Hence, by principle of induction, there is a playing strategy for Hobbes that will not allow both coins of any block to show tails after each move of Hobbes.
Claim: Calvin may not win the game if $k\geq N+2$.
Proof. Now we know that there are $N$ blocks and a coin in total and after each move of Hobbes there is at most one coin showing tails in each block by the above strategy and hence there is at most $N+1$ coins showing tails in total after Hobbes move with the above strategy and hence Calvin will not win the game if $k\geq N+2$ and if Hobbes adopts the above strategy.
Remark: If the total number of coins were even, say $2N$, then Calvin wins the game if $k\in \{1, 2, \ldots, N\}$, which can be easily proved using an argument similar to the above procedure.
Problem 5
Euler marks $n$ different points in the Euclidean plane. For each pair of marked points, Gauss writes down the number $\left\lfloor\log _2 d\right\rfloor$ where $d$ is the distance between the two points. Prove that Gauss writes down less than $2n$ distinct values.
Note: For any $d>0,\left\lfloor\log _2 d\right\rfloor$ is the unique integer $k$ such that $2^k \leq d<2^{k+1}$.
Solution :
We will prove that Gauss writes down at most $n-1$ distinct even numbers and $n-1$ distinct odd numbers and hence, in total at most $2n-2$ distinct numbers which is less than $2n$.
Definition:
• A connected graph is a graph in which for every pair of two distinct vertices, there exist a path connecting them.
• Cycle is a path in a graph whose starting and ending vertices are the same.
Claim 1: Consider a simple graph $G$ with $m$ vertices, where $m\in\mathbb{N}$. If $G$ is connected with no cycle in it, then $G$ has exactly $m-1$ edges.
Proof. If $m=1$, then the claim is trivially true. For $m>1$, let the $m$ vertices be $x_1, x_2, \ldots, x_m$. Consider any vertex $x_i$ and an edge, say $E_1$, that is connecting it to $x_j$ for some $i, j\in\{1, 2, \ldots, m\}$ (since $G$ is connected, there must be such an edge). Consider the subgraph containing $x_i$ and $x_j$ and the edge $E_1$ to be $G_2$, which is a connected subgraph. For $i\geq2$, considering $G_i$ to be a connected subgraph of $G$ with $i$ vertices and $i-1$ edges, let us inductively define that $G_{i+1}$ is a union of the graph $G_i$ with a new edge $E_i$ that is connecting some vertex of $G_i$ to a vertex not in $G_i$, say $V_i$ and the vertex $V_i$ as shown below. Notice that $G_{i+1}$ is a connected graph because there exists such an edge $E_i$ (since $G$ is connected) and $G_i$ is also connected. So, by induction, the graph $G_m$ is connected with all the $m$ vertices and $m-1$ edges in it. Now, if there is any more edge that is in $G$ but not in $G_m$, then it must form a cycle in $G$ because there are would be two paths connecting the endpoints of this edge, one in $G_m$ and other is this edge, which contradicts our assumption in the claim. Hence, there are exactly $m-1$ edges in the graph $G$.
Claim 2: If $G$ is a simple graph with $m$ vertices, $m-1$ edges and no cycle in it, then $G$ is connected.
Proof. Let the graph $G$ have $t$ connected components in it. We know that any two connected components are each connected and disjoint so that there is no path connecting them, because they form a partition for both vertex and edge sets. Since each of the connected component is connected with no cycle in it, the number of edges in it is exactly one less than number of vertices in it. So, the total number of edges in $G$ will be $m-t$, because components partition all the edges. Hence, $t=1$ as there are $m-1$ edges. So, there must be only one connected component, which means that $G$ is connected.
Claim 3: If $G$ is a simple graph with $m$ vertices and at least $m$ edges, then $G$ must have a cycle.
Proof. Assume that there is no cycle in $G$. Now, remove few edges from the graph to get a subgraph $G_0$ with the same $m$ vertices and $m-1$ edges and no cycle in it. By the Claim 2, $G_0$ should be connected. Now, if you insert back any one of the removed edge, say $E$, into $G_0$, it forms a cycle because there are now two paths connecting the endpoints of $E$, one in $G_0$ and other is $E$, which is a contradiction to our assumption that there was no cycle in $G$. Hence, $G$ must have a cycle.
Now assume that Gauss writes at least $n$ distinct even integers and we will derive a contradiction. For every distinct even integer that Gauss writes down, construct an edge between the two points whose logarithmic distance corresponds to this value. Since there are at least $n$ distinct values, there are at least $n$ edges constructed with distinct even numbers, which forms a simple graph with vertices as the $n$ points.
Thus, we have a graph with $n$ vertices and $n$ edges and so by the Claim 3, there must be a cycle in the graph. Suppose say that the distinct even integers are $2k_1, 2k_2, 2k_3, \ldots, 2k_r$, where $r\in\mathbb{N}$ is the cycle length. Without loss of generality, let $2k_r$ be the largest and let $2k_i$ be the second largest integer. Then sum of distances of the edges corresponding to the logarithmic values $2k_1, 2k_2, \ldots, 2k_{r-1}$ is less than,
$2^{2k_1+1}+2^{2k_2+1}+\cdots+2^{2k_{r-1}+1} \leq 2^{2k_i+1}\left(1+2^{-2}+2^{-4}+\cdots+2^{-2(r-2)}\right)$
since $k_i$ is the second largest and the integers are distinct. So we get,
$2^{2k_i+1}\left(1+2^{-2}+2^{-4}+\cdots+2^{-2(r-2)}\right) = 2^{2k_i+1}\left(\frac{1-\left(2^{-2}\right)^{r-1}}{1-2^{-2}}\right) = \frac{4}{3}2^{2k_i+1}\left(1-2^{-(2r-2)}\right)$
$\Rightarrow 2^{2k_1+1}+2^{2k_2+1}+\cdots+2^{2k_{r-1}+1} \leq \frac{4}{3}2^{2k_i+1}\left(1-2^{-(2r-2)}\right) < \frac{4}{3}2^{2k_i+1} < 2^{2k_i+2} \leq 2^{2k_r}$
Triangle Inequality states that for any $r\geq 3$ points in the Euclidean plane, say $x_1, x_2, \ldots, x_r$, we have, $d(x_1, x_2) + d(x_2, x_3) + d(x_3, x_4) + \cdots + d(x_{r-1}, x_r) \geq d(x_r, x_1)$, which is obtained by addition of all the triangle inequalities in the triangles $(x_1, x_2, x_3)\ ;\ (x_1, x_3, x_4)\ ;\ (x_1, x_4, x_5)\ ;\ \ldots\ ;\ (x_1, x_{r-1}, x_r)$. But the above $r$ points violate this inequality as the largest distance is greater than the sum of other $r-1$ distances. Hence, we get a contradiction.
So, Gauss writes at most $n-1$ distinct even integers and by a similar argument, he writes at most $n-1$ distinct odd integers and hence, Gauss writes at most $2n-2$ distinct integers in total which is less than $2n$.
Problem 6
Euclid has a tool called cyclos which allows him to do the following:
• Given three non-collinear marked points, draw the circle passing through them.
• Given two marked points, draw the circle with them as endpoints of a diameter.
• Mark any intersection points of two drawn circles or mark a new point on a drawn circle.
Show that given two marked points, Euclid can draw a circle centered at one of them and passing through the other, using only the cyclos.
Solution :
For any given non-collinear points $X, Y, Z,$ let $\left(XY\right)$ represent the circle drawn with $XY$ as the diameter, $\left(XYZ\right)$ represent the circumcircle of $\triangle XYZ$ drawn using the cyclos.
Claim 1: Given any 3 non-collinear points that forms a non-right angled triangle, one can draw the nine-point circle of this triangle using only the cyclos.
Proof. Let $A, B, C$ be the three non-collinear points, then the intersection of $(AB)$ and $(AC)$ is the foot of perpendicular from $A$ to $BC$, say $P_A$. Similarly one can construct all the foot of perpendiculars, $P_B, P_C$. The points $P_A, P_B, P_C$ must be distinct as $\triangle ABC$ is non-right angled triangle. Now the circle $\left(P_AP_BP_C\right)$ is the nine-point circle of $\triangle ABC$ as it is the unique circle passing through all the foot of perpendiculars.
Claim 2: Given two points in the plane, one can mark its mid-point using only the cyclos.
Proof. Let $A, B$ be the two points, then draw the circle $(AB)$. Choose a new point $X$ on the circumference of $(AB)$. Mark the other intersection of $(XA)$ and $(XB)$ as $D$, which is the foot of perpendicular from $X$ to $\overline{AB}$. Mark the intersections of $(AD)$ and $(XD)$ to be $E$ and $(BD)$ and $(XD)$ to be $F$, which are the feet of perpendiculars from $D$ to $XA$ and $XB$ respectively. Since, $\angle AED = \angle BFD = 90^{\circ}$ and $D$ lies inside the segment $\overline{AB}\Rightarrow \angle AEB$ and $\angle AFB >90^{\circ}$, i.e., obtuse and hence $\triangle AEB$ and $\triangle AFB$ are non-right angled triangles and so we can construct the nine-point circles of $\triangle AEB$ and $\triangle AFB$ by the Claim 1. Observe that $X$ is the foot of perpendiculars from $A$ to $BF$ and $B$ to $AE$, so both the nine-point circles must pass through $X$ and also they both must pass through the midpoint of $\overline{AB}$ by definition and hence we mark the other intersection point of the two nine-point circles to get the midpoint of $\overline{AB}$, say $M$, which is illustrated in the diagram below,
Remark: Dotted red circles represent the nine-point circles of $\triangle AEB$ and $\triangle AFB$. The two nine-point circles are different, since the feet of perpendiculars from $E$ and $F$ to $AB$ are different and each of them do not lie on the other nine-point circle. And two different circles intersect at most at two points and hence we can uniquely mark the midpoint of $\overline{AB}$ without any dilemma, since the intersection point $X$ is already marked.
Construction:
Suppose say that the two marked points as per the question to be $A$ and $B$. Consider point $C$ as the reflection of $B$ with respect to the point $A$, so that $A,B,C$ are collinear with $AC=AB$, but you need not mark $C$. Mark a new point $Y$ on the circumference of $(AB)$. Mark the intersection of $(YA)$ and $(YB)$ to be $U$, which is the foot of perpendicular from $Y$ to $AB$. By the Claim 2, we can mark the midpoint of $\overline{YB}$ as $W$ using only the cyclos. Now, construct $(AUW)$ and mark the other intersection of $(AUW)$ and $(YB)$ to be $S$. Note that $(AUW)$ is the nine-point circle of $\triangle CYB$ and thus feet of perpendiculars from $Y$ to $CB$ (which is $U$) and $B$ to $CY$ lie on both $(AUW)$ and $(YB)$. So, other intersection $S$ is the foot of the perpendicular from $B$ to $CY \Rightarrow \angle CSB=90^{\circ}$. Observe that $A$ is the midpoint of hypotenuse, which is the circumcenter of $\triangle CSB \Rightarrow AB = AS$.
Repeat the same procedure by marking a new point $Z$ on the circumference of $(AB)$ to obtain the foot of perpendicular from $B$ to $CZ$, say $T$, so that $AB=AT$. Most probably, $S$ and $T$ will be distinct because $CY$ and $CZ$ cannot intersect at more than 1 point. But it might happen that $C, Y, Z$ are collinear, in which case the lines $CY$ and $CZ$ coincide and so the points $S$ and $T$ coincide. In that case, we take some other new point $Z'$ on the circumference of $(AB)$ and repeat the procedure to obtain a new point $T'$, which must be different from $S$ because now $CY$ and $CZ'$ are two different lines.
Now, Construct $(BST)$ using cyclos, which is the required circle passing through $B$ and centered at $A$ because $AB = AS = AT$ and there is such a unique point which should be the circumcenter of $\triangle BST$. The final construction diagram is provided below with dotted red circle as the required circle,
Remark: This proves that the cyclos can perform everything that a compass can perform in constructions because now cyclos can construct a circle passing through a given point centered at a given point.
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+ Learn a new musical skill in ten minutes. Explore Soundfly’s wide array of free online courses and gain a musical edge during your lunch break. Or sign up for the Soundfly Weekly newsletter and learn something new every Tuesday!
One of the most valuable skills a musician can have is being able to identify musical intervals by ear. Once you can comfortably pick out musical intervals as you hear them, you are officially on your way towards the upper echelons of musical ability. But, let’s step back for a moment. What is an interval?
An interval is simply the distance between two notes. We classify intervals in two ways— by quantity and by quality. Quantity tells us roughly how far apart the notes are on the scale; and quality tells us more about the unique sound of the interval or which scale it’s pulling from.
Whether you’re a vocalist learning to sight sing from sheet music or an instrumentalist working on transcribing songs, arranging for an ensemble, or even composing new musical material, learning to identify intervals by ear can be useful across almost any situation. Plus, if you can identify intervals by ear, who needs perfect pitch?
The best way we’ve found to practice identifying note intervals by ear is to associate each interval with a familiar song or melody that you could likely sing in your head already. All you have to do is commit a piece of the melody to memory, and voilà, you’re on your way to interval recall!
We’ve already covered this with ascending intervals, so let’s look at how to identify descending intervals using popular songs and melodies. Wherever possible, we’ve offered more than one example, in case you’re not familiar with the first. Ready?
Ludwig van Beethoven’s “Für Elise” provides us with the first example of a minor second, and it can be found in the first “seconds” of the piece — in other words, the first two notes are a descending minor second apart.
We have George Frideric Handel to thank for the next example, in a melody that would later be adapted as the holiday standard, “Joy to the World.” The minor second interval here appears with the lyrics “joy” and “to” at 0:04 in the version below. Easy enough!
The Beatles’ classic “Yesterday” is where we can find a handy example of the major second. Just listen to the syllables “yes” and “ter” around 0:05.
Another simple-to-remember example is the nursery rhyme, “Three Blind Mice.” The first two notes in the melody are what you’ll want to focus on, at 0:00.
Everyone can sing this tune off the top of their heads! Listen to “Frosty the Snowman,” and focus on when the lyrical syllables move from “fros” to “ty” at around 0:12.
And since we’re here, we may as well bring it back to the Fab Four again. The Beatles‘ “Hey Jude” starts on a minor third interval, once again at 0:00.
It seems we can always turn to a popular old melody for some familiar sing-along action. The lyrics of this tune have a bit of an icky past, harking back to the American Civil War, but like most public domain songs to stand the test of time, the simplistic melody often reminds us of childhood. In “Shoo Fly, Don’t Bother Me” you’ll find the major third at 0:04, on the lyrics “Shoo” and “fly.”
But don’t get too comfortable with that one, here’s perhaps the most well-known melody of the bunch: the first two unique notes of Beethoven’s 5th Symphony. The most classic-est of classic major thirds.
For a descending perfect fourth, look no further than the traditional “I’ve Been Working on the Railroad,” as the lyrics “I’ve” and “been” at 0:00 feature this interval.
The tritone is a mainstay interval of heavy, dissonant rock riffage. In a most classic example, Black Sabbath’s self-titled song “Black Sabbath” (off the self-titled record, Black Sabbath) hits us with this massively dissonant tritone as soon as the band enters at 0:36, first jumping an octave before descending a gnarly diminished fifth, aiming to invoke the unequivocal power of the devil. The first time features a fast trill on the guitar, with a cleaner example of the interval at 0:47.
Pyotr Ilyich Tchaikovsky’s ballet, “Swan Lake, Opus 20” contains a very distinct melody that you might not think you’d recognize until you hear it. At 0:03 of Act II, the oboe solo clearly establishes the perfect fifth in a beautiful descending line.
Alright, this one is another melody you’ve definitely heard before, whether you think you have or not. Francis Lai’s theme song to the film Love Story actually opens with two notes ascending a minor sixth, but if you skip ahead to 0:24, just after a short horn fanfare, those two notes reverse briefly before entering the meat of the run. Either way you play those two notes back, it’s a minor sixth interval, but train your ears to hear it as a descending melody to recognize the gap as such.
This minor sixth interval wastes no time, jumping in right at the top of the tune. It’s a bit tough to pick out of the dissonant, bluesy context that D’Angelo sets up with the rest of his opening riff, but if you can isolate just those first two notes, you’ll have a perfectly handy minor sixth to memorize.
Here’s one of my favorite on the list! One of Michael Jackson’s most uplifting melodies ever, his lyrical ramp up to the title refrain in “Man in the Mirror,” actually gets a jump start with a downward interval of an enormous major sixth gap! At 1:06, when he sings, “I’m” and “start,” you can hear how energetically it leads into the ascending melody. Top-line composers, take note!
Herbie Hancock’s funky anthem, “Watermelon Man,” from Head Hunters, was originally painted with lighter colors as a piano-driven jazz tune in 1962. While it opens with some rocking back and forth on two chords, at 0:15, the first two notes of the lead melody take us a minor seventh down.
Cole Porter’s standard, “I Love You,” sung below by Sarah Vaughan, provides a broad, sweeping plunge into the major seventh interval. Listen as Vaughan sings “Love” and “You” after briefly holding on the “I” around 0:21.
Lastly, for the television-savvy listener, the sort-of-famous theme music to the sort-of-famous ’90s sitcom, Doogie Howser, M.D. (written by Mike Post) can also serve as a handy reminder of the descending octave interval. Bop along with the synth at 0:00.
New to Soundfly?
All of our mentored online courses come with six weeks of one-on-one professional coaching, guidance, and feedback on your work. It’s like having a personal trainer, but for music! Whether you’re interested to dive deep into a topic covered by one of our courses, like Unlocking the Emotional Power of Chords, The Creative Power of Advanced Harmony, and Orchestration for Strings. Here’s a snippet of what to expect!<|endoftext|>
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A blood transfusion is a common procedure in which a donor’s blood is infused into the patient’s blood to help replace missing blood components. Patients can receive red blood cells, white blood cells, platelets or plasma. Before receiving a transfusion, donor blood undergoes a rigorous screening process and medical staff must perform many safety measures to ensure that the patient’s transfusion is administered safely.
It is imperative that the donor blood a person receives matches the patient’s own. Laboratories determine a person’s blood type based on the presence or absence of A or B antigens. According to the National Institutes of Health, antigens are a chemical response that produces an immune response in the body. For example, B blood types will have only B antigens. Each blood type also has anti-antigens that will cause the body to have a severe immune reaction. Blood types A have a B anti-antigen, meaning that if they receive type B blood, they will suffer from a hemolytic reaction, which causes blood cells to burst and leak toxins into the bloodstream. Blood typing also examines the Rh factor. People are Rh positive or Rh negative. Giving positive blood to a person with a negative blood type will not initially cause a reaction but could result in a severe or fatal transfusion reaction in later transfusions.
Crossmatching is the second step in the blood transfusion process. After the donor’s blood and recipient’s blood are typed and Rh factor is determined, both blood samples will undergo screening for specific chemicals known as antibodies to avoid a reaction. Laboratories must repeat this process for every blood transfusion the patient receives.
Informed Consent and Verification
According to the University of Michigan, before a blood transfusion, patients must sign a piece of paper known as an informed consent. An informed consent form ensures that the patient is aware of how the procedure works and the risks, the benefits and any alternatives. A physician, a nurse or a physician’s assistant must obtain the informed consent. Before the transfusion, the nurse will verify through a series of blood component and patient identifiers that the right blood is being given to the right patient.
Before the transfusion, a set of vital signs will be taken determining the patient’s temperature, heart rate, breathing rate and blood pressure. This will serve as a baseline during the transfusion. The nurse administers the transfusion via an intravenous catheter placed into the vein. Nurses will monitor the patient closely for the first 15 minutes and then take another series of vital signs looking for any symptoms of a reaction to the blood components. Reaction symptoms include rash, itching, fever, bloody urine, chills and back or chest pain. The nurse will collect vital signs again after 30 minutes and every hour after until the transfusion is complete.
In the event that a patient experiences a minor allergic reaction including itching, hives or a mild fever, physicians may choose to treat with an antihistamine known as diphenhydramine rather than stop the transfusion. In the event of a severe reaction, the nurse will immediately stop the transfusion and blood from both the patient and the donor will be sent to the lab for further analysis. Physicians will often prescribe antihistamines to control itching and rash, corticosteroids to control the immune response and possibly fluids and diuretics to prevent kidney failure.<|endoftext|>
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# 1987 AJHSME Problems/Problem 22
## Problem
$\text{ABCD}$ is a rectangle, $\text{D}$ is the center of the circle, and $\text{B}$ is on the circle. If $\text{AD}=4$ and $\text{CD}=3$, then the area of the shaded region is between
$[asy] pair A,B,C,D; A=(0,4); B=(3,4); C=(3,0); D=origin; draw(circle(D,5)); fill((0,5)..(1.5,4.7697)..B--A--cycle,black); fill(B..(4,3)..(5,0)--C--cycle,black); draw((0,5)--D--(5,0)); label("A",A,NW); label("B",B,NE); label("C",C,S); label("D",D,SW); [/asy]$
$\text{(A)}\ 4\text{ and }5 \qquad \text{(B)}\ 5\text{ and }6 \qquad \text{(C)}\ 6\text{ and }7 \qquad \text{(D)}\ 7\text{ and }8 \qquad \text{(E)}\ 8\text{ and }9$
## Solution
The area of the shaded region is equal to the area of the quarter circle with the area of the rectangle taken away. The area of the rectangle is $4\cdot 3=12$, so we just need the quarter circle.
Applying the Pythagorean Theorem to $\triangle ADC$, we have $$(AC)^2=4^2+3^2\Rightarrow AC=5$$ Since $ABCD$ is a rectangle, $$BD=AC=5$$
Clearly $BD$ is a radius of the circle, so the area of the whole circle is $5^2\pi =25\pi$ and the area of the quarter circle is $\frac{25\pi }{4}$.
Finally, the shaded region is $$\frac{25\pi }{4}-12 \approx 7.6$$ so the answer is $\boxed{\text{D}}$<|endoftext|>
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# How do you factor r^3 - 1?
Dec 20, 2015
Use the difference of cubes identity to find:
${r}^{3} - 1 = \left(r - 1\right) \left({r}^{2} + r + 1\right)$
#### Explanation:
The difference of cubes identity can be written:
${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$
Use this with $a = r$ and $b = 1$ as follows:
${r}^{3} - 1$
$= {r}^{3} - {1}^{3}$
$= \left(r - 1\right) \left({r}^{2} + \left(r\right) \left(1\right) + {1}^{2}\right)$
$= \left(r - 1\right) \left({r}^{2} + r + 1\right)$
If you allow Complex coefficients then this can be factored a little further:
$= \left(r - 1\right) \left(r - \omega\right) \left(r - {\omega}^{2}\right)$
where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.<|endoftext|>
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# AB is a a diameter and AC is a chord of a circle with centre O such that angle BAC =30 the tangent at C intersects extended AB at a point D . prove that BC =BD.
Given : A circle with AB as diameter having chord AC.
∠BAC = 30°
Tangent at C meets AB produced at D.
To prove : BC = BD
Construction : Join OC
Proof : In Δ AOC,
OA = OC (radii of same circle)
⇒ ∠1 = ∠BAC (angles opposite to equal sides are equal)
⇒∠1 = 30°
By angle sum property of Δ,
We have, ∠2 = 180° – (30° + 30°)
= 180° – 60°
= 120°
Now, ∠2 + ∠3 = 180° (linear pair)
⇒ 120° + ∠3 = 180°
⇒ ∠3= 60°
AB is diameter of the circle.
We know that angle in a semi circle is 90°.
⇒ ∠ACB = 90°
⇒ ∠1 + ∠4 = 90°
⇒ 30° + ∠4 = 90°
⇒ ∠4 = 60°
Consider OC is radius and CD is tangent to circle at C.
We have OC ⊥ CD
⇒ ∠OCD = 90°
⇒ ∠4 + ∠5 (=∠BCD) = 90°
⇒ 60° + ∠5 = 90°
⇒ ∠5 = 30°
In ΔOCD, by angle sum property of Δ
∠5 + ∠OCD + ∠6 = 180°
⇒ 60° + 90° + ∠6 = 18°
⇒ ∠6 + 15° = 180°
⇒ ∠6 = 30
In ΔBCD , ∠5 = ∠6 (= 30°)
⇒ BC = BD (sides opposite to equal angles are equal)
• 35
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