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1. What is hepatitis A?
Hepatitis A, also known as infectious hepatitis, is a viral disease of the liver. The hepatitis A virus is relatively stable in the environment, and is easily spread to close contacts who are not immune.
2. Who can get hepatitis A?
All people are at risk of hepatitis A infection. Hepatitis A can spread among young children in daycare because many are in nappies and cannot wash their own hands. These children are usually asymptomatic, and so prevention is difficult. People who share the same household, and sexual partners of persons with hepatitis A are at risk of becoming infected. Persons who develop hepatitis A, or persons who are vaccinated are protected for life against subsequent infections.
3. Where does hepatitis A occur in South Africa?
The distribution of hepatitis A infection varies according to socio-economic conditions. Persons in low socio-economic conditions tend to become infected with hepatitis A very early in life, as infants or young children. Consequently, the disease in this population is largely silent, as children tend to have asymptomatic disease with no long term consequences. Amongst persons from higher socio-economic conditions, the disease tends to infect persons of older age. Paradoxically therefore, the disease is more apparent amongst these groups, as jaundice is more frequent. In studies done in the early 1990’s in South Africa, 90% of black and 60% of white South African adults had evidence of previous hepatitis A infection.
4. How is hepatitis A transmitted?
Hepatitis A virus is found in the stools (faeces) of persons with active disease. It is spread by the ‘faeco-oral’ route – for example - when people don’t wash their hands after using the toilet, or changing a nappy or soiled sheets, and then touch their mouths, prepare food for others, or touch others with their contaminated hands. It may also be spread by contaminated food or water. A person with hepatitis A is most contagious 2 weeks before illness starts until 1 week after the onset of jaundice. Amongst household contacts of persons with hepatitis A, only 20-50% of contacts develop hepatitis A.
5. What are the signs and symptoms of hepatitis A?
The incubation period of hepatitis A ranges from 15-50 days. Symptoms are usually age dependent. Adults are more likely to have symptoms of fever, fatigue, loss of appetite, nausea, diarrhoea and jaundice. With the onset of jaundice persons will also notice dark brown urine, pale stools (faeces) and yellow discolouration of skin and eyes. Young children may have mild flu-like symptoms or no symptoms, and seldom become jaundiced. Symptoms normally last 1-2 weeks but may last longer.
6. How is hepatitis A diagnosed?
Hepatitis A is diagnosed through a blood test. The laboratory will look for antibodies to the hepatitis A virus. Laboratory tests can identify persons who have active disease, or persons who are immune to hepatitis A – either through past infection, or through vaccination
7. How is hepatitis A treated?
There is no treatment for the disease and most people do not need any, as the condition resolves on its own. Amongst the elderly, and persons with other underlying illness, hepatitis A can cause severe illness and rarely, death from liver failure.
8. How can hepatitis A be prevented?
Hepatitis A can be effectively prevented by vaccination. However, vaccination against hepatitis A is not part of the South African Expanded Programme of Immunisation (EPI). Persons who wish to receive hepatitis A vaccine are required to purchase it privately. Two vaccines are licensed for use in South Africa – Havirx®and Avaxim®. Twinrix® which provides vaccination against hepatitis A and B in a single infection is also available. Both require an initial injection with a booster after 6-12 months. Even in its mildest form Hepatitis A is not a pleasant disease as an older child or adult. Vaccination is advised.
When an adult is diagnosed with hepatitis A, this person should not handle food, nor care for patients, nor work with young children or the elderly until 14 days after the onset of illness. Scholars/learners infected with hepatitis A should be excluded from school for 1 week following onset of jaundice or, if they did not become jaundiced, for 14 days after onset of their illness.
Persons who are susceptible to hepatitis A (not vaccinated and never had infection) and who have been in close contact with someone with hepatitis A up to two weeks before onset of their illness, and up to 1 week after the onset of jaundice can receive hepatitis A vaccine to prevent infection. To be most effective, the hepatitis A vaccine should be given within 2 weeks of exposure to the case. The hepatitis A vaccine may not be effective as post- exposure prevention in those who have underlying illness, and there is not much data regarding its effectiveness in those over 40 years of age. These people require an injection of immunoglobulin (antibodies) to prevent infection after exposure. To be most effective, immunoglobulin (pooled human immune globulin) should be given within 2 weeks of exposure to the infected person.
General preventative efforts for hepatitis A include good hand hygiene, especially after use of the toilet, and before food preparation.<|endoftext|>
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Three Kinds of “Average”
There are three different statistics that are commonly taught as “averages”, or “measures of central tendency”, of a set of numbers: mean, median, and mode. (There are others as well, which we will get to later.) What are they? How do they differ? How do you use them? We’ll look into questions like these as we start a series on averages. To some extent, this will be an expansion of the post Averages, Probability, and Reality.
Defining the terms
Our first question is from 1997:
Mean, Median, Mode, Range
What are definitions for mean, median, mode, and range?
Several of the questions here will mention range; I’m going to omit that from most answers, to narrow our focus.
Doctor Scott answered with straightforward definitions by means of an example:
Hi Jamie!
Let's define each of the words and give an example.
Consider the set of numbers 80, 90, 90, 100, 85, 90. They could be math grades, for example.
The MEAN is the arithmetic average, the average you are probably used to finding for a set of numbers - add up the numbers and divide by how many there are: (80 + 90 + 90 + 100 + 85 + 90) / 6 = 89 1/6.
The MEDIAN is the number in the middle. In order to find the median, you have to put the values in order from lowest to highest, then find the number that is exactly in the middle:
80 85 90 90 90 100
^
Since there is an even number of values, the MEDIAN is between these two, or it is 90. Notice that there is exactly the same number of values ABOVE the median as BELOW it!
The MODE is the value that occurs most often. In this case, since there are 3 90's, the mode is 90. A set of data can have more than one mode.
Observe that the mean is a “center of gravity” takes into account each individual value, so it is pulled strongly toward the extreme:
I represent it with a triangle representing a fulcrum on which the data are balanced.
The median ignores the actual values, and just takes into account their order, so that it is “in the middle” in a different sense:
The mode pays attention only to the single most common value (if there is one):
When we use the word “average” by itself, we typically mean the mean. Next week, we’ll look into some details about it; we’ll also see that just as there are different “averages”, there are different “means”!
We have discussed elsewhere some subtleties in the definition of the median. We will eventually have a similar in-depth look at the mode.
We have previously looked at differences between these statistics in Mean, Median, Mode: Which is Best?, which focused on choosing the most appropriate measure for a given situation.
A single number to represent the whole
For more about what each of the “averages” means, and how they differ, we can turn to this question from 1998:
Mean, Median, and Mode
I understand the median, but I am having trouble understanding the mode. Can you please explain?
Doctor Anthony answered, explaining all three:
If you have a set of numbers, say the ages of pupils in a group, then there are 3 different ways of finding a single number to represent the whole group. The most common is the 'mean' or average. For the mean you add up all the ages and divide this total by the number of pupils.
The second way you could find a single representative number is to arrange all the pupils in a line in ascending order of age, with the youngest on the left and oldest on the right. You then go to the person standing in the exact middle of this line and find his/her age. This will be the median age. If there is an even number of pupils you will not have a single person at the midpoint, so you will take the middle pair and give the average of their ages as the median age.
The third way to find a single representative number is to group the pupils by age, so you could have 5 pupils of age 10, 8 pupils of age 11, 14 pupils of age 12, 7 pupils of age 13, 2 pupils of age 14 and 3 pupils of age 15. Looking at this distribution of ages you see that the biggest group is those of age 12, so you say the 'mode' of the distribution is 12. In short, the mode is the most frequently occurring value. If no value occurs more than once then you don't have a mode. Sometimes two values will occur at an equal but greater frequency than other values, and in this case we say that the distribution is bi-modal.
As we’ll see, the main weakness of the mode is that there may not be a mode at all, or there may be more than one.
Working through an example, in detail
Here’s another question from 1998 that asked about one example, and got a little more:
Finding Mean, Median, Mode
I have been out of school for 8 years and I am trying to get my diploma. On one of my practice tests it said to get the mean, median, and mode of scores. However in my textbook there are no examples to show me how to do it. The scores are: 100, 78, 93, 84, 91, 100, 82, 79. I would greatly appreciate if someone would show me how to do them.
Here are Adam’s data, placed on a number line:
Adam,
The mean of the scores is another name for their average. Just add them up and divide by the number of scores:
mean = (100+78+93+84+91+100+82+79)/8
= 88.375
The median of the scores is "the number in the middle" when the scores are sorted in order. In your example:
100 100 93 91 84 82 79 78
If there is an even number of scores (as in your example) there is no number in the middle so the two numbers in the middle are averaged:
median = (91 + 84)/2 = 87.5
The mean of the numbers can be misleading. If I tell you the mean income in my neighborhood is 1 million dollars a year you might think that I am wealthy. But maybe the neighborhood is a poor one with one very rich person making many millions of dollars a year. The average will be large because one number is very large.
The median and the mean together give a better idea of the spread of the numbers. If there were one really wealthy person then the median income would be quite low. In your case the median is quite near to the mean, which suggests that the spread is probably evenly balanced.
The comments here are that the mean is highly affected by “outliers”, values that are far from the others. This is revealed when the median is very different (as in the first example in this post); in the present example, the numbers are distributed fairly evenly.
Here are the three averages on the number line:
The mode of a set of numbers is the number that occurs most often. In your example:
mode = 100
The mode may help to correct false impressions if you know the mean and the median but don't actually see the data.
For example, if I tell you the mean of four numbers is 50.5 and the median is 50 you may think that the four numbers are close to one another, like the numbers in your example. In fact, however, I am thinking of:
101 99 1 1
The mean is (101 + 99 + 1 + 1)/4 = 202/4 = 50.5 and the median = (99+1)/2 = 50. But the mode here is 1. If you know that, you will know that 1 is repeated more than once in the data and, since the mean is near 50 with only four numbers, that might give you the idea that the data include several large numbers and several small numbers rather than four numbers close to one another.
As we’ve seen, the mode doesn’t always even exist, and therefore is not very useful; yet there are times when it can be very revealing. In Adam’s problem, however, the mode really tells us nothing! It just happens that two values are the same, and if one of the 100’s were changed to 99, there would be no mode at all.
Why do we need three “averages”?
The next question, also from 1998, is not about “what”, but about “why”:
Range, Mean, Median, and Mode
I have some questions that you may want to answer for me:
1. Why do we have to study range, mean, median, and mode?
2. Could you help me understand them more?
3. How is it going to help me later in life?
Doctor Stacey took this one:
Hi Stephanie! Thanks for writing Dr. Math.
Mean, median, and mode are all types of averages, although the mean is the most common type of average and usually refers to the _arithmetic mean_ (There are other kinds of means that are more difficult).
The arithmetic mean is a simple type of average. Suppose you want to know what your numerical average is in your math class. Let's say your grades so far are 80, 90, 92, and 78 on the four quizzes you have had. To find your quiz average, add up the four grades:
80 + 90 + 92 + 78 = 340
Then divide that answer by the number of grades that you started with, four: 340 / 4 = 85. So, your quiz average is 85! Whenever you want to find a mean, just add up all the numbers and divide by however many numbers you started with.
Here is the mean; you may notice something special about the data:
The term “arithmetic mean” is used to distinguish “the” mean from other means we’ll be looking at next week.
But sometimes the arithmetic mean doesn't give you all the information you want, and here is where your first and third questions come in. Suppose you are an adult looking for a job. You interview with a company that has ten employees, and the interviewer tells you that the average salary is $200 per day. Wow, that's a lot of money! But that's not what you would be making. For this particular company, you would make half of that. Each employee makes$100 per day, except for the owner, who makes $1100 per day. What? How do they get$200 for average then?!
Well, let's take a look:
Nine employees make $100, so adding those up is 9 x 100 = 900. Then the owner makes$1100, so the total is $1100 +$900 = $2000. Divide by the total number of employees, ten, and we have$2000/10 = $200. Because the owner makes so much more than everyone else, her salary "pulls" the average up. The owner’s salary is an “outlier”, which was mentioned before: Only the mean is affected by the outlier! The single outlier is 9 times as far from the mean as the others. Although the outlier doesn’t pull the mean too far, it does make the mean deviate from what is typical. A better question to ask is, "What is the _median_ salary?" The median is the number in the middle, when the numbers are listed in order. For example, suppose you wanted to find the median of the numbers 6, 4, 67, 23, 6, 98, 8, 16, 37. First, list them in order: 4, 6, 6, 8, 16, 23, 37, 67, 98. Now, which one is in the middle? Well, there are nine numbers, so the middle one is the fifth, which is 16, so 16 is the median. We don’t need a number line to show the median: By the way, the way I commonly find the middle is to divide the count by 2 and round down. Then I know there will be that many on each side. In this case, $$9\div 2 = 4.5$$, which rounds down to 4, so there are 4 below and 4 above the median. Now, what about when there is an even number of numbers? Look at the quiz grade example again: 90, 80, 92, 78. First list the numbers in order: 78, 80, 90, 92. The two middle ones are 80 and 90. So do we have two medians? No, we find the mean of those two: 80 + 90 = 170, and 170 / 2 = 85. So 85 is the median (and in this case the same as the mean)! Here we see that the mean and median are the same because the data are symmetrical: Here when I divide 4 by 2 I get 2 exactly, so I don’t need to round down; and again, there are 2 numbers on each side of the median, which this time is between the two sets of two. What about outliers? Now look at those salaries again. To find the median salary, we look at the salaries in order: 100, 100, 100, 100, 100, 100, 100, 100, 100, 1100. This is an even number of x salaries, so we look at the middle two. They are both 100, so the median is$100. That's much better at telling you how much you'll make if you accept the job.
We saw this above:
Again, I divide 10 by 2 to get exactly 5, so there are 5 on each side of the middle; the fifth and sixth numbers have to be averaged.
But the median doesn't always give you the best information either. Suppose you interview with a company that has 10 general employees, 7 assistants, 3 managers, and 1 owner. For this company, the mean salary is $400, and the median is also$400. But you are applying for the position of general employee, whose starting salary is $100! Why are the mean and median so far away? Well, the 10 general employees each make$100. The 7 assistants each make $400, the 3 managers each make$900, and the owner makes $1900. If you do the math to find the median or mean,$400 is the answer (try it!). So what can you do?
Here are these employees’ incomes on a number line:
And here is the calculation for the median:
and for the mean: $$\frac{10\times 100+7\times 400+3\times 900+1900}{21} = \frac{8400}{21} = 400$$
(Here rather than list all the numbers as I did for the median, I used multiplication as a shortcut for that long addition. This leads, as we’ll see later, to the idea of the “weighted mean”.)
The mode is the type of average you want to know in this situation. The mode is the number that occurs most frequently. In the example for median, 6 would be the mode because it occurs twice, while the other numbers each occur once. In our employee example, the mode is \$100 because that number occurs ten times, which is more than any other number occurs.
Here is a similar question about the need for different averages, giving a researcher’s perspective on the various statistics:
Using Mean, Mode, and Range
Which is which?
I’ll close with this question, from 2010:
Remember the Mean -- and the Median, and the Mode, and the Range, and the Outlier
I need help with mean, median, mode, range, and outlier. I just can't remember which one's which -- especially the first four (I kinda already know "outlier").
All of them are rather hard. I don't have the first idea of how to do them, so I can't show any work. And since most of them begin with the letter "m," it gets confusing.
My math teacher hasn't give us any memory strategies (or if she has, they weren't memorable).
Also, I have a b-i-i-i-g-g-g AIMS test tomorrow about data analysis. I want to do well, especially since math is my worst subject. I'm sure you get that a lot.
So if you could respond today, that would be awesome. But p-p-l-l-e-e-a-a-s-s-e, by tomorrow.
Thank you, Dr. Math!
Hi, Riley.
The main way to learn which word has which meaning is just to use them a lot. Do you have some trick to remember which of your friends has which name? Probably not (unless they are twins and you don't know them well as individuals). Once you've spent enough time with someone, you know who they are! The same is true with vocabulary like this.
Having said that, let's see if we can make the meanings more memorable, so you can get to know them as individuals, keeping their meaning in mind as you use them.
An outlier is a number that doesn't fit in with others in its group -- sort of a loner. It "lies outside" of the main group.
The range of a set of numbers is how far they are spread out -- just as the range of a cell phone is how far you can go and still talk, or the range of an animal is how far it travels.
Everyday usages are often not as far from the technical usage as you think, when you look at them right!
The mean of a set of numbers is just another word for the average -- add them up and divide by how many.
The median is the "middle" -- the number in the middle when they are lined up in order. If there are two middle numbers, calculate the median by taking their average (mean).
The mode is the "most common." You may hear the word "mode" as in "a la mode" (originally meaning "fashionable") or "modish" (also meaning "stylish"). These mean that something is, in a sense, popular. The mode is the most "popular" value.
One last tip: of the three terms that start with the letter "m," "median" is the one that sounds most like "middle."
We could also say that “mean” is what we usually mean by “average”; “median” is in the middle alphabetically; and “mode” is … the other one.
5 thoughts on “Three Kinds of “Average””
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Let’s role play!
Who is/was your favorite superhero? Batman? Spider-Man? Superman? or maybe Sentry?
When I was a child, I’d love imaginary play. I’d love to role play as a superhero (usually Wolverine or Spider-Man), one of the Teenage Mutant Ninja Turtles (Raphael), or some other popular character from my childhood. I would play out entire story lines with my siblings or friends. We would have a lot of fun fighting and pretending to be our favorite heroes. Those were some great times!
Role play activities not only help develop imagination, but also speech and acting skills. You can read about how role play influences child development here and here. Lev Vygotsky, a major influence on current child developmental theory, said that when children participate in imaginary play.
They are developing the following skills:
- social skills – learning to play with others
- speech and language skills – produce words and sentences, imitating speech from another person
- and also problem solving skills – able to make props or assist playmates
How does this relate to the ESL classroom?
Role play activities help the development of language production. Students are able to practice pronunciation and speaking fluency, expand their vocabulary base, develop listening comprehension, and interpersonal communication skills. Drama activities also incorporate movement into the lesson, which help kids who cannot sit still for long periods of time.
Drama not only encourages ESL beginners to communicate at the “pre-production” stage of language acquisition, it’s also an engaging way of teaching any language learner.
Drama offers opportunities to simulate real-life situations, draws on the creativity of students and introduces them to the cultural significance of various gestures and body language familiar to native English speakers.
Acting out a story transcends language barriers. It can be used to encourage students to get an authentic “feel” for the weight of the words they are learning. It’s also great for the many students who prefer physical engagement in their learning and can provide an excellent jumping-off point to bridge from oral activities into reading and writing.
Teaching dramatic performance to ESL students may seem like a somewhat challenging task. It’s important to find a play script that’s at an appropriate skill level, and not too time-consuming. If you cannot find a good script on your own, then perhaps try your hand at writing one yourself (I’ve done it a number of times).
When choosing a script, you should look at what objective the students should meet.
- Should your students practice using target vocabulary?
- Should they practice improvisational language?
- Should they practice situational language?
- Do you want them to develop speaking fluency regardless of grammar mistakes or content?
These are but some things to consider when planning to use drama activities for ESL students. One of the main points is for students to practice producing language, rather than acquiring new language.
What are some benefits to using drama in the ESL classroom?
A quote from this article, sums it up well:
- the acquisition of meaningful, fluent interaction in the target language;
- the assimilation of a whole range of pronunciation and prosodic features in a fully contextualized and interactional manner;
- the fully contextualized acquisition of new vocabulary and structure;
- an improved sense of confidence in the student in his or her ability to learn the target language.” (Wessels, p.10).
That same article goes on to say this:
Drama for second language learners can provide an opportunity to develop the imagination of the students. The students can go beyond the here and now and even ‘walk in the shoes’ of another. It provides an opportunity for independent thinking (McCaslin 1996).
Drama also helps kids build self-confidence, team work, organizational and management skills, authentic language production and listening skills, taking pride in their work, and self-awareness (among other important skills). These skills can help students throughout their entire lives as they continue their education, and even into their job.
Tips for teaching ESL students
It’s great to know the benefits to teaching drama, and now here are some useful tips.
- Pre-learn the vocabulary first – You can use songs, flashcards, or memory games to help students get used to saying key words from the play.
- Start practicing simple lines – Once the students have learned the key words, have them start speaking lines from the play. It doesn’t matter if they say it perfectly, it’s about vocabulary in-context, as well as whole language production.
- Use props – Using props, only after the students have learned their lines, allow kids to develop their character. Props add a level of authenticity to the play/dramatic performance, and it’s also fun!
- Get them moving – Good storytelling not only comes from well-written lines of dialogue and prose, but also through movement. Students should incorporate eye contact, hand gestures, and various facial expressions into their performance.
- Get them talking – That’s one of the reasons to use drama for ESL students, isn’t it? Students have the opportunity to talk, and while some may be more confident in their abilities, less confident students are also able to participate. All students should work on their pace and volume – not speaking too quickly or slowly, nor too loud or too soft. They should also work on using a variety of speaking styles – like different accents, tones, and inflections, to better express their lines.
Examples of different kinds of dramatic activities
There are many ways to incorporate drama into ESL classrooms, and also for all ages – from kindergarten to adult ELLs. Here are just a few I’ve seen (source).
- Improvisation – use Theatresports games. The rules create great boundaries but encourage problem solving.
- Role Play – Acting out short scripted dialogue, making up dialogue.
- Reader’s Theatre – performing written stories. The writer reads the stories aloud whilst the others perform in mime.
- Masked Theatre
- Puppet Shows
One thing that’s important to remember, is to keep the language simple enough for students to work with. If the dialogue has too many difficult words above the students’ level, they won’t be able to develop their skills as easily. Instead, you should scale the language accordingly: simple sentences for kindergarten/young learners, and steadily increase the difficulty towards adult learners.
Speech and drama activities are not only fun, but they also allow students to develop a wide variety of skills. Drama imitates life and help students understand different situations more easily. They also develop empathy as they experience life through someone else’s eyes. As students grow and develop into adults, they may experience similar situations in their own lives someday, and these activities may give them the tools they need to resolve conflict, or deal with an important issue they face.
I’d like to share this quote,
Therefore, it makes sense that dramatic skills can help us become the person we want to be. In this way, drama has a wider reach than simply making us more fluent in a second language. It has the potential of making our lives better as we will be better understood and may help us become the people we want to be. Drama is all about how we present ourselves. If the student can communicate better, the more likely others will see him/her as he/she wishes to be seen.
As teachers, we should encourage our students to try new things, and challenge themselves. We should also not discourage them from speaking their minds. Dramatic play can help students develop their own ideas and challenge themselves to set goals.
Don’t be afraid to use dramatic play in your ESL classroom!<|endoftext|>
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Since the day King Croesus of Lydia ordered the first gold coins to be struck around 550 BC, civilizations around the world have been fascinated by gold's monetary qualities. International debts were settled with gold. Paper money could be exchanged for gold. And under what’s referred to as “the gold standard,” many world currencies were defined by a set quantity of gold. As simple as it sounds, the gold standard was a truly complex system. Read more to find out, specifically, what historians, gold buyers, and economists are referring to when they reference the gold standard, including the many ways its demise could be affecting the U.S. economy even today.
How did the gold standard work?
“The principle behind a gold standard,” writes Forbes contributor Nathan Lewis, “is to make a currency that is as stable in value, neutral, unchanging, and reliable as possible—the monetary equivalent of a standard measure like a kilogram or meter.” In short, the goal is to prevent the monetary distortion that results from a currency's market volatility.
Under the gold standard, a country’s money supply was tied to gold. Fiat currency—or paper money—like cash, could be converted into gold on demand. This naturally limited the supply of currency in circulation since it had to stay within a certain ratio to the supply of gold. In addition to functioning as a tool for regulating the quantity and growth of a country’s domestic currency supply, the gold standard was also used on an international level. Under the standard, two countries with different currencies could easily determine the market value of one currency in terms of the other, and vice versa. Since participating countries maintained a fixed price for gold, the rates of exchange between currencies tied to gold were also fixed.
The dollar, for instance, “was defined as roughly one-twentieth of an ounce of gold, the pound sterling as roughly one-fifth of an ounce of gold,” writes financial market veteran Jim Grant. “The dollar-sterling exchange rate was therefore fixed at roughly $5,” or $4.867 to be exact. This predictability and fixedness, says Grant, encouraged trade and limited the aggressive money-policy interventions that we see today. “As there was only so much gold, there could be only so many dollars, pounds, French francs or German marks.”
The rise and fall of the gold standard
By 1900, all countries, except for China and some countries in Central America, had adopted some form of gold standard. According to the World Gold Council, the widespread adoption of the gold standard was highly successful for the world economy. World trade expanded and many countries benefited from rapid growth and low instability. As such, conditions were favorable to the degree that there were substantial flows of “direct investment that opened up the 'emerging markets' of the era, such as the Americas, Australia, New Zealand, and South Africa.”
This lasted up until WWI, when the cost of war weighed so much on countries that they abandoned the gold standard and resorted to inflationary policies to finance the demands of war, and later, reconstruction. (Remember, under a true gold standard system, paper money is limited.) Despite periodic attempts in various countries to return to a gold standard after the war, most did not survive the Great Depression of the 1930s.
The U.S., in particular, severed its last ties between the dollar and gold in 1971 when President Richard Nixon announced that the U.S. would officially no longer back the dollar with gold reserves. Gold prices then rose 2,330 percent that decade from $35 per ounce to $850, reports Kitco Contributor and CEO of U.S. Global Investors Frank Holmes.
Likewise, all debt (both public and private), reports Forbes, “began to soar past economic growth once the gold standard ended,” as is illustrated in the graph below.
The purchasing power of the dollar also fell severely, declining by 85 percent since the U.S. moved away from the gold standard in 1971, according to calculations by Peter Ferrara, former Associate Deputy Attorney General of the United States under President George H.W. Bush. His estimates further show that a dollar saved in 1971 was worth just 12 cents by 2012.
While governments have always had tools with which to steer the economy, the fall of the gold standard opened up wider avenues for them to do so.
A return to the “gold period of the gold standard”
For many, the advantages of the gold standard remain strong. With it, central banks have less unchecked power, world price levels are effectively anchored, and international exchange rates are more predictable. Even former Fed Chairman, Alan Greenspan, has returned to being a proponent of the gold standard after speaking out otherwise for more than a decade. In a July 2016 interview discussing U.K.’s Brexit referendum, U.S. economic growth, wage stagnation, and more, he suggests that the world may have been better off if the U.S. had never abandoned the gold standard. In the wake of Brexit especially, Greenspan believes a “debt crisis will be inevitable,” as he has no faith that political leaders will “restrain entitlement spending” when another crisis breaks out. “[I] would not be surprised to see the next unexpected move to be on the inflation side,” he says. His solution to high inflation? Good old gold.
“Now if we went back on the gold standard and we adhered to the actual structure of the gold standard as it exists let’s say, prior to 1913, we’d be fine. Remember that the period 1870 to 1913 was one of the most aggressive periods economically that we’ve had in the U.S., and that was a gold period of the gold standard.”
As economies all over the world continue to struggle with a variety of problems, gold could be a solution to protecting your money and increasing your wealth potential. Unlike the dollar, its supply cannot be increased to suit the day's political agenda. In fact, “physical gold is very scarce,” says author and gold market veteran Jim Rickards, and no one is sure how much gold remains embedded in the earth or how much it would cost to mine it.
Take your first steps towards gold ownership and a healthier financial future. Get free access to U.S. Money Reserve's exclusive Gold Information Kit with special offers, diversification strategies, and a precious metals guidebook. Sign up now to gain access to this exclusive resource!
Whether the U.S. returns to a variation of the gold standard in the future or not, you can work to safeguard your wealth by owning gold. “It’s scarce, durable…beautiful and portable. At a glance, anyone can tell it’s valuable,” says financial markets analyst Jim Grant. Bring home a hard asset. Bring home a tangible means of safeguarding your wealth. Call 1-844-307-1589 for your free one-on-one consultation today.<|endoftext|>
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## Engage NY Eureka Math 6th Grade Module 1 Lesson 4 Answer Key
### Eureka Math Grade 6 Module 1 Lesson 4 Example Answer Key
Example 1.
The morning announcements said that two out of every seven sixth-grade students In the school have an overdue library book. Jasmine said, “That would mean 24 of us have overdue books!” Grace argued, “No way. That is way too high.” How can you determine who is right?
You would have to know the total number of sixth-grade students, and then see if the ratio 24: total is equivalent to 2: 7.
### Eureka Math Grade 6 Module 1 Lesson 4 Exercise Answer Key
Exercise 1.
Decide whether or not each of the following pairs of ratios is equivalent.
→ If the ratios are not equivalent, find a ratio that is equivalent to the first ratio.
→ If the ratios are equivalent, identify the nonzero number, c, that could be used to multiply each number of the first ratio by in order to get the numbers for the second ratio.
a. 6: 11 and 42: 88
________ Yes, the value, c, is ________
________ No, an equivalent ratio would be ________
________ Yes, the value, c, is ________
x No, an equivalent ratio would be 42: 77
b. 0: 5 and 0: 20
________ Yes, the value, c, is ________
________ No, an equivalent ratio would be ________
x _ Yes, the value, c, is 4
________ No, an equivalent ratio would be ________
Exercise 2.
In a bag of mixed walnuts and cashews, the ratio of the number of walnuts to the number of cashews is 5: 6. Determine the number of walnuts that are in the bag if there are 54 cashews. Use a tape diagram to support your work. Justify your answer by showing that the new ratio you created of the number of walnuts to the number of cashews is equivalent to 5: 6.
54 divided by 6 equals 9.
5 times 9 equals 45.
There are 45 walnuts in the bag.
The ratio of the number of walnuts to the number of cashews is 45: 54. That ratio is equivalent to 5: 6.
### Eureka Math Grade 6 Module 1 Lesson 4 Problem Set Answer Key
Question 1.
Use diagrams or the description of equivalent ratios to show that the ratios 2: 3, 4: 6, and 8: 12 are equivalent.
Question 2.
Prove that 3: 8 is equivalent to 12: 32.
The ratio of Isabella’s money to Shane’s money is 3: 11. If Isabella has $33, how much money do Shane and Isabella have together? Use diagrams to illustrate your answer. Answer: Isabella has$33, and Shane has $121.$33 + $121 =$154. Together, Isabella and Shane have \$154. 00.<|endoftext|>
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With the ever-increasing epidemic of obesity in North America, almost 20 percent of children and more than 70 percent of adults are overweight, which increases risk for heart attacks, diabetes, strokes, arthritis and certain cancers.
A new study of 84 children and teens shows that the heavier ones have more of the types of gut bacteria that convert carbohydrates into fats so readily that they absorb more calories from their food than the thinner children do (Journal of Clinical Endocrinology & Metabolism, Sept. 20, 2016). The participants ranged from ages 7 to 19 and from normal weight to obese, with MRIs used to measure how much fat they had in their bodies. The colons of the heavier children were far more likely than those of the skinny ones to have the eight groups of gut bacteria that have been previously associated with carbohydrate fermentation to short chain fatty acids that markedly increases the number of calories absorbed from food (Nutr Clin Pract, April 2012;27(2):201–214).
Since bacteria in your colon eat the same food that you do, what you eat determines which types of bacteria thrive in your colon. These bacteria are a prime driver of how high your blood sugar rises after meals and how many calories you absorb from the food you eat. Many epidemiological studies show that people who take in lots of sugared drinks and sugar-added foods, foods made with flour (ground whole grains) and other refined carbohydrates are more likely to be overweight and that those who eat lots of fruits, vegetables, nuts and grains that have not been ground into flour tend to be thin.
Gut Bacteria and Carbohydrate Absorption Carbohydrates are sugars alone, in pairs and in chains of up to millions of sugars bound together. Only single sugars can be absorbed into your bloodstream. You can't even absorb two sugars bound together, so enzymes in your intestines and bacteria in your colon break them apart. For example, milk contains a double sugar called lactose. To split the double sugar into single sugars that can be absorbed, your intestines are supposed to have an enzyme called lactase. However, many people have intestines that do not make lactase, so they cannot break down the lactose into single sugars. Since the lactose cannot be absorbed in the upper intestines, it passes to the colon where bacteria ferment the double sugar, which can cause gas, cramping and diarrhea.
Fiber is the type of carbohydrate formed from long chains of sugars that human intestines cannot split into single sugars. Since fiber cannot be absorbed in the intestines, it passes to the colon. There are two types of fiber: soluble and insoluble fiber. Insoluble fiber cannot be absorbed at all so it passes unchanged from your body, but soluble fiber is readily broken down by certain bacteria in the colon. These bacteria break the soluble fiber into short-chain fatty acids that are easily absorbed through the colon to provide extra calories. The colons of obese people contain a higher concentration of the types of bacteria that break down soluble fiber so that they absorb more calories from the food that they eat. When these bacteria are present, you can gain extra calories not only from the soluble fiber but also from the other sugars and undigested food particles that are bound up with the soluble fiber and that would have passed through undigested.
How Processing Foods Contributes to Obesity The plants we eat contain lots of fiber, much of which cannot be absorbed and passes from your body contributing no calories whatever. When food manufacturers strip fiber away from plant parts, the resulting foods are more easily broken down in the intestines and are quickly absorbed. For example, whole grains are seeds that have carbohydrates bound in a tight fiber capsule that slows the breakdown of the carbohydrates into absorbable sugars. After you eat cooked but unground whole grains, blood sugar levels barely rise. On the other hand, when you eat foods made from whole grains that have been ground into flour (bread, pasta, many dry cereals and so forth), you have no capsule to keep you from breaking down the carbohydrates and your blood sugar can rise very quickly.
Can Changing Colon Bacteria Help to Control Weight? A review of several studies showed that eating a diet based on plant foods and restricting refined carbohydrates and most other foods appears to be a safe and effective way of changing colon bacteria to help control weight and high blood sugar (Gut Microbes, Jan-Feb, 2012;3(1):29-34).
Currently, the best way to change your colon bacteria to favor the types that do not increase absorption of calories is to eat a diet that includes lots of foods that are not quickly absorbed. However, many people are unwilling or unable to lose weight or control weight just by restricting refined carbohydrates and eating a diet based on lots of plants that have not been processed.
The Future Potential of Stool Transplants Two studies are currently being conducted to see if taking pills containing freeze-dried feces from skinny people will cause obese people to have fewer high-calorie-absorbing bacteria in their colons. Dr. Elaine Yu of the Massachusetts General Hospital will test adults and Dr. Nikhil Pay of McMaster Children's Hospital in Ontario, Canada will test children.
Animal studies have shown that germ-free mice fed stool from obese mice gain more weight than those fed bacteria from the guts of lean mice. * Transferring the feces of skinny mice to fat mice changed their colon bacteria to ones that are less likely to convert soluble fiber to acetate and other short-chain fatty acids, and decreased calorie absorption and lowered blood levels of acetate and insulin (Nature, October 9, 2016; 534 (7606):213-217). Blood levels of acetate are increased by absorbing more calories from food and eating a lot of fat. Insulin levels are increased by high rises in blood sugar and high insulin levels are associated with increased risk for obesity and early diabetes.
* Jeffrey Gordon of Washington University in St. Louis showed that feeding mice a diet high in fat and low in fruits, vegetables, and fiber changed their colon bacteria to a more fattening type and made them fat. Then he showed that placing skinny mice in the same cages as the fat mice fed this terrible diet made them fat, presumably because mice eat each other's stool.
* Giving baby mice antibiotics makes them 15 percent fatter than mice not given antibiotics, presumably from changing their colonic bacteria.
* Rob Knight of the University of Colorado Boulder and Maria Gloria Dominguez-Bello of N.Y.U. have shown that human babies delivered by C-section are fatter and more likely to suffer diabetes than those delivered vaginally, probably because vaginally-delivered babies acquire bacteria from their mother's vaginas, while C-section delivered babies do not have that opportunity.
My Recommendations North Americans are in the midst of an epidemic of obesity and diabetes that is still increasing. Emerging evidence suggests that obesity and diabetes are driven to a significant degree by the types of bacteria in the colon. To a large extent, a person's diet determines the type of bacteria that live in his or her colon. * A diet that encourages the growth of bacteria that help to prevent and treat obesity and diabetes has plenty of vegetables, fruits and seeds (beans, nuts and whole grains that have not been ground into flour). * Foods that encourage growth of the types of colon bacteria associated with obesity and diabetes include sugared drinks, sugar-added foods, foods made from flour (ground whole grains) and all other refined carbohydrates.
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Research at Lucent Technologies’ Bell Lab-oratories in Murray Hill, N. J., may be hastening the onset of a new era in electronic materials. Scientists have long known that electrons travel much faster in gallium arsenide (GaAs) than in silicon. But GaAs has found limited use in computing devices, partly because of the difficulty of fabricating suitable transistors.
Silicon chips use metal-oxide-semiconductor field-effect transistors, or MOSFETs. Most GaAs devices now in use (principally in wireless communications) are MESFETs, lacking the oxide. To tap the advantages of GaAs fully will require MOSFETs, which use less power. Bell Labs took the first step two years ago, but its prototypes were woefully inadequate-current fluctuated by 20 percent over a few hours. In the new GaAs devices, however, current drifts only about 1.5 percent after 200 hours. One key benefit: longer time between cell phone rechargings.<|endoftext|>
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Severity: 1st worst
Cause: Eruption of Siberian Traps
Climate: Cold to extremely warm; ocean acidification and anoxia, ozone destruction
Aftermath: Permanent ecosystem reorganization; low O2 for >106 years
There’s good reason why the End-Permian extinction is referred to as “The Great Dying”; 95% of all marine families, 53% of all marine families, 84% of marine genera, and 70% of known land species went extinct,
The extinction likely occurred in three stages:
1. Land extinctions over ~40,000 yrs
2. Very abrupt marine extinctions
3. Second phase of land extinctions
Calcifying marine organisms such as brachiopods and bryozoa were the hardest hit, representative of ocean acidification. The last of the Cambrian fauna also died off, and this was the only known mass extinction of insects
So what exactly made the End-Permian extinction so severe? There truly was a perfect storm to make this the deadliest million years in Earth’s history.
Earth had been emerging from a moderate ice age when the largest flood basalt event in history (the Siberian Traps) occurred, which released vast amounts of CO2. The oceans then became increasingly warm, acidic, stratified, and euxinic from decaying organic matter. The atmosphere also became flooded with light (biogenically fixed) C, possibly from seafloor methane hydrates or from coal gas released as a result of heating from the Siberian Traps. Greenhouse gases soon caused global temperatures to spike, leading to massive extinction. Global euxinia in the oceans then became a severe problem, with sulfate reducing bacteria releasing large amounts of H2S, poisoning the oceans and atmosphere and thinning the ozone layer. These systems then created a cycle of positive feedbacks:
more die-offs → more euxinia → more H2S → more die-offs.
Marine ecosystems were forever changed after the extinction. Land ecosystems didn’t recover for ~5 My, and O2 levels remained low throughout much of Triassic time.
Click HERE to see all Mass Extinction Monday posts<|endoftext|>
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Here's another useful reference article.
A lot of beginner car enthusiasts have trouble understanding differentials, both regular and limited-slip ones. In fact it is a very simple device. So simple in fact, that the first vehicle that is considered in modern terms a car, the Mercedes Simplex of 1903, was distinguishable from other self-propelling carriages by the fact that it had one. (Among other things.)
The differential is a part that splits torque, usually into two equal parts. It has an input shaft and two output shafts, one per each wheel (or axle if it's the central diff in an AWD car). Now, when a car turns, the outer wheels take a wider curve than the inner ones, covering more distance and thus traveling at a higher speed. This happens because the axles have differentials, as opposed to just solid pieces of metal connecting the two wheels.
The differential is made up of gears. The input shaft has a cone-shaped gear at the end. The output shafts are just simple gears. (Their teeth are cut as curved, not straight - this helps them interlock properly and cuts down on noise; straight-cut gears on old offroaders and such tend to produce a loud whine.)
Now, the cone works just like a set of gears, each smaller than the previous one. At the base of the cone, the gear ratio is high and the torque is low. At the tip of the cone, the gear ratio is low and the torque is high. While driving straight on a dry surface with good grip, the output gears sit at the same height on the cone. The gear ratio of the spin, transferred from the input shaft to the output shafts, is the same. Both wheels spin at the same speed.
When a car turns, the outside wheel spins more quickly, so its gear travels towards the base of the cone; the gear ratio gets taller, with the output gear making more revolutions per one cone revolution. The inside wheel travels away from the base of the cone, the gear ratio gets shorter, and its output gear makes less revolutions. Effectively this is the same as the outside wheel shifting up a gear and the inside wheel shifting down a gear. This is what happens when both wheels have the same amount of grip.
But the differential is primarily meant to transfer torque, and torque goes down the path of least resistance. The more grip the wheel has on the ground, the more resistance to spinning it feeds back to its gear - and the less grip it has, the less resistance it gives.
Now, here is the counter-intuitive bit that took me a while to figure out, pondering over a simple drawing of a diff. Resistance makes the gear away from the base of the cone, looking for a shorter gear ratio and less spin. In normal circumstances both wheels have more grip than torque, so when you feed the grip in, the wheels simply start turning and you go forward. But torque really just wants to make the wheel spin freely, so if one wheel has significantly less grip, it will be as if the other wheel is stuck. Its gear will move so far up the cone, looking for less revolutions and more torque to move it, that it will simply pop off the tip of the cone. On the other hand, the wheel with no grip will use less torque to start slipping and its gear will be happy at the base of the cone, spinning quickly but getting little torque.
So you see, torque does travel down the path of least resistance - but not because more torque can be used there, but because less torque is required there. Torque is lazy like that.
To combat this situation, where one wheel is spinning and the other is stationary, engineers use a limited-slip differential. It has a sort of bypass gearing setup, where a certain amount of torque from one output shaft is fed to the other. As long as they are both spinning at the same speed and getting the same amount of torque, it doesn't matter; but if one wheel pops off the tip of the cone and loses torque, it will still get some from other wheel. Limited-slip diffs are rated in percent, as in, how much difference in percent can there be between the amount of torque the wheels are getting.
Active differentials, like the ones on the Mitsubishi Evo or Subaru WRX STI, are limited-slip diffs that have electronics controlling the rate of torque transfer through the bypass.
Locking diffs are much simples. You push a button (or, much more satisfyingly, pull a lever) and the gears are simply held in place, not allowed to travel up and down the cone. Both wheels get half of the torque, but they can't move at different speeds, so turning is tricky. Useful for offroaders though.
And torque steer? Oh, quite simple. In a FWD car, for packaging reasons, the output shafts are different lengths (because the gearbox hooks up right to the differential, it usually can't be in the middle of the axle). Because it takes a bit of torque to spin the shaft itself, more if it's longer, one wheel will get more torque than the other. More torque on one side means that the car will pull in the opposite direction. Besides, differentials are made so that the torque actually travels to one side before the other - normally it makes no difference, but if there's a lot of torque to put down, it does; which is why very powerful RWD cars will throw the tail end to one side under hard acceleration.
The exact sciences
4 weeks ago<|endoftext|>
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# GRE Math : How to multiply exponents
## Example Questions
← Previous 1
### Example Question #1 : How To Multiply Exponents
(b * b* b7)1/2/(b3 * bx) = b5
If b is not negative then x = ?
–1
1
–2
7
–2
Explanation:
Simplifying the equation gives b6/(b3+x) = b5.
In order to satisfy this case, x must be equal to –2.
### Example Question #2 : How To Multiply Exponents
If〖7/8〗n= √(〖7/8〗5),then what is the value of n?
2/5
√5
5/2
1/5
25
5/2
Explanation:
7/8 is being raised to the 5th power and to the 1/2 power at the same time. We multiply these to find n.
### Example Question #3 : How To Multiply Exponents
Quantity A:
(0.5)3(0.5)3
Quantity B:
(0.5)7
Quantity A is greater.
The relationship cannot be determined from the information given.
The two quantities are equal.
Quantity B is greater.
Quantity A is greater.
Explanation:
When we have two identical numbers, each raised to an exponent, and multiplied together, we add the exponents together:
xaxb = xa+b
This means that (0.5)3(0.5)3 = (0.5)3+3 = (0.5)6
Because 0.5 is between 0 and 1, we know that when it is multipled by itself, it decreases in value. Example: 0.5 * 0.5 = 0.25. 0.5 * 0.5 * 0.5 = 0.125. Etc.
Thus, (0.5)6 > (0.5)7
### Example Question #21 : Exponential Operations
For the quantities below, x<y and x and y are both integers.
Please elect the answer that describes the relationship between the two quantities below:
Quantity A
x5y3
Quantity B
x4y4
The quantities are equal.
The relationship cannot be determined from the information provided.
Quantity B is greater.
Quantity A is greater.
The relationship cannot be determined from the information provided.
Explanation:
Answer: The relationship cannot be determined from the information provided.
Explanation: The best thing to do here is to notice that quantity A is composed of two complex terms with odd exponents. Odd powers result in negative results when their base is negative. Thus quantity A will be negative when either x or y (but not both) is negative. Otherwise, quantity A will be positive. Quantity B, however, has two even exponents, meaning that it will always be positive. Thus, sometimes Quantity A will be greater and sometimes Quantity B will be greater. Thus the answer is that the relationship cannot be determined.
### Example Question #1561 : Gre Quantitative Reasoning
Simplify: (x3 * 2x4 * 5y + 4y2 + 3y2)/y
10x7y + 7y2
10x7 + 7y3
10x7 + 7y
10x11 + 7y3
10x7 + 7y
Explanation:
Let's do each of these separately:
x3 * 2x4 * 5y = 2 * 5 * x* x* y = 10 * x7 * y = 10x7y
4y2 + 3y2 = 7y2
Now, rewrite what we have so far:
(10x7y + 7y2)/y
There are several options for reducing this. Remember that when we divide, we can "distribute" the denominator through to each member. That means we can rewrite this as:
(10x7y)/y + (7y2)/y
Subtract the y exponents values in each term to get:
10x7 + 7y
### Example Question #1 : How To Multiply Exponents
Quantitative Comparison
Quantity A: x3/3
Quantity B: (x/3)3
Quantity B is greater.
Quantity A is greater.
The relationship cannot be determined from the information given.
The two quantities are equal.
The relationship cannot be determined from the information given.
Explanation:
First let's look at Quantity B:
(x/3)3 = x3/27. Now both columns have an xso we can cancel it from both terms. Therefore we're now comparing 1/3 in Quantity A to 1/27 in Quantity B. 1/3 is the larger fraction so Quantity A is greater.
However, if , then the two quantities would both equal 0. Thus, since the two quantities can have different relationships based on the value of , we cannot determine the relationship from the information given.
### Example Question #7 : How To Multiply Exponents
Quantitative Comparison: Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given.
Quantity A Quantity B
(23 )2 (22 )3
Quantity A is greater.
Quantity B is greater.
The relationship cannot be determined from the information given.
The two quantities are equal.
The two quantities are equal.
Explanation:
The two quantites are equal. To take the exponent of an exponent, the two exponents should be multiplied.
(2)or 23*2 = 64
(2)or 22*3 = 64
Both quantities equal 64, so the two quantities are equal.
### Example Question #21 : Exponential Operations
Compare and .
The answer cannot be determined from the information given.
Explanation:
To compare these expressions more easily, we'll change the first expression to have in front. We'll do this by factoring out 25 (that is, ) from 850, then using the fact that .
When we combine like terms, we can see that . The two terms are therefore both equal to the same value.
### Example Question #9 : How To Multiply Exponents
Which of the following is equal to ?
Explanation:
is always equal to ; therefore, 5 raised to 4 times 5 raised to 5 must equal 5 raised to 9.
is always equal to . Therefore, 5 raised to 9, raised to 20 must equal 5 raised to 180.
### Example Question #10 : How To Multiply Exponents
Which of the following is equal to ?<|endoftext|>
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# ISEE Lower Level Quantitative : Geometry
## Example Questions
### Example Question #852 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #853 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #854 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #855 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #856 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #857 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #858 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #859 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #860 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Select the graph that displays the polygon created using the following coordinates:
Explanation:
When we are given coordinate points, it's important to know the difference between the x-axis and the y-axis, and which order these points are given. The x-axis is the axis that runs left to right and the y-axis is the axis the runs up and down. When coordinate points are written, the x value goes first, followed by the y value .
Knowing this information, we can plot the points and use straight lines to connect them in a counter-clockwise or clockwise direction. The provided coordinate points should create the following graph:
### Example Question #861 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Select the graph that displays the polygon created using the following coordinates:<|endoftext|>
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Eric R. Jackson, Guest Editor of our Spotlight on Cincinnati, is an Associate Professor of History at Northern Kentucky University. Dr. Jackson is the author of several books including Cincinnati's Underground Railroad, with Richard Cooper, and Northern Kentucky, a part of the Black America Series. Here he discusses the often overlooked African American history of Cincinnati.
African Americans through Cincinnati's History
African Americans in Cincinnati have played a vital role in the history of the Queen City. Struggles for racial equality, social justice, and economic opportunities have taken place in the city's streets, homes, churches, schools, governments, and workplaces, and these efforts been woven into every fabric of Cincinnati's rich historical tapestry. However, the role of African Americans in the region's history remains largely neglected by scholars and writers, an issue that many modern historians are trying to correct.
1788–1860: Racial Tensions and Racial Segregation
The village of Cincinnati—located on the Ohio River—linked the North, South, and West, presenting a wealth of economic opportunity for thousands of immigrants to the region during the nineteenth century. By the mid-1820s the city had begun to emerge as a force in the manufacturing industry, becoming the nation's leader in pork packing (giving the city its other nickname, "Porkopolis") and steamboat construction. Individuals seeking better jobs and economic opportunities left the Northeast and upper South for Cincinnati in large numbers. In about a forty-five year period, the Queen City had been transformed from a small, relatively unknown village to a booming metropolis, rivaling more established Eastern and Midwestern cities. By the 1850s, Cincinnati had become the sixth largest city in the United States, with a population of about 115,000. Its African American population also increased to about 3,200, making it one of the largest communities of Black Americans in the nation during the antebellum era. (Taylor, 2005, pp. 1–2)
For many of its new residents, Cincinnati was the Promised Land. However, for many African American Cincinnatians that promise never materialized. Although Cincinnati was located in a free state, its African American residents had very limited rights and freedoms. Despite the Ohio state constitution prohibiting slavery in 1802, the passage of a series of legal codes, known as the Black Laws, enacted in 1804 and 1807 respectively, left little doubt about the real views of most of the state's white politicians about the presence of persons of African descent. These laws prohibited the migration of African Americans into the state of Ohio without a $500 "bond guaranteeing good behavior." (Taylor, 2005, pp. 2–3). These laws also required persons of African descent to register with the local state clerk office as well as produce a certificate of freedom upon of the request of any white Ohio citizen. (Taylor, 2005, pp. 32–33)
During the decades that followed a series of additional laws were enacted that denied African Americans the right to testify against whites, serve on juries, and vote in several Ohio counties. Indeed, most African Americans were relegated to an inferior status throughout the state. In Cincinnati, African Americans were plagued by the frequency of racial violence throughout most of the 1800s. In 1829 an urban race riot erupted when a gang of whites targeted a group of African Americans for walking down the street. This event occurred, in part, as result of the growing population of African Americans in the city and the perceived competition for various semi-skilled labor jobs by the majority of white middle-class Cincinnatians. In the end, the attacks resulted in some Black Americans moving to Canada where they created the free Wilberforce settlement Several years later, in 1841, one of the worst urban race riots in American history broke out when a group of dockworkers, mostly Irish, attacked a group of African Americans in downtown Cincinnati. Fifty-six people were killed and two hundred were injured. More race riots occurred in Cincinnati throughout the 1840s. Only Philadelphia experienced more urban upheavals during the same period in American history. As a result, Cincinnati began to be known by many as "Queen City of Mobs" (Taylor, 2005, pp. 2–3).
Despite such a harsh racial climate, African Americans in Cincinnati ultimately established several thriving communities during the antebellum years. The impact of the Underground Railroad in the city helped to build valuable bonds between both fugitive and free Americans within various communities. Compared to the development of African American communities in other cities at the time, where the church was typically the most important institution, in Cincinnati African American schools were the center of Black American life and community growth. Though its schools were segregated, Cincinnati nevertheless led the way in establishing educational institutions for black students. African American public schools not only provided the community with a political, social, and an educational space to thrive. They also became the centers of protests and activism throughout the antebellum period. Several private all-Black high schools were formed during these years, including Gilmore High School for Negroes, whose graduates included John I. Gaines, an African American abolitionist and a highly regarded intellectual in Cincinnati during the antebellum years, and Peter H. Clark, a vocal African American politician, educator, and community activist in the "Queen City" during the 1800s. (Tenkotte, 2014, pp. 41–42). In 1852 the Independent Colored School System was established, which eventually led to the creation of an all-Black public school, Gaines High School.
1861–1954: The Development of Several Local Civil Rights Organizations and Community Leaders
Cincinnati's history of race riots, which started during the antebellum period, but continued during the 1860s, demonstrated the resiliency of many African Americans, including John M. Langston, William Ellison, and George Cary, who helped to maintain and mobilize various African American communities despite the continuous episodes of racial violence. The 1862 race riot in particular marked a major turning point in the history of African American life in the Queen City. This incident was a result of the perceived loss of jobs for Irish and German immigrants in favor of African Americans in Cincinnati. More importantly, after this event, the economic gains that many African American workers had made during the previous decades nearly vanished. The thirty percent employment rate that African American laborers in Cincinnati experienced during the 1860s, in various commercial industries, was almost cut-in-half by 1870. Despite this economic downturn, and the elimination of the Independent Colored School System Board in 1874, the all-Black Gaines High School had already graduated more than sixty students per year by 1878, which helped to fortify the emerging African American middle class in the city. (Tenkotte, 2014, pp. 41 - 42)
As in many American cities, racial segregation became codified in Cincinnati following the 1896 Plessy v. Ferguson case. Moreover, the local government's clumsy dismantling of the Independent Colored School System during the 1880s' clearly demonstrated that the city's white leaders aimed to not only separate the races, but to undermine the collective political power of the city's black population. By 1910 the only black high school that was still in operation was the Elm Street School in Walnut Hills, which had been renamed the Douglass School. All of the other schools were eliminated. By 1912 only seven African American teachers were employed by the Cincinnati Public School System to teach less than 300 African American students, which was a major decline from eighty-seven African American teachers and almost 3,800 African American students in 1870. (Taylor, 2005, pp. 2–3)
In 1914 a young African American teacher named Jennie D. Porter convinced the Cincinnati School Board to allow her to operate an all-Black school in the old Hughes High School building in the West End. It was renamed the Harriet Beecher Stowe School. As an admirer of Booker T. Washington, Porter contended that segregated schools would offer a greater cultural awareness and job opportunities to both African American teachers and students. However, some local African American leaders, such as newspaper publisher Wendell P. Dabney, disagreed with Porter's educational philosophy. The first president of the local branch of the NAACP, Dabney published several editorials and articles in his paper titled The Union that criticized Washington's emphasis on vocational training as the chief means of political empowerment (Tenkotte, 2014, pp. 73–74).
Within this highly segregated environment several local African American-led civil rights organizations were founded. One of the first was the above-mentioned Cincinnati branch of the National Association for the Advancement of Colored People (NAACP). In 1915, with only about twenty members, the Cincinnati NAACP chapter was formed; established six years after the national organization. One of its earliest victories was the elimination of segregated public schools in the city. Several prominent local African Americans served as president of the organization, including Dabney; Theodore "Ted" Berry, who was president of the organization from 1932 to 1946, as well as later the first African American mayor of Cincinnati; and Reverend L. Venchel, who served as president and chaired a highly successful membership campaign that started in 1954.
Another African American-led civil rights organization that started in a similar fashion was the Greater Cincinnati Urban League (GCUL). Known by several different names, such as the Negro Civic Welfare Committee of the Council of Social Agencies and the Negro Civic Welfare Association Department of the Council of Social Agencies, the League, was founded in 1948 as one of Cincinnati's Community Chest Agencies (Mjagkij, 1993, pp. 280–281). A major focus of the League during its initial years was the development of employment opportunities for Cincinnati's continuously expanding African American workforce. In 1948, no African Americans were employed by any of the Cincinnati companies employing at least 1,500 workers. To correct this, the League developed several industrial relations and job training programs and fostered numerous relationships with leading company executives, all of which led to the hiring of hundreds of African Americans in several local businesses such as Shillito, Ben's, and Max Department Stores over the next few years. (Mjagkij, 1993, pp. 282–283)
During these same years, a small but influential group of African American community leaders and political activists were emerging. Several of the individuals were directly involved in various local efforts to fully integrate the city's public school system. They were especially active after the 1954 Brown v. Board of Education decision, which called for an end to the use of segregated public schools across the country. Some of the most significant people active in this period were Virginia Coffey, Theodore "Ted" Berry, Frank A.B. Hall, William L. Mallory, Sr., Wilber A. Page, Marjorie Parham, Fred L. Shuttlesworth, and Marian Spencer. With the assistance of the local NAACP, they helped to organize a campaign, to end segregation at Coney Island, a local amusement just outside of downtown Cincinnati. All these individuals would play a vital role in the budding Modern Civil rights movement in Cincinnati during the next few decades.
1955–2015: Triumphs and Tragedies
African Americans in Cincinnati experienced both triumphs and tragedies as they organized various political, social, and economic campaigns to improve their daily lives. Thousands of African American Cincinnatians, especially those who lived in the West End, had to deal with being relocated to other communities, particularly to Mt. Auburn, Walnut Hills, and Over the Rhine, as a result of the city's 1948 urban renewal plan. After numerous budgetary and management issues, the plan was enforced in1960, and did not include an adequate relocation component for the over 40,000 people that were displaced. This plan also did away with numerous African American institutions in the West End, such the Ninth Street YMCA and the Cotton Club, a dance and music hall, as well as several African Americans churches. (Tenkotte, 2014, pp. 120–121)
African American Cincinnatians also had to deal with the city's lasting racial segregationist tactics. While Cincinnati's numerous public facilities were not legally segregated, most local African Americans knew that there were customs that unofficially reinforced the wishes of white private property owners who sought to prevent African Africans from entering their hotels, restaurants, and amusement parks. For instance, not a single African American worked as a manager at any of the city's major hotels or restaurants. In July 1965, the local Congress of Racial Equality (CORE) chapter began a work stoppage and a sit-in in the Federal Building in downtown Cincinnati that drew national attention on the issue of racial discrimination. As a result, the city started a federal assistance job training program for African American males. However, very few individuals took advantage of the program.
During the 1970s and 1980s most individuals who were involved in the local Civil rights movement concentrated their efforts on the areas of business development, access to quality public education, and the creation of integrated neighborhoods. During the 1970s Virginia A. Coffey led several community groups in the fight to create more integrated neighborhoods throughout the city. Several years later, during late 1970s and early 1980s, a group of local African American parents, backed by the local NAACP, pushed for the city to finally end the use of the remaining unofficially segregated public schools. The 1981 Bronson case forced the complete integration of the city's public school system immediately as a result of several law suits by a number of local African American families. These types of community-based Civil rights activities continued throughout the rest of the 1980s and into the 1990s. By this time, the city's racial problems began to be linked a clear class divide within Cincinnati and the surrounding county which made trying to mobilizing groups of citizens much more difficult.
Despite these limitations, some progress was made during the 1990s. Specifically, various neighborhoods took the lead in the improvement of race relations and the plight of African Americans in the city. For example, Avondale attempted to rebuild itself starting with the transformation of many empty, former Jewish synagogues into active Protestant churches. Although this part of the city lost more residents than any other area of the city, non-profit companies, such as the Uptown Consortium, invested about the $20 million into redevelopment of the neighborhood. In addition, the corporation financed extra police protection throughout the neighborhood.
In 2001 another riot took place, this time in response to the shooting of an unarmed African American teenager, Timothy Thomas, by a white police officer in the Over-the-Rhine neighborhood. Following the riots, black activists led an economic boycott of the city. The ACLU and the Black United Front continued to pursue a lawsuit filed before the shooting alleging racial profiling, and the settlement led to The Cincinnati Collaborative Agreement, a plan to improve police relations with the community. The agreement took several years to enact, following intense negations between the police, black leaders, business owners, and community members. Though tensions still exist, the years since the riot led to a more transparent relationship between the law enforcement and the community.
Today the city continues to change with regions of the city, such as Over-the-Rhine and the riverfront (now known as "The Banks") in a new redevelopment phase. Though some believe this is an "'urban renaissance'", others have raised concerns of gentrification and the displacement of poor, often African American, neighborhoods.
Mjagkij, Nina. 1993. "Behind the Scenes: The Cincinnati Urban League, 1948 - 63." In Race and the City: Work, Community, and Protest in Cincinnati, 1820 - 1970, by Jr Henry Louis Taylor, 280 - 294. Urbana: University of Illinois Press
Taylor, Nikki N. 2005. Frontiers of Freedom: Cincinnati's Black Community, 1802 - 1868. Athens: Ohio University Press.
Tenkotte, Daniel Hurley and Paul A. 2014. Cincinnati: The Queen City - 225th Annnersary Edition. San Antonio: HPN Books.<|endoftext|>
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Table of Contents
In this post we will study about Continent – Continent Convergence. Understanding Continent – Continent Convergence is important to understand the Formation of the Himalayas, the Alps, the Urals and the Atlas mountains.
We have studied in See Floor Spreading how convectional currents in the mantle drive the lithospheric plates. Rising vertical limbs of the convection currents in the mantle create a divergent plate boundary and falling limbs create a convergent plate boundary.
In convergence there are sub-types namely:
- Collision of oceanic plates or ocean – ocean convergence. [Explained in the previous posts]
- Collision of continental and oceanic plates or ocean – continent convergence [Explained in the previous post].
- Collision of continental plates or continent – continent convergence [This Post].
- Collision of continent and arc or continent – arc convergence [This Post].
In all types of convergence, denser plate subducts and the less denser plate is either up thrust or folded or both [up thrust and folded].
Continent – Continent Convergence or The Himalayan Convergence
- In ocean – ocean convergence and continent – ocean convergence, at least one of the plates is denser and hence the subduction zone is quite deep [few hundred kilometers].
- At continental – continental convergent margins, due to lower density, both of the continental crustal plates are too light [too buoyant] to be carried downward (subduct) into a trench. In most cases, neither plate subducts or even if one of the plates subducts, the subduction zone will not go deeper than 40 – 50 km.
- The two plates converge, buckle up [The subduction of the continental crust is not possible beyond 40 km because of the normal buoyancy of the continental crust. Thus, the fragments of oceanic crust are plastered against the plates causing welding of two plates known as suture zone. Example: The- Indus-Tsangpo suture zone], fold, and fault.
- Geoclinal sediments are found along the continental margins. As the continental plates converge, the ocean basin (geosynclinical basin) is squeezed between the two converging plates. Huge slivers of rock, many kilometers wide are thrust on top of one another, forming a towering mountain ranges.
- With the building up of resistance, convergence comes to an end. The mountain belt erodes and this is followed by isostatic adjustment.
- As two massive continents weld, a single large continental mass joined by a mountain range is produced.
- Examples: The Himalayas, Alps, Urals, Appalachians and the Atlas mountains.
Volcanism and Earthquakes in Continent – Continent Convergence
- Oceanic crust is only 5 – 30 km thick. But the continental crust is 50 – 70 km thick. Magma cannot penetrate this thick crust, so there are no volcanoes, although the magma stays in the crust.
- Metamorphic rocks are common because of the stress the continental crust experiences.
- With enormous slabs of crust smashing together, continent – continent collisions bring on numerous and large earthquakes. [Earth Quakes in Himalayan and North Indian Region]
Convergent boundary = More deep focus earthquakes. Example: Kachchh region, Himalayan region.
Formation of Himalayas and Tibet
- The Himalayan mountains are also known as the Himadri, Himavan or Himachal.
- The Himalayas are a part of Alpine mountain Chain.
- The Himalayas are the youngest mountain chain in the world.
- Indo – Australian plate è Indian plate + Australian plate + Some parts of Indian Ocean.
Indo – Australian Plate boundary
- North ==> Himalayas
- East ==> Purvanchal, Rakinyoma Mountains, Arakan coast, Andaman & Nicobar islands and Java Trench, South western Pacific plate.
- West ==> Suleiman and Kirthar ranges, Makrana coast, western margin of Red Sea rift, Spreading site between Indio – Australian plate and African plate
- South ==> Spreading site between Indio – Australian plate and Antarctic plate
Explain the formation of Himalayas
- Himalayan mountains have come out of a great geosyncline called the Tethys Sea and that the uplift has taken place in different phases.
- During Permian Period (250) million years ago, there was a super continent known as
- Its northern part consisted of the present day North America and Eurasia (Europe and Asia) which was called Laurasia or Angaraland or Laurentia.
- The southern part of Pangaea consisted of present day South America, Africa, South India, Australia and Antarctica. This landmass was called
- In between Laurasia and Gondwanaland, there was a long, narrow and shallow sea known as the Tethys Sea (All this was explained in detail in Continental Drift Theory).
- There were many rivers which were flowing into the Tethys Sea (Older than Himalayas. We will see this in detail while studying Antecedent and Subsequent Drainage).
- Sediments were brought by these rivers and were deposited on the floor of the Tethys Sea.
- These sediments were subjected to powerful compression due to the northward movement of the Indian Plate. This resulted in the folding of sediments.
- Once the Indian plate started plunging below the Eurasian plate, these sediments were further folded and raised. This process is still continuing (India is moving northwards at the rate of about five cm per year and crashing into rest of the Asia).
- And the folded sediments, after a lot of erosional activity, appear as present day Himalayas.
- Tibetan plateau was formed due to up thrusting of the Eurasian Plate. And the Indo-Gangetic plain was formed due to consolidation of alluvium brought down by the rivers flowing from Himalayas.
- The curved shape of the Himalayas convex to the south, is attributed to the maximum push offered at two ends of the Indian Peninsula during its northward drift.
- Himalayas do not comprise a single range but a series of at least three ranges running more or less parallel to one another.
- Therefore, the Himalayas are supposed to have emerged out of the Himalayan Geosyncline i.e. the Tethys Sea in three different phases following one after the other.
- The first phase commenced about 50-40 million years ago, when the Great Himalayas were formed. The formation of the Great Himalayas was completed about 30 million years ago.
- The second phase took place about 25 to 30 million years ago when the Middle Himalayas were formed.
- The Shiwaliks were formed in the last phase of the Himalayan orogeny — say about two million to twenty million years ago.
- Some of the fossil formations found in the Shiwalik hills are also available in the Tibet plateau. It indicates that the past climate of the Tibet plateau was somewhat similar to the climate of the Shiwalik hills.
- There are evidences to show that the process of uplift of the Himalayas is not yet complete and they are still rising.
- [Recent studies have shown that convergence of the Indian plate and the Asian plate has caused a crustal shortening of about 500 km in the Himalayan region. This shortening has been compensated by sea floor spreading along the oceanic ridge in the Indian Ocean]
Formation of Himalayas in Short
- Pangea’s breakup starts in Permian period [225 million years ago].
- India started her northward journey about 200 million years
- It travelled some 6,000 kilometres before it finally collided with Asia.
- India collided with Asia about 40-50 million years ago.
- Convergent boundary gave rise to Himalayas 40 – 50 million years ago [Tertiary Period] [Formation of Deccan Traps began 70-60 million years ago]
- Scientists believe that the process is still continuing and the height of the Himalayas is rising even to this date.
Evidences for the rising Himalayas
- Today’s satellites that use high precision atomic clocks can measure accurately even a small rise of one cm. The heights of various places as determined by satellites indicate that the Himalayas rise by few centimeters every year. The present rate of uplift of the Himalayas has been calculated at 5 to 10 cm per year.
- Due to uplifting, lakes in Tibet are desiccated (lose water) keeping the gravel terraces at much higher levels above the present water level. This could be possible only in the event of uplift of the region.
- The frequent tectonic activity (occurrence of earthquakes) in the Himalayan region shows that the Indian plate is moving further northwards and plunging into Eurasian plate. This means that the Himalayas are still being raised due to compression and have not yet attained isostatic equilibrium.
- The Himalayan rivers are in their youthful stage and have been rejuvenated [make or cause to appear younger or more vital] in recent times. This shows that the Himalayan Landmass is rising keep the rivers in youth stage since a long time.
Formation of Alps, Urals, Appalachians and the Atlas mountains
- The formation of each of these mountains is similar to the formation of the Himalayas.
- Alps are young fold mountains which were formed due to collision between African Plate and the Eurasian Plate.
- Atlas mountains are also young folded mountains which are still in the process of formation. They are formed due to collision between African Plate and the North American, Eurasian Plates.
- Urals are very old fold mountains which were formed even before the breakup of Pangaea. They were formed due to collision between Europe and Asia.
- Appalachians are also very old fold mountains which were formed even before the breakup of Pangaea. They were formed due to collision between North America and Europe.
Mains Question on Fold Mountains
Why are the world’s fold mountain systems located along the margins of continents? Bring out the association between the global distribution of Fold Mountains and the earthquakes and volcanoes.
Why fold mountains at continental margin?
- Fold mountains are formed due to convergence between two continental plates (Himalayas) or between an oceanic and a continental plate (Rockies. Explained in previous post).
- In Continent – Continent (C-C) convergence, oceanic sediments are squeezed and up thrust between the plates and these squeezed sediments appear as fold mountains along the plate margins.
- In Continent – Ocean (C-O) convergence, the continental volcanic arc formed along the continental plate margin is compressed and is uplifted by the colliding oceanic plate giving rise to fold mountains along the continental plate margin.
- In both C-C convergence and C-O convergence, there is formation of fold mountains and frequent occurrence of earthquakes.
- This is because of sudden release of friction between the subducting plate and up thrust plate. In C-C convergence, the denser plate pushes in to the less denser plate creating a fault zone along the margin. Further collision leads to sudden release of energy along this fault zone generating disastrous earthquakes (Himalayan Region).
- In C-O regions the subducting oceanic plate grinds against the surrounding denser medium producing mostly deep focus earthquakes.
- Volcanism is observed only in C-O convergence and is almost absent in C-C convergence. This is because of the thick continental crust in C-C convergence which prevents the outflow of magma. Magma lies stocked within the crust.
- In C-O convergence, metamorphosed sediments and melting of the subducting plate form magma which escapes to the surface through the less thicker continental crust.
Continent – Arc Convergence or New Guinea Convergence
- New Guinea came into being about 20 million years ago as a result of continent – arc collision.
- The continental plate pushes the island arc towards the oceanic crust. The oceanic plate plunges under the island arc.
- A trench occurs on the ocean side of the island arc and, ultimately, the continental margin is firmly welded against the island arc.
If you have any questions, doubts, corrections etc., leave a comment below.<|endoftext|>
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# Solve for: sqrt(7+4\sqrt{3)}+sqrt(7-4\sqrt{3)}
## Expression: $\sqrt{ 7+4\sqrt{ 3 } }+\sqrt{ 7-4\sqrt{ 3 } }$
Use ${a}^{2}+2ab+{b}^{2}={\left( a+b \right)}^{2}$ to factor the expression
$\sqrt{ {\left( 2+\sqrt{ 3 } \right)}^{2} }+\sqrt{ 7-4\sqrt{ 3 } }$
Use ${a}^{2}-2ab+{b}^{2}={\left( a-b \right)}^{2}$ to factor the expression
$\sqrt{ {\left( 2+\sqrt{ 3 } \right)}^{2} }+\sqrt{ {\left( 2-\sqrt{ 3 } \right)}^{2} }$
Reduce the index of the radical and exponent with $2$
$2+\sqrt{ 3 }+\sqrt{ {\left( 2-\sqrt{ 3 } \right)}^{2} }$
Reduce the index of the radical and exponent with $2$
$2+\sqrt{ 3 }+2-\sqrt{ 3 }$
Since two opposites add up to $0$, remove them from the expression
$2+2$
$4$<|endoftext|>
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# 28.2 Inverting matrices
## 28.2-1
Let $M(n)$ be the time to multiply two $n \times n$ matrices, and let $S(n)$ denote the time required to square an $n \times n$ matrix. Show that multiplying and squaring matrices have essentially the same difficulty: an $M(n)$-time matrix-multiplication algorithm implies an $O(M(n))$-time squaring algorithm, and an $S(n)$-time squaring algorithm implies an $O(S(n))$-time matrix-multiplication algorithm.
Showing that being able to multiply matrices in time $M(n)$ implies being able to square matrices in time $M(n)$ is trivial because squaring a matrix is just multiplying it by itself.
The more tricky direction is showing that being able to square matrices in time $S(n)$ implies being able to multiply matrices in time $O(S(n))$.
As we do this, we apply the same regularity condition that $S(2n) \in O(S(n))$. Suppose that we are trying to multiply the matrices, $A$ and $B$, that is, find $AB$. Then, define the matrix
$$C = \begin{pmatrix} I & A \\ 0 & B \end{pmatrix}$$
Then, we can find $C^2$ in time $S(2n) \in O(S(n))$. Since
$$C^2 = \begin{pmatrix} I & A + AB \\ 0 & B \end{pmatrix}$$
Then we can just take the upper right quarter of $C^2$ and subtract $A$ from it to obtain the desired result. Apart from the squaring, we've only done work that is $O(n^2)$. Since $S(n)$ is $\Omega(n^2)$ anyways, we have that the total amount of work we've done is $O(n^2)$.
## 28.2-2
Let $M(n)$ be the time to multiply two $n \times n$ matrices, and let $L(n)$ be the time to compute the LUP decomposition of an $n \times n$ matrix. Show that multiplying matrices and computing LUP decompositions of matrices have essentially the same difficulty: an $M(n)$-time matrix-multiplication algorithm implies an $O(M(n))$-time LUP-decomposition algorithm, and an $L(n)$-time LUP-decomposition algorithm implies an $O(L(n))$-time matrix-multiplication algorithm.
Let $A$ be an $n \times n$ matrix. Without loss of generality we'll assume $n = 2^k$, and impose the regularity condition that $L(n / 2) \le cL(n)$ where $c < 1 / 2$ and $L(n)$ is the time it takes to find an LUP decomposition of an $n \times n$ matrix. First, decompose $A$ as
$$A = \begin{pmatrix} A_1 \\ A_2 \end{pmatrix}$$
where $A_1$ is $n / 2$ by $n$. Let $A_1 = L_1U_1P_1$ be an LUP decomposition of $A_1$, where $L_1$ is $/ 2$ by $n / 2$, $U_1$ is $n / 2$ by $n$, and $P_1$ is $n$ by $n$. Perform a block decomposition of $U_1$ and $A_2P_1^{-1}$ as $U_1 = [\overline U_1|B]$ and $A_2P_1^{-1} = [C|D]$ where $\overline U_1$ and $C$ are $n / 2$ by $n / 2$ matrices. Since we assume that $A$ is nonsingular, $\overline U_1$ must also be nonsingular.
Set $F = D - C\overline U_1^{-1}B$. Then we have
$$A = \begin{pmatrix} L_1 & 0 \\ C\overline U_1^{-1} & I_{n / 2} \end{pmatrix} \begin{pmatrix} \overline U_1 & B \\ 0 & F \end{pmatrix} P_1.$$
Now let $F = L_2U_2P_2$ be an LUP decomposition of $F$, and let $\overline P = \begin{pmatrix} I_{n / 2} & 0 \\ 0 & P_2 \end{pmatrix}$. Then we may write
$$A = \begin{pmatrix} L_1 & 0 \\ C\overline U_1^{-1} & L_2 \end{pmatrix} \begin{pmatrix} \overline U_1 & BP_2^{-1} \\ 0 & U_2 \end{pmatrix} \overline P P_1.$$
This is an LUP decomposition of $A$. To achieve it, we computed two LUP decompositions of half size, a constant number of matrix multiplications, and a constant number of matrix inversions. Since matrix inversion and multiplication are computationally equivalent, we conclude that the runtime is $O(M(n))$.
## 28.2-3
Let $M(n)$ be the time to multiply two $n \times n$ matrices, and let $D(n)$ denote the time required to find the determinant of an $n \times n$ matrix. Show that multiplying matrices and computing the determinant have essentially the same difficulty: an $M(n)$-time matrix-multiplication algorithm implies an $O(M(n))$-time determinant algorithm, and a $D(n)$-time determinant algorithm implies an $O(D(n))$-time matrix-multiplication algorithm.
(Omit!)
## 28.2-4
Let $M(n)$ be the time to multiply two $n \times n$ boolean matrices, and let $T(n)$ be the time to find the transitive closure of an $n \times n$ boolean matrix. (See Section 25.2.) Show that an $M(n)$-time boolean matrix-multiplication algorithm implies an $O(M(n)\lg n)$-time transitive-closure algorithm, and a $T(n)$-time transitive-closure algorithm implies an $O(T(n))$-time boolean matrix-multiplication algorithm.
Suppose we can multiply boolean matrices in $M(n)$ time, where we assume this means that if we're multiplying boolean matrices $A$ and $B$, then $(AB)_{ij} = (a_{i1} \wedge b_{1j}) \vee \dots \vee (a_{in} \wedge b_{nj})$. To find the transitive closure of a boolean matrix $A$ we just need to find the $n^{\text{th}}$ power of $A$. We can do this by computing $A^2$, then $(A^2)^2$, then $((A^2)^2)^2$ and so on. This requires only $\lg n$ multiplications, so the transitive closure can be computed in $O(M(n)\lg n)$.
For the other direction, first view $A$ and $B$ as adjacency matrices, and impose the regularity condition $T(3n) = O(T(n))$, where $T(n)$ is the time to compute the transitive closure of a graph on $n$ vertices. We will define a new graph whose transitive closure matrix contains the boolean product of $A$ and $B$. Start by placing $3n$ vertices down, labeling them $1, 2, \dots, n, 1', 2', \dots, n', 1'', 2'', \dots, n''$.
Connect vertex $i$ to vertex $j'$ if and only if $A_{ij} = 1$. Connect vertex $j'$ to vertex $k''$ if and only if $B_{jk} = 1$. In the resulting graph, the only way to get from the first set of $n$ vertices to the third set is to first take an edge which "looks like" an edge in $A$, then take an edge which "looks like" an edge in $B$. In particular, the transitive closure of this graph is:
$$\begin{pmatrix} I & A & AB \\ 0 & I & B \\ 0 & 0 & I \end{pmatrix} .$$
Since the graph is only of size $3n$, computing its transitive closure can be done in $O(T(3n)) = O(T(n))$ by the regularity condition. Therefore multiplying matrices and finding transitive closure are are equally hard.
## 28.2-5
Does the matrix-inversion algorithm based on Theorem 28.2 work when matrix elements are drawn from the field of integers modulo $2$? Explain.
It does not work necessarily over the field of two elements. The problem comes in in applying theorem D.6 to conclude that $A^{\text T}A$ is positive definite. In the proof of that theorem they obtain that $||Ax||^2 \ge 0$ and only zero if every entry of $Ax$ is zero. This second part is not true over the field with two elements, all that would be required is that there is an even number of ones in $Ax$. This means that we can only say that $A^{\text T}A$ is positive semi-definite instead of the positive definiteness that the algorithm requires.
## 28.2-6 $\star$
Generalize the matrix-inversion algorithm of Theorem 28.2 to handle matrices of complex numbers, and prove that your generalization works correctly. ($\textit{Hint:}$ Instead of the transpose of $A$, use the conjugate transpose $A^*$, which you obtain from the transpose of $A$ by replacing every entry with its complex conjugate. Instead of symmetric matrices, consider Hermitian matrices, which are matrices $A$ such that $A = A^*$.)
We may again assume that our matrix is a power of $2$, this time with complex entries. For the moment we assume our matrix $A$ is Hermitian and positivedefinite. The proof goes through exactly as before, with matrix transposes replaced by conjugate transposes, and using the fact that Hermitian positivedefinite matrices are invertible. Finally, we need to justify that we can obtain the same asymptotic running time for matrix multiplication as for matrix inversion when $A$ is invertible, but not Hermitian positive-definite.
For any nonsingular matrix $A$, the matrix $A^*A$ is Hermitian and positive definite, since for any $x$ we have $x^*A^*Ax = \langle Ax, Ax \rangle > 0$ by the definition of inner product. To invert $A$, we first compute $(A^*A)^{-1} = A^{-1}(A^*)^{-1}$. Then we need only multiply this result on the right by $A^*$. Each of these steps takes $O(M(n))$ time, so we can invert any nonsingluar matrix with complex entries in $O(M(n))$ time.<|endoftext|>
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Frederick Douglass on Slavery and the Civil War: Selections from His Writings
A former slave, self-taught writer, editor, and public servant, Frederick Douglass was also among the foremost leaders of the abolitionist movement. Recognized as one of the first great African-American speakers in the United States, Douglass was an advisor to President Lincoln during the Civil War and fought for the adoption of constitutional amendments that guaranteed voting rights and other civil liberties for blacks. This book includes representative selections from the speeches and writings of this great statesman, with topics focusing on the slave trade, the Civil War, suffrage for African-Americans, reconstruction in the South, and other vital issues. A powerful voice for human rights throughout much of the 19th century, Douglass remains highly respected today for his fight against racial injustice.<|endoftext|>
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Soil drainage, or how well the soil holds water, is determined by soil texture (see previous section), as well as other elements in the soil.
Soil Drainage terms you should know:
(from Colorado State University Extension- GardenNotes #261 Soil Water Holding Capacity and Irrigation Management)
- Water Holding Capacity: Soil texture and structure (actually the pore space created by soil texture and structure) primarily determine a soil’s ability to hold water. Water coats the soil particles and organic matter and is held in the small pore spaces by cohesion (the chemical forces by which water molecules stick together). Air fills the large pore spaces. Water readily moves downward by gravitational pull through the large pore spaces. In small pore spaces, water moves slowly in all directions by capillary action.
- Saturation refers to the situation when the soil’s pore spaces are filled with water. With water replacing air in the large pore spaces, root functions temporarily stop (since roots require oxygen for water and nutrient up-take). Prolonged periods without root oxygen will cause most plants to wilt (due to a lack of water uptake), to show general symptoms of stress, to decline (due to a lack of root function and possible root dieback), and to die.
- Field capacity refers to the situation when excess water has drained out due to gravitational pull. Air occupies the large pore spaces; and water coats the soil particles and organic matter, and fills the small pore spaces. A handful of soil at or above field capacity will glisten in the sunlight. In clayey and/or compacted soils, the lack of large pore space slows or prohibits water movement down through the soil profile, keeping soils above field capacity and limiting plant growth.
- Permanent wilting point refers to the situation when a plant wilts beyond recovery due to a lack of water in the soil. At this point the soil feels dry to the touch. However, it still holds about half of its water; the plant just does not have the ability to extract it. Plants vary in their ability to extract water from the soil.
- Available water is the amount of the water held in a soil between field capacity and the permanent wilting point. This represents the quantity of water “available” or usable by the plant. Note from the illustration below that the amount of available water is low in a sandy soil. Loamy soils have the largest quantity of available water. In clayey soils, the amount of available water decreases slightly as capillary action holds the water so tightly that plants cannot extract it.
Soil Compaction, compression of soil by heavy machinery or animal traffic, can decrease the soil pore space leading to decreased soil air and increase water and bulk density (harder soil). This can decrease the quality of the soil and make it a more stressful environment for plants-whether crops or forage-to live in.<|endoftext|>
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Metals are one of the most common elements we see in our daily life. Aluminum foil, gold and silver ornaments are made up from metals only.
There are different types of metal placed separately in periodic table.
1. Alkali metals
These metals are placed in 1st group of periodic table. They are highly reactive and soft metals. They form basic oxide in their aqueous solution, therefore known as alkali metal. Most common examples of alkali metals are sodium, potassium, lithium etc.
2. Alkaline Earth Metals
These metals are located in 2nd group of periodic table like calcium, magnesium etc. They are found in earth crust so called as alkaline earth metals.
3. Transition Metals
These metals are located from 3rd to 12th group of periodic table. They are most common metals used in daily life like iron, cooper, zinc, gold, silver etc. These metals are very hard, malleable and ductile in nature, therefore can be convert in sheets or wires for various uses. They are good conductor of heat and electricity, therefore also used in electrical circuits and equipment.
These metals are placed separately at the bottom of periodic table. Due to less abundance in nature, they are also known as rare elements. Compare to transition metals, they are silvery soft metals with high melting point. These metals and their compounds are mainly used in lasers, lamps, magnets and to improve the properties of other metals.
These metals are placed below lanthanides in periodic table. They are reactive and radioactive metals, mainly used in medicines and nuclear devices.
Similarly German silver is an alloy of copper, zinc and nickel and non-magnetic in nature. When copper and gold are mixed together, it forms pink alloy which looks quite different from its constituent metals.
- Uses of metals are not new for human beings as many human civilizations are related to metals. For a long time, we are using metals to meet our needs.
- But metals do not have unlimited characteristics, they also have some limits. Like some metals; aluminum, zinc are not so strong metals.
- Iron gets rusted due to atmospheric oxidation and cooper metal also gets a green color layer of carbonate and oxide over it.
- All most all pure metals have weak strength and easily bends. This is due to weak metallic bonds between same size atoms which become weaker under stress due to sliding of atoms over each other.
- To overcome these all limitations, a new concept of alloys introduced. Alloy is a mixture of metals in certain composition. They show different physical and chemical properties compare to metals.
- For example; bronze is an alloy of tin and copper. It is composed of 12% tin and 88% of copper. Bronze is harder than its constituent metals and easier to use for various applications.
- Since alloys melt at a range of temperature instead of a sharp melting point like metals, they can be used for different purposes.
- The appearance of alloy depends upon their composition. Like white gold is an alloy of nickel and gold, looks like nickel.
For the improvement of properties of metals, alloys produced. Alloys are prepared by mixing of metals on their molten state. The molten mixture of metals is poured in sand moulds to get solidify. Alloys must contain those metals which are miscible into each other in their molten state, otherwise after casting; metals would separate into two layers.Alloys can be a combination of metals and non-metals.
Each constituent element of alloy is known to have its own specific set of properties. Various combinations of metals can create different alloys which show different chemical and physical properties. The constituent metals of alloy cannot be separated by physical processes.Some common examples of alloys are as follow.
Alloys are not a new material for us as we have been used them since pre-historic times. Man used alloys soon after he worked with metals and have created many alloys which are suitable for particular objects. Some alloys like White Gold (gold with nickel or palladium) has limited uses while other alloys such as bronze, steel, and brass have changed our history and our modern world.
It is an alloy of copper and tin. It was first alloy to be made by alloying copper with arsenic and used by human civilization. Since arsenic shows toxic effects, therefore later it replaced by tin. Bronze is mainly used to make utensils like jugs, pans, pots, vases as well as to make weapons and armor. It also uses to make tools like hammers, mallets, wrenches and instruments such as bells, saxophones and cymbals.
It is a combination of metals and non-metals. As we know, iron metals get rust in atmospheric air, hence to create resistance to corrosion in iron; it gets mix with some other metals like chromium, manganese, molybdenum and non-metals like carbon. The percentage of carbon changes the hardness of metal. While addition of chromium provides shine to metal surface and molybdenum makes steel stronger.
The trace of tungsten makes steel more durable. In general steel is composed of 95% iron, up to 2% carbon, and trace quantities of other metals. This composition may vary in different types of steel.Composition of various grades of steel is as follow.
| Chemical Composition of Steel
|| C % Max
|| Mn % Max.
|| P%Max .
In the manufacturing of steel, first refine the pig iron in blast furnace to remove all impurities except little carbon. Than other trace metals are added to refine metal. Steel is mainly used to make cutlery,armor, utensils, weapons and tools.
It is a bright, shiny alloy composed of copper and zinc. It is more malleable compare to its constituent metals and has low melting point. Therefore it can be easily melt and mold in various shapes. It mainly used to make musical instruments like the trombone, the tuba, the trumpet and the French horn.
Due to shiny gold like appearance, it may also use to make decorative objects. Since it is more malleable and ductile compare to copper and zinc, it is widely used into pipes and electrical wires.
Alloys show quite different properties compare to their constituent metals. They show quite batter properties than metals. For example; steel is an alloy of iron and used as building material due to high strength. Similarly brass and bronze can easily mold in various shapes compare to constituent metals.
Bronze which is an alloy of copper and tin has more resistance to corrosion and widely used is ship-building. Titanium alloys are less dense and stronger compare to other metals. Therefore they mainly used in ship, aircraft, and spacecraft. Generally pure metals are less useful sue to more softness. For example; Gold is very soft in pure state and cannot use to manufacture ornaments. Therefore gold is mixed with other metals like silver, copper, or zinc.
These additional metals impart various properties and change appearance of gold. The properties of alloys are influenced by the manner of formation and treatment of alloy. The post-heat treatment and physical process can also modify the properties of alloys. Alloys are chemically more stable than their constituent metals; therefore alloys are designed for specific resistance to actions like corrosion, fatigue, wear, and temperature.
Some other properties of alloys are as follow -
- Steel is an alloy of iron, chromium and nickel with carbon. Each of elements imparts some unique characteristic to steel like resistance to corrosion and shiny surface.
- Alloy of beryllium and copper are stronger and good conductor of electricity.
- Alloy of gallium and arsenic used as superconductor in laser-beam technology.
- Due to their corrosion- and heat-resistance, alloys of nickel and cobalt are used in aircraft engines.
- Alloys of aluminum with silicon, iron, copper, manganese, magnesium and zinc are specifically designed for the manufacture of beverage cans.
- An alloy of tin with minor amounts of antimony and copper is known as pewter.
- Alloys of copper with zinc brass used for a variety of fittings, and copper with tin; bronze is used for plumbing fixtures.
- The combination of more than one metal or other element is formed metallic material called an alloy.
- Alloys are formed by mixing of metals through different physical processes like fusion, electrolytic deposition, etc.
- Generally metals like zinc, copper, tin, chromium, nickel, silver are used to prepare alloys with other metals and non-metals like carbon, boron, sulfur, zinc, etc.
- The composition of metals in alloy determines the chemical and physical properties of alloys.
- On the basis number of metals present in alloys, they can be classified as binary, ternary alloys and so on.
- Binary alloys created by the combination of two metals while ternary alloys are formed by three elements.
Some other common types of alloys are as follow -
1. Aluminium Alloys
Due to good conductivity, aluminium metal is useful for a variety of applications. The main disadvantage of this metal is less strength which can be overcome by addition of other metals to form alloys. For example, alnico is composed of aluminium, magnalium is composed of nickel, iron and cobalt with base metal aluminium and alloy with copper is called as duralumin.
2. Copper Alloys
Like iron, copper is also prone to oxidation and turns surface dull and pale-greenish. Therefore the formation of alloys of copper prevents the oxidation and increases its strength. For example; brass contains 80% copper with 20% zinc. Another alloy of copper is bronze contains 10% tin.
3. Iron Alloys
Steel is the most common alloy of iron which is composed of 95% iron with some content of carbon and other metals. The presence of carbon contents prevents iron metal from rusting and increase its strength. Other example of iron alloys are Invar, Fermico and Kovar.
4. Gold Alloys
Due to high malleability of pure gold, it cannot use to make jewellery. For example; alloy of gold with copper, silver and cobalt is known as yellow gold. Similarly white god is composed of gold, copper, zinc, nickel and trace amount of palladium.
There are many possible combinations of different metals to form alloys. Each alloy has its own specific properties and uses. The uses of alloys depend upon the composition of metal and complexity of alloy. Some common uses of alloys are as follow.
- Uses of aluminum alloys: Compare to aluminum, alloys of aluminum are good conductor and more strong. Due to these properties, they are widely used in the manufacturing of automotive engine part, electrical equipment and in furniture. Due to high strength they are also used in high altitude flying to bear huge pressure and stress.
- Uses of copper alloys: Like copper metal, alloys are copper are good conductor of heat and electricity, have high ductility and corrosion resistance. They are mainly used in automotive heat exchanger, sleeve bearing and in the manufacturing of electrical equipments.
- Uses of Nickel alloys: Alloys of nickel have good corrosion resistance and heat resistance. Therefore they are used for aircraft gas turbines, petrochemical industries, control equipment and nuclear power systems.
- Uses of Iron alloys: Stainless steel is one of the most common alloys of iron which is used for many commercial purposes. Iron alloys are mainly used in agriculture electronic and rail industries.
- Uses of Titanium alloys: Alloys of titanium have high strength, stiffness and toughness, therefore mainly used in aerospace and high-performance applications like for spacecraft parts. Jet engines and airframes. They have wide applications in bio-material and petrochemical industries also.<|endoftext|>
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The Lancet: Diet and food production must radically change to improve health and avoid potentially catastrophic damage to the planet
Credit: The Lancet
- Feeding a growing population of 10 billion people by 2050 with a healthy and sustainable diet will be impossible without transforming eating habits, improving food production, and reducing food waste. First scientific targets for a healthy diet that places healthy food consumption within the boundaries of our planet will require significant change, but are within reach.
- The daily dietary pattern of a planetary health diet consists of approximately 35% of calories as whole grains and tubers, protein sources mainly from plants – but including approximately 14g of red meat per day – and 500g per day of vegetables and fruits.
- Moving to this new dietary pattern will require global consumption of foods such as red meat and sugar to decrease by about 50%, while consumption of nuts, fruits, vegetables, and legumes must double.
- Unhealthy diets are the leading cause of ill-health worldwide and following the diet could avoid approximately 11 million premature deaths per year.
- A shift towards the planetary health diet would ensure the global food system The diet can exists within planetary boundariess for food production such as those for climate change, biodiversity loss, land and freshwater use, as well as nutrient cycles.
Transformation of the global food system is urgently needed as more than 3 billion people are malnourished (including people who are undernourished and overnourished), and food production is exceeding planetary boundaries – driving climate change, biodiversity loss, pollution due to over-application of nitrogen and phosphorus fertilizers, and unsustainable changes in water and land use.
The findings are from the EAT-Lancet Commission which provides the first scientific targets for a healthy diet from a sustainable food production system that operates within planetary boundaries for food. The report promotes diets consisting of a variety of plant-based foods, with low amounts of animal-based foods, refined grains, highly processed foods, and added sugars, and with unsaturated rather than saturated fats.
Human diets inextricably link health and environmental sustainability, and have the potential to nurture both. However, current diets are pushing the Earth beyond its planetary boundaries, while causing ill health. This puts both people and the planet at risk. Providing healthy diets from sustainable food systems is an immediate challenge as the population continues to grow – projected to reach 10 billion people by 2050 – and get wealthier (with the expectation of higher consumption of animal-based foods).
To meet this challenge, dietary changes must be combined with improved food production and reduced food waste. The authors stress that unprecedented global collaboration and commitment will be needed, alongside immediate changes such as refocussing agriculture to produce varied nutrient-rich crops, and increased governance of land and ocean use.
“The food we eat and how we produce it determines the health of people and the planet, and we are currently getting this seriously wrong,” says one of the commission authors Professor Tim Lang, City, University of London, UK. “We need a significant overhaul, changing the global food system on a scale not seen before in ways appropriate to each country’s circumstances. While this is unchartered policy territory and these problems are not easily fixed, this goal is within reach and there are opportunities to adapt international, local and business policies. The scientific targets we have devised for a healthy, sustainable diet are an important foundation which will underpin and drive this change.”
The Commission is a 3-year project that brings together 37 experts from 16 countries with expertise in health, nutrition, environmental sustainability, food systems, economics and political governance.
Scientific targets for a healthy diet – the planetary health diet
Despite increased food production contributing to improved life expectancy and reductions in hunger, infant and child mortality rates, and global poverty over the past 50 years, these benefits are now being offset by global shifts towards unhealthy diets high in calories, sugar, refined starches and animal-based foods and low in fruits, vegetables, whole grains, legumes, nuts and seeds, and fish.
The authors argue that the lack of scientific targets for a healthy diet have hindered efforts to transform the food system. Based on the best available evidence, the Commission proposes a dietary pattern that meets nutritional requirements, promotes health, and allows the world to stay within planetary boundaries.
Compared with current diets, global adoption of the new recommendations by 2050 will require global consumption of foods such as red meat and sugar to decrease by more than 50%, while consumption of nuts, fruits, vegetables, and legumes must increase more than two-fold. Global targets will need to be applied locally – for example, countries in North America eat almost 6.5 times the recommended amount of red meat, while countries in South Asia eat only half the recommended amount. All countries are eating more starchy vegetables (potatoes and cassava) than recommended with intakes ranging from between 1.5 times above the recommendation in South Asia and by 7.5 times in sub-Saharan Africa.
“The world’s diets must change dramatically. More than 800 million people have insufficient food, while many more consume an unhealthy diet that contributes to premature death and disease,” says co-lead Commissioner Dr Walter Willett, Harvard University, USA. “To be healthy, diets must have an appropriate calorie intake and consist of a variety of plant-based foods, low amounts of animal-based foods, unsaturated rather than saturated fats, and few refined grains, highly processed foods, and added sugars. The food group intake ranges that we suggest allow flexibility to accommodate various food types, agricultural systems, cultural traditions, and individual dietary preferences – including numerous omnivore, vegetarian, and vegan diets.”
Please view the image to see the dietary targets based on a 2,500 kcal/day diet .
The authors estimate that widespread adoption of such a diet would improve intakes of most nutrients – increasing intake of healthy mono and polyunsaturated fatty acids and reducing consumption of unhealthy saturated fats. It would also increase essential micronutrient intake (such as iron, zinc, folate, and vitamin A, as well as calcium in low-income countries), except for vitamin B12 where supplementation or fortification might be necessary in some circumstances.
They also modelled the potential effects of global adoption of the diet on deaths from diet-related diseases. Three models each showed major health benefits, suggesting that adopting the new diet globally could avert between 10.9-11.6 million premature deaths per year – reducing adult deaths by between 19-23.6%.
The authors highlight that evidence about diet, human health, and environmental sustainability is continually evolving and includes uncertainty, so they include ranges in their estimates, but are confident of the overall picture. Professor Lang says: “While major transformations to the food system occurred in China, Brazil, Vietnam, and Finland in the 20th century, and illustrate that diets can change rapidly, humanity has never aimed to change the food system this radically at such speed or scale. People might warn of unintended consequences or argue that the case for action is premature, however, the evidence is sufficient and strong enough to warrant action, and any delay will increase the likelihood of not achieving crucial health and climate goals.”
Since the mid-1950s, the pace and scale of environmental change has grown exponentially. Food production is the largest source of environmental degradation. To be sustainable, food production must occur within food-related planetary boundaries for climate change, biodiversity loss, land and water use, as well as for nitrogen and phosphorus cycles. However, production must also be sustainably intensified to meet the global population’s growing food demands.
This will require decarbonising agricultural production by eliminating the use of fossil fuels and land use change losses of CO2 in agriculture. In addition, zero loss of biodiversity, net zero expansion of agricultural land into natural ecosystems, and drastic improvements in fertiliser and water use efficiencies are needed.
The authors estimate the minimum, unavoidable emissions of greenhouse gases if we are to provide healthy food for 10 billion people by 2050 . They conclude that non-CO2 greenhouse gas emissions of methane and nitrous oxide will remain between 4.7-5.4 gigatonnes in 2050, with current emissions already at an estimated 5.2 gigatonnes in 2010. This suggests that the decarbonisation of the world energy system must progress faster than anticipated, to accommodate the need to healthily feed humans without further damaging the planet.
Phosphorus use must also be reduced (from 17.9 to between 6-16 teragrams), as must biodiversity loss (from 100 to between 1-80 extinctions per million species each year).
Based on their estimates, current levels of nitrogen, land and water use may be within the projected 2050 boundary (from 131.8 teragrams in 2010 to between 65-140 in 2050, from 12.6 M km2 in 2010 vs 11-15 M km2 in 2050, and from 1.8 M km3 in 2010 vs 1-4 M km3, respectively) but will require continued efforts to sustain this level. The boundary estimates are subject to uncertainty, and will require continuous update and refinement.
Using these boundary targets, the authors modelled various scenarios to develop a sustainable food system and deliver healthy diets by 2050. To stay within planetary boundaries, a combination of major dietary change, improved food production through enhanced agriculture and technology changes , and reduced food waste during production and at the point of consumption will be needed, and no single measure is enough to stay within all of the limits.
“Designing and operationalising sustainable food systems that can deliver healthy diets for a growing and wealthier world population presents a formidable challenge. Nothing less than a new global agricultural revolution. The good news is that it is not only doable, we have increasing evidence that it can be achieved through sustainable intensification that benefits both farmer, consumer and planet,” says co-lead Commissioner Professor Johan Rockström, Stockholm Resilience Centre, Sweden and Potsdam Institute for Climate Impact Research, Germany.
“Humanity now poses a threat to the stability of the planet. Sustainability of the food system must therefore be defined from a planetary perspective. Five key environmental processes regulate the state of the planet. Our definition of sustainable food production requires that we use no additional land, safeguard existing biodiversity, reduce consumptive water use and manage water responsibly, substantially reduce nitrogen and phosphorus pollution, produce zero carbon dioxide emissions, and cause no further increase in methane and nitrous oxide emissions. There is no silver bullet for combatting harmful food production practices, but by defining and quantifying a safe operating space for food systems, diets can be identified that will nurture human health and support environmental sustainability.” Professor Rockström continues.
Transforming the global food system
The Commission proposes five strategies to adjust what people eat and how it is produced.
Firstly, policies to encourage people to choose healthy diets are needed, including improving availability and accessibility to healthy food through improved logistics and storage, increased food security, and policies that promote buying from sustainable sources. Alongside advertising restrictions and education campaigns, affordability is also crucial, and food prices must reflect production and environmental costs. As this may increase costs to consumers, social protection for vulnerable groups may be required to avoid continued poor nutrition in low-income groups.
Strategies to refocus agriculture from producing high volumes of crops to producing varied nutrient-rich crops are needed. Currently, small and medium farms supply more than 50% of the essential nutrients in the global food supply. Global agriculture policies should incentivise producers to grow nutritious, plant-based foods, develop programmes that support diverse production systems, and increase research funding for ways to increase nutrition and sustainability. In some contexts, animal farming is important to nutrition and the ecosystem and the benefits and risks of animal farming should be considered on a case-by-case basis.
Sustainably intensifying agriculture will also be key, and must take into account local conditions to help apply appropriate agricultural practices and generate sustainable, high quality crops.
Equally, effective governance of land and ocean use will be important to preserve natural ecosystems and ensure continued food supplies. This could be achieved through protecting intact natural areas on land (potentially through incentives), prohibiting land clearing, restoring degraded land, removing harmful fishing subsidies, and closing at least 10% of marine areas to fishing (including the high seas to create fish banks).
Lastly, food waste must be at least halved. The majority of food waste occurs in low- and middle-income countries during food production due to poor harvest planning, lack of access to markets preventing produce from being sold, and lack of infrastructure to store and process foods. Improved investment in technology and education for farmers is needed. Food waste is also an issue in high-income countries, where it is primarily caused by consumers and can be resolved through campaigns to improve shopping habits, help understand ‘best before’ and ‘use by’ dates, and improve food storage, preparation, portion sizes and use of leftovers.
Dr Richard Horton, Editor-in-Chief at The Lancet, says: “Poor nutrition is a key driver and risk factor for disease. However, there has been a global failure to address this. It is everyone’s and no-one’s problem.”
He continues: “The transformation that this Commission calls for is not superficial or simple, and requires a focus on complex systems, incentives, and regulations, with communities and governments at multiple levels having a part to play in redefining how we eat. Our connection with nature holds the answer, and if we can eat in a way that works for our planet as well as our bodies, the natural balance of the planet’s resources will be restored. The very nature that is disappearing holds the key to human and planetary survival.”
The EAT-Lancet Commission is one of several reports on nutrition being published by The Lancet in 2019. The next Commission – The Global Syndemic of Obesity, Undernutrition, and Climate Change – will publish later this month.
Peer-reviewed / Review, modelling, opinion
NOTES TO EDITORS
This study was funded by the Wellcome Trust and EAT (specifically funding from the Wellcome Trust and Stordalen Foundation). The Stockholm Resilience Centre was the scientific coordinator of the report.
The labels have been added to this press release as part of a project run by the Academy of Medical Sciences seeking to improve the communication of evidence. For more information, please see: http://www.sciencemediacentre.org/wp-content/uploads/2018/01/AMS-press-release-labelling-system-GUIDANCE.pdf if you have any questions or feedback, please contact The Lancet press office [email protected]
Quote direct from author and cannot be found in the text of the Article.
This takes into account the average global energy intake being around 2,370 kcal/day (with some countries being even higher than this) based on country-specific body weights. The diet corresponds to the average energy needs of a 70-kg man aged 30 years and a 60-kg woman aged 30 years whose level of physical activity is moderate to high. It is designed to meet nutritional requirements of healthy individuals over 2 years old (with energy intake depending on age, body size, and physical activity), but the authors note that there are special considerations for young children, adolescents and pregnant and breastfeeding women.
This is based on the expectation that commitments to decarbonise the energy system by 2050 (no fossil-fuels for tractors, electricity, heat) will be met globally, there will be net-zero CO2 emissions from land-use change (through sustainable land management), and there will be improved nitrogen use efficiency and reduced methane emissions from ruminant livestock.
The study focusses on methane and nitrous oxide and does not include carbon dioxide. This is because food production is a prime source of methane, and nitrous oxide, which have 56 times and 280 times the global warming potential (over 20 years) of carbon dioxide, respectively, and because it is assumed there are no net inputs of carbon dioxide from fossil fuels to agriculture by 2050.
These estimates only include technologies that are currently available and proven at scale.
Lancet Press Office
Related Journal Article<|endoftext|>
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There’s a certain charm to the fact that such vast equipment has to be constructed to study the smallest known bits of matter. The tunnel of the Large Hadron Collider has a circumference of almost 27 kilometers. And KATRIN, an experiment in Karlsruhe, Germany, described by physics writer Emily Conover in this issue, requires a blimplike metal tank that’s wider than some of the neighborhood streets.
Conover knows firsthand the exacting work of building a physics experiment. While a graduate student at the University of Chicago, Conover toiled away on Double Chooz. The experiment was designed to detect antineutrino oscillations, which occur when antineutrinos change from one type into another.
That endeavor required a source of antineutrinos. They’re produced during certain nuclear reactions, so the experiment was located in a tunnel near a nuclear power plant in Chooz, France. It also required detectors to spot heavy relatives of electrons called muons, which constantly rain down on Earth and can cause reactions that mimic antineutrinos. The detectors, which included scintillators that would light up when a muon zipped through, made sure muons didn’t get counted as antineutrinos by mistake.
Conover and her colleagues used an extruder to form hundreds of thin, 3-meter-long strips of plastic, then assembled them to make the detectors. “They were kind of floppy and covered with aluminum,” Conover says of the detectors. “We were always terrified that we would break one.”
The project involved taking several trips to France to install and test the detectors in the subterranean tunnel. All told, Conover worked on Double Chooz for six years. In the end, the experiment was a success: It was one of the first to measure a particular type of neutrino oscillation.
Scientists come in many flavors, and we’re delighted to provide a behind-the-scenes look at their work across the disciplines.<|endoftext|>
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This week we finished off our Science unit of 'Rocks and Fossils'. We started the week by testing the hardness of a selection of rocks by investigating to see if we could scratch them with (1) our fingernail, (2) sandpaper and (3) an iron nail. We recorded what we found on a chart. Our results varied but our overall conclusion was that metamorphic rocks, such as slate and marble, are the hardest and sedimentary rocks, such as limestone and sandstone, are the softest.
In Art this week, we were all very excited (including the teachers!) because we used clay. On Thursday afternoon all of Year 3 worked together in the dining hall. We used a grey buff clay and tried out a range of clay skills. We learned how to make a lump of clay into a sphere and flatten it out; how to attach pieces of clay together by scratching the surface and wetting; how to roll a coil and how to impress designs using a selection of tools. Our practice pieces looked like miniature sculptures. We then used our skills to decorate thumb pots we had made.
On Friday, it was FOSSIL DAY!
We started the day by sorting a set of picture cards into two groups; fossils or not fossils. After that, we watched a video clip about how fossils are formed and had to use the information to create a storyboard explaining the process.
A really enjoyable part of the day was the fossil dig. In groups, we went outside and used trowels and brushes to search for fossils and rocks at our own archaeological dig site. With guidance from Mrs Sherwood and Mrs Wardell, we found a variety of items including small rocks, bones and fossils. It was all very exciting!
Finally, after lunch, we had another clay session. This time, we used red terracotta clay, which was messier than the buff clay. We used our new skills to make an ammonite and then our very own dinosaurs. All of our clay work will need time to air-dry and then it will be fired (to a temperature of 1,200°C) in our kiln. A huge thank you to all our parent helpers over both days. We couldn't have done it without you .
We can hardly believe that we have reached the end of another half term and are halfway thorough Year 3; where has the time gone?! We are looking forward to next half term with plenty of exciting work, activities and events to look forward to.<|endoftext|>
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Children with autism are often described as robotic: They are emotionless. They engage in obsessive, repetitive behavior and have trouble communicating and socializing.
Now, a humanoid robot designed to teach autistic children social skills has begun testing in British schools.
Known as KASPAR (Kinesics and Synchronisation in Personal Assistant Robotics), the $4.33 million bot smiles, simulates surprise and sadness, gesticulates and, the researchers hope, will encourage social interaction amongst autistic children.
Children with autism have difficulty understanding and interpreting people's facial expressions and body language, says Dr. Ben Robins, a senior research fellow at the University of Hertfordshire's Adaptive Systems Research Group, who leads the multi-national team behind KASPAR.
"Human interaction can be very subtle, with even the smallest eyebrow raise, for example, having different meanings in different contexts," Robins said. "It is thought that autistic children cut themselves off from interacting with other humans because, for them, this is too much information and it is too confusing for them to understand."
With this in mind, the team designed KASPAR to express emotion consistently and with the minimum of complexity.
KASPAR's face is made of silicon-rubber supported on an aluminum frame. Eight degrees of freedom in the head and neck and six in the arms and hands enable movement.
The researchers hope that the end result is a human-like robot that can act as a "social mediator" for autistic children, a steppingstone to improved social interaction with other children and adults.
"KASPAR provides autistic children with reliability and predictability. Since there are no surprises, they feel safe and secure," Robins said, adding that the purpose is not to replace human interaction and contact but to enhance it.
Robins has already tested some imitation and turn-taking games with the children and his preliminary findings are positive.
"When I first started testing, the children treated me like a fly on the wall," he said. "But each one of them, in their own time, started to open themselves up to me. One child in particular, after weeks on end of ignoring me, came and sat in my lap and then took my hand and brought me to the robot, to share the experience of KASPAR with me."
Using robots to interact with children is nothing new, although there's been a lot of new research lately into this kind of work. The Robota dolls, a series of mini humanoid bots developed as part of the AURORA project, have been in use as educational toys since 1997.
The Social Robotics Lab at Yale is collaborating with a robotics team from the university’s department of computer science to develop Nico, a humanoid robot designed to detect vulnerabilities for autism in the first year of life.
Relying on a robot to teach human social skills might seem counterintuitive, but autism presents a special case, said Dr. Cathy Pratt, director of the Indiana Resource Center for Autism at Indiana University.
"Autistic kids often interact better with inanimate objects than with other people, so a project like this makes sense and might lead to a safe way for these kids to learn social skills," she said.
However, autistic children often don’t make the connection between what they have learned in a training situation and the outside world, said Dr. Gary Mesibov, a professor of psychiatry at the University of North Carolina and editor of the Journal of Autism and Developmental Disorders.
"I think this project will still be worthwhile, even if the children don’t fully generalize what they have learned to the real world," Mesibov said. "But the key question facing the researchers is whether the autistic children will be able to apply what they have learned from KASPAR in different situations and contexts."
Face recognition and emotion processing is a major area of deficit for autistic children and hampers their social development, said Dr. Jennifer Pinto-Martin, director of the Center for Autism and Developmental Disabilities Research and Epidemiology at the University of Pennsylvania.
Although autistic children often respond well to training, the process can be very labor intensive and the quality of the trainer is paramount, Pinto-Martin said. "People who work in this area need more creative ways to train around the deficits of autism. The quality and consistency of the trainer can be hard to control, but that's not the case with a robot.
"There is interactive computer software and video out there for testing and interaction, but the idea of using a robot trainer like KASPAR is a creative and wonderful step beyond current technologies and techniques," she said.
The project is due to end in October 2009.<|endoftext|>
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### Chap05 discrete probability distributions
• 1. Statistics for Managers Using Microsoft® Excel 4th Edition Chapter 5 Some Important Discrete Probability Distributions Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-1
• 2. Chapter Goals After completing this chapter, you should be able to: Interpret the mean and standard deviation for a discrete probability distribution Explain covariance and its application in finance Use the binomial probability distribution to find probabilities Describe when to apply the binomial distribution Use the hypergeometric and Poisson discrete probability distributions to find probabilities Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-2
• 3. Introduction to Probability Distributions Random Variable Represents a possible numerical value from an uncertain event Random Variables Ch. 5 Discrete Random Variable Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Continuous Random Variable Ch. 6 Chap 5-3
• 4. Discrete Random Variables Can only assume a countable number of values Examples: Roll a die twice Let X be the number of times 4 comes up (then X could be 0, 1, or 2 times) Toss a coin 5 times. Let X be the number of heads (then X = Using Statistics for Managers0, 1, 2, 3, 4, or 5) Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-4
• 5. Discrete Probability Distribution Experiment: Toss 2 Coins. 4 possible outcomes T Let X = # heads. Probability Distribution T T H H Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. 1 2/4 = .50 1/4 = .25 Probability H 1/4 = .25 2 H Probability 0 T X Value .50 .25 0 1 2 X Chap 5-5
• 6. Discrete Random Variable Summary Measures Expected Value (or mean) of a discrete distribution (Weighted Average) N µ = E(X) = ∑ Xi P( Xi ) i=1 Example: Toss 2 coins, X = # of heads, compute expected value of X: X P(X) 0 .25 1 .50 2 .25 E(X) = (0 x .25) + (1 x .50) + (2 x .25) Statistics for Managers Using = 1.0 Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-6
• 7. Discrete Random Variable Summary Measures (continued) Variance of a discrete random variable N σ 2 = ∑ [Xi − E(X)]2 P(Xi ) i=1 Standard Deviation of a discrete random variable σ = σ2 = N [Xi − E(X)]2 P(Xi ) ∑ i=1 where: E(X) = Expected value of the Statistics for Managers Using discrete random variable X Xi = the ith outcome of X Microsoft Excel, 4e © 2004ith occurrence of X P(Xi) = Probability of the Prentice-Hall, Inc. Chap 5-7
• 8. Discrete Random Variable Summary Measures (continued) Example: Toss 2 coins, X = # heads, compute standard deviation (recall E(X) = 1) σ= ∑ [X − E(X)] P(X ) 2 i i σ = (0 − 1)2 (.25) + (1 − 1)2 (.50) + (2 − 1)2 (.25) = .50 = .707 Possible number of heads = 0, 1, or 2 Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-8
• 9. The Covariance The covariance measures the strength of the linear relationship between two variables The covariance: N σ XY = ∑ [ Xi − E( X)][( Yi − E( Y )] P( Xi Yi ) i=1 where: X = discrete variable X Xi = the ith outcome of X Y = discrete variable Y Yi = the ith outcome of Y Statistics for Managers Using P(XiYi) = probability of occurrence of the condition affecting Microsoft Excel, 4e ©ith2004 of X and the ith outcome of Y the outcome Prentice-Hall, Inc. Chap 5-9
• 10. Computing the Mean for Investment Returns Return per \$1,000 for two types of investments P(XiYi) Economic condition Investment Passive Fund X Aggressive Fund Y .2 Recession - \$ 25 - \$200 .5 Stable Economy + 50 + 60 .3 Expanding Economy + 100 + 350 E(X) = μX = (-25)(.2) +(50)(.5) + (100)(.3) = 50 Statistics for Managers Using E(Y) = μY = (-200)(.2) +(60)(.5) + (350)(.3) = 95 Microsoft Excel, 4e © 2004 Chap 5-10 Prentice-Hall, Inc.
• 11. Computing the Standard Deviation for Investment Returns P(XiYi) Economic condition Investment Passive Fund X Aggressive Fund Y .2 Recession - \$ 25 - \$200 .5 Stable Economy + 50 + 60 .3 Expanding Economy + 100 + 350 σ X = (-25 − 50)2 (.2) + (50 − 50)2 (.5) + (100 − 50)2 (.3) = 43.30 σ Y = (-200 − 95)2 (.2) + (60 − 95)2 (.5) + (350 − 95)2 (.3) Statistics for Managers Using = 193 71 Microsoft Excel, .4e © 2004 Chap 5-11 Prentice-Hall, Inc.
• 12. Computing the Covariance for Investment Returns P(XiYi) Economic condition Investment Passive Fund X Aggressive Fund Y .2 Recession - \$ 25 - \$200 .5 Stable Economy + 50 + 60 .3 Expanding Economy + 100 + 350 σ X,Y = (-25 − 50)(-200 − 95)(.2) + (50 − 50)(60 − 95)(.5) + (100 − 50)(350 − 95)(.3) = 8250 Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-12
• 13. Interpreting the Results for Investment Returns The aggressive fund has a higher expected return, but much more risk μY = 95 > μX = 50 but σY = 193.21 > σX = 43.30 The Covariance of 8250 indicates that the two investments are positively related and will vary Statistics for Managers Using in the same 2004 Microsoft Excel, 4e © direction Prentice-Hall, Inc. Chap 5-13
• 14. The Sum of Two Random Variables Expected Value of the sum of two random variables: E(X + Y) = E( X) + E( Y ) Variance of the sum of two random variables: Var(X + Y) = σ 2 + Y = σ 2 + σ 2 + 2σ XY X X Y Standard deviation of the sum of two random variables: Statistics for Managers Using σ Microsoft Excel, 4e © 2004 X + Y = Prentice-Hall, Inc. σ2 +Y X Chap 5-14
• 15. Portfolio Expected Return and Portfolio Risk Portfolio expected return (weighted average return): E(P) = w E( X) + (1 − w ) E( Y ) Portfolio risk (weighted variability) σ P = w 2σ 2 + (1 − w )2 σ 2 + 2w(1 - w)σ XY X Y Where w = portion of portfolio Statistics for Managers Using value in asset X (1 - w) = portion of portfolio value in asset Y Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-15
• 16. Portfolio Example Investment X: Investment Y: μX = 50 σX = 43.30 μY = 95 σY = 193.21 σXY = 8250 Suppose 40% of the portfolio is in Investment X and 60% is in Investment Y: E(P) = .4 (50) + (.6) (95) = 77 σ P = (.4)2 (43.30)2 + (.6)2 (193.21) 2 + 2(.4)(.6)(8250) = 133.04 Statistics for Managers Using The Excel, return and portfolio variability are between the values Microsoft portfolio 4e © 2004 for investments X and Y considered individually Chap 5-16 Prentice-Hall, Inc.
• 17. Probability Distributions Probability Distributions Ch. 5 Discrete Probability Distributions Continuous Probability Distributions Binomial Normal Hypergeometric Ch. 6 Uniform Statistics for Poisson Using Managers Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Exponential Chap 5-17
• 18. The Binomial Distribution Probability Distributions Discrete Probability Distributions Binomial Hypergeometric Statistics for Poisson Using Managers Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-18
• 19. Binomial Probability Distribution A fixed number of observations, n Two mutually exclusive and collectively exhaustive categories e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse e.g., head or tail in each toss of a coin; defective or not defective light bulb Generally called “success” and “failure” Probability of success is p, probability of failure is 1 – p Constant probability for each observation e.g., Probability of getting a tail is the same each time we toss the coin Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-19
• 20. Binomial Probability Distribution (continued) Observations are independent The outcome of one observation does not affect the outcome of the other Two sampling methods Infinite population without replacement Finite population with replacement Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-20
• 21. Possible Binomial Distribution Settings A manufacturing plant labels items as either defective or acceptable A firm bidding for contracts will either get a contract or not A marketing research firm receives survey responses of “yes I will buy” or “no I will not” New job applicants either accept the offer Statistics for Managers Using or reject it Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-21
• 22. Rule of Combinations The number of combinations of selecting X objects out of n objects is n n! = X X! (n − X)! where: n! =n(n - 1)(n - 2) . . . (2)(1) X! = X(X - 1)(X - 2) . . . (2)(1) Statistics for Managers 0! = 1 (by definition) Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-22
• 23. Binomial Distribution Formula n! X n−X P(X) = p (1-p) X ! (n − X)! P(X) = probability of X successes in n trials, with probability of success p on each trial X = number of ‘successes’ in sample, (X = 0, 1, 2, ..., n) n = sample size (number of trials or observations) p = probability of “success” Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Example: Flip a coin four times, let x = # heads: n=4 p = 0.5 1 - p = (1 - .5) = .5 X = 0, 1, 2, 3, 4 Chap 5-23
• 24. Example: Calculating a Binomial Probability What is the probability of one success in five observations if the probability of success is .1? X = 1, n = 5, and p = .1 P( X = 1) = n! p X (1 − p)n− X X! (n − X)! 5! = (.1)1(1 − .1)5−1 1 (5 − 1)! ! = (5)(.1)(.9)4 Statistics for Managers Using Microsoft Excel, 4e © 2004 = .32805 Prentice-Hall, Inc. Chap 5-24
• 25. Binomial Distribution The shape of the binomial distribution depends on the values of p and n Mean Here, n = 5 and p = .1 .6 .4 .2 0 P(X) X 0 Here, n = 5 and p = .5 Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. .6 .4 .2 0 n = 5 p = 0.1 P(X) 1 2 3 4 5 n = 5 p = 0.5 X 0 1 2 3 4 5 Chap 5-25
• 26. Binomial Distribution Characteristics Mean μ = E(x) = np Variance and Standard Deviation σ = np(1 - p) 2 σ = np(1 - p) Where n = sample size p = probability Statistics for Managers of success Using (1 – p) = probability of failure Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-26
• 27. Binomial Characteristics Examples μ = np Mean = (5)(.1) = 0.5 σ = np(1 - p) = (5)(.1)(1 − .1) = 0.6708 μ = np = (5)(.5) = 2.5 σ = np(1 - p) = (5)(.5)(1 − .5) Statistics for Managers Using = 1.118 Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. .6 .4 .2 0 P(X) X 0 .6 .4 .2 0 n = 5 p = 0.1 P(X) 1 2 3 4 5 n = 5 p = 0.5 X 0 1 2 3 4 5 Chap 5-27
• 28. Using Binomial Tables n = 10 x … p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.50 0 1 2 3 4 5 6 7 8 9 10 … … … … … … … … … … … 0.1074 0.2684 0.3020 0.2013 0.0881 0.0264 0.0055 0.0008 0.0001 0.0000 0.0000 0.0563 0.1877 0.2816 0.2503 0.1460 0.0584 0.0162 0.0031 0.0004 0.0000 0.0000 0.0282 0.1211 0.2335 0.2668 0.2001 0.1029 0.0368 0.0090 0.0014 0.0001 0.0000 0.0135 0.0725 0.1757 0.2522 0.2377 0.1536 0.0689 0.0212 0.0043 0.0005 0.0000 0.0060 0.0403 0.1209 0.2150 0.2508 0.2007 0.1115 0.0425 0.0106 0.0016 0.0001 0.0025 0.0207 0.0763 0.1665 0.2384 0.2340 0.1596 0.0746 0.0229 0.0042 0.0003 0.0010 0.0098 0.0439 0.1172 0.2051 0.2461 0.2051 0.1172 0.0439 0.0098 0.0010 10 9 8 7 6 5 4 3 2 1 0 … p=.80 p=.75 p=.70 p=.65 p=.60 p=.55 p=.50 x Examples: n = 10, p = .35, x = 3: P(x = 3|n =10, p = .35) = .2522 Statistics for Managers Using n= Microsoft 10, p = .75, x =2004 P(x = 2|n =10, p = .75) = .0004 Excel, 4e © 2: Chap 5-28 Prentice-Hall, Inc.
• 29. Using PHStat Select PHStat / Probability & Prob. Distributions / Binomial… Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-29
• 30. Using PHStat (continued) Enter desired values in dialog box Here: n = 10 p = .35 Output for X = 0 to X = 10 will be generated by PHStat Optional check boxes Statistics for Managers Using for additional output Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-30
• 31. PHStat Output P(X = 3 | n = 10, p = .35) = .2522 Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. P(X > 5 | n = 10, p = .35) = .0949 Chap 5-31
• 32. The Hypergeometric Distribution Probability Distributions Discrete Probability Distributions Binomial Hypergeometric Statistics for Poisson Using Managers Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-32
• 33. The Hypergeometric Distribution “n” trials in a sample taken from a finite population of size N Sample taken without replacement Outcomes of trials are dependent Concerned with finding the probability of “X” successes in the sample where there are “A” successes in the population Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-33
• 34. Hypergeometric Distribution Formula A N − A X n − X P( X) = N n Where N = population size A = number of successes in the population N – A = number of failures in the population n = sample size Statistics for Managers Using X = number of successes in the sample Microsoft Excel, = number of failures in the sample n – X 4e © 2004 Prentice-Hall, Inc. Chap 5-34
• 35. Properties of the Hypergeometric Distribution The mean of the hypergeometric distribution is nA μ = E(x) = N The standard deviation is nA(N - A) N - n σ= ⋅ 2 N N -1 Where N-n is called the “Finite Population Correction Factor” N -1 from Using Statistics for Managers sampling without replacement from a Microsoft Excel, 4efinite population © 2004 Chap 5-35 Prentice-Hall, Inc.
• 36. Using the Hypergeometric Distribution ■ Example: 3 different computers are checked from 10 in the department. 4 of the 10 computers have illegal software loaded. What is the probability that 2 of the 3 selected computers have illegal software loaded? N = 10 A=4 n=3 X=2 A N − A 4 6 X n − X 2 1 (6)(6) = = P(X = 2) = = 0.3 120 N 10 n 3 Statistics for Managers2Using 3 selected computers have illegal The probability that of the Microsoft Excel, 4e © .30, or 30%. software loaded is 2004 Chap 5-36 Prentice-Hall, Inc.
• 37. Hypergeometric Distribution in PHStat Select: PHStat / Probability & Prob. Distributions / Hypergeometric … Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-37
• 38. Hypergeometric Distribution in PHStat (continued) Complete dialog box entries and get output … N = 10 A=4 n=3 X=2 Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. P(X = 2) = 0.3 Chap 5-38
• 39. The Poisson Distribution Probability Distributions Discrete Probability Distributions Binomial Hypergeometric Statistics for Poisson Using Managers Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-39
• 40. The Poisson Distribution Apply the Poisson Distribution when: You wish to count the number of times an event occurs in a given area of opportunity The probability that an event occurs in one area of opportunity is the same for all areas of opportunity The number of events that occur in one area of opportunity is independent of the number of events that occur in the other areas of opportunity The probability that two or more events occur in an area of opportunity approaches zero as the area of Statisticsopportunity becomes smaller for Managers Using MicrosoftThe average 2004 Excel, 4e © number of events per unit is λ (lambda) Chap 5-40 Prentice-Hall, Inc.
• 41. Poisson Distribution Formula −λ x e λ P( X) = X! where: X = number of successes per unit λ = expected number of successes per unit e = base of the natural logarithm system (2.71828...) Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-41
• 42. Poisson Distribution Characteristics Mean μ=λ Variance and Standard Deviation σ2 = λ σ= λ where λ = expected number of successes per unit Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-42
• 44. Graph of Poisson Probabilities 0.70 Graphically: 0.60 λ = .50 0 1 2 3 4 5 6 Statistics 7 λ= 0.50 P(x) X 0.50 0.40 0.30 0.6065 0.20 0.3033 0.10 0.0758 0.0126 0.00 0 1 0.0016 0.0002 0.0000 for Managers UsingP(X 0.0000 Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. 2 3 4 5 6 7 x = 2) = .0758 Chap 5-44
• 45. Poisson Distribution Shape The shape of the Poisson Distribution depends on the parameter λ : λ = 0.50 λ = 3.00 0.70 0.25 0.60 0.20 0.15 0.40 P(x) P(x) 0.50 0.30 0.10 0.20 0.05 0.10 0.00 0.00 0 1 2 3 4 5 6 7 x Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. 1 2 3 4 5 6 7 8 9 10 x Chap 5-45 11 12
• 46. Poisson Distribution in PHStat Select: PHStat / Probability & Prob. Distributions / Poisson… Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-46
• 47. Poisson Distribution in PHStat (continued) Complete dialog box entries and get output … Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. P(X = 2) = 0.0758 Chap 5-47
• 48. Chapter Summary Addressed the probability of a discrete random variable Defined covariance and discussed its application in finance Discussed the Binomial distribution Discussed the Hypergeometric distribution Reviewed the Poisson distribution Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 5-48
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Polynomials Applied to Linear Operators Examples 1
Polynomials Applied to Linear Operators Examples 1
Recall from the Polynomials Applied to Linear Operators page that we can apply a polynomial to a linear operator $T \in \mathcal L(V)$ by defining $T^m = \underbrace{T \circ T \circ ... \circ T}_{\mathrm{m \: times}}$ where $m$ is a positive integer. Furthermore, if $T$ is an invertible linear operator then we define $T^{-m} = \underbrace{T^{-1} \circ T^{-1} \circ ... \circ T^{-1}}_{\mathrm{m \: times}}$.
We will now look at some examples regarding polynomials applied to linear operators.
Example 1
Let $T$ be a linear operator on $V$. Suppose that $T^3 - 6T^2 + 11T - 6I = 0$. Prove that then if $\lambda$ is an eigenvalue of $T$ then $\lambda = 1$ or $\lambda = 2$ or $\lambda = 3$.
Let $\lambda$ be an eigenvalue of $T$. Then for some nonzero vector $u \in V$ we have that $T(u) = \lambda u$. Note that:
(1)
\begin{align} \quad T^2(u) = T(T(u)) = T(\lambda u) = \lambda T(u) = \lambda^2 u \\ \quad T^3(u) = T(T^2(u)) = T(\lambda^2 u) = \lambda^2 T(u) = \lambda^3 u \\ \quad \quad \quad \vdots \quad \quad \quad \\ \quad T^n(u) = T(T^{n-1}(u)) = T(\lambda^{n-1} u) = \lambda^{n-1} T(u) = \lambda^n u \end{align}
Substituting this into $T^3 - 6T^2 + 11T - 6I = 0$ and we have that:
(2)
\begin{align} \quad T^3 - 6T^2 + 11T - 6I = 0 \\ \quad \lambda^3 - 6\lambda^2 + 11 \lambda - 6 \lambda = 0 \\ \quad (\lambda - 1)(\lambda - 2)(\lambda - 3) = 0 \end{align}
Therefore we have that $\lambda = 1$ or $\lambda = 2$ or $\lambda = 3$.
Example 2
Let $T$ be a linear operator on $V$ such that $T = T^2$. Show that then $V = \mathrm{null} (T) \oplus \mathrm{range} (T)$.<|endoftext|>
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The if statement is probably the most well-known flow control statement in Python. To show you how that works, let's create a variable called X, and X is going to convert to an integer, the input it receives from a console prompt.
We'll evaluate X with an if statement and say if X is less than zero, we'll set the value to zero and then print negative number change to zero. We can run that with the Python 3 command and then provide the name of the file.
I'm prompted for the integer, so if I enter the value negative 1, it evaluates to true in the if statement, sets the value to zero and then prints out our statement. If we run it again and I give it something that won't evaluate to true such as the number 2, whenever it runs, the if statement doesn't evaluate to true and there's nothing left in the program to do, so the program exits.
Let's expand on it a little bit and give it something else to do. If statements can have optional elif and else statements, so let's give it an elif which stands for else if, and we'll say if X is equal to zero, then we'll just print the word zero.
Notice that the if and elif conditional statements aren't indented, but the code blocks that execute when they evaluate to true are indented. This is important because in Python everything is space sensitive so you have to get it right or your program won't execute.
Next, we'll say elif X equal to 1, then we're going print out the word 1 and we'll give it a final else clause that's going to execute if none of the other statements evaluate to true. Let's run it again and if I give it a negative 1, that's the condition, we originally had, so that still works, and run it one more time.
If I give it the integer of 1, the 1 is printed out, and if I give it a number that we didn't code for such as 3, that triggers the else statement which prints out some other number and then exits.
An if statement constructed like this with multiple elifs is a substitute in Python for a switch or case statements that you'll find in other languages. A for statement can be used to iterate over any sequence which can be a list or a string.
To show you how those work, I'm just going to go into the Python console. I'm going to create a list called pets and inside of the list we'll have cat, dog and elephant. I'm going to start our for loop by typing for pet in pets.
We've got the keyword for and it's going to iterate over the list object pets and for each item in the pets list, it's going to create a temporary variable called pet with that value. Inside the for block, we're going to print out I have a, and then use the format method to and the positional operator to print out the name of the pet.
As the for loop iterates through the list, the variable pet is set to the list element from that iteration. We can also create a for loop in combination with the range operators. We can say for i in range, and then specify the end of the range. It'll increment up to but not including the endpoint number specified in the range.
You can also specify the starting point of the range in the range method if you supply two arguments. The first one is the starting point, the second one is the endpoint and as that runs, it starts as 10 and then goes up to but doesn't include the endpoint of 15.
You can specify the increment or the step for it as well. The increment or step is provided as the third argument to the range method, and as this executes it starts at zero, goes up to but doesn't include 10 counting by two's.
You can increment with negative numbers as well. This time if we start the range at 10 and then go down to zero with a step of negative 1, when it runs, it counts from 10 all the way down to 1. You can use a while statement to keep a loop that continues running as long as a given condition is true.
I've got a variable of X set to 1 and I can say while X is less than 10, print X and then increment X by 1. Remember that the code inside of your loop block has to be indented, and when this runs, it continues to print in an increment the value of X until the statement X less than 10 no longer evaluates to true.
All of the loop statements with the exception of while will continue to iterate and loop until they've reached the end of their given sequence. Our original for statement iterated over the list of pets, cat, dog and elephant, and once it started, it continued iterating until it got to elephant, the last item in the list.
You can terminate a loop with a break statement. If we do for pet in pets, and then evaluate if the pet is equal to the word dog, we'll print no dogs allowed and then break. Otherwise or in Python lingo else, we'll print out we love and use our positional operator with the format method to print out the name of the pet.
If we look at what happened here, the first item in the list, cat, doesn't evaluate to true in the if statement, so it triggered the else and printed we love cat. The next item dog does evaluate to true, so it printed no dogs allowed and then broke the loop and the word elephant was never reached in iteration.<|endoftext|>
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Home | | Maths 8th Std | Applications of Percentage in Word Problems
# Applications of Percentage in Word Problems
We know that Per Cent means per hundred or out of a hundred. It is denoted by the symbol %. x% denotes the fraction x/100.
Applications of Percentage in Word Problems
We know that Per Cent means per hundred or out of a hundred. It is denoted by the symbol %. x% denotes the fraction x/100. It is very useful in comparing quantities easily. Let us see its uses in the following word problems.
Example 4.1
If x % of 600 is 450, then find the value of x.
Solution:
Given, x % of 600 = 450
(x/100) × 600 = 450
x = 450/6
x = 75
Example 4.2
When a number is decreased by 25 % , it becomes 120. Find the number.
Solution:
Let the number be x .
x = 160
Aliter: Using the above formula,
If we start with a quantity A and then decrease that quantity by x% , we will get the decreased quantity as,
D = (1 − x/100) A
Example 4.3
Akila scored 80 % of marks in an examination. If her score was 576 marks, then find the maximum marks of the examination.
Solution:
Let the maximum marks be x .
Now, 80 % of x = 576
80/100 × x = 576
x = 576 × (100/80) = 720 marks
Therefore, the maximum marks of the examination = 720.
Example 4.4
If the price of Orid dhall after 20% increase is ₹96 per kg, then find the original price of Orid dhall per kg.
Solution:
Let the original price of Orid dhall be ₹ x .
New price after 20% increase
Aliter: Using the above formula,
A=₹80
If we start with a quantity A and then increase that quantity by x% , we will get the increased quantity as,
D = (1 + x/100) A
Try these
1. What percentage of a day is 10 hours?
Solution:
In a day, there are 24 hours
10 hrs out of 24 hrs is 10/24
As a percentage, we need to multiply by 100
∴ Percentage = [10/24] × 100 = 41.67%
2. Divide ₹350 among P, Q and R such that P gets 50% of what Q gets and Q gets 50% of what R gets.
Solution:
Let R get x, Q gets 50% of what R gets
Q gets = (50/100) × x = x/2
P gets 50% of what Q gets
P gets = 50/100 × x/2 = x/4
Since 350 is divided among the three
∴ 350 = x + x/2 + x/4
350 = [4x + 2x + x ] / 4 = 7x / 4 = 350
x = [350 × 4] / 7 = 200
Q gets = x/2 = 200/2 = 100, P gets = x/4 = 200/4 = 50
P = 50, Q = 100, R = 200
Think
With a lot of pride, the traffic police commissioner of a city reported that the accidents had decreased by 200% in one year. He came up with this number by stating that the increase in accidents from 200 to 600 is clearly a 200% rise and now that it had gone down from 600 last year to 200 this year should be a 200% fall. Is this decrease from 600 to 200, the same 200% as reported by him? Justify.
Solution:
Increase from original value 200 to 600
% increase = [ Change in value / original value ] × 100
= [(600 – 200) / 600] × 100 = [400 / 200] × 100 = 200 % increase
Decrease from original value 600 to 200
% decrease = [ Change in value / original value ] × 100
here original value is 600
% decrease = [ (600 – 200) / 600 ] × 100 = [ 400 / 600 ] × 100 = 66.67 % increase
Increase from 200 → 600 and % decrease from 600 → 200 are not the same
Example 4.5
The income of a person is increased by 10% and then decreased by 10 % . Find the chang in his income.
Solution:
Let his initial income be ₹ x .
Income after 10% increase is
That is, income of the person is reduced by 1%.
Aliter:
Let his income be ₹100
Income after 10% increase is
100 + 100 × (10/100) = ₹110.
Now, income after 10% decrease is,
110 – 110 × (10/100) = 110–11= ₹99
Net change in his income = 100 – 99 = 1
Percentage change = 1/100 × 100% = 1% .
That is, income of the person is reduced by 1%.
Note
For any given number, when it is increased first by x % and then decreased by y %, then the value of the number is increased or decreased by ( x + y + [xy/100] )% . Use ‘negative’ sign for decrease and also assume ‘decrease’ if the sign is negative. Use this note to check the answer for the Example 4.5.
Tags : Questions with Answers, Solution | Life Mathematics | Chapter 4 | 8th Maths , 8th Maths : Chapter 4 : Life Mathematics
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
8th Maths : Chapter 4 : Life Mathematics : Applications of Percentage in Word Problems | Questions with Answers, Solution | Life Mathematics | Chapter 4 | 8th Maths<|endoftext|>
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# The standard form of the equation of a circle with its center at the origin is Notice that both the x and y terms are squared. Linear equations don’t.
## Presentation on theme: "The standard form of the equation of a circle with its center at the origin is Notice that both the x and y terms are squared. Linear equations don’t."— Presentation transcript:
The standard form of the equation of a circle with its center at the origin is Notice that both the x and y terms are squared. Linear equations don’t have either the x or y terms squared. Parabolas have only the x term was squared (or only the y term, but NOT both). r is the radius of the circle so if we take the square root of the right hand side, we'll know how big the radius is.
Let's look at the equation The center of the circle is at the origin and the radius is 3. Let's graph this circle. This is r 2 so r = 3 2-7-6-5-4-3-21573 0468 Center at (0, 0) Count out 3 in all directions since that is the radius
If the center of the circle is NOT at the origin then the equation for the standard form of a circle looks like this: The center of the circle is at (h, k). The center of the circle is at (h, k) which is (3,1). Find the center and radius and graph this circle. The radius is 4 This is r 2 so r = 4 2 -7 -7 -6 -6 -5 -5 -4 -4 -3 -3 -2 -2 -1 1 5 7 3 0 4 6 8
If you take the equation of a circle in standard form for example: You can find the center and radius easily. The center is at (-2, 4) and the radius is 2. Remember center is at (h, k) with (x - h) and (y - k) since the x is plus something and not minus, (x + 2) can be written as (x - (-2)) This is r 2 so r = 2 (x - (-2)) But what if it was not in standard form but multiplied out (FOILED) Moving everything to one side in descending order and combining like terms we'd have:
If we'd have started with it like this, we'd have to complete the square on both the x's and y's to get in standard form. Group x terms and a place to complete the square Group y terms and a place to complete the square Move constant to the other side 4416 Write factored and wahlah! back in standard form. Complete the square
Now let's work some examples: Find an equation of the circle with center at (0, 0) and radius 7. Let's sub in center and radius values in the standard form 00 7
Find an equation of the circle with center at (0, 0) that passes through the point (-1, -4). The point (-1, -4) is on the circle so should work when we plug it in the equation: Since the center is at (0, 0) we'll have Subbing this in for r 2 we have:
Find an equation of the circle with center at (-2, 5) and radius 6 Subbing in the values in standard form we have: -256
Find an equation of the circle with center at (8, 2) and passes through the point (8, 0). Subbing in the center values in standard form we have: 82 Since it passes through the point (8, 0) we can plug this point in for x and y to find r 2.
Identify the center and radius and sketch the graph: To get in standard form we don't want coefficients on the squared terms so let's divide everything by 9. So the center is at (0, 0) and the radius is 8/3. 999 Remember to square root this to get the radius. 2 -7 -7 -6 -6 -5 -5 -4 -4 -3 -3 -2 -2 -1 1 5 7 3 0 4 6 8
Identify the center and radius and sketch the graph: Remember the center values end up being the opposite sign of what is with the x and y and the right hand side is the radius squared. So the center is at (-4, 3) and the radius is 5. 2 -7-7 -6-6 -5-5 -4-4 -3-3 -2-2 -1 1 5 73 046 8
We have to complete the square on both the x's and y's to get in standard form. Group x terms and a place to complete the square Group y terms and a place to complete the square Move constant to the other side 9944 Write factored for standard form. Find the center and radius of the circle: So the center is at (-3, 2) and the radius is 4.
Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. www.slcc.edu Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.www.mathxtc.com Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www.ststephens.wa.edu.au
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ramjet propulsion Current Affairs - 2019
Category Wise PDF Compilations available at This Link
India has successfully undertaken the demonstration of the indigenously developed propulsion technology. The successful demonstration of indigenous technology is a step forward in the development of long-range air-to-air missiles.
A Long Journey
India had embarked on the journey to indigenously develop the Solid Fuel Ducted Ramjet technology in 2013 with estimated funding of nearly Rs 500 cr to develop the technology and demonstrate it in 5 years.
The Defence Research Development Laboratory (DRDL) at Hyderabad was the lead agency for the collaborative mission project. The first test of this indigenous technology was carried out in May 2018. The recent one was a second trail and it established a series of technologies including the ground booster, separation of the ground booster and nozzle-less booster.
During the test flight, the missile was guided to high altitude to simulate aircraft release conditions and subsequently the nozzle-less. All the mission parameters were successfully met.
How Solid Fuel Ducted Ramjet propelled missile is different?
The Solid Fuel Ducted Ramjet propelled missile uses the air-breathing ramjet propulsion technology, which helps propel the missile at high supersonic speeds (above Mach 2) for engaging targets at long ranges.<|endoftext|>
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Geschichte Europas/ Die Zeit der Erforschungen und Entdeckungen
- 1 Einleitung
- 2 Ursachen für die Zeit der Erforschungen und Entdeckungen
- 3 Portugiesische Rollen in den frühen Entdeckungen
- 4 Wichtige portugiesische Entdecker
- 5 Frühe spanische Entdecker
- 6 Englische Entdecker
- 7 Französische Entdecker
- 8 Niederländische Entdecker
- 9 Folgen aus der Zeit der Entdeckungen
During the fifteenth and the sixteenth century the states of Europe began their modern exploration of the world with a series of sea voyages. The Atlantic states of Spain and Portugal were foremost in this enterprise though other countries, notably England and the Netherlands, also took part.
These explorations increased European knowledge of the wider world, particularly in relation to sub-Saharan Africa and the Americas. These explorations were frequently connected to conquest and missionary work, as the states of Europe attempted to increase their influence, both in political and religious terms, throughout the world.
Ursachen für die Zeit der Erforschungen und Entdeckungen[Bearbeiten]
The explorers of the fifteenth and sixteenth centuries had a variety of motivations, but were frequently motivated by the prospects of trade and wealth. The earliest explorations, round the coast of West Africa, were designed to bypass the trade routes that brought gold across the Sahara Desert. The improved naval techniques that developed then allowed Europeans to travel further afield, to India and, ultimately, to the Americas.
The early explorations of Spain and Portugal were particularly aided by new ship designs. Prior to the fifteenth century Spain and Portugal largely relied on a ship known as the galley. Although gallies were fast and maneuverable, they were designed for use in the confined waters of the Mediterranean, and were unstable and inefficient in the open ocean. To cope with oceanic voyages, European sailors adapted a ship known as the cog, largely used in the Baltic and North Sea, which they improved by adding sail designs used in the Islamic world. These new ships, known as caravels, had deep keels, which gave them stability, combined with lateen sails, which allowed them to best exploit oceanic winds.
The astrolabe was a new navigational instrument in Europe that borrowed from the Islamic world, which used it in deserts. Using coordinates via the sky, one rotation of the astrolabe's plate, called a tympan, represented the passage of one day, allowing sailors to approximate the time, direction in which they were sailing, and the number of days passed. The astrolabe was replaced by the sextant as the chief navigational instrument in the 18th century. The sextant measured celestial objects in relation to the horizon, as opposed to measuring them in relation to the instrument. As a result, explorers were now able to sight the sun at noon and determine their latitude, which made this instrument more accurate than the astrolabe.
Portugiesische Rollen in den frühen Entdeckungen[Bearbeiten]
In 1415, the Portuguese established a claim to some cities (Ceuta, Tangiers) on what is today the Kingdom of Morocco, and in 1433 they began the systematic exploration of the west African coast. In August 1492, Christopher Columbus, whose nationality is still today subject to much debate, set sail on behalf of Ferdinand and Isabella whose marriage had united their crowns forming what is still today the Kingdom of Spain, and on October 12 of that same year, he eventually reached the Bahamas thinking it was the East Indies. In his mind he had reached the eastern end of the rich lands of India and China described in the thirteenth century by the Venetian explorer Marco Polo.
As a result, a race for more land, especially in the so-called "East Indies" arose. In 1481, a papal decree granted all land south of the Canary Islands to Portugal, however, and the areas explored by Columbus were thus Portuguese territories. In 1493, the Spanish-descendant Pope Alexander VI, declared that all lands west of the longitude of the Cape Verde Islands should belong to Spain while new lands discovered east of that line would belong to Portugal. These events led to increasing tension between the two powers given the fact that the king of Portugal saw the role of Pope Alexander VI Borgia as biased towards Spain. His role in the matter is still today a matter of strong controversy between European historians of that period.
The resolution to this occurred in 1494 at the Treaty of Tordesillas, creating, after long and tense diplomatic negotiations between the Kingdoms of Spain and Portugal, a dividing line 370 leagues west of the Cape Verde Islands. Portugal received the west African Coast and the Indian Ocean route to India, as well as part of the Pacific Ocean waterways, while Spain gained the Western Atlantic Ocean and the lands further to the west. «Unknowingly», Portugal received Brazil. King John II of Portugal, however, seems to have had prior knowledge of the location of that Brazilian territory, for in the difficult negotiations of the Treaty of Tordesillas he managed, in a move still open for debate amongst historians of the period today, to push the dividing line further to the west, making it possible to celebrate the official discovery of Brazil and the reclaiming of the land only in 1500, already under the auspices of the treaty.
Wichtige portugiesische Entdecker[Bearbeiten]
Prince Henry (1394–1460)[Bearbeiten]
Prince Henry "the Navigator" financially supported various voyages. He created a school for the advancement of navigation, laying the groundwork for Portugal to become a leader in the Age of Exploration.
Bartolomeu Dias (1450-1500)[Bearbeiten]
Bartolomeu Dias, the first European to sail around the Cape of Good Hope, also found that India was reachable by sailing around the coast of the continent. As a result, trade with Asia and India was made considerably easier because travellers would no longer have to travel through the Middle East. Thus, there was a rise in Atlantic trading countries and a decline in Middle East and Mediterranean countries.
Vasco da Gama (1460–1524)[Bearbeiten]
Vasco da Gama was the first to successfully sail directly from Europe to India in 1498. This was an important step for Europe because it created a sea route from Europe that would allow trade with the Far East instead of using the Silk Road Caravan route.
Pedro Álvares Cabral (1467–1520)[Bearbeiten]
On April 21, 1500, Pedro Álvares Cabral accidentally discovered Brazil while seeking a western route to the Indes. He first landed in modern-day Bahia.
Ferdinand Magellan (1480–1521)[Bearbeiten]
Magellan was a Portuguese explorer sailing in a Spanish expedition, and was the first person to sail the Pacific Ocean and around South America. He attempted to circumnavigate the globe but died in the Philippines, although his crew successfully completed the voyage. One of his ships led by Juan Sebastian Elcano, who took over after Magellan died, made all the way around the globe!
Francis Xavier (1506 –1552)[Bearbeiten]
Francis Xavier was a Spanish missionary, born in the castle of Xavier, a village near the city of Pamplona, from where he has his name. He was a member of the nobility and during his student years in Paris he became friends with Ignacio de Loiola with whom he would found the Jesuit Order He travelled extensively around Africa, India, the South Pacific, and even Japan and China.
Frühe spanische Entdecker[Bearbeiten]
There were a number of other important explorers that were involved in the Age of Exploration.
Pizarro was a Spanish explorer who militarily fought and conquered the Incan people and culture, claiming most of South America for Spain. He gained immense gold and riches for Spain from the defeat of the Incan empire.
Christopher Columbus (1451-1506)[Bearbeiten]
Columbus, an explorer thought to be of Genoa (Italy), who after many unsuccessful attempts at finding patronship, explored the possibility of a western passage to the East Indies for the Spanish crown. Due to miscalculations on the circumference of the world Columbus did not account for the possibility of another series of continents between Europe and Asia, Columbus discovered the Caribbean in 1492. He introduced Spanish trade with the Americas which allowed for an exchange of cultures, diseases and trade goods, known as The Grand Exchange, whose consequences, good and bad, are still being experienced today.
Magellan was a Portuguese explorer who served the King of Spain, and was the first person to sail the Pacific Ocean and around South America. He attempted to be the first to circumnavigate the globe but was killed in the Philippines. His crew managed to successfully complete the voyage under the leadership of the Spanish Juan Sebastian del Cano.
Vasco Nuñez de Balboa (1475-1519)[Bearbeiten]
Cortés was a Spanish conquistador who assembled an army from the Spanish Colonies consisting of 600 men, 15 horsemen and 15 cannons. Using the assistance of a translator, Doña Marina, he assembled alliances with discontented subdued tribes in the Aztec empire. Through decisive use of superior weapons and native assistance, also the help of European disease which had already wrecked native populations, successfully conquered the Aztecs capturing Montezuma II, the current emperor, the city of Tenochtitlan and looting large amounts of Aztec gold.
Bartolomé de las Casas (1484-1566)[Bearbeiten]
Las Casas was a Spanish priest who advocated civil rights for Native Americans and strongly protested the way they were enslaved and badly treated. He wrote A Short Account of the Destruction of the Indies and De thesauris in Peru.
Juan Ponce de León[Bearbeiten]
Juan Ponce de León was a Spanish conquistador hailing from Valladolid, Spain. He had served as the Governor of Puerto Rico when he started his own expedition in 1513, discovering Florida on March 27 of the same year and reaching its eastern coast on April 2. He called the land Florida (Spanish for flowery), either because of the vegetation he saw there, or it was Easter (Spanish: Pascua Florida) that time. De Leon then organized subsequent voyages to Florida; the last one occurring in 1521 when he died.
Sir Francis Drake (1540-1596)[Bearbeiten]
Sir Francis Drake was a letter of marque, one who is a privateer, for Britain during the time of Queen Elizabeth I. Though he is most remembered for helping command the English fleet against the Spanish Armada, he also spent many years in the Caribbean and successfully circumnavigated the world between 1577-1580.
John Cabot (1450-1499)[Bearbeiten]
John Cabot, originally Giovanni Caboto was born in Genoa, Italy.
Rene-Robert de La Salle[Bearbeiten]
Father Jacques Marquette[Bearbeiten]
Jacques Cartier (1491-1551)[Bearbeiten]
Jacques Cartier was an explorer who claimed Canada for France. He was born in Saint Malo, France in 1491. He was also the first European, not just the first Frenchman to describe and chart Saint Lawrence River and Gulf of Saint Lawrence. He made three important voyages. He died in Saint Malo, in 1551, aged 65.
Samuel de Champlain[Bearbeiten]
Samuel de Champlain was the "Father of New France". Founded Quebec City and today Lake Champlain is named in his honor.
In the late 16th century Dutch explorers began to head out all over the world.
On June 5, 1594 Barentsz left the island of Texel aboard the small ship Mercury, as part of a group of three ships sent out in separate directions to try and enter the Kara Sea, with the hopes of finding the Northeast passage above Siberia. During this journey he discovered what is today Bjørnøya, also known as Bear Island.
Later in the journey, Barentsz reached the west coast of Novaya Zemlya, and followed it northward before being forced to turn back in the face of large icebergs. Although they did not reach their ultimate goal, the trip was considered a success.
Setting out on June 2 1595, the voyage went between the Siberian coast and Vaygach Island. On August 30, the party came across approximately 20 Samoyed "wilde men" with whom they were able to speak, due to a crewmember speaking their language. September 4 saw a small crew sent to States Island to search for a type of crystal that had been noticed earlier. The party was attacked by a polar bear, and two sailors were killed.
Eventually, the expedition turned back upon discovering that unexpected weather had left the Kara Sea frozen. This expedition was largely considered to be a failure.
In May of 1596, he set off once agian, returning to Bear Island. Barentsz reached Novaya Zemlya on July 17. Anxious to avoid becoming entrapped in the surrounding ice, he intended to head for the Vaigatch Strait, but became stuck within the many icebergs and floes. Stranded, the 16-man crew was forced to spend the winter on the ice, along with their young cabin boy.
Proving successful at hunting, the group caught 26 arctic foxes in primitive traps, as well as killing a number of polar bears. When June arrived, and the ice had still not loosened its grip on the ship, the scurvy-ridden survivors took two small boats out into the sea on June 13. Barentsz died while studying charts only seven days after starting out, but it took seven more weeks for the boats to reach Kola where they were rescued.
In 1609, Hudson was chosen by the Dutch East India Company to find an easterly passage to Asia. He was told to sail around the Arctic Ocean north of Russia, into the Pacific and to the Far East. Hudson could not continue his voyage due to the ice that had plagued his previous voyages, and many others before him. Having heard rumors by way of Jamestown and John Smith, he and his crew decided to try to seek out a Southwest Passage through North America.
After crossing the Atlantic Ocean, his ship, the Halve Maen (Half Moon), sailed around briefly in the Chesapeake and Delaware Bays, but Hudson concluded that these waterways did not lead to the Pacific.
He then sailed up to the river that today bears his name, the Hudson River. He made it as far as present-day Albany, New York, where the river narrows, before he was forced to turn around, realizing that it was not the Southwest Passage.
Along the way, Hudson traded with numerous native tribes and obtained different shells, beads and furs. His voyage established Dutch claims to the region and the fur trade that prospered there. New Amsterdam in Manhattan became the capital of New Netherland in 1625
Early in Willem's life,1601 and 1602, he set out on two trips to the Dutch possessions in the East Indies. On November 18, 1605, he sailed from Bantam to the coast of western New Guinea. He then crossed the eastern end of the Arafura Sea, without seeing the Torres Strait, into the Gulf of Carpentaria, and on February 26 1606 made landfall at the Pennefather River on the western shore of Cape York in Queensland, near the modern town of Weipa. This is the first recorded European landfall on the Australian continent. Willem Janszoon proceeded to chart some 320 km of the coastline, which he thought to be a southerly extension of New Guinea.
Willem Janszoon returned to the Netherlands in the belief that the south coast of New Guinea was joined to the land along which he coasted, and Dutch maps reproduced this error for many years to come.
Janszoon reported that on July 31, 1618 he had landed on an island at 22° South with a length of 22 miles and 240 miles SSE of the Sunda Strait. This is generally interpreted as a description of the peninsula from Point Cloate to North West Cape on the Western Australian coast, which Janszoon presumed was an island without fully circumnavigating it.
In 1634 Tasman was sent as second in command of an exploring expedition in the north Pacific. His fleet included the ships Heemskerck and Zeehaen. After many hardships Formosa (now Taiwan) was reached in November, 40 out of the crew of 90 having died. Other voyages followed, to Japan in 1640 and 1641 and to Palembang in the south of Sumatra in 1642, where he made a friendly trading treaty with the Sultan. In August 1642 Tasman was sent in command of an expedition for the discovery of the "Unknown Southland", which was believed to be in the south Pacific but which had not been seen by Europeans
On November 24, 1642 Tasman sighted the west coast of Tasmania near Macquarie Harbour. He named his discovery Van Diemen's Land after Anthony van Diemen, Governor-General of the Dutch East Indies. Proceeding south he skirted the southern end of Tasmania and turned north-east until he was off Cape Frederick Hendrick on the Forestier Peninsula. An attempt at landing was made but the sea was too rough; however, the carpenter swam through the surf, and, planting a flag, Tasman claimed formal possession of the land on December 3, 1642.
Tasman had intended to proceed in a northerly direction but as the wind was unfavourable he steered east. On December 13 they sighted land on the north-west coast of the South Island, New Zealand. After some exploration he sailed further east, and nine days later was the first European known to sight New Zealand, which he named Staten Landt on the assumption that it was connected to an island (Staten Island, Argentina) at the south of the tip of South America. Proceeding north and then east one of his boats was attacked by Māori in waka, and four of his men were killed.
En route back to Batavia, Tasman came across the Tongan archipelago on January 21, 1643. While passing the Fiji Islands Tasman's ships came close to being wrecked on the dangerous reefs of the north-eastern part of the Fiji group. He charted the eastern tip of Vanua Levu and Cikobia before making his way back into the open sea. He eventually turned north-west to New Guinea, and arrived at Batavia on June 15, 1643.
With three ships on his second voyage (Limmen, Zeemeeuw and the tender Braek) in 1644, he followed the south coast of New Guinea eastward. He missed the Torres Strait between New Guinea and Australia, and continued his voyage along the Australian coast. He mapped the north coast of Australia making observations on the land and its people.
From the point of view of the Dutch East India Company Tasman's explorations were a disappointment: he had neither found a promising area for trade nor a useful new shipping route. For over a century, until the era of James Cook, Tasmania and New Zealand were not visited by Europeans - mainland Australia was visited, but usually only by accident
Folgen aus der Zeit der Entdeckungen[Bearbeiten]
The Age of Exploration led, directly to new communication and trade routes being established and the first truly global businesses to be established. Tea, several exotic fruits and new technologies were also introduced into Europe.
It also led to the decimation and extinction of Natives in other nations due to European diseases and poor working conditions. It also led indirectly to an increase in slavery (which was already widely practised throughout the world), as the explorations led to a rise in supply and thus demand for cotton, indigo, and tobacco.
Finally, as a result of the Age of Exploration, Spain dominated the end of the sixteenth century. The Age of Exploration provided the foundation for the European political and commercial worldwide imperialism of the late 1800s. From 1580 to 1640 Spain would inherit the right to reign over Portugal, whose interests where now in the hands of its political and geographical neighbour. Spain's power, under Spanish leader Philip II, was bigger than ever before and renewed and financed the power of the Papacy to fight against Protestant Reformation. However, in the seventeenth century, as the explorations were coming to an end and money was becoming scarcer, other countries began to openly challenge the spirit of the Tordesilas treaty and the power of Spain, which began to lapse and lose its former power.<|endoftext|>
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# Three consecutive sums of two squares
$0, 1, 2$ is an example of three consecutive non-negative integers $n, n+1, n+2$ which are each the sum of two integer squares. Using modular arithmetic you can prove that in all of these triplets $n \equiv 0 \pmod 8$ .
Now I'm wondering, is there a way to find all such triplets? Below are two ways I've found which generate infinite triplets, but not all:
Because $(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2$, you can take any triplet (except $0, 1, 2$ which will give itself) and generate a new triplet. If $n, n+1, n+2$ is the chosen triplet then the following is also a triplet:
$$n(n+2) = n^2 + 2n$$
$$(n+1)^2 + 0^2 = n^2 + 2n + 1$$
$$(n+1)^2 + 1^2 = n^2 + 2n + 2$$
Assume that there exists a triplet in the following form (letting $b^2 = 2a^2 + 1$):
$$a^2 + a^2 = 2a^2$$
$$2a^2 + 1 = b^2 + 0^2$$
$$2a^2 + 2 = b^2 + 1$$
Solving $b^2 = 2a^2 + 1$ using the convergents for $\sqrt 2$ you'll find that every second convergent ($3/2$, $17/12$, $99/70$, etc.) will be in the form $b/a$.
For example:
$$70^2 + 70^2 = 9800$$
$$99^2 + 0^2 = 9801$$
$$99^2 + 1^2 = 9802$$
• The sequence of numbers n such that n, n+1, n+2 are all sums of two squares is given at oeis.org/A082982 . (As you'd expect, they're all multiples of 8.) This sequence doesn't look to me like there's a nice formula. Of course this doesn't prove there isn't one. – Michael Lugo Jun 20 '11 at 15:07
• I believe you can find a complete solution. Here's one possible attack: 1. Let $n = n_1^2+n_2^2$ for integers $n_1,n_2 \ge 0$. 2. Rearrange the double relation $$n_1^2+n_2^2+2=n_3^2+n_4^2+1=n_5^2+n_6^2$$ so that you can [twice] apply Cauchy’s classical solution to the linear equation $$rX+sY+tZ=0.$$ 3. “Simplify”. Obviously, Step 3 is the hardest. ;) – Kieren MacMillan Aug 4 '16 at 16:53
The numbers $4n^4+4n^2,4n^4+4n^2+1,4n^4+4n^2+2$ are all sums of two squares, so there's another infinite family. I don't think you're going to find a useful formula that will give all solutions, but I don't see how to prove this.<|endoftext|>
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# STA291 Fall 2009 day 17
Document Sample
``` STA 291
Fall 2009
1
LECTURE 17
THURSDAY, 22 OCTOBER
Le Menu
2
• 5 Probability (Review, mostly)
• 21 Random Variables and Discrete
Probability Distributions
Suggested problems
3
• Suggested problems from the textbook:
21.1 to 21.4,21.7 , 21.8, 21.10, and 21.11
Conditional Probability & the Multiplication Rule
4
P A B
P A | B , provided PB 0
P B
• Note: P(A|B) is read as “the probability that A
occurs given that B has occurred.”
• Multiplied out, this gives the multiplication rule:
P A B P B P A | B
Multiplication Rule Example
5
The multiplication rule:
P A B P B P A | B
Ex.: A disease which occurs in .001 of the population is
tested using a method with a false-positive rate of .05 and a
false-negative rate of .05. If someone is selected and tested
at random, what is the probability they are positive, and the
method shows it?
Independence
6
• If events A and B are independent, then the events A
and B have no influence on each other.
• So, the probability of A is unaffected by whether B
has occurred.
• Mathematically, if A is independent of B, we write:
P(A|B) = P(A)
Multiplication Rule and Independent Events
7
Multiplication Rule for Independent Events: Let A and B
be two independent events, then
P(AB)=P(A)P(B).
Examples:
• Flip a coin twice. What is the probability of observing two
heads?
• Flip a coin twice. What is the probability of getting a head
and then a tail? A tail and then a head? One head?
• Three computers are ordered. If the probability of getting
a “working” computer is 0.7, what is the probability that
all three are “working” ?
Conditional Probabilities—Another Perspective
8
Conditional Probabilities—Another Perspective
9
P A B
P A | B
P B
Conditional Probabilities—Another Perspective
10
P A B
P A | B
P B
Terminology
11
• P(A B) = P(A,B) joint probability of A and B (of the
intersection of A and B)
• P(A|B) conditional probability of A given B
• P(A) marginal probability of A
Chapter 21: Random Variables
12
• A variable X is a random variable if the value that
X assumes at the conclusion of an experiment cannot
be predicted with certainty in advance.
• There are two types of random variables:
– Discrete: the random variable can only assume a
finite or countably infinite number of different
values (almost always a count)
– Continuous: the random variable can assume all
the values in some interval (almost always a physical
measure, like distance, time, area, volume, weight,
…)
Examples
13
Which of the following random variables are discrete
and which are continuous?
a. X = Number of houses sold by real estate developer per
week?
b. X = Number of heads in ten tosses of a coin?
c. X = Weight of a child at birth?
d. X = Time required to run a marathon?
Properties of Discrete Probability
Distributions
14
Definition: A Discrete probability distribution is just
a list of the possible values of a r.v. X, say (xi) and the
probability associated with each P(X=xi).
Properties:
1.All probabilities non-negative.
2.Probabilities sum to _______ .
P xi 0
Px 1
i
Example
15
The table below gives the # of days of sick leave for
200 employees in a year.
Days 0 1 2 3 4 5 6 7
Number of 20 40 40 30 20 10 10 30
Employees
An employee is to be selected at random and let
X = # days of sick leave.
a.) Construct and graph the probability distribution of X
b.) Find P ( X 3 )
c.) Find P ( X > 3 )
d.) Find P ( 3 X 6 )
Population Distribution vs.
Probability Distribution
16
• If you select a subject randomly from the population,
then the probability distribution for the value of the
random variable X is the relative frequency
(population, if you have it, but usually approximated
by the sample version) of that value
Cumulative Distribution Function
17
Definition: The cumulative distribution function, or
CDF is
F(x) = P(X x).
Motivation: Some parts of the previous example
would have been easier with this tool.
Properties:
1. For any value x, 0 F(x) 1.
2. If x1 < x2, then F(x1) F(x2)
3. F(- ) = 0 and F() = 1.
Example
18
Let X have the following probability distribution:
X 2 4 6 8 10
P(x) .05 .20 .35 .30 .10
a.) Find P ( X 6 )
b.) Graph the cumulative probability distribution of X
c.) Find P ( X > 6)
Attendance Question #17
19
Write your name and section number on your index
card.
Today’s question:
```
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Calculus Of One Real Variable – By Pheng Kim Ving Chapter 5: Applications Of The Derivative Part 1 – Section 5.6: Sketching Graphs Of Functions 5.6 Sketching Graphs Of Functions
1. Asymptotes
# Horizontal Asymptotes.
The horizontal line y = L is called a horizontal asymptote of the graph of y = f(x) if:
# Vertical Asymptotes.
The vertical line x = a is called a vertical asymptote of the graph of y = f(x) if one of the following is satisfied:
## Oblique Asymptotes
The graph of f(x) = (x2 + 5x – 4)/(2x – 2) is sketched in Fig. 1.4. Long division of the numerator x2 + 5x – 4 by the
denominator 2x – 2 gives: quotient = (1/2)x + 3 and remainder = 2. So:
# Oblique asymptote of graph of y = f(x) =(x2 + 5x – 4)/(2x – 2) is line y = (1/2)x + 3.
Remark 1.1
The graph of a function can intersect a horizontal or oblique asymptote, but can never intersect a vertical asymptote (why?
hint: definition of a function).
## One- And Two-Sided Asymptotes
iii. The graph of a function f may have two one-sided horizontal asymptotes. For example, see Fig. 1.5.
iv. The notions of one- and two-sided asymptotes also apply to vertical and oblique asymptotes.
# The graph of y = f(x) has two one-sided horizontal asymptotes.
2. Asymptotes Of Rational Functions
Let f(x) = P(x)/Q(x) be a rational function, ie, a ratio or fraction of two polynomials P(x) and Q(x). Let's denote the
degree of a polynomial p by deg( p). Then:
The graph of f has a vertical asymptote at every point x where Q(x) = 0.
If deg(P) < deg(Q), then the line y = 0 (the x-axis) is a horizontal asymptote of the graph of f.
iii. Suppose deg(P) = deg(Q). Let a and b be the coefficients of the dominating terms of P and Q respectively; see
If deg(P) = deg(Q), then the line y = a/b, where a and b are the coefficients of the dominating terms of P and Q respectively, is a horizontal asymptote of the graph of f. The line y = a/b cannot be the x-axis.
iv. Suppose deg(P) = deg(Q) + 1. For an example, see the discussion of the function f(x) = (x2 + 5x – 4)/(2x – 2)
above and its graph sketched in Fig. 1.4. Divide P by Q using long division to get:
3. Sketching Graphs Of Functions
When sketching the graph of a function y = f(x), we have three sources of useful information: f itself, f ', and f ''. We'll
follow these suggested steps:
i. Information From f: Determine each of the following if any:
a. The domain of f.
b. Intercepts: x-intercepts and y-intercept.
c. Symmetry of the graph (is f even, odd, or neither?).
d. Limits at points of discontinuity and at positive and negative infinity.
e. Asymptotes: determine vertical and horizontal ones from the limits above, and oblique ones by long division if
appropriate.
ii. Information From f ':
a. Find points x where f '(x) = 0 or f '(x) doesn't exist (critical points). Calculate the value of f at each of them if
defined.
b. Determine the signs of f '; if it's not obvious to do so by simply examining the expression of f ' , draw a chart for
f '; see Section 5.3 Part 3 and Part 5.
iii. Information From f '':
a. Find points x where f ''(x) = 0 or f ''(x) doesn't exist. Calculate the value of f at each of them if defined.
b. Determine the signs of f ''; if it's not obvious to do so by simply examining the expression of f '' , draw a chart for
f ''; see Section 5.4 Part 4.
iv. Draw the chart for f, where the intervals of increase, decrease, and concavity are determined; see Section 5.4 Part 4. Indicate local maxima, local minima, and inflection points if any.
v. Determine additional points of the graph if helpful.
vi. Sketch the graph using all the information obtained above.
Example 3.1
Sketch the graph of the function:
### Solution
Asymptotes:
long division yields:
vertical: lines x = –1 and x = 1,
horizontal: none,
oblique: line y = x.
First Derivative:
The chart for y'' is shown in Fig. 3.1. The chart for y is shown in Fig. 3.2. The graph of y is shown in Fig. 3.3.
Fig. 3.1 Chart For y''.
# Chart For y.
Fig. 3.3 Graph of:
EOS
Remark 3.1
In the charts, the double vertical line below a point x and on a row means that the function on that row isn't defined or
doesn't exist at that point.
Problems & Solutions
1. Sketch the graph of the function y = x2(x2 – 1), making use of any suitable information you can obtain from the
function and its first and second derivatives.
Solution
Domain: R.
y = x2(x2 – 1) = x2(x + 1)(x – 1);
y = x2(x2 – 1) = x4x2.
Intercepts:
x-intercepts: x = 0, –1, and 1,
y-intercept: y = 0.
Symmetry:
y(–x) = (–x)4 – (–x)2 = x4x2 = y(x);
function is even; graph is symmetric about y-axis.
Limits:
Second Derivative:
from y' = 4x3 – 2x we get:
2. Sketch the graph of the function:
using information from the function and its first and second derivatives.
## Solution
Domain: R – {–1, 1}.
Intercepts:
x-Intercepts: x = 0,
y-Intercept: y = 0.
Symmetry:
Second Derivative:
3. Sketch the graph of the function:
using information from the function and its first and second derivatives.
Solution
Domain: R – {1}.
Intercepts:
x-intercepts: x = – 2,
y-intercept: y = – 4.
Symmetry:
y' is never 0;
y' doesn't exist at x = 1 where y doesn't exist either;
4. Sketch the graph of the function:
utilizing any useful information you can get from f , f ', and f ''.
Solution
Domain: R – {–1}.
Asymptotes:
long division yields:
vertical: line x = –1,
horizontal: none,
oblique: line y = x – 1.
First Derivative:
5. Sketch the graph of the function:
employing any useful information you can get from y , y ', and y ''.
Solution
Domain: R – {1}.
Intercepts:
x-Intercepts: x = 0,
y-Intercept: y = 0.
Symmetry:
Asymptotes:
vertical: line x = 1,
horizontal: none,
oblique: none.
First Derivative:
y'' = 0 at x = 0,
y'' doesn't exist at x = 1 where y doesn't exist either;
if x = 0 then y = 0, point (0, 0);
sign of y'' is same as that of x/(x – 1)3;
if x < 0 then x < 1 and so (x – 1)3 < 0, thus y'' > 0,
if 0 < x < 1 then y'' < 0,
if 1 < x then y'' > 0.<|endoftext|>
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# What comes after calculus: from zero to hero
In this article, we’re going, to begin with, what students study after calculus in high school till the end at the unsolved mathematics topics.
So keep reading to know how much is math after calculus?
## What comes after calculus in high school?
After studying the introductory calculus in 11 or 12th high school classes, in college students start to study calculus in 3 versions:
• calculus 1
• calculus 2
• calculus 3
### calculus 1 and 2
in calculus 1 and 2 students will study these principal 7 subjects:
• Limits and continuity
• Derivatives
• Analyzing functions
• theorems
• Integrals
• Differential equations
• Series
#### 1 – Limits and continuity
The first thing that students focus on in college calculus is limits and continuity. To simplify and make short definition limits are the method that allows you to determine in which value the function approach.
the limits and continuity of real-life applications are in many industries like economics and engineering.
#### 2 – Derivatives
the other interesting thing in calculus that comes after is derivatives. The derivative is a study of a change, it is a field that allows you to determine at which speed your car is going or how a company can study the revenues over the year.
derivatives are used hugely in engineering fields and physics including many aspects and areas of life.
#### 3 – Analyzing functions
in calculus, the most other important thing is to be able and know how to analyze functions. In general, you will need to apply and calculate limits to the four famous following functions:
• polynomial function
• rational function
• trigonometric function
• exponential function
#### 4 – theorems
in calculus 1 and 2 you will learn some theorems that are principally needed in this phase which are:
#### 5 – Integrals
integrals are the functions that allow mathematicians to calculate surfaces and volumes of objects. Integrals are used hugely in physics and engineering it a principal subject.
integrals have a solid relation to derivatives. It is impossible to understand integrals without studying derivatives.
#### 6 – Differential equations
differential equations are equations in the first and second-degree derivatives. in other words, in differential equations, you will solve equations that include derivative objects as variables.
differential equations are hugely used in physics and mechanical engineering.
#### 7 – Series
Series are function that allows making complex and commutative operations. For instance, series allows you to calculate how much money you will have after retiring. also many banking applications.
series are used hugely in finance domains.
The subject that we noticed above are calculus 1 and 2 so were going to talk about calculus 3 which is the hardest subject in calculus 3
### Calculus 3
calculus 3 is the hardest subject in calculus so this is the beginning where math becomes serious and tends to be harder. In calculus 3 students study the following subjects:
• multivariable functions
• studying vector fields
• discover Green’s, Stokes’,
• know what is the divergence theorems
• tensor
• Laplace transforms
you might study in some university programs calculus 4 which is normally just a part of calculus 3. In other words, some universities programs split calculus 3 into 2 parts the:
• The first part is called calculus 3
• The second part is called calculus 4
But they have the same subject, so in some universities, you might study until calculus 3 while others up to calculus 4. you could read this article if you ask if there is a calculus 4 or 5.
#### 1 – multivariable functions
the first subject you will study in calculus 3 is multivariable. In other words, you will study a function that has more than one variable like f(x,y) or f(x,y,z) where you have to make many operations to these functions such as:
• calculate a derivative of a multivariable function
• calculate an integral of a multivariable function
#### 2 – Studying vector fields
a vector field is a domain where we represent a function in two parts:
• a real part
• imaginary part
if we take an example of this kind of function we can write f(x)=X+iY where:
#### 3 – discover Green’s, Stokes’,
green and theorem stocks based on how to calculate integrals of a curve. before learning this theorem you need to be aware of what is:
• double integral
• triple integral
#### 4 – know what is the divergence theorems
the divergence theorem is based on how to calculate a surface integral for a close surface. Also, in this theorem, you will need to use a lot of integrals.
#### 5 – tensor
tensor is a mathematical representation of scalar is a complex subject to understand. But you can watch this video below explaining in a fun way what is a tensor.
#### 6 – Laplace transforms
Laplace transform is a mathematic method that we use to transform exponential functions. it ti like a Fourier transform that we apply to transform trigonometric functions like
• sin
• cos
• tang
There is an amazing video that talks about Laplace’s transformation which can explain in detail what is Laplace transform. you could find it at this link.
## What comes after calculus advanced math: level 1
At this level we’re going to ding into advanced and complex mathematics topics, the most explanations are going to be by videos. So you can understand a little what we’re talking about.
### group theory
is a domain that studies symmetry, group theory is a field that is used in physics especially studying atoms, and in computer science domains like robotics and computer vision.
you could watch this video below to see and understand what is group theory.
### game theory
Game theory is a study of the outcomes that affect a game It has applications in all fields of social science, as well as in logic, systems science, and computer science.
game theory is teaching the ability to solve issues with minimal requirements. For example, if have been caught by the police how you’re going to solve your issues with minimal costs.
### complex analysis
It is a mathematical method that helps to solve complex differential equations. this subject is an extension of Fourier transformation and Laplace transformation.
complex analysis is used hugely in physics and engineering fields such as electrical and mechanical engineering.
### measure theory
measure theory is a domain that bases on how to calculate the intervales of functions. if you are interested to know more about measure theory you could watch this video.
### Reiman surfaces
it is hard to explain Reiman’s surface in writing words. So for this reason you found to you this video to watch below.
## What comes after calculus advanced math: level 2
In these 2 videos, we will explain each topic by a video. Because it is impossible to explain these heavy math subjects in writing form. So the video will be more comfortable and deep about each subject.
## Conclusion
So as you can see after calculus that math starts to become serious. at the highest level, exactly the level where we stopped. Today mathematicians use ai to solve more complex problems and issues.
#### yassin.ajanif
Yassin ajanif is a physics graduate and electromechanical engineer width more than 5 years in the field. My goal and my team are to share our experience to help you succeed in your career as a stem major. we talk about all tips, problems, and struggle STEM students face in their career and how to overcome them.<|endoftext|>
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Home | Teacher | Parents | Glossary | About Us
When an expression contains more than one operation, you can get different answers depending on the order in which you solve the expression. Mathematicians have agreed on a certain order for evaluating expressions, so we all arrive at the same answers. We often use grouping symbols, like parentheses, to help us organize complicated expressions into simpler ones. Here's the order we use:
1. First, do all operations that lie inside parentheses.
2. Next, do any work with exponents or roots.
3. Working from left to right, do all multiplication and division.
4. Finally, working from left to right, do all addition and subtraction.
In Example 1, without any parentheses, the problem is solved by working from left to right and performing all the addition and subtraction. When parentheses are used, you first perform the operations inside the parentheses, and you'll get a different answer!
Example 1 - Parenthesis
Without Parenthesis With Parenthesis 8 - 7 + 3 = 1 + 3 = 4 8 - (7 + 3) = 8 - 10 = -2
Example 2
Order of Operations Explanation 22 x 20/4 - 7 x 3 + 55 = Calculate the exponent 4 x 20/4 - 7 x 3 + 55 = 4 x 5 - 21 + 55 = OR 80/4 - 21 + 55 = (4 x 5 and 80/4 both = 20) Working from left to right, do all multiplications and divisions. When there are several of these operations in the same term, the order within the term doesn't matter 20 - 21 + 55 = Add and subtract from left to right 54 The correct answer!<|endoftext|>
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# Proving properties of Quadrilaterals Wrksht
```Team Parallelogram:
Definition of a Parallelogram:
A parallelogram is a quadrilateral with opposite sides parallel.
Today you will prove the following properties of parallelograms:
1. The opposite sides are congruent
2. The opposite angles are congruent
3. The diagonals bisect each other
4. Any pair of consecutive angles are supplementary
Once you have proved a property, you may use it in proving subsequent
properties.
Property 1:
Property 2:
Property 3:
Property 4:
Team Rectangle
Definition of a Rectangle
A rectangle is a parallelogram in which at least one angle is right. Thus a
rectangle has four right angles.
Today you will prove the following Properties of Rectangles:
1. All angles of a rectangle are right angles.
2. The diagonals are congruent.
Remember the following are properties of parallelograms:
1. The opposite sides are congruent
2. The opposite angles are congruent
3. The diagonals bisect each other
4. Any pair of consecutive angles are supplementary
Once you have proved a property, you may use it in proving subsequent
properties.
Property 1:
Team Kite
Definition of a Kite
A kite is a quadrilateral with two sets of consecutive sides congruent. (those two
sets cannot share a side, they must be disjoint).
Today you will prove the properties of a kite:
1. The diagonals are perpendicular
2. One diagonal bisects the other.
3. One diagonal bisects a pair of opposite angles
4. One pair of opposite angles is congruent
Once you have proved a property you may use it to prove subsequent
properties.
Property 1
Property 2
Property 3
Property 4
Team Rhombus
Definition of a Rhombus
A rhombus is a parallelogram with at least one set of consecutive sides
congruent.
Today you will prove the following properties of a Rhombus
1. All four sides of a rhombus are congruent
2. The diagonals are perpendicular bisectors of each other
3. The diagonals bisect opposite angles
4. The diagonals create right isosceles triangles
Remember that rhombuses are parallelograms and kites, so the following
properties apply:
Properties of a kite:
1. The diagonals are perpendicular
2. One diagonal bisects the other.
3. One diagonal bisects a pair of opposite angles
4. One pair of opposite angles is congruent
Properties of parallelograms:
1. The opposite sides are congruent
2. The opposite angles are congruent
3. The diagonals bisect each other
4. Any pair of consecutive angles are supplementary
Property 1
Property 2
Property 3
Properties of Isosceles Trapezoid
Definition of an Isosceles trapezoid
An isosceles trapezoid is a quadrilateral with one set of parallel sides, with the
non-parallel sides congruent.
Today you will prove the properties of an Isosceles Trapezoid:
1. The lower base angles are congruent
2. The upper base angles are congruent
3. The diagonals are congruent
4. Any lower base angle and upper base angle are supplementary
Property 1
Property 2
Property 3
Property 4
```<|endoftext|>
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# NCERT Solutions for Class 8 Maths Chapter 13 : Direct and Inverse Proportions
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## Chapter 13 : Direct and Inverse Proportions
### 13.1 Introduction
Mohan prepares tea for himself and his sister. He uses 300 mL of water, 2 spoons of sugar, 1 spoon of tea leaves and 50 mL of milk. How much quantity of each item will he need, if he has to make tea for five persons? If two students take 20 minutes to arrange chairs for an assembly, then how much time would five students take to do the same job? We come across many such situations in our day-to-day life, where we need to see variation in one quantity bringing in variation in the other quantity.
For example:
(1) If the number of articles purchased increases, the total cost also increases.
(2) More the money deposited in a bank, more is the interest earned.
(3) As the speed of a vehicle increases, the time taken to cover the same distance decreases.
(4) For a given job, more the number of workers, less will be the time taken to complete the work.
### 13.2 Direct Proportion
If the cost of 1 kg of sugar is Rs 18, then what would be the cost of 3 kg sugar? It is Rs 54. Similarly, we can find the cost of 5 kg or 8 kg of sugar.
Observe that as weight of sugar increases, cost also increases in such a manner that their ratio remains constant. Take one more example. Suppose a car uses 4 litres of petrol to travel a distance of 60 km. How far will it travel using 12 litres? The answer is 180 km. How did we calculate it? Since petrol consumed in the second instance is 12 litres, i.e., three times of 4 litres, the distance travelled will also be three times of 60 km. In other words, when the petrol consumption becomes three-fold, the distance travelled is also three fold the previous one. Let the consumption of petrol be x litres and the corresponding distance travelled be y km .
### 13.3 Inverse Proportion
Two quantities may change in such a manner that if one quantity increases, the other quantity decreases and vice versa. For example, as the number of workers increases, time taken to finish the job decreases. Similarly, if we increase the speed, the time taken to cover a given distance decreases. To understand this, let us look into the following situation. Zaheeda can go to her school in four different ways. She can walk, run, cycle or go by car.Observe that as the speed increases, time taken to cover the same distance decreases. As Zaheeda doubles her speed by running, time reduces to half. As she increases her speed to three times by cycling, time decreases to one third. Similarly, as she increases her speed to 15 times, time decreases to one fifteenth. (Or, in other words the ratio by which time decreases is inverse of the ratio by which the corresponding speed increases). Can we say that speed and time change inversely in proportion? Let us consider another example. A school wants to spend Rs 6000 on mathematics textbooks. How many books could be bought at Rs 40 each? Clearly 150 books can be bought. If the price of a textbook is more than Rs 40, then the number of books which could be purchased with the same amount of money would be less than 150. Observe the following table.
### Smartur
Learning is (Super) rewarding!<|endoftext|>
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One day something will outgrow the blue whale – but it won't be another whale. EPA
When life on Earth began around 3.6 billion years ago, all organisms were small.
Indeed, it took some 2.5 billion years to evolve any organism that grows larger than a single cell.
Since then, things have accelerated a bit and – along with the great diversification of body forms – animals have tended to get bigger. Indeed, the largest animal ever to live, the blue whale, is still very much with us, and has been swimming the world’s oceans for only a couple of million years – a mere blink of the eye in the long, long history of life in the sea.
This trend towards larger body sizes through evolutionary time has become known as Cope’s Rule, after the American paleontologist Edward Drinker Cope. Cope’s rule has been documented or disputed in hundreds of studies of numerous animal lineages over the last century, but a new study in the journal Science provides perhaps the most comprehensive test yet of its existence.
Sea creatures are getting bigger
The team, led by Noel Heim from Stanford University, delved into the fossil record to compile information on the body sizes of more than 17,000 kinds of marine animals that have existed since the start of the Cambrian period, 542 million years ago. The results are clear: both the average and maximum sizes of marine organisms have increased substantially over this period, whereas the minimum size has remained reasonably constant.
To some extent this may seem inevitable: if life starts small, the only way to go is bigger. And although evolutionary biologists are always wary of narratives of “progress”, many innovations in evolution require a large body size – for example, the smallest vertebrates are inevitably larger than the smallest invertebrates, because it takes a certain size of organism to pack in all the stuff that vertebrates have.
Likewise, warm-blooded marine animals like whales can only stave off hypothermia if they are more than about a meter long. So the re-invasion of the seas by the ancestors of today’s marine mammals imposed a new hard boundary on the minimum size within this group, which in turn affects the average size across groups.
In the new study, Heim and colleagues tested whether the observed increase in size could be explained by a simple evolutionary random walk, where body size is allowed to change randomly at each branching in the tree of life. They also modified this to impose a minimum possible size, such that the evolution of body sizes proceeded randomly but “bounced back” if a lineage hit this lower size limit.
Neither of these models fitted the observed data well. Instead, they show that only persistent directional selection for larger body sizes – due to the many advantages to being large – can explain the observed trends.
Age of the giants
Does this mean that sea creatures are all inexorably getting bigger, and will continue to do so until the oceans are full of behemoths? Not really. First, the minimum size has not changed, and – moving for a moment from evolution to ecology – it is well known that most species are small. In the seas this is especially pronounced, because marine food webs are typically highly size structured – that is, big things eat small things.
It takes a lot of small fish to meet the energetic demands of a big fish, and so the only way these food webs can work is if small organisms substantially outnumber their larger predators.
Pieter Bruegel the Elder, 1556
Second, Heim and colleagues show that most of the overall increase in body size across all marine animals is explained by the evolution of major new groups, with all of the anatomical and physiological innovation that implies. There is rather less of a drive towards larger sizes within any existing group. In fact, many ocean giants are already more or less as big as they could be, given physical and physiological limits.
In their fascinating study Sizing Ocean Giants published in the journal PeerJ earlier this year, marine biologist Craig McClain and colleagues document the factors limiting size in many of the most conspicuous large marine species. These include the risk of tentacle tangling in jellyfish, metabolic constraints on giant clams, physiological limitations of pumping water over gills in large bony fish, or the reliance of blue whales on dense concentrations of their crustacean prey.
In the case of most groups of marine animals, then, it is unlikely that significantly larger members will evolve any time soon. So, even if Cope continues to rule unchallenged, a visitor to our future oceans is less likely to find them populated with fish the size of whales and whales the size of supertankers than with some new giants whose blueprints we do not yet know.
The human factor
However, Cope has an important rival now as an evolutionary force, and that is you, me, and everyone who directly or indirectly exploits our seas. The attitude of people down the ages when confronted with large marine creatures is encapsulated by my reaction when I first saw pictures of newly discovered giant deep sea amphipods: “Barbecue!”.
Oceanlab, University of Aberdeen
As a species we’ve been pretty effective at removing large animals wherever we roam. As McClain and colleagues say of manta rays (although this is equally applicable to most exploited marine creatures), “In the face of fishing pressure and other anthropogenic threats, it is likely that individuals in many populations may not be near their maximum possible ages or sizes.”
In some species, such as plaice and cod, fisheries appear to have driven selection for smaller body sizes, and our evolving understanding of extinction risk in the seas suggests we should not take for granted the continued existence of the ocean’s giants.
Of course, we have been around for too short a time to know if human-driven selection will remain a true competitor to Cope’s rule in the longer term. Indeed, as this new research shows, previous mass extinctions have led to sharp increases in body size among survivors. So who knows? Maybe Cope will again rule the waves following the current human-driven extinction crisis.<|endoftext|>
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# Question: How To Divide By Devi?
## How do you divide any number?
When we divide any number, divisor must be smaller than the number of dividend. Subtract the product from the number of dividend every time and then take the next number until all number finished. Quotient is always written above the line and product below the number of dividend.
## What is the trick to Division?
More Divide by Number Tricks Divide by 1 – Anytime you divide by 1, the answer is the same as the dividend. Divide by 2 – If the last digit in the number is even, then the entire number is divisible by 2. Remember that divide by 2 is the same as cutting something in half.
## How do you multiply by Shakuntala Devi?
* Long multiplication: This is the skill that got her into the Guinness Book of Records in 1982. At Imperial College on June 18, 1980, Shakuntala Devi was asked to multiply two 13-digit numbers: 7,686,369,774,870 × 2,465,099,745,779.
## What number do you divide first?
1) Dividing by 1: When dividing something by 1, the answer is the original number. In other words, if the divisor is 1 then the quotient equals the dividend. 2 ) Dividing by 0: You cannot divide a number by 0.
## What is short division method?
Short division is a formal written method of dividing numbers. It’s often used when dividing numbers with up to four digits by a one-digit number.
You might be interested: When Is Sri Devi Funeral?
## How was Shakuntala Devi so good?
Shakuntala Devi attained universal fame when she demonstrated her ability to multiply two random numbers of 13 digits. She could mentally multiple 7,686,369,774,870 × 2,465,099,745,779 on 18 June 1980 and gave the correct answer as 18,947,668,177,995,426,462,773,730 within just 28 seconds.
## How Shakuntala Devi mind works?
Shakuntala Devi had just calculated the 23rd root of a 201-digit number. At three, her parents noticed her flair for calculation as she played with cards. By five, she could compute cube roots in her mind.
## Why Shakuntala Devi is so intelligent?
He discovered his daughter’s ability to memorise numbers while teaching her a card trick when she was about three years old. Her father left the circus and took her on road shows that displayed her ability at calculation. She did this without any formal education.<|endoftext|>
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Home | Mathematics | * Calculus Primer | * Calculus Primer 02. A Moving Experience 03. Derivation 04. Integration 05. Volume 1 06. Volume 2 07. Differential Equations 08. Terminal Velocity 09. Nature's Math 10. Tools Share This Page
Calculus Primer 3: Derivation
P. Lutus Message Page
(double-click any word to see its definition)
Now that we have presented an overview, we can look more closely at some of the elements of Calculus. This page discusses derivation, the operation meant to find a rate of change.
In the previous page we showed a table of values for position and velocity. What this meant was the data points lay on integer seconds, something not usually seen in nature. In real life, data are rarely so coöperative, and neither are mathematical functions. I think this may be an appropriate moment to discuss functions.
Functions
A function is a named abstraction that refers to an arbitrary mathematical operation. A function accepts one or more arguments and returns a result. Here is an example:
(1) p(t) = t2
We have defined a function p(t) and we have assigned it the operation "t2". But, just as with a letter variable in algebra, this assignment is arbitrary, and the function name may appear in a context for which any function definition is acceptable. In the same way that letters symbolize numbers in algebra, functions symbolize operations in Calculus and other fields of mathematics. In both cases, the substitution of a general symbol for a specific value greatly increases the power and flexibility of the expression in which you find it.
Finding a rate of change
Graph 1: The function Graph 2: An average rate of change Graph 3: A specific rate of change
In this example, instead of examining a table of data, let's explore the properties of the function described above. A graph of this function appears as Graph 1.
Let us say the function p(t) describes the position of the car from the previous example with respect to time. Our job is to find out how its position changes as time passes, and use that information to derive its velocity and acceleration.
In the previous section we worked with a table of numbers. We could pick any two numbers and obtain their difference, thus easily deriving a velocity. What happens if we try that same method using a function instead of numbers? Let's apply two time arguments to our function and see what happens. You may remember from the previous page that a rate of change with respect to time is equal to:
(2) p2 - p1 = v t2 - t1
But in this problem we are using a function, not a variable, so we might write our equation like this:
(3) p(t2) - p(t1) = v t2 - t1
We need one more refinement to our equation. We are no longer using times that represent indices into a table of values, and we could simplify this equation if we used a single variable to represent the chosen difference in time. Like this:
(4) p(t+Δt) - p(t) = v Δt
In this new equation, Δt, spoken as "delta t", represents a small increment in time.
Let's start by taking a large time interval, as we did in the previous section using a table of numbers, and see if this will produce something useful. We set Δt = 12, t = 0, and apply equation (4) above:
(5) p(12) - p(0) = 12 12
This result, shown in Graph 2 as a green line with two measured endpoints in blue, creates a single velocity result of 12t (12 times t). But it is clear that there are many other velocities present during the time interval, which we can see by examining the red position curve.
Let's experiment with setting a smaller Δt value and see what happens. An example of this is shown as Graph 3 on this page.
We know from consulting the table from the previous page that the correct velocity value for t = 4 is 8 meters per second. Let's see if we can obtain an approximation for this value by applying equation (4) above using different time increments. Let's set t = 4 and Δt = .1:
(6) p(4+.1) - p(4) = 8.1 .1
Okay, that looks promising. Let's try a smaller value for Δt:
(7) p(4+.001) - p(4) = 8.001 .001
A pattern is emerging. It seems the smaller we make Δt, the more accurate our result becomes, but because Δt is not zero, our result never exactly equals the correct value. How can we make the result agree with the true velocity value of 8?
Let's consider our options. Why not set Δt = 0? Wouldn't that solve the problem? As it turns out, we cannot use a value of zero because that would require us to divide by zero, an operation that is undefined in mathematics. This introduces an important topic in Calculus, the idea of a limit.
The Limit
Although we cannot solve our problem by setting Δt = 0, we can make a statement in which we approach zero. Using words we would say "The limit of a function f(x), as x approaches zero, is equal to y", or using symbols:
(8) limΔx -> 0 f(x) = y
The limit allows us to make statements about functions we would not otherwise be allowed to make. In this case we cannot say "if Δt equals zero, then our velocity result is equal to 8". But we can say "As Δt approaches zero, our result approaches 8 (or 'is equal to 8 in the limit')."
Deriving a new function
So far, we have used the method of derivatives and the notion of a limit to obtain a numerical result that represents a specific outcome for a specific argument. But a version of the method presented above can be used to obtain a new function that represents the derivative of our original function. The advantages of this should be obvious — we can obtain a function, then use that function to obtain any desired numerical result, for any argument value. This approach is much more useful than acquiring specific numerical results as we did above.
Because this is an introductory Calculus tutorial, we will not dwell very long on this relatively technical topic, and those who don't care to follow an algebraic derivation may wish to skip the next section. But it is important to understand that producing a function as a derivative is much more useful than producing a numerical result.
Let's review what we know. Based on the numerical result we acquired above (using the idea of a limit, at t = 4 our result was approximately 8), we might guess that the derivative of t2 with respect to t is 2t, but if we only examine numerical results, this can never be more than a guess. How do we prove this guess, and how can we create a systematic approach to obtaining a function as a derivative?
We begin with the position function we have been discussing above, p(t) = t2. Our goal is to obtain a new function that is the derivative of this one, a function that provides velocities. We will use the notation p'(t) (note the apostrophe) to stand for the derivative we seek. This same notation can be extended to higher derivatives, for example p''(t) represents the second derivative of p(t), or acceleration.
Let's begin by asserting something we will then prove:
(9) limΔt -> 0 p(t+Δt) - p(t) = p'(t) Δt
Using words to describe equation (9), we might say "as the limit of Δt approaches 0, the expression at the right approaches the derivative of p(t)", indicated above as p'(t). Note that the assertion above applies to any function, and our use of functional notation allows us to make a general statement about all functions, not just one.
We will now perform a step-by-step process of algebraic manipulation to obtain the derivative of p(t) = t2. As a first step, we temporarily abandon the functional notation, so that this:
(10) p(t+Δt) - p(t) Δt
Becomes this:
(11) (t + Δt)2 - t2 Δt
For our next step, using high school algebra, we can show that (t + Δt)2 is equivalent to t2 + 2tΔt + Δt2, so we now have:
(12) t2 + 2tΔt + Δt2 - t2 Δt
Simplifying the numerator (where both t2 and -t2 are present), we get:
(13) 2tΔt + Δt2 Δt
Now dividing through by Δt we get:
(14) 2t + Δt
Taking the limit of this expression, we have:
(15) limΔt -> 0 2t + Δt = 2t
And, as they say, Q.E.D.. As it turns out, derivatives for most functions can be obtained using methods like this.
Home | Mathematics | * Calculus Primer | * Calculus Primer 02. A Moving Experience 03. Derivation 04. Integration 05. Volume 1 06. Volume 2 07. Differential Equations 08. Terminal Velocity 09. Nature's Math 10. Tools Share This Page<|endoftext|>
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Before we can talk about hybrid engine power, let's first explain how hybrid engines work. Hybrid engines combine two different sources of power to move the vehicle. The first source is the traditional internal combustion engine, which produces power by burning fuel, usually gasoline. The second source is usually an electric motor that gets its power from a battery pack within the vehicle. The engine and the electric motor work together to produce the power the vehicle needs to operate. However, the internal combustion engine in a hybrid car is typically much smaller than usual for efficiency and to accommodate the electric motor. This dichotomy raises a big question for green driving enthusiasts wanting to combine engine performance and fuel economy: How can a hybrid engine create more power?
One way to boost hybrid power is to update the batteries. For instance, the battery pack used in the third-generation Toyota Prius is smaller and more efficient [source: Garrett] than those in previous versions of the car, which gives it a slightly higher power output [Source: Voelcker]. The second generation's battery pack was rated for 28 horsepower, compared to the third generation's 36 horsepower [source: Toyota]. Although it's only a slight power improvement over earlier generations, battery technology is moving towards lithium-ion batteries that can potentially produce even more power in the near future.
These new batteries have plenty of upside: Lithium-ion batteries can produce more power in the same amount of space, because lithium has a greater energy density than the nickel metal hydride batteries used in most hybrids -- and it weighs less as well. Due to these advantages, you may start to see lithium-ion batteries used in hybrid vehicles relatively soon.
In addition to installing more powerful batteries, the electric motors can increase their power by upping the voltage. The 2010 Prius increases its power from a 500-volt system in the previous version, to 650 volts in the redesigned model [Source: Toyota].
It's easy to see that an increase in electric motor power, combined with a powerful combustion engine, is just the recipe needed to create a hybrid that can rival traditional vehicles.
Go on to the next page to see how hybrids can attain serious horsepower output while managing to keep their fuel efficiency up.<|endoftext|>
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# Trig Equations
A trig equation is similar to a regular equation in that we need to find a number to replace the letter that makes the equation correct. For example:
We solve this to find that . We check it by replacing with :
The solution to is .
A trig equation is different from a regular equation because there are generally many solutions, sometimes infinitely many.
The solution in the case of a simple trig equation refers to an angle. Trigonometry is useful to us in the context of something that changes but that repeats itself over time – such as the amount of moon we see each night which ranges from 0% to 100%. In applied trig equations the solution often refers to a time value – a trig equation is what we might use to calculate what day of the month we see a certain amount of the moon. We can apply trig to a context like this in the grade 12 precalculus course.
# Using the unit circle
Solve:
This equation is asking for all the angles that give a sine value of 0.77, where those angles are between 0 and 360 degrees. That is, the first rotation of the unit circle.
We can find the answers by simply scrolling through angle on this applet until the green line (the coordinate of the point on the circumference) measures 0.77. You can find two solutions:
By scrolling through angle we find that and .
The solutions to are and . Or, since there is usually more than one solution we can write the solutions in a set:
We don’t ordinarily have access to this applet and so we need an alternative approach using a scientific calculator only.
# Example 1
Solve
First, we note that 0.31 is a positive number. Referring to the CAST diagram, we see that sine is positive in the first and second quadrant (marked ‘A’ and ‘S’ for ‘all’ and ‘sine’), that is, the upper half where the coordinate is positive.
Second, let’s find the reference angle – the angle that lies in the first quadrant. On the calculator, we find
Now let’s draw a ‘bow-tie’ or a ‘stick-diagram’ on a circle, showing 18.1 degrees above and below the horizontal axis. Put a checkmark on the values in the first and second quadrant, because that is where sine is positive:
We now calculate the number of degrees represented by the points at those two checkmarks. The calculation required in the second quadrant is .
We have now found the two angles that have a sine value of 0.31.
Finally, let’s check those two solutions:
# Example 2
Solve
First, we note that -0.8 is a negative number. Referring to the CAST diagram, we see that cosine is negative in the second and third quadrants, marked ‘S’ and ‘T’. In other words, where the coordinate is negative.
Second, let’s find the reference angle – the angle that lies in the first quadrant. We use positive 0.8 to find the first quadrant angle. There are other ways to proceed from here, be creative if you feel confident.
Now let’s draw a ‘bow-tie’ or a ‘stick-diagram’ on a circle, showing 36.9 degrees above and below the horizontal axis. Put a checkmark on the values in the second and third quadrants, because that is where cosine is negative:
We now calculate the number of degrees represented by the points at those two checkmarks. The calculations required are and .
We have now found the two angles that have a cosine value of -0.8.
Finally, let’s check those two solutions:
# Practice
practice simple trig equation applet
bowtie applet<|endoftext|>
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I was watching a flock of birds fly by my house this morning and thinking of all the wild animals caught in the way of Hurricane Irma, currently barreling down on nations in the Caribbean. What happens to animals during such a mammoth storm?
I did a little digging and found that the impact of a hurricane affects creatures way before it does humans. The destruction starts deep in the sea and in the air before the storm even reaches land. And it doesn’t end when the storm passes either.
In the ocean
As hurricanes move over the ocean, warm surface waters mix with cooler, deeper layers. In shallow areas along the coast, cool rain can mix with cool surface waters, which makes water less saline. Waves from the storm destroy objects beneath the surface and shift tons of sand from the seabed. All of this chaos can wreak havoc on animals beneath the surface of the water.
Other creatures have to take the full insult. Fish and invertebrates can be driven to land by the storm surge and become stranded. When Hurricane Andrew hit Louisiana, for example, officials estimated that more than nine million fish were killed offshore. During the same storm, experts estimated that 182 million fish were killed in the Everglades Basin.
Coral reefs can also be destroyed. If any animals do survive, it takes years for them to recover. Violent storms can kill fish, sea turtles, crabs and other marine organisms. Dolphins and manatees have even been blown ashore during rough storms.
FACT: Sharks are the only creatures under the water that can sense the change in barometric pressure associated with the storm and depart from the area.
On the beach
Shorebirds often move further inland for protection before the onslaught of a heavy storm. Winds, storm surges and waves can cause both beach and dune erosion, which can heavily impact many species of animals who make their homes in ecological niches in coastal dunes and sandy areas. In some cases, the beach can even disappear. Sea turtle nests can be washed out entirely or flooded by water surges, along with the nests of terns and plovers.
On the ground and in the trees
Animals that make their home in trees are better off than those that are on the ground. Tree mammals and birds brace themselves and ride out storms. The poor creatures on the land, however, are often drowned. Even if they survive, their food sources may be severely compromised.
Strong gusts from tropical storms and hurricanes can send birds flying off course and push them far away from their home habitat. During one severe storm, a North Carolina brown pelican was found on a nightclub roof in Halifax, Nova Scotia. Younger birds and those that are sick and weak may become separated from their flock and not be able to find their way back.
In 1992, Hurricane Andrew brought massive wind velocities to the shore and knocked down 80 percent of all trees on several coastal Louisiana basins. The loss of coastal forests and trees during Hurricane Katrina in 2005 displaced many species that were dependent on those forests for shelter and food.
Sometimes hurricanes and strong storms can disrupt the ecosystem for good. For example, the Hawaiian island of Kauai is overrun by feral chickens. Apparently, the population went crazy when a hurricane blasted through chicken coops.
Massive amounts of rain in upstream areas can wash soil, sediment and dangerous pollutants into coastal and marine environments. When Hurricane Agnes passed, the cloudiness of the water was so severe in the Chesapeake Bay that the native grasses died. These grasses provided homes and spawning areas for crabs, fish and other species. It took a very long time for the bay to recover. Coral reefs also suffer when sediment blocks sunlight and causes algae to grow.
Salt water and fresh water balance
In coastal habitats, there is a fine balance between saltwater and freshwater. With powerful waves crashing on the shore, massive amounts of salt water are spilled into freshwater and brackish areas. Vegetation and animals that are not tolerant to salt can be severely harmed and killed by the onslaught of the salty water. Grasses, crabs, minnows, insects and other creatures living in the freshwater and estuarine areas are all at risk. Because the salt water stays around for a while, some bottomland forests and trees can also be killed.
It is equally devastating when heavy rains dump fresh water in coastal area river basins, called watersheds. This causes massive amounts of fresh water to surge downstream into coastal bays and estuaries. This also turns the fresh water/salt water balance upside down. Normally brackish water can remain fresh water for months after a hurricane, placing great pressure on all species living there.
In the eye of the storm
Interestingly enough, the eye of the storm — which has fast-moving walls of intense wind — can form a massive “bird cage” that holds birds inside. This eye effect often displaces birds as much, or more, than strong winds.
The good news
Yes, hurricanes have a massive impact on wildlife but there is some good news in the midst of all this destruction. Just like wildfires can create opportune conditions for species to colonize, hurricanes can open up new opportunities for some animals as well.
Storms may strip foliage off and knock down trees but now these areas are more open to sunlight and new seeds can germinate. There is also evidence that hurricanes allow new species to colonize. In 1996, for example, 15 green iguanas floated on a tangled raft of trees from Guadeloupe to their new home on Anguilla, across the Caribbean Sea. Scientists say this is a way that species can be naturally introduced into new areas.
What you can do to help
— Allen Reid (@Allen_Reid) August 27, 2017
After a storm, be on the lookout for unusual or rare species that may show up in your area. Alert your local wildlife agencies if you see anything unusual. Also, wildlife rescue operations should be contacted if you find a wild animal that has been injured. Avoid trying to handle any wild animal on your own — this could cause injury!
After the hurricane in Texas, it was amazing how many reports of wild animals were being shared on social media. People were reporting floating carpets of fire ants, snakes swimming, orphaned squirrels and alligators in backyards. All of these animals had been displaced by the teeming rains and wind.
David Steen, a reptile expert and assistant research professor at the Auburn University Museum of Natural History, has this to say about wildlife and storms:
“Be conscious of where you put your hands and feet and do not try to mess with animals,” he said. “Getting in a fight with you is really low on the list of a snake or alligator’s priorities right now. They’re trying to get through the storm, too.”
Once the storm has passed, if you have bird feeders in your yard, be sure to fill them to the top — the hungry birds will thank you.
— Rich Everfail<|endoftext|>
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The Hanford Nuclear Site, located in Eastern Washington, is home to nine nuclear reactors that produced 60% of the plutonium that fueled the US’s nuclear weapons arsenal, including plutonium used in the bomb dropped on Nagasaki on August 9, 1945. Though the reactors are no longer in operation, the radioactive waste they produced continues to pose a major challenge to public health. The nuclear waste created at Hanford poses a threat to the Columbia River and beyond.
Hanford is currently the largest environmental cleanup project in the world. Educating the public about the environmental and public health threats posed by Hanford has been a priority for WPSR for decades.
Our educational resources include:
- A free online curriculum
- Down by the River: Stories of Hanford, a podcast series devoted to unpacking the long and controversial history of Hanford
- Recorded seminars on the worldwide impacts of Hanford
Educating for Activism
Educating for Activism is a curriculum focused on the area that became the Hanford nuclear reservation, as well as the Columbia River and the Tri-Cities. It is divided into six modules, each of which highlights a different period of Hanford history: Geology, Native American Tribes, Home to New Settlers, The Manhattan Project, The Cold War, and Cleanup and Restoration. This curriculum has an interdisciplinary focus, using history, society, culture, and the arts to provide a holistic understanding of Hanford and its impacts on the world. It is designed to appeal to multiple learning styles, and each module can be used independently. The goal is to provide users with a greater understanding of not only the nuclear history of Washington State but also the ripples it has left in its wake. In addition to the free online sources, there is a bibliography of relevant texts with discussion questions to promote critical thinking in students.
The curriculum is most appropriate for use in college and high school courses.
Down by the River: Stories of Hanford
Down by the River: Stories of Hanford is an audio podcast devoted to unpacking the long and nuanced history of the world's first industrial nuclear reactor. Built along the Columbia River, the Hanford Nuclear Site produced 60% of the plutonium used in the US nuclear arsenal. At the end of the Cold War, the mission of Hanford transitioned from production to remediation of the area. When it shut down production, Hanford had 1,700 identified waste sites, 500 contaminated buildings, and 56 million gallons of radioactive and chemical waste stored underground. Cleanup at Hanford – which has been ongoing for 26 years – will last for decades to come, cost billions of taxpayer dollars, and require technology not yet invented. On each podcast, we will explore the science and the personal stories of Hanford to give listeners a deeper and broader understanding of a place that is a significant part of our country's past, and a difficult challenge for our future.
Particles on the Wall
Particles on the Wall (POTW) is an interdisciplinary exhibit that fuses art, science, and history to explore Hanford, the Columbia River and the Atomic Age. The exhibit features visual art, literary works, memorabilia, and scientific information to explorer the complicated issues that surround Hanford’s legacy of nuclear weapons production and its current cleanup challenges. The full exhibit includes over 100 pieces and continues to evolve.
From 2010 - June 2016, WPSR through grants from WA Department of Ecology provided support to Particles on the Wall, working with the exhibit’s founders and curators: Nancy Dickeman, Dianne Dickeman, and Steven Gilbert. WPSR continues to recommend this project. As an extension of POTW, the Atomic Footprints poster series shares excerpts from POTW as a portable display for venues that would not be able to host the physical exhibit. The series of seven posters features art and poetry from POTW with relevant historical and scientific information. POTW is now an ongoing project of the curators.
For more information: www.particlesonthewall.org
Washington Nuclear Museum and Education Center
The Washington Nuclear Museum and Education Center (WANMEC) provides extensive information on the history of Hanford and the current cleanup, as well as the effects of activities at Hanford on the local residents and the surrounding environment. Through historical information, firsthand accounts, art, and poetry, WANMEC educates the public about nuclear issues related to Hanford, generates public interest and informed perspective on the cleanup effort, and conveys the diverse emotions that accompany these issues.
nuclear power at hanford
The Columbia Generating Station (CGS), located on the Hanford Nuclear Reservation, is the only nuclear power plant operating in the Pacific Northwest. Previously known as the Washington Public Power Supply System (WPPSS) plant #2 (WNP-2), CGS has a GE Boiling Water Reactor with a Mark II containment system. The plant began operating in 1984 along the Columbia River, north of Richland, Washington. The plant produces around 4% of the Pacific Northwest’s electricity on an average annual basis.
CGS was the only nuclear plant completed by Washington public power utilities out of five under construction, leading to what was at the time the largest municipal bond default in US history. WPPSS (pronounced “whoops”) has since changed its name to Energy Northwest.
Ten years in advance of its license expiration, Energy Northwest went before the Nuclear Regulatory Commission and extended the plant’s license until 2043, a full 20 years beyond its designed life.
In May, 2016, the Seattle City Council unanimously voted in favor of a resolution, drafted in part by activists from Physicians for Social Responsibility and others from the anti-nuclear movement. It calls on Seattle City Light to oppose the Columbia Generating Station, the sole nuclear power plant in the Pacific Northwest, and to support replacing it with green energy. Read more about the resolution here.
Due to economic concerns and risks to human health and the environment, Washington PSR has partnered with Oregon PSR to work towards shutting down CGS. Both groups are also part of Nuclear Free Northwest coalition.<|endoftext|>
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Question Video: Finding the Moment of a Couple Equivalent to a System of Three Forces Acting on a Triangle | Nagwa Question Video: Finding the Moment of a Couple Equivalent to a System of Three Forces Acting on a Triangle | Nagwa
# Question Video: Finding the Moment of a Couple Equivalent to a System of Three Forces Acting on a Triangle Mathematics • Third Year of Secondary School
## Join Nagwa Classes
π΄π΅πΆ is a triangle, where π΅πΆ = 48 cm, and three forces of magnitudes 13, 13, and 24 newtons are acting along lines πΆπ΄, π΄π΅, and π΅πΆ respectively. If the system of forces is equivalent to a couple, determine the magnitude of its moment.
05:03
### Video Transcript
π΄π΅πΆ is a triangle, where π΅πΆ equals 48 centimeters and three forces of magnitudes 13, 13, and 24 newtons are acting along lines πΆπ΄, π΄π΅, and π΅πΆ, respectively. If the system of forces is equivalent to a couple, determine the magnitude of its moment.
Letβs begin by sketching out our triangle. We have our triangle π΄π΅πΆ, where the length π΅πΆ is 48 centimeters. Weβre going to add the forces of magnitude 13, 13, and 24 newtons to the diagram. A 13-newton force acts along the line πΆπ΄, as shown. Another 13-newton force acts along the line π΄π΅. And then our 24-newton acts from π΅ to πΆ.
Now, in fact, we can use the relationship between the length of line segment π΅πΆ and the force that acts along this line to calculate the length of lines π΄π΅ and π΄πΆ. We might spot that the ratio between 48 and 24 is one-half. And so line segments π΄π΅ and π΄πΆ will need to be 26 centimeters each for the ratio between the lengths and the forces in these sides to be the same.
And this is really useful because weβre told that the system of forces is equivalent to a couple. And so if we take the sum of the moments about any point on this triangle, weβll get the same value. So weβll determine the magnitude of the moment by taking the moments about a given point. We could choose any point on our triangle. Letβs take point π΄ and take the clockwise direction to be positive.
The moment, of course, is the product of the force and the perpendicular distance from the pivot. Now the only force that this applies to really is the force acting from π΅ to πΆ. We see this acts at a point 26 centimeters away from π΄, but the 24-newton force does not necessarily act at an angle of 90 degrees. So we need to calculate the component of this force thatβs perpendicular to line segment π΄π΅.
To do so, weβll begin by using the law of cosines to find the angle π΄π΅πΆ that weβve labeled π. The law of cosines tells us that cos of π΅ is equal to π squared plus π squared minus π squared over two ππ. In this case then, cos of π is 48 squared plus 26 squared minus 26 squared over two times 48 times 26. And this means we can evaluate cos of π exactly. 26 squared minus 26 squared is zero. Then we can simplify by dividing through by 48.
So weβre left with 48 over two times 26. And then we can divide the numerator and denominator of this fraction by four. And so cos of π simplifies to 12 over 13. And whilst we could take the inverse cos or arccos of both sides of this equation to find the value of π, we can actually use the Pythagorean theorem to find exact values for sin of π as well. Since the cosine ratio links the adjacent and hypotenuse in a right triangle and since five squared plus 12 squared equals 13 squared, we know that the third side in this triangle, the opposite side, must be five units. And so in this triangle, sin of π is five over 13.
Now the reason thatβs going to be useful will become apparent later on. For now, we need to work out the component of the 24-newton force that is perpendicular to line segment π΄π΅. Since they are perpendicular, they meet at an angle of 90 degrees. And so the angle on the outside of the triangle here we can label as being 90 minus π. And so the component of our 24-newton force thatβs perpendicular to π΄π΅ is found by considering the adjacent side in this right triangle. And so itβs 24 cos of 90 minus π. This means that the moment is the product of 24 cos of 90 minus π and the distance from the pivot, 26.
And now we notice that we can evaluate cos of 90 minus π using the identity sin π. And of course, we worked out that sin π was five over 13. And so the moment is going to be 24 times five over 13 times 26. Letβs simplify this expression by dividing through by 13. And we get 24 times five over one times two. But of course, five over one times two is 10, so this becomes 24 times 10, which is 240. And that is the magnitude of our moment. Since weβre working in newtons and centimeters, the moment is given in newton centimeters. And so given the information about our system of forces, the magnitude of its moment is 240 newton centimeters.
## Join Nagwa Classes
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# Two part Question - Calculus (a) Verify that a rectangle with one side x and perimeter k has an area A = x (k-2x/2) (b) In the early days of settlement , settlers were permitted...
Two part Question - Calculus
(a) Verify that a rectangle with one side x and perimeter k has an area
A = x (k-2x/2)
(b) In the early days of settlement , settlers were permitted to select a homestead block, a rectangular area of crown land that had a perimeter of approximately 1.6 km. Use calculus to find the shape of the block that gave the greatest area to the selection and give the maximum size of this selection in square metres.
krishna-agrawala | Student
A rectangle has four sides with measures of opposite sides being equal. Thus the perimeter of a rectangle (k) is given by:
k = 2*(Measure of one side + Measure of other side)
If measure of one side is x, then this equation becomes:
k = 2*(x + Measure of other side)
Therefore:
Measure of other side = (k/2 - x) = (k - 2x)/2
Area of rectangle = A = (Measure of one side)*(Measure of other side)
Substituting value of the two sides in the above equation we get:
A = x*[(k - 2x)/2]
We can simplify this further as:
A = kx/2 - x^2
When value of k is 1.6 km this equation becomes:
A = (1.6x)/2 - x^2 = 0.8x - x^2
To find the value of x when the area (A) is maximum we find the differential of the right hand side of above equation. A will be maximum when this is zero.
Therefore: 0.8 - 2x = 0
Or: x = 0.4
Thus one side of the rectangle will measure 0.4 km.
And the other side will measure (k/2 - x) = 1.6/2 - 0.4 = 0.4
Area of this rectangle
A = (Measure of one side)*(Measure of other side) = 0.4*0.4 = 0.16 km^2
Or = 0.16*100,000 m^2 = 160.000 m^2
Block that gives maximum area will be a square with each side equal to 0.4 km.
The total area of this block is 160.000 m^2
neela | Student
To verify the formula, A=x(k-2x/2), where A=area, k = perimeter and x is one side of the rectangle.
Verification:
If x=2 and the other side of the rectangle is 3, the area =side* other side =2*3 = 6....................................(1)
k=(2+3)2=10
k = 10 . Then the area by the given formula, A = 2(10-2*2/2) = 2*8=16, which does not tally with the result side*other side result at (1).
But , when x=2, and k=10, the other side is (k-2*2)/2 = (10-2*2)/ = 6/2=3. And area 2*3=6. So the given formula does not hold good. It needs a correction.
The other side should be (k-2x)/2 and not (k-2x/2). The formula for area should be A = x(k-2x)/2. Now substitute x=2 and k=10 in A = x(k-2x)/2 and you get A = 2(10-2*2)/2 =2*(10-4)/2 = 2*3 = 6.
b)
Shape of the plot is rectangular. Therefore area of the plot is given by:
A(x) = x*(k-2x)/2 , k is the perimeter fixed and x is variable.
A(x) is maximum for for that value of x for which A'(x) = 0 and A''(x) = negative.
Therefore, A'(a)=0 gives: [x*(k-2x)/2]'= 0 or{(kx-2x^2)/2}' = 0 or
(k-4x)/2=0 or x= k/4.
A"(k/4) = {( k-4x)/2}' at x=k/4
=-4/2 = -2, which is negative.
Therefore, x= k/4. Here k is given 1.6m =1600 m
Therefore, x= 1600/4=400m and A = 400^2 sq m=160000 sq m<|endoftext|>
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As the world looks towards a continuous move far from fossil fuel powered cars and trucks, greener alternative advances are being investigated, for example, electric battery controlled vehicles. Hydrogen power is another green technology with great potential. However, they require high pressure and/or complex heat management systems to achieve acceptable storage densities.
Now, scientists at Lancaster University now have discovered a new material that could potentially unlock the potential of hydrogen-powered vehicles. Dubbed as KMH-1 (Kubas Manganese Hydride-1), the material is made from manganese hydride and expected to use in making molecular sieves within fuel tanks—which store the hydrogen and work alongside fuel cells in a hydrogen-powered ‘system’.
The sieve works by absorbing hydrogen under around 120 atmospheres of pressure, which is less than a typical scuba tank. It then releases hydrogen from the tank into the fuel cell when the pressure is released.
The material uses Kubas binding as the principal mechanism, a process that enables the storage of hydrogen by distancing the hydrogen atoms within an H2 molecule and works at room temperature. This eliminates the need to split, and bind, the bonds between atoms, processes that require high energies and extremes of temperature and need complex equipment to deliver.
A storage material with these properties will allow the DOE system targets for storage and delivery to be achieved, providing a practical alternative to incumbents such as 700 bar systems, which generally provide volumetric storage values of 40 kgH2 m−3 or less, while retaining advantages over batteries such as full time and energy density.
What’s more, the material could excellently absorb and stores any excess energy so external heat and cooling are not needed. This is crucial because it implies cooling and heating equipment shouldn’t be utilized in vehicles, resulting in systems with the possibility to be undeniably more proficient than existing designs.
Professor David Antonelli, Chair in Physical Chemistry at Lancaster University and who has been researching this area for more than 15 years, said: “The cost of manufacturing our material is so low, and the energy density it can store is so much higher than a lithium-ion battery, that we could see hydrogen fuel cell systems that cost five times less than lithium-ion batteries as well as providing a much longer range—potentially enabling journeys up to around four or five times longer between fill-ups.”
Scientists experiments show that the material could enable the storage of four times as much hydrogen in the same volume as existing hydrogen fuel technologies. This is great for vehicle manufacturers as it provides them with the flexibility to design vehicles with increased range of up to four times, or allowing them to reduce the size of the tanks by up to a factor of four.
Professor Antonelli said, “Although vehicles, including cars and heavy goods vehicles, are the most obvious application, we believe there are many other applications for KMH-1.”
“This material can also be used in portable devices such as drones or within mobile chargers so people could go on week-long camping trips without having to recharge their devices. The real advantage this brings is in situations where you anticipate being off-grid for long periods of time, such as long haul truck journeys, drones, and robotics. It could also be used to run a house or a remote neighborhood off a fuel cell.”<|endoftext|>
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# Pie Charts
GCSEKS3Level 4-5AQACambridge iGCSEEdexcelEdexcel iGCSEOCRWJEC
## Pie Charts
A pie chart is a circular chart used to compare the sizes of data sets in relation to one another.
Make sure you are happy with the following topics before continuing.
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
## How to Draw a Pie Chart
Start by drawing a circle with a compass, and then measure the angles with a protractor to make sure we get them right.
To measure the size of each slice, we use the size of the angle that it will take up, out of the total $360\degree$ of the circle. To calculate the angle, we use the following formula:
$\textcolor{black}{\text{angle }}=\dfrac{\textcolor{black}{\text{number of things in that category}}}{\textcolor{black}{\text{total number of things}}}\times \textcolor{black}{360\degree}$
Note: Knowing how to use a compass and protractor will come in handy for other topics and will save time in the exam so make sure you get plenty of practise.
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
@mmerevise
## Example 1: Drawing a Pie Chart
Brian asks $60$ people what their favourite colour is and separates the answers into $5$ categories. His results are shown in the table below. Draw a pie chart to display Brian’s results.
[4 marks]
To draw the pie chart, we need to calculate the size of the angle needed for each category using the formula.
$\text{‘Red’ angle }=\dfrac{10}{60}\times 360=60\degree$
$\text{‘Blue’ angle }=\dfrac{13}{60}\times 360=78\degree$
$\text{‘Green’ angle }=\dfrac{24}{60}\times 360=144\degree$
$\text{‘Yellow’ angle }=\dfrac{5}{60}\times 360=30\degree$
$\text{‘Other’ angle }=\dfrac{8}{60}\times 360=48\degree$
Now we have all the angles, we can draw the chart.
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
## Example 2: Interpreting Pie Charts
A survey was done asking $90$ people how many bathrooms were in their home. Freddie drew a pie chart to display the results of this survey.
a) What number of bathrooms was most common in this survey?
[1 mark]
b) Calculate the number of people who have $1$ bathroom in their house.
[2 marks]
c) If Freddie picks someone at random from the group surveyed, what is the probability that he chooses a person with exactly $1$ bathroom in their home?
[2 marks]
a) Clearly, ‘$2$ bathrooms’ has the biggest slice, so the most common number of bathrooms is $2$.
b) To work out how many people had exactly $1$ bathroom in their house, we will have to use a protractor to measure the size of the angle in the ‘$1$ bathroom’ slice.
We can see that the angle for this sector is $100\degree$. Using the pie chart formula,
$\textcolor{black}{\text{number of things in category}}= \textcolor{black}{\dfrac{\text{angle}}{360\degree}} \times \textcolor{black}{\text{total number of things}}$
Subbing in the numbers we get,
Number of people with $1$ bathroom $=\dfrac{100}{360}\times 90=25$
c) To find the probability of randomly picking someone with $1$ bathroom, we just find the proportion of people with $1$ bathroom as a fraction of the total. Doing this, we get
probability of picking someone with $1$ bathroom $=\dfrac{100}{360}=\dfrac{5}{18}$
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
## Pie Charts Example Questions
From the pie chart, we can see that the slice representing red jumpers is a right angle which tells us that this slice is exactly one quarter of the circle.
Therefore, one quarter of the class wore a red jumper on this non-uniform day:
$32\text{ students}\div4 = 8\text{ students}$
Gold Standard Education
We have to work out the size of the angle that each grade ‘slice’ should have. We know that John recorded the grades of 24 pieces of homework $(6+5+10+3=24)$.If $6$ of the $24$ pieces of homework were awarded a grade A, we can write this as a fraction :$\dfrac{6}{24}$ or $\dfrac{1}{4}$.
Therefore $\frac{1}{4}$ of the pie will be the slice for grade A so one quarter of the whole pie chart will be a slice with a right-angle $(90\degree)$ since,
$360\degree \div 4 = 90\degree$
Homework that was awarded a grade B can be expressed as the following fraction:$\dfrac{5}{24}$. Since we know that there are $360\degree$ in a circle, then the angle for the grade B slice must be $\frac{5}{24}$ of $360\degree$:
$\dfrac{5}{24} \times 360\degree = 75\degree$
The angle for the grade C slice must be $\frac{10}{24}$ of $360\degree$:
$\dfrac{10}{24} \times 360\degree = 150\degree$
The angle for the grade D slice must be $\frac{3}{24}$ of $360\degree$:
$\dfrac{3}{24} \times 360\degree = 45\degree$
Drawing the circle with a compass, and measuring the angles with a protractor, we should have a pie chart like the one shown below:
Gold Standard Education
We know that $510$ cars passed underneath the bridge and that the number of yellow cars is represented by a slice of $60$ degrees.
We know that in a circle there is a total of $360\degree$, thus the proportion of yellow cars can be expressed as the following fraction:
$\dfrac{60\degree}{360\degree}$
If $\frac{60}{360}$ of the total number of cars were yellow, then the number of yellow cars can be calculated as follows:
$\dfrac{60}{360}\times 510\text{ cars} = 85\text{ yellow cars}$
Gold Standard Education
Question 4: The pie chart shows different makes of cars in an airport car park.
There are $600$ Kia cars, which is represented by an angle of $40\degree$.
a) How many Renault cars are there if they are represented by an angle of $85\degree$?
[2 marks]
b) How many cars are there in total in the airport car park?
[2 marks]
Gold Standard Education
The first thing we need to work out in this question is the amount of time that Oliver spends playing golf.
We are told that he has $12$ hours of leisure time in total and that this is represented by a $60\degree$ slice of the pie chart. Since we know that there are $360\degree$ in a circle, then the proportion of the time that Oliver spends playing golf can be expressed as follows:
$\dfrac{60\degree}{360\degree}$
We can therefore calculate the amount of time he playing golf as follows:
$\dfrac{60}{360}\times12\text{ hours} = 2 \text{ hours}$
If Lewis spends $2$ hours of the available $9$ playing golf, this will be represented by $\frac{2}{9}$ of the pie chart. Since the pie chart is a full circle $(360\degree)$, then the slice of the pie chart can be calculated as follows:
$\dfrac{2}{9} \times 360\degree = 80\degree$
Gold Standard Education
## Pie Charts Worksheet and Example Questions
### (NEW) Pie Charts Exam Style Questions - MME
Level 4-5GCSENewOfficial MME
Product
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Learn an entire GCSE course for maths, English and science on the most comprehensive online learning platform. With revision explainer videos & notes, practice questions, topic tests and full mock exams for each topic on every course, it’s easy to Learn and Revise with the MME Learning Portal.
Level 1-3GCSE
Level 1-3GCSE<|endoftext|>
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# Fast Multiplication
In aptitude examination, we need to save time and be quick in our calculations. Multiplication is something that you will definitely have to do in any Aptitude or Maths exam.
So, in this article we will explore some of the alternate methods that allow us to know the value of a product in a fast manner. We hope that you already know the basic traditional method of multiplication (where we start multiplying from the rightmost side, i.e. from unit digits and then work our way to the left hand side).
## Trick Method
Are you one of those people who find it hard to remember all different numbers, e.g. tables. Let me tell you one secret of mine. I never remembered table in my school days. It was just too much to remember.
What I did instead was that I evolved a new method. Though there’s nothing very extraordinary in this method, and many other people use it too.
In this method, we do not multiply, but rather add or subtract numbers. Afterall multiplication is just a faster way to do addition. 4 × 3 means that we need to add 4 three times (4 + 4 + 4), or add 3 four times (3 + 3 + 3 + 3).
### Find multiples till 3
If a number is multiplied by 1, 2 or 3, then it is very easy to find the product. No rocket science here.
For example, 13 × 2 = 13 + 13 = 26 (add tens digit first then unit digits, i.e. 10 + 10 = 20, 3 + 3 = 6; So, 20 + 6 = 26)
17 × 3 = 51 (add tens digit first then unit digits, i.e. 10 + 10 + 10 = 30, 7 + 7 + 7 = 21; So, 30 + 21 = 51)
### Find multiples from 4 to 8
If a number is multiplied by 4, 5, 6, 7 or 8, then we will use a trick.
The given number multiplied by 5 will form our hinge point. You can remember it for various numbers, or you can also calculate it easily.
For example, 13 × 5 = (13 × 10)/2 = 130/2 = 65
11 × 5 = (11 × 10)/2 = 110/2 = 55
How can we use it to find other multiples of a number?
Say, we need to find 13 × 4. It is nothing but (13 × 5) - 13 = 65 - 13 = 52
Similarly, 13 × 6 = (13 × 5) + 13 = 65 + 13 = 78
17 × 7 = (17 × 5) + (17 × 2) = 85 + 34 = 119
17 × 8 = (17 × 5) + (17 × 3) = 85 + 51 = 136 (we can also calculate it as 17 × 10 - 17 × 2 = 170 - 34 = 136)
37 × 6 = (37 × 5) + 37 = [(37 × 10)/2] + 37 = 370/2 + 37 = [300/2 + 70/2] + 37 = 150 + 35 + 37 = 150 + 35 + 35 + 2 = 150 + 70 + 2 = 220 + 2 = 222
All the above calculation needs to be done in your mind, not on paper. It is much faster when done mentally, than it seems on paper.
### Find multiples from 9 to 13
If a number is multiplied by 9, 10, 11, 12 or 13, then we will use a similar trick.
The given number multiplied by 10 will form our hinge point.
For example, 13 × 9. It is nothing but (13 × 10) - 13 = 130 - 13 = 117
Similarly, 13 × 11 = (13 × 10) + 13 = 130 + 13 = 143
17 × 12 = (17 × 10) + (17 × 2) = 170 + 34 = 204
17 × 13 = (17 × 10) + (17 × 3) = 170 + 51 = 221
37 × 11 = (37 × 10) + 37 = 370 + 37 = 407
We can find higher multiples of a number in a similar manner. For example, 44 × 31 = (44 × 30) + (44 × 1) = (440 + 440 + 440) + 44 = (400 + 400 + 400) + (40 + 40 + 40) + (40 + 4) = 1200 + 160 + 4 = 1364
For the sake of fast calculations, you should remember tables till 20. That is, tables of numbers from 2 to 20. You need to remember their 20 multiples (not just 10). It will boost your calculation speed a lot.
For example, you should know what is 13 × 17, 11 × 13, 19 × 7 etc.
However, if you cannot, or you are not very confident regarding your memory, then use the method mentioned above. I still use that method in many cases. A beauty of that method is that we can use it even for higher numbers, e.g. 23 × 31, 123 × 102 etc.
## Specific Cases
### Multiplication of numbers near hundred
If the numbers being multiplied are near 100, then there’s another fast method to find their product.
Let our two numbers be 100 + a and 100 + b. Then our product can be found in two parts: 2nd part | 1st part
• 1st part will have 2 digits: a × b (if there are more than 2 digits, then we will carry it to the left)
• 2nd part: (100 + a) + b
For example, 102 × 106 = (100 + 2) × (100 + 6)
• 1st part will have 2 digits: 2 × 6 = 12
• 2nd part: (100 + a) + b = 102 + 6 = 108
So, our product will be 10812
113 × 109 = (100 + 13) × (100 + 9)
• 1st part will have 2 digits: 13 × 9 = 117 (we will keep 17, and carry over 1 to the left)
• 2nd part: (100 + a) + b = 113 + 9 = 122
So, our product will be 122 | 117 = 123 | 17 = 12317
What if the numbers are below 100?
96 × 92 = (100 - 4) × (100 - 8)
• 1st part will have 2 digits: (-4) × (-8) = 32
• 2nd part: (100 + a) + b = 96 + (-8) = 88
So, our product will be 8832
87 × 91 = (100 - 13) × (100 - 9)
• 1st part will have 2 digits: (-13) × (-9) = 117 (we will keep 17, and carry over 1 to the left)
• 2nd part: (100 + a) + b = 87 + (-9) = 78
So, our product will be 78 | 117 = 79 | 17 = 7917
What if one of the number is above 100, and the other is below 100?
104 × 92 = (100 + 4) × (100 - 8)
• 1st part will have 2 digits: 4 × (-8) = -32 (we will make it positive by taking 100 from the left hand side)
• 2nd part: (100 + a) + b = 104 + (-8) = 96
So, our product will be 96 | (-32) = 95 | (100 - 32) = 95 | 68 = 9568
What if the numbers are near 200 or 300?
For example, 202 × 206 = (200 + 2) × (200 + 6)
• 1st part will have 2 digits: 2 × 6 = 12
• 2nd part: n [(200 + a) + b] = 2 [202 + 6] = 2 × 208 = 416 (value of n will be 2, if numbers are close to 200, 3 if they are close to 300 and so on.)
So, our product will be 41612
### Unit digits add upto 10 and Ten’s digits are same
If two 2-digit numbers being multiplied are such that:
• There ten’s digit is the same, say t.
• The sum of their unit digits is 10, say a + b = 10.
Then, their product will have two parts: 2nd part | 1st part
1st part (it will have 2 digits) = a × b
2nd part = t (t + 1)
For example, 44 × 46 = 4 (4 + 1) | 4 × 6 = 4 × 5 | 4 × 6 = 20 | 24 = 2024
31 × 39 = 3 (3 + 1) | 1 × 9 = 3 × 4 | 1 × 9 = 12 | 09 = 1209
### Difference of the numbers is 10 and Unit digits are 5
If two numbers being multiplied are such that:
• There unit’s digit of both is 5. So, let the numbers be a5 and b5
• The difference between the numbers is 10.
Then, their product will have two parts: 2nd part | 1st part
1st part = 75 (it will always be 75)
2nd part = a (a + 2), such that a ⩽ b
For example, 35 × 45 = 3 (3 + 2) | 75 = 3 × 5 | 75 = 15 | 75 = 1575
### Multiplication by a number close to 10, 100, 1000
To multiply with such numbers, convert them into the form of (10 ± a), (100 ± a), (1000 ± a).
For example, 43 × 98 = 43 × (100 – 2) = 43 × 100 – 43 × 2 = 4300 – 86 = 4214
### Multiplication by 5 or powers of 5
To multiply with 5 or powers of 5, convert them into 10 or powers of 10, and then divide by 2 or its powers.
For example, 67 × 25 = 67 × $5^2$ = 67 × $(\frac{10}{2})^2$ = 67 × $\frac{100}{4}$ = $\frac{6700}{4}$ = 1675
Share on:<|endoftext|>
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#### • Class 11 Physics Demo
Explore Related Concepts
#### • example of descriptive and inferential statistics
From Wikipedia
Descriptive statistics
Descriptive statistics describe the main features of a collection of data quantitatively. Descriptive statistics are distinguished from inferential statistics (or inductive statistics), in that descriptive statistics aim to summarize a data set quantitatively without employing a probabilistic formulation, rather than use the data to make inferences about the population that the data are thought to represent. Even when a data analysis draws its main conclusions using inferential statistics, descriptive statistics are generally also presented. For example in a paper reporting on a study involving human subjects, there typically appears a table giving the overall sample size, sample sizes in important subgroups (e.g., for each treatment or exposure group), and demographic or clinical characteristics such as the average age, the proportion of subjects of each sex, and the proportion of subjects with related comorbidities.
## Inferential statistics
Inferential statistics tries to make inferences about a population from the sample data. We also use inferential statistics to make judgments of the probability that an observed difference between groups is a dependable one, or that it might have happened by chance in this study. Thus, we use inferential statistics to make inferences from our data to more general conditions; we use descriptive statistics simply to describe what's going on in our data.
## Use in statistical analyses
Descriptive statistics provide simple summaries about the sample and the measures. Together with simple graphics analysis, they form the basis of quantitative analysis of data.
Descriptive statistics summarize data. For example, the shooting percentage in basketball is a descriptive statistic that summarizes the performance of a player or a team. This number is the number of shots made divided by the number of shots taken. A player who shoots 33% is making approximately one shot in every three. One making 25% is hitting once in four. The percentage summarizes or describes multiple discrete events. Or, consider the scourge of many students, the grade point average. This single number describes the general performance of a student across the range of their course experiences.
Describing a large set of observations with a single indicator risks distorting the original data or losing important detail. For example, the shooting percentage doesn't tell you whether the shots are three-pointers or lay-ups, and GPA doesn't tell you whether the student was in difficult or easy courses. Despite these limitations, descriptive statistics provide a powerful summary that may enable comparisons across people or other units.
### Univariate analysis
Univariate analysis involves the examination across cases of a single variable, focusing on three characteristics: the distribution; the central tendency; and the dispersion. It is common to compute all three for each study variable.
#### Distribution
The distribution is a summary of the frequency of individual or ranges of values for a variable. The simplest distribution would list every value of a variable and the number of cases who had that value. For instance, computing the distribution of gender in the study population means computing the percentages that are male and female. The gender variable has only two, making it possible and meaningful to list each one. However, this does not work for a variable such as income that has many possible values. Typically, specific values are not particularly meaningful (income of 50,000 is typically not meaningfully different from 51,000). Grouping the raw scores using ranges of values reduces the number of categories to something for meaningful. For instance, we might group incomes into ranges of 0-10,000, 10,001-30,000, etc.
Frequency distributions are depicted as a table or as a graph. Table 1 shows an age frequency distribution with five categories of age ranges defined. The same frequency distribution can be depicted in a graph as shown in Figure 2. This type of graph is often referred to as a histogram or bar chart.
#### Central tendency
The central tendency of a distribution locates the "center" of a distribution of values. The three major types of estimates of central tendency are the mean, themedian, and themode.
The mean is the most commonly used method of describing central tendency. To compute the mean, take the sum of the values and divide by the count. For example, the mean quiz score is determined by summing all the scores and dividing by the number of students taking the exam. For example, consider the test score values:
15, 20, 21, 36, 15, 25, 15
The sum of these 7 values is 147, so the mean is 147/7 =21.
The median is the score found at the middle of the set of values, i.e., that has as many cases with a larger value as have a smaller value. One way to compute the median is to sort the values in numerical order, and then locate the value in the middle of the list. For example, if there are 500 values, the median is the average of the two values in 250th and 251st positions. If there are 501 values, the value in 250th position is the median. Sorting the 7 scores above produces:
15, 15, 15, 20, 21, 25, 36
There are 7 scores and score #4 represents the halfway point. The median is 20. If there are an even number of observations, then the median is the mean of the two middle scores. In the example, if there were an 8th observation, with a value of 25, the median becomes the average of the 4th and 5th scores, in this case 20.5.
The mode is the most frequently occurring value in the set. To determine the mode, compute the distribution as above. The mode is the value with the greatest frequency. In the example, the modal value 15, occurs three times. In some distributions there is a "tie" for the highest frequency, i.e., there are multiple modal values. These are called multi-modal distributions.
Notice that the three measures typically produce different results. The term "average" obscures the difference between them and is better avoided. The three values are equal if the distribution is perfectly "normal" (i.e., bell-shaped).
#### Dispersion
Dispersion is the spread of values around the central tendency. There are two common measures of dispersion, the range and the standard deviation. The range is simply the highest value minus the lowest value. In our example distribution, the high value is 36 and the low is 15, so the range is 36 − 15 = 21.
Th
Mathematical statistics
Mathematical statistics is the study of statistics from a mathematical standpoint, using probability theory as well as other branches of mathematics such as linear algebra and analysis. The term "mathematical statistics" is closely related to the term "statistical theory" but also embraces modelling for actuarial science and non-statistical probability theory, particularly in Scandinavia.
Statistics deals with gaining information from data. In practice, data often contain some randomness or uncertainty. Statistics handles such data using methods of probability theory.
## Introduction
Statistical science is concerned with the planning of studies, especially with the design of randomized experiments and with the planning of surveys using random sampling. The initial analysis of the data from properly randomized studies often follows the study protocol.
Of course, the data from a randomized study can be analyzed to consider secondary hypotheses or to suggest new ideas. A secondary analysis of the data from a planned study uses tools from data analysis.
Data analysis is divided into:
• descriptive statistics - the part of statistics that describes data, i.e. summarises the data and their typical properties.
• inferential statistics - the part of statistics that draws conclusions from data (using some model for the data): For example, inferential statistics involves selecting a model for the data, checking whether the data fulfill the conditions of a particular model, and with quantifying the involved uncertainty (e.g. using confidence intervals).
While the tools of data analysis work best on data from randomized studies, they are also applied to other kinds of data --- for example, from natural experiments and observational studies, in which case the inference is dependent on the model chosen by the statistician, and so subjective.
Mathematical statistics has been inspired by and has extended many procedures in applied statistics.
## Statistics, mathematics, and mathematical statistics
Mathematical statistics has substantial overlap with the discipline of statistics. Statistical theorists study and improve statistical procedures with mathematics, and statistical research often raises mathematical questions. Statistical theory relies on probability and decision theory. Mathematicians and statisticians like Gauss, Laplace, and C. S. Peirce used decision theory with probability distributions and loss functions (or utility functions). The decision-theoretic approach to statistical inference was reinvigorated by Abraham Wald and his successors., and makes extensive use of scientific computing, analysis, and optimization; for the design of experiments, statisticians use algebra and combinatorics.
Advanced Placement Statistics (AP Statistics, AP Stats or AP Stat) is a college-level high schoolstatistics course offered in the United States through the College Board's Advanced Placement program. This course is equivalent to a one semester, non-calculus-based introductory college statistics course and is normally offered to juniors and seniors in high school.
One of the College Board's more recent additions, the AP Statistics exam was first administered in May 1997 to supplement the AP program's math offerings, which had previously consisted of only AP Calculus AB and BC. In the United States, enrollment in AP Statistics classes has increased at a higher rate than in any other AP class.
Students may receive college credit or upper-level college course placement upon the successful completion of a three-hour exam ordinarily administered in May. The exam consists of a multiple choice section and a free response section that are both 90 minutes long. Each section is weighted equally in determining the students' composite scores.
## History
The Advanced Placement program has offered students the opportunity to pursue college-level courses while in high school. Along with the Educational Testing Service, the College Board administered the first AP Statistics exam in May 1997. The course was first taught to students in the 1996-1997 academic year. Prior to that, the only mathematics courses offered in the AP program included AP Calculus AB and BC. Students who didn't have a strong background in college-level math, however, found the AP Calculus program inaccessible and sometimes declined to take a math course in their senior year. Since the number of students required to take statistics in college is almost as large as the number of students required to take calculus, the College Board decided to add an introductory statistics course to the AP program. Since the prerequisites for such a program doesn't require mathematical concepts beyond those typically taught in a second-year algebra course, the AP program's math offerings became accessible to a much wider audience of high school students. The AP Statistics program addressed a practical need as well, since the number of students enrolling in majors that use statistics has grown. A total of 7,667 students took the exam during the first administration, which is the highest number of students to take an AP exam in its first year. Since then, the number of students taking the exam rapidly grew to 98,033 in 2007, making it one of the 10 largest AP exams.
## Course
If the course is provided by their school, students normally take AP Statistics in their junior or senior year and may decide to take it concurrently with a pre-calculus course. This offering is intended to imitate a one-semester, non-calculus based college statistics course, but high schools can decide to offer the course over one semester, two trimesters, or a full academic year.
The six-member AP Statistics Test Development Committee is responsible for developing the curriculum. Appointed by the College Board, the committee consists of three college statistics teachers and three high school statistics teachers who are typically asked to serve for terms of three years.
### Curriculum
Emphasis is placed not on actual arithmetic computation, but rather on conceptual understanding and interpretation. The course curriculum is organized around four basic themes; the first involves exploring data and covers 20–30% of the exam. Students are expected to use graphical and numerical techniques to analyze distributions of data, including univariate, bivariate, and categorical data. The second theme involves planning and conducting a study and covers 10–15% of the exam. Students must be aware of the various methods of data collection through sampling or experimentation and the sorts of conclusions that can be drawn from the results. The third theme involves probability and its role in anticipating patterns in distributions of data. This theme covers 20–30% of the exam. The fourth theme, which covers 30–40% of the exam, involves statistical inference using point estimation, confidence intervals, and significance tests.
## Exam
Along with the course curriculum, the exam is developed by the AP Statistics Test Development Committee as well. With the help of other college professors, the committee creates a large pool of possible questions that is pre-tested with college students taking statistics courses. The test is then refined to an appropriate level of difficulty and clarity. Afterwards, the Educational Testing Service is responsible for printing and administering the exam.
### Structure
The exam is offered every year in May. Students are not expected to memorize any formulas. Therefore, a list of common statistical formulas related to descriptive statistics, probability, and inferential statistics are provided. Moreover, tables for the normal, Student's t and chi-square distributions are given as well. Students are also expected to use graphing calculators with statistical capabilities. The exam is three hours long w
Non-parametric statistics
In statistics, the term non-parametric statistics has at least two different meanings:
1. The first meaning of non-parametric covers techniques that do not rely on data belonging to any particular distribution. These include, among others:
*distribution free methods, which do not rely on assumptions that the data are drawn from a given probability distribution. As such it is the opposite of parametric statistics. It includes non-parametric statistical models, inference and statistical tests.
*non-parametric statistics (in the sense of a statistic over data, which is defined to be a function on a sample that has no dependency on a parameter), whose interpretation does not depend on the population fitting any parametrized distributions. Statistics based on the ranks of observations are one example of such statistics and these play a central role in many non-parametric approaches.
2. The second meaning of non-parametric covers techniques that do not assume that the structure of a model is fixed. Typically, the model grows in size to accommodate the complexity of the data. In these techniques, individual variables are typically assumed to belong to parametric distributions, and assumptions about the types of connections among variables are also made. These techniques include, among others:
*non-parametric regression, which refers to modeling where the structure of the relationship between variables is treated non-parametrically, but where nevertheless there may be parametric assumptions about the distribution of model residuals.
*non-parametric hierarchical Bayesian models, such as models based on the Dirichlet process, which allow the number of latent variables to grow as necessary to fit the data, but where individual variables still follow parametric distributions and even the process controlling the rate of growth of latent variables follows a parametric distribution.
## Applications and purpose
Non-parametric methods are widely used for studying populations that take on a ranked order (such as movie reviews receiving one to four stars). The use of non-parametric methods may be necessary when data have a ranking but no clear numerical interpretation, such as when assessing preferences; in terms of levels of measurement, for data on an ordinal scale.
As non-parametric methods make fewer assumptions, their applicability is much wider than the corresponding parametric methods. In particular, they may be applied in situations where less is known about the application in question. Also, due to the reliance on fewer assumptions, non-parametric methods are more robust.
Another justification for the use of non-parametric methods is simplicity. In certain cases, even when the use of parametric methods is justified, non-parametric methods may be easier to use. Due both to this simplicity and to their greater robustness, non-parametric methods are seen by some statisticians as leaving less room for improper use and misunderstanding.
The wider applicability and increased robustness of non-parametric tests comes at a cost: in cases where a parametric test would be appropriate, non-parametric tests have less power. In other words, a larger sample size can be required to draw conclusions with the same degree of confidence.
## Non-parametric models
Non-parametric models differ from parametric models in that the model structure is not specified a priori but is instead determined from data. The term non-parametric is not meant to imply that such models completely lack parameters but that the number and nature of the parameters are flexible and not fixed in advance.
## Methods
Non-parametric (or distribution-free) inferential statistical methods are mathematical procedures for statistical hypothesis testing which, unlike parametric statistics, make no assumptions about the probability distributions of the variables being assessed. The most frequently used tests include<|endoftext|>
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# AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2
AP State Syllabus AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2 Textbook Questions and Answers.
## AP State Syllabus 9th Class Maths Solutions 5th Lesson Co-Ordinate Geometry Exercise 5.2
Question 1.
Write the quadrant in which the following points lie.
i) (- 2, 3)
ii) (5, – 3)
iii) (4, 2)
iv) (- 7, – 6)
v) (0, 8)
vi) (3, 0)
vii) (-4,0)
viii) (0, – 6)
Solution:
i) (- 2, 3) – Q2 (second quadrant)
ii) (5, – 3) – Q4 (fourth quadrant)
iii) (4, 2) – Q1 (Iirst quadrant)
iv) (- 7, – 6) – Q3 (third quadrant)
v) (0, 8) – on Y-axis
vi) (3, 0) – on X-axis
vii) (-4,0) – on X’ – axis
viii) (0, – 6) – on Y’: axis
Question 2.
Write the abscissae and ordinates of the following points.
i)(4,-8)
ii)(-5,3)
iii)(0,0)
iv)(5, 0)
v)(0, -8)
Note: Plural of abscissa is abscissae.
Solution:
Point abscissa ordinate i) (4, – 8) 4 -8 ii) (-5, 3) -5 3 iii) (0,0) 0 0 iv) (5,0) 5 0 v) (0,-8) . 0 -8
Question 3.
Which of the following points lie on the axes ? Also name the axis.
i) (-5,-8)
ii) (0, 13)
iii) (4, – 2)
iv) (- 2,0)
v) (0, – 8)
vi) (7,0)
vii) (0,0)
Solution:
The points (ii) (0,13) ; (v) (0,- 8) lie on Y – axis.
The points (iv) (- 2, 0), (vi) (7, 0) lie on X – axis.
The point (vii) (0, 0) lie on both X – axis and Y – axis.
The points (i) (- 5, – 8); (iii) (4, – 2) do not lie on any axis.
Question 4.
Write the following based on the graph.
i) The ordinate of L
ii) The ordinate of Q
iii) The point denoted by (- 2,-2)
iv) The point denoted by (5, – 4)
v) The abscissa of N
vi) The abscissa of M
Solution:
i) – 7
ii) 7
iii) The point ’R’
iv) The point P
v) 4
vi) – 3
Question 5.
State true or false and write correct statement.
i) In the Cartesian plane the horizon-tal line is called Y – axis, if) in the Cartesian plane, the vertical line is called Y – axis.
iii) The point which lies on both the axes is called origin.
iv) The point (2, – 3) lies in the third quadrant.
v) (-5, -8) lies in the fourth quadrant.
vi) The point (- x, – y) lies in the first quadrant where x < 0; y < 0.
Solution:
i) False
Correct statement: In the Cartesian plane the horizontal line is called X – axis.
ii) True
iii) True
iv) False
Correct statement: The point (2, -3) lies in the fourth quadrant.
v) False
Correct statement: (- 5, – 8) lies in the third quadrant.
vi) True
Question 6.
Plot the following ordered pairs on a graph sheet. What do you observe ?
i) (1, 0), (3, 0), (- 2, 0), (- 5, 0), (0, 0), (5, 0), (- 6, 0)
ii) (0, 1), (0, 3), (0, – 2), (0, – 5), (0, 0), (0,5), (0,-6)
Solution:
i) All points lie on X – axis,
ii) All points lie on Y – axis.<|endoftext|>
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# Converting a Fraction With a Denominator of 100 or 1000 to a Decimal Online Quiz
Following quiz provides Multiple Choice Questions (MCQs) related to Converting a Fraction With a Denominator of 100 or 1000 to a Decimal. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - Write $\mathbf {\frac{43}{100}}$ as a decimal.
### Explanation
Step 1:
We rewrite 43 as 43.0
Step 2:
We shift the decimal two places to the left in 43.0 as there are two zeros in 100.
Step 3:
So, $\frac{43}{100} = 0.43$
Q 2 - Write $\mathbf {\frac{539}{1000}}$ as a decimal.
### Explanation
Step 1:
We rewrite 539 as 539.0
Step 2:
We shift the decimal three places to the left in 539.0 as there are three zeros in 1000.
Step 3:
So, $\frac{539}{1000} = 0.539$
Q 3 - Write $\mathbf {\frac{113}{100}}$ as a decimal.
### Explanation
Step 1:
We rewrite 113 as 113.0
Step 2:
We shift the decimal two places to the left in 113.0 as there are two zeros in 100.
Step 3:
So, $\frac{113}{100} = 1.13$
Q 4 - Write $\mathbf {\frac{217}{1000}}$ as a decimal.
### Explanation
Step 1:
We rewrite 217 as 217.0
Step 2:
We shift the decimal three places to the left in 217.0 as there are three zeros in 1000.
Step 3:
So, $\frac{217}{1000} = 0.217$
Q 5 - Write $\mathbf {\frac{391}{100}}$ as a decimal.
### Explanation
Step 1:
We rewrite 391 as 391.0
Step 2:
We shift the decimal two places to the left in 391.0 as there are two zeros in 100.
Step 3:
$\frac{391}{100} = 3.91$
Q 6 - Write $\mathbf {\frac{441}{1000}}$ as a decimal.
### Explanation
Step 1:
We rewrite 441 as 441.0
Step 2:
We shift the decimal three places to the left in 441.0 as there are three zeros in 1000.
Step 3:
So, $\frac{441}{1000} = 0.441$
Q 7 - Write $\mathbf {\frac{191}{100}}$ as a decimal.
### Explanation
Step 1:
We rewrite 191 as 191.0
Step 2:
We shift the decimal two places to the left in 191.0 as there are two zeros in 100.
Step 3:
So, $\frac{191}{100} = 1.91$
Q 8 - Write $\mathbf {\frac{1162}{1000}}$ as a decimal.
### Explanation
Step 1:
We rewrite 1162 as 1162.0
Step 2:
We shift the decimal three places to the left in 1162.0 as there are three zeros in 1000.
Step 3:
So, $\frac{1162}{1000} = 1.162$
Q 9 - Write $\mathbf {\frac{972}{100}}$ as a decimal.
### Explanation
Step 1:
We rewrite 972 as 972.0
Step 2:
We shift the decimal two places to the left in 972.0 as there are two zeros in 100.
Step 3:
So, $\frac{972}{100} = 9.72$
Q 10 - Write $\mathbf {\frac{2384}{1000}}$ as a decimal.
### Explanation
Step 1:
We rewrite 2384 as 2384.0
Step 2:
We shift the decimal three places to the left in 2384.0 as there are three zeros in 1000.
Step 3:
So, $\frac{2384}{1000} = 2.384$
converting_fraction_with_denominator_100_1000_decimal.htm<|endoftext|>
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The Greek Philosopher Aristotle devised a way of classifying animals and plants. He grouped plants and animals into tree: the plants which have vegetative souls, animals with both vegetative and animals souls, and human beings, who have vegetative, animal, and radional souls, as manifested by their reasoning ability and intellectual pursuits. Among organisms, those that could move by themselves ( animals ) and those that could not move by themselves (plants) where grouped separately. Those that could move by themselves were further grouped according to their habitat, whether they were land-dwellers, or air-dwellers.
Those in the second group were classified according to size. In the 16th and 17th centuries, geographical studies led to the discovery of new species of plants and animals. Scholars then classified plants into three groups: trees, shrubs, and herbs.
Aristotle's scheme of classification was not always applicable. Take the case of the sea turtles. They stay underwater to get food, but stay on land to lay their eggs. Are turtles water-dwellers or land-dwellers? Baby frogs or baby toads stay in water.
As they mature, they can live on land. Aristotle's scheme, therefore, did not encompass all organisms.
Aristotle's attempts to classify organisms prompted other scientist to find better ways of classifying organisms. Some worked on determining more reliable ways of looking at how organisms are related to each other.<|endoftext|>
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Newton's Laws - Lesson 3 - Newton's Second Law of Motion
Finding Acceleration
As learned earlier in Lesson 3 (as well as in Lesson 2), the net force is the vector sum of all the individual forces. In Lesson 2, we learned how to determine the net force if the magnitudes of all the individual forces are known. In this lesson, we will learn how to determine the acceleration of an object if the magnitudes of all the individual forces are known. The three major equations that will be useful are the equation for net force (Fnet = m•a), the equation for gravitational force (Fgrav = m•g), and the equation for frictional force (Ffrict = μ • Fnorm).
The process of determining the acceleration of an object demands that the mass and the net force are known. If mass (m) and net force (Fnet) are known, then the acceleration is determined by use of the equation.
a = Fnet / m
Your Turn to Practice
Thus, the task involves using the above equations, the given information, and your understanding of Newton's laws to determine the acceleration. To gain a feel for how this method is applied, try the following practice problems. Once you have solved the problems, click the button to check your answers.
Practice #1
An applied force of 50 N is used to accelerate an object to the right across a frictional surface. The object encounters 10 N of friction. Use the diagram to determine the normal force, the net force, the mass, and the acceleration of the object. (Neglect air resistance.)
Practice #2
An applied force of 20 N is used to accelerate an object to the right across a frictional surface. The object encounters 10 N of friction. Use the diagram to determine the normal force, the net force, the coefficient of friction (μ) between the object and the surface, the mass, and the acceleration of the object. (Neglect air resistance.)
Practice #3
A 5-kg object is sliding to the right and encountering a friction force that slows it down. The coefficient of friction (μ) between the object and the surface is 0.1. Determine the force of gravity, the normal force, the force of friction, the net force, and the acceleration. (Neglect air resistance.)
A couple more practice problems are provided below. You should make an effort to solve as many problems as you can without the assistance of notes, solutions, teachers, and other students. Commit yourself to individually solving the problems. In the meantime, an important caution is worth mentioning:
Avoid forcing a problem into the form of a previously solved problem. Problems in physics will seldom look the same. Instead of solving problems by rote or by mimicry of a previously solved problem, utilize your conceptual understanding of Newton's laws to work towards solutions to problems. Use your understanding of weight and mass to find the m or the Fgrav in a problem. Use your conceptual understanding of net force (vector sum of all the forces) to find the value of Fnet or the value of an individual force. Do not divorce the solving of physics problems from your understanding of physics concepts. If you are unable to solve physics problems like those above, it is does not necessarily mean that you are having math difficulties. It is likely that you are having a physics concepts difficulty.
We Would Like to Suggest ...
Sometimes it isn't enough to just read about it. You have to interact with it! And that's exactly what you do when you use one of The Physics Classroom's Interactives. We would like to suggest that you combine the reading of this page with the use of our Force Interactive. You can find it in the Physics Interactives section of our website. The Force Interactive allows a learner to explore the effect of variations in applied force, net force, mass, and friction upon the acceleration of an object.
Visit: Force
1. Edwardo applies a 4.25-N rightward force to a 0.765-kg book to accelerate it across a tabletop. The coefficient of friction between the book and the tabletop is 0.410. Determine the acceleration of the book.
2. In a physics lab, Kate and Rob use a hanging mass and pulley system to exert a 2.45 N rightward force on a 0.500-kg cart to accelerate it across a low-friction track. If the total resistance force to the motion of the cart is 0.72 N, then what is the cart's acceleration?
Next Section:<|endoftext|>
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Download the Epic! App and create an account using the classroom code on the picture to the left.
Select a nonfiction book about an animal that interests you. Keep in mind your reading levels, but you are not limited to your level. Read the book and create a thinking map using Popplet. Model your map after the one below. Yours doesn’t have to be the same, you can organize your thinking however works for you.
Description: This lesson is student centered due to students selecting their topic based on interest. They also chose their research tools and select books based not just on reading levels, but their desire to read. Once they have finished reading, they create their movie to teach others. Their tool is also self selected based on their comfort level and what they feel works best for their topic.
Students will be using the Popplet brain mapping program to record their learning and organize it to make sense. Recording their information in like categories helps them to develop ideas, which will make it easier to make their script in the next step. Popplet is an easy program to use, simply touching the bubbles adds more, you can change colors, drag to new positions, and add images as well. It allows creativity, while keeping students on target with recordingneeded information. The sample shown can be recreated by students, or they can make their own. By giving them an example they are able to have a concrete starting point, and make it easier to build their own important information from this starting point.<|endoftext|>
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Math = Love: Linear vs Non-Linear Functions INB Pages
Sunday, February 23, 2014
Linear vs Non-Linear Functions INB Pages
Daily, I am learning how to be a better teacher. I know what worked well for my students last year. And, I know what didn't work so well. Last year, I assumed that if I taught my students the definition of a linear function, they would be able to apply that definition and use it to find the missing value of a function. I assumed incorrectly. Last year, I taught my students the vocabulary, but I didn't give them the practice they needed to be successful with this type of question on our end-of-instruction exams.
I decided this year was going to be different. Instead of spending 15 minutes discussing linear vs. non-linear and then moving on, we spent an entire class period working with linear and non-linear functions.
We started out by creating a frayer model for linear functions. I provided students with the definition and characteristics. They worked together, as a class, to create their own examples and non-examples. In almost every class, one student would suggest that a vertical line was an example of a linear function. It was a happy teacher moment for me when another student would realize and explain to the class why the vertical line belonged in the non-example box. Student propelled discussions are awesome! I need to have more of them in my classroom.
Linear Function Frayer Model for Interactive Notebook
Next, I gave my students a sheet of problems to cut apart. The first thing we looked at was a set of tables. Students were tasked with determining if the table represented a linear function or a non-linear function.
This was the perfect opportunity to use color with a purpose. Students used one color of marker to find the rate of change between the first two lines in the table. Students used a different color of marker for the rate of change between the second and third lines in the table. And, they used a third color to find the rate of change between the third and fourth lines.
In retrospect, I should have given my students more of these tables to practice with. But, we moved on next to some practice EOI questions. These are the questions that I had just assumed in the past that my students would be able to solve on their own. We practiced finding the pattern / rate of change. And, we extended the tables as necessary to find the requested values.
I'm really glad I printed these questions off and had my students glue them in their notebooks. This allowed students to write all over the questions without having to worry about copying down the questions.
Interactive Notebook Pages 1-2: Linear vs Non-Linear Functions
Is this teaching to the test? Yes. But, I think it's more than that. This is me equipping my students with tools that they need to be successful. Students have to be taught to organize their work. They need to see different ways to approach problems. As we worked through these problems together, discussion naturally happened. In some classes, a few students would realize that we could take a short-cut that would keep us from having to draw extra lines on our tables. Part of the class would use the short-cut. The rest of us would write out the process, step-by-step. We would compare answers and discuss why the short cut worked.
Last year, my students would panic when they would see problems like this. This year, they view them as a sort of puzzle that they have the tools to figure out. I view that as a win!
Interactive Notebook Page 3: Linear vs Non-Linear Functions<|endoftext|>
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How did the war in Vietnam reflect social and political tensions that had already existed in American life since 1900?
The Vietnam War produced in the United States one of the most comprehensive efforts in social conditioning [adjusting of public opinion] in modern times. Nothing was spared in the attempt to make the war seem necessary and acceptable to the American public.In the war of protest, students may have taken the lead, but they were not alone. A measure of success was achieved when the U.S. congress responded with what has been called “the greatest outpouring of human rights legislation in this century.
Young people also showed their rejection of the system by rejecting its conventional dress and grooming standards.
For many women it was the day of miniskirts and hot pants; for men, of beards and long hair; disheveled appearance that later became known as the hippie look.
Some of the music of the day also promoted the spirit of protest by encouraging the use of drugs and by condoning permissive sex and homosexuality. Rock stars and pop singers became idols, dictating both fashions and conduct. Communal living became popular. This and other life-styles formerly considered unacceptable were now viewed as being acceptable alternatives.
For more information on this subject and others, please go to jw.org "Online Library" Also for free downloads, publications or read online..<|endoftext|>
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This chapter introduces two types of cell divisions. First it explains mitosis and then meiosis. This chapter also explains why cells divide and how the divisions are regulated. The errors in division may lead to diseases, such as leukemia.
- 7.1: Case Study: Genetic Similarities and Differences
- Steve asked his doctor about Pharmocogenomics. The doctor explains to Steve that Pharmacogenomics is the tailoring of drug treatments to people’s genetic makeup, a form of ‘personalized medicine’.
- 7.2: Cell Cycle and Cell Division
- This baby girl has a lot of growing to do before she's as big as her mom.
- 7.3: Mitotic Phase - Mitosis and Cytokinesis
- Can you guess what this colorful image represents? It shows a eukaryotic cell during the process of cell division.
- 7.4: Mutations and Cancer
- Your cells may grow and divide without performing their necessary functions, or without fully replicating their DNA, or without copying their organelles. Probably not much good could come of that. So the cell cycle needs to be highly regulated and tightly controlled. And it is.
- 7.5: Sexual Reproduction: Meiosis and gametogenesis
- This self-portrait of an 18th century artist and his family clearly illustrates an important point.
- 7.6: Genetic Variation
- Genetic variation. It is this variation that is the essence of evolution. Without genetic differences among individuals, "survival of the fittest" would not be likely. Either all survive, or all perish.
- 7.7: Mitosis vs. Meiosis and disorders
- Both mitosis and meiosis result in eukaryotic cells dividing. So what is the difference between mitosis and meiosis? The primary difference is the differing goals of each process. The goal of mitosis is to produce two daughter cells that are genetically identical to the parent cell, meaning the new cells have exactly the same DNA as the parent cell. Mitosis happens when you want to grow, for example. You want all your new cells to have the same DNA as the previous cells.
- 7.8: Case Study Genes Conclusion and Chapter Summary
- Humans are much more genetically similar to each other than they are different.
Thumbnail: Image of the mitotic spindle in a human cell showing microtubules in green, chromosomes (DNA) in blue, and kinetochores in red. Image used with permission (Public Domain; Afunguy).<|endoftext|>
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Question Video: Pascal’s Principle | Nagwa Question Video: Pascal’s Principle | Nagwa
# Question Video: Pascal’s Principle Physics • Second Year of Secondary School
A solid object falls through water of density 1000 kg/m³. At the instant that the top of the object is 25 cm below the water’s surface, the water exerts a pressure on the top, bottom, and side of the object, as shown in the diagram. Find 𝑃₁, 𝑃₂, and 𝑃₃.
07:16
### Video Transcript
A solid object falls through water of density 1000 kilograms per cubic meter. At the instant that the top of the object is 25 centimeters below the water’s surface, the water exerts a pressure on the top, bottom, and side of the object, as shown in the diagram. Find 𝑃 one. Find 𝑃 two. Find 𝑃 three.
Looking at our diagram, we see this object submerged a complete distance of 25 centimeters below the surface of this water. Along with this, we can see these three pressures exerted on the top, side, and bottom of this object respectively, 𝑃 one on the top, 𝑃 two on the side, and 𝑃 three on the bottom. It’s these three pressures that we want to calculate. To do this, it will be helpful to recall a relationship telling us the pressure created by a liquid a certain depth below the surface of that liquid.
If we have a liquid of density 𝜌 that’s filling a container and we want to know the pressure 𝑃 created by the liquid at some point within the liquid, then the way we do that is we multiply the liquid density 𝜌 by the acceleration due to gravity 𝑔. And then, if this point where we want to solve for the pressure is a distance, we can call it ℎ, below the surface of the liquid. Then we multiply the liquid density and gravity by that term as well. The pressure a distance ℎ below the surface of a liquid of density 𝜌 is equal to 𝜌 times 𝑔 times ℎ.
Now, in our scenario, the liquid we’re working with is water. And we’re told its density, 1000 kilograms per cubic meter. Knowing the density 𝜌, we can now recall what 𝑔, the acceleration due to gravity near the surface of the Earth, is. We can approximate that acceleration as 9.8 meters per second squared. Now, that we know density 𝜌 and the gravitational acceleration 𝑔, all we need to know is ℎ the distance below the surface of the liquid to solve for the pressure at that point.
This is where our diagram helps us. We know that we want to calculate 𝑃 one, which is the pressure on the top side of our object, the pressure right here at this depth below the surface of our liquid. And our diagram shows us what that depth is. It’s 25 centimeters. This means we can write an equation for the pressure on the top of our object 𝑃 one. It’s equal to the density of our liquid multiplied by 𝑔 multiplied by the depth at that point, 25 centimeters below surface level. And it’s at this point that we can plug in values for 𝜌 and 𝑔. We know that the density is 1000 kilograms per cubic meter and that 𝑔 is 9.8 meters per second squared.
At this point, we can notice that the units of distance in our expressions for 𝜌 and in our expression for 𝑔 are meters but for our ℎ value, we have a distance measured in centimeters. So, we’ll want to convert that to meters so our units are consistent all through. If we recall that 100 centimeters is equal to one meter. That means, to convert 25 centimeters to meters, we’ll move the decimal point two spots to the left. And in doing so, we find that 25 centimeters is equal to 0.25 meters.
Before we multiply these three terms together and solve for 𝑃 one, take a look again at the units involved in these terms. We have kilograms per meter cubed multiplied by meters per second squared multiplied by meters. Considering our distance units, we have two powers of the unit meter in our numerator, here and here. And we have three powers in our denominator here. This means that two of those meters units cancel out from both top and bottom, numerator and denominator. And when we do this cancellation, we find that in our denominator we simply have units of meters.
And again, this is after we multiply all three of these values together. If we gathered all the remaining units and put them to the far right of this expression, we would see our final units end up being kilograms per meter second squared. If we were to rearrange these units, we would find that this is equivalent to a newton per square meter. That’s the same thing as a kilogram per meter second squared.
But then, if we consider what a newton per meter squared is, that’s equivalent to the SI unit for pressure, the pascal. Since a newton per square meter is a pascal, and a newton per square meter is the same as the units we’ll end up with in our expression. What we’re saying is the overall units we’ll have, once we calculate 𝑃 one, are pascals, the SI base unit of pressure. All that being said, let’s now multiply these three numbers together and figure out just how many pascals of pressure 𝑃 one is. When we do this, we find a result of 2450 pascals. That’s the pressure being exerted on the top side of our object.
Next, looking back at our diagram, we want to solve for the pressure 𝑃 two, the pressure exerted on the side of the object. Just like before, we’ll write an equation for this pressure, In this case 𝑃 two, in terms of density, acceleration due to gravity, and the height of this particular point below the surface level. And since this height is different from our previous height, let’s give it a subscript. We’ll call it a ℎ two.
We already know what density and 𝑔 are; they’re the same as they were before. But it’s ℎ two that we want to solve for. Taking a look at our diagram, we see that 𝑃 two, this pressure exerted on the side of our object, is exerted exactly 7.5 centimeters below the elevation of 𝑃 one. In other words, if we take the depth value of our pressure point 𝑃 one, 25 centimeters, and add to it seven and a half centimeters, then we’ll have the accurate height, or, in this case, depth, of this pressure point 𝑃 two. So, ℎ two is 25 centimeters plus 7.5 centimeters, or, in other words, 32.5 centimeters.
Now, just like before, we’ll want to convert this value from units of centimeters to units of meters. And to do that, we’ll shift the decimal point two spots to the left and find that this depth of pressure point 𝑃 two in units of meters is 0.325 meters. Plugging in again for density and 𝑔, when we multiply these three numbers together, once again, we’ll get a result in units of pascals. This time, though, we calculate a pressure of 3185 pascals. The fact that this pressure 𝑃 two is greater than the pressure we calculated for 𝑃 one makes sense because we’re now at a lower point below the surface than 𝑃 one.
And now, having solved for the pressures 𝑃 one and 𝑃 two on the top and sides of our object, respectively, we’ll solve for the pressure on the bottom of the object, 𝑃 three. Once again, we’ll do this by multiplying the density of our liquid, the water, by the acceleration due to gravity multiplied by the height below the surface of this point 𝑃 three.
Looking at our diagram, we see that the underside of our object is located a distance of 7.5 centimeters below the side of the object where we calculated the depth of 𝑃 two. So, the depth of the bottom of our object is equal to 25 centimeters plus seven and a half centimeters plus seven and a half centimeters again. Or, in other words, 25 centimeters plus 15 centimeters, which adds up to 40 centimeters. And then, just like we’ve done before, we’ll convert this from a number in centimeters to a number in meters. It’s equal to 0.40 meters.
Lastly, we’ll once more plug in our values for 𝜌 and 𝑔. And having done that, we’ll multiply these three numbers together. When we do, the result we find is 3920 pascals. That’s the pressure we calculate acting on the bottom of this object. And having solved for that, we’ve now calculated the pressure on the top, side, and bottom of our object as shown in the diagram.
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Home | | Maths 7th Std | Area of Pathways
# Area of Pathways
We come across pathways in different shapes. Here we restrict ourselves to two kinds of pathways namely circular and rectangular.
Area of Pathways
We come across pathways in different shapes. Here we restrict ourselves to two kinds of pathways namely circular and rectangular.
1. Circular Pathways
We observe around us circular shapes where we need to find the area of the pathway. The area of pathway is the difference between the area of outer circle and inner circle. Let ‘R’ be the radius of the outer circle and ‘r’ be the radius of inner circle.
Therefore, the area of the circular pathway = πR2 πr2
= π(R2 r2 ) sq. units.
2. Rectangular Pathways
Consider a rectangular park as shown in Fig 2.23. A uniform path is to be laid outside the park. How do we find the area of the path? The uniform path including the park is also a rectangle. If we consider the path as outer rectangle, then the park will be the inner rectangle. Let l and b be the length and breadth of the park. Area of the park (inner rectangle) = l b sq.units . Let w be the width of the path. If L, B are the length and breadth of the outer rectangle, then L = l + 2w and B = b + 2w .
Here, the area of the rectangular pathway
= Area of the outer rectangle – Area of the inner rectangle
= (LBlb) sq. units
Example 2.15
A park is circular in shape. The central portion has playthings for kids surrounded by a circular walking pathway. Find the walking area whose outer radius is 10 m and inner radius is 3 m.
Solution
The radius of the outer circle, R = 10 m
The radius of the inner circle, r = 3 m
The area of the circular path = Area of outer circleArea of inner circle
= π R 2 πr2
= π( R2 r2 ) sq.units
= 22/7 × (10232 )
= 22/7 × ((10 ×10)(3 × 3))
= 227 × (1009)
= 22/7 × 91
= 286 m2
Try these
(i) If the outer radius and inner radius of the circles are respectively 9 cm and 6 cm, find the width of the circular pathway.
Outer radius of the circle R = 9 cm
Inner radius r = 6 cm
The width of the pathway w = R – r
= 9 – 6 = 3 cm.
(ii) If the area of the circular pathway is 352 sq.cm and the outer radius is 16 cm, find the inner radius.
Area of the pathway = π (R2 – r2) = 352 sq.cm
Outer radius R = 16 cm
22/7 (162 – r2) = 352
256 – r2 = 352 / 22 7 = 112
– r2 = 112 – 256 = – 144
r2 = 144 = 122
r = 12 cm
Inner radius = 12 cm
(iii) If the area of the inner rectangular region is 15 sq.cm and the area covered by the outer rectangular region is 48 sq.cm, find the area of the rectangular pathway.
Area of the outer rectangular region πR2 = 48 sq.cm
Area ofthe inner rectangular region πr2 =15 sq.cm
Area of the rectangular pathway = πR2 – πr2
= (48–15) sq.cm
= 33 sq.cm
Example 2.16
The radius of a circular flower garden is 21 m. A circular path of 14 m wide is laid around the garden. Find the area of the circular path.
Solution
The radius of the inner circle r = 21 m
The path is around the inner circle.
Therefore, the radius of the outer circle, R = 21 + 14 = 35 m
The area of the circular path = π ( R 2 r2 ) sq.units
= 22 /7 × (352 −212)
= 22/7 × ((35 × 35) − (21 × 21))
= 22/7 × (1225 − 441)
= 22/7 × 784
= 22 × 112 = 2464 m2
Example 2.17
The radius of a circular cricket ground is 76 m. A drainage 2 m wide has to be constructed around the cricket ground for the purpose of draining the rain water. Find the cost of constructing the drainage at the rate of ₹180/- per sq.m.
Solution
The radius of the inner circle (cricket ground), r = 76 m
A drainage is constructed around the cricket ground.
Therefore, the radius of the outer circle, R = 76 + 2 = 78 m
We have, area of the circular path = π( R 2r2 ) sq. units
= 22/7 × (782 − 762 )
= 22/7 × (6084 − 5776)
= 22/7 × 308
= 22 × 44 = 968 m2
Given, the cost of constructing the drainage per sq.m is ₹180.
Therefore, the cost of constructing the drainage = 968 × 180 = ₹1,74,240.
Example 2.18
A floor is 10 m long and 8 m wide. A carpet of size 7 m long and 5 m wide is laid on the floor. Find the area of the floor that is not covered by the carpet.
Solution
Here, L = 10 m
B = 8 m
Area of the floor = L × B
=10×8
=80 m2
Area of the carpet = l × b
= 7 × 5
= 35 m2
Therefore, the total area of the floor not covered by the carpet = 80 35
= 45 m2
Example 2.19
A picture of length 23 cm and breadth 11 cm is painted on a chart, such that there is a margin of 3 cm along each of its sides. Find the total area of the margin.
Solution
Here L = 23 cm B = 11 cm
Area of the chart = L × B
= 23 ×11
= 253 cm2
l = L 2w = 23 2(3) = 23 6 = 17 cm
b = B 2w = 11 2(3) = 11 6 = 5 cm2
Area of the picture 17× 5 = 85cm2
Therefore, the area of the marin = 235 85
= 168 cm2 .
Example 2.20
A verandah of width 3 m is constructed along the outside of a room of length 9 m and width 7 m. Find (a) the area of the verandah (b) the cost of cementing the floor of the verandah at the rate of ₹15 per sq.m.
Solution
Here, l = 9 m , b = 7 m
Area of the Room = l × b
=9×7
= 63 m2
L = l + 2w = 8 + 2(3) = 8 + 6 = 14 m
B = b + 2w = 5 + 2(3) = 5 + 6 = 11 m
Area of the room including verandah = L × B
= 14 ×11
= 154 m2
The area of the verandah = Area of the room including verandah Area of the room
=15463
= 91 m2
The cost of cementing the floor for 1sq.m = ₹15
Therfore, the cost of cementing the floor of the verandah = 91 × 15 = ₹1365.
Example 2.21
A Kho-Kho ground has dimensions 30 m × 19 m which includes a lobby on all of its sides. The dimensions of the playing area is 27 m × 16 m . Find the area of the lobby.
Solution
From the dimensions of the ground we have,
L = 30 m ; B = 19 m ; l = 27 m ; b = 16 m
Area of the kho kho ground = L × B
= 30 ×19
= 570 m2
Area of the play field = l × b
= 27 ×16
= 432 m2
Area of the lobby = Area of Kho-Kho ground Area of the play field
= 570432
= 148 m2
Tags : Measurements | Term 2 Chapter 2 | 7th Maths , 7th Maths : Term 2 Unit 2 : Measurements
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
7th Maths : Term 2 Unit 2 : Measurements : Area of Pathways | Measurements | Term 2 Chapter 2 | 7th Maths<|endoftext|>
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What is balanced power?
When 120-volt AC power is balanced, one side of the circuit has +60 Volts to ground while the other has -60 Volts to ground. (Across the circuit, the usual 120 Volts is still present. Fig. 1) A European 230 Volt balanced power system has +115 Volts and -115 Volts to ground on the conductors.
Standard unbalanced AC power systems have a "hot" conductor and a "neutral" conductor. In the US, the "hot" conductor nominally has 120 Volts to ground and the "neutral" conductor has 0 Volts to ground. (Fig. 2) Europe has a similar system but with 230 volts on the "hot" and 0 Volts on the "neutral."
In a balanced power system, the voltages on the system's two output terminals are 180 degrees out of phase to each other with respect to ground. The system reference (ground) originates at the output center tap of an AC isolation transformer. In other words, the system's grounding reference (zero position) is located at the system's mean voltage differential or zero crossing point of the AC sinewave. This is a far more effective way to establish a reference potential for an AC system. The center tap is then grounded to Earth for electrical safety and for referencing shields.
There is never any voltage or current present on the ground reference in a balanced power system. Transient voltages and reactive currents which normally would appear on the neutral and ground wires are also out of phase and likewise, sum to zero at the ground reference thereby canceling out AC hum and noise.
A balanced AC Power system works the same way as a balanced audio circuit but with a higher amplitude. Both balanced audio and balanced AC incorporate phase cancellation or common mode rejection to eliminate noise.<|endoftext|>
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Math 115 Spring 2014 Written Homework 10-SOLUTIONS Due Friday, April 25
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1 Math 115 Spring 014 Written Homework 10-SOLUTIONS Due Friday, April 5 1. Use the following graph of y = g(x to answer the questions below (this is NOT the graph of a rational function: (a State the domain and range of g in interval notation. Solution: There is a point on the graph corresponding to every x value except x =. Hence, the domain of g is (, (, 5]. The range is (, 1 (1, ] (b What is g(? What is g(? What is g(4? Solution: We are looking for the y-values corresponding to the points where x =, x = and x = 4. We see that g( =, g( is undefined (as x = is not in the domain of g, and g(4 =. (c For what values of x does (i g(x =?, (ii g(x =?, (iii g(x = 1? Solution: We start by looking at the set of points on the graph of g that intersect the horizontal lines, y =, y = and y = 1, respectively. For y = we see that y = intersects the graph at the points (1,, (, and the line interval from (, to (0,. Hence, the solutions to the equation g(x = are the set {x R x 0, x = 1 or x = }. For y =, there is only one point of intersection with the graph; (,. g(x = when x =. Hence, For y = 1, there is no point of intersection with the graph. Hence, there are no solutions to the equation g(x = 1.
2 (d Determine all of the following limits from the graph of y = g(x: x g(x x g(x x + g(x x 0 g(x x 4 g(x x 4 + g(x x g(x Solution: g(x = 1 x g(x = x x + g(x = x 0 g(x does not exist (since the limit from the right of x = 0 is 1 and the limit from the left of x = 0 is x 4 g(x = 1 g(x = x 4 + x g(x =
3 . Express the lengths a and b in the figure below in terms of θ. Solution: Use the right-triangle definitions of the trig ratios and then solve for a and b: sin θ = a 4 a = 4 sin θ and cos θ = b 4 b = 4 cos θ. A boat approaches a 0-ft lighthouse whose base is at sea level. Let b be the distance between the boat and the base of the lighthouse. Let L be the distance between the boat and the top of the lighthouse. Let θ be the angle of elevation between the boat and the top of the lighthouse. (a Express b as a function of θ. Solution: To model this situation, we can use a right triangle: Using the right-triangle definitions, we know tan θ = 0 0. Thus, b = b tan θ.
4 (b Express L as a function of θ. Solution: Using the same right triangle from above, sin θ = 0 0. Thus, L = L sin θ. 4. (a In a circle of radius r, an arc of length 10 is swept out by an angle of radians. What is the exact radius of the circle? Solution: The radian was defined in such a way that the arc length, s, is equal to the radius times the angle, when the angle is in radians. i.e. Here we have that s = 10 and θ =. Thus, s = rθ. 10 = r r = 10. (b In a circle of radius r =, what is the exact length of the arc swept by an angle of 10? Solution: Before we can apply the arc length formula, we must have an angle in radians. Thus, here we must convert the angle from degree measure into radian measure. 10 ( π 180 = 10π 180 = 1π 18 = 7π. Now we have r = and θ = 7π. s = rθ = ( 7π = 7π. The length of the are subtended by the angle 10 on the circle is 7π. 5. Suppose θ is an angle in standard position (meaning it starts on the positive x-axis whose radian measure is an integer value between 0 and π. (a If the angle falls in the third quadrant, what is the radian measure θ? Solution: The angles in the third quadrant are between the angles π = and = Thus, the only integer value between these is 4. So, θ = 4. π (b If the terminal side of the angle had fallen in any other quadrant, could you still answer the question with a single value? Explain. Solution: If the terminal side had fallen in quadrant I, we would be able to answer this question because those angles fall between 0 and π = Hence, in quadrant I, θ would be equal to 1. In quadrants II and IV, there would be two possible values for θ; and in II and 5 and in IV.
5 . For each angle θ given below, when sketched in standard position, determine the quadrant in which it lies. (Hint: in some cases, it might help to determine a coterminal angle that is between 0 and π. You should justify your answer in some way - either with a brief explanation or a picture. (a θ = 7π Solution: θ = 7π = π + π. π is full revolutions and then π is less than π terminal side of θ will end up in the first quadrant. so the (b θ = 11π 4 Solution: θ = 11π 4 = π π 4 clockwise π which is more than π 4 in the third quadrant.. π is 1 full revolution clockwise an then we continue but less than π. Thus, the terminal side of θ will end up (c θ = 5.5 Solution: θ = 5.5 is between the quadrantal angles π = and π = Thus, the terminal side of θ will end up in the fourth quadrant. 7. (a Suppose θ is an angle in standard position which intersects the unit circle at the point (x, y. If y = 1, what are the possible values of cos θ? If you know that cos θ is positive, in 8 which quadrant does the angle lie? Solution: The equation of the unit circle is x + y = 1. If we plug y = 1 8 we get two possible values of x and thus two possible values of cos θ: into the equation, x + y = 1 x + ( 1 8 = 1 x = 1 x = 4 x = ± 8 The possible values are cos θ = and cos θ =. If we know that cos θ is positive, then 8 8 the angle must lie in quadrant I because both the x and the y coordinate of the corresponding point on the unit circle are positive.
6 (b Suppose sin α = 4 5 and π < α < π. Find cos α, tan α, cot α, sec α, and csc α. Solution: Since π < α < π, we know that the angle lies in quadrant III. One method would be to draw a right triangle in quadrant III with the opposite side length 4 and the hypotenuse 5. This is a -4-5 triangle so the adjacent side is. Then we can use the right triangle trig definitions to determine the other trig values - remembering that the angle is in the third quadrant. Another method would be to use the equation for the unit circle: x + y = 1 or the equivalent identity cos (x + sin (x = 1. By plugging in sin α = y = 4, we can then solve for 5 cos α = y = 5. Either way, we get the following solutions: cos α = 5, tan α = 4, cot α = 4, sec α = 5, csc α = 5 4 (c Suppose cos β = and β lies in quadrant II. Find sin β, tan β, cot β, sec β, and csc β. Solution: One method would be to draw a right triangle in quadrant II with the adjacent side length and the hypotenuse. This is a triangle so the opposite side is 1. Then we can use the right triangle trig definitions to determine the other trig values - remembering that the angle is in the second quadrant. Another method would be to recognize the point on the Unit Circle here. The point is in ( the second quadrant with an x-coordinate of. Thus, the point is, 1. Either way, we get the following solutions: sin β = 1, tan β = 1 = 1, cot β =, sec β =, csc β = 8. Find( the exact value of each of the following. You must justify your answer in some way. 9π (a cos Solution: 9π = 8π + π = 4π + π. Thus the angle is coterminal to π. Thus, cos ( 9π ( π = cos = 0
7 (b sin ( 4π Solution: The terminal side of 4π ends up in quadrant so it s sine value will be negative. The reference angle here is π. Using either a special right triangle or the unit circle, we find that ( 4π ( π sin = sin = ( 11π (c tan Solution: The terminal side of 11π ends up in quadrant 4 so it s tangent value will be negative. The reference angle is π. Using either a special right triangle or the unit circle, we find that (d ( csc 5π 4 ( 11π tan ( π = tan = sin π cos π = 1 = 1 Solution: The angle 5π 4 will fall in quadrant with a reference angle of θ r = π 4. Using either a special right triangle or the unit circle, we find that ( csc 5π ( π = csc = sin π = 1 1 = 4 (e ( 48π sec Solution: The angle 48π = 1π is a quadrantal angle that is coterminal to π. Thus, ( 48π 1 sec = sec(1π = cos 1π = 1 cos π = 1 1 = 1 ( (f cos 11π Solution: The angle 11π = 9π π = π π will fall in quadrant 1 with a reference angle of θ r = π. Using either a special right triangle or the unit circle, we find that ( cos 11π ( π = cos = 1
8 9. As we did with the sine function in lecture, state the properties of the cosine function. (a State the domain and range of f(x := cos x. Solution: The domain is x R. The range is f(x [ 1, 1]. (b Determine all values of x where f(x is zero (i.e. the x-intercepts. Solution: The zeros occur when x = π + kπ where k is any integer. (c Determine the y-intercept for the graph y = cos x. Solution: The y-intercept occurs at (0, Use your knowledge of transformations to sketch the graph of at least two periods/cycles of each function. ( (a f(x := sin x + π 4 Solution: We begin with the graph of y = sin x: Next we reflect the graph over the x axis to get the graph of y = sin x: Finally, we shift the previous graph to the left π ( 4 to get the graph of y = sin x + π : 4
9 (b g(x := cos(x + 4 Solution: We begin with a graph of y = cos x: Then we shift the graph up 4 to get the graph of y = cos(x + 4:
10 (c ( h(x := cos x π + 4 Solution: We begin with a graph of y = cos x. Then we shift the graph to the right π 4 to get the graph of y = cos(x π 4 Now applying the vertical stretch cause by A = yields the graph y = cos ( x π 4. Lastly, we apply the vertical shift up. This results in the final graph y = cos ( x π 4 +.
11 Remember - there is more than correct answer for each of these graphing questions. ANY complete cycles is correct. Some of you may have used a different method on part (c and ended up with the following graph:
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# Rate of Return Learn How to Calculate Rates of Return ROR
In many chemical reactions, the molecules must be in very specific orientations for a chemical reaction to occur. Understanding the rates of chemical reactions, and why they are fast or slow, is a field called kinetics. In this lesson, we will learn methods for computing the rates of chemical reactions based on how quickly chemicals react or products form.
To calculate a 1-year annual return, take the end-of-year investment value, deduct the value from the beginning of the year, and then divide it also by the beginning-of-year value. What if we wanted the rate of reaction at any given instant, say, at 3 s, rather than a specific time interval? The name for that perfect (black) line is a tangent line because it passes tangent to (3, 4.3) but does not cross the data. Once we have drawn the tangent line and have extended it in both directions, we can pick any two points on the line to calculate the slope, that is, the reaction rate. Note that the regular rate of return describes the gain or loss, expressed in a percentage, of an investment over an arbitrary time period.
## Video Explanation of Rate of Return
Any graph that has time as the horizontal axis can be used to determine a rate. In these cases, the rate is the slope of the line on the graph (many of you will know this as “rise over run”) or the change in the vertical axis variable divided by the change in time (on the horizontal axis). Notice in the example above that we rounded the unit price to the nearest cent. Sometimes we may need to carry the division to one more place to see the difference between the unit prices. With this, we can determine that the annual interest rate for this loan is 5.42%. You will notice that cell C7 is set to negative in the formula.
• However, this equation is not very useful if one is just looking at a graph and does not know the rate constant or the values of n and o.
• The % discount rate represents the time value of money of capital that is tied up in a project, and reflects the minimum rate of return needed to produce an acceptable investment result for a given level of risk.
• The IRR calculation would take these interim cashflows into consideration.
• Similar to the simple rate of return, any gains made during the holding period of this investment should be included in the formula.
• In this lesson, we will learn methods for computing the rates of chemical reactions based on how quickly chemicals react or products form.
• When the ROR is positive, it is considered a gain, and when the ROR is negative, it reflects a loss on the investment.
• In this lesson we defined the meaning of a rate and its units.
To find the unit price, divide the total price by the number of items. The RATE Function[1] is an Excel Financial function that is used to calculate the interest rate charged on a loan or the rate of return needed to reach a specified amount on an investment over a given period. A Because O2 has the smallest coefficient in the balanced chemical equation for the Calculating the Rate reaction, define the reaction rate as the rate of change in the concentration of O2 and write that expression. Write expressions for the reaction rate in terms of the rates of change in the concentrations of the reactant and each product with time. The Internal Rate of Return (IRR) is the annual rate of growth that an investment or project generates over time.
## Rate Of Return: Formula, Calculation & Examples
Reaction rates generally decrease with time as reactant concentrations decrease. For example, if an investment is worth \$70 at the end of the year and was purchased for \$60 at the beginning of the year, the annual rate of return would be 16.66%. For example, if a share price was initially \$100 and then increased to a current value of \$130, the rate of return would be 30%. Companies can use rates of return to measure the performance of various business segments or assets which can assist them in making future decisions about how to best invest their capital.
### How to calculate the rate of change?
Rate of change problems can generally be approached using the formula R = D/T, or rate of change equals the distance traveled divided by the time it takes to do so.
The line shown is the concentration of the reactant, but lines to show the products could also be included in the graph. For the chemist, the rate of a reaction is a window into what is happening as molecules and atoms collide with each other. Sometimes, two colliding molecules immediately react, but not always.
## Sciencing_Icons_Functions Functions
A rate of return calculates the percentage change in value for any investment, regardless of whether it continues to be held, or was sold. In this Module, the quantitative determination of a reaction rate is demonstrated. Reaction rates can be determined over particular time intervals or at a given point in time. A rate law describes the relationship between reactant rates and reactant concentrations. Reaction rates are reported as either the average rate over a period of time or as the instantaneous rate at a single time.
What rate of return are Seeking Alpha contributors expecting from the S&P 500 in 2023? Change and time are two of the main themes in the geosciences. Because geologists are interested in how rapidly changes can occur, we also look at the time over which those changes took place. Combined, this creates a rate, the rapidity in which change occurs.
## Interpreting a Kinetics Graph
We can also draw a tangent line to determine the instantaneous rate of reaction at any point between the two endpoints. The average speed on the trip may be only 50 mph, whereas the instantaneous speed on the interstate at a given moment may be 65 mph. Whether the car can be stopped in time to avoid an accident https://accounting-services.net/uncollectible-accounts-expense/ depends on its instantaneous speed, not its average speed. There are important differences between the speed of a car during a trip and the speed of a chemical reaction, however. The speed of a car may vary unpredictably over the length of a trip, and the initial part of a trip is often one of the slowest.
• The rate of reaction is used to know how long a mixture need to sit before the next step of an experiment can happen.
• In that time frame, Company A paid yearly dividends of \$1 per share.
• You could instead monitor the reaction progress by measuring the volume of the balloon over time, so the rate would be measured in cm3/s.
• The Internal Rate of Return (IRR) is the annual rate of growth that an investment or project generates over time.
• When rates are simplified, the units remain in the numerator and denominator.
In other words, the rate of return is the gain (or loss) compared to the cost of an initial investment, typically expressed in the form of a percentage. When the ROR is positive, it is considered a gain, and when the ROR is negative, it reflects a loss on the investment. The instantaneous rate of a reaction is the reaction rate at any given point in time. As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate.
You could instead monitor the reaction progress by measuring the volume of the balloon over time, so the rate would be measured in cm3/s. Since the volume of the balloon should go up over time, the rate of reaction should have a positive slope. Sometimes we buy common household items ‘in bulk’, where several items are packaged together and sold for one price. To compare the prices of different sized packages, we need to find the unit price.
To calculate rate of reaction from a graph, the general formula change in concentration/change in time is used. To find the average rate, find the change in concentration/change in time from the beginning to the end of the reaction. In this lesson we defined the meaning of a rate and its units. By measuring the amount of reactant depleted or the amount of product formed over time, and plotting this data, we can compute the average rate of a chemical reaction by using the two endpoints in the data.<|endoftext|>
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# What is 51/177 as a decimal?
## Solution and how to convert 51 / 177 into a decimal
51 / 177 = 0.288
51/177 converted into 0.288 begins with understanding long division and which variation brings more clarity to a situation. Both represent numbers between integers, in some cases defining portions of whole numbers But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
## 51/177 is 51 divided by 177
Converting fractions to decimals is as simple as long division. 51 is being divided by 177. For some, this could be mental math. For others, we should set the equation. The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. We use this as our equation: numerator(51) / denominator (177) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. Here's 51/177 as our equation:
### Numerator: 51
• Numerators are the parts to the equation, represented above the fraction bar or vinculum. 51 is one of the largest two-digit numbers you'll have to convert. 51 is an odd number so it might be harder to convert without a calculator. Values closer to one-hundred make converting to fractions more complex. Now let's explore the denominator of the fraction.
### Denominator: 177
• Denominators differ from numerators because they represent the total number of parts which can be found below the vinculum. 177 is one of the largest two-digit numbers to deal with. But 177 is an odd number. Having an odd denominator like 177 could sometimes be more difficult. Overall, two-digit denominators are no problem with long division. So grab a pen and pencil. Let's convert 51/177 by hand.
## Converting 51/177 to 0.288
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 177 \enclose{longdiv}{ 51 }$$
We will be using the left-to-right method of calculation. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 177 \enclose{longdiv}{ 51.0 }$$
Because 177 into 51 will equal less than one, we can’t divide less than a whole number. So that means we must add a decimal point and extend our equation with a zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 177 into 51 + 0 or 510.
### Step 3: Solve for how many whole groups you can divide 177 into 510
$$\require{enclose} 00.2 \\ 177 \enclose{longdiv}{ 51.0 }$$
Since we've extended our equation we can now divide our numbers, 177 into 510 (remember, we inserted a decimal point into our equation so we we're not accidentally increasing our solution) Multiply by the left of our equation (177) to get the first number in our solution.
### Step 4: Subtract the remainder
$$\require{enclose} 00.2 \\ 177 \enclose{longdiv}{ 51.0 } \\ \underline{ 354 \phantom{00} } \\ 156 \phantom{0}$$
If you don't have a remainder, congrats! You've solved the problem and converted 51/177 into 0.288 If you still have a remainder, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 51/177 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. Same goes for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But they all represent how numbers show us value in the real world. Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 51/177 into a decimal
Pay & Salary - Anything to do with finance or salary will leverage decimal format. If you look at your pay check, you will see your labor is worth $20.28 per hour and not$20 and 51/177.
### When to convert 0.288 to 51/177 as a fraction
Distance - Any type of travel, running, walking will leverage fractions. Distance is usually measured by the quarter mile and car travel is usually spoken the same.
### Practice Decimal Conversion with your Classroom
• If 51/177 = 0.288 what would it be as a percentage?
• What is 1 + 51/177 in decimal form?
• What is 1 - 51/177 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.288 + 1/2?<|endoftext|>
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Most teachers welcome student participation in class discussions. When the school year starts, only one or two brave students seem to have the confidence to volunteer to speak. After a few weeks when more students want to participate, how does that once brave student who talks too much become so annoying? Beep. Beep. Beep. Back it up and start over.
STRATEGY-anticipate the problem before it happens
Prior to that first classroom discussion, ask students if they know any details about what makes a good discussion. In small groups, ask students to think of the best and worst discussions they remember and write down the reasons why it was the best or worst. Then each group shares their answers while the teacher writes results on the board with extra marks beside repeated characteristics. At the end of this exchange, a few informal rules can be developed based on student input. Most likely, one person who tries to take over or talks too much will be in the examples of a bad discussion. If it isn’t, then the teacher can contribute the idea.
STRATEGY-test the first strategy with fun activity
A strategy that appears to be more fun than serious should help keep Mr. Bossy from dominating the class discussion without a verbal scolding. This activity will help determine the student’s extensive talking as a harmless habit or something more serious.
Before you begin place student desks or chairs in a circle, then hold up a tennis ball or a similar object for the class to view. Explain that you will moderate. Go over brief rules. For example, students may not make negative personal comments about other students or teachers, stay on topic, no profanity and the only person allowed to talk is the one holding the tennis ball! After the initial comments and a beginning statement, the teacher tosses the ball to a student.
After a few exchanges, if Mr. Bossy hasn’t raised his hand, toss the ball to him. Allow him the same time as the other speakers and when he’s in the middle of his filibuster, BEEP. The buzzer, horn, whistle or your sound effect goes off. Motion for the ball. This is that defining moment most students were anticipating. If MB tosses the ball back with minimal grumbling, let it go. Continue with the process. If he tosses the ball back and whines about how it’s not fair or a similar negative response, then you need to remind him of the rules determined by the entire class. Ask the other students to shout out the rule. Then reinforce the fairness of the rule and restate its importance in a good discussion. Most of the time, the issue is resolved and the group continues. Problem solved. Spend some extra time with the student for extra support.
Under the unfortunate circumstance that Mr. Bossy can’t handle giving up the spotlight, then you should already have plan B prepared and ready to roll. First, keep in mind one of the objectives was to address this very issue without using a confrontational manner. Try to use humor; try to keep it positive. If Mr. Bossy’s actions exhibit completely unacceptable behavior such as hurling the ball at someone’s head or using the special ‘F’ word reserved for such instances, then the teacher must quickly and effectively remove the student from the room. Consequences for violating classroom rules must be enforced as well as additional consequences for his poor choices during the discussion. If possible diffuse the tension with humor. “He should try out for the baseball team,” or a similar comment might work. Do not make negative comments about the student or dwell on what happened. Simply restate the rule and how it was violated. Then regroup and move on. All students are weighing your reaction. Handle it professionally.
Variations of this activity can continue for the discussion format after the rules become familiar. Students should be able to handle throwing the ball to another student instead of back to you. They will get to the point that they don’t need a prompt to give someone else a turn. This strategy allows all students to participate in a discussion and keeps one person from dominating the discussion with a simple method using a tennis ball.
STRATEGY- Other techniques for participation
Assign roles for 2-3 students per groups to conduct discussion within that group with each member serving a certain role. After everyone has contributed, discuss results.
Another way to keep participation opportunities equal is a similar activity. Students speak at a podium with a timer buzzing after the indicated time limit. The teacher gives each student a card or another object used like a ticket to hand in when the student comes up to speak. Once the card is handed in, that student is done speaking.
If the need arises, use participation points for credit or non-credit to enforce no talking from those who don’t have the floor or to encourage more participation from quieter students. Depending on the particular class, there are many ways to tweak these activities to fit a variety of group dynamics.
STRATEGY-Make the student important
Take the perpetrator under your advisement. Remind him that you both want the same thing. Allow him to act as your personal assistant or classroom aide the next time there is a class discussion. For example, tell him he already knows how to speak to the group and you need him to take the role of recorder. Remove him completely from the next discussion. Following the discussion, point out strengths and weaknesses of the activity and listen to student feedback. Ask MB to stay a few minutes to organize notes. See if he noticed or will admit to a smoother flow in the discussion. Continue to give him extra support and attention and allow him back into the discussion group. Hopefully, your relationship has improved to the point that more respect and less insecurity exist for MB. The extra attention might provide the turning point in changing his attitude. If he can see the situation from another person’s perspective, you helped him overcome a potential roadblock in his education.
There are many useful strategies teachers can use to make sure the same students don’t dominate class discussions. Many of them involve common sense. Most importantly, avoid being negative or talking to the student in front of the class. If you have a student with the urge to dominate class discussions, have him sign up for the debate team! Try some activities, role playing, assigning roles in small groups or listening to feedback as ways to address appropriate student participation.<|endoftext|>
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OLFACTORY SENSE ANATOMYThe olfactory portion of the cerebral cortex receives the necessary information to determine and discern smell via olfactory receptors. These are the dendritic ends that belong to the olfactory nerve and respond to chemical stimuli. This information is then sent directly to the brain for interpretation.
Nature has determined that the human olfactory sense is the least important sense of the human body. Nature determined this when she created much stronger olfactory senses in other creatures, and enabled them to find food, water, and shelter through the act of sniffing.
OLFACTORY SENSE FACTSHumans rely on sight, sound, and dexterity more than they rely on smell for basic survival skills. The human sense of smell detects odor, but does not discern its actual intensity.
Accommodation occurs quickly when considering the sense of smell. Olfaction and gestation (the sense of taste) are rather similar to each other as they each need the chemical stimuli to dissolve in order for them to react. This, by definition, categorizes them as chemoreceptors.
OLFACTORY SENSE DIAGRAM
Image: Olfactory Sense
The nasal mucosa rests within the roof of the nasal cavity. It can be found along both sides of the nasal septum. This is where the olfactory cells are located. The glandular goblet cells which surround the olfactory cells help to provide the necessary moisture for effective olfactory cell function. In between the supporting columnar cells there are the cell bodies of the bipolar olfactory cells. Olfactory hairs are the sensitive segment of the receptor cell, which is created by the dendrite ends of the free ends of each olfactory cell. These ends are unmyelinated and are quite responsive each time airborne molecules make their way into the nasal cavity.
OLFACTORY SENSE DEFINITIONThere are in fact, numerous neural segments which in turn create the sensory pathway of olfaction. It begins with the unmyelinated axons of the olfactory cells, which conjoin to create the olfactory nerves. The olfactory nerves travel along, through the foramina of the cribriform plate where they finally come to and end in the olfactory bulbs, which are created by grey and white matter.
On either side of the crista galli of the ethmoid bone, the olfactory bulbs can be located just beneath the frontal lobes of the cerebrum. Neurons which belong to the olfactory nerves synapse with dendrites which are just beginning to form the beginning of the olfactory tract. This happens inside the olfactory bulbs. The olfactory tract is responsible for directed the sensory impulses to the cerebral cortex. Within the cerebral cortex, the impulses can then be discerned as smell, aroma, or scent.
OLFACTORY SENSE FUNCTIONSOlfactory senses can be heightened to be able to discern thousands of distinct smells. People who work as perfume testers are trained to understand their fine sense of olfaction. This varies from the senses, like the sense of taste, which can only be distinguished as four basic modalities. There is still a great deal to be studied and learned in regards to the molecular basis of olfaction. What is known however, is that it only takes one odorant molecule to enter the nasal cavity for an olfactory receptor to respond.<|endoftext|>
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The arctic tundra biome is characterized by low-growing vegetation composed of low shrubs, sedges, grasses, forbs, lichens and mosses.
As this map illustrates, the continental portion of the Arctic tundra occupies a thin strip of land between the Arctic Ocean and the boreal forest. Within this strip of land, differences in average July air temperatures lead to important variations in the total amount of summer warmth available for growth. This results in major differences in vegetation that help to distinguish five distinct bioclimate subzones, as marked by different color schemes on this map.
The tundra biome is extremely vulnerable to climate warming due to three of its distinct features: 1) the strong climatic influence of the nearby sea, 2) narrow bioclimate zone associated with the coastline, and 3) extensive lowland plains near most of the Arctic coast.
Click here to read more from the “Terrestrial Ecosystems” chapter of the ABA.<|endoftext|>
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What is the social and/or ecological challenge?
In Guatemala, the worst human health and environmental problems stem from housing conditions. The country's housing deficit is more than 1.5 million units, affecting close to eight million people (50% of the population) and causing massive human suffering and environmental degradation, on several fronts. The above figure includes people living in inadequate dwellings without one or more basic services (water, energy and sanitation), in unsanitary conditions and with a harmful, wasteful impact on the environment. Socially, this is one of the main barriers to human development. Guatemala has extensive forest cover and ample water resources, although both are decreasing dramatically and are at great risk. Most people cook with firewood in open fires, causing severe deforestation, which in turn depletes aquifers while respiratory ailments from smoke inhalation have become the leading cause of death. Many people dig unregulated wells and defecate in open-pit latrines, and most piped sewage in the country is discharged untreated into rivers and waterways. Over 95% of rivers are now polluted. Deforestation, water pollution and lack of sanitation are all interrelated, and most households are part of the problem. Much of it can be fixed with proper housing infrastructure that uses natural resources in clean and sustainable ways. There are ample historical and economic reasons for the current state of affairs. Colonial-era governments ignored the development needs of rural areas and left the indigenous populations to fend for themselves, which explains the vast, underdeveloped regions where most people live today. Culturally, there is a tendency to view nature as a provider of free and inexhaustible resources and a public waste dump, with no concern for limits or best practices. This attitude translates into individual and collective behaviours that degrade shared natural resources. Culturally and economically appropriate technological solutions are the best approach to overcoming these obstacles.
The social entrepreneurial approach
CASSA provides sustainable social housing with clean water, clean energy and sanitation. The organisation helps its customers acquire vital services, save money and protect the environment, specifically Guatemala's forests and water bodies. The financial opportunity for CASSA is huge. With some 500 000 households able to afford a low monthly payment on a home, and houses selling for USD 10 000 on average, Guatemala is a USD 5 billion-plus market for CASSA's product alone – never mind the enormous positive externalities and the value of services to save and reclaim ecosystems, which would also amount to billions of US dollars. CASSA customers benefit substantially from their homes – economically, socially, health-wise and psychologically. Economically, monthly savings average USD 50 or more a month, for life, which for CASSA's target group is significant. Socially, many CASSA customers previously without water service or sanitation services benefit from the mobility and integration that comes with having basic, modern infrastructure. CASSA homes eliminate most of the disease vectors that are among the leading causes of death in Guatemala (indoor smoke, lack of safe drinking water). Then, there are the psychological benefits of living in a renewable, safe and healthy home, and they are profound. Our target customers – whether rural or urban – are at the bottom of the population pyramid. They lack a home of their own and live in cramped conditions with extended family, or they rent a place or live in makeshift and inadequate structures that lack basic services. CASSA's target customers make their living in every type of trade. They include agricultural workers, teachers, public servants and micro-entrepreneurs, to name a few. The average income per household (with more than one earner and on average five people) is between USD 300-600. CASSA has established key partnerships with financial institutions, technology providers and customers. Each customer becomes a loyal CASSA ambassador, allowing live tours of their homes and participating in marketing activities (see here and here). On the financial side, CASSA has an agreement with Oikocredit, an international cooperative, to provide customers with long-term micro-mortgages, eliminating a key obstacle for customers to access capital. CASSA also maintains partnerships with all its providers of water filters, water pumps, solar panels, smokeless stoves, and building materials, ensuring that CASSA can deliver its product at all times. Success will depend on the strength of CASSA's organisation. That includes human capital, robust marketing channels, strong partnerships and efficient construction processes. There is ample evidence of need and demand for CASSA's product. To take it to scale, CASSA, having demonstrated the efficacy and impact of its product, now needs additional capital to hire more personnel and launch a wide-ranging and targeted marketing campaign.
Goals and expected impact
We aim to transform the way social housing is built in such a way that the entire market begins integrating renewable and clean water and energy systems into each house. Long term, we want to build 75 000 new homes and retrofit 50 000, which would directly benefit 700 000 customers whose lives will be transformed, create 50 000 BoP jobs, and indirectly benefit millions of people as other market participants replicate our model. Economically, we will create average monthly savings of USD 50 per household (on saved electricity, firewood and water bills), which at 75 000 homes equals USD 3.75 million monthly savings across our projected customer base. The environmental benefits are equally large and probably multiples of the economic savings, since precious ecosystems, waterways will be preserved. Deforestation will halt in the regions where our homes are built, as will pollution of the water table and local streams. We will reduce carbon emissions by 700 000 tonnes per annum.<|endoftext|>
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# Match the geometrical shapes which form the object.List $1$List $2$Ice-cream coneCylinder + CylinderIglooCylinderLampCone + SpherePoleHemisphere + Cylinder
Last updated date: 24th Jul 2024
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Hint: We have studied shapes in geometry, let us recollect them. Basic 2D-Shapes are Square, Rectangle, Triangle and Circle and the basic 3D-Shapes are Cube, Cuboid, Cone, Sphere and Cylinder. We also have some other 3D-Shapes they are, Hemisphere and Conical frustum. Hemisphere is nothing but the half sphere and Conical frustum is half of the cone when its upper part is removed.
Complete step-by-step solution:
Our aim is to match the object to its relevant 3D-shapes. First let us draw the 3D-shapes that have been given in the option.
The given 3D-shapes are Cylinder, Cone, Sphere and Hemisphere.
Now let us draw the given objects using these 3D-shapes. First let’s draw the ice-cream cone using these 3D-shapes.
Thus, ice cream cones can be drawn using a cone and a sphere that is Cone + Sphere.
Now let us draw the igloo using these 3D-shapes.
Thus, igloo can be drawn using a hemisphere and a cylinder that is Hemisphere + Cylinder.
Let us now draw the lamp using these 3D-shapes.
Thus, a light lamp can be drawn using a cylinder alone.
Now let us draw the pole using these 3D-shapes.
Thus, a pole can be drawn using two cylinders that are Cylinder + Cylinder.
Hence, we have found the shapes of the objects given, now let us match them with the respective shapes.
Ice-cream cone = Cone + Sphere
Igloo = Hemisphere + Cylinder
Lamp = Cylinder
Pole = Cylinder + Cylinder
Thus, we have matched all the objects to their respective 3D-shapes.
Note: Any objects can be represented by using 2D or 3D shapes. We need to know all the basic 2D shapes and 3D shapes. We have studied shapes in smaller classes. The thing is that we need to have some basic knowledge of which shape can be attached perfectly to which shape. Even the lamp can be drawn using a frustum and cylinder, but here there is no other option thus we have drawn that kind of lamp. It all depends upon the option given to us.<|endoftext|>
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# Acceleration
After the air resistance becomes large enough to balance out a skydiver’s weight, they will have no net force. From Newton's First Law we already know that an object’s inertia prevents a change in velocity unless it experience a net force, so from that point when the forces are balanced and onward, the skydiver continues at a constant velocity until they open their parachute.
During the initial part of a skydive, before the drag force is large enough to balance out the weight, there is a net force so their velocity changes. The rate at which the velocity changes is known as the acceleration. Note that students often confuse velocity and acceleration because they are both rates of change, so to be specific: velocity defines the rate at which the position is changing and acceleration defines the rate at which the velocity is changing. We can calculate the average acceleration (a) during a certain time interval (Δt) by subtracting the initial velocity (vi) from the final velocity (vf) to get the change in velocity (Δv) and then dividing by the time interval (Δt):
(1)
### Everyday Example
Let’s calculate the average acceleration during the roughly 2 seconds it takes a parachute to fully open and slow a skydiver from 120 MPH to 6.0 MPH. First let’s remember that the skydiver is moving in our negative direction so the initial and final velocities should be negative. Also, lets convert to meters per second: and .
Starting with our definition of acceleration:
Inserting our values:
The two negatives in front of the 54 m/s make a positive, and then we calculate a value.
We now get a chance to see that the units of acceleration are m/s/s or equivalently m/s2
# Acceleration Direction
The direction of acceleration depends on the direction of the change in velocity. If the velocity becomes more negative, then acceleration must be negative. This is the case for our skydiver during the first part of the jump; their speed is increasing in the negative direction, so their velocity is becoming more negative and therefore acceleration is negative. Conversely, if an object moves in the negative direction, but slows down, the acceleration is positive, even though the velocity is still negative! This was the case for our skydiver just after opening their parachute, when they still moved downward, but were slowing down. Slowing down in the negative direction means the velocity is becoming less negative, so the acceleration must be positive. All of the possible combinations of velocity direction and speed change and the resulting acceleration are summarized in the following chart:
Initial direction of motion ( initial velocity direction) Speed change Direction of Acceleration positive speeding up positive positive slowing down negative negative speeding up negative negative slowing down positive
definition<|endoftext|>
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The gas turbine is one of the most widely used forms of propulsion systems for modern aircraft engines. The engine’s core—defined as the compressor, burner, and turbine—is also known as the gas generator, since the output is hot exhaust gas. The compressor and turbine are defined as the turbomachinery, where the energy is added or extracted from the continuous flow by the dynamic and aerodynamic action of rotating blades.
Common Parts of a Turbine Engine
The inlet of the engine brings “free stream air” into the engine. The focus of an inlet is to decelerate the incoming air and convert its kinetic energy into static pressure.
• Subsonic inlets: Subsonic aircraft do not exceed the speed of sound. One can maximize the pressure rise by having either a longer diffuser or a higher diffuser divergence angle (diffuser area ratio).
The flow pattern for a subsonic inlet is divided into external (outside/upstream) and internal segments. External acceleration occurs for low-speed high thrust operation (i.e., takeoff conditions), which raises the inlet velocity and lowers the inlet pressure. Hence, the inlet area is designed to minimize external acceleration during takeoff so that external deceleration occurs during cruising conditions. On a typical subsonic inlet, the surface of the inlet is a continuous smooth curve that has some thickness from inside to outside. The inlet lip or the highlight, the most upstream portion of the inlet, is relatively thick.
• Supersonic inlets: Supersonic aircraft are still required to slow down the flow to subsonic speeds before the air reaches the compressor. The air flow has a Mach number between 0.4-0.7 when it reaches the engine face. Flow diffusion from supersonic to subsonic flow, also known as ram recovery, involves shocks. A normal shock inlet is the simplest supersonic diffuser. The shocks, which have a narrow inlet lip, are used for single normal shock (90° perpendicular to the flow) for Mach values less than 1.6.
Oblique shock inlets achieve higher total pressure recovery. Supersonic flow deceleration is achieved over a series of oblique shocks (a specific angle to the flow) followed by a weak normal shock. In an oblique shock, the supersonic flow is turned into itself; as the number of oblique shocks increase, the shock losses decrease, especially at high Mach numbers.
An axisymmetric external compression inlet is a cone-shaped diffuser that creates a conical shock. Due to the flow over the cone being inherently three-dimensional, the flow field between the shock and cone is no longer uniform. The effect results in a weaker shock wave than for a wedge of the same angle.
Compressors are used to increase the air pressure before it enters the combustor.
• Centrifugal compressors: These compressors were implemented in the first jet engines, and are still used in turbojets and turboshaft engines. They turn the air flow perpendicular to the axis of rotation. The rotating impeller moves the air, which is collected in the scroll or volute. There may be a diffuser between the impeller and the volute.
• Axial compressors: Instead of a perpendicular flow, axial compressors flow the air parallel to the axis of rotation. The compressor consists of several rows of rotors and stators; which are a series of air foils. The rotors are connected to the central shaft and rotate at high speeds, imparting angular momentum to the fluid. Stators are fixed, which connect to the outer casing, increase the pressure while keeping the flow from spiraling around the axis by returning it to the parallel axis (acting as diffusers). Blade length and the annulus area decrease throughout the length of the compressor, reducing the flow area. This compensates for the increase in fluid density as it is compressed.
The burner or combustion chamber sits between the compressor and turbine arranged like an annulus. Here the fuel is mixed with high-pressure air and burned to create high temperature exhaust gas to turn the power turbine and produce thrust. A few of the desired properties of burners are to achieve complete combustion with minimum exhaust emissions, low total pressure loss, low heat loss from walls, and effective cooling. However, many of these properties compete with one another; hence, an optimal burner design is one of compromise.
• Can-annular combustors: Consisting of a series of cylindrical burners arranged around a common annulus, can-annular combustor chambers function independently of each other. At the entrance of each chamber is a diffuser that can reduce the velocity from a typical compressor outlet (100-150 m/s) to the bulk flow average velocity (20-30 m/s) in the combustion zone. It delivers the air to the combustion zone as a stable and uniformed flow field. This is an older design method for a burner.
• Annular combustors: A more modern design is an annular combustor. It is a single burner having an annular cross-section that supplies gas to the turbine. The combustion zone itself occupies an annular space. The improved combustion zone provides uniformity, design simplicity, reduced linear surface area, and a shorter system length.
A turbine is like a compressor in that it consists of several rows of rotors and stators. The turbine stage begins with a stationary blade row called the nozzle guide vane, followed by a rotating blade row. The turbine converts thermal energy to kinetic energy by expanding through nozzles, then into rotational mechanical energy in a spinning rotor.
The flow in a turbine is dominated by favorable pressure gradients. Pressure changes can be quite large, and the boundary layers in a turbine are less susceptible to stall when compared to a compressor. Cooling of turbines is a major challenge; thus, they are designed to handle high temperature and corrosive environments.
The function of the nozzle is to convert the thermal energy into kinetic energy in order to obtain a high exhaust velocity. The nozzle thrust, or gross thrust, is comprised of the momentum and pressure thrust. A maximized gross thrust is when the nozzle is fully expanded or the ambient pressure equals the exhaust pressure.
• Subsonic nozzle: To accelerate a subsonic flow, the cross-section of a duct must decrease in the stream-wise direction. When the duct ends at the smallest cross-section, the result is a converging nozzle. The pressure at the exit of the nozzle is lower than the atmospheric pressure. As a result, the flow accelerates or expands to the atmospheric or local exit pressure. The higher the aircraft flies, the more velocity increases accordingly to the lower ambient atmospheric pressure. A limit is reached when the jet discharges at sonic velocity and the nozzle is said to be choked. Once the choked condition is realized, the nozzle mass flow rate is at its maximum and the conditions remain unchanged regardless of the decreases in ambient pressure. Hence, a converging nozzle can never produce a supersonic flow.
• Supersonic nozzle: For high exhaust velocities required for supersonic flight, a converging-diverging (CD) nozzle is used to create supersonic exhaust velocity. The CD nozzle construction consists of a convergent duct followed by a divergent duct. The cross-sectional area increase in the CD nozzle accelerates supersonic flow. A supersonic or CD nozzle requires a large pressure difference to accelerate the gas to a supersonic speed at the throat and further create supersonic flow in the diverging section of the CD. A significant pressure difference can be created by reducing the back pressure or the exit pressure of the surrounding downstream.
Adjustable nozzles allow a supersonic aircraft to match the varying conditions of ambient pressure and engine power settings for supersonic flight. And altitude adaptive nozzles can change the shape of the nozzle lip angle for optimal performance.
A problem arises when the nozzle is over-expanded or under-expanded. In an under-expanded condition, the pressure falls across the expansion waves and the exhaust plume expands past the nozzle exit, reducing efficiency in high altitudes. For over-expanded nozzles, the pressure rises across the oblique shock waves and a mixture of sub/supersonic flow. The exhaust plume is pinched by high ambient-air pressure, reducing its efficiency in low altitudes. Over-expansion can produce regions of complex wave patterns in the plume, which create a white/yellow luminescent glow as the low exhaust gas pressure tries to match the high ambient pressure.
The turbojet is the simplest type of gas turbine. Large amounts of surrounding air are pulled into the engine inlet due to the compressor. At the rear of the inlet, air enters the compressor. Pressure increases as the air passes the rows of blades. At the exit of the compressor section, the air pressure is higher than the free stream. In the burner section, fuel is combined with the air and ignited. The hot exhaust comes mostly from the surrounding air and passes through the turbine once it leaves the burner. The turbine extracts energy from the hot airflow by making the blades spin in the flow. In a jet engine, the energy extracted by the turbine turns the compressor by linking it and the turbine to the central shaft. The rest of the hot exhaust is used to produce thrust by increasing its velocity through the nozzle. Since the exit velocity is greater than the free stream, thrust is created. Very little fuel is added to the stream, so the exit mass flow is nearly equal to the free-stream mass flow.
The two main parts for a turboprop propulsion system are the core engine and the propeller. The core engine is very similar to a turbojet except for the way it handles the energy from the exhaust. Instead of expanding the hot exhaust through the nozzle to produce thrust, the turboprop uses most of the energy from the exhaust to turn the turbine. An additional turbine stage may be connected to the drive shaft, which in turn is connected to the gearbox. The propeller connects to the gearbox, which produces most of the thrust.
The thrust produced from the exhaust velocity is low because most of the energy from the core exhaust is used to turn the drive shaft. Turboprop (and turbofan) engines usually have a two-spool engine where a separate turbine and shaft powers the fan and gear box, respectively. Turboprops are used only for low-speed aircraft like cargo planes. Propellers become less efficient as aircraft speed increases.
Modern airlines use turbofan engines to propel their airplanes through the air. This is due to their high thrust and fuel efficiency. The turbofan engine is the most modern variation of the basic gas turbine. In the turbofan, two fans surround the core engine. One fan is in the front of the core engine and the other is located in the rear. The fan and fan turbine are connected to an additional fan shaft. The fan shaft passes through the core shaft in a two-spool engine arrangement. To achieve higher efficiency, some engines have additional spools.
The turbofan operates by capturing incoming air in the inlet. Some of the air passes through the fan, into the core compressor, and then the burner. The heat exhaust passes the through the core, fan turbines, and out the nozzle. This process mirrors that of a turbojet. The rest of the incoming air is redirected around the engine after it passes the fan. The air passing through the fan has a slightly higher velocity increased from the free stream.
The ratio of the air redirected around the engine versus the air that passes through the core is known as the bypass ratio. Low-bypass-ratio turbofans are more fuel efficient than the basic turbojet. A turbofan generates more thrust for nearly an equal amount of fuel used by the core because the fuel flow rate is changed by slightly when adding the fan. As a result, the turbofan offers high fuel efficiency.
The air passed through the core as well as the air passing around the engine comprise the thrust. Due to the fact that the inlet encloses the front fan and has many blades, it can operate efficiently at higher speeds than a simple propeller.
Afterburning Turbojet Engine
Afterburners are used in supersonic aircraft like the Concorde, and are turned off after achieving cruising velocity. Many modern fighter planes use a low-bypass-ratio turbofan equipped with afterburners for efficient cruising conditions and to produce high thrusts for dogfights, and on turbojets to fly at supersonic speeds, overcoming the sharp rise in drag near the speed of sound. The afterburner injects fuel directly into the hot exhaust. The nozzle of the basic turbojet becomes extended and a ring of flame holders is installed after the nozzle. Additional fuel is injected through the hoops and into the stream of the hot exhaust. The burning fuel produces extra thrust, but at an inefficient rate.
The burning fuel offers a simple mechanical way to augment thrust, but at an inefficient rate. The thrust calculation is the same as a normal turbojet, except that the exit thrust value is the thrust exiting the afterburner.
FTurbojetor Afterburning Turbojet =á¹e â Ve – á¹FS â VFS
FTurboprop = á¹FS â (VPe – VFS) + á¹e â (Ve – VPe)
FTurbofan = á¹e â Ve – á¹FS â VFS + bpr â á¹c â Vf
á¹FS = mass flow rate of the free stream of air
á¹e = mass flow rate of air at the exit of the core
á¹c = mass flow rate of the hot exhaust passing through the core
á¹f = mass flow rate of the fan flow or bypass flow
Vf = velocity of air at the exit of the fan
Ve = velocity of air at the exit of the core
VPe = velocity of air at the exit of the propeller
VFS = velocity of the free stream of air
Ve = velocity of air at the exit of the core
bpr = bypass ratio which equals á¹f/á¹c
Looking for parts? Go to SourceESB.<|endoftext|>
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Just a little bit of help from evolution allows invasive species to disperse farther and faster, according to Rice University scientists.
Rice graduate student Brad Ochocki and ecologist Tom Miller, using the bean beetle as their model, found that generations born on the leading edge of an invasion are far more able to push on than those born further back. That has implications for agriculture and natural-resource managers who struggle to predict and prepare for the spread of invasive species.
The team's results were reported today in Nature Communications.
The researchers chose to work with common bean beetles for their relative ease of maintenance and their rapid reproduction, Miller said, noting that the species spawns a new generation about once a month.
The researchers bred 10 generations of the insects in their Rice lab and found that beetles at the leading edge of an invasion, where food is more plentiful and competition less fierce, produced more offspring with the right genetic traits to carry on in the same pioneering spirit.
Ochocki called this the "Olympic Village Effect," known by scientists as the assumption that, were Olympians to reproduce during the games, their offspring would likely carry forward evolutionary advantages for strength and agility.
That notion has real-world relevance for those who track invasive plants and animals. "We want to understand this process of spread because it's happening more and more through biological invasions, especially with climate change," Miller said. "Ecologists are under pressure to predict something about this process.
"We obviously want to understand what's happening outside, but it's impossibly hard because we can't replicate the spread of cane toads across Australia, gypsy moths across the Northeast or zebra mussels in the Great Lakes, for example," he said. "To understand the distribution of variability, we created simple landscapes in the laboratory. We realized we could distill this very complicated process to simple ingredients."
Ochocki, who studied biomedical engineering and mechanical engineering as an undergraduate, designed and built game board-like habitats stocked with black-eyed peas. The beetles lay eggs on the surface, and the resulting larvae burrow in and feed on the peas as they mature.
He started each colony with less than 100 insects, evenly split between males and females. Tunnels between the petri dishes allowed the most pioneering among them to spread from one to the next in search of new territory. Ochocki estimates the lab manipulated and observed about 293,000 beetles over 10 months of gathering data.
The researchers ran two sets of experiments: In one, colonies were left to their own devices to expand with the expectation that the insects on the leading edge of a colony would be more prone to advance. Control experiments eliminated this "spatial sorting" by periodically shuffling colony members between different dishes.
The results showed, first, that spatially sorted colonies -- those where pioneers were not reshuffled -- tended to disperse an average of 8.9 percent farther than control groups whose members were shuffled. Second, the experiments demonstrated that spatially sorted populations were more inclined to spread out, with the variance in their patterns of dispersal -- where and how far they traveled -- increasing 41-fold.
Those results are not good news for those who study invasive species or disease dynamics, Miller said.
"Farmers and other people who have an interest in maintaining a natural resource are good at detecting initial outbreaks," he said. "When they detect a new corn pest in a field, they want to know how far might it get the next season and how far ahead of a wave they should warn growers to spray for this new bug.
"What Brad found is that even in this controlled laboratory setting, which is the simplest of all possible worlds, the process of evolution itself can generate uncertainty and variability in how populations move," Miller said. He said that uncertainty means scientists are unlikely to ever perfect the process of predicting invasion dynamics. "That is kind of sobering."
But including the effects of evolution can better define the range of possibilities, Ochocki said. "We suggest in the paper that accounting for evolution won't necessarily enable you to have an accurate point prediction, but it will give you a window of possible scenarios," he said.
Miller and Ochocki said the rapid turnover of beetle generations left little time for genetic mutation to influence spread, but they said gene surfing was a likely contributor. In gene surfing, traits present at the leading edge of an invasion can ride the wave of expansion to become abundant in newly founded populations. In this case, the random genes brought to the edge of the invasion by dispersing beetles likely generated the increase in variance that Ochocki and Miller observed. Shuffling of beetles during invasion appeared to dampen the effect.
Ochocki and Miller aren't the only researchers using insects to investigate invasion. A group at the University of Colorado, Boulder, led by Christopher Weiss-Lehman, performed similar experiments using flour beetles that led to essentially the same results. They were published along with the Rice paper by Nature Communications.
"It's a very nice demonstration that our results and theirs are not flukes," said Miller, who learned of the work by his Colorado counterparts at a meeting several years ago. "That's why the journal was interested in publishing them side by side; they said in their review that they rarely get an opportunity to independently validate results like this."
Miller is the James and Deborah Godwin Assistant Professor of Ecology and Evolutionary Biology at Rice.
The National Science Foundation supported the research. Computing resources were supplied by the NSF-supported DAVinCI supercomputer, which was procured in partnership with Rice's Ken Kennedy Institute for Information Technology and is administered by Rice's Center for Research Computing.
Read the abstract at http://www.
This news release can be found online at http://news.
Read the University of Colorado, Boulder news release at http://www.
Follow Rice News and Media Relations via Twitter @RiceUNews
Miller Lab: http://www.
Rice Department of BioSciences: http://biosciences.<|endoftext|>
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# Given that 2,a,b the terms of a G.P. and 2,17,a the terms of an A.P. calculate a, b.
hala718 | Certified Educator
2, a, b are terms in a geometric progression.
2, 17, a are terms in an arthimatic progression
For the A.P:
17 -2 = 15 = r
Then a = 17 + r = 17+15 = 32
==> a = 32
Now for the G.P
2, a, b
2, 32, b
32/2 = 16 = r
==> b = 32*r = 32*16 = 512
==> b = 512
neela | Student
2,a ,b are in GP............(1)
2,17 , a are in AP..........(2)
Solution:
Since 2 , a and b are in GP,
a/2 = b/a common ratio of consecutive terms in GP.
a^2 = 2b...........(3).
From (2) , 2, 17 and a are in AP.
So the common difference between the consecutive terms are equal.So 17 -2 = a-17.
17+17-2 = a .
32 = a.
a = 32. Substitute a = 32 in the relation at (3):
a^2 = 2b , 32^2 = 2b
b = 32^2/2 = 512.
Therefore a = 32 , b = 512.
I
thewriter | Student
Here I am assuming the they are the consecutive terms of the series.
Using the fact that 2,a,b are the terms of a GP. a/2=b/a or a^2=2b.
Using the fact that 2,17,a are the terms of an AP, 17=(2+a)/2 or 2+a=34 or a=32.
As we know that a^2=32^2=2b, b=32^2/2= 32*32/2=32*16=512
Therefore a=32 and b=512.
giorgiana1976 | Student
If 2,a,b are in geometric progression, then we'll apply the theorem of geometric progressions: each term of a geometric progression, beginning with the second term, is the geometric mean between the 2 joined terms:
a^2=2*b
If 2,17,a are in arithmetic progression, then the middle term is the arithmetic mean between the 2 joined terms:
17 = (a+2)/2
a+2=17*2
a=32
But a^2=2*b => 32^2 = 2*b
We'll divide by 2:
32*16 = b
b=512<|endoftext|>
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The Big 6 is a research process that has 6 steps.
Using the steps in the Big6 will guide you through your research project for ELA.
(Big 6 is a process model of how people of all ages solve an information problem.)
STEPS 1 & 2: TASK DEFINITION & INFORMATION SEEKING STRATEGIES
Ask, "What's the task? What types of information do I need?"
Think about the final product you will need to create, and consider what special requirements you need to follow!
Brainstorm keywords that will help you find information.
Brainstorm exactly what specific types of information you need.
Next, Ask, "What are all the possible sources to check on my topic? Which sources will have the best information?"
Remember that it is your responsibility to evaluate your sources carefully.
Use this checklist:<|endoftext|>
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# Average of 7 numbers is 6.What is 8th no. to be added if the average increases by 10 ?
Mar 28, 2018
See a solution process below:
#### Explanation:
The formula for the average of a set of numbers is:
$a = \frac{s}{n}$ Where
• $a$ is the average of the numbers. We know from the problem the average of the first 7 numbers is $6$
• $s$ is the sum of the set of numbers. We need to solve for this to answer the questions.
• $n$ is the count of numbers in the set of number we are averaging. We know from the problem the count of numbers in the set is $7$
Substituting and solving for $s$ we can find the some of the numbers:
$6 = \frac{s}{7}$
$\textcolor{red}{7} \times 6 = \textcolor{red}{7} \times \frac{s}{7}$
$42 = \cancel{\textcolor{red}{7}} \times \frac{s}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}}$
$42 = s$
$s = 42$
If the average is to increase by 10, then the new average would be: $6 + 10 = 16$
The count of numbers would increase to $8$
And if we call the value of the new number added in $w$, then the sum of the set of numbers is: $42 + w$
Substituting this into the formula for average and solving for $w$ gives:
$16 = \frac{42 + w}{8}$
$\textcolor{red}{8} \times 16 = \textcolor{red}{8} \times \frac{42 + w}{8}$
$128 = \cancel{\textcolor{red}{8}} \times \frac{42 + w}{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}}$
$128 = 42 + w$
$128 - \textcolor{red}{42} = 42 - \textcolor{red}{42} + w$
$86 = 0 + w$
$86 = w$
$w = 86$
The eight number added in to make an average of 16 would be 86
$a = \frac{42 + 86}{8}$
$a = \frac{128}{8}$
$a = 16$<|endoftext|>
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The human mind is an amazing device. It thinks, it anticipates, and it responds. Studying human psychology has always fascinated people since ages. How the mind reads a certain stimulus and how it registers it to form a particular response to it is an interesting thing to notice and observe. And many researches have been conducted on this very response to stimuli effect that the mind plays around with. One such aspect is the one that deals with 'reinforcement'. As per leading psychologists, the term 'reinforcement' can be said to be a certain act or stimulus which elicits a desired effect or response. Reinforcement can be divided into two types, namely, positive and negative reinforcement.
What is Negative Reinforcement?
Positive reinforcement focuses on granting a favorable stimulus in order to receive the desired response. For instance, a child is permitted to watch television, after he is done with all his homework and chores. In this example, the child is enticed with the positive stimulus of being able to watch TV. Hence, he completes all his work, which was the desired response.
Negative reinforcement is slightly different. In this, an unpleasant stimulus is withdrawn from the equation, once the desired response is generated. For example, a teenager is constantly annoying her brother to let her use his car for a trip to the mall. When he finally lets her use it, she stops annoying him. In this case, the annoying behavior was the stimulus which was withdrawn after the brother gave his car to her, which was the response she expected in the first place. There are many more ways in which negative reinforcement is used, especially in our daily lives.
The reinforcement technique is often used in schools for classroom management. Whether it is positive or negative reinforcement, school authorities and others all around the world, from families to corporates, use this technique to get the expected behavior from people. Here are some classic negative reinforcement examples for children and adults.
Example # 1: A child is told that he will be exempted from doing his chores if he helps out in serving guests at a dinner party.
In this example, the undesirable stimulus of 'doing the chores' is removed, when the child helps in hosting the dinner party. The hosting was the response that was expected of him.
Example # 2: A class is told that they will be kept back for extra classes unless they forgo fifteen minutes of their lunch break for a week.
Here, the stimulus that is taken away is the inconvenience of waiting back after school. The class sacrifices a part of their lunch break, and the syllabus gets completed on time.
Example # 3: An organization decides to declare Saturdays as an official weekly holiday, if the employees complete their targets within the remaining five days of the week.
This is one of the most commonly used negative reinforcement examples at work. With the prospect of avoiding the unpleasant experience of working on a weekend being removed, the employees will make sure that they achieve their targets by Friday. This is what the management required and wanted in the first place.
Example # 4: A woman constantly complaining to her husband about how she needs to take three buses to get to work, and when he buys her a second-hand car, she stops complaining.
In this example, when the man realizes that giving his wife a car will stop her from complaining, it is an occurrence of negative reinforcement. The negative 'complaining' was stopped when the response of getting an easier mode of conveyance was received. Needless to say, that was what the woman was expecting.
There are many more of these that you will now be able to identify and use effectively to get your work done. The technique of using positive and negative reinforcement has proved to be quite efficacious in helping to treat people with anxiety or other mental illnesses. So, do not underestimate its importance. Neither should you misuse it to have your way in everything. Use it wisely, and you will receive what you require.<|endoftext|>
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‘In the beginning, there was no Europe,’ writes Professor Norman Davies in Europe: A History. In the beginning all that existed was an unpopulated peninsula attached to the western edge of the world’s largest landmass (Asia). But after humanoid settlers arrived between 850,000 and 700,000 BC, Europe’s temperate climate and unthreatening environment would make it ripe for agricultural exploitation and the birthplace of great civilisations.
It was in Greece and Rome that the continent’s two great ancient societies arose. Greece (first emerging around 2000 BC) was renowned for its philosophers (Aristotle, Plato, Socrates) and democratic principles. Rome – boasting brilliant politicians, and writers like Cicero, Ovid and Virgil – spread its influence by military might. At its peak, the Roman Empire stretched from England to the Sahara and from Spain to Persia.
By the 4th century AD both empires were in terminal decline. Greece had been swallowed by Macedonia under Alexander the Great, then by Rome itself in AD 146. Although Roman emperors in Constantinople (İstanbul) hung on for another 1000 years, the empire’s western half fell to Germanic tribes in 476.
This marked the start of the Dark Ages in Western Europe. From 768 Charlemagne, King of the Franks, brought together much of Western Europe under his rule into what would later be known as the ‘Holy Roman Empire’. After this territory passed into the hands of Austrian Habsburgs in the 13th century, it became the continent’s dominant political power. Elsewhere, an alliance of Christian nations repeatedly sent troops to reclaim the Holy Land from Islamic control. These unsuccessful ‘Crusades’ (1096–1291) unfortunately set the stage for centuries of skirmishes with the neighbouring Ottoman Empire as it took control of Asia Minor and parts of the Balkans from 1453 onwards.
Europe’s grand reawakening also began in the mid-15th century, and the subsequent Renaissance, Reformation and French Revolution ushered in enormous social upheaval.
The Renaissance fomented mainly artistic expression and ideas. The Reformation was a question of religion. Challenging Catholic ‘corruption’ in 1517, German theologian Martin Luther established a breakaway branch of Christianity, Protestantism. Struggles between Catholics and Protestants for supremacy were behind the bloody Thirty Years War (1618–48).
Forced to pick just one defining era and philosophy to sum up modern Europe, it would be hard to pass over the Enlightenment, or the so-called ‘Age of Reason’. This was the period in the 18th century when science and human logic for the first time took supremacy over religious belief. Heavily intertwined with the rapid scientific advances of the time, it ushered in the modern age with its move away from the church and its emphasis on logic, education, individualism and liberal social values.
Whole university courses are taught on the Enlightenment, its factions (eg rationalism versus empiricism), and its pros and cons. However, two thinkers closely associated with it were the antimonarchist, anti-religious liberal Voltaire (1694–1778) and Immanuel Kant (1724–1804), who believed humans are rational and autonomous beings, so universal moral laws are possible. Kant was also intensely interested in how humans made sense of the surrounding world.
René Descartes (1596–1650), who famously declared ‘I think, therefore I am’, was one of the rationalist forerunners of the Enlightenment. Jean Jacques Rousseau (1712–78) started off as a believer but later fell out with the movement – whose main social consequence was the French Revolution.
The French Revolution in 1789 was about the populace’s attempt to wrest political power from the monarchy. But in the ensuing vacuum, plucky general Napoleon Bonaparte (1769–1821) crowned himself emperor. Napoleon’s efforts to colonise all Europe ended in defeat by the British at Waterloo in 1815, but the civil laws he introduced in France in 1804 would spread the revolutionary ideas of liberty and equality across the globe.
Having vanquished Napoleon, Britain became a major world player itself. With the invention of the steam engine, railways and factories, it unleashed the Industrial Revolution. Needing markets for goods, it and other European powers accelerated their colonisation of countries around the world, bringing new and exotic riches back to Europe.
Meanwhile the death throes of the Habsburg Empire, or the Austro-Hungarian Empire, were about to rock the continent. Serbia was accused of backing the assassination of the heir to the Austro-Hungarian throne in 1914 and the battle between the two states developed into WWI, as allies lined up on each side (Germany and the Ottoman Empire on the Austro-Hungarian side; Britain, France, Russia, Italy and the USA with Serbia).
Crippled by a huge bill for reparations imposed at the war’s end in 1918, Austria’s humbled ally, Germany, proved susceptible to politician Adolf Hitler’s nationalist rhetoric during the 1930s. Other nations watched as Nazi Germany annexed Austria and parts of Czechoslovakia, but its invasion of Poland in 1939 sparked WWII. During the final liberation of Europe in 1945, Allied troops from Britain, France, the USA and the USSR uncovered the full extent of the genocide that had occurred in Hitler’s concentration camps for Jews, Roma (Gypsies), the disabled, homosexuals, communists and other ‘degenerates’.
The Allies carved out spheres of influence, and Germany was divided to avoid its rising up again militarily. Differences in ideology between the Western powers and the communist USSR soon led to a stand-off. The USSR closed off its assigned sectors – East Germany, East Berlin and much of Eastern Europe – behind the figurative Iron Curtain. With the Stasi, Stalinist purges and more, many Eastern European citizens have appalling tales of political repression to relate from these times.
The ‘Cold War’ lasted until 1989 when the Berlin Wall fell. Germany was unified in 1990. A year later the USSR was dissolved. Czechoslovakia, Hungary, Poland, Romania, Bulgaria and Albania all grasped multiparty democracy shortly afterwards.
The downfall of communism had a terrible effect in Yugoslavia, where nationalist leaders seized the chance to stir up political unrest and war: some of the young independent nations there are still recovering. For the most part, however, the end of the Cold War has brought a sense of peace to Europe. A sense of cooperation is proving slightly trickier to locate. The EU was formed in 1957 as a trade alliance and has developed fitfully into a political entity since. At this stage, while 16 members have adopted a common currency, governments are having difficulty pushing through the European Constitution needed.
Europe can be divided into four major physical regions, running from north to south: Western Uplands, North European Plain, Central Uplands, and Alpine Mountains.
The Western Uplands, also known as the Northern Highlands, curve up the western edge of Europe and define the physical landscape of Scandinavia (Norway, Sweden, and Denmark), Finland, Iceland, Scotland, Ireland, the Brittany region of France, Spain, and Portugal.
The Western Uplands is defined by hard, ancient rock that was shaped by glaciation. Glaciation is the process of land being transformed by glaciers or ice sheets. As glaciers receded from the area, they left a number of distinct physical features, including abundant marshlands, lakes, and fjords. A fjord is a long and narrow inlet of the sea that is surrounded by high, rugged cliffs. Many of Europe’s fjords are located in Iceland and Scandinavia.
North European Plain
The North European Plain extends from the southern United Kingdom east to Russia. It includes parts of France, Belgium, the Netherlands, Germany, Denmark, Poland, the Baltic states (Estonia, Latvia, and Lithuania), and Belarus.
Most of the Great European Plain lies below 152 meters (500 feet) in elevation. It is home to many navigable rivers, including the Rhine, Weser, Elbe, Oder, and Vistula. The climate supports a wide variety of seasonal crops. These physical features allowed for early communication, travel, and agricultural development. The North European Plain remains the most densely populated region of Europe.
The Central Uplands extend east-west across central Europe and include western France and Belgium, southern Germany, the Czech Republic, and parts of northern Switzerland and Austria.
The Central Uplands are lower in altitude and less rugged than the Alpine region and are heavily wooded. Important highlands in this region include the Massif Central and the Vosges in France, the Ardennes of Belgium, the Black Forest and the Taunus in Germany, and the Ore and Sudeten in the Czech Republic. This region is sparsely populated except in the Rhine, Rhne, Elbe, and Danube river valleys.
The Alpine Mountains include ranges in the Italian and Balkan peninsulas, northern Spain, and southern France. The region includes the mountains of the Alps, Pyrenees, Apennines, Dinaric Alps, Balkans, and Carpathians.
High elevations, rugged plateau’, and steeply sloping land define the region. Europe’s highest peak, Mount Elbrus (5,642 meters/18,510 feet), is in the Caucasus mountains of Russia. The Alpine region also includes active volcanoes, such as Mount Etna and Mount Vesuvius in Italy.<|endoftext|>
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Pluto and Charon have captured our hearts and imaginations. But how did these adorably strange worlds form, and what consequences could that have on what we see now? Researcher Amy Barr Mlinar chatted with us about catastrophic collisions, subsurface oceans, and Pluto’s lack of craters.
The New Horizons spacecraft is bombarding us with a steady stream of discoveries and realizations about Pluto, Charon, and all the little moons. By far the most startling was when our first close-up look at Pluto contained no craters. The oddly-smooth surface is the hallmark of a freshly resurfaced world, one where the plains of Tombaugh Regio are less than 100 million years old. But how could that happen?
When looking at Pluto at high enough resolution to spot features just 1 kilometer (0.5 miles) in size, we’re seeing a shocking lack of craters. Image credit: NASA/JHUAPL/SwRI.
One of the ideas is that a catastrophic impact between what are now Pluto and Charon could have created a subsurface ocean. That ocean could be the driving force behind icy tectonics, an idea explored by planetary geophysicists Amy Barr Mlinar and Geoff Collins in a paper published last year. In it, they explore how an impact between two equal-sized bodies might have spawned Pluto with Charon starting a piece of shrapnel caught in orbit, and how their tidal push-pull may have formed a subsurface ocean while they were in the process of settling into tidal equilibrium.
I caught up with Barr Mlinar to explain how the new results coming home from the Pluto flyby by the New Horizons probe fit in with her theories.
How did Pluto and Charon Form?
The most popular theory for how Pluto ended up with a moon nearly 1/10th of its mass is through a catastrophic collision between two equal-sized bodies. The collision must have been a glancing blow, knocking the surface off what is now Pluto while more substantially decimating the impactor until Charon was the largest of the intact shards. Some theories look at if the smaller moons—Nix, Hydra, Styx, and Kerberos—could be remnants of debris from the same collision.
A glancing blow between two similarly-sized rocky bodies with ice shells could produce Pluto, Charon, and a debris disk possible of accreting into tiny icy moons with only minimal heating. Image credit: Robin M. Canup.
Barr Mlinar points out this didn’t have to be a huge, high-velocity collision between the proto-worlds to create the system we see today:
The current best model of the collision involves two relatively similar-sized objects hitting at near escape-velocity, which is less than 1 km/s. It is a gentle collision and is thought to heat the young Pluto by a few tens of degrees Celsius.
The object that becomes Charon is essentially a piece of the impactor that remains intact during the impact. This is consistent with the fact that Charon’s mean density is different from that of Pluto.
The timing of this collision is the first place things get interesting. If the two bodies were made of the normal mix of solar system stuff, they’d have silicate rocks with a specific quantity of radioactive material. If they collided long enough ago that they were still heated by radioactive decay at the time of collision, then it’s plausible that being heated by only a few tens of degrees would be sufficient to push the icy worlds into warm enough to create a subsurface ocean.
What Happens After the Collision?
Even if those few degrees aren’t quite enough, the next stage of evolution for the miniature system involve tweaking both dwarf planet and moon as they settled into their new orbital dance.
After the collision, both world would have been tugging on each other, their asynchronous rotations and orbits exerting force slowing the other down to match. They first stage is to become synchronous, where the smaller Charon is yanked enough to always keep its same face towards Pluto. This is equivalent to what we have right now on Earth: while the Earth turns freely so all of us on its surface experience moonrise once a day, the same side of the moon is always facing us. The near side of the moon saw just one Earthrise a long time ago. The second stage, dual synchronous state, is when the larger body falls into the same lock.
Although Barr Mlinar and her co-author Geoff Collins consider this transition from chaos to synchronous as near-instantaneous, the transition to from synchronous to dual synchronous took much longer. According to Barr Mlinar, it probably took Charon about fifty years to reach a synchronous state, but took the much longer timespan of millions of years to reach its current dual synchronous state where both Pluto and Charon are in lockstep with each other. This is part of how we know that the system-forming collision couldn’t have taken place in the recent past: it takes a long time for a system to reach the stability of circular orbits and a full tidal-lock of dual synchronicity.
How Could a Collision Create an Ocean?
One of the things we’ve seen with icy moons around the solar system is that the tidal pull of their massive gas giant parents can be enough to drive geological activity. The same can’t be true on Pluto due to the total lack of nearby giants, but there might be another way. This window of a few million years after Charon settled into gazing steadily at Pluto but when the dwarf planet had not yet mirrored the situation is when Barr Mlinar thinks an ocean could’ve been melted.
To understand why, we’re going to need to talk about mathematics on at least a conceptual level. The tidal impact of one body on another is going to depend on four things:
- The ratio of mass between parent and moon
- The distance between parent and moon
- The eccentricity of the moon’s orbit around its parent
- The difference between the spin period of the parent and the moon’s orbital period
The larger the mass ratio is, the more the parent can throw around any children-moons with impunity. Tidal effects drop off rapidly with distance incredibly quickly: for every unit of distance a moon is farther from its parent, the tides are just one-sixth as strong. The higher the eccentricity of the orbit, the more change in distance between parent and moon takes place over the course of an orbit, then the greater the tidal stress. Finally, as long as the spin period and orbital period are mismatched, then the disequilibrium state of asymmetrical tidal bulge will be enough to keep creative massive tidal pulls deforming the smaller world.
For gas giants with orbiting tiny moons, the mass ratio is 10,000:1 with the huge parent 10,000 times more massive than the insubstantial moons. That means the gas giant is going to be nearly completely unaffected by the moon, while the moon is going to be dramatically impacted every time it gets anywhere near its overwhelming parent. This means the prime component of tidal massage on the moon is going to be the orbital eccentricity of the moon. Barr Mlinar says:
For a moon orbiting a gas giant, the most important effect is its orbital eccentricity. If the moon is in an eccentric orbit, the distance between the planet and moon changes with time, and so the height of the tidal bulge changes with time.
The momentum transfer between Pluto and Charon over millions of years eventually resulted in a tidally-locked system in a dual synchronous state. Image credit: Barr and Collins.
The isn’t the situation for the Pluto-Charon system, where the mass ratio between parent and moon is far more notable. Charon is the largest moon in the solar system compared to its parent. The result is that Pluto is just ten times more massive than Charon, resulting in a nearly unbelievable low mass ratio of just 10:1.
If we look at the time after Pluto and Charon settle into a synchronous state but have not yet reached a dual synchronous state, then the distance between the worlds is small, the mass ratios nearly on-par, and the mismatch between Pluto’s spin period and Charon’s orbital period is going to be severe. The result is a perfect combination to not just subject Charon to a tidal massage, but also Pluto. In theory, this could be enough to push an initially warm and mushy Pluto over the edge into generating a subsurface ocean in a brief bloom of activity.
One possible scenario for heat flux from tidal dissipation in Pluto, with residual viscoelastic tensile stress after orbital evolution and its likelihood of driving tectonics. Image credit: Barr and Collins.
How Long Can a Subsurface Ocean Last?
Once you have a subsurface ocean, you aren’t going back to a boring no-ocean world. Barr Mlinar explains that icy worlds might have a feedback loop when it comes to oceans:
If we find tectonics that could indicate that Pluto had an ocean in the past. It is my personal opinion that if an icy satellite has an ocean ever, it is very difficult to get rid of it. As the ice freezes, all of the anti-freeze materials (salt and ammonia) are excluded. So as the ocean freezes it gets salty and richer in anti-freeze stuff, until eventually you’re left with a thin, deep layer of briny/ammonia-rich liquid.
In other words, as part of the ocean freezes into ice, the salts and ammonias are left behind, reaching higher concentrations in the remaining ocean. Because they act as anti-freeze, they inhibit further freezing. The more water freezes into ice, the more challenging it is for the remaining water to freeze.
The result is that even though the Pluto-Charon collision must’ve happened millions of years ago for it to be in its current tidally-locked dual synchronous state, any subsurface ocean that formed during the collision could potentially still be clinging to liquidity today.
How Could We Tell if Pluto Has a Subsurface Ocean?
If Pluto did have a subsurface ocean, that could be enough to drive icy plate tectonics. If it did, Barr Mlinar is hoping that we might see similar landforms to what we see on Europa, the icy moon of Jupiter with a subsurface ocean and an icy crust. Europa is coated in distinctive linear features and double ridges; although we haven’t seen anything like that on Pluto yet, we could still see a less dramatic version as more photos come back from New Horizons. She’s also hoping that we see any of the usual array of faults and wrinkle ridges we see on any other solid planet with a tectonically active past.
Will we find linear features, double ridges, and faults on Pluto like we see on Europa? Image credit: NASA/JPL.
Finding linear features on Pluto wouldn’t necessary indicate a currently active world with a subsurface ocean; they could be geologic relics from the past, if we can just find a chunk of Pluto’s surface old enough to preserve it.
It’s also possible that we can put some constraints on Pluto’s interior structure through carefully observing its global shape.
Charon is also turning out to be under-cratered for what we’d expect from an ancient, dead moon, leading scientists to speculating if it, too, could have geological activity to some extent. We’ve seen no indication of linear features on Pluto’s largest moon yet, but we’re going to keep looking.
How Does What We’re Seeing at Pluto Impact Our Theories on Icy Worlds?
While Pluto is a dwarf planet, not an icy moon, one of the reasons its smooth surface and implied geological activity are causing so much fuss is because it breaks our current theories of how thermodynamics applies to small, icy worlds. When I asked Barr Mlinar if she could think of other startling discoveries that had a similar impact on trashing the theories, she immediately leapt to how surprising we found Enceladus when we got our first up-close look:
Enceladus was a huge surprise and a big shock. It south polar region has a heat flow something like 10x that of the Earth, yet it’s a moon that is 250 km in radius, the size of an asteroid. A “traditional” tidal model underestimates the amount of heat coming out of Enceladus by a factor of 100, and it also predicts that the heat should be coming out uniformly, so a uniform heat flux across the surface. No tidal model has yet been able to explain either the magnitude or distribution of tidal heat coming out of Enceladus.
Enceladus has a ridiculously hot pole, and we still can’t explain why. Image credit: NASA
Similar statements are true about Jupiter’s moon Io.
So I think it would be safer to say that we don’t understand tidal dissipation at all, either in rocky or icy bodies. We always underestimate.
So yes, Pluto has once again broken our theories for thermodynamics of icy worlds, but it’s not the first time those theories have been royally stomped on by new observations. This time, we’re getting clarification in that Pluto is so far away from anything that could be exerting a substantial, ongoing tidal pull, so any abnormalities cannot be attributed to tidal stress from a gas giant. That means we can’t keep leaning on blaming gas giants as a handy crunch to explain the unexpected. Instead, we need to fundamentally rethink how we’re approaching thermodynamics in these icy worlds.
What Happens Next?
We’ve gotten back the last of the First Look data from the New Horizons probe, the immediate failsafe data downlinked in case something catastrophic happened to the spacecraft. That data is still being released to the public, potentially with all sorts of new surprises including the possibility of finding some of Barr Mlinar’s linear features. Even after that, the probe will be returning data for another 16 months, including lossless, uncompressed versions of existing images so we can crater-hunt with more confidence.
Before we closed out our conversation, I asked Barr Mlinar if she had any final thoughts on the New Horizons mission:
I want to pass along my congratulations to the New Horizons team. The team has done a phenomenal job of planning and executing a very tricky encounter. I am also thrilled for them that Pluto didn’t turn out to be a lifeless cratered ball of ice. We really had no idea what to expect, and I’m so glad that the team’s unwavering interest in Pluto was rewarded with these magnificent images.
[T]here was a time when it looked like budget issues and politics would cause the mission to be cancelled. Alan Stern and the team did a heroic job of fighting to save the mission. I think the message here is, “never give up, never surrender.” If you really want something to happen, you have to go out and fight for it, and not listen to naysayers.
Getting this mission was no sure thing — at one point it involved a postage stamp campaign advertising that the then-ninth planet still hadn’t been explored. Now the spacecraft is out exploring the outer solar system, the extended mission to send New Horizons out to explore a second Kuiper Belt Object is still awaiting funding confirmation. And yet, we’re learning so much already.
The surprises in the Pluto-Charon system are just barely starting to be defined. Scientists on the New Horizons team and those outside it like Barr Mlinar are going to be kept busy sorting how how all these discoveries warp and change their theories of how our solar system functions.
Top image: Pluto and Charon in real colour and appropriate scale, composited from data from the New Horizons probe. Credit: NASA/JHUAPL/SwRI.
Barr Mlinar and I share a peculiar history: she was the scientist who first taught me about asteroids when I was but a wee proto-scientist attending the Summer Science Program. We both highly recommend the program to high school students with a lively interest in science and a desire to be truly pushed to their intellectual limits.<|endoftext|>
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Result:
Detailed Calculation:
Calculation History:
## Introduction
The Long Addition Calculator is a powerful and versatile tool designed to assist individuals in performing complex addition operations with ease and accuracy. This tool is particularly useful for tasks that involve adding large numbers or decimals, making it an essential resource for students, professionals, and anyone who frequently deals with arithmetic calculations.
## Concept and Functionality
The concept behind the Long Addition Calculator is simple yet highly effective. It automates the process of adding long numbers by breaking them down into smaller, manageable parts and performing the addition step by step. This eliminates the risk of making errors due to human miscalculations and enhances the overall efficiency of the addition process.
## Formulae
To understand how the Long Addition Calculator works, let’s take a look at the basic formulae involved in long addition:
Column addition is the fundamental formula used in long addition. It involves adding numbers vertically, one column at a time, starting from the rightmost column (the units column) and moving towards the left.
For example:
`1234 + 567 -------`
In this case, you would start by adding 4 + 7 in the units column, which equals 11. You write down the 1 and carry over the 1 to the next column. Then, you continue adding the numbers in the tens, hundreds, and thousands columns, taking into account the carryovers from the previous columns.
Decimal addition is another important aspect of long addition. When adding numbers with decimal points, it’s essential to align the decimal points correctly and then proceed with column addition as described above.
For example:
`12.34 + 5.67 ---------`
In this case, you would start by adding the numbers to the right of the decimal point, just like in regular column addition. Then, you move to the left of the decimal point and continue the addition process.
## Example Calculations
Let’s illustrate the Long Addition Calculator’s functionality with a few example calculations:
### Example 1: Basic Long Addition
`1234 + 567 --------- 1801`
Here, the Long Addition Calculator breaks down the addition into individual steps, starting with the rightmost column and working towards the left. The result is 1801.
`12.34 + 5.67 --------- 18.01`
In this example, the tool ensures that the decimal points are correctly aligned and then performs the addition, resulting in 18.01.
## Real-World Use Cases
The Long Addition Calculator finds applications in various real-world scenarios:
### Education
In educational settings, students can use this tool to practice and improve their long addition skills. It helps them grasp the concept and gain confidence in performing accurate calculations.
### Accounting and Finance
Professionals working in finance and accounting deal with large numbers and decimals. The Long Addition Calculator simplifies their tasks, reducing the risk of errors in financial calculations.
### Science and Engineering
Scientists and engineers use long addition when working with precise measurements or data analysis. This tool streamlines the process, ensuring the accuracy of results.
### Everyday Life
In everyday life, people encounter situations where they need to add numbers, whether it’s calculating expenses, budgeting, or splitting bills. The Long Addition Calculator makes these tasks quick and error-free.
## Conclusion
The Long Addition Calculator is a valuable tool that simplifies the process of adding long numbers and decimals. Its concept is based on well-established formulae for column addition and decimal addition, making it suitable for a wide range of applications, from education to professional tasks.
## References
1. Larson, R., & Edwards, B. H. (2016). Elementary Linear Algebra. Cengage Learning.
2. Tussy, A. S., & Gustafson, R. D. (2012). Elementary and Intermediate Algebra. Cengage Learning.
What do you think?
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What do you think is the greatest scientific discovery or invention of all time? For some, Benjamin Franklin's discovery of electricity would probably rank near the top of the list. After all, without electricity, our lives would be drastically different than they are today.
Have you ever given much thought to how electricity gets from the power plant to your house? Simply plugging an electronic device into the nearest outlet is a convenience we often take for granted. However, electricity's path to those little plugs in the wall is a fascinating journey.
If you've ever seen objects hanging from the top of utility poles or large boxes sitting near buildings, then you're familiar with some of the most important pieces of machinery in the system that delivers energy to your home. These machines are called transformers. No, they don't change into superhero vehicles when you're not looking, but they are all about change!
Transformers are electrical machines that change electricity from one voltage to a different voltage. Voltage is the measure of the electrical force that pushes electrons around a circuit. In some cases, transformers can take electricity at a lower voltage and change it to a higher voltage. Such transformers are called step-up transformers.
Most transformers, however, are step-down transformers. They take electricity at a high voltage and change it to a lower voltage. This is a critical step in the energy-delivery process, since the electricity that comes from a power plant is at an extremely-high voltage, which is much too high for use in your home.
For example, a power plant transmission line might carry electricity at 400,000 to 750,000 volts. Electricity is sent out at such high voltages because it often has to travel great distances. Using higher voltages helps to minimize energy loss as it travels.
In certain areas, called electrical substations, huge transformers reduce those high voltages to lower voltages to send out to specific areas. Have you ever seen an electrical substation near your home? You can usually identify by them by the presence of lots of electrical lines and equipment, including numerous transformers.
Step-down transformers in substations reduce high voltages to lower voltages in the range of 7,200 volts. When the electricity reaches your neighborhood, transformers on top of utility poles or ground boxes connected to underground wires reduce the voltage of the electricity to 220-240 volts to use in your home. Some major electrical appliances, such as water heaters, stoves, and air conditioners, will use the 220-240 volts, while most other smaller electrical appliances will use 110-120 volts.
So how do transformers work this electrical magic? It all happens because of a couple of simple facts about electricity. Transformers work because the fluctuating current of electricity (known as alternating current or AC) flowing through the wires entering the transformer (the primary current) creates a magnetic field. That fluctuating magnetic field creates a current (the secondary current) in a second set of wires leaving the transformer through a process called electromagnetic induction.
To make this process more efficient, the wires entering and leaving a transformer are coiled into loops or turns around an iron bar called a core. If the primary and secondary coils have the same number of loops or turns, the voltage will be the same in each. If, however, the secondary coil has more or fewer loops or turns, then the secondary current voltage will be greater or less than the primary current.
For example, if the primary coil has 10 turns and the secondary coil has one turn, then the transformer will reduce the primary voltage by a factor of 10. So a current entering the transformer at 1,000 volts will leave the transformer at 100 volts.<|endoftext|>
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# [Maths Class Notes] on Square Root of 6 Pdf for Exam
The square root is a topic that many students find difficult to understand. However, once you understand the concept well, finding square roots will no longer be a challenge for you. So, in simple words square root, in mathematics, is a factor of a number that, when multiplied by itself, gives the original number. For example, both 3 and –3 are square roots of 9.
The square root of a number a is the number b such that b² = a. The square root of any number is represented by the symbol and is also often known as radical. The number or expression given under the square root symbol is known as radicand. The square root is a commonly used function in Mathematics. It is widely used in subjects like Mathematics and Physics. Sometimes it is tedious to find the square root of a number, especially the numbers which are not the perfect square of the number. In this article, we will discuss the square root of 6, and how to calculate the root 6 value using the simplifying square root method.
What is the root 6 value?
The root 6 value is 2.449
### Square Root of 6 Definition
The square root of a number 6 is a number y such that y² = 6. The square root of 6 in radical form is written as √6. The square root of 6 in radical form is expressed as√6
### How to Calculate the Under Root 6 Value?
We can calculate the under root 6 value using different methods of square roots. These methods can be the long division methods, prime factorization method, or simplifying square root method. Let us discuss how to calculate the under root 6 value using the simplifying square root method.
To simplify a square root, make the number under the square root as small as possible while keeping it as a whole number. Mathematically, it can be expressed as: √x.y=√x×√y
To express the square root of 6 in the simplest form, we will make the number 6 as small as possible, ensuring to keep it as a whole number. Hence, the square root of 6 in simplest form is represented as√6=√2×√3. This can be further simplified by substituting the value of √2 and √3
[sqrt{6}=sqrt{2}times sqrt{3}]
[sqrt{6}=1.414times 1.732]
[sqrt{6}=2.449]
Hence, the square root of 6 in simplest form is 2.449. Similarly, we can also calculate the square root of any other whole numbers and their factors. Hence, simplifying the square root method is the simplest method of calculating the square root. We can also calculate the value of under root 6 using the calculator as this will give us the exact value. The exact value of the square root will always be given in a decimal number as it is impossible to determine a positive whole integer as a root for non-rational numbers.
### Simplifying the Square Root Using Perfect Square Method
Following are the steps to simplify the square root using the perfect square method:
1. Find the perfect square that divides the number in the radicand.
2. Express the numbers as a factor of a perfect square.
### Solved Example
1. Simplify \$mathbf{sqrt{300}}\$
Solution:
[sqrt{300}=sqrt{100times 3}]
[sqrt{300}=sqrt{10times 10times 3}]
[sqrt{300}=10sqrt{3}]
Hence,[sqrt{300}] can be simplified as [10sqrt{3}]
2. Simplify the following radical expressions :
1. [sqrt{48}]
2. [sqrt{75}]
Solutions:
i.\$mathbf{sqrt{48}}\$
Step 1: The perfect square 16 will divide the number 48.
Step 2: Express 48 as a factor of 16
48=16×3
Step 3: Reduce the square root of 16 as shown below:
[sqrt{48}=sqrt{16times 3}]
[sqrt{48}=sqrt{4times 4times 3}]
[sqrt{48}=4sqrt{3}]
Hence,[sqrt{48}] can be simplified as[4sqrt{3}]
ii.\$mathbf{sqrt{75}}\$
Step 1: The perfect square 25 will divide the number 75.
Step 2: Express 75 as a factor of 25.
75=25×3
Step 3: Simplify the radicals as shown below:
[sqrt{75}=sqrt{25times 3}]
[sqrt{75}=sqrt{5times 5times 3}]
[sqrt{75}=5sqrt{3}]
Hence,[sqrt{75}] can be simplified as[5sqrt{3}]<|endoftext|>
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By using their body condition as a cue, animals can make good decisions
Gut instinct makes animals appear clever
Animals, including humans, can make surprisingly good decisions just based on the food in their stomach, new research suggests.
The study, led by the University of Exeter and published in Proceedings of the Royal Society of London B, shows that surviving in difficult and dangerous conditions does not necessarily require high brain power.
Instead, animals should be sensitive to their body condition, such as how hungry they are.
Hunger acts as a sort of memory of past food availability, which tells them what conditions are probably like now.
A team led by Dr Andrew Higginson, from the University of Exeter, used computer modelling to predict how animals should behave to maximise their survival when the food supply is unpredictable and the environment contains predators.
According to the model, an animal that bases its decisions only on its current energy reserves can survive almost as long as one that uses its brain to calculate the best thing to do.
Dr Higginson said: “Many of us sometimes get ‘hangry’: when hunger makes us emotional and changes our behaviour. Our model explains why there is link between our gut and our decisions: hunger can act as a memory telling us there’s not been much food around, which it’s important to respond to in the wild.
“The usefulness of such memory means that animals, including humans, may appear to be processing a great deal of information in the brain when in fact they are just following their gut.”
An animal’s body condition tells it how successful it has been in the past, which is a useful guide to how it should behave tomorrow.
This simple, physiological form of memory may have allowed animals to avoid investing in brain tissue, which requires a large amount of energy.
Professor John McNamara, from the University of Bristol’s School of Mathematics and a member of the team, said: “If it costs a lot of resources to be so clever, then natural selection will have found a cheaper way to make decisions.
“The ability to use internal states such as hunger as a memory will have reduced the need to evolve big brains.”
The findings raise the possibility that simple memories may also be encoded in other physiological states, such as emotions.
This might be why it takes a long time to calm down after feeling threatened. Since the threat may come back, the emotion keeps the body ready to fight or flee.
The researchers say it is possible that their usefulness as a ‘memory’ is the reason humans and other animals have emotions.
The research has implications for conservation too.
“By using their body condition as a cue, the animals in our model can still perform well when the environmental conditions change suddenly,” said Dr Higginson.
“This suggests that some species might be able to cope with the effects of climate change better than expected.”
Trust your gut: using physiological states as a source of information is almost as effective as optimal Bayesian learning by Andrew D. Higginson, Tim W. Fawcett, Alasdair I. Houston, and John M. McNamara is published in Proceedings of the Royal Society of London B.
Date: 24 January 2018<|endoftext|>
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Chinese researchers have developed a new Lidar (Light detection and ranging) based camera that can snap photographs of subjects from 45 kilometers away. This is not the first instance of attempts at long-distance photography, but environmental disturbances like pollution hinder capturing subjects beyond a distance.
Zheng-Ping Li and his colleagues from the University of Science and Technology of China have successfully captured subjects from 45 kilometers (28 miles) away in an environment covered with a dense smog.
How Does The Lidar Camera Work?
The Lidar-based system combines single-photon detectors with a breakthrough computational imaging algorithm to create super-high-resolution images by joining the sparsest of data points.
The camera was set up on the 20th floor of a building situated in Chongming Island in Shanghai, and the subjects were captured 45 kilometers away at Pudong Civil Aviation Building across the river.
The Lidar system deployed in the camera illuminates the subject with laser light and then creates an image from the reflected light.
The reflected light is received within a time frame that depends upon the distance. This ensures that photons that are received beyond a specific time duration can be discarded, thus producing noise-free photographs.
To overcome pollution and environmental distortions, the researchers used an infrared laser with a wavelength of 1550 nanometers, a repetition rate of 100 kilohertz and power of 120 milliwatts. Using such an infrared laser makes the system safe for the eyes and also makes it easy to filter out solar photons that would be destructive for the detector.
According to Zheng-Ping and his colleagues, “Our results open a new venue for high-resolution, fast, low-power 3D optical imaging over ultralong ranges.”
Applications of this long-distance imaging technique include target recognition and identification from a distance, airborne surveillance and remote sensing.
Moreover, the entire setup is highly portable and is similar to a shoe box in size. Therefore, you can carry it anywhere and use it for capturing photographs from up to 45 kilometers away.
However, China is infamous for its surveillance activities. Therefore, this breakthrough in technology could also be used for hi-tech surveillance purposes.<|endoftext|>
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# Where is the number 130 archers per meter in attempt 2, in the “Sparta” article in the “What If?” book?
In the article Sparta from the book What If?, the given problem is to know how many archers would be required for the scene of flying arrows blotting out the sun. His method is to calculate the number of arrows in a meter width of the archer battery. Each archer is assumed to have a fire rate of 150 mHz, or 0.15 arrow per second. The arrow flies 3 seconds on air. The number arrows in the air on 1 s for each archer is 0.45.
# Attempt 1
• 2 archers per meter in one row
• A row every 1.5 m
• 20 rows (30 m)
-> number of arrows in the air in 1 s in 1 m width: 2 x 20 x 0.45 = 18 (agrees with the book)
# Attemp 2
• 6 archers per meter width (Since triple the number of archers in square foot essentially means triple the number per square meter, and since each row stands 1.5 m from each other, hence you just simply triple the number in one row.)
• 40 rows (60 meters x 1.5 row/meter)
-> number of arrows in the air in 1 s in 1 m width: 6 x 40 x 0.45 = 6 x 18 = 108
This does not match with the number of 130 in the book. Why? 130 is even indivisible to 40. More importantly, it says "130 archers per meter". The correct number of archers per meter width column is 6 x 40 = 240.
In the end of the attempt 2, there is a bit of information:
I assume that the number 339 is actually the number of arrows fired per meter width per second. But it is $240 \frac{archer}{m} \times \frac{7}{8} \frac{arrow}{archer.s} = 210 \frac{arrow}{m.s}$, not even close to 339. Why?
• Only 0.15 arrows per second? Legolas can do much better than that if you believe the LOTR movies... (Joking aside, given the fantasy-style of the movie 300 with its improbably high fighting quality of various characters, a rate of 1 arrow every >6 seconds seems pretty pessimistic.) – TMM Mar 7 '17 at 11:52
• The only thing that comes to my head is that in one meter wide column there are 240 archers, and this number was misspelled (2 lies next to 1, 4 lies next to 3). Is this number used in further text? By the way, 18*6=108. – Jaroslaw Matlak Mar 8 '17 at 8:52
• @JaroslawMatlak This is a hilarious surprising correct answer. I have added more information in my edit – Ooker Mar 8 '17 at 12:03
It seems, that I've found out 'what author was thinking about'. - In second attempt density of archers from $2/m$ is increased to $1/ft$. Then, if we take the greek foot (as Spartans were Greek), ie. $1ft = 0.3083 m$, we have $\frac{40}{0.3083} \approx 129.74 \approx 130$ archers in one meter firing column.
• hmm, $2/m$ is increased to $6/m$, which equals $0.5/ft$. And using the greek foot seems overcomplicated to me – Ooker Mar 8 '17 at 15:11<|endoftext|>
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# UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1
## Problem
Four siblings BRYAN, BARRY, SARAH and SHANA are having their names monogrammed on their towels. Different letters may cost different amounts to monogram. If it costs $\textdollar{21}$ to monogram BRYAN, $\textdollar{25}$ to monogram BARRY and $\textdollar{18}$ to monogram SARAH, how much does it cost to monogram SHANA?
## Solution
We set up the following equations:
$b+r+y+a+n= 21$
$b+a+2r+y= 25$
$s+2a+r+h= 18$
We are asked to find the price of "Shana", or $s+h+2a+n$. We notice that this expression has no $b$ or $y$, so we subtract the first equation from the second to eliminate those variables:
$r-n=4$
$r=n+4$
Which we substitute into the third equation
$s+2a+n+4+h=18$.
$s+2a+n+h=14$.
This matches the expression we need to solve for, so we are done. $\boxed{14}$.
## Solution 2
Notice that BRYAN is simple BARRY rearranged, and SARAH is simple SHANA rearranged. Let the cost of rearranging SHANA be $x$. We have the following equation: $\[21+18=25+x\]$ Solving for $x$, our answer is $\boxed{14}$.
## Credits
• Credit to Football017 for writing the solution<|endoftext|>
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# Proof of Sum basis Binomial Theorem in One variable for Positive exponent
The sum basis binomial theorem in one variable for positive exponent can be expanded in the following two mathematical forms.
$(1).\,$ $(x+y)^n$ $\,=\,$ $\displaystyle \binom{n}{0} x^n y^0$ $+$ $\displaystyle \binom{n}{1} x^{n-1} y^1$ $+$ $\displaystyle \binom{n}{2} x^{n-2} y^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} x^{n-r} y^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} x^{0} y^n$
$(2).\,$ $(x+y)^n$ $\,=\,$ $^nC_0\,x^n y^0$ $+$ $^nC_1\,x^{n-1} y^1$ $+$ $^nC_2\,x^{n-2} y^2$ $+$ $\cdots$ $+$ $^nC_r\,x^{n-r} y^r$ $+$ $\cdots$ $+$ $^nC_n\,x^{0} y^n$
You can follow any one of the above expansions to prove the binomial theorem in one variable.
### Binomial Theorem in one variable
The Binomial theorem is defined in terms of two variables $x$ and $y$. In this case, the binomial theorem is expressed in one variable. So, substitute $x \,=\, 1$.
$\implies$ $(1+y)^n$ $\,=\,$ $\displaystyle \binom{n}{0} (1)^n y^0$ $+$ $\displaystyle \binom{n}{1} (1)^{n-1} y^1$ $+$ $\displaystyle \binom{n}{2} (1)^{n-2} y^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} (1)^{n-r} y^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} (1)^{0} y^n$
$\implies$ $(1+y)^n$ $\,=\,$ $^nC_0\,(1)^n y^0$ $+$ $^nC_1\,(1)^{n-1} y^1$ $+$ $^nC_2\,(1)^{n-2} y^2$ $+$ $\cdots$ $+$ $^nC_r\,(1)^{n-r} y^r$ $+$ $\cdots$ $+$ $^nC_n\,(1)^{0} y^n$
Now, simplify the expansion of binomial theorem in one variable.
$\implies$ $(1+y)^n$ $\,=\,$ $\displaystyle \binom{n}{0} \times 1 \times y^0$ $+$ $\displaystyle \binom{n}{1} \times 1 \times y^1$ $+$ $\displaystyle \binom{n}{2} \times 1 \times y^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} \times 1 \times y^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} \times 1 \times y^n$
$\implies$ $(1+y)^n$ $\,=\,$ $^nC_0\, \times 1 \times y^0$ $+$ $^nC_1\, \times 1 \times y^1$ $+$ $^nC_2\, \times 1 \times y^2$ $+$ $\cdots$ $+$ $^nC_r\, \times 1 \times y^r$ $+$ $\cdots$ $+$ $^nC_n\, \times 1 \times y^n$
Finally, the binomial theorem in one variable is written in the following forms.
$\implies$ $(1+y)^n$ $\,=\,$ $\displaystyle \binom{n}{0} y^0$ $+$ $\displaystyle \binom{n}{1} y^1$ $+$ $\displaystyle \binom{n}{2} y^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} y^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} y^n$
$\implies$ $(1+y)^n$ $\,=\,$ $^nC_0\, y^0$ $+$ $^nC_1\, y^1$ $+$ $^nC_2\, y^2$ $+$ $\cdots$ $+$ $^nC_r\, y^r$ $+$ $\cdots$ $+$ $^nC_n\, y^n$
### Binomial Theorem in one variable in usual form
The expansion of the Binomial Theorem in one variable is derived in terms of $y$ but we are used to express it in terms of $x$. So, write the binomial theorem in one variable in terms of $x$ by replacing $y$ with $x$.
$(1).\,$ $(1+x)^n$ $\,=\,$ $\displaystyle \binom{n}{0} x^0$ $+$ $\displaystyle \binom{n}{1} x^1$ $+$ $\displaystyle \binom{n}{2} x^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} x^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} x^n$
$(2).\,$ $(1+x)^n$ $\,=\,$ $^nC_0\, x^0$ $+$ $^nC_1\, x^1$ $+$ $^nC_2\, x^2$ $+$ $\cdots$ $+$ $^nC_r\, x^r$ $+$ $\cdots$ $+$ $^nC_n\, x^n$
### Replace the Binomial coefficients by their values
Find the values of binomial coefficients to substitute them in the expansion of the binomial theorem.
$\displaystyle \binom{n}{0}$ $\,=\,$ $^nC_0$ $\,=\,$ $\dfrac{n!}{0!(n-0)!}$ $\,=\,$ $1$
$\displaystyle \binom{n}{1}$ $\,=\,$ $^nC_1$ $\,=\,$ $\dfrac{n!}{1!(n-1)!}$ $\,=\,$ $\dfrac{n}{1!}$ $\,=\,$ $n$
$\displaystyle \binom{n}{2}$ $\,=\,$ $^nC_2$ $\,=\,$ $\dfrac{n!}{2!(n-2)!}$ $\,=\,$ $\dfrac{n(n-1)}{2!}$
$\,\,\,\,\,\vdots$
$\displaystyle \binom{n}{r}$ $\,=\,$ $^nC_r$ $\,=\,$ $\dfrac{n!}{r!(n-r)!}$ $\,=\,$ $\dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!}$
$\,\,\,\,\,\vdots$
$\displaystyle \binom{n}{n}$ $\,=\,$ $^nC_n$ $\,=\,$ $\dfrac{n!}{n!(n-n)!}$ $\,=\,$ $1$
Now, substitute the values of binomial coefficients in the expansion of the binomial theorem.
$\implies$ $(1+x)^n$ $\,=\,$ $1 \times x^0$ $+$ $n \times x^1$ $+$ $\dfrac{n(n-1)}{2!} \times x^2$ $+$ $\cdots$ $+$ $\dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!} \times x^r$ $+$ $\cdots$ $+$ $1 \times x^n$
According to the zero power rule, the $x$ raised to the power zero is equal to one. The $x$ raised to the power one is denoted by $x$ in mathematics.
$\implies$ $(1+x)^n$ $\,=\,$ $1 \times 1$ $+$ $n \times x$ $+$ $\dfrac{n(n-1)}{2!} \times x^2$ $+$ $\cdots$ $+$ $\dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!} \times x^r$ $+$ $\cdots$ $+$ $1 \times x^n$
$\,\,\,\therefore\,\,\,\,\,\,$ $(1+x)^n$ $\,=\,$ $1$ $+$ $nx$ $+$ $\dfrac{n(n-1)}{2!}\,x^2$ $+$ $\cdots$ $+$ $\dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!}\,x^r$ $+$ $\cdots$ $+$ $x^n$
Thus, the sum based binomial theorem in one variable is derived in mathematics and it is also written simply as follows.
$\,\,\,\therefore\,\,\,\,\,\,$ $(1+x)^n$ $\,=\,$ $1$ $+$ $nx$ $+$ $\dfrac{n(n-1)}{2!}\,x^2$ $+$ $\dfrac{n(n-1)(n-2)}{3!}\,x^3$ $+$ $\cdots$
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In a surprising new study, research now suggests that we have hugely underestimated the amount of heat absorbed by the oceans in recent decades.
As first reported in the journal Nature, a team led by Scripps Institution of Oceanography and Princeton University estimated that the ocean has absorbed 60 percent more heat over the past 25 years than the figure that the United Nations IPCC Climate Change report released last month.
The IPCC Climate Change report that was recently released argued that humans must keep global average temperature rise under 1.5℃ below pre-industrial levels. This advisory was after the Paris Climate Accords agreed that the goal would need to be kept under a 2℃ increase. The .5 degree change is extremely significant as once we do get to the new established mark, there is likely no going back for the damage humans have done. To keep within the goal, humans must reduce carbon emissions 25% more than previously believe.
The findings that the amount of heat absorbed by the oceans is significant as we now know that the cooling of our planet is driven even more so by the oceans and without them, our Earth be a lot hotter.
“Imagine if the ocean was only 30 feet deep,” lead author Laure Resplandy, a Princeton assistant professor of geoscience, said in a statement. “Our data shows that it would have warmed by 6.5℃ (11.7℉) every decade since 1991. In comparison, the estimate of the last IPCC assessment report would correspond to a warming of only 4℃ (7.2℉) every decade.”
Scientists work out surface warming with the knowledge that the ocean takes up roughly 90 percent of all the excess energy produced as the Earth warms. If more heat is going into the oceans than we realized, that also means that our greenhouse gas emissions are trapping more heat than we thought. From this, we can conclude that the Earth is more sensitive to fossil-fuel emissions than anticipated.
The paper writes, ” the study suggests that ocean warming is at the high end of previous estimates, with implications for policy-relevant measurements of the Earth response to climate change, such as climate sensitivity to greenhouse gases and the thermal component of sea-level rise.”<|endoftext|>
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Learned, What is the Inductive Method of Economics? Steps, Merits, and Demerits!
The Inductive Method: Induction “is the process of reasoning from a part to the whole, from particulars to generals or from the individual to the universal.” Bacon described it as “an ascending process” in which facts are collected, arranged and then general conclusions are drawn. Also learn, the Methods of Economics, What is the Inductive Method of Economics?
The inductive method was employed in economics by the German Historical School which sought to develop economics wholly from historical research. The historical or inductive method expects the economist to be primarily an economic historian who should first collect material, draw generalizations, and verify the conclusions by applying them to subsequent events. For this, it uses statistical methods. The Engel’s Law of Family Expenditure and the Malthusian Theory of Population have been derived from inductive reasoning.
The inductive method involves the following steps:
1. The Problem:
In order to arrive at a generalization concerning an economic phenomenon, the problem should properly select and clearly stated.
The second step is the collection, enumeration, classification, and analysis of data by using appropriate statistical techniques.
Data are using to make the observation about particular facts concerning the problem.
On the basis of observation, generalization is logically deriving which establishes a general truth from particular facts.
Thus induction is the process in which we arrive at a generalization on the basis of particular observing facts. Also learn, Explain is What is Economics? Meaning and Definition of Criticisms!
The best example of inductive reasoning in economics is the formulation of the generalization of diminishing returns. When a Scottish farmer found that in the cultivation of his field an increase in the amount of labour and capital spent on it was bringing in less than proportionate returns year after year, an economist observing such instances in the case of a number of other farms, and then he is arriving at the generalisation that is known as the Law of Diminishing Returns.
Merits of Inductive Method:
The chief merits of this method are as follows:
The inductive method is realistic because it is based on facts and explains to them as they actually are. It is concrete and synthetic because it deals with the subject as a whole and does not divide it into component parts artificially
(2) Future Inquiries:
Induction helps in future inquiries. By discovering and providing general principles, induction helps future investigations. Once a generalization is establishing, it becomes the starting point of future inquiries.
(3) Statistical Method:
The inductive method makes use of the statistical method. This has made significant improvements in the application of induction for analyzing economic problems of the wide range. In particular, the collection of data by governmental and private agencies or macro variables, like national income, general prices, consumption, saving, total employment, etc., has increased the value of this method and helping governments to formulate economic policies pertaining to the removal of poverty, inequalities, underdevelopment, etc.
The inductive method is dynamic. In this, changing economic phenomena can analyze on the basis of experiences, conclusions can draw, and appropriate remedial measures can take. Thus, induction suggests new problems to pure theory for their solution from time to time.
A generalization drawn under the inductive method is often historical-relative in economics. Since it is drawn from a particular historical situation, it cannot apply to all situations unless they are exactly similar. For instance, India and America differ in their factor endowments. Therefore, it would be wrong to apply the industrial policy which was following in America in the late nineteenth century to present day India. Thus, the inductive method has the merit of applying generalizations only to related situations or phenomena.
Demerits of Inductive Method:
However, the inductive method is not without its weaknesses which are discussing below.
(1) Misinterpretation of Data:
Induction relies on statistical numbers for analysis that “can misuse and misinterpret when the assumptions which are requiring for their use are forgotten.”
(2) Uncertain Conclusions:
Boulding points out that “statistical information can only give us propositions whose truth is more or less probable it can never give us certainty.”
(3) Lacks Concreteness:
Definitions, sources, and methods using in statistical analysis differ from investigator to investigator even for the same problem, as for instance in the case of national income accounts. Thus, statistical techniques lack concreteness.
(4) Costly Method:
The inductive method is not only time-consuming but also costly. It involves detailed and painstaking processes of collection, classification, analyses, and interpretation of data on the part of trained and expert investigators and analysts
(5) Difficult to Prove Hypothesis:
Again the use of statistics in induction cannot prove a hypothesis. It can only show that the hypothesis is not inconsistent with the known facts. In reality, the collection of data is not illuminating unless it is related to a hypothesis.
(6) Controlled Experimentation not Possible in Economics:
Besides the statistical method, the other method used in induction is of controlled experimentation. This method is extremely useful in natural and physical sciences which deal with the matter. But unlike the natural sciences, there is little scope for experimentation in economics because economics deals with human behavior which differs from person to person and from place to place. Also, What is Demand? Meaning and Definition!
Further, economic phenomena are very complex as they relate to the man who does not act rationally. Some of his actions are also bound by the legal and social institutions of the society in which he lives. Thus, the scope of controlled experiments in inductive economics is very little. As pointed Out by Friendman, “The absence of controlled experiments in economics renders the weeding out of unsuccessful hypo-these slow and difficult.”<|endoftext|>
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Meet Casper the Octopus, which scientists discovered 4290 metres below the surface of the Pacific Ocean – an animal unlike anything ever published that looks eerily like a ghost. Researchers from the National Oceanic and Atmospheric Administration (NOAA), part of the US Department of Commerce, said it was the deepest observation on record for this type of cephalopod.
Cephalopods are active predatory mollusks of the large class Cephalopoda, such as octopuses, squid, the chambered nautilus and cuttlefish. They are the most mobile, intelligent and the largest of all mollusks.
Michael Vecchione, of the NOAA National Marine Fisheries Service – National Systematics Laboratory, and colleagues had been conducting the first operation dive of the Okeanos Explorer’s 2016 season, on 27th February, when they came across the fascinating animal, which they are pretty certain is a new species.
When this picture was posted on social media websites, people suggested the little octopus should be called Casper (the friendly cartoon ghost) because of its lack of pigment cells. ( Image courtesy of NOAA Office of Ocean Exploration and Research, Hohonu Moana 2016)
Came across the ghost-like octopus by accident
Okeanos Explorer (R 337) is a converted US Navy ship, now an exploratory vessel for the NOAA. The researchers were exploring depths in excess of 4,000 metres northeast of Mokumanamana (Necker Island) in the Hawaiian Archipelago.
The dive was to collect data on whether a connection exists between Necker Ridge and Necker Island, a narrow feature on the seabed that extends more than 400 miles and protrudes past the current EEZ (exclusive economic zone) of the United States.
The researchers’ main objective was to collect geological samples on the ridge to help them determine whether they have the same composition as the samples that had been previously collected near Necker Island. Deep Discoverer, a remotely-operated vehicle (ROV), was also surveying biological communities in the area.
As the ROV travelled over a flat area of rock and sediment 4,290 metres down, it came a across a fascinating little octopod (octopus) resting on a flat rock dusted with a light coating of sediment.
ROV Deep Discoverer approaches the hitherto unseen octopod at 4,290 meters depth. (Image courtesy of the NOAA Office of Ocean Exploration and Research, Hohonu Moana 2016)
Small octopus unlike anything seen before
The remarkable little animal was unlike anything ever recorded or published before, and was also the deepest observation ever made for this type of mollusk.
There are two types of deap-sea octopods:
1. The Cirrate or finned octopus, also called the ‘dumbo’ octopus. It has fins on the side of its body and finger-like cirri associated with the suckers on its arms.
2. Incirrate octopods. These have neither fins nor cirri and look very similar to the common shallow-water Octopus.
Just one row of suckers per arm
The octopod, which was photographed in detail on this first dive, was found to be an incirrate. The researchers noticed that it had suckers in one row on each arm, rather than the usual two rows.
What made this animal stand out in the world of octopods was its lack of chromatophores – pigment-containing and light-reflecting cells which give an animal colour, typical of most cephalopods. Unlike other octopods, this one did not appear very muscular.
Its lack of chromatophores gave it a ghost-like appearance. When pictures of it were posted on social media sites, some readers suggested it should be called Casper, the friendly cartoon ghost.
Regarding this ghost-like creature, Vecchione wrote:
“It is almost certainly an undescribed species and may not belong to any described genus.”
Cirrate octopods have been seen at depths exceeded 5,000 metres. However, the deepest that incirrates have been observed was considerably less than 4,000 metres, that is, until now.
After seeing this observation, I contacted my colleagues Louise Allcock (currently on a British ship near Antarctica) and Uwe Piatkowski (from Germany) and they agreed that this is something unusual and is a depth record for the incirrate octopods.”
“We are now considering combining this observation with some other very deep incirrate observations by a German cruise in the eastern Pacific into a manuscript for publication in the scientific literature.”
Video – Casper the ghost-like octopus
In this video you can see the fascinating white octopus, probably a new species, moving about on the seabed at a depth of 4,290 metres.<|endoftext|>
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# Linear Systems
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### Linear Systems
1. 1. System’s of Linear Equations Donald Saunders
2. 2. What is it and how do we solve it? <ul><li>Two or more linear equations that are grouped together </li></ul><ul><li>A solution is an ordered pair that is a solution of ALL equations in a system </li></ul><ul><li>There are three methods commonly used to solve a system of linear equations </li></ul>
3. 3. Three methods used to solve <ul><li>Substitution </li></ul><ul><li>Graphing </li></ul><ul><li>Elimination </li></ul>
4. 4. Substitution Method We will be substituting part of one equation into the other equation <ul><li>y = x and x + y = 2 </li></ul><ul><li>x + x = 2 remove the y in the second equation and replace it with x from the first equation </li></ul><ul><li>2x = 2 combine like terms </li></ul><ul><li>x = 1 solve for x </li></ul><ul><li>y = x = 1 plug in what x equals and solve for y in either equation </li></ul><ul><li>(1, 1) identify the solution to our system of equations </li></ul>
5. 5. Graphing Method A linear system of equations are two lines. Let’s graph our lines to see what happens. y = x and x + y = 2
6. 6. Graphing Method Remember our solution was (1,1), so let’s identify this on our graph and deduce how we can identify our solution using the graphing method
7. 7. How can we identify the solution to a system of linear equations if we graph them? <ul><li>The point where the line crosses the x-axis </li></ul><ul><li>The point where the line crosses the y-axis </li></ul><ul><li>The point where the two lines cross each other </li></ul>
8. 8. Elimination method <ul><li>Eliminates one variable from the equations </li></ul><ul><li>Can be either variable </li></ul><ul><li>Key is to get the same number with opposite signs for either variable </li></ul><ul><li>Once you have solved for one variable, you can plug that value in and solve for the second variable </li></ul>
9. 9. Elimination Method <ul><li>y = x and x + y = 2 </li></ul><ul><li>x + y = 0 </li></ul><ul><li>x + y = 2 </li></ul><ul><li>2y = 2 </li></ul><ul><li>y = 1 </li></ul><ul><li>x + (1) = 2 </li></ul><ul><li>x = 1 </li></ul><ul><li>(1, 1) </li></ul>
10. 10. Who prefers which method?
11. 11. More Practice If you need more practice you can go to the J lab website
12. 12. References <ul><li>http://education.jlab.org/solquiz/index.html </li></ul><ul><li>http://player.discoveryeducation.com/index.cfm?guidAssetId=BF5C2DD4-437B-4F0D-B5C6-B90AF853C30E </li></ul>
13. 13. Correct <ul><li>The solution to a system of equations that has been graphed is the point at which they intersect. If they do not intersect there is No Solution, and if the two lines lie on top of each other there are Infinite Solutions. </li></ul><ul><li>Back to presentation </li></ul>
14. 14. Incorrect The y-axis and x-axis have nothing to do with the solution. These are only points used to graph our two lines Back to Quiz<|endoftext|>
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# RD Sharma Solutions Class 12 The Plane Exercise 29.13
RD Sharma Solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.13, is provided here. Students who aim to top their board exams can get access to the RD Sharma Solutions for Class 12 and prepare themselves confidently for the final exams. Experts have formulated the solutions in an easily understandable manner for students to grasp the concepts easily and score good marks in the exams. Students can easily download the pdf consisting of this chapter solutions, which are available in the links provided below.
## Download PDF of RD Sharma Solutions For Class 12 Maths Chapter 29 Exercise 13
### Access RD Sharma Solutions For Class 12 Maths Chapter 29 Exercise 13
EXERCISE 29.13
Q1.
Solution:
Q2.
Solution:
(x + 1) (4 + 3) – (y – 3) (– 6 – 1) + (z + 2) (9 – 2) = 0
7x + 7 + 7y – 21 + 7z + 14 = 0
7x + 7y + 7z = 0
Divide by 7, we get
x + y + z = 0
Hence, the required equation of plane is x + y + z = 0.
Q3.
Solution:
Substitute a, b, c in equation (2), we get
ax + b(y – 7) + c(z + 7) = 0
(– 14λ)x + (– 14λ) (y – 7) + (– 14λ) (z + 7) = 0
Dividing by (– 14λ) we get
x + y – 7 + z + 7 = 0
x + y + z = 0
Q4.
Solution:
Given:
We know that equation of plane passing through (x1, y1, z1) is given by
a(x – x1) + b(y – y1) + c(z – z1) = 0 …… (1)
Thus, required plane passes through points (4, 3, 2) and (3, – 2, 0).
So equation of required plane is
a(x – 4) + b(y – 3) + c(z – 2) = 0 …… (2)
Plane (2) also passes through (3, – 2, 0),
So,
a(3 – 4) + b(– 2 – 3) + c(0 – 2)
– a – 5b – 2c = 0
a + 5b + 2c = 0 …… (3)
Now plane (2) is also parallel to line with direction ratios 1, – 4, 5.
So,
a1a2Â + b1b2Â + c1c2Â = 0
(a) (1) + (b) (– 4) + (c) (5) = 0
a – 4b + 5c = 0 …… (4)
Now substitute the values of a, b, c in equation (2), we get
a(x – 4) + b(y – 3) + c(z – 2) = 0
(11 λ)(x – 4) + (– λ)(y – 3) + (– 3 λ)(z – 2) = 0
11 λx – 44 λ – λy + 3 λ – 3 λz + 6 λ = 0
11 λx – λy – 3 λz – 35 λ = 0
Divide by λ, we get
11x – y – 3z – 35 = 0
Hence, the required equation of plane is 11x – y – 3z – 35 = 0.
Q5.
Solution:
Another equation of line is
3x – 2y + z + 5 = 0
2x + 3y + 4z – 4 = 0
Let a, b, c be the direction ratio of the line so, it will be perpendicular to normal of 3x – 2y + z + 5 = 0 and 2x + 3y + 4z – 4 = 0
So, using a1a2 + b1b2 + c1c2 = 0
(3) (a) + (– 2) (b) + (1) (c) = 0
3a – 2b + c = 0 …… (2)
Again, (2) (a) + (3) (b) + (4) (c) = 0
2a + 3b + 4c = 0 …… (3)<|endoftext|>
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One of the earliest methods of conservation tillage came to be known as contour plowing, or "plowing on the contour." Tilling the soil along the gentle slopes of a piece of cropland, instead of up and down the gradient, prevents fertile topsoil from being carried downhill by flowing rainwater. This preventive measure is most important in areas which are prone to violent storms or heavy rains. Not only is the topsoil kept in place, minerals like salt or additives such as fertilizers, insecticides or weed control agents, as well as bacteria from animal waste are not swept away to pollute bodies of potable water.
In Thomas Jefferson's time, contour plowing was called more simply "horizontal plowing." Jefferson had won a coveted medal from the major agricultural society in France for his design of the moldboard plow, but he began to notice drawbacks to the heavy use of that instrument. One of his relatives, a politically active farmer named Thomas Mann Randolph, was inspired to develop a new plowing technique in order to salvage the hilly areas in Virginia. Instead of funneling water down, like shingles on the roof of a house, it caught the rain in little ridges of upturned earth. Jefferson commented on a noticeable improvement, specifying that the horizontal furrows retained surplus rainwater and allowed it to evaporate back into the soil.
Even after this successful experiment, later versions of the moldboard plow caused damage to the delicate top-soil of the great plains and prairies of the Midwest United States. The most dramatic evidence of soil erosion took the form of huge dust storms and crop failures during the Great Depression. Since then, contour plowing and other forms of conservation tillage have been reinstituted.
Drawbacks to contour plowing have caused it to be less widely used than conventional tillage methods. Some farmers may not have been fully aware of erosion damage and prevention. Lack of access to equipment, funding, or training sometimes take their toll. One of the main limitations of contour plowing results from its contribution of pockets of untilled land. These untended spots eventually develop weeds, which require extra herbicides. Killing off the weeds sometimes destroys surrounding grasses, which in turn leaves another opportunity for rainwater runoff to arise. To combat this possibility, contour plowing is often applied in combination with other soil conservation techniques, such as terracing.<|endoftext|>
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# The Mean Value Theorem
Another application of the derivative is the Mean Value Theorem (MVT). This theorem is very important. One of its most important uses is in proving the Fundamental Theorem of Calculus (FTC), which comes a little later in the year.
See last Fridays post Foreshadowing the MVT for an a series of problems that will get your students ready for the MVT.
Here are some previous post on the MVT:
Fermat’s Penultimate Theorem A lemma for Rolle’s Theorem: Any function extreme value(s) on an open interval must occur where the derivative is zero or undefined.
Rolle’s Theorem A lemma for the MVT: On an interval if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b) and f(a) = f(b), there must exist a number in the open interval (a, b) where ‘(c) = 0.
Mean Value Theorem I Proof
Mean Value Theorem II Graphical Considerations
Darboux’s Theorem The Intermediate Value Theorem for derivatives.
Mean Tables
Revised from a post of October 31, 2017
# Graphing – an Application of the Derivative.
Graphing and the analysis of graphs given (1) the equation, (2) a graph, or (3) a table of values of a function and its derivative(s) makes up the largest group of questions on the AP exams. Most of the other applications of the derivative depend on understanding the relationship between a function and its derivatives.
Here is a list of posts on these topics. Since this list is rather long and the topic takes more than a week to (un)cover,
Tangents and Slopes
Concepts Related to Graphs
Open or Closed? Concerning intervals on which a function increases or decreases.
Extreme Values
Concavity
Joining the Pieces of a Graph
Using the Derivative to Graph the Function
Comparing the Graph of a Function and its Derivative Activities on comparing the graphs using Desmos.
Writing on the AP Calculus Exams Justifying features of the graph of a function is a major point-earner on the AP Exams.<|endoftext|>
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Natural selection is a process of evolution whereby the gene pool for a population is changed over time in favor of traits that are more advantageous to the survival of the population. Less advantageous traits are therefore selected against and are reduced in the population over time. This selective pressure has a population-level effect as a result of the interaction of traits of individuals in their environment. Selection does not take place on the level of the individual for specific traits.
Fitness is a conceptual measure of an individual's potential for reproducing, passing on their traits to progeny and thereby maintaining representation for their underlying genes in the population's gene pool. Many aspects can contribute to fitness, including how fit the individual is for its particular environment. For example, a bird with a mutation creating a long, pointy, awkward beak may have lower fitness than its efficient, short-beak cousins because of a decrease in ability to eat readily available insects. However, if a change in environmental conditions reduces the insect population such that the birds can now only find enough to eat by accessing deep into dirt and the bark of trees, the long-beaked mutant will now have greater fitness, more able to access the available food source.
Differential reproduction refers to unequal capacities for individuals of a population to reproduce due to trait variation. Some individuals will be more likely to reproduce than others. For example, individuals that have a trait for camouflage against predators will be more likely to survive to reproduce.
Natural selection is impactful on the level of a population and does not select for the traits of an individual, but it is the survival and reproduction of a specific individual that passes their traits on to their progeny that plays a role in driving the gene pool of the population. In group selection, traits are selected for not expressly for the survivability and reproduction of an individual but for greater survivability and reproduction of the group. For example, acts of altruism and resource-sharing may not specifically benefit the providing individual but can improve the survivability of a group significant enough to outweigh loss to the individual. Human acts of agriculture and animal breeding produce an artificial selection.
The products of evolution can be seen from many perspectives, from population change to the survivability of individuals to the presence of a particular gene or even its specific alleles. Alleles that produce traits that are selected for will increase in frequency in the gene pool, thereby creating another measure of evolutionary success.
Speciation is the evolutionary process of forming new and distinct species out of a shared ancestral population.
Genetic polymorphism refers to distinct genetic variations present in a population, such as genes which have multiple alleles represented at significant frequencies (e.g. eye color). These variants provide opportunity for selective forces to act on the expressed traits.
Selective pressures can bring about adaptation and specialization within a population. Adaptation is the improved fit of a population for its environment (e.g. mountain goats becoming more capable at traversing steep, rocky terrain). Specialization is the association with a specific ecological niche (e.g. nocturnal behavior that helps to avoid predators).
Inbreeding, reproduction among closely related individuals (usually true for a small population), limits genetic diversity (raw material for selection) and increases likelihood of recessive disorders within a population.
Outbreeding, reproduction among distantly or unrelated individuals, improves chances of genetic diversity and protects against the widespread increase in recessive disorders within a population.
Evolutionary bottlenecks are dramatic reductions in allelic diversity brought about by a sudden reduction in population (e.g. due to environmental catastrophe). This alteration to the gene pool completely eliminates some portion of previous genetic variation while giving opportunity for previously less frequent but surviving traits to increase in frequency. Relatedly, the founder effect, whereby a very small population moves into a new area or is separated off from a larger group, produces a chance for expressed traits based on a very small gene pool and not necessarily selected for.
Genetic drift and other random factors will produce changes over time that are not selected for. By measuring the amount of random change or mutation rates in a genome, it is theorized to be able to estimate the amount of time that passed since the diverging of different species using this so-called molecular evolutionary clock.
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# Sums of Infinite Geometric Series
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Practice Sums of Infinite Geometric Series
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Finding the Sum of an Infinite Geometric Series
Your task as Agent Infinite Geometric Series, should you choose to accept it, is to find the sum of the geometric series $\sum \limits_{n=1}^{\infty}3 \left(\frac{1}{3}\right)^{n-1}$ .
### Guidance
In the previous concept we explored partial sums of various infinite series and observed their behavior as $n$ became large to see if the sum of the infinite series was finite. Now we will focus our attention on geometric series. Look at the partial sums of the infinite geometric series below:
Series $\sum \limits_{n=1}^{\infty}3(1)^{n-1}$ $\sum \limits_{n=1}^{\infty}10 \left(\frac{3}{4}\right)^{n-1}$ $\sum \limits_{n=1}^{\infty}5 \left(\frac{6}{5}\right)^{n-1}$ $\sum \limits_{n=1}^{\infty}(-2)^{n-1}$ $\sum \limits_{n=1}^{\infty}2 \left(- \frac{1}{3}\right)^{n-1}$
$S_5$ 15 30.508 37.208 11 1.506
$S_{10}$ 30 37.747 129.793 $-341$ 1.5
$S_{50}$ 150 40 227485.954 $-3.753 \times 10^{14}$ 1.5
$S_{100}$ 300 40 2070449338 $-4.226 \times 10^{29}$ 1.5
From the table above, we can see that the two infinite geometric series which have a finite sum are $\sum \limits_{n=1}^{\infty}10 \left(\frac{3}{4}\right)^{n-1}$ and $\sum \limits_{n=1}^{\infty}2 \left(- \frac{1}{3}\right)^{n-1}$ . The two series both have a common ratio, $r$ , such that $\left | r \right \vert < 1$ or $-1 .
Take a look at the formula for the sum of a finite geometric series: $S_n=\frac{a_1\left(1-r^n\right)}{1-r}$ . What happens to $r^n$ if we let $n$ get very large for an $r$ such that $\left |r \right \vert <1$ ? Let’s take a look at some examples.
$r$ values $r^5$ $r^{25}$ $r^{50}$ $\ldots$ $r^n$ or $r^{\infty}$
$\frac{5}{6}$ $0.40188$ $0.01048$ $0.00011$ $0$
$- \frac{4}{5}$ $-0.32768$ $-0.00378$ $0.00001$ $0$
$1.1$ $1.61051$ $10.83471$ $117.39085$ keeps growing
$- \frac{1}{3}$ $-0.00412$ $-1.18024 \times 10^{-12}$ $1.39296 \times 10^{-24}$ $0$
This table shows that when $\left | r \right \vert <1$ , $r^n=0$ , for large values of $n$ . Therefore, for the sum of an infinite geometric series in which $\left | r \right \vert <1, S_{\infty}=\frac{a_1 \left(1-r^n\right)}{1-r}=\frac{a_1\left(1-0\right)}{1-r}=\frac{a_1}{1-r}$ .
#### Example A
Find the sum of the geometric series if possible. $\sum \limits_{n=1}^{\infty}100 \left(\frac{8}{9}\right)^{n-1}$ .
Solution: Using the formula with $a_1=100$ , $r=\frac{8}{9}$ , we get $S_{\infty}=\frac{100}{1- \frac{8}{9}}=\frac{100}{\frac{1}{9}}=900$ .
#### Example B
Find the sum of the geometric series if possible. $\sum \limits_{n=1}^{\infty}9 \left(\frac{4}{3}\right)^{n-1}$ .
Solution: In this case, $\left | r \right \vert=\frac{4}{3}>1$ , therefore the sum is infinite and cannot be determined.
#### Example C
Find the sum of the geometric series if possible. $\sum \limits_{n=1}^{\infty}5(0.99)^{n-1}$
Solution: In this case $a_1=5$ and $r=0.99$ , so $S_{\infty}=\frac{5}{1-0.99}=\frac{5}{0.01}=500$ .
Intro Problem Revisit In this case $a_1=3$ and $r=\frac{1}{3}$ , so $S_{\infty}=\frac{3}{1-\frac{1}{3}}=\frac{3}{\frac{2}{3}}=\frac{9}{2}=4.5$ .
### Guided Practice
Find the sums of the following infinite geometric series, if possible.
1. $\sum \limits_{n=1}^{\infty}\frac{1}{9} \left(- \frac{3}{2}\right)^{n-1}$
2. $\sum \limits_{n=1}^{\infty}4 \left(\frac{7}{8}\right)^{n-1}$
3. $\sum \limits_{n=1}^{\infty}3(-1)^{n-1}$
1. $\left | r \right \vert=\left |- \frac{3}{2} \right \vert=\frac{3}{2}>1$ so the infinite sum does not exist.
2. $a_1=4$ and $r=\frac{7}{8}$ so $S_{\infty}=\frac{4}{1- \frac{7}{8}}=\frac{4}{\frac{1}{8}}=32$ .
3. $\left | r \right \vert=\left |-1 \right \vert=1 \ge 1$ , therefore the infinite sum does not converge. If we observe the behavior of the first few partial sums we can see that they oscillate between 0 and 3.
$S_1&=3 \\S_2&=0 \\S_3&=3 \\S_4&=0$
This pattern will continue so there is no determinable sum for the infinite series.
### Practice
Find the sums of the infinite geometric series, if possible.
1. $\sum \limits_{n=1}^{\infty} 5 \left(\frac{2}{3}\right)^{n-1}$
2. $\sum \limits_{n=1}^{\infty} \frac{1}{10} \left(- \frac{4}{3}\right)^{n-1}$
3. $\sum \limits_{n=1}^{\infty} 2 \left(- \frac{1}{3}\right)^{n-1}$
4. $\sum \limits_{n=1}^{\infty} 8(1.1)^{n-1}$
5. $\sum \limits_{n=1}^{\infty} 6(0.4)^{n-1}$
6. $\sum \limits_{n=1}^{\infty} \frac{1}{2} \left(\frac{3}{7}\right)^{n-1}$
7. $\sum \limits_{n=1}^{\infty} \frac{5}{3} \left(\frac{1}{6}\right)^{n-1}$
8. $\sum \limits_{n=1}^{\infty} \frac{1}{5} (1.05)^{n-1}$
9. $\sum \limits_{n=1}^{\infty} \frac{4}{7} \left(\frac{6}{7}\right)^{n-1}$
10. $\sum \limits_{n=1}^{\infty} 15 \left(\frac{11}{12}\right)^{n-1}$
11. $\sum \limits_{n=1}^{\infty} 0.01 \left(\frac{3}{2}\right)^{n-1}$
12. $\sum \limits_{n=1}^{\infty} 100 \left(\frac{1}{5}\right)^{n-1}$
13. $\sum \limits_{n=1}^{\infty} \frac{1}{2} \left(\frac{5}{4}\right)^{n-1}$
14. $\sum \limits_{n=1}^{\infty} 2.5(0.85)^{n-1}$
15. $\sum \limits_{n=1}^{\infty} -3 \left(\frac{9}{16}\right)^{n-1}$
### Vocabulary Language: English
partial sums
partial sums
A partial sum is the sum of the first ''n'' terms in an infinite series, where ''n'' is some positive integer.
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It has four main components: plasma, red blood cells, white blood cells, and platelets. However, while the lack of a nucleus makes a red blood cell more flexible, The percentage of whole blood volume that is made up of red blood cells is. Components of Blood and Blood Disorders - Learn about from the MSD Manuals - Medical Consumer Version. Red blood cells. White blood cells. Platelets. Your blood is made up of liquid and solids. The liquid part, called Bone marrow , the spongy material inside your bones, makes new blood cells. Blood cells.
Blood carries oxygen and nutrients to living cells and takes away their waste products. All of these functions make blood a precious fluid. They circulate around the body for up to days, at which point the old or damaged RBCs are . red blood cells, which carry oxygen throughout the body; white blood cells, which bones — makes the red blood cells, the white blood cells, and the platelets. They have a lot to do with how your body cleans things up and helps wounds. The different components that make up blood. Plasma, white blood cells, red blood cells, platelets.
A blood cell, also called a hematopoietic cell, hemocyte, or hematocyte, is a cell produced Together, these three kinds of blood cells add up to a total 45% of the blood tissue by volume, with the remaining 55% of the volume composed of. Red blood cells play an important role in your health by carrying fresh cells to function, and your bone marrow cannot make enough to keep up with demand. In this latter trait, they are similar to the primitive prokaryotic cells of bacteria. Red cells normally make up % of the total blood volume. They transport. WebMD describes the anatomy of human blood including what makes up our blood and how circulation works.<|endoftext|>
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The symbols are familiar to us—the wreaths, fir trees, light in the darkness of winter, mistletoe and ivy, caroling, the giving of gifts and a big feast with family and friends.
Interestingly enough, these same traditions would also be equally familiar to the pre-Christian denizens of ancient Rome. An ancient holiday, Saturnalia, was the origin to many of the rituals that we observe today during the Christmas season.
Saturnalia was a feast held in honor of Saturn, god of agriculture, plenty and wealth. Ancient Romans celebrated his festival over the course of several days in mid-to-late December, after the harvest was fully completed and at the time of the winter solstice. The holiday was marked by much merrymaking, gambling and drinking. It was a time when slaves could go without punishment and men celebrated free speech. It was a holiday free of work, courts were not in session, schools were closed, exercise went by the wayside and no wars could be declared.
It was also a time when ancient rituals and symbols came into play—many of the same that form the basis for our holiday traditions today.
December 25th is not the real birthday of Jesus. His birth is not mentioned in the Bible at all or in any sacred writings. The more likely reason for this choice of date is to coincide with pagan mid-winter festivals (of which there were many across the vast Empire) in order to unify pagan and Christian worship celebrations. The 25th was the birthday of Saturn and it also happened to coincide with a date important in the cult of Mithras, a religion with striking similarity to Christianity.
The ancient Romans and Greeks both believed that wearing a wreath as a crown, or corona, would help the wearer stave off drunkenness. They were made from flowers and tree branches and during Saturnalia were often made from fir (in honor of Saturn’s grandson, Faunus, known in Greek as Pan) and ivy (in honor of Saturn’s consort, Ops). Roman soldiers gave Jesus a crown of thorns (it may have been holly) in mockery of his Roman “triumph.”
The Romans would also use fir branches as decor inside and outside buildings, to celebrate fertility, abundance and life in the depths of winter.
Holly and Ivy
Like the branches of the fir tree, holly and ivy are plants that stay green throughout the year. Holly was a plant sacred to Saturn and the ancient Romans used it liberally as a decoration during the Saturnalia festivities. Ivy was sacred to Bacchus, the god of wine and revelry, and was used all over Rome in celebration of Saturnalia. It was also a symbol of marriage and friendship and Saturnalia was a time for family and friends to come together.
Ancient Greeks first began the tradition of kissing under the mistletoe. The Romans adopted it as part of the Saturnalia feast and would sometimes settle wartime agreements underneath a mistletoe sprig, making it a symbol of peace. As an evergreen plant, it was also used as a symbol of abundance and fertility in Saturnalia decorations.
During Saturnalia people would go singing door to door, but fortunately for us, caroling of today does not entirely mimic another Saturnalia tradition—running through the streets naked and singing. The latter is the reason that throughout the centuries Christian churches have wavered on their policies about caroling, with the idea that singing coincides with debauchery.
The concept of giving gifts was an important part of Saturnalia. Usually the gifts were small trinkets, dolls, sweets and the like. Masters would also give gifts to their slaves (perhaps a precursor to the idea behind Boxing Day, when employers would give a gift to their servants the day after Christmas).
Taking Care of the Less Fortunate
Saturnalia was also a time of swapping places. Masters would trade places with their slaves, serving them at meals instead of the other way around (although the slaves would still have to cook the meals). This tradition, of masters serving servants, would continue for a time but eventually fell out of favor. Today we look for ways to help those who are needier than ourselves, with many working in soup kitchens on the holiday to feed the poor. In the UK, British officers traditionally serve the holiday meal to junior officers, with Prince Harry recently carrying on the practice.
The winter solstice was the longest day out of the year and light was an important part of the holiday ritual. Romans gave gifts of candles to each other and placed them in the windows of their homes to call the sun back from the darkness.
Saturnalia marked the very end of the harvest. Ancient Romans would dine on the fruits of their labors, with all manner of foods, apples, nuts, honey, fritters, meat and heavily spiced wine. Emperors would shower figs and sweets on crowds during gladiator games. On December 25th, the birthday of Saturn, they would dine on wild boar or the flesh of a goose or an exotic bird such as a peacock. Today we still eat ham and goose in abundance during the Christmas holidays.
While history tells us that many of our Christmas traditions were appropriated from pagan rituals (not just Roman, but many other traditions ranging from the Vikings to Germany and even Mexico where poinsettias originate), the devout should take heart in the underlying meanings of these customs. They symbolize the importance of family, of convivium, of good overcoming evil, of light breaking apart the darkness and life growing once again. All perfectly good reasons to celebrate, most would agree.
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# 2023 AMC 10B Problems/Problem 16
## Problem
Define an $\textit{upno}$ to be a positive integer of $2$ or more digits where the digits are strictly increasing moving left to right. Similarly, define a $\textit{downno}$ to be a positive integer of $2$ or more digits where the digits are strictly decreasing moving left to right. For instance, the number $258$ is an upno and $8620$ is a downno. Let $U$ equal the total number of $upnos$ and let $D$ equal the total number of $downnos$. What is $|U-D|$?
$\textbf{(A)}~512\qquad\textbf{(B)}~10\qquad\textbf{(C)}~0\qquad\textbf{(D)}~9\qquad\textbf{(E)}~511$
## Solution 1
First, we know that $D$ is greater than $U$, since there are less upnos than downnos. To see why, we examine what determines an upno or downno.
We notice that, given any selection of unique digits (notice that "unique" constrains this to be a finite number), we can construct a unique downno. Similarly, we can also construct an upno, but the selection can not include the digit $0$ since that isn't valid.
Thus, there are $2^{10}$ total downnos and $2^9$ total upnos. However, we are told that each upno or downno must be at least $2$ digits, so we subtract out the $0$-digit and $1$-digit cases.
For the downnos, there are $10$ $1$-digit cases, and for the upnos, there are $9$ $1$-digit cases. There is $1$ $0$-digit case for both upnos and downnos.
Thus, the difference is $\left(\left(2^{10}-10-1\right)-\left(2^9-9-1\right)\right)=2^9-1=\boxed{\textbf{(E) }511}.$
~Technodoggo ~minor edits by lucaswujc
## Solution 2
Since Upnos do not allow $0$s to be in their first -- and any other -- digit, there will be no zeros in any digits of an Upno. Thus, Upnos only contain digits $[1,2,3,4,5,6,7,8,9]$.
Upnos are $2$ digits in minimum and $9$ digits maximum (repetition is not allowed). Thus the total number of Upnos will be ${9\choose 2}+{9\choose 3}+ {9\choose 4}+...+{9\choose 9}$, since every selection of distinct numbers from the set $[1,2,3,4,5,6,7,8,9]$ can be arranged so that it is an Upno. There will be ${9\choose 2}$ -$2$ digit Upnos, ${9\choose 3}$ -$3$ digit Upnos and so on.
Thus, the total number of Upnos will be $\sum_{i=2}^{9}{{9 \choose i}}=512-10=502$
Notice that the same combination logic can be done for Downnos, but Downnos DO allow zeros to be in their last digit. Thus, there are $10$ possible digits $[0,1,2,3,4,5,6,7,8,9]$ for Downnos.
Therefore, it is visible that the total number of Downnos are: $$\sum_{i=2}^{10}{{10\choose i}}= 2^{10}-{10\choose 0}-{10\choose 1} = 1024 - 11 = 1013$$.
Thus $|U-D| = |(1013-502)| =\boxed{\textbf{(E) }511}.$
~yxyxyxcxcxcx ~JISHNU4414L (Latex)
## Solution 3
Note that you can obtain a downo by reversing an upno (like $235$ is an upno, and you can obtain $532$). So, we need to find the amount of downos that end with 0 since if you 'flip' the numbers, the upno starts with a 0 which corresponds to a downo. We can find the cases that end with a 0: $$\sum_{n=0}^9 \dbinom{9}{n} = \dbinom{9}{0} + \dbinom{9}{1} + \cdots + \dbinom{9}{9}$$ to get 512. However, 0 itself is not a valid case (since it has 1 digit) so we subtract 1. Our answer is $\boxed{\textbf{(E) }511}$.
-aleyang
-ap246(LaTeX)
## Solution 4 (Educated Guess)
First, note that the only $downnos$ that are not contained by the set of $upnos$ is every $downno$ that ends in $0$.
Next, listing all the two digits $downnos$, we find that the answer is more than 10, since there are more digits to be tested and there are already 9 two digit $downnos$. This leaves us with $512$ or $511$.
Next, we notice that all the possibilities for $2$ through $9$ digit $downnos$ ending in $0$ pair up with one another, as the possibilities are equal (e.g. possibilities for $2$ digits = possibilities for $9$ digits, etc.).
This leaves us with one last possibility, the ten digit $downno$ $9876543210$.
Since all the previous possibilities form an even number, adding one more possibility will make the total odd. Therefore, we need to choose the odd number from the set $[511, 512]$.
Our answer is $\boxed{\textbf{(E) }511}$.
~Stead (a.k.a. Aaron)
## Solution 5
We start by calculating the number of upnos. Suppose we are constructing an upno of $n$ digits such that $n \ge 2$. An upno can't start with a "$0$", so there are $9$ digits to choose from. There are $\dbinom{9}{n}$ ways to choose an upno with $n$ digits. This is because for each combination of digits, only one combination can form an upno. Therefore, for $2 \le n \le 9$, the total number of upnos is $$\binom{9}{2} + \binom{9}{3} + \binom{9}{4} + \cdots + \dbinom{9}{9} = 2^9 - \binom{9}{1} - \binom{9}{0} = 2^9 - 10.$$
Similarly, the digits of a downo of $n$ digits can be chosen among 10 digits to choose from, since $0$ can be a digit of the downo as the last digit. Thus, the number of downos is $$\binom{10}{2} + \binom{10}{3} + \binom{10}{4} + \cdots + \dbinom{10}{9} + \dbinom{10}{10} = 2^{10} - \binom{10}{1} - \binom{10}{0} = 2^{10} - 11.$$ Thus, $$|U - D| = |(2^9 - 10) - (2^{10} - 11)| = (2^{10} - 11) - (2^9 - 10) = 2^{10} - 2^9 - 1 = \boxed{\textbf{(E) }511}$$
~rnatog337
## Solution 6 (1-1 Correspondence)
Notice that the number of upno integers are the number of subsets of the set $(1, 2, 3, 4, 5, 6, 7, 8, 9)$, excluding the empty set ($\emptyset$) and the one-digit integers. So, the number of upno integers is $2^9-1-9=502$.
The number of downo numbers are similar, but with $0$. So, the number of downo integers is the number of subsets of the set $(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)$, excluding the empty set and the one-digit integers. So, the number of downo integers is $2^{10}-1-10=1013$.
Therefore, the difference between the number of downo integers and upno is $1013-502=\boxed{\textbf{(E) }511}$.
~MrThinker
## Solution 7 (Bijection)
Observe that each downno (that doesn't end with a $0$) maps to one upno. It suffices to count the number of downnos that end with $0$.
Given that a downno ends with $0$, we want to choose the remaining digits from the set $\{1, 2, \ldots, 9\}$. We can arrange the elements of any subset so that they are increasing. We may include or exclude any of the $9$ elements, giving $2^{9}$ subsets. Subtracting $1$ to account for the empty set, we have $2^9 - 1 = \boxed{\textbf{(E) }511}$ more downnos than upnos.
-Benedict T (countmath1)
## Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
## See also
2023 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.<|endoftext|>
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Thrush is a condition that is normally associated with newborns and infants. Ask any mother, and the chances are at least one of her children had thrush. It is no big deal for most kids, and usually clears up on its own. However, when thrush appears in adults, it can be harder to get rid of, and may often be an indicator of a more serious problem such as immune system dysfunction due to an underlying illness or certain medical treatments.
What is Thrush?
Thrush, also known as candidiasis, is an infection caused a yeast-like fungus called Candida albicans or Candida for short. This organism can and does cause infections in other parts of the body, but when infection is found in the mouth, it is called thrush. It is the same agent that causes vaginal yeast infections in women, and diaper rash in babies. Candida albicans is found in and on the bodies of almost everyone, and small amounts of it are normal.It is typically found in the digestive tract and on the skin, as well as in the mouth. Thrush occurs when Candida grows out of control and is not held in check by other bacteria and microorganisms that normally provide a balance.
Oral thrush is characterized by creamy, white lesions that are most often found on the tongue or inner cheeks. However, it is known to spread to other areas such as the tonsils, gums, back of the throat, roof of the mouth, or even the esophagus, which can particularly lead to a more serious and critical situation. The lesions are quite unique, and can usually easily be identified as thrush. It is most commonly found in babies and toddlers, and in senior citizens, however it can occur in anyone of any age, especially if they have a compromised immune system. In recent decades, many HIV/AIDS patients have been exhibiting chronic thrush infections, especially in the advanced stages of the illness.
What Are the Symptoms of Thrush?
The lesions associated with thrush are whitish, velvety plaques that often have a red tint due to bleeding of the tissues underneath. They are sometimes described as “cottage cheese-like” in appearance, and may be painful if scraped or aggravated while brushing your teeth, for example. They can show up suddenly, and may slowly increase in number and size. When a thrush infection manifests itself in an infant, it can usually be left to run its course without treatment. Most infections will disappear within about two weeks. The lesions are very identifiable, but in addition to them the baby may have a hard time feeding and be unusually irritable.
It is not unusual for mothers who are breastfeeding to get thrush from their baby if he or she has an infection. Women with thrush infections in their breasts may experience the following signs:
- Very red, sore, and supersensitive nipples.
- Uncharacteristic pain while nursing.
- Tight, shiny areolas.
- Sharp, piercing pains within the breast.
If an individual has a severe case of thrush, or his immune system is weakened, the infection can spread throughout the mouth and throat, and in some instances into the esophagus. Thrush that has spread into the esophagus (esophagitis) can produce a much more serious situation that can result in such signs as:
- Difficulty swallowing
- Painful swallowing
- An uncomfortable sensation of food sticking in the throat or chest.
On rare occasions, usually with immune compromised individuals, Candida infections can spread throughout the body and sometimes create life-threatening situations. Potential complications include infections of the:
- Eyes (endophthalmitis)
- Brain (meningitis)
- Heart (endocarditis)
- Joints (a form of arthritis)
What Are the Causes of Oral Thrush?
There are many different viruses, bacteria, and fungi that exist in our bodies. When in right relationship with each other, they are very beneficial. In exchange for using our bodies as a place to live and breed, these microorganisms protect us against harmful invaders, help to stimulate our immune system when needed, and even aid in the synthesis of essential vitamins and other nutrients.
But there are gangs of other organisms prowling around in the environment that are not so friendly and helpful. In fact, some microbes such as those that cause malaria or certain types of meningitis can be downright deadly. When the balance of “good” and “bad” organisms is in the right proportions, and the immune system is working up to par, the bad guys are usually outnumbered and defeated. But when things are out of whack, unwanted infections can spread like wildfire, and thrush is a good example of that.
The main way that this imbalance occurs is when the body is weakened by other diseases or by the use of certain medications. The following conditions are common ones that can encourage the growth of oral thrush infections:
- Cancer: All types of cancer are very hard on the body’s immune system. Attempting to fight the illness can take a real toll on your ability to fight off other infections as well, and oral thrush is no exception. To complicate matters, cancer treatments like chemotherapy and radiation therapy can further drain the immune system. Candida infections throughout the body can occur alongside many cancers.
- Diabetes is also an illness that is associated with immune system dysfunction. One of side effects of diabetes is that your saliva may have abnormally high amounts of sugar in it. This can easily stimulate outbreaks of oral thrush.
- HIV/AIDS: Oral thrush is only one of several fungal infections that can afflict HIV patients, and it is not the worst scenario. However, the presence of chronic thrush infections can often be an indicator that the condition of the AIDS patient is deteriorating. If an HIV positive person develops chronic thrush, it is not a good sign. It points to the development of full-blown aids. Most folks in this situation will not experience oral thrush until their T-cell count has fallen below 350. T-cells are critical components of the immune system. For HIV positive persons, if the thrush spreads to the esophagus, it is usually considered an “AIDS-defining illness.” Oral thrush (or the lack thereof) is often considered to be a barometer of health for HIV/AIDS patients.
- Vaginal yeast infections: These common infections are caused by the same fungus as oral thrush, and the presence of vaginal yeast infections also increases ones chances of getting oral thrush. Vaginal Candida infections are very common, and the way that most newborns get oral thrush is by being infected by their mother during the birthing process.
- Xerostomia: This is the official name for “dry mouth,” or a condition that occurs when the levels of saliva in the mouth are below normal. Saliva is an amazing substance that has many functions, including cleansing the mouth and helping to provide a balanced environment that keeps microorganisms in check. When saliva is lacking, the risk for oral thrush increases. There are many factors that can cause xerostomia, including many illnesses, nutritional deficiencies, and literally hundreds of medications. Even high levels of stress or anxiety can result in chronic dry mouth.
- Chronic mucocutaneous candidiasis: This is a rare family of disorders that most typically affect kids under the age of three. It is characterized by Candida infections of the mouth, fingernails, and skin on the scalp, hands and feet. It often produces scaly lumps called “granulomas” that appear in the mouth or on the fingernails and skin. Sometimes an adult will develop a tumor of the thymus gland and develop this disorder as well, but that is even more rare than the occurrence in children.
- Certain medications, such as birth control pills, antibiotics, and corticosteroids also increase a person’s risk for oral thrush. The reasons for this are thought to be related to hormonal changes and the negative effects such drugs have on the immune system and/or the balance of flora in the body.
What Treatments Are Available for Thrush and Other Candida Infections?
In otherwise healthy individuals, thrush is not generally a serious condition. It will usually clear up on its own. However, if you suffer from recurring, chronic bouts with thrush, or if you have a compromised immune system, these infections can put you at a greater risk of further complications. In special cases such as these, there are antiviral medications that are available. The problem is that some of them have some very significant side effects, and should probably not be used if you are not in a high-risk group. I would certainly recommend staying away from them if you have any liver problems. Some of them are known to potentially cause liver damage, and patients on these drugs must be regularly monitored for liver function. The use of these medications may be appropriate in the case of someone with a life-threatening immune disorder such as AIDS. Opportunistic infections such as thrush can be very dangerous under such circumstances.
For otherwise healthy people who have thrush, there are some much more “user-friendly” options that are safe and very effective:
- Try eating some organic yogurt that has live cultures. This is a great way to help restore the balance of flora in the body. It works great for adults and even babies who have thrush.
- The use of a quality probiotic product can also help to build up the levels of the good and necessary microorganisms that are typically diminished in folks who have such infections as oral thrush.
Other preventative tips include:
- Don’t smoke. Smoking changes the atmosphere of your mouth and makes it friendlier to infections such as thrush.
- Good oral hygiene: If you allow bacteria from food particles to accumulate in your mouth, it also increases risk for thrush. Be careful about cleanliness if you wear dentures too.
- Limit high-yeast foods such as beer, wine, and bread.
Thrush infections are certainly not the worst health problem a person could have, but they can be quite troublesome for folks in some circumstances. It is a good idea to nip them in the bud if they surface, because it is much more difficult to become free of chronic Candida infections. Thrush is just another way your body talks to you about something that is not quite right. I suggest we get in the habit of listening.<|endoftext|>
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Online abuse is any type of abuse that happens on the web, whether through social networks, playing online games or using mobile phones. Children and young people may experience cyberbullying, grooming, sexual abuse, sexual exploitation or emotional abuse.
Children can be at risk of online abuse from people they know, as well as from strangers. Online abuse may be part of abuse that is taking place in the real world (for example bullying or grooming). Or it may be that the abuse only happens online (for example persuading children to take part in sexual activity online).
Children can feel like there is no escape from online abuse – abusers can contact them at any time of the day or night, the abuse can come into safe places like their bedrooms, and images and videos can be stored and shared with other people.
Developing healthy brains
Every child needs loving, nurturing care from the adults around them if they are to develop the healthy brains they need to be able to thrive and this is even more important in the first few years of life. The toxic stress that abuse causes means that a child's healthy brain development can be impaired.
The good news is our brains aren't fully formed until around the age of 25, so there are plenty of opportunities to help young people develop the skills they need and to overcome the impact of earlier toxic stress.
Watch this short film to find out more about the impact toxic stress has on a child’s healthy brain development.<|endoftext|>
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# What Is The Interval In Math latest 2023
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## Mathematical Modelling of Steps in BharathaNatyam
Mathematical modeling involves creating a model of a real-world system using mathematical techniques such as linear programming, differential equations, etc. When the system model has inherent uncertainty, simulation is used in addition to the mathematical model to represent a stationary or dynamic system. (System in motion).
Adavus in BharathaNatyam (South Indian classical dance art form) represents a set of steps that do not involve expression (nrityam). Thus, Adavus can be studied using mathematical models.
Tattu Adavu involves lifting the feet up and down so that the sound of the tapping can be heard.
The “sollukattu” (Tamil word translated into English as verbal pronunciation of the beats) is rendered at varying tempos. There are also repeated movements of the feet in various counts such as 4, 6 and 8.
The four verb tenses can be pronounced as tai,ya, tai,hi. If the four verb tenses occur at T(1), T(2), T(3) and T(4) where T(I) is the ith instant of the time when the verbal tense is pronounced by the accompanying artist.
The speed or tempo is given by T(2) – T(1) T(3) – T(2) and T(4) – T(3). Ideally, all of these time intervals should be equal. It may be equal if these beats are generated by a machine. But when an artist renders these sounds or beats, the intervals will not be uniform and will vary randomly. These variations can be captured using simulation models.
If the entire step of moving the feet up and down once takes 30 seconds (say) at normal speed. It would take 20 seconds and 10 seconds in the second and third tempos. For example if tai occurs at 0th instant, ya occurs at 13.5 seconds, tai is the waiting time for 3 seconds and hi occurs at 30th second, the upstroke of the feet lasts 13.5 seconds and the downward movement lasts 13.5 seconds and the waiting time lasts 3 seconds. A dancer and a singer cannot render such a uniform movement with exactness as demonstrated by the mathematical model and there may be variations.
The movement of the dancer or performer can be modeled by the position of the torso in space or the x, y, z coordinates and the relative movement of the feet, legs, upper hand, lower hand , head, neck and arm eyes. at the torso.
For a sequence of Tattu Adavu steps beginning at time t = 0 and ending at time t = T the equation of the feet at an instantaneous time t is given by the position of the dancer’s torso and the relative position of the feet with respect to in the Torso.
Since Tattu Adavu involves the tapping of the feet and upward motion, the resulting motion of the toes, for example, can be modeled using algebra using the following discrete time equations, yielding step functions describing movement. Differential equations cannot be used because they would represent a continuous system.
So write these Tattu Adavu equations as y=0 to t=0 y=h to t=T/2 and y=0 to t=T where T is the time period of a beat and h is the maximum pitch reached by one foot. This can be fixed at 30 cm or vary between 25 cm and 50 cm. This is the algebraic model of the 1st Tattu Adavu. In the case where a variation model must be used, the algebraic model used must be replaced by a simulation model.
The second adavu tattu or foot tapping with twice per beat can be modeled as y=0 to t=0 y=h to t=T/4; y=0 to t=T/2; y=h to t=3T/4; y= 0 to t = T.
If the locus of the feet is plotted for a larger number of points along the time interval, the same equation can be described as y = 0 to t = 0; y = h/10 and t = T/10; y = h/9 to t = T/9 etc.
A dancer with natural movement will not be able to reproduce the exact mathematical congruence of the height reached by the moving feet with respect to the divisions in the time period of the Sollukattu.
If one plots the actual movement of a dancer’s feet when performing ‘tattu adavu’ (translated into English as tapping the feet), the resulting equation would be h=0 to t=0, y=0 .6h at t = T/2 and h = 1.1h at t = T etc.
These algebraic equations can be used to write computer programs that use graphics to model the movement of a ballet dancer’s feet. Therefore, some aspects of mechanical steps or adavus can be generated automatically using appropriate models to capture foot movement.
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