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As an introduction Shorrocks surveys savannahs worldwide, describes the climatic patterns of the African savannahs, driven by seasonal shifts in the Intertropical Convergence Zone, and details the different vegetation complexes of "grass and shrub savannah", "tree and shrub savannah", "woodland savannah", and "forest-savannah mosaic".
This is followed by a chapter on vegetation, which considers the relationship between precipitation and biomass or tree cover, offers a typology of the different kinds of savannah plants, and then describes some of the individual grasses and trees. Only the more important species are covered here, but with some sketches of plants and seeds this could serve as a brief field guide. There is also a detailed study of the Serengeti-Mara, looking at local variation in soils and rainfall and the resulting patterns in biomass production and fires.
A chapter on animals passes over insects and birds quickly and focuses on the mammalian megafauna, in something of the manner of a field guide, with descriptions of species accompanied by distribution maps and small black and white photos. Shorrocks also offers some broader biology, however, looking for example at ungulate foot structure and dentition, and at the differences between hindgut fermentation and rumination.
Turning to single species populations, Shorrocks describes some of the methods for estimating species numbers: capture-recapture, identification from neck patterns or ear clipping, transects, aerial photography, etc. Studies of buffalo and wildebeest and modelling of wildebeest and zebra numbers in the Serengeti-Mara, and studies from elsewhere, sugggest that ungulate numbers are limited by rainfall.
Three kinds of species interactions are treated. "Predator-prey" relationships include herbivore-plant, carnivore-herbivore, and parasite-host interactions. Different herbivores are taken by different sized predators; rather than Lotka-Volterra cycles, rainfall appears to be the dominant forcing; and parasites may have the greatest effect on malnourished animals. Competitive interactions include grasses versus trees, herbivore competition and niche differentiation, and kleptoparasitism (kill stealing) between carnivores. And mutualistic interactions include grazing succession, ant-acacias (where Shorrocks provides a guide to identifying the four species of ant that live on whistling thorn acacias) and vulture-mammal scavenging "collaboration".
A final chapter on the savannah community touches on energy flows and food webs, assembly rules (patterns in the species which make up communities), biogeography, and conservation issues (hunting and poaching, habitat destruction, and human-wildlife competition). It's notable that the vast bulk of vegetation is consumed by decomposers such as termites or fires, and that insects (caterpillars and grasshoppers) take about as much as vertebrate herbivores.
All this makes for something of a grab-bag of material: bits and pieces of ecological theory, some mathematical models, field methods, miscellaneous biology, and local detail. The introduction says it is aimed at "senior undergraduate and graduate students" studying savannah or tropical ecology, but it seems a bit unfocused to make a good textbook. There is a clear regional focus, with most of the studies based on work done in East Africa and in particular the Serengeti-Mara.
One obvious audience will be scientists doing fieldwork in the region. Shorrocks includes a good assortment of clearly presented diagrams, however, and doesn't use any statistical tools more sophisticated than regression. So another audience will be travellers on safari who have an interest in ecology and want to go beyond ticking species off checklists and watching animal behaviour — The Biology of African Savannahs is a lot more interesting than the standard wildlife field guides.
Note: The Biology of African Savannahs contains a disconcerting density of "wrong word" spelling errors.
- External links:
- buy from Amazon.com or Amazon.co.uk
- buy from Wordery
- share this review on Facebook or Twitter<|endoftext|>
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In the Black Desert in northeastern Jordan, archaeologists have come across a crispy find — the charred remains of flatbread that was baked 14,400 years ago. This is effectively the oldest evidence of bread making found to date, suggesting that hunter-gatherers were perhaps inspired by their success with wild cereals to set off the agricultural revolution.
The team of researchers at the University of Copenhagen, University College London and the University of Cambridge analyzed 24 charred remains retrieved from fireplaces at a Natufian hunter-gatherer site known as Shubayqa 1.
These remains — which are very similar to flatbreads retrieved at Neolithic and Roman sites in Europe and Turkey — show that our ancestors had been using the wild counterparts of domesticated cereals, such as barley, einkorn, and oat, long before they domesticated the food crops. The wild cereals were ground, sieved, and kneaded prior to being baked into bread.
University of Copenhagen archaeobotanist Amaia Arranz Otaegui, first author of the new study, thinks that the production and cultivation of bread by hunter-gathers may have influenced the domestication of crops — something which the researchers hope to evaluate in the future.
Today, we’re used to plumpy tomatoes and plentiful corn but the wild varieties from whence they came were far less nurturing. For instance, the first bananas that were cultivated in Papua New Guinea used to be stocky and filled with seeds. By contrast, today’s bananas are smooth on the inside and seedless.
It took us more than 10,000 years of selective breeding in order to turn tiny kernels on tall grass into juicy corn on the cob. Imagine the determination, patience, and insight that was required of the first hunter-gatherers that made the huge leap to agriculture.
“Natufian hunter-gatherers are of particular interest to us because they lived through a transitional period when people became more sedentary and their diet began to change. Flint sickle blades as well as ground stone tools found at Natufian sites in the Levant have long led archaeologists to suspect that people had begun to exploit plants in a different and perhaps more effective way. But the flatbread found at Shubayqa 1 is the earliest evidence of bread making recovered so far, and it shows that baking was invented before we had plant cultivation,” explained Tobias Richter, who led the excavations at Shubayqa 1 in Jordan.
“So this evidence confirms some of our ideas. Indeed, it may be that the early and extremely time-consuming production of bread based on wild cereals may have been one of the key driving forces behind the later agricultural revolution where wild cereals were cultivated to provide more convenient sources of food,” he added.
The flatbread was identified after it was analyzed with electronic microscopy at the University College London, a method that allowed the researchers to locate the microstructures and particles of each charred food remain. This was less straightforward than it sounds, however. The researchers had to devise a new set of criteria for identifying flatbread, dough, and porridge-like products.
“Bread involves labor intensive processing which includes dehusking, grinding of cereals and kneading and baking. That it was produced before farming methods suggests it was seen as special, and the desire to make more of this special food probably contributed to the decision to begin to cultivate cereals. All of this relies on new methodological developments that allow us to identify the remains of bread from very small charred fragments using high magnification,” said Professor Dorian Fuller of the UCL Institute of Archaeology.
The findings were reported in the Proceedings of the National Academy of Sciences.<|endoftext|>
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Reader's Theater Methodology
Reader's theater is reading a story aloud, like a play, without memorization, props or a stage. Students are each assigned a character role and read their part with expression, meaning and enthusiasm. It's best done in small non-threatening groups of approximately 6 students so students can become practiced at their roles prior to standing in front of a group or audience. Repeated readings in multi-leveled small groups (high, medium, and low readers together) also provide the practice of "Repeated Guided Oral Reading" which is the only proven method of building reading fluency.
Plays inherently come with built-in strategies to help students read better. The acting out of story dialogue compels readers to work more closely with the text to interpret and project meaning into the experience. As a result, students show improvement in vocabulary, comprehension, and retention.
Reader's theater gives students an outlet for creative expression and a safe platform for building reading confidence which translates into success in many other areas of their lives.
Putting on a single theatrical play in a student environment can require a great deal of planning, preparation, and practice, and it may only provide a spotlight for a handful of students. Reader’s theater, on the other hand, is easy to implement on a daily or weekly basis and provides a balanced platform for "all students to shine."
Strong reading and comprehension skills are the cornerstone for a successful educational experience. Reader’s theater is a well-known fluency building strategy that engages and entertains students while they learn and improve their skills. Because reader’s theater is an approved method of "Repeated Guided Oral Reading," it is a research-based method of reading.
Repeated Guided Oral Reading
Research shows four repeated readings sufficiently improve reading fluency. Most traditional texts cannot hold students' attention for up to four separate readings. Reader's theater truly engages students and is the most popular form of "repeated guided oral reading." Reader's theater provides an easy-to-implement dramatic text that does not require memorization, props, or a stage.<|endoftext|>
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You will translate simple word problems (involving all four basic operations) into one-step equations.
After completing this tutorial, you will be able to complete the following:
Equations and equality.
An equation is a mathematical sentence that shows equality; includes an equal sign, and often includes a variable.
A relationship between mathematical expressions shows equality when the expressions are the same.
In order to solve real-world problems, you can write an equation to help you determine the answer.
Utilizing a step-by-step process when translating problems into one-step equations.
This Activity Objects uses the following step-by-step process when translating problems into one-step equations:
1. Figure out what is being asked for, or identify the unknown.
2. Designate a variable to represent the unknown.
3. Select the words which show the relationship between the unknown and any other information given in the problem.
4. Determine the key word or words which indicate equality.
5. Determine the key word or words which indicate an operation.
6. Form the equation.
|Approximate Time||20 Minutes|
|Pre-requisite Concepts||Learners should be familiar with algebraic expressions, equality, equations, basic operations on whole numbers, and variables.|
|Type of Tutorial||Concept Development|
|Key Vocabulary||algebraic expression, equation, translating problems|<|endoftext|>
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# Thread: Absolute Convergence and the Ratio Test
1. ## Absolute Convergence and the Ratio Test
$\displaystyle \sum(-1)^{1+k}\frac{k^2\cdot4^k}{k!}$
Determine if the series is absolutely convergent, conditionally convergent or divergent. We're using the Ratio and Root Test and absolute convergence in this section.
2. Originally Posted by kl.twilleger
$\displaystyle \sum(-1)^{1+k}\frac{k^2\cdot4^k}{k!}$
Determine if the series is absolutely convergent, conditionally convergent or divergent. We're using the Ratio and Root Test and absolute convergence in this section.
First lets check for absolute convergence. So
$\displaystyle \sum\bigg|\frac{(-1)^{n+1}n^2\cdot{4^n}}{n!}\bigg|$
$\displaystyle =\sum\frac{n^2\cdot{4^n}}{n!}$
So using the root test we see that
$\displaystyle \lim_{n\to\infty}\left(\left|\frac{n^24^n}{n!}\rig ht|\right)^{\frac{1}{n}}$
$\displaystyle =\lim_{n\to\infty}\left(\frac{n^{\frac{2}{n}}4}{(n !)^{\frac{1}{n}}}\right)$
Using the fact that
$\displaystyle n!\sim\sqrt{2\pi{n}}n^ne^{-n}$
We can see that our limit is equal to zero, thus the series converges. And I'm sure you know that if
$\displaystyle \sum|a_n|\text{ converges}\Rightarrow\sum{a_n}\text{ converges}$
$\displaystyle \therefore\sum\frac{(-1)^{n+1}n^2\cdot{4^n}}{n!}$
converges absolutely.<|endoftext|>
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# IBPS PO Quantitative Aptitude Quiz- 12
## IBPS PO Quantitative Aptitude Quiz
Quantitative Aptitude is the most significant part of almost all competitive exams. Candidates appear to be having difficulty with Quantitative Aptitude Questions, so they need to practice more. Even if the portion is difficult, if you practice enough questions and learn the basic concepts, you can surely do well. In this article, we have come up with the IBPS PO Quantitative Aptitude Quiz to enhance your preparation. This IBPS PO Quantitative Aptitude Quiz will assist you to cope with the tough competition. This IBPS PO Quantitative Aptitude Quiz is available to you at no cost. Through this IBPS PO Quantitative Aptitude Quiz, you will learn formulas and short tricks to solve the Quantitative Aptitude Questions. Candidates must solve this IBPS PO Quantitative Aptitude Quiz to improve their exam preparation.
1. Four digits number are formed using numbers 1, 3, 4, 5 without repetition. Find the probability that the number is divisible by 5.
2. In a river the speed of boat in downstream is twice as speed of boat in upstream. If boat takes total 12 hours to go 32 km from A to B and return to A by travelling same distance. Then find speed of boat in still water.
(a) 6 kmph
(b) 8 kmph
(c) 2 kmph
(d) 4 kmph
(e) None of these
3. A invest certain sum on compound interest at 20%, 10% and 25% on 1st year, 2nd year and 3rd year respectively if he receives Rs. 330 as interest in 3rd year. Then how much interest he will receive in all 3 years.
(a) Rs. 720
(b) Rs. 650
(c) Rs. 840
(d) Rs. 780
(e) Rs. 620
4. A is twice efficient as B and together they do the same work in as much time as C and D together. If C and D alone can complete the work in 20 days and 30 days respectively, then in how many days A can complete the same work alone.
(a) 12 days
(b) 18 days
(c) 24 days
(d) 30 days
(e) 32 days
5. The distance between two stations A and B is 497 km. A train starts from A towards B and another from B to A at the same time and they meet after 7 hours. The train travelling from A to B is slower by 7 km/hr compared to other train travelling from B to A then find the speed of slower train.
(a) 22 kmph
(b) 28 kmph
(c) 32 kmph
(d) 40 kmph
(e) 42 kmph
Directions (6 -10): What will come in place of (?) in the following number series?
6. 60, 30, 20, 15, 12, ?
(a) 12
(b) 10
(c) 4
(d) 8
(e) 6
7. 9, 11, 22, 51, 107, ?
(a) 195
(b) 210
(c) 200
(d) 199
(e) None of these
8. 20, 56, 399, 463, 1192, ?
(a) 1280
(b) 1242
(c) 1292
(d) 1442
(e) 1692
9. 112, 141, 172, 209, 250, ?
(a) 301
(b) 287
(c) 282
(d) 297
(e) 293
10. 120, 122, 247, 745, ?, 14931
(a) 2825
(b) 2945
(c) 2725
(d) 2745
(e) 2985
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3<|endoftext|>
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World AIDS Day 2019
World AIDS Day is held on the 1st December each year and is an opportunity for people worldwide to unite in the fight against HIV, show their support for people living with HIV and to commemorate people who have died. World AIDS Day was the first ever global health day, held for the first time in 1988.
Globally there are an estimated 34 million people who have the virus. Despite the virus only being identified in 1984, more than 35 million people have died of HIV or AIDS, making it one of the most destructive pandemics in history.
Today, scientific advances have been made in HIV treatment, there are laws to protect people living with HIV and we understand so much more about the condition. Despite this, people do not know the facts about how to protect themselves and others, and stigma and discrimination remain a reality for many people living with the condition.
World AIDS Day is important because it reminds the public and Government that HIV has not gone away – there is still a vital need to raise money, increase awareness, fight prejudice and improve education.<|endoftext|>
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## Tuesday, September 30, 2014
### Spider Web Math-Art
We used geometry to create a math-art spider web design.
Math often exist within art to levels not detected by the viewer. Learning about the connections not only makes art more interesting, but serves as a very enjoyable way of learning about mathematics.
We have often selected geometric mandalas and used a compass and straight edge to recreate them. The challenge is a math puzzle and an exciting problem solving exercise.
This mandala which looks a bit like a spider web, from the book Das große Mandala-Malbuch für Vorschulkinder was redrawn using only a compass and ruler. Here are the steps.
1. Draw a circle with a 2" radius.
2. Use a ruler to draw a line through the center of the circle.
3. Draw a perpendicular line passing through the center of the circle. Increase compass radius (3 inches or so), place the point of the compass on the line where the circle and straight line intersect and draw tick marks just outside the circle on both sides. Move the compass point to the opposite side and repeat.
Place a ruler to line up with the tick marks and draw the perpendicular line passing through the center.
4. Bisect the perpendicular lines. Place the point of the compass where the circle and line cross. Make a tick mark outside the circle about half way between the perpendicular lines. Repeat on the opposite side of the line. Repeat placing the compass at the other three points where the circle and lines cross.
Place a ruler so that the tick marks are lined up and draw two angle bisectors passing through the center of the circle. (The circle should be divided into eight pieces.)
5. Bisect the eight angles within the circle following the procedure detailed in step 4. Note the bisecting lines are not needed within the circle and are only drawn lightly outside the circle.
6. Draw the outer web arches. Set the compass to the initial radius of 2". Place the pencil part of the compass on the line where the circumference and line passing through the center meet. Place the point on the adjacent line outside the circle. Sketch an arc as shown above. Repeat until the outer web of eight arcs is complete.
7. Draw the second web arc. Increase the compass radius to 2 1/4". Place the point in the same location as in step 6. Draw an arc. Repeat around the entire circle.
8. Increase the compass radius to 2 1/2", then 2 3/4" respectively and repeat step 7 drawing the remaining two web arcs.
9. Draw a small circle in the center.
10. Erase all construction lines.
11. Trace desired pencil lines with a black marker.
12. Color.
For similar projects click the links below.
Tutorial - Four Star Mandala
Math-Art Mandala
## Sunday, September 28, 2014
### Tsunami - Earth Science Unit Study
Week 30 - Tsunami
We built sand castles and then noted the percent destruction when damaging them with varying amounts of water.
Our favorite video series this year by far is How the Earth Was Made. Since there's an episode on tsunamis, we watched it.
We also read a few books about tsunamis.
Tsunami! - When the ocean pulls back, the villagers are curious and walk towards it to investigate. Ojisan is a village elder is not fooled. He's scared, but knows what to do when the Tsunami strikes.
Selvakumar Knew Better - This book tells the true story of how a dog helped save a young boy during the 2004 tsunami.
The Big Wave - This book was about so much more than a tsunami. It told the story of how people dealt with the tsunami years afterwards. I happened to read this book while dealing with my own post traumatic stress disorder due to my daughter's Leukemia and it actually really helped me to cope.
Water Volume verses Destruction
One local park has a nice water/sand play area, so our experiment was done there. First, a sand village was constructed.
Next, one-fourth of a bucket was poured down the channel headed strait for the village.
The water flowed out of the channel and destroyed the structures on the water side of the sea wall.
The village was rebuilt, and the entire process was repeated several times increasing the amount of water in the tsunami.
Each time the percent of damage was noted, and a graph showing water volume vs damage was created with the results.
There are still a few weeks left in our Earth Science study, so be sure to visit next week for more.
## Saturday, September 27, 2014
### Central America and Caribbean Island Songs for Kids
We learned songs to help us remember the countries of Central America and the Islands of the Caribbean.
Learning geography can be confusing to little kids as names referring to locations such as Michigan, United States, North America, Colorado, Europe, Grand Rapids, Rhineland-Pfalz, Traverse City, Denver, Germany and Littleton are thrown around. What is the difference between a continent, country, state and city?
My six year old was so confused at the beginning of the year and so we took a big step back, and now geography is making more sense.
After coloring a map showing the seven continents we learned the continent song.
Next we discussed and learned the names of the four oceans. After working on other projects for several weeks a European country map was colored. In addition, the GeoPuzzle Europe was completed several times over the course of a few weeks.
Proceeding onto North America, rather learning the names of each state, a country map was colored. It was noted that three countries make up the majority of the land.
Since singing worked so well for memorizing the continents, two more songs were learned to aid in remembering the names of the countries in Central America and the Islands of the Caribbean.
Central America Song
Caribbean Island Song
That's two continents down and five more to go. I expect this activity to span the entire year.
Check out these great blog hops. They are filled with activity ideas for kids.
## Friday, September 26, 2014
### Modifying a Crochet Hat Pattern
Do you have difficulty following a pattern? I like to do things my own way, and with knitting and crochet, that tends to involve a lot of math and logical thinking.
I recently completed this Cross-Over Long DC Hat. However, I never seem to have the yarn called for in the pattern, so modifications are always necessary. This hat was made with Boston chunky yarn on a 5.5 size hook. Since both the hook and yarn were bigger than the pattern called for, the number of stitches and rows needed to be reduced. Using chunky yarn the stitches increased at a faster rate than the pattern specified. Instead of 6, 12, 18, 24, 30 and 36 stitches for rounds 1-6, 9, 18, 27, 36, 43, and 52 were used instead. The initial 14 rounds of stitches which defined the top of the hat were reduced to only six rounds.
The next step was the cross-over stitch. Instead of doing the cross-over stitch for rounds 16-23, it was done for rounds 7-14 before moving onto the rim of the hat. Again, more modifications were required at the brim, because I did not want a brim, and only wanted a rim. After six rounds of single crochet at the bottom the hat was finished.
Who said crafts didn't count as math?
## Thursday, September 25, 2014
### Krakow, Poland - Things to See and Do
History from early medieval times through the post World War II period has left its mark on the architecture of Krakow, Poland. The castle area, Jewish quarter and historic district are compact, laid out in a grid and easy to navigate on foot. The people are friendly, the food is good, there's lots to see making Krakow a very affordable and worthwhile place to visit.
Nikolaus Copernicus
Krakow was home to the famous astronomer Nikolaus Copernicus. During his lifetime, astronomers who developed theories of the universe which were contrary to the teachings of the church were severely punished. Copernicus developed a model of the universe which placed the sun in the center, but didn't have his work published until the last days of his life. The University area of Krakow contains a monument memorial to Nikolaus Copernicus.
Barbican
A barbican is a fortified structure which lies just outside a city boundary. Barbicans were built during the middle ages and served as a front line of defense for the city. The city of Krakow has a well preserved barbican which is a rarity as not many are left.
Wawel Castle
The country of Poland and the city of Krakow have a history riddled with occupation by outsiders. The Wawel Castle reflects this history in its architecture. Just looking at the above photo, it is evident from the exterior construction materials and styles that the castle was constructed over a long period of time. The most recent portion, on the right hand side of the photo, was constructed during World War II and severed as a headquarters during the war.
The courtyard portion of the castle dates to the Renaissance period. Tours of the staterooms begin in the courtyard and proceed around the rooms inside this portion of the castle.
St. Mary's Cathedral
The trumpeter still plays every hour from the peak of St. Mary's Gothic Cathedral as a signal that all is well.
The Trumpeter of Krakow is a historical fiction children's book set in the middle ages. After a young boy and his family seek refuge within the city of Krakow after fleeing from the Tarters, he becomes the trumpeter.
The amazingly carved alter within the church stands much higher than this photo indicates. The figures in one of the panels are over 3 feet tall.
Inside, the walls and ceilings are covered with bright and beautiful decoration.
Church of Saints Peter and Paul
Krakow contains many churches including the Renaissance style church of Saints Peter and Paul shown above. This church is most famous for the statues of the apostles in the front and it occasionally gives nighttime orchestral concerts.
Franciscan Church
Pope John Paul is from Poland and spent lots of time in Krakow. His image is visible all over the city; as a statue outside the cathedral at the Wawel castle, as a statue inside the salt mine a 20 minute drive from the city, and on banners, plaques and postcards within the city.
The Franciscan Church contains a plaque on one of the pews stating that John Paul once sat here. Inside the walls were painted with large, bright geometric and floral designs in the Art Nouveau style.
Jewish Quarter
Long before World War II Krakow was a safe haven for Jewish people. When Jews were persecuted in other locations, they not only sought refuge in Krakow, but thrived.
Unfortunately we are all too aware of how that all changed during World War II. The Jewish population of Krakow was virtually wiped off the map. Concentration camps were too close and the horrors of war descended upon the city.
The photo above was taken outside the gates of Oscar Schlinder's factory. Schlinder was a German member of the Nazi party who ran a ceramics factory in Krakow during the war. Once he understood the horrific nature of what was happening around him, he went on a mission to hire and thereby protect as many Jewish people as possible. In comparison to the death toll his efforts were futile. However, he was able to save hundreds of lives and his efforts had lasting impacts on many generations of people. His story is told in the Hollywood movie Schindler's List. (Recommended for adults.)
The Jewish people of Krakow were imprisoned in an area of the city by a concrete wall. The rounded edges of top of the wall were built to resemble the tombstones found in Jewish cemeteries; an especially cruel mental tactic.
Today there is a small Jewish population living in Krakow. The area of the city contains many decorative wrought iron fences featuring Jewish symbols.
Sukiennice (Cloth Hall)
The Sukiennice was built in the Renaissance style in the main square as a place for merchants to sell cloth.
Today the building is still used for sales; only the products now sold are items desired by tourists.
Polish Food
Poland is a northern country and therefore, potatoes, beets, cabbage, pork, and wheat (foods which grow in cold climates) were ingredients in many dishes.
This beet root soup contained beets and dumplings filled with pork.
Stir-fried potatoes and pork were sold in the village square as fast food. In the background, a plate of pierogis is visible. Pierogis are the Polish version of ravioli. Basically they are stuffed with cheese, spinach, or pork and served without tomato sauce.
The city was filled with street vendors selling pretzels. Although they were in rings, they still tasted like pretzels.
Polish Folk Art
Much of Polish folk art and traditions are linked to Christianity. The images contained many bright colors and both orange and blue were prominent in many paintings.
Traditional Polish clothing consisted of a vest, skirt and jacket. Depending on the time frame and region of Poland, the styles were quite different. Some used embroidery, others lace, sequence, beads, or a combination. All were extensively adorned like the beaded vest in the above photo and the hat below.
Hand painting of eggs was perfected in Poland and the Folk Art museum contained a large display.
Wieliczka Salt Mine
About 20 minutes drive outside the city of Krakow there is a salt mine. Inside miners carved over 100 figures and panels into the salt rock.
This figure shows a miner burning off hazardous gas in the morning before other miners arrive for work. As you can imagine, this was the most dangerous job in the mine.
This child-sized elf is one of several which can be found in the mine.
This Bible scene is carved into the wall of one of the three underground cathedrals. The mine contains rooms large enough for banquets and each year it hosts special events such as bungee jumping and weddings.
Traveling is hands-on way to learn. One weekend trip can be packed with enough reading (signs and displays), history, science and PE to easily count as a week of school..... and when it's over, writing about it in a journal makes a nice keepsake.<|endoftext|>
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Fox Going up the food chain, larger nocturnal predators like the red fox and grey fox, are about the size of a small dog weighing nine to 12 pounds, and are typically seen more frequently in the summer. The grey tends to be smaller and stealthier, finding refuge in trees for vantage points and stealing eggs. Foxes are crepuscular, generally hunt at dusk and at dawn, and in midwinter you are more likely to see them out in the daylight if they are cold and hungry, scavenging for food. They are omnivorous and eat mostly rabbits and mice, as well as other small rodents, birds, insects, nuts and fruits.
Bats Bats are the only mammal that can truly fly and Virginia has 17 of the more than 1,000 species of bats, according to the game department. Both tree bats and cave bats reside in the Blue Ridge area, with the big brown bat and little brown bat being the most common cave bats.
Bats generally don’t have much color to them. The ones with the most color are the tree bats like the red and the hoary. Silver-haired bats are more blackish with white tips on the wings that give them their silver look.
Red bats are the only bat species that we have in the East where males and females have distinct coloring. Males are Irish-setter red, while females have a grizzled look to them – white tipping on the fir that gives them a dirtier look – not clean, deep red.
Bats control the mosquito population and can eat as much as 3,000 insects per night.
Bats communicate, hunt and avoid collisions by using ultrasound, creating high-pitched sounds that reverberate off objects in their path. Their calls are too high of a frequency for humans to hear.
A fungus, known as white-nose syndrome, is killing large numbers of bats across the country.It’s found in bats during hibernation and grows on them when their body temperatures drop. It’s typically seen on the face and wings of bats, and it causes them to wake during hibernation and use up their fat reserves that they survive off during the winter. Bats with this fungus can be found flying around midwinter rather than hibernating, causing them to either freeze or starve to death.
“They are seeing a large fatality rate in the northeast and it’s spreading south. In some caves, 90 percent of the bats are dying,” Reynolds says. “It’s now spread south through Virginia and Tennessee.”<|endoftext|>
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# 4.1 Simultaneous Equations
4.1 Simultaneous Equations
(A) Steps in solving simultaneous equations:
1. For the linear equation, arrange so that one of the unknown becomes the subject of the equation.
2. Substitute the linear equation into the non-linear equation.
3. Simplify and expressed the equation in the general form of quadratic equation $$a{x^2} + bx + c = 0$$
$\begin{array}{l} y + x = 9\\ xy = 20 \end{array}$ Solution:
$\begin{array}{l} y + x = 9\\ y = 9 - x \end{array}$ Substitute the linear equation into the non-linear equation.
$\begin{array}{l} xy = 20\\ x(9 - x) = 20\\ 9x - {x^2} = 20 \end{array}$ Simplify and expressed the equation in the general form of quadratic equation $$a{x^2} + bx + c = 0$$
$\begin{array}{l} 9x - {x^2} = 20\\ {x^2} - 9x + 20 = 0 \end{array}$ Solve the quadratic equation.<|endoftext|>
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# Lesson 7
Why Is That Okay?
These materials, when encountered before Algebra 1, Unit 2, Lesson 7 support success in that lesson.
## 7.1: Estimation: Equal Weights (5 minutes)
### Warm-up
The purpose of an Estimation warm-up is to practice the skill of estimating a reasonable answer based on experience and known information, and also help students develop a deeper understanding of the meaning of standard units of measure. It gives students a low-stakes opportunity to share a mathematical claim and the thinking behind it (MP3). Asking yourself “Does this make sense?” is a component of making sense of problems (MP1), and making an estimate or a range of reasonable answers with incomplete information is a part of modeling with mathematics (MP4).
### Student Facing
How many pencils are the same weight as a standard stapler?
1. Record an estimate that is:
too low about right too high
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
Ask a few students to share their estimate and their reasoning. If a student is reluctant to commit to an estimate, ask for a range of values. Display these for all to see in an ordered list or on a number line. Add the least and greatest estimate to the display by asking, “Is anyone’s estimate less than _____? Is anyone’s estimate greater than _____?” If time allows, ask students, “Based on this discussion, does anyone want to revise their estimate?”
Then, reveal the actual value and add it to the display.
Ask students how accurate their estimates were, as a class. Was the actual value inside their range of values? Was it toward the middle? How variable were their estimates? What were the sources of the error?
## 7.2: What’s the Same? What’s Different? (20 minutes)
### Activity
The purpose of this activity is to give students an opportunity to practice checking whether a particular value is a solution to an equation, and to recall properties of operations and equality that preserve the solution set of an equation.
This will be useful when students consider when and why equations have the same solution in the associated Algebra 1 lesson.
Monitor for students who:
• check the value of $$x$$ in only one expression and reason about the second expression
• identify properties by name such as commutative, distributive, and associative
• identify properties informally such as “changing the order doesn’t matter” or “you can add first and then multiply or multiply first and then add” or “you can multiply three or more terms in any order” or “you can add the same thing to each side”
• manipulate the expressions
• graph one or both expressions
Making graphing technology available gives students an opportunity to choose appropriate tools strategically.
### Student Facing
For each pair of equations, decide whether the given value of $$x$$ is a solution to one or both equations:
1. Is $$x = 2$$ a solution to:
1. $$x(2 + 3) = 10$$
2. $$2x + 3x = 10$$
2. Is $$x = 3$$ a solution to:
1. $$x - 4 = 1$$
2. $$4 - x = 1$$
3. Is $$x = \text{-}2$$ a solution to:
1. $$7x = \text{-}14$$
2. $$x \boldcdot 14 = \text{-}28$$
4. Is $$x = \text{-}1$$ a solution to:
1. $$x + 3 = 2$$
2. $$3 + x = 2$$
5. Is $$x = \text{-5}$$ a solution to:
1. $$3 - x = 8$$
2. $$5 - x = 10$$
6. Is $$x = (8 + 1)+3$$ a solution to:
1. $$\frac{12}{2} = \frac12(x)$$
2. $$18 = 2x$$
7. Is $$x = 2$$ a solution to:
1. $$\frac{12}{x} = 6$$
2. $$6x = 12$$
8. Is $$x = \frac{10}{3}$$ a solution to:
1. $$\text{-}1 + 3x = 9$$
2. $$9 = 3x - 1$$
9. Is $$x = \frac12$$ a solution to:
1. $$5(x + 1) = \frac{15}{2}$$
2. $$5x + 1 = \frac{15}{2}$$
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
Ask students to talk to their neighbor about some things they can look for, to see if two equations are likely to have the same solution or different solutions. If needed, draw students’ attention to which equations had the same solution and which had different solutions, perhaps by displaying them for all to see, grouped by whether the equations were equivalent or not.
Call on students to share their thinking and make an informal list displayed for all to see. It’s fine if students do not name all possible properties, or only describe them informally. They will have more opportunities to encounter them formally in their Algebra 1 course.
Remind students that equivalent equations have the same solution set. Use the language of equivalent interchangeably with “have the same solution,” and monitor for students to pick up on this vocabulary.
## 7.3: Generating Equivalent Equations (15 minutes)
### Activity
The purpose of this activity is for students to practice generating equivalent equations. The structure of taking turns and justifying your thinking to your partner supports students to check their work as they go, and the class has a chance to check for equivalence when the total score is recorded.
Students also have to justify, whether formally or informally, why each move keeps the equation equivalent. This will be helpful for students in an associated Algebra 1 lesson when they must determine what moves are acceptable to make to an equation without changing its solution.
Monitor for students who solve the equation first and then work to create new equations with the same answer, and for students who simply create new equivalent equations from the given equation. Monitor for students who add or multiply by the same number on each side of an equation and students who use properties of operations (distributive, commutative, associative).
### Launch
Arrange students in groups of 2. Explain to students that they are going to play a cooperative game in which the class tries to come up with as many different equations with the same solutions as they can. You give them an equation, and their job is to come up with other equations with the same solution as the original equation. The partner’s job is to check that the new equation is equivalent to the original by listening to their partner’s reasoning and making sure they agree. Each partner should create their own equations before moving to the next question.
At the end of each round, you will ask them to share the equations they came up with, and keep track of how many different equations the class came up with. Set a goal for the second round of coming up with 4 more equations than the first round, and continue to set meaningful challenge goals with each round.
Calling on previously identified students to share strategies may help the class be more successful.
Display these equations, one at a time:
1. $$x = 5$$
2. $$x + 1 = 0$$
3. $$2(x + 1) = 10$$
4. $$\frac12x + \frac32 = \frac72$$
### Student Facing
1. Your teacher will display an equation. Take turns with your partner to generate an equivalent equation—an equation with the same solution. Generate as many different equations with the same solution as you can. Keep track of each one you find.
2. For each change that you make, explain to your partner how you know your new equation is equivalent. Ask your partner if they agree with your thinking.
3. For each change that your partner makes, listen carefully to their explanation about why their new equation is equivalent. If you disagree, discuss your thinking and work to reach an agreement.
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
Remind students that today they did a lot of thinking about equivalent equations and moves they could do to equations that wouldn’t change the solution.
Make a semi-permanent display of “moves that won’t change the solutions to equations.” Sort the list by moves that are done to each side, and moves that are done to one side. Use students’ language, adding formal language if that is an emphasis in your school.
Possible list:
• Add the same value to both sides.
• Subtract the same value from both sides.
• Multiply both sides by the same value (but not zero!). Divide both sides by the same value (but not zero!).
• Change the order of terms (on one side) being added or multiplied (commutative property).
• Change the grouping of terms (on one side) being added or multiplied (associative property).
• Distributive property: $$a(b + c) = ab + ac$$.<|endoftext|>
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# Math 12 - Probability Quiz
```Binomial Probability Problems 2
Name:
Solve each of the probability problems below.
1) Another scoundrel alters a coin so that it is a biased coin; each time it’s tossed, there’s a
3
chance of
4
a) Find the probability of tossing exactly 4 tails in 6 tosses.
b) Find the probability of tossing at least 4 tails in 6 tosses.
c) Find the probability of tossing at most 2 tails in 6 tosses
2) You find the empirical probability of winning a game by observing that in the last 100 games played,
you’ve won 76 of them. Based on this, answer the questions below. (Leave answers in expanded form.)
a) Find the probability of winning at most 1 game if you play 5 games in all.
b) Find the probability of winning at least 3 games if you play 5 games in all.
c) Find the probability of losing all games if you play 5 games in all.
3) Mr. Haas was distracted one morning and ended up putting his hard-boiled eggs in a bowl with some
raw eggs. He knows that 5 of the eggs are hard-boiled and 7 of the eggs are raw. Mr. Haas also knows
that if you spin an egg, you can tell if it’s raw or hard-boiled. So he decides to pick up an egg, spin it, and
put it back in the bowl. He’s still a little sleepy, so he forgets to keep track of which eggs he’s already
spun. Given this, answer the questions below. (All answers may be left in expanded form.)
a) If Mr. Haas spins 4 eggs (chosen randomly and replaced), what’s the probability that exactly 3 are
hard-boiled?
b) If Mr. Haas spins 7 eggs (chosen randomly and replaced), what’s the probability that at most 2 are raw?
4) Assume that the figure shown is a dartboard. The dimensions of the large rectangle are 20 by 30 inches
and the dimensions of the inner rectangle are 10 by 15 inches. (Assume that all the darts land
somewhere on the dartboard and that the skill of the dart thrower doesn’t affect the outcome.) Show all
work and find the probability for each question below.
a) If I throw five darts at the dartboard, what’s the probability that exactly 3 of the darts will land in
the smaller rectangle?
b) If I throw ten darts at the dartboard what’s the probability that at most three of the darts will land
in the smaller rectangle? (Leave in expanded form.)
c) If I throw eight darts at the dartboard, what’s the probability that at least six of the darts will land in
the smaller rectangle? (Leave in expanded form.)
```<|endoftext|>
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A threat to security; a liability on the public purse; unwanted and unwelcome. Those sentiments have been applied to today’s refugees, especially this election cycle – but they were also the feelings of many New Englanders to the thousands of Acadians who found themselves placed in towns across Massachusetts (including the District of Maine) in the 1750s and ‘60s. The Acadians weren’t quite refugees, nor were they economic migrants like their French-Canadian cousins who would come to the region in the following centuries in their hundreds of thousands. The Acadians were prisoners and deportees – civilian victims of an undeclared war; described by some as genocide. In 1713, the British had conquered the portion of Acadia that they called Nova Scotia, as well as the northern part of Maine. Although Great Britain and France remained at peace in the following decades, settlers in Maine and Nova Scotia continued to come into conflict with the Acadian inhabitants and, in particular, the Native Americans with whom the Acadians shared close relationships. A series of guerilla conflicts, including Father LeLoutre’s War and Father Rale’s War, not only continued to alienate the English-speaking ruling class from their Acadian subjects, but also increased the desire of land-owners and settlers (many of them from Massachusetts) to gain control of the Acadians’ lands. When France and Britain went to war again in 1744, the British soon found the sympathies of their Acadian subjects to be with the French.
The result was the Grand Dérangement, or “Great Upheaval” – the expulsion of the Acadians from their ancestral lands, and the defining tragedy of the Acadian people. As noted above, some historians refer to the event as a genocide, though the objective was eviction, rather than mass murder. Nonetheless, in their paranoia about the divided loyalties of their French-speaking subjects, and their greed for the Acadians’ fertile land-holdings, the Anglo-Americans caused the death of hundreds. Many Acadians perished in the wrecks of the ships that were transporting them from their homes; unknown numbers surely died of hunger and other deprivations suffered in exile. At the direction of Charles Lawrence, the British governor of Nova Scotia, Acadians across the colony were directed to come to their parish church for an important proclamation. In the presence New England militiamen who had been sent from Boston for the purpose, those who came to the churches were ordered to swear an oath of loyalty to the crown or face deportation. The oath was unacceptable to the Acadians (and designed to be so), because it required them to acknowledge the King of Great Britain as the head of the Church of England – in essence, to recant their faith as Catholics.
For their refusal, thousands of men, women and children were forced onto ships and sent into exile. Their lands would be confiscated, and the Acadians would be scattered across Britain’s colonies, too dispersed to be a security threat any longer. Around two thousand were sent to Massachusetts; many hundreds more made their way there from southern colonies in the years to come. Although the Massachusetts diaspora is fairly well-known, the status of the few dozen Acadians apportioned to the “District of Maine”, then a part of the Commonwealth, has commonly been overlooked. As with many Acadian exiles, records for this group are hard to come by. The new arrivals were destitute; few of them would have been literate, let alone in English, and they were often shuttled from town to town. To add to the difficulty, their names are often Anglicized, so Roy became King and Doucette became Dowset. Nonetheless, what evidence does survive shows that these first Acadian arrivals in Maine were typical of their thousands of fellow-exiles across British North America.
The Committee on the apportionment of Neutrals for the Commonwealth of Massachusetts was assigned the task of dividing the colony’s Acadian population between its various towns. The task was a delicate one, because the Acadians were, frankly, unwelcome. Since they had been required to leave their possessions behind (and, in any case, most had been farmers, whose wealth had been in their now-forfeited land), they were paupers – which made them the responsibility of the various towns to which they were assigned. The board of overseers of the poor in each town was responsible for seeing that the Acadians were clothed, housed and fed, sometimes by entrusting an Acadian family to a wealthy local family. In 1760, the Committee reported to the General Court that it had assigned 61 Acadians to towns in Maine.
The bulk of these individuals appear to have come to Maine from Point Shirley (Winthrop, MA), and for some, this wasn’t the first time they had been shuffled from one town to another. The records of the General Court contain numerous petitions from towns complaining at the number of Acadians (or “French Neutrals”) they were required to support. John King (Jean Roy)’s family of ten were in Billerica in March 1759, but were sent to Dunstable later that year, when the selectmen of Billerica complained there was “not enough available housing.” Just six months later, in June 1760, the Kings were among those moved to Point Shirley for removal to Maine. Similarly, Peter White (Pierre LeBlanc), his wife and five children were supported by the town of Littleton in late 1758, before being sent by the Littleton Selectmen to Point Shirley in 1760.
It’s clear that even before the 1760 reassignments, there were some Acadians already living in York County. Caleb Symmes, a ship’s captain, claimed reimbursement for transporting just thirty-three Acadians from Point Shirley to York that July, so the sixty one individuals listed by the Committee probably include those already in Maine. As early as 1756, Francis (François) LeBlanc petitioned the General Court that he be moved from Point Shirley to the Town of York, “for his more comfortable subsistence among some friends and relations of his who dwell there.” Ultimately, however, the committee determined to assign LeBlanc to Needham – without offering an explanation. Once again, the town of York may have been reluctant to accept the cost of another Acadian family.
Another early Acadian arrival in Maine was Peter Douset (Pierre Doucette) and his family, who came to Wells on 15th January, 1756. The Doucettes were assigned to the care of Colonel Nathaniel Donnell, who put them up at a cost of 30 shillings a week to the town. In 1757, Pierre’s wife [name?] fell pregnant, and Donnell had a new house rented for the family’s use. In a petition to the General Court for additional reimbursements, Donnell wrote that she was “disordered the whole winter and not able to do anything to support the family, and I was obliged to find wood for two fires the most part of the Winter.”
The Doucettes were under the responsibility of Colonel Nathaniel Donnell; another family, the Roys (Kings) were similarly the responsibility of a veteran of the French and Indian Wars – Major Richard Cutts of Kittery. Both Cutts and Donnell had participated in the siege of the Acadian fortress of Louisbourg in 1745, and one can only guess how the Acadians responded to the irony of bring under the “care” of men who had been partly responsible for the conquest of their homeland.
The final fate of most of these early Maine Acadians is hard to determine. Some, no doubt, returned to Nova Scotia, or found their way to countrymen in Quebec or even Northern Maine. One John Mitchell (Jean Michaud) is recorded as having come “to York and thence to Wells about 1760, returned to York, and later made his home in Canada.” One of the ironies of the deportations is that, just a few years later, in 1763, Great Britain all-but evicted France from North America, conquering the final French colony of Canada. The security threat posed by the Acadians was over. In fact, faced with growing dissatisfaction from their American colonies, the British authorities would eventually undertake a dramatic reversal of policy and guarantee freedom of religion for Catholics in Canada in 1774. In 1764, the British allowed Acadians to return to Nova Scotia, in return for an oath of allegiance.
As a result, many Acadians made their way out of exile; some petitioned to leave, while others were repatriated (or re-deported) by their New England towns only too happy to be rid of them. In 1767, the town of Kennebunk voted that “Joseph Denico [Denicourt], a Frenchman, should be Transported to Quebeck in Kannaday [Canada] at the charge of the town.” Despite this, Denicourt appears to have been one of those who chose to remain; it must have been a difficult choice. While most Acadians had no ties to Quebec, and would be no better off there than when they had arrived in Maine, life in New England was tough. As late as 1765, the General Court heard a petition from a John White (Jean LeBlanc) of Falmouth (Portland), begging relief from taxation “which adds greatly to their distress…until they shall get into a way of business to provide for themselves and Families.” Nonetheless, Denicourt was buried in Kennebunk in 1790, leaving behind a number of children.
The spotty history of the Acadian deportees in Maine is emblematic of their history elsewhere – regarded as something between a nuisance and a public burden, and at the whim of local and colonial governments, they were shuttled from place to place as they struggled to find make a living for themselves. It’s no surprise that many chose to leave; it’s a testament to their hardiness that some chose to stay – and it’s unfortunate that we don’t have more records of their struggles and accomplishments.
Appendix – Movements of the Maine Acadians
|Town||Those assigned or enumerated in 1760:||Total number||Number enumerated in 1764 Mass. Census|
|York||DESNET (DOUCETTE) – Francis, his wife, and children Mary, John, Noon, Joseph, Peter, Ann, Francis, Dennis, Charles||11||4 men, 6 women, 6 boys and 5 girls (21 total)|
|Kittery||KING (ROY) – John, his wife, and children Joseph, Margaret, Alexander, Ann, Charles, Paul, Betsy, Sarah||10||2 men, 2 women, 3 boys, 3 girls (10 total)|
|Berwick||WHITE (LEBLANC) – Peter, his wife, and children Mary, Joseph, Sebell, Francis, Charles||7||None|
|Wells||MITCHELL (MICHAUD) – John, his wife, and children Mary, Gregory||6||1 man, 1 woman, 3 boys, 1 girl (6 total)|
|Arundel (Kennebunk)||DENCUR (DENICOURT)– Joseph, his wife, and one child||3||1 man, 1 woman, 1 boy, 1 girl (4 total)|
|Biddeford||BOUDRIN – Claud, his wife, and one child||3||1 man, 1 woman, 1 boy, 3 girls (6 total)|
|Scarbourer [sic]||BOUDRIN – Joseph, John, Mary, Margaret (children of Claud BOUDRIN)||4||Not surveyed|
|Falmouth (Portland)||LEBLANC – Paul, his wife, and children Mary, Joseph, Rose, Tittium, Samuel, Margaret, Madlin, Joseph, Oliver, one “not yet returned”||11||No men, 4 women, 2 boys, 7 girls (13 total)|
|North Yarmouth||Two un-named from Nantucket||2||None|
|Georgetown||Two un-named from Nantucket||2||None|
|Brunswick||Two un-named from Nantucket||2||None|
|Pepperellboro (Saco)||None||0||2 men (2 total)|
Charles Banks History of York Maine, Vol. 1 (n.d.)
Daniel Remich, History of Kennebunk from its earliest settlement to 1890 (n.p, 1911), p526.
Charles Bradbury, History of Kennebunk Port 1602-1837 (Kennebunk , 1837, reprinted 1967), p238. Bradbury lists his children as Joseph, who died in the Revolutionary War; Sally, who married Samuel Anderson; Betsey, who married John Cleaves; Judith, who married (1) John Hall and (2) Joseph Shackley; and Hannah, who married William Green.
Charles Banks History of York Maine, Vol. 1 (nd, np) reports an additional child was born to the Doucettes in 1757. However, he only reports 7 of their 9 children having arrived in York with them in 1756.
Daniel Remich, History of Kennebunk from its earliest settlement to 1890 (n.p, 1911), p526 reports that the Michauds returned from Wells to York at some point, presumably after 1760, and then returned to Canada at an unknown date.
Charles Bradbury, History of Kennebunk Port 1602-1837 (Kennebunk , 1837, reprinted 1967), p238 notes that the Denicourts remained in Arundel/Kennebunk, and that Joseph and his wife eventually had seven children together.
Edward E. Bourne History of Wells and Kennebunk (Portland, 1875), places Margaret and Madeleine in Wells.
A John White of Falmouth is the lead signatory of a petition on behalf of the Acadians of Falmouth (Portland) in 1765. He could be the “not yet returned” child of the 1760 assignments, or this could be an error in the petition.
Since no records exist of Acadians having ever arrived in North Yarmouth, Georgetown or Brunswick, it is quite possible that they were never sent as assigned.<|endoftext|>
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You will be given the length of a string N and the size of the character set allowed for the string K. Also you will be provided with M demands in the format x y (0 ≤ x ≤ y < N), meaning substring from x to y inclusive needs to be a palindrome. Now your job is to count the number of possible strings of length N, the size of the character set K and M demands for palindromic substrings.
Input begins with three integers N, K, M (1 ≤ N, K, M ≤ 10000) representing the length of the string, size of the character set and the number of demands. Next M lines each contains two integers x and y (0 ≤ x ≤ y < N).
Print the number of strings possible according to the above constraints. As the answer can be big, print it modulo 1000000007.
3 2 1 0 2
2 3 1 0 1<|endoftext|>
| 4.09375 |
870 |
# How to Convert a Linear Equation in Standard Form to Slope-Intercept Form?
The standard form and slope-intercept form are ways of writing linear equations. In this guide, you learn more about converting standard form to slope-intercept form.
The equation with the highest degree $$1$$ is known as the linear equation. There are different formulas available to find the equation of a straight line.
## Step by step guide toconverting standard form to slope-intercept form
The equation of a line is the equation that is satisfied by each point that lies on that line. There are several ways to find this equation in a straight line, as follows:
• Slope-intercept form
• Point slope form
• Two-point form
• Intercept form
### Standard form
The standard form of linear equations is also known as the general form and is represented as:
$$Ax+By=C$$
where $$A, B$$, and $$C$$ are integers, and the letters $$x$$ and $$y$$ are the variables.
### Slope-intercept form
The slope-intercept form of a straight line is used to find the equation of a line. For the slope-intercept formula, we need to know the slope of the line and the intercept cut by the line with the $$y$$-axis. Let’s consider a straight line of slope $$m$$ and $$y$$-intercept $$b$$. The slope-intercept form equation for a straight line with a slope, $$m$$, and $$b$$ as the $$y$$-intercept can be given as $$y=mx + b$$.
### Converting standard form to slope-intercept form
By rearranging and comparing, we can convert the equation of a line given in the standard form to slope-intercept form. We know that the standard form of the equation of a straight line represents as follows:
$$Ax + By + C = 0$$
Rearranging the terms to find the value of $$y$$, we get,
$$B×y=-Ax – C$$
$$y = (-\frac{A}{B})x + (-\frac{C}{B})$$
where, $$(-\frac{A}{B})$$ makes the slope of the line and $$(-\frac{C}{B})$$ is the $$y$$-intercept.
### Converting Standard Form to Slope-Intercept Form – Example 1:
Write the following standard form equation of a line in slope-intercept form. $$x-2y=-6$$
Solution:
Subtract $$x$$ from each side.
$$-2y = -x-6$$
Multiply each side by $$-1$$.
$$2y = x + 6$$
Divide each side by $$2$$.
$$y = \frac{(x + 6)}{2}$$
$$y=\frac{x}{2}+\frac{6}{2}$$
$$y=\frac{x}{2}+ 3$$
## Exercises forConverting Standard Form to Slope-Intercept Form
### Write the standard form equation of a line in slope-intercept form.
1. $$\color{blue}{7x-2y=5}$$
2. $$\color{blue}{2x-6y=-11}$$
3. $$\color{blue}{3x-3y=12}$$
4. $$\color{blue}{12x-12y=5}$$
1. $$\color{blue}{y=\frac{7}{2}x-\frac{5}{2}}$$
2. $$\color{blue}{y=\frac{1}{3}x+\frac{11}{6}}$$
3. $$\color{blue}{y=x-4}$$
4. $$\color{blue}{y=x-\frac{5}{12}}$$
### What people say about "How to Convert a Linear Equation in Standard Form to Slope-Intercept Form?"?
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BBC Bitesize, Human body for KS2, has a comprehensive section that would help support a classroom science project on the human body. It provides four topics to study: Skeletons and Muscles, Digestive System, Circulatory System and Health.
Skeletons and Muscles
This topic offers two learners guides:
It also has three class clips:
- Muscles needed for movement – How muscles work in pairs to allow for smooth movement.
- Muscles needed in exercise – How judo players use all their muscles and need to be very strong and flexible.
- Skeletons – An introduction to the skeletons of humans and other animals.
The Digestive System
This topic has five learners guides:
- What is the digestive system? Discover how the food gets into the body, through the digestive system.
- What happens to food in your mouth? Find out how the teeth and tongue help break down food before it is swallowed.
- What are the types of teeth? Find out the functions of different teeth types including canines, incisors and molars.
- What happens in your stomach? Find out about the role that the stomach plays in the digestive system.
- What happens in your intestines? Find out about the role of the small and large intestines in the digestive system.
The Circulatory System
This topic has four learners guides:
- What is the circulatory system? Find out how the circulatory system takes oxygen around the body.
- What is in your blood? Discover how blood vessels and blood cells help keep your body working.
- What are blood vessels? Inside every person is an amazing network of blood vessels. Find out how they carry blood around the body.
- How does your heart work? Find out how the heart keeps all the blood in your circulatory system flowing.
This topic has three learner guides:
- What are medicines and drugs? Discover why it’s important to be responsible around drugs.
- Why is a healthy lifestyle important? Find out why plenty of exercise, sleep and a good diet are really important for a healthy lifestyle.
- What is a balanced diet? Find out the principles of a balanced diet and learn about the four main food groups.
It also has twenty one class clips:
- Developing a new toothpaste – Developing new toothpastes means existing recipes are changed and it needs testing.
- Does eating breakfast affect concentration? – Does eating breakfast affect concentration? A survey aims to find out.
- Eating a varied diet – Why it’s important to eat different foods in order to have a varied and healthy diet.
- Edward Jenner, the discovery of the smallpox vaccine – Edward Jenner describes his discovery of the smallpox vaccine.
- Five types of food – An athlete explains the importance of healthy eating and exercise.
- Food needed by the human body – A bit like a car, the human body also needs fuel in the form of food and a balanced diet.
- Is there iron in breakfast cereal? – An experiment to prove that some breakfast cereals contain iron.
- Keeping healthy (clip compilation – no narration) – A compilation of clips based on the theme of keeping healthy.
- Lungs and keeping them healthy – The function of the lungs and how they can be damaged by smoking cigarettes.
- Sir Alexander Fleming, the discovery of penicillin – Sir Alexander Fleming describes his discovery of penicillin.
- Sir Alexander Fleming, the discovery of penicillin (signed) – Sir Alexander Fleming describes his discovery of penicillin.
- The effects of different drinks on teeth – A series of experiments looking at the effects of different drinks on teeth.
- The effects of different drinks on teeth (signed) – A series of experiments looking at the effects of different drinks on teeth.
- The importance of fitness – A look at the importance of regular exercise to stay fit and healthy.
- The importance of hand washing in food hygiene – An investigation into the best way to wash hands to reduce the risk of food poisoning.
- What do humans need to stay healthy? – An animation about how we can stay healthy and keep our bodies working at their best.
- Why animals need a healthy diet – Zookeepers talk about the diet needed to keep a gorilla and a penguin healthy.
- Why do we take medicine? – An interview with children who need to take asthma medicine regularly.
The Human Body for KS2 is a great set of videos and activities for children studying the human body.
human body in KS2 human body in KS2<|endoftext|>
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# Toronto Math Forum
## MAT244-2014F => MAT244 Math--Tests => Quiz 1 => Topic started by: Victor Ivrii on September 29, 2014, 02:21:19 AM
Title: Q1 problem 1 (day section)
Post by: Victor Ivrii on September 29, 2014, 02:21:19 AM
Title: Re: Q1 problem 1 (day section)
Post by: Rhoda Lam on October 04, 2014, 09:42:22 PM
2.1 p. 40, #18
\begin{gather}
ty' + 2y = \sin t,\\
y(Ï€/2) = 1
\end{gather}
First, change the equation by dividing every term by $t$.
y' + (2/t)y = (\sin (t))/t
Then, solve for the integrating factor.
μ = e^{\int (2/t)\,dt} = e^{2\ln(t)\,} = t^2
Multiply every term in equation by μ.
t^2y' + 2ty = t\sin(t)
Integrate both sides of the equation.
\begin{gather}
\int[t^2y]' = \int{t\sin(t)}dt\implies
t^2y = −t\cos(t) − \int{-\cos(t)}dt= −t\cos(t) + \sin(t) + C\implies y = (−t \cos(t) +\sin (t) + C)/t^2
\end{gather}
Substitute the initial value to solve for C.
\begin{gather}
C = (Ï€/2)^2-1
\end{gather}
Therefore, the solution is:
\begin{gather}
y = (−t \cos(t) +\sin (t) + (π/2)^2-1)/t^2
\end{gather}<|endoftext|>
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Introduction to Graph Theory
What is Graph Theory?
Graph Theory studies how things are connected, through a network of points and lines.
A graph looks like this:
An Example Graph
Yes, it is called a "graph"... but it is NOT this kind of graph:
They are both called "graphs". But they are different things. Just how it is.
This subject explores how these points and lines relate to each other, and how to best travel around them.
You can think of it as places and roads, or as stations and trains ... or as a map of friendships where each person is a node and each friendship is an edge.
There are many ways to think about graphs!
Graphs have applications in many areas such as computer science, biology, social sciences, transport and more.
Let's learn some special words:
A point is called a vertex (plural vertices) A line is called an edge. The whole diagram is called a graph.
• The number of edges that lead to a vertex is called the degree.
• A route around a graph that visits every vertex once is called a simple path.
• A route around a graph that visits every edge once is called an Euler path.
A Simple Path is like finding a way to visit every friend's house exactly once.
An Euler Path is like a mail carrier's route that goes down every street once.
Some Examples:
This Graph has:
• 5 vertices: A, B, C, D and E
• 8 edges: AB, BC, CD, DA, AE, BE, AC and BD
• Vertices A and B have degree 4
• Vertices C and D have degree 3
• Vertex E has degree 2
This Graph has:
• 6 vertices: A, B, C, D, E and F
• 10 edges: AB, BC, CD, DA, AF, BF, CF, DF, AE and BE
• Vertices A, B and F have degree 4
• Vertices C and D have degree 3
• Vertex E has degree 2
Real-World Uses
In the real world, we can find many situations that can be represented using graph theory:
• Transportation networks: Vertices can represent stations, and edges can represent the routes connecting them.
• Social networks: Vertices can be people, and edges can denote friendships or connections.
• Internet: Web pages can be seen as vertices, and hyperlinks as edges connecting them.
Graph theory can help in planning the most efficient public transportation routes or managing network traffic on the internet.
Types
There are many types of graphs such as directed and undirected graphs, weighted graphs, and algorithms that can solve fascinating problems like finding the shortest path between two points.<|endoftext|>
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(HealthDay News) -- A child can learn a great deal online, but the Internet can be a dangerous place if the child isn't prepared.
The University of Michigan Health System suggests these Internet safety tips:
- Teach children about websites that are appropriate and those that are inappropriate. A child should know to leave any site that doesn't seem appropriate or makes the child feel uncomfortable.
- Help your child create a screen name that reveals nothing about the child's identity. Make sure a child knows to never reveal any personal information.
- Your child should avoid chat rooms.
- Make sure your child knows never to reveal a password to anyone, even to a friend.
- Your child should never agree to meet anyone met online in person.<|endoftext|>
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Plants are Nature’s pharmacy. Most drugs are derived from plant compounds, but the original plants chemicals tend to be more effective and have fewer side effects than their pharmaceutical derivatives. Herbalism is often seen as a quack science by mainstream medicine, but the pharmacological effects of many herbs have been scientifically observed and reported. Many plants posses cardioprotective effects if consumed regularly.
Hawthorn is found in hedgerows throughout Europe. The flower of the hawthorn tree contains the substance tyramine which is thought to be protective of heart function. Tyramine may increase catecholamine release, which explains the ability of hawthorne to strengthen the heat beat. The flowers and berries are rich sources of flavonoids, including anthocyanins, which may have cardioprotective properties through the prevention of endothelial dysfunction.
2. Horse Chestnut
The horse chestnut tree is best known for producing seeds called conkers. The conkers from the horse chestnut tree contain aescin which is able to cause constriction of blood vessels and thus prevent oedema and inflammation. Both the bark and the conkers are a rich source of flavonoids such as aesculetin, which may improve circulation. Evidence suggests that horse chestnut may be an effective treatment to strengthen blood vessels and increase blood flow.
3. Olive Leaf
Olive groves are found throughout the Mediterranean regions, where the olive leaves are used medicinally in tinctures and infusions. In particular olive leaf is used in conjunction with other medicinal plants to lower blood pressure. The leaves contain secoiridoids such as oleuropein which may be the active chemical that lowers blood pressure. Oleuropein may also dilate the coronary artery, modulate heartbeat and help regulate blood sugar levels.
4. Butcher’s Broom
Butcher’s broom is found in southern England and Wales. It has spiky leaves and possesses red berries. The plant is a rich source of steroidal saponins that may have cardioprotective effects. In particular, butcher’s broom may be effective at preventing oedema, poor circulation, varicose veins and may protect veins and capillaries. Studies in Europe attest to the effectiveness of butcher’s broom through published papers from clinical trials.
Celery is a plant possessing a ridged shiny stem. The wild variety has an unpleasant taste, but cultivated varieties are often used in soups, salads and other foods. The stem contains a compound called 3-n-butylphthalide that may have blood pressure lowering effects. Evidence suggests that celery also possesses anti-inflammatory effects which may also provide cardioprotective activity. Blood pressure lowering effects may be seen by eating only a few stems per day.<|endoftext|>
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Cornell researchers have developed a "one-pot" process to create porous films of crystalline metal oxides that could lead to more efficient fuel cells and solar cells.
In a fuel cell, a material with nanoscale pores offers more surface area over which a fuel can interact with a catalyst. Similarly in solar cells, a porous material offers more surface area over which light can be absorbed, so more of it is converted to electricity.
Previously such porous materials have been made on hard templates of carbon or silica, or by using soft polymers that self-assemble into a foamy structure. Making a hard porous template and getting the metal oxides to distribute evenly through it is tedious. The polymer approach is easier and makes a good structure, but the metal oxides must be heated to high temperatures to fully crystallize, and this causes the polymer pores to collapse.
The Cornell researchers have combined what Ulrich Wiesner, Cornell professor of materials science and engineering, calls "the best of the two approaches," using a soft block copolymer called poly(isoprene-block-ethylene oxide) or PI-b-PEO that carbonizes when heated to high temperatures in an inert gas, providing a hard framework around which the metal oxide crystallizes. Subsequent heating in air burns away the carbon. Wiesner calls this "combined assembly by soft and hard chemistries," or CASH.
The research is described in an online paper in the journal Nature Materials by Wiesner, Francis DiSalvo, the J.A. Newman Professor of Chemistry and Chemical Biology, and colleagues.
The researchers created porous films of titanium oxide, used in solar cells, and niobium oxide, a potential fuel-cell catalyst support. Chemicals that will react to form the metal oxides and a solution of PI-b-PEO are combined. As the reaction proceeds, the PI portion of the copolymer forms cylinders some 20 nanometers across surrounded by metal oxides, and subsequent heat treatments leave uniform, highly crystalline metal oxide with cylindrical pores. The pores are neatly ordered in hexagonal patterns, which creates a larger surface area than if the pores were randomly distributed. "When the pores are ordered, you can get more of them into the same space," Wiesner explains.
The resulting materials were examined by electron microscopy, X-ray diffraction and a variety of other techniques, all of which confirmed a highly crystalline structure and a uniform porosity, the researchers reported.
The next step, Wiesner said, is to apply the CASH process to the creation of porous metals.
Co-authors of the Nature Materials paper are postdoctoral researcher Jinwoo Lee and graduate research assistants M. Christopher Orilall, Scott Warren and Marleen Kampeman.<|endoftext|>
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Cultural interactions result in both progressive and aggressive interactions due to the evolution of those cultures being uninfluenced by one another. What may be considered good etiquette in one culture may be considered an offensive gesture in another. As this occurs constantly, cultures push each other to change.The biological variations between humans are summarized in the ideas of natural selection and evolution. Human variation is based on the principle that there is variation in traits that result for recombination of genes from sexual reproduction. These traits are variable and can be passed down generation to generation. It also relies on differential reproduction, the idea that the environment can't support unlimited population growth because not all individuals get to reproduce to their full potential.
An example of human variation can be found with a cline. A cline is a genetic variation between populations of species that are isolated in their reproduction (such as skin color variation in humans). Human skin color variation is a selective adaptation that relates to the populations' proximity to the equator. Because of pigmentation characteristics within the human population, a system and term emerged to categorize the differing variations. This category is recognized as race. Populations of humans in equatorial regions have selective advantages as a result of their darker skin pigmentation, whereas populations in more northern environments have less selective pressure to evolve darker pigmentation and have lighter skin. Other clines include differences in stature and hair type.<|endoftext|>
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First of all, something that is “difficult” that is hard to do or understand. The word “difficulty” is used without restrictions as to the nature or intensity. As a result, there seems to be “a slight difficulty” and “a great difficulty”, for instance. “Hardship” is stronger in connotation. When dealing with writing difficulties, it is preferable to define the degree and intensity of the “difficulty” under question.
To plan support for struggling writers, it is necessary first to identify their specific difficulties. The research work of Graves (1983) emphasized the value of viewing students’ written work diagnostically from beginning writers to proficient writers.
It is also necessary to observe the strategies a student is already using, and what he or she can do. Romeo (2008) strongly supports the view that learners’ daily writing should be used in a formative assessment way to determine their learners’ strengths and weaknesses.
Research studies investigating the characteristics of weak writers have suggested that the main areas of concern are those listed below (Westwood, 2008) that weak writers:
Produce a much smaller amount of work than more proficient writers.
Spend a little time thinking and planning before they start to write.
Are usually reluctant to review, edit and polish a first draft.
Tend to be preoccupied with the mechanics of writing.
Have problems with spelling
For various reasons writing appears to be a difficult task. Learners try to write the composition all at once: ideas do not get a chance to form. Therefore, Tribble (1996) claims that writing can be an extremely daunting task when the main focus of a writing task is the final product from the very beginning.
The need to produce a coherent and well-written text can be a great source of stress to the writer if the intervening stages in the process of creating this text are overlooked. Furthermore, Cerbin and Beck (2001) argue that in many classes learners are expected to write well, but are not taught to do so. Courses do not try to develop learners’ writing: they simply require it and learners are left to use whatever strategies and competencies they have.
Actually, unless learners are given feedback and helped with their composing processes, they will not get better by simply writing a lot. Mooney (2004) expressed a similar view that learners should understand that writing has an explicit structure and process in order to empower them to become good writers. Many learners see writing as a magical act, when in fact there is a specific structure and process. Learners should understand that writing consists of idea generation, outlining, getting the words on the page and rewriting.
The frequent lack of a clear purpose or audience for writing resulting from the artificial nature of many EFL writing assignments, makes writing difficult for the learners. This will make learners lose interest in writing (Berkenkotter, 2000).
Moreover, incomplete understanding of the subject matter makes writing hard. Cerbin and Beck (2001) point out that learners often have to write about topics that are unfamiliar to them. Thus, it is very common that their writing lacks coherence and structure, reflecting their fragmented understanding of the topic, not necessarily their incompetence as writers.
Writing has little to do with spelling and grammar, but is more about ideas, emotions and finding a way to express oneself. Most learners are seldom taught to see writing this way (Mooney, 2004).
Unfortunately, what is emphasized at an early age with writing is not ideas or creativity, but one’s ability to master the technical elements like handwriting and spelling. In some classes, writing may be treated entirely as a list of rules governing the use of language (such as grammar, spelling and punctuation) rather than as a focused communication of ideas.
As a result, the frequently limited and often purely grammatically focused nature of teacher feedback on the completed piece of writing can contribute to a strong lack of learner motivation and a distinct reluctance to complete writing assignments both inside and outside the classroom. Agreeing to this, Cerbin and Beck (2001) note that learners perceive writing as an unpleasant task rather than as a meaningful learning experience. Consequently, learners are more likely to be interested in their work when they have some control over the selection of the topic and the work has an authentic purpose beyond getting a grade.
It is not hard to perceive learners’ writing problems; but what about the problems caused by teachers? Some of the problems are teacher-centered rather than learner-centered. According to Fulwiler (2000), vaguely explained directions on a writing assignment; examination questions which make false assumptions about what learners know or should know, assignments which do not challenge learners and are perceived as dull, incomplete responses by teachers to learn writing, and poor planning, timing, or sequencing of assignments, all of these are but some of the ways that teachers, with good intentions, may affect the quality of learner writing.
Additionally, the learners’ task of completing a writing assignment has been made yet more complex by the lack of provision for practice of the writing skill in class. Writing often becomes a low priority for the teacher when time and syllabus constraints come to the fore (Holmes, 2003).
Hence, writing is always tested and rarely taught. Cerbin and Beck (2001) drew attention to the fact that lack of explicit criteria and standards is considered a writing difficulty for learners. In some courses, learners have little or no information about what constitutes appropriate or good writing.
Learners are taught to dumb down the writing. They lose their authentic voice in school because they silence their writing in fear of making mistakes. Mooney (2004) mentions that:
Most learners have a huge discrepancy between their verbal skills and their writing ability. They have words in their minds that they could never spell. But in school, correct spelling is valued more than getting one’s true vocabulary on the page. As a result, many Learners do not write what is actually in their minds, but dumb-downed versions so they do not make mistakes. Over time, learners learn that this dumb-downed version is all they can do, and so they stop trying to write what is in their heads and lose their voice as writers.
Therefore, it is important to accept mistakes made by learners in order to encourage them to write more.
Writing has psychological and cognitive problems. Byrne (1991, cited in Ahmed, 2003) mentions these problems:
Psychological problems: writing is a solitary activity and learners are required to write on their own without the possibility of interaction such as found in speech.
Cognitive problems: people seem to speak without much conscious effort or thought. Writing, however, is learned through a process of instruction that has to be mastered.
The problem of deficiencies in learner writing or learner underachievement in writing can be attributed to many factors: one, learners have a poor attitude towards writing in addition to attitudes from previous writing failure experiences (Cumberworth & Hunt, 1998; Buhrke and others, 2002); two, learners are unmotivated to use the writing process and lack a cognitive awareness of the purpose for the writing process (Cumberworth & Hunt, 1998); three, reluctant writers experience difficulties due to the following factors: spelling and handwriting problems; poor mechanical skills; or a fear of exposing their feelings (Pierce and others, 1997); four, inadequate teacher training and reliance on ineffective past practices, daily time constraints, as well as a lack of immediate and positive feedback (Adams and others, 1996); and five, an insistence by many teachers that writing be accomplished in a silent, non-interactive environment (Accomando and others, 1996).
Thus, it is evident that writing appears to be difficult for a lot of reasons: on the part of the learner, the teacher or the instruction. So, efforts should be exerted to tackle these difficulties separately. Besides, difficulties can be minimized or if they are recognized and addressed early. Effective instruction, particularly when it builds confidence and independence in learning, will always increase students’ achievement levels.
I. Accomando, K. et al. (1996). The Development of Writing: A Social Experience among Primary Students. Unpublished M.A. Project, Saint Xavier University, Illinois, U.S.A., ERIC Research Report (ED 399543).
II. Adams, D. et al. (1996). Improving Writing Skills and Related Attitudes among Elementary School Students. Unpublished M.A. Project, Saint Xavier University, Illinois, U.S.A., ERIC Research Report (ED398595).
III. Ahmed, N. (2003). Using School Journalism for Developing some Writing Skills for Secondary Stage Students. Unpublished M.A. Thesis, Faculty of Education, University of Zagazig.
IV. Buhrke, L. et al. (2002). Improving Fourth Grade Students’ Writing Skills and Attitudes. ERIC (ED471788).
V. Cerbin, B. & Beck, T. (2001). Why Learning to Write Well in College is Difficult. University of Wisconsin – La Cross.
VI. Cumberworth, T. & Hunt, J. (1998). Improving Middle School Student Writing Skills and Attitudes toward Writing. Unpublished M.A. Action Research Project, Saint Xavier University and IRI / Skylight, Illinois, U.S.A. ERIC Research Report (ED420865).
VII. Graves, D. H. (1983). Writing: Teachers and children at work. Exeter, NH: Heinemann.
VIII. Mooney, J. (2004). Demystifying the Writing Process. Pearson Education, Inc. Online available at: familyeducation.com
IX. Pierce, J. et al. (1997). Motivating Reluctant Writers. Unpublished M.A. Project, Saint Xavier University, Illinois; USA. ERIC Research Report (ED 408617).
X. Romeo, L. (2008). Informal writing assessment linked to instruction: A continuous process for teachers, students and parents. Reading and Writing Quarterly, 24, 1, 25–51.
XI. Tribble, C. (1996). Writing. OUP.
XII. Westwood, P. (2008). What teachers need to know about reading and writing difficulties. Australia: Australian Council for Educational Research Press.
Mogahed M. Mogahed<|endoftext|>
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Towosahgy State Historic Site preserves the remains of a once
flourishing Native American community. Towosahgy (tah-wah-saw-ge) is a
borrowed Osage Indian word meaning “Old Town.” The site was the location
of a once-fortified village and ceremonial center between 1000 and 1400
A.D. Although other groups of
had lived in this area for 9,000
years prior to the founding of the village, their societies did not
reach the highly organized level of the people at Towosahgy.
The inhabitants of the village were part of the Mississippian cultural
tradition, so named as most of these archaeological sites are located
near the Mississippi River. Although other groups of Native Americans
had lived in this region for 9,000 years prior to the founding of the
village, their societies did not reach the highly organized level of the
Towosahgy. Unlike their predecessors the inhabitants of Towosahgy were
town dwellers. There were small hamlets within a short distance of the
main town and the entire area formed a well-developed cultural and
political system. The well-constructed dwellings, utensils, and other
artifacts uncovered at the site indicate the inhabitants led a
relatively comfortable lifestyle.
Crops of beans, maize, squash and sunflowers were raised in fields
outside the walls that surrounded the village. The inhabitants also
lived off wild game, fish, persimmons, wild plums and a variety of nuts.
They relied on trade to obtain other necessary items such as salt,
paint, chert for tools and ceremonial materials. For transportation,
they may have used large dugout canoes. Their houses were constructed by
digging a shallow pit, which left the floor of the structure below the
surface of the ground. Walls were built of small posts placed in narrow
trenches. A central fire hearth and associated packed clay floor mark
the original floors of these houses.
The first scientific excavations of the site in 1891 were conducted by
Cyrus Thomas of the Smithsonian Institution. In 1894 he published a map
of the village area iwhich showed a ceremonial center and a well defined
by a fortification wall that once encircled most of the village. On the
map of the village are six of the seven earthen mounds, two depressions
or borrow pits where dirt was removed to construct the mounds, and the
house depressions. Thomas knew the site as Beckwith’s Fort, named for
the landowner at the time of his visit.
In 1967, the state purchased the site and since then, the Missouri
Department of Natural Resources has conducted limited archaeological
investigations in conjunction with the University of Missouri-Columbia.
At least three log stockade walls have been unearthed, as evidenced by
dug trenches, post holes and charred fragments of actual posts. These
walls were not straight but had bastions protruding from them. These
bastions were spaced about 90 feet apart and extended about 17 feet out
from the stockade and could have been used as watch towers.
Today, all that is visible are the remains of the earthen mounds within
the village area. Six of the seven mounds still exist and surround the
central plaza where various civic and religious ceremonies were held.
The largest mound, located at the north end of the plaza, is about 180
feet wide by 250 feet long at its base, and about 16 feet high. All of
the earth within the mounds was dug by hand and carried in baskets from
borrow pits such as the large pit west of this mound. A kiosk provides
interpretive information about the village and its inhabitants. A trail
leads visitors to the mounds and other areas of the village where they
can imagine this once-thriving culture found in this area hundreds of<|endoftext|>
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Mount Vesuvius, on the west coast of Italy, is the only active volcano on mainland Europe. It is best known because of the eruption in A.D. 79 that destroyed the cities of Pompeii and Herculaneum, but Vesuvius has erupted more than 50 times.
Mount Vesuvius facts
Vesuvius in 2013 was 4,203 feet (1,281 meters) tall. After each eruption, the size of the cone changes, according to Encyclopedia Britannica. The volcano also has a semicircular ridge called Mount Somma that rises to 3,714 feet (1,132 m). The valley between the cone and Mount Somma is called Valle del Gigante or Giant's Valley.
Mount Vesuvius is considered to be one of the most dangerous volcanoes in the world because of its proximity to the city of Naples and the surrounding towns on the nearby slopes.
The volcano is classed as a complex stratovolcano because its eruptions typically involve explosive eruptions as well as pyroclastic flows. A pyroclastic flow is a high-density mix of hot lava blocks, pumice, ash and volcanic gas, according to the U.S. Geological Survey. Vesuvius and other Italian volcanoes, such as Campi Flegrei and Stromboli, are part of the Campanian volcanic arc. The Campanian arc sits on a tectonic boundary where the African plate is being subducted beneath the Eurasian plate.
Under Vesuvius, scientists have detected a tear in the African plate. This "slab window" allows heat from the Earth's mantle layer to melt the rock of the African plate building up pressure that causes violent explosive eruptions. In the past, Mount Vesuvius has had a roughly 20-year eruption cycle, but the last serious eruption was in 1944.
Mount Vesuvius destroyed the city of Pompeii, a city south of Rome, in A.D. 79 in about 25 hours, according to History. Because the city was buried so quickly by volcanic ash, the site is a well-preserved snapshot of life in a Roman city. There is also a detailed account of the disaster recorded by Pliny the Younger, who interviewed survivors and recorded events in a letter to his friend Tacitus. [Related: Pompeii 'Wall Posts' Reveal Ancient Social Networks]
Pompeii was established in 600 B.C. and was slowly recovering from a major earthquake that rocked the city in February of A.D. 62. The shallow quake, originating beneath Mount Vesuvius, had caused major damage to the springs and piping that provided the city's water. Reconstruction was being carried out on several temples and public buildings. Seneca, a historian, recorded that the quakes lasted for several days and also heavily damaged the town of Herculaneum and did minor damage to the city of Naples before subsiding. The major quake was followed by several minor shakes throughout the following years. [Image Gallery: Pompeii's Toilets]
Because seismic activity was so common in the area, citizens paid little attention in early August of 79 when several quakes shook the earth beneath Herculaneum and Pompeii. People were unprepared for the explosion that took place shortly after noon on the 24th of August. Around 2,000 residents survived the first blast.
Pliny the Elder, a Roman author, described the massive debris cloud. "It resembled a (Mediterranean) pine more than any other tree. Like a very high tree the cloud went high and expanded in different branches … sometimes white, sometimes dark and stained by the sustained sand and ashes." In Pompeii, ash blocked the sun by 1 p.m. and the people tried to clear heavy ash from rooftops as it fell at a rate of about 6 inches (15 centimeters) an hour. [Image Gallery: Preserved Pompeii — Photos Reveal City in Ash]
Shortly after midnight, a wall of volcanic mud engulfed the town of Herculaneum, obliterating the town as its citizens fled toward Pompeii. About 6:30 a.m. on the following morning, a glowing cloud of volcanic gases and debris rolled down Vesuvius' slopes and enveloped the city of Pompeii. Most victims died instantly as the superheated air burned their lungs and contracted their muscles, leaving the bodies in a semi-curled position to be quickly buried in ash and thus preserved in detail for hundreds of years.
Far away in Misenum, approximately 13 miles (21 kilometers) from Pompeii, Pliny the Younger, the 18-year-old nephew of Pliny the Elder, and his mother joined other refugees escaping the earthquakes rocking their city. They observed, "the sea retreating as if pushed by the earthquakes." This was probably caused by a tsunami at the climax of the eruption, which gives us the time frame for historical record. Pliny writes of "black and horrible clouds, broken by sinuous shapes of flaming wind." He describes people wheezing and gasping because of that wind; the same wind that doomed the people of Pompeii.
It is believed that around 30,000 people died from the eruption of Vesuvius in 79.
On March 17, 1944, a two-week-long eruption began with lava from the summit of Mount Vesuvius. In an article by Life Magazine, Giuseppe Imbo, director of the Mt. Vesuvius Observatory, is quoted as saying, "A marvelous thing, my Vesuvius. It covers land with precious ash that makes the earth fertile and grapes grow, and wine. That's why, after every eruption, people rebuild their homes on the slopes of the volcano. That is why they call the slopes of Vesuvius the compania felix — the happy land."
During the eruption, soldiers and airmen of the 340th Bomber Group were stationed at the Pompeii Airfield just a few miles from the base of the volcano. Diaries record the awesome sights and sounds they witnessed in this latest major eruption. Guards wore leather jackets and "steel pot" helmets to protect themselves from rains of hot ash and small rocks. Tents collapsed or caught fire when hot cinders were blown over them.
Sgt. Robert F. McRae wrote in his diary on March 20, 1944, according to the American Geosciences Institute, "As I sit in my tent … I can hear at four- to 10-second intervals the loud rumbling of the volcano on the third day of its present eruption. The noise is like that of bowling balls slapping into the pins on a giant bowling alley. To look above the mountain tonight, one would think that the world was on fire. The thickly clouded sky glows like that above a huge forest fire. Glowing brighter as new spouts of flame and lava are spewn from the crater. As the clouds pass from across the top of the mountain, the flame and lava can be seen shooting high into the sky to spill over the sides and run in red streams down the slopes. ... Today it is estimated that a path of molten lava 1 mile long, half a mile wide, and 8 feet deep is rolling down the mountain. Towns on the slopes are preparing to evacuate. Our location is, apparently, safe. At any rate no one here, civilian or Army authorities, seems too much worried. Lava has not started to flow down this side of the mountain as yet but is flowing on the other side toward Naples."
On March 22, they were forced to evacuate, leaving behind 88 Allied aircraft. After the volcano subsided, they returned on the 30th to find the planes were a total loss. Engines were clogged by ash, control panels were useless tangles of fused wire, canopies had holes from flying rock or were etched to opacity by wind driven ash.
One airman of the 489th Bomber Squadron complained in his diary when Axis Sally broadcast a radio show dedicated to the "survivors" of the Vesuvius eruption (actually the most severe human casualty was a wrist sprained during the evacuation). She told all of Europe that "Colonel Vesuvius" had destroyed all of them. The diarist was justifiably proud of the work he did with his fellows in recovery. By April 15, the planes had been replaced and the 340th Bomber Group was back to full strength and ready to fly missions from their new base.
Though no soldiers were killed, 26 Italian civilians died and nearly 12,000 were displaced by the 1944 eruption, according to the American Geosciences Institute.
Since 1944, there have been hundreds of minor earthquakes in the region around Mount Vesuvius. The most serious earthquake rocked Naples in October 1999. The magnitude-3.6 quake was felt as far as 15 miles (24 km) from the base of the volcano and was of the same magnitude as a quake that occurred 17 years prior to the last truly major explosion that devastated Naples in 1631.
In 2016, excavations on the outskirts of Pompeii revealed more victims of the volcanic eruption. Archaeologists discovered the remains of four people, including one teenage girl, in the ruins of a shop, according to a statement from the Soprintendenza Pompei, the Italian authority in charge of managing the ancient site.
Additional reporting by Alina Bradford, Live Science contributor<|endoftext|>
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# Notebook 16: Expectation Maximization in practice¶
## Learning Goal¶
The goal of this notebook is to gain intuition for Expectation Maximization using a simple example involving coin tosses.
## Overview¶
In Section XIV, we introduce Expectation-Maximization (EM) as a practical way to perform maximum likelihood estimation (MLE) even when some of the data is hidden (i.e in the presence of latent or hidden variables). To better understand EM, in this short notebook we'll explore a very simple coin-tossing example adapted from Do and Batzoglou, Nat. Biotechnol. (2008).
Suppose that we are given two coins A and B with unkown bias $\theta_A$ and $\theta_B$, respectively. Our goal is to estimate the bias vector $\boldsymbol{\theta}= (\theta_A, \theta_B)$ from the outcomes of the following experiment:
First choose one coin at random. Then toss the selected coin 10 times independently and record the number of heads observed. Repeat this procedure 5 times.
Formally, let $z_i\in\{A,B\}$ be the coin selected in experiment $i$ and $x_i\in\{0,1,\cdots 10\}$ be the number heads recorded by tossing $z_i$ 10 times. Since we conduct $n=5$ such experiments, we can summarize the outcomes of these 50 tosses by two vectors: $\boldsymbol{x}=(x_1,x_2\cdots, x_5)$ and $\boldsymbol{z}=(z_1,z_2,\cdots, z_5)$.
### Exercise 1: What if we know everything?¶
• Consider first the case where we have complete knowledge of the experiment, namely, both $\boldsymbol{x}$ and $\boldsymbol{z}$ are known. How would you intuitively estimate the biases of the two coins $\boldsymbol{\theta}= (\theta_A, \theta_B)$ ?
• What's the likelihood of observing the complete outcomes of these experiments? In other words, what is $P(\boldsymbol{x},\boldsymbol{z}| n,\boldsymbol{\theta} )$? You may assume this is a Bernoulli trial. Namely, every time coin A(B) is tossed, we have, with probability $\theta_A$($\theta_B$), that the outcome is heads.
• What's the Maximum Likelihood Estimator (MLE)? Is this consistent with your intuition?
## Comparing MLE and EM¶
To test your answer, let's do some numerics! We will compare the MLE estimates of biases with an Expectation Maximization procedure where we do not know ${\bf z}$. The following code computes our best guess for the biases using MLE -- assuming we know the identity of the coin used -- and compares it estimates arrived at using an EM procedure where we have no knowledge about which coin was being tossed (though we know the same coin was tossed 10 times).
In [1]:
import numpy as np
from scipy.special import comb
import math
def compute_likelihood(obs, n, pheads): # No surprise, it's Binomial!!!
return likelihood
# generate experiments
num_coin_toss = 10 # each experiment contains num_coin_toss tosses
num_exp = 5 # we perform 5 such experiments
theta_A_true = 0.8
theta_B_true = 0.4
coin_choice = np.zeros(num_exp) # initialize: 0 for A and 1 for B
# MLE
MLE_A = 0.0
MLE_B = 0.0
# generate the outcomes of experiment
for i in np.arange(num_exp):
if np.random.randint(2) == 0: # coin A is selected
head_counts[i] = np.random.binomial(num_coin_toss , theta_A_true, 1) # toss coin A num_coin_toss times
else: # coin B is selected
head_counts[i] = np.random.binomial(num_coin_toss , theta_B_true, 1) # toss coin B num_coin_toss times
coin_choice[i] = 1 # record the selection of coin B during experiment i
# MLE is merely the proportion of heads for each coin toss
MLE_A = MLE_A / ((num_exp - np.count_nonzero(coin_choice))*num_coin_toss)
MLE_B = MLE_B / (np.count_nonzero(coin_choice)*num_coin_toss)
pA_heads[0] = 0.60 # initial guess
pB_heads[0] = 0.50 # initial guess
# E-M begins!
epsilon = 0.001 # error threshold
j = 0 # iteration counter
improvement = float('inf')
while (improvement > epsilon):
expectation_A = np.zeros((num_exp,2), dtype=float)
expectation_B = np.zeros((num_exp,2), dtype=float)
eT = tail_counts[i]
# E step:
weightA = lA / (lA + lB)
weightB = lB / (lA + lB)
expectation_A[i] = weightA*np.array([eH, eT])
expectation_B[i] = weightB*np.array([eH, eT])
# M step
theta_A = np.sum(expectation_A, axis = 0)[0] / np.sum(expectation_A)
theta_B = np.sum(expectation_B, axis = 0)[0] / np.sum(expectation_B)
print('At iteration %d, theta_A = %2f, theta_B = %2f' % (j, theta_A, theta_B))
j = j+1
# END of E-M, print the outcome
print('E-M converges at iteration %d' %j)
print('RESULT:')
print('E-M: theta_A = %2f, theta_B = %2f' % (theta_A, theta_B))
print('MLE with complete data: theta_A = %2f, theta_B = %2f' % (MLE_A, MLE_B))
At iteration 0, theta_A = 0.681216, theta_B = 0.503301
At iteration 1, theta_A = 0.732818, theta_B = 0.459266
At iteration 2, theta_A = 0.760207, theta_B = 0.425543
At iteration 3, theta_A = 0.766395, theta_B = 0.409645
At iteration 4, theta_A = 0.766694, theta_B = 0.402593
At iteration 5, theta_A = 0.766301, theta_B = 0.399292
At iteration 6, theta_A = 0.766019, theta_B = 0.397709
At iteration 7, theta_A = 0.765867, theta_B = 0.396945
E-M converges at iteration 8
RESULT:
E-M: theta_A = 0.765867, theta_B = 0.396945
MLE with complete data: theta_A = 0.766667, theta_B = 0.350000
### Exercise 2¶
• How fast does EM converge? Is the converged result close to what you'd get from MLE?
• Following Exercise 1, what's the objective function we're optimizing in the E-step? Does this function have a unique global maximum?
• Compare both the results of MLE and EM to the actual bias (i.e. theta_A_true and theta_B_true in the snippet above), comment on their performance.
## Final remarks: a few practical tricks¶
From Exercise 2 and Section XIV, we know that the E-M algorithm often approximates the MLE even in the presence of latent (hidden variables). Like with most optimization methods for nonconcave functions, E-M only guarantees convergence to a local maximum of the objective function. For this reason, its performance can be boosted by running the EM procedure starting with multiple initial parameters.
### Exercise 3¶
• Now instead of having a fixed initial guess of coin biases (i.e. pA_heads[0] and pB_heads[0] in the snippet), draw these values uniformly at random from $[0,1]$ and run the E-M algorithm. Repeat this twenty times and report what you observed. What's the best initial guess that gives the closest estimate to the true parameters?
• As we discussed in Section X (LinReg), Maximum a posteriori (MAP) estimation differs from MLE in that it employs an augmented objective function which incorporates a prior distribution over the quantities we want to estimate, and the prior distribution can be think of as a regularizer for the objective fuction used in MLE. Here we will explore how to extend E-M to MAP estimation.
(1) First derive the MAP estimate for the one-coin-flipping example, namely, $$\hat{{\theta}}_{MAP}(\boldsymbol{x}) = \arg\max_{\theta\in[0,1]} \log P(\boldsymbol{x}|n,{\theta} ) + \log P({\theta}),$$ where $$P(\boldsymbol{x}|n,{\theta}) = \prod_{i=1}^{10} \text{Binomial} (x_i|n,\theta)$$
$$P({\theta})=\mathcal{N}(\theta|\mu, \sigma)$$
(2) Based on (1), now modify the E-M snippet above to incorporate this prior distribution into the M-step. Comment on the performance. For the prior choice, try $P(\boldsymbol{\theta})=\mathcal{N}(\theta_A|0.83, 1)\mathcal{N}(\theta_B|0.37, 1)$.<|endoftext|>
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# How to Add Three or More Fractions with Like & Unlike Denominators
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Lesson Transcript
Instructor: Dina Albert
Fractions are everywhere. You might use them when you cook or when you share items with friends and family, even if you don't realize it. In this video, we'll explore how to add three or more fractions with unlike denominators.
## Fractions with the Same Denominator
Suppose that you share a pie with a friend. You eat 1/4 of the pie, and your friend eats 3/4 of the pie. How much of the pie did you both eat all together? In this situation, each fraction has the same denominator, which is the bottom number of a fraction. You can also think of the denominator as the 'name' of the fraction. If fractions have the same denominator (name), then we can simply add the numerators together and keep that 'name.'
In this image, you ate one of the four pieces of pie, or ¼. Your friend ate three out of the four pieces of the pie, or ¾. In total, you and your friend ate four out of the four pieces of the pie, or 4/4. Notice that we kept the denominators the same. You and your friend ate from the same pie, and that pie was cut into 4 pieces. You did not change the original number of pie pieces (4) and so the denominator stayed the same.
Now, imagine sharing a pie with two of your friends, not just one. Let's say you had 1/4 of the pie, another friend had 1/4 of the pie, and another friend had 2/4 of the pie. Since the denominators are the same, we will add the numerators together and keep the denominator.
1/4 + 1/4 + 2/4 = (1 + 1 + 2) = 4/4, where both numerator and denominator are 4.
## Fractions with Different Denominators
Let's take the same example of sharing a pie, but this time you eat 1/4 of the pie, while your friend eats 1/2. Since the denominators of the fractions are different, we cannot add them. We need to rewrite each fraction so that the denominators are the same. How many 2s go into 4? Well, 2 x 2 = 4. We can use this information to help us change our fractions, like so:
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As evidenced by my posts so far, I believe effective teaching recognises and embraces individual student cultures. While it is not reasonable to think that a single teacher could produce individual ‘contexts’ for each child, they must recognise that students can learn from the diversity within the classroom (school community); and should therefore aim to provide meaningful learning opportunities which do not cater only to the majority. Zevenbergen, Dole and Wright (2004) suggest that learning environments which match the learning needs, backgrounds and interests of the students; have high expectations for student success, and encourage deep learning, about and through mathematics, ensure all students can learn mathematics. Further; Smith, Ewing and Le Cornu (2007) recognise the importance of providing opportunities for learning in everyday situations; and connecting to student backgrounds and practices in order to build student self esteem.
In this post I will demonstrate some small examples of mathematics ‘taken from systems which are part of students’ realities’ (Rosa and Orey, 2010). Using my interpretation of culture, which is not limited to ethnicity; I will demonstrate Ethnomathematical examples of drawing on knowledge from traditional cultures, sport, and music.
In this ethnomodelling example the teacher has provided an open ended activity which allows the students to demonstrate their understanding of one concept (solids) by incorporating something they are familiar with, and using that knowledge to build on other content knowledge. This example demonstrates how incorporating ethnomathematics does not require the teacher to decontextualize culture (in contrast to my previous understanding), because the children are bringing it to the classroom (rather than the teacher including it in a worksheet or something). Students are able to use the artefact as a foundation for learning, until such time as they can move beyond the concrete and move on to the symbolic and the generic.
By providing opportunities for students to incorporate items which are familiar to them, students are able to: a) develop self esteem, and cultural pride, b) verbalise cultural component, and connections between cultural and mathematical content and c) make meaning through building on prior knowledge.
Talking about these items allows students to strengthen their understanding of social and mathematical concepts in what is a vastly technological and multicultural world.
This is not the whole article – it is just the appendix, but it really demonstrated to me how easy to think about incorporating cultural knowledge without prejudice or stereotype:
Abiam, Abonyi, gama and Okafor (2015) explain that their example was from Africa, where although they do have vernacular terms for circles and curves, they are spacial rather than geometric references (note that Aboriginal Cultures are also highly spatial … see below).
Many times during my degree I have heard the mantra “AFL will get them engaged in maths!” Putting aside any obvious issues with that comment – such as stereotyping the ‘them’ in that statement (gender? race? cultural group? students?); what is it about Aussie Rules/Australian Rules Football/AFL that is engaging, and mathematical?
Before learning about Ethnomathematics, I had thought the engaging part is making connections to lives outside of school – but I recognise not everyone plays AFL. Maybe that’s another mathematical consideration – comparisons between different sports, for example, scoring systems in AFL (6 points for a goal (grouping/multiplication/division), 1 point for each netball goal (tallying), 2 points each for basketball (or one for a foul (vertical number line/addition/subtraction), number of people on a team, times for quarters/halves/breaks (fractions – also field division), average number of times a whistle is blown in a game (did you know netball just changed the rules to reduce the whistle blowing by 30%?)… the list goes on and on. A ‘Google search’ of Aussie rules math brings up a plethora of examples of how to use AFL in the mathematics classroom.
One author, Christine Nicholls (2013), suggests Aboriginal culture is centred around spatial relationships (like the African example above), which is ideal for a game like football which requires a 360 degree perspective of the field. Rather than focussing on left and right, which depends on the way the body is facing; Aboriginal cultures focus on cardinal directions – north/south/east/west; connecting to things like the sun, and the horizon. Spatial awareness is a coveted skill; and directional awareness is cited by many involved in Aboriginal research, but understanding the directionality of Aboriginal culture is beyond the scope of this post. You can read the article for more information, where even Nicholls acknowledges there is more to learn.
I refer to above cultural knowledge to demonstrate possible differences in perspectives about what children may bring to, and get out of, cultural inclusions in mathematics lessons. Another consideration is that children become engaged because it is something they understand and are good at; and feeling successful is something which can encourage further participation. It also helps students to understand that mathematics is all around us in various forms – it is not just a classroom concept.
But did you know Aboriginal people have played games, similar to AFL for many hundreds of years?
By highlighting the connection between traditional games, and AFL; students may also feel their culture is being respected, and consequently more connected to the learning environment. Meanwhile, non-indigenous students learn about other cultures, whilst connecting to something they are already familiar with.
Two examples of traditional Aboriginal ‘football’ games:
Marngrook, Marn Grook: A traditional game from the Gunditjmara people in Victoria. The name comes from a corroboree by the Djabwurrung and Jardwadjali clans in Victoria’s Western District. Marngrook is said to be the Aboriginal game that provided the first lawmakers of football with some of the fundamentals of the game millions know and love as Australian Rules (Aussie Rules) Football, a view which is not totally undisputed
PurljaPurlja: A game like football that the Warlpiri Aboriginal people (north-west of Alice Springs) played for thousands of years.
We often hear that music and mathematics are connected. That makes a lot of sense when you think about it – patterns, fractions, comparisons, categories – all mathematical concepts! And add to that the social and cultural elements of music, and the way people in different communities value and use music, I feel that it is reasonable to include it in this discussion about Ethnomathematics.
According to Johnson and Edelson (2003) music allows students to connect to the symbolism language of mathematics, and to understand that humans use many symbols to express themselves in different ways. Chahine and Montiel (2015) suggest that analysing the mathematical structures of music provides meaningful learning opportunities. And, because ALL of Bishop’s mathematical concepts are present in music, I tend to agree with them!
As noted previously: Alan Bishop (1988) suggests there are six mathematical concepts counting, locating, measuring, designing, playing, explaining, sentimental values – attitudes, feelings, behaviours, idealogical values – beliefs, symbolism, philosophies, sociological values – customs, institutions, rules and patterns, interpersonal behaviours and technological – manufacture and use tools… .
As noted by Pais (2011) there are also opportunities to miss the connections between social and mathematical world views, if the teacher does not understand the cultural relevance (for example, Pais suggests using flute from a given region could just as well be any other instrument if the cultural aspects are missed). Despite this, Pais suggests test results improved through the inclusion of a culturally responsive artefact.
Johnson and Edelson provide a lot of examples of how to incorporate music into mathematics lessons here. Although these activities are not Ethnomathematically centred; they do provide starting points through which culturally responsive mathematics activities could be developed. Some examples might include specific types of music, instruments and songs, comparisons between and discussions about the similarities and differences, and a combination of visual, aural and practical activities which develop socio-cultural, mathematical and cross-disciplinary learning.
Drums to engage students – patterns (Aboriginal school in Alice Springs):
Fractions in musical symbols:
Abiam, P. O., Abonyi, O. S., Ugama, J. O., & Okafor, G. (2015). Effects of Ethnomathematics-based instructional approach on primary school pupils’ achievement in geometry. Journal of Scientific Research & Reports, 9(2), 1-15. doi: 10.9734/JSRR/2016/19079
Aboriginal culture – Sport – Traditional Aboriginal games & activities. Retrieved from http://www.creativespirits.info/aboriginalculture/sport/traditional-aboriginal-games-activities#ixzz4AV2CfYyS
Bishop, A. J. (1988). Mathematics Education in its Cultural Context. Educational Studies in Mathematics, 19(2), 179-191
Chahine, I., & Montiel, M. (2015). Teaching modeling in algebra and geometry using musical rythms: teachers’ perspectives on effectiveness. Journal of mathematics education, 8(2), 126-138.
Groundwater-Smith, S., Ewing, R., & Le Cornu, R. (2007). Teaching: challenges and dilemmas (J. West Ed. 3rd ed.). South Melbourne, VIC: Harcourt Australia.
Johnson, G. L., & Edelson, R. J. (2003). Integrating music and mathematics in the elementay classroom. Retrieved 31st May, 2016, Retreived from lesage.blogs.uoit.ca/wp-uploads/2010/08/Integrating-Math-Music-2003NCTM.pdf
Nicholls, C. (2013). It’s time we draft Aussie Rules to tackle Indigenous mathematics. Retrieved 31st May, 2016, Retrieved from http://theconversation.com/its-time-we-draft-aussie-rules-to-tackle-indigenous-mathematics-15032
Rosa, M., & Orey, D. C. (2010). Ethnomodeling: an ethnomathematical holistic tool. Academic Exchange Quarterly, 14(3), 1-5.
Zevenbergen, R., Dole, S., & Wright, R. J. (2004). Teaching mathematics in primary school. Crows Nest, NSW: Allen & Unwin.<|endoftext|>
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A Guide to Legislative Votes
Deciphering the thousands of votes cast each year can be confusing. Here are some tips to assist you in navigating the votes taken by Members of Congress.
There are two main types of legislation that originate from each house of Congress: bills and resolutions. Bills, if passed by the House and Senate and signed by the President, become binding law and part of the United States Code. Resolutions are not laws; rather, they are expressions of the “sentiments” of either the House or Senate.
H.R. stands for the U.S. House of Representatives, and any legislation with this prefix indicates that the bill originated from the House. If passed by the House, the bill moves on to the Senate for consideration.
H.Res. stands for a resolution of the House of Representatives. House resolutions are not binding law, but rather express the collective sentiment of the House on a particular issue, person, or event. House committees may also be formed through the passage of a House resolution.
S. stands for the Senate, and any legislation with this prefix indicates that the bill has originated from the United States Senate. If passed by the Senate, the bill then moves on to the House for consideration.
S.Res. stands for a resolution of the United States Senate. Senate resolutions are not binding law; rather, they express the collective sentiment of the Senate on a particular issue, person, or event. Senate committees may also be formed through the passage of a Senate resolution.
H.Con.Res. and S.Con.Res. stand for concurrent resolutions, which are taken up simultaneously by both the House and Senate. These resolutions do not become law, but must be passed by both chambers of Congress.
H.J.Res. and S.J.Res. stand for joint resolutions which are taken up simultaneously by both the House and Senate. These resolutions also require the approval of both chambers, but upon passage, they are submitted to the President for approval.
A Rule is a resolution brought forward by the Rules Committee that asks the House to consider a certain bill or bills. A rule will also often put forth the guidelines for debate between Members and governs the ability to bring forward amendments on the specific legislation.
A Motion to Recommit is a legislative tool used for two distinct purposes. It is introduced by the party who has not introduced the current bill on the floor and is used to send a bill back to the committee of jurisdiction. This indefinitely stalls the legislation. It can also be used as a last opportunity to amend the legislation being considered on the House floor.<|endoftext|>
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Some of the most stunning astronomical objects are globular clusters. These spherically-distributed celestial ornaments can pack a million stars within a tiny angle of space as seen from Earth. For example, M13 in Hercules fills the eyepiece of a 14" Schmidt-Cassegrain or 20" Dobsonian telescope. Omega Centauri, better placed for southern-hemisphere observers, is the largest globular of our galaxy and one of the few visible with the naked eye.
Stars in a globular appear densely packed but are actually widely spaced. About 180 of these clusters orbit the center of the Milky Way at various inclinations, and globulars have also been detected around other galaxies. They have played an important role in the development of astronomical thought since the Christian astronomer William Herschel named them in 1789. Few amateurs at star parties may know that they are now centerpieces of a significant upset in astronomy occurring right now, after a near century of consensus.
In 1914, Harlow Shapley noticed their spectra were different from those in the galactic disk. He named the disk stars Population I, and the globular cluster stars Population II. Because they are low in elements heavier than helium, the Population II stars were assumed to be older than the Population I stars. (A third category, Population III, is assumed to comprise the very first stars after the Big Bang, made out of pure hydrogen and helium, but none have been observed.) The distinct spectral signature of globulars, combined with their lack of dust and gas, gave rise to the view that they represent some of the oldest objects in the universe, mostly composed of old red giants in regions where no new stars are forming. The disk, by contrast, was thought to be young and actively engaged in star formation. This had been the textbook orthodoxy for most of the twentieth century.
The story started to unravel three years ago (see Astronomy, Nov. 2003) when the Hubble and other orbiting telescopes found globulars containing mixed populations of stars, with exotic members called "blue stragglers" and even planets. News@Nature last August reported that the findings are "changing our ideas completely" and will require us to "tear up textbooks." New explanations are being considered. Perhaps they formed during galactic mergers. But each new solution breeds new problems: how was there enough material left over to form these densely-packed clusters?
One thing is clear; globulars can no longer be thought of as simple, homogeneous collections of ancient stars. The article said, "In a complex Universe, astronomers thought they had at least one simple system to tell them how stars are born. Turns out they were wrong." Moreover, this upset can have ripple effects on other theories. One astronomer said, "If you have problems reproducing star formation in globular clusters, you will have problems with a galaxy."
Astronomers will undoubtedly come up with new ideas. There's an important lesson here about how science is done in these days of Big Bang-to-man theorizing. It's not that scientists are unable to concoct a story to fit the data, it's that the data require a story to fit a belief.
*David F. Coppedge works in the Cassini program at the Jet Propulsion Laboratory.
Cite this article: Coppedge, D. 2007. The Globular Cluster Bomb. Acts & Facts. 36 (2).<|endoftext|>
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By David Prychitko
More than a century after his death, Karl Marx remains one of the most controversial figures in the Western world. His relentless criticism of capitalism and his corresponding promise of an inevitable, harmonious socialist future inspired a revolution of global proportions. It seemed that—with the Bolshevik revolution in Russia and the spread of communism throughout Eastern Europe—the Marxist dream had firmly taken root during the first half of the twentieth century.
That dream collapsed before the century had ended. The people of Poland, Hungary, Czechoslovakia, East Germany, Romania, Yugoslavia, Bulgaria, Albania, and the USSR rejected Marxist ideology and entered a remarkable transition toward private property rights and the market-exchange system, one that is still occurring. Which aspects of Marxism created such a powerful revolutionary force? And what explains its eventual demise? The answers lie in some general characteristics of Marxism—its economics, social theory, and overall vision.
Labor Theory of Value
The labor theory of value is a major pillar of traditional Marxian economics, which is evident in Marx’s masterpiece, Capital (1867). The theory’s basic claim is simple: the value of a commodity can be objectively measured by the average number of labor hours required to produce that commodity.
If a pair of shoes usually takes twice as long to produce as a pair of pants, for example, then shoes are twice as valuable as pants. In the long run, the competitive price of shoes will be twice the price of pants, regardless of the value of the physical inputs.
Although the labor theory of value is demonstrably false, it prevailed among classical economists through the midnineteenth century. Adam Smith, for instance, flirted with a labor theory of value in his classic defense of capitalism, The Wealth of Nations (1776), and David Ricardo later systematized it in his Principles of Political Economy (1817), a text studied by generations of free-market economists.
So the labor theory of value was not unique to Marxism. Marx did attempt, however, to turn the theory against the champions of capitalism, pushing the theory in a direction that most classical economists hesitated to follow. Marx argued that the theory could explain the value of all commodities, including the commodity that workers sell to capitalists for a wage. Marx called this commodity “labor power.”
Labor power is the worker’s capacity to produce goods and services. Marx, using principles of classical economics, explained that the value of labor power must depend on the number of labor hours it takes society, on average, to feed, clothe, and shelter a worker so that he or she has the capacity to work. In other words, the long-run wage workers receive will depend on the number of labor hours it takes to produce a person who is fit for work. Suppose five hours of labor are needed to feed, clothe, and protect a worker each day so that the worker is fit for work the following morning. If one labor hour equaled one dollar, the correct wage would be five dollars per day.
Marx then asked an apparently devastating question: if all goods and services in a capitalist society tend to be sold at prices (and wages) that reflect their true value (measured by labor hours), how can it be that capitalists enjoy profits—even if only in the short run? How do capitalists manage to squeeze out a residual between total revenue and total costs?
Capitalists, Marx answered, must enjoy a privileged and powerful position as owners of the means of production and are therefore able to ruthlessly exploit workers. Although the capitalist pays workers the correct wage, somehow—Marx was terribly vague here—the capitalist makes workers work more hours than are needed to create the worker’s labor power. If the capitalist pays each worker five dollars per day, he can require workers to work, say, twelve hours per day—a not uncommon workday during Marx’s time. Hence, if one labor hour equals one dollar, workers produce twelve dollars’ worth of products for the capitalist but are paid only five. The bottom line: capitalists extract “surplus value” from the workers and enjoy monetary profits.
Although Marx tried to use the labor theory of value against capitalism by stretching it to its limits, he unintentionally demonstrated the weakness of the theory’s logic and underlying assumptions. Marx was correct when he claimed that classical economists failed to adequately explain capitalist profits. But Marx failed as well. By the late nineteenth century, the economics profession rejected the labor theory of value. Mainstream economists now believe that capitalists do not earn profits by exploiting workers (see profits). Instead, they believe, entrepreneurial capitalists earn profits by forgoing current consumption, by taking risks, and by organizing production.
There is more to Marxism, however, than the labor theory of value and Marx’s criticism of profit seeking. Marx wove economics and philosophy together to construct a grand theory of human history and social change. His concept of alienation, for example, first articulated in his Economic and Philosophic Manuscripts of 1844, plays a key role in his criticism of capitalism.
Marx believed that people, by nature, are free, creative beings who have the potential to totally transform the world. But he observed that the modern, technologically developed world is apparently beyond our full control. Marx condemned the free market, for instance, as being “anarchic,” or ungoverned. He maintained that the way the market economy is coordinated—through the spontaneous purchase and sale of private property dictated by the laws of supply and demand—blocks our ability to take control of our individual and collective destinies.
Marx condemned capitalism as a system that alienates the masses. His reasoning was as follows: although workers produce things for the market, market forces, not workers, control things. People are required to work for capitalists who have full control over the means of production and maintain power in the workplace. Work, he said, becomes degrading, monotonous, and suitable for machines rather than for free, creative people. In the end, people themselves become objects—robotlike mechanisms that have lost touch with human nature, that make decisions based on cold profit-and-loss considerations, with little concern for human worth and need. Marx concluded that capitalism blocks our capacity to create our own humane society.
Marx’s notion of alienation rests on a crucial but shaky assumption. It assumes that people can successfully abolish an advanced, market-based society and replace it with a democratic, comprehensively planned society. Marx claimed that we are alienated not only because many of us toil in tedious, perhaps even degrading, jobs, or because by competing in the marketplace we tend to place profitability above human need. The issue is not about toil versus happiness. We are alienated, he maintained, because we have not yet designed a society that is fully planned and controlled, a society without competition, profits and losses, money, private property, and so on—a society that, Marx predicted, must inevitably appear as the world advances through history.
Here is the greatest problem with Marx’s theory of alienation: even with the latest developments in computer technology, we cannot create a comprehensively planned system that puts an end to scarcity and uncertainty. But for Marxists to speak of alienation under capitalism, they must assume that a successfully planned world is possible. That is, Marx believed that under capitalism we are “alienated” or “separated” from our potential to creatively plan and control our collective fate. But if comprehensive socialist planning fails to work in practice—if, indeed, it is an impossibility, as we have learned from Mises and Hayek—then we cannot be “alienated” in Marx’s use of the term. We cannot really be “separated” from our “potential” to comprehensively plan the economy if comprehensive planning is impossible.
A staunch antiutopian, Marx claimed that his criticism of capitalism was based on the latest developments in science. He called his theory “scientific socialism” to clearly distinguish his approach from that of other socialists (Henri de Saint-Simon and Charles Fourier, for instance), who seemed more content to dream about some future ideal society without comprehending how existing society really worked (see socialism).
Marx’s scientific socialism combined his economics and philosophy—including his theory of value and the concept of alienation—to demonstrate that throughout the course of human history, a profound struggle has developed between the “haves” and the “have-nots.” Specifically, Marx claimed that capitalism has ruptured into a war between two classes: the bourgeoisie (the capitalist class that owns the means of production) and the proletariat (the working class, which is at the mercy of the capitalists). Marx claimed that he had discovered the laws of history, laws that expose the contradictions of capitalism and the necessity of the class struggle.
Marx predicted that competition among capitalists would grow so fierce that, eventually, most capitalists would go bankrupt, leaving only a handful of monopolists controlling nearly all production. This, to Marx, was one of the contradictions of capitalism: competition, instead of creating better products at lower prices for consumers, in the long run creates monopoly, which exploits workers and consumers alike. What happens to the former capitalists? They fall into the ranks of the proletariat, creating a greater supply of labor, a fall in wages, and what Marx called a growing reserve army of the unemployed. Also, thought Marx, the anarchic, unplanned nature of a complex market economy is prone to economic crises as supplies and demands become mismatched, causing huge swings in business activity and, ultimately, severe economic depressions.
The more advanced the capitalist economy becomes, Marx argued, the greater these contradictions and conflicts. The more capitalism creates wealth, the more it sows the seeds of its own destruction. Ultimately, the proletariat will realize that it has the collective power to overthrow the few remaining capitalists and, with them, the whole system.
The entire capitalist system—with its private property, money, market exchange, profit-and-loss accounting, labor markets, and so on—must be abolished, thought Marx, and replaced with a fully planned, self-managed economic system that brings a complete and utter end to exploitation and alienation. A socialist revolution, argued Marx, is inevitable.
Marx was surely a profound thinker who won legions of supporters around the world. But his predictions have not withstood the test of time. Although capitalist markets have changed over the past 150 years, competition has not devolved into monopoly. Real wages have risen and profit rates have not declined. Nor has a reserve army of the unemployed developed. We do have bouts with the business cycle, but more and more economists believe that significant recessions and depressions may be more the unintended result of state intervention (through monetary policy carried out by central banks and government policies on taxation and spending) than an inherent feature of markets as such.
Socialist revolutions, to be sure, have occurred throughout the world, but never where Marx’s theory had predicted—in the most advanced capitalist countries. On the contrary, socialism was forced on poor, so-called Third World countries. And those revolutions unwittingly condemned the masses to systemic poverty and political dictatorship. In practice, socialism absolutely failed to create the nonalienated, self-managed, and fully planned society. It failed to emancipate the masses and instead crushed them with statism, domination, and the terrifying abuse of state power.
Nations that have allowed for private property rights and full-blown market exchange, in contrast to those “democratic socialist republics” of the twentieth century, have enjoyed remarkable levels of long-term economic growth. Free-market economies lift the masses from poverty and create the necessary institutional conditions for overall political freedom.
Marx just didn’t get it. Nor did his followers. Marx’s theory of value, his philosophy of human nature, and his claims to have uncovered the laws of history fit together to offer a complex and grand vision of a new world order. If the first three-quarters of the twentieth century provided a testing ground for that vision, the end of the century demonstrates its truly utopian nature and ultimate unworkability.
In the wake of communism’s collapse, traditional Marxism, which so many mainstream economists criticized relentlessly for decades, is now seriously questioned by a growing number of disillusioned radicals and former Marxists. Today there is a vibrant post-Marxism, associated with the efforts of those active in the scholarly journal Rethinking Marxism, for instance. Rather than trying to solve esoteric puzzles about the labor theory of value or offering new theoretical models of a planned economy, many of today’s sharpest post-Marxists appreciate marginal analysis and the knowledge and incentive problems of collective action. In this new literature, friedrich hayek seems to be getting a more positive reception than Marx himself. Exactly what will come out of these developments is hard to predict, but it is unlikely to look like the Marxism of the past.<|endoftext|>
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# Factors of 13: Negative Factors, Factor Pairs, Sum, Factor Tree!
A factor of a number is any integer that divides evenly into that number. When a division leaves no remainder it means that the divisor is a Factor of the dividend. Furthermore, Factors of 13 helps with solving difficult mathematical problems that you might face. Additionally, in this blog, you will also learn about the Factors of 13, their Negative Factors, Factor Pairs, their Sum, Factor Tree and the Factors of 13 by Division Method.
## What are the Factors of 13?
In simpler terms, factors of a number are those numbers that divide into that number, leaving no remainder. When it comes to 13, things get interesting. The Factors of 13 are:
• 1
• 13
That is right, 13 has only two factors! This unique property is because 13 is a Prime number. Prime numbers are whole numbers greater than 1 that have exactly two distinct positive factors which are 1 and itself.
Here is why 1 and 13 are the only factors of 13. Imagine dividing 13 by different numbers. Dividing by 1 gives us 13, and dividing 13 by 13 gives us 1. Moreover, any other whole number division of 13 will result in a remainder, hence proving that those numbers are not factors of 13.
## What is the Sum of the Factors of 13?
Additionally, as you now know the Factors of 13, finding their Sum is a breeze. To find the Sum of the factors, you simply add up all the factors together.
In the case of 13, the Sum would be:
1 + 13 = 14
Thus, the Sum of the Factors of 13 is 14.
## What are the factors of Negative 13?
Factors apply not only to positive integers but also to negative integers. The Factors of negative 13 are identical to the Factors of 13. Here is why:
• Multiplying any Factor of 13 by -1 results in the corresponding negative factor.
• In this case, -1 x 1 = -1, and -1 x 13 = -13.
Therefore, the factors of negative 13 are:
• -1
• -13
## What are the Factor Pairs of 13?
Factor Pairs are simply the two factors of a number written together as a pair. As you saw earlier, 13 has only two factors. Hence, the factor pairs of 13 are:
• (1, 13)
• (-1, -13) (considering negative factors as well)
## Factors of 13 by Division Method
Furthermore, there are different methods to find the factors of a number. Here, you will get to learn the Division method:
• Start with 2: The first potential factor to check is 2, the smallest Prime number after 1.
• Divide 13 by 2.
• Since 13 divided by 2 gives a remainder of 1, you know 2 is not a factor.
• Continue with Odd Numbers: Since 13 is not divisible by 2, you can focus on odd numbers as potential factors.
• Move on to the next odd number, which is 3.
• Divide 13 by 3.
• Again, you get a remainder of 1, hence signifying that 3 is not a factor.
• Systematic Division: Keep dividing 13 by consecutive odd numbers (5, 7, 11, etc.). You will continue to get remainders until you reach.
• Reaching the Factor: Finally, when you divide 13 by 13, you get a quotient of 1 with no remainder. Therefore, this signifies that 13 is a factor of itself.
Thus, through this systematic division process, you have proved that the Factors of 13 are 1 and 13.
Related Blogs
I hope this helps! Did you like learning about the Factors of 13? You also learn about the Factors of 1 to 25! Also, keep reading our blogs to learn more about the Basic Concepts of Maths!<|endoftext|>
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How the “comparable worth” campaign succeeded
On a hillside in San Francisco a small public school bears the name of one of the pioneers in the movement for workplace equality. Kate Kennedy was born in Ireland, and like so many others, came to the United States during the Great Potato Famine of 1845-49. She was the first San Francisco teacher to join a union. In 1874, she brought a non-discrimination suit that provided the precedent for “equal pay for equal work.” Ultimately a federal law passed in 1963 made it illegal to pay men and women working in the same place different salaries for similar work.
Nonetheless, women’s wages today are between 70 and 80 percent of men’s, though they comprise 58 percent of the workforce. (Among union members, the gap is only 9 percent.) If different subgroups are compared, such as white men and women of color, the gap is much larger. Some of the excuses for this include: “Men need to be paid more because they are heads of the household;” “Women only work to supplement their husbands’ pay;” “Women only work outside the home to earn ‘pin money.’” Of course, employers will always find reasons to pay workers less, women or no.
Employers, both public and private, were able to maintain the “gender gap” by basing their wage structures on “market value,” that is, what workers in similar jobs were paid in different locales. So, if the wages for a job, like that of librarian, were generally depressed in other places, the employer could claim that it would be fair to pay the same wage. This, of course, preserved the gender gap because librarians were predominantly women.
Enter Maxine Jenkins. Born in rural Mississippi, Jenkins arrived in the Bay Area in 1964, and took a clerical job at UC Berkeley while taking classes. At the time, organizers for AFSCME, the American Federation of State, County and Municipal Employees, were organizing clerical workers on campus. Jenkins joined and never looked back. She dropped out of school and worked on organizing campaigns full time. “We must get out of the bind of depending on surveys that compare salaries with already depressed rates elsewhere,” she declared at a California Labor Federation meeting in 1973.
Jenkins and other feminist organizers based their campaign on something called “comparable worth,” which asserted that men and women should be paid equally according to comparable skill, effort, and responsibility under similar working conditions, not by comparing salaries in different locales. In June of 1981, these efforts culminated in a nine-day strike for pay equity by San Jose public sector workers. They won pay increases in more than 60 job categories and brought nationwide attention to the idea of comparable worth.
Fifty-eight percent of the workforce? Working for between 20 to 30 percent less than what the other 42 percent earn? This battle for social justice is far from over.
— By Bill Morgan, a member of the CFT Labor and Climate Justice Education Committee who taught elementary students in San Francisco for 34 years<|endoftext|>
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During the last few lessons in Third Grade, we have been learning about Nested Loops. A nested loop is a loop inside of loop. We’ve used them in our Code.org lessons to move and collect things and to draw shapes and designs.
As an extension, we did the Awesome Sauce Challenge in Scratch. Students were given nine blocks that they had to use in a program. They had a start block (when green flag is clicked), a move block, a turn right and left, a repeat block, a change color, a set pen size, a pen up, and a pen down. Each one must be used at least once. They can re-use any block any amount of times. They also have permission to change the variables in the blocks to make each one unique. They were also encouraged to experiment with using Nested Loops.
The goal was to PLAY and CREATE a unique and amazing digital design — AWESOME SAUCE!! Check out some of our sauciest creations!!<|endoftext|>
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# What is the scalar product of two vectors?
## What is the scalar product of two vectors?
The scalar product of two vectors is defined as the product of the magnitudes of the two vectors and the cosine of the angles between them.
### What is scalar product of two vectors give an example?
Solution – If two vectors are perpendicular to each other then their scalar product is 0. So we get: (-2)(-8) + (-r)(r) = 0 i.e. r2 = 16, hence r = 4 or -4….Solved Examples of Scalar and Vector Product of Two Vectors.
Scalar Quantity Vector Quantity
This is always a positive number This can be positive or negative
How do you find the product of two vectors?
Vector Product of Two Vectors
1. If you have two vectors a and b then the vector product of a and b is c.
2. c = a × b.
3. So this a × b actually means that the magnitude of c = ab sinθ where θ is the angle between a and b and the direction of c is perpendicular to a well as b.
What is the scalar triple product?
The scalar triple product of three vectors a, b, and c is (a×b)⋅c. The scalar triple product is important because its absolute value |(a×b)⋅c| is the volume of the parallelepiped spanned by a, b, and c (i.e., the parallelepiped whose adjacent sides are the vectors a, b, and c).
## What is a scalar product?
Definition of scalar product : a real number that is the product of the lengths of two vectors and the cosine of the angle between them. — called also dot product, inner product.
### Which is the vector triple product?
The cross-product of the vectors such as a × (b × c) and (a × b) × c is known as the vector triple product of a, b, c. The vector triple product a × (b × c) is a linear combination of those two vectors which are within brackets. The ‘r’ vector r=a×(b×c) is perpendicular to a vector and remains in the b and c plane.
What is vector product of two vectors give its four properties?
1) Cross product of two vectors is equal to the area of parallelogram formed by two vectors. 2) Area of triangle formed by two vectors and their resultant is equal to half the magnitude of cross product. 3) Vector product of two vectors is anti commutative.
What are scalar and vector products?
The vector product has the anticommutative property, which means that when we change the order in which two vectors are multiplied, the result acquires a minus sign. The scalar product of two vectors is obtained by multiplying their magnitudes with the cosine of the angle between them.
## What is scalar and vector triple product?
By the name itself, it is evident that the scalar triple product of vectors means the product of three vectors. It means taking the dot product of one of the vectors with the cross product of the remaining two. It is denoted as. [a b c ] = ( a × b) . c.
### Which properties is this ax BxC Ax B xC?
Associative property: For any three whole numbers a, b and c, (a x b) x c = a x (b x c), this means the product is regardless of how grouping is done. This explain the associative property of multiplication.<|endoftext|>
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Reciprocal of 6/6
What is the reciprocal of 6/6? Here we will define reciprocal of 6/6, and show you how to calculate the reciprocal of 6/6 in fraction form and decimal form.
The reciprocal of 6/6 is a fraction or a number that when multiplied by 6/6 is equal to 1. To get the reciprocal of 6/6 based on that definition, we can make the following equation where "R" is the reciprocal of 6/6.
(6/6) × R = 1
When we solve for R, we get the answer to the reciprocal of 6/6 as a fraction as follows:
(6/6) × R = 1
R = 6/6
Reciprocal of 6/6 = 6/6
You can check that the answer is correct by confirming that 6/6 times the reciprocal of 6/6 is equal to 1, like this:
6/6 × 6/6 = 1
To find the decimal answer to the reciprocal of 6/6, you divide the numerator of the reciprocal by the demoninator of the reciprocal, like this:
6 ÷ 6 = 1
Reciprocal of 6/6 = 1
Tip: As you may have inferred from our tutorial above, you can quickly get the reciprocal of any fraction, such as 6/6, by switching the numerator and the denominator.
Reciprocal of a fraction
Do you need the reciprocal of another fraction? No problem! Please enter another fraction below.
Reciprocal of /
Reciprocal of 6/7
Here is the next fraction on our list that we have calculated the reciprocal of.<|endoftext|>
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When the ore had been separated from any rock and vein minerals, the resulting concentrate was taken for smelting. From the mediaeval period to the early twentieth century, three principal methods were used for this. These are bales (or boles), ore-hearths, and reverberatory furnaces.
Bales or Boles
The former in northern England, and the latter in Derbyshire and some other parts of the British Isles, were used to smelt lead until the late sixteenth century. They were basically bonfires, built in relatively exposed and elevated locations by laying repeating layers of timber and lumps of ore on each other. Then the bale was lit and left to burn, apparently with little control, until the molten lead ran out of the base into a collecting area where it cooled.
Bales were inefficient, being ruled by the wind and wasteful of fuel. Moreover, much lead was lost, either through volatilization or into the slags. Nevertheless, from the eleventh to the late-sixteenth century, most lead ore was smelted this way.
Around 1540, a new type of smelting hearth, blown by a foot-operated bellows, was developed on the Mendip, in Somerset. This had evolved into the ore-hearth, with its characteristic work-stone, by 1571 or 72, when the first of them reached Derbyshire. The ore-hearth could be built inside a mill, alongside a stream which turned a waterwheel to drive the bellows. The hearth was normally raised above the ground and placed in an arched recess which supported a chimney. Ore-hearths also burnt less fuel and could smelt smaller pieces of ore, which gave the operators great savings and ensured that the uptake of the new technology was both widespread and rapid.
Early ore-hearths were made entirely from large stones until the early eighteenth century, when they were generally made from iron, although their integral parts were still called stones.
The ore-hearth burnt dried wood, called chopwood, which was branch wood cut into lengths of about 150 mm by 50 mm and dried in a kiln over a slow fire. Chopwood was also supplemented with a little coal at some mills. Smelters in Yorkshire and the north Pennines made further savings on smelting costs by burning peat in their ore-hearths from the 1670s. German copper smelters, at Keswick in Cumbria, had burnt peat in the late sixteenth century. The switch to peat burning never took place in Derbyshire, where smelting had historically been concentrated on the eastern fringe of the orefield, near woods and water supplies, but away from the peat moors. This made the cost of transporting peat to the mills prohibitive even if supplies could be found.
The smelting was done in two basic steps, although secondary processes (for example, roasting the concentrate to drive off moisture and some sulphur) were sometimes done in a separate furnace. The first step took place between 600 and 800 degrees Celsius, and recovered from 40 to 75 per cent of the metal and left a lead-rich ‘grey’ slag.
In the second stage these slags were heated to between 1000 and 1200 degrees in a slag hearth to melt them entirely. In order to get such temperatures coke was used as fuel.
Reverberatory or Cupola Furnaces
In the mid and north Pennines, where coal was relatively expensive, ore-hearths were used until the early twentieth century. Elsewhere, however, where coalfields were nearer, ore-hearths were superseded by reverberatory furnaces or cupolas. They burnt coal on a separate fire grate and the heat was reflected onto the ore, which was in a low-arched compartment, by a brick fire-bridge. This kept the coal smoke away from the lead and allowed more ore to be smelted at once. Because they were built of fire brick and their temperature could be controlled, reverberatory furnaces could also work for much longer than ore-hearths, which had to be put out and allowed to cool. This gave savings in fuel, scale and the furnace’s idle time.
The first successful cupolas in Britain were used for lead and copper smelting near Bristol in the 1680s, and others, for lead, were built in Flintshire during the 1690s, where they were also later used by the London Lead Company at its Gadlys works in Flintshire. In 1701, a cupola was built at Marrick, in Yorkshire, but it only worked for a few years before a dispute amongst the owners forced its closure. It was the only cupola in Yorkshire until 1792, when two were built on Grassington Moor. Here the Duke of Devonshire’s failure to secure a supply of peat forced him to build cupola furnaces and import expensive, high quality coal.
In the north Pennines, the London Lead Company, in particular, had wide experience of the early use of cupolas for lead smelting. In 1705, it acquired the Ryton Smelting Company’s cupola mill, near Newcastle upon Tyne, but the mill was abandoned in 1706 because savings in transport costs could be made if Whitfield smelt mill, which had ore-hearths and was nearer the mines, was leased instead. By 1724, this smelt mill had been fitted with cupolas, but, when the company leased the Nenthead Mill in 1757 the ore-hearths were retained. The London Lead Company’s principal smelting centre in Teesdale, at Eggleston, on a site leased in 1771, eventually had three mills, all with cupolas. Eggleston became a centre of innovation and extensive experiments were conducted on improving the cupola and the condensation of lead fumes.
It was not until 1735 that the London Lead Company introduced them into Derbyshire, at its Bowers Mill in Ashover. Two years later, Richard Bagshawe used them at his Olda Mill, where they initially proved disappointing. Nevertheless, the Bagshawe family and its associates were responsible for spreading the use of cupolas throughout Derbyshire where they superseded the ore-hearth by the 1780s.
Flues and Chimneys
Early smelt mills had stumpy vertical chimneys, directly above the hearth, but a ground-level flue was introduced at the Upper Cupola in Middleton Dale, Derbyshire, in the 1770’s. This took poisonous fumes away from the smelters and prevented pollution of a nearby pasture. It appears to have been a providential discovery that vaporised lead settled on the sides of flues and could be recovered.
Flues got longer during the 19th century and some became very complex. Others were fitted with condensation chambers. Side openings on some flues provided access for workers to enter the flue and scrape condensed ‘fume’ off the sides. These scrapings could be flushed with water down to settling ponds, where they were recovered and taken for smelting.Return to previous page<|endoftext|>
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# 2006 AIME I Problems/Problem 8
(Redirected from 2006 AIME I Problem 8)
## Problem
Hexagon $ABCDEF$ is divided into five rhombuses, $\mathcal{P, Q, R, S,}$ and $\mathcal{T,}$ as shown. Rhombuses $\mathcal{P, Q, R,}$ and $\mathcal{S}$ are congruent, and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $\mathcal{T}$. Given that $K$ is a positive integer, find the number of possible values for $K$.
## Solution 1
Let $x$ denote the common side length of the rhombi. Let $y$ denote one of the smaller interior angles of rhombus $\mathcal{P}$. Then $x^2\sin(y)=\sqrt{2006}$. We also see that $K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y$. Thus $K$ can be any positive integer in the interval $(0, 2\sqrt{2006})$. $2\sqrt{2006} = \sqrt{8024}$ and $89^2 = 7921 < 8024 < 8100 = 90^2$, so $K$ can be any integer between 1 and 89, inclusive. Thus the number of positive values for $K$ is $\boxed{089}$.
## Solution 2
Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where $w*h=\sqrt{2006}$. The height of rhombus T would be 2h, and the width would be $\sqrt{w^2-h^2}$. Substitute the first equation to get $\sqrt{\frac{2006}{h^2}-h^2}$. Then the area of the rhombus would be $2h * \sqrt{\frac{2006}{h^2}-h^2}$. Combine like terms to get $2 * \sqrt{2006-h^4}$. This expression equals an integer K. $2006-h^4$ specifically must be in the form $n^2/2$. There is no restriction on h as long as it is a positive real number, so all we have to do is find all the positive possible values of $n^2$ for $2006-h^4$. Now, quick testing shows that $44^2 < 2006$ and $45^2>2006$, but we must also test $44.5^2$, because the product of two will make it an integer. $44.5^2$ is also less than $2006$, so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5 is also a valid value. (notice 0 is not valid because the height must be a positive number) That gives us $44*2+1=$ $\boxed{089}$
-jackshi2006<|endoftext|>
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## Algebra - Rings and fields
### 2. Characteristic of a ring
• Definition 2.1:The characteristic of a ring $R$ is a smallest positve integer $n$ such that $na=0$ $\forall$ $a \in R$.
i.e., $na=\smash{\underbrace{a+a+...+a}_{n\hspace{0.1cm}times}}=0$ $\forall$ $a \in R$ denoted by $ch(R)=n$.
• Examples: $ch(\mathbb{Z})=0$, $ch(\mathbb{Q})=0$, $ch(\mathbb{R})=0$, $ch(\mathbb{C})=0$, $ch(\mathbb{Z}_n)=n$
• Theorem 2.1: The characteristic of an integral domain is either $0$ or prime number.
Proof: Let $D$ be an integral domain. To prove $ch(D)=0$ or $ch(D)=n$ where $n$ is a prime number.
If $ch(D)=0$ then nothing to prove.
Let $ch(D)=n$ where $n$ is a smallest positive integer such that $na=0$ $\forall$ $a \in D$.
Suppose $n$ is a prime. Then we can have $n=n_1n_2$ where $n_1\neq 1$, $n_2\neq 1$ and $n_1<n$, $n_2<n$
Let $a \in D$ $\Rightarrow$ $a^2 \in D$. Then $na^2=0$
$\Rightarrow$ $(n_1n_2)a^2=0$
$\Rightarrow$ $\smash{\underbrace{a^2+a^2+...+a^2}_\text{$n_1n_2$times}}=0$
$\Rightarrow$ $\smash{\underbrace{(a+a+...+a)}_\text{$n_1$times}}$ $\smash{\underbrace{(a+a+...+a)}_\text{$n_2$times}}$ $=0$
$\Rightarrow$ $(n_1a)(n_2a)=0$
$\Rightarrow$ $n_1a=0$ or $n_2a=0$ which is a contradiction to the fact $ch(D)=n$ [$\because$ $n_1<n$ and $n_2<n$] Hence $n$ must be prime. $\blacksquare$
• Definiton 2.2: An element $a$ of a ring $R$ is said to nilpotent if $\exists$ a positive integer $n$ such that $a^n=0$.
• Definition 2.3: An element $a$ of a ring $R$ is said to be idempotent if $a^2=a$.
• Definition 2.4: A ring $R$ is said to be boolean ring if $a^2=a$ $\forall$ $a \in R$.
• Problem 2.1: In a ring $R$, if $(ab)^2=a^2b^2$ $\forall$ $a,b \in R$. Then show that $R$ is commutative.
Solution: Given $(ab)^2=a^2b^2$ $\forall$ $a,b \in R$
$\Rightarrow$ $(ab)(ab)=(aa)(bb)$
$\Rightarrow$ $a[b(ab)]=a[a(bb)]$
$\Rightarrow$ $b(ab)=a(bb)$ [by left cancellation law in ring $R$]
$\Rightarrow$ $(ba)b=(ab)b$ [by associative law in $R$]
$\Rightarrow$ $ba=ab$ [by right cancellation in $R$]
$\therefore$ $ab=ba$ $\forall$ $a,b \in R$ $\spadesuit$
• Problem 2.2: If $a^2=a$ $\forall$ $a \in R$, then show that
(1.) $a+a=0$
(2.) $a+b=0$ $\Rightarrow$ $a=b$
(3.) $R$ is commutative
Solution: (1.) Given, $a^2=a$ $\forall$ $a \in R$ $\Rightarrow$ $a+a \in R$
So $(a+a)^2=a+a$
$\Rightarrow$ $(a+a)(a+a)=a+a$
$\Rightarrow$ $a^2+a^2+a^2+a^2=a+a$ [by distributive law in ring $R$]
$\Rightarrow$ $a+a+a+a=a+a$ [$\because$ Given, $a^2=a$ $\forall$ $a \in R$]
$\Rightarrow$ $a+a+a=a$ [by left cancellation in $R$]
$\Rightarrow$ $a+a+a=a+0$
$\Rightarrow$ $a+a=0$ [by left cancellation in $R$]
(2.) Given $a+b=0$ and from (1.) we have $a+a=0$
$\Rightarrow$ $a+b=a+a$
$\Rightarrow$ $b=a$ [by right cancellation in $R$]
(3.) Given $a^2=a$ $\forall$ $a \in R$
Let $a,b \in R$ $\Rightarrow$ $a^2=a$ and $b^2=b$ also $a,b \in R$ $\Rightarrow$ $ab \in R$
So we have $(ab)^2=ab$
$\Rightarrow$ $(ab)^2=a^2b^2$
$\Rightarrow$ $(ab)(ab)=(aa)(bb)$
$\Rightarrow$ $a[b(ab)]=a[a(bb)]$
$\Rightarrow$ $b(ab)=a(bb)$
$\Rightarrow$ $(ba)b=(ab)b$ $\Rightarrow$ $ba=ab$ $\forall$ $a,b \in R$
$\therefore$ $R$ is commutative. $\spadesuit$
• Problem 2.3: Find all the idempotent elements of $(\mathbb{Z}_6,\oplus_6,\otimes_6)$
Solution: $\mathbb{Z}_6$ $=\{0,1,2,3,4,5\}$
$0^2=0\otimes_60=0$
$1^2=1\otimes_61=1$
$2^2=2\otimes_62=4$
$3^2=3\otimes_63=3$
$4^2=4\otimes_64=4$
$5^2=5\otimes_65=1$
$\therefore$ $0,1,3,4$ are idempotent elements in $\mathbb{Z}_6$. $\spadesuit$
• Exercise 2.1: Prove that a skew field has no zero divisors.
• Exercise 2.2: Find all the idempotents and nilpotents of $(\mathbb{Z}_4,\oplus_4,\otimes_4)$.
• Exercise 2.3: Show that the only idempotents of an integral domain is $0$ and $1$.
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The body needs insulin to transport glucose into cells, where it is needed for producing energy. Without insulin-producing beta cells, the glucose builds up in the blood stream, and the body cannot use it for energy.
It is not yet known what causes type 1 diabetes, and there is currently no cure for the illness. The most common treatment option is administering insulin, but the ultimate goal of the medical research community is to stop the body's immune system from attacking its own beta cells, or reversing this process.
Omega-3s are a class of polyunsaturated fatty acids (PUFAs). They are typically found in fish, seafood, and some vegetable oils, as well as in dietary supplements.
These fatty acids are important for the good functioning of several organs, as these beneficial fats improve the activity of muscles, prevent blood clotting, help digestion, and aid the division and growth of cells.
The health benefits of omega-3s have been investigated in several studies. Some research has suggested that omega-3s protect against cardiovascular events and rheumatoid arthritis, but other studies seem to contest these positive effects.
What is known, however, is that omega-3s stop inflammatory processes in the body. This has led the authors of the new research - led by Allan Zhao at Guangdong University of Technology in Guangdong, China - to think that these fatty acids might be able to either prevent or improve the negative outcomes of autoimmune diseases.
Omega-3s may regenerate beta cells
Zhao and team used non-obese diabetic (NOD) mice, to which they fed a regular diet and a diet enriched with PUFAs. They also increased the levels of omega-3s in these mice through genetic modification.
The team tested the mice every 3 months for glucose tolerance and insulin tolerance.
The researchers also examined the pancreas of mice for insulitis - an infiltration of lymphocytes in the islets of the pancreas, which is a phenomenon typical of type 1 diabetes.
Additionally, the researchers collected blood from the mice and measured their levels of serum insulin.
The study revealed that adding omega-3s to the diet of NOD mice significantly improved the metabolism of glucose and decreased the incidence of type 1 diabetes.
The researchers noted a decrease in pro-inflammatory cell-signaling proteins, as well as a considerable decrease in insulitis. Namely, they noticed that omega-3s lowered the levels of interferon gamma, interleukin 17, interleukin 6, and tumor necrosis factor alpha, or TNF-α.
Moreover, Zhao and colleagues noticed signs of beta cell regeneration in the mice that had been treated with omega-3.
Both nutritional supplementation and genetic therapy normalized blood sugar and insulin levels for a minimum of 182 days, stopped the development of autoimmunity, blocked the lymphocytes from entering the regenerated islets in the pancreas, and drastically increased the levels of beta cell markers.
These results suggest that omega-3 PUFAs may serve as a new therapy for type 1 diabetes. Referring to the potential therapeutic effects of nutritional supplementation of omega-3s or the increase of omega-3s through gene therapy, the authors conclude:
"Our observations may also offer clinical guidance, in that those patients who are either at the early-onset stage of [type 1 diabetes] or have consistently had good management of their blood glucose levels may benefit the most from these interventions. These treatment modalities, if cleared in safety evaluations, may potentially be helpful in the treatment of other types of autoimmune diseases as well."<|endoftext|>
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asked in category: General Last Updated: 30th May, 2020
# How do you find the focal point of an ellipse?
Each ellipse has two foci (plural of focus) as shown in the picture here: As you can see, c is the distance from the center to a focus. We can find the value of c by using the formula c2 = a2 - b2. Notice that this formula has a negative sign, not a positive sign like the formula for a hyperbola.
In this regard, how do you find the foci of an ellipse?
actually an ellipse is determine by its foci. But if you want to determine the foci you can use the lengths of the major and minor axes to find its coordinates. Lets call half the length of the major axis a and of the minor axis b. Then the distance of the foci from the centre will be equal to a^2-b^2.
Also, how do you find focal points? To find the focal point of a parabola, follow these steps: Step 1: Measure the longest diameter (width) of the parabola at its rim. Step 2: Divide the diameter by two to determine the radius (x) and square the result (x ). Step 3: Measure the depth of the parabola (a) at its vertex and multiply it by 4 (4a).
Thereof, what is the focal point of an ellipse?
An ellipse is defined as follows: For two given points, the foci, an ellipse is the locus of points such that the sum of the distance to each focus is constant. Note: If the interior of an ellipse is a mirror, all rays of light emitting from one focus reflect off the inside and pass through the other focus.
How do you find the equation of an ellipse with foci and points?
Use the standard form (x−h)2a2+(y−k)2b2=1 ( x − h ) 2 a 2 + ( y − k ) 2 b 2 = 1 . If the x-coordinates of the given vertices and foci are the same, then the major axis is parallel to the y-axis. Use the standard form (x−h)2b2+(y−k)2a2=1 ( x − h ) 2 b 2 + ( y − k ) 2 a 2 = 1 .
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30th May, 2020
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Mrs. Joans Robinson
Model of Economic Growth:
Mrs. Robinson includes the issue of
population growth in her model. She shows the effects of Population on the rate
of Capital Accumulation and rate of Growth of Output.
Her model is based upon two
(i) Capital accumulation depends upon distribution of income.
(ii) The utilization of labor depends upon supply of labor and capital.
Assumptions of the Model:
(i) National income is distributed between two classes of the economy, i.e.,
labor and producers.
(ii) The laborers entirely spend their wages on consumption and do not make
(iii) The producers who earn profits make the savings and reinvest
such savings and do not make any consumption. If they do not earn profits, they will not be able to
invest. And if they do not invest, they will not be able to earn profits.
(iv) There are
no technological changes and ratio of K to L remains constant.
(v) There is no change in price level.
(vi) There is closed economy and no intervention by the state.
(vii) There is no shortage of supply of labor in the economy.
As told above that model assumes the existence of two classes in the economy.
The labor class consumes all of its income while the entrepreneurs reinvest
their profits. Accordingly, the real savings remain equal to real investment.
But how much money producers invest, depends upon the real wages demanded by.
the labor. Moreover, the productive efficiency and BOP, etc., are obstacles in
the way of promoting investment. As the economy grows, such obstacles become
high and high. Thus, in the presence of such obstacles, economic growth depends
upon the role of entrepreneurs. If they make inventions and innovations then the
obstacles in the way of economic growth will be surmounted. In this way, the
economy will enter in a stage which is known as Golden Age by Mrs.
Robinson. Here, capital accumulation is at a higher rate. Here, capital
accumulation is maintained through the constant rate of profit - which is
known as Golden Rule of economic growth by Mrs. Robinson.
The growth rate during golden age will be similar to Harrod's Natural Growth
rate. We now explain the model symbolically.
The equation (3) shows that the rate of profit (π) depends upon the
efficiency of labor = p = Y/N, real wages (w/P) and the ratio of K to N. i.e., θ
As producers wish to maximize their profits. Accordingly, we will take first
derivative of profit function and keep it equal to zero. As P.F. is:
Y = f (N, K)
Equation (11) shows the growth rate of capital which the entrepreneurs can
attain by depending upon capitalistic principle. This equation also shows that
growth rate of capital may grow if net return of capital (p - w/P) increases
more than capital labor ratio (9). As Ricardo said, "Capital accumulation will
be strengthened if the level of real wages is low". This means that Mrs.
Robinson wishes to take us towards Ricardian growth theory through Keynesian
In Robinson's golden age, at
equilibrium, labor will be fully employed and full use of capital will become
possible. But such will happen only if θ = K/N
remains constant. At Yf, the rate of change of labor and rate of change of
capital will be equalized. It is as:
K/N = θ
K = Nθ
K/θ = N
N = K/θ
In case of change, we have:
ΔN = ΔK/θ
From this relation we can find the rate of change of labor force having
compatibility with the rate of change of capital.
Dividing both sides by N, we get:
The eq.(12) shows that the labor will be fully employed if labor and capital
grow at the same rate or if capital grows at the same rate to that of growth of
labor, Yf of labor will become possible. Because in this situation the ratio of K to N, i.e.,
remain constant. When labor and capital grow in the same ratio - the equilibrium
position of the economy is known as Golden Age. It is shown in
Here, OW* is minimum wage rate. ON is growth of labor force. While OY* is
expansion path. At C, we draw a tangent. Here, K/N = OK, while growth rate of
labor is ON*. The wages given to labor are OW*, then the rate of surplus
(profits) will be equal to HC. Such profits will be able to absorb the
growth of labor which is ON*. This is the golden age, because profits absorb the
Showing the effect of capital accumulation on growth of the economy, we show
the effects of growth of labor force on economic growth. She says if in an
economy, population and labor grow but capital does not grow. As a result, the
MPL will decrease. If real wages remain constant then profits of producers will
decline - badly affect the capital accumulation.
This will result in unemployment, as it is happening in LDC's. Thus
Yf ia possible only if increase in labor force and increase in capital are equal. This is
the golden age. It is as:
ΔN/N = ΔK/K
In golden age, the proportion of profits and wages remain the same. All the
variables in the economy grow at the same rate. Under the proper rate of capital
accumulation - there is a neutral technical progress and there is a constant
ratio between capital and output.<|endoftext|>
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Any material that’s under pressure can be dangerous if it’s not handled properly. If the material is a compressed gas, it may be flammable, explosive, reactive, toxic or a combination of these. Because of the hazards of compressed gases, it’s very important to know what you’re working with, what its hazardous properties are and how to safely handle its container-the compressed-gas cylinder.
Use Compressed Gases Safely
Compressed gases are hazardous because of the high pressure at which they are stored in cylinders and pressure tanks. The compressed gases can be flammable, poisonous, corrosive, or any combination of these.
How can they hurt me?
Mishandling of compressed gases has been responsible for fatalities, serious injuries, and property damage that has amounted to millions of dollars.
Flammable compressed gases:
- Explode if handled roughly or exposed to heat.
- Ignite by heat, sparks, or flames.
- Flash back if vapors travel to a source of ignition.
- Produce irritating or poisonous gas when burning.
Non-flammable compressed gases:
- Explode when in a mixture with fuels.
Health Effects of Compressed Gases
- Are harmful if inhaled.
- Have extremely irritating vapors.
- Can cause cryogenic burns to skin and eyes.
- Produces irritating or poisonous gas when burning.
- Causes dizziness, unconsciousness, or suffocation.
Handling compressed gas cylinders
Compressed gas cylinders require careful handling to prevent damage. When handling cylinders:
- Move cylinders (securely fastened, in as near an upright position as possible) on special hand trucks.
- Don’t drop or bang cylinders together.
- Don’t roll, drag, or slide cylinders and never use cylinders as rollers or supports.
- Don’t lift cylinders by their caps.
- Don’t use magnets to lift cylinders.
- Cradles or platforms can be used to lift cylinders only if the cylinder was manufactured with lifting attachments.
Compressed Gas Storage
Some general guidelines in storing compressed gas cylinders include:
- Store cylinders in an upright position.
- Storing the cylinders in a safe, dry, well-ventilated place that is clean and free of combustible material.
- Avoiding areas where cylinders can be knocked down or damaged.
- Storing the cylinders in a position that ensures that the safety relief device is always in direct contact with the cylinder’s vapor space.
- Store oxygen CGCs at least 20 feet from flammables or combustibles, or separate them by a 5 foot, fire-resistant barrier.
Tips for Compressed Gas Safety
- Before handling any compressed-gas cylinder, identify the type of gas it houses by its identification and hazard labels, not its color. Different manufacturers use different color codes.
- Check the cylinder’s label for hazards, and read the material safety data sheet (MSDS) for instructions on protective equipment and handling.
- Look for the maximum approved pressure label and make sure a current test date is indicated. If the cylinder is missing this information, it should not be handled.
- Only trained personnel should unload compressed-gas cylinders.
- Inspect cylinders for damage or leaks.
- Move defective cylinders to an isolated storage area; a ruptured cylinder can become a rocket with the force to blast through a concrete wall.
- When moving cylinders, use special cylinder hand trucks, keeping the cylinder lashed to the cradle and standing as upright as possible.
- Avoid dropping, banging or rolling cylinders.
- Keep compressed-gas cylinders away from fire, heat and sparks.
- When using a cylinder, open the valve slowly, with the cylinder pointed away from people.
- Make sure the hoses and connections are clean and in good condition each time you use the cylinder.
- When a cylinder is not in use, screw down the protective metal cap to the last thread.
- Label empty cylinders with “MT” and keep them separate from full ones.
- Store compressed-gas cylinders upright, secured with a chain or cable, in a safe, well-ventilated, fire-resistant area with a controlled temperature below 125° F (51.7° C).
- Keep cylinders out of direct sunlight and away from heat sources, combustible materials and electrical wiring.·
- Group cylinders with others housing the same contents.
- Rotate stock, using older cylinders first.
- Avoid using cylinders in confined spaces.
- Keep oxygen cylinders at least 20 feet away from flammable-gas containers, combustible materials, oil and grease.
Compressed Gases That Need Special Handling
- Acetylene and hydrogen: Both of these gases are highly explosive and must be handled with extreme caution. Hydrogen escapes easily from threaded fittings that aren’t completely tight. and such leaks can ignite spontaneously from the friction of the escaping gas. Hydrogen has no odor to warn of a leak.
- Oxygen: While not flammable itself, oxygen increases the tendency of things around it to bum or explode. Keep oxygen cylinders away from combustible or flammable materials and fire hazards, including grease and oil on your clothes, hands and work area. Oxygen should not be used in place of compressed air.
- Chlorine and fluorine: These gases are highly corrosive and irritating. When mixed with acetylene and exposed to light, they may explode. Chlorine will form corrosive hydrochloric acid in water, eating into iron or steel equipment. The proper respirator and other protective equipment should be available in case of a leak.
- Ammonia: This is a highly corrosive gas. When using it, make sure you have quick access to the proper respirator and other protective equipment.<|endoftext|>
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Thread: Create a decomposition of functions
1. Create a decomposition of functions
Decomposing a function as a composition
Find f(x) and g(x) such that h(x)=(fog)(x)
h(x)=x2-3x/square root of x+5
I tried to do it but I do not think is right. this is the answer I get: f(x)= x2-3x g(x)= 1/square root of x+5
2. Re: Create a decomposition of functions
Your answer was a quotient, not a composition. You found f and g so that h(x) = f(x)/g(x).
There are an unlimited number of choices given the way the problem is stated.
For instance, there are always these trivial choices:
Choice #1) f(x) = h(x) and g(x) = x
Choice #2) f(x) = x and g(x) = h(x)
Maybe more interesting is letting g(x) = x+5.
Then $\displaystyle h(x) = \frac{x^2-3x}{\sqrt{x+5}}$ and $\displaystyle x = g(x)-5$, so
$\displaystyle h(x) = \frac{(g(x)-5)^2-3(g(x)-5)}{\sqrt{(g(x)-5)+5}}$
$\displaystyle = \frac{(g(x)^2 - 10g(x) + 25) + (-3g(x)+15))}{\sqrt{g(x)}}$
$\displaystyle = \frac{g(x)^2 - 13g(x) + 40}{\sqrt{g(x)}}$
$\displaystyle = f(g(x))$, where $\displaystyle f(t) = \frac{t^2 - 13t + 40}{\sqrt{t}}$
Choice #3) $\displaystyle f(t) = \frac{t^2 - 13t + 40}{\sqrt{t}}, g(x) = x+5$
Now try $\displaystyle g(x) = \sqrt{x+5}$. Note that $\displaystyle g(x) \ge 0$ (actually, it can't be allowed to be 0). Have $\displaystyle x = g(x)^2 - 5$.
Thus $\displaystyle h(x) = \frac{x^2-3x}{\sqrt{x+5}}$ and $\displaystyle x = g(x)^2-5$, so
Thus $\displaystyle h(x) = \frac{(g(x)^2-5)^2-3(g(x)^2-5)}{\sqrt{(g(x)^2-5)+5}}$
$\displaystyle = \frac{(g(x)^4-10g(x)^2+25) +(-3g(x)^2 + 15)}{\sqrt{g(x)^2}}$
$\displaystyle = \frac{g(x)^4 -13 g(x)^2 + 40}{|g(x)|}$ (now use $\displaystyle g(x) \ge 0$ to know that $\displaystyle |g(x)| = g(x)$)
$\displaystyle = \frac{g(x)^4 -13 g(x)^2 + 40}{g(x)}$
$\displaystyle = g(x)^3 - 13g(x) + \frac{40}{g(x)}$
$\displaystyle = f(g(x))$, where $\displaystyle f(t) = t^3 - 13t + \frac{40}{t}$
Choice #4) $\displaystyle f(t) = t^3 - 13t + \frac{40}{t}, g(x) = \sqrt{x+5}$
3. Re: Create a decomposition of functions
Hi! Thanks for answering my question. However my professor said that is not the way he wants it. He said that the answer g(x) should be g(x)=1/square root of x+5 and I only need to find f(x). I guess he only wants a quotient? Mine is wrong.Can you help me please?? So when I find (fog) I will end up with: x^2-3x/ square root of x+5 back again. Thanks!!
4. Re: Create a decomposition of functions
Review what I did for Choice #3 and Choice #4. In each case, I began with choice of g(x) (one time was x+5, the other it was sqrt(x+5)). Then I did some algebra and eventually found f(t) such that h(x) = f(g(x)). So if you repeat that process, you'll solve the problem. Repeat the process I did, only now your choice of g(x) will be:
$\displaystyle g(x) = \frac{1}{\sqrt{x+5}}$.<|endoftext|>
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Lipids are organic molecules found in all living organism. Some lipids are hydrophobic molecules (triglycerides, sterol esters) while others are hydrophilic molecules (phospholipids, short chain fatty acids).
Lipids are grouped into fats, phospholipids, steroids, and waxes.
Fats are made of three fatty acid chains and glycerol, and are also called triglycerides. They are either solid or liquid (oil) at room temperature. The fatty acids consist of long carboxyl group at one end, which may be saturated or unsaturated.
Saturated fats are solids at room temperature, and do not have double bonds between carbon molecules. Instead they are saturated with hydrogen molecules. Unsaturated fats contain one or more double bonds and are liquids at room temperature. Unsaturated fats with one double bond are called monounsaturated fats, while those with more than one double bond are called polyunsaturated fats.
Body fat is stored in adipose tissue, and used as source of energy. Fats act as insulators in our body, and help in maintaining body temperature. They are also important constituents of a cell.
Phospholipids have of two fatty acids, a glycerol unit, and a phosphate group, and are amphipathic molecules. The phosphate group is water soluble (hydrophilic) while the fatty acid tail is insoluble in water (hydrophobic). So, when phospholipids are placed in water they form a bilayer in which the non-polar tails face away from the water, and the polar ends face outward and interact with water.
Phospholipids are a major structural component of cell membrane and other contents of the cell.
Steroids contain 17 carbon atoms which are bonded in 4 fused ring-like structures of which 3 are cyclohexane rings and 1 is a cyclopentane ring. Cholesterol and sex hormones are included under steroids.
Waxes are long chain polar lipids comprising ester of alcohol (n-alkanes, ketones, primary and secondary alcohols) and fatty acids. Plant leaves and animal fur are coated with wax to repel water and to prevent water loss. Unlike most waxes, ear wax contains phospholipids and esters of cholesterol.
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In science there are many key concepts and terms that are crucial for students to know and understand. Often it can be hard to determine what the most important science concepts and terms are, and even once you’ve identified them you still need to understand what they mean. To help you learn and understand key science terms and concepts, we’ve identified some of the most important ones and provided detailed definitions for them, written and compiled by Chegg experts.<|endoftext|>
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# Centroid: Definition, Formula, Properties, Practice Problems and FAQs
In a triangle, there are many centres, unlike a circle which has only one centre. Some of the most common centres of a triangle are centroid, incentre,ex-centre, circumcentre and orthocentre. In this article, we are going to learn about the centroid of a triangle in detail. The centroid, also known as the geometrical centre is basically the point of intersection of the medians i.e the line joining the mid-point of sides and the opposite vertices.
• Definition of centroid
• Properties of Centroid
• Centroid Formula
• Practice Problems
• FAQs
## Definition of centroid
The centroid of a triangle is defined as a point of concurrency of the medians i.e, lines joining the mid-point of sides and the opposite vertices of a triangle. It is generally denoted by G. Referring to the figure above, ABC is a triangle AD, BE and CF are the medians of the triangle which intersect at a common point ‘G’ known as the centroid of the triangle ABC . Here points D, E, and F are the mid-points of the side BC, CA and AB respectively.
## Properties of Centroid
Following are the important properties of the centroid of a triangle -
• It is located at the point of intersection of the medians
• Centroid always divides the median internally in the ratio of 2:1 from the vertex.
• Centroid divides the triangle into three parts of equal areas
• The centroid of a triangle always lies inside the triangle
## Centroid Formula
Given the coordinates of a triangle as A(x1,y1), B(x2,y2)and C(x3,y3), then the coordinates of the centroid is given by-
$=\frac{{\sum }_{i=1}^{3}{x}_{i}}{3},\frac{{\sum }_{i=1}^{3}{y}_{i}}{3}$
## Derivation for finding the centroid of a triangle (Proof)
Let ABC be a triangle having coordinates A(x1,y1), B(x2,y2) and C(x3,y3) and G(x,y) be the centroid of the triangle.
Since, AD, BE and CF are the medians of the triangle then D, E and F become the mid-point of the side BC, AC and AB respectively.
As D is the mid-point of BC then using mid-point formula co-ordinates of D is given by -
From the properties of centroid, we know that G divides the median in the ratio of 2:1
Therefore, on applying section formula on the median AD we have,
Therefore, co-ordinates of the centroid (G) is given by
## Practice Problems
1. If are the roots of the equation then, find the centroid of Δ ABC having coordinates as A(x 1, 1/ x 1), B(x2 ,1/ x2 ),C( x3 , 1/ x3 ) as shown in the figure given below
Solution- Given, are the roots of the equation then ,
Sum of roots
Sum of product of roots taken two at a time
Product of roots
We know that, centroid of a triangle G(x,y) is given by
Therefore coordinates of centroid are
1. In a ABC, coordinates of B = (0,0) , AB= 2 units , ∠ABC = 600 and mid-point of BC = (2,0) , then find the centroid of ABC.
Solution - Let D be the mid-point of BC
∴ coordinates of D=(2,0)
Hence, coordinates of C(x3,0) becomes (4,0)[∵BD = DC]
Now, in Δ ABD, we have
AB=BD= 2 units
∴ ∠ADB = ∠BAD = θ (say) [Angles opposite to equal sides are equal]
Also, in Δ ABD
60+ = θ + θ = 1800 [Angle sum property of a triangle]
θ = 600
Therefore, Δ ABD becomes an equilateral triangle
Now, applying distance formula between A and B, we have
Squaring both sides
Again applying distance formula between A and D
Squaring both sides
Solving equation (1) and (2)
___________________
Putting = 1 in equation (1)
Therefore, coordinates of )
Note- ) is neglected because in the problem ABC is given as 600 so , the coordinate of A is going to lie in the first quadrant.
Now, the centroid of Δ ABC i.e, G(x,y) =
=
=
1. Find the centroid of the triangle made by the lines x + y = 2 and x y = 0 .
Solution: Given
x . y = 0
∴ either x = 0 or y = 0
Therefore, the lines are x = 0 , y = 0 and x + y = 2
The centroid of the triangle G(x,y) =
=
=
1. If (1, 1) are the co-ordinates of the centroid of a triangle whose vertices are (0, 4), (-2,0) and (p, q). Determine the value of p,q
Solution:
Given (x1,y1)=(0,4)
(x2,y2) = (-2,0)
(x3,y3) = (p,q)
We know that, Centroid of a triangle is given by G(x,y) =
Hence, and
5 q= -1
## FAQs
Ques 1) If ‘P is any internal point of a triangle such that area of △APB, △BPC and △APC are equal then P must be a ………………..
As ‘P’ is an internal point of the triangle and it is dividing the triangle into three parts of equal area then P must be the centroid of the triangle.
Ques 2) What will be the location of the centroid in a right isosceles triangle? (inside/outside/on the hypotenuse/ at right-angled vertex)<|endoftext|>
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The Mayan Calendar system was a collection of multiple calendars that were used not only by the Mayans but also by various other civilisations of Mesoamerica.
A lot of features of the Mayan calendar were laid down as early as 5th century BC and developments took place over the subsequent centuries.
Broadly speaking, there were two main parts of the calendar. One was the religious calendar while the other one was the solar calendar. Additionally, there was a third calendar known as the Long Count.
Mayans had a very advanced knowledge of astronomy and keenly observed the movements of the celestial bodies Read more about the Calendar Predictions >>
The Mayan apocalypse refers to a number of different theories which ties the end of the Mayan Long Count calendar to an apocalypse Read more about the Mayan Apocalypse >>
Essential features of the Mayan calendar were laid down as early as 5th century BC. Several features of the calendar are similar to the systems that were already in use by earlier Mesoamerican civilisations including the Zapotec and the Olmec. It was also used by subsequent civilisations such as the Aztecs. Thus the history of the calendar system extends over most of the history of the Mesoamerican civilisations.
There were two main purposes of the Mayan calendar system. The first one, the sacred calendar was used to keep track of religious festivals and calculate days for them. This function was performed by the 260-day calendar called Tzolkin. The other function was the calculation of ordinary days and years and this was called the solar calendar, known as Haab.
There were different types of calendars which combined to form the system of Mayan calendar. However, the two most important types were Tzolkin and Haab, the Mayan sacred calendar and the solar calendar. There is another type of Mayan calendar that records the longest span of time. This one is called the Long Count and marks the number of days that have passed since a mythological creation date of August 11 or 13 of 3114BC.
The Mayan solar calendar was called Haab and it was made up of 18 months of 20 days each. In addition to this, there were 5 extra days which were considered ominous by the Mayans. According to the Mayan beliefs, ill-willing gods could unleash disastrous events during these five days and thus they remained inside their homes. A special number was associated with each day in the calendar and a symbol was used to signify that day.
The Mayan sacred calendar was called Tzolkin and it consisted of 260 days in total. The system of Tzolkin consisted of 20 day names combined with 13 day numbers. This calendar was used to keep track of the time of religious festivals and ceremonies. Just like Haab, this calendar also had symbols associated with each day. The Aztecs who came to power after the Mayans had a similar religious calendar which was called Tonalpohualli in the Nahuatl language.
The Long Count was an important calendar type in the Mayan calendar system. The most important aspect of this calendar was that it identified a day by counting the number of days passed since an ancient mythological creation date of August 11 or 13 of 3114 BC. Mesoamerican numerals were used for dates on the Long Count and the calendar was widely used on Mayan monuments.
Date setting on Mayan calendar was a rather complex task and involved Tzolkin and Haab as well as the Long Count. Date is identified by counting the number of days since the creation date. A typical long count date format in the Mayan calendar system would be: Baktun.Katun.Tun.Uinal.Kin. Here, kin refers to one day, uinal refers to 20 days, tun refers to 360 days, katun refers to 7,200 days, and baktun refers to 144,000 days.
Various prophecies of apocalypse are often associated with Mayan calendar due to its division of time into various world eras. According to Mayan literature, the world is destroyed after every era and newly created by the gods for the next era. The first three eras were composed of 5,125 years. The fourth era started on August 11 or 13 of 3,114BC and ended on 21 December 2012 which according to some Mayan prophecies marked the end of the world and the Doomsday.
There were three Mayan calendars known as the Haab, the Tzolkin, and the Long Count. The first one was the usual 365-day calendar which kept track of ordinary days and was based on the rotation of earth around the sun. It consisted of 18 months of 20 days each plus 5 extra days. The second one, Tzolkin, was the scared Mayan calendar composed to keep track of religious festivals and ceremonies. It consisted of 20 periods of 13 days each. The Long Count was an astronomical calendar used for tracking of long time periods. It tracked total days in what the Mayans called “universal cycle”.
The Mayan Calendar round was made from the interweaving of the sacred and the solar calendars, Tzolkin and Haab. Any combination of day from one calendar with day from the other calendar did not repeat itself until 52 periods of 365 days had passed. The figure of 52 was considered important and according to the Mayan beliefs, any person who reached the age of 52 attained special wisdom. Thus the two calendars on the calendar system were related with each other and completed a Mayan “century” of 52 years. The same mechanism was used to form the Aztec calendar system which was heavily influenced by the Mayans.
Mayan Calendar system was a combination of various calendars that were used by different civilisations of Mesoamerican region. Various elements of this calendar system was already being used by previous civilisations and continued to be used by civilisations to come such as the Aztecs. The calendar system had three calendars known as Haab, Tzolkin, and the Long Count. Haab was the solar calendar for the tracking of ordinary days while Tzolkin was the sacred calendar for tracking of religious festivals and ceremonies. Both these calendars were similar to the ones used by the Aztecs, with the difference of symbols used for the days. The third calendar measured the number of days in each “universal cycle”. The current universal cycle, according to this calendar, began in August 11 or 13 of 3,114BC and ended on 21 December 2012.
Some Great Resources to help you learn more about the Mayan Calendar:<|endoftext|>
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Autism is a complex developmental disability that affects a person’s ability to communicate and interact with others, with a wide range of behavioral, social, and language problems. Autism usually appears during the first three years of life.
Autism is characterized by a collection of neurobehavioral, neurological, gastrointestinal and immunological dysfunctions that include a loss of eye contact, deficiencies in socialization and communication, abnormal theory of mind function, language dysfunction, restrictive, repetitive, and stereotypical behaviors, food allergies, constipation, yeast infections and other behavioral and medical conditions.
Autism is called a “spectrum disorder” since it affects individuals differently and to varying degrees.
It is estimated that one in every 150 American children has some degree of autism with males being affected three to four times more frequently than females. Autism Spectrum Disorders (ASD) are complicated conditions that may require an integrative treatment protocol involving many factors including behavioral and social therapy, pharmacotherapy, environmental control, dietary supplements, nutritional, alternative and biomedical therapies. Many are sick with gastrointestinal, immunologic, and metabolic problems that significantly affect their behavior and their physical and emotional health. Treating the medical problems often leads to improvement in clinical signs and symptoms and, in some cases, to recovery.
Causes of Autism
There has been an exponential increase in the number of new diagnoses of autism. And there is continuing debate and controversy as to what the diagnosis ‘autism’ actually constitutes. There is even more debate and controversy as to its causes, and potential for treatment /amelioration.
The fundamental cause of Autism, based in our experience, is severe intestinal flora imbalances resulting into immune imbalances mainly due to gastrointestinal infections, antibiotics and vaccinations that in turn affect the brain. We have daily phone and e-mail contacts with parents of ASD children where parents detail causes and symptoms of Autism and treatments they undertake for their children. Our evaluation of causes of Autism is as follows.
Autism usually appears during the first three years of life. The reason is, and it is well documented, that between the ages of 0-3, the intestinal microflora of a child is not well established yet. The same can be said about the brain and nervous system. They are in a stage of formation and are fragile. During this time, if a lot of antibiotics and/or vaccinations are given, and if the immune system of a child is somewhat low, the intestinal microflora being fragile is affected negatively. As a result a toxic condition is produced in the gut that will affect the brain and the nervous system to various extents. The more toxicity in the gut the more the effect on the brain. As a result, the majority of children with Autism reveal abnormal gastrointestinal symptoms including food allergies, yeast infections, constipation.
Factors affecting Gut/Brain/Immune dysfunction and Autism are indicated in Fig. 1. These factors result into at least two types of ASD with regard to disease development: abnormal cognitive development evident from birth (classical autism); and in the majority cases developmental regression, usually between 18-36 months of age, following apparent normal development (regressive autism).
Gut/Brain/Immune Dysfunction and Autism
It is very important to understand the direct relationship between the gut and brain, especially during the first three years of life, when both are in a stage of formation. The above factors make the immune system of each child different and unique. Therefore physicians, especially pediatricians, should not treat all children alike, but on a case by case basis after careful evaluation with the parents. A detailed history should be taken that can determine, at least qualitatively, the immune system of each child. Hence the standard vaccination schedule can not be applicable, if the above factors are not evaluated and taken into consideration properly.
The gastrointestinal tract is a complex ecosystem in which there is a delicate balance between the intestinal microflora and the host. A healthy gastrointestinal tract should contain a high percentage of Lactobacilli and Bifidobacteria beneficial bacteria to prevent the over colonization of disease causing pathogenic micro-organisms such as E. Coli, Clostridia, Salmonella and Candida.
Bidirectional interactions between the brain and intestinal microflora might have an important role in modulating gut and brain function and may be involved in the modulation of emotions, pain perception, mucosal immune activity and general well-being. The reduction of Lactobacilli and Bifidobacteria and overgrowth of pathogenic bacteria will be stressful to this brain/gut interactions especially for infants aged 0-3 producing neurological and immune imbalances and affecting their development. The neurological and immune systems are inextricably intertwined beginning in the embryonic stage of life. Disruption of the bidirectional interactions between the intestinal microflora and the nervous system may be involved in the pathophysiology of acute and chronic gastrointestinal and neurological disease states.
Abbreviations: ANS, autonomic nervous system; CNS, central nervous system; EMS,
emotional motor system; GI, gastrointestinal; HPA, hypothalamus–pituitary–adrenal. Fig. 2 Source: Rhee, S. H. et al. Nat. Rev. Gastroenterol. Hepatol. 6, 306–314 (2009)
D-Lactate Free Probiotic powder formulation have been used specifically for children with Autism Spectrum Disorder (ASD), to improving their intestinal microflora and digestive processes. D-lactate is a result of fermentation of probiotic bacteria in the digestive system.
Probiotics and Autism
Probiotics are defined as “live microorganisms which, when administered in adequate amounts, confer a health benefit to the host”. They possess the ability to transiently colonize the GI tract, increase the concentration of beneficial microbes, and thereby create a balance in the gut microbiota to the ultimate benefit of the host, in a natural and safe way.
Potential or known mechanisms whereby probiotic bacteria might impact on the microbiota include:
competition for dietary ingredients as growth substrates
bioconversion of, for example, sugars into fermentation products with inhibitory properties
production of growth substrates
direct effect on pathogens
competitive exclusion for binding sites
reduction of inflammation, thus altering intestinal properties for colonization and persistence within, and (8) stimulation of innate immune response.
Probiotics can therefore have been used to prevent or reduce the risk of ASD for infants aged 0-3. The aim is to protect their digestive and immune systems by using probiotics at appropriate dosages prior, during and after any intervention, such as antibiotics or vaccinations, that affect negatively the intestinal microflora and immune system of the mother and infant, as indicated in Fig. 1. Choosing the correct probiotic formulation and dosages are considerations that also must be understood and followed by pediatricians and parents alike.<|endoftext|>
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Surface runoff (also known as overland flow) is the flow of water that occurs when excess storm water, melt water, or other sources flows over the Earth's surface. This might occur because soil is saturated to full capacity, because rain arrives more quickly than soil can absorb it, or because impervious areas (roofs and pavement) send their runoff to surrounding soil that cannot absorb all of it. Surface runoff is a major component of the water cycle. It is the primary agent in soil erosion by water.
Runoff that occurs on the ground surface before reaching a channel is also called a non point source. If a nonpoint source contains man-made contaminants, or natural forms of pollution (such as rotting leaves) the runoff is called nonpoint source pollution. A land area which produces runoff that drains to a common point is called a drainage basin. When runoff flows along the ground, it can pick up soil contaminants including petroleum, pesticides, or fertilizers that become discharge or nonpoint source pollution.
In addition to causing water erosion and pollution, surface runoff in urban areas is a primary cause of urban flooding which can result in property damage, damp and mold in basements, and street flooding.
Surface runoff can be generated either by rainfall, snowfall or by the melting of snow, or glaciers.
Snow and glacier melt occur only in areas cold enough for these to form permanently. Typically snowmelt will peak in the spring and glacier melt in the summer, leading to pronounced flow maxima in rivers affected by them. The determining factor of the rate of melting of snow or glaciers is both air temperature and the duration of sunlight. In high mountain regions, streams frequently rise on sunny days and fall on cloudy ones for this reason.
In areas where there is no snow, runoff will come from rainfall. However, not all rainfall will produce runoff because storage from soils can absorb light showers. On the extremely ancient soils of Australia and Southern Africa, proteoid roots with their extremely dense networks of root hairs can absorb so much rainwater as to prevent runoff even when substantial amounts of rain fall. In these regions, even on less infertile cracking clay soils, high amounts of rainfall and potential evaporation are needed to generate any surface runoff, leading to specialised adaptations to extremely variable (usually ephemeral) streams.
This occurs when the rate of rainfall on a surface exceeds the rate at which water can infiltrate the ground, and any depression storage has already been filled. This is called flooding excess overland flow, Hortonian overland flow (after Robert E. Horton), or unsaturated overland flow. This more commonly occurs in arid and semi-arid regions, where rainfall intensities are high and the soil infiltration capacity is reduced because of surface sealing, or in paved areas. This occurs largely in city areas where pavements prevent water from flooding.
When the soil is saturated and the depression storage filled, and rain continues to fall, the rainfall will immediately produce surface runoff. The level of antecedent soil moisture is one factor affecting the time until soil becomes saturated. This runoff is called saturation excess overland flow, saturated overland flow or Dunne runoff.
Soil retains a degree of moisture after a rainfall. This residual water moisture affects the soil's infiltration capacity. During the next rainfall event, the infiltration capacity will cause the soil to be saturated at a different rate. The higher the level of antecedent soil moisture, the more quickly the soil becomes saturated. Once the soil is saturated, runoff occurs.
After water infiltrates the soil on an up-slope portion of a hill, the water may flow laterally through the soil, and exfiltrate (flow out of the soil) closer to a channel. This is called subsurface return flow or throughflow.
As it flows, the amount of runoff may be reduced in a number of possible ways: a small portion of it may evapotranspire; water may become temporarily stored in microtopographic depressions; and a portion of it may infiltrate as it flows overland. Any remaining surface water eventually flows into a receiving water body such as a river, lake, estuary or ocean.
Urbanization increases surface runoff by creating more impervious surfaces such as pavement and buildings that do not allow percolation of the water down through the soil to the aquifer. It is instead forced directly into streams or storm water runoff drains, where erosion and siltation can be major problems, even when flooding is not. Increased runoff reduces groundwater recharge, thus lowering the water table and making droughts worse, especially for agricultural farmers and others who depend on the water wells.
When anthropogenic contaminants are dissolved or suspended in runoff, the human impact is expanded to create water pollution. This pollutant load can reach various receiving waters such as streams, rivers, lakes, estuaries and oceans with resultant water chemistry changes to these water systems and their related ecosystems.
As humans continue to alter the climate through the addition of greenhouse gases to the atmosphere, precipitation patterns are expected to change as the atmospheric capacity for water vapor increases. This will have direct consequences on runoff amounts.
Surface runoff can cause erosion of the Earth's surface; eroded material may be deposited a considerable distance away. There are four main types of soil erosion by water: splash erosion, sheet erosion, rill erosion and gully erosion. Splash erosion is the result of mechanical collision of raindrops with the soil surface: soil particles which are dislodged by the impact then move with the surface runoff. Sheet erosion is the overland transport of sediment by runoff without a well defined channel. Soil surface roughness causes may cause runoff to become concentrated into narrower flow paths: as these incise, the small but well-defined channels which are formed are known as rills. These channels can be as small as one centimeter wide or as large as several meters. If runoff continue to incise and enlarge rills, they may eventually grow to become gullies. Gully erosion can transport large amounts of eroded material in a small time period.
Reduced crop productivity usually results from erosion, and these effects are studied in the field of soil conservation. The soil particles carried in runoff vary in size from about .001 millimeter to 1.0 millimeter in diameter. Larger particles settle over short transport distances, whereas small particles can be carried over long distances suspended in the water column. Erosion of silty soils that contain smaller particles generates turbidity and diminishes light transmission, which disrupts aquatic ecosystems.
Entire sections of countries have been rendered unproductive by erosion. On the high central plateau of Madagascar, approximately ten percent of that country's land area, virtually the entire landscape is devoid of vegetation, with erosive gully furrows typically in excess of 50 meters deep and one kilometer wide. Shifting cultivation is a farming system which sometimes incorporates the slash and burn method in some regions of the world. Erosion causes loss of the fertile top soil and reduces its fertility and quality of the agricultural produce.
The principal environmental issues associated with runoff are the impacts to surface water, groundwater and soil through transport of water pollutants to these systems. Ultimately these consequences translate into human health risk, ecosystem disturbance and aesthetic impact to water resources. Some of the contaminants that create the greatest impact to surface waters arising from runoff are petroleum substances, herbicides and fertilizers. Quantitative uptake by surface runoff of pesticides and other contaminants has been studied since the 1960s, and early on contact of pesticides with water was known to enhance phytotoxicity. In the case of surface waters, the impacts translate to water pollution, since the streams and rivers have received runoff carrying various chemicals or sediments. When surface waters are used as potable water supplies, they can be compromised regarding health risks and drinking water aesthetics (that is, odor, color and turbidity effects). Contaminated surface waters risk altering the metabolic processes of the aquatic species that they host; these alterations can lead to death, such as fish kills, or alter the balance of populations present. Other specific impacts are on animal mating, spawning, egg and larvae viability, juvenile survival and plant productivity. Some researches show surface runoff of pesticides, such as DDT, can alter the gender of fish species genetically, which transforms male into female fish.
Surface runoff occurring within forests can supply lakes with high loads of mineral nitrogen and phosphorus leading to eutrophication. Runoff waters within coniferous forests are also enriched with humic acids and can lead to humification of water bodies Additionally, high standing and young islands in the tropics and subtropics can undergo high soil erosion rates and also contribute large material fluxes to the coastal ocean. Such land derived runoff of sediment nutrients, carbon, and contaminants can have large impacts on global biogeochemical cycles and marine and coastal ecosystems.
In the case of groundwater, the main issue is contamination of drinking water, if the aquifer is abstracted for human use. Regarding soil contamination, runoff waters can have two important pathways of concern. Firstly, runoff water can extract soil contaminants and carry them in the form of water pollution to even more sensitive aquatic habitats. Secondly, runoff can deposit contaminants on pristine soils, creating health or ecological consequences.
The other context of agricultural issues involves the transport of agricultural chemicals (nitrates, phosphates, pesticides, herbicides etc.) via surface runoff. This result occurs when chemical use is excessive or poorly timed with respect to high precipitation. The resulting contaminated runoff represents not only a waste of agricultural chemicals, but also an environmental threat to downstream ecosystems.
Flooding occurs when a watercourse is unable to convey the quantity of runoff flowing downstream. The frequency with which this occurs is described by a return period. Flooding is a natural process, which maintains ecosystem composition and processes, but it can also be altered by land use changes such as river engineering. Floods can be both beneficial to societies or cause damage. Agriculture along the Nile floodplain took advantage of the seasonal flooding that deposited nutrients beneficial for crops. However, as the number and susceptibility of settlements increase, flooding increasingly becomes a natural hazard. In urban areas, surface runoff is the primary cause of urban flooding, known for its repetitive and costly impact on communities. Adverse impacts span loss of life, property damage, contamination of water supplies, loss of crops, and social dislocation and temporary homelessness. Floods are among the most devastating of natural disasters.
Mitigation of adverse impacts of runoff can take several forms:
Land use controls. Many world regulatory agencies have encouraged research on methods of minimizing total surface runoff by avoiding unnecessary hardscape. Many municipalities have produced guidelines and codes (zoning and related ordinances) for land developers that encourage minimum width sidewalks, use of pavers set in earth for driveways and walkways and other design techniques to allow maximum water infiltration in urban settings. An example land use control program can be seen in the city of Santa Monica, California.
Erosion controls have appeared since medieval times when farmers realized the importance of contour farming to protect soil resources. Beginning in the 1950s these agricultural methods became increasingly more sophisticated. In the 1960s some state and local governments began to focus their efforts on mitigation of construction runoff by requiring builders to implement erosion and sediment controls (ESCs). This included such techniques as: use of straw bales and barriers to slow runoff on slopes, installation of silt fences, programming construction for months that have less rainfall and minimizing extent and duration of exposed graded areas. Montgomery County, Maryland implemented the first local government sediment control program in 1965, and this was followed by a statewide program in Maryland in 1970.
Flood control programs as early as the first half of the twentieth century became quantitative in predicting peak flows of riverine systems. Progressively strategies have been developed to minimize peak flows and also to reduce channel velocities. Some of the techniques commonly applied are: provision of holding ponds (also called detention basins) to buffer riverine peak flows, use of energy dissipators in channels to reduce stream velocity and land use controls to minimize runoff.
Chemical use and handling. Following enactment of the U.S. Resource Conservation and Recovery Act (RCRA) in 1976, and later the Water Quality Act of 1987, states and cities have become more vigilant in controlling the containment and storage of toxic chemicals, thus preventing releases and leakage. Methods commonly applied are: requirements for double containment of underground storage tanks, registration of hazardous materials usage, reduction in numbers of allowed pesticides and more stringent regulation of fertilizers and herbicides in landscape maintenance. In many industrial cases, pretreatment of wastes is required, to minimize escape of pollutants into sanitary or stormwater sewers.
The U.S. Clean Water Act (CWA) requires that local governments in urbanized areas (as defined by the Census Bureau) obtain stormwater discharge permits for their drainage systems. Essentially this means that the locality must operate a stormwater management program for all surface runoff that enters the municipal separate storm sewer system ("MS4"). EPA and state regulations and related publications outline six basic components that each local program must contain:
Other property owners which operate storm drain systems similar to municipalities, such as state highway systems, universities, military bases and prisons, are also subject to the MS4 permit requirements.
Runoff is analyzed by using mathematical models in combination with various water quality sampling methods. Measurements can be made using continuous automated water quality analysis instruments targeted on pollutants such as specific organic or inorganic chemicals, pH, turbidity etc. or targeted on secondary indicators such as dissolved oxygen. Measurements can also be made in batch form by extracting a single water sample and conducting any number of chemical or physical tests on that sample.
In the 1950s or earlier hydrology transport models appeared to calculate quantities of runoff, primarily for flood forecasting. Beginning in the early 1970s computer models were developed to analyze the transport of runoff carrying water pollutants, which considered dissolution rates of various chemicals, infiltration into soils and ultimate pollutant load delivered to receiving waters. One of the earliest models addressing chemical dissolution in runoff and resulting transport was developed in the early 1970s under contract to the United States Environmental Protection Agency (EPA). This computer model formed the basis of much of the mitigation study that led to strategies for land use and chemical handling controls.
Other computer models have been developed (such as the DSSAM Model) that allow surface runoff to be tracked through a river course as reactive water pollutants. In this case the surface runoff may be considered to be a line source of water pollution to the receiving waters.
Agricultural wastewater treatment is a farm management agenda for controlling pollution from surface runoff that may be contaminated by chemicals in fertiliser, pesticides, animal slurry, crop residues or irrigation water.Colluvium
Colluvium (also colluvial material or colluvial soil) is a general name for loose, unconsolidated sediments that have been deposited at the base of hillslopes by either rainwash, sheetwash, slow continuous downslope creep, or a variable combination of these processes. Colluvium is typically composed of a heterogeneous range of rock types and sediments ranging from silt to rock fragments of various sizes. This term is also used to specifically refer to sediment deposited at the base of a hillslope by unconcentrated surface runoff or sheet erosion.Combined sewer
A combined sewer is a sewage collection system of pipes and tunnels designed to simultaneously collect surface runoff and sewage water in a shared system. This type of gravity sewer design is no longer used in almost every instance worldwide when constructing new sewer systems. Modern-day sewer designs exclude surface runoff from sanitary sewers, but many older cities and towns continue to operate previously constructed combined sewer systems.Combined sewers can cause serious water pollution problems during combined sewer overflow (CSO) events when combined sewage and surface runoff flows exceed the capacity of the sewage treatment plant, or of the maximum flow rate of the system which transmits the combined sources. In instances where exceptionally high surface runoff occurs (such as large rainstorms), the load on individual tributary branches of the sewer system may cause a back-up to a point where raw sewage flows out of input sources such a toilets, causing inhabited buildings to be flooded with a toxic sewage-runoff mixture, incurring massive financial burdens for cleanup and repair. When combined sewer systems experience these higher than normal throughputs, relief systems cause discharges containing human and industrial waste to flow into rivers, streams, or other bodies of water. Such events frequently cause both negative environmental and lifestyle consequences, including beach closures, contaminated shellfish unsafe for consumption, and contamination of drinking water sources, rendering them temporarily unsafe for drinking and requiring boiling before uses such as bathing or washing dishes.Environmental impact of concrete
The environmental impact of concrete, its manufacture and applications, are complex. Some effects are harmful; others welcome. Many depend on circumstances. A major component of concrete is cement, which has its own environmental and social impacts and contributes largely to those of concrete.
The cement industry is one of the primary producers of carbon dioxide, a potent greenhouse gas. Concrete causes damage to the most fertile layer of the earth, the topsoil.
Concrete is used to create hard surfaces which contribute to surface runoff that may cause soil erosion, water pollution and flooding. Conversely, concrete is one of the most powerful tools for proper flood control, by means of damming, diversion, and deflection of flood waters, mud flows, and the like. Light-colored concrete can reduce the urban heat island effect, due to its higher albedo. Concrete dust released by building demolition and natural disasters can be a major source of dangerous air pollution. The presence of some substances in concrete, including useful and unwanted additives, can cause health concerns due to toxicity and (usually naturally occurring) radioactivity. Wet concrete is highly alkaline and should always be handled with proper protective equipment. Concrete recycling is increasing in response to improved environmental awareness, legislation, and economic considerations. Conversely, the use of concrete mitigates the use of alternative building materials such as wood, which is a carbon sink. Concrete structures also last much longer than wood structures.Erosion control
Erosion control is the practice of preventing or controlling wind or water erosion in agriculture, land development, coastal areas, river banks and construction. Effective erosion controls handle surface runoff and are important techniques in preventing water pollution, soil loss, wildlife habitat loss and human property loss.Freshwater marsh
A freshwater marsh is a marsh that contains fresh water. Freshwater marshes are usually found near the mouths of rivers and are present in areas with low drainage. It is the counterpart to the salt marsh, an upper coastal intertidal zone of bio-habitat which is regularly flushed with sea water.
Freshwater marshes are non-tidal biomes containing little or no peat (unlike bogs and fens, both a kind of mire and mires consisting heavily of moist, biologically active peat). They are most common in the Gulf Coast region, specifically in Florida. They can be one of two principal types: either fresh water mineralized marshes, which derive their water from groundwater, streams and surface runoff; or poorly mineralized fresh water marshes whose moisture comes mostly from regular direct precipitation. Freshwater marshes support an independent pH-neutral ecosystem which encourages biodiversity. Common species include ducks, geese, swans, songbirds, swallows, coots, and black ducks. Although shallow marshes do not tend to support many fish, deeper ones are home to many species, including such large fish as the northern pike and carp. Some of the most common plants in these areas are cattails, water lilies, arrowheads, and rushes.The Florida Everglades represent the largest contiguous freshwater marsh in the entire world. This immense marsh covers 4,200 square miles (11,000 km2) and is located in the southern tip of Florida. Continued human development, including drainage for development and polluted agriculture runoff, as well as alterations in the water cycle threaten the existence of the Everglades. The remaining parts of the Everglades are grasses, sedges and other emergent hydrophytes.Great Lakes Basin
The Great Lakes Basin consists of the Great Lakes and the surrounding lands of the states of Illinois, Indiana, Michigan, Minnesota, New York, Ohio, Pennsylvania, and Wisconsin in the United States, and the province of Ontario in Canada, whose direct surface runoff and watersheds form a large drainage basin that feeds into the lakes. It is generally considered to also include a small area around and beyond Wolfe Island, Ontario, at the east end of Lake Ontario, which does not directly drain into the Great lakes, but into the Saint Lawrence River.
The Basin is at the center of the Great Lakes region.Hydrological transport model
An hydrological transport model is a mathematical model used to simulate river or stream flow and calculate water quality parameters. These models generally came into use in the 1960s and 1970s when demand for numerical forecasting of water quality was driven by environmental legislation, and at a similar time widespread access to significant computer power became available. Much of the original model development took place in the United States and United Kingdom, but today these models are refined and used worldwide.
There are dozens of different transport models that can be generally grouped by pollutants addressed, complexity of pollutant sources, whether the model is steady state or dynamic, and time period modeled. Another important designation is whether the model is distributed (i.e. capable of predicting multiple points within a river) or lumped. In a basic model, for example, only one pollutant might be addressed from a simple point discharge into the receiving waters. In the most complex of models, various line source inputs from surface runoff might be added to multiple point sources, treating a variety of chemicals plus sediment in a dynamic environment including vertical river stratification and interactions of pollutants with in-stream biota. In addition watershed groundwater may also be included. The model is termed "physically based" if its parameters can be measured in the field.
Often models have separate modules to address individual steps in the simulation process. The most common module is a subroutine for calculation of surface runoff, allowing variation in land use type, topography, soil type, vegetative cover, precipitation and land management practice (such as the application rate of a fertilizer). The concept of hydrological modeling can be extended to other environments such as the oceans, but most commonly (and in this article) the subject of a river watershed is generally implied.Hydrophobic soil
Hydrophobic soil – soil that is hydrophobic – causes water to collect on the soil surface rather than infiltrate into the ground. Wild fires generally cause soils to be hydrophobic temporarily, which increases water repellency, surface runoff and erosion in post-burn sites. Soil dispersion due to sodification causes similar problems.
Hydrophobic soils are created when hydrocarbon residue is created after organic material is burnt and soaks into empty pore spaces in the soils, making it impervious to water.
Dryness, plant chemicals, aromatic oils, and other chemicals also cause hydrophobicity.Libya Montes
The Libya Montes are a highland terrain on Mars up-lifted by the giant impact that created the Isidis basin to the north.
During 1999, this region became one of the top two that were being considered for the canceled Mars Surveyor 2001 Lander. The Isidis basin is very ancient. Thus, the Libya Montes that form the southern Isidis basin rim contain some of the oldest rocks available at the Martian surface, and a landing in this region might potentially provide information about conditions on early Mars.
After they formed by the Isidis impact, the Libya Montes were subsequently modified by a large variety of processes, including fluvial activity, wind erosion and impact cratering. In particular, precipitation induced surface runoff and groundwater seepage resulted in the formation of fluvial landforms, i.e., dense valley networks, broad and elongated valleys, delta deposits, alluvial fans, open-basin paleolakes and coastlines. Crater size - frequency distribution measurements ("crater counting") revealed that the majority of valleys were formed early in Martian history (more than 3.7 billion years ago, Late Noachian). However, recent studies show that the formation of valleys continued throughout the Middle Ages of Mars (Hesperian period) and stopped 3.1 billion years ago in the Late Hesperian.Park River (Connecticut)
The Park River, sometimes called the Hog River, flows through and under the city of Hartford, Connecticut. Between 1940 and the 1980s, the 2.3-mile (3.7 km) river was buried by the Army Corps of Engineers to prevent the spring floods regularly caused by increased surface runoff from urban development.Runoff
Runoff, run-off or RUNOFF may refer to:
Runoff (program), the Multics operating system type-setting program
Runoff or run-off, another name for bleed (printing), printing that lies beyond the edges to which a printed sheet is trimmed
Runoff or run-off, a stock market term
RUNOFF, the first computer text-formatting programRunoff curve number, an empirical parameter used in hydrology
Runoff model (reservoir), a mathematical model describing the relationship between rainfall and surface runoff (see below) in a rainfall catchment area or watershed
Runoff voting system, also known as the two-round system, a voting system where a second round of voting is used to elect one of the two candidates receiving the most votes in the first round
Instant-runoff voting, an extension or variation of runoff voting where a second round can be rendered unnecessary by voters ranking candidates in order of preference
Run-off area, a racetrack safety featureSurface runoff, the flow of water over land as a consequence of rain, melting snow, etc.Runoff model (reservoir)
A runoff model is a mathematical model describing the rainfall–runoff relations of a rainfall catchment area, drainage basin or watershed. More precisely, it produces a surface runoff hydrograph in response to a rainfall event, represented by and input as a hyetograph. In other words, the model calculates the conversion of rainfall into runoff.
A well known runoff model is the linear reservoir, but in practice it has limited applicability.
The runoff model with a non-linear reservoir is more universally applicable, but still it holds only for catchments whose surface area is limited by the condition that the rainfall can be considered more or less uniformly distributed over the area. The maximum size of the watershed then depends on the rainfall characteristics of the region. When the study area is too large, it can be divided into sub-catchments and the various runoff hydrographs may be combined using flood routing techniques.
Rainfall-runoff models need to be calibrated before they can be used.Sanitary sewer
A sanitary sewer or foul sewer is an underground pipe or tunnel system for transporting sewage from houses and commercial buildings (but not stormwater) to treatment facilities or disposal. Sanitary sewers are part of an overall system called a sewage system or sewerage.
Sewage may be treated to control water pollution before discharge to surface waters. Sanitary sewers serving industrial areas also carry industrial wastewater.
Separate sanitary sewer systems are designed to transport sewage alone. In municipalities served by sanitary sewers, separate storm drains may convey surface runoff directly to surface waters. Sanitary sewers are distinguished from combined sewers, which combine sewage with stormwater runoff in one pipe. Sanitary sewer systems are beneficial because they avoid combined sewer overflows.Sediment
Sediment is a naturally occurring material that is broken down by processes of weathering and erosion, and is subsequently transported by the action of wind, water, or ice or by the force of gravity acting on the particles. For example, sand and silt can be carried in suspension in river water and on reaching the sea bed deposited by sedimentation. If buried, they may eventually become sandstone and siltstone (sedimentary rocks) through lithification.
Sediments are most often transported by water (fluvial processes), but also wind (aeolian processes) and glaciers. Beach sands and river channel deposits are examples of fluvial transport and deposition, though sediment also often settles out of slow-moving or standing water in lakes and oceans. Desert sand dunes and loess are examples of aeolian transport and deposition. Glacial moraine deposits and till are ice-transported sediments.Sewage
Sewage (or domestic wastewater or municipal wastewater) is a type of wastewater that is produced by a community of people. It is characterized by volume or rate of flow, physical condition, chemical and toxic constituents, and its bacteriologic status (which organisms it contains and in what quantities). It consists mostly of greywater (from sinks, tubs, showers, dishwashers, and clothes washers), blackwater (the water used to flush toilets, combined with the human waste that it flushes away); soaps and detergents; and toilet paper (less so in regions where bidets are widely used instead of paper).
Sewage usually travels from a building's plumbing either into a sewer, which will carry it elsewhere, or into an onsite sewage facility (of which there are many kinds). Whether it is combined with surface runoff in the sewer depends on the sewer design (sanitary sewer or combined sewer). The reality is, however, that most wastewater produced globally remains untreated causing widespread water pollution, especially in low-income countries: A global estimate by UNDP and UN-Habitat is that 90% of all wastewater generated is released into the environment untreated. In many developing countries the bulk of domestic and industrial wastewater is discharged without any treatment or after primary treatment only.
The term sewage is nowadays regarded as an older term and is being more and more replaced by "wastewater". In general American English usage, the terms "sewage" and "sewerage" mean the same thing. In common British usage, and in American technical and professional English usage, "sewerage" refers to the infrastructure that conveys sewage.Sewerage
Sewerage is the infrastructure that conveys sewage or surface runoff (stormwater, meltwater, rainwater) using sewers. It encompasses components such as receiving drains, manholes, pumping stations, storm overflows, and screening chambers of the combined sewer or sanitary sewer. Sewerage ends at the entry to a sewage treatment plant or at the point of discharge into the environment. It is the system of pipes, chambers, manholes, etc. that conveys the sewage or storm water.
It is also an alternate noun for the word sewage.
In American colloquial English, "sewer system" is applied more frequently to the large infrastructure of sewers that British speakers more often refer to as "sewerage".Sump
A sump (American English and some parts of Canada: oil pan) is a low space that collects often undesirable liquids such as water or chemicals. A sump can also be an infiltration basin used to manage surface runoff water and recharge underground aquifers. Sump can also refer to an area in a cave where an underground flow of water exits the cave into the earth.Urban runoff
Urban runoff is surface runoff of rainwater created by urbanization. This runoff is a major source of flooding and water pollution in urban communities worldwide.
Impervious surfaces (roads, parking lots and sidewalks) are constructed during land development. During rain storms and other precipitation events, these surfaces (built from materials such as asphalt and concrete), along with rooftops, carry polluted stormwater to storm drains, instead of allowing the water to percolate through soil. This causes lowering of the water table (because groundwater recharge is lessened) and flooding since the amount of water that remains on the surface is greater. Most municipal storm sewer systems discharge stormwater, untreated, to streams, rivers and bays. This excess water can also make its way into people's properties through basement backups and seepage through building wall and floors.
|Other types of pollution|<|endoftext|>
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Generally if one of the genes' biochemical functions becomes knocked out completely, the other copy will fill in for it, making the trait recessive - requiring both copies being knocked out.
An example of such a recessive trait is Albinism - if both copies of the enzyme participating in melanin biosynthesis are ineffective, the result is someone with no pigment.
Dominant genes are often variant genes which convey a new ability (phenotype) and as such the trait can show up with just one copy has this variant. Phenylthiocarbamide tasting is an example of this dominance. If both copies of the gene were the variant, the original ability might disappear - making the original trait dominant as well. On the molecular level, genes most often encode proteins which perform some function for the cell: For example, they could be enzymes and catalyze chemical reactions. They could also have some structural function, such as make up the "muscle" part of your muscle cells... You get the idea.
In the most simple case, the dominant allele encodes a protein that can perform its function. For example, the dominant allele for the CFTR gene encodes a channel that can let chloride into and out of the cells. The recessive allele, on the other hand encodes a protein that cannot do its job correctly (this also called a loss-of-function mutation). So if you inherit a functional copy from one parent and a non-functional copy from the other parent, you will still have one copy of the protein that can do its job. Only if you get a nonfunctional copy from both parents will you have a recessive condition called cystic fybrosis.<|endoftext|>
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# Solving Timoshenko beam equation for cantilever beam
Update with solution
Solution 1
Problem: Beam clamped at left side, free end on right side, point load pointing downwards. x is defined positive from the clamped end towards the free end.
Create FBD from a section x from the free end gives:
$$M(x) = -P(L-x)$$
$$Q(x) = -P$$
Differential equation for equilibrium is:
$$-EI\frac{d\alpha}{dx} = M(x) = -P(L-x)$$
Integrating this function results in:
$$EI\alpha(x) = \frac{-Px(2L-x)}{2}+C_{1}$$
Using the boundary condition that the beam is clamped at $$x=0$$:
$$\alpha(0) = 0 \rightarrow 0 + C_{1} = 0 \rightarrow C_{1} = 0$$
Now using the second differential equation for equilibrium:
$$\frac{d}{dx}\left(\frac{dw}{dx}-\alpha(x)\right)+\frac{p(x)}{\kappa G A} = 0$$
Integration of this equation leads to:
$$\frac{dw}{dx} = \frac{Q(x)}{\kappa A G}+\alpha(x)$$
Substituting in the previous found equation for $$\alpha(x)$$ gives:
$$\frac{dw}{dx} = -\frac{P}{\kappa A G} - \frac{Px(2L-x)}{2EI}$$
Integrating this once more to find the equation for $$w(x)$$ leads to:
$$w(x) = -\frac{Px}{\kappa A G} - \frac{Px^2(3L-x)}{6EI} + C_{2}$$
Using the boundary condition $$w(0) = 0$$ gives:
$$w(0) = 0 = -0 - 0 + C_{2} \rightarrow C_{2} = 0$$
Which gives:
$$w(x) = -\frac{Px}{\kappa A G} - \frac{P}{EI}\left(-\frac{x^3}{6}+\frac{x^2L}{2}\right)$$
Solution 2
Problem: Beam clamped at right side, free end on left side, point load pointing downwards.\
Create FBD from a section x from the free end gives:
$$M(x) = -Px$$
$$Q(x) = -P$$
Differential equation for equilibrium is:
$$-EI\frac{d\alpha}{dx} = M(x) = -Px$$
Integrating this function results in:
$$EI\alpha(x) = \frac{Px^2}{2}+C_{1}$$
Using the boundary condition that the beam is clamped at $$x=L$$:
$$\alpha(L) = 0 \rightarrow \frac{PL^2}{2} + C_{1} = 0 \rightarrow C_{1} = -\frac{PL^2}{2}$$
Which gives:
$$\alpha(x) = \frac{P\left(x^2-L^2\right)}{2EI}$$
Now using the second differential equation for equilibrium:
$$\frac{d}{dx}\left(\frac{dw}{dx}-\alpha(x)\right)+\frac{p(x)}{\kappa G A} = 0$$
Integration of this equation leads to:
$$\frac{dw}{dx} = \frac{Q(x)}{\kappa A G}+\alpha(x)$$
Substituting in the previous found equation for $$\alpha(x)$$ gives:
$$\frac{dw}{dx} = -\frac{P}{\kappa A G} - \frac{P\left(x^2-L^2\right)}{2EI}$$
Integrating this once more to find the equation for $$w(x)$$ leads to:
$$w(x) = -\frac{Px}{\kappa A G} + \frac{P}{2EI}\left(\frac{x^3}{3}-L^2x\right) + C_{2}$$
Using the boundary condition $$w(L) = 0$$ gives:
$$w(L) = 0 = -\frac{PL}{\kappa A G} + \frac{P}{2EI}\left(\frac{L^3}{3}-L^3\right) + C_{2} \rightarrow C_{2} = \frac{PL}{\kappa A G} + \frac{PL^3}{3EI}$$
Which gives:
$$w(x) = \frac{P}{\kappa A G}\left(L-x\right) + \frac{P}{EI}\left(\frac{x^3}{6}-\frac{L^2x}{2}+\frac{L^3}{3}\right)$$
Note that both solution now produce the same result, but in reverse.
Old message
I am trying to solve the simple problem of a left side supported beam, with a point load at the free end. Lets say the free end is located at L and the force is pointing downwards (negative z-direction). x is defined as positive towards the right (and thus positive towards the free end). I would arrive at equations:
$$Q = -P = \kappa AG(-\phi + \frac{\partial w}{\partial x})$$
$$M = -P(L-x) = EI \frac{\partial \phi}{\partial x}$$
After which I integrate the second equation, which results in:
$$\frac{-P(L-x)}{EI} = \frac{\partial \phi}{\partial x} \rightarrow \phi(x) = \frac{Px(x-2L)}{2EI} + C_{1}$$
Next I apply the boundary condition: $$\phi = 0$$ ax $$x=0$$, which gives:
$$\phi(0) = 0 = \frac{P\cdot 0 (0-2L)}{2EI} + C_{1} \rightarrow C_{1} = 0$$
Next I integrate the other equation and substitute the found solution and the other boundary condition into it which is $$w(0) = 0$$, which results in:
$$-P = \kappa AG(-\phi + \frac{\partial w}{\partial x}) = \kappa AG(-\frac{Px(x-2L)}{2EI} + \frac{\partial w}{\partial x}) \rightarrow \frac{\partial w}{\partial x} = \frac{-P}{\kappa AG} + \frac{Px(x-2L)}{2EI}$$
$$\rightarrow w(x) = \frac{Px^{2}(x-3L)}{6EI} - \frac{Px}{\kappa AG} + C_{2}$$
$$w(0) = 0 = \frac{P \cdot 0^{2}(0-3L)}{6EI} - \frac{P \cdot 0}{\kappa AG} + C_{2} \rightarrow C_{2} = 0$$
However when taking a look at the solution provide on Wikipedia: https://en.wikipedia.org/wiki/Timoshenko%E2%80%93Ehrenfest_beam_theory#Example:_Cantilever_beam, I would expect a solution that provides the same values, except in opposite direction, which is not the case. The difference between my solution and the one provided seems to be created in the first part, which finds the solution for $$\phi$$, but I am unsure what I did wrong. The solution at the extremities is the same, but not the solution throughout the beam itself.
Any help would be highly appreciated.
• Create FBD from a section x from the free end gives: M(x)=−P(L−x) ,---- How? I rather you deselect my answer, as I don't think we are on the same page. Now you have followed the wiki's pattern - mixing, jumping, and disorganized. Sorry, the change is worse than the original.
– r13
Commented Jan 2, 2022 at 15:08
• I have deselected your answer as per your wish. I have not followed the wiki example but actually this paper: iieta.org/journals/ti-ijes/paper/10.18280/ti-ijes.630105, but their approach is indeed very similar. I arrived at the moment, because for that solution the problem is defined differently where x is defined from the clamped end towards the free end, instead of the other way around. Which I now realize I did not specify in my previous post
– Dion
Commented Jan 2, 2022 at 16:01
• Note that steps 171 & 172 are in line with the Timoshenko solution shown in my answer. Case closed.
– r13
Commented Jan 2, 2022 at 17:38
• Yes I never disagreed with the solution you provided. I just wanted to post the solution using a little more detail for other people that might stumble onto this thread and wanted to also show that you can indeed also arrive at the solution using x going from the fixed end towards the free end. Once again thank you for your help.
– Dion
Commented Jan 2, 2022 at 18:02
• Note that we've never said you can't work out from the fixed end, but provided that there is a simple way to do it why bother to go around, then make unnecessary complications that potentially cause mistakes. Hope you will keep in mind in your future practices that, oftentimes, simple is the true beauty, and that's made Timoshenko one of the best. Wish you doing well. Good luck.
– r13
Commented Jan 2, 2022 at 18:30
As addressed before, you should formulate the problem with the x-axis pointing from the free end to the fixed end, which is the conventional method, and was used by the wiki example as well.
On wiki example:
I've carefully reviewed the wiki example, the solution seems incorrect. However, I've failed to pinpoint the root problems, as the steps jump around quite a lot, with many steps containing problematic/questionable terms. Rather than make the line-by-line correction, which could lead to more confusion, the deflection, based on Timoshenko Beam Theory, of a cantilever beam with concentrate load at the free end is provided below for your information. (Per the textbook of Timoshenko & Gere)
Revised per updated info:
Total curvature of an elastic beam (per Timoshenko):
$$\dfrac{d^2w}{dx^2} = \dfrac{d^2w_b}{dx^2} + \dfrac{d^2w_s}{dx^2}$$ = $$-\dfrac{M}{EI} - \dfrac{\kappa}{GA}\dfrac{dV}{dx}$$
Where $$M = Px$$, $$V = P$$ , $$\kappa =$$ Shear (form) Factor ($$\dfrac{_3}{_2}$$ for rectangle shape; $$\dfrac{_4}{_3}$$ for circular shape)
Integrate the above equation twice to get the deflection due to $$P$$, thus
$$w = \dfrac{PL^3}{3EI} + \dfrac{\kappa PL}{AG}$$
• That would indeed solve the problem. I am however curious why you have to define x from the free end towards the fixed end and why I would not be able to do this the other way around? Similar to en.wikipedia.org/wiki/…, where x is defined from the fixed end towards the free end.
– Dion
Commented Dec 29, 2021 at 17:30
• I have edited my original post, to now include the integration steps that I took as well as applying the boundary conditions which I believe should be $\phi(0) = 0$ and $w(0) = 0$, please correct me if I am wrong. And if you notice any mistake please let me know. Thank you for your help already
– Dion
Commented Dec 29, 2021 at 20:25
• I do not agree with this simplyfication, neglecting the $q$ term, simplifies the equation to the euler-bernoulli equation. Furthermore as was shown in the equation on the wiki is that it should be possible to solve the beam equation for a point load.
– Dion
Commented Dec 30, 2021 at 11:05
• This is not a simplification, it is responding to your own statement "...left side supported beam, with a point load at the free end. ". If you wish, you can integrate the shear deformation term ("q") too with q = 0, why don't you, but sit and complain?
– r13
Commented Dec 30, 2021 at 12:26
• I am not trying to complain or to be rude. However the solution you posted above does not account shear deformation which is the added term from timoshenko compared to the euler-bernoulli beam equation. Also the wiki solution is clearly different from the one you provided above, which confuses me, as the problem is the same, with a different reference frame and a change of the frame of reference should not provided a different solution for a physical problem. Thank you for your help anyways.
– Dion
Commented Dec 30, 2021 at 12:37
Your moment equation is missing the positive moment, $$PL$$ at the left support.
$$M= PL-P(L-x)$$<|endoftext|>
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# Lesson 4
Mismo tamaño, tamaños relacionados
### Lesson Purpose
The purpose of this lesson is for students to use visual representations to reason about the fractions that have the same size and to locate them on the number line.
### Lesson Narrative
In grade 3, students reasoned about equivalent fractions, using fraction strips, tape diagrams, and number lines. Here, they begin to revisit the idea of equivalence. Students examine fractions that have the same size but are expressed with different numerators and denominators. They use diagrams of fraction strips, now expanded to include fractions with denominator 10 and 12, and then transition to using number lines to support their reasoning.
The relationships between fractions such as $$\frac{1}{4}$$ and $$\frac{1}{8}$$, $$\frac{1}{5}$$ and $$\frac{1}{10}$$, and $$\frac{1}{6}$$ and $$\frac{1}{12}$$, in which one denominator is a multiple of the other, continue to be highlighted and offer many opportunities for students to look for and make use of structure (MP7).
Later in the unit, students will take a closer look at equivalence and investigate new ways to reason about equivalence.
As in earlier activities, rulers can be provided to help students draw, extend, or align partition lines, but should not be used to measure the location of a fraction.
• Engagement
Activity 2: Fracciones en rectas numéricas
### Learning Goals
Teacher Facing
• Use the relationships between fractions whose denominators are multiples of one another (for instance $\frac{1}{2}$, $\frac{1}{4}$, and $\frac{1}{8}$) to locate fractions on the number line.
• Use visual representations to reason about fractions that have the same size. Recall that these fractions are equivalent.
### Student Facing
• Encontremos algunas fracciones que tengan el mismo tamaño.
### Required Materials
Materials to Gather
Building On
Building Towards
### Lesson Timeline
Warm-up 10 min Activity 1 20 min Activity 2 15 min Lesson Synthesis 10 min Cool-down 5 min
### Teacher Reflection Questions
This lesson is students’ first experience with the number line in grade 4. What understandings or misunderstandings about the number line did you observe today as students worked? Did you see students relating the idea of partitioning a tape diagram to partitioning a number line?
### Suggested Centers
• Get Your Numbers in Order (1–5), Stage 3: Denominators 2, 3, 4, or 6 (Addressing)
• Number Line Scoot (2–3), Stage 3: Halves, Thirds, Fourths, Sixths and Eighths (Supporting)<|endoftext|>
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## Integral Calculus, The Volume of the Hypersphere
### The Volume of the Hypersphere
The sphere in n dimensions is the set of points that are 1 unit away from the origin. In 3 space the sphere has the equation x2+y2+z2 = 1. In the previous section we calculated the volume of this sphere. Is there a formula for the volume of the unit sphere in n dimensions?
Before diving into integral calculus, let's develop some intuition. Enclose the sphere inside a cube whose coordinates run from -1 to 1. In 2 dimensions the circle takes up almost all of the square's area, π out of 4 to be exact. Even in 3 dimensions the sphere takes up most of the cube. This is not true in 100 dimensions. The sphere touches the cube along the axes, but watch what happens as we follow the diagonal line from the origin to the far corner of the cube. When all coordinates are 0.1 we reach the boundary of the sphere. That's only a tenth of the distance to the corner of the cube. The remaining 90% is all empty space. We will not be surprised to see the volume of the sphere drop to 0 as n approaches infinity.
In one dimension the "sphere" is a line segment of length 2. In two dimensions the sphere is a circle, having area π. Proceed by induction on n.
Use polar coordinates to describe the first two dimensions of an n dimensional sphere. For every point r,θ, there is a sphere in n-2 dimensions with radius sqrt(1-r2). Let v be the volume of the unit sphere in n-2 dimensions. Scale by the radius, and the volume of the sphere at a distance r from the origin is v×(1-r2)½(n-2). Integrate with respect to r, and then θ. Don't forget the extra factor of r, for polar coordinates.
∫∫ vr(1-r2)½(n-2)
We are integrating r times a power of 1-r2. The result is -v(1-r2)½n/n. Evaluate at 0 and 1 and get v/n. Then integrate with respect to θ and bring in a factor of 2π. Therefore vn = 2π/n×vn-2.
Set n = 3 and confirm the volume of the 3-sphere = 4π/3. The volume of the 4-sphere is π2/2, and the volume of the 5-sphere is 8π2/15. For larger values of n, n exceeds 2π, hence the volume of the n-sphere approaches zero.
The volume of an ellipsoid is the volume of the corresponding unit sphere, vn, multiplied by the lengths of the semi-axes. If the sphere has radius r it's volume is vnrn.
As the radius increases, the sphere encloses an ever-larger volume. The surface of the sphere is always perpendicular to its outward motion. Therefore the surface area is the derivative of volume. Differentiate rn to get an extra factor of n. If an is the surface area then an = nvn. In other words, the surface area of the unit hypersphere is volume times dimension. The circumference of the unit circle is 2 times π, or 2π, as expected. The surface area of the unit sphere is 3 times 4π/3, or 4π. Here is the volume and surface area for the first 10 dimensions. As you can see, the greatest volume occurs at dimension 5, but the greatest surface area occurs at dimension 7. Like volume, surface area approaches 0 as n approaches infinity.
Dimension Volume Area
1 2.0000 2.0000
2 3.1416 6.2832
3 4.1888 12.5664
4 4.9348 19.7392
5 5.2638 26.3189
6 5.1677 31.0063
7 4.7248 33.0734
8 4.0587 32.4697
9 3.2985 29.6866
10 2.5502 25.5016
Calculus, Medical Research, and Peptide BPC 157:
Experimental Peptides | Medical Research Calculus Concepts | BPC 157 5mg<|endoftext|>
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# Section 8.6. You probably know that a lighter tree climber can crawl farther out on a branch than a heavier climber can, before the branch is in danger.
## Presentation on theme: "Section 8.6. You probably know that a lighter tree climber can crawl farther out on a branch than a heavier climber can, before the branch is in danger."— Presentation transcript:
Section 8.6
You probably know that a lighter tree climber can crawl farther out on a branch than a heavier climber can, before the branch is in danger of breaking. What do you think the graph of (length, mass) data will look like when mass is added to a length of pole until it breaks? Is the relationship linear, like line A, or does it resemble one of the curves, B or C?
Procedural Note 1. Lay a piece of linguine on a table so that its length is perpendicular to one side of the table and the end extends over the edge of the table. 2. Tie the string to the film canister so that you can hang it from the end of the linguine. (You may need to use tape to hold the string in place.) 3. Measure the length of the linguine between the edge of the table and the string. Record this information in a table of (length, mass) data. 4. Place mass units into the container one at a time until the linguine breaks. Record the maximum number of weights that the length of linguine was able to support.
Step 1:Work with a partner. Follow the Procedure Note to record at least five data points, and then compile your results with those of other group members. Step 2: Make a graph of your data with length as the independent variable, x, and mass as the dependent variable, y. Does the relationship appear to be linear? If not, describe the appearance of the graph. The relationship between length and mass is an inverse variation. The parent function for an inverse variation curve, f(x)=1/x, is the simplest rational function. Step 3: Your data should fit a dilated version of the parent function f (x) = 1/x. Write an equation that is a good fit for the plotted data.
Graph the function f(x) =1/x on your calculator and observe some of its special characteristics. The graph is made up of two branches. One part occurs where x is negative and the other where x is positive. There is no value for this function when x =0. What happens when you try to evaluate f(0)?
This graph is a hyperbola. Its like the hyperbolas you studied in Lesson 8.4, but it has been rotated 45°. It has vertices (1, 1) and (-1, 1), and its asymptotes are the x- and y-axes.
To understand the behavior of the graph close to the axes, make a table with values of x very close to zero and very far from zero and examine the corresponding y-values. Consider these values of the function f (x) =1/x.
The behavior of the y-values as x gets closer to zero shows that the y-axis is a vertical asymptote for this function.
As x approaches the extreme values at the left and right ends of the x-axis, the curve approaches the x-axis. The horizontal line y = 0, then, is a horizontal asymptote. This asymptote is called an end behavior model of the function. In general, the end behavior of a function is its behavior for x-values that are large in absolute value.
Example A You can change the form of the equation so that the transformations are more obvious. Because the numerator and denominator both have degree 1, you can use division to rewrite the expression.
Example B
To find when the solution is 60% acid, substitute 0.6 for P and solve the equation.
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Asteroid smelter. Image: Bryan Versteeg, spacehabs.com
- Roadmap Table of Contents
- Part 1: General Milestones
- Part 2: Utilization and Development of Cislunar Space
- Part 3: To the Moon
- Part 4: To Mars
- Part 5: Asteroid Mining and Orbital Space Settlements (this page)
- Part 6: Additional Expansion and Greater Sustainability of Human Civilization
Remote or robotic characterization of near-Earth (and other) asteroid orbits, compositions and structures.
Osiris Rex asteroid sample return mission launched September 2016. Image: NASA
Telescopic observations will initially identify asteroids as Near Earth Objects (NEO’s), Earth threatening NEOs, main belt asteroids and other orbital groupings. Initial robotic missions to NEO asteroids of commercial interest will confirm the size and composition of different types of asteroids as being rocky, metallic or carbonaceous, and identify the actual abundances of minerals on each one. Metallic asteroids contain iron, nickel and platinum group metals, and carbonaceous ones contain carbon compounds and water. The probes will also estimate the structure of the asteroids, as being apparent “rubble piles” of loose fragments, or made of solid, non-fractured rock and metal. Some missions may bring back actual samples of asteroid material for analysis. All this information will assist governments in planning planetary defense against threatening NEOs and will assist mining companies to decide which asteroids to focus on. Earth-threatening NEO’s that are composed of useful minerals could be put on a list of objects to be totally mined away so there is no remaining risk. Radio beacons may be placed on NEO’s to make tracking them easier.
- Lack of information on the composition and physical structure of individual asteroids.
- Lack of telescopes dedicated to spectral analysis of asteroid composition.
- Lack of telescopes dedicated to finding and tracking small Earth-threatening asteroids.
- Lack of inexpensive robotic probes that can rendezvous with and analyze asteroids.
The investigation of asteroids will be ongoing as activity expands into the main belt with thousands of objects, so finding a completion point would be difficult. Partial completion would be indicated when sufficient knowledge of asteroids exists to plan planetary defense or actual mining of surveyed asteroids begins.
After robotic identification of suitable asteroids, robotic and human crews following to establish mining bases and habitats for transient occupation, and eventually building permanent human settlements nearby.
The eventual construction of rotating space settlements from minable asteroids. Image: Bryan Versteeg, spacehabs.com
Asteroids have huge mineral wealth if it can be accessed, including iron, nickel, platinum group metals, other non-volatile materials, and also volatiles like water ice. There are different classes of asteroids with varying amounts of these materials that would be useful both on Earth and in space. That potential value may be the primary driver for asteroid exploration and mining.
As is expected to be the case on the Moon and Mars, deposits of volatiles can be converted to rocket fuel and oxygen, thus enabling further space operations. The metals in asteroids can be refined and turned into construction materials for building large structures in space. Smelting and fabrication of parts from asteroids will require either the development of new techniques for doing this in microgravity, or the use of rotating structures to provide gravity so that existing methods can be used.
The practicality of returning asteroidal materials to Earth or other locations would depend on (1) transport costs, (2) value of the materials as delivered and (3) the extent to which those materials can be separated and purified to reduce the total mass before transport. Asteroid resources may greatly improve life on Earth. Economic return of materials to Earth is expected to involve the use of fuel obtained from asteroid mining.
In time, asteroids will see more permanent rotating habitats created nearby, probably made of asteroid derived materials, and using unprocessed asteroid materials for radiation shielding. These habitats will house either visiting crews or, if there is sufficient mining to be done, permanent occupants. With an appropriate asteroid, these mining stations may go through the same processes of growth as settlements on the Moon and Mars, and evolve into permanent settlements where people will raise their children and live out their lives, as on any far frontier. Proposals have also been made for hollowing out an asteroid and building a rotating space settlement inside it.
The eventual construction and location of rotating space settlements in the orbits of minable asteroids would reduce the materials transport costs of asteroid resources and derived materials for the construction of such settlements. This would create a synergism which should accelerate the asteroid mining industry.
- Economic barriers (transport costs) to moving asteroidal products to space settlement construction sites and to the Earth and vicinity.
- Lack of knowledge of methods for mining, refining, and fabricating asteroidal products such as building materials in microgravity.
- Lack of detailed planning and design for use of the fabricated building materials to build large space structures.
This milestone will be considered achieved when asteroid mines and smelters are regularly sending ores or refined products to the Earth, Moon or Mars or are contributing substantial structural mass to rotating space settlements in any location.
Orbital “cities in space” built from asteroid or lunar materials.
Image: Alexander Preuss
Orbital space settlements are large pressurized structures that constitute cities or villages with residential, commercial and/or governmental functions, built in space from asteroid or lunar materials, where families live. The settlements would rotate to provide artificial gravity.
In 1974, Princeton physicist Gerard O’Neill proposed the construction of orbital space settlements. An orbital settlement (sometimes called an “O’Neill Settlement”) is a giant rotating space structure, large enough and rotating fast enough so that people standing on the inner surface would experience a centrifugal force equivalent to gravity on the surface of the Earth. Thus, children on orbital space settlements would be raised in Earth-normal gravity, which is important for normal bone and muscle development. Three proposed types of orbital settlements are Bernal Spheres (and a variation called Kalpana), Stanford Tori and O’Neill Cylinders.
Since orbital space settlements must rotate, only a few basic shapes work well: sphere, torus, cylinder, disk, or some combination. Current materials are strong enough for habitats many kilometers in extent, big enough for a moderately large city. The inner surface of the hull is real estate, i.e., land on which crops could be grown and homes and businesses could be constructed. While the outer hull will experience one gravity, interior structures can be positioned for fractional gravity, and even zero gravity at the axis of rotation. People and their families can live there indefinitely, in communities ranging in size from villages to cities which have their own internal economies as well as external imports and exports.
The Equatorial Low Earth Orbit (ELEO) settlements discussed in Milestone 16, which can act as precursors for later orbital settlements, use the Earth’s geomagnetic field to shield them from space radiation. All other settlements would need to use extensive radiation shielding.
Unlike Lunar or Martian surface settlements, radiation shielding is required on all sides, so roughly twice as much shielding mass is necessary. Shielding can consist of a substantial mass of asteroidal rubble, water, waste material, or some other mass.
Orbital settlements could be built in, or moved to, a variety of orbits, including Earth, solar or other orbits, including special locations such as Lagrange points. Most of these orbits would be selected to have continuous solar energy available. The choice of orbit may be driven by access to materials, such as sites co-orbiting near an asteroid mine. For use in cislunar space, lunar material could be launched into space using electromagnetic launchers (mass drivers). Material mined from an asteroid could be utilized either in an orbit close to the asteroid or moved to some other desired location. There are thousands of candidate asteroids among the Near Earth Objects, some requiring less energy to reach than the Moon.
Eventually such cities in space could be located throughout the solar system, orbiting around planets or moons, co-orbiting with asteroids, at Lagrange points, or in solar orbit. These settlements may be very different from each other, each reflecting the particular tastes and cultures of those who built, financed and settled it. Such diversity could provide a new flowering of human creativity. The result would also be to disperse humankind throughout the solar system, enabling survival even if some disaster were to befall the Earth.
Orbital settlements may be built by private companies, governments, or consortiums. It will be financially possible to build them only when the cost of construction and support is less than the expected value of the settlement, financial or otherwise.
The Potential Scale of Orbital Settlement
Orbital settlements could be built in virtually unlimited numbers. NASA Publication SP-413 (Space Settlements: A Design Study) states: “If the asteroids are ultimately used as the material resource for the building of new colonies, and … assuming 13 km of total area per person, it appears that space habitats might be constructed that would provide new lands with a total area some 3,000 times that of the Earth.”
COMPONENTS (required capacities)
Habitats in space beyond Low Earth Orbit designed as space settlements must be able to:
- Provide redundant life support systems for the residents that will last for many decades.
- Store sufficient reserves of food and water for the residents, recycle water and grow food.
- Provide a high level of protection and redundancy against loss of air pressure accidents.
- Protect residents against constant cosmic radiation (heavy, fast nuclei from outside the solar system) to a level consistent with the presence of children and pregnant women.
- Protect against intermittent radiation from solar mass ejections and general solar radiation to a level consistent with the presence of families and children.
- Provide artificial (centrifugal) gravity with a sufficient level of gravity to maintain health.
- Provide for permanent residency by making the habitats suitable (large enough, etc,) for comfortable living.
- Provide employment and recreation for the residents.
- Lack of asteroidal or lunar-derived materials and parts in space for construction of orbital settlements.
- Lack of detailed planning, design and methods for creation of orbital space settlements, including use of materials to build large, pressurized, rotating structures in space.
- Lack of immediate economic incentive to work toward orbital settlement construction.
- Inadequate understanding of human physical adaptation and the psychology of individuals and large groups of people living in space.
- Lack of information on all phases of the cost to construct orbital space settlements.
- Planetary chauvinism: the idea that people should only live on planetary surfaces.
This milestone can be considered achieved when a rotating space settlement built primarily from non-terrestrial materials has a population of at least 1000 including families and children. Another milestone will be achieved when the total population in orbital space settlements exceeds the population on Earth.
ORBITAL SPACE SETTLEMENT DESIGNS (click for larger image)<|endoftext|>
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# NCERT Solutions for Class 8 Maths chapter 10 Exercise 10.2 Exponents and Powers - PDF Download
NCERT solutions class 8 maths chapter 10 Exercise 10.2 in Exponents and Powers is composed of questions that are based on expressing numbers in standard form. This exercise also covers the topic of comparing very large and very small numbers. Class 8 maths chapter 10 exercise 10.2 NCERT solutions has examples and questions that go into detail about these topics so that students can understand them quickly.
Ex 10.2 class 8 maths solutions consists of four questions with its sub-parts. With regular practice of questions and examples in these solutions, you will be able to easily represent any number. Ex 10.2 class 8 maths chapter 10 NCERT solutions focuses on covering the topic with appropriate examples. The solutions are prepared by eSaral's subject experts with easy to understand language. Students can use these solutions to prepare for exams and achieve good marks. NCERT solutions are also available here in PDF format to download for free from the link provided here. You can download the free PDF and practice all the questions.
## Topics Covered in Exercise 10.2 Class 8 Mathematics Questions
NCERT solutions for ex 10.2 class 8 maths ch 10 is based on the topics of use of exponents to express small numbers in standard form and comparing very large and very small numbers. Questions of ex 10.2 can be solved if you comprehend these topics with proper understanding. For better understanding of these topics our experienced faculty of maths has explained them in detailed manner.
1 Use of Exponents to Express Small Numbers in Standard Form Comparing very large and very small numbers
1. Use of Exponents to Express Small Numbers in Standard Form
Sometimes, we need to write the numbers in very large or very small form, and we can use exponents to write them in very small numbers.
A. Standard form to write the natural numbers like xyz000000......
Step 1- The first step is to count the number of digits from left, leaving only the first number.
Step 2- To write it in exponent or standard form, write down the first digit.
Step 3- If there are more digits in the number then put a decimal after the first digit and then write down the other digits until the zero comes. And if there are no digits after the first digit then skip this step.
Step 4- Now place a multiplication sign and then write down the counted digits in the first step as the exponent to the base number 10.
For example - we can write the number 149000000000 in standard form 1.49×1011
B. Standard form to write decimal numbers like 0.00000.....xyz.
Step 1- We will count the number of digits from the decimal point to the last digit.
Step 2- If there is only one digit after the zeros then simply write down that digit. Now, place a multiplication sign and write down the counted digits in step-1 with a negative sign as the exponent to base number 10.
Step 3- If there are two or more non-zero digits at the end of the number. Then, write down the digits followed by a decimal point after the first digit and the other non-zero digits.
Step 4- Calculate the number of digits in step 1 and subtract the number of digits appearing after the decimal point.
Step 5- Place a multiplication sign and write down the counted digits in step-4 with a negative sign as an exponent to base number 10.
For ex. - we can write the number 0.000000564 in standard form 5.64×10-8
• Comparing very large and very small numbers
In order to compare the very large numbers and the very small numbers, we need to equalize their exponents. When the exponents are equal, we can compare them and see which number is large and which number is small.
To also add and subtract, we have to equalize their exponents and then we can easily add or subtract them.
## Tips for Solving Exercise 10.2 Class 8 Chapter 10 Exponents and Powers
NCERT solutions for ex 10.2 class 8 maths chapter 10 are well-structured resources written in an understandable language. Here, we provide some helpful tips to solve the questions of ex 10.2.
1. NCERT solutions for class 8 maths chapter 10 ex 10.2 are reliable and will provide you with the accurate solutions to all the important questions about the ex 10.2. Learning well with these solutions will help you find the right and step by step solutions to all questions in a systematic way.
2. Students are encouraged to practice all issues and examples to gain a thorough understanding of concepts of comparing very large and very small numbers.
3. By carefully studying all the topics provided in NCERT solutions class 8 Maths chapter 10 ex 10.2, students can master this exercise easily.
## Importance of Solving Ex 10.2 Class 8 Maths Chapter 10 Exponents and Powers
To excel in chapter 10 ex 10.2, students must practice a lot of questions from this exercise. Students can practice the questions of class 8 chapter 10 ex 10.2 from the NCERT solutions provided by eSaral. Here are some important benefits of solving questions from NCERT solutions class 8 maths chapter 10 ex 10.2.
1. Questions in ex 10.2 class 8 maths are solved step by step and with easy methods by subject experts of eSaral which will provide you a clear idea on how to solve the questions in exams.
2. Concepts of use of exponents to express small numbers in standard form and comparing very large and very small numbers are described in most understandable language so that students can comprehend the topics and solve all the questions of ex 10.2.
3. By practicing and revising questions in NCERT solutions, students can also improve their time management.
4. These solutions are also available in PDF, you can download the PDF for future reference.<|endoftext|>
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# Intermediate Geometry : How to find the diagonal of a prism
## Example Questions
### Example Question #1 : How To Find The Diagonal Of A Prism
What is the length of the diagonal of a rectangular box with the dimensions of ?
Explanation:
To solve this problem we need an extension of the Pythagorean Theorem:
So the equation to solve becomes
So the distance of the diagonal is .
### Example Question #2 : How To Find The Diagonal Of A Prism
Find the diagonal of the prism. The diagonal is represented by the dashed line.
Explanation:
The length of the diagonal is from the bottom left hand corner closest to us to the top right hand corner that's farthest away from us.
This kind of a problem may seem to be a little more complicated than it really is.
In order to solve for the diagonal length, all that's required is the Pythagorean Theorem. This equation will be used twice to solve for the dashed line.
For the first step of this problem, it's helpful to imagine a triangle "slice" that's being taken inside the prism.
, where the diagonal of interest is D2, and D1 is the diagonal that cuts from corner to corner of the bottom face of the prism. Of this triangle that's outlined in pink dashed lines, the given information (the dimensions of the prism) provides a length for one of the legs (16).
We can already "map out" that D2 (the hypotenuse of the dashed triangle) can be solved by using the Pythagorean Theorem if we can obtain the length of the other leg (D1).
The next step of this problem is to solve for D1. This will be the first use of the Pythagorean theorem. D1 is the diagonal of the base and is limited to a 2D face. This can be represented as:
The hypotenuse of the base, or the mystery length leg of the dashed triangle, can be solved by using the Pythagorean Theorem:
Now that we calculated the length of D1, D2 can be solved for by using the Pythagorean Theorem a second time:
### Example Question #3 : How To Find The Diagonal Of A Prism
A right rectangular prism has a width of cm, a length of cm, and a height of cm. Find the diagonal distance of the prism.
None of the other answers.
Explanation:
To find the diagonal distance of a prism, you can use the formula:
, where = height; = width, and = length.
So, in this problem
.
### Example Question #4 : How To Find The Diagonal Of A Prism
A right rectangular prism has a height of ft, a width of feet and a length that is twice its width. The volume of the prism is cubic feet. What is the diagonal of the prism?
None of the other answers.
Explanation:
First, given the volume, you need to find the width and length. The volume of a right, rectangular prism can be found using
, so , where represents the length and represents the width.
Solving for , you get
So, the width of the prism is 3 feet.
Remember that the length is twice the width, so the length is 6 feet.
Now you may use the formula for finding the diagonal:
. So, .
### Example Question #5 : How To Find The Diagonal Of A Prism
A right, rectangular prism has a length of meters, width that is meters longer than the length, and a height of meters. The volume of the prism is cubic meters. Find the diagonal of the prism.
None of the other answers.
Explanation:
First, use the volume formula,
to find the missing length and width.
Since cannot be a negative value is it represents a length of a prism, we know . So the length is 2 meters, and therefore the width is 5 meters.
Now you can plug in the length, width and height into the formula for finding the diagonal of the prism.
.
### Example Question #6 : How To Find The Diagonal Of A Prism
A right, rectangular prism has a volume of cubic centimeters. Its width is cm and its length is three times its height. Find the length of the diagonal of the prism.
None of the other answers.
Explanation:
First, use the volume to find the missing height and length.
Since the length is three times the height, use to represent the length and to represent the height.
So, .
So the height of the prism is 2 centimeters, and the length is 6 centimeters.
Use these values to now solve for the diagonal distance.
.
### Example Question #7 : How To Find The Diagonal Of A Prism
The surface area of a right, rectangular prism is square inches. The height is inches and the length is times the width. Find the diagonal distance of the prism.
None of the other choices.
Explanation:
Use the surface area, 280 square inches, and the formula for finding the surface area of a right, rectangular prsim to find the missing length and width measurements.
So,
Since represents the length of a solid figure, we must assume , rather than the negative value.
So, the width of the figure is 2 inches and the length is 10 inches.
Now, use the formula for finding the diagonal of a right, rectangular prism:
.
### Example Question #8 : How To Find The Diagonal Of A Prism
A right, rectangular prism has a surface area of square meters. Its width is twice its length, and its height is four times its length. Find the diagonal distance of the prism.
None of the other answers.
Explanation:
Use the surface area of the prism to find the missing length, width and height.
So, the prism's length is 1 meter, the width is 2 meters and the height is 4 meters.
Now you can find the diagonal distance using those values.
.
### Example Question #9 : How To Find The Diagonal Of A Prism
The above cube has edges of length 1. True or false: The dashed line has length .
True
False
False
Explanation:
Examine the diagram below.
is a right triangle with legs of length , so it is an isosceles - or 45-45-90 - right triangle. By the 45-45-90 Triangle Theorem its hypotenuse measures
is a right triangle with legs of lengths and , so the length of its hypotenuse is
.
, the diagonal in question, has length , not .<|endoftext|>
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Teachers and tutors, do you want to save time and get double or triple use from the same source? Use your students’ vocabulary workbook to teach writing.
Many of my students use the Wordly Wise 3000 series (which I recommend). It has 20 lessons per booklet, one booklet per grade, first through twelfth. In each lesson is an annotated list of new vocabulary words plus exercises using the words.
Like other vocabulary building series, each lesson also has a reading selection in which each new vocabulary word is used. These reading selections are followed by many questions asking the student to use one of the new vocabulary words in a complete sentence answer.
But other ways to use the vocabulary and reading selections augment their original purpose and make them valuable as writing tools. Here are some I have used.
- Summarizing. I teach students to underline the most important or key words in each paragraph. Next, I show how to analyze each paragraph and to write an identification in the margin next to the paragraph. Those phrases might be “dodo bird’s appearance,” “raising $ for Statue of Liberty base,” or “Renaissance dates and definition.” Then, using the underlines and margin information, I teach the student to write a summary of each paragraph in about one or two sentences. When he is done, he has a good summary of the reading selection.
- Paraphrasing. Taking one sentence at a time, I ask students to rewrite the sentence, keeping the meaning but changing the sentence structure and, where possible, the vocabulary.
- Writing RACE responses. I write a question based on the article. Then I ask the student to respond using the RACE format (Repeat the question, Answer the question, Cite part of the article used as evidence, and Elaborate on that evidence with more evidence).
- Writing sentences using new vocabulary words. So many times students can define a word but they cannot use it properly in a sentence. I ask them to write sentences using vocabulary words. This shows their weakness in understanding certain words and helps me to explain the words better to them.
- Writing paragraphs using new vocabulary words. I ask students to write each new word in a coherent paragraph or two. Writing a paragraph takes more skill than writing independent sentences. Not only does the student need to know how to use the word, but he needs to know its noun, adjective and verb forms and whether it is the best word in a given situation. Forming a coherent whole takes imagination and hard work.
- Writing narratives. Put a person or animal into the nonfiction situation in the reading passage and write about it. What if you were a dodo bird encountering your first human being? What if you were a Cherokee forced to say good-bye to your land in North Carolina and trek toward the unknown? What if you were Leonardo’s apprentice, entrusted to carry the rolled up canvas of the Mona Lisa from Florence to France?
If you are teaching children to write, you know that coming up with a writing topic is tedious. But by using the reading selections from the vocabulary workbooks, the subject matter is identified, the student has prior knowledge, and the vocabulary words are identified.
There is no need to reinvent the wheel.<|endoftext|>
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The Legionnaires' rebellion and the Bucharest pogrom occurred in Bucharest, Romania, between 21–23 January 1941. As the privileges of the Iron Guard paramilitary organization were being cut off gradually by the Conducător Ion Antonescu, its members, also known as the Legionnaires, revolted. During the rebellion and pogrom the Iron Guard killed 125 Jews, and 30 soldiers died in the confrontation with the rebels. Following this, the Iron Guard movement was banned and 9,000 of its members were imprisoned.
|Part of World War II|
The Sephardic Temple in Bucharest after it was looted and set on fire
|Kingdom of Romania||Iron Guard|
|Commanders and leaders|
|Ion Antonescu||Horia Sima|
4 armored vehicles
|Casualties and losses|
200–800 killed or wounded|
|125+ Jews killed during the pogrom|
Following World War I Romania gained many new territories, thus becoming "Greater Romania". However, the international recognition of the formal union with these territories came with the condition of granting civil rights to ethnic minorities in those regions. The new territories, especially Bessarabia and Bukovina, included large numbers of Jews, whose presence stood out because of their distinctive clothing, customs, and language. Intellectuals together with a wide array of political parties and the clergy led an anti-semitic campaign; many of these eventually came to cast their political lot with Nazi Germany.
The Molotov-Ribbentrop Pact (August 1939) gave the Soviet Union a green light to take back Bessarabia and northern Bukovina in June 1940, leading to the June 1940 Soviet Ultimatum and Soviet occupation of Bessarabia and Northern Bukovina. In August 1940 Germany and Italy mediated Romanian disputes with Hungary about Transylvania (resulting in the Second Vienna Award) and with Bulgaria regarding Dobruja (resulting in the Treaty of Craiova). Large areas of Romania were ceded to Hungary and Bulgaria.
During the Romanian army's withdrawal from Bessarabia, some local residents celebrated. Attacks on soldiers by locals are also documented. Various reports speak of attacks on the retreating soldiers by Jews—though the veracity of those reports is disputed—and some have been proven to be fabrications. Additionally, although the reports defined all of the celebrators and attackers as "Jews", some were Ukrainians, Russians, pro-Communists, newly released criminals, and ethnic Romanians. These reports, regardless of veracity, did much to incite many Romanians against Jews, strengthening existing anti-Semitic sentiment.
The Romanians were traumatized and frustrated by giving up these areas without a war, and the regime's position weakened significantly. The government scapegoated the Jews, with the support of the press:
Confronted with an extremely serious crisis and doubting their regime could survive, Romanian government officials turned the Jews into a political "lightning rod", channeling popular discontent toward the minority. Notable in this report is the reaction of the Romanian press, whose rage was directed more toward Jews than the Soviets, the real aggressors. Given that the Romanian press was censored in 1940, the government must have played a role in creating this bias. A typical form of anticipatory scapegoating was to let Jewish leaders know that the Romanian authorities might launch acts of repression against the Jews.
The anti-semitic legislation that began with the "Jewish Codex" in Romania, and the establishment of the National Legionary State government, set in motion the laws of Romanianization, which deprived Jewish people of their property and distributed it among supporters of the new regime. This created an atmosphere in which anti-semitism was seen as legitimate, and even sanctioned.
Politically, control was in the hands of the Conducător Ion Antonescu, heading the anti-semitic fascist coalition government, together with Horia Sima. The latter commanded the paramilitary Legionnaire militia known as the Iron Guard (originally called "The Legion of the Archangel Michael", hence the name "Legionnaires"). There was a great deal of tension between the two leaders due to Iron Guard seizures of Jewish property. Antonescu thought the robbery was done in a fashion detrimental to the Romanian economy, and the stolen property did not benefit the government, only the Legionnaires and their associates. Besides the Jewish issue, the Legionnaires, achieving power after many years of persecution by the former regime of King Carol II (which killed their first leader and founder Corneliu Zelea Codreanu, "the Captain"), were vengeful toward anyone associated with the regime.
The disagreement between Antonescu and the Iron Guard about the robbery of the Jews was not about the robbery itself but about the method, and the final destination of the stolen property. Antonescu held that the robbery should be done by way of expropriation, gradually, through an orderly process of passing anti-semitic laws.
... the Legionnaires wanted everything, and they wanted it immediately; Antonescu, while sharing the same goal, intended to achieve it gradually, using different methods. The leader stated this clearly in an address to Legion-appointed ministers: "Do you really think that we can replace all Yids immediately? Government challenges are addressed one by one, like in a game of chess.
The Legionnaires, on the other hand, wanted to rob as much as possible, as quickly as possible, utilizing methods based not in law but in terror, murder and torture. The Legionnaires had an additional quarrel with the German minority in Romania.
According to the laws of Romanianization, Jews were forced to sell many of their businesses, a fact used by many Romanians to purchase those businesses for close to nothing. The German minority introduced a level of competition by offering the Jews a better price than the one offered by the Legionnaires (on average, about one-fifth of the real worth). The local Germans had capital received as a loan from Germany, Romanian money paid to the Germans for keeping military units in their territory (to protect them from the Soviets). Antonescu demanded that the Legionnaires cease their terror tactics, and the Legionnaires began plotting to usurp Antonescu and take over sole control of the country.
Initially, the Legionnaires began "defaming" Antonescu, mentioning his family relation to Jews (his stepmother and his ex-wife, whom he had married when on a diplomatic mission to France, were Jews). They also accused him of being linked to Freemasonry. According to Nazi propaganda, the Freemasons were enemies of humanity, second only to Jews in wickedness.
In the 20 days preceding the rebellion, the level of anti-Semitic propaganda greatly increased, using all the tools at the Legionnaires' disposal. The propaganda emphasized the need for solving the "Jewish problem". Horia Sima and his comrades sought the sympathy of the Nazi regime in Germany, and built upon the ideological similarities between their movement and the Nazi movement, and had quite a few supporters within the Nazi establishment.
Antonescu, who had the support of Romania's military, met with Adolf Hitler on 14 January 1941, in Germany. During this meeting he promised Hitler the cooperation of Romania in any future German conflict with the Soviet Union, and gained Hitler's silent agreement to eliminate Antonescu's opponents in the Legionnaire Movement. Between 17–19 January the Legionnaire movement conducted a series of "lectures" throughout Romania, designed to demonstrate the national socialist nature of their movement and to show their loyalty to Hitler.
Antonescu took measures to curb the actions of the Legionnaires, and on 19 January issued an order canceling the position of Romanization commissars: well-paying jobs, held by Legionnaires. Additionally, he fired the persons responsible for terror acts committed by Legionnaires, from Minister of the Interior Constantin Petrovicescu to the commanders of the Security Police and the Bucharest police. He appointed loyal military men in their place. The military also took control of strategic installations, such as telephone exchanges, police stations and hospitals. The district officers of the Legionnaires were called to the capital for an important economic consultation, but found themselves arrested in the middle of the meeting.
As a paramilitary force, the Iron Guard had no shortage of firearms while it was in power. At the start of 1941, in Bucharest alone, the Legionnaires had 5,000 guns (rifles, revolvers and machine guns) as well as numerous hand grenades. The Legion also possessed a small, mostly symbolic armored force of four vehicles: two police armored cars and two Renault UE Chenillettes from the Malaxa factory. The Malaxa factory had been licence-producing these French armored vehicles since mid-1939, and aside from the two such machines, the factory also supplied the Legion with machine guns and rifles. For transport, the Legion possessed almost 200 trucks in Bucharest alone.
On 20 January 1941, a German officer was killed in Bucharest by a Greek citizen. This murder remains unsolved to this day, but it was the spark that lit the Legionnaire Rebellion. Antonescu had replaced the commanders of the Security Police and the Bucharest police, but their subordinates, who received their orders from Horia Sima, refused to allow the new commanders to take their place. Armed Legionnaires captured the ministry of the interior, police stations and other government and municipal buildings, and opened fire on soldiers trying to regain these buildings.
Antonescu's public addresses, intended to calm the public, were not published or broadcast, as the media was under Legionnaire control. The Legionnaires called the people to rise up against the Freemasons and the Jews (hinting at Antonescu's relations). The people who were possible targets for assassination by the Legionnaires were held, for their own protection, at the ministry of the interior. The Legionnaires' leaders, headed by Horia Sima, went underground. The Legionnaires held mass drafts at neighboring villages, and masses of peasants flooded the streets of Bucharest, answering the call to defend the country against the Jews and Freemasons. The Legionnaires took over gas stations and tankers, and used burning oil cans as weapons against the soldiers. Only 15 loyal officers remained with Antonescu in his palace. For two days the Romanian military defended itself and tried to besiege the Legionnaires' strongholds, but did not initiate attacks and gave them a free hand. During this time the Legionnaires published announcements claiming that the Jews had revolted. During the days of the rebellion, the Legionnaires' newspapers (the only ones active during this time) engaged in vicious propaganda against the Jews. At the end of the articles would appear the motto "You know whom to shoot".
The Bucharest pogrom was not a side effect of the rebellion, but a parallel event, purposefully organized to give legitimacy to the rebellion and to equate the Legionnaires' opponents with Jewish sympathizers. Many parties took part in the riots against the Jews: police officers loyal to the Legionnaires, various Legionnaire organizations, the workers' union, student union, high-school students, Roma and Sinti and criminals. The attacks on the two Jewish boroughs (Dudeşti and Văcăreşti) began a few hours before the rebellion. Minister Vasile Iasinschi gave the order to set the Jewish neighborhoods on fire, and mobs stormed Jewish homes, synagogues and other institutions. The Legionnaires' headquarters became torture centers, and Jews kidnapped from their homes were brought to them. Jews' homes were set on fire and the Jews themselves were concentrated in places where they could be tortured to take their property and their women raped. Jews were murdered at random, but also in planned executions. Some Jews were thrown from the top floors of the police headquarters building, and others killed in the slaughterhouse. Soldiers did not take part in the pogrom, nor did police officers loyal to Antonescu. Those officers were forced to surrender their weapons and uniforms, and were put under arrest.
Besides extorting the Jews for their hidden property, sadistic youth (including teenagers) took part in the torture, for their own pleasure. It continued for hours and even days and nights, the torturers taking turns. Jews were robbed of any possessions on their person, and sometimes even their clothes. They were made to turn over property hidden elsewhere, private or communal, and were often shot afterwards, as happened to the community's treasurer. Some Jews were coerced into writing suicide notes before being killed.
The persecutors were headed by Mircea Petrovicescu, the son of the minister of the interior who was deposed by Antonescu. Petrovicescu tied Jews to targets and shot them, aiming not to hit them but to draw a line around them. He also used Jewish women stripped naked and tied with their backs to the target. After he was done shooting, they bore into the women's breasts with a drill, or cut them. Only one woman survived this treatment, but she was later executed with other Jews. Legionnaire women took part in the pogrom; all survivors noted their involvement in the torture, and some of the worst acts of abuse were at their hands. According to the witnesses, Legionnaire women stripped Jewish men and hit their genitalia.
On 23 January, a few hours before the rebellion was quelled, a group of Legionnaires selected 15 Jews at random. They took them in trucks to the local slaughterhouse, where they were shot. Five of the Jews, including a five-year-old girl, were hung on the slaughterhouse's hooks, still alive. They were tortured, their bellies cut and their entrails hung around their necks in a parody of shehita, kosher slaughter of cattle. The bodies were labeled "kosher". The slaughterhouse was closed for a week to purge and clean the house of the results. When Antonescu appointed a military prosecutor to investigate the events at the slaughterhouse, he reported that
he recognized three of his acquaintances among the "professionally tortured" bodies (lawyer Millo Beiler and the Rauch brothers). He added, "The bodies of the dead were hanged on the hooks used by slaughterers.
American minister to Romania, Franklin Mott Gunther, toured the meat-packing plant where the Jews were slaughtered with the placards reading "Kosher meat" on them. He reported back to Washington: "Sixty Jewish corpses were discovered on the hooks used for carcasses. They were all skinned . . . and the quantity of blood about was evidence that they had been skinned alive". Gunther wrote he was especially shocked that one of the Jewish victims hanging on the meat hooks was a five-year-old girl, saying that he could not imagine such cruelty was possible until he saw the evidence of it firsthand.
Of the slaughterhouse episode, Romanian author Virgil Gheorghiu later wrote:
In the big hall of the slaughterhouse, where cattle are hanged up in order to be cut, were now human naked corpses . . . On some of the corpses was the inscription "kosher". There were Jewish corpses. … My soul was stained. I was ashamed of myself. Ashamed being Romanian, like criminals of the Iron Guard.
During the pogrom 125 Bucharest Jews were murdered: 120 bodies were eventually counted, and five never found. Other Jews, not from the Bucharest community, who happened to be in Bucharest at the time may have also been killed. The Legionnaires ignited the Jewish synagogues and danced around the flames, roaring with joy. To accomplish their mission they used a fuel tanker, sprayed the walls of Kahal Grande (the great Sephardic synagogue) and lit it. It was completely burnt. In the various synagogues the Legionnaires robbed the worshipers, abused them, took all their valuables and tore up the holy scriptures and ancient documents. They destroyed everything, even the lavatories.
During the riots 1,274 businesses, shops, workshops and homes were badly damaged or destroyed. After the suppression of the rebellion, the army took the Legionnaires' loot in 200 trucks (not including money and jewelry). Some synagogues were partly saved. The large Choral Temple (Heichal Hakorali) synagogue was saved from burning completely, because the Legionnaires didn't bring enough fuel. In the large synagogue was a Christian, Lucreţia Canjia. She begged the rioters not to burn the synagogue, reminding them of their Christian teachings. The synagogue was saved.
In Turda, Buhuși and Ploiești, hundreds of legionnaires marched down the streets while singing Legionary songs, but they eventually dispersed quietly. Two gangs of unarmed legionnaires in Vrata patrolled the main street of the village, interrogating anyone who tried to enter it. In Piatra Neamț, 600 Legionnaires gathered to support Sima, but they were peacefully dispersed by the intervention of local police. Nevertheless, a small group of legionnaires later vandalized Jewish homes in the town. In Buzău, legionnaires gathered at the police station, but they were surrounded by soldiers and trapped inside. In Târgu Frumos, the mayor deployed groups of teenage legionnaires by train to Iași on 20 January. He soon resigned, however, when situation deteriorated on the evening of 21 January. By far the most active spot of the legionnaire rebellion outside Bucharest was Brașov. Better organized than in other places outside the capital, the legionnaires occupied the gendarmerie, the council chambers, municipal offices, the treasury, the post office and telephone exchange, the radio station, as well as other gendarmerie posts in nearby villages. Five armed legionnaires seized a bus and held its passengers hostage for several hours.
During the days of the rebellion, Antonescu avoided direct confrontation with the Legionnaires but brought military units, including 100 tanks, into Bucharest from other cities. As the chaos spread—worrying even Hitler, who was interested in Romania as an ally—the horrific picture of the pogrom became clear. As stories spread, the military's fury against the Legionnaires grew (the Legionnaires had assaulted captured soldiers, stripped them of their uniforms, and even burned several of them). When Antonescu thought the moment was most appropriate, he gave the order to crush the rebellion. The military, led by Gen. Ilie Șteflea, quelled the rebellion in a matter of hours with little difficulty. The Legionnaires could not defend against the military's superior firepower. As soldiers stormed their strongholds, the Legionnaires fled. During the skirmishes 30 soldiers were killed and 100 were wounded. The number of legionnaires killed during the rebellion was approximately 200, although in later years Horia Sima would claim there had been 800 legionnaire casualties. After the rebellion was suppressed Antonescu addressed the public on the radio, telling them "the truth", but never mentioning the pogrom. He asked the German garrison, which had sat idly by throughout the rebellion, to show their support. German troops were sent marching through the streets of Bucharest, ending in front of the Prime Minister's building, where they cheered Antonescu.
After the Legionnaires' fall the trend reversed, and those who had joined them fled. The press stopped supporting the Legionnaires, but remained anti-Semitic and nationalistic. Some of the Legionnaires' leaders, including Horia Sima, fled to Germany. Around 9,000 members of the Legionnaires' movement were sentenced to prison. The Legionnaires who led the anti-Semitic movement in Romania had fallen and never regained power. However, the movement continued even without them, although it was set back for a while, as the atrocities of the Bucharest pogrom gradually became known to the Romanian public. A few months later those atrocities paled in severity compared to those of the Iaşi pogrom, initiated at the orders of Antonescu. One leader of the pogrom, Valerian Trifa, became a cleric and emigrated to the US, where he became a citizen, but he was stripped of his citizenship in 1982 and left the US rather than be deported.
Events from the year 1941 in Romania.Corpul Muncitoresc Legionar
Corpul Muncitoresc Legionar or Corpul Muncitorilor Legionari (CML, the Legionary Worker Corps or Legionary Workers' Corps) was a fascist association of workers in Romania, created inside the Iron Guard (which was originally known as the Legionary Movement) and having a rigid hierarchical structure. From its creation until September 1940, the CML was led by Gheorghe Clime; afterwards, the position was filled by Dumitru Groza, who oversaw the Corps during the period when the Iron Guard was in power — the National Legionary State —, and involved it in the 1941 Rebellion and Pogrom. The CML had its headquarters in Bucharest, on Calea Călăraşilor.Together with the Iron Guard, it was outlawed by Conducător Ion Antonescu during the Rebellion, and dissolved itself. In time, the group formed around Dumitru Groza was drawn into collaboration with Antonescu, and later refused to become involved in talks with the Romanian Communist Party over the possibility of a political truce.Cuvântul
Cuvântul (Romanian pronunciation: [kuˈvɨntul], meaning "The Word") was a daily newspaper, published by philosopher Nae Ionescu in Bucharest, Romania, from 1926 to 1934, and again in 1938. It was primarily noted for progressively adopting a far right and fascist agenda, and for supporting, during the 1930s, the revolutionary fascist Iron Guard.For My Legionaries
For My Legionaries (Romanian: Pentru legionari) is an autobiographical book by Iron Guard leader Corneliu Zelea Codreanu first published in 1936. The book has been described by historian Irina Livezeanu as being to Codreanu what Mein Kampf was to Adolf Hitler. It was first published in Sibiu, as it was not allowed to pass censorship in Bucharest.The book is a first-person narrative describing Codreanu's leadership role in a series of political movements, "The Guard of the National Conscience", "League of National Christian Defence", "the Legion of the Archangel Michael", and finally, the Iron Guard. His goal within these movements was to defend the newly established Greater Romania against a set of demonised enemies, particularly, the Soviet Union and the Jewish people. The narratives are interspersed with quotations from Romanian intellectuals, as well as clippings from contemporary newspapers.Codreanu makes clear in his book that his ideology is not compatible with the liberal democratic institutions. He loathed the elections and the parliamentary system and he considered his movement to be part of a greater family of ultra-nationalist ideologies, which included Italy's Fascism and Germany's National Socialism.German People's Party (Romania)
The German People's Party (German: Deutsche Volkspartei in Rumänien; Romanian: Partidul Poporului German din România, PPGR) was a political party that operated in Romania between 1935 and 1938, claiming to represent the ethnic German community.
Alfred Bonfert founded the PPGR on April 22, 1935, in a split with the Nazi-oriented German Party, whose president he accused of having a conciliatory attitude toward the party's democratic leaders, and which he denounced as Judeo-Communist during the next few years. The party's base was the Volksdeutsche bourgeoisie, influenced by Nazism. It was organised on the Hitler-created model, with a paramilitary system in which the cadres were named by superior hierarchical organs. It ran three official newspapers: Der Stürmer (Timișoara), Ost-deutscher beobachter (Sibiu) and Sachsenburg (Brașov).
In its programme of 1935, the PPGR asked for the 1923 Constitution to be respected, as well as for cultural autonomy for the local German community. Besides its programme, the party's practical activity entailed cultivating a German (in this case Nazi) spirit among the Germans of Romania, and implanting each one of these with the idea that he represented an element, living abroad, of the Great Reich, whose interests he had to serve. The PPGR was hostile to Romanians and tried to isolate ethnic Germans from the general population. It adopted an intransigent attitude toward the country's governments, disavowing collaboration and pursuing a policy of confrontation toward them. A veritable fifth column for the Reich, it was never very popular, gaining under 1% of the vote at the 1937 election despite Germans forming over 4% of the population.
The German People's Party, along with all other parties extant in Romania, was dissolved on March 30, 1938. On October 27, 1938, following orders from Berlin, the remnants of PPGR were merged with the German Party.Horia Sima
Horia Sima (3 July 1907 – 25 May 1993) was a Romanian fascist politician. After 1938, he was the second and last leader of the fascist para-military movement known as the Iron Guard.Ion Moța
Ion I. Moța [or Motza] (5 July 1902, Orăștie, Austria-Hungary—13 January 1937, Majadahonda, Spain) was the Romanian nationalist deputy leader of the Iron Guard killed in battle during the Spanish Civil War.Iron Guard death squads
During the 1930s, three notable death squads emerged from Romania's Iron Guard: the Nicadori, the Decemviri and the Răzbunători. Motivated by a combination of fascist political ideology and religious-nationalist mysticism, they carried out several high-level political assassinations in the inter-war period.Jilava massacre
The Jilava Massacre took place during the night of November 26, 1940 at Jilava penitentiary, near Bucharest, Romania. Sixty-four political detainees were killed by the Iron Guard (Legion), with further high-profile assassinations in the immediate aftermath. It came about halfway through the fascist National Legionary State and led to the first open clash between the Guard and conducător Ion Antonescu, who ousted the Legion from power in January 1941.National-Christian Defense League
The National-Christian Defense League (Romanian: Liga Apărării Național Creștine, LANC) was a far-right political party of Romania formed by A. C. Cuza.National Christian Party
The National Christian Party (Romanian: Partidul Național Creștin) was a radical-right authoritarian and strongly antisemitic political party in Romania active between 1935 and 1938. It was formed by a merger of Octavian Goga's National Agrarian Party and A. C. Cuza's National-Christian Defense League (LANC); a prominent member of the party was the philosopher Nichifor Crainic. Goga was chosen in December 1937 by King Carol II to form a government which included Cuza. The government lasted for only 45 days and was followed by a royal dictatorship by Carol.National Fascist Movement
The National Fascist Movement (Romanian: Mișcarea Națională Fascistă, MNF) was a Romanian political movement formed in 1923 by the merger of the National Romanian Fascia and the National Italo-Romanian Cultural and Economic Movement.
With its roots in an avowedly pro-Italian group, the MNF also became close supporters of the Italian model of Fascism - although the movement also admired the methods of Action Française. The movement did not enjoy the success that it had hoped for, largely because of its attachment to foreign influences, and it was ultimately superseded by the Iron Guard, which offered a more domestic form of fascism.National Italo-Romanian Cultural and Economic Movement
The National Italo-Romanian Cultural and Economic Movement (Romanian: Mișcarea Națională Culturală și Economică Italo-Română) or National Italo-Romanian Fascist Movement (Mișcarea Națională Fascistă Italo-Română) was a short-lived Fascist movement active in Romania during the early 1920s.
The movement was formed in 1921 by Elena Bacaloglu, a female journalist who had an Italian husband at the time, and was an acquaintance of Benito Mussolini (she had been briefly the wife of Ovid Densusianu). The group deliberately mimicked Italian fascism and stressed the close ethnic bonds between the Italians and the Romanians. The group attracted only around 100 members. The group was based in Cluj, where it was initially established. It was wound up in 1923, when it merged with the National Romanian Fascia to form the National Fascist Movement.National Romanian Fascio
The National Romanian Fascio (Romanian: Fascia Națională Română) was a small fascist group that was active in Romania for a short time during the 1920s.
Led by Titus Panaitescu Vifor, the group emerged from the short-lived National Fascist Party in 1921 and, at its peak, had around 1,500 members. It defined itself as national socialist, although generally it pursued a policy of corporatism, land reform and support for the creation of agricultural cooperatives. It was critical of capitalism and also espoused antisemitism. The movement's main areas of influence were Western Moldavia, Bukovina, and Banat.The party merged with the National Italo-Romanian Cultural and Economical Movement in 1923 to form the National Fascist Movement, although a small rump movement carried on, with little significance. Both groups shared a close affinity to Italian fascism which facilitated their merger.Radu Gyr
Radu Gyr (Romanian pronunciation: [ˈradu ˈd͡ʒir]; pen name of Radu Ștefan Demetrescu [ˈradu ʃteˈfan demeˈtresku]; March 2, 1905, Câmpulung-Muscel – 29 April 1975, Bucharest) was a Romanian poet, essayist, playwright and journalist.Relationship between the Romanian Orthodox Church and the Iron Guard
The relationship between the Romanian Orthodox Church and the Iron Guard was one of ambivalence: while the Romanian Orthodox Church supported much of the fascist organization's ideology, it did not outright support the movement. Nevertheless, many individual Orthodox clerics supported the Iron Guard and spread their propaganda.
The Orthodox Church promoted its own version of nationalism which highlighted the role of Orthodoxy in preserving the Romanian identity. Starting with the 1920s, Orthodoxy became entangled with fascist politics and antisemitism: the most popular Orthodox theologian at the time, Nichifor Crainic, advocated in his magazine Gândirea a mix of Orthodoxy and nationalism, while philosopher Nae Ionescu argued that Orthodoxy is inseparable from the Romanian identity.Romanian Youth Labour
The Romanian Youth Labor (Munca Tineretului Român – MTR) was a paramilitary movement present in Romania during 1942-1944.Vasile Marin
Vasile Marin (January 29, 1904 in Bucharest – January 13, 1937 in Majadahonda) was a Romanian politician, public servant and lawyer. A member of the National Peasants' Party until 1932, Vasile Marin become a prominent member of the Iron Guard. His death in the Spanish Civil War after volunteering to fight for the Nationalists along with the subsequent death of Ion Moța is credited with contributing to the growth of the Iron Guard.Văcărești, Bucharest
Văcărești (Romanian pronunciation: [vəkəˈreʃtʲ]) is a neighbourhood in south-eastern Bucharest, located near Dâmboviţa River and the Văcăreşti Lake. Nearby neighbourhoods include Vitan, Olteniței, and Berceni. Originally a village, it was incorporated into Bucharest as it expanded. Its name is related to the Wallachian aristocratic Văcărescu family, with an etymology leading back to the Romanian văcar, "cow-herder," and the suffix -eşti.
|1st – 11th century|
|12th – 19th century|<|endoftext|>
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# The sum of three consecutive integers is 75, what are the integers?
Jun 3, 2015
Three consecutive integers can be written as $\left(x\right) , \left(x + 1\right) , \left(x + 2\right)$
$\left(x\right) + \left(x + 1\right) + \left(x + 2\right) = 75$
$3 x + 3 = 75$
$3 x = 75 - 3 = 72$
color(purple)(x = 72/ 3 = 24
so ,$x + 1 = 25 , x + 2 = 26$
the numbers are color(purple)(24,25,26
Jun 7, 2017
This is the type of puzzle that we should try to perform as mental math (in your head). The answer is $24 , 25 , 26.$
#### Explanation:
To arrive at this answer, we can avoid a lot of math formulas, and the use of a calculator by investigating the given statement and quantities.
Three consecutive integers means three numbers that follow each other like $1 , 2 , 3.$ So if we add these up we get $6$ as a sum.
But in this case the desired sum of three consecutive integers is $75$.
That means we can start by dividing the target sum $75$ by $3$.
$\frac{75}{3} = 25 \to$ you have probably memorized that one already.
That means $3 \times 25 = 75 \mathmr{and} 25 + 25 + 25 = 75$
But if we take a $1$ from the first $25$ and move it to the third $25$, we arrive at:
$24 , 25 , 26 \to$ which are three consecutive integers that sum to 75#.
If you can understand this method and are careful in using it you will save a lot of time when these types of puzzles appear.<|endoftext|>
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# Which Word Describes the Slope of the Line?
Which Word Describes the Slope of the Line?
When studying linear equations, one important concept to understand is the slope of a line. The slope tells us how steep or gradual the line is, and it plays a significant role in various real-life applications, such as calculating rates of change or determining the direction of a relationship between two variables. In this article, we will explore different words that can be used to describe the slope of a line and answer some frequently asked questions about this topic.
Slope describes the ratio of the vertical change (rise) to the horizontal change (run) between two points on a line. It tells us how much the y-value changes for each unit increase in the x-value. The slope can be positive, negative, zero, or undefined, and each of these cases has a specific word associated with it to describe the line’s slope.
1. Positive Slope:
A line with a positive slope rises from left to right. It suggests that as the x-values increase, the y-values also increase. Words to describe a positive slope include “upward,” “ascending,” “increasing,” or “rising.”
2. Negative Slope:
A line with a negative slope falls from left to right. It indicates that as the x-values increase, the y-values decrease. Words to describe a negative slope include “downward,” “descending,” “decreasing,” or “falling.”
3. Zero Slope:
A line with a zero slope is horizontal and has no steepness. It suggests that the y-values do not change as the x-values increase or decrease. The line is completely flat. Words to describe a zero slope include “horizontal,” “flat,” “level,” or “unchanging.”
See also What Do Flamingos Say
4. Undefined Slope:
An undefined slope occurs when the line is vertical. It indicates that the x-values do not change, and the y-values can be any value. Words to describe an undefined slope include “vertical” or “infinite.”
Frequently Asked Questions (FAQs):
Q1: How do I calculate the slope of a line?
To calculate the slope of a line, you need two points on the line. Let’s call them (x1, y1) and (x2, y2). The slope (m) is calculated using the formula: m = (y2 – y1) / (x2 – x1).
Q2: Can the slope be a fraction or a decimal?
Yes, the slope can be a fraction or a decimal. The slope’s value represents the ratio between the vertical and horizontal changes. Therefore, it can be any numerical value, including fractions or decimals.
Q3: What does a slope of zero mean?
A slope of zero indicates that the line is horizontal. It suggests that the y-values do not change as the x-values increase or decrease.
Q4: Is there a difference between zero slope and no slope?
No, there is no difference between zero slope and no slope. Both terms refer to a horizontal line where the y-values remain constant.
Q5: Can a line have more than one slope?
No, a line can only have one slope. The slope represents a unique characteristic of the line, describing its steepness or direction.
Q6: What does an undefined slope mean?
An undefined slope occurs when the line is vertical. It suggests that the x-values do not change, and the y-values can be any value.
Q7: How can I identify the slope from an equation?
In a linear equation in the form y = mx + b, the coefficient of x (m) represents the slope of the line. If the equation is not in this form, you may need to rearrange it to identify the slope.
See also What to Say When You Light a Candle for Someone
Understanding the slope of a line is fundamental when working with linear equations. By using the appropriate words to describe the slope, we can easily communicate the direction, steepness, and behavior of the line. Whether it is positive, negative, zero, or undefined, the slope provides valuable information about the relationship between variables and helps us interpret real-world situations with mathematical precision.
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How we discovered a new species of the 'missing link' between dinosaurs and birds
Perhaps one of the world's best known fossils is Archaeopteryx. With its beautifully preserved feathers, it has long been regarded as the first bird in the fossil record, and is often called "the icon of evolution". Only a handful of specimens have ever been found, its elusivity adding to its fascination.
But was it really the first bird – and could it really fly? Given that we now know birds descended from dinosaurs, was Archaeopteryx actually just another small dinosaur with a feathery covering?
My colleagues and I had the rare opportunity to examine an Archaeopteryx skeleton using one of the world's most powerful synchrotrons, a type of particle accelerator rather like an X-ray machine but ten thousand billion times brighter than those in a hospital. We were able to see inside the rock and uncover fragments of bone that had never been seen before. What we discovered surprised us: this was an entirely new species of Archaeopteryx.
The first Archaeopteryx fossil was uncovered in the Jurassic limestones of Bavaria, in the summer of 1861, just two years after the publication of Darwin's Origin of Species. It appeared to be one of Darwin's predicted "missing links," the link between reptiles and birds, specifically between dinosaurs and birds.
It certainly looked like a bird, with delicately preserved feathers on its wings and a fan-shaped tail. It also had a wishbone or "furcula," just like you find in a roast chicken. And both of these features were thought at the time only to occur in birds. But Archaeopteryx also had some very reptilian features – a long, bony tail, and a jaw filled with very sharp teeth – and seemed to be part way between the two groups of animals.
A second specimen was found in 1876 but Archaeopteryx fossils have generally remained elusive and we still have discovered only 12 specimens. One of these has been lost and several others are fragments.
Despite more than 150 years of study, we still have much to discover about this primitive bird. Much of the controversy surrounds the question of whether Archaeopteryx could fly, the consensus being that it was at best a "feeble flapper."
The eighth specimen to be discovered was one of the least well-known. It was found by a private collector in a quarry near Daiting in Bavaria, Germany, in the early 1990s, and changed hands several times before being sold on cheaply to another private collector in the belief that it was a common pterosaur. It is rumoured that the finder, on realising what he had done, threw himself off the top of the quarry, as these specimens can change hands for millions of dollars.
The new owner, perhaps feeling guilty, was very secretive about his purchase, and the scientific community was unaware of this specimen until 1996 (hence its nickname, "the phantom") when a cast of the specimen was shown briefly at the Naturkundemuseum in Bamberg, Germany. Then, in 2009, dinosaur hunter Raimund Albersdörfer, also from Bavaria, purchased the specimen and in 2011 made it available to our research team for scientific study.
It is not well preserved and we nicknamed it "the ugly bird," for the bones had become disconnected after it died and were jumbled together and crushed. The lower half of the body was missing completely. But we took it to Grenoble, to the European Synchrotron Radiation Facility (ESRF), where we were able to perform a virtual dissection of the skeleton, and piece together fragments of bones that had been broken when the animal died 150m years ago.
All of the specimens of Archaeopteryx had been classified as a single species, Archaeopteryx lithographica, but this one seemed to be different. It had also been collected from a different layer in the rock record than the others, which had all come from the well-known Solnhofen Formation. Our specimen was collected from the Moernsheim Formation, which lies above the Solnhofen Formation and is considered to be around half a million years younger.
Closer to birds
We first noticed that the skull bones were fused together. Then we spotted that the furcula had a knob-like extension, which in modern birds is an attachment for the powerful flight muscles. Neither of these features were seen in Archaeopteryx lithographica. An expanded opening in one of the bones of the shoulder-girdle (called the coracoid) suggested that a large nerve passed through this bone, again suggesting a more powerful wing stroke.
But our biggest surprise was when we noticed that the bones of the wrist were fused together – another feature of modern birds known as the "carpo-metacarpus." This provided a rigid structure that would have allowed the wings to make a powerful downbeat.
Archaeopteryx lithographica possessed instead the flexible wrist seen in its dinosaur ancestors. It was obvious that our specimen was showing many of the characteristics seen in modern flying birds that were not seen in the "feebly flapping" Archaeopteryx lithographica. We had discovered a species new to science, and we named it Archaeopteryx albersdoerferi after its owner who had rescued it from obscurity.
Whenever a missing link is discovered, this merely creates two further missing links: what came before, and what came after. In this case, what came before Archaeopteryx was discovered in 1996 with the description of feathered dinosaurs in China. Our new species is what came after. It confirms Archaeopteryx as the first bird and not just one of a number of feathered theropod dinosaurs, as some researchers have suggested recently. Our research puts Archaeopteryx back on its perch as the first bird.<|endoftext|>
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# Acceleration Vs. Deceleration: 7 Facts You Should Know
Let’s see the difference between acceleration ad deceleration.
Acceleration is a rate of change of velocity within a required time along a positive or negative direction, but what is deceleration? Whenever an object slows down during its motion, it decelerates. So we can say that in acceleration and deceleration, velocity is changing with respect to time, but What is the difference between Acceleration Vs Deceleration?
## Acceleration: Detailed analysis
Acceleration is a vector quantity. When the body changes its velocity by changing magnitude (Speed) or direction, it accelerates. SI unit of acceleration is m/s2. Acceleration is always in the order of applied force on a body. It can be positive or negative. Whether the acceleration of a body is positive or negative is depends upon the following two factors,
1. the coordinate system used to describe the motion
2. weather the body speeding up or slowing down
To learn more about negative acceleration, see the post on negative acceleration
## Vector form of acceleration
We know that acceleration is a rate of change of velocity to time,
If we resolve the velocity vector, i.e.
Where
[latex]v_{x},v_{y},v_{z}, are component of velocity along x, y, z respectively [/latex] and i, j, k are the unit vector in the x, y, z-axis, respectively.
Then,
We can find out the average acceleration or an instantaneous acceleration of a moving body. Now, what is average acceleration? And what is fast acceleration?
## Average acceleration
Consider a car moving with a continuously changing velocity. Its average acceleration is the ratio of a total speed change during a motion to the required time to complete that motion. Mathematically it is represented as,
where aave – Averege acceleration
vf – final velocity
vi – initial velocity
## Instantaneous acceleration
Instantaneous acceleration is acceleration in a particular time instant during motion. Mathematically it is denoted as,
## Deceleration
A moving object slows down by losing its speed, and the rate by which it loses its velocity is called deceleration. Deceleration is always in the opposite direction of the velocity of the moving object. The SI unit of deceleration is m/s2. The following formulae give the magnitude of deceleration,
Therefore,
v- final velocity
u- initial velocity
t- time of motion
Here, the negative sign indicates that deceleration is opposite to the direction of acceleration.
## Sign of Acceleration & Deceleration
As we discussed earlier, a sign of acceleration depends upon two factors, namely, the coordinate system chosen to describe the motion and whether the object is speeding up or slowing down. Deceleration is also dependent upon these two factors.
To understand this correctly, consider the motion of a car in one dimension. We can divide the motion of a car into four cases; these are as follows,
Case1-A car is moving in a forward direction and speeding up
As the vehicle is speeding up in a forward direction, deceleration is zero, and acceleration is in the direction of velocity. Both velocity and acceleration are positive in this condition.
Case2– A car is moving in positive x-direction and slowing down.
As we know, deceleration is always associated with the slowing down of motion. In this case, the car is slowing down while moving in + x-direction. The friction force is responsible for the slowing down of the car and is in – x-direction; therefore, the acceleration of a car is also in – x-direction. As the car is slowing down, it means a car is decelerating in the negative x-direction. Therefore, in this case, we can say that negative acceleration and deceleration are the same.
Case3– A car is moving in a negative X- direction and slowing down.
In this case, the car is decelerating, and its direction is along the positive x-axis. Hence the deceleration of the car is positive; this proves that slowdown can be positive.
Case4– A car is moving in negative x-direction and speeding up
In this case, the vehicle is speeding up in –ve x-direction, so it has no deceleration. As the direction of motion is negative, so the acceleration of the car is also negative.
Hence deceleration and negative acceleration are not necessarily equal.
## A man starts to walk on a road with a velocity of 0.5 m/s. After 4 minutes, its velocity is 2 m/s. What is the acceleration of that man during motion?
Solution-
Given, initial velocity, u = 0.5 m/s
Final velocity, v = 2 m/s
Time, t = 4 minutes = 4×60 = 240 sec
To find: acceleration of a man
By using the formulae for average acceleration,
0.00625 m/s2 is the acceleration of a man within the motion.
## A car moves with a uniform velocity of 40 km/hr from point A and stops at point B. A car completes its motion from point A to B within 5 hours. What will be the deceleration of the car in km/hr2?
Solution,
Given the initial velocity of the car, u = 40 km/hr
Final velocity of car, v = 0 km/h
Time to complete the motion, t = 5 hr
To find: deceleration of a car
By using the formulae of deceleration
Here, the negative sign indicates the direction of deceleration
= – 8 km/hr2
-8 km/ hr2 is the deceleration of the car within motion.
Shambhu Patil
I am Shambhu Patil, a physics enthusiast. Physics always intrigues me and makes me think about, how this universe works? I have an interest in nuclear physics, quantum mechanics, thermodynamics. I am very good at problem solving, explaining complex physical phenomenon in simple language. My articles will walk you through each and every concept in detail. Join me over LinkedIn to https://www.linkedin.com/in/shambhu-patil-96012b1a1 . E-mail :- [email protected]<|endoftext|>
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Though language development varies from child to child, most children follow predictable patterns of language development as they grow up. Language development progresses rapidly at the age of 6, so your child’s language skills will grow more advanced during the year. If you think your child has fallen behind in language development, consult a physician who specializes in child development.
By the age of 6, your child should be able to pronounce most of the sounds in her native language. According to the Child Development Institute, a 6-year-old should be able to pronounce all vowels and diphthongs (complex vowel sounds) and almost all consonants. Most 6-year-olds have mastered the consonant sounds of b, d, f, g, h, k, m, n, p, t, v, w, y, ng, sh, zh and th. Six-year-olds might still have trouble pronouncing s, z, th, ch, wh and the soft g.
Time and Numbers
Most 6-year-olds have a strong understanding of the concept of numbers and can count up to seven. Six-year-olds also understand simple time concepts, such as different times of day. They can express relative time with words such as before and after or yesterday, today and tomorrow.
Six-year-olds can usually put together complex sentences with multiple clauses. Six-year-olds should also be able to form grammatically correct sentences. Most 6-year-olds can speak fluently with clear, intelligible enunciation. By the age of 6, children should have the confidence to communicate with people in a variety of social settings.
Story Telling and Literacy
Most 6-year-olds have developed the ability to provide stories, descriptions or explanations relating to a specific picture or illustration. They can describe the connections and relationships between separate events and objects. Most 6-year-olds cannot yet read or write independently, though they often can memorize songs and nursery rhymes. Some 6-year-olds might begin to read on their own.
Encouraging Language Development
Encourage your child’s language development by reading poems or stories to her on a daily basis. Ask engaging questions about the story and illustrations. Encourage your child’s use of descriptive language by playing guessing games or “I Spy” games or by looking at pictures and artwork together.<|endoftext|>
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# How do you find the area of a rectangle and circle?
Category: science space and astronomy
5/5 (1,509 Views . 28 Votes)
The formula to find the area of a rectangle is egin{align*}A = lwend{align*}. The formula to find the area of a circle is egin{align*}A= pi r^2end{align*}. To find the area of a composite shape, simply find the area of each individual shape and add them together.
Accordingly, how do you find the area of shapes?
To find the area of a square or rectangle, just multiply the width of the shape by its height. To find the area of a circle, start by measuring the distance between the middle of the circle to the edge, which will give you the radius. Then, square the radius and multiply it by pi to find the area.
Also, how do you find the maximum area of a rectangle inscribed in a circle? Since x must be positive, then x = r/√2. Thus, x = r/√2 and y = r/√2. Solving for the width and height and noting 2r is equal to the diameter d we have: The width and height have the same length; therefore, the rectangle with the largest area that can be inscribed in a circle is a square.
Keeping this in consideration, what is the formula for area of a sector of a circle?
The area of a sector of a circle is ½ r² ∅, where r is the radius and ∅ the angle in radians subtended by the arc at the centre of the circle. So in the below diagram, the shaded area is equal to ½ r² ∅ .
What is the area of an irregular shape?
There are times when you need to find the area of a shape that is not a regular shape. One method for finding the area of an irregular shape is to divide the shape into smaller shapes which you do have the formula for. Then you find the area of all of the smaller shapes and add all of your areas together.
### What is the area of this trapezoid?
Explanation: To find the area of a trapezoid, multiply the sum of the bases (the parallel sides) by the height (the perpendicular distance between the bases), and then divide by 2.
### How do you find the area calculator?
Calculate the Area as Square Footage
1. If you are measuring a square or rectangle area, multiply length times width; Length x Width = Area.
2. For other area shapes, see formulas below to calculate Area (ft2) = Square Footage.
### What is the formula for circumference?
The circumference = π x the diameter of the circle (Pi multiplied by the diameter of the circle). Simply divide the circumference by π and you will have the length of the diameter. The diameter is just the radius times two, so divide the diameter by two and you will have the radius of the circle!
### What is arc of a circle?
Arc Of A Circle. The arc of a circle is a portion of the circumference of a circle. The formula for finding arc length in radians is where r is the radius of the circle and θ is the measure of the central angle in radians.
### How do I find the length of an arc?
To find arc length, start by dividing the arc's central angle in degrees by 360. Then, multiply that number by the radius of the circle. Finally, multiply that number by 2 × pi to find the arc length.
### Can a rectangle fit in a circle?
Benneth, Actually - every rectangle can be inscribed in a (unique circle) so the key point is that the radius of the circle is R (I think). One of the properties of a rectangle is that the diagonals bisect in the 'center' of the rectangle, which will also be the center of the circumscribing circle.
### What is a circle inside a triangle called?
In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. The center of the incircle is a triangle center called the triangle's incenter.
### How do you find the largest area of a rectangle?
A rectangle will have the maximum possible area for a given perimeter when all the sides are the same length. Since every rectangle has four sides, if you know the perimeter, divide it by four to find the length of each side. Then find the area by multiplying the length times the width.
### What is the area of the largest rectangle that can be inscribed in the ellipse?
Thus the maximum area of a rectangle that can be inscribed in an ellipse is 2ab sq. units. Here a = 1,b=1/2. Thus area = 2.1.
### Can a parallelogram be inscribed in a circle?
For a quadrilateral to be inscribed in a circle, its opposite angles must be supplementary. For a quadrilateral to be a parallelogram, it's opposite angles must be equal. Therefore, for a parallelogram to be inscribed in a circle, it must have four right angles, i.e. be a rectangle.
### How many circles can fit in a rectangle?
If we make 257 the vertical dimension, then the rectangle is a bit over 144⋅r units tall, and a bit over (2+49√3)⋅r units wide. So we can arrange the circles in 50 columns that alternate between 72 and 71 circles, for 25⋅72+25⋅71=3575 circles.<|endoftext|>
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# How do you graph -10x+15y=60 using intercepts?
Aug 19, 2017
See a solution process below:
#### Explanation:
x-intercept
To find the $x$-intercept we set $y$ to $0$ and solve for $x$:
$- 10 x + \left(15 \cdot 0\right) = 60$
$- 10 x + 0 = 60$
$- 10 x = 60$
$\frac{- 10 x}{\textcolor{red}{- 10}} = \frac{60}{\textcolor{red}{- 10}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 10}}} x}{\cancel{\textcolor{red}{- 10}}} = - 6$
$x = - 6$ or $\left(- 6 , 0\right)$
y-intercept
To find the $y$-intercept we set $x$ to $0$ and solve for $y$:
$\left(- 10 \times 0\right) + 15 y = 60$
$0 + 15 y = 60$
$15 y = 60$
$\frac{15 y}{\textcolor{red}{15}} = \frac{60}{\textcolor{red}{15}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{15}}} y}{\cancel{\textcolor{red}{15}}} = 4$
$y = 4$ or $\left(0 , 4\right)$
We can now plot the two points and draw a line through them to graph the equation:
graph{((x+6)^2+y^2-0.025)(x^2+(y-4)^2-0.025)(-10x+15y-60)=0}<|endoftext|>
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# Missing-digit sum
Missing-digit sums are integer numbers that are equal to the sum of numbers created by deleting one or more digits at a time from the original number. For example, the OEIS lists these two integers as missing-digit sums in base ten:
1,729,404 = 729404 (missing 1) + 129404 (missing 7) + 179404 (missing 2) + 172404 + 172904 + 172944 + 172940
1,800,000 = 800000 (missing 1) + 100000 (missing 8) + 180000 (missing first 0) + 180000 + 180000 + 180000 + 180000[1]
Missing-digit sums are therefore a subset of narcissistic numbers, when these are defined as numbers that are equal to some manipulation of their own digits (for example, 153 and 132 are narcissistic numbers in base ten because 153 = 13 + 53 + 33 and 132 = 13 + 32 + 12 + 31 + 23 + 21).
## Dropping two and more digits
When one digit is dropped from a d-digit integer, there are d integers in the sum and each is d-1 digits long. In general, when n digits are dropped from a d-digit integer, the number of integers in the sum is equal to d! / (n!(d - n)!), or the combination of n digits taken 2, 3, 4... at a time. For example, when d = 20 and n = 3, there are 20! / (3!(20 - 3)!) = 1,140 integers in the sum. In base ten, the integers31171093 1523163197662495253514 and 47989422298181591480943 are equal to their missing-digit sums when dropping two, three and four digits, respectively. Here is the delete-2 sum, containing 12! / (2!(12 - 2)!) = 66 integers:
183477122641 = 3477122641 (missing 1 and 8) + 8477122641 (missing 1 and 3) + 8377122641 (missing 1 and 4) + 8347122641 (missing 1 and first 7) + 8347122641 (missing 1 and second 7) + 8347722641 (missing 1 and second 1) + 8347712641 + 8347712641 + 8347712241 + 8347712261 + 8347712264 + 1477122641 + 1377122641 + 1347122641 + 1347122641 + 1347722641 + 1347712641 + 1347712641 + 1347712241 + 1347712261 + 1347712264 + 1877122641 + 1847122641 + 1847122641 + 1847722641 + 1847712641 + 1847712641 + 1847712241 + 1847712261 + 1847712264 + 1837122641 + 1837122641 + 1837722641 + 1837712641 + 1837712641 + 1837712241 + 1837712261 + 1837712264 + 1834122641 + 1834722641 + 1834712641 + 1834712641 + 1834712241 + 1834712261 + 1834712264 + 1834722641 + 1834712641 + 1834712641 + 1834712241 + 1834712261 + 1834712264 + 1834772641 + 1834772641 + 1834772241 + 1834772261 + 1834772264 + 1834771641 + 1834771241 + 1834771261 + 1834771264 + 1834771241 + 1834771261 + 1834771264 + 1834771221 + 1834771224 + 1834771226
## Trivial missing-digit sums
In any base b, there will be a set of delete-1 missing-digit sums with b+1 digits and first digits in the range 1...b-1, followed by b zeroes. Consider base-2 and base-3:
100b=2 = 4b=10 = 00b=2 (deleting 1) + 10b=2 (deleting first 0) + 10b=2 (deleting second 0)
= 0 + 10b=2 + 10b=2
= 0 + 10b=2 x 10b=2 = 100b=2
= 0 + 2 x 2 = 4b=10
1000b=3 = 27b=10 = 000b=3 (deleting 1) + 100b=3 (deleting first 0) + 100b=3 (deleting second 0) + 100b=3 (deleting third 0)
= 0 + 100b=3 + 100b=3 + 100b=3
= 0 + 10b=3 x 100b=3 = 1000b=3
= 0 + 3 x 9 = 27b=10
2000b=3 = 54b=10 = 000b=3 (deleting 2) + 200b=3 (deleting first 0) + 200b=3 (deleting second 0) + 200b=3 (deleting third 0)
= 0 + 200b=3 + 200b=3 + 200b=3
= 0 + 10b=3 x 200b=3 = 2000b=3
= 0 + 3 x 18 = 54b=10
Accordingly, the eleven-digit numbers 10,000,000,000 through 90,000,000,000 are trivial missing-digit sums in base ten, because their sums take this form:
10,000,000,000 = 0,000,000,000 (deleting 1) + 1,000,000,000 (deleting first 0) + 1,000,000,000 (deleting second 0) + 1,000,000,000 (deleting third 0) + 1,000,000,000 + 1,000,000,000 + 1,000,000,000 + 1,000,000,000 + 1,000,000,000 + 1,000,000,000 + 1,000,000,000
= 0 + (10 x 1,000,000,000) = 10,000,000,000
And in hexadecimal or base sixteen, the seventeen-digit numbers 10,000,000,000,000,000 through F0,000,000,000,000,000 are missing-digit sums of the same form.
## Searching for missing-digit sums
Searching for delete-1 missing-digit sums is simplified when one notes that the final two digits of n determine the final digit of its missing-digit sum. One can therefore test simply the final two digits of a given n to determine whether or not it is a potential missing-digit sum. In this way, the search-space is considerably reduced. For example, consider the set of seven-digit base-ten numbers ending in ...01. For these numbers, the final digit of the sum is equal to (digit-0 x 1 + digit-1 x 6) modulo 10 = (0 + 6) mod 10 = 6 mod 10 = 6. Therefore no seven-digit number ending in ...01 is equal to its own missing-digit-sum in base ten.
Now consider the set of seven-digit numbers ending in ...04. For these numbers, the final digit of the sum is equal to (0 x 1 + 4 x 6) modulo 10 = (0 + 24) mod 10 = 24 mod 10 = 4. This set may therefore contain one or more missing-digit sums. Next consider seven-digit numbers ending ...404. The penultimate (last-but-one) digit of the sum is equal to (2 + 4 x 2 + 0 x 4) modulo 10 = (2 + 8 + 0) mod 10 = 10 mod 10 = 0 (where the 2 is the tens digit of 24 from the sum for the final digit). This set of numbers ending ...404 may therefore contain one or more missing-digit sums. Similar reasoning can be applied to sums in which two, three and more digits are deleted from the original number.
## Incomplete list of missing-digit sums in base ten
### Delete-1 sums
1729404, 1800000, 13758846, 13800000, 14358846, 14400000, 15000000, 28758846, 28800000, 29358846, 29400000, 1107488889, 1107489042, 1111088889, 1111089042, 3277800000, 3281400000, 4388888889, 4388889042, 4392488889, 4392489042, 4500000000,[2] 5607488889, 5607489042, 5611088889, 5611089042, 7777800000, 7781400000, 8888888889, 8888889042, 8892488889, 8892489042, 10000000000, 20000000000, 30000000000, 40000000000, 50000000000, 60000000000, 70000000000, 80000000000, 90000000000
### Delete-2 sums
167564622641, 174977122641, 175543159858, 175543162247, 183477122641, 183518142444, 191500000000, 2779888721787, 2784986175699, 212148288981849, 212148288982006, 315131893491390, 321400000000000, 417586822240846, 417586822241003, 418112649991390, 424299754499265, 424341665637682, 526796569137682, 527322398999265, 533548288981849, 533548288982006, 636493411120423, 636531893491390, 642800000000000, 650000000000000, 738986822240846, 738986822241003, 739474144481849, 739474144482006, 739474144500000, 739512649991390, 745699754499265, 745741665637682, 746186822240846, 746186822241003, 751967555620423, 848722398999265, 849167555620423, 854948288981849, 854948288982006, 855396569137682, 862148288981849, 862148288982006, 957893411120423, 957931893491390, 965131893491390, 971400000000000
### Delete-3 sums
124611932292235425, 257559932292235425, 273161719965897657, 2159824675153518576, 5751345323192555691
### Delete-4 sums
1523163197662495253514, 47989422298181591480943, 423579919359414921365511, 737978887988727574986738
## References
1. ^ Jon Ayres. "Sequence A131639". Neil Sloane. Retrieved 10 March 2014.
2. ^ Jon Ayres. "Sequence A131639". Neil Sloane. Retrieved 10 March 2014.<|endoftext|>
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What is Obsessive Compulsive Disorder (OCD)?
Obsessive compulsive disorder (OCD) is a condition in which a person’s life is interrupted or ruled by a cycle of obsessions and compulsions. It’s a common, yet misunderstood condition. Many people self-diagnose themselves as “OCD” when casually remarking on preferences concerning things like cleanliness. An official OCD diagnosis is characterized by severe distress and an inability to function normally if conditions imposed by their illness are not met.
Compulsions are defined as repetitive behaviors that an individual with OCD ritualizes, such as turning a light switch on and off or washing their hands until they are raw. Obsessions, on the other hand, are clinically defined as the irrational thoughts or fears that someone with OCD could experience. These obsessions could include wanting everything to be symmetrical or having an irrational fear of dirt.
OCD affects around 2% of the population, and it is equally prevalent among males and females. However, specific types of obsessive compulsive disorders, such as pulling out hair, may be more common in females.
People suffering from obsessive compulsive disorder can experience obsessions and compulsions, either concurrently or separately. An effort to try and abstain from their habits or compulsions can cause an individual anxiety. This often makes it “easier” for someone affected with OCD to continue participating in compulsive behavior, rather than struggling with the bouts of intense anxiety that often follow attempts to return to normal functioning without professional help.
Getting a Diagnosis for Obsessive Compulsive Disorder (OCD)
Getting a diagnosis for obsessive compulsive disorder (OCD) can sometimes be a lengthy process. Medical providers often run a series of tests and evaluations to ensure that there isn’t a physical reason for OCD-like symptoms to manifest. Similarly, OCD can be masked within a number of psychiatric conditions, such as depression, anxiety, and schizophrenia, making an accurate diagnosis established by a qualified professional crucial to successful treatment.
Psychological evaluation often involves a patient undergoing therapy so that their psychiatrist or therapist can better understand the context in which their obsessions and compulsions exist.
Treatments for Obsessive Compulsive Disorder (OCD)
Once a patient has been given a professional diagnosis of having obsessive compulsive disorder, there are a number of treatments available that may help manage their condition.
As with most psychiatric conditions, therapy is often a first-line treatment option. One particular method, cognitive behavioral therapy (CBT), is highly-effective in people with OCD as it uses “exposure and response prevention therapy” in combination with “cognitive therapy” to effectively shift the way an individual reacts to their triggers and compulsions.
There are several options when it comes to medications that can help patients with OCD by minimizing their obsessive compulsive behaviors. In most cases, antidepressants are also FDA-approved to treat OCD, so doctors may try a series of antidepressant prescriptions to find the right option for their patient.
Transcranial Magnetic Stimulation (TMS) for Obsessive Compulsive Disorder
As of late 2018, the FDA has approved transcranial magnetic stimulation as a treatment option for those diagnosed with OCD. TMS can modulate the anterior cingulate cortex, a region of the brain that is correlated with OCD symptoms.
Patients who have not responded to other treatment options, like medication and or therapy, may benefit from TMS as it is helpful in stimulating the brain, often resulting in significant symptom relief.
For more information about how transcranial magnetic stimulation could help you manage your OCD when other treatments have failed, or to schedule your consult at TMS & Brain Health, please contact us.<|endoftext|>
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ELA Learning Targets:
Students will distinguish their own point of view from that of the narrator or those of the characters. Students will determine how their point of view is similar or different from the characters in the story.
During centers students will be practicing on IXL, I-Station, reviewing text features. Students will learn how to determine their point of view as well as the narrator’s point of view in a story. Students will be reading “Wolf!” during center time. IXL: X.1, X.2, U.1, U.5, V.2, Z.1, Y.3
Science Learning Targets: What are some physical properties? How are mass and volume measured? What are the states of matter? How can the state of matter change?
Science Learning Activities: Unit 3 Science Fusion. Students will work on inquiry science learning activities and sort through different types of matter, measure and compare mass and volume and describe the changes of states of matter. Students will be participating in virtual labs and inquiry experiments.
Math Learning Targets: Students will identify regions that have been divided into equal-sized parts and divide regions into equal-sized parts. They will be able to describe a fractional part of a set.
Math Learning Activities: Students will be practicing diving shapes into equal parts, naming the fractional parts within a set, and using fraction strips and number lines to represent fractions.
IXL: W.1-6, W.9, W.11
Social Studies: Students will learn about landforms on maps and create their own topographic maps. We will also discuss what population means and will read an article about China’s population.
Passion: A very strong feeling or liking.
Admire: Respect or think well of.
Concentrate: Pay attention or think carefully about something.
Splendid: Very good or beautiful.
Bothering: Giving someone trouble or annoying them.
Dangerous: Likely to cause harm.
Ached: Felt a dull and steady pain.
Challenge words: Catcher, Sandwich<|endoftext|>
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Home
# Event probability
### Probability of events (Pre-Algebra, Probability and
• Types of Events in Probability: Impossible and Sure Events. If the probability of occurrence of an event is 0, such an event is called an impossible... Simple Events. Any event consisting of a single point of the sample space is known as a simple event in probability. Compound Events. Contrary to.
• Probability of events Probability is a type of ratio where we compare how many times an outcome can occur compared to all possible outcomes. Probability=\frac{The\, number\, of\, wanted \, outcomes}{The\, number \,of\, possible\, outcomes}\$
• g part
### Event probability - Minita
• Mutually Exclusive Events More Lessons On Probability. What Is An Event In Probability? In an experiment, an event is the result that we are interested in. The probability of an event A, written P(A), is defined as. Example: When a fair dice is thrown, what is the probability of getting a) the number 5 b) a number that is a multiple of
• Probability: Types of Events Events. When we say Event we mean one (or more) outcomes. Rolling a 5 is an event. Let's look at each of those types. Independent Events. Events can be Independent, meaning each event is not affected by any other events. This is an... Dependent Events. But some.
• The probability of an event is the number of favorable outcomes divided by the total number of outcomes
• Probability of an Event Very simply put, a probability is the chance of something happening. Probability is a measure of the likelihood of a given event's occurrence. There are events which we cannot predict with certainty, so we find out the probability of their occurrence
• Probability is the likliehood that a given event will occur and we can find the probability of an event using the ratio number of favorable outcomes / total number of outcomes
• Another event, event B, could be to throw an odd number with the die, so B ={1,3,5}. Probability. Referring to the event A and the event B defined above, we can denote the following probabilities: The probability of event A occurring is 0.5 is denoted P(A)=0.5 The probability of event B occurring is 0.5 is denoted P(B)=0.5 Venn diagram
• e the total number of outcomes for the first event. Find the probability of the first event. Deter
Tutorial on how to solve for the probability of a simple event. About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features. Probability is simply how likely something is to happen, probability theory applies precise calculations to quantify uncertain measures of random events. Use our online probability calculator to calculate the single and multiple event probability based on number of possible outcomes
### Probability Of An Event (video lessons, examples and
• Chapter 3 Probability and random events. Probability is a method of mathematically modeling a random process so that we can understand it and/or make predictions about its future results. Probability is an essential tool for casinos, as well as for banks, insurance companies, and any other businesses that manage risks
• What is an Event? In probability, the set of outcomes from an experiment is known as an Event. So say for example you conduct an experiment by tossing a coin. The outcome of this experiment is the coin landing 'heads' or 'tails'. These can be said to be the events connected with the experiment. So when the coin lands tails, an event can be said to have occurred
• The event, E, is any subset of the sample space, S that is a set of outcomes which does not necessarily need to have all outcomes for random phenomena
• Probability is the chance that a certain event or series of events will occur
• Outcome and event are not synonymous. Yes, an outcome is the result of a random experiment, like a rolling a die has six possible outcomes (say). However, an event is a set of outcomes to which a probability is assigned. One possible event is rolling a number less than 3
• Probability is chance to happen of something/events. Probability equals with number of something happen divide by total numbers of outcomes. Probability of independent event means probability that is not related or affected with previous or next events. Some examples of independent events are: tossing coins, rolling dices, deck of cards, and combined between them. If there are two independent.
• e the probability that an event does not happen. As in the previous section, consider the situation of rolling a six-sided die and first compute the probability of rolling a.
### Probability: Types of Events - mathsisfun
You can use the following steps to calculate the probability: Step 1: Identify an event with one result. Step 2: Identify the total number of results that can occur. Step 3: Divide the number of favourable events by the total number of possible outcomes Home > Probability > Exhaustive Events. Exhaustive Events. Jalal Afsar September 6, 2012 Probability No Comments. Definitions. When a sample space is distributed down into some mutually exclusive events such that their union forms the sample space itself, then such events are called exhaustive events. OR. When two or more events form the sample space collectively than it is known as. Event Examples. If a single face is considered when a die is rolled, then it will be simple event. For example suppose getting 5 or 6 or 3 or 2 etc on the die when it is thrown, is called as simple event. If the event is any even number on the die, then the event is consist of points {2, 4, 6}, which is known as compound event
### Finding the Probability of an Event Prealgebr
1. Probability Questions with Solutions. Tutorial on finding the probability of an event. In what follows, S is the sample space of the experiment in question and E is the event of interest. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E
2. Posted in Probability, Statistics and Probability Tagged Probability of a single event, Probability of combined events Post navigation Greater than, less than or equal to 0.
3. The Single Event Probability Calculator uses the following formulas: P(E) = n(E) / n(T) = (number of outcomes in the event) / (total number of possible outcomes) P(E') = P(not E) = 1 - P(E) Where: P(E) is the probability that the event will occur, P(E') is the probability that the event will not occur, n(E) is the number of outcomes in the event E, n(T) is the total number of possible outcomes.
Probability explained | Independent and dependent events | Probability and Statistics | Khan Academy - YouTube. Probability explained | Independent and dependent events | Probability and. In general, the revisedprobability that an event Ahas occurred, taking into account the additional information that another event Bhas definitely occurred on this trial of the experiment, is called the conditional probability ofAgivenBand is denoted by P(A|B) Conditional Probability for Mutually Exclusive Events. In probability theory, mutually exclusive events Mutually Exclusive Events In statistics and probability theory, two events are mutually exclusive if they cannot occur at the same time. The simplest example of mutually exclusive are events that cannot occur simultaneously. In other words, if one event has already occurred, another can. Dependent and Independent Events - Probability. Probability theory is an important topic for those who study mathematics in higher classes. For example, Weather forecast of some areas says that there is a fifty percent probability that it will rain today. The probability is a chance of some event to happen In an experiment, an event is the result that we are interested in. The probability of an event A, written P (A), is defined as. Example: When a fair dice is thrown, what is the probability of getting. a) the number 5. b) a number that is a multiple of 3. c) a number that is greater than 6
The probability of an event tells us how likely that event is to occur. We usually write probabilities as fractions or decimals. For example, picture a fruit bowl that contains five pieces of fruit - three bananas and two apples. If you want to choose one piece of fruit to eat for a snack and don't care what it is, there is a ${\Large\frac{3}{5}}$ probability you will choose a. In probability theory, an event is a set of outcomes (a subset of the sample space) to which a probability is assigned.Typically, any subset of the sample space is an event (i.e. all elements of the power set of the sample space are events), but when defining a probability space it is possible to exclude certain subsets of the sample space from being events (see §2, below) Simple Probability expresses the probability of one event occurring, and is often visually expressed using coins, dice, marbles, or spinner. Compound Probability describes the chances of more than. Independent and dependent events. Independent probability. Up to this point, we've been focusing on independent events, which are events that don't effect one another.For example, if I flip a coin two times in a row, the result of the first flip doesn't effect the second flip, so those flips are independent events
Probability of Two Events Complement of A and B. Given a probability A, denoted by P (A), it is simple to calculate the complement, or the... Intersection of A and B. The intersection of events A and B, written as P (A ∩ B) or P (A AND B) is the joint... Union of A and B. In probability, the union. Best online Probability Calculator. Probability is simply how likely something is to happen, probability theory applies precise calculations to quantify uncertain measures of random events. Use our online probability calculator to calculate the single and multiple event probability based on number of possible outcomes The probability that a student is taking art or English is 0.833 or 83.3%. When we calculate the probability for compound events connected by the word or we need to be careful not to count the same thing twice. If we want the probability of drawing a red card or a five we cannot count the red fives twice. If we want the probability a.
### Probability of an Event : Videos, Formulae, Methods with
Probability. Probability implies 'likelihood' or 'chance'. When an event is certain to happen then the probability of occurrence of that event is 1 and when it is certain that the event cannot happen then the probability of that event is 0. Hence the value of probability ranges from 0 to 1. Probability has been defined in a varied manner by. Probability Definition. The probability of any event is defined as the chance of occurrence of the events to the total possible outcomes. If there are 'n' exhaustive, mutually exclusive and equally likely outcomes of a random experiment.Out of which, 'm' are favorable to the occurrence of an event E Because the total probability for a sample space must be equal to 1, the probabilities of complementary events must sum to 1. In symbols, p(A)+p(A C)=1. As a result, p(A C)=1-p(A). In words, the probability that an event does not happen is equal to one minus the probability that it does Probability is: (Number of ways it can happen) / (Total number of outcomes) Dependent Events (such as removing marbles from a bag) are affected by previous events. Independent events (such as a coin toss) are not affected by previous events. We can calculate the probability of two or more Independent events by multiplying
Outcome and event are not synonymous. Yes, an outcome is the result of a random experiment, like a rolling a die has six possible outcomes (say). However, an event is a set of outcomes to which a probability is assigned. One possible event is rolling a number less than 3. See the Wikipedia page for probability theory and probability space. In probability, two events are independent if the incidence of one event does not affect the probability of the other event. If the incidence of one event does affect the probability of the other event, then the events are dependent. Determining the independence of events is important because it informs whether to apply the rule of product to calculate probabilities The probability of an event is the chance that the event will occur in a given situation. The probability of getting tails on a single toss of a coin, for example, is 50 percent, although in statistics such a probability value would normally be written in decimal format as 0.50. The individual probability values of multiple events can be combined to determine the probability of a specific. Definition of Probability using Sample Spaces . When an experiment is performed, we set up a sample space of all possible outcomes.. In a sample of N equally likely outcomes we assign a chance (or weight) of 1/N to each outcome.. We define the probability of an event for such a sample as follows:. The probability of an event E is defined as the number of outcomes favourable to E divided by. The comment by Dilip Sarwate points to conditioning on the level of densities which can be interpreted as conditioning on a family of events of probability zero. It is a probabilistic version of Radon-Nikodym derivative.. One can also condition on an individual event of probability zero, if that event admits a natural approximation by events of positive probability
Probability is the study of events and how likely they are to happen. This likelihood is usually expressed as a fraction. The denominator expresses the total number of possible events in a given situation while the numerator expresses the number of ways that the indicated event can happen. Sometimes this fraction is converted to a decimal or a percentage, depending on the situation. The two. In general, the revised probability that an event A has occurred, taking into account the additional information that another event $$B$$ has definitely occurred on this trial of the experiment, is called the conditional probability of $$A$$ given $$B$$ and is denoted by $$P(A\mid B)$$. The reasoning employed in this example can be generalized to yield the computational formula in the. Name: Probability of events union. Explanation: Used to represent the probability of event A or event B. P (A | B) Name: Conditional probability function. Explanation: Used to represent the probability of Event A. f (x) Name: Probability density function. Probability is both theoretical and practical in terms of its applications Probability Questions with Solutions. Tutorial on finding the probability of an event. In what follows, S is the sample space of the experiment in question and E is the event of interest. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E Single-event probability is used to find the probability for a single event that occurs for an experiment. For example, consider tossing a coin, we will get single event (either head or tail) as expected result. Formula: Probability that event A occurs P(A) = n(A) / n(S). Probability that event A does not occur P(A') = 1 - P(A). where, n(A) - number of event occurs n(S) - number of possible.
Conversely, for conditional probability of event B with respect to event A, probability of event A can never be zero. If two events can never occur simultaneously, they are termed as mutually exhaustive events, that is A∩B= ф. The formula for finding the probability of two events occurring simultaneously is derived from the multiplication theorem of probability. A∩B is represented by the. The probability of an event A, denoted by P(A), is the sum of the probabilities of the corresponding elements in the sample space. For rolling an even number, we have P(A) = p(x 2) + p(x 4) + p(x 6) = 1 2 Given an event Aof our sample space, there is a complementary event which consists of all points in our sample space that are not in A. We denote this event by :A. Since all the points in a. Probability: random experiments - exhaustive events. When a sample space is divided into multiple mutually exclusive events where their union forms the sample space itself, then these events are called exhaustive events.A collectively exhaustive event contains all the possible elementary events for a certain experiment under consideration
The probability that both events happen is the product of each if they're independent. If they're not, the probability of the second must be modified based on the results of the first. The probability that either one or the other happens is the sum of their probabilities, less the product of both if they overlap. It may be easier to calculate 1 - the opposite of the desired probability. Be. Probability of Event P(E) = No. of times that event occurs/ Total number of trials. Axiomatic Probability; One of the ways to define probability is through axiomatic probability. Here, few axioms are predefined before predicting the outcome of any event. The event is quantified that makes it easy to calculate the expected outcome. Probability Tree . It helps to apprehend and visualize various. Complementary Events Two events are said to be complementary when one event occurs if and only if the other does not. The probabilities of two complimentary events add up to 1.. For example, rolling a 5 or greater and rolling a 4 or less on a die are complementary events, because a roll is 5 or greater if and only if it is not 4 or less. The probability of rolling a 5 or greater is = , and the. Probability is the measure of the likelihood that an event will occur in a Random Experiment. Probability is quantified as a number between 0 and 1, where, loosely speaking, 0 indicates impossibility and 1 indicates certainty. The higher the probability of an event, the more likely it is that the event will occur. Example
### 4 Ways to Calculate Probability - wikiHo
1. ator gives us two hundred twenty one. So the probability of pulling two kings in a.
2. The probability that a coin will show head when you toss only one coin is a simple event. However, if you toss two coins, the probability of getting 2 heads is a compound event because once again it combines two simple events. Suppose you say to a friend, I will give you 10 dollars if both coins land on head
3. A general statement of the chain rule for n events is as follows: Chain rule for conditional probability: P ( A 1 ∩ A 2 ∩ ⋯ ∩ A n) = P ( A 1) P ( A 2 | A 1) P ( A 3 | A 2, A 1) ⋯ P ( A n | A n − 1 A n − 2 ⋯ A 1) Example. In a factory there are 100 units of a certain product, 5 of which are defective
4. Probability = Event Outcomes \text{Probability} = \dfrac{\text{Event}}{\text{Outcomes}} Probability = Outcomes Event To understand this formula in a better manner, we can go through another example. Consider that you have a bottle filled with 7 peanuts, 4 pistachios and 6 almonds. What is the probability that when you randomly pick one dry.
5. Probability is the likelihood or chance of an event occurring. For example, the probability of flipping a coin and it being heads is ½, because there is 1 way of getting a head and the total number of possible outcomes is 2 (a head or tail). We write P (heads) = ½ . The probability of something which is certain to happen is 1
6. PROBABILITY 259 13.1.4 Independent Events Let E and F be two events associated with a sample space S. If the probability of occurrence of one of them is not affected by the occurrence of the other, then we sa
7. Independent Events and Conditional Probability. Remember that conditional probability is the probability of an event A occurring given that event B has already occurred. If two events are independent, the probabilities of their outcomes are not dependent on each other. Therefore, the conditional probability of two independent events A and B is
### Sample Space, Events And Probabilities - Statistical Data
Joint Probability: A joint probability is a statistical measure where the likelihood of two events occurring together and at the same point in time are calculated. Joint probability is the. Event C is an intersection of event A & B. Probabilities are then defined as follows. P (C) = P (A ꓵ B) We can now say that the shaded region is the probability of both events A and B occurring together. 1.4 Disjoint Events. What if, you come across a case when any two particular events cannot occur at the same time. For example: Let's say you have a fair die and you have only one throw. Probability calculator is free and easy to use. You just need to follow below steps. Step #1: Define the probabilities of single or multiple events you want to calculate. Probabilities must have two separate events. Probability of A: P (A) and. Probability of B: P (B) Step #2: Find the Probability of an event The axioms of probability are mathematical rules that probability must satisfy. Let A and B be events. Let P(A) denote the probability of the event A.The axioms of probability are these three conditions on the function P: . The probability of every event is at least zero. (For every event A, P(A) ≥ 0.There is no such thing as a negative probability.
### Probability for Multiple Events College Algebr
1. 2 Events and Probabilities 2.1 What is probability and why do we care? Probability theory is a branch of mathematics that allows us to reason about events that are inherently random. However, it can be surprisingly difficult to define what probability is with respect to the real world, without self-referential definitions. For example, you might try to define probability as follows.
2. Probability - Independent & Dependent Events 1. Using Probability 2. Independent Events <ul><li>Result of the first draw does not effect the outcome of the second draw
3. Poker Probability and Statistics with Python. Tackle probability and statistics in Python: learn more about combinations and permutations, dependent and independent events, and expected value. Data scientists create machine learning models to make predictions and optimize decisions. In online poker, the options are whether to bet, call, or fold
4. The probability of the event corresponding to any node on a tree is the product of the numbers on the unique path of branches that leads to that node from the start. If an event corresponds to several final nodes, then its probability is obtained by adding the numbers next to those nodes. Key Takeaways . A conditional probability is the probability that an event has occurred, taking into.
5. Math ↠ Probability ↠ Probability of an event. PDF Export Premium; Note Premium; Report an issue; Probability. Although the outcome of an experiment is unpredictable, the relative frequency approaches more and more a fixed value as the number of trials increases.! Remember. The stabilization of the relative frequency with increasing number of experiments is called the empirical law of large.
6. e the ratio of the number of favorable outcomes to the number of total outcomes. Table 1 lists all possible outcomes. There are eight different outcomes, only one of which is favorable.
7. development of probability theory. For example, the event the sum of the faces showing on the two dice equals six consists of the five outcomes (1, 5), (2, 4), (3, 3), (4, 2), and (5, 1). Oftentimes probabilities need to be computed for related events. For instance, advertisements are developed for the purpose of increasing sales of a.
The probability of one event occurring is quantified as a number between 0 and 1, with 1 representing certainty, and 0 representing that the event cannot happen. Therefore, the probability of an event lies between 0 ≤ P(A) ≤ 1. If we plot the likelihood of rolling a 6 on a dice in the probability line, it would look something like this: What's the formula for an event that will not occur. Probability of an event happening = Number of ways it can happen Total number of outcomes . Example: there are 4 Kings in a deck of 52 cards. What is the probability of picking a King? Number of ways it can happen: 4 (there are 4 Kings) Total number of outcomes: 52 (there are 52 cards in total) So the probability = 4 52 = 1 13. Mutually Exclusive. When two events (call them A and B) are. We saw that the probability of an event (for example, the event that a randomly chosen person has blood type O) can be estimated by the relative frequency with which the event occurs in a long series of trials. So we would collect data from lots of individuals to estimate the probability of someone having blood type O. In this section, we will establish the basic methods and principles for. Probability of Complementary Events and At Least One Probabilities At least one is equivalent to one or more The complement of getting at least one item of a particular type is that you get no items of that type. Examples: 1. Find the probability of couple having at least 1 boy among 4 children. 2. A unprepared student makes random.
The probability, or likelihood, of an event is also commonly referred to as the odds of the event or the chance of the event. These all generally refer to the same notion, although odds often has its own notation of wins to losses, written as w:l; e.g. 1:3 for a 1 win and 3 losses or 1/4 (25%) probability of a win Adding up these probabilities of disjoint events, the desired total probability is P(at least 7 blue in 9) 12 = P(exactly 7)+P(exactly 8)+P(exactly 9) = 9 7 3 5 7 2 5 9−7 + 9 8 3 5 8 2 5 9−8 + 9 9 3 5 9 2 5 9−9 13. Conditional probability The conditional probability that an event A will occur given that an event B occurs is defined to be P(A|B) = P(A∩B) P(B) [0.7] Example: The. Independent events and probability can be defined as those occurrences that are not dependent on any specific event. A good example will be if an individual flips a coin, then he/she has the chance of getting head or tail. In both outcomes, the occurrences are independent of each other, which makes an event of probability. This theory can be understood with the Venn diagram, which gives. Events with positive probability can happen, even if they don't. Some authors also insist on the converse condition that only events with positive probability can happen, although this is more controversial — see our discussion of 'regularity' in Section 3.3.4. (Indeed, in uncountable probability spaces this condition will require the employment of infinitesimals, and will thus take us.
Probabilities help us measure how likely it is that two events occur together. If two events are closely related, then their probabilities will show it. As soon as one event occurs, the other becomes much more likely (or perhaps much less likely). If two events are completely unrelated, their probabilities will also show it. It won't matter whether the first event has occurred or not occurred. Zero-probability events are of paramount importance in probability and statistics. Often, we want to prove that some property is almost always satisfied, or something happens almost always. Almost always means that the property is satisfied for all sample points, except possibly for a negligible set of sample points. The concept of zero-probability event is used to determine which sets are. The probability of an event cannot be - 1.5 because Probability of an event can never be negative. The probability of happening of an event always lies between 0 to 1 (0 and 1 inclusive) i.e 0 ≤ P(E) ≤ 1 . Also in percentage, it lies between 0 % to 100 %(0 and 100 inclusive) Independent events give us no information about one another; the probability of one event occurring does not affect the probability of the other events occurring. Independent events. Two events, $$A$$ and $$B$$ are independent if and only if $P(A \text{ and } B) = P(A) \times P(B)$ At first it might not be clear why we should call events that satisfy the equation above independent. We will. Find probability of an event occurring lesson plans and teaching resources. Quickly find that inspire student learning
### Probability of Simple Event - YouTub
The probability that an event does not occur is 1 minus the probability that it does occur. (also called the complement of A) 18. Probability- General Rules(contd.) Probability of a sure event is 1. Probability of an impossible eventis 0. 19. Possible outcomes and countingtechniques If you can do one task in A ways and a second task in B ways, then both tasks can be done in A x B ways. Flip a. = Probability is the measure of how likely an event is. the ratio of the number of favourable cases to the number of all the cases or P(E) = Number of outcomes favourable to E Number of all possible outcomes of the experiment Generally the word probability is used in our day to day conversations by coming across following statements such as :1) Probably it may rain today. 2) He may. Unconditional Probability: The probability that an event will occur, not contingent on any prior or related results. An unconditional probability is the independent chance that a single outcome. Sample Spaces and Events. Rolling an ordinary six-sided die is a familiar example of a random experiment, an action for which all possible outcomes can be listed, but for which the actual outcome on any given trial of the experiment cannot be predicted with certainty.In such a situation we wish to assign to each outcome, such as rolling a two, a number, called the probability of the outcome.
### Multiple Event Probability Calculator Online Calculato
To determine the probability of one event or another occurring, you first need to determine if the events are overlapping or non-overlapping. If they do not overlap, then you just need to add the. The probability of an event, like rolling an even number, is the number of outcomes that constitute the event divided by the total number of possible outcomes. We call the outcomes in an event its favorable outcomes. If a die is rolled once, determine the probability of rolling a 4: Rolling a 4 is an event with 1 favorable outcome (a roll of. Determine the probability of the second event. To do this, set up the ratio, just like you did for the first event. For example, if the second event is throwing a 4 with one die, the probability is the same as the first event: p r o b a b i l i t y = 1 6 {\displaystyle probability= {\frac {1} {6}}} Combined events. Listing or counting all the possible outcomes for two or more combined events enables you to calculate the probability of any particular event occurring The probability of an event cannot be: (a) Equal to zero (b) Greater than zero (c) Equal to one (d) Less than zero MCQ 6.7 A measure of the chance that an uncertain event will occur: (a) An experiment (b) An event (c) A probability (d) A trial MCQ 6.8 A graphical device used to list all possibilities of a sequence of outcomes in systematic way is called: (a) Probability histogram (b) Venn.
### Chapter 3 Probability and random events Introductory
1. Probability[pred, x \[Distributed] dist] gives the probability for an event that satisfies the predicate pred under the assumption that x follows the probability distribution dist. Probability[pred, x \[Distributed] data] gives the probability for an event that satisfies the predicate pred under the assumption that x follows the probability distribution given by data
2. Using the Binomial Probability Calculator. You can use this tool to solve either for the exact probability of observing exactly x events in n trials, or the cumulative probability of observing X ≤ x, or the cumulative probabilities of observing X < x or X ≥ x or X > x.Simply enter the probability of observing an event (outcome of interest, success) on a single trial (e.g. as 0.5 or 1/2, 1.
3. The probability that both die rolls are 6 is 1 36 \boxed{\dfrac{1}{36}} 3 6 1 . An important requirement of the rule of product is that the events are independent. If one were to calculate the probability of an intersection of dependent events, then a different approach involving conditional probability would be needed
### Event and its Types : Probability, Concepts and Videos
1. The probability of an event is calculated by adding up the probabilities of all the outcomes comprising that event. So, if all outcomes are equally likely, we have P(A)= |A| |S|. In our example, both A and B have probability 4/8=1/2. An event is simple if it consists of just a single outcome, and is compound otherwise. In the example, A and B are compound events, while the event 'heads on.
2. Probability - A risk is an event that may occur. The probability of it occurring can range anywhere from just above 0 percent to just below 100 percent. (Note: It can't be exactly 100 percent, because then it would be a certainty, not a risk. And it can't be exactly 0 percent, or it wouldn't be a risk.) Impact - A risk, by its very nature, always has a negative impact. However, the size.
3. Probability calculation for a single event is a process in which we find the probability for a single event that can occur. Like, in the example of tossing a coin, whether it will be a head or it will be a tail. The formula for calculating probability of a single event is as follows. Theoretical Probability Formula : ⇒ P(A) = n(A)/n(S
4. dict.cc | Übersetzungen für 'event [probability theory]dingsbums' im Rumänisch-Deutsch-Wörterbuch, mit echten Sprachaufnahmen, Illustrationen, Beugungsformen,.
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But in Native American cultures served quite a different function. Native Americans viewed music and bodily movement as ways to achieve connection with the spiritual realm. As an early Jesuit priest, Father Paul La Jeune, wrote of the Hurons: "All their religion consists mainly in singing."
Among Native Americans, music occupied a conspicuous place in a wide variety of ceremonies and activities. Native American music was primarily vocal, usually accompanied by percussive instruments (such as drums, rattles, or sticks) and by flutes and whistles. Music was typically performed by a single singer or by a group of singers singing in unison. Native American music tended to involve the repetition of brief fragments of music or sound—and often the sounds had no specific meaning. Technically, such sounds are known as vocables.<|endoftext|>
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Dung beetles use the Milky Way to roll their dung balls in a straight line.
Dung beetles, looking up. I’m Bob Hirshon and this is Science Update.
The humble dung beetle, best known for rolling around balls of poop, now has a more lofty distinction. It’s the first creature known to use the Milky Way as a compass. Biologist Marie Dacke, of Lund University in Sweden, says her team wanted to know how the beetles roll their dung balls in a straight line at night. So they brought the bugs to a planetarium.
And here, we could now dictate and say, okay, we want the eighteen brightest stars, or we want the four thousand dimmest stars, or we want only the Milky Way. And if we showed them the whole starry sky or the Milky Way, the beetles did equally well. And this is what told us the beetles are using the Milky Way for their orientation.
Although they’re the first animal proven to do so, Dacke says there are probably many others out there that we just haven’t studied closely enough. I’m Bob Hirshon for AAAS, the Science Society.
Making Sense of the Research
From a human perspective, dung beetles have a pretty disgusting lifestyle. Their diet consists almost exclusively of animal feces. To make sure they have a steady supply, they collect it and roll it into balls, which they can bury for later or keep with them for snacking. When they're feeling romantic, they'll also mate underground with a dung ball, so the female can lay her eggs in it.
A less icky but impressive characteristic of dung beetles is their ability to roll dung balls in a straight line. Many live in deserts, grasslands, or other environments with relatively few landmarks to navigate by. It was already known that the beetles could use the night sky to navigate, since they need clear skies to roll their balls in a straight line by night.
The question Dacke's team asked is: which component of the night sky are the dung beetles using? The obvious answer would be either the moon or the configuration of the stars, and that's what's been assumed. But to find out for sure, they took the dung beetles into a model environment where they could control the night sky: a planetarium.
The results were surprising. Although the beetles, as expected, were able to navigate by the entire starry sky, they were also equally successful using just the Milky Way. The Milky Way is the galaxy that the sun, Earth, and our solar system are located in. We can see part of it because the entire Milky Way is shaped like a disk, and our solar system is located relatively close to one edge of it. So when we look toward the center, we can see a flat cross-section of the rest of the disk.
In the night sky, the Milky Way appears as a blurry, narrow, pale band. It's blurry and relatively dim because we're looking at a thick layer of stars, many of which are too far to see individually. In fact, we can only see any light at all from just part of the Milky Way—the actual center is too far away, let alone the opposite edge. For this reason, the Milky Way is also visible only on clear nights, in places far from artificial light. So it's not visible from major cities or nearby areas.
Back to the dung beetles: When the researchers adjusted the planetarium so that it showed only the pale band of the Milky Way, and not any stars that were close enough to be seen as individual points of light, the beetles still rolled the ball in a straight line. That proves that they can use, to some extent, the alignment of the Milky Way to keep their dung ball on course. It's the first creature ever shown to use the Milky Way itself, but the researchers suspect there are many others out there. We just haven't identified them yet, or we've discovered that they use the night sky to navigate and left it at that.
Now try and answer these questions:
- How do we know dung beetles use the night sky to navigate?
- Why did the researchers go to a planetarium to study this more closely?
- How did they prove that the beetles use the Milky Way, itself, to get their bearings?
- Why is it reasonable to assume that other animals can do this?
You may want to check out the GoSkyWatch Planetarium App, which allows you to identify and locate stars, planets, constellations, and more by touching the screen or by pointing it to the sky.
- Ultra-White Beetle
- 2011 BioBlitz BobCast 3: Bugs in the Night
- Looking at the Night Sky
- Sizing Up the Universe
The GoSkyWatch Planetarium App allows you to identify and locate stars, planets, constellations, and more by touching the screen or by pointing it to the sky.<|endoftext|>
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Use these resources to introduce students to how the American people elect national leaders, the laws that govern the nation, and the three branches of government.
What Do Political Parties Stand For?
Are you a Democrat or a Republican? If you can answer that question, you're ahead of the game. For many people, political parties are a puzzle. The dictionary defines them as groups of people "who control or seek to control a government." So much for music and dancing.
The U.S. Constitution doesn't even mention political parties. President George Washington warned about "the danger of parties." But even then, our leaders didn't always agree. In the 1790s, a quarrel broke out between Thomas Jefferson and Alexander Hamilton over how much power to grant the federal government. The argument split their followers into two separate groups — the beginning of U.S. political parties.
The argument still forms the underlying dispute between our two main political parties: the Democrats and the Republicans. What does each group stand for?
The Party of the People
The Democratic Party is the oldest existing political party in the United States. Some scholars say that it began when Jefferson founded the Democratic-Republicans in 1792. Jefferson opposed a strong central government.
The party later split. Some scholars say that the Democrat Party grew from a branch headed by Andrew Jackson. Jackson, elected President in 1828, believed in a strict interpretation of the Constitution and a limitation of the government's powers.
Today's Democratic Party takes a different stand. Democrats are sometimes referred to as "the Party of the People," attracting immigrants, blue-collar workers, women, and minorities. Democrats tend to take a more liberal stand on important issues. They believe that the federal government should take a more active role in people's lives, particularly those who are in need.
One example is Franklin D. Roosevelt's presidency (1933–1945). To pull the United States out of an economic depression, Roosevelt started a slew of government programs to create jobs.
The Grand Old Party
The Republican Party was formed in 1854, when a man named Alvan E. Bovay brought together antislavery leaders. These leaders opposed the Kansas-Nebraska Act, which would permit slavery in these new territories if the people voted for it.
The party's candidate lost in 1856. The Republicans realized they needed more than one issue to win. In 1860, they still opposed slavery in the territories, but also called for a transcontinental railroad and free land to settlers. The candidate that year, Abraham Lincoln, won.
What do Republicans stand for today? In general, Republicans tend to take a more conservative stand on issues. They believe that the federal government should not play a big role in people's lives. Most Republicans favor lower taxes and less government spending on social programs. They believe in less government intervention in business and the economy.
Who Would You Choose?
Not everyone agrees with everything his or her party stands for. Most voters choose a party that most closely represents their values, concerns, and ideals. Which party would you support?
Other Voices in the Electoral Crowd
Over the last 40 years, Americans have seldom granted their Presidents much freedom to enact their proposals. For example, when Republican Presidents have been in the White House, while Democrats have controlled both houses of Congress or vice versa. The result is what we call a divided government.
Some experts say that this situation can spark trouble. Each party blames the other when things go wrong. Voters, in turn, have grown disillusioned with a government that is constantly deadlocked.
While we have a two-party system, there are hundreds of political parties in the U.S. There are also people who are not linked to any party and run as independents. No third-party or independent candidate has ever been elected President. But several third-party proposals have gained such widespread support that the major parties were forced to adopt them. The direct election of U.S. senators and primary elections, which are part of selecting presidential candidates, both began as third-party ideas.
Adapted from Junior Scholastic.<|endoftext|>
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Fossil fuels have long been the precursor to plastic, but new research from the University of Nebraska–Lincoln and European collaborators could help send that era up in smoke — carbon dioxide, to be exact.
Produced almost entirely from burning fossil fuels, carbon dioxide concentrations in the atmosphere have risen from 280 parts per million in the pre-industrial era to about 410 PPM today. That trend, combined with the finite supply of fossil fuels, has pushed researchers to explore methods for producing plastic from CO2 rather than petroleum or natural gas — recycling CO2 just as plastic is now.
Nebraska’s Vitaly Alexandrov and colleagues have now detailed a catalyst-based technique that can double the amount of carbon dioxide converted to ethylene, an essential component of the world’s most common plastic, polyethylene.
“The conversion of CO2 is very important to help offset the emissions that lead to global warming and other detrimental processes in the environment,” said Alexandrov, assistant professor of chemical and biomolecular engineering.
Copper has emerged as the prime candidate for catalyzing chemical reactions that convert carbon dioxide to plastic-forming polymer molecules, which it does when voltage is applied to it. But some copper-based setups have failed to convert more than about 15 percent of CO2 into ethylene, a yield too small to meet the needs of industry.
So researchers at Swansea University in Wales decided to try coating copper with different polymers in the hope of increasing that efficiency. After overlaying it with a polymer called polyacrylamide, they found that their copper foam’s conversion rate rose from 13 to 26 percent.
Alexandrov and postdoctoral researcher Konstantin Klyukin then ran quantum mechanics-based simulations through Nebraska’s Holland Computing Center to help explain why polyacrylamide managed to outperform its polymeric cousins. They discovered that the polyacrylamide breaks up CO2 and reassembles it into a pair of bonded C-O compounds, then stabilizes that new molecule as it drives further chemical reactions – those ultimately producing ethylene.
“CO2 is a very stubborn molecule because it has double bonds that are very difficult to break,” Alexandrov said. “That’s the most challenging part of trying to convert it to something else. You don’t want to spend too much energy converting it; otherwise, it’s a trade-off that becomes inefficient.”
Even as researchers look to further improve that efficiency, Alexandrov said, they have an eye toward a larger goal: turning CO2 directly into the polyethylene that makes up plastic bags, containers and films.
“One of the things that experimentalists want is to go from synthesizing simple molecules, like ethylene, to very complicated molecules in one batch reaction,” Alexandrov said. “You put in CO2 catalysts, and you end up with polymer structures that you can sell in a store. But those molecules have very complicated structures. This is a first step toward understanding how we can (create them).”
Alexandrov and Klyukin authored the study with Swansea’s Enrico Andreoli, Sunyhik Ahn, Russell Wakeham, Jennifer Rudd, Aled Lewis and Shirin Alexander, along with Francesco Carla of the European Synchrotron Radiation Facility. The team’s paper appeared in ACS Catalysis, a journal published by the American Chemical Society.
The researchers received support from the Engineering and Physical Sciences Research Council and the Welsh Government Sêr Cymru Program.<|endoftext|>
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# In the Given Figure, Rs is a Diameter of the Circle. Nm is Parallel to Rs and ∠Mrs = 29°. Calculate: (I) ∠Rnm, (Ii) ∠Nrm - ICSE Class 10 - Mathematics
ConceptArc and Chord Properties - Angle in a Semi-circle is a Right Angle
#### Question
In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°.
Calculate:
(i) ∠RNM,
(ii) ∠NRM
#### Solution
(i) Join RN and MS.
∴ ∠RMS = 90°
(Angle in a semicircle is a right angle)
∴ ∠RSM = 90° - 29° = 61°
(By angle sum property of triangle RMS)
∴ ∠RNM =180° ∠RSM =180° - 61° = 119°
(pair of opposite angles in a cyclic quadrilateral are supplementary)
(ii) Also, RS || NM
∴ ∠NMR = ∠MRS = 29° (Alternate angles)
∴ ∠NMS = 90° + 29° = 119°
Also, ∠NRS + ∠MS = 180°
(pair of opposite angles in a cyclic quadrilateral are supplementary)
⇒ ∠NMR + 29° +119° = 180°
⇒ ∠NRM = 180° -148°
∴ ∠NRM = 32°
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution In the Given Figure, Rs is a Diameter of the Circle. Nm is Parallel to Rs and ∠Mrs = 29°. Calculate: (I) ∠Rnm, (Ii) ∠Nrm Concept: Arc and Chord Properties - Angle in a Semi-circle is a Right Angle.
S<|endoftext|>
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Scientists have discovered a new type of bacteria that they believe could be unique to fracking sites.
Researchers examining micro-organisms living in shale oil and gas wells found a remarkable similarly in the make-up of the microbes present at each site – including a never-before-seen genus of bacteria they have dubbed Frackibacter.
The word Frackibacter – or Candidatus Frackibacter, to give its official name – gives a nod to the method of extraction used to produce oil and gas from shale rocks. The process, known as fracking, releases the hydrocarbons by blasting a mixture of sand, water and chemicals into the rock.
The process is widespread in the US but only in the early stages in the UK, where commercial fracking is expected but has not yet begun.
Frackibacter is one of 31 microbes found living inside two separate fracking wells in the US, according to a new report by Ohio State University researchers, published in the journal Nature Microbiology.
Even though the wells were hundreds of miles apart and drilled in different kinds of shale formations, the microbes in each were nearly identical, they said.
“We thought we might get some of the same types of bacteria, but the level of similarity was so high it was striking”
“We thought we might get some of the same types of bacteria, but the level of similarity was so high it was striking,” said Kelly Wrighton, assistant professor of microbiology and biophysics at Ohio State.
“That suggests that whatever’s happening in these ecosystems is more influenzed by the fracturing than the inherent differences in the shale,” she added.
The microorganisms living in the shale must be able to tolerate high temperatures, pressure and salinity and one bacteria in particular – known as Halanaerobium – was found to dominate both wells. Halanaerobium can be converted into hydrogen.
The researchers took fluid samples from each well over a period of 328 days.
Shale energy companies typically formulate their own recipes for the fluid they pump into the wells to break up the rock. They all start with water and add other chemicals. Once the fluid is inside the well, salt within the shale leaches into it, making it briny.
It is unknown how much, if any, of the UK’s considerable shale gas reserves it will be commercially viable to extract. This is because production costs can be high and planning permission can be difficult to get because of environmental concerns about fracking.
These include fears that the process could pollute the local groundwater, release methane into the atmosphere and cause earth tremors.<|endoftext|>
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Begin With the Five W’s and H
Students can begin their skills in second language acquisition and students writing performance simply by writing on separate lines on a sheet of paper the words who, what, where, why and how. From here have each student write an appropriate or relevant detail of his or her life next to each word. For example, students can write their own name next to whom, or the name of a relative or political figure or celebrity. They then add additional details to each word, eventually creating a complete story. In the beginning, have students write their story in parts, mark the beginning; then after you have corrected this they move on to the middle, and once again this is corrected before they write the end.
Another approach to writing skills for second language learners is to have students draw pictures before they write. The teacher can then integrate the use of the Present Progressive Tense by asking the students questions about their drawings. In this way, the teacher produces a model to help the student to write. For example, the teacher can ask “What building is this?” “I see a bench with two people sitting. I also see some cars. What is happening here?" The student can then explain before starting to write that he and his classmate went outside of the building during break, and sat on a bench near the car park. He can add further details; that they ate some cookies and drank some coffee there. These are very simple sentences with very appropriate and relevant details. The teacher then asks the student to put this in writing, and corrects the student’s grammar and spelling mistakes, if any. The teacher also helps with vocabulary as necessary.
Index cards are handy tools for teaching. Use index cards for writing projects. Have students make invitations, greeting cards or announcements on them. Encourage them to engage in writing for real life situations. This will motivate them and pique their interest, in addition to increasing their vocabulary.
Reading and Writing
Impress on your students how closely reading and writing are connected. Have them read as much as possible in class, and ask them to observe the different styles of writing. Next, have them discuss the styles they like. Then tell them try to emulate those styles in their writing. Like speaking, the imitation of good models is the key to good writing in a language. While speaking should be kept as the primary focus in the student’s acquisition of the new language, teachers should encourage their students to write frequently. They should also make students aware that writing needs to be edited. The student will already know this from writing in their own language, so it should not make them think the editing is mainly because of their language mistakes. In any case, the student should accept that mistakes are expected, and mistakes help me the learning process.
As students get deeper in the writing process, have them write about subjects that are of interest and meaningful to them. Also have them practice writing letters, book reviews, and anything else that may be useful to them if they need to get a job using the language. Their writing skills will be very beneficial to them in the future.<|endoftext|>
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# How do you find the vertex of f(x)= -3x^2-6x-7?
Jan 17, 2016
Vertex$\to \left(x , y\right) \to \left(- 1 , - 4\right)$
Solved using Calculus and by not using Calculus
#### Explanation:
As you are using $f \left(x\right)$ I am assuming you are at a higher level of Mathematics.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Using Calculus with shortcuts}}$
$\textcolor{b l u e}{\text{To find "x_("vertex}}$
Given: $f \left(x\right) = - 3 {x}^{2} - 6 x - 7$..............(1)
$f ' \left(x\right) = - 6 x - 6$
Equating to zero gives $\textcolor{b l u e}{{x}_{\text{vertex}} = - 1}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{To find "y_("vertex}}$
Substitute $x = - 1$ into equation (1) giving
${y}_{\text{vertex}} = - 3 {\left(- 1\right)}^{2} - 6 \left(- 1\right) - 7$
$\textcolor{b l u e}{{y}_{\text{vertex}} = - 4}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Vertex$\to \left(x , y\right) \to \left(- 1 , - 4\right)$
'~~~~~~~~~~~~~~~~~~~~~~~~
This is a maximum as $- 3 {x}^{2}$ being negative is indicative of an inverted U shape. Also the 2nd differential is negative which is also indicative of a maximum
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{NOT using Calculus}}$
$\textcolor{b l u e}{\text{I am going to show you a really cool trick!}}$
Write as:$y = \left(- 3 {x}^{2} - 6 x\right) - 7$
Factor the -3 out
$y = - 3 \left({x}^{2} + 2 x\right) - 7$............................(1)
Now consider the $+ 2 \text{ from } 2 x$ inside the bracket
Multiply this by $\left(- \frac{1}{2}\right)$
$\left(- \frac{1}{2}\right) \times \left(+ 2\right) = - 1$
This is the value you are after for $x$ so:
$\textcolor{b l u e}{{x}_{\text{vertex}} = - 1}$
Substitute this x value into the original equation to find the value of ${y}_{\text{vertex}}$<|endoftext|>
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# Parallel Perpendicular Lines Slope of Parallel Lines Two
• Slides: 16
Parallel & Perpendicular Lines
Slope of Parallel Lines • Two non-vertical lines with the same slope are parallel.
Parallel Lines The equations of two parallel lines have the same slope, but two different y-intercepts.
Parallel lines have the same slope If Line A has a slope of 4 (m = 4), then which of these four lines is parallel to Line A? Line B, m = 16 Line C, m = 12 Line D, m = 4 Line E, m = 2/3
y = 3 x + 4 and y = 3 x - 2 y = 3 x + 4 y = 3 x -2 BACK
Are the two lines: L 1, through (-2, 1) and (4, 5) L 2, through (3, 0) and (0, -2), parallel?
Slopes of Perpendicular Lines • If neither line is vertical, then the slopes of perpendicular lines are negative reciprocals. • If the product of the slopes of two lines is -1 then the lines are perpendicular. • Horizontal lines are perpendicular to vertical lines.
Perpendicular Slopes y-axis x-axis What can we say about the intersection of the two lines? 4 3
Write parallel perpendicular or neither for the pair of lines that passes through (5, -9) and (3, 7) and the line through (0, 2) and (8, 3).
Perpendicular lines have the ‘opposite reciprocal’ slope To get the opposite reciprocal… Express the slope as a fraction Then flip it and give it the opposite sign!
Perpendicular lines have the ‘opposite reciprocal’ slope So, a line with slope 3 is perpendicular to a line with slope – 1/3
Perpendicular lines have the ‘opposite reciprocal’ slope If Line A has a slope of – 4/5 (m = -4/5), then which of these four lines is perpendicular to Line A? Line B, m = 9 Line C, m = 5/4 Line D, m = 20 Line E, m = -5/4
Slope-Intercept Form • Useful for graphing since m is the slope and b is the y-intercept Point-Slope Form • Use this form when you know a point on the line and the slope • Also can use this version if you have two points on the line because you can first find the slope using the slope formula and then use one of the points and the slope in this equation. Standard Form • Commonly used to write linear equation problems or express answers
Remember parallel lines have the same slopes so if you need the slope of a line parallel to a given line, simply find the slope of the given line and the slope you want for a parallel line will be the same. Perpendicular lines have negative reciprocal slopes so if you need the slope of a line perpendicular to a given line, simply find the slope of the given line, take its reciprocal (flip it over) and make it negative.
Rules and Properties Parallel Lines Parallel lines have the same slope.
Parallel & Perpendicular Lines Algebra 1 & Algebra 1 Honors<|endoftext|>
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"In spite of difference of soil and climate, of language and manners, of laws and customs, in spite of things silently gone out of mind and things violently destroyed, the Poet binds together by passion and knowledge the vast empire of human society, as it is spread over the whole earth, and over all time. The objects of the Poet's thoughts are everywhere; though the eyes and senses of man are, it is true, his favorite guides, yet he will follow wheresoever he can find an atmosphere of sensation in which to move his wings. Poetry is the first and last of all knowledge—it is as immortal as the heart of man."
—William Wordsworth, "Preface to Lyrical Ballads"
Romanticism was arguably the largest artistic movement of the late 1700s. Its influence was felt across continents and through every artistic discipline into the mid-nineteenth century, and many of its values and beliefs can still be seen in contemporary poetry.
It is difficult to pinpoint the exact start of the romantic movement, as its beginnings can be traced to many events of the time: a surge of interest in folklore in the early to mid-nineteenth century with the work of the brothers Grimm, reactions against neoclassicism and the Augustan poets in England, and political events and uprisings that fostered nationalistic pride.
Romantic poets cultivated individualism, reverence for the natural world, idealism, physical and emotional passion, and an interest in the mystic and supernatural. Romantics set themselves in opposition to the order and rationality of classical and neoclassical artistic precepts to embrace freedom and revolution in their art and politics. German romantic poets included Fredrich Schiller and Johann Wolfgang von Goethe, and British poets such as Wordsworth, Samuel Taylor Coleridge, Percy Bysshe Shelley, George Gordon Lord Byron, and John Keats propelled the English romantic movement. Victor Hugo was a noted French romantic poet as well, and romanticism crossed the Atlantic through the work of American poets like Walt Whitman and Edgar Allan Poe. The romantic era produced many of the stereotypes of poets and poetry that exist to this day (i.e., the poet as a tortured and melancholy visionary).
Romantic ideals never died out in poetry, but were largely absorbed into the precepts of many other movements. Traces of romanticism lived on in French symbolism and surrealism and in the work of prominent poets such as Charles Baudelaire and Rainer Maria Rilke.<|endoftext|>
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# Solve $5a^2 - 4ab - b^2 + 9 = 0$, $- 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0$
Solve $\left\{\begin{matrix} 5a^2 - 4ab - b^2 + 9 = 0\\ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0. \end{matrix}\right.$
I know that we can use quadratic equation twice, but then we'll get some very complicated steps. Are there any elegant way to solve this? Thank you.
-
You can notice that many terms of $$(b+5a-4)^2=b^2+10ab-8b-40a+25a^2+16$$ appear in the first equation. Similarly, in the first one, you can notice $(b+2a)^2$.
By algebraic manipulation you get that the original equations are equivalent to \begin{align} (b+5a-4)^2&=4(a^2+1)\\ (b+2a)^2&=9(a^2+1) \end{align} which implies $4(b+2a)^2=9(b+5a-4)^2$ and $2(b+2a)=\pm 3(b+5a-4)$. This should simplify things a little. (In each of the two possibilities you can express $b$ using $a$ as a linear expression. Then you will get a quadratic equation in $a$. Or you can start by eliminating $a$.)
-
I've corrected $(b+2)^a$. Are there other typos? – Martin Sleziak Oct 10 '12 at 15:47
Subtract the two equation (thereby eliminating the $b^2$ term) and solve for $b$, which you can plug into either equation and get an equation in $a$; so $a$ is a solution to $(8a^2-24a+15)(3+4a)^2$. The roots are $a=-3/4, 3/2\pm \sqrt{6}/4$.
-
By arrange the terms, you can note that
$$\begin{cases} 5a^{2}-4ab-b^2+9=0\\ -21a^{2}-10ab+40a-b^{2}+8b-12=0 \end{cases} \Leftrightarrow \begin{cases} 9(a^{2}+1)=(2a+b)^{2}\\ 4(a^{2}+1)=(5a+b-4)^{2} \end{cases}$$
So we get
$$\begin{cases} 9(a^{2}+1)=(2a+b)^{2}\\ 4(2a+b)^{2}=9(5a+b-4)^{2} \end{cases}$$
According to the second equation, we can get two cases of first relation between $a$ and $b$, then substitute them into the first equation respectively, we can get all the cases of values of pairs $(a,b)$.
-
According to maple, $a=-3/4, b=21/4$ is one solution, and if $r$ is a solution of $2x^2-12x+15=0$ then $a=r/2, b=12-(11/2)r$ is another solution. As the quadratic here has two real zeros, there are three pairs $(a,b)$ of real numbers for your system.
Of course this is not an elegant solution, and even worse it relies on maple. But it seemed odd to me since when the equations are subtracted and the result solved for $b$, we get $b=(26a^2-40a+21)/(8-6a)$, which may then be put into the first equation, so I would have expected a fourth degree equation for a.
A comment by @zyx prompted another look at this without use of maple. When the above expression for $b$ is put into the first equation, the factored equation for $a$ is $$\frac{(4a+3)^2(8a^2-24a+15)}{4(3a-4)^2}=0.$$ The zero denominator at $a=4/3$ doesn't lead to a solution to both equations, and each zero of the linear and quadratic factors in the numerator leads to only one $b$ so that the pair $(a,b)$ satisfies both equations. So geometrically the two conics meet at exactly three points in the plane. Suspecting that one must be a point of tangency led me to compute the two gradients of the functions $f,g$ at the simplest of the points, namely the $(-3/4,21/4)$, and both gradients turned out to be parallel to the vector $(19,5)$. The other two intersections coming from the quadratic roots then (it would seem to me) are geometrically transverse intersections of the two conics.
-
I admit the above is not elegant. But I thought it interesting that, if you subtract the two equations and solve for b, and then plug that into the first one, you expect a fourth degree equation for a, yet there are apparently only 3 values for a. – coffeemath Oct 10 '12 at 15:26
Is one of the four solutions a double point? This is an intersection of two affine conics. – zyx Aug 8 '13 at 5:01
@zyx -- Yes, the $a=-3/4$ is a double root of the quartic suggested in the post. And at the corresponding point the two conics are tangent to each other. – coffeemath Aug 8 '13 at 6:18
@zyx Note I just added something about the double root to the answer. Thanks for the question. – coffeemath Aug 8 '13 at 6:30
Note that \begin{equation*} 4(5a^2 - 4ab - b^2 + 9) - 9(-21a^2-10ab+40a-b^2+8b-12) = (19a+5b-12)(11a+b-12). \end{equation*}
-
How did you note that ? – Belgi Oct 11 '12 at 6:11
$21a^2+10ab+b^2-10a-8b+12=0$
$(7a+b)(3a+b)-4(7a+b+3a+b)+12=0$
$(7a+b-4)(3a+b-4)=4--->(1)$
$5a^2-4ab-b^2=(5a+b)(a-b)=-9--->(2)$
If $7a+b-4=\frac 2 k, 3a+b-4=2k$,
Express $a,b$ in terms of $k,$ replace their values in (2).
-<|endoftext|>
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Suggested languages for you:
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Q18.
Expert-verified
Found in: Page 266
### Precalculus Mathematics for Calculus
Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508
# Radicals and Exponents Evaluate each expression.a. ${\mathbf{\left(}\mathbf{-}\mathbf{5}\mathbf{\right)}}^{\mathbf{3}}$b. role="math" localid="1648704247043" $\mathbf{-}{\mathbf{5}}^{\mathbf{3}}$c.
a) The value of ${\left(-5\right)}^{3}$ is $\mathbf{-}\mathbf{125}$.
b) The value of $-{5}^{3}$ is $\mathbf{-}\mathbf{125}$.
c) The value of is 4.
See the step by step solution
## Part a Step 1. Given information.
The given expression is ${\left(-5\right)}^{3}$.
## Part a Step 2. Write the concept.
The exponential expression “$a$ to the power $n$” is defined as:
${a}^{n}=a×a×a×\cdots ×a\text{\hspace{0.17em}}\left(n\text{\hspace{0.17em}times}\right)$
## Part a Step 3. Determine the value of the expression.
The given expression can be written as:
$\begin{array}{l}{\left(-5\right)}^{3}=\left(-5\right)×\left(-5\right)×\left(-5\right)\\ =-125\end{array}$
Thus, the value of ${\left(-5\right)}^{3}$ is $\mathbf{-}\mathbf{125}$.
## Part b Step 1. Given information.
The given expression is $-{5}^{3}$.
## Part b Step 2. Write the concept.
The exponential expression “$a$ to the power $n$” is defined as:
${a}^{n}=a×a×a×\cdots ×a\text{\hspace{0.17em}}\left(n\text{\hspace{0.17em}times}\right)$
## Part b Step 3. Determine the value of the expression.
The given expression can be written as:
$\begin{array}{l}-{5}^{3}=-\left(5×5×5\right)\\ =-125\end{array}$
Thus, the value of $-{5}^{3}$ is $\mathbf{-}\mathbf{125}$.
## Part c Step 1. Given information.
The given expression is .
## Part c Step 2. Write the concept.
The exponential expression “$a$ to the power $n$” is defined as:
${a}^{n}=a×a×a×\cdots ×a\text{\hspace{0.17em}}\left(n\text{\hspace{0.17em}times}\right)$
## Part c Step 3. Determine the value of the expression.
The given expression can be written as:
$\begin{array}{l}{\left(-5\right)}^{2}\cdot {\left(\frac{2}{5}\right)}^{2}=\left(-5\right)×\left(-5\right)×\left(\frac{2}{5}\right)×\left(\frac{2}{5}\right)\\ =\left(25\right)×\left(\frac{4}{25}\right)\\ =4\end{array}$
Thus, the value of is 4.<|endoftext|>
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# Chapter 1 Calculus-Integration.
author: Hertzberg, R. H.
## 1.1 PREFACE
Frequently in the application of calculus to ecological problems, relationships between measured quantities involve rates of change of the quantities of interest, rather than the quantities themselves. Stated in mathematical terms, it is possible to determine derivatives of functions instead of the functions themselves. It is then necessary to use an inverse operation on the derivative to determine the desired function. This operation is called integration. Each function’s behavior used in this module is comparable to characteristics of the ecosystem it represents. Topics are introduced and extended in the problem sets.
## 1.2 INTRODUCTION
Most applications of calculus to ecological problems involve the determination of specific relationships between measured quantities. Frequently, the obvious relations involve rates of change of the quantities of interest, and not the quantities themselves. In mathematical terms, we often can determine equations involving the derivatives of functions instead of the functions themselves. We then need to use an inverse operation on the derivative to determine the desired function. This operation is called integration and its field of mathematical study is called integral calculus.
For example, we may wish to know the blood level of a toxic material that is absorbed through the skin during a certain time period. However, the mathematical model may be based on the rate of absorption through the skin and rate of excretion via the urine. Thus we wish to know the magnitude of a quantity (body level of pollutant) when we know only the rate of change of that quantity (rate of increase by absorption and rate of decrease by excretion).
## 1.3 FUNDAMENTALS
The inverse operation to differentiation is called antidifferentiation or, more commonly, integration. In models of exponential growth we assume that the growth rate is proportional to the population size.
$$$\frac{dN}{dt} = kN \tag{1.1}$$$
where $$N$$ is the population size, $$t$$ is time and $$k$$ is a proportionality constant. We could solve for $$N$$ as a function of time by integrating the derivative. However, we know that the exponential function is the only function which equals its derivative.
$\frac{d(e^t)}{dt} = e^t$
Thus, we can modify the function to give the solution
$N = N_0 e^{kt}$
since differentiating $$N_0 e^{kt}$$ gives eqn. (1.1).
$\frac{dN}{dt} = \frac{d(N_0 e^{kt})}{dt} = kN_0 e^{kt} = kN$
If we write eqn. (1.1) as
$$$\frac{dN}{dt} = kN_0 e^{kt} \tag{1.2}$$$
then the “antidifferentiation” becomes more obvious: find $$N$$ so that its derivative equals $$kN_0 e^{kt}$$. If we define
$g(t) = kN_0 e^{kt}, \;\;\;\;\;f(t) = N_0 e^{kt}$
then eqn. (1.2) becomes
$g(t) = df/dt$
Thus $$g(t)$$ is the derivative of $$f(t)$$ and, conversely, $$f(t)$$ is an antiderivative of $$g(t)$$. But there is a slight problem. The antiderivative is not unique. If we define
$x(t) = N_0 e^{kt} + 10, \;\;\;\;y(t) = N_ e^{kt} +354$
then $$x(t)$$ and $$y(t)$$ are also antiderivatives of $$g(t)$$, as can be checked by differentiating: $$dx/dt = g(t), dy/dt = g(t)$$. One interpretation of this is that the graphs of $$x(t)$$ and $$y(t)$$ have the same slope (derivative) for any given value of $$t$$. Since the functions differ by only a constant, we can write the general form of the antiderivative as $$f(t) + C$$ where $$C$$ can be any constant. We usually write the antiderivative as the indefinite integral
$$$\int g(t)dt = f(t) + C \tag{1.3}$$$
and call $$C$$ the integration constant. Recall that the term “$$dt$$” identifies the variable of integration just as it does the variable of differentiation in the derivative $$df/dt$$.
## 1.4 THE DEFINITE INTEGRAL
The integration constant does not appear if the integral is a definite integral, written as
$A = \int^b_a g(x)dx$ where $$a$$, $$g$$ are called the limits of the integral. In this case, the integral is determined by evaluating the antiderivative of $$g(x)$$ at the values $$x = b$$, $$x = a$$ and subtracting. Thus, as with eqn. (1.3), if
$\frac{df}{dx} = g(x)$
then
\begin{align*} \int^b_a g(x)dx &= [f(x)]^b_a \\ &= f(b) - f(a) \end{align*}
Note that the integration constant is not written since it is eliminated through subtraction. If $$g(x) = 2x^2$$, then $$\int g(x)dx = f(x) + C = 2x^3/3 + C$$ and
\begin{align*} \int^b_ag(x)dx = \int^b_a 2x^2dx &= (2b^3 / 3 + C ) - (2a^3/3 + C) \\ &= 2b^3/3 - 2a^3/3 + C - C \\ &= f(b) - f(a) \\ \end{align*}
### 1.4.1 Area Under the Curve
Whereas the derivative can be used to represent the slope of a curve, the definite integral can represent the area under the curve (specifically, the area between the curve and the horizontal axis). For a curve given by $$y = g(x)$$, the area ($$A$$) under the curve from $$x = a$$ to $$x = b$$ is written (see fig. 1.1).
$A = \int^b_a g(x) dx$
This use of the definite integral has strong intuitive appeal. Consider the case where $$g(x) = 1$$ as in figure 1.2.
The area under $$g(x)$$ is then
$A = \int^b_a g(x)dx = \int^b_a 2 dx = 2 \int^b_a dx = 2(b-a)$
Thus $$\int^b_a dx$$ represents the width and $$g(x)$$ is the height. Thus the definite a integral as area has intuitive meaning: area = height $$\times$$ width. In general, the same visual identification applies:
Example 1
Diffusion of a gas across a membrane is often described by Fick’s Law which states that the rate of transport is proportional to the product of the surface area of the membrane and the concentration gradient. Thus knowledge of the surface area is important. Certain leaves have nearly parabolic edges (fig. 1.3). The area of the top surface available for transport of water is then found by integrating the parabola function. If the dimensions of the leaf are length = 2a, width = 2b, then the parabola describing one edge is written
$y(x) = -bx^2/a^2 + b$
The area of half the leaf is
$\frac{A}{2} = \int^a_{-a} (-bx^2/a^2 + b)dx$
i.e., the area between the x-axis and the upper curve. Thus we calculate
\begin{align*} A &= 2 \int^a_{-a} (-bx^2/a^2 + b)dx \\ &= 2[-(b/a^2)x^3/3+bx]^a_{-a} \\ &= 2[(-ab/3 + ab ) - (ab/3 - ab)] \\ &= 8ab/3 \\ \end{align*}
### 1.4.2 Improper Integrals
Occasionally we are interested in long term behavior of a measured quantity. For example, when raw sewage and other chemical pollutants are continually dumped into a lake, one effect is the rapid increase in the number of microorganisms. The resultant high level of organic oxidation from metabolism of the sewage can lead to extreme oxygen depletion of the lake, with obvious detrimental effects on other aquatic life (Dugan 1972). If we know something about the rate of oxygen consumption by the microorganisms, then we obtain the amount of oxygen consumed during a time period $$T$$ by integrating the rate over that time period, i.e., using, a definite integral with limits 0, $$T$$. We then estimate the maximum oxygen depletion by integrating over an infinite time interval. Such a definite integral with at least one infinite limit is called an improper integral.
Example 2
Assume that self-inhibition by the microorganism population causes the oxygen depletion to taper off as time becomes large. One model might be
$$$\frac{dC}{dt} = te^{-t^2} + e^{-t} \tag{1.4}$$$
where $$C$$ represents the quantity of oxygen consumed. A graph of $$dC/dt$$ looks like figure 1.4.
The total amount of oxygen consumed by time $$T$$ is then found by integrating eqn. (1.4).
$C(T) = \int^T_0 (te^{-t^2} + e^{-t}) dt$
The maximum depletion is then found by letting $$T$$ increase to $$\infty$$.
$C(\infty) = \lim_{T \rightarrow \infty} \int^T_0 (te^{-t^2} + e^{-t}) dt$ With most nicely behaved models, we can evaluate $$C(\infty)$$ directly. The above integral is then written
$$$C(\infty) = \int^\infty_0 (te^{-t^2} + e^{-t}) dt \tag{1.5}$$$
and is evaluated as with previous definite integrals.
Some functions are not “nicely behaved” and certain integrals cannot be evaluated directly. If we desired to evaluate the circumference $$C$$ of a tube with circular cross-section by using.an integral (instead of the well-known formula $$C = 2 \pi r$$) we can encounter difficulties. For one-quarter of the circle (fig. 1.5) the arc-length is given by (Schwartz 1974, p. 634)
$L = \int^r_0 \frac{r}{\sqrt{r^2-x^2}} dx$
where $$r$$ is the radius. The integrand is infinite at $$x=r$$ (fig. 1.6). However, we can evaluate the integral by taking the limit of another integral:
\begin{align*} L &= \lim_{a \rightarrow r} \int^q_0 \frac{r}{\sqrt{r^2-x^2}} dx \\ &= \lim_{a \rightarrow r} [r \sin^{-1} (x/r) ]^a_0 \\ &= \lim_{a \rightarrow r} [r \sin^{-1} (a/r) ] = r\pi/2 \\ \end{align*}
## 1.5 METHODS OF INTEGRATION
Since there are no foolproof formulae that can be used for all integrals (as there are for derivatives), the following integration methods all attempt to change a difficult integral into a simpler one. All examples use definite integrals.
### 1.5.1 Substitution
This method substitutes each part of a definite integral with a counterpart so that the result is kept the same. With the substitution $$u = h(x)$$, the integral
$\int^b_a f(x)dx$
becomes transformed into
$\int^{h(b)}_{h(a)}g(u)du$ where $$g(u)du = f(x)dx$$. Thus the integrand, $$dx$$, and the limits have been transformed into equivalent counterparts. We obtain $$g(u)$$ by finding the inverse function $$h^{-1} (u)$$ so that
\begin{align*} x &= h^{-1}(u) \\ dx &= \frac{d[h^{-1}(u)]}{du} du \\ g(u)du &= [f(x)][dx] \\ g(u) &= [f(h^{-1}(u))][\frac{dh^{-1}(u)}{du}] \\ \end{align*}
Often, $$h(x)$$ appears explicitly in the integrand so that $$u$$ is substituted directly.
Example 3
The problem presented above on the oxygen depletion due to sewage dumping can now be solved. The integral in eqn. (1.5) is first separated.
$$$\int^{\infty}_0(te^{-t^2} +e^{-t}) dt = \int^{\infty}_0 te^{-t^2}dt + \int^{\infty}_0|e^{-t}dt \tag{1.6}$$$
The antiderivative of $$te^{-t^2}$$ is not obvious, so we substitute a new function into the integral. Let $$u = t^2$$. Then the differential $$du$$ is
$du = (du/dt)dt = 2tdt$ so that $\frac{du}{2} = tdt$ The limits remain, since $$u = 0$$ when $$t = 0$$ and $$u = \infty$$ when $$t = \infty$$. Direct substitution then gives
\begin{align*} \int^{\infty}_0 te^{-t^2}dt &= \int^{\infty}_0e^{-t^2}(tdt) \\ &= \int^{\infty}_0 e^{-u} (\frac{du}{2}) = \frac{1}{2} \int^{\infty}_0 e^{-u}du \\ &= \frac{1}{2} [-e^{-u}]^{\infty}_0 = \frac{1}{2}[0-(-1)] \\ &= \frac{1}{2} \\ \end{align*}
Since the second term in eqn. (1.6) is now obvious $$\int^{\infty}_0 e^{-u}du = \int^{\infty}_0 e^{-t}dt=1$$, we obtain the solution to eqn. (1.5) of $C(\infty) = 1/2 + 1 = 3/2$
### 1.5.2 Integration by Parts
This method utilizes a relation from differential calculus concerning the total differential. Recall that the differential of a product of two functions $$u(x)$$, $$v(x)$$ can be written $d(uv) = udv + vdu$ If we evaluate antiderivatives, we obtain
$$$uv = \int udv + \int vdu \tag{1.7}$$$
Rearrangement of eqn. (1.7) gives the integration by parts formula $\int udv = uv - \int vdu$ With definite integrals, we usually write this formula as
$$$\int^b_a [u(x)\frac{dv}{dx}]dx = [uv]^b_a - \int^b_a [v(x)\frac{du}{dx}]dx \tag{1.8}$$$
Again, the goal is to change a difficult integral, the left side of eqn. (1.8), into a simpler one, the right side of eqn. (1.8). An example is presented later.
### 1.5.3 Partial Fractions
When the integrand is a ratio of two polynomials, it often can be decomposed into a sum of simpler terms. Only the case of non-repeated linear terms in the denominator is treated here. For more complicated cases in an ecological setting, see Clow and Urquhart (1974) p. 559.
The integrand is assumed to be of the form $$F(x)/G(x)$$ where $$F(x)$$ and $$G(x)$$ are polynomial functions and where $$G(x)$$ is the product of linear factors. For example, $G(x) = (1 + 2x)(2 + 2x)(1+x)$ is a polynomial composed of factors linear in $$x$$. The partial fraction technique replaces the single rational expression by a sum of terms where each denomination is one of the linear factors of $$G(x)$$. If we have two linear factors, $G(x) = A(x)B(x)$ then a partial fractions decomposition gives $\frac{F(x)}{G(x)} = \frac{F(x)}{A(x)B(x)} = \frac{C_1}{A(x)} + \frac{C_2}{B(x)}$ where $$C_1$$ and $$C_2$$ are constants. Multiplication by $$A(x)B(x)$$ gives
$$$F(x) = C_1B(x) + C_2A(x) \tag{1.9}$$$
Let $$r_1, r_2$$ be zeros of $$A(x), B(x)$$, respectively, i.e. $$A(r_1 ) = B(r_2 ) = 0$$. Then, with $$x = r_1$$, eqn. (1.9) is $F(r_1) = C_1 B(r_1)$
and with $$x = r_2$$, eqn. (1.9) becomes $F(r_2) = C_2 A(r_2)$ so that $$C_1,C_2$$ can be easily determined.
Example 4
The logistic growth model for animal populations is represented as a differential equation $\frac{dN}{dt} = rN(1 - N/K), \;\;\;\text{where }\; r,K = \mbox{constant}$
Separating variables (see the section, Differential Equations) gives, using the differentials $$dN$$ and $$dt$$, $\frac{dN}{N(1 - N/K)} = rdt$ which is integrated to yield the equation
$$$\int \frac{dN}{N(1 - N/K)} = \int rdt \tag{1.10}$$$
The right side of eqn. (1.10) is easily integrated. The left side, however, must be reworked. First rewrite the integral as $\frac{1}{N(1 - N/K)} = \frac{K}{N(K-N)}$
Now expand in partial fractions as $\frac{K}{N(K-N)} = \frac{a_1}{N} + \frac{a_2}{K-N}$ Multiplying both sides by $$N(K - N)$$ gives
$$$K = a_1(K - N) + a_2N \tag{1.11}$$$
Since (1.11) must hold for all values of $$N$$ (Clow and Urquhart 1974, p. 562), then setting $$N = 0$$ gives $a_1 = 1$ and $$N = K$$ gives
$a_2 = 1$ so that
$\frac{K}{N(K-N)} = \frac{1}{N} + \frac{1}{K-N}$ Then eqn. (1.10) is $\int \frac{dN}{N} + \int \frac{dN}{K-N} = \int rdt$ which integrates to
$\ln N - \ln (K-N) = rt + C$
## 1.6 APPLICATIONS OF DEFINITE INTEGRALS
Direct applications of integrals generally fall into discrete categories in contrast to applications of derivatives which usually are based on slopes. The first group discussed below uses the integral as the accumulation of changes in the function. The second category uses the integral as an area or generalized volume. The last application is more mathematical, although it actually relies on the accumulation concept, and uses the integral to estimate the error in a given approximation.
### 1.6.1 Accumulation of Changes in the Function
The integral as a total accumulation has been presented before in example 2 on oxygen depletion. This use of the integral is actually fairly intuitive. Let us call our quantity of interest $$F(x)$$. Then $$F'(x) = dF/dx$$ is certainly the rate of change of $$F(x)$$ and $$F(x)$$ is certainly the antiderivative of $$F'(x)$$. Then integrating the rate of change of $$F$$ gives the total change in $$F$$.
$\int^b_a F'(x)dx = F(x)]^b_a = F(b) - F(a)$ Thus the definite integral
$\int^b_a F'(x)dx$ is the total change in $$F(x)$$ as $$x$$ changes from $$a$$ to $$b$$.
### 1.6.2 Average Change
The average change in $$F(x)$$ is then found by dividing by the change in $$x$$, since the average is the change in $$F$$ per unit change in $$x$$. Note that this formula can be shown graphically as the average height of the function. For a given curve, the area under the curve equals the average height multiplied by the width. Thus the average height $$\overline y$$ of a curve $$y = f(x)$$ is the area $$A$$ divided by the width. $A = \int^b_a f(x)dx = (b-a)\overline y$ $\overline y = \frac{A}{b-a} = \frac{1}{b-a} \int^b_a f(x)dx$
Example 5
An interesting example is a study (Fisher 1963 in Warren 1971, pp. 161-163) of the effects of dissolved oxygen content and food ration on the growth rate of Coho salmon. The data appear in figure 1.7. The upper curve is well approximated by
$$$y = 7.3 (x + 3.5) e^{-0.05x} \tag{1.12}$$$
The lower curve is the straight line $y = 28$ here $$y$$ = growth rate, $$x$$ = dissolved oxygen. A simple comparison of the effect of diet (restricted vs. unrestricted ration) on growth rate is to compare the average growth rates ($$\overline y$$) for the two diets. Since the lower curve has a constant growth rate, we must have $$\overline y = 28$$.
\begin{align*} \overline y = \frac{1}{27} \int^{30}_3 28dx &= \frac{1}{27}[28x]^{30}_3 = \frac{1}{27}[(28)(30)-(28(3))] \\ &= \frac{1}{27}(28)(27)=28 \\ \end{align*}
For the upper curve, \begin{align*} \overline y &= \frac{1}{27} \int^{30}_3 7.3(x + 3.5)e^{-0.5x}dx \\ &= \frac{7.3}{27} \int^{30}_3 3.5e^{-0.5x}dx + {\frac{7.3}{27}}\int^{30}_3 xe^{-0.05x}dx \\ &= 0.95 \int^{30}_3 e^{-0.5x}dx + 0.27\int^{30}_3 xe^{-0.05x}dx \\ \end{align*}
The first integral is the same form as in previous examples. $0.95 \int^{30}_3 3.5e^{-0.5x}dx = 0.95[-\frac{1}{0.05} e^{-.05x}]^{30}_3$ $= 19(-e^{-0.05(30)}+e^{-0.05(3)}) =12.07$
Rather than finding the second integral in tables, we will evaluate it using integration by parts.
With $$dv(x)$$ representing the differential of $$v(x)$$, the second integral can be written $\int^{30}_3 xe^{-0.05x}dx = \int^{30}_3 u(x)dv(x)$ where $$u(x) = x, dv(x) = e^{-0.05x}dx$$.
Recall that integration by parts uses the formula $\int udv = uv - \int vdu$ Since $$du = dx$$, and $$v = \int dv = -e^{-0.05x}/0.05$$, we get
\begin{align*} \int^{30}_3 xe^{-0.05x}dx &= [{\frac{-xe^{-0.05x}}{0.05}}]^{30}_3 - \int^{30}_3 \frac{-e^{-0.05x}}{0.05}dx \\ &= \frac{1}{0.05} \bigg([-30e^{-1.5}+3e^{-0.15}]-[{\frac {e^{-0.05x}}{0.05}}]^{30}_3\bigg) \\ &= 20[-30e^{-1.5}+3e^{-0.15} + \frac{e^{-0.15}}{0.05} - \frac{e^{-1.5}}{0.05}] \\ &= 173.04 \\ \end{align*}
Thus, $\overline y = 12.07 + 0.27(173.04) = 58.8$ In summary, we have the averages:
ration average growth rate
restricted 29.3
unrestricted 58.8
It is somewhat surprising that the average unrestricted ration rate is over twice that of the rate for the restricted ration. The data is deceptive visually due to the close values near $$x = 3 mg/l$$ and the distorted logarithmic scale.
### 1.6.3 Distance
Velocity is defined as the rate of change of position. Since the distance covered is the total change in position, it must equal the integral of the velocity.
For a free-falling object, the velocity is given by $$v(t) = g \cdot t$$ where $$g$$ is the acceleration due to gravity and $$t$$ is time elapsed. The distance $$S$$ covered after $$T$$ seconds is given by $$S = \frac{1}{2} gT^2$$ which is merely the integral of velocity.
\begin{align*} S &= \int^T_0 v(t)dt = \int^T_0 (gt)dt \\ &= g \int^T_0 t \, dt = g[t^2/2]^T_0 \\ &= \frac{1}{2}gT^2 \\ \end{align*}
### 1.6.4 Volumes
Volumes and areas of complicated regions are also evaluated using the definite integral. Previously, the area under a curve was bounded by three straight perpendicular lines. When the bottom is not the base axis, the integration is still simple. For a region shown in figure 1.8 the area is the difference between the area under curve $$f$$ and the area under curve $$g$$. Thus
\begin{align*} A &= \int^b_a f(x)dx - \int^b_a g(x)dx \\ &= \int^a_b [f(x)-g(x)]dx \\ \end{align*}
Note that $$[f(x)-g(x)]$$ is merely the height of the region at the point $$x$$, so the height times width interpretation is still applicable.
### 1.6.5 Surface Area of Revolution
When the region is not planar, the evaluation of its area must take into account the changes in the third dimension. If the surface is obtained by revolving a curve around a straight line, the evaluation needs only a single integral. The following example illustrates the method.
Example 6
One study of temperature regulation in mammals requires knowledge of the surface area exposed to the sun. The model views the torso of the animal as symmetric with respect to a longitudinal axis. Each vertical cross-section is then a circle. The simplest such approximation is a cylinder:
The cylinder can be described by revolving a straight line around the axis, as in figure 1.9.
Unrolling the surface gives a rectangle whose width equals the circumference of the circular end face of the cylinder. This surface area of revolution equals the line length multiplied by the width, thus $$A=2\pi r\:l$$ (fig. 1.10)
As with the area under a curve, the general formula for a surface area of revolution must be intuitive, i.e., must visually appear as length times circumference. Let the torso have a profile of varying radius:
Assume the line to be revolved is graphed as follows,
and is represented by $$r = a - bx^2$$, $$a$$ and $$b$$ positive constants. The radius then changes with $$x$$, and the integral must be used:
In this example, let $$x_0 = 0.5m$$, $$a = 0.28$$, $$b = 0.24$$. The total surface area is
\begin{align*} A &= \int^{0.5}_{-0.5} 2 \pi(0.28-0.24x^2)dx \\ &= 2 \pi [0.28x - \frac{0.24}{3} x^3]^{0.5}_{-0.5} \\ &= 1.63 m^2 \\ \end{align*}
Note that any function will work in the formula, as long as the area desired is a surface area of revolution. The only problem might be in using a function which is difficult to integrate.
### 1.6.6 Volume of Revolution
The volume of a solid generated by revolving a curve around an axis can be derived as an intuitive extension of the surface area of revolution. The area and volume formulae for the cylinder and the general revolved solid (figure 1.11) are seen to be analogous.
Cylinder Revolved solid
Surface area $$2\pi r^2 l$$ $$\int^l_0 2\pi f(y)dy$$
Volume $$\pi r^2 l$$ $$\int^l_0 \pi[f(y)]^2 dy$$
Now we develop the general volume formula by expressing the integral as a limit of sums of pieces of the solid. Consider a curve $$z = f(y)$$ and the region $$R$$ under the curve (figure 1.12). We revolve the region $$R$$ around the y-axis (figure 1.13) to obtain the solid.
Suppose we divide the interval $$[a,b]$$ into many subintervals, each of width $$dy$$. Then, if $$dy$$ is sufficiently small, the area of the subregion $$R_i$$ is well approximated by a rectangle of width $$dy$$ and height $$f(y_i)$$, as figure 1.14 indicates. By revolving $$R_i$$ about the y-axis, we sweep out a circular slab with radius $$f(y_i)$$ and thickness $$dy$$ (figure 1.15).
The volume of each slab is $$\pi (radius)^2(length) = \pi f(y_i)^2dy$$. Thus we have volume $$V$$ of the solid as
\begin{align*} V &= \lim_{dy \rightarrow 0} \sum^N_{i=1} \pi f(y_i)^2dy, \;\;\;\;\;\; N= (b-a)/dy \\ &= \int^b_a \pi f(y)^2 dy \\ \end{align*}
### 1.6.7 General Surface Areas
When the surface is more irregular and is not axially symmetric, its area can still be found. The surface must now, however, be described by a three dimensional function which gives the height as a function of the length and width coordinate: $$z = f(x,y)$$.
The dependence on two variables requires two integrals, and the method used is called double integration.
Given a function of two variables, say $$z = f(x,y)$$, we can write a double integral of $$z$$ over a region $$R$$ as: $F = \int_R \int f(x,y)dA$ with $$A$$ representing the area coordinates from the region $$R$$. This integral is more often evaluated by writing it as an iterated double integral:
$F = \int^b_a \int^{h(x)}_{g(x)} f(x,y) dy dx = \int^b_a \bigg[ \int^{h(x)}_{g(x)} f(x,y)dy \bigg] dx$
Evaluation of the “inner” integral yields a function of $$x$$, which becomes the integrand of the “outer” integral.1
When the surface is described by $$z = f(x,y)$$, its area is found using an iterated integral. The limits of integration are found by projecting the boundary of the surface onto the x,y plane. The formula for the surface area is2 $A = \int^b_a \int^{h(x)}_{g(x)} [1 + (\frac{\partial z}{\partial x})^2 +(\frac{\partial z}{\partial y})^2]^{\frac{1}{2}} dy dx$
and is best illustrated by example.
Example 7
If the animal is again considered to look like a cylinder, one improvement would be to account for the neck rising at an angle from the shoulder. To keep the calculations simple, we assume the neck rises vertically, and is also cylindrical (figure 1.16).
We determine the area of the torso by subtracting the “back” area inside the vertical cylinder, $$A$$, from the area of the horizontal cylinder, which we found was $$0.25 \pi$$. The equation for the “back” surface is $$x^2 + z^2 = (0.25)^2$$. The vertical cylinder is defined by $$x^2 + y^2 = (0.10)^2$$. The projection is then half a circle of radius 0.10 (figure 1.17), and is given by $$y = \sqrt {0.10^x - x^2}$$.
For a given $$x$$, $$y$$ varies from 0 up to $$\sqrt {0.10^x - x^2}$$ . The limits on $$x$$ are -0.10 to 0.10. The equation for the surface of the back yields the required partial derivatives: $\frac{\partial z}{\partial x} = - \frac{x}{\sqrt{0.25^2 - x^2}},\;\;\; \frac{\partial z}{\partial y} = 0$ $A = \int^{0.10}_{-0.10} \int^{\sqrt{0.10^2-x^2}}_0 [1 + \frac{x^2}{0.25^2 - x^2} + 0] dy \, dx$ This strange formula still has visual intuitive appeal since the limits on $$y$$ are obtained from the width in the $$y$$ direction (for a given $$x$$) and the limits on $$x$$ are from the length in the $$x$$ direction. The quantity in brackets accounts for the changing height of the surface. The remaining parts of the problem are left as an exercise.
### 1.6.8 Error Estimation
The integral can be used to develop approximate solutions to certain equations or approximate models of given data. The integral provides a qualitative error estimate for the approximation. The most well known application is the least squares fit of a line through a set of data points. One of the most recent applications is the residual norm as an error indicator for approximate solutions to partial differential equations. In each example discussed below, a function is integrated over a domain of interest. If this function represents the difference between the approximate and true solution, then the value of the integral decreases as the approximation improves. The integral is then minimized to provide the “best” approximation.
#### 1.6.8.1 Least Squares
Many experiments produce data as pairs of numbers $(x_1,y_1),(x_2,y_2), \ldots, (x_n,y_n)$ The underlying relationship is often assumed to be linear, that is, the model is assumed to be $y = Ax + B, \text{ where }A,B = \text{constants}$ The points ($$x_i,y_i$$) are usually not collinear (see figure 1.18) due to experimental error, inaccuracies in the model, round-off error during measurement, etc. Thus the problem is to choose the constants $$A,B$$ so the line matches the data points as closely as possible. The method of least squares uses the sum of squared deviations for the error function, $$E^2$$ (see figure 1.18):
$$$E^2 = \sum^n_{i=1} [y(x_i)-y_i]^2 \tag{1.13}$$$
where $$y(x_i) = Ax_i + B$$. The goal is then to choose $$A$$, $$B$$ so that $$E^2$$ is minimal. By writing $$E^2$$ as a function of the parameters $$A$$, $$B$$, we minimize $$E^2$$ by setting the partial derivatives equal to zero: $\frac{\partial E^2}{\partial A} = 0,\;\; \frac{\partial E^2}{\partial B} = 0$
We then obtain two linear equations in $$A$$, $$B$$ which are easily solved (see problem 5) and can be shown to give the line with the least sum of squared deviations (see problem 6).
#### 1.6.8.2 Norms
The squared deviation used in the least squares method is an example of a norm. A norm of a function $$f(x)$$, denoted $$\|f(x)\|$$, is a function (Pearson 1974, p. 946) such that, for any scalar $$\alpha$$, and any functions $$f(x), g(x)$$, $\|f(x)\| \ge 0$ $\|f(x)\| = 0$ if and only if $$f(x) = 0$$ $\| f (x) \| = |\alpha| \cdot \| f (x) \|$ $\| f (x) + g(x) \| \le \| f (x) \| + \|g(x) \|$
When a curve $$y = f(x)$$ is approximated by a least-squares straight line $$y = Ax + B$$ over the interval $$[a,b]$$, we choose $$A$$, $$B$$ to minimize the error norm $$E$$ defined by
$$$E^2 = \int^b_a [Ax + B - f(x)]^2 dx \tag{1.14}$$$
Note that equation (1.13) is a discrete analog of equation (1.14).
The norm can be used for evaluating the closeness of fit of one curve to another, or for obtaining a qualitative estimate of the accuracy of an approximate solution to an equation. The $$L_p$$ norm of a function $$f(x)$$ is defined as $L_p (f(x)) = \| f(x)\|_\rho = \bigg[\int^b_a |f(x)|^P dx\bigg]^{1/p}$
The least squares norm $$E$$ is in fact $$L_2 (f(x))$$, since $\| f(x) \|_2 = \bigg[ \int^b_a |f(x)|^2 dx\bigg]^{1/2}$ and thus $E^2 = [\|f(x)\|_2]^2 = [L_2(f(x))]^2$
Example 8
An interesting comparison is between the line fitted to the data points of example 5 and the line fitted to the smooth exponential curve of eqn. (1.12), rewritten as follows.
$$$y_1 = 7.3 (x + 3.5) e^{-0.05x} \tag{1.15}$$$
We can approximate this function with a linear function,
$$$y_2 = Ax + B \tag{1.16}$$$
by choosing $$A$$ and $$B$$ to minimize the $$L_2$$ (least squares) norm of the difference between them, $$y_1 - y_2$$. Denote the norm by $E = \|y_1 - y_2\|_2 = [\int^{30}_3 (y_1-y_2) dx]^{1/2}$ Substituting eqns. (1.15) and (1.16) yields
$$$E = [\int^{30}_{3} (7.3[x+3.5]e^{-0.05x} - Ax - B)^2dx]^{1/2} \tag{1.17}$$$
We now minimize $$E$$ by taking partial derivatives with respect to $$A$$ and $$B$$ (see Hertzberg 1977 for a review of differentiation). One first derivative will now be calculated, with the remainder of the minimization left as additional problems 2.
First, note that the minimum of $$E$$ occurs at the same values of $$A$$ and $$B$$ which minimize $$E_2$$, since $$E$$ is positive. The derivative of $$E_2$$ with respect to $$A$$ is now attempted. First, we write $\frac{\partial(E^2)}{\partial A} = \frac{\partial}{\partial A} \bigg[ \int^{30}_3 f(A,B,x)dx\bigg]$ where $f(A,B,x) = (7.3[x+3.5]e^{-0.05x}-Ax - B)^2$ Then the derivative is calculated after the integration is completed, which is not a simple task. It may be easier to differentiate under the integral sign first, and then integrate the result.
Theorem (Pearson 1974, p. 100).
Let $$f(x,y)$$ be an integrable function of $$x$$ for each $$y$$, and let $$\partial f / \partial y$$ be continuous over $$a \le x \le b, c \le y \le d$$. Then $\frac{d}{dy} \int^b_a f(x,y)dx = \int^b_a (\partial f / \partial y)dx$
Thus we evaluate:
$$$\frac{\partial (E^2)}{\partial A} = \int^{30}_3 \frac{\partial f}{\partial A} dx \\ = \int^{30}_3 2(7.3[x+3.5]e^{-0.05x}-Ax-B)(-x)dx \tag{1.18}$$$
which can now be integrated.
No justification has yet been given for the constants in eqn. (1.15). These, too, may be determined by a least squares norm. See additional problems 1 for details.
The norm can also be used to estimate qualitatively the relative accuracies of approximate solutions to a differential equation. The limits are determined as above by the interval in which the accuracy is to be judged. An example is presented in the next section.
## 1.7 DIFFERENTIAL EQUATIONS
As was previously mentioned, the integral is also used to solve a differential equation. In the simplest form, if an equation is written as $\frac{dx}{dt} = f(t)$ then the solution (general solution) is $x(t) = \int f(t)dt$ If we know the value of $$x$$ at $$t = 0$$, the initial condition, then the differential equation has a unique solution given by $x(t) = \int^t_0 f(\tau)d \tau + x(0)$
### 1.7.1 Separation of Variables
When the differential equation also involves $$x$$ on the right-hand side, the solution is not so easily represented. Other techniques must then be employed to represent the problem by a set of integrals whose integrands are functions of only the variable of integration. One such technique is separation of variables.
Example 9
$$$\frac{dN}{dt} = rN(1 - N/K) \tag{1.19}$$$
The simple representation given above fails here: $N(t) = \int rN(1 - N/K)dt$ To integrate, we need to know $$N$$ as an explicit function of $$t$$, which of course we do not know. However, by multiplying both sides of eqn. (1.19) by the differential $$dt$$, we obtain $dN = rN(1 - N/K)dt$
and by dividing by $$N(1 - N/K)$$ we successfully separate the variables $$N$$, $$t$$ on either side of the equality: $\frac{dN}{N(1- N/K)} = rdt$
Integration is now possible since each integrand is a function of just the variable of integration. $\int \frac{1}{N(1- N/K)} dN = \int rdt$
How are initial conditions incorporated now? Since the indefinite integral produces an integration constant, the constant is determined by the initial condition, making the general solution unique.
Example 10
A recent study on water purification (Dickson, et al, 1977) included the mortality of fish due to intermittent chlorination. The following data were obtained (Table A) where frequency is the number of doses per 24 hours.
Table A. Fish Mortality Due to Intermittent Chlorination
Mortality Frequency Duration (hr)
0.01 1 1.5
2 0.7
4 0.3
0.10 1 3.2
2 1.4
4 0.7
0.20 1 4.3
2 2.0
4 0.9
0.50 1 7.5
2 3.4
4 1.6
A mathematical model has been devised which shows the effect of duration ($$T$$) and frequency ($$F$$) on mortality ($$M$$) using two differential equations:
$$$\partial M / \partial T = a_1T^{1.5} \tag{1.20}$$$
$$$\partial M / \partial F = a_2 F^{1.8} \tag{1.21}$$$
where $$a_1$$ is a function of $$F$$ and $$a_2$$ is a function of $$T$$. Another model was derived by curve fitting, using Table A and is quite different. For a given $$M$$,
$$$\ln(T)=b_1 - b_2 \ln F \tag{1.22}$$$
where it is now assumed that $$b_1$$ is a function of $$M$$: $b_1 = 2.16 + 0.40 \ln M$
Can both of these models hold? Are they consistent? We can answer this by integrating the first model to derive a single expression for $$M$$ as a function of $$F, T$$. Each equation (1.20), (1.21) involves a single partial derivative, so that one of the variables can be assumed constant. By holding $$F$$ constant, we integrate eqn. (1.20) to obtain $M = (a_1/1.5)T^{2.5} + C_1$ which is more precisely written, with $$a_1$$ as a function of $$F$$, as
$$$M = \frac{a_1(F)}{1.5}T^{2.5}+C_1 \tag{1.23}$$$
Similarly, we integrate eqn. (1.21) with respect to $$F$$, holding $$T$$ constant.
\begin{align*} M &= \int a_2 F^{1.8} dF = (a_2 / 1.8) F^{2.8} + C_2 \\ &= \frac{a_2(T)}{1.8} F^{2.8} +C_2 \\ \end{align*}
Thus $\frac{a_1(F)}{1.5} T^{2.5} +C_1 = \frac{a_2(T)}{1.8}F^{2.8} +C_2$
We assume no observable mortality if no chlorination occurs, so that, if $$F = 0$$ or $$T = 0$$, then $$M = 0$$. Thus $$C_1= C_2 =0$$. $\frac{a_1(F)}{1.5} T^{2.5} = \frac{a_2(T)}{1.8} F^{2.8}$ Therefore, we must conclude that $a_1(F) = 1.5C F^{2.8}$ $a_2(T) = 1.8C T^{2.5}$ where $$C$$ is some proportionality constant. From eqn. (1.23) we have
$$$M = C F^{2.8} T^{2.5} \tag{1.24}$$$
It is now straightforward to manipulate eqn. (1.22) to look like eqn: (1.24). $\ln T = (2.16 + 0.40 \ln M) - b_2 \ln F$ $\frac{\ln T + b_2 \ln F - 2.16}{0.40} = \ln M$ $\frac{\ln T + ln(F^{b_2}) - \ln(e^{2.16})}{0.40} = \ln M$ $2.5 \ln (TF^{b_2}/8.7) = lnM$ $\ln[(TF^{b_2} /8.7)^{2.5} ] = \ln M$ $T^{2.5} F^{2.5b_2} /(8.7^{2.5} ) = M$ Thus, $$b_2 = 2.8/2.5 = 1.1, \;\;C = 1/(8.7)^{2.5} = 0.0045$$ and the models agree.
### 1.7.2 Residual Norm
Often a differential equation cannot be solved by any standard methods and an approximate solution is developed. In some cases, bounds on the error in the approximation can be derived. Too often, however, calculating the error bounds is as complicated as determining the solution itself. A recent idea is to use the least-squares norm (or occasionally some other norm) as a gauge of the accuracy of the approximation. With only the differential equation at hand, a new type of norm, the residual norm, is used.
Consider a differential equation written as $\frac{dx}{dt} = f(x,t)$
with initial condition $$x(0) = x_0$$. If the equation is rewritten as $dx/dt - f(x,t) = 0$
then we obtain an expression which vanishes when the exact solution is used. In operator notation the equation is $L(x,t) = 0$ where $$L$$ is the differential operator defined by $L(x,t) = dx/dt - f(x,t)$ If $$L$$ operates on an approximate solution, $$\overset\sim x (t)$$, then $$L(\overset\sim x) \neq 0$$, and its value is the residual. Yet we would expect $$L(\overset\sim x)$$ to be small if $$x$$ is a good approximation. The final step is to decide on some interval over which the approximation’s accuracy is to be judged, say $$[a,b]$$. Then the least-squares residual norm for the approximate solution $$\overset\sim x (t)$$ is $\| R(\overset\sim x) \|_2 = [\int^b_a (L(\overset\sim x,t))^2 dt]^{1/2}$
Note that $$\| R(\overset\sim x) \|_2 = 0$$ if the exact solution is used, since $$L$$ is then 0. This limiting behavior suggests the residual norm as a qualitative indicator of the accuracy of an approximate solution. No theory exists, however, for estimating the actual error from the value of the residual norm.
Example 11
A deer population is threatened by severe storms as well as reduced grazing area due to a small fire. Consequently, a moratorium on deer hunting is imposed until the deer population returns to 60% of the carrying capacity for the area. How long should the moratorium be imposed?
Assumptions are now presented which allow us to model the population dynamics of the deer and estimate the duration of the moratorium. Let the logistic growth model be used: $\frac{dn}{dt} = Rn(1-n/K)$
Assume that the current population size is 400, the carrying capacity ($$K$$) is 2000, the intrinsic growth rate ($$R$$) is 1.0 and that the approximate solutions should be fairly accurate for the first three years. Assume also that the moratorium is not obeyed immediately, so that hunting pressure gradually tapers off. Let the growth model with the “harvesting” (i.e. hunting) term be
$$$\frac{dn}{dt} = Rn(1-n/K) - e^{-t}n \tag{1.25}$$$
with initial condition $$n(0)-n_0$$.
Assume also that you do not have access to either a programmable calculator or a computer. This equation cannot be solved by separation of variables (try it), so some approximation must be made. From previous studies, you learn that the following two models have been used fairly successfully to model short term population growth.
$$$n_1(t) = (K-n_0)(1-e^{-Rt/5} ) + n_0 \tag{1.26}$$$
$$$n_2(t) = n_0 e^{Rt/3} \tag{1.27}$$$
Note that $$n_1(t)$$ tapers off at the carrying capacity, but $$n_2(t)$$ is exponential growth for $$n_2 \approx 0$$; both are properties the exact solution must possess. Which approximation is better? We compare $$n_1$$ and $$n_2$$ by evaluating their least-squares residual norms. Rewrite eqn.(1.25) as $$L(n) = 0$$, where $$L(n)$$ represents the differential operator acting on $$n$$: $L(n) = dn/dt - Rn(1-n/K) + e^{-t}n = 0$
We expect a good approximate solution, $$n_a$$, to give $L(n_a) \approx 0$ and thus its residual norm should also be small: $\| R(n_a) \|_2 \equiv [\int^T_0 (L(n_a))^2 dt]^{1/2} \approx 0$ Intuitively, the smaller $$\| R \|_2$$ is, the better the approximation should be. For this example, assume the following values: $K = 2000, \; R=1, \; n_0=400, \; T=3$
The integrands in $$\| R(n_1) \|_2$$ and $$\| R(n_2) \|_2$$ are now calculated. From eqn. (1.26) we obtain, after a few steps,
$$$[L(n_1)]^2 = (2000 e^{-t}- 1600 e^{-1.2t} + 1280 e^{-0.4t} -2800 e^{-0.2t} +1600)^2 \tag{1.28}$$$
From eqn. (1.27) we obtain
$$$[L(n_2)]^2=[ (400/3) e^{t/3} - (1-400 e^{t/3}/2000 -e^{-t}) 400 e^{t/3}]2 \tag{1.29}$$$
After multiplying out the square in each equation and collecting terms, we can write both (1.28) and (1.29) as sums of exponential functions, and thus both can be integrated by hand easily (but tediously). The results are, upon integrating over the interval $$[0,3]$$: $\| R(n_1) \|_2 = 661.8$ $\| R(n_2) \|_2 = 172.3$ Figure 1.19 shows the accuracy of the two approximate solutions as compared to a numerical solution to the original differential equation (1.25). It appears that, for the chosen values of $$R$$, $$K$$ and $$n_0$$, the approximation $$n_2(t)$$ is better, and the residual norm for $$n_2$$ supports this evaluation. The 60% of carrying capacity, i.e. 1200, is reached in 3.30 years using $$n_2(t)$$, and in 3.08 years according to the numerical solution $$n(t)$$. It is interesting that, without the term for hunting (curve $$n_0(t)$$ in fig. 1.19), the population reaches the 60% level in only 1.79 years.
## 1.9 PROBLEM SET
General
1. An application of both the definite integral and exponential growth is modelling demand for natural resources. Assume $$P(t)$$ represents the rate of use of a resource at time $$t\geq0$$, and $$P(t)$$ represents the rate of consumption at time $$t = 0$$. The exponential model then gives $$P(t) = P_0 e^{kt}$$. The constant $$k$$ can be defined as the rate of increase in the use of this resource. Let $$A(T)$$ represent the amount of resource used during the interval $$[0,T]$$, $$T\geq0$$.
1. Write the definite integral representing $$A(T)$$ in terms of a general function $$P(t)$$.
2. Evaluate the integral for general $$T\geq0$$ by substituting the above exponential function for $$P(t)$$.
3. In 1973 (say $$t = 0$$), the world use of copper was estimated to be 6.99 x 109 kg, and the demand was increasing at an exponential rate of 8% per year. How many tons of copper will then be used from 1973 to 1983? (Hint: First find $$P_0$$, $$k$$, $$T$$ and then use your answer to b.)
4. If the rate of growth in demand for copper remains at 8% and no new reserves are discovered, when will the world supply of copper be exhausted? Assume the reserves in 1973 are 336 x 109 kg, and no recycling occurs.
1. Five thousand trout, each weighing 80 grams, are planted in a lake. The population size, $$N(t)$$, declines exponentially according to the equation $$N(t) = 5000e^{-0.5t}$$ where $$t$$ is measured in months. Each fish grows according to the formula $$W(t) = 10000 (1 - 0.8e^{-0.05t})^3$$ where $$W$$ is the weight in grams.
1. Check that both $$N(t)$$ and $$W(t)$$ satisfy the given initial conditions.
2. Write the formula for the biomass (total weight of all the planted trout) in the lake as a function of $$t$$, and calculate the biomass at 12 months.
3. What is the average biomass for the first 12 months? What was the initial biomass? Sketch how you think the graph of biomass versus time would appear. Mark on the vertical axis the initial, final and average biomass values.
1. Assume that a radioactive element disintegrates so that the number ($$N$$) of atoms present at a given time ($$t$$) decreases at a rate proportional to $$N$$ itself. Let $$k$$ be the constant of proportionality and $$N_0$$ be the number of atoms present at $$t = 0$$ (thus $$N(0) = N_0$$).
1. With the derivative representing the rate of change of $$N$$, use the above defined parameters and variables to write an equation for the rate of decrease in the number of radioactive atoms with passing time.
2. Solve this differential equation for $$N(t)$$ so that the solution satisfies the initial condition, $$N(0) = N_0$$. (Hint: first separate variables. Do not forget the integration constant!)
3. Determine the “half-life” : the time ($$t_1$$) such that $$N(t_1) = 0.5 N_0$$. At $$t =t_1$$, half of the original number of radioactive atoms have disintegrated (decayed).
4. We now make certain assumptions which will allow us to estimate the age of a piece of wood by radiocarbon dating:
1. living organic material contains the carbon isotopes $$C^{12}$$ and in a fixed proportion independent of time,
2. the $$C^{12}$$ atoms do not disintegrate,
3. the $$C^{14}$$ atoms disintegrate radioactively with a half-life of 5568 years, and
4. the $$C^{12}$$ and $$C^{14}$$ atoms are not replaced once the organism dies, even though the $$C^{14}$$ atoms are being lost through disintegration. A sample of wood in an American Indian cliff dwelling is measured to have 87% of the $$C^{14}$$ isotope expected in living wood. Estimate the age of the sample.
1. The survival and health of intertidal organisms often depend on the amount of time they are exposed, i.e. not covered by sea water. A given position on the beach can be identified by its tidal height: the height of the tide when the water’s edge just touches that spot. A model of tidal height as a function of time can then be used to approximate the length of time between successive high and low tides (or vice versa) during which a position on the beach is exposed.
The following data were taken from tide tables for Port Townsend, Washington for the day of maximum tide difference for daylight low tides.
Time Height (m)
4:35 2.47
11:49 -0.82
19:40 2.77
For each of the two times intervals 4:35-11:49, 11:49-19:40, the model uses the following cosine function fitted to the tide data: $H = a \cos (b(T-T_h)) + H_a,\: H_a=\frac{high\:tide + low\:tide}{2}$
1. Sketch by hand a graph of the cosine function with the maximum at the high tide level and a minimum at the low tide level for the first time interval to aid in interpreting the equation. Consider the general form of a cosine function: $$y=a \cos(bx+c)+d$$, where the amplitude is $$|a|$$; the period is $$2\pi/b$$; the phase shift in $$-c/b$$; and the vertical shift in $$d$$. Use this information to determine what the constants a, b, $$T_h$$ represent in the provides tidal cosine function? Calculate these constants and $$H_a$$ for each of the two time intervals.
2. The constant $$H_a$$ seems to be the average tide height. Verify this for the first time interval by evaluating the appropriate definite integral. Prove that the average height $$\overline H$$ for the two intervals combined can be written as the weighted average of the average heights $$\overline H_1$$, $$\overline H_2$$ for both subintervals. That is, show that (using decimal hours)
$\overline H= \frac{1}{15.09} \int^{19.67}_{4.58} H(t)dt= \frac{7.24}{15.09} \overline {H_1} + \frac{7.85}{15.09} \overline H_2$
1. For a point on the beach at the 0.0 meter tide level, how long is it exposed between these successive high tides? (Hint: find the inverse function T=g(H), but be careful with the second interval.)
Least Squares
1. Write equation (1.13) using $$y(x_i) = Ax_i + B$$. Evaluate the two first partial derivatives and, by setting both equal to zero, obtain two linear equations in A and B.
1. For the following data, derive the “least squares” line and prove that for your choices for A and B, the sum of squares is at a true minimum: $f(x,y) \text{ is at a minimum at } x_o, y_o \text{ if, for } x=x_0, y=y_0$
1. $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$$,
2. $$\frac{\partial^2 f}{\partial x^2}-\frac{\partial^2 f}{\partial y^2}>0$$, and
3. $$\frac{\partial^2 f}{\partial^2 x}>0$$.
Data from figure 1.8 in text
x=c(3,5.3,9.6,18,30)
y=c(39,55,57.5,60,57)
1. Now determine a “best” exponential fit to the above data. First transform the desired function $$y = Axe^{Bx}$$ into a linear function (with respect to $$x$$): $$h(x,y) = f(A,B) + g(A,B)x$$. A neater format is $$h= f + gx$$. Make a table of values for $$x$$ and $$h$$. Determine the least squares line through this new data set and, in the process, calculate $$f$$, $$g$$, and thus $$A$$, $$B$$.
## 1.10 ANSWERS TO THE PROBLEM SET
General
1. $$A(T)$$ represents the total change in the resource. Thus the rate of change, $$P(t)$$, is integrated: $A(T)=\int^T_0 P(t)dt$
1. With $$P(t) = P_0 e^{kt}$$, $A(T)=\int^T_0 P_0 e^{kt} dt= (P_0/k)(e^{kT}-1)$
2. The initial use rate in $$P_0$$, the rate of increase is $$k$$, and the time interval ends at 1983-1973 = 10 years. Thus
P0= 6.99 * 10^9 #kg
k=0.08
T=10.0 #years
#Estimate A(T) in kg
(P0/k)*(exp(k*T)-1)
## [1] 107081638627
1. We solve for $$T$$. The total change equals the current reserves (at $$t=0$$); thus, the parameters are
require("stats")
library(stats)
P0= 6.99 * 10^9 #kg
k=0.08
A_T= 336 * 10^9 #kg
#Using A(T)= (P0/k)*(exp(k*T)-1)
#Solve T in years
Tsolve= function(T) (P0/k)*(exp(k*T)-1) - A_T
uniroot(Tsolve, interval=c(0, 100))\$root
## [1] 19.72562
2a. Check the initial conditions: N(0)=5000, W(0)=80.
N_t= function(time) 5000*exp(-0.5*time)
N_t(0)
## [1] 5000
W_t= function(time) 10000*(1-0.8*exp(-0.05*time))^3
W_t(0)
## [1] 80
#Biomass B(t)=N(t)*W(t)
#Calculate biomass at 12 months
B_t= function(time) 5000*exp(-0.5*time) * 10000*
(1-0.8*exp(-0.05*time))^3
B_t(12)
## [1] 21876.47
1. Average biomass (g) for the first 12 months, $$\overline B(12)$$ uses the definite integral. $\overline B(12)= (1/12) \int^{12}_0 B(t)dt$ Substitution for $$B(T)$$ gives $\overline B(12)= (1/12) \int^{12}_0 5.0\times10^7 e^{-0.5t} (1-0.8e^{-0.05t})^3 dt$ $\overline B(12)= 4.17 \times 10^6 \int^{12}_0 e^{-0.5t}(1-0.8e^{-0.05t})^3 dt$ Integrate by parts where $$u=(1-0.8e^{-0.05t})^3$$ and $$dv= e^{-0.5t} dt$$ $\int^{12}_0 e^{-0.5t}(1-0.8e^{-0.05t})^3 dt$
$= \frac{e^{-0.5t}}{-0.5}(1-0.8e^{-0.05t})^3 \Biggr|^{12}_0 - \int^{12}_0 \frac{e^{-0.5t}}{-0.5}(3)(1-0.8e^{-0.05t})^2(0.04)e^{-0.05t}dt$ $=(-0.00496)(0.177)+0.016+0.24 \int^{12}_0 e^{-0.55t}(1-0.8e^{-0.05t})^2 dt$ Integrate by parts again with $$u=(1-0.8e^{-0.05t})^2$$ and $$dv= e^{-0.55t} dt$$ An intermediate stage is (two steps skipped), with the left side included, $\int^{12}_0 e^{-0.5t}(1-0.8e^{-0.05t})^3 dt= 0.0324 +0.0349 \int^{12}_0 e^{-0.60t}(1-0.8e^{-0.05t}) dt$ $= 0.0324 +0.0349 \int^{12}_0 e^{-0.60t}-0.8e^{-0.65t} dt = 0.0324 + 0.0349(0.4353)=0.0476$
#Biomass B(t)=N(t)*W(t)
#Calculate average biomass (g) for the first 12 months
(4.17*10^6)*(0.0476)
## [1] 198492
B_t= function(time){5000 * exp(-0.5 * time) * 10000 *
(1 - 0.8 * exp(-0.05 * time))^3}
#Initial biomass in g
B_t(0)
## [1] 4e+05
#Plot the graph of biomass over time.
plot(0:12, B_t(0:12)/10^3, ylab="Production (kg)",
xlab= "Time (months)", type='l')
3a. The rate of change of $$N$$ is a decrease (i.e. negative) and proportional to $$N$$. For $$k > 0$$, we have $$\frac{dN}{dt}=-kN$$.
1. Separate variables $\frac{dN}{N}=-kdt$ $\int \frac{dN}{N}= \int -kdt$ $$\ln N= -kt+C$$, $$C$$ = integration constant $N= e^{-kt}e^{C}$ But $$N(0)=N_0$$. Thus $$N(0)=N_0=e^0 e^C=e^C$$ and the solution is $$N(t)= N_0e^{-kt}$$.
2. Substitute into the solution equation. \begin{align*} N(t_1)=N_0e^{-kt_1} \\ 0.5 N_0= N_0e^{-kt_1} \\ \ln(0.5)=-kt_1 \\ t_1= -(\ln(0.5))/k \\ = (\ln 2)/k \end{align*}
3. Let N(t) be the number of $$C^{14}$$ atoms in the wood. Then $$N(t)/N_0 = 0.87$$. The half life of 5568 gives $$(\ln2)/k=5568$$, $$k= (\ln 2)/5568=0.0001245$$. The equation for N(t) then gives \begin{align*} N(t)/N_0=e^{-kt} \\ 0.87 = \exp(-0.0001245t)\\ \ln(0.87) = -0.0001245t\\ t=1119 years \end{align*}
1. a is the amplitude (half the total change in tidal height); b is $$2\pi/p$$ where p=period=$$2|T_{high}-T_{low}|$$; and $$T_h$$ is the phase shift or lag. The simple cosine function begins at $$t=0$$ so the lag is the time of high tide. $$H_a$$ is the vertical shift.
#first interval
h_h= 2.47
h_l= -0.82
time_h= 4+35/60
time_l= 11+40/60
#second interval
h_h= 2.77
h_l= -0.82
time_h= 19+40/60
time_l= 11+40/60
a= (h_h-h_l)/2
p= 2*abs(time_h-time_l)
b=2*pi/p
time_h= time_h #phase lag, time of high tide
Ha= (h_h+h_l)/2
a; p; b; time_h; Ha
## [1] 1.795
## [1] 16
## [1] 0.3926991
## [1] 19.66667
## [1] 0.975
H=function(a,b,time, time_h, Ha) a*cos(b*(time-time_h))+Ha
plot(1:24, H(a=a,b=b,time=1:24, time_h=time_h, Ha=Ha), type='l',
xlab="Tide height (m)", ylab="Time (hours)")
b. $$\overline H$$= average height, first interval = 7.24 hours. \begin{align*} \overline H&= \frac{1}{7.24} \int^{11.82}_{4.58}a \cos(bT-bT_h)+H_a dt \\ &= \frac{1}{7.24} (-a/b)(\sin(11.82b-4.58b)-\sin(4.58b-4.58b)) + \frac{1}{7.24} (H_a)(11.82-4.58)\\ &=-\frac{a}{7.24b}(\sin(7.24(2\pi/2(7.24)))-0)+H_a\\ &=H_a \text{, since } \sin \pi=0. \end{align*}
For the two intervals combined, we must weight the average height in each interval by the length of the time interval. \begin{align*} \overline H&= \frac{1}{15.09} \int^{19.67}_{4.58}H(t)dt=\frac{1}{15.09}\int^{11.82}_{4.58}H(t)dt +\frac{1}{15.09}\int^{19.67}_{11.82}H(t)dt\\ &=\frac{7.24}{15.09}\frac{1}{7.24}\int^{11.82}_{4.58}H(t)dt +\frac{7.85}{15.09}\frac{1}{7.85}\int^{19.67}_{11.82}H(t)dt\\ &=\frac{7.24}{15.09} \overline H_1 +\frac{7.85}{15.09} \overline H_2\\ \end{align*}
1. We need the times of day when the 0.0 tide level is reached. The easiest way is to invert the function $$H(t)$$ for each subinterval. \begin{align*} &H= a \cos(b(T-T_h))+H_a\\ &(H-H_a)/a= \cos(b(T-T_h))\\ &\cos^{-1}((H-H_a)/a)=b(T-T_h)\\ &(1/b) \cos^{-1}((H-H_a)/a)+T_h=T\\ \end{align*} Now substitute the appropriate constants and 0 for $$H$$.
#first interval
h_h= 2.47
h_l= -0.82
time_h= 4+35/60
time_l= 11+40/60
#second interval
h_h= 2.77
h_l= -0.82
time_h= 19+40/60
time_l= 11+40/60
a= (h_h-h_l)/2
p= 2*abs(time_h-time_l)
b=2*pi/p
time_h= time_h #phase lag, time of high tide
Ha= (h_h+h_l)/2
H=0
#Estimate T
(1/b)*acos((H-Ha)/a)+time_h
## [1] 25.12889
This answer is obviously wrong since the 0.0 height must be attained sometime between low tide (11.82 hours) and the next high tide (19.67 hours). The fallacy is in treating the “arcos x” as an inverse function. It is one-to-one only when the domain of “cos x” is restricted to $$[0,\pi]$$; i.e., arcos x always has $$[0,\pi]$$ as its range. Thus the first term is $\frac{1}{0.400}\cos^{-1}(-0.975/1.795)$ correctly tells how far the maximum ($$T_h$$) is from the desired time $$T$$, but now whether $$T$$ lies above or below $$T_h$$. With the second subinterval, $$T$$ is obviously below $$T_h$$ and hence we subtract from $$T_h$$:
19.67-1/0.400*acos(-0.975/1.795)
## [1] 14.30747
\begin{align*} \frac{\partial E}{\partial B} &=\sum^n_{i=1}2[Ax_i+B-y_i] \\ E&= \sum^n_{i=1}[Ax_i+B-y_i]^2\\ \frac{\partial E}{\partial A} &= \sum^n_{i=1}[Ax_i+B-y_i]x_i\\ \end{align*} Set equal to zero: \begin{align} \sum^n_{i=1}2[Ax_i+B-y_i]x_i=0 \notag \\ \sum^n_{i=1}[Ax_i+B-y_i]x_i=0 \notag \\ \frac{\partial E}{\partial B} =\sum^n_{i=1}2[Ax_i+B-y_i] \tag{1.30} \end{align}
Equate to 0 and divide by 2: $$$\sum^n_{i=1}[Ax_i+B-y_i]=0 \tag{1.31}$$$ Thus we have two equations which are linear in A and B. This is more easily seen if we rework them. From (1.30)
$$$A(\sum^n_{i=1}x_i^2)+B(\sum^n_{i=1}x_i)-(\sum^n_{i=1}x_iy_i)=0 \tag{1.32}$$$
And (1.31) becomes $$$A(\sum^n_{i=1}x_i^2)+nB-(\sum^n_{i=1}y_i)=0 \tag{1.33}$$$
1. Eq. (1.32) is: $A(1353.25) + B(65.90) - (3750.50) = 0$ Eq. (1.33) is: $A(65.90) + 5B - (268.50) = 0$
We solve for A and B.
\begin{align*} B &= (268.50 - 65.90A)/5 \\ A &= (3750.50 - 65.90B)/(1353.25) \\ &= 2.771 - 2.615 + 0.642A = 0.156 + 0.642A \\ \end{align*}
Thus
\begin{align*} A &= 0.156/(1 - 0.642) = 0.436 \\ B &= (268.50 - 65.90(0.436))/5 = 47.95 \\ \end{align*}
To prove that these values for $$A$$, $$B$$ do give a minimum sum of squares, we must show that, for $$A$$ = 0.44, $$B$$ = 47.95,
1. $$\partial f/\partial A=\partial f/\partial B = 0$$
2. $$(\partial^2f/\partial A^2)(\partial^2f/\partial B^2)-(\partial^2f/\partial A\partial B)^2 > 0$$
3. $$\partial^2f/\partial A^2 > 0$$,
where $$f$$ is the sum of squares $f=\sum_i(y_i-Ax_i-B)^2$ We have \begin{align*} \partial f/\partial A &= -2\sum_i(y_i-Ax_i-B)x_i \\ \partial f/\partial B &= -2\sum_i(y_i-Ax_i-B) \\ \end{align*}
Since $$A$$, $$B$$ were calculated by setting these derivatives to zero, we do not need to check (i). \begin{align*} \partial^2f/\partial A^2 &= 2\sum_ix^2_i,\;\;\partial^2f/\partial B^2 = 2n \\ \partial^2f/\partial A\partial B &= 2 \sum_ix_i \\ \end{align*}
Substituting into (ii) gives $(2\sum_ix^2_i)(2n)-4(\sum_ix_i)^2=20\sum_ix^2_i-4(\sum_ix_i)^2$ $= 20(1353.25) - 4(65.90)^2$ $=9693.76 > 0$ Note that (iii) is obviously satisfied so we have a minimum for $$f$$.
1. Begin with $y = Axe^{Bx}$ Divide by $$x$$ to isolate the exponential function. $y/x=Ae^{Bx}$ Now take logarithms of both sides. $\ln(y/x)=\ln A+Bx$ Thus, in the form $$h=f+gx$$, we have \begin{align*} h &= \ln(y/x) \\ f &= \ln A \\ g &= B \\ \end{align*}
The transformed table is below.
$$x$$ $$h$$
3.0 2.565
5.3 2.340
9.6 1.790
18.0 1.204
30.0 0.642
The least squares calculations give the transformed equation
$h = 2.642 - 0.071x$
and thus
$y = 14.041x e^{-0.071x}$
1. The following graph 1.20 compares the least squares line and exponential curve of parts (a) and (b) with the data and the exponential curve in the text eqn. (1.12). The fit of eqn. (1.12) seems superior to that of the curve from (b). However, the fit of
$h = 2.642 - 0.071x$
to the transformed data is quite good, as shown by figure 1.21. Thus we conclude that a good least squares fit to transformed data does not necessarily imply a good curve fit to the original data.
1. One way to use least-squares is to approximate one function by another. Here we approximate the curve $y = 6.1(x+5.0)e^{-0.045x}$ by a straight line $y = Ax + B$
Evaluate both $$\partial (E^2)/\partial A$$ and $$\partial (E^2)/\partial B$$, then set both equal to zero to obtain two equations in A and B. The goal is to minimize: $E^2 = \int^{30}_3[6.1(x+5.0)e^{-0.045x}-Ax-B]^2dx$ (Hint: when evaluating $$\partial E^2/\partial A$$ and $$\partial E^2/\partial B$$, differentiate under the integral sign.)
Solve your equations for A and B and the least squares integral.
1. The stomates in leaves are small pores which permit the exchange of gases and water vapor. The stomates change shape from a near circle to a slit in order to adjust this exchange in response to varying environmental conditions. Each stomate is roughly elliptical in shape with constant perimeter throughout the shape change. The area of the opening is then given by $A= \frac{2b}{a}\int^a_{-a}(a^2-x^2)^{1/2}dx$ where $$2a$$, $$2b$$ are the lengths of the major and minor axes, respectively, of the ellipse.
1. First evaluate this integral using either of two trigonometric substitutions: let $$x = a \cos \theta$$ or let $$x = a \sin \theta$$. Be sure to check the limits on $$\theta$$ and the sign of each trig function in the transformed integrand. Your result must be non-negative since the area is non-negative. For a fixed perimeter, say $$35 \mu$$, the area should be maximal when the hole is a circle, i.e. $$a=b$$. Demonstrate this by evaluating the area when $$a=b=5.57$$, and when $$a=5$$, $$b=6.09$$. (The perimeter formula is $$P = 2\pi\sqrt{(a^2+b^2)/2}$$.)
2. Use your area formula (the answer to part a) to investigate the dilution of the sun’s energy flux due to the angle at which the rays strike the earth’s surface. The dilution is caused by a fixed amount of solar energy being spread over a larger area (as the rays become more horizontal) and thus the energy per unit area decreases as compared to the energy density of vertical rays. The situation is pictured below, giving a top and side view for two inclination angles. The sunlight passes through a circular hole in an opaque sheet and strikes the earth at an angle $$R$$ from the perpendicular.
The illuminated area changes from a circle ($$R=0$$) to an elongated ellipse (as $$R$$ increases). For a given diameter $$B$$, write the illuminated area $$A$$ as a function of the angle of inclination $$R$$. Check your result by choosing values for $$B$$ and $$R$$ and calculating the dilution $$D$$, which equals the circle’s area divided by the ellipse’s area. Does $$D = 0.6$$ mean more dilution (less energy density) than $$D = 0.3$$ or vice versa? The inclination angle in Minnesota changes from about 20° in summer to 65° in winter. What is the resulting change in energy density, measured by $$D$$?
## 1.12 SOLUTION TO THE ADDITIONAL PROBLEMS
1. First partial derivatives: \begin{align*} \frac{\partial E^2}{\partial A} &= \int^{30}_3\frac{\partial}{\partial A}(6.1(x+5.0)e^{-0.045x}-Ax-B)^2dx \\ &= 2 \int^{30}_3(6.1(x+5.0)e^{-0.045x}-Ax-B)(-x)dx \end{align*} Equating to zero and dividing by 2 gives $0 = -\int^{30}_36.1(x-2+5x)e^{-0.045x}dx +\int^{30}_3Ax^2+Bxdx$ Successive integration by parts (for the first integral) and direct integration (for the second integral) yield the first equation: $0 = 8991.0A + 445.5B - 25984.9$ \begin{align*} \frac{\partial E^2}{\partial B} &= \int^{30}_3\frac{\partial}{\partial B}(6.1(x+5.0)e^{-0.045x}-Ax-B)^2dx \\ &= -2 \int^{30}_3(6.1(x+5.0)e^{-0.045x}-Ax-B)dx \end{align*}
Equating to zero, dividing by 2 and solving yields the second equation: $0 = 445.5A + 27.0B - 1568.8$
The solutions are $$A= 0.061$$, $$B = 57.1$$. The least squares integral using the approximation $$Y = 0.061x + 57.1$$ is 546.452.
1. Let $$x=a\sin\theta$$. The limits of integration are then changed as follows $-a \le x \le a$ $-a \le a\sin\theta \le a$ $-1 \le \sin\theta \le 1$
There are many options for the range of $$\theta$$. We first change the integrand to see which range is suitable. We know the integrand is non-negative since $$x^2 \le a^2$$. Substitution gives $x = a\sin\theta$ $dx = a\cos\theta\:d\theta$ $\sqrt{a^2-x^2}dx = \sqrt{a^2 - a^2\sin^2\theta}\cdot a\cos\theta \:d\theta$ The integrand is then non-negative when the positive square root is used and when $$\cos \theta$$ is non-negative. Thus, we choose the limits of integration to be $$-\frac{\pi}{2}$$, $$\frac{\pi}{2}$$ since $$\sin(-\frac{\pi}{2}) = 1, \sin(\frac{\pi}{2}=1)$$ and $$\cos\theta \le0$$ if $$-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$$.
The integral is then $\int^{\pi/2}_{-\pi/2}\sqrt{a^2-a^2\sin^2\theta}a\cos\theta \:d\theta = a^2\int^{\pi/2}_{-\pi/2}\sqrt{1-\sin^2\theta}\cos\theta\:d\theta$ \begin{align*} &= a^2\int^{\pi/2}_{-\pi/2}\cos^2\theta\:d\theta=a^2[\frac{\theta}{2}+\frac{\sin2\theta}{4}]^{\pi/2}_{-\pi/2} \\ &= a^2[\pi/4+(1/4)(0)-(-\pi/4)-(1/4)(0)] =\pi a^2/2 \end{align*} Thus $A=\frac{2b}{a}\int^a_{-a}\sqrt{a^2-x^2}dx = \pi ab$ When $$a = b = 5.57$$, $$A = 97.47$$.
When $$a = 5$$, $$b = 6.09$$, $$A = 95.66$$.
From the right-hand diagram, we have $\cos R = B/x$ and thus $x=B/\cos R$ The minor axis is then the circle’s diameter, $$B$$. In the answer to part “a”, we replace $$2a$$ by $$B/\cos R$$ and $$2b$$ by $$B$$.
\begin{align*} A &= \pi ab = \pi(B/2\cos R)(B/2) \\ &= \pi B^2 / (4 \cos R) \end{align*}
Since the area of the circle is $$\pi B^2/4$$,the dilution coefficient is $D = \frac{\pi B^2/4}{\pi B^2/4\cos R} = \cos R$ Thus $$D=0.6$$ means less dilution than $$D=0.3$$, and $$D=1.0$$ means no dilution at all. For Minnesota, the coefficients are
$$D=0.94$$ for $$R=20°$$
$$D=0.42$$ for $$R=65°$$
1. Note that the order of integration can be reversed when $$f(x,y)$$ is continuous in $$x$$ and $$y$$.↩︎
2. see Ayres (1964), p.319.↩︎<|endoftext|>
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limate change policies will fail unless women have greater influence over decisions, from where to build wells to how to negotiate a global deal
The impacts of climate change will disproportionately affect people living in poverty in poorer countries, notwithstanding their minimal per-capita contributions to greenhouse-gas emissions. It’s here where the damage will be greatest and where people have the lowest capacity to cope.
According to UN Women, some 70% of people in the developing world living below the poverty threshold are women, yet gender issues receive little attention in the climate-change debate. People are vulnerable to the hazards of climate change to a greater or lesser degree depending on factors such as their wealth, education, skills, management capability and access to technology, infrastructure and information. Women’s access to these resources is often inferior to that of men, and this increases their vulnerability and limits their ability to cope with the advent of climate shocks and to recover when they have passed.
These gender-related inequalities are particularly pervasive in the developing world. Women and children are 14 times more likely to die than men during natural disasters. More than 70% of the dead from the Asian tsunami were women; roughly 87% of unmarried women and 100% of married women lost their main source of income when Cyclone Nargis hit the Ayeyarwaddy Delta in Myanmar in 2008.
In Bangladesh, social prejudice keeps girls and women from learning to swim and climb trees. Many women cannot leave their homes without accompaniment or consent from their husband or one of their male relatives. Men can more easily warn each other as they meet in public spaces, but they rarely communicate information to the rest of the family. As a result, far more women than men perish in the major floods which characterise life in the coastal areas of Bangladesh. When a cyclone and floods hit Bangladesh in 1991, the death rate for women was almost five times higher than for men.
Research by the London School of Economics found that in a sample of 141 countries over the period 1981–2002, natural disasters (and their subsequent impact) killed on average more women than men, and killed women at an earlier age than men. Boys were more likely to receive preferential treatment in rescue efforts, and women and children suffered more from food shortages and a lack of privacy and safety in the aftermath of disasters.
In India, various studies have shown that over the past decade more women than men have suffered from premature deaths on account of heatwaves and cold snaps and other climate-related extreme events. Following extreme events such as storms and floods, the burden of devastation also falls primarily on women, who must keep the family together.
Changing weather patterns could affect farming activities such as paddy cultivation in Asia, and cash crops such as cotton and tea, the cultivation of which employs many women. In Africa, for example, women contribute to 70% of food production; they account for nearly half of all farm labour and 80–90% of food processing. Wangari Maathai acknowledged this when she received the Nobel Peace Prize in Oslo on 10 December, 2004, describing Africa’s women as “the primary caretakers, holding significant responsibility for tilling the land and feeding their families”.
In Bangladesh, women and children provide nearly all household water in rural areas, both for domestic use such as drinking, cooking, bathing and washing, and for irrigating gardens and watering livestock. The intrusion of saltwater into freshwater resources in the districts of Satkhira, Khulna and Bagerhat on the south-western coast of Bangladesh is already having a disproportionate impact on women, particularly during the dry season. Because nearly all local water sources have high salinity, women must travel long distances by foot every day to find drinking water, even if they are in poor health.
Ensuring greater gender equality will benefit society as a whole and help promote sustainable development. However, getting gender issues into debates on climate change and sustainable development is happening piecemeal, extremely slowly and often as an afterthought.
This is in part due to the lack of participation by women in decision-making at all levels. In the Kilombera district in Tanzania, for example, a newly constructed well dried up. Its location had been determined by an all-male local committee, despite the fact that it was the task of local women to dig for water by hand as they know the most likely places to find water. By the same token, at the international climate-change negotiations in 2010, women made up only 30% of negotiators and just 10% of heads of delegations.
This lack of representation must change, because climate-change policies will be unsuccessful if women have no opportunity to influence decision-making, build their capacity, lower their vulnerability and diversify their income sources. In India, for example, women have played a huge role in improving the public service health sector. A 1992 constitutional amendment mandated the reservation of one-third of panchayat (local government) seats for women. Since then, relative spending on public water and latrines for low-caste communities has increased.
Special attention needs to be paid to the opportunities arising from international climate-change negotiations. Many emerging solutions intended to help people cope with climate change involve land use and agriculture in rural areas, a key sector for women.
The dependence of women on biomass energy – for example, burning wood for household cooking – means that they should also be involved in projects promoting renewable-energy resources. In El Salvador and Guatemala the primary source of fuel is wood, and it is the job of women and girls to gather it. Many spend as much as three or four hours, three to five times a week, searching for wood. And when they cook food for their households, they are exposed to toxic cooking smoke.
Agencies promoting clean, renewable energy, such as solar ovens, have found it vital to target groups of women, who can learn from one another whilst practising the new technologies. Despite this, men often influence the uptake of new-energy technologies in this domain too. In one case in Zimbabwe, men are reported to have rejected the use of solar cookers by their wives, since technology and its development are seen traditionally as a male preserve.
This piece was first published by www.chinadialogue.net. It is an edited extract from the book Climate Change and Human Development (Zed Books, 2014) by Hannah Reid<|endoftext|>
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# Difference between revisions of "2012 AIME II Problems/Problem 8"
## Problem 8
The complex numbers $z$ and $w$ satisfy the system $$z + \frac{20i}w = 5+i$$ $$w+\frac{12i}z = -4+10i$$ Find the smallest possible value of $\vert zw\vert^2$.
## Solution
Multiplying the two equations together gives us $$zw + 32i - \frac{240}{zw} = -30 + 46i$$ and multiplying by $zw$ then gives us a quadratic in $zw$: $$(zw)^2 + (30-14i)zw - 240 =0.$$ Using the quadratic formula, we find the two possible values of $zw$ to be $7i-15 \pm \sqrt{(15-7i)^2 + 240}$ = $6+2i,$ $12i - 36.$ The smallest possible value of $\vert zw\vert^2$ is then obviously $6^2 + 2^2 = \boxed{040}$.
2012 AIME II (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions<|endoftext|>
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Over the past week, dead fish have been pulled out of Germany's rivers and lakes by the ton — the extreme heat and a lack of rain have proven too much for them. Global warming is among factors altering fish habitat.
Although the Dreisam River is less than 30 kilometers (19 miles) long, the residents of Freiburg are still proud of their waterway. It runs right through the southern German city, carrying snow melt and providing a lovely opportunity to the residents of Freiburg to cool off in summer.
But these days, the river has turned into a rocky desert wash (pictured above).
The local fishing club had preemptively caught fish and released them in other bodies of water.
Peter Rudolph, an expert for aquatic biology and environmental issues in Freiburg, has been at the ready. "We regularly measure the temperatures of lakes and rivers," Rudolph told DW.
If the weather gets too dry or temperatures rise precariously, he informs authorities. Then he and other fish-savers take their electric dip nets and go to work.
Such nets create a guiding current and stun fish at the same time. With this method, large quantities of fish can be removed and transferred to other areas — without damaging the aquatic animals.
"You have to continuously measure water temperature and oxygen content," Rudolph explained. "When you see the first dead fish, it is actually too late to save the stock. All you can do then is leave the matter over to nature."
Control and timely action necessary
In the Rhine River, which begins in the Swiss Alps, runs through Germany, and empties into the North Sea, the situation is more dire.
Although the river is wider and contains more water than the Dreisam, fish particularly on the Swiss side of the Rhine have been dying by the ton.
When river water temperature exceeds 27 degrees Celsius (80.6 degrees Fahrenheit), oxygen concentration in the water drops precariously, particularly endangering fish that prefer cold water.
Over the course of the heat wave in Germany, which lasted all of July and the first two weeks of August (with some very warm weeks in May and June as well), water in sections of the Rhine warmed to 28 degrees Celsius.
In addition, unfavorable wind conditions and low water levels placed grayling and trout in particular under long-term stress.
The Federal Institute of Hydrology (BfG) put daily average temperature for the Upper Rhine at 27.2 degrees Celsius. Also the number of consecutive days with water temperature above 25 degrees Celsius is significant — according to the BfG, that's currently at 15.
Water temperature in the Upper Rhine hovered around 27 degrees Celsius this summer — due to heat and lack of rain
Climate change alters fish population
Firefighters and the Federal Agency for Technical Relief have pumped millions of liters of water into lakes for oxygen enrichment. In Switzerland, measures included digging out deeper pools to create cool zones and opening tributaries to fish.
Nevertheless, 20 tons of fish carcasses were pulled out of a reservoir near Ellwangen in Baden-Württemberg.
The mass die-off recalls memories of the summer of 2003, when 50,000 fish died in the Rhine alone. At that time, 90 percent of the grayling population in that region was lost.
Like trout, grayling occurs in low mountain regions and needs clear, cool water. Graylings thrive in fast-flowing rivers with a hard sand or stone bottom, and well-oxygenated water.
"These are typical mountain stream-dwellers, readily adapted to cold water," Rudolph explained.
Carp, chub and bream are more robust in warmer temperatures, and tolerate the lower oxygen concentration better. As climate change progresses, such species have greater chances of survival, biologists say.
Rudolph also believes that invasive species such as the topmouth gudgeon — imported from Asia — and the sunfish, originally native to the USA, will better withstand climate change. But declines of local species will reverberate through the ecosystem.
Although 2003 saw similar high temperatures, this summer, the drought has also been a major problem for the fish.
"Heat and lack of precipitation make for a disastrous combination," explained Rudolph.
Land use changes also come into play
Because in Germany many riverbanks have been deforested, water along the water's edge offers no shade. At low water levels, the fish are exposed to full sunlight, and can find no protective shelter.
Floodplains are crowded out by settlements or industrial areas, or are used by agriculture. Rivers have also been channelized to make their paths fit with human wishes.
"The river needs more space, and more trees along its banks," Rudolph said, calling for restoration. "Trees would provide shade, and underground roots that protrude into the water could provide shelter."
"But the problem is, who's going to pay for compensation if owners of the shore areas have to give up some of their land?"
A 1-degree change in water temperature can have devastating consequences for graylings that prefer a cool 23 degrees Celsius
The agricultural, shipping, traffic and mining lobbies are too powerful, complain environmental groups such as BUND (German Federation for the Environment and Nature Conservation).
The EU has been exerting pressure on Germany for years, as its Water Framework Directive prescribes largely natural plants and fish in the waters, free flow of streams and rivers to benefit all living organisms, and near-natural and unspoiled shore zones.
Lakes in climate stress
Lakes in Central Europe have also lost their ecological balance, as their water temperature has increased to up to 30 degrees Celsius.
Read more: Can we save our lakes from global warming?
Light and heat promote the growth of algae and plants. Especially treacherous is blue-green algae, which can produce toxins, endangering bathers. This can even contaminate drinking water.
Moreover, in lakes, the different layers of water don't easily mix. As a result, oxygen from the air does not reach deeper water. If there is no oxygen, the fish swim to the warmer water surface. This speeds up circulation in these cold-blooded creatures, so that they have to eat more — which leads to further stresses.
Since the 1970s, researchers from the Leibniz Institute of Freshwater Ecology and Inland Fisheries have been documenting changes in the Müggel Lake near Berlin.
Researchers have been alarmed at the 0.34-degree Celsius increase in water temperature per decade over recent decades, as the shorter ice duration in winter and indirect effects such as changes in light and oxygen, as well as runoff from over-fertilization by agriculture, are majorly impacting lake ecosystems.
Long-term climate impact research will be indispensable in order to develop adaptation strategies based on the observations, researchers think.<|endoftext|>
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Temporal range: Early Oligocene–recent
|Siberian musk deer|
Musk deer (Sanskrit: कस्तूरी मृग, and Hindi: कस्तूरी हिरन) can refer to any one, or all seven, of the species that make up Moschus, the only extant genus of the family Moschidae. The musk deer family differs from cervids, or true deer, by lacking antlers and facial glands and by possessing only a single pair of teats, a gallbladder, a caudal gland, a pair of tusk-like teeth and—of particular economic importance to humans—a musk gland.
Musk deer live mainly in forested and alpine scrub habitats in the mountains of southern Asia, notably the Himalayas. Moschids, the proper term when referring to this type of deer rather than one/multiple species of musk deer, are entirely Asian in their present distribution, being extinct in Europe where the earliest musk deer are known to have existed from Oligocene deposits.
Musk deer resemble small deer with a stocky build, and hind legs longer than their front legs. They are about 80 to 100 cm (31 to 39 in) long, 50 to 70 cm (20 to 28 in) high at the shoulder, and weigh between 7 and 17 kg (15 and 37 lb). The feet of musk deer are adapted for climbing in rough terrain. Like the Chinese water deer, a cervid, they have no antlers, but the males do have enlarged upper canines, forming sabre-like tusks. The dental formula is similar to that of true deer: 0.1.3.3
Musk deer are herbivores, living in hilly, forested environments, generally far from human habitation. Like true deer, they eat mainly leaves, flowers, and grasses, with some mosses and lichens. They are solitary animals, and maintain well-defined territories, which they scent mark with their caudal glands. Musk deer are generally shy, and either nocturnal, or crepuscular.
Males leave their territories during the rutting season, and compete for mates, using their tusks as weapons. Female musk deer give birth to a single fawn after about 150–180 days. The newborn young are very small, and essentially motionless for the first month of their lives, a feature that helps them remain hidden from predators.
Musk deer have been hunted for their scent glands, which are commonly used in perfumes. The glands can fetch up to $45,000/kg on the black market. It is rumored that ancient royalty wore the scent of the musk deer and that it is an aphrodisiac.
Musk deer may be a surviving representative of the Palaeomerycidae, a family of ruminants that is probably ancestral to deer. They originated in the early Oligocene epoch and disappeared in the Pliocene. Most species lacked antlers, though some were found in later species. The musk deer are, however, still placed in a separate family.
While they have been traditionally classified as members of the deer family (as the subfamily "Moschinae") and all the species were classified as one species (under Moschus moschiferus), recent studies have indicated that moschids are more closely related to bovids (antelope, goat-antelope and wild cattle). The following taxonomy is after Prothero (2007)
- Hydropotopsis lemanensis
- Hispanomeryx aragonensis
- Hispanomeryx daamsi
- Hispanomeryx duriensis
- Hispanomeryx andrewsi
- Oriomeryx major
- Oriomeryx willii
- Friburgomeryx wallenriedensis
- Bedenomeryx truyolsi
- Bedenomeryx milloquensis
- Bedenomeryx paulhiacensis
- "Moschus (musk deer) Classification". Animal Diversity Web. University of Michigan Museum of Zoology.
- Frädrich H (1984). "Deer". In Macdonald D (ed.). The Encyclopedia of Mammals. New York: Facts on File. pp. 518–9. ISBN 978-0-87196-871-5.
- Wild Russia, Discovery Channel
- Hassanin A, Douzery EJ (April 2003). "Molecular and morphological phylogenies of ruminantia and the alternative position of the moschidae". Systematic Biology. 52 (2): 206–28. doi:10.1080/10635150390192726. PMID 12746147.
- Prothero DR (November 2007). Evolution: what the fossils say and why it matters. Columbia University Press. pp. 221–226.
- Aiglstorfer M, Costeur L, Mennecart B, Heizmann EP (2017). "Micromeryx? eiselei-A new moschid species from Steinheim am Albuch, Germany, and the first comprehensive description of moschid cranial material from the Miocene of Central Europe". PLOS ONE. 12 (10): e0185679. doi:10.1371/journal.pone.0185679. PMC 5642927. PMID 29036194.
- Guha S, Goyal SP, Kashyap VK (March 2007). "Molecular phylogeny of musk deer: a genomic view with mitochondrial 16S rRNA and cytochrome b gene". Molecular Phylogenetics and Evolution. 42 (3): 585–97. doi:10.1016/j.ympev.2006.06.020. PMID 17158073.<|endoftext|>
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# 10.05 Volume and capacity
Lesson
## Are you ready?
We have looked at units of volume and units of capacity . It is important that we remember the difference.
### Examples
#### Example 1
What is the most appropriate unit for measuring the volume of a box?
A
\text{ mm}^3
B
\text{ cm}^3
Worked Solution
Create a strategy
Think about what type of objects would be measured by these units.
Apply the idea
\text {mm}^3 is useful to measure a drop of liquid, or the volume of a coin and other tiny objects. \text {cm}^3 is useful to measure bigger objects like a ball or a box.
So, the answer is option B.
#### Example 2
Choose the most appropriate unit of measure for finding the amount of water in a dam.
A
Millilitres
B
Megalitres
C
Litres
Worked Solution
Create a strategy
Use the table of examples below to help you.
Apply the idea
Dams can store massive amounts of water, enough to supply entire cities. Cities have thousands of people, that each use 200\text{ L} of water per day. So it would need to store millions of litres.
A megalitre is equal to a million of litres. So this is the most appropriate unit.
The correct answer is option B.
Idea summary
Volume measures the amount of space an object takes up. Some units are: \text{mm}^3, \,\text{cm}^3, \,\text{m}^3.
Capacity measures the amount of liquid an object can hold. Some units are: \text{mL}, \,\text{L}, \,\text{kL}, \,\text{ML}.
## Volume and capacity
Let's look at how to convert between millilitres (\text{mL}) and cubic centimetres (\text{cm}^3).
### Examples
#### Example 3
Convert 40 millilitres (\text{mL}) to cubic centimetres (\text{cm}^3).
Worked Solution
Create a strategy
Use the conversion rule 1 \text{ mL} = 1 \text{ cm}^3.
Apply the idea
Since 1 \text{ mL} = 1 \text{ cm}^3 we don't need to change the number, we just change the units:40 \text{ mL} = 40 \text{ cm}^3
Idea summary
Below are the conversions between volume and capacity:
\begin{array}{c} &\text{Volume} & &\text{Capacity} \\ &1 \text{ cm}^3 &= &1 \text{ mL} \\ &1000 \text{ cm}^3 &= &1 \text{ L} \\ &1 \text{ m}^3 &= &1000 \text{ L} \\ &1 \text{ m}^3 &= &1 \text{ kL} \\ \end{array}
### Outcomes
#### VCMMG225
Connect volume and capacity and their units of measurement<|endoftext|>
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# What is the slope of the line passing through the following points: (5, -6) , (-8, 1)?
Feb 7, 2016
$- \frac{7}{13}$
#### Explanation:
To find the gradient (slope ) of a line passing through 2 points use the $\textcolor{b l u e}{\text{ gradient formula}}$
$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$
where $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \textcolor{b l a c k}{\text{ are coords of 2 points }}$
let $\left({x}_{1} , {y}_{1}\right) = \left(5 , - 6\right) \textcolor{b l a c k}{\text{ and }} \left({x}_{2} , {y}_{2}\right) = \left(- 8 , 1\right)$
substitute into gradient formula :
$\Rightarrow m = \frac{1 - \left(- 6\right)}{- 8 - 5} = \frac{7}{-} 13 = - \frac{7}{13}$<|endoftext|>
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# Math Assignment Help With Decimal Numbers Ordering
## 2.2 Ordering
The method of placing numbers in increasing or decreasing format is called ordering.
2.2.1 Ordering of number
Numbers are arranged in either Ascending order or Descending order. In case of Ascending order numbers are placed from lowest to highest. The number being the lowest is placed first and the higher number is placed.
For eg: if we asked to place 12, 5, 10, 4, and 3 in ascending order then,
3,4,5,10,12
Similarly, in Descending order numbers are placed from highest to lowest. The number having highest value is placed first and the lower valued number is placed
For eg: taking the similar example, if we are asked to place 12, 5, 10, 4, and 3 in descending order then,
12, 10, 5, 4, 3
2.2.2 Ordering of Decimals
Ordering of decimal is not same as of integers, it can be confusing sometimes. If you take two numbers
0.24 And 0.234
You may say 0.234 is greater than 0.24 as it contains more number of digits. But this is wrong.
0.24 is greater than 0.234. Now we will explain how.
Let us follow a stepwise method.
Take few decimal numbers, say,
0.235, 0.456, 0.237, 0.346, 1.765
Step1: First compare the digit at Units place of all the given decimal numbers. One of them has 1 and rest have 0. So 1.765 is the highest. Put it aside.
Step2: now compare the tenth place of the remaining decimal numbers that is,
0.235, 0.456, 0.237, 0.346
We see 2, 3, and 4 at tenth place, since 4 is greater of them therefore 0.456 is the next highest, the decimals left are
0.235, 0.237, 0.346
Comparing the tenth place again we find 0.346 is the next highest. Here we see two decimals having the same digit at tenth place that is, 2 so we will have to compare the hundredth place as well.
Step3: compare the hundredth place of the remaining two decimals that is,
0.235,0.237
Here digit at hundredth place is also same so compare the thousandth place, 7>5
Therefore 0.237 is the next highest and 0.237 is the lowest.
Hence the correct order is
1.765, 0.456, 0.346, 0.237, 0.235
Note: In decimal number system decimals must be from highest to lowest.
### Email Based Assignment Help in Decimal Numbers Ordering
To Schedule a Decimal Numbers Ordering tutoring session<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 3.1: Even and Odd Identities
Difficulty Level: At Grade Created by: CK-12
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You and your friend are in math class together. You enjoy talking a lot outside of class about all of the interesting topics you cover in class. Lately you've been covering trig functions and the unit circle. As it turns out, trig functions of certain angles are pretty easy to remember. However, you and your friend are wishing there was an easy way to "shortcut" calculations so that if you knew a trig function for an angle you could relate it to the trig function for another angle; in effect giving you more reward for knowing the first trig function.
You're examining some notes and starting writing down trig functions at random. You eventually write down:
cos(π18)\begin{align*}\cos \left( \frac{\pi}{18} \right)\end{align*}
Is there any way that if you knew how to compute this, you'd automatically know the answer for a different angle?
As it turns out, there is. Read on, and by the time you've finished this Concept, you'll know what other angle's value of cosine you already know, just by knowing the answer above.
### Guidance
An even function is a function where the value of the function acting on an argument is the same as the value of the function when acting on the negative of the argument. Or, in short:
\begin{align*}f(x) = f(-x)\end{align*}
So, for example, if f(x) is some function that is even, then f(2) has the same answer as f(-2). f(5) has the same answer as f(-5), and so on.
In contrast, an odd function is a function where the negative of the function's answer is the same as the function acting on the negative argument. In math terms, this is:
\begin{align*}-f(x) = f(-x)\end{align*}
If a function were negative, then f(-2) = -f(2), f(-5) = -f(5), and so on.
Functions are even or odd depending on how the end behavior of the graphical representation looks. For example, \begin{align*}y = x^2\end{align*} is considered an even function because the ends of the parabola both point in the same direction and the parabola is symmetric about the \begin{align*}y-\end{align*}axis. \begin{align*}y = x^3\end{align*} is considered an odd function for the opposite reason. The ends of a cubic function point in opposite directions and therefore the parabola is not symmetric about the \begin{align*}y-\end{align*}axis. What about the trig functions? They do not have exponents to give us the even or odd clue (when the degree is even, a function is even, when the degree is odd, a function is odd).
\begin{align*}& \underline{\text{Even Function}} && \underline{\text{Odd Function}} \\ & y = (-x)^2 = x^2 && y = (-x)^3 = -x^3\end{align*}
Let’s consider sine. Start with \begin{align*}\sin(-x)\end{align*}. Will it equal \begin{align*}\sin x\end{align*} or \begin{align*}-\sin x\end{align*}? Plug in a couple of values to see.
\begin{align*}\sin(-30^\circ) & = \sin 330^\circ = - \frac{1}{2} = - \sin 30^\circ \\ \sin (-135^\circ) & = \sin 225^\circ = - \frac{\sqrt{2}}{2} = - \sin 135^\circ\end{align*}
From this we see that sine is odd. Therefore, \begin{align*}\sin(-x) = -\sin x\end{align*}, for any value of \begin{align*}x\end{align*}. For cosine, we will plug in a couple of values to determine if it’s even or odd.
\begin{align*}\cos(-30^\circ) & = \cos 330^\circ = \frac{\sqrt{3}}{2} = \cos 30^\circ \\ \cos (-135^\circ) & = \cos 225^\circ = - \frac{\sqrt{2}}{2} = \cos 135^\circ\end{align*}
This tells us that the cosine is even. Therefore, \begin{align*}\cos(-x) = \cos x\end{align*}, for any value of \begin{align*}x\end{align*}. The other four trigonometric functions are as follows:
\begin{align*}\tan(-x) & = - \tan x \\ \csc (-x) & = - \csc x \\ \sec(-x) & = \sec x \\ \cot (-x) & = - \cot x\end{align*}
Notice that cosecant is odd like sine and secant is even like cosine.
#### Example A
If \begin{align*}\cos (-x) = \frac{3}{4}\end{align*} and \begin{align*}\tan(-x) = - \frac{\sqrt{7}}{3}\end{align*}, find \begin{align*}\sin x\end{align*}.
Solution: We know that sine is odd. Cosine is even, so \begin{align*}\cos x = \frac{3}{4}\end{align*}. Tangent is odd, so \begin{align*}\tan x = \frac{\sqrt{7}}{3}\end{align*}. Therefore, sine is positive and \begin{align*}\sin x = \frac{\sqrt{7}}{4}\end{align*}.
#### Example B
If \begin{align*}\sin(x) = .25\end{align*}, find \begin{align*}\sin(-x)\end{align*}
Solution: Since sine is an odd function, \begin{align*}\sin(-\theta) = -\sin(\theta)\end{align*}.
Therefore, \begin{align*}\sin(-x) = -\sin(x) = -.25\end{align*}
#### Example C
If \begin{align*}\cos(x) = .75\end{align*}, find \begin{align*}\cos(-x)\end{align*}
Solution:
Since cosine is an even function, \begin{align*}\cos(x)= \cos(-x)\end{align*}.
Therefore, \begin{align*}\cos(-x) = .75\end{align*}
### Vocabulary
Even Function: An even function is a function with a graph that is symmetric with respect to the 'y' axis and has the property that f(-x) = f(x)
Odd Function: An odd function is a function with the property that f(-x) = -f(x)
### Guided Practice
1. What two angles have a value for cosine of \begin{align*}\frac{\sqrt{3}}{2}\end{align*}?
2. If \begin{align*}\cos \theta = \frac{\sqrt{3}}{2}\end{align*}, find \begin{align*}\sec(-\theta)\end{align*}
3. If \begin{align*}\cot \theta = -\sqrt{3}\end{align*} find \begin{align*}\cot -\theta\end{align*}
Solutions:
1. On the unit circle, the angles \begin{align*}30^\circ\end{align*} and \begin{align*}330^\circ\end{align*} both have \begin{align*}\frac{\sqrt{3}}{2}\end{align*} as their value for cosine. \begin{align*}330^\circ\end{align*} can be rewritten as \begin{align*}-30^\circ\end{align*}
2. There are 2 ways to think about this problem. Since \begin{align*}\cos \theta = \cos -\theta\end{align*}, you could say \begin{align*}\sec \left(-\theta \right) = \frac{1}{\cos \left( -\theta \right)} = \frac{1}{\cos \left( \theta \right)}\end{align*} Or you could leave the cosine function the way it is and say that \begin{align*}\sec \left( -\theta \right) = \sec \left( \theta \right) = \frac{1}{\cos \theta}\end{align*}. But either way, the answer is \begin{align*}\frac{2}{\sqrt{3}}\end{align*}
3. Since \begin{align*}\cot(-\theta) = -\cot(\theta)\end{align*}, if \begin{align*}\cot \theta = -\sqrt{3}\end{align*} then \begin{align*}-\cot(-\theta) = -\sqrt{3}\end{align*}. Therefore, \begin{align*}\cot(-\theta) = \sqrt{3}\end{align*}.
### Concept Problem Solution
Since you now know that cosine is an even function, you get to know the cosine of the negative of an angle automatically if you know the cosine of the positive of the angle.
Therefore, since \begin{align*}\cos \left( \frac{\pi}{18} \right) = .9848\end{align*}, you automatically know that \begin{align*}\cos \left( -\frac{\pi}{18} \right) = \cos \left( \frac{17\pi}{18} \right)= .9848\end{align*}.
### Practice
Identify whether each function is even or odd.
1. \begin{align*}y=\sin(x)\end{align*}
2. \begin{align*}y=\cos(x)\end{align*}
3. \begin{align*}y=\cot(x)\end{align*}
4. \begin{align*}y=x^4\end{align*}
5. \begin{align*}y=x\end{align*}
6. If \begin{align*}\sin(x)=.3\end{align*}, what is \begin{align*}\sin(-x)\end{align*}?
7. If \begin{align*}\cos(x)=.5\end{align*}, what is \begin{align*}\cos(-x)\end{align*}?
8. If \begin{align*}\tan(x)=.1\end{align*}, what is \begin{align*}\tan(-x)\end{align*}?
9. If \begin{align*}\cot(x)=.3\end{align*}, what is \begin{align*}\cot(-x)\end{align*}?
10. If \begin{align*}\csc(x)=.3\end{align*}, what is \begin{align*}\csc(-x)\end{align*}?
11. If \begin{align*}\sec(x)=2\end{align*}, what is \begin{align*}\sec(-x)\end{align*}?
12. If \begin{align*}\sin(x)=-.2\end{align*}, what is \begin{align*}\sin(-x)\end{align*}?
13. If \begin{align*}\cos(x)=-.25\end{align*}, what is \begin{align*}\sec(-x)\end{align*}?
14. If \begin{align*}\csc(x)=4\end{align*}, what is \begin{align*}\sin(-x)\end{align*}?
15. If \begin{align*}\tan(x)=-.2\end{align*}, what is \begin{align*}\cot(-x)\end{align*}?
16. If \begin{align*}\sin(x)=-.5\end{align*} and \begin{align*}\cos(x)=-\frac{\sqrt{3}}{2}\end{align*}, what is \begin{align*}\cot(-x)\end{align*}?
17. If \begin{align*}\cos(x)=-.5\end{align*} and \begin{align*}\sin(x)=\frac{\sqrt{3}}{2}\end{align*}, what is \begin{align*}\tan(-x)\end{align*}?
18. If \begin{align*}\cos(x)=-\frac{\sqrt{2}}{2}\end{align*} and \begin{align*}\tan(x)=-1\end{align*}, what is \begin{align*}\sin(-x)\end{align*}?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Even Function An even function is a function with a graph that is symmetric with respect to the $y$-axis and has the property that $f(-x) = f(x)$.
Odd Function An odd function is a function with the property that $f(-x) = -f(x)$. Odd functions have rotational symmetry about the origin.
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# Prime Factors of 5132
Prime Factors of 5132 are 2, 2, and 1283
#### How to find prime factors of a number
1. Prime Factorization of 5132 by Division Method 2. Prime Factorization of 5132 by Factor Tree Method 3. Definition of Prime Factors 4. Frequently Asked Questions
#### Steps to find Prime Factors of 5132 by Division Method
To find the primefactors of 5132 using the division method, follow these steps:
• Step 1. Start dividing 5132 by the smallest prime number, i.e., 2, 3, 5, and so on. Find the smallest prime factor of the number.
• Step 2. After finding the smallest prime factor of the number 5132, which is 2. Divide 5132 by 2 to obtain the quotient (2566).
5132 ÷ 2 = 2566
• Step 3. Repeat step 1 with the obtained quotient (2566).
2566 ÷ 2 = 1283
1283 ÷ 1283 = 1
So, the prime factorization of 5132 is, 5132 = 2 x 2 x 1283.
#### Steps to find Prime Factors of 5132 by Factor Tree Method
We can follow the same procedure using the factor tree of 5132 as shown below:
So, the prime factorization of 5132 is, 5132 = 2 x 2 x 1283.
#### How do you define Prime Factors of a number?
Prime numbers, in mathematics are all those natural numbers greater than 1 that have exactly 2 factors which are 1 and the number itself. When we express any number as the product of these prime numbers than these prime numbers become prime factors of that number. Eg- Prime Factors of 5132 are 2 x 2 x 1283.
#### Properties of Prime Factors
• For any given number there is one and only one set of unique prime factors.
• Any number can have only 1 even prime factor and that is number 2.
• Two prime factors are always coprime to each other.
• 1 is the factor of any given number, but 1 is neither a prime number nor a composite number. So, 1 is the factor of 5132 but not a prime factor of 5132.
#### Frequently Asked Questions
• Which is the smallest prime factor of 5132?
Smallest prime factor of 5132 is 2.
• What is the prime factorization of 5132?
Prime factorization of 5132 is 2 x 2 x 1283.
• What are the factors of 5132?
Factors of 5132 are 1 , 2 , 4 , 1283 , 2566 , 5132.
• Is 5132 a prime number or a composite number?
5132 is a composite number.
• Is 5132 a prime number?
false, 5132 is not a prime number.
• Which is the largest prime factors of 5132?
The largest prime factor of 5132 is 1283.
• What is the sum of all prime factors of 5132?
Prime factors of 5132 are 2 x 2 x 1283. Therefore, their sum is 1287.
• What numbers are the prime factors of 5132?
Prime factors of 5132 are 2 , 2 , 1283.
• What is the sum of all odd prime factors of 5132?
Prime factors of 5132 are 2 , 2 , 1283, out of which 1283 are odd numbers. So, the sum of odd prime factors of 5132 is 1283 = 1283.
Share your expierence with Math-World<|endoftext|>
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Andromeda’s once and future stars
Two ESA observatories have combined forces to show the Andromeda Galaxy in a new light. Herschel sees rings of star formation in this, the most detailed image of the Andromeda Galaxy ever taken at infrared wavelengths, and XMM-Newton shows dying stars shining X-rays into space.
During Christmas 2010, ESA’s Herschel and XMM-Newton space observatories targeted the nearest large spiral galaxy M31. This is a galaxy similar to our own Milky Way – both contain several hundred billion stars. This is the most detailed far-infrared image of the Andromeda Galaxy ever taken and shows clearly that more stars are on their way.
Sensitive to far-infrared light, Herschel sees clouds of cool dust and gas where stars can form. Inside these clouds are many dusty cocoons containing forming stars, each star pulling itself together in a slow gravitational process that can last for hundreds of millions of years. Once a star reaches a high enough density, it will begin to shine at optical wavelengths. It will emerge from its birth cloud and become visible to ordinary telescopes.
Many galaxies are spiral in shape but Andromeda is interesting because it shows a large ring of dust about 75 000 light-years across encircling the centre of the galaxy. Some astronomers speculate that this dust ring may have been formed in a recent collision with another galaxy. This new Herschel image reveals yet more intricate details, with at least five concentric rings of star-forming dust visible.
Superimposed on the infrared image is an X-ray view taken almost simultaneously by ESA’s XMM-Newton observatory. Whereas the infrared shows the beginnings of star formation, X-rays usually show the endpoints of stellar evolution.
XMM-Newton highlights hundreds of X-ray sources within Andromeda, many of them clustered around the centre, where the stars are naturally found to be more crowded together. Some of these are shockwaves and debris rolling through space from exploded stars, others are pairs of stars locked in a gravitational fight to the death.
In these deadly embraces, one star has already died and is pulling gas from its still-living companion. As the gas falls through space, it heats up and gives off X-rays. The living star will eventually be greatly depleted, having much of its mass torn from it by the stronger gravity of its denser partner. As the stellar corpse wraps itself in this stolen gas, it could explode.
Both the infrared and X-ray images show information that is impossible to collect from the ground because these wavelengths are absorbed by Earth’s atmosphere. The twinkling starlight seen from Earth is indeed a beautiful sight but in reality contains less than half the story. Visible light shows us the adult stars, whereas infrared gives us the youngsters and X-rays show those in their death throes.
To chart the lives of stars, we need to see it all and that is where Herschel and XMM-Newton contribute so much.<|endoftext|>
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So how is a threat detected?
Well, there are these Pattern Recognition Receptors
- PRRs, that can recognize infectious agents.
And what they recognize on the
infectious pathogen are structures
that are called Pathogen-Associated
Molecular Patterns or PAMPs, P-A-M-P.
Also, Pattern Recognition Receptors, as well
as recognizing PAMPs on foreign infectious
agents can recognize structures associated
with damage to our own body cells.
And we call these structures
Damage-Associated Molecular Patterns.
So Pattern Recognition Receptors
can recognize both PAMPs and DAMPs.
And these can be present on the cell
surface or sometimes inside cells,
and these may be on pathogen cells or
they may be on our own body cells.
What is important to appreciate, is that although this
recognition is often described as being broadly
specific, what is actually being recognized is
recognized in a very, very highly specific way.
So for example, we’ll mention a few Pathogen-Associated
Molecular Patterns in a few moments.
One of them is called
lipopolysaccharide or LPS.
LPS is found on
And there are lots of different
types of Gram-negative bacteria.
So LPS is shared between several different bacteria but
the recognition of LPS is very, very highly specific.
So recognition is structurally-specific but what is recognized
is common to whole groups of organisms or host cells.
These Pattern Recognition Receptors can be
inside cells, in other words, intracellular.
And if they’re intracellular, if they’re
inside a cell, they may be present on the
endosomes within the cell or they may be
present within the cytosol of the cell.
Alternatively, they may be present on the surface of
cells, cell surface Pattern Recognition Receptors.
Or indeed, they may be released or secreted from
cells as soluble Pattern Recognition Receptors.
You can now look at a number of different
Pattern Recognition Receptors and the PAMPs that they recognize.
Let’s start with endosomal
Pattern Recognition Receptors.
There is a group of 10 or so Pattern Recognition
Receptors that are called toll-like receptors.
Couple of examples for you now; TLR3, toll-like
receptor recognizes the PAMP viral double-stranded RNA.
TLR7 and TLR8 recognize
viral single-stranded RNA.
Whereas TLR9 recognizes a particular nucleotide sequence
within the DNA of bacteria, called bacterial unmethylated CpG.
Let’s now turn to cytosolic
Pattern Recognition Receptors.
NOD-1 (nucleotide-binding oligomerization
domain-containing protein-1) and NOD-2.
These recognize bacterial
These structures are found
on Gram-positive bacteria.
So again, shared between many different
bacteria but the recognition of
peptidoglycan by NOD-1 and NOD-2 is highly
specific for that particular structure.
RIG-1 (retinoic acid-inducible gene 1)
recognizes viral double-stranded RNA.
And as a third example of a cytosolic Pattern
Recognition Receptor, NLRP3 (NOD-like receptor family,
pyrin domain containing 3) which is part of the
inflammasome which we’ll discuss in a few seconds.
This recognizes bacterial
Cell surface Pattern Recognition Receptors; TLR2,
again recognizes bacterial structures that are shared
between many different bacterial species, various
bacterial lipopeptides and lipoproteins.
TLR4 recognises bacterial
And TLR5 recognises
And finally, soluble Pattern
Mannose binding lectin that recognizes
the sugar mannose as its name suggests.
And Ficolin, which recognizes
N-acetylglucosamine, another sugar.
Let’s now turn to Pattern Recognition Receptors
which recognize DAMPs - Damage-Associated
Molecular Patterns or sometimes called
Danger-Associated Molecular Patterns.
These are Pattern Recognition Receptors that recognize
structures produced by our own body cells following damage.
A couple of examples of Pattern
Recognition Receptors present on cell
surfaces that recognize DAMPS: RAGE,
the receptor for advanced glycation
end products as its name suggests
recognizes advanced glycation end products
that are produced by our
own cells in response to damage.
And RAGE, TLR2 and TLR4 recognize
HMGB1 (high mobility group box 1).
So if you were paying attention you’ll see
that RAGE recognizes two
different DAMPs - advanced glycation end products and HMGB1.
Whereas TLR2 and TLR4 are specific not for
advanced glycation end products, but for HMGB1.
And then finally, cytosolic Pattern Recognition
Receptors, involved with a structure called the
inflammasome; NLRP3 recognizes the Damage or
Danger-Associated Molecular Pattern, uric acid.<|endoftext|>
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Findings could help feline preservationists implement breeding strategies for rare species
Whole genome sequencing (WGS), which is the process of determining an organism’s complete DNA sequence, can be used to identify DNA anomalies that cause disease. Identifying disease-causing DNA abnormalities allows clinicians to better predict an effective course of treatment for the patient. Now, in a series of recent studies, scientists at the University of Missouri are using whole genome sequencing through the 99 Lives Cat Genome Sequencing Consortium to identify genetic variants that cause rare diseases, such as progressive retinal atrophy and Niemann-Pick type 1, a fatal disorder in domestic cats. Findings from the study could help feline preservationists implement breeding strategies in captivity for rare and endangered species such as the African black-footed cat.
The 99 Lives project was established at Mizzou by Leslie Lyons, the Gilbreath-McLorn Endowed Professor of Comparative Medicine in the College of Veterinary Medicine, to improve health care for cats through research. The database has genetically sequenced more than 50 felines and includes DNA from cats with and without known genetic health problems. The goal of the database is to identify DNA that causes genetic disorders and have a better understanding of how to treat diseases.
In the first study, Lyons and her team used the 99 Lives consortium to identify a genetic mutation that causes blindness in the African black-footed cat, an endangered species often found in U.S. zoos. The team sequenced three cats ― two unaffected parents and an affected offspring ― to determine if the mutation was inherited or spontaneous. The genetic mutation identified was located the IQCB1 gene and is associated with progressive retinal atrophy, an inherited degenerative retinal disorder that leads to blindness. The affected cat had two copies of the genetic mutation, indicating that it was an inherited disorder.
“African black-footed cats are closely related to domestic cats, so it was a good opportunity to use the 99 Lives database,” Lyons said. “When sequencing DNA, we are looking for the high priority variants, or genetic mutations that result in disease. Variants in the IQCB1 gene are known to cause retinal degeneration in humans. We evaluated each gene of the African black-footed cat, one at a time, to look for the genetic mutation that is associated with vision loss.”
In another study representing the first time precision medicine has been applied to feline health, Lyons and her team used whole genome sequencing and the 99 Lives consortium to identify a lysosomal disorder in a 36-week-old silver tabby kitten that was referred to the MU Veterinary Health Center. The kitten was found to have two copies of a mutation in the NPC1 gene, which causes Niemman-Pick type 1, a fatal disorder. The NCP1 gene identified is not a known variant in humans; it is a rare mutation to the feline population.
“Genetics of the patient is a critical aspect of an individual’s health care for some diseases,” Lyons said. “Continued collaboration with geneticists and veterinarians could lead to the rapid discovery of undiagnosed genetic conditions in cats. The goal of genetic testing is to identify disease early, so that effective and proactive treatment can be administered to patients.”
Identification of both the IQCB1 gene in the African black-footed cat and the NCP1 in the silver tabby will help to diagnose other cats and allow them to receive appropriate treatment. Using results of the black-footed cat study, zookeepers will be implementing species survival plans to help manage the cats in captivity in North America.
The study, “Early-Onset Progressive Retina Atrophy Associated with an IQCB1 Variant in the African Black-Footed Cates (Felis nigripes),” recently was published in Scientific Reports. Funding was provided by the University of Missouri, College of Veterinary Medicine Clinician Scientific Grant. The study, “Precision Medicine in Cats: Novel Niemann-Pick Type C1 Diagnosed by Whole-Genome sequencing,” recently was published in the Journal of Veterinary Internal Medicine.
Published by Mizzou News, 329 Jesse Hall, Columbia, MO 65211<|endoftext|>
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Lesson 4-4
1 / 4
# Lesson 4-4 - PowerPoint PPT Presentation
Proving Triangles Congruent. Lesson 4-4. (AAS, HL). A. D. D. A. B. C. F. E. B. C. F. E. Postulates. If two angles and a non included side of one triangle are congruent to the corresponding two angles and side of a second triangle, then the two triangles are congruent. AAS.
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## PowerPoint Slideshow about 'Lesson 4-4' - pearlie
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Proving Triangles CongruentLesson 4-4
(AAS, HL)
Lesson 4-4: AAS & HL Postulate
A
D
D
A
B
C
F
E
B
C
F
E
Postulates
If two angles and a non included side of one triangle are congruent to the corresponding two angles and side of a second triangle, then the two triangles are congruent.
AAS
If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and corresponding leg of another right triangle, then the triangles are congruent.
HL
Lesson 4-4: AAS & HL Postulate
Problem 1
Step 1: Mark the Given
Step 2: Mark vertical angles
AAS
Step 3: Choose a Method (SSS /SAS/ASA/AAS/ HL )
Step 4: List the Parts in the order of the method
Step 5: Fill in the reasons
Step 6: Is there more?
Given
Vertical Angle Thm
Given
AAS Postulate
Lesson 4-4: AAS & HL Postulate
Problem 2
Step 1: Mark the Given
Step 2: Mark reflexive sides
HL
Step 3: Choose a Method (SSS /SAS/ASA/AAS/ HL )
Step 4: List the Parts in the order of the method
Step 5: Fill in the reasons
Step 6: Is there more?
Given
Given
Reflexive Property
HL Postulate
Lesson 4-4: AAS & HL Postulate<|endoftext|>
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a physicochemical quantity that by convention is used to characterize the equilibrium potential difference between an electrode and a solution when the substances participating in the electrode reaction are in the standard state, that is, when the activities of the substances are equal to unity. Since there is no way to measure the actual potential difference between the electrode and the solution, quantities that characterize the potentials of different electrodes with respect to some reference electrode are used. A normal hydrogen electrode, whose potential is assumed to be zero at any temperature, is usually used as the reference electrode.
The potential of an electrode that is negatively charged with respect to the normal hydrogen electrode has a minus sign, while the potential of a positively charged electrode has a plus sign. An arrangement of the standard potentials of discharge-ionization reactions of metals and hydrogen, placed in ascending order, is called the electromotive-force series. One element will displace a second element from a solution containing the second element’s cations if the standard potential of the first element is less positive than that of the second. Standard potentials are calculated from the results of measurements of the electromotive force of voltaic cells and from the standard values of the change in the Gibbs free energy ΔG° during the reaction. The values of the standard potential can be used to calculate ΔG° and the equilibrium constant of chemical reactions. Such calculations are necessary for thermodynamic calculations and for assessing the possibility that a given chemical reaction will take place.
REFERENCESKireev, V. A. Kratkii kursfizicheskoi khimii. Moscow, 1963. Chapter 13, section 175.
Spravochnik khimika, vol. 3. Moscow-Leningrad, 1965.
Perel’man, V. I. Kratkii spravochnik khimika, 6th ed. Moscow, 1963. Goronovskii, I. T., Iu. P. Nazarenko, and E. F. Nekriach. Kratkii spravochnik po khimii, 3rd ed. Kiev, 1965.<|endoftext|>
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# SOLUTION: how do you simplify 15x^2+75x-1260
Algebra -> Algebra -> Radicals -> SOLUTION: how do you simplify 15x^2+75x-1260 Log On
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Algebra: Radicals -- complicated equations involving roots Solvers Lessons Answers archive Quiz In Depth
Question 347687: how do you simplify 15x^2+75x-1260
Found 2 solutions by jim_thompson5910, Fombitz:
You can put this solution on YOUR website!
Factor out the GCF
Now let's focus on the inner expression
------------------------------------------------------------
Looking at we can see that the first term is and the last term is where the coefficients are 1 and -84 respectively.
Now multiply the first coefficient 1 and the last coefficient -84 to get -84. Now what two numbers multiply to -84 and add to the middle coefficient 5? Let's list all of the factors of -84:
Factors of -84:
1,2,3,4,6,7,12,14,21,28,42,84
-1,-2,-3,-4,-6,-7,-12,-14,-21,-28,-42,-84 ...List the negative factors as well. This will allow us to find all possible combinations
These factors pair up and multiply to -84
(1)*(-84)
(2)*(-42)
(3)*(-28)
(4)*(-21)
(6)*(-14)
(7)*(-12)
(-1)*(84)
(-2)*(42)
(-3)*(28)
(-4)*(21)
(-6)*(14)
(-7)*(12)
note: remember, the product of a negative and a positive number is a negative number
Now which of these pairs add to 5? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 5
First NumberSecond NumberSum
1-841+(-84)=-83
2-422+(-42)=-40
3-283+(-28)=-25
4-214+(-21)=-17
6-146+(-14)=-8
7-127+(-12)=-5
-184-1+84=83
-242-2+42=40
-328-3+28=25
-421-4+21=17
-614-6+14=8
-712-7+12=5
From this list we can see that -7 and 12 add up to 5 and multiply to -84
Now looking at the expression , replace with (notice adds up to . So it is equivalent to )
Now let's factor by grouping:
Group like terms
Factor out the GCF of out of the first group. Factor out the GCF of out of the second group
Since we have a common term of , we can combine like terms
So factors to
So this also means that factors to (since is equivalent to )
------------------------------------------------------------
So our expression goes from and factors further to
------------------<|endoftext|>
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### Prompt Cards
These two group activities use mathematical reasoning - one is numerical, one geometric.
### Consecutive Numbers
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Exploring Wild & Wonderful Number Patterns
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
##### Age 7 to 11Challenge Level
I was with a class of children in Bromley near to London, when I suddenly came up with this idea, and I put it to the youngsters at the school. They did a lot of work on it so I thought I'd share it with you.
It's all about $64$!
Lots of you know that $64$ is $8$ times $8$. So if you were asked to write down all the numbers up to $64$ you might decide to do eight lots of $8$ . [It's a bit like $100$ in that you may well write ten lots of $10$ to get up to $100$ and produce a $100$ square.]
I suggested to them that they tried writing the numbers up to $64$ in an interesting way so that the shape they made at the end would be interesting, different, more exciting ... than just a square. Here are the ones that some of them came up with to show that the numbers could be arranged in an interesting way.
Most of them, as you see, ended up with shapes that were not squares. Those that did end up with an $8$ by $8$ square put the numbers in an interesting order into the shape.
When they did that they were then asked to made a tile [or frame] that was made up of four squares.
Here are some examples:-
The idea now was to place one these tiles/frames somewhere on the table of $64$ so that it covered four numbers. [The tiles were made so that the squares were the same size as the squares on each of the numbers in the $64$ table.]
The numbers underneath the tile/frame were added up and recorded. The tile/frame was then moved around the table of $64$ to different positions and each time the total of the four numbers underneath was recorded.
Well that's what you need to do. It's fun creating new $64$ tables in different shapes.
Now comes the investigative part ...
Explore by looking at the totals that you've found and and think about any relationships that you notice.
You then need to think about why these sets of answers are occurring. The youngsters at the Bromley school found lots of things out ... now it is your turn to do the same.
Lastly of course you need to ask, "I wonder what would happen if ...?"<|endoftext|>
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Decimal Vs Duodecimal System
I think that duodecimal (dozenal) system would be much more appropriate for architecture and let me explain you why. A big chunk of the design procedure is composed of dividing objects and distances into fractions, which is a mess in the decimal system. For example, we need to divide 20 m long Facade into thirds, which results in those funny numbers with infinite dividers: 6,666 and 13,333. Those numbers are not very useful on the construction site and no producer will make prefabricates in that sizes. We are forced to either use approximations (in this case 6 and 13) or to avoid certain dividers. First option creates some residueg and the second option is not really feasible, as simply too much of architectural design is composed of dividing geometric forms into fractions: halves, thirds, quarters, fifths, tenths, and so on.
Duodecimal system solves this problem by elegantly adding two extra numbers to the counting system. Instead of grouping numbers into tenths, we group them into twelfths.
1 2 3 4 5 6 7 8 9 10 (11) (12)
1 2 3 4 5 6 7 8 9 ᘔ Ɛ 10
Technically, number 10 gets a new symbol ᘔ (rotated number 2) and 11 gets Ɛ (rotated number 3). It is actually the oldest counting system, invented in the ancient Sumeria. At that time with the basis of 60 (4 x 12), which is still in use in measuring time. Ancient Sumerians found that this way they could simply divide a unit of crops into thirds and fourths for the purpose of taxation. The Germanic languages still have separate words for 11 and 12, and until the end of 19th century many traders used dozens as their preferred units of measurement.
The argument of having 10 fingers to count, well, it is only partially true. We have 3 bones in each finger, with 4 fingers it makes 12. Index finger is used for counting. Try it, it does work :).
Let’s take a look at dividing into fractions. In the decimal system 1 meter is composed of 10 dm. Let’s divide it:
DECIMAL
1/1 of a meter = 10 dm
1/2 of a meter = 5 dm
1/3 of a meter = 3,333333 dm
1/4 of a meter = 2,5 dm
1/5 of a meter = 2 dm
1/6 of a meter = 1,6666666 dm
1/8 of a meter = 1,25 dm
1/10 of a meter = 1 dm
In the duodecimal system 1 meter is composed of 12 dm:
DUODECIMAL
1/1 of a meter = 12 dm
1/2 of a meter = 6 dm
1/3 of a meter = 4 dm
1/4 of a meter = 3 dm
1/5 of a meter = 2,4 dm
1/6 of a meter = 2 dm
1/8 of a meter = 1,5 dm
1/10 of a meter = 1,2 dm
1/12 of a meter = 1 dm
It is quite obvious that we got rid of those infinite dividers (1,333, etc.) and that almost all numbers are natural numbers. Calculations are much easier, and have almost no residue.
That is why I almost always use as the basis for my plans 1,2 m (and not 1 m), as it is so easy to divide it into fractions. Do I have to extend a road for 1/3? I just add 0,4 m. And I try to avoid the number 5, as it is a prime number and so non-dividable (numbers as 1’5, 0’5, etc.).
Am I a proponent of the duodecimal system? Yes. Do I think it is realistic to implement it? Of course not.
More on this here and here<|endoftext|>
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