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# Limitation and Propositional Logic and Predicates
In this section, we will learn about the limitations of Propositional logic and predicates. For this, we will cover the following topics:
• Limitation of Propositional logic
• Predicate Logic and Predicates
## Limitations of Propositional Logic
As we know that the propositional logic contains the statements. In case of propositional logic, we are not allowed to conclude the truth of some or ALL statements. Hence, it is not possible to translate or conclude some valid arguments of the propositional logic into purely propositional logic. In case of propositional logic, there is no possibility to describe properties that apply to the object's category. It is also impossible to describe the relationship between those properties.
### Examples of Propositional logic
There are various examples of propositional logic, and some of them are shown below:
Example 1:
• All the chemicals and equipment of the chemistry lab are functioning properly.
• Chemistry lab of my college is functioning properly.
• However, we are not able to determine the truth related to whether the business lab is functioning.
Example 2:
• Harry is playing.
• If Harry is playing, then she will not watch the movie.
• So, Harry will not watch the movie.
Example 3:
• A virus is used to infiltrate computer system A.
• A virus is used to infiltrate the computer system B.
• However, a virus has been used by someone to infiltrate the city network of the organization.
So, in order to infer the statements, we use propositional logic from general rules.
## Predicate Logic
Suppose there is a statement that contains variables a and b. If there is a variable that is not specified by any value, then that type of statement will neither be true nor false.
Propositions can be made with the help of predicate logic from statements that have variables. If there is a statement that has a variable, then it will have two parts, which are described as follows:
Suppose there is a statement "a is equal to 5".
• The first part of this statement is "the variable a", which is used to indicate the subject of the statement.
• The second part of this statement is "is equal to 5", which is used to indicate the property that the subject of the statement can have.
• With the help of symbol P(a), we can indicate the statement "a is equal to 5", where P is used to indicate the predicate "is equal to 5", and a is used to indicate the variable.
• Once the variable x is assigned, in this case, statement P(a) becomes the proposition and truth table.
### Examples of Predicate
A statement can have one variable or more than one variable. Now we will explain the one variable statement and two variable statements one by one with the help of their examples, which are shown below:
One variable
Here we will explain those types of statements that have only one variable. The examples of statements with one variable are described as follows:
Example 1: Suppose there is a statement P(x) = x>3. Now we have to determine the truth values of p(4) and p(2).
Solution: From the question, we have a statement P(x) = x>3
When we put 2 in place of x, then we will get the following:
• P(2) has a statement "2>;3". This statement is false.
• P(4) has a statement "4>3". This statement is true.
Hence, the truth value of P(2) is false, and the truth value of P(4) is true.
Example 2: Suppose there is a statement P(x) = "A virus is used to infiltrate our computer network". Suppose a virus is used to infiltrate the CS20 and Business. Now we have to determine the truth values of A(CS10), A(CS20), and A(Business).
Solution: From the question, we have a statement:
P(x) = "A virus is used to infiltrate our computer network".
• As we can see that CS10 is not on the infiltrate list. So we can say that A(CS10) will be false.
• The CS20 and Business are on the infiltrate list. So we can say that A(CS20) and A(Business) will be true.
Hence, the truth value of A(CS10) is false, and the truth value of A(CS20) and A(Business) are true.
Two Variables
There can be those types of statements that are related to more than one variable. The examples of statements with two variables are described as follows:
Example 1: Suppose we have a proposition Q(a, b) that has a statement "a = b+6". Now we have to determine the truth value of Q(3, 6) and Q(6, 0).
Solution: From the question, we have a statement
Q(a, b) = "a = b+6".
• Q(3, 6) has a statement "3 = 6 + 6". This statement is false because 3 is not equal to 12.
• Q(6, 0) has a statement "6 = 0 + 6". This statement is true because 6 = 6.
Hence, the truth value of Q(3, 6) is false, and the truth value of Q(6, 0) is true.
Example 2: Suppose we have a proposition Q(a, b) that has a statement "a = b-5". Now we have to determine the truth value of Q(7, 4) and Q(0, 5).
Solution: From the question, we have a statement
Q(a, b) = "a = b-5".
• Q(7, 4) has a statement "7 = 4 - 5". This statement is false because 7 is not equal to -1.
• Q(0, 5) has a statement "0 = 5 - 5". This statement is true because 0 = 0.
Hence, the truth value of Q(7, 4) is false, and the truth value of Q(0, 5) is true.
### n-ary Predicate
In general, if there is a statement that is used to contain the n number of variables x1, x2, x3, ...., xn, in this case, it will be denoted in the following way:
P(x1, x2, x3, ...., xn).
Suppose there is a statement in the form P(x1, x2, x3, ...., xn). This statement is used to indicate the value of P at the n-tuple (x1, x2, x3, ..., xn). Here P is used to indicate the propositional function, and P can also be known as the n-ary predicate.<|endoftext|>
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This free course, Gene testing, looks at three different uses of genetic testing: pre-natal diagnosis, childhood testing and adult testing. Such tests provide genetic information in the form of a predictive diagnosis, and as such are described as predictive tests. Pre-natal diagnosis uses techniques such as amniocentesis to test fetuses in the womb. For example, it is commonly offered to women over 35 to test for Down's syndrome. Childhood testing involves testing children for genetic diseases that may not become a problem until they grow up, and adult testing is aimed at people at risk of late-onset disorders, which do not appear until middle age. In addition, we address some of the issues involved in carrier testing, another predictive test. This involves the testing of people from families with a history of genetic disease, to find out who carries the gene, and who therefore might pass the disease onto their children even though they themselves are unaffected. Here the aim is to enable couples to make informed choices about whether or not to have children, and if so whether they might have a genetic disease studies 'proteins'. Starting with a simple analysis of the molecular make up, the course moves on to look at the importance of protein and how they are digested and absorbed.
Course learning outcomes
After studying this course, you should be able to:
- understand something of the role of a genetic counsellor and its non-directiveness
- understand the difference between pre-natal diagnosis, childhood testing and adult testing and give some examples of diseases that may be tested for
- understand the ethical and moral difficulties involved in making decisions on whether or not to carry out such tests.
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- Learning outcomes
- 1 Genetic testing
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The Roman Province Moesia, mainly modern day Bulgaria and Serbia, was a stretch of mountainous terrain in the west and more fertile plains in the east. It bordered Macedonia and Thracia to the south, the Black Sea (Pontus Euxinus) to the east and shared the Danube as a border with Dacia in the North.
The region was initially inhabited by Dacian tribes, but were mostly supplanted by migrating Thracian Moesi tribe people in the mid 1st millenium BC. These Thracians would eventually come to dominate the territory.
Roman interest in Moesia was prompted by the realization of the rich mining and fertile fields that it had to offer. In 75 BC, C. Scribonius Curio, proconsul of Macedonia, in expanding territory and protecting Roman interests, penetrated as far as the Danube. The Romans gained victories as they went, but would only stay a short time, leaving the area out of Roman control. Little is really known about these early expeditions or the formal declaration of Moesia as a province, but after the ascension of Augustus the conquest was completed.
During his reign, Augustus commissioned Marcus Licinius Crassus, grandson of the original Caesarian Triumvir, serving as proconsul of Macedonia, to subdue the local tribes and brings its wealth under Roman influence. By 29 BC the campaign commenced and the Moesi and Dacian residents were defeated, seemingly by 6 AD. It is this year, according to Dio Cassius, that the first actual governor of Moesia is recorded, Caecina Severus.
Moesia was initially developed as a single province, but the different challenges in terrain and size prompted Domitian (85 AD) to the divide it into two: Upper (superior) and Lower (Inferior, also called Ripa Thracia) Moesia. Moesia was not only rich in mineral wealth in the west, and grain in the east, but served as an all important buffer between the Greek provinces and various potential invaders.
The primary settlements of the provinces were Singidunum (Belgrade), Viminacium (sometimes called municipium Aelium, both major legionary outposts on the Danube, Bononia (Widdin), Ratiaria (Artcher): Naissus (birthplace of Constantine), Oescus (colonia Ulpia, Gigen), Novae (near Sistova), Nicopolis ad Istrum (Nikup), Odessus (Varna), and Tomis (Constanta). Of these, Tomis was interesting in that it became the place of banishment for the Roman Poet Ovid. He was banished by Augustus in 9 AD, for seemingly disagreeable poetry, where he lived until his death in 17 AD. Ovid painted a bleak picture of Tomi and Moesia, claiming it was a violent and uncivilized place, obviously far different than what a highly cultured artist would be accustomed to.
Romanization of the Moesian people, being a volatile border province, was far less observed than in other more peaceful concerns, but the people seemed to settle in under Roman rule without a great deal of trouble over the years. While Ovid's descriptions must be measured by his prejudice, it is evident that the province was under constant pressure from Germanic Goths, Scythians, Sarmatians and other regional migrating tribes. The permanent legionary garrisons may have contributed to the lack of trouble from within.
In the early 2nd century AD, with the conquest of Dacia by Trajan, Moesia likely got some relief from these external pressures. Dacia, however would not remain long within Roman control and the pressure would once again fall to Moesia. Aurelian (270 - 275) AD) abandoned the Dacian conquest to the Germanic tribes, moving Romanized Dacians south of the Danube. At this point the province was renamed Dacia Aureliani to reflect this, and was again divided between upper and lower territories.
The Goths, who were the major source of conflict for the Romans in the area, crossed the Danube, for the countless and final time, under their King Fritigern in 376 AD, under pressure from the Huns themselves. The Emperor Valens, seeking to maintain the peace and avoid losing the province to the Goths, allowed them to settle the region in exchange for defending it from the Huns. Terms were disagreeable, however, and the Goths, feeling as fodder for Rome's legions, soon broke the yoke of Roman control. In 378 AD, at the battle of Adrianople, in modern day Turkey, Fritigern and his Goths destroyed the army of Valens, and Moesia passed forever out of Roman dominion.
As already mentioned, Moesia was a province of volatile activity. A frontier province bordered by the Goths, the region required the constant presence of several legions. For the bulk of the Roman occupation, Moesia was home to 4 legions-- Legio I Italica at Novae and Legio XI Claudia at Durostorum in Inferior, and Legio IV Flavia at Singidunum (Belgrade) and Legio VII Claudia at Viminacium in Superior. The province was also home to the fleet of the Euxinus (Black Sea), the Classis Moesica at Noviodunum.
The Danube, thanks to legionary engineers, became a a vast network of fortifications stretching from the Rhein in Germania all the way to the Black Sea. This province not only served as a border protecting interior provinces, but was the source of vast mineral wealth requiring important occupation of force.
Economy of Moesia
The mountainous stretches of Moesia Superior, to the west, were rich in several valuable metals. Gold, especially, was the key mineral resource mined by the Romans, and its high content was a key source of Roman coinage. In Moesia Inferior, to the east, the plains provided rich and fertile farmland. Under the governing of Ti. Plautius Silvanus Aelianus, in 57 - 67 AD, the province first added Moesian wheat to the grain supply. The area also contained a vast amount of pasture land for livestock, bountiful orchards and timber. Its position bordering the Black Sea and the Danube also made it an important trade hub for west to east caravans and shipments.<|endoftext|>
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It’s a familiar fall scene: you hear the honking first, then see the v-shaped flock fly over – geese heading to some exotic locale for the winter. It’s obvious why “snow birds” – ironically human – head to warmer climes to avoid the frigid Northern winters, but what about these birds? Why do they migrate back and forth each year, and how do they even know where they’re going?
Approximately 1,800 of the world’s 10,000 bird species take this annual, large-scale movement between their breeding or nesting (summer) homes to their non-breeding (winter) homes each year. (1) Like many things in life, food is the main motivating factor. Birds that nest in the northern hemisphere hang out in the spring to take advantage of the plentiful insect populations, budding plants, and large quantity of places to set up “nest”. As winter sets in and the availability of insects and other food options declines, the birds head south – simple as that. (2) Many of these birds that breed in North America migrate to areas south of the Tropic of Cancer (Southern Mexico, Central and South America and the Lesser and Greater Antilles in the Caribbean Sea) in the fall (August-October) and then winter there until April when they head back to their old stomping grounds up North to breed and raise young. (3)
There are three different types of bird migration: short (moving from a higher to lower elevation on a mountainside), medium (moving a distance that spans several states) and long-distance (generally moving from the northern hemisphere to the southern hemisphere). For short-distance migrants, the reason really is as simple as a need for available food. But the origins of long-distance migration are a little more complicated. What “tips” the birds off that it’s time to get moving varies – days getting shorter and colder, dwindling food supplies, or it even something in their genetic predisposition. (2)
In the period before migration, many birds display higher activity or Zugunruhe – German for “migratory restlessness”. Even cage-raised birds with no environmental cues (e.g. shortening of day and falling temperature) show signs of Zugunruhe, further leading scientists to believe migratory tendencies might be genetically predisposed. (1)
Birds also eat more food pre-migration, storing it as fat. Fat is normally three to five percent of the bird’s mass, but some will almost double their body weights as they pack it on for the trip! The ruby-throated hummingbird, for example, weighs only 4.8 grams but can use stored fat to fuel a non-stop, 24-hour flight across a 600-mile stretch of open water from the U.S. Gulf coast to the Yucatan Peninsula of Mexico! But most songbirds don’t fly to their non-breeding grounds non-stop. They stop a number of times to rest and feed in places called stopover sites. Some birds stop only one day to rest and feed, and then continue their migration. Others will remain at stopover areas for weeks, storing up more fat. The arctic tern may hold the longest distance migration, made possible because they stop over various places to eat fish and feed along the way. The tern migrates about 18,600 miles each year! Amazing! (3)
It’s not just the distance traveled that is amazing, though – the “how” of the travel is pretty wild too. They don’t come equipped with GPS, but somehow migrating birds can cover thousands of miles, often on the same exact “bird highway”, year after year. Even first year birds may migrate – without a guide – to a winter home they have never before seen and return in the spring to their birth land. (2) They use the age-old compasses in the sky to navigate the way – the sun and stars – and also something really cool: the Earth’s magnetic field! Yes, you read that right! Birds apparently have tiny grains of the mineral magnetite just above their nostrils, which helps them find what direction is true north by using the Earth’s magnetic field.
Beyond that, day flyers navigate using the positions of the sun and night flyers find their way by following the patterns of the stars. And – get this: In their very first year of life, those birds memorize the position of the constellations in relation to the North Star! Some birds can also use their sense of small to help find the way. (3)
Many, if not most, birds migrate in flocks, which for larger birds, can conserve energy. Geese save from 12 to 20 percent of the energy they would need to fly alone – and some even fly faster in flock formation! (1)
In the spring, we’ll see them heading back up the “road” home, and now, when you gaze up at the first honk, know it might even be the same exact birds you saw this fall!
Aryn Henning Nichols was constantly amazed as she researched this Science, You’re Super. How cool are migrating birds?!?<|endoftext|>
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The case for Planet Nine is growing. Two new findings presented at a planetary science meeting in Pasadena, California, have uncovered hints for the existence of this distant, mysterious world in the motions of known solar system objects.
The results could help astronomers hone in on their target, which – if it really is out there – could fundamentally alter our understanding of the solar system.
The hunt for Planet Nine began in earnest in 2014 after astronomers Scott Sheppard and Chadwick Trujillo found 2012 VP113, a planetoid nicknamed “Biden.” Its closest point to the sun in its orbit is 80 astronomical units – that is, 80 times the Earth-Sun distance of 93 million miles.
Objects like 2012 VP113 exist far beyond the typical denizens of the Kuiper belt, the icy ring of debris that stretches from Neptune’s orbit at 30 AU out to 50 AU (and whose largest member is distant Pluto, sitting around 49 AU).
The scientists also noticed that 2012 VP113 and another far-out mini-world named Sedna were making their closest approaches to the sun at similar angles – which could mean that something massive but unseen was tugging on both their orbits. Since the scientists could not directly see a planet in the darkness of our solar system, they would have to keep searching for its gravitational fingerprint on the motions of other bodies.
Then, early this year, California Institute of Technology scientists Konstantin Batygin and Mike Brown found that the paths of half a dozen extremely distant objects seemed to be similarly tilted relative to the plane of the solar system, and that their perihelia – their closest points to the sun – seemed to cluster together. They estimated that a Planet Nine would weigh about 10 Earth masses and take somewhere from 10,000 to 20,000 years to orbit the sun.
Now, a team led by Renu Malhotra, a planetary scientist at the University of Arizona, has examined the orbits of four extreme Kuiper belt objects with the longest-known orbital periods and found an elegant relationship among their orbits: They can be described essentially in simple, whole-number ratios. This suggests that they’re pulled into these resonances by the gravity from an unseen massive object.
Malhotra’s team calculated that such a planetoid would be 10 times the mass of Earth and would orbit the sun roughly every 17,000 years – which fits with the Caltech scientists’ range estimate of 10,000 to 20,000 years. At its farthest point, this planet would lie a whopping 665 astronomical units from the sun. The results were published in the Astrophysical Journal Letters.
Meanwhile, Brown and Batygin have found more potential evidence of Planet Nine’s influence. Their calculations, accepted for publication in the Astrophysical Journal, suggest that the solar system’s slight tilt relative to the sun might have been caused by the massive world’s pull.
Since the mid-1800s, scientists have wondered why the plane of the solar system – the plane in which all the planets orbit – is tilted 6 degrees relative to the spin axis of the sun, said lead author Elizabeth Bailey, an astronomer and Ph.D. student at Caltech. Given that Planet Nine is suspected to be circling the sun at an even more extreme angle, the scientists calculated that it could indeed have pulled the planets out of alignment with the sun, causing the 6-degree mismatch.
“From our vantage point, it looks like it’s the sun that’s tilted; but really it’s the plane of the planets precessing around the total angular momentum of the solar system, just like a top,” she said.
Neither study is a slam-dunk case for the planet’s existence, scientists said, but the evidence continues to mount.
The results from both teams were presented at the joint 48th meeting of the Division for Planetary Sciences of the American Astronomical Society and 11th European Planetary Science Congress in Pasadena.<|endoftext|>
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Lesson 3
A Gallery of Data
• Let’s make, compare, and interpret data displays.
3.1: Notice and Wonder: Dot Plots
The dot plots represent the distribution of the amount of tips, in dollars, left at 2 different restaurants on the same night.
What do you notice? What do you wonder?
3.2: Data Displays
Your teacher will assign your group a statistical question. As a group:
1. Create a dot plot, histogram, and box plot to display the distribution of the data.
2. Write 3 comments that interpret the data.
As you visit each display, write a sentence or two summarizing the information in the display.
Choose one of the more interesting questions you or a classmate asked and collect data from a larger group, such as more students from the school. Create a data display and compare results from the data collected in class.
Summary
We can represent a distribution of data in several different forms, including lists, dot plots, histograms, and box plots. A list displays all of the values in a data set and can be organized in different ways. This list shows the pH for 30 different water samples.
• 5.9
• 7.6
• 7.5
• 8.2
• 7.6
• 8.6
• 8.1
• 7.9
• 6.1
• 6.3
• 6.9
• 7.1
• 8.4
• 6.5
• 7.2
• 6.8
• 7.3
• 8.1
• 5.8
• 7.5
• 7.1
• 8.4
• 8.0
• 7.2
• 7.4
• 6.5
• 6.8
• 7.0
• 7.4
• 7.6
Here is the same list organized in order from least to greatest.
• 5.8
• 5.9
• 6.1
• 6.3
• 6.5
• 6.5
• 6.8
• 6.8
• 6.9
• 7.0
• 7.1
• 7.1
• 7.2
• 7.2
• 7.3
• 7.4
• 7.4
• 7.5
• 7.5
• 7.6
• 7.6
• 7.6
• 7.9
• 8.0
• 8.1
• 8.1
• 8.2
• 8.4
• 8.4
• 8.6
With the list organized, you can more easily:
• interpret the data
• calculate the values of the five-number summary
• estimate or calculate the mean
• create a dot plot, box plot, or histogram
Here is a dot plot and histogram representing the distribution of the data in the list.
A dot plot is created by putting a dot for each value above the position on a number line. For the pH dot plot, there are 2 water samples with a pH of 6.5 and 1 water sample with a pH of 7. A histogram is made by counting the number of values from the data set in a certain interval and drawing a bar over that interval at a height that matches the count. In the pH histogram, there are 5 water samples that have a pH between 6.5 and 7 (including 6.5, but not 7). Here is a box plot representing the distribution of the same data as the dot plot and histogram.
To create a box plot, you need to find the minimum, first quartile, median, third quartile, and maximum values for the data set. These 5 values are sometimes called the five-number summary. Drawing a vertical mark and then connecting the pieces as in the example creates the box plot. For the pH box plot, we can see that the minimum is about 5.8, the median is about 7.4, and the third quartile is around 7.9.
Glossary Entries
• categorical data
Categorical data are data where the values are categories. For example, the breeds of 10 different dogs are categorical data. Another example is the colors of 100 different flowers.
• distribution
For a numerical or categorical data set, the distribution tells you how many of each value or each category there are in the data set.
• five-number summary
The five-number summary of a data set consists of the minimum, the three quartiles, and the maximum. It is often indicated by a box plot like the one shown, where the minimum is 2, the three quartiles are 4, 4.5, and 6.5, and the maximum is 9.
• non-statistical question
A non-statistical question is a question which can be answered by a specific measurement or procedure where no variability is anticipated, for example:
• How high is that building?
• If I run at 2 meters per second, how long will it take me to run 100 meters?
• numerical data
Numerical data, also called measurement or quantitative data, are data where the values are numbers, measurements, or quantities. For example, the weights of 10 different dogs are numerical data.
• statistical question
A statistical question is a question that can only be answered by using data and where we expect the data to have variability, for example:
• Who is the most popular musical artist at your school?
• When do students in your class typically eat dinner?
• Which classroom in your school has the most books?<|endoftext|>
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The Greenland ice sheet, a vast body of ice covering approximately 80 percent of the country, was discovered recently by scientists as hiding beneath its icy surface billions of gallons of fresh water. This finding was revealed as scientists and researchers are monitoring the ice sheet’s continuous melting as it relates to climate change and the rise of the sea level.
What baffles scientists is that the vast reservoir of fresh water, located at the southeastern part of Greenland, remains liquid even if the temperature above is always below freezing point. The reservoir of water was discovered in 2011 by researchers from the University of Utah while drilling for samples. Upon further inspections, the area covered by the liquid water was estimated to be 27,000 square miles, or half the size of New York State. According to the study published in the scientific journal Nature Geoscience, on the first drill, the water was found at a depth of 33 feet and on the second drill it was at about 82 feet.
The discovery has significant implications in understanding climate change and global sea-level rise. In terms of contributions to the global sea-level rise, the Greenland ice sheet is the largest contributor and it is continuously melting at a record pace. According to Rick Forster, professor of geography at the University of Utah and lead author of the study published in Nature, the water reservoir can help researchers better understand the relationship between meltwater runoff and sea levels.
An earlier National Geographic report indicates that the global sea levels have risen from 0.07 inches in the 1980s to o.14 inches in the 1990s. And, if the Greenland ice sheet is lost all together due to rapid melting, it is estimated that an additional six meters of water would be added to the global sea level and this will be catastrophic. In the last 20 years, the ice losses have already increased the average sea levels by 0.34 inches.
Earlier studies have indicated three significant contributory factors to rising sea levels: first is thermal expansion, which pertains to water heating up; second is the continuous melting of ice caps and glaciers; and, lastly, the rapid ice loss from Greenland and West Antarctica.
Between 1992 and 2001, the Greenland ice sheet lost about 34 billion tons of ice annually; and, between the years 2002 and 2011, that figure increased to 215 billion tons. With the new discovery of fresh water stored underneath the vast ice sheet, this could suggest that a possible large amount of melted ice over the years goes to this place.
What is causing concern among the researchers is the fact that if these vast amounts of water trapped under the Greenland ice sheets escaped, this could greatly hasten global sea-level rise. According to Forster, “Most models assume water runs off or refreezes.” Yet, with this latest finding, this could either mean that the water reservoir is buffering the sea-level rise and delaying the whole process or the water acts as a lubricant to moving glaciers and hastening their movement into the global ocean. He also added, “We don’t know the answer to this right now. It’s massive, it’s a new system we haven’t seen before.”
The discovery that the Greenland ice sheet is hiding billions of gallons of fresh water underneath is indeed a great scientific find. Understanding this and its impact on the global sea-level rise would help researchers accurately predict sea-level rise and advise people on any disaster preparedness plans to undertake.
By Roberto I. Belda<|endoftext|>
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Growing fruit trees in the home garden can be an enjoyable, relaxing and rewarding experience. However, success takes careful planning and hard work.
When planting fruit trees, it is important to consider site selection. Fruit trees should be planted in fertile, well-drained soil in a site that receives full sun. It is also important to consider fertility and soil pH. For fruit trees, the soil pH should be between 6.0 and 6.5.
Take a soil sample to determine the soil’s fertility needs and pH. Do not put fertilizer in the planting hole or fertilize immediately after planting. Fertilizers in direct contact with young roots can burn and cause damage.
Good root growth is essential
If purchased bare root, fruit trees are best planted in late fall or winter. If plants are purchased in containers, they can be planted any time of the year. Just take care to add the needed water when there is not enough rain. University of Georgia Cooperative Extension experts recommends planting fruit trees during the winter months to achieve the best success of establishment.
Whether trees are purchased bare root or containerized, the planting hole should at least two to three times the width of the root system. Prune any damaged roots on bare root plants and open up the root ball of container grown plants before planting. This prevents root circling.
Weeds and grasses that grow in the immediate area around fruit trees compete for both moisture and nutrients. One of the best methods of weed control is mulch. Place mulch in a ring 2 to 4 inches deep out around the tree to the drip line. Be sure to keep mulch several inches away from the trunk.
Avoid using hoes or string trimmers around trees. Hoeing can damage the root system and string trimmers can damage tree trunks.
Without proper management, insects and diseases can seriously damage fruit trees and reduce production. Both inorganic and organic control products are available at most garden centers. Always read and follow label directions carefully.
Proper sanitation can reduce insect and disease problems. Apples, nectarines, peaches and pears should be thinned approximately four weeks after they bloom. Remaining fruit should be 4 to 6 inches apart. Remove dead, diseased or damaged wood, fruit and leaves from around fruit trees. This debris can provide overwintering sites for insects and diseases.
Train and prune
Fruit trees need proper training and pruning for improved production. Pruning should be done during the winter months. Dormant or winter pruning consists of removing upright branches and any dead, diseased or damaged wood.
Gardeners who preplan should be rewarded with a bountiful harvest of fruit in coming years.
For more information on installing a backyard orchard, contact your local UGA Extension office at 1-800-ASK-UGA1 or see the website www.caes.uga.edu/publications/.<|endoftext|>
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# 3.1 Complex numbers (Page 2/8)
Page 2 / 8
## Plotting a complex number on the complex plane
We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number we need to address the two components of the number. We use the complex plane , which is a coordinate system in which the horizontal axis represents the real component and the vertical axis represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs $\text{\hspace{0.17em}}\left(a,b\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ represents the coordinate for the horizontal axis and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ represents the coordinate for the vertical axis.
Let’s consider the number $-2+3i.\text{\hspace{0.17em}}$ The real part of the complex number is $-2\text{\hspace{0.17em}}$ and the imaginary part is $\text{\hspace{0.17em}}3i.\text{\hspace{0.17em}}$ We plot the ordered pair $\text{\hspace{0.17em}}\left(-2,3\right)\text{\hspace{0.17em}}$ to represent the complex number $-2+3i\text{\hspace{0.17em}}$ as shown in [link] .
## Complex plane
In the complex plane , the horizontal axis is the real axis, and the vertical axis is the imaginary axis as shown in [link] .
Given a complex number, represent its components on the complex plane.
1. Determine the real part and the imaginary part of the complex number.
2. Move along the horizontal axis to show the real part of the number.
3. Move parallel to the vertical axis to show the imaginary part of the number.
4. Plot the point.
## Plotting a complex number on the complex plane
Plot the complex number $\text{\hspace{0.17em}}3-4i\text{\hspace{0.17em}}$ on the complex plane.
The real part of the complex number is $\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}$ and the imaginary part is $\text{\hspace{0.17em}}-4i.\text{\hspace{0.17em}}$ We plot the ordered pair $\text{\hspace{0.17em}}\left(3,-4\right)\text{\hspace{0.17em}}$ as shown in [link] .
Plot the complex number $\text{\hspace{0.17em}}-4-i\text{\hspace{0.17em}}$ on the complex plane.
## Adding and subtracting complex numbers
Just as with real numbers, we can perform arithmetic operations on complex numbers. To add or subtract complex numbers, we combine the real parts and combine the imaginary parts.
## Complex numbers: addition and subtraction
$\left(a+bi\right)+\left(c+di\right)=\left(a+c\right)+\left(b+d\right)i$
Subtracting complex numbers:
$\left(a+bi\right)-\left(c+di\right)=\left(a-c\right)+\left(b-d\right)i$
Given two complex numbers, find the sum or difference.
1. Identify the real and imaginary parts of each number.
2. Add or subtract the real parts.
3. Add or subtract the imaginary parts.
Add $\text{\hspace{0.17em}}3-4i\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}2+5i.\text{\hspace{0.17em}}$
Subtract $\text{\hspace{0.17em}}2+5i\text{\hspace{0.17em}}$ from $\text{\hspace{0.17em}}3–4i.\text{\hspace{0.17em}}$
$\left(3-4i\right)-\left(2+5i\right)=1-9i$
## Multiplying complex numbers
Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.
## Multiplying a complex numbers by a real number
Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So, for example,
Given a complex number and a real number, multiply to find the product.
1. Use the distributive property.
2. Simplify.
## Multiplying a complex number by a real number
Find the product $\text{\hspace{0.17em}}4\left(2+5i\right).$
Distribute the 4.
$\begin{array}{l}4\left(2+5i\right)=\left(4\cdot 2\right)+\left(4\cdot 5i\right)\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=8+20i\hfill \end{array}$
Find the product $\text{\hspace{0.17em}}-4\left(2+6i\right).$
$-8-24i$
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
what is foci?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations<|endoftext|>
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Class Teacher- Mrs Collins
Classroom Assistants: Miss Hackett and Mr Franklin
Welcome to Primary 6
The Areas of Learning
- Language and Literacy
- Mathematics and Numeracy
- The Arts
- The World Around Us
- Physical Education
- Religious Education
Cross Curricular Skills
Thinking Skills and Personal Capabilities
-Thinking, Problem Solving, Decision Making
-Working with Others
Children have the opportunity to acquire, develop and demonstrate the Cross Curricular Skills and Thinking skills and Personal Capabilities in all areas of the Curriculum. Children work individually, in pairs, in groups and as a whole class.
This year is a very busy one for us. We work hard every day in class at a range of oral, practical and recording activities through which we are actively engaged in our learning.
We understand how important attendance is to learning and progressing.
Language and Literacy
Every day we engage in Reading, writing and Talking and Listening activities.
We participate in guided reading, accelerated reading and paired reading.
We use Accelerated Reading to develop our reading skills. When we have finished reading a book we can take a test online and can find out our score. We can even find out how many words we have read.
I wonder who will be our first Millionaire!!
We keep a wide selection of books in our classroom and change them regularly.
Our class Novel is called 'The Water Horse' by Dick King-Smith.
Throughout the year we will be looking at many different writing genres and learning how to become great writers.
We need to keep checking over our work and think of ways to extend our sentences and improve our vocabulary. This will make our writing really interesting and, who knows, one day the next big author could be telling the world that they learnt all their writing skills in John Paul II Primary School.
We use drama strategies to help us understand and explore the text in more detail.
We are also improving our handwriting skills through practice of the Nelson Handwriting Scheme.
Mathematics and Numeracy
We have been learning how to develop a range of mental skills to help us problem solve. It is important to keep learning tables and mathematical facts every night and challenge ourselves to become better mathematicians.
We know that maths is not just about recording work in books so we also engage in lots of oral and practical activities.
We are exposed to ‘real life’ mathematical problems that encourage us to discuss and explore many different ways to find a solution. We are encouraged to be mathematicians, not just in school, but at home; while playing with our friends; shopping with our families or even planning a family trip. We recognise that maths is all around us and we aim to make connections using the skills we are taught.
Through structured activities we apply all of these skills to number, measure, shape and space and handling data. Some areas that are new to us include: angles; decimal fractions; percentages; 24 hour clock. We are also now working hard to learn all of our times tables off by heart – help me learn at home.
Everyone in Primary 2 to Primary 7 took part in an exciting Mental Maths competition. Next term I am going to work even harder to try and win a prize.
Some websites that will help me learn more:
It is fair to say that we all love our ICT lessons and we are given lots of time to investigate and use it to create our own stories, newsletters, poems etc. We like the way we are given a topic and can use the internet to help us to investigate it. The skills that we work hard to improve in ICT are: desktop publishing; managing data; online communication; presenting and working with images. In the classroom ICT is an important tool in all areas of our learning.
World around us
In Primary 6 we will discover so much about the world around us. In Term One we looked at the STEM Topic of Water and had great fun discovering new facts and experimenting. We discovered the importance of water and how lucky we are to have fresh water at our fingertips. We couldn't believe how people throughout the world have to cope with no clean water. We hope to do something about this in Term 2 and 3 so watch this space!
In Term 2 we we will be learning all about The Victorians.
Above and beyond all of this we even manage to fit in singing, swimming, basketball, Spanish, assemblies, golden time, circle time and art.
Alive-O 6 is our religion programme where we learn about God’s love for us and how we can share this love with others.<|endoftext|>
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Claude-Achille Debussy (French: ; 22 August 1862 – 25 March 1918) was a French composer. Along with Maurice Ravel, he was one of the most prominent figures associated with Impressionist music, though he himself disliked the term when applied to his compositions. He was made Chevalier of the Legion of Honour in his native France in 1903. Debussy was among the most influential composers of the late 19th and early 20th centuries, and his use of non-traditional scales and chromaticism influenced many composers who followed. Debussy's music is noted for its sensory content and frequent usage of atonality. The prominent French literary style of his period was known as Symbolism, and this movement directly inspired Debussy both as a composer and as an active cultural participant.<|endoftext|>
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Comparing Fractions is to compare the two fractions and find greatest and smallest fraction among the given two fractions. To compare fractions, we need to see that all the fractions are greater than zero. And, also check that the fraction is a proper fraction or an improper fraction.
## How to Compare Fractions
Rule 1: If the denominators of the two fractions are same, then the fraction with larger numerator is greater.
Example:
Compare
Solution: The fractions $\frac{3}{4}$ and $\frac{1}{4}$ have same denominator. So, $\frac{3}{4}$ is greater than $\frac{1}{4}$
$\frac{3}{4}$ > $\frac{1}{4}$
Rule 2: If the numerators of the two fractions are same, then the fraction with smaller denominator is greater.
Example:
Compare
Solution: The fractions $\frac{2}{4}$ and $\frac{2}{8}$ have same numerator. So, $\frac{2}{4}$ is greater than $\frac{2}{8}$
$\frac{2}{4}$ > $\frac{2}{8}$
Rule 3: If both numerator and denominator of the two fractions are different, then multiply the numerator and denominator of both the fractions with same numbers so as to have same denominator. Then, compare the numerator of the two fractions.
Example: Compare
Solution: The fractions $\frac{2}{3}$ and $\frac{3}{4}$ are unlike fractions. We need to multiply $\frac{2}{3}$ by $\frac{4}{4}$ and multiply $\frac{3}{4}$ by $\frac{3}{3}$ to get same denominator.
$\frac{2}{3} \times \frac{4}{4} = \frac{8}{12}$ and $\frac{3}{4} \times \frac{3}{3} = \frac{9}{12}$
Now, on comparing the numerators. The fraction $\frac{3}{4}$ which is equal to $\frac{9}{12}$ is greater than $\frac{2}{3}$ which is equal to $\frac{8}{12}$.
$\frac{2}{3}$ < $\frac{3}{4}$
### Ordering Fractions Calculator
Comparing Fractions Comparing Unlike Fractions Compare Numbers how to compare integers Comparing and Ordering Rational Numbers What are Like Fractions?
Compare and Order Fractions Calculator Compare Decimals Calculator Comparing Proportions Calculator A Fraction Calculator Add 3 Fractions Calculator Add Fraction Calculator<|endoftext|>
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# How calculate the shaded area in this picture?
Let the centers of four circles with the radius $R=a$ be on 4 vertexs a square with edge size $a$. How calculate the shaded area in this picture?
Let $(APD)$ be the area of the figure $APD$.
And let $x,y,z$ be $(KEPM),(PAD),(MPD)$ respectively.
First, we have $$(\text{square}\ ABCD)=a^2=x+4y+4z.\tag1$$
Second, we have $$(\text{sector}\ BDA)=\frac{\pi a^2}{4}=x+2y+3z.\tag2$$
Third, note that $KA=KD=a$ and that $(\triangle KAD)=\frac{\sqrt 3}{4}a^2$ since $\triangle KAD$ is a equilateral triangle.
So, since we have \begin{align}(K(E)AD(M))&=(\text{sector}\ AKD)+(\text{sector}\ DKA)-(\triangle KAD)\\&=\frac{\pi}{6}a^2+\frac{\pi}{6}a^2-\frac{\sqrt 3}{4}a^2\\&=\frac{\pi}{3}a^2-\frac{\sqrt 3}{4}a^2,\end{align} we have $$\frac{\pi}{3}a^2-\frac{\sqrt 3}{4}a^2=x+y+2z.\tag3$$
Solving $(1),(2),(3)$ gives us $$(KEPM)=x=\left(1+\frac{\pi}{3}-\sqrt 3\right)a^2.$$
Here is a geometric solution.
Let's find the angle $EDK$. Since $KA=KD=a=AD$ hence the triangle $AKD$ is equiliteral and the angle $KDA$ is $\pi/3$. Therefore angle $KDC$ is $\pi/6$. In the same way the angle $EDA$ is also $\pi/6$ and we get that the angle $EDK$ is $\pi/6$.
Now the shaded area $$(EPMK) = ({\rm square\,} EPMK) + 4\times ({\rm segment\,}EKE).$$
The edge of the square $EPMK$ (from the triangle $EDK$) is $$EK=2a\sin\frac{\pi}{12}=a\frac{\sqrt{3}-1}{\sqrt{2}},$$ hence the area $$({\rm square\,} EPMK) =(2-\sqrt{3})a^2.$$
And the area $$({\rm segment\,}EKE) = \frac{1}{2}\left(\frac{\pi}{6}-\sin\frac{\pi}{6}\right)a^2=\frac{1}{4}\left(\frac{\pi}{3}-1 \right)a^2.$$
So $$(EPMK) = (2-\sqrt{3})a^2 + \left(\frac{\pi}{3}-1 \right)a^2=\left(\frac{\pi}{3}+1-\sqrt{3}\right)a^2.$$
Well this one is very easy to solve. We put a coordinate system directly in the middle of your square. The upper arc from the top left to bottom right is a quater circle and can be considered the graph of a function $g$. Now, since it is on a circle that must have its middle point at $(-a/2,-a/2)$, $g$ satisfies $$(g(x)+a/2)^2 + (x + a/2)^2 = a^2,$$ because $a$ is its radius. Now we solve this equation for $g(x)$, keeping in mind, that it is the upper half of the circe and get $$g(x) = \sqrt{a^2 - (x+a/2)^2} - a/2.$$ Due to the symmetry along the line $x=y$, it must be $g(g(0)) = 0$. Due to further symmetries, we get the area as $$A = 4\int_0^{g(0)} g(x)\,dx.$$ We can easily see, by standard integration techniques, that $$G(x) = \frac{1}{8} \left(4 a^2 \tan ^{-1}\left(\frac{a+2 x}{\sqrt{(a-2 x) (3 a+2 x)}}\right)-4 a x+(a+2 x) \sqrt{(a-2 x) (3 a+2 x)}\right)$$ is an antiderivative. Now we use the fundamental theorem of calculus and finally get $$A = 4(G(g(0)) - G(0)) = 4 \frac1{12} a^2 (3 - 3\sqrt 3 + \pi) = a^2(1 - \sqrt 3 + \pi/3).$$
• Sorry for sounding so cocky. I wrote this for someone who asked the same question, but was very rude in formulating it. It was closed before I could post the answer but I didn't want it to go to waste. :) Nov 8, 2017 at 13:38<|endoftext|>
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Architectural acoustics is the science that studies the behaviour of sound within a building. The design of a conference room, a hall, a theatre, etc., requires the analysis of all parameters related to sound and its propagation in the air.
In this first lesson we will address the first key concept related to the topic of acoustics: the sound wave.
The sound wave
In physics, the sound is defined as an oscillation (movement in space) done by the particles (atoms and molecules) in a given medium. This simple oscillatory motion propagates mechanically, originating a sound wave (or acoustic wave). The sound source can be any device, appliance, etc., that causes pressure variations, and the portion of space concerned by these variations is called sound field.
The propagation of sound occurs spherically, i.e. in all directions, and the speed of propagation depends on the nature of the elastic medium on which it diffuses.
In the acoustic design of a room you must take into account certain factors related to the sound wave such as magnitude, intensity, speed of propagation, and wavelength. When a sound wave comes into contact with the surface of an object, it can be reflected totally, partially, or not be reflected at all; it depends on the wavelength.
In the following illustrations, we see that if the size of the object placed in front of the sound source is greater than three times the wavelength, there will be total reflection; if the object is of equal size, there will be 50% reflection and 50% diffraction; if the object instead is smaller than 1/3 of the wavelength, there will be no reflection.
In the next lesson, we will study the concept of acoustic reverberation time and how it is measured during phonometric analyses.Architectural acoustics is the science that studies the behaviour of sound within a building. The design of a conference room, a hall, a theatre, etc., requires the analysis of all parameters related to sound and its propagation in the air.<|endoftext|>
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At the surface, Antarctica is a motionless and frozen landscape. Yet hundreds of miles down the Earth is moving at a rapid rate, new research has shown.
The study, led by Newcastle University, UK, and published this week in Earth and Planetary Science Letters, explains for the first time why the upward motion of the Earth's crust in the Northern Antarctic Peninsula is currently taking place so quickly.
Previous studies have shown the earth is 'rebounding' due to the overlying ice sheet shrinking in response to climate change. This movement of the land was understood to be due to an instantaneous, elastic response followed by a very slow uplift over thousands of years.
But GPS data collected by the international research team, involving experts from Newcastle University, UK; Durham University; DTU, Denmark; University of Tasmania, Australia; Hamilton College, New York; the University of Colorado and the University of Toulouse, France, has revealed that the land in this region is actually rising at a phenomenal rate of 15mm a year - much greater than can be accounted for by the present-day elastic response alone.
And they have shown for the first time how the mantle below the Earth's crust in the Antarctic Peninsula is flowing much faster than expected, probably due to subtle changes in temperature or chemical composition. This means it can flow more easily and so responds much more quickly to the lightening load hundreds of miles above it, changing the shape of the land.
Lead researcher, PhD student Grace Nield, based in the School of Civil Engineering and Geosciences at Newcastle University, explains: "You would expect this rebound to happen over thousands of years and instead we have been able to measure it in just over a decade. You can almost see it happening which is just incredible.
"Because the mantle is 'runnier' below the Northern Antarctic Peninsula it responds much more quickly to what's happening on the surface. So as the glaciers thin and the load in that localised area reduces, the mantle pushes up the crust.
"At the moment we have only studied the vertical deformation so the next step is to look at horizontal motion caused by the ice unloading to get more of a 3-D picture of how the Earth is deforming, and to use other geophysical data to understand the mechanism of the flow."
Since 1995 several ice shelves in the Northern Antarctic Peninsula have collapsed and triggered ice-mass unloading, causing the solid Earth to 'bounce back'.
"Think of it a bit like a stretched piece of elastic," says Nield, whose project is funded by the Natural Environment Research Council (NERC). "The ice is pressing down on the Earth and as this weight reduces the crust bounces back. But what we found when we compared the ice loss to the uplift was that they didn't tally - something else had to be happening to be pushing the solid Earth up at such a phenomenal rate."
Collating data from seven GPS stations situated across the Northern Peninsula, the team found the rebound was so fast that the upper mantle viscosity - or resistance to flow - had to be at least ten times lower than previously thought for the region and much lower than the rest of Antarctica.
Professor Peter Clarke, Professor of Geophysical Geodesy at Newcastle University and one of the authors of the paper, adds: "Seeing this sort of deformation of the earth at such a rate is unprecedented in Antarctica. What is particularly interesting here is that we can actually see the impact that glacier thinning is having on the rocks 250 miles down."<|endoftext|>
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Rule Britannia! The English Empire, 1660–1763
By the end of this section, you will be able to:
- Analyze the causes and consequences of the Restoration
- Identify the Restoration colonies and their role in the expansion of the Empire
When Charles II ascended the throne in 1660, English subjects on both sides of the Atlantic celebrated the restoration of the English monarchy after a decade of living without a king as a result of the English Civil Wars. Charles II lost little time in strengthening England’s global power. From the 1660s to the 1680s, Charles II added more possessions to England’s North American holdings by establishing the Restoration colonies of New York and New Jersey (taking these areas from the Dutch) as well as Pennsylvania and the Carolinas. In order to reap the greatest economic benefit from England’s overseas possessions, Charles II enacted the mercantilist Navigation Acts, although many colonial merchants ignored them because enforcement remained lax.
The chronicle of Charles II begins with his father, Charles I. Charles I ascended the English throne in 1625 and soon married a French Catholic princess, Henrietta Maria, who was not well liked by English Protestants because she openly practiced Catholicism during her husband’s reign. The most outspoken Protestants, the Puritans, had a strong voice in Parliament in the 1620s, and they strongly opposed the king’s marriage and his ties to Catholicism. When Parliament tried to contest his edicts, including the king’s efforts to impose taxes without Parliament’s consent, Charles I suspended Parliament in 1629 and ruled without one for the next eleven years.
The ensuing struggle between the king and Parliament led to the outbreak of war. The English Civil War lasted from 1642 to 1649 and pitted the king and his Royalist supporters against Oliver Cromwell and his Parliamentary forces. After years of fighting, the Parliamentary forces gained the upper hand, and in 1649, they charged Charles I with treason and beheaded him. The monarchy was dissolved, and England became a republic: a state without a king. Oliver Cromwell headed the new English Commonwealth, and the period known as the English interregnum, or the time between kings, began.
Though Cromwell enjoyed widespread popularity at first, over time he appeared to many in England to be taking on the powers of a military dictator. Dissatisfaction with Cromwell grew. When he died in 1658 and control passed to his son Richard, who lacked the political skills of his father, a majority of the English people feared an alternate hereditary monarchy in the making. They had had enough and asked Charles II to be king. In 1660, they welcomed the son of the executed king Charles I back to the throne to resume the English monarchy and bring the interregnum to an end ([link]). The return of Charles II is known as the Restoration.
Charles II was committed to expanding England’s overseas possessions. His policies in the 1660s through the 1680s established and supported the Restoration colonies: the Carolinas, New Jersey, New York, and Pennsylvania. All the Restoration colonies started as proprietary colonies, that is, the king gave each colony to a trusted individual, family, or group.
Charles II hoped to establish English control of the area between Virginia and Spanish Florida. To that end, he issued a royal charter in 1663 to eight trusted and loyal supporters, each of whom was to be a feudal-style proprietor of a region of the province of Carolina.
These proprietors did not relocate to the colonies, however. Instead, English plantation owners from the tiny Caribbean island of Barbados, already a well-established English sugar colony fueled by slave labor, migrated to the southern part of Carolina to settle there. In 1670, they established Charles Town (later Charleston), named in honor of Charles II, at the junction of the Ashley and Cooper Rivers ([link]). As the settlement around Charles Town grew, it began to produce livestock for export to the West Indies. In the northern part of Carolina, settlers turned sap from pine trees into turpentine used to waterproof wooden ships. Political disagreements between settlers in the northern and southern parts of Carolina escalated in the 1710s through the 1720s and led to the creation, in 1729, of two colonies, North and South Carolina. The southern part of Carolina had been producing rice and indigo (a plant that yields a dark blue dye used by English royalty) since the 1700s, and South Carolina continued to depend on these main crops. North Carolina continued to produce items for ships, especially turpentine and tar, and its population increased as Virginians moved there to expand their tobacco holdings. Tobacco was the primary export of both Virginia and North Carolina, which also traded in deerskins and slaves from Africa.
Slavery developed quickly in the Carolinas, largely because so many of the early migrants came from Barbados, where slavery was well established. By the end of the 1600s, a very wealthy class of rice planters who relied on slaves had attained dominance in the southern part of the Carolinas, especially around Charles Town. By 1715, South Carolina had a black majority because of the number of slaves in the colony. The legal basis for slavery was established in the early 1700s as the Carolinas began to pass slave laws based on the Barbados slave codes of the late 1600s. These laws reduced Africans to the status of property to be bought and sold as other commodities.
Visit the Charleston Museum’s interactive exhibit The Walled City to learn more about the history of Charleston.
As in other areas of English settlement, native peoples in the Carolinas suffered tremendously from the introduction of European diseases. Despite the effects of disease, Indians in the area endured and, following the pattern elsewhere in the colonies, grew dependent on European goods. Local Yamasee and Creek tribes built up a trade deficit with the English, trading deerskins and captive slaves for European guns. English settlers exacerbated tensions with local Indian tribes, especially the Yamasee, by expanding their rice and tobacco fields into Indian lands. Worse still, English traders took native women captive as payment for debts.
The outrages committed by traders, combined with the seemingly unstoppable expansion of English settlement onto native land, led to the outbreak of the Yamasee War (1715–1718), an effort by a coalition of local tribes to drive away the European invaders. This native effort to force the newcomers back across the Atlantic nearly succeeded in annihilating the Carolina colonies. Only when the Cherokee allied themselves with the English did the coalition’s goal of eliminating the English from the region falter. The Yamasee War demonstrates the key role native peoples played in shaping the outcome of colonial struggles and, perhaps most important, the disunity that existed between different native groups.
NEW YORK AND NEW JERSEY
Charles II also set his sights on the Dutch colony of New Netherland. The English takeover of New Netherland originated in the imperial rivalry between the Dutch and the English. During the Anglo-Dutch wars of the 1650s and 1660s, the two powers attempted to gain commercial advantages in the Atlantic World. During the Second Anglo-Dutch War (1664–1667), English forces gained control of the Dutch fur trading colony of New Netherland, and in 1664, Charles II gave this colony (including present-day New Jersey) to his brother James, Duke of York (later James II). The colony and city were renamed New York in his honor. The Dutch in New York chafed under English rule. In 1673, during the Third Anglo-Dutch War (1672–1674), the Dutch recaptured the colony. However, at the end of the conflict, the English had regained control ([link]).
The Duke of York had no desire to govern locally or listen to the wishes of local colonists. It wasn’t until 1683, therefore, almost 20 years after the English took control of the colony, that colonists were able to convene a local representative legislature. The assembly’s 1683 Charter of Liberties and Privileges set out the traditional rights of Englishmen, like the right to trial by jury and the right to representative government.
The English continued the Dutch patroonship system, granting large estates to a favored few families. The largest of these estates, at 160,000 acres, was given to Robert Livingston in 1686. The Livingstons and the other manorial families who controlled the Hudson River Valley formed a formidable political and economic force. Eighteenth-century New York City, meanwhile, contained a variety of people and religions—as well as Dutch and English people, it held French Protestants (Huguenots), Jews, Puritans, Quakers, Anglicans, and a large population of slaves. As they did in other zones of colonization, native peoples played a key role in shaping the history of colonial New York. After decades of war in the 1600s, the powerful Five Nations of the Iroquois, composed of the Mohawk, Oneida, Onondaga, Cayuga, and Seneca, successfully pursued a policy of neutrality with both the English and, to the north, the French in Canada during the first half of the 1700s. This native policy meant that the Iroquois continued to live in their own villages under their own government while enjoying the benefits of trade with both the French and the English.
The Restoration colonies also included Pennsylvania, which became the geographic center of British colonial America. Pennsylvania (which means “Penn’s Woods” in Latin) was created in 1681, when Charles II bestowed the largest proprietary colony in the Americas on William Penn ([link]) to settle the large debt he owed the Penn family. William Penn’s father, Admiral William Penn, had served the English crown by helping take Jamaica from the Spanish in 1655. The king personally owed the Admiral money as well.
Like early settlers of the New England colonies, Pennsylvania’s first colonists migrated mostly for religious reasons. William Penn himself was a Quaker, a member of a new Protestant denomination called the Society of Friends. George Fox had founded the Society of Friends in England in the late 1640s, having grown dissatisfied with Puritanism and the idea of predestination. Rather, Fox and his followers stressed that everyone had an “inner light” inside him or her, a spark of divinity. They gained the name Quakers because they were said to quake when the inner light moved them. Quakers rejected the idea of worldly rank, believing instead in a new and radical form of social equality. Their speech reflected this belief in that they addressed all others as equals, using “thee” and “thou” rather than terms like “your lordship” or “my lady” that were customary for privileged individuals of the hereditary elite.
The English crown persecuted Quakers in England, and colonial governments were equally harsh; Massachusetts even executed several early Quakers who had gone to proselytize there. To avoid such persecution, Quakers and their families at first created a community on the sugar island of Barbados. Soon after its founding, however, Pennsylvania became the destination of choice. Quakers flocked to Pennsylvania as well as New Jersey, where they could preach and practice their religion in peace. Unlike New England, whose official religion was Puritanism, Pennsylvania did not establish an official church. Indeed, the colony allowed a degree of religious tolerance found nowhere else in English America. To help encourage immigration to his colony, Penn promised fifty acres of land to people who agreed to come to Pennsylvania and completed their term of service. Not surprisingly, those seeking a better life came in large numbers, so much so that Pennsylvania relied on indentured servants more than any other colony.
One of the primary tenets of Quakerism is pacifism, leading William Penn to establish friendly relationships with local native peoples. He formed a covenant of friendship with the Lenni Lenape (Delaware) tribe, buying their land for a fair price instead of taking it by force. In 1701, he also signed a treaty with the Susquehannocks to avoid war. Unlike other colonies, Pennsylvania did not experience war on the frontier with native peoples during its early history.
As an important port city, Philadelphia grew rapidly. Quaker merchants there established contacts throughout the Atlantic world and participated in the thriving African slave trade. Some Quakers, who were deeply troubled by the contradiction between their belief in the “inner light” and the practice of slavery, rejected the practice and engaged in efforts to abolish it altogether. Philadelphia also acted as a magnet for immigrants, who came not only from England, but from all over Europe by the hundreds of thousands. The city, and indeed all of Pennsylvania, appeared to be the best country for poor men and women, many of whom arrived as servants and dreamed of owning land. A very few, like the fortunate Benjamin Franklin, a runaway from Puritan Boston, did extraordinarily well. Other immigrant groups in the colony, most notably Germans and Scotch-Irish (families from Scotland and England who had first lived in Ireland before moving to British America), greatly improved their lot in Pennsylvania. Of course, Africans imported into the colony to labor for white masters fared far worse.
The American Weekly Mercury, published by William Bradford, was Philadelphia’s first newspaper. This advertisement from “John Wilson, Goaler” (jailer) offers a reward for anyone capturing several men who escaped from the jail.
BROKE out of the Common Goal of Philadelphia, the 15th of this Instant February, 1721, the following Persons:
John Palmer, also Plumly, alias Paine, Servant to Joseph Jones, run away and was lately taken up at New-York. He is fully described in the American Mercury, Novem. 23, 1721. He has a Cinnamon coloured Coat on, a middle sized fresh coloured Man. His Master will give a Pistole Reward to any who Shall Secure him, besides what is here offered.
Daniel Oughtopay, A Dutchman, aged about 24 Years, Servant to Dr. Johnston in Amboy. He is a thin Spare man, grey Drugget Waistcoat and Breeches and a light-coloured Coat on.
Ebenezor Mallary, a New-England, aged about 24 Years, is a middle-sized thin Man, having on a Snuff colour’d Coat, and ordinary Ticking Waistcoat and Breeches. He has dark brown strait Hair.
Matthew Dulany, an Irish Man, down-look’d Swarthy Complexion, and has on an Olive-coloured Cloth Coat and Waistcoat with Cloth Buttons.
John Flemming, an Irish Lad, aged about 18, belonging to Mr. Miranda, Merchant in this City. He has no Coat, a grey Drugget Waistcoat, and a narrow brim’d Hat on.
John Corbet, a Shropshire Man, a Runaway Servant from Alexander Faulkner of Maryland, broke out on the 12th Instant. He has got a double-breasted Sailor’s Jacket on lined with red Bays, pretends to be a Sailor, and once taught School at Josephs Collings’s in the Jerseys.
Whoever takes up and secures all, or any One of these Felons, shall have a Pistole Reward for each of them and reasonable Charges, paid them by John Wilson, Goaler
—Advertisement from the American Weekly Mercury, 1722
What do the descriptions of the men tell you about life in colonial Philadelphia?
Browse a number of issues of the American Weekly Mercury that were digitized by New Jersey’s Stockton University. Read through several to get a remarkable flavor of life in early eighteenth-century Philadelphia.
THE NAVIGATION ACTS
Creating wealth for the Empire remained a primary goal, and in the second half of the seventeenth century, especially during the Restoration, England attempted to gain better control of trade with the American colonies. The mercantilist policies by which it tried to achieve this control are known as the Navigation Acts.
The 1651 Navigation Ordnance, a product of Cromwell’s England, required that only English ships carry goods between England and the colonies, and that the captain and three-fourths of the crew had to be English. The ordnance further listed “enumerated articles” that could be transported only to England or to English colonies, including the most lucrative commodities like sugar and tobacco as well as indigo, rice, molasses, and naval stores such as turpentine. All were valuable goods not produced in England or in demand by the British navy. After ascending the throne, Charles II approved the 1660 Navigation Act, which restated the 1651 act to ensure a monopoly on imports from the colonies.
Other Navigation Acts included the 1663 Staple Act and the 1673 Plantation Duties Act. The Staple Act barred colonists from importing goods that had not been made in England, creating a profitable monopoly for English exporters and manufacturers. The Plantation Duties Act taxed enumerated articles exported from one colony to another, a measure aimed principally at New Englanders, who transported great quantities of molasses from the West Indies, including smuggled molasses from French-held islands, to make into rum.
In 1675, Charles II organized the Lords of Trade and Plantation, commonly known as the Lords of Trade, an administrative body intended to create stronger ties between the colonial governments and the crown. However, the 1696 Navigation Act created the Board of Trade, replacing the Lords of Trade. This act, meant to strengthen enforcement of customs laws, also established vice-admiralty courts where the crown could prosecute customs violators without a jury. Under this act, customs officials were empowered with warrants known as “writs of assistance” to board and search vessels suspected of containing smuggled goods.
Despite the Navigation Acts, however, Great Britain exercised lax control over the English colonies during most of the eighteenth century because of the policies of Prime Minister Robert Walpole. During his long term (1721–1742), Walpole governed according to his belief that commerce flourished best when it was not encumbered with restrictions. Historians have described this lack of strict enforcement of the Navigation Acts as salutary neglect. In addition, nothing prevented colonists from building their own fleet of ships to engage in trade. New England especially benefited from both salutary neglect and a vibrant maritime culture made possible by the scores of trading vessels built in the northern colonies. The case of the 1733 Molasses Act illustrates the weaknesses of British mercantilist policy. The 1733 act placed a sixpence-per-gallon duty on raw sugar, rum, and molasses from Britain’s competitors, the French and the Dutch, in order to give an advantage to British West Indian producers. Because the British did not enforce the 1733 law, however, New England mariners routinely smuggled these items from the French and Dutch West Indies more cheaply than they could buy them on English islands.
After the English Civil War and interregnum, England began to fashion a stronger and larger empire in North America. In addition to wresting control of New York and New Jersey from the Dutch, Charles II established the Carolinas and Pennsylvania as proprietary colonies. Each of these colonies added immensely to the Empire, supplying goods not produced in England, such as rice and indigo. The Restoration colonies also contributed to the rise in population in English America as many thousands of Europeans made their way to the colonies. Their numbers were further augmented by the forced migration of African slaves. Starting in 1651, England pursued mercantilist policies through a series of Navigation Acts designed to make the most of England’s overseas possessions. Nonetheless, without proper enforcement of Parliament’s acts and with nothing to prevent colonial traders from commanding their own fleets of ships, the Navigation Acts did not control trade as intended.
To what does the term “Restoration” refer?
What was the predominant religion in Pennsylvania?
What sorts of labor systems were used in the Restoration colonies?
Since the proprietors of the Carolina colonies were absent, English planters from Barbados moved in and gained political power, establishing slave labor as the predominant form of labor. In Pennsylvania, where prospective servants were offered a bounty of fifty acres of land for emigrating and finishing their term of labor, indentured servitude abounded.
- English interregnum
- the period from 1649 to 1660 when England had no king
- Navigation Acts
- a series of English mercantilist laws enacted between 1651 and 1696 in order to control trade with the colonies
- proprietary colonies
- colonies granted by the king to a trusted individual, family, or group
- Restoration colonies
- the colonies King Charles II established or supported during the Restoration (the Carolinas, New York, New Jersey, and Pennsylvania)
- salutary neglect
- the laxness with which the English crown enforced the Navigation Acts in the eighteenth century<|endoftext|>
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# A Fractions Number Line Project for Fourth Graders
I’ve finished up this fractions/number-line project that I’ve been thinking about. I worked with a class of fourth graders who were just starting their fractions unit. My plan from the start was to try to present a project that was dynamic enough to capture their interest. The center piece of the project was to make a “magic wallet,” which is a shortened variation of a Jacob’s Ladder. I’ve been using the name “Li’l Jacob” instead of “magic wallet” because, originally, I couldn’t remember the magic-wallet name, and Li’l Jacob seems properly descriptive. This structure opens in two different ways, to reveal two different visuals. It’s tricky, and seems magical. I am happy to report that these students were over-the-moon happy to learn how make this.
After showing the students what the finished book project would look like we dove right into making the L’il Jacobs. Making this requires a completely non-intuitive sequence of precise folding and gluing. The students have to keep track of where they are in the sequence in order to get the folding to work. I was nervous about how I could get them to see for themselves what was going on. A great surprise was that they offered me the best description I could hope for: they saw the arrangement of papers and immediately recognized it a human figure, legs and torso. Perfect! Now I was completely convinced that I would continue calling this a Little Jacob.
The students decided that this next step was best described by saying that they were covering Jacob’s innards.
As the sequence of folding continues, the Jacob becomes smaller. (Notice the paper that student is using to protect the desk from getting mucked up with glue.)
The last fold reduces the paper into a square.
Each student made four Li’l Jacobs. Each of these had a set of equivalent fractions written on and in it. But we didn’t even start with the fraction labeling until the third class. Our second class was about making the book that was going to hold our fraction cards.
We folded a 33″ x 4.5″ paper int halves, then quarters, then eighths, to make an eight page accordion. I know that most people don’t have access to paper this size, but with a some thought this can be created by combining smaller sheets of paper.
Students then made origami pockets out of 5.5″ squares of paper. Starting with the second page, these were glued on to every other page of the book.
Next came the cover. The book needed an extension so that the number line could start at zero. To accomplish this we attached an extra long cover piece then folded it over. I know I’ve explained that badly, so I hope the pictures above are adequate explanation.
Finally we were ready to label the Little Jacobs with equivalent fractions. I talked to students about how fractions could be a way of counting to one: one-fourth, two-fourths, three-fourths, four-fourths(one). I showed them my animated zero-to-one gif and some static images of equivalent fractions but they seems to like the image above the best. We circled the columns of fractions equivalent to eighths. The really seemed to get the concept, and kept referring to it as they did their labeling.
The picture above is my sample that shows the labeling, with different ways to write equivalent fractions, as well as a simple addition problem, using the fractions.
Here are some images of the students finished books.
The even pages hold the fractions in the pockets.
On the odd pages, students wrote out the fractions. These fractions had no equivalents on our chart. Forth graders don’t do fractions beyond the twelfths, so 1/8, 3/8. 5/8, and 7/8 stand alone.
Some students were more inclined than other to do decorations on the books.
One thing that was wonderful about this class was that the students were incredibly helpful to each other. I could have never gotten this far with this project if I had to problem shoot with each child individually. The students who grasped each step were enthusiastic about working with a classmate that didn’t quite get a step.
We lined up the fractions so the eighths showed, thus showing the 1, 2, 3, 4, 5, 6, 7 and 8 in the numerator. By the end of my third meeting with these students, just about everyone had finished with their books. This is one class that I can say is excited about equivalent fractions.
## 6 thoughts on “A Fractions Number Line Project for Fourth Graders”
1. ivasallay says:
Li’l Jacob might just help these kids love fractions!
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1. It’s hard to convey just how much they loved making these containers for their fractions. They were so open and excited: I hope they carry this positive energy through the unit of studying equivalents!
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1. Sorry to hear it was tedious, but glad to hear it was fun and worth it. I can’t tell you how overwhelmed with pleasure that I was to get these photos from you! The one with the little books falling out, did that happen a lot? When I did these book with the students I ended up punching a little hole at the top of the last page, and threaded then looped a rubber band so that the book had a way of securing the Li”l jacabs in when the book was closed. I know exactly how much time it takes to gather the right materials and prep this project, and I really admire that you took on this challange. Those smiles are priceless.
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2. I totally needed you as a teacher when I was younger. I had such trouble understanding fractions back then. This project would have made it so clear and I would have LOVED making these books since I loved (and still do) making anything 🙂
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practice problem 1
The diagrams below represent generic objects before a collision followed by a set of outcomes to be considered. Comment on the outcomes, paying attention to the energy and momentum before and after each collision. Does the outcome describe a completely inelastic, partially inelastic, completely elastic, or impossible collision? Provide a brief explanation to accompany each answer.
Here we are given the initial conditions of the two colliding objects.
∑p = m1v1 + m2v2 ∑p = (8 kg)(+6 m/s) + (4 kg)(−9 m/s) ∑p = +12 kg m/s ∑K = ½m1v12 + ½m2v22 ∑K = ½(8 kg)(6 m/s)2 + ½(4 kg)(9 m/s)2 ∑K = 306 J
Momentum is a vector quantity, so the total momentum is found by a vector sum. Since the momentums of the two objects are in opposite directions one of them is going to be negative. Since positive answers are preferred over negative ones, let's choose right as the positive direction. This gives us a total momentum of +12 kg m/s. Energy being a scalar is much easier to handle, especially here since the only energy that matters is kinetic (which is always positive). The total mechanical energy of this system in 306 J.
The first outcome we'll be examining has the two objects sticking together and moving off to the right.
∑p′ = (m1 + m2)v′ ∑p′ = (8 + 4 kg)(+1 m/s) ∑p′ = +12 kg m/s ∑K′ = ½(m1 + m2)v′2 ∑K′ = ½(8 + 4 kg)(1 m/s)2 ∑K′ = 6 J
This is a sensible outcome since the initial total momentum is positive (to the right). Calculations show that the final total momentum is still +12 kg m/s, but the final total energy has dropped significantly to a relatively low value of 6 J. Momentum was conserved, but mechanical energy was lost. This a classic example of an inelastic collision. (Some would even call this a completely inelastic collision.) The lost energy has likely gone into plastic deformation of the two objects (given the distorted edges shown in the diagram).
Here the two objects have separated after collision and are moving in opposite directions. Each is moving more slowly than it was before the collision. This hints at a loss of mechanical energy.
∑p′ = m1v1′ + m2v2′ ∑p′ = (8 kg)(−1 m/s) + (4 kg)(+5 m/s) ∑p′ = +12 kg m/s ∑K′ = ½m1v12 + ½m2v22 ∑K′ = ½(8 kg)(1 m/s)2 + ½(4 kg)(5 m/s)2 ∑K′ = 54 J
Momentum was conserved as it should be, but mechanical energy was lost making this an inelastic collision. Since more energy was retained than in the previous outcome, some would call this a partially inelastic collision. Lost energy is not a big deal and does not violate the conservation of energy. The energy wasn't destroyed in this outcome. It just turned into a form that isn't easy to see with our eyes — internal energy. When kinetic energy transforms into internal energy, either the temperature of the system increases or it experience a phase change (melting, for example). Internal energy will be dealt with in more detail later in this book.
This outcome is similar to the previous one only now the objects are moving a bit more quickly. Still, their speeds after the collision are slower than before.
∑p′ = m1v1′ + m2v2′ ∑p′ = (8 kg)(−3 m/s) + (4 kg)(+9 m/s) ∑p′ = +12 kg m/s ∑K′ = ½m1v12 + ½m2v22 ∑K′ = ½(8 kg)(3 m/s)2 + ½(4 kg)(9 m/s)2 ∑K′ = 198 J
Momentum was conserved and energy was lost, but to a lesser extent than in the previous outcome. Of all the outcomes so far, this inelastic collision is the least inelastic. Whether one would call it partially inelastic or partially elastic doesn't really matter.
Here we see an elastic collision. (Some would even call this a perfectly elastic collision.)
∑p′ = m1v1′ + m2v2′ ∑p′ = (8 kg)(−4 m/s) + (4 kg)(+11 m/s) ∑p′ = +12 kg m/s ∑K′ = ½m1v12 + ½m2v22 ∑K′ = ½(8 kg)(4 m/s)2 + ½(4 kg)(11 m/s)2 ∑K′ = 306 J
Momentum and total mechanical energy of the system were both conserved. The total kinetic energy of the two objects after the collision is the same as it was before. In the macroscopic world such an outcome would never happen. The results above show the limit of what could happen. That is, macroscopic objects will always have less total mechanical energy after a collision than before. Never equal to or greater than.
This outcome is difficult to explain.
∑p′ = m1v1′ + m2v2′ ∑p′ = (8 kg)(−6 m/s) + (4 kg)(+15 m/s) ∑p′ = +12 kg m/s ∑K′ = ½m1v12 + ½m2v22 ∑K′ = ½(8 kg)(−6 m/s)2 + ½(4 kg)(15 m/s)2 ∑K′ = 594 J
Momentum was conserved, but mechanical energy increased. How could this happen? Where did the extra energy come from? Since this is a problem about generic objects, we are free to contrive all sorts of explanations. Perhaps there was a compressed spring on one of the objects, or a chemical explosive, or the two objects were small mammals that kicked off of each other after they collided. I think "contrive" was an appropriate word choice on my part to describe what we're doing here. This outcome seems improbable as it violates the conservation of mechanical energy in a way different from the previous outcomes.
This outcome is also perplexing.
∑p′ = m1v1′ + m2v2′ ∑p′ = (8 kg)(−8 m/s) + (4 kg)(+5 m/s) ∑p′ = −44 kg m/s ∑K′ = ½m1v12 + ½m2v22 ∑K′ = ½(8 kg)(8 m/s)2 + ½(4 kg)(5 m/s)2 ∑K′ = 306 J
Mechanical energy was conserved in this outcome (which is rare, but not impossible), but momentum was not. This would require an impulse applied from outside of the system. Since the description of this problem makes no mention of the existence or even the possibility of a third object, we would have to conclude that this outcome is impossible as it violates the conservation of linear momentum.
practice problem 2
- A 5 kg bowling ball moving at 8 m/s approaches a row of stationary balls lined up end to end in a ball return. Comment on the likelihood of the following outcomes.
- The incoming ball stops and one 5 kg ball leaves the row of stationary balls at a speed of 8 m/s.
- The incoming ball stops and two 5 kg balls leave the row of stationary balls at a speed of 4 m/s.
- Two 5 kg bowling balls moving at 8 m/s approach a row of stationary balls lined up end to end in a ball return. Comment on the likelihood of the following outcomes.
- The incoming balls stop and two 5 kg balls leave the row of stationary balls at a speed of 8 m/s.
- The incoming balls stop and one 5 kg ball leaves the row of stationary balls at a speed of 16 m/s.
Compute the total linear momentum and mechanical energy of the bowling balls before and after each collision has occurred. Compare the values of these quantities to answer this question.
In the case of one bowling ball the initial momentum and energy are…
In the first hypothetical outcome shown below, both momentum and energy are completely conserved. Such a collision is said to be perfectly elastic. While totally fine from a theoretical perspective, such an outcome is practically impossible. In the macroscopic real world, momentum and energy are always dissipated. (In the microscopic world of atoms and molecules collisions are always elastic, but that is another story.)
This outcome, while highly improbable, is not theoretically impossible.
In this hypothetical outcome, momentum is conserved but mechanical energy is lost. Such a collision is said to be inelastic. While totally fine from a theoretical standpoint I think, from personal observations of bowling balls, that such an outcome is highly unlikely.
Of the two outcomes presented neither will occur. The real world doesn't work out according to hypothetical ideals. The real questions should be, "Is the real outcome of this collision more like the first hypothetical outcome or the second?" Experimental observation confirms that the collisions between bowling balls are more like elastic collisions than inelastic collisions. If one bowling ball comes into a row of stationary balls, the most likely outcome is that one bowling ball will leave the other end.
In the case of two bowling balls the initial momentum and energy are…
This is a trivial solution to the problem. Obviously both momentum and energy are conserved. This is another example of a perfectly elastic collision.
This outcome is possible, but not probable.
This last possible outcome makes no sense. The momentum after collision is the same as before, but the mechanical energy has somehow increased. Miraculously, it doubled.
While energy can neither be created nor destroyed, it certainly can become "lost". This is why I have no problem with the second outcome of the first collision. However, this outcome is surely impossible. The kinetic energy of a system cannot increase without work being done by some outside agent. For a row of bowling balls sitting in a ball return I cannot see anyway for positive work to be done. Thus if two bowling balls approach a row of stationary balls the most likely outcome is that two will emerge from the other end.
In general, the collision between bowling balls is more like an elastic collision than an inelastic one. If one ball approaches a row of stationary balls, one ball will leave from the other side. If two balls approach, then two will leave. If three balls approach, three will leave. And so on. This rule is the basis for a popular desktop ornament — and when I say "desktop" I mean the top of a real desk, not the pattern on a computer monitor that one sees when no applications are running or documents are open. Often called "Newton's cradle" or (more hilariously) "Newton's balls" and identified by a whole host of trade names, it is standard issue for the executive desktop in movies and television. In fact, I once heard it called "the executive intelligence tester". The dialog below illustrates this application.
Human ResourcesWatch. One goes in. One comes out. Two go in. Two come out. Three go in. Three come out.ExecutiveWow.Human ResourcesKeep watching. Four go in. What happens next?ExecutiveUh… Four come out?Human ResourcesGood. Now five go in.ExecutiveOo, oo, I know. Five come out!Human ResourcesCongratulations, you're our new vice president.
practice problem 3
- the kinetic energy of the bullet
- the recoil velocity of the pistol
- the kinetic energy of the pistol
- the fraction of the total energy delivered to the bullet
- the velocity of the man and bullet together
- the kinetic energy of the man and bullet together
- the fraction of the total kinetic energy lost in the collision
practice problem 4
- the final velocity of the two objects stuck together (easy)
- the kinetic energy lost as a result of the collision (hard)
The first part is the easy part.
∑p = ∑p′ m1v1 − m2v2 = (m1 + m2)v′ v′ = m1v1 − m2v2 (m1 + m2)
The second part is the hard part. Start with the basics.
∆K = Kf − Ki
Both terms on the right side of this equation are huge vomit piles of symbols. Work on each term separately. Start with the final total kinetic energy.
Kf = 1 (m1 + m2) ⎛
m1v1 − m2v2 ⎞2
2 m1 + m2
Simplify, if that's the right word.
Kf = m12v12 − 2m1v1m2v2 + m22v22 2(m1 + m2)
Go back to the initial total kinetic energy.
Ki = 1 m1v12 + 1 m2v22 2 2
We can't subtract until we have a common denominator with the final total energy.
Ki = m1v12(m1 + m2) + m2v22(m1 + m2) 2(m1 + m2)
Ki = m12v12 + m1m2v12 + m1m2v22 + m22v22 2(m1 + m2)
We now have two monster fractions than need to be subtracted. Instead of writing the whole mess out, let's just work on the numerator.
(m12v12 − 2m1v1m2v2 + m22v22) −
(m12v12 + m1m2v12 + m1m2v22 + m22v22)
A few terms cancel and the masses can be factored out along with the minus sign.
− m1m2(v12 + 2v1v2 + v22)
The term in parentheses is the square of the sum of the two speeds (their absolute values in this case)
− m1m2(v1 + v2)2
Stack numerator on denominator and call it quits.
∆K = − m1m2(v1 + v2)2 2(m1 + m2)
Of what use is this? I forgot why I wrote this question.<|endoftext|>
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Stereolithography is one of the most detailed methods of rapid prototyping and 3D printing currently available. It is also known as SLA or SL with the file extension .stl being drawn from this abbreviation.
This process is an additive manufacturing procedure which uses a receptacle of liquid ‘stereolithography resin’ photopolymers (a substance which solidifies on exposure to an ultra violet laser. The 3D model is built layer by layer, on a mobile platform. The laser touches the container solidifying the parts required to achieve the creation of the SLA prototype. source: materialgeeza via Wikipedia
When a layer is completed the mobile platform upon which the solid layers are placed is lowered by a distance which is equal to the thickness of each layer of material being used (usually between 0.05 mm and 0.15 mm).
This exposes a liquid surface once more, enabling the laser to begin tracing it again in order to create a new layer which will be placed directly on top of the one below it. Once the process has been completed, the printed 3D object is then rinsed in a chemical solution in order to cleanse it of excess resin. Following this, it is then ‘baked’ in an ultra violet oven in order to harden it.
|Materials used||Liquid polymer|
|Available Materials||Thermoplastic (Elastomers)|
|Maximum printing size||150 cm x 75 cm x 50 cm|
|Maximum resolution||0,00254 cm|
|Tolerance||+/- 0,0127 cm|
|Applications||- Form/fit testing - Functional testing - Rapid tooling patterns, Snap fits, Very detailed parts, Presentation models, High heat applications|
Stereolithography is a manufacturing procedure which takes a short time to complete and it is possible to create functional products within a day. The manufacturing periods for procedures involving stereolithography vary according to the size and complexity of the object being produced with differences ranging from a few hours to slightly over 24 hours. The average dimensions of 3D printing platforms used in Stereolithography are 50 cm x 50 cm x 60cm; there are however large printers capable of printing objects up to two metres in height
Items created using stereolithography 3d printers are sufficiently durable for processing or for use as plastic injection molds, thermoforming and in casting processes.
Stereolithography can be used for: - Adjustment and size tests. - Obtaining finished products (painting, texture) for market tests - Producing molds for processing purposes or mold patterns for vacuum casting
source: fdecomite via Flickr
It is a relatively expensive process. The liquid ‘stereolithography resin’ photopolymers required may range from 70 to 220 euros, while stereolithography printers cost 100 000 euros and above.
Was invented in 1986 by Charles W. Hull. He also founded the 3d society which is responsible for patenting the procedure.<|endoftext|>
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# Simplifying Expression 3 Success
• Jul 9th 2009, 12:19 PM
Sooda
Simplifying Expression 3 Success
I solved this by my own, using things learned from my previous two threads. I am so proud. I post it here so maybe others can learn from it if they decide to.
Expression to simplify:
$\displaystyle (\frac{5}{2a+3}+\frac{2}{3-2a}+\frac{2a+9}{4a^2-9})\div\frac{8}{4a^2+12a+9}$
$\displaystyle \frac{2a+3}{2}$
Step by step:
$\displaystyle \frac{5}{2a+3}+\frac{2}{3-2a}+\frac{2a+9}{4a^2-9}=\frac{5}{2a+3}-\frac{2}{2a-3}+\frac{2a+9}{4a^2-9}$
$\displaystyle {4a^2-9}$ is formula: $\displaystyle a^2-b^2=(a+b)(a-b)$
$\displaystyle {4a^2-9}=(2a+3)(2a-3)$
$\displaystyle \frac{5}{2a+3}-\frac{2}{2a-3}+\frac{2a+9}{(2a+3)(2a-3)}$
$\displaystyle \frac{5(2a-3)-2(2a+3)+(2a+9)}{(2a+3)(2a-3)}$
$\displaystyle \frac{10a-15-4a-6+2a+9}{(2a+3)(2a-3)}=\frac{8a-12}{(2a+3)(2a-3)}$
Time to divide,
$\displaystyle 4a^2+12a+9$ is formula: $\displaystyle a^2+2ab+b^2=(a+b)^2$
$\displaystyle 4a^2+12a+9=(2a+3)^2$
$\displaystyle \frac{8a-12}{(2a+3)(2a-3)}\div\frac{8}{(2a+3)^2}$
$\displaystyle \frac{8a-12}{(2a+3)(2a-3)}*\frac{(2a+3)^2}{8}$
$\displaystyle \frac{8a-12}{(2a-3)}*\frac{(2a+3)}{8}$
$\displaystyle \frac{4(2a-3)}{(2a-3)}*\frac{(2a+3)}{8}$
$\displaystyle \frac{4}{1}*\frac{(2a+3)}{8}$
$\displaystyle \frac{2a+3}{2}$
So good to give something back.
• Jul 9th 2009, 04:41 PM
AlephZero
Excellent job. Just one tip:
noticing that
$\displaystyle \frac{8a-12}{(2a+3)(2a-3)}=\frac{4(2a-3)}{(2a+3)(2a-3)}=\frac{4}{2a+3}$
would have saved you some trouble.
Cheers.<|endoftext|>
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What is a Neuron
A neuron is a very specialized cell that forms the basis of the nervous system. It is also referred to as a nerve cells. Neurons are unique in that they transmit signals throughout the body. These signals are referred to as action potentials, nerve impulses or spikes.
Lets make this very practical. If you need to send a signal very quickly to a part of the body, the neuron is the cell for the job. Your brain needs to tell your hand to move? It sends that signal via neurons. You touch a hot stove, the signal needs to go to your brain quickly, so that you can realize that it hurts and pull your hand away before you do some serious damage, neurons send that signal faster than you can say OUCH!
The parts of a neuron:
- Dendrites: these are the points of connections between neurons (synaptic connections). When a connection is made with the soma of another cell (as opposed to another dendrite), this is a stronger connection.
- Soma: The cell body of a neuron. This is the part of the neuron that is most like other cells. It has the nucleus, genetic machinery and is where many of the metabolic processes happen.
- Axon: This is where the magic happens. The Axon has a base where it is continuous with the soma. This swollen section is called the axon hillock, or the “spike” initiating zone (where the nerve impulse originates). The nerve impulse then travels down the axon to the Axon Terminals.
- Axon Terminals: These are the ends of axons. They contain neurotransmitters. When the nerve impulse reaches the axon terminals, the axon terminals release the neurotransmitter, which may then result in an action potential in the next cell.
There are three basic types of neurons: Sensory neurons, motor neurons and interneurons. Sensory neurons bring signals to the central nervous system (CNS) and Motor neurons send signals from the CNS to the rest of the body. Interneurons are in the CNS and allow the sensory neurons to communicate with the motor neurons.
The take home message is that neurons are what it’s all about in the nervous system.<|endoftext|>
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1. /
2. CBSE
3. /
4. Class 12
5. /
6. Mathematics
7. /
8. NCERT Solutions class 12...
# NCERT Solutions class 12 Maths Exercise 13.3
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## NCERT Solutions class 12 Maths Probability
1. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Ans. Case (i) : S1 = {5 red balls, 5 black balls}
= 10
Let us draw a red balls first, i.e., A1 = {5 red balls}
= 5
P (A1) =
Now after adding 2 balls of the same colour, i.e.,
S2 = {7 red balls, 5 black balls}
= 12
Let us draw a red balls first, i.e., A2 = {7 red balls}
= 7
P (A2) =
P (a red ball is drawn) =
Case (ii) : When a black ball is drawn, i.e., A2 = {5 red balls}
= 5
P (A1) =
Now after adding 2 balls of the same colour, i.e.,
S2 = {5 red balls, 7 black balls}
= 12
Let us draw a red balls first, i.e., A2 = {5 red balls}
= 5
P (A2) =
P (a red ball is drawn) =
Therefore, required probability in both cases =
2. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
Ans. Let A be the event hat ball drawn is red and let E1 and E2 be the events that the ball drawn is from the first bag and second bag respectively.
P (E1) = , P (E2) = ,
P = P (drawing a red ball from bag I) =
P = P (drawing a red ball from bag II) =
Therefore, by Bayes’ theorem,
P = P (red ball drawn from bag I) =
=
3. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostler?
Ans. Let E1 = the examinee knows the answer, E2 = the examinee guesses the answer and
A = student who attain grade A, P (E1) = , P (E2) = ,
P = P =
Therefore, by Bayes’ theorem,
P =
=
4. In answering a question on a multiple choice test a student either knows the answer or guesses. Let be the probability that he knows the answer and be the probability that he guesses. Assuming that a student who guesses the answer, will be correct with probability . What is the probability that a student knows the answer given that he answered it correctly?
Ans. Let E1 = students residing in the hostel, E2 = day scholars (not residing in the hostel) and
A = the examinee answers correctly
Now P (E1) = , P (E2) = ,
Since E1 and E2 are mutually exclusive events and exhaustive events, and if E2 has already occurred, then the examinee guesses, therefore the probability that he answers correctly given that he has made a guess is i.e., P =
And P = P (answers correctly given that he knew the answer) = 1
Therefore, by Bayes’ theorem,
P =
=
5. A laboratory blood test is 99% effective in detecting a certain disease when it is, in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e., if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Ans. Let E1 = The person selected is suffering from certain disease, E2 = The person selected is not suffering from certain disease and A = The doctor diagnoses correctly
Now P (E1) = 0.1% = = 0.001, P (E2) = ,
P = 99% = P = 0.005%
Therefore, by Bayes’ theorem,
P =
=
#### 6. There are three coins. One is a two headed coin, another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows head, what is the probability that it was the two headed coin?
Ans. Let E1 = a two headed coin, E2 = a biased coin, E3 = an unbiased coin and A = A head is shown
Now P (E1) = , P (E2) = , P (E3) =
P = 1, P = and P =
Therefore, by Bayes’ theorem,
P =
=
#### 7. An insurance company insured 2000 scooter driver, 4000 car drivers and 6000 truck drivers. The probability of accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Ans. Let E1 = Person chosen is a scooter driver, E2 = Person chosen is a car driver, E3 = Person chosen is a truck driver and A = Person meets with an accident
Since there are 12000 persons, therefore,
Now P (E1) = , P (E2) = , P (E3) =
It is given that P= P (a person meets with an accident, he is a scooter driver) = 0.01
Similarly, P = 0.03 and P = 0.15
To find: P (person meets with an accident that he was a scooter driver)
Therefore, by Bayes’ theorem,
P =
=
#### 8. A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?
Ans. Given: P (A) = P (B) =
Let D denotes a defective item:
P = and P =
P = =
### 9. Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability if 0.3, if the second group wins. Find the probability that the new product introduced was by the second group.
Ans. Given: P (G1) = 0.6, P (G2) = 0.4
Let P denotes the launching of new product.
P (PG1) = 0.7, P (PG2) = 0.3
P = =
### 10. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4 she tosses a coin once and noted whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 and 4 with the die?
Ans. Let E1 = 5 or 6 appears on a die, E2 = 1, 2, 3 or 4 appears on a die and A = A head appears on the coin.
Now P (E1) = , P (E2) =
Now P Probability of getting a head on tossing a coin three times,
when E1 has already occurs = P (HTT) or P (THT) or P (TTH)
= =
P = Probability of getting a head on tossing a coin once,
when E2 has already occurred =
P (there is exactly one head given that 1, 2, 3 or 4 appears on a die)
P =
=
#### 11. A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job for 30% of the time and C on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?
Ans. Let E1 = the item is manufactured by the operator A, E2 = the item is manufactured by the operator B, E3 = the item is manufactured by the operator C and A = the item is defective
Now P (E1) = , P (E2) = , P (E3) =
Now P= P (item drawn is manufactured by operator A) =
Similarly, P = and P =
Now Required probability = Probability that the item is manufactured by operator A given that the item drawn is defective
P =
=
#### 12. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Ans. Let E1 = the missing card is a diamond, E2 = the missing card is a spade, E3 = the missing card is a club, E4 = the missing card is a heart and A = drawing of two heart cards from the remaining cards.
Now P (E1) = , P (E2) = , P (E3) = , P (E4) =
P= P (drawing 2 heart cards given that one diamond card is missing) =
Similarly, P = , P = and P =
By Bayes’ theorem,
P =
=
=
#### 13. Probability that A speaks truth is A coin is tossed. A report that a head appears. The probability that actually there was head is:
(A)
(B)
(C)
(D)
Ans. Let A be the event that the man reports that head occurs in tossing a coin and let E1 be the event that head occurs and E2 be the event head does not occur.
P (E1) = , P (E2) =
P= P (A reports that head occurs when head had actually occur red on the coin) =
P= P (A reports that head occurs when head had not occur red on the coin) =
By Bayes’ theorem,
P = =
Hence, option (A) is correct.
#### 14. If A and B are two events such that A B and P (B) 0, then which of the following is correct:
(A) P =
(B) P < P (A)
(C) P P (A)
(D) None of these
Ans. A B A B = A P and P (B) 0 P =
Since P (B) 0
< 1 P (A) < P (B) P P (A)
Hence, option (C) is correct.
## NCERT Solutions class 12 Maths Exercise 13.3
NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 13 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide
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### 2 thoughts on “NCERT Solutions class 12 Maths Exercise 13.3”
1. Answer to question 1 case (2) part 1 is wrong. It should be 1/2 not 5/12.
2. Q1 coorect s1 sample space as 10
You have done 5/12<|endoftext|>
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# Curves MCQ Quiz in বাংলা - Objective Question with Answer for Curves - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Oct 31, 2022
পাওয়া Curves उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). এই বিনামূল্যে ডাউনলোড করুন Curves MCQ কুইজ পিডিএফ এবং আপনার আসন্ন পরীক্ষার জন্য প্রস্তুত করুন যেমন ব্যাঙ্কিং, এসএসসি, রেলওয়ে, ইউপিএসসি, রাজ্য পিএসসি।
## Top Curves MCQ Objective Questions
#### Curves Question 1:
If R is the radius of the curve in metres and C is the chord length in metres what would be the expression to denote versine ‘V’ in millimetres?
1. V = 127 C2/R
2. V = 127 R2/C
3. V = 125 C2/R
4. V = 125 R2/C
Option 3 : V = 125 C2/R
#### Curves Question 1 Detailed Solution
Concept:
Versine of curve
The versine is the perpendicular distance of the midpoint of a chord from the arc of a circle.
The relationship between radius and versine of a curve in various units is shown below:
$$V = \frac{{{C^2}}}{{8R}}$$ .............(V, C, R are in same units, say, m or cm)
$$V = \frac{{125{C^2}}}{R}$$ ......... (V in mm, C in m, and R in m)
$$V = \frac{{1.5{C^2}}}{R}$$...........(C and R in feet and V in inches)
where,
V = Versine of curve
C = Length of chord
#### Curves Question 2:
Calculate the length (m) of the longer chord of 250 m radius curve having deflection angle of 90 degree.
1. 250
2. 353.6
3. 392.7
4. 500
Option 2 : 353.6
#### Curves Question 2 Detailed Solution
Concept:
For the given curve:
Tangent length $$\left( \text{T} \right)=\text{R}\tan \frac{\text{ }\!\!\Delta\!\!\text{ }}{2}$$
Length of curve $$\left( \text{l} \right)=\frac{\text{ }\!\!\pi\!\!\text{ R }\!\!\Delta\!\!\text{ }}{180}$$
Long chord $$\left( \text{L} \right)=2\text{R}\sin \frac{\text{ }\!\!\Delta\!\!\text{ }}{2}$$
External distance $$\left( \text{E} \right)=\text{R}\left( \sec \frac{\text{ }\!\!\Delta\!\!\text{ }}{2}\text{ }\!\!~\!\!\text{ }-1 \right)$$
Mid-ordinate $$\left( \text{M} \right)=\text{R}\left( 1-\cos \frac{\text{ }\!\!\Delta\!\!\text{ }}{2}\text{ }\!\!~\!\!\text{ } \right)$$
Chainage of T1 = Chainage of P.I. – T, and Chainage of T2 = Chainage of T1 + l
Calculations:
Length of long chord (L) = 2 × R × sin Δ/2 = 2 × 250 × sin 45° = 353.55 m
#### Curves Question 3:
Calculate the length (m) of tangent of a 5-degree curve, if the deflection angle is 60 degree.
1. 172.5
2. 198.6
3. 360
4. 596
Option 2 : 198.6
#### Curves Question 3 Detailed Solution
The tangent length of a simple curve of radius R and deflection angle D is $$R \times \tan \left( {\frac{D}{2}} \right)$$
Tangent length VT = $$R \times \tan \left( {\frac{D}{2}} \right)$$
Radius of the curve $$= \frac{{1720}}{5} = 344\;m$$
Tangent length = $$344 \times \tan \left( {\frac{{60}}{2}} \right) = 198.6084\;m$$
#### Curves Question 4:
The relationship between the length l and radius r of an ideal transition curve is stated as :
1. $$l \propto r\;$$
2. $$l \propto \frac{1}{r}\;$$
3. $$l \propto r^2\;$$
4. $$l \propto \frac{1}{r^2}\;$$
Option 2 : $$l \propto \frac{1}{r}\;$$
#### Curves Question 4 Detailed Solution
Explanation:
Transition curve is provided at the junction of the straight and curved portion of a railway track to serve the following purposes:
i) to provide the superelevation in a gradual manner
ii) to avoid the sudden jerk to the passengers
The requirement of an ideal transition curve:
i) The length of the transition curve is inversely proportional to the radius of curvature ( $$l \propto \frac{1}{r}\;$$).
ii) The curve should be tangential to its junction points.
iii) The rate of change of superelevation provided should be equal to the rate of change of curvature so that full superelevation can be provided within the length of the transition curve.
iv) The rate of change of centrifugal acceleration should be consistent, this implies the radius of the transition curve should be consistently decreased from infinity at the tangent point to the radius R of the circular curve.
#### Curves Question 5:
A railway curve of 1350 m radius is to be set out to connect two tangents. If the design speed is 110 kmph and the rate of change of acceleration is 0.3 m/s3, the shift of the circular curve will be nearly
1. 0.18 m
2. 0.16 m
3. 0.14 m
4. 0.12 m
Option 2 : 0.16 m
#### Curves Question 5 Detailed Solution
Concept:
Length of transition curve, Ls is given as:
$${L_s} = \;\frac{{{V^3}}}{{CR}}$$
The shift of the circular curve, S is given as:
$$S = \frac{{L_s^2}}{{24\;R}}$$
Where, V = speed,
C = rate of change of acceleration,
Calculation:
V = 110 kmph = 110× 5/18 = 30.55 m/s
C = 0.3 m/s3
R = 1350 m
$${L_s} = \frac{{{{30.55}^3}}}{{0.3\; \times \;1350}}$$
∴ Ls = 70.4 m
$$S = \frac{{\;{{70.4}^2}}}{{24\; \times \;1350}}$$ = 0.153 m
S ≈ 0.16 m
#### Curves Question 6:
On doubling the radius of the ideal transition curve, the length of the curve will tend to:
1. Increased by 100%
2. Remain same
3. Increased by 50%
4. Reduced by 50%
Option 4 : Reduced by 50%
#### Curves Question 6 Detailed Solution
Concept:
Length of Ideal transition curve is given by:
$${{\rm{L}}_{\rm{C}}} = \frac{{{{\rm{V}}^3}}}{{{\rm{CR}}}}$$
C → rate of change of radial acceleration
V → design speed
Explanation:
$${{\rm{L}}_{\rm{C}}}_1 = \frac{{{{\rm{V}}^3}}}{{{\rm{CR}}}}$$
$${{\rm{L}}_{\rm{C}}} \propto \frac{1}{{\rm{R}}}$$
R2 = 2R1
$${{\rm{L}}_{{{\rm{C}}_2}}} \propto \frac{1}{{{{\rm{R}}_2}}}$$
$${{\rm{L}}_{{{\rm{C}}_2}}} \propto \frac{1}{{2{{\rm{R}}_1}}}$$
$$\frac{{{{\rm{L}}_{{{\rm{C}}_2}}}}}{{{{\rm{L}}_{{{\rm{C}}_1}}}}} = \frac{{1 \times {{\rm{R}}_1}}}{{2{{\rm{R}}_1}}} = \frac{1}{2}$$
$${{\rm{L}}_{{{\rm{C}}_2}}} = \frac{{{{\rm{L}}_{{{\rm{c}}_1}}}}}{2} = 0.5{{\rm{L}}_{{{\rm{C}}_1}}}$$
i.e. the length of the curve will become half or reduce by 50%
#### Curves Question 7:
The radial offset at a distance x from the beginning of circular curve of radius R is given by
1. $$\sqrt{R^2-x^2} - R$$
2. $$R-\sqrt{R^2-x^2}$$
3. $$R-\sqrt{R^2+x^2}$$
4. $$\sqrt{R^2+x^2}-R$$
Option 4 : $$\sqrt{R^2+x^2}-R$$
#### Curves Question 7 Detailed Solution
Concept:
The methods of setting out curves:
The methods of setting out curves may be divided into two classes according to the instrument employed for setting out of curves as follows:
1. Linear or chain and tape method:- Linear methods are those in which the curve is set out with chain and tape only. It contains sub-methods as follows:
• By offsets or ordinates from the long chord
• By successive bisections of arcs
• By offsets from the tangents
• By offsets from chord produced
2. Angular or instrumental method:- Instrumental methods are those in which theodolite with or without a chain is employed to set out a curve. It contains sub-methods as follows:
• Rankine’s method of tangential angle
• Two theodolite method
By offsets from the tangents: In this method, the offsets are set out either radially or perpendicular to the tangents. It contains sub-methods as follows:
• By offsets perpendicular to tangents
Explanation:
The radial offset at a distance X from the point of commencement of curve of radius R
Let, T1 be the first tangent point and R is the radius of the curve
EE1 = Ox= the radial offset at E at a distance of X from T1 along with the tangent AB.
Now in the ∆OT1E,
OT1 = R and T1E = X
OE = OE1+ EE1 =R + Ox
Now, OE2 = (OT1) 2 + (T1E)2
(R + Ox)2 = R2 + x 2
$$O_X = \sqrt {R^2 + X^2} - R$$
The radial offset at a distance X from the point of commencement of curve of radius R is given by $$O_X = \sqrt {R^2 + X^2} - R$$
#### Curves Question 8:
If g1 = + 1.2% and g2 = + 0.8% and rate of change of grade = 0.1% per 20 m chain, then the length of the vertical curve is
1. 10 m
2. 15 m
3. 30 m
4. 80 m
Option 4 : 80 m
#### Curves Question 8 Detailed Solution
Concept:
Length of curve = (Total grade/Rate of change of grade per chain length) × Length of chain
Calculation:
Grade = + 1.2 – (+ 0.8) = 0.4 % (Upward)
Change of grade is 0.1% per 20 m chain.
L = (0.4/0.1) × 20 = 80 m
#### Curves Question 9:
Calculate the apex distance, if the deflection angle is 60 degree and the degree of curve is 8 degree.
1. 33.26
2. 124.13
3. 215
4. 262.8
Option 1 : 33.26
#### Curves Question 9 Detailed Solution
Concept:
Apex Distance: It is the distance between the point of Intersection and the apex (highest point) of the curve.
Apex distance = R ($$\sec \frac{\Delta }{2}$$ - 1)
Where, R: Radius of the curve, ∆: Deflection angle in degree
Arc Definition: If R is the radius of the curve and D is its degree for 30 m arc, then
× D × $$\frac{\pi }{{180}}$$ = 30
Where, R: Radius of the curve, D: Degree of the curve
Calculation:
The radius of the curve, R = $$\frac{{30 \times 180^\circ }}{{D \times {\rm{\pi }}}}$$ = $$\frac{{30 \times 180^\circ }}{{8 \times {\rm{\pi }}}}$$ = 214.9 m
Apex distance = R ($$\sec \frac{\Delta }{2}$$ - 1) = 214.9 ($$\sec \frac{{60}}{2}$$ - 1) = 33.255 m
#### Curves Question 10:
In a simple curve, external distance is the distance between:
1. Vertex and middle point of the curve
2. Vertex and point of the curve
3. Point of the curve and point of tangency
4. Vertex and the center of the curve
Option 1 : Vertex and middle point of the curve
#### Curves Question 10 Detailed Solution
Concept:
Simple circular curve:
• A simple curve consists of a single arc of a circle connecting two straights.
• It has a radius of the same magnitude throughout.
• Point of the intersection of the tangents and also called the vertex.
Point of curvature(T1) - It is the beginning of the curve.
Point of tangency(T2) - It is the end of the curve.
Figure: Simple curve
The length of the curve (l):
the length of curve T1 K T2 is given by
$${\rm{l}} = \frac{{2{\rm{\pi R}}}}{{360}} \times {\rm{D }}$$
Where, R = Radius of the curve, D = Deflection angle, and l = Length of the curve
Tangent Length (T) = R tan (D/2)
Length of long Chord (L) = 2R sin (D/2)
Mid-ordinate (M):
The middle ordinate is the distance from the midpoint(K) of the curve to the midpoint(C) of the chord.
M= R {1 - cos (D/2)}
External distance (E):
External distance is the distance from the vertex to the midpoint(K) of the curve.
E= R {sec(D/2) - 1}<|endoftext|>
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Sharks ruled the waters of the Mediterranean for hundreds of thousands of years before humans ever so much as dipped a toe into the waves. Before overfishing and fear-driven persecution put the run on them, some sharks were the Mediterranean’s apex predators, weeding out sick and weak individuals and ultimately enforcing a balanced and robust ecosystem through the virtue of their jagged teeth.
But when humans began to take to the water, things quickly went sour. Soon, writers and artists began portraying sharks in the role of villain.
In an effort to better understand the way our relationship with sharks has changed throughout history, Massimiliano Bottaro, a marine scientist at the Anton Dohrn Zoological Station in Naples, Italy, and his colleagues went through thousands of years of historical and archaeological evidence and artistic depictions. Their results, published in Regional Studies in Marine Science, found several dozen records of interactions between humans and sharks, as well as rays, a close relative.
The earliest evidence of interaction was gustatory—shark bones left over from late Stone Age meals in caves in southern Italy and Spain. Then, over the years, images of sharks appeared on bowls, vases, and in mosaics; they were written into histories and epic poems; and, in one case, part of a ray became a religious relic affixed to the wall of a church. These accounts provide some biological insight into species’ presence and abundance before overfishing and intense environmental degradation brought on by the Industrial Revolution precipitated a cascade of changes in the Mediterranean ecosystem. The records also reveal that fear and misunderstanding have complicated our relationship with sharks since some of our earliest encounters.
In 492 BCE, the first Persian invasion of Greece was proceeding well for military commander Mardonius. According to Histories, Greek historian Herodotus’s classic, Mardonius had already “deposed all the despots” of the Ionian Islands, replacing them with popular governments and forcing the Macedonians into servitude. Mardonius’s luck didn’t last. His fleet was caught in a nasty north wind, sinking some 300 ships against the cliffs of Mount Athos. In Herodotus’s account of the event, “sea monsters” moved in to “seize and devour” many of the 20,000 sailors in the water, marking what’s likely the first infamous appearance of sharks in literature.
While Herodotus could be taking artistic liberties in describing so many shark deaths, Bottaro speculates the tidbit could reveal a snapshot of historical shark abundance. “In that time … the Mediterranean Sea was so full of sharks that after an incident in a ship, sharks came immediately, [following] the olfactory trace of blood,” he says, adding that the noise of drowning, flailing sailors could have also been a draw.
Herodotus’s account that shark attacks could be a potential hazard to mariners is reflected in the work of visual artists. One of the earliest known images of sharks in the Mediterranean comes from an eighth-century-BCE ceramic bowl found on the island of Ischia in the Gulf of Naples—an area that saw intense maritime commerce at the time, according to Bottaro. The bowl shows sailors fighting off a large marine creature reminiscent of a great white shark.
About five centuries after the Persian campaign, Roman philosopher and naturalist Pliny the Elder illustrated shark abundance when describing the swarms of dogfish that plagued sponge divers in Natural History. “Divers have fierce encounters with sharks, which make for their groin, heels and all the pale parts of their body. The only safe course is to turn on the sharks and frighten them. For sharks fear men just as much as men fear them,” he wrote, adding that divers searched for places with a particular species of flatfish, the presence of which indicated that sharks—“noxious monsters” in some translations—were likely absent.
Pliny is the first known author to describe the interactions between sharks and humans in some detail, but one of the earliest accounts that broke the shark-as-man-eater rhetoric was written about 500 years earlier. In History of Animals, Greek philosopher and scientist Aristotle detailed shark biology and ecology in a way rarely seen before the Renaissance. In one passage, for instance, Aristotle described sharks’ use of shallow water nurseries: “Cartilaginous fishes come out from the main seas and deep waters towards the shore and there bring forth their young, and they do so for the sake of warmth and by way of protection for their young.”
Mark Bond, a shark expert at Florida International University, says Aristotle’s descriptions of nurseries and his observation that some sharks hatch from internal eggs before exiting their mothers is fascinating, and provides further evidence of the abundance of the species at the time. “You would have thought that [sharks] must have been that much more abundant for [Aristotle] to pick up on stuff that we are still investigating centuries later,” he says.
Art from these early periods also sheds light on the species present in the Mediterranean. A fourth-century mosaic from the ancient Roman city Aquileia shows a variety of fish, including a type of electric ray, while another mosaic from roughly 200 years earlier in Pompeii shows two sharks and an electric ray. The great graphic detail in these pieces allows modern researchers to clearly identify most of the species, Bottaro says.
Despite these efforts to rationally study and observe sharks, in the years leading up to the Enlightenment, science and accurate depictions fell by the wayside. The Middle Ages were a dark period for the knowledge of sharks and science in general, says Bottaro. Superstition seemed to take over from the more studied approach taken by Pliny and Aristotle, and sharks were depicted as sea monsters or devils.
The 16th-century French epic poem translated roughly to History and True Novel of the Duke of Lyon de Bourges describes an episode in which crusader Olivier de Bourges is held up on his maritime journey toward the Holy Land by a “devil” in the sea. An accompanying illustration shows de Bourges about to hack into a sharkish-looking devil from a rather improbable-looking boat with a halberd. The creature, red-eyed with giant scales and a collar, already has its ignoble teeth on the heroic de Bourges’s helmet.
Humans’ relationship with sharks moved from superstitious to miraculous in one case the authors recount in their research. In 1573, an Italian ship traveling between Naples and Sicily was caught in a storm and began taking on water. Fearing imminent death, the legend goes, the crew prayed to the Virgin of Carmine for salvation, and the virgin responded. The storm promptly ended and the ship made it safely to land, where the crew discovered a large sawfish, a type of shark-like ray, stuck in a crack in the ship that had been letting in water. The sawfish didn’t enjoy the same fortune, as the sailors cut off its serrated rostrum and donated it to the Santa Maria del Carmine church in Naples as a votive offering, according to Bottaro.
The long, spiky rostrum is still revered in the church today as a symbol of an ancient miracle and has also fueled an ongoing debate among ecologists. Sawfish are not found in the region, but the story behind the relic seems to suggest they were once there. “If this rostrum was of a species [living] close to the Mediterranean Sea, it would be the first proof of the possible presence,” Bottaro says, adding that they are conducting work, including genetic analysis, to confirm the origin of the specimen.
Even if the sawfish wasn’t a resident of the Mediterranean, Bond says the relic’s existence could still reveal new information about trade routes at the time, considering the species would likely have come from southern Asia.
Tony Pitcher, a fisheries professor at the University of British Columbia who has been looking at the history of fisheries for years, says studies like those of Bottaro and his colleagues give more profile to the value of historical ecology to researchers.
“[The study] has utility in terms of understanding what we have lost and what we should really be seeking to restore,” he says.
As the artistic renderings and stories have accumulated through the centuries, shark numbers have dwindled. Ocean environments across the globe have taken a drastic change for the worse since the Industrial Revolution—a 2008 paper estimated that shark numbers in the Mediterranean Sea had dropped by 97 percent over the past 200 years.
“The Mediterranean has very few shark species and very few that are doing well,” Bond says. “We ecologists give it the nickname of the Deaditerranean.”
Bond says one of the biggest takeaways from these recorded interactions is the persistence of our fearmongering and how little has changed in the relationship between humans and sharks over time. But despite the ongoing challenge of convincing the public to accept the presence of these predators, a number of groups are working to improve the plight of sharks in the Mediterranean. If they succeed, future artists and writers depicting scenes of the Mediterranean in their work will have a robust ecosystem to draw on—one that includes its apex predators.<|endoftext|>
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Calculating Areas of Geometric Figures
Area of geometric figures are very common in Civil Service Exams and also other types of examinations. Area is basically the number of square units that can fit inside a closed region. In a closed region, if all the unit squares fit exactly, you can just count them and the number of squares is the area. For example, the areas of the figures below are 4, 10, 8 and 20 square units.
The figures blow are rectangles (yes, a square is a rectangle!). Counting the figures and observing the relationship between their side lengths and their areas, it is easy to see that the area is equal to the product of the length and the width (Why?).
The blue rectangle has length 5 and width 2, and counting the number of squares, we have 10. Of course, it is easy to see that we can group the squares into two groups of 5, or five groups of 2. From this grouping, we can justify why the formula for the area of a rectangle is described by the formula
$A_R = l \times w$
where $A_R$ is the area of the rectangle, $l$ is the length and $w$ is the width. Since the square has the same side length, we can say that
$A_S = s \times s = s^2$
where $A_S$ is its area and $s$ is its side length.
There are also certain figures whose areas are difficult to calculate intuitively such as the area of a circle, but mathematicians have already found ways to calculate the areas for these figures.
Challenge: Find the area of the green and blue figure below and estimate the area of the circle.
Below are some formulas for the most common shapes used in examinations. Don’t worry because we will discuss them one by one.
Triangle: $A = \frac{1}{2}bh$, $b$ is base, $h$ is height.
Parallelogram: $A = bh$, $b$ is base, $h$ is height
Trapezoid: $A = \frac{1}{2}h(b_1 + b_2)$, $b_1$ and $b_2$ are the base, $h$ is the height
Circle: $A = \pi r^2$ r is radius
In this series, we are going to discuss the areas of the most commonly used figure in examinations and we will discuss various problems in calculating areas of geometric figures. We are also going to discuss word problems about them. Questions like the number of tiles that can be used to tile a room is actually an area problem.
NEXT in this series: How to Solve Rectangle Area Problems Part 1<|endoftext|>
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How do you simplify (8/27)^(-2/3)?
Feb 14, 2017
$\frac{9}{4}$
Explanation:
${\left(\frac{8}{27}\right)}^{- \frac{2}{3}}$
$\text{ }$
$= {\left(\frac{27}{8}\right)}^{\frac{2}{3}}$
$\text{ }$
The prime factorization of 27 and 8 is:
$\text{ }$
$27 = 9 \times 3 = 3 \times 3 \times 3 = {3}^{3}$
$\text{ }$
$8 = 4 \times 2 = 2 \times 2 \times 2 = {2}^{3}$
Substituting the factorization on the above fraction we have:
$\text{ }$
${\left(\frac{27}{8}\right)}^{\frac{2}{3}}$
$\text{ }$
$= {\left({3}^{3} / {2}^{3}\right)}^{\frac{2}{3}}$
$\text{ }$
$= {\left({\left(\frac{3}{2}\right)}^{3}\right)}^{\frac{2}{3}}$
$\text{ }$
Applying the property of power of a power that says:
$\text{ }$
$\textcolor{red}{{\left({a}^{n}\right)}^{m} = {a}^{m \times n}}$
$\text{ }$
${\left({\left(\frac{3}{2}\right)}^{3}\right)}^{\frac{2}{3}}$
$\text{ }$
$= {\left(\frac{3}{2}\right)}^{3 \times \left(\frac{2}{3}\right)}$
$\text{ }$
$= {\left(\frac{3}{2}\right)}^{2}$
$\text{ }$
$= \frac{9}{4}$
Feb 14, 2017
$= \frac{9}{4}$
Explanation:
There are 3 different processes indicated in this expression.
Laws of indices:
${x}^{-} m = \frac{1}{x} ^ m \text{ "and" } {\left(\frac{a}{b}\right)}^{-} m = {\left(\frac{b}{a}\right)}^{m}$
The second law is the one we will apply.
Also ${x}^{\frac{p}{q}} = {\sqrt[q]{x}}^{p}$
${\left(\frac{8}{27}\right)}^{- \frac{2}{3}} = {\left(\frac{27}{8}\right)}^{+ \frac{2}{3}}$
$= {\sqrt[3]{\frac{27}{8}}}^{2} \text{ } \leftarrow$ find the cube roots first
$= {\left(\frac{3}{2}\right)}^{2}$
$= \frac{9}{4}$<|endoftext|>
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Geographos is classified as an Earth-crossing asteroid because its orbit can evolve to intersect Earth's orbit. Scientists have found fewer than 300 Earth-crossing asteroids; however, they believe hundreds of thousands of objects might exist. The asteroids probably include several hundred objects larger than Geographos, thousands larger than half a mile across, and a few hundred thousand that are larger than a football field.
Geographos is an irregular body with dimensions of about 5.1 kilometers by 1.8 kilometers (3.2 miles by 1.2 miles). It has the largest length-to-width ratio of any solar system object ever imaged to date. Scientists do not know whether the asteroid is a single coherent body or consists of several distinct pieces. Geographos was discovered at Palomar Observatory near San Diego, California, in 1951. The asteroid's name, which means geographer, was chosen to honor the National Geographic Society for its support of the Palomar Mountain Sky Survey.
The above image shows the outline of asteroid Geographos viewed from above its north pole. Researchers obtained radar images of the asteroid on August 30, 1994, when the asteroid was 7.2 million kilometers (4.5 million miles) from Earth. They used a planetary radar instrument to image the asteroid from the Deep Space Network's facility at Goldstone, California. The tick marks on the borders are 1 kilometer apart. The central white pixel locates the asteroid's pole. The gray scale is arbitrary and no meaning is attached to brightness variations inside the silhouette. Scientists conducted the radar observations a few days after Geographos passed 5 million kilometers (3.1 million miles) from Earth, its closest approach in at least two centuries. (Courtesy Dr. Steven J. Ostro, JPL/NASA).
Asteroids Vesta Mathilde
Copyright © 1997-2000 by Calvin J. Hamilton. All rights reserved. Privacy Statement.<|endoftext|>
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Assessment is a huge topic that encompasses everything from national or international accountability tests to everyday classroom observations and recording. It involves diagnostic, formative and summative approaches, and includes the use of standards as a means of benchmarking across large cohorts.
Examples and links
In order to grapple with what seems to be an over use of testing, we need to think of assessment as information. The more information we have about students, the clearer the picture we have about achievement or where gaps may occur, and the areas we may need to direct extra effort and guidance.
When we think of assessment as information it becomes easier to think of the myriad of ways in which that information can be gathered, stored and reported.
Technology offers teachers a broad range of tools to collect and analyze data, and richer sets of student data to guide instructional decisions.
Increasingly, digital technologies are being used to assist with assessment practices, including:
With an increasing emphasis on accountability through assessment in schools, there is a need to focus on strategies that support assessment for and of learning that are consistent with the philosophical frameworks adopted by schools, and with the intent of the New Zealand Curriculum and Te Whariki.<|endoftext|>
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## Mathematics and Applied Statistics Lesson of the Day – The Harmonic Mean
The harmonic mean, H, for $n$ positive real numbers $x_1, x_2, ..., x_n$ is defined as
$H = n \div (1/x_1 + 1/x_2 + .. + 1/x_n) = n \div \sum_{i = 1}^{n}x_i^{-1}$.
This type of mean is useful for measuring the average of rates. For example, consider a car travelling for 240 kilometres at 2 different speeds:
1. 60 km/hr for 120 km
2. 40 km/hr for another 120 km
Then its average speed for this trip is
$S_{avg} = 2 \div (1/60 + 1/40) = 48 \text{ km/hr}$
Notice that the speed for the 2 trips have equal weight in the calculation of the harmonic mean – this is valid because of the equal distance travelled at the 2 speeds. If the distances were not equal, then use a weighted harmonic mean instead – I will cover this in a later lesson.
To confirm the formulaic calculation above, let’s use the definition of average speed from physics. The average speed is defined as
$S_{avg} = \Delta \text{distance} \div \Delta \text{time}$
We already have the elapsed distance – it’s 240 km. Let’s find the time elapsed for this trip.
$\Delta \text{ time} = 120 \text{ km} \times (1 \text{ hr}/60 \text{ km}) + 120 \text{ km} \times (1 \text{ hr}/40 \text{ km})$
$\Delta \text{time} = 5 \text{ hours}$
Thus,
$S_{avg} = 240 \text{ km} \div 5 \text{ hours} = 48 \text { km/hr}$
Notice that this explicit calculation of the average speed by the definition from kinematics is the same as the average speed that we calculated from the harmonic mean!<|endoftext|>
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# Solve the following question:When x=2011, the value of $\dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{x}}}}}}}$ is ____.
Last updated date: 24th Jun 2024
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Hint: An expression is given and it has four divisions. It has a division within a division with a division within a division. So first solve the lowest division and after solving it go to the next one. Solve the equation by putting x as it is and then substitute the value of x.
We are given an expression $\dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{x}}}}}}}$ . We have to find its value when the value of x is 2011
So the given expression has four divisions.
So the first division to be solved is $1 + \dfrac{1}{x}$
$1 + \dfrac{1}{x} = \dfrac{{x + 1}}{x}$
After substituting the division value, the given expression will become
$\Rightarrow \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{x}}}}}}} = \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{\left( {\dfrac{{x + 1}}{x}} \right)}}}}}} = \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{x}{{x + 1}}}}}}$
The next division is $1 + \dfrac{x}{{x + 1}}$
$1 + \dfrac{x}{{x + 1}} = \dfrac{{\left( {x + 1} \right) + x}}{{x + 1}} = \dfrac{{2x + 1}}{{x + 1}}$
On substituting the above division value, we get
$\Rightarrow \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{x}{{x + 1}}}}}} = \dfrac{1}{{1 + \dfrac{1}{{\left( {\dfrac{{2x + 1}}{{x + 1}}} \right)}}}} = \dfrac{1}{{\left( {1 + \dfrac{{x + 1}}{{2x + 1}}} \right)}}$
The next division to be solved is $1 + \dfrac{{x + 1}}{{2x + 1}}$
$\Rightarrow 1 + \dfrac{{x + 1}}{{2x + 1}} = \dfrac{{\left( {2x + 1} \right) + x + 1}}{{2x + 1}} = \dfrac{{3x + 2}}{{2x + 1}}$
Substituting the above division value in the expression, we get
$\Rightarrow \dfrac{1}{{\left( {1 + \dfrac{{x + 1}}{{2x + 1}}} \right)}} = \dfrac{1}{{\left( {\dfrac{{3x + 2}}{{2x + 1}}} \right)}} = \dfrac{{2x + 1}}{{3x + 2}}$
Therefore, the value of the expression $\dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{x}}}}}}}$ is $\dfrac{{2x + 1}}{{3x + 2}}$
We are asked to find its value when x is equal to 2011.
So we are substituting 2011 in the place of x.
$\Rightarrow \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{2011}}}}}}}} = \dfrac{{2\left( {2011} \right) + 1}}{{3\left( {2011} \right) + 2}} \\ = \dfrac{{4022 + 1}}{{6033 + 2}} \\ = \dfrac{{4023}}{{6035}} \\$
Therefore, the value of $\dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{x}}}}}}}$ when x is equal to 2011 is $\dfrac{{4023}}{{6035}}$ .
So, the correct answer is “ $\dfrac{{4023}}{{6035}}$ ”.
Additional Information: If the denominator of a fraction is zero, then the value of that fraction is infinity.
Note: The reciprocal of a number is the fraction in which the numerator is 1 and the denominator is the number itself. The reciprocal of a reciprocal of a number is the number itself. That is what we have done in the above solution. Let a fraction be $\dfrac{a}{b}$ , then the reciprocal of this fraction will be $\dfrac{b}{a}$ . The numerator of the original fraction will become the denominator in its reciprocal and the denominator of the original fraction will become the numerator in its reciprocal.<|endoftext|>
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# Write and interpret numerical expressions 5th grade worksheets
Construct viable arguments and critique the reasoning of others. Students will be eager to solve this, rather than re-write it in another way. This is also building up their endurance to solve longer, more complicated math problems.
You can further model this if needed giving out counters to students and comparing numbers between students. With 5 to 8 minutes left in the class period, distribute an index card to each student.
Car Rider Line Guided Practice 15 minutes This lesson focuses on the use of grouping symbols in expressions. Writing and Interpreting Numerical Expressions Independent Practice 15 minutes Using MP3, students construct viable arguments and critique the reasoning of others in the next few problems.
Ask students to write a mathematical expression to represent this phrase. The matching game consists of 10 cards. Then we write and solve an equation together. In the second problem, students explain why two verbal expressions represent the same numerical calculations.
Parentheses are not needed in the last two expressions because multiplication and division are done before addition and subtraction. A third Expression Matching Game is also provided.
I also have students write as an exit ticket to the following: I have each student divide their paper into quarters and label each box. Tracie waits 21 minutes for mom to pick her up from school.
Students ready for an additional challenge will find this Web site helpful. You can use sentence frames as well. If possible, copy each game set onto a different color of paper. I point out to my students examples of the key words and phrases that indicate that parentheses should be used, such as "the result" and "twice the sum".
If the two cards they uncover are a match, they keep that pair of cards. The students work with their table partner to attempt to solve these expressions. For example, when students ask someone for "three more chips," they are expressing the operation "add 3".
Use the following strategies and activities to meet the needs of your students during the lesson and throughout the year. Students may then take turns flipping over two cards.
Ask students to work in pairs. To begin, put all 10 cards on your table with the writing showing.
Students sometimes confuse the divisor and the dividend when interpreting statements involving division. How long does Gwen wait for the bus? Explain why or why not. Each pair of students will need a copy of each of the three versions of the game.
Independent Practice Questions Closure 10 minutes Using cold calling, I call upon random students to discuss how they came upon their answers in the independent practice. Explain why or why not?This freebie is for all the 5th grade teachers out there that are as frustrated as I am.
practice sheets contain Algebraic Expression which are not covered in the 5th grade Common Core. 19, Downloads. Numerical Expressions Worksheet (killarney10mile.com2) Subject. Math, Other (Math So here is a two sided worksheet that just contains numerical 4/5(59). Write numerical expressions killarney10mile.comA.2 - Write simple expressions that record calculations with numbers, and interpret numerical expressions without.
Students learn to write and interpret numerical expressions. Students will: The lesson deals with writing and interpreting numerical expressions. Now distribute the Writing Numerical Expressions practice worksheet (M_Writing Numerical Expressions Practice Worksheet and killarney10mile.com).
Ask students to work in pairs to write expressions. 5th Grade Math (Jessica Field) Unit 2. Unit 1: We then walk through writing numerical expressions for the following: a) twice the sum of five and seven.
b) three times the result of subtracting four from nine. Writing and Interpreting Numerical Expressions.
Independent Practice. Write simple expressions that record calculations with numbers, and interpret numerical expressions without evaluating them. For example, express the calculation "add 8 and 7, then multiply by 2" as 2 x (8 + 7).
5th Grade Worksheets - Write and interpret numerical expressions.
Use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols. Write simple expressions that record calculations with numbers, and interpret numerical expressions without evaluating them.
Write and interpret numerical expressions 5th grade worksheets
Rated 3/5 based on 61 review<|endoftext|>
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# probability using permutations/combinations
• Jul 10th 2007, 05:03 PM
sam48
probability using permutations/combinations
A bag contains 4 red and 3 green blocks. 3 blocks are randomly selected from the bag. Determine the probability that all three are red given that
i.) replacement occurs between the second and third selections
ii.) no replacement occurs
• Jul 10th 2007, 05:59 PM
Soroban
Hello, Sam!
Quote:
A bag contains 4 red and 3 green blocks.
Three blocks are randomly selected from the bag.
Determine the probability that all three are red given that:
a) replacement occurs between the second and third selections
First block is red: $\displaystyle P(\text{\#1 red}) = \frac{4}{7}$
Second block is red: $\displaystyle P(\text{\#2 red}) = \frac{3}{6}$ . . . . The second block replaced.
Third block is red: $\displaystyle P(\text{\#3 red}) = \frac{3}{6}$
Therefore: $\displaystyle P(\text{3 red}) \;=\;\frac{4}{7}\cdot\frac{3}{6}\cdot\frac{3}{6} \;=\;\frac{1}{7}$
Quote:
b) no replacement occurs
First block is red: $\displaystyle P(\text{\#1 red}) = \frac{4}{7}$
Second block is red: $\displaystyle P(\text{\#2 red}) = \frac{3}{6}$
Third block is red: $\displaystyle P(\text{\#3 red}) = \frac{2}{5}$
Therefore: $\displaystyle P(\text{3 red}) \;=\;\frac{4}{7}\cdot\frac{3}{6}\cdot\frac{2}{5} \;=\;\frac{4}{35}$<|endoftext|>
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September 2017 saw a spate of solar activity, with the Sun emitting 27 M-class and four X-class flares and releasing several powerful coronal mass ejections, or CMEs, between Sept. 6-10. Solar flares are powerful bursts of radiation, while coronal mass ejections are massive clouds of solar material and magnetic fields that erupt from the Sun at incredible speeds.
The activity originated from one fast-growing active region — an area of intense and complex magnetic fields — as it travelled across the Sun’s Earth-facing side in concert with the star’s normal rotation. As always, NASA and its partners had many instruments observing the Sun from both Earth and space, enabling scientists to study these events from multiple perspectives.
With multiple views of solar activity, scientists can better track the evolution and propagation of solar eruptions, with the goal of improving our understanding of space weather. Harmful radiation from a flare cannot pass through Earth’s atmosphere to physically affect humans on the ground, however — when intense enough — they can disturb the atmosphere in the layer where GPS and communications signals travel. On the other hand, depending on the direction they’re traveling in, CMEs can spark powerful geomagnetic storms in Earth’s magnetic field.
To better understand the fundamental processes that drive these events, and ultimately improve space weather forecasts, many observatories watch the Sun around the clock in dozens of different wavelengths of light. Each can reveal unique structures and dynamics in the Sun’s surface and lower atmosphere, giving researchers an integrated picture of the conditions driving space weather.
Scientists also have their eyes on the Sun’s influence on Earth and even other planets. Effects from September’s solar activity were observed as Martian aurora and across the globe on Earth, in the form of events known as ground-level enhancements — showers of neutrons detected on the ground, produced when energetic particles accelerated by a solar eruption stream along Earth’s magnetic field lines and flood the atmosphere.
The imagery below shows the wide swath of views available to researchers as they use these recent space weather events to learn more and more about the star we live with.
NOAA’s Geostationary Operational Environmental Satellite-16, or GOES-16, watches the Sun’s upper atmosphere — called the corona — at six different wavelengths, allowing it to observe a wide range of solar phenomena. GOES-16 caught this footage of an X9.3 flare on Sept. 6, 2017. This was the most intense flare recorded during the current 11-year solar cycle. X-class denotes the most intense flares, while the number provides more information about its strength. An X2 is twice as intense as an X1, an X3 is three times as intense, etc. GOES also detected solar energetic particles associated with this activity.
NASA’s Solar Dynamics Observatory watches the corona at 10 different wavelengths on a 12-second cadence, enabling scientists to track highly dynamic events on the Sun such as these X2.2 and X9.3 solar flares. These images were captured on Sept. 6, 2017, in a wavelength of extreme ultraviolet light that shows solar material heated to over one million degrees Fahrenheit. The X9.3 flare was the most intense flare recorded during the current solar cycle.
JAXA/NASA’s Hinode caught this video of an X8.2 flare on Sept. 10, 2017, the second largest flare of this solar cycle, with its X-ray Telescope. The instrument captures X-ray images of the corona to help scientists link changes in the Sun’s magnetic field to explosive solar events like this flare. The flare originated from an extremely active region on the Sun’s surface — the same region from which the cycle’s largest flare came.
Key instruments aboard NASA’s Solar and Terrestrial Relations Observatory, or STEREO, include a pair of coronagraphs — instruments that use a metal disk called an occulting disk to study the corona. The occulting disk blocks the Sun’s bright light, making it possible to discern the detailed features of the Sun’s outer atmosphere and track coronal mass ejections as they erupt from the Sun.
On Sept. 9, 2017, STEREO watched a CME erupt from the Sun. The next day, STEREO observed an even bigger CME, which was associated with the X8.2 flare of the same day. The Sept. 10 CME traveled away from the Sun at calculated speeds as high as 7 million mph, and was one of the fastest CMEs ever recorded. The CME was not Earth-directed. It side-swiped Earth’s magnetic field, and therefore did not cause significant geomagnetic activity. Mercury is in view as the bright white dot moving leftwards in the frame.
Like STEREO, ESA/NASA’s Solar and Heliospheric Observatory, or SOHO, uses a coronagraph to track solar storms. SOHO also observed the CMEs that occurred during Sept. 9-10, 2017; multiple views provide more information for space weather models. As the CME expands beyond SOHO’s field of view, a flurry of what looks like snow floods the frame. These are high-energy particles flung out ahead of the CME at near-light speeds that struck SOHO’s imager.
NASA’s Interface Region Imaging Spectrometer, or IRIS, peers into a lower level of the Sun’s atmosphere — called the interface region — to determine how this area drives constant changes in the Sun’s outer atmosphere. The interface region feeds solar material into the corona and solar wind: In this video, captured on Sept. 10, 2017, jets of solar material appear like tadpoles swimming down toward the Sun’s surface. These structures — called supra-arcade downflows — are sometimes observed in the corona during solar flares, and this particular set was associated with the X8.2 flare of the same day.
NASA’s Solar Radiation and Climate Experiment, or SORCE, collected this data on total solar irradiance, the total amount of the Sun’s radiant energy, throughout Sept. 2017. While the Sun produced high levels of extreme ultraviolet light, SORCE actually detected a dip in total irradiance during the month’s intense solar activity. A possible explanation for this observation is that over the active regions — where solar flares originate — the darkening effect of sunspots is greater than the brightening effect of the flare’s extreme ultraviolet emissions. As a result, the total solar irradiance suddenly dropped during the flare events. Scientists gather long-term solar irradiance data in order to understand not only our dynamic star, but also its relationship to Earth’s environment and climate. NASA is ready to launch the Total Spectral solar Irradiance Sensor-1, or TSIS-1, this December to continue making total solar irradiance measurements.
The intense solar activity also sparked global aurora on Mars more than 25 times brighter than any previously seen by NASA’s Mars Atmosphere and Volatile Evolution, or MAVEN, mission. MAVEN studies the Martian atmosphere’s interaction with the solar wind, the constant flow of charged particles from the Sun. These images from MAVEN’s Imaging Ultraviolet Spectrograph show the appearance of bright aurora on Mars during the September solar storm. The purple-white colors show the intensity of ultraviolet light on Mars’ night side before (left) and during (right) the event.<|endoftext|>
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Electrophysiology (or EP) is the name of the branch of cardiology that deals with heart rhythm disorders. Catheter ablations are performed by an electrophysiologist (also sometimes called an EP). The University of Michigan electrophysiology program has been a national and international leader in catheter ablation of arrhythmias (irregular heart rhythms) over the last three decades. We’ve been at the forefront of catheter ablation of atrial fibrillation during the last eight years and played a key role in understanding the mechanisms of atrial fibrillation and developing new treatment strategies.
Doctors at the University of Michigan have performed crucial studies that have helped to perfect the tools used in ablation. We handle large volumes of patients and have conducted extensive studies to prove that ablation is effective long term.
What is Catheter Ablation?
Catheter ablation is a minimally invasive technique intended to treat atrial fibrillation (Afib) without major surgery. Using a specially designed catheter that is positioned in the left atrium, radiofrequency energy is applied to the heart muscle to cauterize the “short circuits” that are triggering atrial fibrillation.
For the ablation procedure, catheters are inserted with a needle into a vein that runs up to the heart from the groin. The procedure requires the insertion of a catheter into the left atrium. This is accomplished by “transeptal catheterization,” in which a small hole is purposely made with a needle that is pushed through a thin membrane that separates the two top chambers of the heart.
With a computerized, three-dimensional mapping system to guide the procedure, the doctor can see the catheter and the left atrium on the computer screen, which makes it possible to guide the catheter very precisely and cuts down on the amount of X-ray needed during the procedure.
By recording the electrical activity inside the heart, we can identify the short circuits that are generating the atrial fibrillation. These spots are cauterized with the radiofrequency energy. To eliminate the atrial fibrillation, a typical patient requires between 150 and 250 different spots to be ablated.
The entire procedure usually takes 2 to 4 hours and the patient spends one night in the hospital. Patients are put on blood thinners for at least three months before and after the procedure, until the inner lining of the heart has healed from the effects of the radiofrequency energy.
Another type of ablation, known as cryoballoon ablation, uses liquid nitrogen instead of radiofrequency energy. Cryoballoon ablation precisely targets the multiple nerve connections around the opening of the pulmonary veins, which deliver oxygenated blood back to the left side of the heart. Pulmonary vein connections contribute to all types of atrial fibrillation — including paroxysmal and persistent — so it is important to target them during any ablation procedure.
A cryoballoon ablation procedure can be shorter than radiofrequency ablation, although both approaches are equally effective. With cryoballoon ablation, however, only pulmonary vein connections can be targeted. Radiofrequency ablation enables all types of arrhythmias (other than atrial fibrillation), which can coexist in up to 30 percent of patients, to be targeted.
During cryoballoon ablation, a catheter is threaded from the femoral vein in the groin to the pulmonary veins of the heart’s left atrium, followed by insertion of the cryoballoon catheter. The balloon, filled with liquid nitrogen, is positioned against the opening of the pulmonary veins and inflated. This causes the tissue touched by liquid nitrogen to become frozen, or scarred. This ablated tissue then can no longer trigger the electrical currents responsible for atrial fibrillation.
Success Rates for Catheter Ablation at the University of Michigan
We have performed thousands of ablation procedures aimed at curing atrial fibrillation over the past few years. Success rates of left atrial ablation in patients with atrial fibrillation depend on whether the atrial fibrillation is paroxysmal (the kind that comes and goes on its own) or persistent (the kind that has been present consistently for several months to years).
Paroxysmal atrial fibrillation can be eliminated in 70-75 percent of patients with a single procedure. When the procedure is repeated in patients who still have atrial fibrillation after the first procedure, the overall success rate is approximately 85-90 percent.
Persistent atrial fibrillation can be eliminated in approximately 50 percent of patients with a single procedure. In about 30 percent of patients who undergo ablation of chronic atrial fibrillation, the atrial fibrillation is replaced by a different kind of short circuit referred to as “left atrial flutter.” These patients are treated temporarily with medications and the left atrial flutter sometimes goes away on its own within a few months. If it does not, you may need a second catheter ablation procedure to eliminate the flutter. In these cases, the overall success rate is approximately 75-85 percent. If the atrial fibrillation has been persistent for more than 1-2 years, almost all patients will require more than one ablation procedure before a normal heart rhythm is restored.
Make an Appointment
The Arrhythmia Program at the University of Michigan has been a national and international leader in the treatment of arrhythmias for more than 30 years. To schedule an appointment to discuss catheter ablation or any other cardiovascular condition or treatment, call us at 888-287-1082 or visit our Make a Cardiovascular Appointment page, where you may fill out a Patient Appointment Request Form and view other details about scheduling an appointment.<|endoftext|>
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# Difference between revisions of "2008 USAMO Problems/Problem 2"
## Problem
(Zuming Feng) Let $ABC$ be an acute, scalene triangle, and let $M$, $N$, and $P$ be the midpoints of $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Let the perpendicular bisectors of $\overline{AB}$ and $\overline{AC}$ intersect ray $AM$ in points $D$ and $E$ respectively, and let lines $BD$ and $CE$ intersect in point $F$, inside of triangle $ABC$. Prove that points $A$, $N$, $F$, and $P$ all lie on one circle.
## Solution
### Solution 1
$[asy] /* setup and variables */ size(280); pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8); pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */ /* construction and drawing */ pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C); D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s)); D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,s))); D(D(MP("D",D,SE,s))--MP("P",P,W,s)); D(B--D(MP("F",F,s))); D(O--F--A,linetype("4 4")+linewidth(0.7)); D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5)); picture p = new picture; draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7)); draw(p,circumcircle(A,B,C),linetype("1 4")+linewidth(0.7)); clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p); [/asy]$
Construct $T$ on $AM$ such that $\angle BCT = \angle ACF$. Then $\angle BCT = \angle CAM$. Then $\triangle AMC\sim\triangle CMT$, so $\frac {AM}{CM} = \frac {CM}{TM}$, or $\frac {AM}{BM} = \frac {BM}{TM}$. Then $\triangle AMB\sim\triangle BMT$, so $\angle CBT = \angle BAM = \angle FBA$. Then we have
$\angle CBT = \angle ABF$ and $\angle BCT = \angle ACF$. So $T$ and $F$ are isogonally conjugate. Thus $\angle BAF = \angle CAM$. Then
$\angle AFB = 180 - \angle ABF - \angle BAF = 180 - \angle BAM - \angle CAM = 180 - \angle BAC$.
If $O$ is the circumcenter of $\triangle ABC$ then $\angle BFC = 2\angle BAC = \angle BOC$ so $BFOC$ is cyclic. Then $\angle BFO = 180 - \angle BOC = 180 - (90 - \angle BAC) = 90 + \angle BAC$.
Then $\angle AFO = 360 - \angle AFB - \angle BFO = 360 - (180 - \angle BAC) - (90 + \angle BAC) = 90$. Then $\triangle AFO$ is a right triangle.
Now by the homothety centered at $A$ with ratio $\frac {1}{2}$, $B$ is taken to $P$ and $C$ is taken to $N$. Thus $O$ is taken to the circumcenter of $\triangle APN$ and is the midpoint of $AO$, which is also the circumcenter of $\triangle AFO$, so $A,P,N,F,O$ all lie on a circle.
### Solution 2
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
## Resources
2008 USAMO (Problems • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 All USAMO Problems and Solutions
• <url>viewtopic.php?t=202907 Discussion on AoPS/MathLinks</url>
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# “Mathematics in Our World”: Project 1 & Project 2
## Introduction
Generally, a quadratic equation has the following form: ax2+bx+c; with, c, b, and a being constant that are referred to as quadratic efficient, and x is a variable. In numerical stipulations, a quadratic equation is simply a polynomial-statement of a subsequent degree (Katz, & Barton, 2007). There are manifold methods of solving a quadratic equation, such as completing-the-square method; factoring; graphical method; employing quadratic formula; and Newton’s method. Superlatively, the rudimentary task of a prime numeral is generally set up by the arithmetic theorem.
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In mathematical terms, a prime-number is simply a natural numeral that has two distinct numeral divisors, i.e. the number itself and one (Stoppard, 1993). In this paper, someone will complete the weekly reading, i.e. up to Chapter 7 in the textbook entitled “Mathematics in Our World” that was written by Bluman G Allan; Thus, this paper aims at completing Project #1 and Project #2. In project 1, the paper will follow the example given by completing the six steps. In project 2, the paper will choose not less than five numbers, which can be odd, even numbers, or even zero, then substitute them in the formula to determine whether it yields a number that is composite or prime.
## Project #1
To answer this project, we ought to follow a fascinating method to solve quadratic equations. This method is believed to have originated from India, and it is referred to as the completing-the-square method (Bluman, 2005). Thus, this method will be employed in solving these equations:
1. X2 – 2x-13=0;
2. 4x2-4x+3=0;
3. X2+12x-64=0; and
4. 2x2-3x-5=0
### Project #1 Solution
We will follow the six steps as per the example given
1. x2 – 2x = 13; Step 1, 4x2 – 8x = 52; then 4x2 – 8x + 4 = 56; from this,
We get (2x – 2)2 = 56; this implies that 2x-2 = √56 = 2√14, i.e. 2(x – 1) = 2√14
Simplifying we get x – 1 = √14; thus x = 1 + √14; Then 2x -2 = √56 = -2√14
From this 2(x – 1) = -2√14, which can be expressed as x – 1 = -√14;
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Therefore, x = 1 – √14
1. X2– 4x = -3; Step 1, 4×2-16x =-12; then 4(x2– 4x) =-12; from this.
We get x2-4x =-3; this implies that x2-4x + 4 =1, i.e. (x – 2)2 = 1; simplifying we get
X-2 =1; thus x = 2 + √1 = 2 + 1 = 3; then x – 2 = -√1; from this x = 2 – √1, which can be expressed as x = 2 – 1 = 1
1. x2 + 12x = 64; Step 1, 4x2+48x=256; then 4x2+48x+144 = 400; from this,
We get (2x + 12)2 = 400; this implies that 2x+12=√400; simplifying we get
2x + 12 = 20, x=4; or 2x + 12 = -20, and x= -16
1. 2x2-3x=5; Step 1, x2-1.5x=2.5; then 4×2 – 6x = 10; step 3, 4x2 – 6x + 2.25 = 12.25;
Step 4, (2x – 1.5)2 = 12.25, step 5, 2x -1.5 = √12.25; and finally, step 6, 2x-1.5 = 3.5;
X= 2.5, or 2x – 1.5 = -3.5; x= -1
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## Project #2
The searched formula that will be capable of yielding prime numbers is x2-x+41. This formula will be useful in this project, whereby at least 5 numerals for x will be selected and then substituted in this formula, so as to determine whether there is an occurrence of a prime number. Ideally, in this project, a numeral x that yields a composite numeral when substituted in the formula will also be found (Bluman, 2005). Thus;
### Project #2 Solution
Let, P (x) =x2-x+ 41, lets plug-in the following values of x: x=1; x=2; x=5; x=10; x=12; x=20.
Then
P (1) =12 -1 +41= 41, this is a prime number; P (2) = 22-2 + 41 = 43, this is a prime number;
P (5) =52-5+4 = 61, this is a prime number; P (7) = 72-7 41= 83, this is a prime number;
P (10) = 102-10+41=131, this is a prime number; P (12) =122-12+41=173, this is a prime number; and P (20) =202-20+41=421, this is also a prime number. In fact, this list will continue to yield prime numbers until the value of x equals to 40. For instance, if x=41 then
P (41) =412-41+41=1681, this is a composite numeral, but not a prime number.
## Conclusion
The completing-the-square method can be employed in deriving the quadratic formula, to be capable of using the algebraic identity. Ideally, the majority of general notions that apply to the algebraic structure, in which multiplication, adding up and subtraction are defined, arise from prime numbers.
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## References
Bluman, A. G. (2005). Mathematics In Our World. College of Allegheny: McGraw-Hill Publishers.
Katz, V. J., & Barton, B. (2007). History of Algebra. Educational and Mathematical Studies, 65 (2), 180-199.
Stoppard, T. (1993). Arcadia. London: Faber & Faber.
## Cite this paper
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Reference
StudyCorgi. (2022, February 25). “Mathematics in Our World”: Project 1 & Project 2. Retrieved from https://studycorgi.com/mathematics-in-our-world-project-1-and-amp-project-2/
Reference
StudyCorgi. (2022, February 25). “Mathematics in Our World”: Project 1 & Project 2. https://studycorgi.com/mathematics-in-our-world-project-1-and-amp-project-2/
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StudyCorgi. "“Mathematics in Our World”: Project 1 & Project 2." February 25, 2022. https://studycorgi.com/mathematics-in-our-world-project-1-and-amp-project-2/.
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StudyCorgi. 2022. "“Mathematics in Our World”: Project 1 & Project 2." February 25, 2022. https://studycorgi.com/mathematics-in-our-world-project-1-and-amp-project-2/.
References
StudyCorgi. (2022) '“Mathematics in Our World”: Project 1 & Project 2'. 25 February.
This paper was written and submitted to our database by a student to assist your with your own studies. You are free to use it to write your own assignment, however you must reference it properly.
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Find the value
Question:
Find the value
$(x+2)\left(x^{2}+25\right)-10 x^{2}-20 x$
Solution:
$(x+2)\left(x^{2}+25\right)-10 x(x+2)$
Taking (x + 2) common in both the terms
$=(x+2)\left(x^{2}+25-10 x\right)$
$=(x+2)\left(x^{2}-10 x+25\right)$
Splitting the middle term of $\left(x^{2}-10 x+25\right)$
$=(x+2)\left(x^{2}-5 x-5 x+25\right)$
$=(x+2)\{x(x-5)-5(x-5)\}$
$=(x+2)(x-5)(x-5)$
$\therefore(x+2)\left(x^{2}+25\right)-10 x^{2}-20 x=(x+2)(x-5)(x-5)$<|endoftext|>
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FutureStarr
A How to Find 75 Percent of a Number
# How to Find 75 Percent of a Number
Sharing this tip may help you or a friend solve a problem. Imagine you want to divide a number by 75 percent. Hopefully, this article will provide a clear and accurate answer to this problem.
### Calculate
It is valuable to know the origin of the term ​percent​ if you want to truly understand how to calculate a percentage. The word ​percent​ comes from the phrase per cent. ​Cent​ is a root that means one hundred, so per cent literally means per one hundred. For example, if you know that 30 percent of the students in a school are boys, that means that there are 30 boys per one hundred students. Another way to say this is that 30 out of 100 students are boys.
Generally, the way to figure out any percentage is to multiply the number of items in question, or ​X​, by the ​decimal​ form of the percent. To figure out the decimal form of a percent, simply move the decimal two places to the left. For example, the decimal form of 10 percent is 0.1. Then, to calculate what 10 percent of is, say, 250 students, simply multiply the number of students by 0.1. (Source: sciencing.com)
### Divide
To calculate percentages, start by writing the number you want to turn into a percentage over the total value so you end up with a fraction. Then, turn the fraction into a decimal by dividing the top number by the bottom number. Finally, multiply the decimal by 100 to find the percentage.
To calculate percentages, start by writing the number you want to turn into a percentage over the total value so you end up with a fraction. Then, turn the fraction into a decimal by dividing the top number by the bottom number. Finally, multiply the decimal by 100 to find the percentage. (Source: percentagecalculator.guru)
### Number
Remember that in decimal multiplication, you multiply as if there were no decimal points, and the answer will have as many “decimal digits” to the right of the decimal point as the total number of decimal digits of all of the factors. So when you multiply 0.7 × 80, think of multiplying 7 × 80 = 560. Since 0.7 has one decimal digit, and 80 has none, the answer has one decimal digit: 56.0 Thus, 0.7 × 80 = 56.
More percentage problems: 150% of what number is 15 75% of what number is 30 225% of what number is 15 75% of what number is 45 375% of what number is 15 75% of what number is 75 525% of what number is 15 75% of what number is 105 75% of 15 What percent is 75 of 15 (Source: answers.everydaycalculation.com)
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Science is essential to understand the natural world. It encompasses many topics, such as astrophysics, astronomy, biology, mathematics, history and latest technologies. With science fairs, students will understand how the universe works. During the science fairs, students are able to collect information, ask questions, make observations and perform experiments. There are new discoveries that can bring understanding and improvement in our quality of life. With science fairs, it should be possible for students to understand the whole and they could have a desire to become scientists. It is human nature to make a lot of questions and we should be aware of things that happen around us. Science exercises should be essential to ensure that students can implement when they have learned in the classroom. There are laboratory works and hands-on activities that students could do during science fairs. Science fairs should also offer interactions between pure science with technology and society. If students want to participate in the science fairs, they should choose topics that they are familiar with. As an example, students may choose insects and other animals as the primary focus. Students may show in an interesting way interesting things about animals, such as how the communicate and eat to survive. There are different characteristics and traits of an animal that people will find it interesting. Students may choose a few exotic animals with unique life cycles and behaviors. People could be impressed with how these animals live and students should provide enough information.
Astronomy is also an interesting thing that we may share to others. There could be specific stars, planets, asteroids, planetoids, galaxies, nebula and other distant objects that students want to focus on. Students may also provide interesting facts about how extraterrestrial life may exist based on facts on Earth. We may also show how solar energy works and how it can be harnessed. Human body is a complex biological machine and it could become a subject of science fair. We should show people how our body reacts to external factors and good. We may show how nutrients could be beneficial for our body to sustain growth and prevent disease. Earth is a planet where we live in with so many interesting things. Volcanoes, streams and oceans are interesting subjects for science fairs. We should be able to provide people with an information how the ocean and weather interacts with one another. Water cycle is a commonly known fact, but it can still be shown in an interesting way. We could show how the Earth surface would be like when the North and South poles melt partially and completely. We may also show how the continents would look like in two billion years or more from today. Gardening and plants are another subjects that we can choose for science fairs. We could show others how photosynthesis works and why it is critical to life on Earth, even to carnivores and human. Another idea is to show how types of soil works with a specific type of plant. We could also show the percentage of water in different fruits, including showing fresh fruits and completely sun-dried fruits.<|endoftext|>
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The Addition Snake Game is a fun way for children to begin memorising facts, and of course, boost their adding skills. The goal of the activity is to turn a colourful snake into a golden snake by counting to ten.
The game has three components: a black and white bead stair, multicoloured beads, golden beads and a special notched card for the purpose of counting and marking – organised by coloured boxes in a bead tray.
The directress will begin the exercise by introducing the child to the black and white beads, which are then placed in an inverted triangle on the right side of the surface (the point of which will become evident in due course); the coloured bars are then introduced into the game. During this first presentation, the coloured beads (which come in varying sequences from 1-9) are prepared in the shape of a snake, in sets of 10: so nine beads plus one bead; seven beads plus three…and so on. The child is then asked to count up to ten, using the coloured beads as a guide. Once the number 10 is reached, the coloured bead set is replaced with a golden bar (all of which constitute 10 beads).
During the second presentation, the directress will demonstrate what to do if the numbers don’t come out to exact sets of 10, which may happen when the child is creating his/her own snake. If the child gets to 10 in the middle of a set of beads, the remaining beads on that bar are replaced with the appropriate quantity from the black and white beads, otherwise known as the Black and White Bead Stair. These help the child with any leftovers they may need to fill in to add up to 10.
Have a look at the below tutorial:
Montessori materials can often seem arbitrary outside of their application but once used within the context of an activity, they begin to make sense…as is the case with the conjoined beads that form the crux of the Addition Snake Game. Once the game is understood, it becomes clear how this exercise helps children identify sets of they become acquainted with the possible number combinations that make up 10.
If you’d like further information about this tutorial, feel free to contact us at [email protected]. We’re happy to answer any questions.
Source: Montessorium.com – Addition Snake Game<|endoftext|>
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Time to hand in those open responses! I will start class by collecting both versions: the one they were going to hand in yesterday and the edited version from last night's homework.
In the chapter we are reading today, we learn the convict's name, both real and assumed. Early on in our reading, we discussed the importance of names in this text; this handout will help us understand the significance of the convict's name.
Before reading chapter 40, students will read this handout to themselves quietly and write a two-sentence summary at the bottom to prove their understanding. The handout is a copy of Genesis 4, the story of how Cain killed his brother Abel. I'm not going to be looking for an in depth analysis of the passage; I merely want the students to understand which brother is which and generally what happened. After reading and learning more about the convict we will return to this story, which juxtaposes generosity and selfishness, much like in Great Expectations (RL.9-10.7). The convict, Abel Magwich, embodies generosity, much like the original Abel, even if Pip doesn't realize it yet (RL.9-10.9).
We won't read all of chapter 40 in class, but we will start it together. In this section, we learn more about the convict, including his name. As we read, the students will bullet point adjectives that describe Magwich in their notebook and explain their choices. Ultimately, I want to highlight how different Magwich, the lowly convict, is from Miss Havisham. The best way to do that is to analyze the interaction between him and Pip. The nicer that Magwich is to Pip, the more Pip seems to hate him, and yet Magwich loves him still. Like the Abel of the Old Testament, Abel Magwich offers everything and gives it to Pip; he is complete generosity, whereas Miss Havisham is totally selfish (RL.9-10.3). Understanding these key elements of their personalities will help us understand theme, which will discuss more as we near the end of the text.
In this clip, student tries to imitate me reading aloud and walking around the room, although I hope that I read with more life!
Once we learn the convict's real and assumed name, I will ask the class why Dickens made these choices. It will be important to remember that there is no "correct" answer; instead, we get to interpret and decide what makes sense, given what we know. I will direct them to think about both the synonym "able" and the allusion "Abel" and to write a thesis in the notebooks about what they think. We will approach the assumed name, Provis, similarly. What does the name sound like? What might it connote? This will be a full-class discussion that will be mostly student directed. Although I plan to pose the questions, their answers and interpretations will guide the discussion overall (SL.9-10.1c).
In the last few minutes of class, I will assign homework. Tonight students will finish chapter 40 (there will probably be three pages left to read) and read chapter 41.<|endoftext|>
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# Primary Mathematics/Fractions
Primary Mathematics ← Negative numbers Fractions Working with fractions →
## Learning to use fractions (Visually)
Fractions or rational numbers are in essence the same as division, however we use them more often to express numbers less than one - for instance a half or a quarter. Fractions have a numerator (on the top) and a denominator (on the bottom). If a fraction is larger than 1 then the numerator will be larger.
### Modern methods to teach fractions
Today's modern methods of teaching math and fractions are drastically different than how they were taught just 10 years ago. The difference between these methods is that the later method explores the visual evidence for certain ways of manipulating fractions and whereas the earlier approach simply used variables from the beginning. Tiles of different colors sorted into groups can be useful in representing fractions visually.
### Origami and Fractions
There is perhaps little emphasis these days which shows the elaborate "visualization" that math requires. Students used to be taught merely by equation, but to my understanding, by teaching them such methods they tend to take the "cooking approach" to problems in that they have an inadequate sense of "visualizing" the concept the problem in their mind. What I want to do is emphasize the creative aspect of fractions while at the same time exploring the richness of why such problems are true.
Why I call this section Origami and math is because they are very much related to each other. The thing about Origami which is rich in math is that essentially folding a piece of paper proves that such a fraction exists!
In order to do this experiment, you need the following materials:
1. Square piece of paper
2. pencil
Steps to seeing some nice fractions
First, we have to say to ourselves, "This piece of paper is 1 piece of paper"
Now, we will explore fractions by seeing how "much" of the remaining paper we see as we fold. Every time we see the paper, we will write down the fraction of the paper on the front of it. By the time we get done, we'll have lots of fractions written on a piece of paper.
Step 1: Obtain square piece of paper. Write "1" on it.
Step 2: fold piece of paper in half. Write "1/2" on it.
Step 3: Fold again in half. write "1/4"
Step 4: Fold again in half. write "1/8"
Step 5: Unfold the piece of paper and write lines in where there are folds in the paper
Notice: if there is a 1 on top, then whatever number is on the bottom, it takes that many "pieces" to make a "whole". For example, 1/8 needs 8 pieces to become 1 whole.
### The square model vs. the circle model
A square divided into fractions A circle divided into fractions
### The money approach: 1, 1/4, 1/2, 1/10, 1/20, 1/100
A very practical way to learn fractions is the use of money, as we use it everyday.
Questions:
1. How many quarters are in a dollar?
2. How many dimes are in a dollar?
3. How many nickels are in a dollar?
4. How many pennies are in a dollar?
As said from the previous section, if there is a one on top and some number on bottom, that means it needs that much pieces to make it a whole (a whole means 1 by the way)
For example, 1/10 is the value of a dime (10 cents). You need 10 dimes (10 x 10 = 1 dollar) to get one full dollar.
### Whole number fractions
First of all, as always, instead of looking at complicated variable jargon, we will instead look at certain ways to "view" certain types of number. Just like art, you don't need to be an accomplished artist to draw, but rather you just need to know how to look at things better (in this case, numbers)
Since fractions have both a top (called a numerator. think "topinator") and a bottom (denominator, which "downominator" which is divided by a bar, we have to "adjust our thinking" so that we can recognize what our friendly fractions might look like.)
For example,
Q: 5 ÷ 5 reads "5 divided by 5". What does that look in Fraction form?
Well, since we are prospective mathematicians (and artists...) we will look at the magic of what the divided sign actually means:
o === o (my divided sign) What it actually means is that dots tell you to "Make me a number!" (because Zeroes often become lonely...they want to have value in life...) and you see that bridge which devides them too says that, "In order to separate such ZEROES in life and making more zeroes, we say always divide your numbers from each other."
A: ${\displaystyle {\frac {5}{5}}=1}$, which also reads "5 divided by 5".
Since we know how to recognize some fractions, we have to "see" with our naked eyes what some fractions actually mean. The best preparation that you will have in our "artist" training is that whole numbers actually have "1" on the bottom. You may ask yourself, "How can that be? Is that even possible?"
Well Just for fun, we will go through this little exercise to show that there is indeed ones on the bottom when you have whole numbers.
Example: express "5" in a fraction form.
Look at our previous example about the placement of things. We now know what a denominator is and a numerator is. Since the whole number is the number of top, all we have to do is see what number is on bottom, i wonder what it is.
${\displaystyle {\frac {5}{?}}}$ In order to figure out what goes on bottom, we can figure it out many ways. Fractions are in ways actually like ratios. Such as if there were two ice creams for me and none for you, my ratio would be 2:1 and yours would be 0:1...Sad isn't it? Getting back to the point, we must think about why and how it becomes a whole number. ${\displaystyle {\frac {(X)(X)(X)(X)(X)}{(X)}}}$ The upper part represents five oranges. the lower represents 1. ${\displaystyle ={\frac {5}{1}}}$
So our lesson is, if there is any whole number, "1" will always go on the bottom. So if I say, "What's the number that goes on the bottom if the number is 1,2,3,4,5,6,7, or any other whole number?"
Of course, your answer should be "1", "1", "1", "1", "1", "1", "1" and always "1." I want you to remember that!
## Multiplying fractions
In general, multiplying fractions involves a simple formula:
${\displaystyle {\frac {A}{B}}*{\frac {C}{D}}={\frac {A*C}{B*D}}}$
Where A, B, C, and D represent the numbers within the fractions. If this seems intimidating, there are methods used to tackle the multiplication more easily.
1. Know what the fractions are. Write it out.
2. If you wrote it out correctly, ignore the fractions part and look at it this way.
For example:
${\displaystyle {\frac {5}{3}}*{\frac {5}{3}}={\frac {?}{?}}}$
If you wrote it correctly, this is what it should look like. Also, it would be helpful if you wrote it equal to the mysteriously equal question marks, "?/?." They're there so you can see where to write your answers.
3. Using your finger or card, cover or hide the bottom (called the denominator) so that you can concentrate on multiplying the top. Focus solely on multipling the upper portion:
${\displaystyle {\frac {5}{}}*{\frac {5}{}}={\frac {?}{}}}$
5 x 5 = ? (what is 5 times 5? what is the value (in cents) of 5 nickels? why 25 of course!) So if 25 is equals ?, then simply "erase" the "?" and write "25." Its that simple!
Again, since you know whats on top, do the same on the bottom:
${\displaystyle {\frac {}{3}}*{\frac {}{3}}={\frac {}{?}}}$
Again, what is 3 x 3? Of course, it is 9. So, again, erase the "?" and write down "9"
Once you remove the finger or card, you are presented with the answer: ${\displaystyle {\frac {5}{3}}*{\frac {5}{3}}={\frac {25}{9}}}$
### Multiplying Fractions with nice pictures
Another good way to prove that answers are actually correct is to use a diagram or tiles to prove that it is correct. So how we do that is first write out one of our fractions on one side and the other fraction on the other side. To see what i mean, here is an example:
Here is our little lesson. We will multiply 3/4 x 1/2 = ?.
1/4 1/4 1/4 1/4
So just imagine the figure to the right is one whole 4 story "chocobar" (i will remind you that one chocobar that has 4 stories still is one chcobar) You will see that each story represents a fraction. 1/4 means its needs 4 pieces to make a whole.
0000 __________________
000000/ |
OOOO Mr. Anaconda | "CHOBARS ARE YUMMY!"
\ _______ | "I hope I eat a lot
| / \ | of them!"
| / \ |
| / \ |
| / \ /
\-/ \-/
But the thing is, Mr. Anaconda only has enough energy to eat once a week. His mouth is 1/2-full with chocobar and the deepest he can eat is 3/4 a chocobar. How much of the chocobar has he eaten?
The fraction is set up like this:
${\displaystyle {\frac {3}{4}}*{\frac {1}{2}}={\frac {?}{?}}}$
Visually, it will look like this:
1/2 1/2
1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8
1/4
1/4
1/4
1/4
So, simplified from our little illustration is our little chocobar broken up into nice convenient blocks so that Mr. Anaconda can eat it.
1/2 1/2
1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8
1/4
1/4
1/4
1/4
From the image, you can count 3/8ths of the chocobar was eaten.
### Dividing fractions
As mentioned earlier, a fraction is one number divided by another. Because of textual limitations, fractions here will be represented using numbers and slashes. For example, the fraction one half will be represented as 1/2. The fraction two thirds will be represented as 2/3.
Now, on to some specifics. You may already know how to multiply fractions. Remember all you have to do is multiply the two top numbers together and put them above your two bottom numbers multiplied together. If know how to multiply fractions, dividing fractions will also be easy, and only requires one extra step.
In order to divide two fractions, you invert the second fraction and multiply. This means that you switch around the numerator (top) and denominator (bottom) of the fraction before multiplication.
Let's look at examples using our two fractions. First, let's divide the fraction 1/2 by the fraction 1/2. Since we are dividing a number by itself, we should expect to get an answer of 1.
Here is our problem.
${\displaystyle {\frac {1}{2}}\div {\frac {1}{2}}={\frac {1}{2}}*{\frac {2}{1}}={\frac {1*2}{2*1}}={\frac {2}{2}}=1}$
Now let's try another example.
${\displaystyle {\frac {1}{2}}\div {\frac {2}{3}}={\frac {1}{2}}*{\frac {3}{2}}={\frac {1*3}{2*2}}={\frac {3}{4}}}$
Following these simple steps will let you solve any fraction division problem.
There is also an alternate way to divide fractions using an equivalent fraction approach. You obtain this by multiplying the numerator and denominator so that they're equivalent, cancelling out the denominator, and creating a fraction from the two remaining numerators.
${\displaystyle {\frac {1}{2}}\div {\frac {2}{3}}={\frac {3}{6}}\div {\frac {4}{6}}={\frac {3}{4}}}$
Here is another example:
${\displaystyle {\frac {2}{5}}\div {\frac {3}{4}}={\frac {8}{20}}\div {\frac {15}{20}}={\frac {8}{15}}}$
It works because the common denominator will always be ${\displaystyle {\frac {n}{n}}=1}$
Fractions that have the same or "Common" denominator are called "Like" fractions.
1/3, 2/3
(one-third, two-thirds)
To add Like fractions together such as these:
${\displaystyle {\frac {1}{3}}+{\frac {2}{3}}=?}$
1. Add the numerators (the top numbers):
• 1 + 2 = 3 (one plus two equals three)
2. Use the common denominator (the bottom numbers):
• /3
All together it looks like this:
${\displaystyle {\frac {1}{3}}+{\frac {2}{3}}={\frac {3}{3}}}$
3. Simplify the answer as much as you can by dividing the denominator into the numerator.
• ${\displaystyle {\frac {3}{3}}=1}$
If the numerator is now larger than the denominator it might look like this:
4/5 + 3/5 = 7/5 (four-fifths plus three-fifths equals seven-fifths)
This is simplified by creating a mixed number:
• 5 goes into 7 one time, with 2 left over. Put the remainder (2) over the denominator (5).
• ${\displaystyle {\frac {7}{5}}=1{\frac {2}{5}}}$
To add or subtract fractions with different denominators, you must first convert them to equivalent fractions with common denominators.
### Multiplying to get equivalent fractions
In some cases, you may need to add 1/2 to 1/3:
${\displaystyle {\frac {1}{2}}+{\frac {1}{3}}=?}$
You will first have to convert the fractions into like fractions. The easiest means of doing so is to multiply the top and bottom of the left fraction by the bottom on the right, and the top and bottom of the right fraction by the bottom on the left:
${\displaystyle {\frac {1*3}{2*3}}+{\frac {1*2}{3*2}}=?}$
${\displaystyle {\frac {3}{6}}+{\frac {2}{6}}=?}$
The addition can now take place as expected:
${\displaystyle {\frac {3}{6}}+{\frac {2}{6}}={\frac {5}{6}}}$
Primary Mathematics ← Negative numbers Fractions Working with fractions →<|endoftext|>
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This site is devoted to mathematics and its applications. Created and run by Peter Saveliev.
## Holes
Let's review the following exercise. A $3\times 3 \times 3$ cube is made from $27$ smaller cubes and then $7$ cubes are removed: one from the middle of each face and the one in the middle of the cube. How many holes does it have now?
Below is a paraphrase of how we were trying to answer the question and, at the same time, figuring out what "hole" means.
-- Is it 1 hole?
-- Hmm, no... Don't think "rooms", think "tunnels".
-- 3? We can just drill 3 tunnels...
-- Hmm, no... Don't think how many tunnels you drill but how many "windows" you create.
-- 6? We have 6 openings to the outside...
-- Hmm, not 6... but close, it's 5. You see, one of the windows doesn't count!
-- Why not?!
-- Hmm... Try to flatten the whole thing.
-- What for?
-- Well, you should have the same number of holes after you flatten it.
-- The bottom "window" stretches out and become the outer border...
-- Right.
Exercise. Once we know how many holes we have, can you point them out?
In case this seems arbitrary, keep in mind that there is another way to arrive at this result. We can take the data about all the vertices, edges, faces, and how they are attached to each other and feed it into a machine. This machine, called “homology”, will produce the same answer, $5$, every time even if you start with a $5\times 5\times 5$ cube...
Exercise. Count the holes below. Then think about the pattern.
Another look at counting holes: how many holes does a pair of pants have? A pair. No, not three. One can see this by creating a pair of “topological pants”: take a piece of fabric, puncture two(!) holes for the legs, you then realize that there is no need for another hole because the outer edge goes around your waist.<|endoftext|>
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# 8th Class Mathematics Squares and Square Roots Square Root Methods
## Square Root Methods
Category : 8th Class
### Square Root Methods
We know that addition is inverse of subtraction, multiplication is inverse of division, and similarly square root is the inverse of square of a number. There are different methods to find the square root of a number. We can find the square root either by finding the factors or by repeated subtraction or by long division method.
By Prime Factorization Methods
In this method we find the prime factors of the given number and then pair them up. We multiply the numbers taking one from each pair and get the square root of the required number. It is useful only in case the numbers have perfect pairs of prime factors.
$\sqrt{256}=\sqrt{(2\times 2)\times (2\times 2)\times (2\times 2)\times (2\times 2)}$
There are four such order pairs within the square root of the number, therefore, the square root of the above number is $\text{2}\times \text{2}\times \text{2}\times \text{2}=\text{16}$
Repeated Subtraction Method
In this method we subtract the successive odd numbers from the given number starting from 1 till we get the result zero. The number of steps required to reduce the given number to zero will be the square root of the given number.
Take the number 64
Solution:
$64-1=63\Rightarrow 63-3=60\Rightarrow 60-5=55\Rightarrow 55-7=48$
$\Rightarrow 48-9=39\Rightarrow 39-11=28\Rightarrow 28-13=15\Rightarrow 15-15=0$
There are eight steps required to reduce the number to 0. Therefore, square root of 64 is 8.
Here are some more Squares and Square Roots
$\,Square\xrightarrow[{}]{{}}$ $\xleftarrow[{}]{{}}\,Square\,Root$ 4 16 5 25 6 36
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Students will watch selected video segments from THE VIETNAM WAR and examine the impact of French colonial rule of Vietnam on both the French and the Vietnamese people. Then students will conduct a cost/benefit analysis for each country and discuss ways to find a middle ground.
- Examine the different types of colonialism experienced by the Vietnamese.
- Explore the effects of the French pacification program.
- Conduct a cost/benefit analysis of French colonialism on the Vietnamese.
- Begin by showing students the video segments from THE VIETNAM WAR. Then, discuss the following questions:
- How did colonialism, first under the Chinese and then the French, impact the Vietnamese people? What did they gain, and what did they lose?
- Under the conditions of colonialism, why was it easy for the French to feel superior to the Vietnamese, and for the Vietnamese to hate the French?
- How did the French policy of pacification set the tone for relations between the Vietnamese and the French? How did it set the standard for brutality among the combatants?
- Describe how the Vietnamese fight for independence was similar to other colonies’ struggles for independence from more powerful and domineering countries. For example, what similarities and differences do you see in Vietnam’s quest for independence from France and the American colonies’ break with Great Britain?
- Form students into small groups and have each group do a quick cost/benefit analysis of colonialism for the French and the Vietnamese. Distribute two copies of the graphic organizer handout to each group and ask them to complete an analysis for each country, France and Vietnam.
- When students have completed their handouts, debrief the class with the following questions:
- Which country benefited the most from colonialism? In what way?
- What were the political, economic, and social costs of colonialism for both countries?
- Which country seemed to “pay” more?
- What arguments could each country have made to convince the other that either the benefits were good for both France and Vietnam if Vietnam remained under French rule, or the costs were too high for the French to maintain colonial rule over Vietnam?
National Standards for History
9.2C.1 ( U.S. History Grades 5-12 ): Assess the Vietnam policy of the Kennedy, Johnson, and Nixon administrations and the shifts of public opinion about the war. [Analyze multiple causation]
9.2C.2 ( U.S. History Grades 5-12 ): Explain the composition of the American forces recruited to fight the war. [Interrogate historical data]
9.2C.3 ( U.S. History Grades 5-12 ): Evaluate how Vietnamese and Americans experienced the war and how the war continued to affect postwar politics and culture. [Appreciate historical perspectives]
9.2C.4 ( U.S. History Grades 5-12 ): Explain the provisions of the Paris Peace Accord of 1973 and evaluate the role of the Nixon administration. [Differentiate between historical facts and historical interpretations]
9.2C.5 ( U.S. History Grades 5-12 ): Analyze the constitutional issues involved in the war and explore the legacy of the Vietnam war. [Formulate a position or course of action on an issue]
College, Career, and Civic Life (C3) Framework
D2.Civ.10.9-12 ( By the end of Grade 12 ): Analyze the impact and the appropriate roles of personal interests and perspectives on the application of civic virtues, democratic principles, constitutional rights, and human rights.
Handout: Colonialism: A Cost/Benefit Analysis Graphic Organizer
Directions: After discussing the questions in the corresponding activity, fill out the chart below.
______________________________________________________ (Name of Country)<|endoftext|>
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# Applications of Slope
Slope can be interpreted in many real-world relationships, such as speed, unit cost, and rates of change.
In real-world situations, a rate of change is often represented using the word per, as in kilometers per hour, dollars per hour, or liters per minute. The first quantity indicates the dependent variable, which is often represented by the variable $y$. The second quantity indicates the independent variable, which is often represented by the variable $x$, or by the variable $t$ if it represents time. The y-intercept, which is the value of $y$ where a graph touches or crosses the $y$-axis, represents the value of the independent variable at the start of the problem, where $x=0$ or $t=0$.
Step-By-Step Example
Interpreting Slope and $y$-Intercept from a Graph
Leila is walking to the library at a constant speed. The graph represents her distance, or $d$, in kilometers from the library after a number of hours, or $t$. Identify the slope and $y$-intercept.
Step 1
The graph goes through the point $(0, 6)$, so the $y$-intercept is 6.
Step 2
The rate of change is the slope of the line. To determine the slope, choose two points on the graph.
At time $t=0$, Leila is 6 kilometers from the library. This is shown on the graph as $(0, 6)$.
After 2 hours, at $t=2$, her distance from the library is zero kilometers. This is shown on the graph as $(2, 0)$.
Step 3
Use the two chosen points on the graph to calculate the slope.
\begin{aligned}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{0-6}{2-0}\\&=\frac{-6}{2}\\&=-3\end{aligned}
Solution
The $y$-intercept is 6. This means that when Leila starts walking and zero hours have passed, she is 6 kilometers from the library.
Leila's rate of change is –3 kilometers per hour. This means she is walking at a speed of 3 kilometers per hour. The rate is negative because her distance from the library is decreasing as she gets closer.
Step-By-Step Example
Interpreting Slope and $y$-Intercept from an Equation
A 20-liter bucket is being filled with water. The equation describes the volume of water in the bucket in liters, $y$, after $x$ minutes:
$y=2x+5$
Identify the slope and $y$-intercept of the line described by the equation, and describe their meanings. At what time will the bucket be full?
Step 1
The equation is in slope-intercept form, so the $y$-intercept is the value of $b$, or 5.
Step 2
The rate of change is the slope of the equation. The equation is in slope-intercept form, so the slope is the value of $m$, or 2.
Step 3
Determine when the bucket will be full. The bucket holds 20 liters. So, substitute 20 for $y$ in the equation and solve for $x$.
\begin{aligned}y&=2x+5\\20&=2x+5\\15&=2x\\7.5&=x\end{aligned}
Solution
The $y$-intercept is 5. This means that there were 5 liters of water already in the bucket at $x=0$ minutes.
The slope represents the number of liters added to the bucket each minute, which is 2 liters per minute.
When $y=20$, the value of $x$ is 7.5. So, the bucket will be full after 7.5 minutes.<|endoftext|>
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# What two forces affect the gravitational attraction between two objects give an example?
Contents
Mass and distance are the two factors that affect the gravitational force between any two objects. The force due to gravity (Fg) between any two…
## What two forces affect the gravitational attraction between two objects?
When dealing with the force of gravity between two objects, there are only two things that are important – mass, and distance. The force of gravity depends directly upon the masses of the two objects, and inversely on the square of the distance between them.
## What is an example of gravitational attraction?
Some examples of the force of gravity include: The force that holds the gases in the sun. The force that causes a ball you throw in the air to come down again. … The force that causes the moon to revolve around the Earth.
## What force is the force of attraction between two objects?
Gravitational force -an attractive force that exists between all objects with mass; an object with mass attracts another object with mass; the magnitude of the force is directly proportional to the masses of the two objects and inversely proportional to the square of the distance between the two objects.
## What affects gravitational acceleration?
The acceleration due to gravity depends on the terms as the following: Mass of the body, Distance from the center of mass, Constant G i.e. Universal gravitational constant.
## How do you find the gravitational force between two objects?
Find out how to calculate gravitational forces
We can do this quite simply by using Newton’s equation: forcegravity = G × M × mseparation2 . Suppose: your mass, m, is 60 kilogram; the mass of your colleague, M, is 70 kg; your centre-to-centre separation, r, is 1 m; and G is 6.67 × 10 11 newton square metre kilogram2.
## What is the effect of gravitational force on an object?
Although the gravitational force the Earth exerts on the objects is different, their masses are just as different, so the effect we observe (acceleration) is the same for each. The Earth’s gravitational force accelerates objects when they fall. It constantly pulls, and the objects constantly speed up.
## How does gravitational force affect the motion of an object?
When objects fall to the ground, gravity causes them to accelerate. Acceleration is a change in velocity, and velocity, in turn, is a measure of the speed and direction of motion. Gravity causes an object to fall toward the ground at a faster and faster velocity the longer the object falls.
## What is meant by gravitational force or force of gravity )? Give its one example?
Gravitational force is the force that attracts to objects with mass. … This forces tries to pull masses together, that is why attractive in nature. For example: When ball is thrown upwards it always come downwards after a certain height because the ball is pulled or attracted by the Earth’s gravitation la force.
IMPORTANT: Your question: What visa do I need for masters in USA?
## Which statement correctly describes the factors affecting gravitational force between two objects?
Newton’s law also states that the strength of gravity between any two objects depends on two factors: the masses of the objects and the distance between them. Objects with greater mass have a stronger force of gravity between them.
## What is the relation between the gravitational force acting between any two objects with their masses and the distance between their Centres?
The Law of Universal Gravitation
The force of gravitational attraction between any two massive bodies is proportional to their masses and inversely proportional to the square of the distance between their centers.
## Which has more gravitational force?
The greater an object’s mass, the more gravitational force it exerts. So, to begin answering your question, Earth has a greater gravitational pull than the moon simply because the Earth is more massive.<|endoftext|>
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# How do you evaluate log_[3]sqrt(18) + log_[3]sqrt(24) - log_[3](12)?
Apr 9, 2016
${\log}_{3} \sqrt{18} + {\log}_{3} \sqrt{24} - {\log}_{2} \left(12\right) = \frac{1}{2}$
#### Explanation:
In general:
$\textcolor{w h i t e}{\text{XXX}} {\log}_{b} p + {\log}_{b} q = {\log}_{b} \left(p q\right)$
and
$\textcolor{w h i t e}{\text{XXX}} {\log}_{b} r - {\log}_{b} s = {\log}_{b} \left(\frac{r}{s}\right)$
So
$\textcolor{w h i t e}{\text{XXX}} {\log}_{3} \sqrt{18} + {\log}_{2} \sqrt{24} - {\log}_{3} \left(12\right)$
$\textcolor{w h i t e}{\text{XXX}} = {\log}_{3} \left(\frac{\sqrt{18} \cdot \sqrt{24}}{12}\right)$
$\textcolor{w h i t e}{\text{XXX}} = {\log}_{3} \sqrt{3}$
${\log}_{3} \sqrt{3} = k$ means ${3}^{k} = \sqrt{3}$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow k = \frac{1}{2}$
So
$\textcolor{w h i t e}{\text{XXX}} {\log}_{3} \sqrt{18} + {\log}_{2} \sqrt{24} - {\log}_{3} \left(12\right) = {\log}_{3} \sqrt{3} = \frac{1}{2}$<|endoftext|>
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People from all over the world gather nutritional products from trees including fruits, nuts, seeds, leaves, bark and even sap.
Tree products have been an important part of diets for thousands of years, from early humans gathering fruits and nuts (there is evidence of humans eating apples in the Neolithic period) to the first cultivation of important trees, such as mango (Mangifera indica) which has been grown in India for over 4,000 years.
Today, products such as apples, oranges, pistachios and brazil nuts are routinely eaten the world over and form the basis for multi-million dollar industries – the apple industry is estimated to be worth US $10 billion a year, for example.
At the local level, edible tree products are often highly valued by local communities as a core part of their diet, as an important supplement or to sustain them when food is seasonally scarce or when harvests are poor. This role as ‘emergency’ food sources is particularly important and there are several examples of entire communities surviving periods of famine by collecting food from trees.
There are thousands of tree species that provide important food products, but some good examples include:
The Brazil nut tree (Bertholletia excelsa)
This species takes a long time to reach maturity and cannot be grown in plantations. As a result all Brazil nuts are collected from wild populations. One of the most valuable forest products in international trade, the collection of the brazil nut has resulted in the species becoming threatened. It is currently classified as Vulnerable.
Maple trees (Acer sp.)
Trees from this genus produce a sugary sap that, for certain species, is eaten, used for baking or as a flavouring and sweetener. This sap – commonly known as maple syrup – can only be produced in large enough quantities from two species: the sugar maple Acer saccharum and the black maple Acer nigrum.
Holes are bored in the trunk and a single tree can produce up to 100 litres of sap in one season. Canada is the largest producer of maple syrup and in 2008 the value of all maple products sold was estimated to be CAD$212 million. You can read about threatened maple species in the Red List of Maples, published by GTC in 2009.
Baobabs (Adansonia sp.)
Baobabs are found in mainland Africa, Australia and Madagascar. Both Grandidier’s baobab (Adansonia grandidieri) and Adansonia digitata are extremely important food sources. The fruit of both species is rich in vitamins, very nutritious and is eaten raw and made into juice. The seeds can also be turned into oil and the leaves of A. digitata are eaten. Baobab fruit has been called the new ‘superfood’ on the international market after it was approved for use in smoothies by the European Union in 2008.
Apples (Malus sp.)
Apples are one of the world’s most cultivated fruit trees, with over 7,000 different varieties in existence.
Despite their great diversity, most domesticated apples can be traced back to a common ancestor, the wild apple – Malus sieversii. The wild species still exists in China, Kazakhstan, Kyrgyzstan and Tajikistan but is now, alongside a host of other fruit and nut species from Central Asia, threatened due to the loss of its forest habitat.
This species represents an important genetic storehouse of natural variation, which could include disease and pest resistance that may be important in the future.<|endoftext|>
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What is astigmatism?
Astigmatism is a structural abnormality of the eye that bends light as it passes through the eye in such a way that makes it difficult to focus properly, causing blurred vision. It is due to an irregularity in the shape of the cornea, which in mostly spherical in a normal eye, but more egg shaped in an eye with astigmatism.
The causes of astigmatism
There is some evidence that suggests that regular astigmatism may be hereditary in nature, passing from parents to their children, but there's not enough information available to know for certain. Irregular astigmatism is often caused by damage to the eye, resulting in corneal scarring, or scattering within the crystalline lens inside the eye.
In some cases, astigmatism can be present at the time of birth, although detecting vision problems in very young children is very difficult. In other cases, astigmatism can develop later in life. It's also possible for the condition to either improve or worsen as time passes. Everyone's own experience will be different.
Symptoms of Astigmatism
For people with very minor levels of astigmatism, there may not be any obvious symptoms at all, aside from requiring a slight change in prescription to match the shape of the eye. Otherwise, there may not be any noticeable issues. For others, a severe case of astigmatism may cause blurred vision and squinting, which can then also lead to headache and eye fatigue.
If you're experiencing any of these symptoms, it's recommended that you see you eye doctor as soon as possible.
Types of Astigmatism
The measurements of the shape and curvature of the eye, using an instrument called a Keratometer, are used to determine the type of and severity of the astigmatism. In addition to regular astigmatism, which develops naturally, and irregular astigmatism, which is the result of damage to the eye, further categorization of astigmatism includes:
- Simple Myopic Astigmatism – The shape of the eye causes light to refract onto two separate focal points, one on the retina, and the other just in front of it.
- Simple Hyperopic Astigmatism – The inverse of simple myopic astigmatism. Also causing two focal points of light, with one being on the retina, but the other behind it.
- Compound Myopic Astigmatism – Similar to simple myopic astigmatism, but with both focal points in front of the retina, but distinct from one another.
- Compound Hyperopic Astigmatism – Following the same pattern, both focal points are found behind the retina, separate from one another.
- Mixed Astigmatism – A hybrid of the other forms of astigmatism, where one focal point exists in front of the retina, and the other behind it.
How is Astigmatism Diagnosed?
A standard comprehensive eye exam is going to be enough to detect astigmatism. The eye exam will consist of several different tests, each designed to measure a different aspect of vision and eye health. The primary tests performed during the exam are:
- The Visual Acuity Test – Probably the most recognizable eye test in the world, where lines of random letters that decrease in size are viewed from a fixed distance of 20 feet (or 6 metres). A person with normal “20/20” vision (or 6/6 in the UK) can see exactly what their expected to see from 20 feet away. A person with 20/40 vision (6/12 in the UK), for example, can see letters clearly at 20 feet that a person with normal vision could see from 40 feet away.
- The Refraction Test – A vision testing device called a phoropter is used to determine the exact numerical value of someone’s prescription. The phoropter, which looks like a large mask with many dials and lenses, is put in front of the patient's face. While looking at the same lettered chart, different lenses are changed out with one another to find which one provides a clearer image. Through a process of elimination, the best lenses are determined, and the results used to write a prescription for eyeglass lenses.
- The Keratometry Test – The tool most relevant to diagnosing the cause of astigmatism in an individual, is a Keratometer. This is a device that shines a light across the surface of the cornea, to measure the exact curvature of the surface of the eye. This is used not only to detect astigmatism, but also for the proper fitting of contact lenses. In some cases, a more complex variation of this test called corneal topography may be used for a more detailed mapping of the cornea. The keratometer can give an insight as to the location of the astigmatism - often it is corneal in nature, but sometimes it can be caused by the lens inside the eye (lenticular astigmatism). This is very important information to determine the best means of correcting it with contact lenses.
How to treat astigmatism
The recommended treatment will depend largely on the type and severity of astigmatism in question.
Regular astigmatism can be corrected with any sort of standard corrective lens - contacts or spectacles - while irregular astigmatism can only be rectified with contact lenses. Laser refractive surgery, such as LASIK or PRK, can be beneficial, but can cost quite a bit more than the alternatives, and require a complicated surgical process.
The type of contact lenses used to correct astigmatism, known as toric lenses, can come in both soft and rigid gas permeable varieties. Which lens is the right one for you will depend on both the strength of the prescription, the nature of the astigmatism, as well as personal preference.
Small amounts of astigmatism (< 0.75D) can often be ignored with soft lenses - most people will experience vision only slightly less clear than their glasses.
However if you want any astigmatism corrected up to 3.00D there are now soft lenses capable of that - even daily ones. Focus Dailies Aquacomfort Plus Toric can correct up to 2.00D of astigmatism easily and 1 Day Acuvue Moist for Astigmatrism can correct up to 3.00D and may be just what you need to sharpen up your vision.
Correct your astigmatism with contact lenses
If you are concerned about problems with your vision, the best thing to do is get an eye examination. Your eye care professional will carry out various tests to determine the nature and severity of your condition. They will ask you questions about your lifestyle and working habits to determine the best way to get you seeing clearly again.
If you have been diagnosed with astigmatism, a common solution is the use of corrective contact lenses. At Contactlenses.co.uk, we believe that visual problems should not get in the way of your normal lifestyle. That's why we've made it simple, fast and affordable to get contact lenses to correct your vision.
We can also help you to keep up with the latest contact lenses news, enabling you to find out about the most recent research into vision correction and treatment options.
Author: John Dreyer Optometrist Bsc(Hons), MCOPTOM, DipCLP
Created: 24 Apr 2015, Last modified: 18 Apr 2019<|endoftext|>
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The First Map of North America
Exceedingly rare map of North America first published by Paolo Forlani in 1565. This is an example of the second state of the map, published in 1566. It is the first separately-published map of North America and only the second to include the Strait of Anian and is thus a landmark in the history of cartography.
In the early 1560s, Forlani also published a map of South America and the West Indies, La descrittione de tutto il Peru , the only map of South America ever completed by the Lafreri school. With this North America map, Forlani completed his coverage of the New World. The map stretches from Greenland down the coast of Canada and the Atlantic Seaboard to the West Indies, including a corner of South America, and from the coast of China in the west to the Azores and Cape Verde in the east.
Forlani's scarce and finely engraved map of North America is one of the most significant early maps of America. It is the earliest printed map devoted solely to North America, the first to portray that landmass as a separate continent and the second to show the so-called Strait of Anian separating America from Asia.
The geography of Forlani’s North America
The geographic features Forlani chose to include reveal how much, and little, was known about the continent in the late-sixteenth century. Certain places are familiar to the modern eye. Labrador is included, near Terra de Baccalos, a reference to the cod fisheries that made the area one of the first to see European ships. La Florida covers the entire southeast, with a nascent peninsula jutting into the Gulf of Mexico. California is not labeled, but it is readily recognizable.
Another familiar name is Apalchen which, although nearly in the center of the continent, denotes the Appalachian Mountains. This reveals a source for Forlani, for this feature first appeared on maps with the Gutiérrez-Cock map of 1562. Another source is revealed in the use of the city name Ochelaga, or Montreal. It is placed on a river other than the St. Lawrence, which is itself confused with a Gamas River. This misplacement of Montreal appeared in Ramusio’s important travel collection.
Other representations appear strange to the modern reader, such as the placement of a squat Japan in the middle of a horizontally-shrunk Pacific, or Mare del Sur. Two unfamiliar mountain ranges cut east to west, while there are no lakes beyond an un-named body of water in Canada and a large lake in the center of Mexico, most likely meant to be Lake Texcoco.
Canada Pro and La Nova Franza mark the presence of French colonists, a process that began in 1534. In the northeast is Larcadia, a reference to Acadia, which was not permanently settled by the French until early in the seventeenth century. Just to the west is Terra de Norumbega, which first appeared as Oranbega on Giovanni da Verrazzano’s 1529 map. Soon after this map was made, the place would gain a mythic reputation based on the stories of David Ingram, a marooned English sailor. He described silver thrones and vast cities, but his story was doubted by both Richard Hakluyt and Samuel Purchas.
Farther west still is Quivira Pro. Quivira refers to the Seven Cities of Gold sought by the Spanish explorer Francisco Vasquez de Coronado in 1541. In 1539, Coronado wandered over what today is Arizona and New Mexico, eventually heading to what is now Kansas to find the supposedly rich city of Quivira. Although he never found the cities or the gold, the name stuck on maps of southwest North America, wandering from east to west. Here it is used to describe the entire western half of the North America.
Strait of Anian
One of the most important features of the map is the Streto de Anian, which is the name of the thin channel that separates Parte di Asia and North America. This is only the second map to use the place name. Forlani based his map on Giacomo Gastaldi's 1561 and 1546 world maps. Gastaldi had been the first to formulate the concept of the Strait of Anian and included it in his nine-sheet world map of 1561 that survives in only one example, at the British Library.
Anian derives from Ania, a Chinese province on a large gulf mentioned in Marco Polo’s travels (ch. 5, book 3). The gulf Polo described was actually the Gulf of Tonkin, but the province’s description was transposed from Vietnam to the northwest coast of North America. After Gastaldi and Forlani included the feature, it became common as a shorthand for a passage to China, i.e. a Northwest Passage. It appeared on maps until the mid-eighteenth century.
States and rarity
Until the late-twentieth century, this map was attributed to Venetian publisher Bolognino Zaltieri, whose name and imprint appear on the second state, like this example, and was published in 1566. As David Woodward has demonstrated, however, authorship should be ascribed to Forlani, who sold some copperplates—including, presumably, the one used to print this map—to Zaltieri sometime around late 1565 or early 1566.
The present example is a second state, as evidenced by the publisher’s imprint in the upper left corner. Both states are seldom found in institutional collections and are rare on the market. This rarity, and the importance of the map as the first stand-alone map of the North American continent, mark it as an extremely important document for the history of cartography and for the histories of exploration and of the Americas.
Paolo Forlani (fl. ca. 1560-1571) was a prolific map engraver based in Venice. All that is known of his life are his surviving maps and prints, of which there are almost 100 (185 with later states included in the total). He also produced a globe and two town books. It is likely he came from Verona and that he died in Venice in the mid-1570s, possibly of the plague.<|endoftext|>
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Children count reliably with numbers from 1 to 20, place them in
order and say which number is one more or one less than a given number.
Using quantities and objects, they add and subtract two single-digit numbers
and count on or back to find the answer. They solve problems, including
doubling, halving and sharing.
The latest Numbers teaching resources
Select Your Location
Choosing your country and state helps us to provide you with the most relevant teaching resources for your students.<|endoftext|>
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# Three taps A, B, C fill up a tank independently in 10 hr, 20 hr, 30 hr respectively. Initially, the tank is empty and exactly one pair of taps is open during each hour and every pair of taps is open at least for an hour. What is the minimum number of hours required to fill the tank?(a) 8(b) 9(c) 10(d) 11
Last updated date: 12th Sep 2024
Total views: 468k
Views today: 7.68k
Answer
Verified
468k+ views
Hint: Find the capacity of the tank filled by each pair of taps AB, BC and CA in 1 hour and then use the pair of taps which fills the highest capacity of tank in 1 hour to fill the remaining tank.
Given that three taps A, B, C fill up the tank independently in 10 hr, 20 hr, 30 hr respectively. We have to find the minimum number of hours required to fill the tank if each pair of taps is open for at least an hour.
Since 60 is LCM of 10, 20 and 30, therefore let us take the total capacity of the tank as L = 60 liters.
So, we are given that A can fill the full tank in 10 hours.
That is, in 10 hours, A can fill 60 liters.
So, in 1 hour, A can fill
$\dfrac{60}{10}=6\text{ liters}$
Similarly, in 20 hours B can fill 60 liters.
So, in 1 hour, B can fill
$\dfrac{60}{20}=3\text{ liters}$
Similarly, in 30 hours C can fill 60 liters.
So, in 1 hour, C can fill
$\dfrac{60}{30}=2\text{ liters}$
Therefore, in 1 hour, a pair of A and B can fill = 6 + 3 = 9 liters.
Similarly, in 1 hour, pair of B and C can fill = 3 + 2 = 5 liters
Similarly, in 1 hour, pair of C and A can fill = 6 + 2 = 8 liters
It is given in the question that each pair is open for at least an hour. Therefore, if we open AB for first hour, BC for second hour and CA for third hour,
Then in total 3 hours, they can fill = 9 + 5 + 8 = 22 liters.
Since we know that total capacity is 60 liters, we still have 60 – 22 = 38 liters to be filled.
It is given in question that the tank must be filled in minimum hours. We have already found that the pair AB fills the highest capacity of the tank in 1 hour. This capacity is 9 liters. Therefore, we will use the pair AB to fill the remaining part of the tank. This remaining part of the tank, as we have found already is 38 liters.
Let us consider that AB takes x hours to fill the remaining tank,
Then, we get 9x = 38 liters. Or,
$x=\dfrac{38}{9}\text{ hours}$
$x=4.22\text{ hours}$
Therefore, total time taken to fill the tank
\begin{align} & =3+4.22 \\ & =7.22\text{ hours} \\ \end{align}
As the tank gets filled in the 8th hour, therefore option (a) is correct.
Note: In these types of questions, students take the total capacity of the tank to be x and get confused with fractions. So it is advisable to take the total capacity of the tank to be a number preferably the one which is the LCM of the given hours so that, easily it gets cancelled. Also, students must remember that each pair of taps must get opened for at least 1 hour.<|endoftext|>
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# Some puzzles from Cav
A couple of puzzles that came my way via @srcav today:
Cav’s solutions to this one are here; mine are below the line further down.
And to this one, here
Have a go yourself before you read on!
I’ve mentioned before about @solvemymaths and @cshearer41 puzzles: they usually consist of two puzzles. First, finding an answer; secondly, finding a method that makes you go ‘oo!’.
### Area 48
• The white triangle at top-left is $\frac{1}{4}$ of the smaller square.
• The white triangle at top-right is $\frac{1}{16}$ of the smaller square (similar triangles)
• So the pink shape is $\frac{11}{16}$ of the smaller square.
• The larger square has side length $\frac{\sqrt{5}}{2}$ as long as the smaller square (via Pythagoras)
• So its area is $\frac{5}{4}$ as large.
• Therefore, the pink shape is $\frac{11}{16} \div \frac{5}{4} = \frac{11}{20}$ of the larger square.
### Angle puzzle
My first thought, faced with this, was “I really can’t be bothered chasing all those angles. Surely there’s some symmetry to exploit?”
Surely there is.
• All of the points on the outside of the squares lie on a circle (by symmetry). This circle has the same centre as the one going through the points of the pentagon.
• The red lines divide the angle at the centre of the circles into fifths - so the angle between the two outer red lines is thus $\frac{4}{5}\pi$.
• The angle subtended at the circumference yadda yadda - the angle between the two black lines is $\frac{2}{5}\pi$.
• So the original shaded angle is $\frac{8}{5}\pi$ (or 144 degrees, if the Ninja isn’t looking).
Nice puzzles!
## Colin
Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.
#### Share
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0
Q:
What is the probabiltiy of getting a mutliple of 2 or 5 when an unbiased cubic die is thrown?
A) 1/3 B) 1/2 C) 2/3 D) 1/6
Explanation:
Total numbers in die=6
P(multiple of 2)=1/2
P(multiple of 5)=1/6
P(multiple of 2 or 5)=2/3
Q:
When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 12?
A) 35/36 B) 17/36 C) 15/36 D) 1/36
Explanation:
When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36
Required, the sum of the two numbers that turn up is less than 12
That can be done as n(E)
= { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5) }
= 35
Hence, required probability = n(E)/n(S) = 35/36.
3 638
Q:
In a purse there are 30 coins, twenty one-rupee and remaining 50-paise coins. Eleven coins are picked simultaneously at random and are placed in a box. If a coin is now picked from the box, find the probability of it being a rupee coin?
A) 4/7 B) 2/3 C) 1/2 D) 5/6
Explanation:
Total coins 30
In that,
1 rupee coins 20
50 paise coins 10
Probability of total 1 rupee coins = 20C11
Probability that 11 coins are picked = 30C11
Required probability of a coin now picked from the box is 1 rupee = 20C11/30C11 = 2/3.
7 1044
Q:
In a box, there are 9 blue, 6 white and some black stones. A stone is randomly selected and the probability that the stone is black is ¼. Find the total number of stones in the box?
A) 15 B) 18 C) 20 D) 24
Explanation:
We know that, Total probability = 1
Given probability of black stones = 1/4
=> Probability of blue and white stones = 1 - 1/4 = 3/4
But, given blue + white stones = 9 + 6 = 15
Hence,
3/4 ----- 15
1 ----- ?
=> 15 x 4/3 = 20.
Hence, total number of stones in the box = 20.
10 1079
Q:
What is the probability of an impossible event?
A) 0 B) -1 C) 0.1 D) 1
Explanation:
The probability of an impossible event is 0.
The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.
The probability of a certain event is 1.
9 1499
Q:
In a box, there are four marbles of white color and five marbles of black color. Two marbles are chosen randomly. What is the probability that both are of the same color?
A) 2/9 B) 5/9 C) 4/9 D) 0
Explanation:
Number of white marbles = 4
Number of Black marbles = 5
Total number of marbles = 9
Number of ways, two marbles picked randomly = 9C2
Now, the required probability of picked marbles are to be of same color = 4C2/9C2 + 5C2/9C2
= 1/6 + 5/18
= 4/9.
9 1824
Q:
A bag contains 3 red balls, 5 yellow balls and 7 pink balls. If one ball is drawn at random from the bag, what is the probability that it is either pink or red?
A) 2/3 B) 1/8 C) 3/8 D) 3/4
Explanation:
Given number of balls = 3 + 5 + 7 = 15
One ball is drawn randomly = 15C1
probability that it is either pink or red =
14 1677
Q:
Two letters are randomly chosen from the word TIME. Find the probability that the letters are T and M?
A) 1/4 B) 1/6 C) 1/8 D) 4
Explanation:
Required probability is given by P(E) =
19 2408
Q:
14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together.
A) 11/379 B) 21/628 C) 24/625 D) 26/247
Explanation:
Total no of ways = (14 – 1)! = 13!
Number of favorable ways = (12 – 1)! = 11!
So, required probability = $11!×3!13!$ = $39916800×66227020800$ = $24625$<|endoftext|>
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# Mathematics.
## Presentation on theme: "Mathematics."— Presentation transcript:
Mathematics
Trigonometric ratios and Identities
Session 1 Trigonometric ratios and Identities
Topics Measurement of Angles Definition and Domain and Range
of Trigonometric Function Compound Angles Transformation of Angles
Measurement of Angles J001 B O A
Angle is considered as the figure obtained by rotating initial ray about its end point.
Measure and Sign of an Angle
J001 Measure of an Angle :- Amount of rotation from initial side to terminal side. Sign of an Angle :- B Rotation anticlockwise – Angle positive O A B’ Rotation clockwise – Angle negative
Right Angle O Y X J001 Revolving ray describes one – quarter of a circle then we say that measure of angle is right angle Angle < Right angle Acute Angle Angle > Right angle Obtuse Angle
Quadrants J001 Y O X’ X X’OX – x - axis Y’OY – y - axis Y’ II Quadrant
III Quadrant IV Quadrant X’OX – x - axis Y’OY – y - axis
System of Measurement of Angle
J001 Measurement of Angle Sexagesimal System or British System Centesimal System or French System Circular System or Radian Measure
System of Measurement of Angles
J001 Sexagesimal System (British System) 1 right angle = 90 degrees (=90o) 1 degree = 60 minutes (=60’) 1 minute = 60 seconds (=60”) Is 1 minute of sexagesimal 1 minute of centesimal ? = Centesimal System (French System) 1 right angle = 100 grades (=100g) 1 grade = 100 minutes (=100’) 1 minute = 100 Seconds (=100”) NO
System of Measurement of Angle
J001 Circular System O r A B 1c If OA = OB = arc AB
System of Measurement of Angle
J001 Circular System O A C B 1c
J002 OR
Illustrative Problem J002
Illustrative Problem J002 Solution
Relation Between Angle Subtended by an Arc At The Center of Circle
B J002 Arc AC = r and Arc ACB =
Illustrative Problem J002 Solution B Arc AB = 88 m and AP = ? 72o P A
A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope tight and describes 88 meters when it has traced out 72o at the center. Find the length of rope. [ Take = 22/7 approx.]. Solution P A B 72o Arc AB = 88 m and AP = ?
Definition of Trigonometric Ratios
x O Y X P (x,y) M y r J003
Some Basic Identities
Illustrative Problem J003 Solution
Signs of Trigonometric Function In All Quadrants
J004 In First Quadrant x O Y X P (x,y) M y r Here x >0, y>0, >0
Signs of Trigonometric Function In All Quadrants
J004 In Second Quadrant X X’ Y Y’ P (x,y) x y r Here x <0, y>0, >0
Signs of Trigonometric Function In All Quadrants
J004 X’ X P (x,y) O Y’ Y M In Third Quadrant Here x <0, y<0, >0
Signs of Trigonometric Function In All Quadrants
J004 In Fourth Quadrant X O P (x,y) Y’ M Here x >0, y<0, >0
Signs of Trigonometric Function In All Quadrants
J004 X Y’ X’ Y O II Quadrant sin & cosec are Positive I Quadrant All Positive III Quadrant tan & cot are Positive IV Quadrant cos & sec are Positive ASTC :- All Sin Tan Cos
Illustrative Problem J004 lies in second If cot =
quadrant, find the values of other five trigonometric function Solution Method : 1
Illustrative Problem J004 Solution Y P (-12,5) r 5 -12 X X’ Y’
lies in second If cot = quadrant, find the values of other five trigonometric function Solution Method : 2 Y X X’ Y’ P (-12,5) -12 5 r Here x = -12, y = 5 and r = 13
Domain and Range of Trigonometric Function
J005 Functions Domain Range sin R [-1,1] cos R [-1,1] tan R cot R Faculty should explain domain of any one of trigonometric function in the class with the help off funda book. sec R-(-1,1) cosec R-(-1,1)
Illustrative problem Prove that J005
is possible for real values of x and y only when x=y J005 Solution But for real values of x and y is not less than zero
Trigonometric Function For Allied Angles
If angle is multiple of 900 then sin cos;tan cot; sec cosec If angle is multiple of 1800 then sin sin;cos cos; tan tan etc. Trig. ratio - 90o- 90o+ 180o- 180o+ 360o- 360o+ sin - sin cos cos sin - sin - cos tan - tan cot - cot -tan
Trigonometric Function For Allied Angles
Trig. ratio - 90o- 90o+ 180o- 180o+ 360o- 360o+ cot - cot tan -tan -cot sec cosec - cosec - sec cosec - cosec sec -cosec
Periodicity of Trigonometric Function
J005 If f(x+T) = f(x) x,then T is called period of f(x) if T is the smallest possible positive number Periodicity : After certain value of x the functional values repeats itself Period of basic trigonometric functions sin (360o+) = sin period of sin is 360o or 2 cos (360o+) = cos period of cos is 360o or 2 tan (180o+) = tan period of tan is 180o or
Trigonometric Ratio of Compound Angle
J006 Angles of the form of A+B, A-B, A+B+C, A-B+C etc. are called compound angles (I) The Addition Formula sin (A+B) = sinAcosB + cosAsinB cos (A+B) = cosAcosB - sinAsinB
Trigonometric Ratio of Compound Angle
J006 We get Proved Ask the students to attempt cot(A+B) themselves
Illustrative problem Find the value of (i) sin 75o (ii) tan 105o
Solution (i) Sin 75o = sin (45o + 30o) = sin 45o cos 30o + cos 45o sin 30o
Trigonometric Ratio of Compound Angle
(I) The Difference Formula sin (A - B) = sinAcosB - cosAsinB cos (A - B) = cosAcosB + sinAsinB Note :- by replacing B to -B in addition formula we get difference formula
Illustrative problem If tan (+) = a and tan ( - ) = b Prove that
Solution
Some Important Deductions
sin (A+B) sin (A-B) = sin2A - sin2B = cos2B - cos2A cos (A+B) cos (A-B) = cos2A - sin2B = cos2B - sin2A Ask the students to find the expressions for Sin ( A+B+C) and Cos ( A+B+C)
To Express acos + bsin in the form kcos or sin
Similarly we get acos + bsin = sin
Illustrative problem Find the maximum and minimum values of 7cos + 24sin Solution 7cos +24sin
Illustrative problem Solution Max. value =25, Min. value = -25 Ans.
Find the maximum and minimum value of 7cos + 24sin Solution Max. value =25, Min. value = Ans. Ask them to try using the Sin (A+B ) form
Transformation Formulae
Transformation of product into sum and difference 2 sinAcosB = sin(A+B) + sin(A - B) 2 cosAsinB = sin(A+B) - sin(A - B) 2 cosAcosB = cos(A+B) + cos(A - B) Proof :- R.H.S = cos(A+B) + cos(A - B) = cosAcosB - sinAsinB+cosAcosB+sinAsinB = 2cosAcosB =L.H.S Ask student to do 2 sinAsinB = cos(A - B) - cos(A+B) 2 sinAsinB = cos(A - B) - cos(A+B) [Note]
Transformation Formulae
Transformation of sums or difference into products By putting A+B = C and A-B = D in the previous formula we get this result Note or
Illustrative problem Prove that Solution Proved
Class Exercise - 1 If the angular diameter of the moon be 30´, how far from the eye can a coin of diameter 2.2 cm be kept to hide the moon? (Take p = approximately)
Class Exercise - 1 Solution :-
If the angular diameter of the moon be 30´, how far from the eye can a coin of diameter 2.2 cm be kept to hide the moon? (Take p = approximately) Solution :- Let the coin is kept at a distance r from the eye to hide the moon completely. Let AB = Diameter of the coin. Then arc AB = Diameter AB = 2.2 cm
Class Exercise - 2 Prove that tan3A tan2A tanA = tan3A – tan2A – tanA.
Solution :- We have 3A = 2A + A Þ tan3A = tan(2A + A) Þ tan3A = Þ tan3A – tan3A tan2A tanA = tan2A + tanA Þ tan3A – tan2A – tanA = tan3A tan2A tanA (Proved)
Class Exercise - 3 If sina = sinb and cosa = cosb, then (b) (a) (d)
Solution :- and
Class Exercise - 4 Prove that Solution:-
LHS = sin20° sin40° sin60° sin80°
Class Exercise - 4 Prove that Solution:- Proved.
Class Exercise - 5 Prove that Solution :-
Class Exercise - 5 Prove that Solution :-
Class Exercise - 6 The maximum value of 3 cosx + 4 sinx + 5 is
(d) None of these (a) 5 (b) 9 (c) 7 Solution :-
Class Exercise - 6 Solution :-
The maximum value of 3 cosx + 4 sinx + 5 is Solution :- \ Maximum value of the given expression = 10.
Class Exercise - 7 If a and b are the solutions of a
cosq + b sinq = c, then show that Solution :- We have … (i) are roots of equatoin (i),
Class Exercise - 7 Solution :- sin and sin are roots of equ. (ii).
If a and b are the solutions of acosq + bsinq = c, then show that Solution :- sin and sin are roots of equ. (ii). Hence Again from (i),
Class Exercise - 7 Solution :- (iv)
If a and b are the solutions of acosq + bsinq = c, then show that Solution :- (iv) and be the roots of equation (i), cos and cos are the roots of equation (iv). Now
Class Exercise - 8 If a seca – c tana = d and b seca + d tana = c, then (a) a2 + b2 = c2 + d2 + cd (b) (c) a2 + b2 = c2 + d2 (d) ab = cd
Class Exercise - 8 If a seca – c tana = d and b seca + d tana = c, then Solution :- ….. (I) Again Squaring and adding (i) and (ii), we get ….. ii
Class Exercise -9 The value of (a) 2 sinA (c) 2 cosA (b) (d)
Class Exercise -9 The value of Solution :-
Class Exercise -10 If , , and b lie between0 and , then value
of tan2a is (a) 1 (c) 0 (b) (d) and between 0 and , Solution :- Consequently, cos(a - b) and sin(a + b) are positive.
Class Exercise -10 Solution :- If , ,
and b lie between0 and , then value of tan2a is Solution :-<|endoftext|>
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English
# How to Calculate Conditional Probability
An easy tutorial to calculate the probability of intersection of more than two events(Conditional Probability) with formula and examples.
## Calculating Conditional Probability
Conditional Probability is a probability of one event occurring based on the another event (s) that has already occurred. To put it simple, we have already given the probability of event A and we need to find the probability of event B based on the event A. So the probability obtained is the conditional probability of B given A. It can be written as P(B/A). Learn here how to calculate conditional probability with simple example problem.
Calculating conditional probability is essential if we are calculating the probability of intersection of more than two elements.
Conditional Probability Formula:
P (A and B) = P (A) x p (B/A)
So,
P(B/A) = P (A and B) / P(A)
Conditional Probability Example:
Consider that a interviewer is conducting the online aptitude and software test for a set of candidates. 45 % of people passed both the test and 65% passed the aptitude test. Find the percentage of candidates who passed the first test, also passed the second one?
Step 1:
Consider,
Aptitude test (A) = 65 %
Aptitude and Software test (A and B) = 45%
P(B|A) = ?
Step 2: Substitute the given values in the formula.
P(B/A) = P (A and B) / P(A)
= 45 / 65 = 0.6923 = 69 %
P (B/A) = 69%
The percentage of candidates who passed the first test, also passed the second one is 69%<|endoftext|>
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Greenland is melting faster than previously believed, with its ice retracting four times more quickly than in 2003.
Researchers concerned about rising sea levels have studied melting glaciers in Greenland for many years, specifically from the southeast and northwest regions of the country.
Glaciers here are responsible for pushing enormous iceberg-sized chunks of ice into the Atlantic Ocean, which eventually float away and melt.
However, a new study into Greenland's southwest region - which is mostly devoid of large glaciers - found that the rate of ice loss was even higher than previously believed.
Here, instead of glaciers pushing icebergs into the sea, ice is actually melting inland and entering the ocean as melt-water.
"Whatever this was, it couldn't be explained by glaciers, because there aren't many there," said Professor Michael Bevis, lead author of the paper. "It had to be the surface mass - the ice was melting inland from the coastline."
The work from researchers at Ohio State University was published in the Proceedings of the National Academy of Sciences.
"We knew we had one big problem with increasing rates of ice discharge by some large outlet glaciers," said Professor Bevis.
"But now we recognise a second serious problem: increasingly, large amounts of ice mass are going to leave as melt-water, as rivers that flow into the sea."
The researchers warn that their findings could have serious implications for coastal cities in the US, as well as island nations such as the UK which are vulnerable to rising sea levels.
Professor Bevis added: "The only thing we can do is adapt and mitigate further global warming - it's too late for there to be no effect.
"This is going to cause additional sea level rise. We are watching the ice sheet hit a tipping point."
Scientists already understood Greenland to be one of the biggest contributors to sea-level rise due to its glaciers and ice sheets which cover 80% of its surface.
According to Professor Bevis, the new findings indicate that scientists may not have been watching the right parts of the island to detect where potential sea-level rise would come from.
There are GPS systems in place which monitor the movements of the ice sheet around most of Greenland's perimeter, but this has not had good coverage in the southwest region of the country.
The researchers say it is necessary to make the network there more dense, given these new findings, to monitor melt-water rates.
"We're going to see faster and faster sea level rise for the foreseeable future," Professor Bevis said.
"Once you hit that tipping point, the only question is: how severe does it get?"<|endoftext|>
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Divisibility rules
This lesson presents divisibility rules for the numbers 2, 3, 4, 5, 6, 7, 8, 9, and 10.
Divisibility rules of whole numbers are very useful because they help us to quickly determine if a number can be divided by 2, 3, 4, 5, 9, and 10 without doing long division.
Divisibility means that you are able to divide a number evenly
For instance, 8 can be divided evenly by 4 because 8/4 = 2. However, 8 cannot be divided evenly by 3.
To illustrate the concept, let's say you have a cake and your cake has 8 slices, you can share that cake between you and 3 more people evenly. Each person will get 2 slices.
However,if you are trying to share those 8 slices between you and 2 more people, there is no way you can do this evenly. One person will end up with less cake
In general, a whole number x divides another whole number y if and only if you can find a whole number n such that
x times n = y
For instance, 12 can be divided by 3 because 3 times 4 = 12
When the numbers are large, use the following divisibility rules:
Rule #1: divisibility by 2
A number is divisible by 2 if it's last digit is 0,2,4,6,or 8.
For instance, 8596742 is divisible by 2 because the las t digit is 2.
Rule # 2: divisibility by 3:
A number is divisible by 3 if the sum of its digits is divisible by 3
For instance, 3141 is divisible by 3 because 3+1+4+1 = 9 and 9 is divisible by 3.
Rule # 3: divisibility by 4
A number is divisible by 4 if the number represented by its last two digits is divisible by 4.
For instance, 8920 is divisible by 4 because 20 is divisible by 4.
Rule #4: divisibility by 5
A number is divisible by 5 if its last digit is 0 ot 5.
For instance, 9564655 is divisible by 5 because the last digit is 5.
Rule # 5: divisibility by 6
A number is divisible by 6 if it is divisible by 2 and 3. Be careful! it is not one or the other. The number must be divisible by both 2 and 3 before you can conclude that it is divisible by 6.
Rule # 6: divisibility by 7
To ckeck divisibility rules for 7, study carefully the following two examples:
Is 348 divisible by 7?
Remove the last digit, which is 8. The number becomes 34. Then, Double 8 to get 16 and subtract 16 from 34.
34 − 16 = 18 and 18 is not divisible by 7. Therefore, 348 is not divisible by 7
Is 37961 divisible by 7?
Remove the last digit, which is 1. The number becomes 3796. Then, Double 1 to get 2 and subtract 2 from 3796.
3796 − 2 = 3794, so still too big? Thus repeat the process.
Remove the last digit, which is 4. The number becomes 379. Then, Double 4 to get 8 and subtract 8 from 379.
379 − 8 = 371, so still too big? Thus repeat the process.
Remove the last digit, which is 1. The number becomes 37. Then, Double 1 to get 2 and subtract 2 from 37.
37 − 2 = 35 and 35 is divisible by 7. Therefore, 37961 is divisible by 7.
Rule #7:divisibility by 8
A number is divisible by 8 if the number represented by its last three digits is divisible by 8.
For instance, 587320 is divisible by 8 because 320 is divisible by 8.
Rule #8: divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9.
For instance, 3141 is divisible by 9 because the sum of its digits is divisible by 9.
Rule # 9: divisibility by 10
A number is divisible by 10 if its last digits is 0
For instance, 522480 is divisible by 10 because the last digit is 0.
Divisibility rules quiz
Take the quiz below to see how well you understand the lesson on this page.
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# How to Calculate the Circumference of a Circle
Author Info
Updated: September 19, 2019
Whether you're doing craft work, putting fencing around your hot tub, or just solving a math problem for school, knowing how to find the circumference of a circle will come in handy in a variety of circle-related problems.
### Method 1 of 2: Using the Diameter
1. 1
Write down the formula for finding the circumference of a circle using the diameter. The formula is simply this: C = πd. In this equation, "C" represents the circumference of the circle, and "d" represents its diameter. That is to say, you can find the circumference of a circle just by multiplying the diameter by pi. Plugging π into your calculator will give you its numerical value, which is a closer approximation of 3.14 or 22/7.[1]
2. 2
Plug the given value of the diameter into the formula and solve.[2]
• Example problem: You have a circle tub with a diameter of 8 feet, and you want to build a white fence that creates a 6-foot wide space around the tub. To find the circumference of the fence that has to be created, you should first find the diameter of the tub and the fence which will be 8 feet + 6 feet + 6 feet, which will account for the entire diameter of the tub and fence. The diameter is 8 + 6 + 6, or 20 feet. Now plug it into the formula, plug π into your calculator for its numerical value, and solve for the circumference:
• C = πd
• C = π x 20
• C = 62.8 feet
### Method 2 of 2: Using the Radius
1. 1
Write down the formula for finding the circumference of a circle using the radius. The radius is half as long as the diameter, so the diameter can be thought of as 2r. Keeping this in mind, you can write down the formula for finding the circumference of a circle given the radius: C = 2πr. In this formula, "r" represents the radius of the circle. Again, you can plug π into your calculator to get its numeral value, which is a closer approximation of 3.14.[3]
2. 2
Plug the given radius into the equation and solve. For this example, let's say you're cutting out a decorative strip of paper to wrap around the edge of a pie you've just made. The radius of the pie is 5 inches. To find the circumference that you need, just plug the radius into the equation:[4]
• C = 2πr
• C = 2π x 5
• C = 10π
• C = 31.4 inches
## Community Q&A
Search
• Question
What is the perimeter of a circle?
wikiHow Staff Editor
The perimeter of a circle is the same as its circumference, the distance around it. The term "perimeter" refers to the distance around any closed shape, and “circumference” applies specifically to a circle or arc.
• Question
What’s the difference between the circumference and the diameter?
wikiHow Staff Editor
The diameter is the length of a straight line drawn through the center of a circle from one side to the other. The circumference is the length all the way around the outside of the circle.
• Question
What is a circumference?
Circumference is the distance around the perimeter of a circle. It is calculated by multiplying the distance across the center (diameter) by Pi (3.1416).
• Question
If I know the radius of a circle, how I can find the circumference?
Donagan
Double the radius to get the diameter. Then multiply by pi to get the circumference.
• Question
How do I find the radius when I get the 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!
• Question
How can I find the diameter of the circle, once I know the circumference?
Circumference (C) = pi * diameter(d). To get the diameter of a circle take the circumference (C) and divide it by pi (3.14).
• Question
How do I find the area of a circle when only given the circumference?
Donagan
Divide the circumference by pi. That's the diameter. Divide by 2. That's the radius. Square the radius, and multiply by pi. That's the area.
• Question
What is the formula used to calculate diameter?
The diameter is twice the radius. The circumference is diameter x pi, or 2 x radius x pi.
• Question
How do I change the mixed number into radius?
If your radius is a mixed number, turn the number into an improper fraction. To do this, simply multiply the whole number part by the denominator and add that number to the numerator. The denominator should remain the same throughout the process. You can then use the improper fraction in your formula.
• Question
What is the answer of the circumference of a circle?
200 characters left
## Tips
• Consider buying a scientific or graphing calculator that already has π as one of the buttons. This will mean less typing for you and a more accurate answer because the π button produces an approximation to π that is much more accurate than 3.14.
Thanks!
• Remember: some worksheets will ask to replace pi with a subside, such as 3.14 or 22/7.
Thanks!
• To find circumference from the diameter, just multiply pi by the diameter.
Thanks!
• The radius is always going to be half of the diameter.[5]
Thanks!
## Warnings
• If you are stuck, ask a friend, family member, or teacher for help.
Thanks!
Thanks!
• Remember to always double-check your work because one mistake will set off all your data.
Thanks!
wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 217 people, some anonymous, worked to edit and improve it over time. Together, they cited 5 references. This article has also been viewed 8,360,129 times.<|endoftext|>
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Following a ballistic entry, parachute deceleration and a large number of airbags being deployed which protected the sensitive onboard equipment, the “twin” NASA rovers made a 480 million kilometer hole in one landing on MARS! After deploying their solar arrays and other scientific instruments, the rovers took numerous panoramic images of the landing sites and began their preliminary 90 day plus mission trek over the Martian surface, performing onsite scientific investigations every where they went. The rovers have already investigated the promising geological locations where they landed, i.e., the Gusev Crater and Meridiani Planum. One of the main reasons for landing in Meridiani Planum was to look for hematite, an iron oxide mineral that usually forms in the presence of water.
Before the eyes of the world, we were once again able to see panoramic camera shots of Mars. But this time scientists were well equipped to identify rock and soil targets and then command the vehicles to traverse bedrock and search out suspected water locations, creating a vast amount of information regarding Mars on a daily basis, and showing that the planet is still the most similar to our own in the solar system.
The realities of futuristic robots, equipped with basic logic programs, capable of conducting complex scientific explorations at ever greater distances from Earth are secondary to finding something more important: evidence that Earth and Mars shared the same biosphere of water, maybe even salt water! After the first problems with the landing gear and complex “Flash” command signals with “Spirit” (the first rover which landed), the “Opportunity” rover proved to the world that Mars was once very soggy and wet-having an environment that may have supported life. Even more exciting is this is just the beginning. Scientists are becoming more psychologically prepared to accept the fact that life may have once lived upon the red planet. The missions will continue to carry out high_level monitoring and explorations, formulating and executing complex mission plans, but the initial objective of finding evidence of past water is done, the new objective awaits: finding the fingerprint of life.
Investigations into biological mechanisms of self-organization and other evolutionary processes are leading to possible models made by new computing paradigms of self-designing hardware and software. All this is providing insight into the origins and evolution of life and the possible evolution of our universe. The opening of the magnificent picture of Mars begins with the actual landing on the anniversary of the blueprint of The Keys-January 3rd-now coming into interplanetary fulfillment as revealed. The House of Many Mansions stands before our eyes awaiting our exploration and citizenship.
— J.J. Hurtak, Ph.D., Ph.D.
When ordering from the Periodicals area of the online catalog, please add the date of the issue you are ordering (e.g., Spring 2012). Write that into the end of purchase comment area to ensure you receive the issue you want.<|endoftext|>
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The Spanish Conquistador
Born around 1485, Hernán Cortés was a Spanish conquistador and explorer who defeated the Aztec empire and claimed Mexico for Spain. He first set sail to the New World at the age of 19. Cortés later joined an expedition to Cuba. In 1518, he set off to explore Mexico. There he strategically aligned some native peoples against others to overthrow them. King Charles I appointed him governor of New Spain in 1522. Cortés died in Spain in 1547.<|endoftext|>
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# Area and Perimeter Activities
I love teaching area and perimeter activities and lessons. There are so many ways to make the lessons engaging and relevant to students.
This post if full of ideas for teaching area and perimeter activities that students will LOVE. These lessons were created for my fourth grade class but can be adapted to any grade level. If you want to see the area and perimeter post with lessons I used when I was teaching third grade, check out this post. The lessons below are a small part of my 4th Grade Measurement Unit.
## Pattern Block Perimeter and Area
Even though area and perimeter is introduced in third grade, most students need a review of the concepts through hands-on activities. In the first lesson(s), it is not necessary to explain how to add the sides to find the perimeter or multiply the sides to find the area. Instead, make sure students understand the general concept of perimeter and area. In this pattern block activity, students use the side of the square as one unit. Using the square, they find the area and perimeter of each of the pattern blocks.
## Edible Perimeter and Area
Food always brings in an extra element of fun, especially in math! In this area and perimeter activity students are given a fixed area of a fixed perimeter and use square crackers to build polygons with those areas or perimeters. If you’re not allowed to use food, you can use square tiles. Even though this is fun for students, it is quite challenging, and it allows students to see how there are many different ways to build a set number of square units.
## Area and Perimeter Carnival
After the hands-on lessons that incorporate some pictorial representations, I move to a lesson that is primarily pictorial. Students use grid paper with an included set of directions to build a carnival with fixed areas and perimeters. I have students focus on the math portion first and use their extra time to color or decorate the rides and booths in the carnival. It’s important to let students know that the math is the priority of the lesson.
## Pentomino Area and Perimeter
This is another highly pictorial lesson. In each square, students combine three pentomino pieces to form a polygon. Make sure students line the squares up and don’t try to create diagonal designs. Then, students find the area and perimeter of each polygon. This leads to great conversations about how the areas remain fairly constant but the perimeters change significantly.
## Ordering Rectangles
If you feel that students need additional work with pictorial models, you can have them complete this ordering rectangles area and perimeter activity. Students use grid paper to draw all the rectangles they can with a given perimeter (you can decide or let the class decide). After they build the rectangles, they find the area of each rectangle and write the area inside the rectangle. You can have students cut out the rectangles and glue them in order from smallest area to largest area which helps students visualize the differences in measurements.
## Fixed Area and Perimeter
When you feel that students are ready to move toward abstract thinking, this fixed area and perimeter is a good starting area and perimeter activity. Students draw and label all possible rectangles with a perimeter of 24 units (or different set perimeter). Then, they draw and label all possible rectangles with an area of 24 units.
## Construction Decisions
After a time working with abstract perimeter and area, I like to step up the problem solving into more complex problems. This is where students must draw from their conceptual understanding and problem solving experiences to find solutions.
## Area and Perimeter Activities – Problem Solving
Through problem solving area and perimeter activities, I also teach students how area is additive. This means you can add sections together to find the total area, which is typically not too challenging for students. However, it also means students can subtract to find area, which is often a surprise. You can download three FREE problem solving activities here!
## Integrating Art
I love to integrate art whenever possible, and this was a lot of fun. I used it as an early finisher activity, so it wasn’t an actual lesson. If students finished their task early, they could work on their name. Students wrote their name in block letters on grid paper (we tape it together to make a long sheet of paper). Then the partitioned each letter of their name into rectangles and found the area for each section of each letter. I had students color-code each section and the corresponding equation.
I hope you were able to find exciting area and perimeter activities that your students will love!
### 3 thoughts on “Area and Perimeter Activities”
1. Hi,
I love your materials! Can you update the pool problem-solving page by making the last column of the pool white for the walkway? Right now my copy shows the pool as 11 ft wide and the walkway as 4 ft on the right. Thank you!
Scroll to Top<|endoftext|>
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# Thread: Calculate X in a circle
1. ## Calculate X in a circle
I'm having some difficulties trying to solve this problem seen in the picture below. I just don't know where to begin. Any help would be very appreciated, thanks.
Calculate the angel X in the circle
2. ## Re: Calculate X in a circle
Originally Posted by gingerale
I'm having some difficulties trying to solve this problem seen in the picture below.
Calculate the angel X in the circle
.
Don't you know that $x+58=180~?$
Why is that true?
3. ## Re: Calculate X in a circle
So the opposite angel is a reflection that together is equal to 180? X = 122, and the angel that is unmarked would then be 77, all together 360. Makes sense. Why is the circle there then, just for confusing?
4. ## Re: Calculate X in a circle
Hello, gingerale!
Plato is absolutely correct.
Calculate the angel X in the circle.
Code:
B
o * *
* *
* *
* o C
A o *
* + *
* *
* *
* *
D o *
* * *
We have cyclic quadrilateral $ABCD.$
Draw chords $AB, BC, CD, DA.$
$\text{We have: }\:\begin{Bmatrix}\angle A &=& \frac{1}{2}\overarc{BCD} \\ \angle C &=& \frac{1}{2}\overarc{DAB} \end{Bmatrix}$
$\text{Hence: }\:\angle A + \angle C \;=\;\tfrac{1}{2}\overarc{BCD} + \tfrac{1}{2}\overarc{DAB} \;=\; \tfrac{1}{2}(\overarc{BCD} + \overarc{DAB}) \;=\; \tfrac{1}{2}(360^o)$
$\text{Therefore: }\:\angle A + \angle C \;=\;180^o$
$\text{Opposite angles of a cyclic quadrilateral are supplementary.}$
5. ## Re: Calculate X in a circle
Generally speaking it is true that the measure of an angle, with vertex on a circle, is half the angle measure of the arc it subtends. Here, the angle opposite x has measure 58 degrees and so subtends an arc of angle 116 degrees. Then entire circle has 360 degrees so that leaves 360- 116= 144 degrees for the arc subtended by angle x.<|endoftext|>
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Ginkgo is a genus of highly unusual non-flowering plants. The scientific name is also used as the English name. The order to which it belongs, Ginkgoales, first appeared in the Permian, 270 million years ago, possibly derived from "seed ferns" of the order Peltaspermales, and now only contains this single genus and species. The rate of evolution within the genus has been slow, and almost all its species had become extinct by the end of the Pliocene; the exception is the sole living species, Ginkgo biloba, which is only found in the wild in China, but is cultivated across the world. The relationships between ginkgos and other groups of plants are not fully resolved.
|Ginkgo biloba Eocene, McAbee, B.C., Canada|
The ginkgo (Ginkgoales) is a living fossil, with fossils similar to modern ginkgo from the Permian, dating back 270 million years. The most plausible ancestral group for the order Ginkgoales is the Pteridospermatophyta, also known as the "seed ferns", specifically the order Peltaspermales. The closest living relatives of the clade are the cycads, which share with the extant G. biloba the characteristic of motile sperm. Fossils attributable to the genus Ginkgo first appeared in the Early Jurassic, and the genus diversified and spread throughout Laurasia during the middle Jurassic and Early Cretaceous. It declined in diversity as the Cretaceous progressed with the extinction of species such as Ginkgo huolinhensis, and by the Palaeocene, only a few Ginkgo species, Ginkgo cranei and Ginkgo adiantoides, remained in the Northern Hemisphere, while a markedly different (and poorly documented) form persisted in the Southern Hemisphere. At the end of the Pliocene, Ginkgo fossils disappeared from the fossil record everywhere except in a small area of central China, where the modern species survived. It is doubtful whether the Northern Hemisphere fossil species of Ginkgo can be reliably distinguished. Given the slow pace of evolution and morphological similarity between members of the genus, there may have been only one or two species existing in the Northern Hemisphere through the entirety of the Cenozoic: present-day G. biloba (including G. adiantoides) and G. gardneri from the Palaeocene of Scotland.
At least morphologically, G. gardneri and the Southern Hemisphere species are the only known post-Jurassic taxa that can be unequivocally recognised. The remainder may have been ecotypes or subspecies. The implications would be that G. biloba had occurred over an extremely wide range, had remarkable genetic flexibility and, though evolving genetically, never showed much speciation. While it may seem improbable that a species may exist as a contiguous entity for many millions of years, many of the ginkgo's life-history parameters fit. These are: extreme longevity; slow reproduction rate; (in Cenozoic and later times) a wide, apparently contiguous, but steadily contracting distribution coupled with, as far as can be demonstrated from the fossil record, extreme ecological conservatism (restriction to disturbed streamside environments).
Modern-day G. biloba grows best in well-watered and drained environments, and the extremely similar fossil Ginkgo favoured similar environments; the sediment records at the majority of fossil Ginkgo localities indicate it grew primarily in disturbed environments along streams and levees. Ginkgo therefore presents an "ecological paradox" because, while it possesses some favourable traits for living in disturbed environments (clonal reproduction), many of its other life-history traits (slow growth, large seed size, late reproductive maturity) are the opposite of those exhibited by modern plants that thrive in disturbed settings.
Given the slow rate of evolution of the genus, it is possible that Ginkgo represents a pre-angiosperm strategy for survival in disturbed streamside environments. Ginkgo evolved in an era before flowering plants, when ferns, cycads, and cycadeoids dominated disturbed streamside environments, forming a low, open, shrubby canopy. The large seeds of Ginkgo and its habit of "bolting"—growing to a height of 10 metres (33 ft) before elongating its side branches—may be adaptations to such an environment. Diversity in the genus Ginkgo dropped through the Cretaceous (along with that of ferns, cycads, and cycadeoids) at the same time the flowering plants were on the rise, which supports the notion that flowering plants, with their better adaptations to disturbance, displaced Ginkgo and its associates over time.
As of February 2013[update], molecular phylogenetic studies have produced at least six different placements of Ginkgo relative to cycads, conifers, gnetophytes and angiosperms. The two most common are that Ginkgo is a sister to a clade composed of conifers and gnetophytes or that Ginkgo and cycads form a clade within the gymnosperms. A 2013 study examined the reasons for the discrepant results, and concluded that the best support was for the monophyly of Ginkgo and cycads.
- Taylor & Taylor (1993), pp. 138, 197.
- R. Govaerts. "Ginkgo L., Mant. Pl. 2: 313 (1771)". World Checklist of Selected Plant Families. Royal Botanic Gardens, Kew. Retrieved June 8, 2013.
- "Genus: Ginkgo L." Germplasm Resources Information Network. United States Department of Agriculture. Retrieved June 8, 2013.
- Royer et al. (2003)
- Royer et al. (2003), p. 84.
- Royer et al. (2003), p. 85.
- Royer et al. (2003), p. 91.
- Royer et al. (2003), p. 87.
- Royer et al. (2003), p. 92.
- Royer et al. (2003), p. 93.
- Wu et al. (2013)
- Dana L. Royer, Leo J. Hickey & Scott L. Wing (2003). "Ecological conservatism in the 'living fossil' Ginkgo". Paleobiology. 29 (1): 84–104. doi:10.1666/0094-8373(2003)029<0084:ECITLF>2.0.CO;2.
- Thomas N. Taylor & Edith L. Taylor (1993). The Biology and Evolution of Fossil Plants. Englewood Cliffs, NJ: Prentice Hall. ISBN 0-13-651589-4.
- Chung-Shien Wu, Shu-Miaw Chaw & Ya-Yi Huang (2013). "Chloroplast phylogenomics indicates that Ginkgo biloba is sister to cycads". Genome Biology and Evolution. 5 (1): 243–254. doi:10.1093/gbe/evt001. PMC 3595029. PMID 23315384.<|endoftext|>
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Exponential Functions
• Jan 11th 2009, 08:07 AM
Macleef
Exponential Functions
Quote:
Once the initial publicity surrounding the release of a new book is over, sales of the hardcover edition tend to decrease exponentially. At the time publicity was discontinued, a certain book was experiencing sales of 25,000 copies per month. One month later, sales of the book had dropped to 10,000 copies per month. What will the sales be after 1 more month?
I have no idea on how to solve this problem. Please show me a step by step solution! The answer is 4000.
And also need assistance with the following question...
Find $\displaystyle f(2)$ if $\displaystyle f(x) = e^{kx}$ and $\displaystyle f(1) = 20$.
Here's what I did...
$\displaystyle 20 = e^{k \times 1}$
$\displaystyle e^{20} = e^{k \times 1}$
$\displaystyle 20 = e$
$\displaystyle f(2) = e^{20 \times 2}$
$\displaystyle f(2) = 2.3539$ x $\displaystyle 10^{17}$
• Jan 11th 2009, 08:13 AM
james_bond
For your second question: $\displaystyle 20=f(1)=e^k\rightarrow k=\log 20$ so $\displaystyle f(2)=e^{2\cdot \log 20}=\left(e^{\log 20}\right)^2=20^2=\boxed{400}$.
• Jan 11th 2009, 08:13 AM
Prove It
Quote:
Originally Posted by Macleef
I have no idea on how to solve this problem. Please show me a step by step solution! The answer is 4000.
And also need assistance with the following question...
Find $\displaystyle f(2)$ if $\displaystyle f(x) = e^{kx}$ and $\displaystyle f(1) = 20$.
Here's what I did...
$\displaystyle 20 = e^{k \times 1}$
$\displaystyle e^{20} = e^{k \times 1}$
$\displaystyle 20 = e$
$\displaystyle f(2) = e^{20 \times 2}$
$\displaystyle f(2) = 2.3539$ x $\displaystyle 10^{17}$
Step 1 to Step 2 is wrong. How have you turned 20 into $\displaystyle e^{20}$?
You should have
$\displaystyle 20 = e^k$
$\displaystyle k = \ln{20}$.
Therefore $\displaystyle f(x) = e^{x\ln{20}} = e^{\ln{20^x}} = 20^x$.
Thus $\displaystyle f(2) = 20^2 = 400$.
• Jan 11th 2009, 08:19 AM
james_bond
Quote:
Originally Posted by Macleef
$\displaystyle 20 = e^{k \times 1}$
$\displaystyle e^{20} = e^{k \times 1}$
$\displaystyle 20 = e$
$\displaystyle f(2) = e^{20 \times 2}$
$\displaystyle f(2) = 2.3539$ x $\displaystyle 10^{17}$
How did you get that?? If you take a look at the first two lines you can see that it just can be true. (It would mean that $\displaystyle 20=e^{20}$..) I suppose that you meant $\displaystyle 20 = k$ in the third line, then it would at least come from the line above and the other two lines would come from that...<|endoftext|>
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Global warming is an aspect of climate change, referring to the long-term rise of the planet's temperatures. It is caused by increased concentrations of greenhouse gases in the atmosphere, mainly from human activities such as burning fossil fuels, deforestation and farming.
1. Burning fossil fuels
When we burn fossil fuels like coal, oil and gas to create electricity or power our cars, we release CO2 pollution into the atmosphere.
Australians are big producers of CO2 pollution compared to the rest of the world. Our level of CO2 pollution per person is nearly double the average of other developed nations and more than four times the world average.
Electricity generation is the main cause of carbon pollution in Australia as 73% of our electricity comes from burning coal and 13% from burning gas. The remaining 14% comes from renewable energy sources such as hydro, solar and wind, which do not emit carbon.
Reducing the amount of electricity generated from coal and gas
Increasing the amount of electricity from clean, renewable energy sources like solar and wind
Join the movement for stronger action on climate change and urge key Australian politicians to get us back on track to meeting our Paris Agreement targets.
2. Deforestation & Tree-Clearing
Plants and trees play an important role in regulating the climate because they absorb carbon dioxide from the air and release oxygen back into it. Forests and bushland act as carbon sinks and are a valuable means of keeping global warming to 1.5°C.
But humans clear vast areas of vegetation around the world for farming, urban and infrastructure development or to sell tree products such as timber and palm oil. When vegetation is removed or burnt, the stored carbon is released back into the atmosphere as CO2, contributing to global warming.
Up to one-fifth of global greenhouse gas pollution comes from deforestation and forest degradation.
3. Agriculture & Farming
Animals, particularly livestock like sheep and cattle, produce methane, a greenhouse gas. When livestock are grazed at a large scale, as in Australia, the amount of methane produced is a big contributor to global warming.
Some fertilisers that farmers use also release nitrous oxide, which is another greenhouse gas.
Australian farming contributes 16% of our total greenhouse gas emissions.
Use different stock feeds can help to reduce farming's contribution to climate change
WWF is working with leading beef producers through ‘Project Pioneer’ to develop, trial and validate improved livestock and pasture management that can deliver significant economic, social and environmental gains. Find out more...<|endoftext|>
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How do you write the quadratic function in vertex form given vertex (-4,6) and point (-1,9)?
Apr 24, 2017
$3 y - 18 = {\left(x + 4\right)}^{2}$ or ${\left(y - 6\right)}^{2} = 3 \left(x + 4\right)$
Explanation:
A vertex form of equation is of the type
$\left(y - k\right) = a {\left(x - h\right)}^{2}$ or $\left(x - h\right) = a {\left(y - k\right)}^{2}$, where $\left(h , k\right)$ is the vertex.
As the vertex is $\left(- 4 , 6\right)$, it would be either
$\left(y - 6\right) = a {\left(x + 4\right)}^{2}$ or $\left(x + 4\right) = a {\left(y - 6\right)}^{2}$
Case 1
If it is $\left(y - 6\right) = a {\left(x + 4\right)}^{2}$, as it passes through $\left(- 1 , 9\right)$, we have
$\left(9 - 6\right) = a {\left(- 1 + 4\right)}^{2}$ i.e. $9 a = 3$ and $a = \frac{1}{3}$ and equation is
$\left(y - 6\right) = \frac{1}{3} {\left(x + 4\right)}^{2}$ i.e. $3 y - 18 = {\left(x + 4\right)}^{2}$
Case 2
If it is $\left(x + 4\right) = a {\left(y - 6\right)}^{2}$, as it passes through $\left(- 1 , 9\right)$, we have
$\left(- 1 + 4\right) = a {\left(9 - 6\right)}^{2}$ or $3 = 9 a$ and $a = \frac{1}{3}$ and equation is
$\left(x + 4\right) = \frac{1}{3} {\left(y - 6\right)}^{2}$ or ${\left(y - 6\right)}^{2} = 3 \left(x + 4\right)$
graph{(3y-18-(x+4)^2)((y-6)^2-3(x+4))=0 [-10.96, 9.04, 0.04, 10.04]}<|endoftext|>
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In treaties such as the Fort Laramie Treaty of 1851, the lands of Native peoples in the west were legally recognized and guaranteed. That such agreements carry the full weight of the United States Constitution belies the casual nature of their neglect.
The Sand Creek Massacre of Nov. 29, 1864, represents just one example of our nation’s failure to live up to the ideals and laws sanctified in the documents of civilization.
Less than a decade after the Fort Laramie Treaty was signed, gold was discovered in land covered by the agreement. For some settlers, this news made the Fort Laramie Treaty a troublesome nuisance. The influx of prospectors and settlers into the Rocky Mountain region hastened the establishment of a territorial government and an urgent necessity to deal with Native nations living there. As more settlers arrived, the resulting violence demanded attention. But the territorial government was unwilling and perhaps unable to secure the treaty boundaries against such violation, so instead they set out to create a new treaty greatly reducing the land of the Arapaho and Cheyenne people within Colorado territory.
Known as the Treaty of Fort Wise, this new agreement was signed in 1861 by a smaller group of Arapaho and Cheyenne chiefs and representatives. Given that many of the Native people affected were unaware of the treaty or opposed the stipulations, it was widely rejected. Even so, land in Colorado’s central and eastern plains and in Kansas—including the area around Sand Creek—was encompassed within the treaty boundaries.
These places, by rights of occupation, treaty and law, were Indian land.
By 1862, John Evans, a physician from Ohio, would be appointed by President Lincoln as Governor of Colorado Territory. Born into the cauldron of war that was the old Northwest where the Shawnee leader, Tecumseh, and his Native allies had faced down General William Henry Harrison and the United States Army, it is not surprising that he harbored unsympathetic views of Native people. His arrival in Denver was marked by escalating violence and cries for the wholesale annihilation of Native people.
There was strong pressure for driving Native people from the territory by the time Arapaho and Cheyenne chiefs, accompanied by Major Edward “Ned” Wyncoop and Captain Silas Soule, arrived at Denver’s Camp Weld on Sept. 28, 1864, to meet with Evans for peace. Cheyenne Chief Black Kettle reportedly said that the trip had been “like coming through the fire,” only hoping for peace—but Evans, who’d first rejected the meeting, stated that he was “in no condition” to make a treaty, and his soldiers were “preparing for the fight.”
When Wyncoop returned to his command at Fort Lyon—some 20 miles from Sand Creek—he assured the assembled Arapaho and Cheyenne that he would protect them. But his superiors were displeased with his humane treatment of these people, and charges were made that forced him to relinquish his command on November 25 of that year.
The Sand Creek Massacre itself would take place a mere four days later. According to reports by Lt. Joseph Cramer—who, along with Capt. Silas Soule, defied orders by refusing to take part in the massacre—“scouts” were placed “around the Post, with instructions to let no one out” to prevent news of the plan from getting out.
Early in the morning of Nov. 29, Cheyenne chiefs Black Kettle and White Antelope came out from their lodges as the troops neared. Black Kettle carried an American Flag and a white flag of truce presented to him by President Lincoln; White Antelope wore a peace medal, also from Lincoln. The soldiers opened fire, wounding Black Kettle and killing White Antelope as he sang his death song: “nothing lives long . . . only the earth and mountains . . .” The massacre would not end until the following day, by which time approximately 200 people, mostly women and children, lay dead, with as many wounded. In the report of his subsequent investigation into what had happened, Wyncoop characterizes Col. John Chivington, who led the attack, as an “inhuman monster,” stating that, “the most fearful atrocities were committed that ever was heard of.”
After military and congressional investigations determined Sand Creek to be a massacre, the U.S. government negotiated the Little Arkansas Treaty in October 1865 as a gesture of conciliation, accepting full responsibility and providing reparations. But the terms of this treaty would once again be forgotten within a few years, and Arapaho and Cheyenne people confined to reservations in Wyoming, Montana and Oklahoma.
Now, more than 150 years after the Sand Creek Massacre, its anniversary is still commemorated each year by a days-long Spiritual Healing Run that retraces the route of taken by Chivington and his troops as they made their way back to Denver, there to be welcomed as conquering heroes. But the runners also make their way to the grave of Silas Soule, one of the Army officers who defied Chivington. As we consider the moral and ethical dimensions of these events on this, the 152nd anniversary of the massacre, that aspect of the history should be remembered as well: for this is not just a story about the status of treaties and the question of justice in our American past, but also a reminder that it is always possible to stand up for what is right in the face of hatred and violence.
Billy J. Stratton is an associate professor of English at the University of Denver and the author of Buried in Shades of Night: Contested Voices, Indian Captivity, and the Legacy of King Philip’s War.<|endoftext|>
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All about Words: Increasing Vocabulary in the Common Core Classroom, PreK-2 (Common Core State Standards in Literacy)De (autor) Susan B. Neuman, Tanya S. Wright Timothy Shanahan
en Limba Engleză Carte Paperback – 19 Apr 2013
Vocabulary forms a relentless divide between children who succeed and those who do not. This divide is often between poor children and their privileged counterparts. Without vocabulary knowledge, children cannot interpret text meaningfully or respond in ways that enable them to fully participate in classroom discussions.
All About Words is a practical guide designed to help early childhood teachers take advantage of the unique opportunity provided by the Common Core State Standards. It offers strategies for planning and presenting vocabulary instruction and for monitoring children's word learning progress, along with helpful appendices that provide specific guidance on which words to teach. Each chapter includes ideas to support home-school connections, recognizing the important role parents play in children's vocabulary development. Throughout, the authors encourage readers to examine day-to-day classroom issues, making it an ideal resource for professional development.
- Helpful summary of key CCSS features and how they play out for PreK-2 teachers.
- Accessible reviews of key research.
- Evidence-based, developmentally appropriate teaching strategies.
- Classroom examples demonstrating how strategies look in action.
- "Think About It" sections to help teachers and coaches facilitate discussion.
- Helpful lists of vocabulary words children should be able to use, as well as student texts that support vocabulary growth and development.<|endoftext|>
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4
Q:
# In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal ?
A) 40 B) 30 C) 50 D) 25
Explanation:
Here given 120 men = 200 children
therefore 1 child = 3/5 of Man------ step I
Now given that the food was consumed by 150 children
From step I we get 150 children =90 Men ( 1 child = 3/5 Man ie , 150 X 3/5 = 90)
So the food is for (120-90) men = 30 Men
Q:
"Compounding frequency" refers to
A) What type of interest your account earns B) How often your interest is calculated and added back into your account C) What interest rate you can expect from your account D) How easily you can add money into your account
Explanation:
Compounding frequency :
The time periods when interest will be calculated on top of the original loan amount (or) The number of compounding periods in one year.
The greater the compounding frequency, the more often your interest is calculated and added back into your account.
0 113
Q:
How many inches is 5' 4"?
A) 66 B) 64 C) 62 D) 60
Explanation:
We know 1 feet = 12 inches.
According to the question,
5' 4'' = 5 feet 4 inches = 12 x 5 + 4 = 64 inches.
Hence, 5' 4'' = 5 feet 4 inches = 64 inches.
0 80
Q:
Multiplicative inverse of 0
A) 0 B) 1 C) infinity D) None of the above
Explanation:
In the real numbers, zero does not have a reciprocal because no real number multiplied by 0 produces 1 (the product of any number with zero is zero).
Hence, 0 doesn't have any multiplicative inverse.
0 126
Q:
6 divided by 2(1 + 2) =
A) 1 B) 0 C) 9 D) Can't be determined
Explanation:
The given expression can be simplified as
$÷$ 2 (1 + 2)
The expression can be simplified further by the order of operations, using BODMAS rule.
First evaluate Parentheses/Brackets, then evaluate Exponents/Orders, then evaluate Multiplication-Division, and finally evaluate Addition-Subtraction.
Now, the expression becomes $÷$ 2 (3)
According to the order of operations, division and multiplication have the same precedence, so the correct order is to evaluate from left to right. First take 6 and divide it by 2, and then multiply by 3.
6 ÷ 2 × 3
= 3 × 3
= 9
But not 6÷2×3 = 6 ÷ 6 = 1
Hence, 6 ÷ 2 (1 + 2) = 9.
1 141
Q:
Multiplicative inverse of 7?
A) 1 B) -1/7 C) 0 D) 1/7
Explanation:
Multiplicative inverse is nothing but a reciprocal of a number.
It is defined as one of a pair of numbers that when multiplied with another number equals the number 1.
Multiplicative inverse or reciprocal of 7 is
7 x n = 1
=> n = 1/7.
Hence, Multiplicative inverse or reciprocal of 7 is 1/7.
2 119
Q:
What is the product of a number and its reciprocal?
A) 0 B) 1 C) -ve of the number D) the number itself
Explanation:
The product of a number and its reciprocal is always equals to 1.
For Example : Let the number be 4.
Now, its reciprocal is 1/4
Hence, required product = 4 x 1/4 = 1.
Now, take the number as -15.
Then its reciprocal is -1/15
Required product = -15 x -1/15 = 1.
Hence, the product of a number and its reciprocal is 1.
0 149
Q:
A plane is an undefined term because it
A) is a flat surface that extends indefinitely in all directions B) is where other geometric shapes can be constructed C) is described generally, not using a formal definition D) can be named using three noncollinear points
Answer & Explanation Answer: C) is described generally, not using a formal definition
Explanation:
In geometry, we can define plane as a flat surface with no thickness.
The surface extends with no ends.
A plane does not have any edges even if we draw it on paper with edges, it does't mean it has edges.
Hence, A plane is an undefined term because it is described generally, not using a formal definition.
0 203
Q:
Absolute value of 9?
A) 0 B) 9 C) 8 D) -9
Explanation:
Absolute value of a number :
It means that the distance of a number from 0 on a number line.
Here absolute value of 9 is that on a number line 9 is 9 units away from 0. Hence its absolute value is 9. Similarly, absolute value of -9 means -9 is also 9 units away from the 0 on number line. Hence, absolute value of -9 is also 9.
Hence, the absolute value of 9 is 9.<|endoftext|>
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The Common Core State Standards place a high value on informational text, emphasizing it more in the early grades than ever before. We've looked at how to work with narrative text in the classroom and how to deepen interest in literature by tying in facts (and vice versa); now, let's take a look at how informational text can function on its own in the classroom. For easy reference, I've reproduced here the standards that students are expected to meet with regards to prose literature in the sidebar.*
Helping Students to Recognize Features of Informational Texts
One of the main ways that informational texts differ from narrative literature texts, aside from the veracity of their content, is that informational texts have key recognizable features that can clue readers in to the fact that they are reading something that is intended to inform. We are all familiar with these features: table of contents, glossaries, indices, photos and realistic illustrations, etc., but it is likely that our students are completely new to the medium. As early as the first grade, the Common Core State Standards expect children to be able to recognize these features and know how to use them.
One way to familiarize students with the features is to scan the pertinent pages of an informational text that show the different features, and have students read along with their own copies of the book. You can present the pages in diagram form, labeling things such as headings, illustrations, and the index, and test the students on their knowledge with a worksheet with blank labels. An example of a bare-bones version of this is below.
Informational Text Standards:
Students should be able to do the following:
~ Ask and answer questions about key details in a text.
~ Identify the main topic and retell key details of a text.
~ Describe the connection between two individuals, events, ideas, or pieces of information in a text.
~ Ask and answer questions to help determine or clarify the meaning of words and phrases in a text.
~ Know and use various text features (e.g., headings, tables of contents, glossaries, electronic menus, icons) to locate key facts or information in a text.
~ Distinguish between information provided by pictures or other illustrations and information provided by the words in a text.
~ Use the illustrations and details in a text to describe its key ideas.
~ Identify the reasons an author gives to support points in a text.
~ Identify basic similarities in and differences between two texts on the same topic (e.g., in illustrations, descriptions, or procedures).
Students should be able to do the following:
~ Ask and answer such questions as who, what, where, when, why, and how to demonstrate understanding of key details in a text.
~ Identify the main topic of a multiparagraph text as well as the focus of specific paragraphs within the text.
~ Describe the connection between a series of historical events, scientific ideas or concepts, or steps in technical procedures in a text.
~ Determine the meaning of words and phrases in a text relevant to a grade 2 topic or subject area.
~ Know and use various text features (e.g., captions, bold print, subheadings, glossaries, indexes, electronic menus, icons) to locate key facts or information in a text efficiently.
~ Identify the main purpose of a text, including what the author wants to answer, explain, or describe.
~ Explain how specific images (e.g., a diagram showing how a machine works) contribute to and clarify a text.
~ Describe how reasons support specific points the author makes in a text.
~ Compare and contrast the most important points presented by two texts on the same topic.
Choosing Your Informational Text
When introducing your students to informational texts, you'll want to choose topics that both have all or most of the informational text features and will also be interesting to students. One way of finding topics that will interest them is to choose books that draw from and expand upon things the students are already familiar with. Another way is to find topics that they may have recently heard or read about through exposure to narrative texts.
These first informational texts should be bright and captivating, with pictures—illustrations and especially photographs—that draw students in. Where possible, the photographs should feature children, though this won't be applicable to every topic, of course. The informational texts should be at the correct reading level, with lines of text broken up into bite-sized chunks at sentence or phrase boundaries, with enough space between the letters and words that fledgling readers will not have trouble making them out.
While Hameray's new series Real World was designed to pair with the Story World series (narrative texts), they can stand just as well on their own as terrific informational texts. The topics are diverse enough that there will be something of interest for almost every young reader, from animals to the water cycle, and from baking to ball games. This also allows them to be worked into various teaching units throughout the school year. Disciplines covered span from the sciences to the arts and beyond.
The Story World Real World series was specifically tailored to help students meet these Common Core standards, and the free Teacher's Guide (available for download in June) will include many ideas to support teaching informational texts.
Below are example lessons extracted from the Story World-Real World Teacher's Guide for one of the books in the set of informational texts intended to support the traditional tale Cinderella.** The book is called What's the Time?, and it is about different ways of telling time and where you can find clocks. There are many more activities and lessons in the teacher's guide; this is just a sample. There's a flip book of the title below the lesson.
Example Lesson: What's the Time?
Before reading the book:
Look at the cover with the children. Ask what kind of book is this going to be—a story (like Cinderella) or an informational text? (CCSS Reading Standards for Literature Grade 1 #5)
During the reading of the book:
If you are doing a shared reading, read the text with the children, spending time on the photographs and illustrations as well as the text (CCSS Reading Standards for Informational Text Grade 1 #7). For guided reading, decide on a specific learning focus: for example, vocabulary (reading, saying and understanding new words), or informational text features, or science information and concepts.
After reading the book:
Discuss Informational Text features (table of contents, headings, photographs, captions, glossary, index). How do they help us read and write about a nonfiction or informational subject? (CCSS Reading Standards for Informational Texts Grade 1 #5)
Reread pages 4–5: Can we see our own shadows? Use a light to act as the sun. Have a student stand in one place. Shine the light on the student so he or she casts a shadow. What happens if we move the light around the student? (The student’s shadow moves.) What have we made? (A human sundial!) Talk about the students’ shadows when they are out in the sunlight on a sunny day. Why is it that sometimes our shadow is front of us and sometimes it is behind us? Why do shadows seem to get smaller towards the middle of the day and then longer in the afternoon?
Reread pages 12–13. Ask the children to count how many clocks they have at home. Are there any clocks on public buildings in your town or city? How many clocks can they find in the classroom? (Don’t forget computers.) In the school? Why do we need so many clocks?
This was a sample lesson from the Story World-Real World Teacher's Guide. The suggested text is meant to help guide the discussion and facilitate interactions, but is in no way meant to dictate exactly how a lesson is taught.
Features of What's the Time?
- Informational text type: Description (Text type description: “informs the reader about the subject being described”.)
- Informational text features: Table of contents, headings, glossary, index. (CCSS Reading Standards for Informational Texts Grade 1 #5)
- Visual information: photographs with captions and labels. (CCSS Reading Standards for Informational Texts Grade 1 #6,7)
- Vocabulary: (egg timer, sundial)
- Supports integrated curriculum learning – literacy learning (reading informational texts: (CCSS Reading Standards for Informational Text Grade 1), plus science (Physical science: measurement, light and shadows, gravity).
To see all titles available in the series, you can visit our website or download the series highlights by clicking below.
- Tara Rodriquez
*All information regarding the CCSS comes directly from the Common Core website and can be found in the downloadable PDFs available on that site.
** Because this is a sneak-preview of the forthcoming Teacher's Guide, which is still in draft form, the final downloadable Teacher's Guide soon available may deviate slightly from what is presented here.
Photo credit: Kim Gunkel<|endoftext|>
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In the early 1900s black bears were nearly eliminated from Missouri until reintroduction efforts in the 1960s led to population increases across the southern portion of the state. Today, there is an estimated population size between 250 to 300 individual black bears within the known breeding range. Black bears have been documented throughout the Ozark Highland Plateau and human-bear conflict is an increasing concern for wildlife managers. The Missouri Department of Conservation (MDC) and Mississippi State University studied the expansion patterns of black bears and the potential impacts expansion has on the number of human-bear conflicts.
The study used historical occurrence data and forest cover to characterize broad patterns of black bear re-colonization. Researchers estimated the bear population and analyzed bear incident reports to understand their distribution, frequency, and type of incident occurrence over time. Statewide public bear occurrence data from 1989 to 2010 was obtained through MDC’s “Report a Bear Sighting system.” It was used to examine current distribution and potential range expansion of bears throughout Missouri.
Data collected during this study suggests the overall population size is likely less than the 500 individuals, the population size considered necessary to implement a harvest season. Final reports show the spatial distribution of black bear occurrences, both adults and dependent young, closely reflect the distribution of forest cover. Bears, likely from Arkansas, have contributed to the distribution of bears throughout Missouri. The current distribution of dependent young suggests reproducing bears primarily occur in the Ozark Highlands, reinforcing the possibility of bears expanding northward from Arkansas. Further observation is needed.
The information gathered to date is basic, but important. Information on statewide bear occurrences and human–bear incident patterns will be useful for establishing future research and management plans in Missouri. As black bears in Missouri continue to re-colonize their historical range, understanding large-scale spatial and temporal changes in their distribution and interactions with humans will help managers better assess research and management objectives.
Twice a week, SCI Foundation informs readers about conservation initiatives happening worldwide and updates them on SCI Foundation’s news, projects and events. Tuesdays are dedicated to an Issue of the Week and Thursday’s Weekly Updates will provide an inside look into research and our other science-based conservation efforts. Follow us on Facebook and Twitter for more SCI Foundation news.<|endoftext|>
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California Condor Population Hits 100
One of the most endangered birds in the world celebrates a happy milestone this week: On Wednesday, the US Fish and Wildlife Service announced that for the first time in half a century, 100 wild California condors now fly free in California.
In 1987, the last 22 condors remaining in the wild were captured for a captive breeding program. The program, run jointly by the Los Angeles Zoo and the San Diego Wild Animal Park, began slowly; scientists were frustrated by the birds’ unusually slow reproductive habits (see below). But eventually the condors began to reproduce reliably, and in 1991, the first juveniles were reintroduced to the wild. Since then, young birds have been released each fall, and a handful of chicks have also hatched naturally in wild nests.
Small populations of the birds also survive in the southwestern U.S. and in Baja California, but scientists estimate that the entire population, wild and captive, remains below 400.
Anyone who has ever seen a condor knows that these are extraordinary birds. They’re huge, with the largest wingspan of any North American bird: more than 9 feet. Their heads and necks are nearly featherless. Condors can soar for miles without flapping their wings, which gives them a grace in flight that is at odds with their reputation as homely birds.
These long-lived scavengers are picky about their mates (as befits any animal that mates for life and lives as long as 50 years) and do not breed until the relatively advanced age of six, making them especially prone to population loss. Throughout the 20th Century, condors suffered terribly from various human-related causes: lead poisoning (from consuming animals contaminated with lead shot), DDT poisoning, poaching and the destruction of their habitat.
By the 1960s, fewer than 100 condors survived in the state, making this week’s announcement very good news for anyone who wants to see condors once again become a significant part of the ecosystem of the American West.
How Our Work Helps the Condor
The National Wildlife Federation works to strengthen and defend the Endangered Species Act, the law which has helped protect the California condor. Our California affiliate, the Planning and Conservation League, also works to protect important habitat for the condor, including the establishment of the Tejon Preserve.<|endoftext|>
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Calculating an average rate shows the amount of change of one variable with respect to another. The other variable is commonly time and could describe the average change in distance (speed) or chemical concentrations (reaction rate). You can replace time with any correlated variable, however. For example, you might calculate the change in a local bird population with respect to the number of bird feeders you place. These variables could be plotted against one another, or you could use a function curve to extrapolate data from one variable.
Measure the variables at two points. As an example, you might measure 50 grams of a reactant at time zero and 10 grams after 15 seconds. If you're looking at a graph, you could reference data at two plot points. If you have a function, such as y = x^2 + 4, plug in two values of "x" to extract the respective values of "y." In this example, x-values of 10 and 20 produce y-values of 104 and 404.
Subtract the first value of each variable from the second. Continuing with the reactant example, subtract 50 from 10 to get the concentration change of -40 grams. Likewise, subtract zero from 15 to get a change in time of 15 seconds. In the function example, the changes in x and y are 10 and 300, respectively.
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Divide the primary variable's change by the influencing variable's change to get the average rate. In the reactant example, dividing -40 by 15 gets an average rate of change of -2.67 grams per second. But reaction rates are typically expressed as positive numbers, so drop the negative sign to get just 2.67 grams per second. In the function example, dividing 300 by 10 produces a "y" average rate of change of 30 between x-values of 10 and 20.
A negative rate describes a decrease, whereas a positive figure describes an increase. Therefore, always keep the negative sign, unless you're calculating chemical reaction rates, which are expressed as positive figures.
The primary variable is the one that's changing with respect to the other variable. In the examples, the chemical concentration changed over time and y changed with respect to x.<|endoftext|>
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When a person stutters, we sometimes say that their speech is “dysfluent” because the flow of speech is interrupted. These interruptions may include one or more of the following characteristics:
Repetitions – involve repeating a sound, syllable or phrase (e.g. m-m-m-my)
Prolongations – involve stretching out a sound (e.g. mmmmmy)
Blocks – involve stoppage of the airflow so no sound comes out
These characteristics are often caused by increased tension somewhere within the lungs, throat, mouth, tongue or lips. When people stutter, they often struggle to push the words out as they feel the words are getting stuck.
Unfortunately, this strategy is often ineffective as it results in more tension which can then result in even more struggle. This struggle may be heard in the individual’s speech (e.g. change in pitch or loudness), seen throughout their body (e.g. body movements, eye blinking) or it may come out in other behaviours (e.g. rapid breathing).
Over time, people who stutter may experience negative emotional reactions to their speech (e.g. teasing, bullying, embarrassment, frustration).
Other people’s reactions or the individual’s own negative thoughts may result in feelings of anxiousness or worry about speaking as they fear getting caught in a stutter again. For many, the result is often a cycle of tension, struggle and inevitably more stuttering.
This loop is very difficult to break without intervention. If these feelings are not addressed, the individual may begin to avoid speaking situations in an attempt to keep themselves from stuttering.
Is there a Cure?
People often wonder if stuttering can be cured. Although there is no easy “cure” for stuttering, speech-language therapy can be very effective in helping the individual learn strategies or techniques to help him modify his speech. Speech-language intervention for stuttering often involves the following:
Increasing awareness of stuttering and its related behaviours;
Reducing how often someone stutters;
Decreasing the tension and struggle of stuttering moments;
Working to decrease word or situation avoidances;
Using effective communication skills; and,
Increasing overall communication confidence.
If you have concerns about your speech or your child’s speech, it is always best to get it checked out by a trained Speech-Language Pathologist.
After reviewing the assessment results, the treating Speech-Language Pathologist will work with the person who stutters and/or their family members to set individual treatment goals that will help them become the best communicators they can be.<|endoftext|>
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1721 – John Hanson, first U.S. President under the Articles of Confederation, was born in Maryland. He was the heir of one of the greatest family traditions in the colonies and became the patriarch of a long line of American patriots – his great-grandfather died at Lutzen beside the great King Gustavus Aldophus of Sweden; his grandfather was one of the founders of New Sweden along the Delaware River in Maryland; one of his nephews was the military secretary to George Washington; another was a signer of the Declaration; still another was a signer of the Constitution; yet another was Governor of Maryland during the Revolution; and still another was a member of the first Congress; two sons were killed in action with the Continental Army; a grandson served as a member of Congress under the new Constitution; and another grandson was a Maryland Senator. Thus, even if Hanson had not served as President himself, he would have greatly contributed to the life of the nation through his ancestry and progeny. As a youngster he began a self-guided reading of classics and rather quickly became an acknowledged expert in the juridicalism of Anselm and the practical philosophy of Seneca – both of which were influential in the development of the political philosophy of the great leaders of the Reformation. It was based upon these legal and theological studies that the young planter – his farm, Mulberry Grove was just across the Potomac from Mount Vernon – began to espouse the cause of the patriots. In 1775 he was elected to the Provincial Legislature of Maryland. Then in 1777, he became a member of Congress where he distinguished himself as a brilliant administrator. Thus, he was elected President in 1781. The new country was actually formed on March 1, 1781 with the adoption of The Articles of Confederation. This document was actually proposed on June 11, 1776, but not agreed upon by Congress until November 15, 1777. Maryland refused to sign this document until Virginia and New York ceded their western lands (Maryland was afraid that these states would gain too much power in the new government from such large amounts of land). Once the signing took place in 1781, a President was needed to run the country. John Hanson was chosen unanimously by Congress (which included George Washington). In fact, all the other potential candidates refused to run against him, as he was a major player in the Revolution and an extremely influential member of Congress. As the first President, Hanson had quite the shoes to fill. No one had ever been President and the role was poorly defined. His actions in office would set precedent for all future Presidents. He took office just as the Revolutionary War ended. Almost immediately, the troops demanded to be paid. As would be expected after any long war, there were no funds to meet the salaries. As a result, the soldiers threatened to overthrow the new government and put Washington on the throne as a monarch. All the members of Congress ran for their lives, leaving Hanson running the government. He somehow managed to calm the troops and hold the country together. If he had failed, the government would have fallen almost immediately and everyone would have been bowing to King Washington. Hanson, as President, ordered all foreign troops off American soil, as well as the removal of all foreign flags. This was quite a feat, considering the fact that so many European countries had a stake in the United States since the days following Columbus. Hanson established the Great Seal of the United States, which all Presidents have since been required to use on all official documents. President Hanson also established the first Treasury Department, the first Secretary of War, and the first Foreign Affairs Department. Lastly, he declared that the fourth Thursday of every November was to be Thanksgiving Day, which is still true today. The Articles of Confederation only allowed a President to serve a one-year term during any three-year period, so Hanson actually accomplished quite a bit in such little time. He served in that office from November 5, 1781 until November 3, 1782. He was the first President to serve a full term after the full ratification of the Articles of Confederation – and like so many of the Southern and New England Founders, he was strongly opposed to the Constitution when it was first discussed. He remained a confirmed anti-federalist until his untimely death. Six other presidents were elected after him – Elias Boudinot (1783), Thomas Mifflin (1784), Richard Henry Lee (1785), Nathan Gorman (1786), Arthur St. Clair (1787), and Cyrus Griffin (1788) – all prior to Washington taking office. George Washington was the first President of the United States under the Constitution we follow today. And the first seven Presidents are forgotten in history.
1743 – Thomas Jefferson was born at Shadwell in Albemarle county, Virginia. He was tutored by the Reverend James Maury, a learned man, in the finest classical tradition. He began the study of Latin, Greek, and French at the age of nine. He attended William and Mary College in Williamsburg at sixteen years old, then continued his education in the Law under George Wythe, the first professor of law in America (who later would sign Jefferson’s Declaration in 1776). Thomas Jefferson attended the House of Burgesses as a student in 1765 when he witnessed Patrick Henry’s defiant stand against the Stamp Act. He gained the Virginia bar and began practice in 1769, and was elected to the House of Burgesses in 1769. It was there that his involvement in revolutionary politics began. He was never a very vocal member, but his writing, his quiet work in committee, and his ability to distill large volumes of information to essence, made him an invaluable member in any deliberative body. In 1775 when a Virginia convention selected delegates to the Continental Congress, Jefferson was selected as an alternate. It was expected that Payton Randolph, (then Speaker of the Virginia House and president of the Continental Congress too,) would be recalled by the Royal Governor. This did happen and Jefferson went in his place. Thomas Jefferson had a theory about self governance and the rights of people who established habitat in new lands. Before attending the Congress in Philadelphia he codified these thoughts in an article called A Summary View of the Rights of British America. This paper he sent on ahead of him. He fell ill on the road and was delayed for several days. By the time he arrived, his paper had been published as a pamphlet and sent throughout the colonies & on to England where Edmund Burke, sympathetic to the colonial condition, had it reprinted and circulated widely. In 1776 Jefferson, then a member of the committee to draft a declaration of independence was chosen by the committee to write the draft. This he did, with some minor corrections from James Madison and an embellishment from Franklin, the document was offered to the Congress on the first day of July. The congress modified it somewhat, abbreviating certain wording and removing points that were outside of general agreement. The Declaration was adopted on the Fourth of July. Jefferson returned to his home not long afterward. His wife and two of his children were very ill, he was tired of being remote from his home, and he was anxious about the development of a new government for his native state. In June of 1779 he succeeded Patrick Henry as Governor of Virginia. The nation was still at war, and the southern colonies were under heavy attack. Jefferson’s Governorship was clouded with hesitation. He himself concluded that the state would be better served by a military man. He declined re-election after his first term and was succeeded by General Nelson of Yorktown. In 1781 he retired to Montecello, the estate he inherited, to write, work on improved agriculture, and attend his wife. It was during this time that he wrote Notes on the State of Virginia, a work that he never completed. Martha Jefferson died in September of 1782. This event threw Jefferson into a depression that, according to his eldest daughter he might never have recovered from. Except that Washington called on him in November of 1782 to again serve his country as Minister Plenipotentiary to negotiate peace with Gr. Britain. He accepted the post, however it was aborted when the peace was secured before he could sail from Philadelphia. In 1784 Jefferson went to France as an associate Diplomat with Franklin and Adams. It was in that year that wrote an article establishing the standard weights, measures, and currency units for the Untied States. He succeeded Franklin as Minster to France the following year. When he returned home in 1789, he joined the Continental Congress for a while, and was then appointed Secretary of State under George Washington. This placed him in a very difficult position. The character of the executive was being established during the first few terms. Jefferson and many others were critical of the form it was taking under the first Federalist administration. Jefferson was sharply at odds with fellow cabinet members John Adams and Alexander Hamilton, both of whom he found to be too authoritarian and too quick to assume overwhelming power for the part of the executive. He resigned from the cabinet in 1793 and formed the Democrat-republican party. Heated competition continued. Jefferson ran for president in 1796, lost to John Adams, and, most uncomfortably, this made him vice president under a man whom he could no longer abide. After a single meeting, on the street, the two never communicated directly during the whole administration. Jefferson again ran for the presidency in 1801 & this time he won. He served for two terms & he did ultimately play a deciding role in forming the character of the American Presidency. The 12th amendment to the Constitution changed the manner in which the vice president was selected, so as to prevent arch enemies from occupying the first and second positions of the executive. Jefferson also found the State of the Union address to be too magisterial when delivered in person. He performed one and afterwards delivered them, as required by the constitution, only in writing. He also undertook the Louisiana Purchase, extending the boundaries of the country and establishing the doctrine of manifest destiny. Thomas Jefferson retired from office in 1808. He continued the private portion of his life’s work, and sometime later re-engaged his dearest & longest friend James Madison, in the work of establishing the University of Virginia. In 1815 one of his projects, a Library of Congress, finally bore fruit, when he sold his own personal library to the congress as a basis for the collection. Shortly before his death in 1826, Jefferson told Madison that he wished to be remembered for two things only; as the Author of the Declaration of Independence, and as the founder of the University of Virginia. Jefferson died on the 4th of July, as the nation celebrated the fiftieth anniversary of his splendid Declaration.
1775 – Lord North extended the New England Restraining Act to South Carolina, Virginia, Pennsylvania, New Jersey and Maryland. The act forbade trade with any country other than Britain and Ireland.
1777 – American forces are ambushed and defeated in the Battle of Bound Brook, New Jersey. The Battle of Bound Brook was a surprise attack conducted by British and Hessian forces against a Continental Army outpost at Bound Brook, New Jersey. The British objective of capturing the entire garrison was not met, although prisoners were taken. The American commander, Major General Benjamin Lincoln, left in great haste, abandoning papers and personal effects. Late on the evening of April 12, 1777, four thousand British and Hessian troops under the command of Lieutenant General Charles Cornwallis marched from the British stronghold of New Brunswick. All but one detachment reached positions surrounding the outpost before the battle began near daybreak the next morning. During the battle, most of the 500-man garrison escaped by the unblocked route. American reinforcements arrived in the afternoon, but not before the British plundered the outpost and began the return march to New Brunswick.
1847 – Naval Forces begin 5 day battle to capture several towns in Mexico.
1847 – Marines captured LaPaz, California, during the Mexican War.
1860 – 1st Pony Express reached Sacramento, Calif.
1861 – After a thirty-three hour bombardment by Confederate cannon, Fort Sumter in Charleston Harbor surrenders. The first engagement of the war ended in Rebel victory. The surrender concluded a standoff that began with South Carolina’s secession from the Union on December 20, 1860. When President Lincoln sent word to Charleston in early April that he planned to send food to the beleaguered garrison, the Confederates took action. They opened fire on Sumter in the predawn of April 12. Over the next day, nearly 4,000 rounds were hurled toward the black silhouette of Fort Sumter. Inside Sumter was its commander, Major Robert Anderson, 9 officers, 68 enlisted men, 8 musicians, and 43 construction workers who were still putting the finishing touches on the fort. Captain Abner Doubleday, the man often inaccurately credited with inventing the game of baseball, returned fire nearly two hours after the barrage began. By the morning of April 13, the garrison in Sumter was in dire straits. The soldiers had sustained only minor injuries, but they could not hold out much longer. The fort was badly damaged, and the Confederate’s shots were becoming more precise. Around noon, the flagstaff was shot away. Louis Wigfall, a former U.S. senator from Texas, rowed out without permission to see if the garrison was trying to surrender. Anderson decided that further resistance was futile, and he ran a white flag up a makeshift flagpole. The first engagement of the war was over, and the only casualty had been a Confederate horse. The Union force was allowed to leave for the north; before leaving, the soldiers fired a 100-gun salute. During the salute, one soldier was killed and another mortally wounded by a prematurely exploding cartridge. The Civil War had officially begun.
1862 – In the Washington area volunteers led by Sarah J. Evans paid homage to the graves of Civil War soldiers. Villagers in Waterloo, NY, held their 1st Memorial Day service on May 5, 1866. In 1966 Pres. Johnson gave Waterloo, NY, the distinction of holding the 1st Memorial Day.
1863 – Battle of Irish Bend, LA (Ft. Bisland).
1863 – Hospital for Ruptured and Crippled in NY became the 1st orthopedic hospital.
1865 – Union forces under Gen. Sherman began their devastating march through Georgia. Sherman’s troops took Raleigh, NC.
1873 – In the Colfax Massacre in Grant Parish, Louisiana, 60 blacks were killed. The dispute over the government of Louisiana continued to escalate. Republican officers of Grants Parrish were holed up in the city of Colfax. Blacks from the surrounding area feared an attack, so they entrenched themselves in front of the court house. A huge white mob attacked. The day was a massacre, as somewhere between 60 and 100 local blacks were killed even as they tried to surrender. The white mob suffered only 3 casualties. The battle for the courthouse of Colfax, Louisiana has been renamed the Colfax Massacre. All of the blacks in the area and governor Kellogg were spared only because the President ordered the federal troops to intercede and stop the white mob before they moved to another area, killing all the blacks and their white sympathizers. The New Orleans Times’ headline the next day read, “War at Last!!” They also warned other white sympathizers to beware. The majority of the white people in Louisiana supported the “Colfax Massacre,” and the systematic annihilation of blacks and the white sympathizer governments.
1885 – Marines guarded the rail line to Panama City.
1919 – Eugene V. Debs is imprisoned at the Atlanta Federal Penitentiary in Atlanta, Georgia, for speaking out against the draft during World War I.
1939 – USS Astoria arrives in Japan under the command of Richard Kelly Turner in an attempt to photograph the Japanese battleships Yamato and Musashi. U.S. Navy Rear Admiral Turner, whose motto was “If you don’t have losses, you’re not doing enough,” saw the cruiser Astoria through many assignments, from assessing Japanese naval strength before U.S. entry in the war, to returning the ashes of a Japanese ambassador to Japan, to the amphibious assault at Guadalcanal. The Astoria was unfortunately sunk, along with the Quincy and the Vincennes, during Operation Watchtower, the landing of 16,000 troops on Guadalcanal, in the Solomon Islands, in August 1942.
1941 – The USSR and Japan sign a five year Neutrality Agreement. For Stalin this is an invaluable piece of diplomacy which, backed by secret information from Soviet spies in Tokyo, will allow him to transfer forces from Siberia to face a possible German attack. These moves begin now. The agreement represents a complete change in Japanese policy and marks the growing concern of the Japanese military leaders and statesmen to look south to the resources of the East Indies. The agreement has been negotiated almost alone by Foreign Minister Matsuoka, in Moscow on the way back from a European visit.
1943 – President Roosevelt dedicated the Jefferson Memorial. It was designed by John Russell Pope. Located in Washington, DC, the Jefferson Memorial honors Thomas Jefferson — author of the Declaration of Independence, first Secretary of State, and third President of the United States. The memorial is built along the Tidal Basin, in line with the White House, other memorials and the Capitol. The Tidal Basin is surrounded by cherry blossom trees. The trees were a gift from the city of Tokyo, Japan, to the city of Washington, DC. The structure of the building is based on the classic style of architecture Jefferson introduced into this country. In the center of the memorial is a standing statue of Jefferson. On the inside walls are four inscriptions based upon Jefferson’s writings. They describe his beliefs in freedom, education of all people, and the need for change in the laws and institutions of a democracy.
1943 – The Jefferson Memorial is dedicated in Washington, D.C., on the 200th anniversary of President Thomas Jefferson’s birth.
1943 – US bombers conduct day and night raids on Kiska Island.
1944 – American and British tactical air forces conduct numerous attacks on German coastal batteries in Normandy.
1945 – On Okinawa, elements of the US 6th Marine Division not engaged on the Motobu Peninsula continue to advance up the west coast of the island and reach the northwest tip at Hedo Point. Japanese Kamikaze attacks hit a destroyer. British carriers attack Sakashima Gunto.
1945 – The Nazi concentration camps at Belsen and Buchenwald are liberated by British and American forces respectively. Jena is captured by US 3rd Army units. To the south, US 7th Army forces take Bamberg.
1945 – In Manila Bay, American forces land on Fort Drum, known as “the Concrete Battleship”, and begin to pour 5,000 gallons of oil fuel into the fortifications. This is then set on fire and burns for five days, eliminating the Japanese garrison.
1945 – Some 327 American B-29 Superfortress bombers attack Tokyo during the night, dropping some 2139 tons of incendiaries. The nominal target area is the arms manufacturing district.
1945 – Adolf Hitler proclaims from his underground bunker that deliverance was at hand from encroaching Russian troops–Berlin would remain German. A “mighty artillery is waiting to greet the enemy,” proclaims Der Fuhrer. This as Germans loyal to the Nazi creed continue the mass slaughter of Jews. As Hitler attempted to inflate his troops’ morale, German soldiers, Hitler Youth, and local police chased 5,000 to 6,000 Jewish prisoners into a large barn, setting it on fire, in hopes of concealing the evidence of their monstrous war crimes as the end of the Reich quickly became a reality. As the Jewish victims attempted to burrow their way out of the blazing barn, Germans surrounding the conflagration shot them. “Several thousand people were burned alive,” reported one survivor. The tragic irony is that President Roosevelt, had he lived, intended to give an address at the annual Jefferson Day dinner in Washington, D.C., on that very day, proclaiming his desire for “an end to the beginnings of all wars–yes, an end to this brutal, inhuman, and thoroughly impractical method of settling the differences between governments.”
1951 – As Lieutenant General Ridgway was winding up affairs as Eighth Army commander prior to assuming command of the U.N. Command, he put the final touches on plans developed during his term of command for rotating Army troops. Over 70,000 soldiers already were eligible under the length of service criteria of six months in combat units or one year in a support unit. The backlog of eligible troops would leave in monthly quotas based on the replacement flow. Since replacements currently exceeded casualty losses by more than 50 percent, the first quota of troops would leave Korea beginning on April 22.
1953 – CIA director Allen Dulles launches the mind-control program Project MKULTRA. Project MKUltra — sometimes referred to as the CIA’s mind control program — MKULTRA was the code name given to an illegal and clandestine program of experiments on human beings, made by the CIA – the Intelligence Service of the United States of America. Experiments on humans were intended to identify and develop drugs and procedures to be used in interrogations and torture, in order to weaken the individual to force confessions through mind control. Organized through the Scientific Intelligence Division of the Central Intelligence Agency (CIA), the project coordinated with the Special Operations Division of the U.S. Army’s Chemical Corps. The program began in the early 1950s, was officially sanctioned in 1953, was reduced in scope in 1964, further curtailed in 1967 and officially halted in 1973. The program engaged in many illegal activities; in particular it used unwitting U.S. and Canadian citizens as its test subjects, which led to controversy regarding its legitimacy. MKUltra used numerous methodologies to manipulate people’s mental states and alter brain functions, including the surreptitious administration of drugs (especially LSD) and other chemicals, hypnosis, sensory deprivation, isolation, and verbal abuse. The scope of Project MKUltra was broad, with research undertaken at 80 institutions, including 44 colleges and universities, as well as hospitals, prisons and pharmaceutical companies. The CIA operated through these institutions using front organizations, although sometimes top officials at these institutions were aware of the CIA’s involvement. As the Supreme Court later noted, MKULTRA was:
concerned with “the research and development of chemical, biological, and radiological materials capable of employment in clandestine operations to control human behavior.” The program consisted of some 149 subprojects which the Agency contracted out to various universities, research foundations, and similar institutions. At least 80 institutions and 185 private researchers participated. Because the Agency funded MKULTRA indirectly, many of the participating individuals were unaware that they were dealing with the Agency.
Project MKUltra was first brought to public attention in 1975 by the Church Committee of the U.S. Congress, and a Gerald Ford commission to investigate CIA activities within the United States. Investigative efforts were hampered by the fact that CIA Director Richard Helms ordered all MKUltra files destroyed in 1973; the Church Committee and Rockefeller Commission investigations relied on the sworn testimony of direct participants and on the relatively small number of documents that survived Helms’ destruction order. In 1977, a Freedom of Information Act request uncovered a cache of 20,000 documents relating to project MKUltra, which led to Senate hearings later that same year. In July 2001 some surviving information regarding MKUltra was officially declassified.
1960 – Navy’s navigation satellite, Transit 1-B, placed into orbit from Cape Canaveral, FL and demonstrates ability to launch another satellite.
1970 – Disaster strikes 200,000 miles from Earth when oxygen tank No. 2 blows up on Apollo 13, the third manned lunar landing mission. Astronauts James A. Lovell, John L. Swigert, and Fred W. Haise had left Earth two days before for the Fra Mauro highlands of the moon but were forced to turn their attention to simply making it home alive. Mission commander Lovell reported to mission control on Earth: “Houston, we’ve had a problem here,” and it was discovered that the normal supply of oxygen, electricity, light, and water had been disrupted. The landing mission was aborted, and the astronauts and controllers on Earth scrambled to come up with emergency procedures. The crippled spacecraft continued to the moon, circled it, and began a long, cold journey back to Earth. The astronauts and mission control were faced with enormous logistical problems in stabilizing the spacecraft and its air supply, and providing enough energy to the damaged fuel cells to allow successful reentry into Earth’s atmosphere. Navigation was another problem, and Apollo 13’s course was repeatedly corrected with dramatic and untested maneuvers. On April 17, with the world anxiously watching, tragedy turned to triumph as the Apollo 13 astronauts touched down safely in the Pacific Ocean.
1972 – Three North Vietnamese divisions attack An Loc with infantry, tanks, heavy artillery and rockets, taking half the city after a day of close combat. An Loc, the capital of Binh Long Province, was located 65 miles northwest of Saigon. This attack was the southernmost thrust of the three-pronged Nguyen Hue Offensive (later more commonly known as the “Easter Offensive”), a massive invasion by North Vietnamese forces designed to strike the knockout blow that would win the war for the communists. The attacking force included 14 infantry divisions and 26 separate regiments, with more than 120,000 troops and approximately 1,200 tanks and other armored vehicles. The main North Vietnamese objectives, in addition to An Loc in the south, were Quang Tri in the north, and Kontum in the Central Highlands. Initially, the South Vietnamese defenders in each case were almost overwhelmed, particularly in the northernmost provinces, where the South Vietnamese abandoned their positions in Quang Tri and fled south in the face of the enemy onslaught. In Binh Long, the North Vietnamese forces crossed into South Vietnam from Cambodia to strike first at Loc Ninh on April 5, then quickly encircled An Loc, holding it under siege for almost three months while they made repeated attempts to take the city. The defenders suffered heavy casualties, including 2,300 dead or missing, but with the aid of U.S. advisors and American airpower, they managed to hold An Loc against vastly superior odds until the siege was lifted on June 18. Fighting continued all over South Vietnam throughout the summer months, but eventually the South Vietnamese forces prevailed against the invaders, even retaking Quang Tri in September. With the communist invasion blunted, President Nixon declared that the South Vietnamese victory proved the viability of his Vietnamization program, instituted in 1969 to increase the combat capability of the South Vietnamese armed forces.
1974 – Western Union (in cooperation with NASA and Hughes Aircraft) launches the United States’ first commercial geosynchronous communications satellite, Westar 1.
1980 – US boycotted the Summer Olympics in Moscow.
1990 – The Soviet government officially accepts blame for the Katyn Massacre of World War II, when nearly 5,000 Polish military officers were murdered and buried in mass graves in the Katyn Forest. The admission was part of Soviet leader Mikhail Gorbachev’s promise to be more forthcoming and candid concerning Soviet history. In 1939, Poland had been invaded from the west by Nazi forces and from the east by Soviet troops. Sometime in the spring of 1940, thousands of Polish military officers were rounded up by Soviet secret police forces, taken to the Katyn Forest outside of Smolensk, massacred, and buried in a mass grave. In 1941, Germany attacked the Soviet Union and pushed into the Polish territory once held by the Russians. In 1943, with the war against Russia going badly, the Germans announced that they had unearthed thousands of corpses in the Katyn Forest. Representatives from the Polish government-in-exile (situated in London) visited the site and decided that the Soviets, not the Nazis, were responsible for the killings. These representatives, however, were pressured by U.S. and British officials to keep their report secret for the time being, since they did not want to risk a diplomatic rupture with the Soviets. As World War II came to an end, German propaganda lashed out at the Soviets, using the Katyn Massacre as an example of Russian atrocities. Soviet leader Joseph Stalin flatly denied the charges and claimed that the Nazis were responsible for the slaughter. The matter was not revisited for 40 years. By 1990, however, two factors pushed the Soviets to admit their culpability. First was Gorbachev’s much publicized policy of “openness” in Soviet politics. This included a more candid appraisal of Soviet history, particularly concerning the Stalin period. Second was the state of Polish-Soviet relations in 1990. The Soviet Union was losing much of its power to hold onto its satellites in Eastern Europe, but the Russians hoped to retain as much influence as possible. In Poland, Lech Walesa’s Solidarity movement was steadily eroding the power of the communist regime. The Katyn Massacre issue had been a sore spot in relations with Poland for over four decades, and it is possible that Soviet officials believed that a frank admission and apology would help ease the increasing diplomatic tensions. The Soviet government issued the following statement: “The Soviet side expresses deep regret over the tragedy, and assesses it as one of the worst Stalinist outrages.” Whether the Soviet admission had any impact is difficult to ascertain. The communist regime in Poland crumbled by the end of 1990, and Lech Walesa was elected president of Poland in December of that year. Gorbachev resigned in December 1991, which brought an effective end to the Soviet Union.
1991 – Speaking at Maxwell Air Force Base in Montgomery, Alabama, President Bush warned Iraq the United States would “not tolerate any interference” with the international relief effort for Kurdish refugees.
1993 – The day before a visit by Pres. Bush, fourteen people were arrested in Kuwait for plotting to assassinate him. Kuwaiti officials said the plot was organized by Iraqi intelligence.
1993 – NATO forces began combat patrols over Bosnia to enforce a UN ban on flights.
1996 – The US agreed to close the Futenma Air Station at Okinawa, Japan. The 1200 acre base is surrounded by the densely populated city of Ginowan. Marine Corps Air Station, Futenma began in 1945 as a bomber base. Construction of hangars and barracks began in 1958. The airfield was commissioned as a “Marine Corps Air Facility” in 1960 and became an Air Station in 1976. Located within Ginowan City, Okinawa, the Air Station is home to approximately 4,000 Marines and Sailors. It is capable of supporting most aircraft and serves as the base for Marine Aircraft Group 36, Marine Air Control Group 18, and Marine Wing Support Squadron 172. The Air Station provides support for the III Marine Expeditionary Force and for Marine Corps Base, Camp Butler. Since 15 January 1969 MCAS Futenma serves as a United Nations air facility and a divert base for Air Force and Naval aircraft operating in the vicinity of Okinawa.
1999 – NATO bombs were dropped on Pristina. Yugoslav infantry troops crossed into northeastern Albania for a short time and clashed with Albanian border police. Refugees in Albania reported gang-rapes and murders by Serbian soldiers.
2001 – With the crew of a U.S. spy plane safely back in the United States, American officials gave their detailed version of what happened when the plane collided with a Chinese fighter on April 1; the United States said its plane was struck by the jet. China maintained that the U.S. plane rammed the fighter.
2002 – Yasser Arafat issued a statement condemning terrorism and planned to meet with Colin Powell the next day. Hamas declared it had no intention of halting attacks.
2003 – In the 26th day of Operation Iraqi Freedom US troops pushed into Tikrit. Marines found 7 missing US troops, including Army Specialist Shoshana Johnson, on the road between Baghdad and Tikrit. Army engineers worked to help restore electricity in Baghdad.
2003 – U.S.-led forces announced the capture of Watban Ibrahim Hasan, a half-brother of and adviser to Saddam Hussein. Watban is one of three of Saddam’s half-brothers (both share the same mother). During his tenure as Interior Minister, he is accused of having overseen the deportations, torture, and executions of hundred of prisoners. Some of those executions were reportedly taped with copies kept at the ministry. Despite his family ties to Saddam and his position, Watban was not thought to be fully trusted by Saddam Hussein. Watban is believed to have been shot in the leg by Uday Husein during a party in 1995. British SAS troops were believed to have orchestrated the capture of Watban Ibrahim Hasan Al-Tikriti near the town of Rabia, on the road between the city of Mosul and the Syrian border, as he was attempting to flee to Syria. He is the 5 of Spades on the Most-Wanted-Deck-of-Cards.
2004 – Pres. Bush defended his Iraq policy, vowed no retreat and conceded the need for UN help in a televised press conference.
2004 – Cuba agreed to buy $13 million in food from American companies and reached a tentative deal for up to $10 million in farm goods from California.
2004 – A 2,500-strong U.S. force, backed by tanks and artillery, pushed to the outskirts of the Shiite holy city of Najaf for a showdown with a radical cleric. One soldier was killed enroute. US forces in Fallujah killed over 100 insurgents.
2009 – The U.S. federal government rescinds travel and gift restrictions to Cuba.
Congressional Medal of Honor Citations for Actions Taken This Day
WESTON, JOHN F.
Rank and organization: Major, 4th Kentucky Cavalry. Place and date: Near Wetumpka, Ala., 13 April 1865. Entered service at: Kentucky. Birth: Kentucky. Date of issue: 9 April 1898. Citation: This officer, with a small detachment, while en route to destroy steamboats loaded with supplies for the enemy, was stopped by an unfordable river, but with 5 of his men swam the river, captured 2 leaky canoes, and ferried his men across. He then encountered and defeated the enemy, and on reaching Wetumpka found the steamers anchored in midstream. By a ruse obtained possession of a boat, with which he reached the steamers and demanded and received their surrender.
COX, ROBERT EDWARD
Rank and organizarion: Chief Gunner’s Mate, U.S. Navy. Born: 22 December 1855, St. Albans, W. Va. Accredited to: West Virginia. G.O. No.: 43, 14 April 1921. (Medal presented by President Harding.) Citation: For extraordinary heroism on U.S.S. Missouri 13 April, 1904. While at target practice off Pensacola, Fla., an accident occurred in the after turret of the Missouri whereby the lives of 5 officers and 28 men were lost. The ship was in imminent danger of destruction by explosion, and the prompt action of C.G. Cox and 2 gunners’ mates caused the fire to be brought under control, and the loss of the Missouri, together with her crew, was averted.
Rank and organization. Chief Gunner’s Mate, U.S. Navy. Born: 20 January 1867, Norway. G.O. No.: 160, 26 May 1904. Citation: Serving on board the U.S.S. Missouri, for extraordinary heroism in entering a burning magazine through the scuttle and endeavoring to extinguish the fire by throwing water with his hands until a hose was passed to him, 13 April 1904.
Rank and organization: Chief Boatswain, U.S. Navy. Born: 24 May 1876, Goteborg, Sweden. Accredited to: New York. G.O. No.: 142, 4 December 1924. Citation: For gallant conduct upon the occasion of the disastrous fire of accidentally ignited powder charges, which occurred in the forward turret of the U.S.S. Kearsage during target practice on 13 April 1906. Chief Boatswain Nordstrom, then chief boatswain’s mate, was among the first to enter the turret in order to assist in bringing out the injured.
SCHEPKE, CHARLES S.
Rank and organization: Gunner’s Mate First Class, U.S. Navy. Born: 26 December 1878, New York, N.Y. Accredited to: New York. G.O. No.: 160, 26 May 1904. Citation: For extraordinary heroism while serving on the U.S.S. Missouri in remaining by a burning magazine and assisting to extinguish the fire, 13 April 1904.
GAUJOT, JULIEN E.
Rank and organization: Captain, Troop K, 1st U.S. Cavalry. Place and date: At Aqua Prieta, Mexico, 13 April 1911. Entered service at: Williamson, W. Va. Birth: Keweenaw, Mich. Date of issue: 23 November 1912. Citation: Crossed the field of fire to obtain the permission of the rebel commander to receive the surrender of the surrounded forces of Mexican Federals and escort such forces, together with 5 Americans held as prisoners, to the American line.
ANDERSON, BEAUFORD T.
Rank and organization: Technical Sergeant, U.S. Army, 381st Infantry, 96th Infantry Division. Place and date: Okinawa, 13 April 1945. Entered service at: Soldiers Grove, Wis. Birth: Eagle, Wis. G.O. No.: 63, 27 June 1946. Citation: He displayed conspicuous gallantry and intrepidity above and beyond the call of duty. When a powerfully conducted predawn Japanese counterattack struck his unit’s flank, he ordered his men to take cover in an old tomb, and then, armed only with a carbine, faced the onslaught alone. After emptying 1 magazine at pointblank range into the screaming attackers, he seized an enemy mortar dud and threw it back among the charging Japs, killing several as it burst. Securing a box of mortar shells, he extracted the safety pins, banged the bases upon a rock to arm them and proceeded alternately to hurl shells and fire his piece among the fanatical foe, finally forcing them to withdraw. Despite the protests of his comrades, and bleeding profusely from a severe shrapnel wound, he made his way to his company commander to report the action. T/Sgt. Anderson’s intrepid conduct in the face of overwhelming odds accounted for 25 enemy killed and several machineguns and knee mortars destroyed, thus single-handedly removing a serious threat to the company’s flank.
KERSTETTER, DEXTER J.
Rank and organization: Private First Class, U.S. Army, Company C, 130th Infantry, 33d Infantry Division. Place and date: Near Galiano, Luzon, Philippine Islands, 13 April 1945. Entered service at: Centralia, Wash. Birth: Centralia, Wash. G.O. No.: 97,1 November 1945. Citation: He was with his unit in a dawn attack against hill positions approachable only along a narrow ridge paralleled on each side by steep cliffs which were heavily defended by enemy mortars, machineguns, and rifles in well-camouflaged spider holes and tunnels leading to caves. When the leading element was halted by intense fire that inflicted 5 casualties, Pfc. Kerstetter passed through the American line with his squad. Placing himself well in advance of his men, he grimly worked his way up the narrow steep hogback, meeting the brunt of enemy action. With well-aimed shots and rifle-grenade fire, he forced the Japs to take cover. He left the trail and moving down a cliff that offered only precarious footholds, dropped among 4 Japs at the entrance to a cave, fired his rifle from his hip and killed them all. Climbing back to the trail, he advanced against heavy enemy machinegun, rifle, and mortar fire to silence a heavy machinegun by killing its crew of 4 with rifle fire and grenades. He expended his remaining ammunition and grenades on a group of approximately 20 Japs, scattering them, and returned to his squad for more ammunition and first aid for his left hand, which had been blistered by the heat from his rifle. Resupplied, he guided a fresh platoon into a position from which a concerted attack could be launched, killing 3 hostile soldiers on the way. In all, he dispatched 16 Japs that day. The hill was taken and held against the enemy’s counterattacks, which continued for 3 days. Pfc. Kerstetter’s dauntless and gallant heroism was largely responsible for the capture of this key enemy position, and his fearless attack in the face of great odds was an inspiration to his comrades in their dangerous task.
NORRIS, THOMAS R.
Rank and organization: Lieutenant, U.S. Navy, SEAL Advisor, Strategic Technical Directorate Assistance Team, Headquarters, U.S. Military Assistance Command. Place and date: Quang Tri Province, Republic of Vietnam, 10 to 13 April 1972. Entered service at: Silver Spring, Md. Born: 14 January 1944, Jacksonville, Fla. Citation: Lt. Norris completed an unprecedented ground rescue of 2 downed pilots deep within heavily controlled enemy territory in Quang Tri Province. Lt. Norris, on the night of 10 April, led a 5-man patrol through 2,000 meters of heavily controlled enemy territory, located 1 of the downed pilots at daybreak, and returned to the Forward Operating Base (FOB). On 11 April, after a devastating mortar and rocket attack on the small FOB, Lt. Norris led a 3-man team on 2 unsuccessful rescue attempts for the second pilot. On the afternoon of the 12th, a forward air controller located the pilot and notified Lt. Norris. Dressed in fishermen disguises and using a sampan, Lt. Norris and 1 Vietnamese traveled throughout that night and found the injured pilot at dawn. Covering the pilot with bamboo and vegetation, they began the return journey, successfully evading a North Vietnamese patrol. Approaching the FOB, they came under heavy machinegun fire. Lt. Norris called in an air strike which provided suppression fire and a smoke screen, allowing the rescue party to reach the FOB. By his outstanding display of decisive leadership, undaunted courage, and selfless dedication in the face of extreme danger, Lt. Norris enhanced the finest traditions of the U.S. Naval Service.<|endoftext|>
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PDF chapter test
In the previous sections, we learned about linear momentum.
In this section, we will discuss the conservation of linear momentum law and prove them with an example.
Principle of conservation of linear momentum:
There is no change in the linear momentum of a system of bodies as long as no net external force acts on them.
Let us prove the conservation of linear momentum law with the following example.
Proof:
Consider two bodies, A and B, having masses ${m}_{1}$ and ${m}_{2}$, move with an initial velocity ${u}_{1}$ and ${u}_{2}$ in a straight line.
Let the initial velocity of the body, A be higher than that of the body, B. i.e., ${u}_{1}\phantom{\rule{0.147em}{0ex}}\succ \phantom{\rule{0.147em}{0ex}}{u}_{2}$.
During a period of time $$t second$$, they tend to have a collision. After the impact, both bodies move along the same straight line with a velocity ${v}_{1}$ and ${v}_{2}$, respectively.
Force on body B due to A,
${F}_{B}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{{m}_{2}\left({v}_{2}-{u}_{2}\right)}{t}$
Force on body A due to B,
${F}_{A}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{{m}_{1}\left({v}_{1}-{u}_{1}\right)}{t}$
By Newton’s III law of motion,
$$Action\ force\ =\ Reaction\ force$$
$\begin{array}{l}{F}_{A}=-{F}_{B}\\ \\ \phantom{\rule{0.147em}{0ex}}\frac{{m}_{1}\left({v}_{1}-{u}_{1}\right)}{t}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}-\left(\phantom{\rule{0.147em}{0ex}}\frac{{m}_{2}\left({v}_{2}-{u}_{2}\right)}{t}\right)\\ \\ \frac{{\left(m}_{1}{v}_{1}\right)-\left({m}_{1}{u}_{1}\right)}{t}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}\left(\frac{{\left(m}_{2}{v}_{2}\right)-\left({m}_{2}{u}_{2}\right)}{t}\right)\\ \\ \mathit{Remove}\phantom{\rule{0.147em}{0ex}}"t"\phantom{\rule{0.147em}{0ex}}\mathit{from}\phantom{\rule{0.147em}{0ex}}\mathit{bothsides},\\ \\ {\left(m}_{1}{v}_{1}\right)-\left({m}_{1}{u}_{1}\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}\left({\left(m}_{2}{v}_{2}\right)-\left({m}_{2}{u}_{2}\right)\right)\\ \\ {\left(m}_{1}{v}_{1}\right)-\left({m}_{1}{u}_{1}\right)\phantom{\rule{0.147em}{0ex}}=-\phantom{\rule{0.147em}{0ex}}{\left(m}_{2}{v}_{2}\right)+\left({m}_{2}{u}_{2}\right)\\ \\ {\left(m}_{1}{v}_{1}\right)+\phantom{\rule{0.147em}{0ex}}{\left(m}_{2}{v}_{2}\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\left({m}_{1}{u}_{1}\right)+\left({m}_{2}{u}_{2}\right)\phantom{\rule{0.147em}{0ex}}-\left(\mathit{eq}.\phantom{\rule{0.147em}{0ex}}1\right)\end{array}$
The above equation (eq. 1) confirms in the absence of an external force, the algebraic sum of the momentum after collision is numerically equal to the algebraic sum of the momentum before collision.
Hence, the conservation of linear momentum law is proved.<|endoftext|>
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The next mission that Nasa sends to Mars will use methods originally developed to find the the oldest life on Earth to hunt down measurements that could indicate signs of ancient life once existing on the planet.
The techniques, known as 'spatially resolved biosignature analysis', involve procedures like X-ray fluorescence and Raman spectroscopy. They're capable of mapping the elements, minerals and organic components of bits of rock as small as the width of a human hair.
Major step up
They're also major step up from tools available on the rovers currently exploring our second-closest neighbour in the solar system.
"Previous missions to Mars have used a relatively broad brush - analyzing average chemistry over roughly the size of a postage stamp - to 'follow the water' and seek ancient habitable environments," said Ken Williford, from the Jet Propulsion Laboratory'.
"Mars 2020 takes the next natural step in its direct search for evidence of ancient microbial life, focusing measurements to the microbial scale and producing high-resolution maps over similarly postage stamp-sized analytical areas."
As well as hunting for signs of ancient life, the rover will collect samples that will hopefully one day be examined in labs back home. It'll collect about 30 to 40 rock and sediment samples, seal them in titanium tubes, and then leave them in a safe place for future pickup.
"Mars 2020 represents a crucial first step towards a possible Mars sample return. Our objective is to collect a diverse set of samples from our landing site with the best potential to preserve records of the evolution of Mars - including the presence of life if it was there," added Williford.
"We'll use our onboard instruments to provide the critical field context that future scientists would need to understand the measurements made back on Earth."
Via Science Daily<|endoftext|>
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# Solve the equation t^2+10=6t by completing square method?
Oct 27, 2016
$t = 3 - i$ or $t = 3 + i$
#### Explanation:
${t}^{2} + 10 = 6 t$ can be written as
${t}^{2} - 6 t + 10 = 0$
or ${t}^{2} - 6 t + 9 + 1 = 0$
Now we use the identity ${\left(x + 1\right)}^{2} = {x}^{2} + 2 x + 1$ and imaginary number $i$ defined by ${i}^{2} = - 1$
or ${t}^{2} - 2 \times 3 t + {3}^{2} - \left(- 1\right) = 0$
or ${\left(t - 3\right)}^{2} - {i}^{2} = 0$
Now using identity ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, this becomes
$\left(t - 3 + i\right) \left(t - 3 - i\right) = 0$
Hence, either $t - 3 + i = 0$ i.e. $t = 3 - i$
or $t - 3 - i = 0$ i.e. $t = 3 + i$<|endoftext|>
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Gene therapy refers to the alteration of the existing genes to cure diseases. This is done by inserting genetically modified or normal genetic material into the cell’s genome to correct the abnormality.
Example of a gene abnormality is the brain tumor. The reason behind the formation of this tumor is the rapid and abnormal growth of the cells due to a mutation in a gene. Gene therapy uses a harmless virus to insert a normal gene into the target cells.
Some other theoretical uses of gene therapy include treatment of detrimental conditions such as hemophilia, cystic fibrosis, muscular dystrophy, and sickle cell anemia. There are three basic approaches used in gene therapy:
- Replacing a mutated gene (the gene which causes the disease) with a healthy copy of the gene
- Deactivating or “knocking out” a mutated gene that does not function properly
- Introducing a new gene into the human body to help fight against certain diseases
The following sections explore some of the most popular gene therapy approaches and how is it used for the treatment of specific disease.
Types of Gene Therapy
Gene therapy is of two types depending on the type of cells being treated:
Somatic gene therapy: This type of gene therapy involves the transfer of DNA into the cells that do not produce eggs or sperms. With this type of therapy, the only person whose cells are altered will be affected. The effects will not be passed on to the patient’s children.
Germline gene therapy: This type of gene therapy involves the transfer of DNA to the cells that produce eggs and sperms. The effects of this therapy will be passed on to the patient’s children and further generations.
Gene Therapy Techniques
Gene Augmentation Therapy
This technique is used for the treatment of diseases, which are caused by a mutation that stops a gene from producing a functional product such as a protein. In this therapy, DNA containing a functional copy of the lost gene is added back into the target cell. The new gene produces the functional protein, which was not being produced earlier. For example, this technique can be used for the treatment of cystic fibrosis by introducing a functional copy of the gene to correct the disease.
Gene Inhibition Therapy
This gene therapy technique is used for treating diseases due to abnormal gene activity including certain infectious diseases, cancer, and inherited diseases. In this technique, a gene is introduced whose functional protein will either inhibit the expression of another gene or interfere with the activity of the product of another gene.
The main aim of this technique is to eliminate the activity of the gene that causes the growth of diseased cells. For example, cancer is caused by overexpression of oncogenes (genes which cause cancer). So, by the elimination of the activity of such oncogenes by gene inhibition therapy, further cell growth, and cancer can be treated or prevented. Gene therapy in cancer has advanced a lot in the last few years.
Killing Specific Cells
This type of gene therapy is used for diseases such as cancer. It works by the destruction of certain groups of cells. A DNA is inserted into the diseased cells that cause them to die. This is done by inserting the DNA that contains a “suicide gene” and produces a highly toxic product that kills the diseased cell. It also works by inserting DNA, which produces a protein that marks the cells so that the diseased cells are attacked by the body’s own immune system. It is important to target DNA properly to avoid the death of normally functioning cells.
How does gene therapy work?
Gene therapy works by introducing or inserting a gene with the use of carriers known as vectors. Most of the times, the vectors used are viruses because they can deliver the new gene by entering the cell.
These viruses are modified so that they do not cause any disease when introduced in patients. Some examples of viral vectors used are DNA virus, retrovirus, and adenovirus. This vector can be either directly injected or given intravenously into a specific tissue inside the body, where it can be taken up by the cells. This is known as in situ somatic gene therapy.
Another way is to take a sample of the patient’s cells and expose them to the vector in a laboratory setting. The cells containing the vector are put back into the patient’s body. This is known as ex vivo somatic gene therapy. If the treatment becomes successful, the newly delivered gene by the vector will start making a functional protein.
Potential Applications of Gene Therapy in Medicine
Research is being carried out to find out the potential applications of gene therapy in the treatment of a number of diseases such as cancer, infectious diseases, cardiac diseases, neurological disorders and inherited conditions.
Gene Therapy in Cancer Treatment
Gene therapy techniques used against cancer include the introduction of tumor suppressor genes, the introduction of genes that cause apoptosis of cells (cause them to die), the introduction of genes which inhibit tumor angiogenesis, genes which code for an enzyme to convert prodrugs to active drugs and immunotherapy.
Gene therapy is being tested in clinical trials for cancers such as glioblastoma, metastatic melanoma, head and neck cancer, prostate cancer, and colorectal cancer. Cancers such as melanomas are highly sensitive to immunotherapy and hence gene therapy vaccines are under development for such cancer.
Some gene therapies are being developed to target mutated genes in cancer patients, an example of which is the use of p53 gene for the treatment of neck and head cancer.
Gene Therapy in Neurological Disorders Treatment
Examples of the neurological disorder that can be treated with gene therapy include Parkinson’s disease, Alzheimer’s disease, and motor neuron disease. ProSavin is currently under trial for the treatment of Parkinson’s disease which is caused by the deficiency of dopamine in the brain.
ProSavin delivers genes that encode for enzymes required for dopamine synthesis in the brain through a lentivirus vector. The aim of this gene therapy is to start the production of dopamine and restore its levels in the brain. ProSavin is under Phase I and Phase II clinical trials and has shown some clinical efficacy. Further trials are being conducted to analyze the safety and efficacy of higher doses of ProSavin.
Gene Therapy in Infectious Disease Treatment
Gene therapy vaccines are currently being developed and tested for treating infectious diseases such as tuberculosis, malaria, HIV, and influenza. A genetically modified vaccine comprising a recombinant fowlpox virus and a recombinant MVA virus (each containing tuberculosis antigen) is being tested for treatment of tuberculosis.
A Phase II study of gene therapy is being conducted in 74 patients with HIV-1 infection. The patients are being given an anti-HIV ribozyme (OZI) or placebo. Results have shown reduced viral load in patients receiving OZI as compared to placebo. Such kind of therapeutic vaccines can make it possible to preserve the immune system and reduce the need for lifelong treatment with antiretroviral drugs in patients with HIV.
Gene Therapy in Inherited Disorders
In patients with inherited disorders, gene therapy can be used to replace a defective gene or missing gene, or start the expression of a missing biological factor or enzyme or protein. In patients with Hemophilia, therapies to deliver genes that express the missing factors VIII and IX could be designed. With such a treatment, individuals with hemophilia will no longer need injections of exogenous clotting factors. Research is ongoing with the use of adeno-associated virus and lentivirus vectors. Another inherited disorder, cystic fibrosis, is caused by a mutation in the gene that encodes for a cellular protein called “cystic fibrosis transmembrane conductance regulator” (CFTR).
This mutation causes the abnormal transportation of ions in lungs and other cells. Researchers have cloned the CFTR gene and are now developing therapies to enable the expression of CFTR. Non-viral vectors are being tested to deliver gene therapy locally into the lungs through a nebulizer.
Gene Therapy in Cardiac Disease Treatment
Gene therapy is currently being investigated to target angiogenesis (formation of new blood vessels) during cardiac surgery and to improve calcium-handling mechanisms in heart failure.
Challenges and Future Perspective
The clinical application of gene therapy has shown promising results in the first phases of clinical trials. These early phase results suggest such trials would further be continued to correct the abnormalities or diseases caused by gene mutations.
Some challenges in the path of gene therapy are delivering the gene into the right place and activating it, avoiding immune response against the inserted genes, and making sure that the new gene does not disrupt the functioning of other genes.
Scientists and researchers are working hard to overcome these challenges and hope that gene therapy and modification will prove to be a cure for many diseases and successful clinical trials will create new opportunities.<|endoftext|>
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## Question 500:
1
a) Find the probability that a student had a score higher than 325?
a. Find the z-score for the score of 325 by subtracting the mean from this and dividing the result by the standard deviation:
b. (325-276.1)/34.4 = a z-score of 1.42. Use the percentile to z-score calculator to find the area under the normal curve above 1.42, you should get 7.78% probability of getting a score higher than 325
b) Find the probability that a student had a score between 250 and 305?
a. For between area questions we need to find the larger then subtract the smaller area.
b. 1st z-score is (305-276.1)/34.4 = .84 and the 2nd z-score is (250-276.1)/34.4 = -.758 (notice the negative sign).
c. Looking up the area for both using the z-score to percentile calculator gets us areas of 79.95% and 22.42%
d. Subtracting the two gets us an area of 79.95-22.42 = 57.53 meaning that 57.53% probability a students score is between 250 and 305
c) What percentage of the students had a test score that is greater than 250?
a. The z-score is (250-276.1)/34.4 = -.758 and the area above this is 77.57%
d) If 2000 students are randomly selected, how many will have a test score that is less than 300?
a. The z-score is (300-276.1)/34.4 = .6947 and has the area of 75.64% below a score of 300. If 2000 are selected, we'd expect .7564*2000 = 1512.8 of them to have scores below 300.
e) What is the lowest score that would place a student in the top 5% of the scores?
a. Construct an equation and solve for the unknown score
b. First find the z-score for the area above 5% by entering .95 and in the percentile to z-score calculator and you should get 1.642. Now use this z-score to solve for
c. ( x -276.1 )/34.4 = 1.642
d. x-276.1 = 56.48
e. x = 332.58 or you'd need a score of at least 332.58 or rounding up to 333 to be in the top 5% of scorers.
f) What is the highest score that would place a student in the bottom 25% of the scores?
a. Same process as in e, the z-score for the bottom 25% is -.6742.
b. ( x -276.1 )/34.4 = -.6742
c. x-276.1 = -23.19
d. x = 252.91
e. So a score of 252.91 is the highest a student can score to still be scoring in the lowest 25%.
g) A random sample of 60 students is drawn from this population. What is the probability that the mean test score is greater than 300?
a. We would use the z-score for a sample mean formula here. The only difference is that we use the standard error of the mean (SEM) instead of the standard deviation to calculate z. The SEM is the population standard deviation divided by the square root of the sample size= 34.4/SQRT(60)= 4.441021. We then substute the SEM into the formula we get (300-276.1)/4.441021 =5.38165. This z-score has a 1-sided above probability of less than .00001. In other words, it would be extremely rare given a sample of 60 students for the mean to be higher than 300. The reason it is so rare is that 60 is a large enough sample that it should be pretty close to the population mean of 276.1.
h) Are you more likely to randomly select one student with a test score greater than 300, or are you more likely to select a sample of students with a mean test score greater than 300? Explain?
a. The way to find out is compare the z-score we got in the last portion, which is from a mean, with a z-score from just 1 score. That z-score is (300-276.1)/34.4 = .69476. This z-score has a 1-sided probability of .2436 or 24.36%. So comparing that to the probability of getting a MEAN score above 300 of less than 1%, it's clear that it is more common to see an individual score higher. The reason is that as your sample size increases the mean of that sample will converge onto the population mean, which is 276.1. With just one student, its like having a sample size of 1, and it is more likely for a single point to vary further from the population mean.<|endoftext|>
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Karl Marx is a renowned philosopher, social scientist, historian and a revolutionist who influenced people and social thinkers in the 19th century. His social, economic and political ideas however gained acceptance after his death in 1883. And since then, almost a half of the world population claimed to be under Marxist regime until recently. As a member of the Karl Heinrich group, Karl Marx produced a radical critique to Christianity by implicating a liberal opposition to the Prussian autocracy. He consistently opposed the divisions amongst people and the exploitation of the poor by the rich. He advocated for equitable allocation of resources a system which he termed it as communism. Karl Marx jointly participated with Friedrich Engels in the people’s revolution movement, made the two close partners. The paper seeks to analyze the role played by Engels in the revolutionary struggle.
His challenging criticism led to his dismissal from the university by the Prussian government. He later migrated to France at the end of the 1843. While in Paris Marx contacted a group of German workers and started assisting them to air their view to the public in the series of his writings. Although this series of writings were never published until in 1930s he was able to influence and bring out the contrast between the alienated nature of labor under communist and capitalist regimes. It was during this time that Marx partnered with Friedrich Engels. The two revolutionists advocated for a communist regime since they argued that the capitalists continued to accumulate wealth at the expense of the poor (Marx & Engels Para 3).
Marx influenced people through his writings, and although his eminent influence caused him to be expelled from Paris in 1844, he never backed down. It is also believed that Engels was the major financier of Marx throughout his exile period. Having come from a poor background, Marx felt the need for enhancing and bridging the growing gap between the rich and the poor.
In the German ideology, Marx and Engels argued that human beings are generally productive and that their material and social needs triggers them into production. According to them, there was an eminent need to form a society and a state which enhances a collective human production. But as production progresses, people will start developing mode of production which will consequently lead to the formation of a communist regime. And once the plight of workers and their awareness starts motivating them, they establish a revolution. He once said that “Political Economy regards the proletarian ... like a horse, he must receive enough to enable him to work. It does not consider him, during the time when he is not working, as a human being. It leaves this to criminal law, doctors, religion, statistical tables, politics, and the beadle” (Marx & Engels Para 5).
Since Marx lived in a capitalist era, he argued that the continued hold on to power and accumulation of capital by the ruling class or the bourgeoisie will eventually lead to scarcity of need and as a result the mass will rebel to overturn this. He developed a strong dislike towards the bourgeois capitalist society since he always argued that they maximized their wealth at the expense of the poor. Instead he praised communism by saying, “Communism is the riddle of history solved, and it knows itself to be this solution.”
Marx observed that capitalism was unjust and that it aimed at exploiting the worker. He argued that capitalism used some dirty secrets that are not in the realm of mutual benefit to extract profit from others. And although he was never quoted concluding that capitalism was unjust, his arguments portrayed exactly that. He argued that the proletariat was being exploited by the wealthy class in their bid to accumulate more wealth. He was also noted to classify the capitalist as corrupt and unjust system which failed to effectively accommodate the needs of all people within the society. Some of the notable quotes which help us understand Marx feeling towards capitalism were; “capital is dead labor, which, vampire-like, lives only by sucking living labor, and lives the more, the more labor it sucks” (Marx & Engels Para 16). Although the later theorists did not justify the moral ground of capitalist as it’s argued by Marx, some supported his argument while others deferred with it. He therefore advocated for a revolution since according to him only the society had power to check the excess of capitalism. In his book he was noted saying, “Capital is reckless of the health or length of life of the laborer, unless under compulsion from society” (Marx & Engels Para 13).
Marx also classified people into different classes, depending with their capacity and influence in the society. He therefore argued that the society is divided into two major classes, the ruling also termed as the bourgeoisie and the working also known as the proletariat class. According to Engels, the bureaucrats manipulated the world in order to attain their goals, but they did so at the expense of their subjects. The bourgeoisie therefore triggered revolution as they seek to maximize wealth. However, he advocated that people should never allow this to happen since they should be guided by their conscious state of mind, social needs and satisfaction. To Marx, alienation happens whenever the human interaction is subordinated with other things such as money.
However, he noted that the reason why the mass continued to live under the oppressive regime of the rich is due to lack of unity. That is why he dedicated his entire life for the benefit of all. He said, “If we have chosen the position in life in which we can most of all work for mankind, no burdens can bow us down, because they are sacrifices for the benefit of all; then we shall experience no petty, limited, selfish joy, but our happiness will belong to millions, our deeds will live on quietly but perpetually at work, and over our ashes will be shed the hot tears of noble people”. But he still had some hope that one day, the mass will eventually unit against the rich and claims their position and stake in their land. Over the years, his prophesy have partly come true to the nations of the world. The French Revolution was a notable mass revolution triggered by the continued oppression of the mass by the rich few. Recently, the same was realized by the mass revolution in Egypt and Tunisia and also in many other Arab countries. We should therefore take note of Marxist argument seriously and let people have their stake in the country<|endoftext|>
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# Whats u need to graphing polynomials
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### Transcript of "Whats u need to graphing polynomials"
1. 1. TARUNGEHLOTSGraphing PolynomialsGraphing polynomials can be easy if you know what all the x-intercepts are. Or graphingpolynomials, by hand without a graphing calculator, can be only accomplished usingcalculus. We will look at methods of graphing more “manageable” polynomials as wellas some methods of quickly predicting behavior of the less-manageable.Using Function Shift Rules to Plot Even PowersYou can easily plot even powers of x if they are in a function-shift form since all evenpowers of x like y=x2, y=x4, y=x6, y=x8, etc have a similar “U” shape containing points(0,0), (1,1), and (-1,1), as shown below. The higher the power on x, the more “flattened”out the curve will be between x=1 and x=-1 and the steeper the curve will be for x>1 andx<-1.Example: Graph y = (x –3)10 + 1 by using function shift rules.This will be a shift of y=x10 right 3 and up 1. So, we get a flattened out “U” shaped curvewith the points (0,0), (1,1), and (-1,1) shifted right 3 and up 1 as shown below.
2. 2. Using Function Shift Rules to Plot Odd PowersYou can easily plot odd powers of x if they are in a function-shift form since all evenpowers of x like y=x3, y=x5, y=x7, y=x9, etc have a similar “S” shape containing points(0,0), (1,1), and (-1,-1), as shown below. The higher the power on x, the more “flattened”out the curve will be between x=1 and x=-1 and the steeper the curve will be for x>1 andx<-1, as was the case with the even powers.Example: Graph y = (x + 2)11 - 1 by using function shift rules.This will be a shift of y=x11 left 2 and down 1. So, we get a flattened out “S” shapedcurve with the points (0,0), (1,1), and (-1,-1) shifted left 2 and down 1 as shown below.Leading Term Test For PolynomialsOften, your polynomial does not fit the simple form of the previous two examples, andfunction shift rules do not apply. We can still make predictions about the behavior of thegraph by using the Leading Term Test, which in summary, states:The leading term of a polynomial will predict the behavior of the graph on the farright side and far left side.
3. 3. Example: Use the Leading Term Test to predict the behavior of the graph off(x) = -x5 + 2x + 1.The leading term is –x5. For positive x-values, this term is negative. So the graph willfall on the far right. Why does this happen? The leading term has the highest power ofx and thus “dominates” the function in determining what the graph does as x gets larger.For negative values of x, –x5 is positive. So the graph will rise on the far left. Again,the leading term has the highest power of x and thus “dominates” the function indetermining what the graph does as x takes on larger negative values.The leading term test tells us what happens on the far right and far left, and ourconclusions are verified below in the actual graph.Graphing Polynomials With Known ZerosIf you know the zeros of a polynomial, or they may be determined by factoring, then youcan use the procedure covered back in graphs of functions. The method and exampleare given below.Intercepts Method For Graphing Functions 1. Find and plot all intercepts. To find y-intercepts, let x=0 and solve for y. To find x-intercepts, let y=0 and solve for x. 2. Plot points on each side of each x-intercept. Find and plot at least one point between each two x-intercepts and one point on each side of the largest and smallest x-intercept. 3. Draw a smooth curve through the points from left to right.Example: Graph f(x) = x3 – x.First, rewrite as y = x3 – x. Now, find intercepts.When x = 0, we get y = 03 – 0 = 0. So our y-intercept is (0,0)When y=0, we get 0 = x3 – x. We solve this equation.0 = x3 – x Given0 = x(x2 – 1) Use Distributive Property to factor out x.0 = x(x + 1)(x – 1) Use Distributive Property to factor x2 – 1. CONTINUED ON NEXT PAGE
4. 4. x=0 The Zero Product Law allows us to let each factor = 0x+1=0x–1=0x = -1, x = 1, x = 0 Solve each equation using Addition Property of EqualitySo the x-intercepts are (-1,0), (1,0), and (0,0)We plot these points to getNow, find and plot points on each side of each of these x-intercepts by letting x= -2, x=-1/2, x=1/2, and x=2. Draw a smooth curve through the points. Note that you could alsouse the Leading Term Test to predict that the graph rises to the far right since x3 ispositive for x>0 and falls to the far left since x3 is negative for x<0. The Leading TermTest would allow us to skip plotting the points (2,6) and (-2,-6). We get the points x y -2 -6 2 6 -1/2 3/8 1/2 -3/8
5. 5. Multiplicity Rules For Polynomial ZerosThe multiplicity of a zero x=c refers to the power N on the factor (x – c)N where (x – c)N ispart of the factorization of the polynomial. 3 2For example, if f(x) = (x – 2) (x – 5) , then f(x) will have zero x=2 with multiplicity 3 andzero x = 5 with multiplicity 2. We know x=2 and x=5 are zeros since f(2) = 0 andf(5) = 0.If a real zero has even multiplicity, then the graph will “bounce off” the x-interceptcorresponding to the zero. (Remember that x-intercepts occur at real zeros).If a real zero has odd multiplicity, then the graph will “pass through” the x-interceptcorresponding to the zero. (Remember that x-intercepts occur at real zeros). 2 (x , the graph will (bounceoff) the point (5,0) since x=5 has even multiplicity and the graph will pass through thepoint (2,0) since x=2 has odd multiplicity. These conclusions are verified below in thegraph of f(x).Why Do These Multiplicity Rules Work?When a factor is raised to an even power, then values above and below the value of the 2 (x ,the x-value x = 4.9 results in 3 2f(4.9) = (4.9 – 2) (4.9 – 5) = 2.93 (-.1)2 = 24.389 0.01 = 0.24389and when x=5.1, we get 3 2f(5.1) = (5.1 – 2) (5.1 – 5) = 3.13 (.1)2 = 29.791 0.01 = 0.29791The even power resulted in a positive result each time, so the graph “bounces off”.When a factor is raised to an odd power, then values above and below the value of the 2 (x , thex-value x = 2.1 results in 3 2f(2.1) = (2.1 – 2) (2.1 – 5) = 0.13 (-2.9)2 = 0.001 8.41 = 0.00841and when x=1.9, we get 3 2f(1.9) = (1.9 – 2) (1.9 – 5) = (-0.1)3 (-3.1)2 = -0.001 9.61 = -0.00961The odd power resulted in a negative result at x=1.9 and a positive result at x=2.1.
6. 6. Example: Given f(x) = x3 – x2 – 5x – 3, use the fact that x=3 is a zero to write f(x) incompletely factored form. Then use this form to find all the zeros and then graph. Showall x-intercepts and make use of the Multiplicity Rules.Given that x=3 is a zero, this means that (x – 3) is a factor. So if we divide f(x) by(x – 3), we are able to write f(x) as a product of (x-3) and this result.We can divide f(x) using synthetic division. Synthetic division results in a zero remainder(as it should be!)So this means f(x) = (x – 3)(x2 + 2x + 1). We can factor this some more to getf(x) = (x – 3)(x +1)(x + 1) orf(x) = (x – 3)(x+1)2The factor (x-3) is raised to the 1st power so it has odd multiplicity 1.The factor (x+1) is raised to the 2nd power so is has even multiplicity 2.So, with this information, we know that : ƒ The graph has x-intercepts (3,0) and (-1,0). ƒ The graph “bounces off” the point (-1,0) and passes through the point (3,0). ƒ Also, the Leading Term test predicts that the graph rises up to the right since leading term x3 is positive for x>0 and falls to the left since leading term x3 is negative for x<0.Putting all this together, we conclude that the graph bounces off (-1,0) from below since itis negative on the far left. Also, we conclude that it must rise up again and pass through(3,0) in order to rise up on the far right. Finding a few more points like (0,-3) and (1,-8)also help to give us a little better picture of the graph. The graph is shown below.Students often ask, “How low will the graph go in between the x-intercepts? Is theresomething like the vertex (of a parabola) that we can calculate for higher orderpolynomials?” The answer is: Yes, you can find this “low point” exactly, but you need touse calculus in order to do so.<|endoftext|>
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The style of music, with which we are familiar in Western societies, has not long been in this form. Music is constantly evolving, since it represents the technology, culture and beliefs of society in the moment.
Early musicians probably used simple woodwind and plucked string instruments. In some societies, it was believed that the gods inspired music. Music and instruments were also thought to create different moods, and so were chosen to sooth when relaxing or to encourage a warrior spirit before battle. Single note melodies were played for ceremonies, dramas and other important occasions.
During the Middle Ages, music became increasingly more structured and complex. Until around 800, instrumental music was monophonic – a single melody line. After this, two or more different melody lines were played at the same time – it was polyphonic. Music was often used to accompany single voice lines or poems.
The Renaissance was a period of fast cultural development in Europe. Composers in this time experimented with new combinations of tone and rhythm. This was the beginning of counterpoint – different instruments imitated the first, with slight variations in melody, creating harmonies.
The Baroque Period brought counterpoint into an elaborate and expressive musical form. At this time, the modern system of Well Tempering was invented – the allocation of pitch as the 12 half steps that are used on the keyboard. Italy led the way in music of this era. Important composers of the period include JS Bach, Handel, Montverdi, Lully and Purcell.
During this era, the middle class had increasing access to written music and instruments. Music became a more popular form of entertainment. The lighter music, written simply for public enjoyment, became the Classical Style. It sought balance and expression of ideas and experiences. Well-known composers of this time include Haydn, Mozart and Beethoven.
Composers of the Romantic Period believed music should express the deepest of emotions and imaginations. Shorter, pleasing forms were created – such as the German Leider and character pieces. The piano underwent rapid invention and the symphony orchestra was at its highest point. Famous composers of this time include Schubert, Schumann, Berlioz, Mendelssohn, Liszt and Chopin.
Some composers sought to express the character of the people of their own countries: Nationalism. Among these composers were Grieg (Norway), Mussorgsky, Rinsky-Korsakov and Tchaikovsky (all from Russia), and Smetana and Dvorak (from Czechoslovakia). Some of Chopin’s pieces show how much he missed his homeland, Poland.
Impressionism was one of the significant art movements through the 1900s. Non-traditional harmonies, structures and ideas were included in composition. The movement originated with artists and poets in France. Debussy was one of the first to include such forms of imagery in his music.
Serialism, another 20th century style, used all 12 notes (all half steps), with no distinction of key and tone.
Music of the modern era has its roots in many of the earlier styles. The harmonies are usually more moving, and some styles have no discernible structure in form or harmony. Very popular in the early 21st Century is the NeoClassical and Symphonic forms definitely experiencing revival and renewal across all age groups.<|endoftext|>
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Two teams of researchers have apparently gone on record as saying they plan on cloning the mammoth. In 2008, when the mammoth genome was announced in the journal Nature, we took at look at that possibility, and concluded it wouldn't work. Given the recent press attention, we thought we'd rerun an updated version of the relevant section from our original report.
Given that the genome is often called the blueprint for an organism, Nature took the liberty of commissioning an evaluation of what it would take to rebuild the mammoth using that blueprint. The challenge is enormous: each one of the mammoth's chromosomes are likely to be over 100 million base pairs long; the average surviving fragment of DNA we've obtained from mammoth remains is under 200 bases long.
That means the sort of cloning technique that we use on currently living mammals wouldn't work, since it relies on a genome that's largely intact. The cloned cells can undoubtedly repair some DNA damage, but nothing like the scrambled fragments we have from mammoths. There's always the chance that some mammoth remains contain larger fragments of DNA, but basic chemistry indicates that we're unlikely to ever find anything close to an intact chromosome.
So the piece suggests starting from scratch, using a process similar to the one that constructed the first artificial genome. Unfortunately, that bacterial genome is about three orders of magnitude smaller than a single mammoth chromosome, and the techniques used are simply unlikely to scale. Mammoths also had dozens of chromosomes, and we'd need to get two copies of each into a single cell, safely encapsulated in a nucleus. We've only got techniques that work for some of this, and we've never tried any of the ones that work on a task approaching this scale.
Assuming we have two full sets of mammoth chromosomes together in a single nucleus, advances in stem cell research suggest we could reset them to an embryonic stem cell state using molecular tools. Unfortunately, we still don't know how to get these stem cells to develop into adult organisms without implanting them into a viable egg or embryo. That would mean we'd need the embryo of a closely related species to work with.
It would obviously be best to do this with elephants (as the teams of researchers have realized), both as egg donors and surrogates. But, apparently thanks to an aquatic lifestyle in the elephant's evolutionary past, they have a baroque reproductive tract and an internal organ arrangement that makes laparoscopy to harvest eggs a non-starter. So, the elephant represents yet another technical hurdle.
There are a host of other issues that are relatively minor in scale—we'd need a Y chromosome and sequence from enough individuals to create a diverse breeding population—but resurrecting the mammoth faces some technological obstacles that we haven't yet even started to try to overcome. A more likely solution, Nature concludes, would be to identify the regions of the genome that have diverged most significantly between elephants and mammoths, and engineer the mammoth equivalent back into an elephant's DNA. Depending how well we can identify these, the mammophant that we produce may be at least physically indistinguishable from artists' renderings we're all familiar with.
Overall, Nature's analysis is pretty persuasive. Given the technology we have now, it's tough to imagine putting a mammoth together, even given the complete genome sequence.
But it's difficult to predict how technology advances will proceed. The article quotes one of the researchers who lead the efforts to sequence the Neanderthal and Denisovan genomes, Svante Paabo, as saying he doesn't expect to see anything more than a mammophant in his lifetime. Of course, Paabo's in his 50s, and I'd imagine that, in his 20s, he wouldn't have expected to see a Neanderthal genome completed in his lifetime. He has done just that.
Nature, 2008. DOI: 10.1038/456310a<|endoftext|>
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# Year 5&6 Maths Championships Semi-Finals
Our Year 5 and 6 team travelled all the way to Priory Primary School to take part in the Year 5 and 6 Maths Championships
For Target 24, we were given 4 numbers and told we had to use all of them to try and make 24. To make 24 out of these numbers: 2, 6, 12 and 16.
12-6=6
6+2=8
16+8=24
To make it harder, there were some easy questions that you got 1 point for - 1, 5, 14 and 14 - or some hard ones that gave you 3 points! 10, 10, 12 and 20
Which could you do?
Star Battle had children working hard to put a star in every column, row and shape on a grid. The catch was, none of the stars could touch. Have a go yourself at Star Battle.
We were then given 16 statements about maths and had to decide if the statement was always true, sometimes true or never true.
One of the following statements is always true, one is sometimes true and one is never true. Can you work out which is which?
A multiple of 5 is always a multiple of 10.
You can find one more than a number.
An odd number add an odd number always equals an odd number.
We also had to give reasons why!
Our next task was to make a hexomino. This is where 6 squares are joined together to make a shape. There are 35 altogether and we had to find as many as we could. Once we had found them all, we had to work out the rotational symmetry and perimeter of each hexomino.
The penultimate round was all about mathematical language. We were given 20 phrases such as, “What S is another word for minus?” and had to come up with the correct mathematical word (subtract or subtraction).
Finally it was countdown. We had a set of numbers and a target number. The aim was to get as close to the target number as we possibly could in 3 minutes using the numbers only once.
There was a lot of teamwork involved as well of lots of serious brain work needed. Unfortunately, unlike our Year 3 and 4 team, the 5 and 6 team did not make it through to the finals in the summer.
Well done to all children taking part and thanks to Miss Owens for organising and hosting it.<|endoftext|>
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# Distance Problem 5 - Uniform Motion rt=d
Taught by YourMathGal
• Currently 4.0/5 Stars.
2775 views | 1 rating
Part of video series
Meets NCTM Standards:
Lesson Summary:
This lesson uses the rt=d formula to solve a word problem involving a car and a bus traveling in the same direction. The car is twice as fast as the bus and after 2 hours, it is 68 miles ahead of the bus. The lesson breaks down the problem step-by-step using a chart and visual aids to explain the equation. The answer is revealed at the end of the lesson: the bus speed is 34 mph and the car speed is 68 mph.
Lesson Description:
Solves this word problem using rt=d formula: A car and a bus set out at 2 pm from the same spot, headed in the same direction. The average speed of the car is twice the average speed of the bus. After 2 hours, the car is 68 miles ahead of the bus. Find the rate of the bus and the car.
Answer: Bus speed: 34 mph; Car speed: 68 mph
More free YouTube videos by Julie Harland are organized at http://yourmathgal.com
• How do you solve uniform motion problems?
• How do you solve equations using the formula rate * time = distance?
• How can you draw and use pictures to solve word problems using the formula d = rt?
• How can you come up with an equation to solve a uniform motion problem?
• How can you use a chart to solve a uniform motion problem?
• How can you check your solutions to a uniform motion problem?
• If a car and a bus set out at 2pm from the same spot headed in the same direction, the average speed of the car is twice as fast as the bus, and after 2 hours, the car is 68 miles ahead of the bus, how do you find the rate of the bus and the car?
• If you know the time of the bus and car, how can you write an expression for their rate and distance?
• #### Staff Review
• Currently 4.0/5 Stars.
This video shows another good thinking problem. It is another uniform motion word problem in which a sketch and a chart are critical to forming an equation and solving the problem. This is yet another scenario you might encounter in your class.<|endoftext|>
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Johannes Kepler is a famous mathematician, astronomer, and astrologer of the Scientific Revolution during the seventeenth century. Kepler has made some very important contribution to the fields of astronomy and mathematics. Without him we might not have made some discoveries until much later. He is one of the most important scientists of the Scientific Revolution. Johannes Kepler made some important contributions to astronomy and had some incredible works and accomplishments all due to his early developed love for science.
To begin, the early life of Johannes Kepler plays a big role in his development to love science. Kepler was born on December 27, 1571 in Weil der Stadt, Württemberg in Germany. He had two brothers and one sister. His mother was Katharina Guldenmann and his father was Heinrich Kepler, who is believed to have died in the Eighty Years’ War. His mother was an innkeeper's daughter, was a healer and herbalist, and she was tried for witchcraft. Kepler was born prematurely and so as a child he was very sickly. When he was young he got chicken pox and that gave him weak vision and crippled hands. On top of this his family was very poor. He was introduced to astronomy early and developed a love for it. He had observed a few astronomical events at an early age, which surely contributed to his love for it. He naturally understood mathematics very well starting at a young age and impressed many people with his mathematical abilities. For example, “he often impressed travelers at his grandfather's inn with his phenomenal mathematical faculty” (Johannes kepler: Early Years, n.d.). His intelligence earned him a scholarship to the University of Tübingen. There he studied theology and philosophy and was introduced to the ideas of Copernicus. In school, he was introduced to the Ptolemaic system and the Copernican system of planetary motion. He strongly believed and defended heliocentrism. He also became the chancellor of Tübingen in 1590. Kepler wanted to be a minister but instead took a job as a
mathematics and astronomy teacher at a Protestant school in Graz at the age of twenty-three. All these things are what made him one of the most influential and important scientist/mathematician in the Scientific Revolution.
In addition, Kepler has made many contributions to the scientific field. One of the most popular discoveries Kepler made was the elliptical orbit of the planets around the sun. This discovery is part of the three laws of planetary orbit. The first law states, “The path of the planets about the sun is elliptical in shape, with the center of the sun being located at one focus” (Kepler’s Three Laws, n.d.). The second law states that an imaginary line connected from the center of the sun to the center of the planet will sweep out in equal areas in equal amounts of time. The third law states that the ratio of the squares of the periods of two planets is equal to the ratio of the cubes of their distances from the sun. He also...<|endoftext|>
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A microgravity environment is a perfect place to demonstrate basic physics, i.e. Newton’s laws of motion. In the videos published by the NASA Johnson channel, astronauts aboard the International Space Station (ISS) just do that.
Newton’s laws of motion
In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.
Newton’s first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia. The key point here is that if there is no net force acting on an object (if all the external forces cancel each other out) then the object will maintain a constant velocity. If that velocity is zero, then the object remains at rest. If an external force is applied, the velocity will change because of the force.
In an inertial frame of reference, the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration (a) of the object: F = ma. It is assumed here that the mass m is constant.
The second law explains how the velocity of an object changes when it is subjected to an external force. The law defines a force to be equal to change in momentum (mass times velocity) per change in time. Newton also developed the calculus of mathematics, and the “changes” expressed in the second law are most accurately defined in differential forms. (Calculus can also be used to determine the velocity and location variations experienced by an object subjected to an external force.) For an object with a constant mass m, the second law states that the force F is the product of an object’s mass and its acceleration a:
F = m * a
For an external applied force, the change in velocity depends on the mass of the object. A force will cause a change in velocity; and likewise, a change in velocity will generate a force. The equation works both ways.
When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
The third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal force on object A. Notice that the forces are exerted on different objects. The third law can be used to explain the generation of lift by a wing and the production of thrust by a jet engine.<|endoftext|>
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Polar bears are members of bear family and are mostly found in the arctic regions. They do require extreme cold climates for survival and these Polar Regions are their only habitat. Baby polar bears are born as cute and healthy youngsters. They grow in to very big size if they survive maturity.
Female polar bears do give birth to the young ones between the age of four and eight. In most cases a female would give birth to the first set of cubes at the age of five Baby polar bears are mostly born in pairs. A female would give birth to at least three cubs. Only in rare cases will there be a one cub. The period from November to January is known as the maternity period for a polar bear. It gives birth to its young one in this period. The den were the cubs are born is known as ‘maternity dens’ as these are made specially to protect the new born babies from the freezing polar climates.
A baby polar bear is born small and helpless. Their body would have thick hair which helps them to survive the cold of the arctic. These hairs would make the body warm and helps in slowly adapting with the extreme cold climate of the polar region. At birth, a baby polar bear would weigh about a pound. These cubs would be 30 to 35 centimeters in length and starts growing very fast. New born polar bears will not have any senses so they are very much protected by their mothers from all enemies. The mothers keep their babies close to their body so as to keep the body of the cubs warm. A female polar bear do have special crevices within its body which can help to keep the body of cub warm as well.
A baby polar bear grows very fast. Like all other animals the mother does protect the child up to a certain period. During these times it is the calorie rich milk of their mothers which survives the life of a cub. Scientists say that the milk of a mother polar bear is rich in fat content. It is the mother which helps a cub to adapt itself with the climate of the arctic. She would emerge out of her den at the end of March or April so as to teach her babies the basic skills for hunting food. The cubs would be first treated with the food that is hunted by mothers and so these babies start eating solid food when they are four or five months old.
The baby polar bears stay with their mothers till the age of two and a half years. But the bears of the low arctic region leave their child even at the age of one and a half years. Within these years the baby polar bears would be taught how to catch the prey of their own. When these cubs have grown and attain a stage from which they can survive of their own, the mothers would chase them away to go and lead their own life.<|endoftext|>
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The Apache tribe is noted among Native Americans for the bravery with which they initially faced the arrival of the “white man.”
As a nomadic group of people, the Apache were the first of the tribespeople to master horse riding, according to legend. They snaked through Alaska and Canada, and down into North America.
The Apache eventually reached the southern part of the continent’s plateau, then split into two large groups. If there was a border of sorts between the groups, it was the Rio Grande.
At the dawn of the 18th century, Apache Native Americans largely populated the vast plains of Kansas. While they grew beans and other crops, the Apache’s favorite dish was buffalo meat. The tribespeople also used the animal’s skin to manufacture garments.
When the Apache were on the move, they rested in temporary homes called teepee’s, easily movable, particularly in times of some serious buffalo-hunting. They were constructed with long poles and buffalo hide.
The wikiups, also known as wigwams, were the Apaches’ more durable homes. Tree saplings were used to form the dome of the wigwam structure, and natural materials like tree bark was used to cover the small but comfortable home.
The Apache outfits in the portraits were frequently made of leather. Men opted for loincloths and shirts, while women wore dresses made of buckskin. Leather moccasins were the most common footwear, especially among men.
Clothing would be decorated with vegetable-pigment paints, beads and feathers. Strings of beads made from materials such as turquoise and jet, as well as shells and seeds, were worn in abundance by Apache women as bracelets, necklaces and chokers. Other jewelry included stitched or woven necklaces, pouches, and hair ornaments.
Despite their being more military-minded than some other Native American tribes, over time the Apache lost the lands they traditionally held. The Apaches’ first clashes were with the Spaniards, then with the U.S. government–both exhausting and costly.
Though for a brief period during the 18th century they lived in peace with the newcomers on the continent, things took a turn as their assets became subject to perpetual raiding, and their people were murdered.
The population significantly shrank over time, and the last of the great warriors who led the Apache into battle were eventually taken as prisoners. As of the 21st century, it is believed that descendants of this fierce Native American peoples number about 100,000.
We hope you are enjoying The Vintage News. Please consider helping us with our journey to bring popular historical content to everyone by becoming a supporter today. Thanks.Become a Supporter<|endoftext|>
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Angles are all over our world. Without the knowledge of how to measure and calculate between angles, we would have a difficult time engineering and using them in our everyday lives. One of the most important shapes in the engineering world is the triangle. Just the name itself is made up of tri (three) and angle. It’s fairly self-descriptive. But, moving on, let’s look at how angles are measured.
The first measurement of an angle is in regards to slope. If you look at the roof on your house you’ll notice that it probably has an angle to it. This angle is called slope and is measured in the amount of rise for a given amount of run. In other words, for every unit of lenght you have horizontally, how many units of height are there? A slope can also be described as pitch or percentage of slope. The percentage is the amount of rise divided by the amount of run. If your roof slope goes up 6 inches for every 12 inches of horizontal run, you have a 50% slope.
Another way of calculating an angle is with radians. If you take the radius of a circle and apply that length distance around the circumference of the circle, it will make a pie shape. The angle formed by that shape is one radian. Because it is related to the circle, you can also look at it this way. There are 2 x Pi radians in a full circle. That’s approximately 6.28 radians. It’s easiest to look at an illustration.
The final, and most common way of measuring an angle is with degrees. Again, we refer to a circle. Every circle has 360 degrees. An angle that is 90 degrees takes up on quarter of a circle and is also known as a right angle.
When you measure angles it is important to know what type of measurement is being used and how to decipher between them. After all if someone gives you a slope of 45 degrees, what would that convert to in a percentage? For more on how to convert between different angle measurements try the online conversion calculators below.
More Geometry Calculators<|endoftext|>
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# CHAPTER 10 CIRCLES
## (A) To Identify Parts of a Circle Centre Circumference
Diameter
Chord
Arc
Sector
Segment
(B) To Draw a Circle and Parts of a Circle Draw the circles below: 1. Draw a circle with center O and radius 2 cm.
## 3. Draw a diameter of 3cm passing through a point C on a circle with centre O.
4. Draw two chords of length 4cm passing through a point E on the circumference of the circle with radius 3cm.
5. Draw the sector of a circle given that the angle at the centre is 60o and its radius is 4cm.
(C) To Determine The Centre Or Radius Of A Circle By Construction The perpendicular bisector of chords of a circle intersect at the centre of the circle.
1. Determine the centre and radius of the circles given below: (a) (b)
## (D) To Understand and Use the Concept of Circumference to Solve Problems.
= =
Formula :
22 7 3.142
Circumference = 2r
(i)
To find the circumference of a circle given its diameter or radius 1. Find the circumference of a circle of (a) diameter 14 cm (b) radius 2.1 cm
2. Take = 3.142, find the circumference of the circles below: (a) (b)
3 cm 10 cm
3. Circle P has a radius of 21cm and circle Q has a diameter of 5.6cm. Find the difference in the circumference of circle P and Q. 22 (Take = ) 7
(ii)
To find the diameter or radius of a circle given its circumference 1. Take = 17.6cm. 22 , find the diameter of the circle with circumference 7
2. Take = 3.142, find the radius of a circle given its circumference is 314.2 cm.
3. A circle and a square with length 5 cm have the same perimeter. Find the radius of the circle, correct to 2 decimal places.
(iii) To solve problems involving the circumference of circles. 1. A circle is inscribed inside a square of length 1.4 cm as shown. Find (a) the circumference of the circle. 1.4cm (b) the difference in the perimeter of the square and the circle.
2. The tyre of a bicycle rotates 50 times per minutes. Given the diameter of the tyre is 42 cm, find the distance traveled by the bicycle in 5 minutes. Give your answer in meters.
3. XYZ is a right-angled triangle. Semicircles are drawn on all 3 sides of the triangle. Find the perimeter of the diagram. X
13 cm cm 5 cm
## (E) Arc of a Circle 7
The length of arc is proportional to the angle at the centre Formula: Length of Arc Angle at centre = Circumference 360o
Length of arc Angle At Centre = Circumference 360
(i) To find the length of the arc given the angle at the centre and the radius 1. Find the length of the arc of a circle with radius 4.9cm and an angle of 56o at the centre.
2. Find the value of x for each of the following circles: (a) (b) x
7 cm 72o
300o 3.5cm cm
3. The figure shows a circle of diameter 12.6cm and and arc AB that subtends an angle of 80o at the centre. Find the length of arc AB. A O
80
o
(ii) To find the angle at the centre given the length of the arc and the radius of the circle
1. Find the angle subtended at the centre of the circle, given (a) radius = 35mm, length of arc = 143mm (b) radius = 28cm, length of arc = 22cm 22 (Take = ) 7
2. Find the angle subtended at the centre of a circle by an arc of 22 length 12.1 cm, given that its radius is 3.5cm. (Take = ) 7
3. Find the value of x given that the circle with centre O has a radius of 14m and the minor arc AB is 44m long. O A x B
(iii) To find the radius of a circle given the length of the arc and the angle at the centre
1. Find the radius of the circle, given (a) angle at the centre = 36o, length of arc = 8.8cm (b) angle at the centre = 45o, length of arc = 121mm 22 (Take = ) 7
2. Find the radius of a circle given that an arc of length 19.8m subtends an angle of 180o at the centre of the circle.
3. Find the radius of the circle with centre O given that the major arc PQ is 121cm. Q O
30o
10
## (iv) To solve problems involving arcs of a circle
4. The minute hand of a clock has traveled one quarter of a revolution. Find the distance traveled by the tip of the minute hand given that the minute hand is 12.1cm.
5. The hour hand of a clock moves from 12am to 8am, covering a 2 length of 50 cm. Find the length of the hour hand. 7
6. The figure shows a right-angled triangle joined to a semicircle. find the perimeter of the figure.
6cm 8cm
11
7. The figure given shows a rectangle PQRS overlapped by two quadrants PAD and RCB of radius 7cm. If PS = 10cm and PQ = 14cm, calculate the perimeter of the shaded area.
## (F) Area of a Circle Formula: Area of circle = r2
(i) To find the area of a circle given the radius or diameter 1. Take =
22 , find the area of a circle with radius 7 cm. 7
2. Take =3.14, find the area of a circle with radius 5.6 cm.
3. Take =
## (ii) To find the radius or diameter given the area of a circle 12
1. The area of a circle is 154 cm2. Find its radius and diameter. (Take =
22 ) 7
2. A circle has an area of 78.55cm2. Determine the diameter of the circle given =3.142.
## 4. Determine the circumference of a circle which has an area of 616cm2.
13
5. The figure shows two concentric circles with centre O. Find the area of the annulus. Important! 3 cm O 4 cm Concentric Circles
= circles having the same centre
Annulus
18cm
## 22 , find the area of the shaded regions below. 7
(b)
14cm
O
24cm
42cm
14
2. The diameter of the circle has the same length as the sides of the square. Take =3.142, find the area of the figure below. 28 cm
3. Mr Lingam bought a pizza which is circular in shape. His daughter ate quarter of the cake. Given the area of the top surface of the remaining portion is 462 cm2, find the diameter of the pizza.
15
## (G) Area of a Sector of a Circle
1 circle Formula:
semicircle = circle
## Area of Sector Angle At Centre = = Area of Area of Circle 360 Sector
Area of Circle
(i) To find the area of a sector given the radius and angle at the centre 1. Find the area of the sector of a circle, given (a) Radius = 14cm (b) Radius = 24m o Angle at centre = 45 Angle subtended at centre = 210o (Take =
22 ) 7
2. Find the area of each of the shaded regions where O is the centre of the circle. (a) (b) O
72o
7cm
4.9m
16
(ii) To find the angle at the centre given the radius and area of a sector 1. Find the angle at the centre for the circles shown below. (a) (b) O
21 m 18.75cm2 Area of sector = 77 m2 2.8cm
2. The area of the sector of a circle, with the radius 4.2cm, is 15.28cm2. Find the angle at the centre.
17
(iii) To find the radius given the area of a sector and the angle at the centre 1. The area of the sector of a circle, with an angle of 70o at the centre is 22 m2. Find the radius of the circle. (Take =
22 ) 7
2. The sector of a circle subtends an angle of 60o at the centre of the circle. Given the area of the sector is 231cm2, find the diameter of the circle.
18
(iv) To solve problems involving area of sectors and area of circles 1. The diagram below shows a circle with centre O and a radius of 14mm. OAB is a right-angled triangle. Find the area of the shaded part.
O B A
2. Find the area of the shaded part for the diagram below. (Take =
22 ) 7
10cm 36o
6cm
19
PMR Past Year Questions (Chapter 10 Circles) 2004 Paper 1 Question 19 Diagram 12 shows a circle with centre O.
## The radius of the circle is 10 minor arc MN. (Use = A
55 2 1617 8 22 ) 7
## 1 cm. Calculate the length, in cm, of the 2
77 2
1155 8
2004 Paper 1 Question 20 In Diagram 13, POR is the diameter of circle PQR.
Given that PQ = QR and PR = 14cm, calculate the area, in cm2, of the shaded region. (Use =
22 ) 7
20
A 56
B 93
C 105
D 142
2004 Paper 1 Question 39 In Diagram 25, PQR is an arc of a circle with centre O and PRO is an equilateral triangle.
## The perimeter, in cm, of the whole diagram is (Use = A 42
22 ) 7
B 63
C 64
D 85
2005 Paper 1 Question 23 In Diagram 18, PQR is a straight line and PQS is a quadrant of a circle with centre Q.
The area if triangle QRS is 21cm2. Calculate the area, in cm2, of quadrant PQS. (Use = A 11
22 ) 7
B 19.25
C 38.5
D 154
21
2005 Paper 1 Question 24 In Diagram 19, PQR is an arc of a circle with centre O.
The radius of the circle is 14 cm and PT = 2 TS. Calculate the perimeter, in cm, of the whole diagram. (Use = A 66 B 94
22 ) 7
C 98
D 126
2005 Paper 1 Question 40 Diagram 25 shows a circle with centre O. The length of the minor arc PQ is 3.3 cm.
Calculate the radius, in cm, of the circle. (Use = A 2.42 B 4.50 C 14.85
22 ) 7
D 29.70
22
## Calculate the length, in cm, of the minor arc PQ. (Use = A 22 B 44 C 88
22 ) 7
D 176
2006 Paper 1 Question 21 Diagram 16 shows the surface of a table PQRSTU. PQST is a rectangle. PUT and QRS are semicircles with centres X and Y respectively.
It is given that PT = 1.4 m and PQ = 3 m. Calculate the perimeter, in m, of the surface of the table. (Use =
22 ) 7
23
A 10.4
B 13.2
C 14.8
D 17.6
2006 Paper 1 Question 22 Diagram 17 shows two sectors, QSR and UST, with the common centre S. RSU and QST are straight lines.
It is given that ST = 2QS. Calculate the area, in cm2, of the sector UST. (Use = A 11
22 ) 7
B 19.25
C 38.50
D 77
2007 Paper 1 Question 23 Diagram 16 shows a circle with centre O. PT, QU, RV and SW are diameters of the circle.
## Which of the following minor arcs is the longest? A TU B UV C WP
D RS 24
2007 Paper 1 Question 24 Diagram 17 shows a circle with centre O and radius 6 cm.
## Calculate the area, in cm2, of the shaded region. A 18 B 5.4 C 10.2
D 30.6
2007 Paper 1 Question 31 Diagram 22 shows a square OPQR and an arc RSP with centre O.
25
## 2008 Paper 1 Question 21 Diagram 12 shows a circle with centre O.
Calculate the area, in cm2, of the coloured region. (Use = A 105 B 116 C 539
22 ) 7
D 616
2008 Paper 1 Question 22 Diagram 13, O is the centre of the circle and POR is a diameter of the circle. PQRS is a rectangle.
26
It is given that PR = 10 cm and PQ = 6 cm. Find the area, in cm2, of the coloured region. A 25 48 B 25 96 C 100 48 D 100 96 2008 Paper 1 Question 30 In Diagram 19, PQRS is a square and STU is an arc of a circle with centre P.
The area of PQRS is 576 cm2. Calculate the length, in cm, of the arc STU. A 16 B 28 C 32 D 48
27
28<|endoftext|>
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The ‘basic’ school curriculum includes the ‘national curriculum’, religious education and sex education.
The national curriculum is a set of subjects and standards used by primary and secondary schools so children learn the same things. It covers what subjects are taught and the standards children should reach in each subject.
At Holly House, all pupils follow the National Curriculum, but adapted to meet their needs and their previous learning experiences. There is a 7 period day with lessons lasting 40 minutes. All classes start their day with Spelling, Punctuation and Grammar in their form groups. Apart from the main subjects there are opportunities to experience a broader curriculum based on developing social skills and improving behaviour. Holly House also offers Life Skills and Outdoor Education mainly with our KS3 pupils. This is aimed at preparing pupils for life after Holly House, you can find further information about Outdoor Education in our ‘Outdoor Ed’ section of the website.
The national curriculum is organised into blocks of years called ‘key stages’ (KS). At the end of each key stage, your child’s teacher will formally assess their performance to measure your child’s progress.
|3 to 4||Early years|
|4 to 5||Reception||Early years|
|5 to 6||Year 1||KS1||Phonics screening check|
|6 to 7||Year 2||KS1||Teacher assessments in English, maths and science|
|7 to 8||Year 3||KS2|
|8 to 9||Year 4||KS2|
|9 to 10||Year 5||KS2|
|10 to 11||Year 6||KS2||National tests and teacher assessments in English and Maths|
|11 to 12||Year 7||KS3||Teacher assessments|
|12 to 13||Year 8||KS3||Teacher assessments|
|13 to 14||Year 9||KS3||Teacher assessments|
|14 to 15||Year 10||KS4||Some children take GCSEs|
|15 to 16||Year 11||KS4||Most children take GCSEs or other national qualifications|
Assessment of progress
There are no longer levels of attainment attached to the National Curriculum. At Holly House we recognise that pupils are admitted at a fairly low starting point. We assess on entry their reading, comprehension and spelling age. We also assess their skills in maths. In Year 7 we also use Cognitive ability testing as an indicator set against national norms. To measure progress at Key Stage 2 we use the Age Appropriate National Curriculum content as a guide and use ‘ages’ to show where they are in comparison. At Key Stage 3 we have a learning objective based system using the 3 E’s. For each learning objective pupils can be ‘Engaged’ (orange – they have been experienced the learning) , ‘Embedded’ (green – they have been able demonstrate they have learnt the objective) or ‘Exceeded’ (blue – they have gone beyond the objective and extended their learning). This is then summarised for parents using a 5 point scale.
Special Educational Needs
All pupils at Holly House have a Statement/EHC plan. Their primary need will be SEMH (Social, Emotional or Mental Health) with Behaviour being the main priority. Pupils may well have additional needs and diagnosed conditions. For more information on Special Educational Needs you can visit the Derbyshire Website or follow this link to the SEND Local Offer.
To view our current Curriculum Policy, click the link below:
We also have Curriculum Statements for English and Maths. Please click the links below to find out more<|endoftext|>
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Using a computer to predict an infectious disease outbreak before it starts may sound like a bit of Philip K. Dick sci-fi, but scientists are coming close. In a new study, researchers have used machine learning—teaching computers to recognize patterns in large data sets—to make accurate forecasts about which animals might harbor dangerous viruses, bacteria, and fungi. Better predictions could help experts improve how they prevent and respond to disease outbreaks.
“I can’t emphasize enough how exciting a paper this is. I think it’s really going to resonate with the scientific community,” says Lynn Martin, a disease ecologist at University of South Florida, Tampa, who was not involved in the study.
Nearly all new infectious disease outbreaks occur when a virus, bacterium, or fungus jumps from an animal to a human. Accurately predicting when and where these infections—called zoonotic diseases—cross species could squelch outbreaks before they become epidemics. But maintaining active disease surveillance around the world is costly and time-consuming.
To help narrow the search, a team of scientists built a computer program to analyze a massive database of mammalian habits and habitats, including the geographic range and reproductive strategies for hundreds of species. Their program evaluated 86 different variables, like body size, life span, and population density, to hunt for patterns common among animals known to carry zoonotic diseases. “I was actually surprised that nobody had done it. It seemed like a natural approach,” says team leader Barbara Han, a disease ecologist at the Cary Institute of Ecosystem Studies in Millbrook, New York.
To simplify the results, Han and colleagues restricted their analysis to rodents—a group that carries a disproportionately high number of zoonotic diseases. Rodents carrying zoonotic infections tend to have “live fast, die young” lifestyles, Han says. They have large geographic ranges, early reproductive maturity, large litter sizes, and shorter gestation periods. Scientists aren’t certain why this lifestyle is common among rodents carrying zoonotic diseases, but they suggest that the fast-paced reproductive cycle may allow the animals to pass on their genes successfully before the disease kills them. “Even if it kills you in 6 months, you’re going to be able to put out like 5 litters,” Han says.
Han and her team first used their program to identify lifestyle patterns common to rodents harboring diseases like black plague, rabies, and hanta virus and found that their model had an accuracy rate of 90%. After the machine had “learned” the telltale signs, the researchers searched for new rodents that fit the profile but were not previously thought to be carriers. So far, the model has identified more than 150 new animal species that could harbor zoonotic diseases, the researchers report online today in the Proceedings of the National Academy of Sciences. The computer program also predicted 58 new infections in rodents that were already known to carry one zoonotic disease.
Based on its findings, the team was also able to identify hot spots where a disease was more likely to jump from rodent to human in the Middle East, Central Asia, and the American Midwest. Rodents living near humans were also more likely to be carriers of zoonotic diseases. But even population densities as low as 50 people per square kilometer were associated with increased chance for zoonotic diseases in the animals—a threshold that was surprising to Han. “I expected the reservoirs to be in areas where there are a lot of humans,” she says, “but it seems that if you have an average level of humans in your range that that’s enough.”
As exciting as it is to predict new zoonotic diseases, it’s another thing entirely to prevent them. “I think all the hard work is ahead of us,” Han says. It’s hard to know which rodents singled out by the study to focus on; many rarely interact with humans, if at all. Human behaviors like deforestation, urbanization, and increased carbon emissions could all influence where and when a disease makes the jump from animal to person, but precisely when and how this could happen is hard to predict. Finally, once a disease does cross species, health care workers and epidemiologists will have to face the challenge of containing and treating never-before-seen infections. “They can really only tell us about whether a species carries a parasite,” Martin says. Actually understanding how likely it is to cross to humans will require more study.
But Han says her team’s work should still make that job easier. It also speaks to the importance of basic research, she says. Without the information from the database—much of which had no obvious application as it was being compiled—her study would never have been possible. “I kind of worry if we stop appreciating the contribution of basic science then we won’t be able to make these types of breakthroughs in the future,” she says. “We’re not going to be able to predict much of anything if we don’t have the back story.”<|endoftext|>
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Overview: Conic Sections
## Introduction
• A curve, generated by intersecting a right circular cone with a plane is termed as ‘conic’.
• It has distinguished properties in Euclidean geometry.
• The vertex of the cone divides it into two nappes referred to as the upper nappe and the lower nappe.
• The initials as mentioned in the above figure A carry the following meaning,
(i) V is the vertex of the cone.
(ii) l is the axis of the cone.
(iii) m, the rotating line is a generator of the cone.
• In figure B, the cone is intersected by a plane and the section so obtained is known as a conic section.
Depending upon the position of the plane which intersects the cone and the angle of intersection β, different types of conic sections are obtained. i.e.
(i) Circle
(ii) Ellipse
(iii) Parabola
(iv) Hyperbola
## Terminology
A conic section or conic is the locus of a point that moves in a plane so that its distance from a fixed point is in a constant ratio to its perpendicular distance from a fixed straight line.
• The fixed point is called the Focus.
• The fixed straight line is called the Directrix.
• The constant ratio is called the Eccentricity denoted by e.
• The line passing through the focus & perpendicular to the directrix is called the Axis.
• A point of intersection of a conic with its axis is called a Vertex.
• Eccentricity is the factor related to conic sections which show how circular the conic sections.
More the eccentricity less circular the shape is. The eccentricity of the line is ∞.
The two conic sections will be of the same shape if they have the same eccentricity.
Suppose, the angle formed between the surface of the cone and its axis is β and the angle formed between the cutting plane and the axis is α, then
eccentricity, e = cos α/cos β
Note:
(i) If eccentricity, e = 0, the conic is a circle
(ii) If 0<e<1, the conic is an ellipse
(iii) If e=1, the conic is a parabola
(iv) If e>1, it is a hyperbola
• Principal Axis: Line joining the two focal points or foci of ellipse or hyperbola. Its midpoint is the center of the curve.
• Linear Eccentricity: Distance between the focus and center of a section.
• Latus Rectum: A chord of section parallel to directrix, which passes through a focus.
• Focal Parameter: Distance from focus to the corresponding directrix.
• Major axis: Chord joining the two vertices. It is the longest chord of an ellipse.
• Minor axis: Shortest chord of an ellipse.
Question for Overview: Conic Sections
Try yourself:
Which conic section has an eccentricity of 0?
## Circle
If β=90o, the conic section formed is a circle as shown below.
The standard form of a circle: (x - h)2 + (y - k)2 = r2
where (h, k) is the center of the circle and r is the radius of the circle.
## Ellipse
If α<β<90o, the conic section so formed is an ellipse as shown in the figure below.
The standard form of an ellipse is:
where (h, k) is the center of the ellipse, a is the horizontal stretch factor and b is the vertical stretch factor.
Ellipse with horizontal major axis:
• Focus: There are 2 focii; (ae, 0) and (-ae, 0)
• Directrix: These foci have corresponding directrices as
• Length of major axis(xx’): 2a
Length of major axis(yy’): 2b
• Vertex: (a, 0) & (-a, 0)
• Centre: (0, 0)
• Latus rectum:
• Length of Latus rectum
Ellipse with vertical major axis:
• Focii: (0, be) & (0, -be)
• Directrices: x = b/e & x = -b/e
• Length of major axis (yy'): 2b
Length of minor axis (xx'): 2a
• Vertex: (0, b) & (0, -b)
• Centre: (0, 0)
• Latus rectum:
• Length of Latus rectum
Note:
(i)
(ii)
(iii) Distance between 2 directices=
(iv) Distance between 2 focii = (major axis) × eccentricity
(v) Distance between focus and directrix =
Question for Overview: Conic Sections
Try yourself:
If ? = 60o and ? = 75o, what type of conic section is formed?
## Parabola
If α = β, the conic section formed is a parabola (represented by the orange curve) as shown below.
The standard form of an ellipse is:
x = a(y - k)2 + h (east to west), where a is the horizontal stretch factor and (h, k) is the vertex.
y = a(x - h)2 + k (north to south), where a is the vertical stretch factor and (h, k) is the vertex.Standard forms of Parabola
• Latus rectum: 4 x (distance between vertex and focus) = 4a
• Latus rectum: 2 x Distance between directrix and focus = 2(2a).
• Point of intersection of Axis and directrix and the focus is bisected by the vertex.
• Focus is the midpoint of the Latus rectum.
• (Distance of any point on parabola from the axis)2 = (LR) (Distance of the same point from tangent at the vertex)
Illustration 1: If extreme points of LR are (11/2, 6) and (13/2, 4). Find the equation of the parabola.
Solution: Mid point of LR = focus = (6, 5)
⇒ 4a = 2 or a = ½
The equation of parabolas are: (y – 5)2 = 2(x – 5.5) and (y – 5)2 = – 2 (x – 6.5)
Question for Overview: Conic Sections
Try yourself:
If the extreme points of the Latus rectum are (7/2, 4) and (9/2, 6), what is the equation of the parabola?
## Hyperbola
If 0 ≤ β < α, then the plane intersects both nappes and the conic section so formed is known as a hyperbola (represented by the orange curves).
The standard form of an ellipse is:
where a is the horizontal stretch factor and b is the vertical stretch factor.
Standard Hyperbola
• Equation:
• Focus: There are 2 focii (ae, 0) and (-ae, 0)
• Directrix: The foci has corresponding directrices as x = +a/e and x = −a/e respectively.
• Length of Transverse axis (xx') = 2a
Length of Conjugate axis (yy') = 2b (Hypothetical)
• Vertex: (0, 0) and (-a, 0)
• Centre: (0, 0)
• Latus rectum: x = +ae
• Length of latus rectum
• Position of point at hyperbola .
Let and
(i) If S1 > 0, point (x1, y1) lies inside the hyperbola
(ii) If S1 = 0, point (x1, y1) lies on the hyperbola
(iii) If S1 < 0, point (x1, y1) lies outside the hyperbola
Illustration 2: Classify the following equations according to their type of conics.
(a) 4x2 + 4y2 - 16x + 4y - 60 = 0
(b) x2 - 4x + 16y + 17 = 0
(c) x2 + 2y2 + 4x + 2y – 27 = 0
(d) x2 – y2 + 3x – 2y – 43 = 0
Solution:
(a) Here A = 4, B = 0, C = 4
Determinant will be B2 - 4AC = 02- 4(4) (4) = -64
This shows that B2 - 4AC < 0, B = 0 and A = C, so this is a circle.
In another way,
Since Both variables are squared, and both squared terms are multiplied by the same number. Therefore, this is a circle
(b) Here A =1,B = 0, C = 0
Determinant will be B2- 4AC = 0- 4(1) (0) = 0
This shows that B2 - 4AC = 0, so this is a parabola.
In another way,
Since only one of the variables is squared. Therefore, this is called a Parabola.
(c) Here A =1, B = 0, C = 2
Determinant will be B2 - 4AC = 02 - 4(1) (2) = -8
This shows that B2 - 4AC < 0, and A ≠ C so this is an ellipse.
In other way,
Since both variables are squared with the same sign, but they aren't multiplied by the same number. Therefore, so this is an ellipse.
(d) Here A =1,B = 0, C = -1
Determinant will be B2- 4AC = 02 - 4(1) (-1) = 8
This shows that B2 - 4AC > 0 so this is a hyperbola.
In another way,
Since both variables are squared, and the squared terms have opposite signs. Therefore, this is a hyperbola.
Question for Overview: Conic Sections
Try yourself:
Which conic section is represented by the equation x^2 - 4x + 16y + 17 = 0?
## Degenerated Conics
A degenerate conic is generated when a plane intersects the vertex of the cone.
There are three types of degenerate conics:
1. If α< β≤90°, then the plane intersects the vertex exactly at a point. This condition is known as the degenerated form of circle or ellipse.
1. If α=β, the plane upon an intersection with a cone forms a straight line containing a generator of the cone. This condition is a degenerated form of a parabola.
1. If 0≤β<α, the section formed is a pair of intersecting straight lines. This condition is a degenerated form of a hyperbola.
The document Overview: Conic Sections | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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## Mathematics (Maths) for JEE Main & Advanced
209 videos|443 docs|143 tests
## FAQs on Overview: Conic Sections - Mathematics (Maths) for JEE Main & Advanced
1. What are the key conic sections covered in the JEE syllabus?
Ans. The key conic sections covered in the JEE syllabus are circle, ellipse, parabola, and hyperbola.
2. How are circle and ellipse different from each other in terms of their equations and properties?
Ans. A circle has a constant distance from its center to any point on the circumference, while an ellipse has two different distances from its center to any point on the ellipse. The equations for a circle and ellipse also differ, with the general equation for a circle being (x-h)^2 + (y-k)^2 = r^2 and for an ellipse being (x-h)^2/a^2 + (y-k)^2/b^2 = 1.
3. What is a degenerated conic in conic sections?
Ans. A degenerated conic is a special case where a conic section loses its defining characteristics and transforms into a simpler shape. For example, a degenerated conic can occur when the eccentricity of an ellipse becomes 0, resulting in a circle.
4. How are parabolas and hyperbolas distinguished from each other in conic sections?
Ans. Parabolas have a single focus point and a directrix, while hyperbolas have two focus points and two branches. The general equation for a parabola is y^2 = 4ax or x^2 = 4ay, while for a hyperbola, it is (x-h)^2/a^2 - (y-k)^2/b^2 = 1 or (y-k)^2/a^2 - (x-h)^2/b^2 = 1.
5. How are conic sections used in real-life applications?
Ans. Conic sections have various real-life applications, such as in astronomy for describing the orbits of planets, in optics for designing lenses and mirrors, in architecture for creating domes and arches, and in engineering for designing antennas and satellite dishes.
## Mathematics (Maths) for JEE Main & Advanced
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Summary and Keywords
The eighty years from 1790 to 1870 were marked by dramatic economic and demographic changes in the United States. Cities in this period grew faster than the country as a whole, drawing migrants from the countryside and immigrants from overseas. This dynamism stemmed from cities’ roles as spearheads of commercial change and sites of new forms of production. Internal improvements such as canals and railroads expanded urban hinterlands in the early republic, while urban institutions such as banks facilitated market exchange. Both of these worked to the advantage of urban manufacturers. By paying low wages to workers performing repetitive tasks, manufacturers enlarged the market for their products but also engendered opposition from a workforce internally divided along lines of sex and race, and at times slavery and freedom. The Civil War affirmed the legitimacy of wage labor and enhanced the power of corporations, setting the stage for the postwar growth of large-scale, mechanized industry.
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(or German socialism), a type of petit bourgeois socialism that gained currency in 1844–47 among the intelligentsia and artisans of Germany and among German émigrés in other countries.
The main representatives of so-called true socialism were K. Grün, M. Hess, O. Liming, and H. Püttmann. G. Kuhl-mann preached true socialism in Switzerland and H. Kriege in the USA. True socialism expressed the sentiments of the German petite bourgeoisie, who were thrown off course by the development of capitalism and frightened by the growth of class antagonisms. The ideology of true socialism was a unique synthesis of the idealist aspects of Feuerbach’s philosophy (such as love for “man” in the abstract, the disregard of social relations, and other features) and an emasculated French Utopian socialism. Under the pretence of criticizing capitalism, the true socialists idealized the precapitalist system, in particular the artisan guilds, and spread illusions that Germany could arrive at socialism without passing through the stage of large-scale capitalist industry. Marx and Engels criticized true socialism in several of their works, including German Ideology (part 2), the Communist Manifesto, (chapter 3), and Circular Against Kriege, and in Engels’ works The True Socialists and German Socialism in Verse and Prose (see K. Marx and F. Engels, Soch., 2nd ed., vols. 3–4).
True socialism disappeared as a separate current during the Revolution of 1848–49, which fully revealed its bankruptcy. In characterizing the true socialists, Lenin wrote in 1906: “The ‘true socialists’ … are something like peaceful Lavrovists, semiuplifters, nonrevolutionaries, heroes of abstruse thought and abstract sermonizing” (Poln. sobr. soch., 5th ed., vol. 13, p. 154).
REFERENCESSerebriakov, M. V. “Nemetskii sotsializm i bor’ba s nim Marksa i En-gel’sa.” Uch. zap. LGU: Ser. filosofskikh nauk, 1948, issue 2.
Kandel’, E. P. “Iz istorii bor’by Marksa i Engel’sa s nemetskim ‘istinnym sotsializmom’.” In the collection Iz istorii formirovaniia i razvitiia marksizma. Moscow, 1959.
Kan, S. B. Istoriia sotsialisticheskikh idei (do vozniknoveniia marksizma), 2nd ed. Moscow, 1967.
L. I. GOL’MAN<|endoftext|>
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Section12.4Solving Quadratic Equations by the Square Root Method
If $k$ is a positive number and $u^2=k\text{,}$ then:
\begin{equation*} u=\sqrt{k}\text{ or } u=-\sqrt{k}\text{.} \end{equation*}
The latter set of equations is frequently abbreviate as
\begin{equation*} u= \pm \sqrt{k} \end{equation*}
which is read aloud as "$u$ is equal to plus or minus the square root of $k\text{.}$"
For example, if $x^2=64\text{,}$ then $x= \pm \sqrt{64}$ which, of course, simplifies to $x= \pm 8\text{.}$ So the original equation has two solutions: $-8$ and $8\text{.}$
Another example follows. Solve $(3x+5)^2=16\text{.}$
\begin{align*} (3x+5)^2\amp=16\\ 3x+5\amp=\pm \sqrt{16}\\ 3x+5\amp=\pm 4 \end{align*}
\begin{align*} 3x+5\amp=-4\amp\amp\text{or}\amp 3x+5\amp=4\\ 3x+5\subtractright{5}\amp=-4\subtractright{5}\amp\amp\text{or}\amp 3x+5\subtractright{5}\amp=4\subtractright{5}\\ 3x\amp=-9\amp\amp\text{or}\amp 3x\amp=-1\\ \divideunder{3x}{3}\amp=\divideunder{-9}{3}\amp\amp\text{or}\amp \divideunder{3x}{3}\amp=\divideunder{-1}{3}\\ x\amp=-3\amp\amp\text{or}\amp x\amp=-\frac{1}{3} \end{align*}
The solutions are $-3$ and $-\frac{1}{3}\text{.}$ The solution set is $\left\{-3, -\frac{1}{3}\right\}\text{.}$
Completing the Square.
We frequently have to do some preliminary work to have an equation of form $u^2=k\text{.}$ The process involves completing the square. Completing the square entails adding a number to an expression of form $x^2+bx$ so that the resultant trinomial factors into a perfect square. The number that complete the square is always the square of half of $b\text{.}$
For example, the number that completes the square for $x^2+10x$ is $5^2$ which is $25\text{.}$ Note:
\begin{align*} x^2+10x+25\amp=(x+5)(x+5)\\ \amp=(x+5)^2 \end{align*}
The steps followed when using the square root method are listed below.
1. Perform any and all manipulations so that the equation has the form $x^2+bx=c\text{.}$
2. Complete the square by adding $\left(\frac{b}{2}\right)^2\text{.}$ Note that to keep the equation in balance $\left(\frac{b}{2}\right)^2$ also needs to be added to the other side of the equation.
3. Factor the non-constant side of the equation. The equation should now have form $u^2=k\text{.}$
4. If the constant, $k\text{,}$ is negative, then the equation has no solution. If $k$ is zero, then the equation has one solution which is determined by solving $u=0\text{.}$ Step 5 is written under the assumption that $k$ is a positive number.
5. Invoke the square root property, i.e. $u=\pm \sqrt{k}\text{.}$ If $k$ is a perfect square, you need to write out and solve two equations.
\begin{equation*} u=-\sqrt{k}\text{ or }u=\sqrt{k}\text{.} \end{equation*}
If $k$ is not a perfect square, you may solve for $u$ maintaining the plus/minus sign. Make sure to simplify the square root should it simplify.
6. State your solutions and/or solution set.
Several examples follow.
Example12.4.1.
Solve $x^2-6x-7=0$ by the square root method after first completing the square.
Solution
The solutions are $-1$ and $7\text{.}$ The solution set is $\{-1, 7\}\text{.}$
Example12.4.2.
Solve $(2x-1)(2x+1)=7-32x$ by the square root method after first completing the square.
Solution
The solutions are $-4-3\sqrt{2}$ and $-4+3\sqrt{2}\text{.}$
The solutions set is $\{-4-3\sqrt{2}, -4+3\sqrt{2}\}\text{.}$
Example12.4.3.
Solve $x^2+9x+25=0$ by the square root method after first completing the square.
Solution
The equation has no real number solutions.
Over the real numbers, the solution set is $\emptyset\text{.}$
Example12.4.4.
Solve $4x^2-28x=-11$ by the square root method after first completing the square.
Solution
The solutions are $\frac{7-\sqrt{38}}{2}$ and $\frac{7+\sqrt{38}}{2}\text{.}$
The solution set is $\left\{\frac{7-\sqrt{38}}{2}, \frac{7+\sqrt{38}}{2}\right\}\text{.}$
You can use Figureย 12.4.5 to generate several more examples/practice problems. You'll need to write the steps down, because only one step is shown at a time.
ExercisesExercises
Use the square root method to solve each of the following quadratic equations over the real numbers. Make sure that all solutions are completely simplified. State the solutions to each equation as well as the solution set to each equation.
1.
$(2t-1)^2=9$
Solution
We begin by applying the square root property.
\begin{align*} (2t-1)^2\amp=9\\ 2t-1\amp=\pm \sqrt{9}\\ 2t-1\amp=\pm 3 \end{align*}
\begin{align*} 2t-1\amp =-3\amp\amp\text{ or }\amp 2t-1\amp=3\\ 2t-1\addright{1}\amp =-3\addright{1}\amp\amp\text{ or }\amp 2t-1\addright{1}\amp=3\addright{1}\\ 2t\amp =-2\amp\amp\text{ or }\amp 2t\amp=4\\ \divideunder{2t}{2}\amp =\divideunder{-2}{2}\amp\amp\text{ or }\amp \divideunder{2t}{2}\amp=\divideunder{4}{2}\\ t\amp =-1\amp\amp\text{ or }\amp t\amp=2 \end{align*}
The solutions are $-1$ and $2\text{.}$
The solution set is $\{-1, 2\}\text{.}$
2.
$(3x-2)^2=72$
Solution
We begin by applying the square root property.
\begin{align*} (3x-2)^2\amp=72\\ 3x-2\amp=\pm \sqrt{72}\\ 3x-2\amp=\pm \sqrt{36 \cdot 2}\\ 3x-2\amp=\pm 6\sqrt{2}\\ 3x-2\addright{2}\amp=\pm 6\sqrt{2}\addright{2}\\ 3x\amp=2\pm 6\sqrt{2}\\ \multiplyleft{\frac{1}{3}}3x\amp=\multiplyleft{\frac{1}{3}}(2\pm 6\sqrt{2})\\ x\amp=\frac{2\pm 6\sqrt{2}}{3} \end{align*}
The solutions are $\frac{2-6\sqrt{2}}{3}$ and $\frac{2+6\sqrt{2}}{3}\text{.}$
The solution set is $\left\{\frac{2-6\sqrt{2}}{3}, \frac{2+6\sqrt{2}}{3}\right\}\text{.}$
3.
$x^2-8x+16=12$
Solution
We begin by factoring the perfect-square trinomial $x^2-8x+16\text{.}$
\begin{align*} x^2-8x+16\amp=12\\ (x-4)^2\amp=12\\ x-4\amp=\pm \sqrt{12}\\ x-4\amp=\pm \sqrt{4 \cdot 3}\\ x-4\amp=\pm 2\sqrt{3}\\ x-4\addright{4}\amp=\pm 2\sqrt{3}\addright{4}\\ x\amp=4\pm 2\sqrt{3} \end{align*}
The solutions are $4-2\sqrt{3}$ and $4+2\sqrt{3}\text{.}$
The solution set is $\{4-2\sqrt{3}, 4+2\sqrt{3}\}\text{.}$
4.
$y^2+20y+80=0$
Solution
We begin by moving the constant term to the right side of the equation and then completing the square on the left side of the equation.
\begin{align*} y^2+20y+80\amp=0\\ y^2+20y+80\subtractright{80}\amp=0\subtractright{80}\\ y^2+20y\amp=-80\\ y^2+20y+100\amp=-80+100\\ (y+10)^2\amp=20\\ y+10\amp=\pm \sqrt{20}\\ y+10\amp=\pm \sqrt{4 \cdot 5}\\ y+10\amp=\pm 2\sqrt{5}\\ y+10\subtractright{10}\amp=\pm 2\sqrt{5}\subtractright{10}\\ y\amp=-10\pm 2\sqrt{5} \end{align*}
The solutions are $-10-2\sqrt{5}$ and $-10+2\sqrt{5}\text{.}$
The solution set is $\{-10-2\sqrt{5}, -10+2\sqrt{5}\}\text{.}$
5.
$x^2+4x+7=0$
Solution
We begin by moving the constant term to the right side of the equation and then completing the square on the left side of the equation.
The equation $x^2+4x+7=0$ has no real number solutions.
Over the real numbers the solution set is $\emptyset\text{.}$
6.
$w^2-7w+7=0$
Solution
We begin by moving the constant term to the right side of the equation and then completing the square on the left side of the equation.
The solutions are $\frac{7-\sqrt{21}}{2}$ and $\frac{7+\sqrt{21}}{2}\text{.}$
The solution set is $\left\{\frac{7-\sqrt{21}}{2}, \frac{7+\sqrt{21}}{2}\right\}\text{.}$<|endoftext|>
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