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Q. What does Reparations mean?
A. It comes from the Latin Word for “repair”. It means redressing a wrong which has been done. In international law it is a recognised principle that those nations or individuals who have wronged other nations or individuals, should make reparation to repair the damage which has been suffered. This is particularly so when the wrongs amount to a crime against humanity. In recent years the term “reparations” has come to be used specifically in the context of the wrongs done to people of African descent in the period of the transatlantic trade in Africans and the system of enslavement in the Americas.
Q. So, is it all about getting money?
A. No, the issue goes far beyond a request for payment in monetary terms. It involves a process of reconciliation. It may include an acknowledgement by those whose countries were enriched by the profits of slavery that a crime against humanity was committed and the making of a solemn apology. And a natural consequence of all this may include monetary compensation in various forms, to repair the damage done.
Q. Even if reparations were paid, how would the people of the Caribbean benefit? Would all the money go to the Government?
A. This is one key question which the CARICOM Reparations Commission will address. This is one of the reasons why we will reaching out to the public of Jamaica in different ways. It is crucial that ordinary people see this cause as one from which they will benefit.
Q. But isn't it too late to be talking about slavery days?
A. No. In principle people are entitled to reparations they are still suffering from the wrongs which were done to their ancestors. The Commission is gathering evidence that the slavery system throughout the Caribbean had traumatic consequences in terms of poverty, destruction of family life and social structured racism, and the use of violence and that these consequences affect our society seriously to this day.
Q. Didn't slavery end with emancipation in 1838?
A. Emancipation meant that all our people were legally free but the former enslaved people received no compensation for decades of forced labour under inhuman conditions, On the contrary, it was the former enslavers who were massively compensated by the British Government for the loss of their “property”.
Q. What about the African chiefs who sold our ancestors into slavery? Weren't they just as much to blame?
A. It is true that many people in Africa collaborated in the transatlantic trade in Africans and made profits from it. This is an issue which the Reparations Commission will address. But it must be remembered that the driving force behind the trade was the insatiable demand of the plantation system for enslaved labourers, and the greatest profits were made by the European nations whose economies boomed because of the slavery system.
Q. What has the movement for reparations achieved so far?
A. Many Africans and Caribbean leaders have called for reparations to be paid by the former enslaving nations. The issue was raised at the United Nations World Conference Against Racism in 2001, which resolved that slavery and the slave trade are a crime against humanity and should always have been so. There are active organisations in the United States which have filed lawsuits against companies which profited from slavery. There has been a debate in the House of Lords in Britain. There has been a growing movement for reparations in countries all over the world where black people are to be found.
Q. Has there been any response from Britain and other European countries?
A. The official response is that reparations are not going to be paid. The UK government has said that there is no evidence that people are still suffering from the effects of the transatlantic trade in Africans. Both Tony Blair and David Cameron, former prime ministers of the UK expressed “our deep sorrow that it happened” but added that Caribbean people should “get over it”, stop dwelling in the past and instead focus on the future. The President of France has admitted France’s responsibility for the tragedy that was the slave trade, and established May 10 as a national day to honour those who suffered from enslavement.
Q. So why should European countries take any notice of the claim for reparations?
A. European societies have also suffered from the history of slavery. Enslavement gave rise to pernicious doctrines of white supremacy which still have their legacy in the shape of racial prejudice and racial violence. Reparations is about reconciliation. Facing up to a shameful past and taking measures to repair the damage. It will be a healing process for black and white alike. Those who commit crimes against humanity must make amends for such crimes and slavery was a crime against humanity.
Q. Have Jamaicans been involved in the movement for reparations?
A. Definitely. Jamaican Rastafari have been forefront of the call for reparations and have petitioned the Queen of England. Jamaica’s High Commissioner to Nigeria, Ambassador Dudley Thompson, was one of the organisers of the First Pan African Conference on Reparations in 1992. Honourable Mike Henry, Minister of Transport, initiated a debate on reparations in the House of Representatives in 2007. And now Jamaica has been the first Caribbean country to set up a National Commission on reparations.
Q. What is the role of the National Commissions on Reparations?
A. They have been set up by the governments of the Caribbean who have recommended the form or forms which reparations may take, and to receive testimony from the public and from experts, with the aim of guiding a national approach to reparations.
Q. Is reparations the same as repatriation?
A. No, but they are linked. One of the forms which reparation could take would be giving assistance to those who so desire to return to the continent from which their ancestors were forcibly removed.<|endoftext|>
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# What is distance between parallel planes?
## What is distance between parallel planes?
The distance between two parallel planes, D=|ax1+by1+cz1+d|√a2+b2+c2. Note: The distance between the given two parallel planes is nothing but the shortest distance between the two plane surfaces.
What is the perpendicular distance between two profile planes?
In geometry, the perpendicular distance between two objects is the distance from one to the other, measured along a line that is perpendicular to one or both. Particular instances include: Distance from a point to a line, for the perpendicular distance from a point to a line in two-dimensional space.
What is the formula of distance between two parallel lines?
The distance between two parallel lines is given by d = |c1-c2|/√(a2+b2).
### What is perpendicular distance formula?
The perpendicular distance is the shortest distance between a point and a line. The perpendicular distance, 𝐷 , between the point 𝑃 ( 𝑥 , 𝑦 ) and the line 𝐿 : 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 = 0 is given by 𝐷 = | 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 | √ 𝑎 + 𝑏 .
What is the perpendicular distance of the point A 6 7 from Y axis?
Step-by-step explanation: the perpendicular distance of a point P (6 , 7) from y axis is -1.
How do you find the perpendicular distance between two lines?
Basically, to find the distance between two perpendicular lines given two points, all we have to do is find the distance from the points to the intersection of the lines, then use the Pythagorean Theorem to find the distance between the two points.
#### How do you find the perpendicular distance between a point and a plane?
We call this the perpendicular distance between the point and the plane because 𝑃 𝑄 is perpendicular to the plane. We could find this distance by finding the coordinates of 𝑄 ; however, there is an easier method. To calculate this distance, we will start by setting ∠ 𝑅 𝑃 𝑄 = 𝜃 and | | 𝑃 𝑄 | | = 𝐷 .
How do you find a perpendicular plane?
If a vector is perpendicular to two vectors in a plane, it must be perpendicular to the plane itself. As the cross product of two vectors produces a vector perpendicular to both, we will use the cross product of →v1 and →v2 to find a vector →u perpendicular to the plane containing them.
How do you find the perpendicular distance from a point to a plane?
## How do you find the perpendicular distance from the origin?
The perpendicular distance from origin to the normal at any point to the curve x=a(cosθ+θsinθ). y=a(sinθ−θcosθ)
What is the definition of parallel planes?
In geometry, parallel lines are lines in a plane which do not meet; that is, two lines in a plane that do not intersect or touch each other at any point are said to be parallel. By extension, a line and a plane, or two planes, in three-dimensional Euclidean space that do not share a point are said to be parallel.
When are planes parallel?
When two planes are perpendicular to the same line, they are parallel planes. When a plane intersects two parallel planes, the intersection is two parallel lines.
### What are parallel lines and angles?
Angles and parallel lines. When two lines intersect they form two pairs of opposite angles, A + C and B + D. Another word for opposite angles are vertical angles. Vertical angles are always congruent, which means that they are equal. Adjacent angles are angles that come out of the same vertex.<|endoftext|>
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By the end of this section, you will be able to do the following:
- Explain the significance of photosynthesis to other living organisms
- Describe the main structures involved in photosynthesis
- Identify the substrates and products of photosynthesis
Photosynthesis is essential to all life on earth; both plants and animals depend on it. It is the only biological process that can capture energy that originates from sunlight and converts it into chemical compounds (carbohydrates) that every organism uses to power its metabolism. It is also a source of oxygen necessary for many living organisms. In brief, the energy of sunlight is “captured” to energize electrons, whose energy is then stored in the covalent bonds of sugar molecules. How long lasting and stable are those covalent bonds? The energy extracted today by the burning of coal and petroleum products represents sunlight energy captured and stored by photosynthesis 350 to 200 million years ago during the Carboniferous Period.
Plants, algae, and a group of bacteria called cyanobacteria are the only organisms capable of performing photosynthesis ((Figure)). Because they use light to manufacture their own food, they are called photoautotrophs (literally, “self-feeders using light”). Other organisms, such as animals, fungi, and most other bacteria, are termed heterotrophs (“other feeders”), because they must rely on the sugars produced by photosynthetic organisms for their energy needs. A third very interesting group of bacteria synthesize sugars, not by using sunlight’s energy, but by extracting energy from inorganic chemical compounds. For this reason, they are referred to as chemoautotrophs.
The importance of photosynthesis is not just that it can capture sunlight’s energy. After all, a lizard sunning itself on a cold day can use the sun’s energy to warm up in a process called behavioral thermoregulation. In contrast, photosynthesis is vital because it evolved as a way to store the energy from solar radiation (the “photo-” part) to energy in the carbon-carbon bonds of carbohydrate molecules (the “-synthesis” part). Those carbohydrates are the energy source that heterotrophs use to power the synthesis of ATP via respiration. Therefore, photosynthesis powers 99 percent of Earth’s ecosystems. When a top predator, such as a wolf, preys on a deer ((Figure)), the wolf is at the end of an energy path that went from nuclear reactions on the surface of the sun, to visible light, to photosynthesis, to vegetation, to deer, and finally to the wolf.
Main Structures and Summary of Photosynthesis
Photosynthesis is a multi-step process that requires specific wavelengths of visible sunlight, carbon dioxide (which is low in energy), and water as substrates ((Figure)). After the process is complete, it releases oxygen and produces glyceraldehyde-3-phosphate (GA3P), as well as simple carbohydrate molecules (high in energy) that can then be converted into glucose, sucrose, or any of dozens of other sugar molecules. These sugar molecules contain energy and the energized carbon that all living things need to survive.
The following is the chemical equation for photosynthesis ((Figure)):
Although the equation looks simple, the many steps that take place during photosynthesis are actually quite complex. Before learning the details of how photoautotrophs turn sunlight into food, it is important to become familiar with the structures involved.
Basic Photosynthetic Structures
In plants, photosynthesis generally takes place in leaves, which consist of several layers of cells. The process of photosynthesis occurs in a middle layer called the mesophyll. The gas exchange of carbon dioxide and oxygen occurs through small, regulated openings called stomata (singular: stoma), which also play roles in the regulation of gas exchange and water balance. The stomata are typically located on the underside of the leaf, which helps to minimize water loss due to high temperatures on the upper surface of the leaf. Each stoma is flanked by guard cells that regulate the opening and closing of the stomata by swelling or shrinking in response to osmotic changes.
In all autotrophic eukaryotes, photosynthesis takes place inside an organelle called a chloroplast. For plants, chloroplast-containing cells exist mostly in the mesophyll. Chloroplasts have a double membrane envelope (composed of an outer membrane and an inner membrane), and are ancestrally derived from ancient free-living cyanobacteria. Within the chloroplast are stacked, disc-shaped structures called thylakoids. Embedded in the thylakoid membrane is chlorophyll, a pigment (molecule that absorbs light) responsible for the initial interaction between light and plant material, and numerous proteins that make up the electron transport chain. The thylakoid membrane encloses an internal space called the thylakoid lumen. As shown in (Figure), a stack of thylakoids is called a granum, and the liquid-filled space surrounding the granum is called stroma or “bed” (not to be confused with stoma or “mouth,” an opening on the leaf epidermis).
On a hot, dry day, the guard cells of plants close their stomata to conserve water. What impact will this have on photosynthesis?
<!–<para> Levels of carbon dioxide (a necessary photosynthetic substrate) will fall. As a result, the rate of photosynthesis will decrease.–>
The Two Parts of Photosynthesis
Photosynthesis takes place in two sequential stages: the light-dependent reactions and the light-independent reactions. In the light-dependent reactions, energy from sunlight is absorbed by chlorophyll and that energy is converted into stored chemical energy. In the light-independent reactions, the chemical energy harvested during the light-dependent reactions drives the assembly of sugar molecules from carbon dioxide. Therefore, although the light-independent reactions do not use light as a reactant, they require the products of the light-dependent reactions to function. In addition, however, several enzymes of the light-independent reactions are activated by light. The light-dependent reactions utilize certain molecules to temporarily store the energy: These are referred to as energy carriers. The energy carriers that move energy from light-dependent reactions to light-independent reactions can be thought of as “full” because they are rich in energy. After the energy is released, the “empty” energy carriers return to the light-dependent reaction to obtain more energy. (Figure) illustrates the components inside the chloroplast where the light-dependent and light-independent reactions take place.
Click the link to learn more about photosynthesis.
Photosynthesis at the Grocery Store
Major grocery stores in the United States are organized into departments, such as dairy, meats, produce, bread, cereals, and so forth. Each aisle ((Figure)) contains hundreds, if not thousands, of different products for customers to buy and consume.
Although there is a large variety, each item ultimately can be linked back to photosynthesis. Meats and dairy link, because the animals were fed plant-based foods. The breads, cereals, and pastas come largely from starchy grains, which are the seeds of photosynthesis-dependent plants. What about desserts and drinks? All of these products contain sugar—sucrose is a plant product, a disaccharide, a carbohydrate molecule, which is built directly from photosynthesis. Moreover, many items are less obviously derived from plants: For instance, paper goods are generally plant products, and many plastics (abundant as products and packaging) are derived from “algae” (unicellular plant-like organisms, and cyanobacteria). Virtually every spice and flavoring in the spice aisle was produced by a plant as a leaf, root, bark, flower, fruit, or stem. Ultimately, photosynthesis connects to every meal and every food a person consumes.
The process of photosynthesis transformed life on Earth. By harnessing energy from the sun, the evolution of photosynthesis allowed living things access to enormous amounts of energy. Because of photosynthesis, living things gained access to sufficient energy that allowed them to build new structures and achieve the biodiversity evident today.
Only certain organisms (photoautotrophs), can perform photosynthesis; they require the presence of chlorophyll, a specialized pigment that absorbs certain wavelengths of the visible spectrum and can capture energy from sunlight. Photosynthesis uses carbon dioxide and water to assemble carbohydrate molecules and release oxygen as a byproduct into the atmosphere. Eukaryotic autotrophs, such as plants and algae, have organelles called chloroplasts in which photosynthesis takes place, and starch accumulates. In prokaryotes, such as cyanobacteria, the process is less localized and occurs within folded membranes, extensions of the plasma membrane, and in the cytoplasm.
Visual Connection Questions
Which of the following components is not used by both plants and cyanobacteria to carry out photosynthesis?
- carbon dioxide
What two main products result from photosynthesis?
- oxygen and carbon dioxide
- chlorophyll and oxygen
- sugars/carbohydrates and oxygen
- sugars/carbohydrates and carbon dioxide
In which compartment of the plant cell do the light-independent reactions of photosynthesis take place?
- outer membrane
Which statement about thylakoids in eukaryotes is not correct?
- Thylakoids are assembled into stacks.
- Thylakoids exist as a maze of folded membranes.
- The space surrounding thylakoids is called stroma.
- Thylakoids contain chlorophyll.
Predict the end result if a chloroplast’s light-independent enzymes developed a mutation that prevented them from activating in response to light.
- GA3P accumulation
- ATP and NADPH accumulation
- Water accumulation
- Carbon dioxide depletion
How are the NADPH and GA3P molecules made during photosynthesis similar?
- They are both end products of photosynthesis.
- They are both substrates for photosynthesis.
- They are both produced from carbon dioxide.
- They both store energy in chemical bonds.
Critical Thinking Questions
What is the overall outcome of the light reactions in photosynthesis?
The outcome of light reactions in photosynthesis is the conversion of solar energy into chemical energy that the chloroplasts can use to do work (mostly anabolic production of carbohydrates from carbon dioxide).
Why are carnivores, such as lions, dependent on photosynthesis to survive?
Because lions eat animals that eat plants.
Why are energy carriers thought of as either “full” or “empty”?
The energy carriers that move from the light-dependent reaction to the light-independent one are “full” because they bring energy. After the energy is released, the “empty” energy carriers return to the light-dependent reaction to obtain more energy. There is not much actual movement involved. Both ATP and NADPH are produced in the stroma where they are also used and reconverted into ADP, Pi, and NADP+.
Describe how the grey wolf population would be impacted by a volcanic eruption that spewed a dense ash cloud that blocked sunlight in a section of Yellowstone National Park.
The grey wolves are apex predators in their food web, meaning they consume smaller prey animals and are not the prey of any other animal. Blocking sunlight would prevent the plants at the bottom of the food web from performing photosynthesis. This would kill many of the plants, reducing the food sources available to smaller animals in Yellowstone. A smaller prey animal population means that fewer wolves can survive in the area, and the population of grey wolves will decrease.
How does the closing of the stomata limit photosynthesis?
The stomata regulate the exchange of gases and water vapor between a leaf and its surrounding environment. When the stomata are closed, the water molecules cannot escape the leaf, but the leaf also cannot acquire new carbon dioxide molecules from the environment. This limits the light-independent reactions to only continuing until the carbon dioxide stores in the leaf are depleted.
- organism that can build organic molecules using energy derived from inorganic chemicals instead of sunlight
- organelle in which photosynthesis takes place
- stack of thylakoids located inside a chloroplast
- organism that consumes organic substances or other organisms for food
- light-dependent reaction
- first stage of photosynthesis where certain wavelengths of the visible light are absorbed to form two energy-carrying molecules (ATP and NADPH)
- light-independent reaction
- second stage of photosynthesis, through which carbon dioxide is used to build carbohydrate molecules using energy from ATP and NADPH
- middle layer of chlorophyll-rich cells in a leaf
- organism capable of producing its own organic compounds from sunlight
- molecule that is capable of absorbing certain wavelengths of light and reflecting others (which accounts for its color)
- opening that regulates gas exchange and water evaporation between leaves and the environment, typically situated on the underside of leaves
- fluid-filled space surrounding the grana inside a chloroplast where the light-independent reactions of photosynthesis take place
- disc-shaped, membrane-bound structure inside a chloroplast where the light-dependent reactions of photosynthesis take place; stacks of thylakoids are called grana
- thylakoid lumen
- aqueous space bound by a thylakoid membrane where protons accumulate during light-driven electron transport<|endoftext|>
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A breakthrough in honeybee immunization has been discovered after a 15-year study on a bee blood protein known as vitellogenin.
Scientists from Arizona State University, University of Helsinki, University of Jyväskylä and Norwegian University of Life Sciences discovered the process of how bees immunize their offspring and what it could mean for vaccinations.
The discovery was published in . The findings explain how immune system priming occurs through vitellogenin because bees lack antibodies.
In honeybee colonies, the queen has food brought to her by worker bees. While gathering pollen and nectar, worker bees can also pick up pathogens. The pollen, nectar and pathogens are transformed into “royal jelly,” which is food that is only eaten by the queen.
Any bacteria picked up from outside that is in the royal jelly is digested and stored in the queen’s “fat body” which functions similarly to a liver. The bacteria are then bound to the protein vitellogenin and carried to the developing larvae through the blood.
Thanks to vitellogenin, baby bees’ immune systems are primed against infection. Moreover, scientists now believe they understand how to vaccinate bees.
“We are patenting a way to produce a harmless vaccine, as well as how to cultivate the vaccines and introduce them to bee hives through a cocktail the bees would eat,” Dalial Freitak, a postdoctoral researcher with University of Helsinki, told . “They would then be able to stave off disease.”
, a disease that is highly contagious and kills honeybee larvae, is one of the dangers that could someday be eliminated by a bee vaccine.
“Because this vaccination process is naturally occurring, this process would be cheap and ultimately simple to implement,” said Gro Amdam, a professor with ASU’s School of Life Sciences. “It has the potential to both improve and secure food production for humans.”<|endoftext|>
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Four of seven mysterious worlds orbiting a nearby star might bear large quantities of water, scientists have revealed, offering a tantalising boost to the possibility of finding life beyond our solar system.
Just 39 light-years away from Earth, the solar system of seven Earth-sized planets orbiting the small, cool star known as Trappist-1 was discovered earlier this year, with the announcement met with excitement by experts.
While scientists said all of the planets could harbour some water, given the right atmosphere, three of the four outermost planets lie within the star’s so-called “habitable zone” – the distance at which temperatures are suitable for liquid water to exist – making them the most likely candidates.
But it remained unclear whether any of the planets were indeed wet. Now scientists say they are a step closer to finding out.
“What we found tells us that probably the [three] inner planets are not good planets to search for life,” said Vincent Bourrier, first author of the study from the University of Geneva.
Writing in the Astronomical Journal, Bourrier and an international team of researchers reveal how they sought to calculate whether the planets could bear water using the Hubble space telescope to probe the levels of ultraviolet radiation emitted by Trappist-1.
“Knowing how much energy is emitted by a star in the ultraviolet is very important because it can strongly affect the atmospheres of the planets,” said Bourrier, adding that the lower energies of ultraviolet radiation can break water molecules into hydrogen and oxygen, while higher energies can be absorbed by the gases, allowing them to escape into space.
The team fed data on the star’s UV radiation and the distances of the planets into computer models which also took into account other factors, including how the star’s radiation and its habitable zone have shifted over millennia.
The results reveal that over 8bn years all but the outermost planet could have lost the equivalent of 20 times the amount of water contained in the Earth’s oceans. But, they add, if water loss drops once a planet enters the habitable zone, the outer four planets could have lost less than three times the water in Earth’s oceans over 8bn years, suggesting it could still be present in substantial amounts. What’s more, they found that the outer planets could have “regained” significant amounts of water through volcanic events that released water from the planets’ interiors.
But, said Bourrier, the calculations assume the planets start with an infinite amount of water, adding that further work on the density of the planets is needed to figure out how much water they actually formed with, and hence the amount of water remaining.
There is more. “When the hydrogen escapes it can form a very extended, comet-like tail around the planet,” said Bourrier. “If you detect such a cloud escaping from the Trappist-1 planets, it could inform us on the presence of water that is being [lost] into the atmosphere.”
To probe the possibility, the team looked at the UV radiation emitted from Trappist-1 as the second planet out passed in front of the star – a dip in such radiation would have suggested the planet had a hydrogen “halo” absorbing energy. Unfortunately the observations were inconclusive.
“[The star’s] energy is changing over time, so that makes it difficult to search for this hydrogen cloud,” he said, adding that the team is planning further observations looking at both UV and infrared wavelengths to hunt for signs of water in the planets’ atmospheres.
Ignas Snellen, an astrophysicist at the Leiden Observatory in the Netherlands who was not involved in the study, described the Trappist-1 system as exciting. “We are all very eager to learn more about the possible climates on these planets, and in particular whether they could host liquid water,” he said.
But while Snellen welcomed the research, he said that further observations were crucial.
“Personally, I think that you can make all the calculations you want, but in the end we need measurements,” he said.
Snellen added that the mystery could soon be solved, pointing out that the James Webb Space Telescope is set to launch next year and is expected to shed light on the atmospheres of the planets.
“If one or more of these planets turn out to contain water it will be an enormous breakthrough,” said Snellen. “We can really start to think seriously about whether these worlds may harbour life.”<|endoftext|>
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Home » Maths » Domain and Range Of Inverse Trigonometric Functions
# Domain and Range Of Inverse Trigonometric Functions
Inverse trigonometric functions are the inverse functions of the basic trigonometric functions: sine, cosine, and tangent. To define these inverse functions, we need to restrict the domain and range of the original functions so that they become one-to-one and onto.
## Here are the domain and range for each of the primary inverse trigonometric functions:
1. Inverse Sine function ($sin^(-1)$ or arcsin):
Domain: $-1 \leq x \leq 1$, as the sine function takes values in this range.
Range: $-\pi/2 \leq y \leq \pi/2$, which corresponds to the interval over which sine is a strictly increasing function.
1. Inverse Cosine function ($cos^(-1)$ or arccos):
Domain: $-1 \leq x \leq 1$, as the cosine function takes values in this range.
Range: $0 \leq y \leq \pi$, which corresponds to the interval over which cosine is a strictly decreasing function.
1. Inverse Tangent function ($tan^(-1)$ or arctan):
Domain: $-\infty < x < \infty$, as the tangent function takes values over the entire real line.
Range: $-\pi/2 < y < \pi/2$, which corresponds to the interval over which tangent is a strictly increasing function.
There are also three other inverse trigonometric functions corresponding to the reciprocal functions of sine, cosine, and tangent: inverse cosecant, inverse secant, and inverse cotangent. Their domain and range are as follows:
1. Inverse Cosecant function ($csc^(-1)$ or arccsc):
Domain: $x \geq -1$ or $x \geq 1$, as the cosecant function takes values in this range.
Range: $-\pi /2 \leq y < 0$ or $0 < y \leq \pi/2$, which corresponds to the interval over which cosecant is a strictly decreasing function.
1. Inverse Secant function ($sec^(-1)$ or arcsec):
Domain: $x \leq -1$ or $x \geq 1$, as the secant function takes values in this range.
Range: $0 \leq y < \pi/2$ or $\pi/2 < y \leq \pi$, which corresponds to the interval over which secant is a strictly decreasing function.
1. Inverse Cotangent function ($cot^(-1)$ or arccot):
Domain: $-\infty < x < \infty$, as the cotangent function takes values over the entire real line.
Range: $0 < y < \pi$, which corresponds to the interval over which cotangent is a strictly decreasing function.
## How to remember it
• The domain of the inverse function is the range of the original trigonometric function
• The range of the Inverse function is the smallest interval in which it becomes one-one and onto.lets see how we can remember it
Lets first rewrite the table in below manner
We can use below tricks to remember the range base on this
• We can clearly see that inverse of sin, cosec and tan have a similar type of range while inverse of cos ,sec and cot having similar type of range
• sin and cosec but it exclude 0 as then it will become undefined, Similar for cos and sec
• inverse of tan and cot are in different interval
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Helvetic Republic hĕlvē´tĭk [key], 1798–1803, Swiss state established under French auspices. In Sept., 1797, several exiled Swiss leaders in France (notably Frédéric César de La Harpe ) formally urged the French Revolutionary government (the Directory) to help in liberating the subject districts of Switzerland and in overthrowing the aristocratic cantonal governments. The Directory, eager to secure the Alpine passes as well as the treasury of Bern , ordered the invasion of Switzerland (Jan., 1798); resistance was brief. A unified state, the Helvetic Republic, was set up. Lack of funds and constant French political and military intervention proved troublesome; finally, the French Revolutionary Wars shifted (1799) into Switzerland. An Austrian army defeated the French at Zürich (June), but Austro-Russian discord led to the victory (Sept.), again at Zürich, of André Masséna over a Russian army under General Korsakov. General Suvorov, who arrived from Italy to aid Korsakov, was obliged to retreat to Lindau in Germany. The survival of the Helvetic Republic until 1803 was largely due to the presence of French troops, since the Swiss were hostile to centralization. In Feb., 1803, Napoleon, imposing the Act of Mediation, established a confederation of 19 cantons, with a federal diet subservient to France.
The Columbia Electronic Encyclopedia, 6th ed. Copyright © 2012, Columbia University Press. All rights reserved.
See more Encyclopedia articles on: Swiss History<|endoftext|>
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Tubes: The "tubes" are medically known as the Fallopian tubes. There are two Fallopian tubes, one on each side, which transport the egg from the ovary to the uterus (the womb). The Fallopian tubes have small hair-like projections called cilia on the cells of the lining.
These tubal cilia are essential to the movement of the egg through the tube into the uterus. If the tubal cilia are damaged by infection, the egg may not get 'pushed along' normally but may stay in the tube.
Infection can also cause partial or complete blockage of the tube with scar tissue, physically preventing the egg from getting to the uterus.
Any process (such as infection, endometriosis, tumors, or scar tissue in the pelvis (pelvic adhesions) that cause twisting or chinking of the tube) that damages the Fallopian tube or narrows its diameter increases the chance of an ectopic pregnancy: a pregnancy developing in the Fallopian tube or another abnormal location outside the uterus.
These tubes bear the name of Gabriele Falloppio (also spelled Falloppia), a 16th-century (c. 1523-62) Italian physician and surgeon who was expert in anatomy, physiology and pharmacology. He was an early expert on syphilis and one of the great surgeons of the age. Of the various works by Falloppio only the "Observationes anatomicae", a work of great originality, was published during his lifetime. In it he made a number of contributions to the knowledge of centers of ossification, to the detailed account of muscles, and to the understanding of the vascular system and the kidneys. His description of the uterine tubes was sufficiently accurate that they bear his name. With Vesalius and Eustachi, Fallopio is often seen as one of the three heroes of anatomy. (Historical information based on the Catalog of the Scientific Community of the 16th and 17th Centuries by Richard S Westfall for the Galileo Project.)<|endoftext|>
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The evolution of computers from ENIAC to the extremely portable laptop has made computing an everyday task. The then used vacuum diodes became extremely heated that diodes had to be replaced after every hour or less. With a little improvement, lifespan was increased to about a day or two but heating was the major obstacle being faced by computer engineers.
Today we have desktops and ultra-thin laptops, which also use processors which are now integrated ones. Every electronic equipment produces a noticeable amount of heat so does the processor. Not only processors but the motherboard along with the GPU(graphic cards) produce a good amount of heat.
Computers are equipped with a variety of cooling equipment, like a fan, to disperse the excess heat generated from electrical energy, and a CPU fan is used exclusively for cooling the processor.
2.Active or Fan cooling.
Mostly it is the fan cooling which is used in most of the computers, where the components like processor get heated at a fast rate. Fan cooling is performed in conjunction with an Aluminium Heat Sink because direct cooling cannot be performed since heat emitted is more conductive than radiative. To understand how this works:
Heat Sink Fan: The aluminum heat sink fan is a cooling device that works on its own to efficiently draw heat away from the components into its large surface area to transfer cooler air into its fin-like aluminum structure. This cooling device works in unison with all the major components of the computer.
The fan is attached to this fin-like structure and improves the transfer of hot air by pulling the hot air from the electrical heat generated by the components and pushing in the cooler air between the aluminum fins, thus keeping the processor cool.
Talking of the passive cooling there is no fan but only a heatsink alone.Passive cooling is widely used in all types of electronic circuits to dissipate component heat, heatsinks thus preventing burnout of crucial components like MOSFETS.
But what happens in a laptop??
Laptop cooling is completely different from desktop CPU cooling. Instead of using big heatsinks with fan. They use heat-tubes which are made of copper which is a good heat conductor.
To increase the conductance of heat from processor to heat-tube thermal grease is used which is a kind of paste. All the major heat producing components are in contact with heat-tube which is then linked to a cooling fan which forces out heated air from the heatsink-tube, making the processor cooler and helps in better performance.
Liquid cooling is also used but it is not in common and used in bigger computers.
Good processors with better operating speeds can withstand up to 100º C. Some can up to 120°C but that is the last limit, for safety reasons your computers gives you a warning and shuts itself when reaching 80° C or more.
Computers being the most important devices in our daily life, it is very important to maintain them properly and check for their proper operation. Temperature is one condition that is quite important for the maintenance of these devices. So one should have proper knowledge of how the technical aspects of these cooling systems in computers and laptops.
Image courtesy: Wikipedia<|endoftext|>
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How many different ways can the alphabet be rearranged?
How many different ways can the alphabet be rearranged?
3,628,800 ways
How many ways can you arrange the alphabet of the English language if you were to form all the words that have three alphabets in them?
Total no of ways = 18*24! + 18*24! = 36*24!
How many ways can the letters in the word world be arranged?
Number of Ways to Arrange n Letters Word Calculator
Number of Ways to Arrance ‘n’ Letters of a Word
‘n’ Letters Words Ways to Arrange
5 Letters Word 120 Distinct Ways
6 Letters Word 720 Distinct Ways
7 Letters Word 5,040 Distinct Ways
How many ways can 10 things be arranged?
How many ways can 3 things be arranged?
6 ways
What is nPr formula?
FAQs on nPr Formula The nPr formula is used to find the number of ways in which r different things can be selected and arranged out of n different things. This is also known as the permutations formula. The nPr formula is, P(n, r) = n! / (n−r)!.
What do N and R mean in combinations?
* (n – r)!, where n represents the total number of items, and r represents the number of items being chosen at a time. To calculate a combination, you will need to calculate a factorial. A factorial is the product of all the positive integers equal to and less than your number.
What does R mean in nPr?
subset size
What does R mean in nCr?
Here “n” stands for total no of items in the given set. “r” means, the no of items required in the subset formed from the main set(n). And as we know, “C” stands for the word “combination”.
Is nPr and nCr same?
Permutation (nPr) is the way of arranging the elements of a group or a set in an order. Combination (nCr) is the selection of elements from a group or a set, where order of the elements does not matter. …
What is the value of 7 \choose 5?
7 CHOOSE 5 = 21 possible combinations. 21 is the total number of all possible combinations for choosing 5 elements at a time from 7 distinct elements without considering the order of elements in statistics & probability surveys or experiments.
How many ways can you pick 3 out of 5?
10 possible
What are N and R in permutations?
n = total items in the set; r = items taken for the permutation; “!”
What does 10 choose 6 mean?
What is 10 CHOOSE 6 or Value of 10C6? 10 CHOOSE 6 = 210 possible combinations. 210 is the total number of all possible combinations for choosing 6 elements at a time from 10 distinct elements without considering the order of elements in statistics & probability surveys or experiments.
How is 5C1 calculated?
What is 5 CHOOSE 1 or Value of 5C1? 5 CHOOSE 1 = 5 possible combinations. 5 is the total number of all possible combinations for choosing 1 elements at a time from 5 distinct elements without considering the order of elements in statistics & probability surveys or experiments.
How many permutations are there in 10 numbers?
10,000,000,000<|endoftext|>
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Literature circles allow students to participate in self-directed discussions about literature. In groups of 4-6, students choose a book for their group to read. The students meet daily, setting goals for completion and choosing jobs for discussion.
With Falcon Apps (Google Apps for Education) and laptop computers, every student collaborates outside of class before meeting with their group. Students monitor their group members’ progress, chat to solve problems before the discussion, and guide each other to better contributions in class. The teacher can monitor student work in progress, providing quick and easy feedback to facilitate better preparation for discussions.
The 1:1 access to laptop computers and Falcon Apps has facilitated:
- increased student collaboration.
- more focused and productive group discussions.
- student goal-setting, teamwork, and accountability.
- faster feedback from peers and teachers.
- extended learning beyond the confines of the classroom walls.
- students seeking knowledge from more sources than just the teacher.
During class students view the shared document. They can view a copy of their partner’s work just by pulling it up on their laptop computer, revising and adding or removing information based on the flow of the discussion. The entire discussion process is facilitated through the shared document, and the students’ responses are improved by the ability to quickly edit their work, provide and receive feedback.
Students use the internet to supplement the text by looking up words and concepts, looking for other interpretations of the text by other readers outside the class, reading comments, finding images and video to help visualize and contextualize, and locating examples of literary terms applied to their text (such as metaphors and themes) used by other internet users.
After the completion of the literature circle discussion, students demonstrate comprehension through a project of their choice. Using a shared Falcon Apps document, students sign up for a project choice and regularly update the document with their progress. The freedom to choose a final product has boosted student engagement. Watch this sample literature circle project by Matt D.
This is a nifty little script I just learned about on the Free Technology for Teachers blog. (By the way, anyone would be wise to subscribe to this blog – lots of useful ideas and tools.)
An effective way to see whether students have grasped some factual content is to set up an assessment using a Google Form (Falcon Apps > Create new > Form). Once students have taken the assessment, you can go to the associated spreadsheet and install the Flubaroo script (Directions are located online at this link.) What will the Flubaroo script do? It will…
- Grade the assessment
- Email feedback to the students
- Print a report for your use
There is an excellent demonstration video available showing how to start using this tool.
Flubaroo is a great example of how technology can make the feedback process (so necessary in teaching and learning) simple and helpful. Both teacher and student receive feedback that can be invaluable in the process of acquiring basic content knowledge.<|endoftext|>
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# Sudha is twice as old as Radha. If 6 years is subtracted from Radha's age and Sudha's age is increased by 4 years, then Sudha's age will be four times that of Radha. What will be the ages of Sudha and Radha, 2 years ago from today?
This question was previously asked in
RRC Group D Previous Paper 62 (Held On: 9 Oct 2018 Shift 1)
View all RRB Group D Papers >
1. 30 years and 14 years
2. 26 years and 12 years
3. 34 years and 20 years
4. 30 years and 16 years
Option 2 : 26 years and 12 years
## Detailed Solution
Given:
Sudha is twice as old as Radha
Calculation:
Let Radha's age be 'x' years
and Sudha's age be 'y' years
y = 2x
⇒ 4(x − 6) = (y + 4)
⇒ 4(x − 6) = 2(2x + 4)
⇒ 4x − 24 = 2x + 4
⇒ 2x = 28
⇒ x =14 years
y = 2x =28 years
2 years ago Radha was 12 years & Sudha was 26 years.
∴ The ages of Sudha and Radha, 2 years ago from today were 12 years and 26 years<|endoftext|>
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# AM-GM Inequality
In mathematics, the inequality of arithmetic and geometric means, or more briefly the AM–GM inequality, states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list; and further, that the two means are equal if and only if every number in the list is the same (in which case they are both that number).
The simplest non-trivial case – i.e., with more than one variable – for two non-negative numbers x and y, is the statement that
${\displaystyle {\frac {x+y}{2}}\geq {\sqrt {xy}}}$
with equality if and only if x = y. This case can be seen from the fact that the square of a real number is always non-negative (greater than or equal to zero) and from the elementary case (a ± b)2 = a2 ± 2ab + b2 of the binomial formula:
{\displaystyle {\begin{aligned}0&\leq (x-y)^{2}\\&=x^{2}-2xy+y^{2}\\&=x^{2}+2xy+y^{2}-4xy\\&=(x+y)^{2}-4xy.\end{aligned}}}
Hence (x + y)2 ≥ 4xy, with equality precisely when (xy)2 = 0, i.e. x = y. The AM–GM inequality then follows from taking the positive square root of both sides and then dividing both sides by 2.
For a geometrical interpretation, consider a rectangle with sides of length x and y, hence it has perimeter 2x + 2y and area xy. Similarly, a square with all sides of length xy has the perimeter 4xy and the same area as the rectangle. The simplest non-trivial case of the AM–GM inequality implies for the perimeters that 2x + 2y ≥ 4xy and that only the square has the smallest perimeter amongst all rectangles of equal area.
Extensions of the AM–GM inequality are available to include weights or generalized means.
## Background
The arithmetic mean, or less precisely the average, of a list of n numbers x1, x2, . . . , xn is the sum of the numbers divided by n:
${\displaystyle {\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}.}$
The geometric mean is similar, except that it is only defined for a list of nonnegative real numbers, and uses multiplication and a root in place of addition and division:
${\displaystyle {\sqrt[{n}]{x_{1}\cdot x_{2}\cdots x_{n}}}.}$
If x1, x2, . . . , xn > 0, this is equal to the exponential of the arithmetic mean of the natural logarithms of the numbers:
${\displaystyle \exp \left({\frac {\ln {x_{1}}+\ln {x_{2}}+\cdots +\ln {x_{n}}}{n}}\right).}$
## The inequality
Restating the inequality using mathematical notation, we have that for any list of n nonnegative real numbers x1, x2, . . . , xn,
${\displaystyle {\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\geq {\sqrt[{n}]{x_{1}\cdot x_{2}\cdots x_{n}}}\,,}$
and that equality holds if and only if x1 = x2 = · · · = xn.
## Geometric interpretation
In two dimensions, 2x1 + 2x2 is the perimeter of a rectangle with sides of length x1 and x2. Similarly, 4x1x2 is the perimeter of a square with the same area, x1x2, as that rectangle. Thus for n = 2 the AM–GM inequality states that a rectangle of a given area has the smallest perimeter if that rectangle is also a square.
The full inequality is an extension of this idea to n dimensions. Every vertex of an n-dimensional box is connected to n edges. If these edges' lengths are x1, x2, . . . , xn, then x1 + x2 + · · · + xn is the total length of edges incident to the vertex. There are 2n vertices, so we multiply this by 2n; since each edge, however, meets two vertices, every edge is counted twice. Therefore, we divide by 2 and conclude that there are 2n−1n edges. There are equally many edges of each length and n lengths; hence there are 2n−1 edges of each length and the total of all edge lengths is 2n−1(x1 + x2 + · · · + xn). On the other hand,
${\displaystyle 2^{n-1}(x_{1}+\ldots +x_{n})=2^{n-1}n{\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}}$
is the total length of edges connected to a vertex on an n-dimensional cube of equal volume, since in this case x1=...=xn. Since the inequality says
${\displaystyle {x_{1}+x_{2}+\cdots +x_{n} \over n}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}},}$
it can be restated by multiplying through by n2n–1 to obtain
${\displaystyle 2^{n-1}(x_{1}+x_{2}+\cdots +x_{n})\geq 2^{n-1}n{\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}}$
with equality if and only if x1 = x2 = · · · = xn.
Thus the AM–GM inequality states that only the n-cube has the smallest sum of lengths of edges connected to each vertex amongst all n-dimensional boxes with the same volume.[2]
## Examples
### Example 1
If ${\displaystyle a,b,c>0}$, then the A.M.-G.M. tells us that
${\displaystyle (1+a)(1+b)(1+c)\geq 8{\sqrt {abc}}}$
### Example 2
A simple upper bound for ${\displaystyle n!}$ can be found. AM-GM tells us
${\displaystyle 1+2+\dots +n\geq n{\sqrt[{n}]{n!}}}$
${\displaystyle {\frac {n(n+1)}{2}}\geq n{\sqrt[{n}]{n!}}}$
and so
${\displaystyle \left({\frac {n+1}{2}}\right)^{n}\geq n!}$
with equality at ${\displaystyle n=1}$.
Equivalently,
${\displaystyle (n+1)^{n}\geq 2^{n}n!}$
### Example 3
Consider the function
${\displaystyle f(x,y,z)={\frac {x}{y}}+{\sqrt {\frac {y}{z}}}+{\sqrt[{3}]{\frac {z}{x}}}}$
for all positive real numbers x, y and z. Suppose we wish to find the minimal value of this function. It can be rewritten as:
{\displaystyle {\begin{aligned}f(x,y,z)&=6\cdot {\frac {{\frac {x}{y}}+{\frac {1}{2}}{\sqrt {\frac {y}{z}}}+{\frac {1}{2}}{\sqrt {\frac {y}{z}}}+{\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}+{\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}+{\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}}{6}}\\&=6\cdot {\frac {x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}}{6}}\end{aligned}}}
with
${\displaystyle x_{1}={\frac {x}{y}},\qquad x_{2}=x_{3}={\frac {1}{2}}{\sqrt {\frac {y}{z}}},\qquad x_{4}=x_{5}=x_{6}={\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}.}$
Applying the AM–GM inequality for n = 6, we get
{\displaystyle {\begin{aligned}f(x,y,z)&\geq 6\cdot {\sqrt[{6}]{{\frac {x}{y}}\cdot {\frac {1}{2}}{\sqrt {\frac {y}{z}}}\cdot {\frac {1}{2}}{\sqrt {\frac {y}{z}}}\cdot {\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}\cdot {\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}\cdot {\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}}}\\&=6\cdot {\sqrt[{6}]{{\frac {1}{2\cdot 2\cdot 3\cdot 3\cdot 3}}{\frac {x}{y}}{\frac {y}{z}}{\frac {z}{x}}}}\\&=2^{2/3}\cdot 3^{1/2}.\end{aligned}}}
Further, we know that the two sides are equal exactly when all the terms of the mean are equal:
${\displaystyle f(x,y,z)=2^{2/3}\cdot 3^{1/2}\quad {\mbox{when}}\quad {\frac {x}{y}}={\frac {1}{2}}{\sqrt {\frac {y}{z}}}={\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}.}$
All the points (x, y, z) satisfying these conditions lie on a half-line starting at the origin and are given by
${\displaystyle (x,y,z)={\biggr (}t,{\sqrt[{3}]{2}}{\sqrt {3}}\,t,{\frac {3{\sqrt {3}}}{2}}\,t{\biggr )}\quad {\mbox{with}}\quad t>0.}$
## Applications
An important practical application in financial mathematics is to computing the rate of return: the annualized return, computed via the geometric mean, is less than the average annual return, computed by the arithmetic mean (or equal if all returns are equal). This is important in analyzing investments, as the average return overstates the cumulative effect.
## Proofs of the AM–GM inequality
### Proof using Jensen's inequality
Jensen's inequality states that the value of a concave function of an arithmetic mean is greater than or equal to the arithmetic mean of the function's values. Since the logarithm function is concave, we have
${\displaystyle \log \left({\frac {\sum _{i}x_{i}}{n}}\right)\geq \sum {\frac {1}{n}}\log x_{i}=\sum \left(\log x_{i}^{1/n}\right)=\log \left(\prod x_{i}^{1/n}\right).}$
Taking antilogs of the far left and far right sides, we have the AM–GM inequality.
### Proof by successive replacement of elements
We have to show that
${\displaystyle \alpha ={\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}=\beta }$
with equality only when all numbers are equal.
If not all numbers are equal, then there exist ${\displaystyle x_{i},x_{j}}$ such that ${\displaystyle x_{i}<\alpha . Replacing xi by ${\displaystyle \alpha }$ and xj by ${\displaystyle (x_{i}+x_{j}-\alpha )}$ will leave the arithmetic mean of the numbers unchanged, but will increase the geometric mean because
${\displaystyle \alpha (x_{j}+x_{i}-\alpha )-x_{i}x_{j}=(\alpha -x_{i})(x_{j}-\alpha )>0}$
If the numbers are still not equal, we continue replacing numbers as above. After at most ${\displaystyle (n-1)}$ such replacement steps all the numbers will have been replaced with ${\displaystyle \alpha }$ while the geometric mean strictly increases at each step. After the last step, the geometric mean will be ${\displaystyle {\sqrt[{n}]{\alpha \alpha \cdots \alpha }}=\alpha }$, proving the inequality.
It may be noted that the replacement strategy works just as well from the right hand side. If any of the numbers is 0 then so will the geometric mean thus proving the inequality trivially. Therefore we may suppose that all the numbers are positive. If they are not all equal, then there exist ${\displaystyle x_{i},x_{j}}$ such that ${\displaystyle 0. Replacing ${\displaystyle x_{i}}$ by ${\displaystyle \beta }$ and ${\displaystyle x_{j}}$ by ${\displaystyle {\frac {x_{i}x_{j}}{\beta }}}$leaves the geometric mean unchanged but strictly decreases the arithmetic mean since
${\displaystyle x_{i}+x_{j}-\beta -{\frac {x_{i}x_{j}}{\beta }}={\frac {(\beta -x_{i})(x_{j}-\beta )}{\beta }}>0}$. The proof then follows along similar lines as in the earlier replacement.
### Induction proofs
#### Proof by induction #1
Of the non-negative real numbers x1, . . . , xn, the AM–GM statement is equivalent to
${\displaystyle \alpha ^{n}\geq x_{1}x_{2}\cdots x_{n}}$
with equality if and only if α = xi for all i ∈ {1, . . . , n}.
For the following proof we apply mathematical induction and only well-known rules of arithmetic.
Induction basis: For n = 1 the statement is true with equality.
Induction hypothesis: Suppose that the AM–GM statement holds for all choices of n non-negative real numbers.
Induction step: Consider n + 1 non-negative real numbers x1, . . . , xn+1, . Their arithmetic mean α satisfies
${\displaystyle (n+1)\alpha =\ x_{1}+\cdots +x_{n}+x_{n+1}.}$
If all the xi are equal to α, then we have equality in the AM–GM statement and we are done. In the case where some are not equal to α, there must exist one number that is greater than the arithmetic mean α, and one that is smaller than α. Without loss of generality, we can reorder our xi in order to place these two particular elements at the end: xn > α and xn+1 < α. Then
${\displaystyle x_{n}-\alpha >0\qquad \alpha -x_{n+1}>0}$
${\displaystyle \implies (x_{n}-\alpha )(\alpha -x_{n+1})>0\,.\qquad (*)}$
Now define y with
${\displaystyle y:=x_{n}+x_{n+1}-\alpha \geq x_{n}-\alpha >0\,,}$
and consider the n numbers x1, . . . , xn–1, y which are all non-negative. Since
${\displaystyle (n+1)\alpha =x_{1}+\cdots +x_{n-1}+x_{n}+x_{n+1}}$
${\displaystyle n\alpha =x_{1}+\cdots +x_{n-1}+\underbrace {x_{n}+x_{n+1}-\alpha } _{=\,y},}$
Thus, α is also the arithmetic mean of n numbers x1, . . . , xn–1, y and the induction hypothesis implies
${\displaystyle \alpha ^{n+1}=\alpha ^{n}\cdot \alpha \geq x_{1}x_{2}\cdots x_{n-1}y\cdot \alpha .\qquad (**)}$
Due to (*) we know that
${\displaystyle (\underbrace {x_{n}+x_{n+1}-\alpha } _{=\,y})\alpha -x_{n}x_{n+1}=(x_{n}-\alpha )(\alpha -x_{n+1})>0,}$
hence
${\displaystyle y\alpha >x_{n}x_{n+1}\,,\qquad ({*}{*}{*})}$
in particular α > 0. Therefore, if at least one of the numbers x1, . . . , xn–1 is zero, then we already have strict inequality in (**). Otherwise the right-hand side of (**) is positive and strict inequality is obtained by using the estimate (***) to get a lower bound of the right-hand side of (**). Thus, in both cases we can substitute (***) into (**) to get
${\displaystyle \alpha ^{n+1}>x_{1}x_{2}\cdots x_{n-1}x_{n}x_{n+1}\,,}$
which completes the proof.
#### Proof by induction #2
First of all we shall prove that for real numbers x1 < 1 and x2 > 1 there follows
${\displaystyle x_{1}+x_{2}>x_{1}x_{2}+1.}$
Indeed, multiplying both sides of the inequality x2 > 1 by 1 – x1, gives
${\displaystyle x_{2}-x_{1}x_{2}>1-x_{1},}$
whence the required inequality is obtained immediately.
Now, we are going to prove that for positive real numbers x1, . . . , xn satisfying x1 . . . xn = 1, there holds
${\displaystyle x_{1}+\cdots +x_{n}\geq n.}$
The equality holds only if x1 = ... = xn = 1.
Induction basis: For n = 2 the statement is true because of the above property.
Induction hypothesis: Suppose that the statement is true for all natural numbers up to n – 1.
Induction step: Consider natural number n, i.e. for positive real numbers x1, . . . , xn, there holds x1 . . . xn = 1. There exists at least one xk < 1, so there must be at least one xj > 1. Without loss of generality, we let k =n – 1 and j = n.
Further, the equality x1 . . . xn = 1 we shall write in the form of (x1 . . . xn–2) (xn–1 xn) = 1. Then, the induction hypothesis implies
${\displaystyle (x_{1}+\cdots +x_{n-2})+(x_{n-1}x_{n})>n-1.}$
However, taking into account the induction basis, we have
{\displaystyle {\begin{aligned}x_{1}+\cdots +x_{n-2}+x_{n-1}+x_{n}&=(x_{1}+\cdots +x_{n-2})+(x_{n-1}+x_{n})\\&>(x_{1}+\cdots +x_{n-2})+x_{n-1}x_{n}+1\\&>n,\end{aligned}}}
which completes the proof.
For positive real numbers a1, . . . , an, let's denote
${\displaystyle x_{1}={\frac {a_{1}}{\sqrt[{n}]{a_{1}\cdots a_{n}}}},...,x_{n}={\frac {a_{n}}{\sqrt[{n}]{a_{1}\cdots a_{n}}}}.}$
The numbers x1, . . . , xn satisfy the condition x1 . . . xn = 1. So we have
${\displaystyle {\frac {a_{1}}{\sqrt[{n}]{a_{1}\cdots a_{n}}}}+\cdots +{\frac {a_{n}}{\sqrt[{n}]{a_{1}\cdots a_{n}}}}\geq n,}$
whence we obtain
${\displaystyle {\frac {a_{1}+\cdots +a_{n}}{n}}\geq {\sqrt[{n}]{a_{1}\cdots a_{n}}},}$
with the equality holding only for a1 = ... = an.
### Proof by Cauchy using forward–backward induction
The following proof by cases relies directly on well-known rules of arithmetic but employs the rarely used technique of forward-backward-induction. It is essentially from Augustin Louis Cauchy and can be found in his Cours d'analyse.[3]
#### The case where all the terms are equal
If all the terms are equal:
${\displaystyle x_{1}=x_{2}=\cdots =x_{n},}$
then their sum is nx1, so their arithmetic mean is x1; and their product is x1n, so their geometric mean is x1; therefore, the arithmetic mean and geometric mean are equal, as desired.
#### The case where not all the terms are equal
It remains to show that if not all the terms are equal, then the arithmetic mean is greater than the geometric mean. Clearly, this is only possible when n > 1.
This case is significantly more complex, and we divide it into subcases.
##### The subcase where n = 2
If n = 2, then we have two terms, x1 and x2, and since (by our assumption) not all terms are equal, we have:
{\displaystyle {\begin{aligned}{\Bigl (}{\frac {x_{1}+x_{2}}{2}}{\Bigr )}^{2}-x_{1}x_{2}&={\frac {1}{4}}(x_{1}^{2}+2x_{1}x_{2}+x_{2}^{2})-x_{1}x_{2}\\&={\frac {1}{4}}(x_{1}^{2}-2x_{1}x_{2}+x_{2}^{2})\\&={\Bigl (}{\frac {x_{1}-x_{2}}{2}}{\Bigr )}^{2}>0,\end{aligned}}}
hence
${\displaystyle {\frac {x_{1}+x_{2}}{2}}\geq {\sqrt {x_{1}x_{2}}}}$
as desired.
##### The subcase where n = 2k
Consider the case where n = 2k, where k is a positive integer. We proceed by mathematical induction.
In the base case, k = 1, so n = 2. We have already shown that the inequality holds when n = 2, so we are done.
Now, suppose that for a given k > 1, we have already shown that the inequality holds for n = 2k−1, and we wish to show that it holds for n = 2k. To do so, we apply the inequality twice for 2k-1 numbers and once for 2 numbers to obtain:
{\displaystyle {\begin{aligned}{\frac {x_{1}+x_{2}+\cdots +x_{2^{k}}}{2^{k}}}&{}={\frac {{\frac {x_{1}+x_{2}+\cdots +x_{2^{k-1}}}{2^{k-1}}}+{\frac {x_{2^{k-1}+1}+x_{2^{k-1}+2}+\cdots +x_{2^{k}}}{2^{k-1}}}}{2}}\\[7pt]&\geq {\frac {{\sqrt[{2^{k-1}}]{x_{1}x_{2}\cdots x_{2^{k-1}}}}+{\sqrt[{2^{k-1}}]{x_{2^{k-1}+1}x_{2^{k-1}+2}\cdots x_{2^{k}}}}}{2}}\\[7pt]&\geq {\sqrt {{\sqrt[{2^{k-1}}]{x_{1}x_{2}\cdots x_{2^{k-1}}}}{\sqrt[{2^{k-1}}]{x_{2^{k-1}+1}x_{2^{k-1}+2}\cdots x_{2^{k}}}}}}\\[7pt]&={\sqrt[{2^{k}}]{x_{1}x_{2}\cdots x_{2^{k}}}}\end{aligned}}}
where in the first inequality, the two sides are equal only if
${\displaystyle x_{1}=x_{2}=\cdots =x_{2^{k-1}}}$
and
${\displaystyle x_{2^{k-1}+1}=x_{2^{k-1}+2}=\cdots =x_{2^{k}}}$
(in which case the first arithmetic mean and first geometric mean are both equal to x1, and similarly with the second arithmetic mean and second geometric mean); and in the second inequality, the two sides are only equal if the two geometric means are equal. Since not all 2k numbers are equal, it is not possible for both inequalities to be equalities, so we know that:
${\displaystyle {\frac {x_{1}+x_{2}+\cdots +x_{2^{k}}}{2^{k}}}\geq {\sqrt[{2^{k}}]{x_{1}x_{2}\cdots x_{2^{k}}}}}$
as desired.
##### The subcase where n < 2k
If n is not a natural power of 2, then it is certainly less than some natural power of 2, since the sequence 2, 4, 8, . . . , 2k, . . . is unbounded above. Therefore, without loss of generality, let m be some natural power of 2 that is greater than n.
So, if we have n terms, then let us denote their arithmetic mean by α, and expand our list of terms thus:
${\displaystyle x_{n+1}=x_{n+2}=\cdots =x_{m}=\alpha .}$
We then have:
{\displaystyle {\begin{aligned}\alpha &={\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\\[6pt]&={\frac {{\frac {m}{n}}\left(x_{1}+x_{2}+\cdots +x_{n}\right)}{m}}\\[6pt]&={\frac {x_{1}+x_{2}+\cdots +x_{n}+{\frac {(m-n)}{n}}\left(x_{1}+x_{2}+\cdots +x_{n}\right)}{m}}\\[6pt]&={\frac {x_{1}+x_{2}+\cdots +x_{n}+\left(m-n\right)\alpha }{m}}\\[6pt]&={\frac {x_{1}+x_{2}+\cdots +x_{n}+x_{n+1}+\cdots +x_{m}}{m}}\\[6pt]&\geq {\sqrt[{m}]{x_{1}x_{2}\cdots x_{n}x_{n+1}\cdots x_{m}}}\\[6pt]&={\sqrt[{m}]{x_{1}x_{2}\cdots x_{n}\alpha ^{m-n}}}\,,\end{aligned}}}
so
${\displaystyle \alpha ^{m}\geq x_{1}x_{2}\cdots x_{n}\alpha ^{m-n}}$
and
${\displaystyle \alpha \geq {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}}$
as desired.
### Proof by induction using basic calculus
The following proof uses mathematical induction and some basic differential calculus.
Induction basis: For n = 1 the statement is true with equality.
Induction hypothesis: Suppose that the AM–GM statement holds for all choices of n non-negative real numbers.
Induction step: In order to prove the statement for n + 1 non-negative real numbers x1, . . . , xn, xn+1, we need to prove that
${\displaystyle {\frac {x_{1}+\cdots +x_{n}+x_{n+1}}{n+1}}-({x_{1}\cdots x_{n}x_{n+1}})^{\frac {1}{n+1}}\geq 0}$
with equality only if all the n + 1 numbers are equal.
If all numbers are zero, the inequality holds with equality. If some but not all numbers are zero, we have strict inequality. Therefore, we may assume in the following, that all n + 1 numbers are positive.
We consider the last number xn+1 as a variable and define the function
${\displaystyle f(t)={\frac {x_{1}+\cdots +x_{n}+t}{n+1}}-({x_{1}\cdots x_{n}t})^{\frac {1}{n+1}},\qquad t>0.}$
Proving the induction step is equivalent to showing that f(t) ≥ 0 for all t > 0, with f(t) = 0 only if x1, . . . , xn and t are all equal. This can be done by analyzing the critical points of f using some basic calculus.
The first derivative of f is given by
${\displaystyle f'(t)={\frac {1}{n+1}}-{\frac {1}{n+1}}({x_{1}\cdots x_{n}})^{\frac {1}{n+1}}t^{-{\frac {n}{n+1}}},\qquad t>0.}$
A critical point t0 has to satisfy f′(t0) = 0, which means
${\displaystyle ({x_{1}\cdots x_{n}})^{\frac {1}{n+1}}t_{0}^{-{\frac {n}{n+1}}}=1.}$
After a small rearrangement we get
${\displaystyle t_{0}^{\frac {n}{n+1}}=({x_{1}\cdots x_{n}})^{\frac {1}{n+1}},}$
and finally
${\displaystyle t_{0}=({x_{1}\cdots x_{n}})^{\frac {1}{n}},}$
which is the geometric mean of x1, . . . , xn. This is the only critical point of f. Since f′′(t) > 0 for all t > 0, the function f is strictly convex and has a strict global minimum at t0. Next we compute the value of the function at this global minimum:
{\displaystyle {\begin{aligned}f(t_{0})&={\frac {x_{1}+\cdots +x_{n}+({x_{1}\cdots x_{n}})^{1/n}}{n+1}}-({x_{1}\cdots x_{n}})^{\frac {1}{n+1}}({x_{1}\cdots x_{n}})^{\frac {1}{n(n+1)}}\\&={\frac {x_{1}+\cdots +x_{n}}{n+1}}+{\frac {1}{n+1}}({x_{1}\cdots x_{n}})^{\frac {1}{n}}-({x_{1}\cdots x_{n}})^{\frac {1}{n}}\\&={\frac {x_{1}+\cdots +x_{n}}{n+1}}-{\frac {n}{n+1}}({x_{1}\cdots x_{n}})^{\frac {1}{n}}\\&={\frac {n}{n+1}}{\Bigl (}{\frac {x_{1}+\cdots +x_{n}}{n}}-({x_{1}\cdots x_{n}})^{\frac {1}{n}}{\Bigr )}\\&\geq 0,\end{aligned}}}
where the final inequality holds due to the induction hypothesis. The hypothesis also says that we can have equality only when x1, . . . , xn are all equal. In this case, their geometric mean t0 has the same value, Hence, unless x1, . . . , xn, xn+1 are all equal, we have f(xn+1) > 0. This completes the proof.
This technique can be used in the same manner to prove the generalized AM–GM inequality and Cauchy–Schwarz inequality in Euclidean space Rn.
### Proof by Pólya using the exponential function
George Pólya provided a proof similar to what follows. Let f(x) = ex–1x for all real x, with first derivative f′(x) = ex–1 – 1 and second derivative f′′(x) = ex–1. Observe that f(1) = 0, f′(1) = 0 and f′′(x) > 0 for all real x, hence f is strictly convex with the absolute minimum at x = 1. Hence x ≤ ex–1 for all real x with equality only for x = 1.
Consider a list of non-negative real numbers x1, x2, . . . , xn. If they are all zero, then the AM–GM inequality holds with equality. Hence we may assume in the following for their arithmetic mean α > 0. By n-fold application of the above inequality, we obtain that
{\displaystyle {\begin{aligned}{{\frac {x_{1}}{\alpha }}{\frac {x_{2}}{\alpha }}\cdots {\frac {x_{n}}{\alpha }}}&\leq {e^{{\frac {x_{1}}{\alpha }}-1}e^{{\frac {x_{2}}{\alpha }}-1}\cdots e^{{\frac {x_{n}}{\alpha }}-1}}\\&=\exp {\Bigl (}{\frac {x_{1}}{\alpha }}-1+{\frac {x_{2}}{\alpha }}-1+\cdots +{\frac {x_{n}}{\alpha }}-1{\Bigr )},\qquad (*)\end{aligned}}}
with equality if and only if xi = α for every i ∈ {1, . . . , n}. The argument of the exponential function can be simplified:
{\displaystyle {\begin{aligned}{\frac {x_{1}}{\alpha }}-1+{\frac {x_{2}}{\alpha }}-1+\cdots +{\frac {x_{n}}{\alpha }}-1&={\frac {x_{1}+x_{2}+\cdots +x_{n}}{\alpha }}-n\\&={\frac {n\alpha }{\alpha }}-n\\&=0.\end{aligned}}}
Returning to (*),
${\displaystyle {\frac {x_{1}x_{2}\cdots x_{n}}{\alpha ^{n}}}\leq e^{0}=1,}$
which produces x1 x2 · · · xnαn, hence the result[4]
${\displaystyle {\sqrt[{n}]{x_{1}x_{2}\cdots x_{n}}}\leq \alpha .}$
### Proof by Lagrangian multipliers
If any of the ${\displaystyle x_{i}}$ are ${\displaystyle 0}$, then there is nothing to prove. So we may assume all the ${\displaystyle x_{i}}$ are strictly positive.
Because the arithmetic and geometric means are homogeneous of degree 1, without loss of generality assume that ${\displaystyle \prod _{i=1}^{n}x_{i}=1}$. Set ${\displaystyle G(x_{1},x_{2},\ldots ,x_{n})=\prod _{i=1}^{n}x_{i}}$, and ${\displaystyle F(x_{1},x_{2},\ldots ,x_{n})={\frac {1}{n}}\sum _{i=1}^{n}x_{i}}$. The inequality will be proved (together with the equality case) if we can show that the minimum of ${\displaystyle F(x_{1},x_{2},...,x_{n}),}$ subject to the constraint ${\displaystyle G(x_{1},x_{2},\ldots ,x_{n})=1,}$ is equal to ${\displaystyle 1}$, and the minimum is only achieved when ${\displaystyle x_{1}=x_{2}=\cdots =x_{n}=1}$. Let us first show that the constrained minimization problem has a global minimum.
Set ${\displaystyle K=\{(x_{1},x_{2},\ldots ,x_{n})\colon 0\leq x_{1},x_{2},\ldots ,x_{n}\leq n\}}$. Since the intersection ${\displaystyle K\cap \{G=1\}}$ is compact, the extreme value theorem guarantees that the minimum of ${\displaystyle F(x_{1},x_{2},...,x_{n})}$ subject to the constraints ${\displaystyle G(x_{1},x_{2},\ldots ,x_{n})=1}$ and ${\displaystyle (x_{1},x_{2},\ldots ,x_{n})\in K}$ is attained at some point inside ${\displaystyle K}$. On the other hand, observe that if any of the ${\displaystyle x_{i}>n}$, then ${\displaystyle F(x_{1},x_{2},\ldots ,x_{n})>1}$, while ${\displaystyle F(1,1,\ldots ,1)=1}$, and ${\displaystyle (1,1,\ldots ,1)\in K\cap \{G=1\}}$. This means that the minimum inside ${\displaystyle K\cap \{G=1\}}$ is in fact a global minimum, since the value of ${\displaystyle F}$ at any point inside ${\displaystyle K\cap \{G=1\}}$ is certainly no smaller than the minimum, and the value of ${\displaystyle F}$ at any point ${\displaystyle (y_{1},y_{2},\ldots ,y_{n})}$ not inside ${\displaystyle K}$ is strictly bigger than the value at ${\displaystyle (1,1,\ldots ,1)}$, which is no smaller than the minimum.
The method of Lagrange multipliers says that the global minimum is attained at a point ${\displaystyle (x_{1},x_{2},\ldots ,x_{n})}$ where the gradient of ${\displaystyle F(x_{1},x_{2},\ldots ,x_{n})}$ is ${\displaystyle \lambda }$ times the gradient of ${\displaystyle G(x_{1},x_{2},\ldots ,x_{n})}$, for some ${\displaystyle \lambda }$. We will show that the only point at which this happens is when ${\displaystyle x_{1}=x_{2}=\cdots =x_{n}=1}$ and ${\displaystyle F(x_{1},x_{2},...,x_{n})=1.}$
Compute ${\displaystyle {\frac {\partial F}{\partial x_{i}}}={\frac {1}{n}}}$ and
${\displaystyle {\frac {\partial G}{\partial x_{i}}}=\prod _{j\neq i}x_{j}={\frac {G(x_{1},x_{2},\ldots ,x_{n})}{x_{i}}}={\frac {1}{x_{i}}}}$
along the constraint. Setting the gradients proportional to one another therefore gives for each ${\displaystyle i}$ that ${\displaystyle {\frac {1}{n}}={\frac {\lambda }{x_{i}}},}$ and so ${\displaystyle n\lambda =x_{i}.}$ Since the left-hand side does not depend on ${\displaystyle i}$, it follows that ${\displaystyle x_{1}=x_{2}=\cdots =x_{n}}$, and since ${\displaystyle G(x_{1},x_{2},\ldots ,x_{n})=1}$, it follows that ${\displaystyle x_{1}=x_{2}=\cdots =x_{n}=1}$ and ${\displaystyle F(x_{1},x_{2},\ldots ,x_{n})=1}$, as desired.
## Generalizations
### Weighted AM–GM inequality
There is a similar inequality for the weighted arithmetic mean and weighted geometric mean. Specifically, let the nonnegative numbers x1, x2, . . . , xn and the nonnegative weights w1, w2, . . . , wn be given. Set w = w1 + w2 + · · · + wn. If w > 0, then the inequality
${\displaystyle {\frac {w_{1}x_{1}+w_{2}x_{2}+\cdots +w_{n}x_{n}}{w}}\geq {\sqrt[{w}]{x_{1}^{w_{1}}x_{2}^{w_{2}}\cdots x_{n}^{w_{n}}}}}$
holds with equality if and only if all the xk with wk > 0 are equal. Here the convention 00 = 1 is used.
If all wk = 1, this reduces to the above inequality of arithmetic and geometric means.
One stronger version of this, which also gives strengthened version of the unweighted version, is due to Aldaz. In particular, There is a similar inequality for the weighted arithmetic mean and weighted geometric mean. Specifically, let the nonnegative numbers x1, x2, . . . , xn and the nonnegative weights w1, w2, . . . , wn be given. Assume further that the sum of the weights is 1. Then ${\displaystyle \sum _{i=1}^{n}w_{i}x_{i}\geq \prod _{i=1}^{n}x_{i}^{w_{i}}+\sum _{i=1}^{n}w_{i}\left(x_{i}^{\frac {1}{2}}-\sum _{i=1}^{n}w_{i}x_{i}^{\frac {1}{2}}\right)^{2}}$. [5]
#### Proof using Jensen's inequality
Using the finite form of Jensen's inequality for the natural logarithm, we can prove the inequality between the weighted arithmetic mean and the weighted geometric mean stated above.
Since an xk with weight wk = 0 has no influence on the inequality, we may assume in the following that all weights are positive. If all xk are equal, then equality holds. Therefore, it remains to prove strict inequality if they are not all equal, which we will assume in the following, too. If at least one xk is zero (but not all), then the weighted geometric mean is zero, while the weighted arithmetic mean is positive, hence strict inequality holds. Therefore, we may assume also that all xk are positive.
Since the natural logarithm is strictly concave, the finite form of Jensen's inequality and the functional equations of the natural logarithm imply
{\displaystyle {\begin{aligned}\ln {\Bigl (}{\frac {w_{1}x_{1}+\cdots +w_{n}x_{n}}{w}}{\Bigr )}&>{\frac {w_{1}}{w}}\ln x_{1}+\cdots +{\frac {w_{n}}{w}}\ln x_{n}\\&=\ln {\sqrt[{w}]{x_{1}^{w_{1}}x_{2}^{w_{2}}\cdots x_{n}^{w_{n}}}}.\end{aligned}}}
Since the natural logarithm is strictly increasing,
${\displaystyle {\frac {w_{1}x_{1}+\cdots +w_{n}x_{n}}{w}}>{\sqrt[{w}]{x_{1}^{w_{1}}x_{2}^{w_{2}}\cdots x_{n}^{w_{n}}}}.}$
### Matrix arithmetic–geometric mean inequality
Most matrix generalizations of the arithmetic geometric mean inequality apply on the level of unitarily invariant norms, owing to the fact that even if the matrices ${\displaystyle A}$ and ${\displaystyle B}$ are positive semi-definite the matrix ${\displaystyle AB}$ may not be positive semi-definite and hence may not have a canonical square root. In [6] Bhatia and Kittaneh proved that for any unitarily invariant norm ${\displaystyle |||\cdot |||}$ and positive semi-definite matrices ${\displaystyle A}$ and ${\displaystyle B}$ it is the case that
${\displaystyle |||AB|||\leq {\frac {1}{2}}|||A^{2}+B^{2}|||}$
Later, in [7] the same authors proved the stronger inequality that
${\displaystyle |||AB|||\leq {\frac {1}{4}}|||(A+B)^{2}|||}$
Finally, it is known for dimension ${\displaystyle n=2}$ that the following strongest possible matrix generalization of the arithmetic-geometric mean inequality holds, and it is conjectured to hold for all ${\displaystyle n}$
${\displaystyle |||(AB)^{\frac {1}{2}}|||\leq {\frac {1}{2}}|||A+B|||}$
This conjectured inequality was shown by Stephen Drury in 2012. Indeed, he proved[8]
${\displaystyle {\sqrt {\sigma _{j}(AB)}}\leq {\frac {1}{2}}\lambda _{j}(A+B),\ j=1,\ldots ,n.}$
### Other generalizations
Other generalizations of the inequality of arithmetic and geometric means include:
## Notes
1. ^ If AC = a and BC = b. OC = AM of a and b, and radius r = QO = OG.
Using Pythagoras' theorem, QC² = QO² + OC² ∴ QC = √QO² + OC² = QM.
Using Pythagoras' theorem, OC² = OG² + GC² ∴ GC = √OC² − OG² = GM.
Using similar triangles, HC/GC = GC/OC ∴ HC = GC²/OC = HM.
## References
1. ^ Hoffman, D. G. (1981), "Packing problems and inequalities", in Klarner, David A. (ed.), The Mathematical Gardner, Springer, pp. 212–225, doi:10.1007/978-1-4684-6686-7_19
2. ^ Steele, J. Michael (2004). The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities. MAA Problem Books Series. Cambridge University Press. ISBN 978-0-521-54677-5. OCLC 54079548.
3. ^ Cauchy, Augustin-Louis (1821). Cours d'analyse de l'École Royale Polytechnique, première partie, Analyse algébrique, Paris. The proof of the inequality of arithmetic and geometric means can be found on pages 457ff.
4. ^ Arnold, Denise; Arnold, Graham (1993). Four unit mathematics. Hodder Arnold H&S. p. 242. ISBN 978-0-340-54335-1. OCLC 38328013.
5. ^ Aldaz, J.M. (2009). "Self-Improvement of the Inequality Between Arithmetic and Geometric Means". Journal of Mathematical Inequalities. 3 (2): 213-216. doi:10.7153/jmi-03-21. Retrieved 11 January 2023.
6. ^ Bhatia, Rajendra; Kittaneh, Fuad (1990). "On the singular values of a product of operators". SIAM Journal on Matrix Analysis and Applications. 11 (2): 272–277. doi:10.1137/0611018.
7. ^ Bhatia, Rajendra; Kittaneh, Fuad (2000). "Notes on matrix arithmetic-geometric mean inequalities". Linear Algebra and Its Applications. 308 (1–3): 203–211. doi:10.1016/S0024-3795(00)00048-3.
8. ^ S.W. Drury, On a question of Bhatia and Kittaneh, Linear Algebra Appl. 437 (2012) 1955–1960.
9. ^ cf. Iordanescu, R.; Nichita, F.F.; Pasarescu, O. Unification Theories: Means and Generalized Euler Formulas. Axioms 2020, 9, 144.<|endoftext|>
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Next: Upper-tailed t-test Up: Testing Hypotheses Previous: Testing Hypotheses
# Test of Significance involving the Sample Average
A bowler is bragging that his average is at least 180''. We observe him play three games, his scores are 125, 155, 140 ( , S=15). Should we accept or reject his claim? We should reject it. Why? Because a sample average as low as 140 is unlikely from a 180 bowler. How unlikely? A 180 bowler will bowl a 3-game average of 140 or lower only 2 percent of the time. Is 2 percent of the time unlikely? In statistics, yes. 5 percent or less is called statistically significant.
The decision making process above is called a test of significance . Here is the way a statistical report would formally present the test, in numbered stages.
1. Hypotheses: versus
2. Test Statistic:
3. P-value: Presuming H0 is true, the likelihood of chance variation yielding a t-statistic as low as -4.62 is .02. (Calculation details later.)
4. Conclusion: Since P-value , the observed sample value is declared significantly unlikely under . Hence, we reject H0 and conclude . The sample provides evidence to reject the bowler's claim.
Here is a more detailed description of each component of the test of significance above.
1.
The null and alternative hypotheses .
H0 and H1 are called the null hypothesis and alternative hypothesis , respectively. The two hypotheses describe the two possibilities: the claim is true ( ), or the claim is false (). Note that
(i) the two hypotheses are statements about the population (ii) the two hypotheses are complementary; if one occurs the other does not (iii) the hypothesis with the equal sign is the null hypothesis
A test of significance rejects (population statement) H0 and concludes H1 if the sample values are significantly far from H0 and inside H1''. Hence, we reject and conclude if is some significant distance below 180. How far below 180 is significant''? The test statistic helps us determine where to draw the line in the sand.
2.
The Test Statistic For tests of hypotheses on , the t-test statistic is a ratio of the form
For the null hypothesis , the t-test statistic is
H0 will be rejected if and only if will be some significant distance below 180, which happens if and only if t is some significant distance below 0. Based on the sample observed scores, the observed t value is
Is t=-4.62 significantly below 0?'' To answer this, we will need the help of the t-curve with n-1 degrees of freedom.
3.
The P-value
Using the t curve with n-1=2 degrees of freedom, the likelihood of chance variation resulting in a t-value as low as -4.62 is .02.
Since this likelihood is less than .05 (the standard for statistical significance), we declare that t=-4.62 is significantly below 0'', or that is significantly below 180'', and reject . In general, the P-value is the total area under the curve more extreme than t in support of H1. If t is deep in H1 territory'', then the P-value is small. If P-value .05, we reject H0 with statistical significance. If P-value .01, we reject H0 with high statistical significance. If P-value is larger than .05, we accept H0.
4.
Conclusion If H0 is rejected, the conclusion is usually stated as there is enough evidence to ...' or there are statistically significant differences...'. If H0 is accepted, the conclusion is usually stated as there is not enough evidence to ...', or there are no statistically significant differences...'. Since P-value=.02 in our example, we conclude that the sample provides enough evidence to reject the bowler's claim of a 180 average'. Or his performance ( ) was much lower than his claimed average (), and the difference is statistically significant.'
Summary of the lower-tailed t-test for :
versus
Test statistic:
P-value: Total area less than t (the direction of H1) under t-curve with n-1 degrees of freedom If t is significantly below 0 (the direction of H1), the P-value will be small.
Conclusion: If P-value .05, we reject H0 with statistical significance. If P-value .01, we reject H0 with high statistical significance. If P-value >.05, we do not reject H0.
Next: Upper-tailed t-test Up: Testing Hypotheses Previous: Testing Hypotheses
2003-09-08<|endoftext|>
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Battle of Miami
Battle of Miami
As settlers pushed further and further into the Pacific Northwest territories, clashes with American Indians increased. On January 2, 1791, the most exposed settlement in Ohio, Big Bottom, was attacked; and all of the settlers, including women and children, were killed. In response, General Washington appointed St. Clair to lead forces to pacify the Indians. The Miami Indians attacked St. Clair and his militia on the banks of the Wabash River. The Indians routed the government forces, who were forced to flee..
The Indians of the Northwest Territory were emboldened at the end of the War Independence by the retention of by the British of their forts in Northern Ohio. The Americans felt that the Indians had lost all their claims to land thanks to their support for the British. The Americans however, attempted to reach agreements with the tribes but not all of them were willing to come to an agreement. Among those who opposed any agreement were the Miami Indians who were determined to drive the White settlers from what they considered their lands. Throughout the spring and summer of 1790 the Miami Indians partook in scattered attacked throughout Northern Ohio. The governor of Ohio sent Major John Hamtramck to investigate and gauge the plans of the Miami and the other tribes. He returned and warned that the tribes were planning to go to war against he settlers.
The governor of the Ohio territory sent out expeditions under the command of Colonel James Hardin and James Trotter. The expeditions achieved little. The Americans showed themselves to be incompetent and the only thing they accomplished was to burn down some of the tribe’s villages further inflaming the tribes.
The one result of the military action was providing the settlers with a false sense of safety. As a result they continued to settle. The most exposed settlement was the settlement of Big Bottom that was 40 miles further upstream then any other settlement. Eleven families had begun the process of settling the area. On the night of January 2, 1791 the Miami attacked the settlers, destroyed the settlement and killed all the settlers; men, woman and children. The massacre became known as the Big Bottom Massacre.
The massacre forced the US government to respond. President Washington appointed St Clair the governor of the territory to take a force and defeat the Native Americans. In late September St Claire set off from Cincinnati. St Claire had twenty three hundred regulars and militia with him. Along the way St Claire first built Fort Hamilton. Then forty miles down the Ohio River he built Fort Jefferson. Finally, after a further ten-day march in which St Claire lost almost half his men through attrition he reached what he thought was the St Mary River. It was in fact just a branch of the Wabash. There he spread out his men in search of the Native Americans. His men reported that the woods were full of warriors. On the morning of November 4, 1781 the warriors attacked. The American regulars who outnumbered the Native Americans when the battle began attempted to use conventional tactics, which were not very successful in a battle with Native Americans in the woods. The Indians were slowly overcoming the Americans and every time they would capture an American they would scalp him in view of those fighting, thus spreading fear among the Americans. Finally St Claire decided the only American option was to fight there way out of the battle. St Claire’s forced managed to fight clear of the Indian warriors leaving behind their equipment and their wounded and began a headlong retreat to Fort Jefferson. They covered in ten hours what had taken them ten days cover the week before. It was an inglorious defeat for the American Army.<|endoftext|>
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Health care, education, appearance, and nutrition all began to get much more attention than previously. Increased production per person eventually led to the drop in prices paid for the thread. As a result, the Spinning Jenny was one of the most important inventions for the advancement of the production of cotton textiles during the early days of the Industrial Revolution. Spinners were unable to keep up with the increased demand for threads. They set up a small mill to supply hosiery makers with suitable yarn. Improved Quality of Life All of the modern inventions and the amount of wealth that was brought into the cities greatly improved the way that people lived.
The Industrial Revolution did not reach America until the 1820s and began with the textile industries in the northeast. In fact, Hargreaves had several daughters, but none named Jenny and neither was his wife. Following methods are used by machine manufacturers to condense the fiber strand. Later versions of the spinning jenny added even more lines which made the machine too large for home use. The technology was already out there and being used in many machines.
This resulted in stress related illnesses, the spread of disease, and death through out the factories. Although this area was known as a major textile center, prior to the Industrial Revolution the production of cloth from raw goods took place within cottage industries. British law of the time was well behind the state of industrial progress. Regardless of how the invention came to be named, it changed forever the way textile manufacturing was accomplished and led the way to the Industrial Revolution. This led the way to factories where these larger machines could be run by fewer workers. With machines and workers concentrated in one place, the transportation costs of raw materials and finished goods were greatly reduced.
Large industrial buildings usually employed one central source of power to drive a whole network of machines. The spinning jenny used eight different spindles that were powered by a single wheel. Increasing production of yarn The spinning Jenny was able to produce up to 8 threads at once, this was later improved to 120 threads The invention used a metal frame with 8 wooden spindles at one end. All the work was done by entire families and men were usually the weavers, women spun the materials into threads and children helped clean the raw materials. Individual thread streams were separated by a vertical wire system referred to as a faller, which dropped down between the lines and ensured they remained unconnected.
The process was time consuming and merchants wanting to meet the demand for textile goods were often frustrated by the huge gap between supply and demand. The metalworks industry producing parts for were also moving to factories at this time. Some sources claim that Hargreaves named the machine after his daughter. This was one of the first labor disputes of the Industrial Revolution. Here some of the advantages of compact spinning are discussed. Moreover, sources claim that whereas in the 1750s Britain imported only 3 million pounds of raw cotton, in 1786, more than 18 million pounds of this material were introduced in the country Allen 2007; citing Crouzet 1985. According to Stahlecker , RoCoS 1 is suitable for 100 % cotton, cotton blends and 100% synthetic fibers with maximum staple length of 60 mm.
A key development in the industrial revolution of England, the spinning jenny allowed yarn to be manufactured in large quantities. Just like a coin, there are two sides to everything. However, it still lowered the price that fabric could be made, making textiles more available to more people. Many people were trying at the time to invent a device to make textile manufacture easier. As the design evolved, the spools increased massively, often exceeding 100 to 120. This meant that the operator or weaver could spin eight threads at once by turning a single wheel.
The Industrial Revolution 1760-1850 began in England and spread throughout Europe and the Americas over the course of the next several decades. The mule was an important development because it could spin thread better than by hand, which led to every finer threads that commanded a better price in the marketplace. But it was the invention of the Spinning Jenny by James Hargreaves that is credited with moving the textile industry from homes to factories. The most important requirement for perfect compact yarn is complete parallel arrangement of fibers and close position before twist is imparted. The supra magnets are equipped with ceramic compactor, which is pressed against front bottom drafting roller by supra magnet Without clearance. The Spinning Jenny: A woolen Revolution. It is said that some of the makers of the spinning wheel sabotaged Hargreaves machines, and Hargreaves fled.
Result of Improvements After huge improvements on the Spinning Jenny; adding more lines, the machines became too large for houses. Model of the spinning jenny in a museum in Wuppertal, Germany The spinning jenny is a multi-spool spinning wheel. The owners of the factories had great power over the laborers and started to divide the workers to be responsible for different sections of the manufacturing process. Some of this include the light bulb, the telephone, and X-rays. The special vacuum element generates the air current underneath the apron.
Hargreaves found a way to ramp up the supply of thread. Moreover, their living conditions were not any better. The Pros of The Industrial Revolution 1. The machine was kept in the cotton industry until 1810 when it was replaced with a more advanced machine named spinning mule, invented by Samuel Crompton. Once perfected, the spinning mule gave the spinner great control over the weaving process, and many different types of yarn could be produced. The spindle continued to revolve, giving Hargreaves the idea that a whole line of spindles could be worked off jus … t one wheel. With the use of water to power later versions of spinning and weaving machinery, quality and strength of the cloth produced was greatly improved.<|endoftext|>
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# Find the value of $\sin(\frac{1}{4}\arcsin\frac{\sqrt{63}}{8})$
Find the value of $\sin(\frac{1}{4}\arcsin\frac{\sqrt{63}}{8})$
Let $\sin(\frac{1}{4}\arcsin\frac{\sqrt{63}}{8})=x$
$\arcsin\frac{\sqrt{63}}{8}=4\arcsin x$
$\arcsin\frac{\sqrt{63}}{8}=\arcsin(4x\sqrt{1-x^2})(2x^2-1)$
$\frac{\sqrt{63}}{8}=(4x\sqrt{1-x^2})(2x^2-1)$
$\frac{63}{64}=16x^2(1-x^2)(2x^2-1)^2$
Let $x^2=t$
$\frac{63}{64}=16t(1-t)(2t-1)^2$ $64t^4-128t^3+80t^2-16t+\frac{63}{64}=0$
Now solving this fourth degree equation is getting really difficult,even after using rational roots theorem.
Is there a better method possible?The answer given is $x=\frac{\sqrt2}{4}$.Please help me.
$$x = \arcsin{\frac{\sqrt{63}}{8}} = \arccos{\frac18}$$
$$\sin{\frac{x}{4}} = \sqrt{\frac{1-\cos{(x/2)}}{2}} = \sqrt{\frac{1-\sqrt{\frac{1+\cos{(x)}}{2}} }{2}}$$
In this case, $\cos{x} = 1/8$ so that
$$\sin{\left (\frac14 \arcsin{\frac{\sqrt{63}}{8}} \right )} = \sqrt{\frac{1-\sqrt{\frac{9/8}{2}} }{2}} = \frac1{2 \sqrt{2}}$$<|endoftext|>
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I’ve been exploring making thinking visible in all curriculum areas. Thinking tools and questioning are being taught and added to a collection of tools the children can choose from to develop perspective, vocabulary and an understanding of their world.
One of the tools I’ve been exploring with my students is Point of View. This tool can be used for all curriculum areas, whether exploring a character from a book, developing a character for a narrative, or to solve and explore a general problem. When children put themselves in the position of others, their empathy and understanding of a problem, situation or character deepens. Our inquiry for the first 8 weeks of the year has been about relationships.
Central Idea: The relationships we have with each other affect how we feel and behave.
What lines of inquiry will define the scope of the inquiry into the central idea?
- Self Awareness (LP Attributes, Attitudes, Skills, Mission Statement, Essential Agreement, School Pledge)
- How we develop relationships (What is relationship? What relationships do you have in your life? What makes it a relationship?
- Roles and behaviours within relationships (Scenarios, Role playing, Photos of LP Attributes, Essential Agreement)
- How relationships affect us (Good, Bod, Reflections task board/Think board- develop their own)
What teacher questions/provocations will drive these inquiries?
How do we develop and maintain healthy relationships?
What makes a supportive relationship?
What/why do actions help to build healthy relationships?
My students have been exploring the relationships they have in their world.
Example: Point of View: Friendship and why we need to have more than one best friend. My role is to record my student’s ideas and not mine, but I think they covered all bases! 🙂
Brainstorm: A good friend…
The Point of View has been a fantastic tool to help sort friendship and playground issues which is part of building healthy relationships. Young children can be egocentric and developing their understanding that there can be other views has led to a very inclusive group of young learners.
The children have also used Point of View to understand the behaviours of a book character and to develop their own characters and plots when writing.<|endoftext|>
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- Outline the distinctive features of the Time of Troubles and how they eventually ended
- The Time of Troubles started with the death of the childless Tsar Feodor Ivanovich, which spurred an ongoing dynastic dispute.
- Famine between 1601 and 1603 caused massive starvation and further strained Russia.
- Two false heirs to the throne, known as False Dmitris, were backed by the Polish-Lithuanian Commonwealth that wanted to grab power in Moscow.
- Rurikid Prince Dmitry Pozharsky and Novgorod merchant Kuzma Minin led the final resistance against Polish invasion that ended the dynastic dispute and reclaimed Moscow in 1613.
The last tsar of the Rurik Dynasty, whose death spurred on a major dynastic dispute.
The Rurikid prince that successfully ousted Polish forces from Moscow.
A form of Russian parliament that met to vote on major state decisions, and was comprised of nobility, Orthodox clergy, and merchant representatives.
The Time of Troubles was an era of Russian history dominated by a dynastic crisis and exacerbated by ongoing wars with Poland and Sweden, as well as a devastating famine. It began with the death of the childless last Russian Tsar of the Rurik Dynasty, Feodor Ivanovich, in 1598 and continued until the establishment of the Romanov Dynasty in 1613. It took another six years to end two of the wars that had started during the Time of Troubles, including the Dymitriads against the Polish-Lithuanian Commonwealth.
Famine and Unrest
At the death of Feodor Ivanovich, the last Rurikid Tsar, in 1598, his brother-in-law and trusted advisor, Boris Godunov, was elected his successor by the Zemsky Sobor (Great National Assembly). Godunov was a leading boyar and had accomplished a great deal under the reign of the mentally-challenged and childless Feodor. However, his position as a boyar caused unrest among the Romanov clan who saw it as an affront to follow a lowly boyar. Due to the political unrest, strained resources, and factions against his rule, he was not able to accomplish much during his short reign, which only lasted until 1605.
While Godunov was attempting to keep the country stitched together, a devastating famine swept across Russian from 1601 to 1603. Most likely caused by a volcanic eruption in Peru in 1600, the temperatures stayed well below normal during the summer months and often went below freezing at night. Crops failed and about two million Russians, a third of the population, perished during this famine. This famine also caused people to flock to Moscow for food supplies, straining the capital both socially and financially.
Dynastic Uncertainty and False Dmitris
The troubles did not cease after the famine subsided. In fact, 1603 brought about new political and dynastic struggles. Feodor Ivanovich’s younger brother was reportedly stabbed to death before the Tsar’s death, but some people still believed he had fled and was alive. The first of the nicknamed False Dmitris appeared in the Polish-Lithuanian Commonwealth in 1603 claiming he was the lost young brother of Ivan the Terrible. Polish forces saw this pretender’s appearance as an opportunity to regain land and influence in Russia and the some 4,000 troops comprised of Russian exiles, Lithuanians, and Cossacks crossed the border and began what is known as the Dymitriad wars.
False Dmitri was supported by enough Polish and Russian rebels hoping for a rich reward that he was married to Marina Mniszech and ascended to the throne in Moscow at Boris Godunov’s death in 1605. Within a year Vasily Shuisky (a Rurikid prince) staged an uprising against False Dmitri, murdered him, and seized control of power in Moscow for himself. He ruled between 1606 and 1610 and was known as Vasili IV. However, the boyars and mercenaries were still displeased with this new ruler. At the same time as Shuisky’s ascent, a new False Dmitri appeared on the scene with the backing of the Polish-Lithuanian magnates.
An Empty Throne and Wars
Shuisky retained power long enough to make a treaty with Sweden, which spurred a worried Poland into officially beginning the Polish-Muscovite War that lasted from 1605 to 1618. The struggle over who would gain control of Moscow became entangled and complex once Poland became an acting participant. Shuisky was still on the throne, both the second False Dmitri and the son of the Polish king, Władysław, were attempting to take control.
None of the three pretenders succeeded, however, when the Polish king himself, Sigismund III, decided he would take the seat in Moscow.
Russia was stretched to its limit by 1611. Within the five years after Boris Godunov’s death powers had shifted considerably:
- The boyars quarreled amongst themselves over who should rule Moscow while the throne remained empty.
- Russian Orthodoxy was imperiled and many Orthodox religious leaders were imprisoned.
- Catholic Polish forces occupied the Kremlin in Moscow and Smolensk.
- Swedish forces had taken over Novgorod in retaliation to Polish forces attempting to ally with Russia.
- Tatar raids continued in the south leaving many people dead and stretched for resources.
The End of Troubles
Two strong leaders arose out of the chaos of the first decade of the 17th century to combat the Polish invasion and settle the dynastic dispute. The powerful Novgordian merchant Kuzma Minin along with the Rurikid Prince Dmitry Pozharsky rallied enough forces to push back the Polish forces in Russia. The new Russian rebellion first pushed Polish forces back to the Kremlin, and between November 3rd and 6th (New Style) Prince Pozharsky had forced the garrison to surrender in Moscow. November 4 is known as National Unity Day, however it fell out of favor during Communism, only to be reinstated in 2005.
The dynastic wars finally came to an end when the Grand National Assembly elected Michael Romanov, the son of the metropolitan Philaret, to the throne in 1613. The new Romanov Tsar, Michael I, quickly had the second False Dmitri’s son and wife killed, to stifle further uprisings.
Despite the end to internal unrest, the wars with Sweden and Poland would last until 1618 and 1619 respectively, when peace treaties were finally enacted. These treaties forced Russia to cede some lands, but the dynastic resolution and the ousting of foreign powers unified most people in Russia behind the new Romanov Tsar and started a new era.<|endoftext|>
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Ground Spiders are a vast and diverse group of spiders found all over Australia from rainforests to deserts.
Ground spiders are rather drably coloured, ranging from charcoal grey to pinkish brown. Some, like Anzacia, have shiny, iridescent hairs. Typically they have large, cylindrical front spinnerets that are well separated from each other. The middle eyes in the rear row are usually angular rather than rounded in shape.
Most Ground Spiders are vagrant hunters that are especially common in forest and woodland habitats, living in leaf litter and under bark. Some are highly specialised.
Ground Spiders are found Australia-wide.
Other behaviours and adaptations
The Slit Spider of inland red sand-dune habitats makes a burrow with a wide, horizontal slit opening on the down-slope of a dune. The lower edge of the slit projects out beyond the upper edge, so that insects running down slope across the slit are guided into the cavity where they are grabbed by the waiting spider. Other Ground Spiders such as Hemicloea, are remarkably flattened for life in narrow spaces under rocks and bark.<|endoftext|>
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The Math Learning Center is committed to offering free tools, materials, and other programs in support of our mission to inspire and enable individuals to discover and develop their mathematical confidence and ability.
The Common Core State Standards for Mathematics prompt teachers to significantly change both what they teach and how they teach it. In a series of posts, we’ll address the three characteristics of the new content standards likely to have the biggest impact on the way you teach math—tighter focus, greater coherence, and increased rigor.
"We must convey, from the very first day, the important message that we will tackle challenging material and do high-quality work in our classroom. But we must also convey that we will tackle this material and do this high-quality work in an atmosphere of support and collaboration." – Denton & Kriete, 2000
"Assessment should be more than merely a test at the end of instruction to see how students perform under special conditions; rather it should be an integral part of instruction that informs and guides teachers as they make instructional decisions." – NCTM<|endoftext|>
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The Savvy Student's Guide to Study Skills—Chapter Seven
- What Should We Memorize?
- Bloom's Taxonomy Revisited
- Kinds of Memory
- The Benefits of Long Term Memory
- How to Memorize
- Approaches to Learning
- Critical Thinking
- Memory and Social Situations
What Should We Memorize?
One of the best parts of the Harry Potter stories was the pensieve, the dish that looks like a large, flat birdbath where the user could store and later view memories. The pensieve plays a pivotal role in tying up many of the subplots in the seven volume series, and more than one Harry Potter fan has talked about how cool it would be to own one, if only they existed.
It's easy to understand how people feel that way, but it's also important to know that I was able to retrieve the video clip of the pensieve by using a device that's very much like a pensieve—a computer. Able to make astronomical calculations (literally) and to show us live events from throughout the globe, as well as memories from days gone by, computers can be far superior to the pensieve. Sure, we have to type or download our memories into the computer, rather than extract them directly from our brains or our tears, but the speed, content, and accessibility of computers has so changed our society, it's fair to ask the question, just what do we need to commit to memory, now that computers can remember everything for us, and bring it up on the screen in a matter of seconds?
This is a question that had predated computers. The great automotive pioneer Henry Ford is known to have said that if something is in a book, there is no need for it to be in his head. This same outlook has been leading to changes in the US educational system, including modifications of Advance Placement exams. The updated AP tests are placing a greater emphasis on the application of knowledge, and not the memorization of facts. It is one thing to know the process of mitosis, but knowing its limits and uses is something very different. That requires more than memory; that requires understanding.
Bloom's Taxonomy Revisited
A quick example of the difference between the two might help here. Once you're ready, take a look at this set of numbers. Study them closely, and when you're ready to take on the question below, go ahead.
Name any four numbers you saw.
There's an excellent chance you were able to answer this question with no problem (and yes, you can now go back to the picture and check your answer if you'd like).
But now, let's ask another question.
What do you think those numbers were used for?
At this very moment, there's an excellent chance you did one of two things. Either you tried to remember what the numbers looked like from memory, or you went back to the picture another time to take a closer look. Either way, what also occurred to you is that you don't need to memorize the numbers to be able to answer this question. Sure, the answer might have something to do with the value of the numbers, but you're probably focusing more on the design of the wood blocks, and thinking about places you've been or games that you've played where you've seen numbers like this before. (And if you have used these before, your parents must be huge fans of the game Bingo, since these are antique Bingo markers, which have were replaced a long time ago by large marking pens.)
The first question asked you to answer a question from memory, while the second question asked you to analyze the markers and create an answer based on your life experiences. These three skills—memorize, analyze, and create—are part of Bloom's Taxonomy, a topic we've previously discussed. Bloom's is designed to label and explain different levels of thinking, with the high end of the pyramid representing more complicated thoughts that require combining many of the lower levels of thinking to develop something new.
When it comes to memory, then, the message from Bloom's is clear: you can't really create something new until you know a great deal about how something already works—and that requires some level of memorization. As someone once said about Henry Ford's quote, you might not have to remember something that's in a book, but you do have to remember what book it's in. So let's talk about memorization.
Kinds of Memory
Neuroscience is the study of how the brain works, and it gives us some clues on how to make the most of memory. Like all sciences, neuroscience is always growing, but right now, it offers these insights in to how memory works.
An idea starts in short-term memory, a place in our brain where we can keep a few ideas for a little while. The real goal of short-term memory is to keep a piece of information long enough for us to use it, and then get rid of it, or store it, so we can hold on to it for a longer period of time and do more things with it. As an example, if you want to call to place a pizza order, you look up the phone number, remember it long enough to make the call, and then forget it. Twenty minutes later, you're savoring the glories of pepperoni and cheese, all thanks to short-term memory. (Want to take your short-term memory for a spin? Take this simple quiz for fun.)
Not surprisingly, long-term memory is the place memories go when we're able to hold on to them for a while. This doesn't mean they all stay there forever; some long-term memories will hang around for days or months, while others will be there for most of our lives. As the link points out, the duration and clarity of a long-term memory depends in part on how strong the initial impression was when it became a memory. Twenty years from now, you probably can't remember much about your tenth grade English teacher, but you'll never forget the face of the student in tenth grade English who asked you to homecoming. You first saw both faces at around the same time, but one meant more to you, so you remember it more clearly.
A relatively new idea in neuroscience is working memory, or the place where you take an idea, either from short- or long-term memory, and do something with it. A great example can be found in this article, where two bloggers talk about walking into a building for the first time. Finding the office you're looking for requires short-term memory to learn where it is; getting back out of the building requires working memory to reverse those directions.
The Benefits of Long-Term Memory
Long-term makes life a lot easier, since it keeps ideas on-hand for us to use in working memory. Remember your first day of school in a new building? It took you forever to find your classes—and it would take that long every day, if it weren't for long-term memory.
And that is exactly why people study—so they can bring up ideas from long-term memory to working memory with little or no problem, and do something with those ideas. Sometimes you just write them down (“What's 5 times 8?”, a question from the Remember section of Bloom's), sometimes you compare more than one idea (“Who used symbolism better, Hemingway or Fitzgerald?”, a question from the Analyze level), and sometimes you put them together in new ways (“Use the qualities from any three US presidents to make the perfect leader of a country”, which comes from the Create level.)
If you're thinking, “But if you can put ideas into working memory from short-term memory, why do you have to study so hard to put things in long-term memory—especially if you aren't going to need those ideas ever again?” That's a fair question— why study? Well:
- If you took the short-term memory quiz, you've likely discovered the space in your short-term memory is mighty small. If you try to put too many ideas in short-term memory the night before the exam, you might not remember all of them, or they may come out a little mixed up. If you're only being tested on four or five basic ideas, this study approach might just work. But if you're being asked for a lot of information, you're likely to find there's too much stuff to stuff in short-term memory (hey, there's a reason they call it cramming.)
- Learning new things now makes it easier to learn new things later. Many of the learning techniques we'll talk about are built on the time-tested idea that once you know a little about something, it's easier to learn more about it later on. To go back to our pizza example, suppose you used your short-term memory to call a pizza place that's new to you. The pizza arrives, you take a bite, and you immediately begin to compare it to the pizzas you've had before (“The crust is a little softer than at Bill's Pizza, and the pepperoni is a little spicier than Uncle Tim's.”)
- You've actually taken pizza memories out of long-term memory, thrown them in working memory, added to them, and refiled them—and that's easier to do than to start a completely new pizza memory. This is one of the reasons students study subjects in high school; it may not seem like you remember much from high school biology, but some of it comes roaring back to mind in life science in college, and makes deeper exploration that much easier (and easier to apply biological concepts to the science of pizza—no, really.)
- It makes the higher end of Bloom's Taxonomy easier to explore. You may have four new ideas in short-term memory about this week's reading of an Emily Dickinson poem, but if the quiz asks you to compare those ideas to the poems of W.E.B. DuBois you read three months ago, short-term memory just isn't going to help you there.
Studying makes your life easier, allows you to do more in less time, and helps you become more creative. Remember the first time you heard that amazing new song, and wanted to learn everything about it? Now you know the words by heart, you can probably even see part of the video in your head, and at least one of your friends thinks it's pretty cool you know all that-- all thanks to long-term memory, and studying.
How to Memorize
First and foremost, this is going to require some time. You don't think twice about how to get to your classes now, because you had to think about it a lot for a long time, and practice it for more than a couple of days.
This means you'll have to build time in to your schedule to study. For high school students, that will likely be 2.5 hours per day; for college students, the rule has long been 2-3 hours of studying for every hour spent in class, but that may be a little too much. Keep in mind that this time includes time for homework assignments, too. Homework requires you to use the ideas you're studying, but it isn't quite the same as studying, so make sure you give yourself enough time for both.
Next, you want to figure out the best way you memorize for each subject. This video gives you a quick look at the difference between verbal memory and visual memory. Many people are surprised to see that visual memory generally works better for the task of memorization than verbal memory, but that goes back to the idea of relating an idea to something you already know—so if you make up a story about a new idea, you're blending both old and new ideas.
Visual memory might help you with the task of memorizing lists, but you may need a different approach if you have to memorize something more complicated. Memorizing state capitols is one thing; explaining where they're located within each state, or how they became the state capitol, is something else. When we talked about writing, we said it was important to keep in mind what the teacher was looking for in the assignment. The same is true for memorizing. If the assignment asks for a recall of facts, verbal or visual memory might do the trick—but what if you need to understand the ideas, and not simply know them?
Approaches to Learning
To figure out how to make the most of our study time, we have to consider the best way to make sure the ideas we are working with have meaning. This is pretty important, and it explains the difference between students who know the material, and students who understand it—or, students who can answer the question “What year did the War of 1812 end?” and the student who can answer the question “What effect does the War of 1812 have on US foreign policy today?” One question asks you to understand a year; the other asks you to understand a number of reasons. There's a big difference, but most students don't understand that.
Since there's usually more involved to understanding the “Why” questions (or the higher end of Bloom's Taxonomy) than the “What” questions (on the bottom of the triangle), it's important to consider how you're going to keep the ideas fresh and interesting as you study. For most students, that means they have to mix up their study approaches, using a variety of activities. This quick summary on memorizing offers some great tips on how to keep study time new and interesting, with approaches based on:
Most classes are still based on the idea of talking about ideas, whether that's through lecture or discussion. It's also the basis for most of your class notes. Listening to a lecture a second time can put you right back in the middle of the classroom when the ideas were brand new, and that can help students learn—so ask you're teacher if recording the lecture or discussion portion of the class is OK.
You can also engage your ears by recording your notes and listening to them while you're working out, doing chores, or eating lunch. Other students will use study time to read their notes out loud while walking around their study area. That works, too—especially if you ask yourself a question out loud that you think might be on the test, then answer that question out loud. (Yes, this means it's really OK to talk to yourself. Young kids do it naturally; we only stop doing it because we think we look uncool.)
A good part of many classes also includes reading books, handouts, or websites, so it's important to keep reading as a vital study skill. We've already talked about the power of rewriting your notes; this allows you to fill in the blanks of the phrases you wrote down in class, which means your second set of notes can actually convey a story that connects the ideas. This is a huge step towards getting meaning out of something.
A second set of notes also allows you to highlight your notes, using different colors to combine different sets of facts. This makes it easy to study when you have a five minute break at work, or at lunch, or if you're waiting for your parents to get ready to go out. This is even easier if you put the important ideas together on note cards you can stick in your backpack or back pocket. A travelling set of note means you'll think about the connections of the big ideas more times during the day, and that can be a huge help. (And remember—you can also use note cards to study for “Why” questions—the answer will just take up way more space.)
A less popular method to study is to “read” a video that has the key facts and ideas you're working with. Videos related to your subject can easily be found online, but you could also record your class lectures (make sure you move the camera around, since 40 minutes of the front of the room can get a little boring), or record yourself summarizing the main ideas, and play it back to yourself. Better yet, think about posting that recording to social media—nothing like a little peer pressure to make sure you know what you're talking about.
Touching and Moving
One way to change things up in your study routine is to use the power of touch to think about the ideas in a different way. Having trouble memorizing a set of words? Write each one on a note card, then cut the word out of the note card, using the shape of the word as your guide. If you run your finger around the edge of the word as you say it, the shape of the word can leave an impression—really.
Other ways to use touch include using note cards for one set of facts (like memorizing basic words) while using sticky notes for a different set of facts (notes you'll use to answer the “Why” questions). And if you're trying to memorize countries, print out some pictures of the countries and cut them out—and leave plenty of time for cutting out Russia.
A closely related strategy is to act out the ideas you're discussing. By pretending you're someone else, you have to think about the ideas in a different way, because you have to see the ideas from their point of view. What Robert E. Lee knew about the Civil War was very different from what Abraham Lincoln knew, so pull out your best southern drawl and give it a shot.
Your teacher has to spend a long time understanding a subject before they can teach it to you, so it makes sense that one way to really know an idea is to teach it to someone else. This is one reason why study groups can—that's can—be a real help as part of your study pattern, if you use them the right way. We'll talk more about this in a chapter on group work.
Over time, you'll develop some “go to” strategies for different kinds of assignments, different kinds of subjects, and different kinds of teachers. Note cards may be your sweet spot when it comes to memorizing terms, but talking out loud may be the best way to shape your analysis answers. Give yourself plenty of time to try a wide variety of approaches, stay with it…
And be sure to remember that each teacher has their own way of teaching and grading. This is especially important to remember, if you're taking a subject where you see yourself as “good at” it. Mrs. Jones may be looking for a completely different level of knowledge in AP Biology than what Mr. Smith wanted in Bio I, so pay close attention in the first few weeks of class, and ask for advice from students who took her class last year who did well in it.
You also want to be careful to avoid the trap of seeing yourself as “only a visual learner”, or “only an auditory learner.” There's ample evidence to suggest that everyone's brain learns in all kinds of ways, and the learning “style” you think best works for you is more about what you're used to doing, what the subject is, or what the teacher expects. As the video points out, the best way to understand the shape and location of a country is to look at a map, not to read about it—but you might not think to try that if you see yourself as not a visual learner. The goal is to do what needs to be done to make sure the ideas have meaning, and that includes a mix of study approaches, and applying them every time you can in the real world.
Another skill that's a must in terms of memory is the ability to think critically. This classic cartoon expresses the concern shared by many teachers and leaders: schools are putting so much emphasis on grades, they aren't showing students how to think, judge, or evaluate—and if it isn't on the test, they just aren't going to do it.
I've seen this in my work as a math teacher, when students are engaged in the story problems they love to hate. “Suzie is seventeen years older than her brother William. Combined, their ages total 13. How old are Suzie and William?”
They set up a nice equation, solve it beautifully, and energetically shout out the answer: “Suzie is 15 years old, and her brother William is negative 2 years old!”
And the thing is, most students don't see a single thing wrong with than answer—which is why I included at least one of them when I taught math.
It's certainly true there are some things you don't have to think about, and probably don't get an opinion on (the atomic mass of Chlorine is a little under 35.5, no matter how you might feel about it), and there are some classes where breaking the rules of what's known is pretty important (particle physics comes to mind). But if you're writing up a lab report involving Sodium Chloride and it acts more like Calcium Chloride, that might be a problem—or it might not. Since we're talking chemicals here, it's best not to guess, and that requires you to use critical thinking.
Applying key critical thinking skills as you study is important at every step:
- In reviewing your notes, you may run across a part that just doesn't make sense. That's critical thinking! You can correct this by reviewing the notes around the one you don't understand (this is called a contextual clue), reviewing the recording or video you may have made of the lecture, asking a fellow student—or even asking the teacher, which gives you the bonus of letting them know you really care about what's going on in class.
- In reading a textbook, you may come across a statement that sounds more like the author's opinion than an objective presentation of fact. This often happens in textbooks, and it happens all the time in literary criticism. This is where critical thinking really comes in handy, since it's likely the reason the teacher picked this particular textbook or reading assignment is because they agree with the author's opinion. If you don't, and you want to present your differing viewpoint, you've got to proceed with objective caution.
- It's vital to apply the same kind of critical thinking to your own finished assignments. This is why rough drafts of papers are so important, especially in high school. If you can see a problem in the facts that you used to create your conclusion, it's pretty likely your teacher will, too—and that just isn't good. Review your papers before you turn them in. Better yet, do that with the teacher a week before they're due.
Key Elements of Critical Thinking
There are many ways to apply critical thinking to school work and real life situations, and many of them are summarized here. They include:
- Using the 5 Ws + How (Who, What, When, Where, Why, and How) to summarize what an author is saying. This is a great way to review something you've just read; answering “Why did the author say this?” gives you a chance to think about why they included this idea in their writing, and if they used good evidence to support the idea or opinion.
- Treat the reading by using the scientific method. This is a great way to analyze for opinion, especially if you love lab sections of science. This method allows students to be objective in their analysis of what's being said, and it helps make sure your response is based more in objective fact that emotional opinion.
- Use the same approach to evaluating all opinions and arguments. This is where The Five Paragraph Essay can come in handy, since many people use some form of it in their writing. By reading for the main idea, supporting points, and conclusion, you can get to the clear parts of their argument, and see if it makes sense. This is especially helpful when you're reading something you agree with, since it requires you to keep your opinion, or bias, out of the way.
- Assume the author has reached the wrong conclusion, and try to present an objective argument that offers a different, and objectively correct, opinion. This not only requires you to know the facts well enough to make your own opinion; it means you have to know the facts well enough to know why the author is wrong.
Memory and Social Situations
Students who are making the most of learning know how to apply their skills both in the classroom and out. When it comes to memory, that skill is best applied in a social situation where you're meeting someone for the first time, where you have to handle the basics—like remembering their name, what they do, and anything they tell you that they think is special. (You don't have to remember what you think is special about them—somehow, that just kind of magically shows up.)
Remembering someone's name boils down to a few easy steps, all applications of what we've talked about earlier:
Focus on that person, and that person only.
Not easy to do at a party or other social situation, but treat it like a study situation, and you'll be amazed what you can block out.
Say their name twice.
This is especially important if the name is one you haven't heard before, so you can show interest in knowing how to say it properly. “Hi Kevin, it's nice to meet you” is a good first use. About three or four minutes later, use their name again when asking a question. “Kevin, I'm getting a soda. Is there something I can get you?”
Link their name to a positive idea you already know.
If there's a Kevin in your life you already like, find a way to link the two Kevins together. If the new Kevin likes Beyonce, picture the two of them standing together. If there's a Kevin in your life you'd rather forget, connect the new Kevin to a different idea.
Use their name when saying goodbye.
This shows them (and you) you've remembered so far, and that's a good thing.
Like all skills, this one takes some practice, and you might think Kevin won't be all that impressed if he can see you're struggling to remember his name. Actually, if Kevin is really worth getting to know, he'll be very impressed, and honored—so if they give you any grief, it's probably pretty OK to forget their name, and them.<|endoftext|>
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## Thursday, 14 December 2017
### On The 'Mystery Calculator' Trick
'The Mystery Calculator' trick is a perennial, favourite 'surprise gift' found in Christmas crackers[1] — along with the (hard) plastic moustache/slug, the fortune-telling fish, and perhaps (if you're lucky) a die.
For a maths teacher, 'The Mystery Calculator' is a potent conceit that piques students' interest and that — when used carefully[2] of course — levers and encourages strong, mathematically mature thinking. After playing the trick out on/with students, it is demystified carefully, and collectively, before students use their newfound knowledge and understanding creatively in devising a different trick, based on the same principles. (A ppt version of the cards, one card per slide, can be downloaded here.)
The Trick:
• Ask someone to choose a number from any of the six cards.
• Show them each card in turn and ask them if their number appears on it.
• Add the numbers in the top left-hand corner of each card that contains their number.
• The total is their number.
The How / Why:
The number 1 in binary is (1 x 2⁰) → (1)10 = (1)2 → Hence the number 1 appears on the 1st card only.
The number 2 in binary is 10... (1 x 2¹) + (0 x 2⁰) → (2)10 = (10)2 → Hence the number 2 appears on the 2nd card, but does not appear on the 1st or any other.
The number 3 in binary is 11... (1 x 2¹) + (1 x 2⁰) → (3)10 = (11)2 → Hence the number 3 appears on the first 2 cards, but does not appear on others.
The number 4 in binary is 100... (1 x 2²) + (0 x 2¹) + (0 x 2⁰) → (4)10 = (100)2 → Hence the number 4 appears on the 3rd card but does not appear on the 1st or 2nd, or any other.
The number 5 in binary is 101... (1 x 2²) + (0 x 2¹) + (1 x 2⁰) → (5)10 = (101)2 → Hence the number 5 appears on the 1st and 3rd cards, but does not appear on the 2nd, or any other.
The number 6 in binary is 110... (1 x 2²) + (1 x 2¹) + (0 x 2⁰) → (6)10 = (110)2 → Hence the number 6 appears on 2nd and 3rd cards but does not appear on the 1st, or any other.
The number 7 in binary is 111... (1 x 2²) + (1 x 2¹) + (1 x 2⁰) → (7)10 = (111)2 → Hence the number 7 appears on on 1st, 2nd and 3rd cards, but does not appear on any other.
The number 41 in binary is 101001... (1 x 2⁵) + (0 x 2⁴) + (1 x 2³) + (0 x 2²) + (0 x 2¹) + (1 x 2⁰) → (41)10 = (101001)2 → Hence the number 41 appears on the 1st, 4th and 6th cards but does not appear on the 2nd, 3rd or 5th.
The number 62 in binary is 111110... (1 x 2⁵) + (1 x 2⁴) + (1 x 2³) + (1 x 2²) + (1 x 2¹) + (0 x 2⁰) → (62)10 = (111110)2 → Hence the number 62 appears on the 1st, 2nd, 3rd, 4th and 5th cards, but does not appear on the 6th.
The number 63 in binary is 111111... (1 x 2⁵) + (1 x 2⁴) + (1 x 2³) + (1 x 2²) + (1 x 2¹) + (1 x 2⁰) → (63)10 = (111111)2 → Hence the number 63 appears on all six cards.
Thinking through some possible starting questions for students to consider (or to consider with students), before moving on perhaps with students creating their own 'Mystery Calculator' trick (see also [3]):
• What do you notice about the numbers on each card?
• Do these patterns matter — what could we do to see if they do?
• Can we have a number larger than 63 on any of the cards?
• What about all of the other numbers on the cards — can they be chosen randomly?
• What number would appear on all cards if there were four, five, seven, ten, ... cards?
• Does the number in the top left-hand corner have to be there?
• Are there any constraints to the numbers we place on the cards?
• Would a similar trick in another base, maybe base-3, be more magical?
• How high can we count in binary on our fingers? (Watch this Ted-Ed video from James Tanton.)
[1] See here for an example, or here.
[2] See 'When Magic Fails in Mathematics,' by Junaid Mubeen.
[3] See this by Katie Steckles in The Aperiodical'On Disreputable numbers'.
For other Christmathsy problems to consider with students, try the 'Santamaths' problem, the '12 Days of Christmas' problem, see this selection from @mathsjem, and have a look at my 'xmaths card'.
## Friday, 8 December 2017
### Problem... Santamaths
#Probability #Combinations #nCr #Binomial #Modelling
## Tuesday, 5 December 2017
### On Mathematics To Listen To
A collection of links to radio and podcast episodes (in English) with a mathematics bent — to inspire teachers to inspire, to feed the minds of our young mathematicians in the making, or, simply, to entertain. Click here or on the 'Audio' tab above for the complete list. Programmes have been (loosely) categorised according to themes below, and arranged alphabetically within each category by programme title.
The collection includes, in no particular order:
1. Select episodes from BBC Radio 4 In Our Time
2. All episodes from BBC Radio 4 Marcus du Sautoy's A Brief History of mathematics
3. All episodes from BBC Radio 4 Marcus du Sautoy's Five Shapes
4. All episodes from BBC Radio 4 Simon Singh's Five Numbers
5. All episodes from BBC Radio 4 Simon Singh's Another Five Numbers
6. All episodes from BBC Radio 4 Simon Singh's A Further Five Numbers
7. Select episodes from BBC Radio 4 The Infinite Monkey Cage
8. All episodes from BBC Radio 4 A History of the Infinite
9. Select episodes from BBC Radio 4 More Or Less
10. Select episodes from BBC Radio 4 Great Lives
11. Select episodes from BBC World Service More Or Less
12. Select episodes from BBC Radio 4 Seriously
13. Select episodes from BBC Radio 4 Incarnations: India in Fifty Lives
14. Select episodes from BBC Radio 4 Start the Week
15. Select episodes from BBC Radio 4 The Life Scientific
16. Other programs from BBC Radio
17. Select episodes from The Partially Examined Life podcast
18. Select episodes from Futility Closet podcast
19. Select episodes from Freakonomics podcast
20. Select episodes from Radiolab podcast
21. Select episodes from Teach Better podcast
22. Select episodes from Modern Learners podcast
23. Select episodes from Plus Magazine podcasts
24. Select episodes from The Guardian's Science Weekly podcast
25. Select episodes from The Naked Scientists podcast
26. Select episodes from The NCETM Maths podcast
27. Select episodes from The James Altucher Show podcast
28. Select episodes from BBC Radio Cambridgeshire and BBC Radio Oxford Funbers
29. Select episodes from Things Fall Apart podcast
30. Select episodes from The Zone of Potential Construction podcast
There are, of course, many other superb podcasts from which episodes have not been included in this selection, purely because as their sole focus is on mathematics or its teaching, I would be listing links to all of their episodes. (Episodes in this selection are from series that don’t concentrate solely on mathematics, or from shorter series that concentrate on one aspect of mathematics.)
I have listed a selection of some other, excellent podcasts solely focused on mathematics and/or its teaching here, but of course this list is not exhaustive. If you are aware of other podcasts or audio to point people in the direction of, do let me know.
## Monday, 4 December 2017
### Problem... The 12 Days of Christmas
#Sequences, #GeneratingSequences, #TriangularNumbers
#Generalising, #WordProblem, #Series, #SummationOfSeries<|endoftext|>
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Chapter 3—Fractions
3-2 Converting Fractions to Higher Terms
When you complete the work for this section, you should be able to: Explain why multiplying the numerator and denominator of a fraction by the same number does not change the value of the fraction. Demonstrate your ability to convert fractions to higher terms.
When performing routine arithmetic operations with fractions, it is often necessary to convert a fraction to higher terms. This means you multiply both the numerator and denominator by a particular integer value. Suppose you have a fraction, 2/3, and you multiply both the numerator and denominator by 7:
2/3 · 7/7 = 14/21
Procedure To convert any fraction to higher terms, multiply both the numerator and denominator by the same integer value.
Examples:
More Examples:
1. 1/2 x 2/2 = 2/4 2. 3/4 · 5/5 = 15/20 3. ( 3/5 ) ( 3/3 ) = 9/15 4. 5/8 x 4/4 = 20/32 5. 7/10 · 12/12 = 84/120
Keeping All Things Equal
Note: Multiplying the numerator and denominator by the same number does not change the actual value of the fraction.
After multiplying the numerator and denominator of 3/4 by 4, we see that 3/4 = 12/16 Equal portions, different expressions.
Examples
1. After multiplying the numerator and denominator of 3/4 by 2, we see that 3/4 = 6/8
2. After multiplying the numerator and denominator of 3/4 by 3, we see that 3/4 = 9/12
3. After multiplying the numerator and denominator of 5/8 by 2, we see that 5/8 = 10/16
4. After multiplying the numerator and denominator of 3/5 by 5, we see that 3/5 = 15/25
Thinking Mathematically Multiplying any number by 1 does not change the value of that number: Examples 2 x 1 = 2 5 x 1 = 5 1 x 120 = 120 When the numerator and denominator of a fraction have the same value, the fraction is equal to 1: Examples 3/3 = 1 12/12 = 1 8/8 = 1 Therefore, multiplying the numerator and denominator of a fraction by the same number does not change the actual value of the fraction.
Determining the Multiplying Factor
You will be doing a lot of these conversions to higher terms, especially when adding and subtracting fractions. In those situations, however, you are not given the value of the common multiplying factor—you must determine it for yourself. How is this done? The clue is that you are given the value of the denominator for the converted fraction. Like this:
3/5 = ?/10
In this problem, the fraction 3/5 is being raised to a higher power. You can see that the denominator is raised from 5 to 10. It is raised by a factor of 2. But what about the numerator? When 5 is raised to 10, 3 is raised to ... . Figure it out. Since the numerator is raised by a factor of 2, the denominator must also be raised by a factor of 2. So:
3/5 = 3/5 · 2/2 = 6/10
3/5 = 6/10
Example
Problem Complete the conversion, 5/8 = ?/32 Procedure Determine the value of the common factor. The denominator is converted from 8 to 32. The factor in this case is 4: 8 x 4 = 32 So the common factor is 4. 5/8 x 4/4 = ?/32 Complete the multiplication. 5/8 x 4/4 = 20/32 Solution 5/8 = 20/32
Examples and Exercises
Raising Fractions to Higher Terms Use these interactive examples and exercises to strengthen your understanding and build your skills:
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Mathematics
Easy
Question
# What is the value of the expression x1 + 5x + 3 where the value of x is 5?
Hint:
## The correct answer is: 28
### The given expression is x1+ 5x + 3There is one variable in the above expression. The variable is: ‘x’. Value assigned to the variable is x = 5. We have to substitute this value and find the final value of the expression.There are two terms in the above expression. The first term is ‘x1'. It means x is taken one time.The second term is ‘5x’. It means variable x is multiplied by constant 5.Third term is constant.We have to first substitute the value and perform the required operations.To substitute the value, we have to replace x by 5. After substituting the value, we will get the following equation:x1+ 5x + 3 = x + 5x + 3x1+ 5x + 3 = (5) + 5(5) + 3x1+ 5x + 3 = 5 + 25 + 3x1+ 5x + 3 = 33Therefore, the value of equation is 33.So option (b), which is ‘33’ is the right option.
We have to be careful about power of variables.<|endoftext|>
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The three bears put their point of view with evidence to Judge Jenny to try to persuade her that they are right. Goldilocks defends her actions and Judge Jenny sums up so that the audience can discuss and decide.
This clip could be used to show the importance of listening as a fundamental part of communication; pupils could partake in a kinesthetic activity whereby they begin the session in the middle of the room. The teacher could place a picture of Goldilocks on one wall and one of the three bears on the opposite one. As the clip progresses, pupils would move in whichever direction they see fit, depending on whom they believe and whom they think has the best argument. At the end, each pupil could be asked why they are standing where they are and this would stimulate further discussion.<|endoftext|>
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What is 15 2?
Discover the answer to the simple math problem “what is 15 2?” in this informative article. Learn how to approach the problem and why it’s important for everyday life.
Introduction
When it comes to solving simple math problems, some people find it challenging to determine the correct answer. One of the most common math problems that people encounter is “what is 15 2?” While the problem may seem simple, it can lead to confusion and incorrect answers if not approached correctly. In this article, we will explore the answer to this problem and provide a step-by-step guide on how to solve it.
Understanding the Problem
Before we dive into the solution, let’s first understand the problem. “What is 15 2?” is a basic math problem that requires the addition of two numbers – 15 and 2. The objective is to find the sum of these two numbers. While the problem may seem simple, it is essential to approach it with the correct methodology to ensure that the answer is accurate.
The Solution
To find the answer to “what is 15 2?”, we need to add 15 and 2. The result of this addition is 17. Therefore, the answer to “what is 15 2?” is 17.
Why is it Important?
While “what is 15 2?” may seem like a trivial question, it is an essential foundation for more complex math problems. Understanding the basics of addition and subtraction is crucial for solving more complex math problems, such as algebra and calculus. Moreover, having a strong foundation in math is essential for everyday life, from calculating tips in a restaurant to budgeting household expenses.
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Conclusion
In conclusion, “what is 15 2?” is a simple math problem that requires the addition of two numbers. The answer to this problem is 17, which can be found by adding 15 and 2. While it may seem like a trivial question, having a strong foundation in math is essential for everyday life and more complex math problems.
Step-by-Step Guide to Solving “What is 15 2?”
Now that we know the answer to “what is 15 2?” let’s take a closer look at how to solve this problem step-by-step.
1. Write down the problem: Write “15 + 2” on a piece of paper or a calculator.
2. Add the numbers: Add 5 and 2 together to get 7.
3. Write down the answer: Write “17” as the answer to the problem.
4. Double-check your work: Make sure that you have correctly added 15 and 2 to get 17. If you are using a calculator, make sure that you have entered the numbers correctly.
5. Practice: To improve your math skills, it’s important to practice solving problems like “what is 15 2?” regularly. You can find similar problems online or in math textbooks.
Tips for Solving Math Problems
While solving “what is 15 2?” is relatively straightforward, tackling more complex math problems can be challenging. Here are some tips that can help you solve math problems more efficiently:
1. Understand the problem: Before you start solving the problem, make sure you understand what it is asking you to do.
2. Break it down: Break the problem down into smaller parts to make it more manageable.
3. Use visual aids: Draw diagrams or use graphs to help you visualize the problem.
4. Practice: The more you practice solving math problems, the more comfortable you will become with the process.
5. Seek help: If you are struggling with a particular problem, don’t be afraid to seek help from a teacher or tutor.
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By following these tips and practicing regularly, you can improve your math skills and solve even the most complex math problems with ease.
Conclusion
In conclusion, solving “what is 15 2?” requires adding 15 and 2 together to get 17. While this problem may seem simple, it is essential to approach it correctly to ensure an accurate answer. By following a step-by-step guide and incorporating tips for solving math problems, you can improve your math skills and tackle even the most complex math problems with confidence.
Common Mistakes and Tips
While “what is 15 2?” may seem like a simple problem, there are a few common mistakes that people make when trying to solve it. One common mistake is to subtract 2 from 15 instead of adding them together. Another mistake is to forget to carry over the one when adding the numbers. To avoid these mistakes, it is essential to approach the problem with the correct methodology.
Here are a few tips to help you solve “what is 15 2?” correctly:
1. Always ensure that you are adding the numbers, not subtracting them.
3. If you are having trouble adding the numbers mentally, try writing them down on paper to help you visualize the problem better.
Real-World Applications
While “what is 15 2?” may seem like a simple math problem with limited real-world applications, it is an essential foundation for more complex math problems. Having a strong foundation in math can open up many career opportunities, from engineering to finance. Moreover, math is essential in everyday life, from calculating grocery bills to determining mortgage payments.
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Here are a few real-world applications of the skills required to solve “what is 15 2?”:
1. Budgeting household expenses
2. Calculating tips in a restaurant
3. Determining the cost of sale items
4. Calculating mortgage payments
5. Analyzing financial data in business
In conclusion, “what is 15 2?” is a simple math problem that requires the addition of two numbers. While it may seem like a trivial question, having a strong foundation in math is essential for everyday life and more complex math problems. By understanding the methodology required to solve this problem, you can apply these skills to real-world situations and open up many career opportunities.
How Many Pounds is 600 kg? – A Comprehensive Guide
Learn how to convert 600 kg into pounds with our comprehensive guide. Discover the factors that impact weight measurement, including gravity, altitude, and temperature.
How Many Pounds is 600 kg? – A Comprehensive Guide
Learn how to convert 600 kg into pounds with our comprehensive guide. Discover the factors that impact weight measurement, including gravity, altitude, and temperature.
What is the Optional DC Cable for the Yaesu FT-70DR?
Learn how to use the optional DC cable for the Yaesu FT-70DR to ensure uninterrupted radio communication in the field. Find out what it is and how to use it!
How Many Ounces in 1.5 Pounds?
Learn how to convert pounds to ounces and vice versa accurately! Discover “how many ounces in 1.5 pounds” and more with this comprehensive guide.
What is Rebirth 2k22? Unlocking the Mysteries of Reincarnation
Discover the mysteries of rebirth 2k22 – the concept of reincarnation in the year 2022. Explore its significance in different beliefs and how to achieve it.
What is a Smoochie Girl?
Discover the truth about what a smoochie girl is and the harmful effects of this behavior. Learn how to break free from the cycle and embrace authenticity.<|endoftext|>
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# Rational Numbers Class- 7th RS Aggarwal Exe-4 I Goyal Brothers ICSE Math Solution
Rational Numbers Class- 7th RS Aggarwal Exe-4 I Goyal Brothers ICSE Math Solution . We provide step by step Solutions of lesson-4 Rational Numbers for ICSE Class-7 Foundation RS Aggarwal Mathematics of Goyal Brothers Prakashan . Our Solutions contain all type Questions of Exe-4 I to develop skill and confidence. Visit official Website for detail information about ICSE Board Class-7 Mathematics. MCQs questions for class 7/8 maths with answers pdf of rational numbers.
## Rational Numbers Class- 7th RS Aggarwal Exe-4 I Goyal Brothers ICSE Math Solution
Board ICSE Publications Goyal brothers Prakashan Subject Maths Class 7th Chapter-4 Rational Numbers Writer RS Aggrawal Book Name Foundation Topics Solution of Exe-4 I Academic Session 2023 – 2024
### MULTIPLE CHOICE QUESTIONS
Rational Numbers Class- 7th RS Aggarwal Exe-4 I Goyal Brothers ICSE Math Solution
#### Choose the correct option in each of the following :
##### 1. The additive inverse of (5/9) is
(a) (9/5)
(b) -(5/9)
(c) (-5/-9)
(d) -(9/5)
Answer: option (b) -(5/9) is correct.
##### 2. The rational number (32/-40) expressed in standard form is
(a) (8/-10)
(b) (4/-5)
(c) (-32/40)
(d) (-4/5)
Answer: option (d) (-4/5) is correct.
##### 3. What should be added to (-3/16) to get (5/8)?
(a) (2/8)
(b) -(1/2)
(c) (3/8)
(d) (13/16)
Answer: option (d) (13/16) is correct.
##### 4. The multiplicative inverse of (-3/7) is
(a) (-7/3)
(b) (3/7)
(c) (4/7)
(d) (7/3)
Answer: option (a) (-7/3) is correct.
##### 5. The sum of -(1/3) and its multiplicative inverse is
(a) 0
(b) -3
(c) -1(2/3)
(d) -3(1/3)
Answer: option (d) -3(1/3) is correct.
##### 6. The product of -(1/3) and its additive inverse is
(a) 0
(b) -3
(c) -(1/9)
(d) -3(1/3)
Answer: option (c) -(1/9) is correct.
##### 7. Which of the following rational numbers is equivalent to (-2/7)?
(a) (-14/21)
(b) (-8/14)
(c) (-14/49)
(d) (-6/28)
Answer: option (c) (-14/49) is correct.
##### 8. If 3(3/4) m of cloth is required for one suit, then how many suits can be prepared from 30 m of cloth?
(a) 4
(b) 5
(c) 8
(d) 9
Answer: option (c) 8 is correct.
### MENTAL MATHS
Rational Numbers Class- 7th RS Aggarwal Exe-4 I Goyal Brothers ICSE Math Solution
#### 1. Fill in the blanks :
(i) The multiplicative inverse of a rational number is also called its reciprocal.
(ii) Every negative rational number is less or smaller than 0.
(iii) A rational number (p/q) is said to be in standard form, if q is positive and p and q have no common divisor other than 1.
(iv) (-5/-9) is a positive rational number.
(v) The additive inverse of a rational number (a/b) is (-(a/b)).
#### 2. Write true (T) or false (F) :
Statement True/False (i) There exists a rational number which is neither positive nor negative. T (ii) Every rational number has a multiplicative inverse. F (iii) Every rational number when expressed in its standard form has its denominator greater than the numerator. F (iv) The sum of a rational number and its additive inverse is always. T (v) The product of a rational number and its multiplicative inverse is always. T (vi) Any tow equivalent rational number have the same standard form. F (vii) The product of any two rational number is also a rational number. T (viii) A rational number when divided by another rational number always given a rational number. T (ix) Every rational number smaller than a given rational number (p/q) lie to the left of (p/q). F
— : end of Rational Numbers Class- 7th RS Aggarwal Exe-4 I Goyal Brothers ICSE Math Solution:–<|endoftext|>
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This item is a tutorial for introductory physics students on the topic of refraction. It contains an introduction to basic concepts, a brief history of refraction experimentation, refractive index values, light dispersion, and applications of Snell’s equation. Also included are links to six interactive Java simulations related to the refraction of light. This item is part of a larger collection of materials on optics and microscopy developed by the National High Magnetic Field Laboratory and Florida State University.
At this site students will learn that in addition to making elements, supernovae scatter them. They learn that the elements that are made both inside the star as well as the ones created in the intense heat of the supernova explosion are spread out into the interstellar medium. These are the elements that make up stars, planets and everything on Earth including ourselves. Except for hydrogen and some helium created in the Big Bang, all of the stuff we, and the Earth around us, are made of, was generated in stars, through sustained fusion or in supernova explosions.
This U.S. Geological Survey site lists and discusses the properties of volcanic ash. The site contains many helpful diagrams, and explains topics including the size of ash particles, the dispersal of ash by wind, and the kind of eruption that produces ash.
This peer reviewed article from Bioscience magazine is about figs and the diversity of tropical rainforests. Ficus (Moraceae) is arguably one of the most important plant genera in lowland tropical rainforests. A brief review of tropical florulas also demonstrates that Ficus is the only ubiquitously diverse genus in lowland rainforests.
This chapter continues the examination and clarification of concepts relating to point objects, for which as argued in Chapter 1, appropriate, complex/ second order, concepts relate to words like ‘distribution’, ‘dispersion’, ‘density’, ‘pattern’ and ‘scale’ and, at higher level still, third order concepts relating to point process models, stationarity and isotropy/anisotropy.<|endoftext|>
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Preserving "Swampeast" Missouri
Early explorers to the "Bootheel" region of southeast Missouri discovered a unique landscape where the Ozark hills dropped off abruptly into flat flood plains covered with giant trees. The forest trees that took root here grew to enormous proportions in the fertile earth. Today, nature-seekers to "swampeast" Missouri will find a forested island, known as Big Oak Tree State Park, in the midst of a vast agricultural area. The park features the state park system's only cypress swamp.
Regular flooding in the area over millions of years caused many of the rich Bootheel forests to convert to swampland, providing temporary protection for many of the giant trees. In 1811, a series of earthquakes known as the New Madrid earthquake altered the topography of the southeast lowlands. Some of the land in the southeast Bootheel area and in northeast Arkansas sank, adding to the existing flooding.
Beginning in the late 1800s, the giant timber and fertile soil of the lowlands began to attract timbermen and farmers. Timbermen cleared the trees leaving bare land. Realizing the potential for agriculture, landowners supported the creation of drainage districts in the early 1900s to enable them to plant crops, such as wheat, soybeans and corn. Over two million acres were converted from forest to cropland.
In the 1930s, citizens of southeast Missouri began to realize that their magnificent lowland forests were about to disappear forever. A campaign to save a large oak tree and 80 acres surrounding it attracted statewide attention. In response, Gov. Lloyd Stark asked the Mississippi Valley Hardwood Co. to spare the tree and surrounding land.
Due to the Great Depression, the state, however, could not afford to purchase the land. Despite the depression, businessmen and local citizens began donating money while area school children gave their nickels and dimes. In 1938, about 1,007 acres were purchased and dedicated as Big Oak Tree State Park.
The big oak tree that the citizens fought to save was located in an 80-acre tract of virgin bottomland hardwood forest, which is now designated as a National Natural Landmark. It stood its ground there for nearly 400 years before dying in 1952 and was cut down. Today, trees in the park are unsurpassed in the state for their size, with a canopy averaging 120 feet and with several trees more than 130 feet tall. Two trees qualify as state champions in their species. Ninety percent of the park is designated as a Missouri natural area because of its rarity and value in preserving this significant representation of Missouri's natural heritage.
In addition to the towering hickories and majestic oaks, the rich soils provide habitat for green ash, swamp cottonwood, American elm, black willow, persimmon, baldcypress and patches of giant cane. An undergrowth of woody vines, such as poison ivy, Virginia creeper, wild grape and peppervine, covers much of the area.
Many swamp plants live in the water-soaked soil, including swamp privet, buttonbush, lizard's tail, swamp leather flower and ladies' eardrop while aquatic liverworts and duckweed float on the surface of the water. The park protects 12 species of rare plants and animals, 250 kinds of plants and 25 mammal, 31 reptile and seven amphibian species.
A boardwalk winds its way through the park past some of the park's largest trees. A walk on the boardwalk also gives visitors a chance to view many common mammals, such as deer, raccoons, squirrels and opossums that call the park home, along with the rare swamp rabbit. Amid the lofty trees live more than 150 species of birds, giving the park a national reputation among bird watchers. Several of the birds are considered rare in the state, including the prothonotory warbler, cerulean warbler, red-shouldered hawk, Mississippi kite and fish crow.<|endoftext|>
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# Algebra Intro 10: Fractions and Multiplication
Many people seem a bit phobic about “fractions”. This anxiety likely has two sources: not really understanding what a fraction represents, and having memorized a bunch of rules way back in elementary school without understanding why they work.
Revisiting fractions using variables as well as constants, with complicated as well as simple problems, helps conquer anxieties about this topic.
### What Is A Fraction?
The fraction
$\dfrac{3}{5}$
can be interpreted as
• The number three divided by the number five
• Three of the quantity one fifth
• A fifth of the quantity three
• Three parts of a five part “whole”
• A rational number
A rational number is a number that can be expressed as the ratio (or quotient) of two integers. Three fifths satisfies this definition, and is therefore a rational number.
“The number three divided by the number five” describes a division problem, with three in the numerator and five in the denominator. This is perhaps the most “rote” and least “intuitive” interpretation of a fraction, but it is also the simplest. Every fraction is a division problem, and every division problem involving two quantities can be expressed as a fraction.
“Three parts of a five part whole” is illustrated above, and is the interpretation that is probably used most often in elementary school. The entire object above represents the quantity one. The denominator instructs us to divide the object into five equal parts, and the numerator instructs us to use three of them.
The above illustration also fits the description “three of the quantity one fifth”. Once again, the entire bar represents the quantity one, and it has been divided into five equal squares, so each square represents the quantity “one fifth”. Using three of the squares therefore represents three “fifths”.
Note that when read the way fractions are usually read (as “three fifths”), the phrase can be translated into algebraic notation as either $(\frac{3}{5})$ (using the first interpretation) or $(3\cdot\frac{1}{5})$ (using the second interpretation). It is useful to be confident rewriting a fraction with the reciprocal of the denominator in either the first or second position:
$\dfrac{3}{5}~~=~~3\cdot\dfrac{1}{5}~~=~~\dfrac{1}{5}\cdot 3$
An illustration of the final fraction above is:
Each row above represents a “whole” quantity divided into five parts. The three rows therefore add up to a total of three “whole” quantities. The left column is one fifth of the entire grid, or one fifth of three. Note that the size of the darker three squares on the left above is exactly the same as the size (after rotating it by 90 degrees) of the darker three squares on the left of the previous illustration, which demonstrates that both interpretations produce the same result.
### Multiplying Fractions
Each of the four arithmetic operations (addition, subtraction, multiplication, and division) work with fractions too. Some operations are easy to compute with fractions, others more complex. I believe that multiplying two fractions is the easiest of the four operations.
To multiply two fractions, multiply their numerators and multiply their denominators. Why does this rule work? What is really going on when we follow this rule? To explore these questions a bit, let’s consider an example:
$\dfrac{2}{3}\cdot \dfrac{4}{5}$
One way of simplifying the above is to rewrite division as multiplication by the reciprocal (scaling by a number between zero and one), then use the commutative property of multiplication to rearrange things a bit:
$2\cdot\dfrac{1}{3}\cdot 4\cdot \dfrac{1}{5}\;\;=\;\;2\cdot 4\cdot \dfrac{1}{5}\cdot \dfrac{1}{3}$
While this now makes it easy to calculate the two times four part, it still requires us to evaluate the product of two fractions. How is the product of these two reciprocals calculated?
Intuitively, if you are asked to divide a whole pizza into three equal parts, you will get something like the illustration below. Each third of the pizza has a different color:
Now, divide each color (third) into five equal parts. Looking at the result of slicing the pizza into thirds, then each slice into fifths, the pizza has been cut into fifteen slices: three large slices (each with a different color above) were the result of the first division, then each large slice became five smaller slices as the second division was carried out, creating a total of fifteen small slices (as numbered above). So, “a fifth of a third” will be one of the above slices, which represents one fifteenth of the original pizza:
$\dfrac{1}{3}\cdot \dfrac{1}{5}\;\;=\;\;\dfrac{1}{3\cdot 5}\;\;=\;\;\dfrac{1}{15}$
Another way to envision the process of taking “a fifth of a third” relies on interpreting multiplication as scaling. The vertical line on the left of the illustration below has a length of one, and the green segment at the bottom of that line has a length of one third.
The horizontal line at the bottom of the above illustration also has a length of one, and the red segment on the left of it has a length of one fifth. Multiplying one quantity by another is done by stretching the first quantity (the green line segment) into a new dimension (to the right) by the length of the second quantity (the red line segment). The result is the yellow rectangle with “1” in its center.
As you can see in the above illustration, the yellow rectangle is identical in size to the other 14 rectangles shown. Together, all fifteen rectangles create a 1 by 1 square, whose area is 1. The yellow rectangle, which is the product of “one third” and “one fifth”, has an area equal to one fifteenth of the square’s area, so the product of these two numbers equals one fifteenth.
A non-geometric approach to the problem can also be taken. Convert both reciprocals into their decimal equivalents, then multiply them:
one third is $\;\;0.\overline{333}$
and one fifth is $\;\;0.20$
and the product of the two is $\;\;0.0\overline{666}$
which is the decimal equivalent of one fifteenth.
And lastly, a physical approach could be based on a room that contains 15 chairs (similar to the square in the illustration above). If you divide them into three equal groups, there will be 5 chairs in each group. If you then divide each group into five equal sub-groups, there will be 1 chair in each sub-group. The result of dividing 15 by three and then five is 1. What proportion (or “fraction”) of the original number of chairs does this represent? One fifteenth… the same answer arrived at via the other approaches.
Some useful insights hopefully came to mind while thinking about the above:
1. Dividing by three then dividing by five will produce the same result as dividing by five then dividing by three. When rewritten as multiplication by the reciprocal, the commutative property of multiplication allows us to rearrange the factors without changing the result.
2. Dividing by one number (three), then dividing by another (five), produces the same results as dividing by their product (fifteen).
Think about how we read the original problem: “two thirds times four fifths”. This could be rephrased as “two of the quantity a third times four of the quantity a fifth”, which not only is equivalent to the first statement grammatically but also mirrors the way division can be rewritten as multiplication by the reciprocal (remember that “of” indicates multiplication).
By rewriting the entire problem as a multiplication problem, and using the commutative property of multiplication to change the order of the factors, the original problem becomes “two, times four, times a third, times a fifth”. Hopefully you are now comfortable with the reasoning that two times four is eight, and a third times a fifth is a fifteenth, so we have now simplified the problem down to the product of eight and a fifteenth. As our last step, we can now reverse the process of turning division into multiplication, and rewrite a number times a reciprocal as the quotient of the two numbers:
$\dfrac{2}{3}\cdot \dfrac{4}{5}$
$=2\cdot\dfrac{1}{3}\cdot 4\cdot \dfrac{1}{5}$
$=2\cdot 4\cdot\dfrac{1}{3}\cdot\dfrac{1}{5}$
$=8\cdot \dfrac{1}{15}$
$=\dfrac{8}{15}$
So, the product of two fractions will always end up being the product of their numerators divided by the product of their denominators.
One note of caution… if any numerator or denominator has more than one term, put the entire numerator or denominator in parentheses before multiplying. The vinculum (the horizontal line indicating division) is also a grouping symbol that indicates that both the numerator and the denominator should be treated as a single factor – as if they were in parentheses:
$\dfrac{4+8}{2}$
$=~~(4+8)\cdot\dfrac{1}{2}$
$=~~4\cdot\dfrac{1}{2}+8\cdot\dfrac{1}{2}$
$=~~2+4$
$=~~6~~=~~\dfrac{12}{2}~~=~~\dfrac{4+8}{2}$
And lastly, I find multiplication by a fraction to be the easiest time to re-introduce and discuss the notion of multiplication as scaling (instead of repeated addition). Since fractions seldom represent whole numbers, the interpretation of multiplication as scaling by a factor is somehow easier to comprehend in examples involving fractions.
Posts that continue this series:
Algebra Intro 11: Dividing Fractions, Equivalent Fractions
Algebra Intro 12: Adding and Subtracting Fractions
## Published by
### Whit Ford
Math teacher, substitute teacher, and tutor (along with other avocations)
## 2 thoughts on “Algebra Intro 10: Fractions and Multiplication”
1. Christopher says:
What? No pictures?
I do appreciate, however, the level of complexity you are willing to allow this topic. Too often, the topic is reduced to “of means multiply and here’s the algorithm for multiplication” (and by the way, kids, don’t try to think too hard about what this might have to do with the multiplication of whole numbers).
Bravo.
Might a diagram or two add something to your text?
GeoGebra (of which I understand you are quite fond) is a lovely tool for good, quick mathematical diagrams.
2. Christopher,
Thank you for the suggestion. I have reworked the posting to include some graphical illustrations, along with some additional verbiage to describe them. See what you think!
Whit
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Many countries and organizations have developed working definitions to identify nanomaterials based on the size of the material, its novel properties, or a combination of both. In fact, each material “exists” at the nanoscale. Materials at nanoscale behave differently than those at macroscale. Nanoparticles present incredible properties based on “quantum effects” and other simple physical effects such as expanded surface area. In summary, “nano” is just a question of size and not of material or composition.
Nanoscience and nanotechnology are the study and application of extremely small things and can be used across all the other science fields, such as chemistry, biology, physics, materials science, and engineering. In the International System of Units, the prefix “nano” means one billionth, (or 10-9) of a meter. The Figure 1 illustrates clearly the nanoscale.
Figure 1. The « Nano » scale.
Nanoparticles are abundant in nature, as they are produced in many natural processes, including photochemical reactions, volcanic eruptions, forest fires, simple erosion and by plants and animals. Any groups of atoms or molecules present in nature have a “nano” size such as amino acids, proteins, DNA, viruses, etc. Those are organic molecules with carbon atoms, hydrogen atoms and oxygen atoms mainly.
In fact, each material “exists” at the nanoscale. During a long time, humans have not been able to produce with a sufficient control material at nanoscale such as nature is doing since millions of years. Other sources of nanoparticles can be identified and originates from the man-made activity such as incidental nanoparticles generated as unwanted by-products from combustion processes (e.g.: charcoal burning, incinerators,…).
Materials at nanoscale behave differently than at macroscale and nanoparticles often presents incredible properties based on “quantum effects” and other simple physical effects such as expanded surface area. The integration of the nanoparticles inside a matrix leads to improve specific properties depending on the nature of the nanoparticles and the matrix in which the nanoparticles are integrated (polycarbonates, polystyrene, polyethylene, natural rubber, polyurethane, epoxies, etc.). Each type of nanoparticle will improve a different set of properties (electrical, mechanical, thermal, etc.).
Figure 2. Sources of nanoparticles.
In summary, “nano” is just a question of size and not of material or composition.
The definition of a nanomaterial does not make the consensus yet. In 2011, the European Commission adopted a recommendation on the definition of a nanomaterial in 3 steps [COMMISSION RECOMMENDATION of 18 October 2011 on the definition of nanomaterial(Text with EEA relevance) (2011/696/EU)]:
- a natural, incidental or manufactured material
- That contains particles, in an unbound state or as an aggregate or as an agglomerate
- and where, for 50% or more of the particles in the number size distribution, one or more external dimensions is in the size range 1 nm – 100 nm.”
It is finally quite recently that humans have developed techniques for manufacturing nanoparticles in a controllable and reproducible way; what are called today engineered nanoparticles (“ENP”).
Figure 3 illustrates that nanoparticles look very different. Spheres, rods, fibers, stars, cups,… are various shapes that have been identified.<|endoftext|>
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# Ideas Made of Light
#### Golden Sections
I often look at the “golden section grid” when analyzing a painting. I’ll describe it below – the further you read the more math will crop up. So, start at the top for the summary and quit reading whenever you feel like it.
## The Not-So-Math Part
When it gets right down to it, certain proportions just look better to us. No one knows why, exactly, but folks as far back as ancient Greece and Egypt recognized the fact and worked out guidelines to help us use those proportions. One of the most important of these guidelines is the so-called “golden section.”
Proportions are all about relationships. How do the parts look compared to each other, and compared to the whole? How long, how tall, how wide? In math, we talk about relationship and proportion by using ratios. The golden section, also called the golden ratio, is:
1 to 1.6180…
So, if you have a rectangle that is 1 foot tall, that rectangle is “golden” if it is 1.6180 feet wide. It’s not the only rectangle that looks good, but it’s one of the most useful. You don’t need to understand the mathematics, but if you want to know I’ll go into some of that below. For some, though, it’s enough to know the golden ratio itself to use it in their art. In the end, even going through the math doesn’t tell you why it works. But it can increase your appreciation.
Before getting to that though, I have one other point to make. If you have a line, like the side of a rectangle, you can divide that line into two sections according to the golden ratio. So, if your line is 100 cm long, you can divide it into two line segments of approximately 38 and 62 cm.
62 : 100 is the ratio of the longer line to the whole thing. It reduces down to 1 : 1.6180.
38: 62 is the ratio of the shorter line to the longer line. It also reduces down to 1 : 1.6180.
If that 100 cm is the side of a rectangle, and that rectangle happens to be your painting canvas, putting things along those lines at 38 and 62 cm will just naturally look right. Then, if you do the same thing along the bottom and top of your rectangle, and draw the lines at those golden section divisions, you’ll end up with a grid of 9 rectangles inside your overall canvas, like this:
A 100 x 150 rectangle divided up according to the golden ratio.
Placing things along those lines, and especially at the intersections of the lines, will result in a more pleasing composition than you’ll get with just about any other method.
## Slightly Mathier Parts
So where did this magic number of 1.6180 come from? For one thing, it’s known as Phi (ϕ) and it’s one of those numbers like Pi where the decimals go on forever. One of its sources is the Fibonacci sequence. If you start with 0 and 1, you get the next number of the sequence by adding the last two. So:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
You can find examples of this sequence all through nature. Tool uses it in their song Lateralus – look at the syllable counts for the first couple parts of the song.
It’s quite approximate at first, but if you take any two consecutive numbers in the sequence and divide the smaller one into the larger, you get a number very close to Phi. So that’s one of the ways that artists and mathematicians got to the golden rectangle.
I’ll have more to say on this soon, but for now I’ll stop here. When I get more time I’ll go into the relationships between the golden rectangle and some of the other rectangles in Dynamic Symmetry. Until then, see the analysis on Maxfield Parrish.<|endoftext|>
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Guided Lessons
# How Many Floors Down?
Students will love incorporating stories and acting as they practise creating their own subtraction equations. This can be used as a stand-alone or support lesson for the Subtracting Numbers lesson plan.
This lesson can be used as a pre-lesson for theSubtracting NumbersLesson plan.
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Which set of standards are you looking for?
This lesson can be used as a pre-lesson for theSubtracting NumbersLesson plan.
Students will be able to solve subtraction problems using numbers 0–10. Students will be able to explain how to subtract items from a group using acting and a number line to represent the problem.
(2 minutes)
• Gather the class together.
• Ask: "Has anyone ever been on an elevator before?"
• Display the ElevatorVocabulary card and use student examples along with the card to provide a relevant definition.
• Hold up the book Elevator MagicAnd go on a brief picture walk to preview the story.
(5 minutes)
• Read the first half of the story aloud, pausing on each page to notice how many floors Ben is going down.
• Display the number line and model how the floors are going down, just like the numbers go down on the number line.
• Introduce the word SubtractionAs taking something away from something else, using the elevator going down as an example (e.g., when they go down one floor, write "10 – 1 = 9" to figure out which floor they will be on next).
• Write up an equation from the beginning of the story 10 – 1 = 9On the board and explain that this is how we write a subtraction number sentence. Model how to read it aloud: "Ten minus one equals nine."
• Model solving the subtraction problem using the number line to show how to take one away, or jump to the smaller number.
• Explain the difference between addition and subtraction using the number line to model an addition problem.
(5 minutes)
• Finish reading the story, inviting the students to act out the scene by pretending to be Ben going down in the elevator.
• Collaboratively write and solve equations to go with several more pages of the story.
• Ask students to turn and talk to answer the question: "If Ben was on the 7th floor and went down 1 floor, which floor would he be on? What might he see?"
• Share out and record the subtraction problem on the board, using the number line (or elevator buttons) to solve the problem.
(20 minutes)
• Explain that now students will get to make their very own elevator and set of elevator buttons.
• Display the cardboard boxes and materials and go over any classroom materials expectations.
• Pair students together and explain that they will have 10 minutes to create their elevator and buttons. After 10 minutes, ring a bell to signal time for them to go down the elevator and practise figuring out which floor will be next.
• Send students to work in pairs.
• Ring the bell halfway through the work period and circulate around the room to ensure students transition to playing "elevator" and working to find out which floor will be next.
Beginning
• Work with students in a smaller group to practise creating and solving subtraction problems.
• Provide students with pre-made number lines 0–10 to support their understanding.
• Ask students to record the subtraction problems using a number sentence.
• Encourage students to explain to a partner how to solve each problem using a number line.
(5 minutes)
• Ask guiding questions throughout the lesson to assess understanding (e.g., "How many floors are you going down?" "What floor will we be on next? How do you know? Can you show me using the elevator button panel?").
• Circulate around the room and take note of how students are interacting with their number line/elevator controls. Check for areas of confusion and/or misunderstanding.
(3 minutes)
• Gather the students back together.
• Share out some of your observations of how students were using their elevator button panel to solve problems.
• Applaud student work.<|endoftext|>
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# How do you solve by substitution 5x-6y=6 and 5x+y=2?
Apr 29, 2018
$y = - \frac{4}{7}$
$x = \frac{18}{35}$
#### Explanation:
$5 x - 6 y = 6$ --- (1)
$5 x + y = 2$ --- (2)
(1) minus (2)
$- 6 y - y = 6 - 2$
$- 7 y = 4$
$y = - \frac{4}{7}$ --- (3)
Sub (3) into (1)
$5 x - 6 \times - \frac{4}{7} = 6$
$5 x + \frac{24}{7} = 6$
$5 x = \frac{18}{7}$
$x = \frac{18}{35}$
Apr 29, 2018
The solution is $\left(\frac{18}{35} , - \frac{4}{7}\right)$ or $\left(0.514 , - 0.571\right)$.
#### Explanation:
Solve the system of equations.
$\text{Equation 1} :$ $5 x - 6 y = 6$
$\text{Equation 2} :$ $5 x + y = 2$
The solution to the system of linear equations is the point they have in common, the point of intersection. The system will be solved using substitution.
Solve Equation 2 for $y$.
$y = - 5 x + 2$
Substitute $- 5 x + 2$ for $y$ in Equation 1. Solve for $x$.
$5 x - 6 \left(- 5 x + 2\right) = 6$
Expand.
$5 x + 30 x - 12 = 6$
Simplify.
$35 x - 12 = 6$
Add $12$ to both sides.
$35 x = 6 + 12$
Simplify.
$35 x = 18$
Divide both sides by $35$.
$x = \frac{18}{35}$ or $0.514$
Substitute $\frac{18}{35}$ for $x$ in Equation 2. Solve for $y$.
$5 \left(\frac{18}{35}\right) + y = 2$
Simplify ${\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}}}^{1} \left(\frac{18}{\textcolor{red}{\cancel{\textcolor{b l a c k}{35}}}} ^ 7\right)$ to $\frac{18}{7}$.
$\frac{18}{7} + y = 2$
Multiply both sides by $7$.
${\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}}^{1} \times \frac{18}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}} ^ 1 + 7 \times y = 2 \times 7$
Simplify.
$18 + 7 y = 14$
Subtract $18$ from both sides.
$7 y = 14 - 18$
Simplify.
$7 y = - 4$
Divide both sides by $7$.
$y = - \frac{4}{7}$ or $- 0.571$
Solution
$\left(\frac{18}{35} , - \frac{4}{7}\right)$ or $\left(0.514 , - 0.571\right)$
graph{(y-5/6x+1)(y+5x-2)=0 [-10, 10, -5, 5]}<|endoftext|>
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A polar chart is essentially an XY plot drawn on a circular grid, showing trends in values on the basis of angles. Like logarithmic scales, polar charts are useful primarily in mathematical and engineering applications.
In a polar chart, the independent variable (X) is charted on the angular axis. By default DPlot draws the origin (zero point) at three o'clock. The dependent variable (Y) is charted on the radial axis, with the origin (by default) at the center of the circle.
As an example we'll walk through recreating the polar chart appearing on the Features page:
This graph plots the function r=abs(sin(3φ)), where φ is the angular value. Select Y=f(X) on the Generate menu, which displays this dialog box:
Enter the values shown above: "abs(sin(3*x))" for the function, from X=0 to 360 degrees.
This initially results in a normal XY plot with linear scaling on both the X and Y axes. Right-click on the plot and select "Polar Coordinates" to get a polar plot:
Note that when using Y=f(X) or similar command, DPlot automatically selects the proper units for the angular values (degrees or radians). If the source of your data is a text file, data pasted from the Clipboard, or data sent to DPlot from another program, the angular values may be in the wrong units. Normally this will be obvious. For example with radians selected, our example looks like:
To instead select degrees, right-click on the plot and select Polar Plot Options. Note that regardless of whether degrees or radians are selected, DPlot will display angular values in degrees with one exception: You can display angular values as multiples of or fractions of π (right-click on any of the angle values and select "Pi Multiples" or "Pi Fractions").
Windows 7, 2008,
Vista, XP, NT,
ME, 2003, 2000,
Windows 98, 95<|endoftext|>
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Illustration by Kimberly Carney
Mice and other mammals — people included — have an exquisite ability to sift out different scents. Humans can recognize at least 10,000 different odors, and some animals may detect far more. The neurons that respond to airborne scent molecules, or odorants, can do more than just recognize scents, such as triggering a visceral memory of childhood Thanksgivings or the desire to get away from a rotten stench.
Some scents, especially those emitted from other animals, can cause instinctive behavior.
How mice respond to social cues — instinctive reactions driven by their superb sense of smell — may involve a unique set of neurons in the nose, according to a new study by Fred Hutch biologist and Nobel Laureate Dr. Linda Buck and colleagues.
The study, published April 20 in the journal Proceedings of the National Academy of Sciences, sheds light on a small but unique set of genes involved in scent recognition, known as trace amine-associated receptor, or TAAR, genes, by capturing the molecular details of how and where these genes are turned on and off in the nose.
Buck, who first discovered TAAR genes in 2006, has spent 25 years delving into the science of smell, launched by her discovery, together with Dr. Richard Axel of Columbia University, of the genes responsible for the majority of scent detection. Those genes comprise mammals’ largest gene family and code for proteins known as odorant receptors. The researchers found that mice have about 1,000 odorant receptors, humans between 350 and 400.
Buck and Axel found these odorant receptors inside olfactory neurons, nerve cells embedded in a small area lining the upper part of the nose. Buck and her colleagues later found evidence that each olfactory neuron produces a protein from only one odorant receptor gene — all the other hundreds of odorant receptor genes remain dormant in that cell.
When airborne scent molecules waft into the nose, those tiny chemicals are captured by copies of the single odorant receptor protein sitting on the surface of nose neurons. Each odorant slots into only specific odorant receptors, like a key fitting in a lock. When that key and lock find each other, the odorant receptor triggers signals in its neuron that travel to the brain.
But even that groundbreaking discovery didn’t explain the full panoply of the sense of smell. Buck and her colleagues later found that odorant receptors are combinatorial — each scent triggers the activation of a small and unique subset of odorant receptor proteins and the neurons that house them, making an “odorant code” in the brain. And that detailed and precise organization finally illuminated how humans and other animals — even fish — can detect so many different odors.
“Even if each odorant is only recognized by three odorant receptors, this combinatorial strategy could generate millions of different odorant codes,” Buck said. “That was a key breakthrough.”
The large number of odorant receptors and their combinatorial use in odor detection also explains how our noses can easily distinguish between even closely related chemicals, Buck said. Odorants with scents of oranges and sweaty socks, for example, are nearly identical molecules, but we perceive them very differently.
A pungent mystery
So Buck’s discovery of the TAAR genes and proteins was a big surprise, she said. The proteins are similar to odorant receptors, although a much smaller family — mice have 14 TAARs, humans six. And a small class of olfactory neurons produce TAAR proteins instead of odorant receptors (as with odorant receptors, neurons that express TAARs seem to choose only one TAAR gene to activate).
“This was really puzzling,” Buck said. “Why would there be TAARs in addition to odorant receptors?”
Then she and her team found that some of the scent molecules that TAARs recognize come from other animals. The researchers discovered that several mouse TAARs recognize compounds in mouse urine, and that was their first hint about the proteins’ special role.
“Urine is a very rich source of social cues in the mouse,” Buck said. “It can stimulate all kinds of things, like maternal behavior, mating behavior or aggression. So we thought, ‘Aha! Maybe the TAARs are involved in sensing social cues.’”
Most responses to smells (for mice or humans) are learned behaviors, said Dr. Zhonghua Lu, a postdoctoral fellow in Buck's group and one of the study’s authors. But some are innate — like attraction to something pleasing or repulsion from something potentially dangerous. Buck and other research teams later showed that TAARs recognized chemicals that trigger innate responses, either attraction or aversion. Buck’s team also found that a human TAAR recognizes spoiled fish, a smell that’s almost universally repugnant to humans, suggesting that these proteins can drive our instinctive reactions, too.
The science of attraction and repulsion
Those behaviors and their link to TAARs are intriguing, but it’s not clear how the neurons that produce individual TAARs drive innate responses.
“If that’s the case, it must be that some neurons are preprogrammed to express TAARs and may be sending signals to different pathways in the brain,” Buck said. “There must be something special about those neurons.”
To try to get at that question, Buck’s team asked how these proteins and their associated neurons are organized in the nose — and how neurons choose among the more than 1,000 different odorant receptors or TAARs available to them in the genome.
Taking advantage of the observations that each olfactory neuron always houses one, and only one, functional scent-sensing protein (either an odorant receptor or a TAAR), the researchers looked at mice with mutations in one of two different TAAR genes.
The Buck group and others found that neurons that chose a mutant TAAR tended to replace that nonfunctioning protein with another TAAR instead of an odorant receptor, even though there are far more odorant receptor genes than TAAR genes. And Buck's group further found two classes of TAARs — a neuron missing a functional TAAR tended to choose another gene from the same subset of TAAR genes.
These two subsets of TAAR neurons are spatially separate in the nose, Buck said. These findings suggest that there’s something unique about neurons that contain TAARs, which could be a clue to help map the cells responsible for TAAR-driven instinctive behaviors.
How the neurons choose their fate
To understand how these special classes of neurons arise, the researchers then looked at TAAR gene organization inside the nucleus, the part of the cell that contains all its genes. Cells often bundle genes together spatially in the nucleus, said Dr. Tobias Ragoczy, one of the study’s authors and a staff scientist in the fundamental research laboratory of Dr. Mark Groudine, Fred Hutch executive vice president and deputy director.
“If you keep them all together, it’s easier to regulate them as a group,” Ragoczy said.
The nucleus contains “dead zones” and “hot zones” for gene activity — physical spots in the cellular compartment where genes are more or less likely to spur production of a protein. Clustering odorant receptor and TAAR genes in such dead zones within neurons makes sense, Ragoczy said, when you consider that only one out of the more than 1,000 genes needs to be activated in any given cell.
It was already known that the genes encoding odorant receptors cluster together in large silenced regions toward the center of the nucleus, so the team expected that TAAR genes might show a similar clustering. But instead they found “a striking difference,” Ragoczy said. TAAR genes are sequestered at the edge of the nucleus instead, a region not thought to be a dead zone for gene activity in olfactory neurons (although it is in many other cell types). They also saw that when a single TAAR gene is activated in a neuron, that gene moves away from its neighbors at the periphery into a different part of the nucleus.
It’s not clear why the cells would choose two different organizational systems to silence the different olfactory receptor genes, Ragoczy said, but their findings indicate that olfactory neurons have another layer of organization than scientists had previously appreciated.
Next up, Buck and her team want to uncover how TAAR-containing neurons trigger instinctive reactions by uncovering the signaling pathways in the brain triggered when a TAAR protein recognizes a scent. They’re already looking at neural pathways that govern other scent-driven behaviors in mice, behaviors like fear, mating and changes in appetite.
“What about those neural pathways causes attraction or aversion?” Buck asked. “Are signals from those TAARs going down different railroad tracks in the brain?”
There’s little known about how the brain prompts such behavior, Buck said. And that’s her team’s next challenge.
Rachel Tompa is a former staff writer at Fred Hutchinson Cancer Research Center. She has a Ph.D. in molecular biology from the University of California, San Francisco and a certificate in science writing from the University of California, Santa Cruz. Follow her on Twitter @Rachel_Tompa.
Are you interested in reprinting or republishing this story? Be our guest! We want to help connect people with the information they need. We just ask that you link back to the original article, preserve the author’s byline and refrain from making edits that alter the original context. Questions? Email us at [email protected]<|endoftext|>
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Cloud forcing (sometimes described as cloud radiative forcing or cloud radiative effect) is, in meteorology, the difference between the radiation budget components for average cloud conditions and cloud-free conditions. Much of the interest in cloud forcing relates to its role as a feedback process in the present period of global warming.
Measuring Cloud ForcingEdit
The following equation calculates this change in the radiation budget at the top of the atmosphere
The net cloud radiative effect can be decomposed into its longwave and shortwave components. This is because net radiation is absorbed solar minus the outgoing longwave radiation shown by the following equations
The first term on the right is the shortwave cloud effect (Qabs ) and the second is the longwave effect (OLR).
The shortwave cloud effect is calculated by the following equation
The longwave effect is calculated by the next following equation
Where σ is the Stefan–Boltzmann constant, T is the temperature at the given height, and F is the upward flux in clear conditions.
Putting all of these pieces together, the final equation becomes
Current Effects of Cloud ForcingEdit
All global climate models used for climate change projections include the effects of water vapor and cloud forcing. The models include the effects of clouds on both incoming (solar) and emitted (terrestrial) radiation.
Clouds increase the global reflection of solar radiation from 15% to 30%, reducing the amount of solar radiation absorbed by the Earth by about 44 W/m². This cooling is offset somewhat by the greenhouse effect of clouds which reduces the outgoing longwave radiation by about 31 W/m². Thus the net cloud forcing of the radiation budget is a loss of about 13 W/m². If the clouds were removed with all else remaining the same, the Earth would gain this last amount in net radiation and begin to warm up.
Without the inclusion of clouds, water vapor alone contributes 36% to 70% of the greenhouse effect on Earth. When water vapor and clouds are considered together, the contribution is 66% to 85%. The ranges come about because there are two ways to compute the influence of water vapor and clouds: the lower bounds are the reduction in the greenhouse effect if water vapor and clouds are removed from the atmosphere leaving all other greenhouse gases unchanged, while the upper bounds are the greenhouse effect introduced if water vapor and clouds are added to an atmosphere with no other greenhouse gases. The two values differ because of overlap in the absorption and emission by the various greenhouse gases. Trapping of the long-wave radiation due to the presence of clouds reduces the radiative forcing of the greenhouse gases compared to the clear-sky forcing. However, the magnitude of the effect due to clouds varies for different greenhouse gases. Relative to clear skies, clouds reduce the global mean radiative forcing due to CO2 by about 15%, that due to CH4 and N2O by about 20%, and that due to the halocarbons by up to 30%. Clouds remain one of the largest uncertainties in future projections of climate change by global climate models, owing to the physical complexity of cloud processes and the small scale of individual clouds relative to the size of the model computational grid.
- NASA (2016). "Clouds & Radiation Fact Sheet : Feature Articles". NASA. Retrieved 2017-05-29.
- Hartmann, Dennis L. (2016). Global Physical Climatology. Amsterdam: Elsevier. ISBN 978-0123285317.
- Intergovernmental Panel on Climate Change (1990). IPCC First Assessment Report.1990. UK: Cambridge University Press.table 3.1
- Schmidt, Gavin A. (2005-04-06). "Water vapour: feedback or forcing?". RealClimate. Retrieved 2008-01-14.
- Pinnock, S.; M.D. Hurley; K.P. Shine; T.J. Wallington; T.J. Smyth (1995). "Radiative forcing of climate by hydrochlorofluorocarbons and hydrofluorocarbons". J. Geophys. Res. 100 (D11): 23227–23238. Bibcode:1995JGR...10023227P. doi:10.1029/95JD02323.
- "Well-mixed Greenhouse Gases". Climate Change 2001: The Scientific Basis. Intergovernmental Panel on Climate Change. 2001. Retrieved 2008-01-14.
- Christidis, N.; M.D. Hurley; S. Pinnock; K.P. Shine; T.J. Wallington (1997). "Radiative forcing of climate change by CFC-11 and possible CFC replacements". J. Geophys. Res. 102 (D16): 19597–19609. Bibcode:1997JGR...10219597C. doi:10.1029/97JD01137.
- Myhre, G.; E.J. Highwood; K.P. Shine; F. Stordal (1998). "New estimates of radiative forcing due to well mixed greenhouse gases". Geophys. Res. Lett. 25 (14): 2715–2718. Bibcode:1998GeoRL..25.2715M. doi:10.1029/98GL01908.<|endoftext|>
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Question of Exercise 2
# Question A rectangular tank is 225 m by 162 m at the base. With what speed must water flow into it through an aperture 60 m by 45 cm that the level may be raised 20 cm in 5 hours?
Option 1 1458 m/hr
Option 2 4505 m/hr
Option 3 2745 m/hr
Option 4 5400 m/hr
Factorize x2+x-6
Solution:
Explanation:
On factoring the equation, we get,
x2+x-6
x2+3x-2x-6
x(x+3)-2(x+3)
(x+3)(x-2)
The simplified form of the equation x2+x-6 is (x+3)(x-2).
Factorize x^2-2x-8
Solution:
Explanation:
We have;
x2-2x-8
x2-4x+2x-8
x(x-4)+2(x-4)
(x-4) (x+2)
x=4,x=-2
Hence, we factorized the given expression as, x=4,x=-2
The opposite angles of a parallelogram are
are (3x - 2) and (x + 48) Find the measure of each angle of the parallelogram.
Solution:
Explanation:
Let the parallelogram be ABCD & the angles of parallelogram be <A,<B,<C & <D
From the given question, opposite angles of a parallelogram are (3x-2) & (x+48).
Let <A=3x-2 & <B=x+48
As we know that, “the opposite angles of a parallelogram are always equal”. Therefore, we can write;
(3x-2)=(x+48)
⇒3x-x=48+2
⇒2x=50
⇒x=50/2
⇒x=25
Substituting the value of x in 3x-2, we get;
3(25)-2
=75-2
=73º
⇒<A=<C=73
Finding the measure of other two angles:
We know that, the sum of adjacent angles of a parallelogram is equal to 180.
∴ <A+<B=180º
⇒73º+<B=180º
⇒<B=180º-73º
⇒<B=107º
∴ <D=107º
Therefore, the measure of each angle of the parallelogram is <A=73º,<B=107º,<C=73º,<D=107º.
Hence, we measured all the angles of parallelogram as;<A=73º,<B=107º,<C=73º,<D=107º.
In the given figure the value of x is
Solution:
Explanation:-
The value of x is 125º.
Estimate the value of square root
square root :√22
Solution:
Final Answer: The estimated square root of √22 is 4.690.<|endoftext|>
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The water level in the Great Lakes is rising quicker than predicted. The outcome of high water levels tends to be erosion.
Water levels tend to rise and fall due to seasonal weather patterns. However, experts indicate that this ebb and flow is occurring at an accelerated rate, causing water to eat away at the shoreline. Erosion is beginning to cause damage.
In fact, the water level in Lake Huron is higher than it usually is this time of the year. Due to high water levels in the Great Lakes, excess water from Lake Superior is spilling over into Lake Huron at an alarming rate.
Water levels in the Great Lakes are influenced by a lot of conditions. Heavy snow in the winter will raise the water levels. Heavy rains in the spring can often cause the lakes to flood the area. Climate changes can also affect weather patterns. It is difficult to pinpoint just one reason for the increased water levels in Lake Huron. It may be the result of a combination of changes.
The biggest fear with higher water levels is the erosion of the land. If high water levels continue, it is likely the shoreline will be eaten away. This is a concern for businesses and homes that are built along the water’s edge.<|endoftext|>
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In this quick tutorial you'll learn how to draw a Tilapia in 8 easy steps - great for kids and novice artists.
The images above represents how your finished drawing is going to look and the steps involved.
Below are the individual steps - you can click on each one for a High Resolution printable PDF version.
At the bottom you can read some interesting facts about the Tilapia.
Make sure you also check out any of the hundreds of drawing tutorials grouped by category.
How to Draw a Tilapia - Step-by-Step Tutorial
Step 1: Let's draw a Tilapia! It is a very common fish. Start with the mouth by drawing a rectangular shape with an opening in the back. Draw a small triangle under that for the bottom of the mouth. Tilapias two sets of jaws that work separately, which most fish do not.
Step 2: Draw the body by making two curved lines that almost form a large oval. Leave space at the back for the tail! The Tilapia is usually a very thin fish.
Step 3: For the tail fins, draw a few lines curving outward and connected them with a straight line. Now add a few horizontal lines inside the tail for detail.
Step 4: For the top fin, make a long teardrop shape that ends at a point. Add a few other small lines at an angle inside the fin for more detail.
Step 5: To add the bottom fin, make another similar teardrop shape, only smaller. Once again, add a few more lines inside for the fin folds.
Step 6: Draw the last bottom fin as shown. One last teardrop shape and a few final lines inside.
Step 7: For the eye, draw two ovals, one inside the other. Then, add a small curve from about the middle of the eye to below the mouth.
Step 8: Add one more fin on the inside, and several small curves to add the scale pattern.
Interesting Facts about the Tilapia
The tilapia is a type of fish that mainly lives in freshwater areas such as shallow streams, rivers, ponds, and lakes as well. These fish aren’t able to survive well in waters that get to temperatures below 21°C, so they have to stay in areas with warm water.
Did you know?
- Tilapias were called Nile Tilapia in times from ancient Egypt and were even depicted in hieroglyphs.
- In biblical times this fish was one of the main types of fish that was caught from the Sea of Galilee
- During the time of the biblical era they were referred to as musht or even more commonly, St. Peter’s fish
- These fish are very thin and have deep bodies.
- Due to the strange way that this fish has its lower pharyngeal bones combined into one large tooth, these fish are able to almost have two sets of jaws that they can work separately
- These fish are very effective eaters because of their “jaws” so they can eat and catch a diverse amount of things.
You wouldn’t want to have one of these fish in your own aquarium. The larger tilapias will dig up the bottom, eat the plants, and even fight with your other fish! However, these fish often do grow up in aquariums as s food source, because they have no problem growing in low qualities of water without having much room.<|endoftext|>
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A new study from the Westmead Institute for Medical Research has uncovered how serious fungal infections grow in humans by conserving phosphate, highlighting a possible target for treatment.
Cryptococcus neoformans is a potentially life-threatening invasive fungal disease that infects people with weakened immune systems, such as cancer patients and organ transplant recipients. The fungus needs phosphate to grow and sustain an infection in its host, as phosphate is essential for functions such as cell division.
Researchers found that, in environments with limited phosphate, Cryptococcus neoformans remodels lipids -- fatty acids in the cell membrane -- to release phosphate.
Lead researcher, associate professor Julie Djordjevic, said, "This is the first time that this strategy of conserving phosphate has been described in a human fungal pathogen. Fungi encounter phosphate starvation when they infect humans. However, fungal infections are quite clever, and have unique strategies to conserve phosphate when it becomes scarce. We found Cryptococcus neoformans conserves phosphate by activating a gene, BTA1, which encodes an enzyme that produces betaine lipids. Production of these betaine lipids allows the fungal pathogen to recycle phosphate from the lipids it normally produces, helping it to survive and spread. Fungi that were missing the BTA1 gene grew more slowly than cells with the gene and were less able to cause disease in animal infection models, indicating the essential role of phosphate conservation in the development of Cryptococcus neoformans infections."
Cryptococcus neoformans initially infects the lungs, where it can spread to the brain, causing the life-threatening condition, Cryptococcal meningitis. Invasive fungal diseases including those caused by Cryptococcus neoformans cause 1.6 million deaths worldwide each year.
Djordjevic said the high rates of mortality are, partly, due to a lack of new treatments. "The growing emergence of drug resistant strains of fungi, as well as a lack of effective existing treatments, means we need new therapies to combat invasive fungal diseases. Now that we understand how Cryptococcus neoformans conserves phosphate, we can investigate how we can prevent this process from occurring to stop the growth and spread of infection."
Source: Westmead Institute for Medical Research<|endoftext|>
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This activity will walk you through the sequencing of several short strings of single-stranded DNA (ssDNA). By understanding how sequencing-by-synthesis works at this small scale, it will be easier to picture what's happening with the billions of much longer strings that get sequenced in a real sequencing machine.
First, you will see an example to show you the process. There will be quick questions so you can make sure you understand how it works. Then, you will sequence a flow cell of unknown strings.<|endoftext|>
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# How do you evaluate cos^-1(cos((17 pi)/5))?
May 16, 2016
$\frac{17 \pi}{5}$
#### Explanation:
An inverse function might be single-valued or many-valued. But applying an inverse function and the function in succession, over an operand, returns the operand.
${O}^{- 1} O \left(c\right) = {O}^{- 1} \left(O \left(c\right)\right) = c$.
$\cos \left(\frac{17 \pi}{5}\right) = \cos \left(3 \pi + \left(\frac{2 \pi}{5}\right)\right) = - \cos \left(\frac{2 \pi}{5}\right) = - - \cos {72}^{o} = - 0.3090$.
You can see the basic difference between evaluating directly
${\cos}^{- 1} \left(- 0.3090\right) = \frac{3 \pi}{5} \mathmr{and} - \frac{3 \pi}{15} \mathmr{and} \frac{7 \pi}{5} \mathmr{and} - \left(7 \pi\right) 5 , \frac{17 \pi}{5} \mathmr{and} - \frac{17 \pi}{5}$..... and
${\cos}^{- 1} \cos \left(\frac{17 \pi}{5}\right) = {\cos}^{- 1} \left(\cos \left(\frac{17 \pi}{5}\right)\right) = \frac{1 7 \pi}{5}$.
For that matter, if the expression is
${\cos}^{- 1} \left(\cos \left(\frac{3 \pi}{5}\right)\right)$, the value is $\frac{3 \pi}{5}$.
Note that cos (+-(17/5)pi) = cos (+-(7/5)pi) = cos (+-(3/5)pi))= -0.3090
Cosine is negative, in 2nd and 3rd quadrants..<|endoftext|>
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Malware (short for “malicious software”) is a file or code, typically delivered over a network, that infects, explores, steals or conducts virtually any behavior an attacker wants. Though varied in type and capabilities, malware usually has one of the following objectives:
Malware is an inclusive term for all types of malicious software, such as:
Viruses – Programs that copy themselves throughout a computer or network. Viruses piggyback on existing programs and can only be activated when a user opens the program. At their worst, viruses can corrupt or delete data, use the user’s email to spread, or erase everything on a hard disk.
Worms – Self-replicating viruses that exploit security vulnerabilities to automatically spread themselves across computers and networks. Unlike many viruses, worms do not attach to existing programs or alter files. They typically go unnoticed until replication reaches a scale that consumes significant system resources or network bandwidth.
Trojans – Malware disguised in what appears to be legitimate software. Once activated, Trojans will conduct whatever action they have been programmed to carry out. Unlike viruses and worms, Trojans do not replicate or reproduce through infection. “Trojan” alludes to the mythological story of Greek soldiers hidden inside a wooden horse that was given to the enemy city of Troy.
Rootkits – Programs that provide privileged (root-level) access to a computer. Rootkits vary and hide themselves in the operating system.
Remote Administration Tools (RATs) – Software that allows a remote operator to control a system. These tools were originally built for legitimate use, but are now used by threat actors. RATs enable administrative control, allowing an attacker to do almost anything on an infected computer. They are difficult to detect, as they don’t typically show up in lists of running programs or tasks, and their actions are often mistaken for the actions of legitimate programs.
Botnets – Short for “robot network,” these are networks of infected computers under the control of single attacking parties using command-and-control servers. Botnets are highly versatile and adaptable, able to maintain resilience through redundant servers and by using infected computers to relay traffic. Botnets are often the armies behind today's distributed denial-of-service (DDoS) attacks.
Spyware – Malware that collects information about the usage of the infected computer and communicates it back to the attacker. The term includes botnets, adware, backdoor behavior, keyloggers, data theft and net-worms.
Polymorphic malware – Any of the above types of malware with the capacity to “morph” regularly, altering the appearance of the code while retaining the algorithm within. The alteration of the surface appearance of the software subverts detection via traditional virus signatures.
Learn how to use Palo Alto Networks® next-generation threat prevention features and WildFire® cloud-based threat analysis service to protect your network from all types of malware, both known and unknown.
More Threat Articles:<|endoftext|>
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# Test yourself now
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# Chapter 14: Constructions
A construction in Mathematics is an accurate drawing of angles and lines. You are only allowed to use a straightedge and a pair of compasses with a pencil in it when you do constructions.
1. A straightedge
A straightedge is an instrument made of plastic, metal or wood that has a straight edge. You use it to draw straight lines. It is a ruler without any markings on it, because when you do constructions you are not allowed to take any measurements.
Most of the time you will use your ruler when you do constructions and have to draw straight lines. The important thing to remember is that although there are markings on the ruler, you should not use them at all. You may only use the ruler to draw a straight line.
2. A pair of compasses is an instrument to draw circles. It can also be called "a compass ".
A compass has two legs:
• One leg has a sharp point so that you can put it firmly into position on a piece of paper or page in a notebook. When you draw a circle the sharp point will stay on the dot where you positioned it. The sharp point will not move.
• The other leg has space to insert a pencil. When you draw a circle, you move this leg around, and the pencil will draw the circle on the page.
The two legs of the compass are joined at the hinge. Make sure that the metal pin at the hinge is fitted tightly so that the two legs of the compass will not open too easily. Also make sure that your pencil is sharpened. A blunt pencil will lead to diagrams that are not accurate.
Do not confuse the compass that we use in mathematical constructions with the compass that we use to find direction.
All the constructions in this chapter are based on the properties of a circle.
In the diagram below, $BC$ is the diameter of the circle. $A$ is the centre of the circle. This means that $AB$ is a radius of the circle and $AC$ is also a radius of the circle. So, $AB$ is equal to $AC$.
When we do constructions we often do not draw a complete circle.
The solid lines drawn at $B$ and $C$ are called arcs. An arc is part of the circumference of a circle.
We will draw arcs in all the constructions that we are going to do. It is important to remember that if we draw two arcs while our compass stays in the same position, we are drawing two radiuses with equal lengths. Even if we do not draw a complete circle, and we only draw an arc at $B$ and at $C$, $AB$ is still a radius of the circle and $AC$ is still a radius of the circle. So, $AB$ is equal to $AC$.
arc An arc is a part of the circumference of a circle.
## 14.1 Perpendicular lines
Lines that are perpendicular intersect each other at right angles (at $90^{\circ}$). When we are asked to construct a perpendicular line, we will be given one of the lines, as well as a point. We then have to construct a line through the given point so that the new line is perpendicular to the given line are perpendicular.
intersect When two lines intersect, they cross each other.
perpendicular line A perpendicular line cuts another line at a right angle.
The point that is given can be on the line, or it can be a small distance away from the line.
In the first diagram, point $P$ is on line $AB$. In the second diagram, point $P$ is some distance away from line $AB$.
### Worked example 14.1: Constructing a perpendicular line from a point on the line
You are given line $AB$ with a point $P$ on the line.
Construct line $PQ$ so that $PQ$ is perpendicular to $AB$.
1. Step 1: Mark equal distances to both sides of point $P$ on line $AB$.
Put the compass on point $P$ (where the blue dot is shown).
Draw an arc to the left of point $P$. Then, while keeping the compass open by exactly the same amount, draw an arc to the right of point $P$. These arcs should intersect (cross) line $AB$.
2. Step 2: Draw an arc from the left.
Put the compass point on the arc at the left (where the orange dot is shown). Draw an arc so that it is above point $P$.
3. Step 3: Draw an arc from the right.
Keep the compass exactly the same width open as for Step 2. Put the compass point on the arc at the right (where the purple dot is shown). Draw an arc so that it intersects the arc from Step 2.
4. Step 4: Draw the perpendicular line.
Point $Q$ is where the two arcs from Step 2 and Step 3 are intersecting each other. Use a ruler and draw a line from point $P$ and through point $Q$.
5. Step 5: Check your construction.
Use a protractor to measure $Q\hat{P}A$ and $Q\hat{P}B$.
If each of them is equal to $90^{\circ}$, you have made an accurate construction.
When copying the lines in order to do constructions, the lines only need to be more or less the same length as given in the questions, and in more or less the same direction. It is a good idea to rotate your workbook if you have to do a construction involving a line that is not horizontal on the page in front of you. Never erase the construction lines after you have completed a construction.
### Exercise 14.1: Construct a perpendicular line from a point on the line
1. Draw line $CD$ as shown below. $P$ is a point on the line.
Construct $PQ$ perpendicular to line $CD$. Remember to check your work.
Follow the Steps as shown in Worked example 14.1.
If you have worked accurately, $Q \hat {P} C$ and $Q \hat{P}D$ will both be equal to $90^{\circ}$.
2. Draw line $EF$ as shown below. $P$ is a point on the line.
Construct $PQ$ perpendicular to line $EF$. Remember to check your work.
Follow the Steps as shown in Worked example 14.1.
If you have worked accurately, $Q\hat{P}E$ and $Q\hat{P}F$ will both be equal to $90^{\circ}$.
3. Draw line $MN$ as shown below. $P$ is a point on the line.
Construct $PQ$ perpendicular to line $MN$. Remember to check your work.
Follow the Steps as shown in Worked example 14.1.
If you have worked accurately, $Q\hat{P}M$ and $Q\hat{P}N$ will both be equal to $90^{\circ}$.
4. Draw line $ST$ as shown below. $P$ is a point on the line.
Construct $PQ$ perpendicular to line $ST$. Remember to check your work.
Follow the Steps as shown in Worked example 14.1.
If you have worked accurately, $Q\hat{P}S$ and $Q\hat{P}T$ will both be equal to $90^{\circ}$.
### Worked example 14.2: Constructing a perpendicular line from a point away from the line
You are given line $AB$ with a point $P$ away from the line.
Construct line $PQ$ so that $PQ$ is perpendicular to $AB$.
1. Step 1: Mark equal distances from point $P$ on line $AB$.
Put the point of the compass on point $P$ (where the blue dot is shown).
Draw an arc to the left of point $P$ on the line. While keeping the compass exactly the same width open, draw an arc on the line to the right of point $P$. These arcs should intersect line $AB$.
2. Step 2: Work from the left and draw an arc below line $AB$.
Put the compass point on the left arc of Step 1 (where the orange dot is shown). Draw an arc so that it is underneath point $P$.
3. Step 3: Work from the right and draw a second arc below line $AB$.
Keep the compass the same width open as for Step 2. Put the compass point on the arc to the right arc of Step 1 (where the purple dot is shown). Draw an arc so that it crosses the arc from Step 2.
4. Step 4: Draw the perpendicular line.
Point $Q$ is where the two arcs from Step 2 and Step 3 cross each other. Use a ruler and draw a line from point $P$ and through point $Q$.
5. Step 5: Check your construction.
Use a protractor and measure the angles where line $AB$ and line $PQ$ intersect. If each angle is equal to $90^{\circ}$, you have made an accurate construction.
Remember to rotate your workbook if you need to, and not to erase your construction lines. Also, always check your work.
### Exercise 14.2: Construct a perpendicular line from a point away from the line
1. Draw line $CD$ as shown below. Point $P$ is a point away from the line.
Construct $PQ$ perpendicular to line $CD$.
Follow the Steps as shown in Worked example 14.2.
If you have worked accurately, the two lines will intersect each other at an angle of $90^{\circ}$.
2. Draw line $EF$ as shown below. Point $P$ is a point away from the line.
Construct $PQ$ perpendicular to line $EF$.
Follow the Steps as shown in Worked example 14.2.
If you have worked accurately, the two lines will intersect each other at an angle of $90^{\circ}$.
3. Draw line $MN$ as shown below. Point $P$ is a point away from the line.
Construct $PQ$ perpendicular to line $MN$.
Follow the Steps as shown in Worked example 14.2.
If you have worked accurately, the two lines will intersect each other at an angle of $90^{\circ}$.
4. Draw line $ST$ as shown below. Point $P$ is a point away from the line.
Construct $PQ$ perpendicular to line $ST$.
Follow the Steps as shown in Worked example 14.2.
If you have worked accurately, the two lines will intersect each other at an angle of $90^{\circ}$.
## 14.2 Parallel lines
Parallel lines are lines that are always the same distance apart from each other. They never meet or cross each other. When we are asked to construct parallel lines, we will be given a line and a point. We then have to construct a line through the given point, and then another line that is parallel to the first line.
parallel lines Parallel lines are lines that always remain the same distance apart, and never meet or cross each other.
You will learn more about parallel lines in Chapter 15, but it will be easier to construct parallel lines if you know the following:
If two parallel lines are crossed by another line, then the angles marked in the diagram below will be equal to each other.
To construct a parallel line, you will be given the line shown in blue in the diagram, as well as the blue dot. The plan is to measure $\hat 1$ and then to construct $\hat 2$ so that the two angles are equal. If the two angles are equal, the lines will be parallel.
### Worked example 14.3: Constructing parallel lines
$P$ is a point away from line $AB$.
Construct line $PQ$ through point $P$ so that $PQ$ is parallel to $AB$.
1. Step 1: Use a ruler (or a straightedge) to draw a line to form an acute angle.
Draw a straight line through $P$ and that also goes through the line $AB$. This forms an angle where the two lines intersect.
2. Step 2: Draw an arc from the vertex of the acute angle.
Put the compass point on the orange dot and draw the arc that goes through both lines. Make sure that the arc is below point $P$.
3. Step 3: Draw the second arc from point $P$.
Keep the compass at the same width to draw the second arc. Put the compass point on point $P$ and draw the second arc above $P$. Make sure that the second arc goes lower than point $P$ on the right hand side.
4. Step 4: Use the compass to measure the distance between the two points where the first arc intersects the lines (the two purple dots).
Put the point of the compass where the first arc intersects the line $AB$ (at the purple point now labelled $R$). Put the pencil end of the compass where this arc intersects the line through $P$ (at the purple point now labelled $S$).
IMPORTANT: Do not draw a line between point $R$ and point $S$. You are only measuring the distance.
5. Step 5: Mark the same length on the second arc.
Keep the compass in exactly the same position. Put the compass point on point $T$, and draw an arc to find point $Q$. Point $Q$ will be where this new arc intersects the second orange arc. The distance between point $T$ and point $Q$ should be exactly the same as the distance between point $R$ and point $S$.
IMPORTANT: Do not draw a line between point $T$ and point $Q$. You are only measuring off the correct distance.
6. Step 6: Draw a line through point $P$ and point $Q$.
When you draw a line through point $P$ and point $Q$, line $PQ$ will be parallel to line $AB$.
7. Step 7: Check your construction.
Use a protractor to measure $\hat 1$ and $\hat 2$. If the two angles are equal to each other, your construction is accurate.
Remember, you can rotate your workbook to make it easier to work with a line that is not horizontal. Do not erase your construction lines after you have completed a construction, and always check your work by measuring the angles.
### Exercise 14.3: Construct a parallel line
1. Draw line $CD$ as shown below. Point $P$ is a point away from the line.
Construct $PQ$ parallel to line $CD$.
Follow the Steps as shown in Worked example 14.3.
If you have worked accurately, the two angles labelled $\hat{1}$ and $\hat{2}$ will be equal.
2. Draw line $EF$ as shown below. Point $P$ is a point away from the line.
Construct $PQ$ parallel to line $EF$.
Follow the Steps as shown in Worked example 14.3.
If you have worked accurately, the two angles labelled $\hat{1}$ and $\hat{2}$ will be equal.
3. Draw line $MN$ as shown below. Point $P$ is a point away from the line.
Construct $PQ$ parallel to line $MN$.
Follow the Steps as shown in Worked example 14.3.
If you have worked accurately, the two angles labelled $\hat{1}$ and $\hat{2}$ will be equal.
4. Draw line $ST$ as shown below. Point $P$ is a point away from the line.
Construct $PQ$ parallel to line $ST$.
Follow the Steps as shown in Worked example 14.3.
If you have worked accurately the two angles labelled $\hat{1}$ and $\hat{2}$ will be equal.
## 14.3 Line bisector
A line segment is part of a straight line. A line segment has a beginning and an end. When we are asked to bisect a line segment, we are asked to divide the line segment into two equal parts.
The bisector is the line segment that divides the given line segment into two equal parts.
bisect When we bisect a line segment or an angle, we divide the line segment or the angle exactly in half.
bisector A bisector is the line segment that divides the given line segment into two equal parts.
### Worked example 14.4: Bisecting a line segment
You are given line segment $AB$ as shown in the diagram below.
Bisect $AB$ and label the bisector $PQ$.
1. Step 1: Draw the first set of two arcs.
Open the compass so that the distance between the two points is more than half of the length of $AB$. Test this on line $AB$.
Put the compass on point $A$ (the blue dot) and draw two arcs as shown. One arc is above $AB$ and one arc is below $AB$.
2. Step 2: Draw the second set of two arcs.
Keep the compass open at exactly the same width as for Step 1.
Put the compass on point $B$ (the orange dot) and draw two arcs as shown. One arc is above $AB$ and one arc is below $AB$.
3. Step 3: Draw the bisector.
Label the points where the arcs of intersect as point $P$ and point $Q$.
Use a ruler and draw a line through point $P$ and point $Q$. $PQ$ is the bisector of line segment $AB$ at point $R$.
4. Step 4: Check your work.
Measure the lengths of the two parts of line segment $AB$. If $AR$ is equal to $RB$, your construction is accurate.
Take note that if you measure the angles at $R$, you will find they are $90^{\circ}$. We say that the bisector is perpendicular to the original line segment. The symbol for perpendicular is $\perp$, so we write: $PQ \perp AB$.
When we do a construction to bisect a line, the bisector is always perpendicular to the line. This means we can call the bisector a "perpendicular line bisector ".
If you find it easier, you may draw two bigger arcs instead of four smaller arcs when you construct the bisector of a line segment, as shown in the diagram below.
### Exercise 14.4: Bisect a line segment
1. Draw line segment $CD$ as shown below.
Bisect the line segment and label the bisector $PQ$.
Follow the Steps as shown in Worked example 14.4.
If you have worked accurately, the length of line segment $CR$ will be the same as the length of line segment $RD$.
2. Draw line segment $EF$ as shown below.
Bisect the line segment and label the bisector $PQ$.
Follow the Steps as shown in Worked example 14.4.
If you have worked accurately, the length of line segment $ER$ will be the same as the length of line segment $RF$.
3. Draw line segment $MN$ as shown below.
Bisect the line segment and label the bisector $PQ$. Remember to check your work.
Follow the Steps as shown in Worked example 14.4.
If you have worked accurately, the length of line segment $MR$ will be the same as the length of line segment $RN$.
4. Draw line segment $ST$ as shown below.
Bisect the line segment and label the bisector $PQ$.
Follow the Steps as shown in Worked example 14.4.
If you have worked accurately, the length of line segment $SR$ will be the same as the length of line segment $RT$.
## 14.4 Sixty-degree angles
Here is a fun activity for you to do: Use a compass and draw a circle. Keep the position of the legs of the compass the same for the whole activity.
Make a dot anywhere on the circumference of the circle.
Put the compass on the dot and start to make small arcs on the circumference of the circle:
You will find that you can fit in exactly six of these small arcs:
You know that a full turn (or revolution) is equal to $360^{\circ}$, so if we join the centre of the circle to each one of the six small arcs, we form six angles of $60^{\circ}$ each.
If you are asked to construct an angle of $60^{\circ}$, you simply do a part of the fun activity that you have done above.
### Worked example 14.5: Constructing an angle of 60 degrees
You are given a line segment $BD$ as shown below. Construct $AB$ so that $C$ is a point on $AB$ and $A\hat{B}C$ = $60^{\circ}$.
1. Step 1: Draw the first big arc.
Put the compass point at point $B$ (on the blue dot) and draw a big arc. Label point $C$ where the arc crosses line segment $BD$. Keep the compass at the same width for Step 2.
2. Step 2: Draw a small arc that crosses the first big arc.
Keep the compass at the same width for Step 1. Put the compass point at point $C$ (on the orange dot) and make an arc to cross the big arc that you drew in Step 1.
Label the point of intersection $A$.
3. Step 3: Complete the construction.
Use a ruler and draw a line segment from point $B$ through point $A$.
4. Step 4: Check your work.
Use a protractor to measure $A\hat{B}C$. The angle should be equal to $60^{\circ}$.
When you construct an angle of $60^{\circ}$ you are actually constructing a triangle with three equal sides. In an equilateral triangle all three sides have equal lengths and each of the angles are equal to $60^{\circ}$.
### Exercise 14.5: Construct an angle of 60 degrees
1. Draw line segment $CD$ as shown below.
Construct $FC$ so that $F\hat{C}D$ = $60^{\circ}$.
Follow the Steps as shown in Worked example 14.5.
If you have worked accurately, $F\hat{C}D$ will be equal to $60^{\circ}$.
2. Draw line segment $EF$ as shown below.
Construct $DE$ so that $D\hat{E}F$ = $60^{\circ}$.
Follow the Steps as shown in Worked example 14.5.
If you have worked accurately, $D\hat{E}F$ will be equal to $60^{\circ}$.
3. Draw line segment $MN$ as shown below.
Construct $LM$ so that $L\hat{M}N$ = $60^{\circ}$.
Follow the Steps as shown in Worked example 14.5.
If you have worked accurately, $L\hat{M}N$ will be equal to $60^{\circ}$.
4. Draw line segment $ST$ as shown below.
Construct $PS$ so that $P\hat{S}T$ = $60^{\circ}$.
Follow the Steps as shown in Worked example 14.5.
If you have worked accurately, $P\hat{S}T$ will be equal to $60^{\circ}$.
## 14.5 Right angle
A right angle is an angle of 90 degrees. In real life we often see right angles. For example, the corners of your workbook are right angles, the corners of a door are right angles, and the corners of most tables are right angles.
When you are asked to construct an angle of $90^{\circ}$, you will again make use of the fun activity in which you made six small arcs on the circumference of a circle.
In this construction you will make only two arcs: you will mark where to form an angle of $60^{\circ}$ and where to form an angle of $120^{\circ}$.
An angle of $90^{\circ}$ is exactly halfway between $60^{\circ}$ and $120^{\circ}$, so you will construct the one arm of the right angle exactly in the middle of the arcs for $60^{\circ}$ and $120^{\circ}$.
### Worked example 14.6: Constructing a right angle
You are given a line segment $BD$ as shown below. Construct $AB$ so that $A\hat{B}C$ = $90^{\circ}$.
1. Step 1: Draw the first big arc.
Put the compass point at point $B$ (on the blue dot) and draw a big arc. Label point $C$ where the arc intersects line segment $BD$.
2. Step 2: Draw the first small arc for an angle of $60^{\circ}$.
Keep the compass at the same width for the following two arcs (Step 2 and Step 3).
Put the compass on point $C$ and make an arc on the big arc that was drawn in Step 1.
Label the point of intersection $P$. (Point $P$ shows the mark for an angle of $60^{\circ}$.)
3. Step 3: Draw the second small arc for an angle of $120^{\circ}$.
The compass should still be at exactly the same width as for the first big arc.
Put the compass on point $P$ and make another arc on the big arc that was drawn in Step 1.
Label the point of intersection point $Q$. (Point $Q$ shows the mark for an angle of $120^{\circ}$.)
4. Step 4: Draw the first arc to find the right angle.
We want to find the line for an angle of $90^{\circ}$ and we know that this line should be exactly in the middle between the arcs for the angles of $60^{\circ}$ and $120^{\circ}$.
The compass can now be less open, but it should stay at the same width for these next two arcs (Step 4 and Step 5).
Put the compass on point $P$ and make an arc as shown below.
5. Step 5: Draw the second arc to find the right angle.
Keep the compass at the same width as for Step 4.
Put the compass on point $Q$ and draw an arc. Label the point of intersection between the two arcs $A$.
6. Step 6: Complete the construction.
Use a ruler and draw a line segment from point $B$ and through point $A$.
7. Step 7: Check your work.
Use a protractor to measure $A\hat{B}C$. The angle should be equal to $90^{\circ}$.
### Exercise 14.6: Construct a right angle
1. Draw line segment $CD$ as shown below.
Construct $FC$ so that $F\hat{C}D$ = $90^{\circ}$.
Follow the Steps as shown in Worked example 14.6.
If you have worked accurately, $F\hat{C}D$ will be equal to $90^{\circ}$.
2. Draw line segment $EF$ as shown below.
Construct $DE$ so that $D\hat{E}F$ = $90^{\circ}$.
Follow the Steps as shown in Worked example 14.6.
If you have worked accurately, $D\hat{E}F$ will be equal to $90^{\circ}$.
3. Draw line segment $MN$ as shown below.
Construct $LM$ so that $L\hat{M}N$ = $90^{\circ}$.
Follow the Steps as shown in Worked example 14.6.
If you have worked accurately, $L\hat{M}N$ will be equal to $90^{\circ}$.
4. Draw line segment $ST$ as shown below.
Construct $PS$ so that $P\hat{S}T$ = $90^{\circ}$.
Follow the Steps as shown in Worked example 14.6.
If you have worked accurately, $P\hat{S}T$ will be equal to $90^{\circ}$.
## 14.6 Summary
• A construction in Mathematics is an accurate drawing of angles and lines.
• You are only allowed to use a straightedge and a pair of compasses with a pencil in it when you do a construction.
• You may use a ruler if you do not have a straightedge, but do not use the markings on the ruler.
• You will draw arcs in all the constructions. An arc is part of the circumference of a circle.
• You have to practise the following constructions:
• Construct a perpendicular line from a point on the line
• Construct a perpendicular line from a point away from the line
• Construct a parallel line through a given point
• Bisect a line segment
• Construct an angle of 60 degrees
• Construct a right angle (an angle of 90 degrees)
• Never erase the construction lines after you have completed a construction.<|endoftext|>
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There are certainly many practical advantages of learning a second language, such as being able to communicate with people from different cultures in their native tongue. It also serves to open up career, academic and relationship opportunities that may otherwise have been closed.
But increasingly it has been recognized that learning a second language at an early age may also give kids an edge academically. Not just as a skill to be put into practise, but it actually works to enhance the learning and cognitive ability of the child in many not so obvious ways.
It’s Easier to Learn Languages at an Early Age
Humans are social beings and use language as a means of communicating simple and complex ideas with each other; a key strategy of our survival as a species. It’s imperative that a child learns to understand the language that they hear and they do it remarkably quickly in comparison to adults.
In fact, the ability to acquire a new language is greatly excelled until children reach between 8 and 12 months old in which they start to hone in on the sounds of their native language and become less attuned to the sounds of other languages.
As we age, our ability for learning a second language greatly reduces following puberty for the reason that by this age, we have a bias towards the tones and speech elements of our first language known as phenomes. It’s much harder at this point to overcome these biases.
This is not to say that there is a point of no return, and no matter our age we can all benefit from learning a new language. Even if we are not looking to increase academic performance, we can still use it to connect with others.
The Links: Language Learning and Cognitive Performance
So, why exactly should you commit the effort to exposing your child to an additional language when they are young? As we touched upon above, there are many cognitive benefits to learning a second language. Overall, learning a second language improves academic performance and enhances learning and there is plenty of research to demonstrate this.
Exercises the Brain to Improve Memory
People often compare your brain to a muscle and while it is not actually a muscle it does perform better the more it is used. When learning a language you are memorizing the rules and the phonetics of that language and so to put it simply: you are exercising your brain.
If a child is learning multiple languages, there are more rules to remember, more variations of how to reference an object, a verb or a noun and so therefore, they are using their brain to remember these rules more than if they were learning only one language.
Jeffrey D. Karpicke, PhD of the American Psychological Association outlines how language relates to the use of memory and how memory can then be improved the more it is exercised which then goes on to benefit the ability to learn more in other subjects.
Learning Second Language Improves Test Scores
It is shown by Armstrong, P. W., & Rogers, J. D. that children who were taking 30 minute lessons in Spanish three times a week for one semester scored higher on standardized mathematic and language tests than children who did not take the lessons.
This supports the idea that polyglots are also better able to perceive relevant information and disregard the irrelevant than their monolingual counterparts.
Tonal Language and Learning Music
The Chinese languages of Mandarin and Cantonese are examples of tonal language in that they incorporate the pitch and certain inflections in the voice to convey different meanings. For example a variation on the way that the word “Ma” is pronounced can dictate its meaning as either being “mother” or “horse” – not a wise idea to get them confused!
Research has recognized that speakers of tonal language are better able to identify pitch and tonal changes giving them an edge when learning music. And just as there are benefits for learning a second language, there are many benefits of playing a musical instrument that interestingly overlap between the two.
Is this because music is considered a language of its own and utilizes similar thought patterns and cognitive functions?
Supports Native Language Learning
It was considered in the past that having a child learn an additional language at the same time as learning their native language would be confusing for them and generally detrimental to their learning.
However, research has indicated that children who learn two languages in early childhood effectively become native speakers of both the languages at once. This helps them to understand the inner workings and the rules of each language, allowing them to draw comparisons and improve comprehension of both simultaneously.
Potentially Guards against Certain types of Dementia
There are different types of dementia, including Alzheimer’s disease. It is described as a group of symptoms which contribute to the decline in cognitive ability which has an impact on daily life. These include impairment in memory, communication, perception and reasoning.
There is evidence to suggest that speaking two languages can delay the progress of three types of dementia by an average of 4 and half years. Alissa Sauer of Alzheimers.net states that it is not yet known why, but supposes that it relates to the repeated use of a specific part of the brain that is responsible for language.
Practical Benefits of Learning Additional Languages
Since, as we have discussed, children are better able to take on additional languages when they are young, there is no doubt that childhood is the best time to teach them. Being bilingual from a young age is a huge advantage that they can reap the rewards from indefinitely.
Greater Career and Academic Opportunities
Employers are attracted to prospective employees who are able to speak multiple languages and it looks fantastic on a resume. Not only does it mean you can communicate with customers and clients from other countries opening up the possibility of higher paying roles but it shows your commitment and ability to learn and understand.
When considering what colleges are looking for in high school students, having a second language under your belt is a factor that impresses them for similar reasons that it would impress an employer. It also opens up the potential to study abroad and be accepted onto international internships, allowing your child to network with peers in their field whilst studying abroad.
Networking and familiarity abroad further opens up opportunities for securing a job in another country too giving them greater flexibility with how they can use their education to secure their future!
Increased Cultural Awareness
Language and culture go hand in hand. A language and the development of languages over time is a reflection of the history of a culture and can demonstrate how the world is perceived by people of a particular location.
It is reasonable to say that the language we learn as our native tongue can have an influence on our thought patterns, (such as we discussed earlier that tonal languages can increase the recognition of pitch and tone in music) does this influence have an effect on the evolution of culture?
Questions like that can be answered by exploration, leaning multiple languages lets us engage in different cultures, communicate with people who come from distinct cultural backgrounds and share an appreciation for how language and culture relate.
It’s a Huge Achievement!
Having learned to speak a new language, whether you are a young child learning to speak for the first time or an adult planning on visiting a new land, it is a huge achievement. And having put all the effort and work into learning this new skill you would be right to feel proud of yourself and confident in your abilities.
For a school age child, this kind of confidence is a great motivator in their academic life. Knowing that they have the capabilities to learn more and build new skills boosts their propensity to carry on learning, to learn different things and to do well at them so that they can be proud in knowing that they are on their way to a successful future.<|endoftext|>
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# Subtraction of Like Algebraic Terms
A mathematical operation of subtracting two like algebraic terms is called the subtraction of like algebraic terms.
## Introduction
In algebra, two like algebraic terms are connected by a minus sign to find the difference between them mathematically. In fact, the like algebraic terms have a common literal coefficient and it is taken common from both the terms to perform the subtraction successfully.
$2x^2y$ and $5x^2y$ are two two like algebraic terms. Take, $2x^2y$ is subtracted from $5x^2y$ and the subtraction can be done in three simple steps.
#### First step
Write $5x^2y$ first and then $2x^2y$ in a row but display a minus sign between them to represent the subtraction.
$5x^2y-2x^2y$
#### Second step
Take the literal coefficient common from both the terms.
$\implies$ $5x^2y-2x^2y$ $\,=\,$ ${(5-2)}x^2y$
#### Third step
Now, find the subtraction of the numbers and multiply the difference by their common literal coefficient.
$\,\,\, \therefore \,\,\,\,\,\,$ $5x^2y-2x^2y$ $\,=\,$ $3x^2y$
It can be observed that the difference of any two like algebraic terms is also a like algebraic term. In this way, the subtraction of any two like algebraic terms can be performed in algebra in three simple steps.
### Examples
For better understanding the subtraction of algebraic terms, obverse the following examples.
$(1)\,\,\,\,\,\,$ $7a-5a$ $\,=\,$ $(7-5)a$ $\,=\,$ $2a$
$(2)\,\,\,\,\,\,$ $2bc-10bc$ $\,=\,$ $(2-10)bc$ $\,=\,$ $-8bc$
$(3)\,\,\,\,\,\,$ $3c^2-2c^2$ $\,=\,$ $(3-2)c^2$ $\,=\,$ $c^2$
$(4)\,\,\,\,\,\,$ $17d^3e^2f-23d^3e^2f$ $\,=\,$ $(17-23)d^3e^2f$ $\,=\,$ $-6d^3e^2f$
$(5)\,\,\,\,\,\,$ $5ghi-ghi$ $\,=\,$ $(5-1)ghi$ $\,=\,$ $4ghi$
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How do I calculate what size basin I need for a pondless waterfall? | Decorative Ponds & Water Gardens Q & A
Q: How do I calculate what size basin I need for a pondless waterfall?
Robert – Waverly, TN
A: With the help of a few friends, installing a pondless waterfall is an easy way to upgrade your land- or waterscape. If you are planning to design and build your own feature, here’s how to do the math to determine your basin size.
Step 1: Calculate total amount of water in your stream
Sound tricky? It’s not. To calculate the water that’s in motion in your stream, first measure your stream’s length, width and depth. Then plug those numbers into this equation: L x W x (0.25 x D of stream, generally 1 to 2 inches) x 7.48.
Here’s an example. If your stream is 25 feet long, 2 feet wide and 1 inch deep, the equation would look like this: 25 x 2 x (0.25 x 1) x 7.48 = 93.5 gallons.
Step 2: Calculate amount of water your basin needs to hold
As a general rule, you will need your basin to hold 2½ times the amount of water in your stream. To find out how much water your basin will need to hold, multiply the gallons of water in your stream by 2.5. In the above example, it would be 93.5 x 2.5 = 233.75 – so your basin would need to hold 233.75 gallons of water.
Step 3: Determine the size basin needed to hold the water
Here’s where things get a little tricky. You’ll likely use boulders, stones and other décor in your design. Well all that rockwork will take up space for water, and so you’ll need to go with a larger vessel.
Try this quick equation that will help you determine what size basin you’ll need with boulders: L x W x D x 2.2 = gallons of water the basin will need to hold. To give you an idea, a 5 foot by 10 foot basin that’s 3 feet deep and contains stone will hold 330 gallons of water.
A space- and cost-saving option is to use a basin matrix in place of some of the boulders. A basin matrix is a strong, 27-by-16-by-17½-inch hollow box that’s perforated with half-inch holes to allow water to collect and pass through. You can stack them together and landscape right over them. Each basin matrix holds 31.5 gallons, adding valuable water volume without increasing the size of the basin. Don’t forget to add a pump vault for easy access to your pump.
Go with a Kit
If you aren’t trying to customize a large waterfall, consider investing in a pond free waterfall kit, like PondBuilder™ Cascading Falls Pondless Kits. They contain all components you’ll need – from underlayment and liner to a waterfall box, pump, basin matrix and plumbing – with predetermined dimensions to help you create your water feature.
Pond Talk: What’s the largest size boulder you’ve installed in your water- or landscape?<|endoftext|>
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# The Ordinal Numbers In Spanish
With ordinal numbers, it is possible to count any number of sets. They can also be used to broaden ordinal numbers.But before you are able to use them, you must comprehend what they are and how they operate.
## 1st
The ordinal number is among the foundational ideas in mathematics. It is a number that identifies the position of an object within a list. Ordinarily, the ordinal numbers fall between one and twenty. While ordinal numbers can serve many functions, they’re most commonly used to show the order of items within the list.
You can use charts, words and numbers to depict ordinal numbers. They are also able to specify how a group of pieces are placed.
The majority of ordinal numbers fall into one of two categories. Transfinite ordinals can be depicted using lowercase Greek letters while finite ordinals are represented using Arabic numbers.
Based on the axioms that govern choice, each set must include at least one ordinal or two. For instance, the highest possible grade would be awarded to the class’s initial member. The contest’s runner-up was the student with a highest score.
## Combinational ordinal numbers
Multidigit numbers are called compound ordinal number. These numbers are created by multiplying an ordinal by its last number. These numbers are typically employed for ranking and dating. They don’t have a distinctive end like cardinal numbers.
Ordinal numbers identify the order of elements that are found in the collection. These numbers can also be used to identify items in collections. Ordinal numbers come in regular and suppletive forms.
Regular ordinals are made by prefixing the cardinal number by a -u suffix. The number is next typed as a word, and then a hyphen is placed after it. There are additional suffixes available.For example, the suffix “-nd” is used for numerals that end with 2, while “-th” is used to refer to numbers that end in 9 and 4.
Suppletive ordinals are made by affixing words with the suffix -u, -ie. This suffix, which is used for counting is broader than the conventional one.
## Ordinal limit
Limits on ordinal numbers are ordinal numbers that aren’t zero. Limit ordinal numeric numeric numbers have one disadvantage, they do not have a maximum element. These can be formed by joining sets of nonempty elements with no element that is larger than.
Limited ordinal numbers are employed in transfinite definitions for the concept of recursion. Each infinite number of cardinals, according to the von Neumann model, is also an ordinal limit.
A limit-ordered ordinal equals the sum of the other ordinals beneath. Limit ordinal numbers can be enumerated using arithmetic, but they are also able to be expressed in a series of natural numbers.
The ordinal numbers are used to organize the information. They provide an explanation of the numerical location of objects. These numbers are frequently used in set theory or math. While they share a common design, they aren’t considered to be in the same class as natural numbers.
In the von Neumann model, a well-ordered collection is used. Consider that fyyfy is an element of g’, a subfunction of a function defined as a single operation. If fy only has one subfunction (ii) then G’ meets the requirements.
The Church-Kleene oral is an limit ordinal in a similar manner. Limit ordinals are properly-ordered collection of smaller or less ordinals. It has a nonzero ordinal.
## Stories that make use of normal numbers as examples
Ordinal numbers are commonly used to show the hierarchy of entities and objects. They are crucial to organize, count, ranking, and other reasons. They are able to show the position of objects in addition to giving the order of things.
The ordinal number is generally indicated by the letter “th”. Sometimes, however, the letter “nd” could be substituted. Titles of books typically contain ordinal numerals as well.
While ordinal numbers are commonly used in list format they can still be expressed in words. They can also be found in the form of numbers or acronyms. Comparatively, these numbers are much simpler to comprehend than the cardinal ones.
Ordinal numbers come in three distinct flavors. Through practice, games and other exercises, you may learn more about the different kinds of numbers. You can increase your math skills by understanding more about the basics of them. You can improve your math abilities by doing coloring activities. Check your work with a handy mark sheet.<|endoftext|>
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To close the literacy and numeracy skills gap between immigrants and Indigenous people and the overall population, more education and training is needed.
Good literacy and numeracy skills will help everyone prepare for the rapidly evolving needs of the modern workplace. These information processing skills are essential to allow workers to perform their jobs, to help them acquire job-specific skills, and to improve their earnings and employment opportunities.
However, international surveys of adults’ levels of literacy and numeracy skills show that Canada’s working-age population has experienced a decline in these skills, even though more Canadians have obtained a post-secondary education over time (see table 1). Some groups, such as immigrants and Indigenous people, have lower average literacy and numeracy skills than the overall population. (In this article, “skills” refers to literacy and numeracy skills unless specified otherwise.)
People born in Canada
Comparing results of the surveys conducted by the OECD in 2003 and 2012, my C.D Howe Institute study shows that both aging and generational differences contribute to the literacy and numeracy skills declines among the Canadian-born population. People get lower test scores as they get older; generational differences are the variations in results among groups of individuals of the same age (e.g., those aged 25 to 34) as measured at different points in time.
The rate of erosion of skills accelerates with age, so with an increase in the proportion of older people in Canada’s population, the average performance has gone down. Age-related skills decline may be caused by people not continuing education or training or not applying existing skills frequently enough. This highlights the importance of obtaining the highest possible skills as early as possible, since skills start declining once people leave school.
In these surveys, recent generations of Canadians had lower scores in literacy and numeracy in 2012 than in 2003, regardless of education level. Although skills declined for all education levels, the drops were substantially larger among those without a university degree. Differing education quality and lower admission requirements for post-secondary education are potential explanations for the lower skills levels of equally educated adults over generations. In addition, the aging effect and generational differences are more profound among low achievers and less educated individuals.
Immigrants come to Canada with various job-specific skills. High literacy skills make immigrants able to use their full skill sets and accelerate their labour market integration. However, there is a large gap in literacy skills between immigrants and non-immigrants, even though a greater proportion of immigrants have post-secondary education.
In a C.D. Howe Institute study on immigrants, I show that language background is a major factor that affects skills gaps between immigrants and non-immigrants. Literacy tests are administered in the official language(s) of the host country. Immigrants to Canada without English or French as a mother tongue would naturally be expected to have lower scores on literacy exams than those who have a greater background in any of these languages.
Another important factor that affects the skills gap between immigrants and non-immigrants is education. Higher levels of education translate into greater skills for both immigrants and non-immigrants, but literacy levels for immigrants across all levels of education, including university-educated immigrants, are persistently lower than those of non-immigrants. However, immigrants who obtained their highest level of education in Canada performed better on literacy tests than those who received degrees in all other countries, regardless of their language background.
There are also notable variations in the literacy gap between immigrants and non-immigrants by immigration program. The literacy gap is much wider for refugees and family-class immigrants, where language and education are not selection factors for immigration.
The poor performance of Indigenous people relative to non-Indigenous Canadians in literacy and numeracy tests (table 1) is not surprising, since according to the 2012 survey only 10 percent of Indigenous people in Canada hold a university degree, compared with 22 percent of non-Indigenous Canadians. Among Indigenous groups, Inuit have the worst performance; only 4 percent of Inuit have a university education.
Skills gaps between off-reserve Indigenous people and non-Indigenous Canadians are wide among those without a high school education. Although the gap in literacy scores between the Indigenous and non-Indigenous populations shrinks with more education, both First Nations and Inuit people experience large skills gaps at every education level.
In work I did with Colin Busby, we showed that differences in education levels and the characteristics of the home environment are able to explain the largest share of the Indigenous skills gap. This is particularly true for Métis.
Since education and training are central to improving skills outcomes for all population groups, they require special policy attention. However, training needs to be tailored to the particular needs of each targeted population.
Ensuring people complete each level of their education with the highest possible skills level, by focusing on education quality, is the best strategy to limit age-related skills decline. Other potential policies include encouraging active learning and offering adult training opportunities for those at higher risk of skills depreciation. Policy-makers also need to review their targeted on-the-job training programs to ensure their effectiveness in slowing the skills decline as people age.
To reduce skills gaps for immigrants, in addition to education, the focus should be on language ability and post-immigration language training, especially for newcomers.
For Indigenous people, attention should be given to improving skills development during primary and secondary education. To do this, more data should be collected. Well-designed training opportunities should also be provided for both unemployed and employed Indigenous people since they experience a greater unemployment rate.
These policy improvements will help develop a more skilled workforce and would drive broader prosperity and economic growth with positive social impacts.
This article is part of the Preparing Citizens for the Future of Work special feature.
Do you have something to say about the article you just read? Be part of the Policy Options discussion, and send in your own submission. Here is a link on how to do it. | Souhaitez-vous réagir à cet article ? Joignez-vous aux débats d’Options politiques et soumettez-nous votre texte en suivant ces directives.<|endoftext|>
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Pablo Picasso: Art primary resource
Explore the life and works of artist Pablo Picasso
This primary resource introduces children to artist Pablo Picasso. Discover the artist’s development over his lifetime. By what age was he a better painter than his father? What was his most famous work? When does Malaga hold a festival in Picasso’s honour?
Pupils will learn about Pablo Picasso’s life and works, and the influence they have had on modern day artistic styles, such as cubism, in our National Geographic Kids’ Art primary resource sheet.
The teaching resource can be used in study group tasks for looking at the significance and influence of Picasso’s works. It can also be used as a printed handout for each pupil to review and annotate, or for display on the interactive whiteboard using the images included in the resource to open a class discussion.
Activity: Ask children to use the information in the resource sheet to make their own Picasso-style portrait. Pupils could be shown further examples of Picasso’s works and discuss which pieces they like, or which pieces they don’t, giving reasons for their answers. They could compare the work of Picasso with other well-known artists (e.g. Leonardo da Vinci), whose style do they find most interesting? Why?
N.B. The following information for mapping the resource documents to the school curriculum is specifically tailored to the English National Curriculum and Scottish Curriculum for Excellence. We are currently working to bring specifically tailored curriculum resource links for our other territories; including South Africa, Australia and New Zealand. If you have any queries about our upcoming curriculum resource links, please email: [email protected]
This Pablo Picasso primary resource assists with teaching the following Key Stage 1 Art objective from the National Curriculum:
Pupils should be taught:
- about the work of a range of artists, craft makers and designers, describing the differences and similarities between different practices and disciplines, and making links to their own work.
National Curriculum Key Stage 2 Art objectives:
Pupils should be taught:
- to create sketch books to record their observations and use them to review and revisit ideas
- about great artists, architects and designers in history.
This Pablo Picasso Art primary resource assists with teaching the following Expressive arts First – Third level objectives from the Scottish Curriculum for Excellence:
- I can respond to the work of artists and designers by discussing my thoughts and feelings. I can give and accept constructive comment on my own and others’ work.
Scottish Curriculum for Excellence Fourth level Expressive arts objectives:
- I can analyse art and design techniques, processes and concepts, make informed judgements and express considered opinions on my own and others’ work
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# Ex.5.2 Q6 Lines and Angles - NCERT Maths Class 7
Go back to 'Ex.5.2'
## Question
In the given figures below, decide whether $$l$$ is parallel to $$m.$$
Video Solution
Lines & Angles
Ex 5.2 | Question 6
## Text Solution
(i) Reasoning:
Let’s visually model this problem. There is one operation that can be done check whether interior angles are supplementary or not. According to this model, the result sum of $$126^\circ + 44^\circ$$ is $$170^\circ.$$ Now, it’s a matter of finding $$l$$ is parallel to $$m$$ or not
Steps:
(i) $$126^\circ + 44^\circ = 170^\circ$$
As the sum of interior angles on the same side of transversal $$n$$ is not $$180^\circ.$$
Therefore, $$l$$ is not parallel to $$m.$$
(ii) Reasoning:
Let’s visually model this problem. There are two operations that can be done in a sequence. First find the value of $$x$$ and then check it is equal to its corresponding angle or not. According to this model, the resultant value of $$x$$ is not equal to its corresponding angle. Now, it’s a matter of finding $$l$$ is parallel to $$m$$ or not
(ii) Steps:
\begin{align}\angle x + 75^\circ &= 180^\circ \text{(Linear pair)}\\\angle x &= 180^\circ - 75^\circ\\\angle x &= 105^\circ\end{align}
For $$l$$ and $$m$$ to be parallel measure of their corresponding angles should be equal but here the measure of $$\angle x$$ is $$105^\circ$$ and its corresponding angle is $$75^\circ.$$
Therefore, the lines $$l$$ and $$m$$ are not parallel.
(iii) Reasoning:
Let’s visually model this problem. There are two operations that can be done in a sequence. First find the value of $$x$$ and then check it is equal to its corresponding angle or not. According to this model, the resultant value of $$x$$ is not equal to its corresponding angle. Now, it’s a matter of finding $$l$$ is parallel to $$m$$ or not
(iii) Steps:
\begin{align}\angle y&=57^\circ \begin{bmatrix}\text{Vertically opposite}\\\text {angles}\end{bmatrix}\\\angle x + 123^\circ &= 180^\circ\text{ (Linear pair)}\\\angle x& = 180^\circ - 123^\circ\\\angle x &= 57^\circ\end{align}
Here, the measure of corresponding angles are equal i.e $$57^\circ.$$
Therefore, lines l and m are parallel to each other.
(iv) Reasoning:
Let’s visually model this problem. There are two operations that can be done in a sequence. First find the value of $$x$$ by using linear pairand then check it is equal to its corresponding angle or not. According to this model, the resultant value of $$x$$ is not equal to its corresponding angle. Now, it’s a matter of finding $$l$$ is parallel to $$m$$ or not
(iv) Steps:
\begin{align}\angle x + 98^\circ &= 180^\circ \text{(Linear pair)}\\\angle x &= 180^\circ - 98^\circ\\\angle x &= 82^\circ\end{align}
For $$l$$ and $$m$$ to be parallel measure of their corresponding angles should be equal but here the measure of corresponding angles are $$82^\circ$$ and $$72^\circ$$ whichn are not equal.
Therefore, $$l$$ and $$m$$ are not parallel to each other.
Learn from the best math teachers and top your exams
• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school<|endoftext|>
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Grade Levels: 9/10, 11/12
Subject Area: Social Studies, History, Civics
Canada has a rich and varied past that can often lie buried in the archives. Too often, the histories of Canada focus on political and military exploits and the lives of the rich or powerful figures. But there are many other stories of Canada’s past to be told. Tales of the struggles and triumphs of Canadian labour, the journey to a new home of people from a multitude of former lands, and the trials and tribulations of growing up in Canada. It has often been said that a picture is worth a thousand words. Just imagine the volumes of history we can uncover through the paintings, symbols, and photographs held by Canadian museums, archives and especially, families.
Three 60 minute class periods
Historical Thinking Concept(s)
This lesson plan uses the following historical thinking concepts: establish historical significance, use primary source evidence, identify continuity and change, and analyze cause and consequence.
- Develop cooperative group work skills
- Enhance their creative skills
- Expand their research and analytical skills
- Deploy critical thinking skills
- Promote a sense of Canadian citizenship and identity
- Create a proper bibliography in an academically recognized format
- Reflect and write critically about their work, experience, and skills developed during the project
Canada is a multicultural country with people from diverse ethnocultural backgrounds who have moved here from all parts of the world. The modern Canadian identity is a conglomerate of diverse customs, traditions and beliefs yet Canadians have a shared sense of common identity and pride in our country that has been fostered by principles that we commonly uphold and cherish. Our diversity and harmony and the way we cherish and respect each other’s values are traits that are admired and respected globally.
The Lesson Activity
- Students will be asked to research sources for information about the migration history of the ethnocultural group that they individually belong to and create a brief list of historical push and pull factors that have impacted the migration history of this particular group to Canada (e.g. why did people from Italy migrate to Canada).
- Students will go to the Government of Canada Justice Laws Website and explore the Multiculturalism Act and identify three reasons how this Act may have impacted the demographic composition of Canada today.
Working in groups, students will create a collage of photographs, symbols, paintings, songs and/or poetry to visually tell the story of a group of Canadians during the past century. The completed collage will consist of 20 images (the equivalent of 20,000 words) that will be creatively placed in an appropriate location in the classroom.
The initial task will foster creativity and cooperative group skills, and introduce students to research and analytical skills. Although students will work cooperatively to complete the project, there will be parts of the assignment that must be completed individually. The collages can focus on any one of the themes listed above, but only one group per topic.
To complete this task, each student must:
- Prepare a bibliography of five to seven sources using the proper format—the bibliography must include at least three primary sources
- Prepare a list of five paintings, symbols, poems, songs of photographs to use in the collage
- Provide a one-paragraph (75 to 100 words) explanation why each of the pieces was chosen to reflect the theme of the collage
- Write a 250-300 word personal reflection which assesses the overall success of the project including the success of the group dynamics, the effectiveness of the collage in expressing the theme, and the overall learning experience
To complete this task, each group must:
- Collectively decide on the focus of the collage
- Share with each other the selections made by individuals to ensure no duplication
- Prepare a sketch/blueprint of the final design with a half to one-page detailed description (including shape, arrangement, colour) which will clearly explain how the collage will appear and how it will capture their theme as it relates to Canada in the twentieth and twenty-first century
- Gather, organize and assemble the hanging collage in the classroom
All of the images can be individually printed and assembled into a collage or can be incorporated into a single slide using Powerpoint or similar software (the slide size can be modified to accommodated all selected images) and then printed. Final collages are to be laminated so that they can be creatively displayed in appropriate locations in the classroom.
Students can hold a Canadian identity and diversity fair where they can invite other students or stakeholders such as teachers and parents to view the collages. Each group can explain their collages to visitors. Alternatively, a virtual fair can be set up using a blog or website with visuals of the collages, themes, and explanations with a comments section for guests to leave their impressions and opinions about the fair.<|endoftext|>
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How Opal Color is produced Visit opalmine shop here
Ok, this article on opal color is not an earth science, geology, or chemistry lesson, but since many folks wonder what makes opals glow in a rainbow of colors, here’s a quick explanation that’s hopefully not too technical. Just in case some of the terms are unfamiliar, there’s Glossary of Terms following the text.
How Opal Color Was Discovered
It took the development of the electron microscope to work this out. Precious opal is made up of tiny uniform spheres of transparent hard silica, which fit together in an orderly three-dimensional frame, sitting in a “bath” of silica solution.
It is the orderliness of the spheres that separates precious opal from common opal.
Light passes through the transparent spheres in a direct line, but when it hits the ‘bath’ of silica, it is bent and deflected at different angles, thus producing a rainbow effect.
Opal Color: Deflection & Diffraction
Depending on the size of the spheres, varying colors of the spectrum are diffracted. So it is a combination of deflection (bending) and diffraction (breaking up) of light rays that creates the color in opal. If you move the stone, light hits the spheres from different angles and bring about a change in color.
The name opal actually means “to see a change in color.” The way in which opal color changes within a particular stone as it is rotated and tilted is called the stone’s play of color as you can see featured in these amazing opal pendants.
How opal color is defined
The size of the spheres has a bearing on the opal color produced. Smaller spheres bring out the blues, from one end of the spectrum. Larger spheres produce the reds from the other end. The more uniform the spheres are placed, the more intense, brilliant and defined thecolorr will be.
Glossary of Terms:
Shapeless. Not consisting of crystals. Non crystalline. Glass is amorphous. Sugar is crystalline.
The bending of rays of light from a straight line.
The Breaking up of a ray of light into either a series of light and dark bands, or into colored bands of the spectrum.
To spread out so as to cover a larger space or surface. To scatter.
A light produced by the electrical stimulation of a gas or vapour. Fluorescent lights have a similar effect on opal as a bright cloudy day–they do not properly bring out the colors in opal.
A compound produced when certain substances chemically combine with water.
Glowing with heat (red or white hot) as in a light bulb which glows white hot, but produces a light that more closely simulates natural sunlight. Sunlight and incandescent lights bring out the natural opal color.
Opal comes from the Latin word opalus which means to see a change in color. (that is, opal color) Chemically, opal is hydrated silica, similar to quartz. (use graphic 1168 to show change of color in the three images)
Opalescence means a play of opal color
A play of color, similar to that of an opal.
Not allowing light to pass through. The opposite of transparent.
Play of Colour
The way in which colors change as an opal is tilted in different directions.
(Silicon Dioxide) A hard, white or colorless substance, that in the form of quartz, enters into the composition of many rocks and is contained in sponges and certain plants. The needle in the mouth of a female mosquito is made of silica. Flint, sand, chalcedony, and opal are examples of silica in different forms.
The band of colours formed when a beam of white light passes through a prism or by some other means (e.g. mist or spray, in the case of a rainbow) The full range of spectrum colors are: red, orange, yellow, green blue, indigo, and violet.
A round three dimensional geometric shape whose surface is equally distant at all points from the centre point.
Letting light through without being transparent. An object that lets some light pass through
Easily seen through.(glass like) it means that you can see through it. if you take a clear glass plate and put it to your face, you can still objects on the other side.
Please leave a comment if you have any further questions about opal color.<|endoftext|>
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Class 8 History
India After Independence
Question 1: Name three problems that the newly independent nation of India faced.
Answer: The three problems that the newly independent nation of India faced are as follows:
- Rehabilitation of a large number of refugees.
- Assimilation of princely states.
- Ensuring the unity of a country which is full of diversity.
Question 2: What was the role of the Planning Commission?
Answer: The Planning Commission was to formulate policies which would guide the economic development. Productivity and employment opportunities were to be increased through proper implementation of those policies.
Question 3: Fill in the blanks:
- Subjects that were placed on the Union List were _________, _________ and _________.
Answer: taxes, defence and foreign affairs
- Subjects on the Concurrent List were _________ and _________.
Answer: forests and agriculture
- Economic planning by which both the state and the private sector played a role in development was called a _________ _________ model.
Answer: Mixed economy
- The death of _________ sparked off such violent protests that the government was forced to give in to the demand for the linguistic state of Andhra.
Answer: Potti Sriramulu
Question 4: State whether true or false:
- At independence, the majority of Indians lived in villages.
- The Constituent Assembly was made up of members of the Congress party.
- In the first national election, only men were allowed to vote.
- The Second Five Year Plan focused on the development of heavy industry.
Answer: (a) True, (b) False, (c) False, (d) True
Question 5: What did Dr Ambedkar mean when he said that “In politics we will have equality, and in social and economic life we will have inequality”?
Answer: By political equality, Dr. Ambedkar meant the universal adult franchise which gave equal rights to all citizens. By socioeconomic inequality, he meant the income disparities among people.
Question 6: After Independence, why was there a reluctance to divide the country on linguistic lines?
Answer: The partition of the country along communal lines changed the mindset of the nationalist leaders. They wanted to prevent further divisions in the country on sectarian lines.
Question 7: Give one reason why English continued to be used in India after Independence.
Answer: Some leaders believed that English should be done away with and Hindi should be promoted as the national language. But this idea was opposed by the leaders from non-Hindi areas. They did not want an imposition on Hindi on the people of those areas. Finally, it was decided that while Hindi would be the ‘official language’; English would be used for communication among various states.
Question 8: How was the economic development of India visualised in the early decades after Independence?
Answer: Removing poverty and building a modern technical and industrial base were important objectives for the new nation. The Planning Commission was set up in 1950 to plan and execute policies for economic development.<|endoftext|>
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|Crane Fly – Epiphragma solatrix|
Family Tipulidae – crane flies, tipules
Flies Main | Flies Index | Tachinidae | Syrphidae | Bee Flies | Blow Flies | Flesh Flies
Tipulidae is the largest family of Diptera with about 1,500 species in North America
|Although adult crane flies are usually sluggish fliers, it's very difficult to see them in flight due to their slim build; there are species that use a very slow wingbeat to propell their body forward and backward, almost as if the body is moving instead of the wings; at the same time the creature flings its hideously long legs to-and-fro inrythm to the whole pulsation – and thereby become all but invisible. Sort of like the vibrating spider. If you ever see one doing it, you'll know what I mean. It's quite a sight!|
They are are often abundant in moist woodlands and around water, usually near places where their larval life is spent. They occur mainly in spring and fall, but species of wingless, snow crane flies (Chionea) appear in the winter. Adult crane flies are most active in the cooler part of the day, usually around dusk. Adult males are more abundant at the beginning of the flight period while females are more numerous toward the end.
Although individual adults have a relatively short life span of 10 to 15 days, the flight period for each species can last from 25-30 days. The main functions of the adult stage are mating and egg-laying. Feeding is less important, and probably water is the most pressing need. Species with elongated rostrum (Geranomyia, Elephantomyia, Toxorhina) have been reported visiting flowers, probably for nectars.
Crane flies serve several important roles in the ecosystem. Most importantly, adult and larval crane flies are food for many animals such as birds, fish, frogs, lizards, spiders and other insects. In addition, the larvae are detritus feeders that break down organic matter in various habitats such as streams and forest floors thereby enriching the soil, renewing and modifying the microhabitat for other invertebrate species. Some crane flies require special habitat conditions, and their presence or absence can be used as an indicator of environmental quality. Fishermen use larvae of some large crane flies as bait. Several species of crane flies are important agricultural pests; their larvae feed on seedlings of field crops and if abundant can be destructive to lawns, rangelands, rice fields, and golf courses.
|Tipulidae, often called crane flies in their adult stage, is the largest family of true flies. Crane flies form a highly diverse group of insects, both in number of species and in larval habitats, which extend from aquatic to terrestrial. The body plan or morphology of crane flies is rather simple. An elongate body, one pair of narrow wings, and long, slender legs characterize them. The body size ranges from 5 to 50 mm and can be described as mosquito-like. They are often mistaken for mosquitoes, but they belong to a group of harmless flies and can be distinguished from all other true flies by the transverse V-shaped groove on the dorsal part of the thorax.|
In North America, more than 1,500 species of crane flies have been described and over 300 species are known from Pennsylvania. This number probably represents only about two-thirds of the estimated actual number for the state, and much more precise taxonomic studies are needed.
Flies of North America – Order Diptera. Flies are prevalent in virtually all habitats, with over 16,000 species in North America. Flies can be distinguished from all other insects in that they only have one pair of normal wings. Most flies have compound eyes and mouthparts adapted for piercing, lapping or sucking fluids.
Insects & Spiders | Flies Index | Syrphidae | Bee Flies | Robber Flies<|endoftext|>
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# Touch x-axis
Find the equations of circles that pass through points A (-2; 4) and B (0; 2) and touch the x-axis.
Result
a = (Correct answer is: a=pow(x+2, 2)+pow(y-2, 2)=4)
b = (Correct answer is: b=pow(x-6, 2)+pow(y-10, 2)=100)
#### Solution:
$(x-m)^2+(y-n)^2 = r^2 \ \\ \ \\ (m+2)^2+(n-4)^2 = r^2 \ \\ m^2+(n-2)^2 = r^2 \ \\ n = r \ \\ \ \\ (m+2)^2+(n-4)^2 = n^2 \ \\ m^2+(n-2)^2 = n^2 \ \\ \ \\ m^2 + 4 \ m - 8 \ n + 20 = 0 \ \\ m^2 - 4 \ n + 4 = 0 \ \\ \ \\ n = (m^2+4)/4 \ \\ \ \\ \ \\ \ \\ m^2 + 4 \ m - 8 \cdot \ ((m^2+4)/4) + 20 = 0 \ \\ -m^2 +4m +12 = 0 \ \\ m^2 -4m -12 = 0 \ \\ \ \\ a = 1; b = -4; c = -12 \ \\ D = b^2 - 4ac = 4^2 - 4\cdot 1 \cdot (-12) = 64 \ \\ D>0 \ \\ \ \\ m_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 4 \pm \sqrt{ 64 } }{ 2 } \ \\ m_{1,2} = \dfrac{ 4 \pm 8 }{ 2 } \ \\ m_{1,2} = 2 \pm 4 \ \\ m_{1} = 6 \ \\ m_{2} = -2 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (m -6) (m +2) = 0 \ \\ n_{ 1 } = (m_{ 1 }^2+4)/4 = (6^2+4)/4 = 10 \ \\ \ \\ n_{ 2 } = (m_{ 2 }^2+4)/4 = ((-2)^2+4)/4 = 2 \ \\ \ \\ r = n \ \\ \ \\ r = 2, m = -2, n = 2 \ \\ r = 10, m = 6, n = 10 \ \\ \ \\ \ \\ a = (x+2)^2+(y-2)^2 = 4$
$b = (x-6)^2+(y-10)^2 = 100$
Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):
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#### Following knowledge from mathematics are needed to solve this word math problem:
For Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc. Looking for help with calculating roots of a quadratic equation? Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? Pythagorean theorem is the base for the right triangle calculator.
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A reservoir contains 28.54 m3 of water when completely full. The diameter of the upper base is 3.5 m while at the lower base is 2.5 m. Determine the height if the reservoir is in the form of a frustum of a right circular cone.
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If the height of a cylinder is 4 times its circumference. What is the volume of the cylinder in terms of its circumference, c?
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In an isosceles triangle, the equal sides are 2/3 of the length of the base. Determine the measure of the base angles.
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# 12.3 The most general applications of bernoulli’s equation
Page 1 / 2
• Calculate using Torricelli’s theorem.
• Calculate power in fluid flow.
## Torricelli’s theorem
[link] shows water gushing from a large tube through a dam. What is its speed as it emerges? Interestingly, if resistance is negligible, the speed is just what it would be if the water fell a distance $h$ from the surface of the reservoir; the water’s speed is independent of the size of the opening. Let us check this out. Bernoulli’s equation must be used since the depth is not constant. We consider water flowing from the surface (point 1) to the tube’s outlet (point 2). Bernoulli’s equation as stated in previously is
${P}_{1}+\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}+\rho {\text{gh}}_{1}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}+\rho {\text{gh}}_{2}\text{.}$
Both ${P}_{1}$ and ${P}_{2}$ equal atmospheric pressure ( ${P}_{1}$ is atmospheric pressure because it is the pressure at the top of the reservoir. ${P}_{2}$ must be atmospheric pressure, since the emerging water is surrounded by the atmosphere and cannot have a pressure different from atmospheric pressure.) and subtract out of the equation, leaving
$\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}+\rho {\text{gh}}_{1}=\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}+\rho {\text{gh}}_{2}\text{.}$
Solving this equation for ${v}_{2}^{2}$ , noting that the density $\rho$ cancels (because the fluid is incompressible), yields
${v}_{2}^{2}={v}_{1}^{2}+2g\left({h}_{1}-{h}_{2}\right)\text{.}$
We let $h={h}_{1}-{h}_{2}$ ; the equation then becomes
${v}_{2}^{2}={v}_{1}^{2}+2\text{gh}$
where $h$ is the height dropped by the water. This is simply a kinematic equation for any object falling a distance $h$ with negligible resistance. In fluids, this last equation is called Torricelli’s theorem . Note that the result is independent of the velocity’s direction, just as we found when applying conservation of energy to falling objects.
All preceding applications of Bernoulli’s equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli’s equation in which pressure, velocity, and height all change. (See [link] .)
## Calculating pressure: a fire hose nozzle
Fire hoses used in major structure fires have inside diameters of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of $1\text{.}\text{62}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}$ . The hose goes 10.0 m up a ladder to a nozzle having an inside diameter of 3.00 cm. Assuming negligible resistance, what is the pressure in the nozzle?
Strategy
Here we must use Bernoulli’s equation to solve for the pressure, since depth is not constant.
Solution
Bernoulli’s equation states
${P}_{1}+\frac{1}{2}{\mathrm{\rho v}}_{1}^{2}+\rho {\text{gh}}_{1}={P}_{2}+\frac{1}{2}{\mathrm{\rho v}}_{2}^{2}+\rho {\text{gh}}_{2}\text{,}$
where the subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds ${v}_{1}$ and ${v}_{2}$ . Since $Q={A}_{1}{v}_{1}$ , we get
${v}_{1}=\frac{Q}{{A}_{1}}=\frac{\text{40}\text{.}0×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\text{/s}}{\pi \left(3\text{.}\text{20}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{m}{\right)}^{2}}=\text{12}\text{.}4\phantom{\rule{0.25em}{0ex}}\text{m/s}\text{.}$
Similarly, we find
${v}_{2}=\text{56.6 m/s}\text{.}$
(This rather large speed is helpful in reaching the fire.) Now, taking ${h}_{1}$ to be zero, we solve Bernoulli’s equation for ${P}_{2}$ :
${P}_{2}={P}_{1}+\frac{1}{2}\rho \left({v}_{1}^{2}-{v}_{2}^{2}\right)-\rho {\text{gh}}_{2}\text{.}$
Substituting known values yields
${P}_{2}=1\text{.}\text{62}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}+\frac{1}{2}\left(\text{1000}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}\right)\left[\left(\text{12}\text{.}4\phantom{\rule{0.25em}{0ex}}\text{m/s}{\right)}^{2}-\left(\text{56}\text{.}6\phantom{\rule{0.25em}{0ex}}\text{m/s}{\right)}^{2}\right]-\left(\text{1000}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}\right)\left(9\text{.}80\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)\left(\text{10}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{m}\right)=0\text{.}$
Discussion
This value is a gauge pressure, since the initial pressure was given as a gauge pressure. Thus the nozzle pressure equals atmospheric pressure, as it must because the water exits into the atmosphere without changes in its conditions.
the meaning of phrase in physics
is the meaning of phrase in physics
Chovwe
write an expression for a plane progressive wave moving from left to right along x axis and having amplitude 0.02m, frequency of 650Hz and speed if 680ms-¹
how does a model differ from a theory
what is vector quantity
Vector quality have both direction and magnitude, such as Force, displacement, acceleration and etc.
Besmellah
Is the force attractive or repulsive between the hot and neutral lines hung from power poles? Why?
what's electromagnetic induction
electromagnetic induction is a process in which conductor is put in a particular position and magnetic field keeps varying.
Lukman
wow great
Salaudeen
what is mutual induction?
je
mutual induction can be define as the current flowing in one coil that induces a voltage in an adjacent coil.
Johnson
how to undergo polarization
show that a particle moving under the influence of an attractive force mu/y³ towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v²k² and distance uk²/√u-vk as origin
show that a particle moving under the influence of an attractive force mu/y^3 towards the axis x. show that if it be projected from the point (0,k) with the component velocities U and V parallel to the axis of x and y, it will not strike the axis of x unless u>v^2k^2 and distance uk^2/√u-k as origin
No idea.... Are you even sure this question exist?
Mavis
I can't even understand the question
yes it was an assignment question "^"represent raise to power pls
Gabriel
Gabriel
An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10kg . Pendulum 2 has a bob with a mass of 100 kg . Describe how the motion of the pendula will differ if the bobs are both displaced by 12º .
no ideas
Augstine
if u at an angle of 12 degrees their period will be same so as their velocity, that means they both move simultaneously since both both hovers at same length meaning they have the same length
Modern cars are made of materials that make them collapsible upon collision. Explain using physics concept (Force and impulse), how these car designs help with the safety of passengers.
calculate the force due to surface tension required to support a column liquid in a capillary tube 5mm. If the capillary tube is dipped into a beaker of water
find the time required for a train Half a Kilometre long to cross a bridge almost kilometre long racing at 100km/h
method of polarization
Ajayi
What is atomic number?
The number of protons in the nucleus of an atom
Deborah
type of thermodynamics
oxygen gas contained in a ccylinder of volume has a temp of 300k and pressure 2.5×10Nm<|endoftext|>
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# Algebra/Completing the Square
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Algebra
← Factoring Polynomials Completing the Square Quadratic Equation →
## Derivation
The purpose of "completing the square" is to either factor a prime quadratic equation or to more easily graph a parabola. The procedure to follow is as follows for a quadratic equation $y = ax^2+bx+c$:
1. Divide everything by a, so that the number in front of $x^2$ is a perfect square (1):
$\frac{y}{a} = x^2 + \frac{b}{a}x + \frac{c}{a}$
2. Now we want to focus on the term in front of the x. Add the quantity $\left(\frac{b}{2a}\right)^2$ to both sides:
$\frac{y}{a} + \left(\frac{b}{2a}\right)^2 = x^2 + \frac{b}{a}x +\left(\frac{b}{2a}\right)^2 + \frac{c}{a}$
3. Now notice that on the right, the first three terms factor into a perfect square:
$x^2 + \frac{b}{a}x +\left(\frac{b}{2a}\right)^2 = \left(x + \frac{b}{2a}\right)^2$
Multiply this back out to convince yourself that this works.
4. Therefore the completed square form of the quadratic is:
$\frac{y}{a} + \left(\frac{b}{2a}\right)^2 = \left(x + \frac{b}{2a}\right)^2 + \frac{c}{a}$ or, multiplying through by a,
$y = a\left(x+\frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}$
## Explanation of Derivation
1. Divide everything by a, so that the number in front of $x^2$ is a perfect square (1):
$x^2 + \frac{b}{a}x + \frac{c}{a} = {a}$
Think of this as expressing your final result in terms of 1 square x. If your initial equation is
2. Now we want to focus on the term in front of the x. Add the quantity $\left(\frac{b}{2a}\right)^2$ to both sides:
$\frac{y}{a} + \left(\frac{b}{2a}\right)^2 = x^2 + \frac{b}{a}x +\left(\frac{b}{2a}\right)^2 + \frac{c}{a}$
3. Now notice that on the right, the first three terms factor into a perfect square:
$x^2 + \frac{b}{a}x +\left(\frac{b}{2a}\right)^2 = \left(x + \frac{b}{2a}\right)^2$
Multiply this back out to convince yourself that this works.
4. Therefore the completed square form of the quadratic is:
$\frac{y}{a} + \left(\frac{b}{2a}\right)^2 = \left(x + \frac{b}{2a}\right)^2 + \frac{c}{a}$ or, multiplying through by a,
## Example
The best way to learn to complete a square is through an example. Suppose you want to solve the following equation for x.
2x2 + 24x + 23 = 0 Does not factor easily, so we complete the square. x2 + 12x + 23/2 = 0 Make coefficient of x2 a 1, by dividing all terms by 2. x2 + 12x = - 23/2 Add – 23/2 to both sides. x2 + 12x + 36 = - 23/2 + 36 Take half of 12 (coefficient of x), and square it. Add to both sides. (x + 6)2 = 49/2 Factor. Now we can take square roots to easily solve this form of the equation. √(x + 6)2 = √49/√2 Take the square root. x + 6 = 7/√2 Simplify. x = -6 + (7√2)/2 Rationalize the denominator.<|endoftext|>
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NASA: Sea Level Rise of Several Feet is UNAVOIDABLE!
[Tide gauges have been used to measure sea level for more than 130 years. Satellite measurements now complement the historical record. (NASA Earth Observatory image by Joshua Stevens, based on data from the Commonwealth Scientific and Industrial Research Organization and NOAA)]
For thousands of years, sea level has remained relatively stable and human communities have settled along the planet’s coastlines. But now Earth’s seas are rising. Globally, sea level has risen about eight inches since the beginning of the 20th century and more than two inches in the last 20 years alone. According to the federal National Aeronautics and Space Administration (NASA), “all signs suggest that this rise is accelerating”.
Seas around the world have risen an average of nearly 3 inches since 1992, with some locations rising more than 9 inches due to natural variation, according to the latest satellite measurements from NASA and its partners. An intensive research effort now underway, aided by NASA observations and analysis, points to “an unavoidable rise of several feet in the future”.
“Sea level rise is a natural [physical] consequence of the warming of our planet”, states NASA’s press release dated August 26, 2015. “We know this from basic physics. When water heats up, it expands. So when the ocean warms, sea level rises. When ice is exposed to heat, it melts. And when ice on land melts and water runs into the ocean [water from unprecedented ice and snow melt off Antarctica, Greenland, land areas north of permafrost region in Northern Hemisphere, mountainous glaciers receding, worldwide] sea level rises”.
As the ocean has warmed, polar ice has melted, and porous landmasses have subsided, global mean sea level has risen by 8 inches (20 centimeters) since 1870. The rate of sea level rise is faster now than at any time in the past 2,000 years, and that rate has doubled in the past two decades.
While NASA and other agencies continue to monitor the warming of the ocean and changes to the planet’s land masses, the biggest concern is what will happen to the ancient ice sheets covering Greenland and Antarctica, which continue to send out alerts that a warming planet is affecting their stability.
NASA has been recording the height of the ocean surface from space since 1992, recording about 2.9 inches (7.4 centimeters) of sea level rise averaged for all the oceans in that 23 year period.
In 2002, NASA and the German space agency launched the Gravity Recovery and Climate Experiment (GRACE) twin satellites, capable of measuring the movement of mass, hence gravity, around Earth at intervals of every 30 days. GRACE has found that earth’s land masses move very little in a month; however, earth’s water masses move through melting, evaporation, precipitation and other processes. GRACE records these movements of water around the planet, while a new NASA network of more than 3,000 floating ocean sensors spread across the entire open ocean supplement that ocean water level data.
Observations from the new NASA ocean level data collection systems have revolutionized scientists’ understanding of contemporary sea level rise and its causes. NASA’s newest release state that: “We know that today’s sea level rise is about one-third the result of the warming of existing ocean water, with the remainder coming from melting land ice”, adding that “currently, regional differences in sea level rise are dominated by the effects of ocean currents and well known cycles, such as the Pacific Ocean’s El Niño phenomenon and the Pacific Decadal Oscillation.” But as the ice sheets located closer to the poles of the planet (which receive less direct solar radiation due to the tilting of the planet as it orbits the Sun), and the once “permanent” ice located at higher elevations around the world continues to melt as a direct result of measurably warming temperatures occurring as a direct consequence of a stronger greenhouse effect that is already unnaturally high due to human activities such as fossil fuel burning, deforestation, and paving over the earth’s still green landscape. NASA scientists now predict that the increasing meltwater resulting from the stronger greenhouse effect will overtake the formerly natural causes of regional variations of the ocean water levels and be the most significant contributor to the overall rise in sea level.
The recent advances in observing the world’s frozen regions using satellite measurements from NASA and its participating organizations have allowed scientists to accurately estimate annual ice losses from Greenland and Antarctica. Not only can they now determine how much sea level around the world is changing – as measured by satellite for the past 23 years – but they can also determine how much of the sea level rise is being caused by our warming of the earth’s biosphere, which includes both the atmosphere at or near the surface, the oceans, the land surface, and the biota (plant and animal kingdoms) that all together comprise earth’s biosphere.
GRACE’s record, spanning over the last decade, shows that the ice loss around the planet is now accelerating in Greenland and West Antarctica. The record shows Greenland has shed on average 303 gigatons of ice every year since 2004, while Antarctica has lost on average 118 gigatons of ice per year. Much of Antarctica’s ice loss has been shown to come from West Antarctica’s ice loss. Greenland’s ice loss has accelerated by 31 gigatons of ice per year, every year since 2004, while West Antarctica’s ice loss has accelerated to 28 gigatons per year.
“Given what we know now about how the ocean expands as it warms and how ice sheets and glaciers are adding water to the seas, it’s pretty certain we are locked into at least 3 feet of sea level rise and probably more,” said Steve Nerem of the University of Colorado, Boulder, and lead of NASA’s new Sea Level Change Team.
The Greenland Ice Sheet, spanning 660,000 square miles (an area almost as big as Alaska), and with a thickness at its highest point of almost 2 miles, has the potential to raise the world’s oceans by more than 20 feet. Situated in the Arctic, which is warming at twice the rate of the rest of the planet, Greenland has been shedding more ice in the summer than it gains back in the winter since 1992.
“In Greenland, everything got warmer at the same time: the air, the ocean surface, the depths of the ocean,” said Ian Joughin, a glaciologist at University of Washington. “We don’t really understand which part of that warming is having the biggest effect on the glaciers.”
What scientists do know is that warming Arctic temperatures – and a darkening surface of the Greenland ice sheet – are causing so much summer melting that it is now the dominant factor in Greenland’s contribution to sea level rise.
NASA has found that Greenland’s summer melt season now lasts 70 days longer than it did in the early 1970s. Every summer, warmer air temperatures cause melt over about half of the surface of the ice sheet – although recently, 2012 saw an extreme event where 97 percent of the ice sheet experienced melt at its top layer.
Greenland’s massive glaciers have sped up, too. Though many of the glaciers in the southeast, west and northwest of the island – an area that experienced quick thinning from 2000 to 2006 – have now slowed down, the melting rate at other areas Greenland’s massive ice sheet has not slowed. A study last year showed that the northeast Greenland ice stream had increased its ice loss rate due to warmer air temperatures.
“The early 2000s was when some big things revealed themselves, such as when we saw the fastest glacier we knew of, the Jakobshavn ice stream in Greenland, double its speed,” said Waleed Abdalati, director of the Cooperative Institute for Research in Environmental Sciences, Boulder, Colorado, and former NASA chief scientist. “The subsequent surprise was that these changes could be sustained for a decade – Jakobshavn is still going fast”, Abdalati said.
The Antarctic Ice Sheet covers nearly 5.4 million square miles, and area larger than the United States and India combined, and contains enough ice to raise the ocean level by about 190 feet. The Transantarctic Mountains split Antarctica in two major regions: West Antarctica and the much larger East Antarctica.
Though Antarctica’s contribution to sea level rise is still at less than 0.02 inches (0.5 millimeters) per year, several events over the past decade and a half have prompted experts to start warning about the possibility of more rapid changes this century.
The mountainous horn of the continent, the Antarctic Peninsula, gave one of the earliest warnings on the impact of a changing climate in Antarctica when warm air and warmer ocean temperatures led to the dramatically fast breakup of the Larsen B ice shelf in 2002. In about a month, 1,250 square miles of floating ice that had been stable for over 10,000 years were gone. In the following years, other ice shelves in the Peninsula, including the last remainder of Larsen B, collapsed, speeding up in the flow of the land lying glaciers that they were buttressing against the warming ocean.
In 2014, two studies focusing on the acceleration of the glaciers in the Amundsen Sea sector of West Antarctica showed that its collapse is currently well underway. And while one of the studies speculated that the demise of the ice overlaying West Antarctica could take as long as 200 to 1,000 years, depending on how rapidly the ocean heats up, both studies concurred that its collapse is already unstoppable, and that when it does collapse, the melt water will add up to 12 feet of sea level rise to the oceans.
The wind is also a factor in determining the timing of West Antarctica’s collapse. The “westerlies”, the winds that spin the ocean waters around Antarctica, are known to have intensified during the last decade, pushing the cold top layer away from the land, and thus allowing the warmer, deeper waters to rise and spill over the border of the continental shelf, flowing all the way back to the base of many of the ice shelves jetting out from the continent. As the ice shelves weaken from underneath, the glaciers behind them are predicted to speed up.
East Antarctica’s massive ice sheet, as vast as the lower continental U.S., remains an unknown in projections of sea level rise. Though it appears to be stable, a recent study on Totten Glacier, East Antarctica’s largest and most rapidly thinning glacier, hints otherwise. Research found two deep troughs that could lead warm ocean water to the base of the glacier and melt it in a similar way to what’s happening to the glaciers in West Antarctica. Other sectors grounded below sea level, such as the Cook Ice Shelf, Ninnis, Mertz and Frost glaciers, have also been found to be losing mass.
For the West Antarctic Ice Sheet, which largely rests on a bed that lies below sea level, the main driver of ice loss is the ocean. The waters of the Southern Ocean are layered: on top and at the bottom, the temperatures are frigid, but the middle layer is warm. The westerlies, the winds that spin the ocean waters around Antarctica, have intensified during the last decade, pushing the cold top layer away from the land. This allows the warmer, deeper waters to rise and spill over the border of the continental shelf, flowing all the way back to the base of many ice shelves. As the ice shelves weaken from underneath, the glaciers behind them speed up.
East Antarctica’s massive ice sheet, as vast as the lower continental U.S., remains the main unknown in projections of sea level rise. Though it appears to be stable, a recent study on Totten Glacier, East Antarctica’s largest and most rapidly thinning glacier, hints otherwise. The research found two deep troughs that could lead warm ocean water to the base of the glacier and melt it in a similar way to what’s happening to the glaciers in West Antarctica. Other sectors grounded below sea level, such as the Cook Ice Shelf, Ninnis, Mertz and Frost glaciers, are also losing mass.<|endoftext|>
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Design a Zoo: A Fun Math and Art Project for Kids
Welcome to the exciting world of zoo design, where math and art come together to create a miniature masterpiece! In this blog, we’ll embark on a thrilling journey to build our very own zoo, complete with animal enclosures, area calculations, and a dash of creativity. So, grab your dice, sharpen your pencils, and let’s dive into the wild world of math and art!
## Rolling the Dice for Enclosure Sizes
To begin our zoo-building adventure, we’ll use a pair of dice to determine the sizes of our animal enclosures. Each roll of the dice will give us a number that corresponds to a specific area measurement. Here’s how it works:
• Roll two dice: Grab a pair of standard six-sided dice and roll them together.
• Add the numbers: Add the numbers shown on the top faces of both dice.
• Determine the area: The sum of the two numbers represents the area of the animal enclosure in square units. For example, if you roll a 3 and a 5, the area of the enclosure will be 3 + 5 = 8 square units.
## Calculating Perimeter: Measuring the Boundaries
Once we’ve determined the area of each enclosure, it’s time to calculate its perimeter. Perimeter is the total length of the boundary surrounding the enclosure. To calculate the perimeter, we’ll use the following formula:
Perimeter = 2 x (Length + Width)
Here’s how to apply the formula:
• Determine the length and width: Based on the area you rolled for each enclosure, decide on the length and width dimensions that fit that area. For example, if the area is 8 square units, you could choose a length of 4 units and a width of 2 units.
• Plug in the values: Substitute the length and width values into the perimeter formula. In our example, the perimeter would be 2 x (4 + 2) = 2 x 6 = 12 units.
## Getting Creative: Designing the Enclosures
Now comes the fun part – unleashing your creativity to design the animal enclosures! Here are some ideas to inspire you:
• Draw the enclosures: Use a pencil and paper to sketch out the shapes of the enclosures based on the area and perimeter you calculated.
• Add details: Draw trees, rocks, ponds, and other features to make the enclosures look realistic and inviting for the animals.
• Color your zoo: Use colored pencils, markers, or crayons to bring your zoo to life with vibrant colors.
• Create animal cutouts: Draw and cut out different animal shapes to place inside the enclosures. You can even label each animal with its name and some fun facts.
## Putting It All Together: Your Miniature Zoo Model
Once you’ve designed and decorated all the enclosures, it’s time to assemble your miniature zoo model. Arrange the enclosures on a large piece of paper or cardboard to create a zoo layout. You can even add a zoo entrance, pathways, and a parking lot.
Bonus Challenge: Calculate the total area and perimeter of your entire zoo model. This will give you an idea of the overall size of your zoo and the amount of space you’ve created for your animal friends.
Remember to share your finished zoo model with us! Take a picture of your creation and send it to us on social media. We’d love to see your unique and creative zoo designs.
So, roll those dice, grab your art supplies, and let’s embark on this exciting journey of designing our very own zoo! Happy building!<|endoftext|>
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Over 250 million years ago the world experienced it’s largest mass extinction. Over 70% of all land species on earth, became extinct. The reason for this event has become clear and concise. Our world’s land masses slammed together into what is known as our ancestral supercontinent named Pangea. The air was rendered poisonous and the land became so vast and arid lacking water and food to sustain life. At this point in time the Appalachians were higher than the present day Himalayas and the winds were much stronger and more sustained than they are anywhere today. Actually the land was so dry and winds so fierce that sediment blew from the Appalachians all the way to an area that occupies to this day Southern Utah, Southwest Colorado, Northern Arizona and North West New Mexico. This sandy sediment mounted up into sky scraper high sand dunes that looked much like today’s Sahara. But this is not where the story starts. The a fore mentioned Dunes lay on top of a series of sediments that occurred before the land became this extreme. To the East of this four state region there was a series of mountains that are now referred to as the Ancestral Rocky Mountains. During times of wetter climate conditions, large deluges of precipitation plundered the ancestral Rockies spilling from it blood red sediment that fanned out into the landscape west of the mountains. Also sewn into this story are a series of low lying seas which ebbed and flowed into the historical land layer cake. This epic tale happened not once but multiple times stacking layer upon layer of variations in rock: sand stone, shale, limestone, more stand stone, lime stone, sand stone, more shale, leaving visible to the naked eye different versions of the landscape. The story of history of this region disappeared deeper and deeper underground giving the impression it would be gone forever.
After almost 200 million years of this North America was now separated from the rest of the super continent and something completely different started to happen. As North America drifted westward the ocean floor of the ancient Pacific became over run and pushed under North America. The Ocean floor, or plate, was pushed under North American at an unusually low angle. This caused mountains to rise as much as 1,000 miles to the east. The huge swath of land from our story that occupies significant parts of present day Utah Colorado, Arizona and New Mexico received an enormous amount of pressure from below ground to rise like the Rocky mountains to the east. Some areas did succumb to this tension and a high mountain was pushed up here or there, land cracked, land folded but the majority of this land did not fold, for the most part it stayed together and was uplifted to one cohesive high country which is now called the Colorado Plateau. Now that this land was elevated with the high Rocky Mountains to the east and the North it was ready to be carved by one last awesome force of Nature: H2O
Today, if you stand at the southern edge of this geologic province you would be looking down a fortress of several thousand foot sandstone and limestone walls that stretch across the entire state of Arizona. These walls reach their climax smack in the middle of the state. Standing here looking down you feel that you are at the edge of the world peering into an alternate reality or dimension, down into the land that has given birth to the New Age, the age of AQUARIUS.
When the moon is in the seventh house
And Jupiter aligns with the Mars
The peace will guide the planets
And love will steer the stars
This is the dawning of the age of Aquarius
The age of Aquarius
Harmony and understanding
Sympathy and trust abounding
No more falsehoods or derisions
Golden living dreams of visions
Mystic crystals revelations
And the minds true liberation
Sedona has become and still is our worldwide capital for communing with the Metaphysical universe. Bits and pieces of spiritual disciplines and religious practices such as Hinduism, Suffism, Yoga and Native American ceremony since as early as the 1950s have slowly been stirred into this small community, allowing it to become a harbor for what is most commonly known as this New Age Spirituality. Along with it came the psychics, the fortune tellers, crystals and the pictures of Auras and a general obsession with the color purple…… By 1980 a self proclaimed psychic with a following named Page Bryant announced that there were sources of positive, negative or neutrally charged energy conveniently spread throughout the Red Rock country of Sedona within close proximity to the road. These were the vortexes or vortices. Belief as well as inspiration from these Vortexes has spread since Page made the designation and today Sedona, a town of just over 10,000 year round residents receives towards 4 million tourists a year. A study conducted by NAU (Northern Arizona University) found that close to 70% of these visitors are here for the vortexes, or “healing properties” of Sedona.
I have no idea how our family would fit into this statistic. Not a believer in Page’s vortexes there is no doubt some incredible energy found in Sedona…or rather in the Red Rock Country that surrounds Sedona. In my personal experience I have found large amounts of positively charged energy on some of the spectacularly exposed climbs and summits of the dozens of steep spires that make up the Red Rock country. Most of the climbs on rock so red you sometimes feel you are climbing the flesh of the earth. To think about climbing up rocks laid down during events that occurred 300 million years ago is certainly sublime. Also need mentioning almost all of Sedona classics have the 15-25 foot high limestone “band” or layer creating an important defining feature of every climb. Deposited by a shallow sea during these ancient times, this band of very different rock always gives a climb a vortex of one sort or another. Dr. Rubo’s Wild ride is still my favorite with multiple pitches of aesthetic hand cracks, interrupted only by a very strenuous and steep lime stone section, followed by exposed pitches of face climbing on more red wine colored stone. All climbs seem to end at a classic Sedona spire summit. I want to carry a level with me at some point to the top of all the Sedona towers. I bet every one has square, level rock summits. During our stay we also had the great opportunity to climb Queen Victoria Spire right above Sedona. The Lime stone band gave us an undercut off-width crack that was a real bear. And then there was Goliath…with more great cracks, wild exposure, positive vortexes.
Walking around downtown Sedona I can’t help to feel that with the overwhelming number of tourists, candy shops, pink jeep tours, fudge shops, souvineer traps etc. it is surely one big negatively charged vortex. It was however easy to ignore while we were there, we remained surrounded by friends and family. Roni and Michael came to be with us while we all stayed at a beautiful hotel removed from down town. Tim joined us with a last hike before he left to Durango and Scott came down from Flagstaff to hike and show us around. Most trails take you to wild, sky high locations with views that give insight to timelessness and vastness. Surrounded by all of this natural, family and friendly goodness the thought of staying here and spending several decades past through our minds more than once. But the thought never stayed for long enough and eventually the time came to move on. From there we moved up onto the top of the Mogollon Rim to our friends Scott and Lindsay Flagstaff, Arizona.<|endoftext|>
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Contents
Cylinder
In mathematics a cylinder is a quadric, i.e. a three-dimensional surface, with the following equation in Cartesian coordinates:
$\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1$
This equation is for an elliptic cylinder. If a = b then the surface is a circular cylinder. The cylinder is a degenerate quadric because at least one of the coordinates (in this case z) does not appear in the equation. By some definitions the cylinder is not considered to be a quadric at all.
In common usage, a cylinder is taken to mean a finite section of a right circular cylinder with its ends closed to form two circular surfaces, as in the figure (right). If the cylinder has a radius r and length h, then its volume is given by
$V = \pi r^2 h$
and its surface area is
$A = 2 \pi r^2 + 2 \pi r h$
For a given volume, the cylinder with the smallest surface area has h = 2r. For a given surface area, the cylinder with the largest volume has h = 2r.
There are other more unusual types of cylinder. These are the imaginary elliptic cylinder:
$\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = -1$
the hyperbolic cylinder:
$\left(\frac{x}{a}\right)^2 - \left(\frac{y}{b}\right)^2 = 1$
and the parabolic cylinder:
$x^2 + 2y = 0$
A cylinder in an engine is the space a piston travels in. The piston is the same size as the two bases of the cylinder (the circular and flat surfaces). In the following drawing, which depicts a cross-section of a steam engine cylinder, the bottom sliding part is the piston, and the top sliding part is a valve that directs steam into the two ends of the cylinder alternately.
The cylinder was also the dominent medium of audio storage from the 1870s to the 1910s, and continued in limited use (such as the dictaphone) through the mid 20th century. See: phonograph cylinder<|endoftext|>
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Doping is the process of adding impurities to intrinsic semiconductors to alter their properties. Normally Trivalent and Pentavalent elements are used to dope Silicon and Germanium. When a intrinsic semiconductor is doped with Trivalent impurity it becomes a P-Type semiconductor. The P stands for Positive, which means the semiconductor is rich in holes or Positive charged ions. When we dope intrinsic material with Pentavalent impurities we get N-Type semiconductor, where N stands for Negative. N-type semiconductors have Negative charged ions or in other words have excess electrons.
How Doping Works
Atoms follow a rule called Octet Rule. According to Octect-rule atoms are stable when there are eight electrons in their valence. If not, atoms readily accept or share neighboring atoms to achieve eight electrons in their valence shell. In the silicon lattice each silicon atom is surrounded by four silicon atoms. Each silicon atom share one of its electron in the valance shell to its neighbor to satisfy the octect-rule. A schematic diagram of an intrinsic semiconductor is shown in image right (Figure : Intrinsic Silicon Lattice).
Now lets see what will happen when we pop in a pentavalent element into the lattice. As you can see the image (Figure : N-type), we have doped the silicon lattice with Phosphorous, a pentavalent element. Now pentavalent element has five electrons, so it shares a electron with each of the four neighboring silicon atoms, hence four atoms are tied up with the silicon atoms in the lattice. This leaves an electron extra. This excess electron is free to move and is responsible conduction. Hence N-type (Negative Type) extrinsic semiconductor (silicon in this case) is made by doping the semiconductor with pentavalent element.
To create a P-type semiconductor, all we must do is to pop in a trivalent element into the lattice. A trivalent element has three electrons in its valence shell. It shares three electrons with three neighboring silicon atoms in the lattice, the fourth silicon atom demands an electron but the trivalent atom has no more electron to share. This creates a void in lattice which we call it has hole. Since the electron is deficient, the hole readily accepts an electron, this makes it a P-type (Positive type) extrinsic semiconductor.
As you can see at image (Figure : P-type), we have poped in boron (trivalent element) in silicon lattice. This has created a hole making the semiconductor a P-type material.
The case is no different in Germanium. Its behaves same as silicon how ever some properties do differ which makes germanium based devices used in certain application and silicon based devices used in other applications.<|endoftext|>
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A child’s early years in life are very important for their health and development. Parents, family, teachers, and others work together so each child can reach their full potential.
Healthy development means that all children can grow up having their social, emotional and educational needs met. A safe and loving home, spending time with family are all very important. Examples of this include playing, singing, reading and talking. Proper diet, exercise, and sleep are foundational.
Positive parenting practices include:
Parents and caregivers who use these practices help children be healthy and safe.
Key learning takes place in a child’s first 6 years. This learning sets the foundation for success in school and in life.
Parents and caregivers play a powerful role in wiring a baby’s brain for learning. From a child’s first breath to the first day of kindergarten, loving relationships are the best teachers.
Raising a child is too big of a job to do alone, and many cultures believe the whole family is involved, as well as communities. Wise families support communities in launching children for success.
Areas of Development
There are different areas of development all children progress through. Here are some key areas and tips.
Social skills help your child build relationships with the people around them: family, friends and neighbours.
Singing songs and telling stories to each other (including make-believe) is a fun way to practice social skills. Children can develop self confidence by engaging you in your stories or learning a song.
- Hold your baby close and talk to her
- Play simple games like peek-a-boo
- Provide opportunities to play with other children. Bring your child to a community playgroup or preschool
- Encourage and model how to share
As your child grows, self-help and independence skills allow your child to be independent and do things for themselves. This eventually leads to their independence in children’s programs, preschool and kindergarten classrooms. Independence also boosts children self esteem and helps them develop life skills.
- Simple things may seem difficult at first, but with help, patience and practice, your child will develop the necessary skills.
- Around 12 months let your child try feeding themselves with a spoon – help when needed.
- When they are ready, encourage your child to put things away, put on their own clothes and other simple things.
Gross Motor Skills
Coordination and balance are a combination of skills called gross motor skills.
- Your child will learn how to coordinate the body by practicing climbing, walking, pushing and pulling things around.
- She will want to do the same thing over and over because she is learning by repetition.
- If she is climbing dangerous things, provide something safe to climb over
- Games where children move their whole bodies help them to learn how to make their bodies do what they want and it feels great!
Fine Motor Skills
The development of fine motor skills eventually leads to writing and drawing, and other activities that require physical precision.
- Babies play with toys in one hand, then as children grow they become more coordinated and can pick up smaller objects, build things, put things together, and use crayons to draw.
- Allow your child to turn the pages of a book you are reading together or to help out with household chores: misting houseplants with water or wringing out the sponge when you do dishes.
All these things can be fun, help them with their fine motor development and help your child feel included too.
Language and Literacy
Language and literacy begins when a baby is born.
- Talk to baby while you change their diaper, bathe them, and cook dinner.
- Talk about things your child sees, hears, feels.
- Include your child in conversations, sing with your child and talk to them about the world that surrounds them.
- Speak in your first language so that children develop an understanding of the patterns of grammar through the stories behind them
-The patterns of language can be applied to a second language – speaking your first language with a child can help them be literate with a second language. Children must hear language spoken fluently to understand how it should sound.
Children develop numeracy skills by sorting objects, pouring water, counting things and adding objects to other objects. As they play, their brains are making connections which they will build on when they go to school.
A child’s emotional development is important for building self confidence and self esteem. When a child feels good about themselves they will learn and grow in productive ways.
- Hold your baby to feed and look into his eyes
- Provide positive guidance
- Talk to your child about feelings and emotions. Help him learn to identify and name them.
Many cultures believe in the holistic development of children and include spiritual development. Many cultures have their own ceremonies, practices and traditions that support the child’s development emotionally, and spiritually that provides the foundation for health and wellness in later years.
Child Development Chart - First Five Years<|endoftext|>
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# Magnetic Flux Formula
## Magnetic Flux Formula - Definitions & Practice Questions
Magnetic flux refers to the total number of magnetic field lines penetrating any surface placed perpendicular to the magnetic field. It is calculated as the product of the average magnetic field strength and the perpendicular area it penetrates.
Magnetic flux is denoted by ΦB where B represents magnetic field, and its unit is tesla-metre2 or weber (Wb).
Mathematically, ${\phi _B} = \vec B\,.\,\vec A$
Or, ${\phi _B} = BA\cos \theta$
Where, B is the magnetic field, A is the surface area and $\theta$is the angle between the normal to the surface and the magnetic field.
If a coil of n turns and area of cross section A is placed in a magnetic field of strength B, then the total flux associated with the coil is:
${\phi _B} = n\,BA\cos \theta$
Example:
A magnetic field of 2.5 T passes perpendicular to a disc of radius 2 cm. Find the magnetic flux associated with the disc.
Solution:
B = 2.5 T, r = 2 cm = 2 × 10–2 m, $\theta$= 0, ${\phi _B}$=?
${\phi _B} = BA\cos \theta = 2.5 \times \pi {(2 \times {10^{--2}})^2} \times \cos 0 = 3.14 \times {10^{--3}}Wb$
Example:
A coil of area of cross section 10–2 m2 and 100 turns is placed in a magnetic field of strength 1 T, with its axis making an angle 60° with the field. Find the total flux associated with the field.
Solution:
B = 1 T, A = 10–2 m2, $\theta$= 60°, n = 100, ${\phi _B}$=?
${\phi _B} = n\,BA\cos \theta = 100 \times 1 \times {10^{--2}} \times \cos 60^\circ = 0.5\,Wb$
Question: When the coil is rotated between the pole pieces of a magnet as shown, during one complete rotation of the coil, how often will the magnetic flux linked with coil be maximum and minimum?
Options:
(a) maximum and minimum once each,
(b) maximum and minimum twice each
(c) maximum once, minimum twice
(d) maximum twice, minimum once<|endoftext|>
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|Voice onset time|
In phonetics, aspiration is the strong burst of breath that accompanies either the release or, in the case of preaspiration, the closure of some obstruents. In English, aspirated consonants are allophones in complementary distribution with their unaspirated counterparts, but in some other languages, notably most Indian and East Asian languages, the difference is contrastive, while in Arabic and Persian, all stops are aspirated.
To feel or see the difference between aspirated and unaspirated sounds, one can put a hand or a lit candle in front of one's mouth, and say spin [spɪn] and then pin [pʰɪn]. One should either feel a puff of air or see a flicker of the candle flame with pin that one does not get with spin.
- 1 Transcription
- 2 Phonetics
- 3 Phonology
- 4 Examples
- 5 Other uses
- 6 See also
- 7 Notes
- 8 References
In the International Phonetic Alphabet (IPA), aspirated consonants are written using the symbols for voiceless consonants followed by the aspiration modifier letter ⟨◌ʰ⟩, a superscript form of the symbol for the voiceless glottal fricative ⟨h⟩. For instance, ⟨p⟩ represents the voiceless bilabial stop, and ⟨pʰ⟩ represents the aspirated bilabial stop.
Voiced consonants are seldom actually aspirated. Symbols for voiced consonants followed by ⟨◌ʰ⟩, such as ⟨bʰ⟩, typically represent consonants with murmured voiced release (see below). In the grammatical tradition of Sanskrit, aspirated consonants are called voiceless aspirated, and breathy-voiced consonants are called voiced aspirated.
There are no dedicated IPA symbols for degrees of aspiration and typically only two degrees are marked: unaspirated ⟨k⟩ and aspirated ⟨kʰ⟩. An old symbol for light aspiration was ⟨ʻ⟩, but this is now obsolete. The aspiration modifier letter may be doubled to indicate especially strong or long aspiration. Hence, the two degrees of aspiration in Korean stops are sometimes transcribed ⟨kʰ kʰʰ⟩ or ⟨kʻ⟩ and ⟨kʰ⟩, but they are usually transcribed [k] and [kʰ], with the details of voice onset time given numerically.
Preaspirated consonants are marked by placing the aspiration modifier letter before the consonant symbol: ⟨ʰp⟩ represents the preaspirated bilabial stop.
Unaspirated or tenuis consonants are occasionally marked with the modifier letter for unaspiration ⟨◌˭⟩, a superscript equals sign: ⟨t˭⟩. Usually, however, unaspirated consonants are left unmarked: ⟨t⟩.
Voiceless consonants are produced with the vocal folds open (spread) and not vibrating, and voiced consonants are produced when the vocal folds are fractionally closed and vibrating (modal voice). Voiceless aspiration occurs when the vocal folds remain open after a consonant is released. An easy way to measure this is by noting the consonant's voice onset time, as the voicing of a following vowel cannot begin until the vocal folds close.
Phonetically in some languages, such as Navajo, aspiration of stops tends to be realised as voiceless velar airflow; aspiration of affricates is realised as an extended length of the frication.
Aspirated consonants are not always followed by vowels or other voiced sounds. For example, in Eastern Armenian, aspiration is contrastive even word-finally, and aspirated consonants occur in consonant clusters. In Wahgi, consonants are aspirated only in final position.
The degree of aspiration varies: the voice onset time of aspirated stops is longer or shorter depending on the language or the place of articulation.
Armenian and Cantonese have aspiration that lasts about as long as English aspirated stops, in addition to unaspirated stops. Korean has lightly aspirated stops that fall between the Armenian and Cantonese unaspirated and aspirated stops as well as strongly aspirated stops whose aspiration lasts longer than that of Armenian or Cantonese. (See voice onset time.)
Aspiration varies with place of articulation. The Spanish voiceless stops /p t k/ have voice onset times (VOTs) of about 5, 10, and 30 milliseconds, whereas English aspirated /p t k/ have VOTs of about 60, 70, and 80 ms. Voice onset time in Korean has been measured at 20, 25, and 50 ms for /p t k/ and 90, 95, and 125 for /pʰ tʰ kʰ/.
When aspirated consonants are doubled or geminated, the stop is held longer and then has an aspirated release. An aspirated affricate consists of a stop, fricative, and aspirated release. A doubled aspirated affricate has a longer hold in the stop portion and then has a release consisting of the fricative and aspiration.
Icelandic and Faroese have consonants with preaspiration [ʰp ʰt ʰk], and some scholars[who?] interpret them as consonant clusters as well. In Icelandic, preaspirated stops contrast with double stops and single stops:
- kapp [kʰɑʰp] or [kʰɑhp] "zeal"
- gabb [kɑpp] "hoax"
- gap [kɑːp] "opening"
Preaspirated stops also occur in most Sami languages. For example, in Northern Sami, the unvoiced stop and affricate phonemes /p/, /t/, /ts/, /tʃ/, /k/ are pronounced preaspirated ([ʰp], [ʰt] [ʰts], [ʰtʃ], [ʰk]) in medial or final position.
Fricatives and sonorants
Although most aspirated obstruents in the world's languages are stops and affricates, aspirated fricatives such as [sʰ], [ɸʷʰ] or [ɕʰ] have been documented in Korean, though these are allophones of other phonemes. Similarly, aspirated fricatives and even aspirated nasals, approximants, and trills occur in a few Tibeto-Burman languages, in some Oto-Manguean languages, in the Hmongic language Hmu, and in the Siouan language Ofo. Some languages, such as Choni Tibetan, have as many as four contrastive aspirated fricatives [sʰ] [ɕʰ], [ʂʰ] and [xʰ].
Voiced consonants with voiceless aspiration
True aspirated voiced consonants, as opposed to murmured (breathy-voice) consonants such as the [bʱ], [dʱ], [ɡʱ] that are common in the languages of India, are extremely rare. They have been documented in Kelabit Taa, and the Kx'a languages. Reported aspirated voiced stops, affricates and clicks are prevoiced [b͡pʰ, d͡tʰ, d͡tsʰ, d͡tʃʰ, ɡ͡kʰ, ɢ͡qʰ, ᶢᵏʘʰ, ᶢᵏǀʰ, ᶢǁʰ, ᶢᵏǃʰ, ᶢᵏǂʰ].
Aspiration has varying significance in different languages. It is either allophonic or phonemic, and may be analyzed as an underlying consonant cluster.
They are unaspirated for almost all speakers when immediately following word-initial s, as in spill, still, skill. After an s elsewhere in a word they are normally unaspirated as well, except sometimes in compound words. When the consonants in a cluster like st are analyzed as belonging to different morphemes (heteromorphemic) the stop is aspirated, but when they are analyzed as belonging to one morpheme the stop is unaspirated. For instance, distend has unaspirated [t] since it is not analyzed as two morphemes, but distaste has an aspirated middle [tʰ] because it is analyzed as dis- + taste and the word taste has an aspirated initial t.
Word-final voiceless stops are sometimes aspirated.
Voiceless stops in Pashto are slightly aspirated prevocalically in a stressed syllable.
In many languages, such as Armenian, Korean, Lakota, Thai, Indo-Aryan languages, Dravidian languages, Icelandic, Ancient Greek, and the varieties of Chinese, tenuis and aspirated consonants are phonemic. Unaspirated consonants like [p˭ s˭] and aspirated consonants like [pʰ ʰp sʰ] are separate phonemes, and words are distinguished by whether they have one or the other.
In Danish and most southern varieties of German, the lenis consonants transcribed for historical reasons as ⟨b d ɡ⟩ are distinguished from their fortis counterparts ⟨p t k⟩, mainly in their lack of aspiration.
Standard Chinese (Mandarin) has stops and affricates distinguished by aspiration: for instance, /t tʰ/, /t͡s t͡sʰ/. In pinyin, tenuis stops are written with letters that represent voiced consonants in English, and aspirated stops with letters that represent voiceless consonants. Thus d represents /t/, and t represents /tʰ/.
Wu Chinese and Southern Min has a three-way distinction in stops and affricates: /p pʰ b/. In addition to aspirated and unaspirated consonants, there is a series of muddy consonants, like /b/. These are pronounced with slack or breathy voice: that is, they are weakly voiced. Muddy consonants as initial cause a syllable to be pronounced with low pitch or light (陽 yáng) tone.
Many Indo-Aryan languages have aspirated stops. Sanskrit, Hindi, Bengali, Marathi, and Gujarati have a four-way distinction in stops: voiceless, aspirated, voiced, and breathy-voiced or voiced aspirated, such as /p pʰ b bʱ/. Punjabi has lost breathy-voiced consonants, which resulted in a tone system, and therefore has a distinction between voiceless, aspirated, and voiced: /p pʰ b/.
Some of the Dravidian languages, such as Telugu, Malayalam, and Kannada, have a distinction between voiced and voiceless, aspirated and unaspirated only in loanwords from Indo-Aryan languages. In native Dravidian words, there is no distinction between these categories and stops are underspecified for voicing and aspiration.
Most dialects of Armenian have aspirated stops, and some have breathy-voiced stops.
Western Armenian has a two-way distinction between aspirated and voiced: /tʰ d/. Western Armenian aspirated /tʰ/ corresponds to Eastern Armenian aspirated /tʰ/ and voiced /d/, and Western voiced /d/ corresponds to Eastern voiceless /t/.
Some forms of Greek before the Koine Greek period are reconstructed as having aspirated stops. The Classical Attic dialect of Ancient Greek had a three-way distinction in stops like Eastern Armenian: /t tʰ d/. These series were called ψιλά, δασέα, μέσα (psilá, daséa, mésa) "smooth, rough, intermediate", respectively, by Koine Greek grammarians.
There were aspirated stops at three places of articulation: labial, coronal, and velar /pʰ tʰ kʰ/. Earlier Greek, represented by Mycenaean Greek, likely had a labialized velar aspirated stop /kʷʰ/, which later became labial, coronal, or velar depending on dialect and phonetic environment.
The other Ancient Greek dialects, Ionic, Doric, Aeolic, and Arcadocypriot, likely had the same three-way distinction at one point, but Doric seems to have had a fricative in place of /tʰ/ in the Classical period, and the Ionic and Aeolic dialects sometimes lost aspiration (psilosis).
So-called voiced aspirated consonants are nearly always pronounced instead with breathy voice, a type of phonation or vibration of the vocal folds. The modifier letter ⟨◌ʰ⟩ after a voiced consonant actually represents a breathy-voiced or murmured dental stop, as with the "voiced aspirated" bilabial stop ⟨bʰ⟩ in the Indo-Aryan languages. This consonant is therefore more accurately transcribed as ⟨b̤⟩, with the diacritic for breathy voice, or with the modifier letter ⟨bʱ⟩, a superscript form of the symbol for the voiced glottal fricative ⟨ɦ⟩.
Some linguists restrict the double-dot subscript ⟨◌̤⟩ to murmured sonorants, such as vowels and nasals, which are murmured throughout their duration, and use the superscript hook-aitch ⟨◌ʱ⟩ for the breathy-voiced release of obstruents.
- Ladefoged, Peter; Barbara Blankenship; Russell G. Schuh, eds. (21 April 2009). "Korean". UCLA Phonetics Archive. Retrieved 20 February 2015. word lists from 1977, 1966, 1975.
- Lisker and Abramson (1964). "A cross-language Study of Voicing in Initial Stops". Word. 20: 384–422.
- Guillaume Jacques 2011. A panchronic study of aspirated fricatives, with new evidence from Pumi, Lingua 121.9:1518–1538
- Robert Blust, 2006, "The Origin of the Kelabit Voiced Aspirates: A Historical Hypothesis Revisited", Oceanic Linguistics 45:311
- Tranel, Bernard (1987). The sounds of French: an introduction (3rd ed.). Cambridge, New York: Cambridge University Press. pp. 129–130. ISBN 0-521-31510-7.
- Frans Hinskens, Johan Taeldeman, Language and space: Dutch, Walter de Gruyter 2014. 3110261332, 9783110261332, p.66
- Cho, T., & Ladefoged, P., "Variations and universals in VOT". In Fieldwork Studies of Targeted Languages V: UCLA Working Papers in Phonetics vol. 95. 1997.<|endoftext|>
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April 26 (UPI) -- Scientists have credited comets and asteroids with delivering water and other elements necessary for life to Earth shortly after the planet was formed. But impact models suggest water should be entirely boiled away by the heat generated by such a collision.
Brown University researchers used a high-powered projectile cannon to investigate the contradiction. The results of their experiments, detailed this week in the journal Science Advances, suggest surprising amounts of water can survive asteroid impacts.
"The origin and transportation of water and volatiles is one of the big questions in planetary science," Terik Daly, now a postdoctoral researcher at Johns Hopkins University, said in a news release. "These experiments reveal a mechanism by which asteroids could deliver water to moons, planets and other asteroids. It's a process that started while the solar system was forming and continues to operate today."
Originally, scientists thought icy comets supplied most of the water found on Earth, but isotopic analysis suggests Earth's early water supply was most similar to water found trapped in carbonaceous asteroids.
How exactly water would survive a rock-on-rock collision, however, has remained a mystery. Computer models suggest all water would be completely vaporized during such a collision.
"But nature has a tendency to be more interesting than our models, which is why we need to do experiments," said Pete Schultz, professor of planetary science at Brown.
Schultz, Daly and their colleagues built miniature asteroid models based on the composition of carbonaceous chondrites, the remains of water-rich asteroid fragments that struck Earth long ago. The research team loaded the mini asteroids into the Vertical Gun Range at the NASA Ames Research Center and shot them into a compacted wall of pumice powder at speeds of approximately 11,000 miles per hour.
Scientists replicated impacts featuring speeds and angles common throughout the solar system. They found as much as 30 percent of the original water content remained trapped in the fragments that survived the collisions.
Though almost all of the water is vaporized, some of that vapor gets trapped in a plume of impact material and reincorporated into debris as melted rocks cool into solid fragments.
"The impact melt and breccias are forming inside that plume," Schultz said. "What we're suggesting is that the water vapor gets ingested into the melts and breccias as they form. So even though the impactor loses its water, some of it is recaptured as the melt rapidly quenches."
The research could not only help explain water on early Earth, but also more recent signs of water on the moon and the planetary bodies.
"The point is that this gives us a mechanism for how water can stick around after these asteroid impacts," Schultz said. "And it shows why experiments are so important because this is something that models have missed."
Earlier this year, researchers in Russia blasted miniature asteroid models with a high-powered laser to better understand the kinds of forces needed to destroy a space rock headed for planet Earth.<|endoftext|>
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The age pyramid describes the age distribution within countries and it is called a pyramid because it closely resembles the shape. There are different stages within the age pyramid as explained on Wikipedia http://en.wikipedia.org/wiki/Population_pyramid.
While all countries’ population pyramids differ, four general types have been identified by the fertility and mortality rates of a country.
A population pyramid showing an unchanging pattern of fertility and mortality.
A population pyramid typical of countries with low fertility and low mortality, very similar to a constrictive pyramid.
A population pyramid showing a broad base, indicating a high proportion of children, a rapid rate of population growth, and a low proportion of older people. This wide base indicates a large number of children. A steady upwards narrowing shows that more people die at each higher age band. This type of pyramid indicates a population in which there is a high birth rate, a high death rate and a short life expectancy. This is the typical pattern for less economically developed countries, due to little access to and incentive to use birth control, negative environmental factors (for example, lack of clean water) and poor access to health care.
A population pyramid showing lower numbers or percentages of younger people. The country will have a greying population which means that people are generally older, as the country has long life expectancy, a low death rate, but also a low birth rate. This pyramid has been occurring more frequently, especially when immigrants are factored out, and is often a typical pattern for a very developed country, a high over-all education and easy access and incentive to use birth control, good health care and few or no negative environmental factors.
Overall 3rd world countries show the first two sages and developed countries show the last two stages.
This development of pyramid shape can be perfectly observed in the population distribution history of the Netherlands.
For more information on the age Pyramid of the Netherlands look to:<|endoftext|>
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Use found objects to make a cool creation.
- Target Age: 3-5
- Subject: engineering, art
- Related Episode: The All-Animal Recycled Band
- building materials like these:
- cardboard boxes
- food and drink containers
- cardboard tubes
- paper plates
- drinking straws
- rubber bands
- tape, glue, or string
- scissors (adult handling or supervision)
Give children (or monkeys!) some simple everyday materials, and their imaginations will help them create all sorts of things. By building and experimenting, your child will also learn some basic engineering concepts.
1. Imagine: Look at the materials together with your child. What possibilities do you see? What could you build? A castle from juice cartons and paper towel rolls? A guitar from a tissue box and rubber bands? A robot? A rocket ship? Let your imaginations go wild!
2. Design: Draw pictures of your ideas on plain paper. Talk about them, then choose one to build.
3. Build: As you build your invention, talk about what's working and what's not. Do you need to change your design? Do you need to scale back, or can you go even further than you thought?
Store all those initial design drawings in a folder or binder and pull them out whenever the mood strikes! When the time comes, talk about those original ideas with your child. Would you change anything? Do you have any new ideas? What will you build this time?
More Ways to Discover and Learn
Look in a Book
- Albert's Alphabet by Leslie Tryon
- Bling Blang by Woody Guthrie
- What Can You Do With a Shoe? by Beatrice de Regniers<|endoftext|>
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April 29, 2019History of Medicine
Asthma was recognized in ancient Egypt and was treated by drinking an incense mixture known as kyphi. Asthma was officially named as a specific respiratory problem by Hippocrates(c. 450 BCE), with the Greek word for panting forming the basis of our modern name. In 200 BCE, asthma was believed to be at least partly related to the emotions. In the 12th century the Jewish physician-philosopher Maimonides wrote a treatise on asthma in Arabic, based partly on Arabic sources, in which he discussed the symptoms, proposed various dietary and other means of treatment, and emphasized the importance of climate and clean air.
In 1873, one of the first papers in modern medicine on the subject tried to explain the pathophysiology of the disease while one in 1872, concluded that asthma can be cured by rubbing the chest with chloroform liniment. Medical treatment in 1880 included the use of intravenous doses of a drug called pilocarpine. In 1886, F. H. Bosworth theorized a connection between asthma and hay fever. Epinephrine was first referred to in the treatment of asthma in 1905. Oral corticosteroids began to be used for this condition in the 1950s while inhaled corticosteroids and selective short acting beta agonist came into wide use in the 1960s.
A notable and well-documented case in the 19th century was that of young Theodore Roosevelt (1858-1919). At that time there was no effective treatment. Roosevelt's youth was in large part shaped by his poor health partly related to his asthma. He experienced recurring nighttime asthma attacks that caused the experience of being smothered to death, terrifying the boy and his parents. Another well-known historic figure, with chronic asthma was Antonio Lucio Vivaldi, (1678-1741), enormously prolific, Italian composer and violinist who left a decisive mark on the form of the concerto and the style of late Baroque instrumental music. Vivaldi was born in Venice, the capital of the Venetian Republic, and is regarded as one of the greatest Baroque composers. Vivaldi's influence during his lifetime was widespread across Europe. He composed many instrumental concertos, for the violin and a variety of other instruments, as well as sacred choral works and more than forty operas. His best-known work is a series of violin concertos known as the Four Seasons.
Vivaldi's health was problematic. One of his symptoms, strettezza di petto (tightness of the chest), has been interpreted as a form of asthma. This did not prevent him from learning to play the violin, composing, or taking part in musical activities, although it did stop him from playing wind instruments. Vivaldi's violinist father taught him to play the violin at an early age. At thirteen, he created the liturgical work Laetatus sum (RV Anh 31), written in 1691. In 1693, at the age of fifteen, Vivaldi began studying to become a priest. He was ordained in 1703, aged 25, and was soon nicknamed il Prete Rosso, The Red Priest. (Rosso is Italian for red, and would have referred to the color of his hair, a family trait.) In September 1703, Vivaldi became maestro di violino (master of violin) at an orphanage called the Pio Ospedale della Pieta (Devout Hospital of Mercy) in Venice. While Vivaldi is most famous as a composer, he was regarded as an exceptional technical violinist as well. The German architect Johann Friedrich Armand von Uffenbach referred to Vivaldi as :
the famous composer and violinist and said that Vivaldi played a solo accompaniment excellently, and at the conclusion he added a free fantasy [an improvised cadenza] which absolutely astounded me, for it is hardly possible that anyone has ever played, or ever will play, in such a fashion.
Not long after his ordination, in 1704, Vivaldi was given a dispensation from celebrating Mass because of his ill health. Vivaldi said Mass as a priest only a few times, and appeared to have withdrawn from liturgical duties, though he formally remained a member of the priesthood. Relieved of priestly duties, Vivaldi was free to utilize his time composing. It's possible that if it weren't for his asthma, Vivaldi might not have become one of the greatest Baroque musicians and composers. Over the next thirty years he composed most of his major works while working at the orphanage, whose purpose was to give shelter and education to children who were abandoned or orphaned, or whose families could not support them. They were financed by funds provided by the Republic. The boys learned a trade and had to leave when they reached the age of fifteen. The girls received a musical education, and the most talented among them stayed and became members of the Ospedale's renowned orchestra and choir. Shortly after Vivaldi's appointment, the orphans began to gain appreciation and esteem abroad, too. Vivaldi wrote concertos, cantatas and sacred vocal music for them. These sacred works, which number over 60, are varied: they included solo motets and large-scale choral works for soloists, double chorus, and orchestra. In 1704, the position of teacher of viola all'inglese was added to his duties as violin instructor. The position of maestro di coro, which was at one time filled by Vivaldi, required a lot of time and work. He had to compose an oratorio or concerto at every feast and teach the orphans both music theory and how to play certain instruments. Vivaldi became responsible for all of the musical activity of the institution when he was promoted to maestro de' concerti (music director) in 1716.
At the height of his career, Vivaldi received commissions from European nobility and royalty. Vivaldi's Opus 9, La cetra, was dedicated to Emperor Charles VI, who gave Vivaldi the title of knight, a gold medal and an invitation to Vienna. Vivaldi gave Charles a manuscript copy of La cetra, a set of concerti almost completely different from the set of the same title published as Opus 9. After the success of his meeting with Emperor Charles VI, Vivaldi wished to take up the position of a composer in the imperial court and to stag operas, which were the most popular musical form at this time. Shortly after his arrival in Vienna, Charles VI died, which left the Vivaldi without any royal protection or a steady source of income. Soon afterwards, Vivaldi became impoverished and died during the night of 27 July 1741, aged 63, of internal infection. On 28 July, Vivaldi was buried in a simple grave in a burial ground that was owned by the public hospital fund. His modest funeral took place at St. Stephen's Cathedral, where no music was performed.
Since readers are probably familiar with the Four Seasons, below for your listening pleasure are five other (of my) favorite Vivaldi pieces. The first two are from Vivaldi operas.
Sposa, son disprezzata Bajazet - Vivaldi, Sung by the great coloratura, Montserrat Caballe
Vivaldi's Most Beautiful Aria, Sovente il sole, Andromeda Liberata, sung by a perfect voice for Baroque, Anne Sophie Von Otter
A. VIVALDI: Concerto for 2 Violins and Cello in D minor Op. 3/11 RV 565, Akademie fur Alte Musik
VIVALDI - A Rain of Tears - ANDERSON & ROE (tears fall down the face of one of the pianists)
A. VIVALDI: Viola d'amore Concerto in A minor RV 397, Accademia Bizantina
All this divine music, (perhaps) because Vivaldi had chronic asthma.<|endoftext|>
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What is a native plant?
A native plant is one that occurs naturally in a particular region, ecosystem or habitat – and occurred prior to European contact.
Native plants can be mosses, ferns, grasses, wildflowers, shrubs, trees and more!
Native plants have co-evolved with animals, fungi, and microbes to form a complex network of relationships. These plants are the foundation of native ecosystems, or natural communities.
Why are native plants important?
They are the basis of our native ecosystems. All forms of native organisms depend on native plants – directly or indirectly.
Insects such as butterflies may use plants for food, shelter or as places to their eggs.
Birds may nest in their branches, eat their berries or hide from predators within them.
Mammals may den in them, eat their roots, shoots and leaves or use them as cover from the weather.
Even fish, amphibians and reptiles need plants!
Where can I see native plants in BC?
Native plants are all around you!
Some, you might already be familiar with – like western red cedar (Thuja plicata) or Pacific dogwood (Cornus nuttallii).
Others you might have seen and not even realized that they’re native – like tiger lilies (Lilium columbianum) and skunk cabbage (Lysichiton americanum).
There are more than 3,000 species of native plants in our province.
How can I find out more?
Check out our Facebook page
Either search for us or try this direct link
There are some good books that will tell you more about native plant species. Click here to find some titles.<|endoftext|>
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Mathematics
Easy
Question
# The father's age is six times his son's age. Four years hence, the age of the father will be four times his son's age. The present ages, in years, of the son and the father respectively are
Hint:
## The correct answer is: 6 and 36
### Here we have to find that age of father and son,Firstly , it is given at first the father age is six times its son .Lets father be a and son be b,So , b = 6a ---(1)And after four year ,Father age is four times to his son , so we can write, b+4 = 4(a+4) [we add 4 in both sides because it is after 4 years so both age ages are added by 4] b + 4 =4a + 16 6a + 4 = 4a + 16 ( using (1)) 6a – 4a = 16 – 4 2a =12 a = 6 put value of a in eq (1),b = 6ab = 6 x 6b = 36Therefore , The present age of son and father is 6 and 36The correct answer is 6 and 36.Or,Let, the present age of father = x yearsLet, the present age of son = y yearsGiven that fathers Age is 6 time of sons age⇒x=6y…(1)After four years,Age of the father will be four times of his son’s age⇒x+4=4(y+4)⇒x+4=4y+16⇒x=4y+16−4⇒x=4y+12…(2) Step2: Solving the equations.Substitute equation (1) in equation (2)⇒6y=4y+12⇒2y=12⇒y=6Substitute y=6 in equation (1)⇒x=36Hence, the solution of the father and his son ages are .
In this question, we have to make equation on the given date about ages. Consider all the character in similar ages question be some variable and make the equation on given details.<|endoftext|>
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People have lived on the outer part of the hook of land that forms Cape Cod for thousands of years. Cape Cod and the islands of Martha's Vineyard and Nantucket were formed as massive terminal moraines at the end of the last glacial period in North America about 15,000 years ago. Ancient artifacts, such as Paleoindian projectile points found at several locations on Cape Cod indicate that humans have occupied this land, or at least traversed it for the last 10,000 years. Thousands of archeological sites exist throughout the Cape. A number of these sites have been scientifically studied and some can be visited. The interpretation of ancient and historic activities at these sites provides a view of human uses of the Cape Cod coastline from earliest times.
The archeology of Cape Cod has been of interest to inhabitants and visitors for hundreds of years. In his travel narrative about the region, Cape Cod, Henry David Thoreau observed that Cape Cod was once "thickly settled" by Indians and that traces of their occupation, in the form of "arrow-heads," and piles of shell, ashes, and deer bones, could be seen around the marsh edges and inlets throughout the Cape. More systematic and concerted archeological studies on the outer Cape by National Park Service archeologists in the 1980s showed concentrations of ancient villages and activities around Nauset Harbor and Wellfleet Harbor, as well as in the High Head area. Other researchers have found concentrations of sites in Truro near the mouth of the Pamet River and in many locations in the western portion of the Cape.
By 5000 years ago, the human presence on Cape Cod was quite extensive. Artifacts, projectile points in particular, dating from this period are found throughout the Cape; however, sites are rare. During this early period of settlement, human groups may have moved seasonally from one part of the Cape to another without establishing permanent settlements. It may also be that the remains of such settlements are buried deeply and are rarely found and investigated by archeologists. By 3000 years ago, people left dense deposits of ancient trash, including discarded stone tools, stone flakes used as tools or from tool sharpening, shell from intensive gathering of shellfish for food, fish and animal bone, and ash and stone from fires for cooking and heat. These are found at sites in the Nauset area and probably exist in other areas where settlement was concentrated. Permanent settlement was probably the norm by this time, with parties of men and women traveling out from the villages to hunt or gather food and raw material for making tools, clothing, and shelter.
Francis P. McManamon, National Park Service
Francis P. McManamon is recently retired as Chief Archeologist for the National Park Service, Department of the Interior, in Washington D.C. His Cape Cod National Seashore Archeological Survey conducted between 1979 and 1981 resulted in a three-volume report entitled "Chapters in the Archeology of Cape Cod", published between 1984 and 1985, and "The Indian Neck Ossuary" published in 1986.
The Nauset Archaeological District, Eastham
Last updated: October 31, 2017<|endoftext|>
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2 Geometry 3 SIMILARITY & CONGRUENCY Congruency: When two figures have same shape and size, then they are said to be congruent figure. The phenomena between these two figures is said to be congruency. CONDITIONS FOR TRIANGLES TO BE CONGRUENT When three sides are equal When two sides and their included angles are equal When two angles and one side is equal CONDITIONS FOR TRIANGLES TO BE CONGRUENT When three sides are equal AB=QR BC =PQ AC = PR ABC RQP. Reference: SSS When two sides and their included angles are equal If AB = PQ BC = QR, and B = Q Then ABC PQR. Reference: SAS
3 When two angles and one side is equal If B = Q A = P and AB = PQ Then ABC PQR. Reference: ASA Similiarity: Two figures are said to be similar when they have same shape. Example : Proto type of a building and the actual building are similar When we deal with two dimension figures we use similarity to prove the given figures are similar. Similarity can be proved by conditions given below CONDITIONS FOR TRIANGLES TO BE SIMILAR Three Angles are Equal If A = X B = Y and C = Z then ABC ~ XYZ. Reference: AAA WHEN TWO TRIANGLE ARE SIMILAR Three Angles are Equal If ABC ~ XYZ. Then, BC = AC = AB YZ XZ XY Reference: By similar Triangles
4 BASIC THEOREM OF PROPORTIONALITY Thales Theorem: If a straight line divides any two sides of a triangle in the same ratio, then the straight line is parallel to the third side of the triangle. If, AD = AE DB EC Then DE BC MID POINT THEOREM If the midpoints of two adjacent sides of a triangle are joined by a line segment, then this segment is parallel and half of the third side, If, AD = BD & AE = CE, Then DE BC and DE = 1 2 BC THEOREM The areas of similar triangles are proportional to the squares of corresponding sides, altitude, or median. Then, ar ( ABC) = AB2 = AC2 = BC2 = AM2 or ( DEF) DE 2 DF 2 EF 2 DN 2 SOME IMPORTANT CONCLUSION APB is a right triangle with P = 90. PN is perpendicular, drawn from P, to the hypotenuse AB. Then, (i) PN 2 = AN.NB (ii) AP 2 = AN.AB (iii) BP 2 = BN.BA
5 Ex: In Fig. AB and DE are perpendiculars to BC. If AB = 9 cm, DE = 3 cm and AC = 24 cm. Calculate AD. Sol: In similar ABC and CDE AB DE = AC DC 9 = 24 3 DC DC = 8 cm AD = AC DC = 24 8 = 16 cm Ex: In Fig. AQ and PB are perpendiculars to AB. If AQ = 14 cm, PB = 3.5 cm and AO = 6 cm. Calculate OP.? (Approx.) Sol: In similar OAQ and OBP, OA = OB AQ PB 6 = OB , OB = 1.5 cm In right triangle OBP, OP 2 = OB 2 + PB 2 OP 2 = = 14.5, OP = 3.8 cm Ex: The sides A, BC of a trapezium ABCD are parallel & the diagonals AC, BD meet at O. The area of the BOB = 5cm 2, and the ratio between OA : OC = 3 :5. Calculate the area of triangle AOD?
6 Sol: In similar AOD and COB Then, ar( AOD) ar( BOC) =AO2 BO 2 ar( AOD) 5 = 9 25 ar( AOD) = 9 = 1.8 cm2 5 Ex: In the given figure, ABC and CEF are two triangles where BA is parallel to CE, and AF: AC = 5: 8. Find AD, if CE = 6 cm Sol: In similar triangles ADF and CEF AD = AF CE CF AD = 5x 6 3x AD = 10 cm (By theorem) (AF = 5x, AC = 8x, CF= 3x) Ex: BE and CF be the two medians of a ABC and G be their intersection. Also let EF cut AG at O. Then find AO: OG? Sol: In similar AOE and ADC, OA AD = AE AC = 1 2 But AD AG = 3 2 OA AD x AD AG = OA = 3 AG 4 OA = 3(OA + OG) OA = 3 OG (E is the mid point of AC) OA : OG = 3 : 1
7 Ex: In PQR, S and T are points on side PR and PQ respectively such that PQR = PST. If PT = 5cm, PS = 3cm and TQ = 3cm, then find the length of SR? Sol: In PQR and PST, PQR = PST and P is common PQ = PR PS PT 8 3 = PR 5 PR = 8 X 5 3 = 40 40, then SR = =31 9 Ex: In ABC it is given that AB = 6cm, AC = 8cm AD is the angle bisector of BAC. Then find the ratio of BD and DC? Sol: In ABC, AD is angle bisector of BAC so BD = AB 6 DC AC 8 BD : AC = 3 : 4 Ex: In ABC D is point on BC in such that AD BC and E is point on AD in such that AE : ED = 5:1, If BAD = 30 0 and tan ( ACB) = 6 tan ( DBE) then find the ( ACB)? Sol. tan ( ACB) = 6 tan ( DBE) ( ACB) = 6 ( DBE) DBA = = 60 0 In ADB & EDB
8 DBA = AD 600 = 6 DBE ED DBE 1 DBE = 10 0 ACB = 6 x 10 = 60 0 Ex: A straight line parallel to BC of ABC intersects AB and AC at point P and Q given. AP = QC, PB =4 units and AQ = 9 units, then find the length of AP? Sol: In ABC, PQ BC AP AB = AQ AC AP+PB AP = AQ+QC AQ PB = QC AP AQ = AP AQ Then AP 2 =PB x AQ = 4 x 9 = 36 AP = 6 Ex: In the given figure BAD = CAD, AB= 4 cm., AC = 5.2 BD = 3 cm then find length of the BC? Sol: BAD = CAD AB AC = BD CD = 3 CD CD= 3.9 BC = = 6.9 cm
9 Ex: In ABC, D and E point onn AB and AC in such that AD = 1 3 AB and AE = 1 AC. If BC = 15cm then find the length of DE? 3 Sol: AD AC = AE AC = DE BC =1 3 DE 15 = 1 3 DE= 5cm Ex: In ABC, ab = 10 cm, BC = 8cm, CA = 6cm and M is mid point of BC is a point on AC in such that MN AB, then find the area of trapezium ABMN? Sol: In ABC MN AB So, CN = NM CA AB 3 = NM 6 10 NM = 5 cm area of ABC = 24 cm 2 area of CNM = 6 So area trapezium ABMN = 24 6 = 18cm 2
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# Statistics in Year 6 (age 10–11)
In Year 6, your child will read and make pie charts and line graphs. They will also calculate and interpret mean averages from sets of data.
The key words for this section are continuous data, discrete data, and pie chart.
## What your child will learn
Take a look at the National Curriculum expectations for statistics in Year 6 (age 10–11):
#### Read and make pie charts and line graphs and use them to solve problems
Your child will connect their work on angles, circles, fractions, and percentages to draw and interpret pie charts. This will involve:
• Illustrating and naming parts of circles, using the language radius, diameter, circumference, and centre.
• Knowing that the diameter is twice the length of the radius.
• Knowing that are 360° in a circle and relate this to pie charts (for example, 360° shows 100% of the data).
Your child will also be able to interpret information from line graphs.
#### Calculate and interpret mean averages
Your child will calculate and interpret the mean average of a set of data. The mean average is the average of adding together the numbers in a data set and then dividing by the number of values in the set.
Your child will know when it is appropriate to find the mean of a data set. For example, it would be appropriate for finding the average daytime temperature over a week. It would not make sense to try to find the mean average of favourite ice cream choice.
## How to help at home
There are lots of everyday ways you can help your child to understand statistics. Here are just a few ideas.
### 1. Create a line graph
Line graphs are used to show information that changes over time. Your child will use them to represent continuous data such as temperature, which can have an unlimited number of possible values within a selected range. Your child will plot the values on a graph from data they have collected or using data given to them, and will join each point together with a continuous line.
Why not measure and record the daily temperature over a period of time? Ask your child to make predictions about what data they expect to collect. When you have collected the data, discuss what their graph may look like and then have a go. Remember, they may have to extend the scales on the graph to include negative numbers.
As an example of what this might look like, here is a graph showing the overnight temperatures recorded between 10pm and 4am one evening during the winter:
What was the highest and lowest temperatures and when did these occur?
How much did the temperature fall between 11pm and 3am?
How long did the temperature stay below zero?
Can you estimate the temperature at 11.30pm?
### 2. Investigate pie charts
A pie chart is a circular chart which is divided into sectors that represent parts of the total. Pie charts are used to represent discrete data (data that can only take certain values) and is used to show different sizes.
At school, your child will use their understanding of circles to interpret pie charts. For example, they will be able to name parts of circles, using the language radius, diameter, and centre.
They will also understand how to calculate fractions from pie charts. For example:
Look at the pie chart below. The blue section represents 42 people who took part in a survey. Can you find the total number of people who took part in the survey?
Looking at the pie chart, we can tell that a quarter of the pie represents 42 people.
Therefore, the whole pie chart represents four groups of 42, which is 168 people (42 × 4).
Your child may need to interpret data from pie charts using percentages to represent each sector. For example:
How many people chose apples?
25% of 200 people = 50 people
How many more people voted for strawberry than apple?
100 – 50 = 50 more people
What was the total number of people who chose grapes, oranges, and strawberries?
20 + 10 + 100 = 130 people
Your child will also construct pie charts by finding the size of each sector in degrees and using a protractor to measure the angles accurately. They will know that angles in a circle total 360° and that this represents 100% of the data in a pie chart. For example, if a sector represents 50% of the pie chart, this is equivalent to 180°.
### 3. Find mean averages
The mean is one type of average. You can find the mean by adding together all values in a set of data, and then dividing your answer by the total number of values.
Try to explore this in real life. For example, you could investigate the mean number of goals per match your child’s favourite football team has scored in the last ten matches. Here is an example:
The number of goals scored per match is: 3, 2, 0, 1, 2, 1, 3, 2, 2, 4. What are the mean average goals per match?
Find the total number of goals first: 3 + 2 + 0 + 1 + 2 + 1 + 3 + 2 + 2 + 4 = 20
Divide 20 (number of goals scored) by 10 (number of matches) to find the mean: 2 goals per match
Alternatively, your child could work out the mean number of minutes it takes to walk to school over a week, or the mean temperature of a given week, or the mean score of their latest tests at school. There are all sorts of places the mean average can be really useful and interesting!
### Video: What are averages?
Find out how to calculate mean, median and mode averages.<|endoftext|>
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# Proportion in a Right Triangle Worksheets
What Proportions Apply in a Right Triangles? Right triangles are pretty interesting shapes having many intricacies and theorems attached to them. A right triangle consists of three sides like every other form of a triangle. All of these sides are unequal in measurements. The longest of these sides is known as the hypotenuse, and the other two sides are called base and perpendicular. The angle between the base and the perpendicular is ninety degrees, and the sum of the other two angles is collectively equal to ninety degrees. A right triangle can be broken down into proportions by dropping a perpendicular from the vertex that is opposite to the hypotenuse and has an angle of ninety degrees, to the hypotenuse. In this way, we break the right triangle in two smaller right triangles that are proportional to the bigger right triangle in size. Following are the ratios in which these dimensions can be divided: The perpendicular of the bigger right triangle to its base. The larger part of the hypotenuse of the bigger right triangle to the length of the perpendicular drawn internally. The internal perpendicular to the smaller part of the hypotenuse of the larger triangle.
• ### Basic Lesson
Guides students through solving Proportion in a Right Triangle.
• ### Intermediate Lesson
Demonstrates the concept of advanced skill while solving Proportion in a Right Triangle. In right triangle ABC, CD is the altitude to the hypotenuse, AB. The segments of the hypotenuse, AB, are in the ratio of 3:7. The altitude is 8. Find the two segments of the hypotenuse.
• ### Independent Practice 1
A really great activity for allowing students to understand the concepts of the Proportion in a Right Triangle. In right triangle ABC, CD is the altitude to the hypotenuse, AB. The segments of the hypotenuse, AB, are in the ratio of 2:5. The altitude is 8. Find the two segments of the hypotenuse.
• ### Independent Practice 2
Students use Proportion in a Right Triangle in 20 assorted problems. The answers can be found below.
• ### Homework Worksheet
Students are provided with 12 problems to achieve the concepts of Proportion in a Right Triangle.
• ### Skill Quiz
This tests the students ability to understand Proportion in a Right Triangle.
• ### Answer Key
Answers for all lessons and independent practice.
#### Eye Color Problem
If you took 100 people and put them in a room, and 63 had blue eyes, than how many people would have brown eyes?
Answer: most people would say 37 people ( 100 minus 63); however, that answer would be wrong, since the odds are that at least a few of the people remaining would have either green or hazel eyes. The real answer is: between 1 and 37.<|endoftext|>
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Origins of World War II and the Holocaust Unit 10 Guiding Questions • What economic and political conditions following World War I encouraged dictatorship? • How did European nations try to prevent war? Terms to Know • • • • • Fascism Totalitarianism Appeasement Exploit Dictator Rise of Dictators • Anti-democratic governments rose in both Europe and Asia following World War I ▫ Aided By: Treaty of Versailles Economic Depression • Countries would soon break the various provisions of the Treaty of Versailles European Dictators ITALY SOVIET UNION GERMANY Leader Benito Mussolini Joseph Stalin Adolph Hitler Type of Government Fascist Communist Nazi (Fascist) Characteristics - Aggressive nationalistic movement Nation more important than individuals Order in society and national greatness comes from dictator who led a strong government Strongly anticommunist Took power by force - - Government owns all means of production Family farms combined into collectives, government-owned farms Target political enemies, artists, and intellectuals Between 15-20 million died under Stalin’s rule - - Called for Germany to expand its borders and to reject terms of Treaty of Versailles Anti-Semitic (AntiJewish) Germans belong to the master Aryan race All Slavic peoples of Eastern Europe were to be made slaves Jews to be punished for causing world’s problems Militarist Gain Control of Japan • Depression weakened Japan’s political system • High tariffs hurt Japanese economy • Military leaders argued expansion was needed to get required resources ▫ Sept. 1931- Japanese invaded Manchuria • Japanese Prime Minister asked Minister of War Hideki Tojo to withdraw troops from China ▫ Wanted to avoid conflict with U.S. ▫ Tojo refused and threatened to bring down government • In October 1941, Tojo took over as prime minister Hitler Breaks Versailles Treaty • In 1935, Hitler began breaking the provisions of the Treaty of Versailles ▫ Began building a new air force ▫ Began a military draft to expand its army • European leaders tried to reason with Hitler instead of threaten war ▫ Did not want to repeat WWI ▫ Thought most of Hitler’s demands were reasonable ▫ Believed Nazis would want peace once they had taken over more land Anschluss and Appeasement • Austrian Anschluss (unification) ▫ Late 1937, Hitler called for German speaking people to be united ▫ Sent troops into Austria and announced the anschluss of Austria and Germany • Munich Conference ▫ Began policy of appeasement- given in to an aggressors demands to prevent conflict ▫ September 29, 1938- Allies agreed that Czechoslovakia must give up the Sudetenland or fight Germany itself ▫ March 1939, German troops invaded the rest of Czechoslovakia Hitler Demands Danzig • Hitler demanded the Polish city of Danzig be returned to German control in Oct 1938 ▫ 90% German ▫ Part of Poland since WWI • March 31, 1939 Britain and France announced it would defend Poland should Germany declare war on them • May 1939, Hitler ordered German army to prepare to invade Poland The Non-Aggression Pact • Signed by Germany and Soviet Union on August 23, 1939 ▫ Agreed not to fight or go to war with each other ▫ Divided Poland amongst them • Allowed Germany to focus on a 1 front war should it invade Poland • Agreement stunned the world ▫ Hate towards communists by Nazis ▫ Stalin thought the best way to protect communism was by having capitalist nations fight each other World War II Begins Sept. 1, 1939: Germany invades Poland Sept. 3, 1939: France and Britain Declare War on Germany June 4, 1940: Miracle at Dunkirk- British and French troops evacuate Belgium May 10 1940: Germany launches blitzkrieg into Netherlands and Belgium June 22, 1940: France surrenders to Germany Aug./Sept. 1940: Battle of Britain Sept. 3, 1939: France and Britain Declare War on Germany Oct. 5, 1939: German blitzkrieg allows Germany to capture Poland Discussion Questions Guiding Questions • Why did many Americans support isolationism and why did President Roosevelt support internationalism? • How did President Roosevelt assist Britain while maintaining U.S. neutrality? • What led to the United States’ involvement in World War II? • How did the United States respond to Japanese Americans? Terms to Know • • • • Internationalism Strategic materials Isolationism Neutral American Neutrality and Isolationism • Many Americans supported isolationism after WWI ▫ Discouraged by rise of dictatorships and militarism in Europe Felt like efforts during WWI were pointless ▫ Depression and inability of European nations to repay U.S. war debts ▫ Nye Committee Report Details about huge profits made by U.S. arms manufacturers Made it appear as if businesses influenced the decision to go to war Legislating Neutrality Neutrality Act of 1935 Neutrality Act of 1936 Neutrality Act of 1937 • Made it illegal for • Made it illegal for Americans to sell Americans to sell arms to any country at arms to any country war in a civil war • Response to the Spanish Civil War • “Cash and Carry” • All countries at war buying nonmilitary goods from the U.S. had to pay in cash and transport the goods on their own ships • Designed to keep U.S. transport ships from being attacked Roosevelt’s Internationalism • Roosevelt did not agree with Neutrality Acts but did not veto the bills ▫ Believed in internationalism Trade between nations created wealth and helped prevent wars ▫ Felt Neutrality Acts would actually force the U.S. into war • Approved sale of weapons to China after the Japanese invasion in 1937 ▫ Japan had not declared war ▫ Did not violate the Neutrality Acts Neutrality Tested Neutrality Act of 1939 • Allowed the sale of weapons to warring nations on the “cash-andcarry” basis Destroyersfor-Bases (Spring 1940) • • Britain asked for destroyers to replace losses FDR sent 50 ships to Britain in exchange for right to build American bases on Britishcontrolled land Lend-Lease Act (Dec. 1940) • • Allowed the U.S. to lend/lease arms to countries that were important to the defense of the U.S. Easily passed Congress Atlantic Charter (Aug. 1941) • FDR and Winston Churchill agreed to a commitment by both nations to a postwar world of democracy Shoot on Sight • • German submarines fired on a U.S. destroyer radioing the U-boats position to British U.S. navy was ordered to shoot German U-boats on sight Deteriorating U.S./Japanese Relations • In July 1940, Congress allowed Roosevelt the power to stop the sale of strategic war materials to Japan (embargo) ▫ Designed to stop Japan’s aggressive expansion in Pacific ▫ Japan signed alliance with Germany and Italy • In 1941, Roosevelt began sending lend-lease aid to China to help stop Japanese invasion ▫ Was not successful • Summer 1941, FDR froze Japanese assets in U.S., reduced oil shipments to Japan, and sent Gen. MacArthur to Philippines Japan Plots Attack • Japanese military planned to attack British and Dutch colonies in Southeast Asia ▫ Needed resources to save war effort in China ▫ U.S. Navy would not allow this expansion • Japan also decided to take over Philippines and attack American fleet at Pearl Harbor ▫ Set out for Hawaii in late Nov. 1941 Japanese Sneak Attack • Japan prepared for attack while continuing negotiations with U.S. ▫ Thought to be in good faith ▫ U.S. intelligence decoded Japanese messages that Japan was preparing for war ▫ U.S. military leaders could not figure out where attack would take place • Dec 7., 1941 – A Day Which Will Live in Infamy ▫ Japanese surprise attack on Pearl Harbor ▫ Great casualties and military losses for U.S. 2,403 Americans dead, 1,178 injured 8 battleships, 3 cruisers, 4 destroyers, 6 other vessels were sunk or damaged U.S. Enters WWII • Dec. 8, 1941- FDR asks Congress for declaration of war against Japan ▫ Overwhelming support 82-0 vote in Senate 388-1 vote in House • December 11, 1941- Germany and Italy declared war on the United States Discussion Questions • What geographic factors might have encouraged the United States to remain isolationist? • What is the cause-and-effect relationship between the embargo on Japan and the Pearl Harbor attack? The Holocaust Guiding Questions • Why did many Jews remain in Nazi Germany within Axis-controlled areas of Europe? • How did the Nazis try to exterminate Europe’s Jewish Population? Terms to Know • • • • Holocaust Concentration Camp Extermination Camp Genocide Nuremberg Laws • Passed in September 1935-1938 ▫ Took citizenship away from Jewish Germans ▫ Banned marriage between Jews and non-Jewish Germans ▫ Barred Jews from holding public office or voting ▫ Jews with German sounding names had to adopt “Jewish” names ▫ Passports were marked with a red J ▫ Banned Jews from practicing medicine, law, and running a business • Many Jews still remained ▫ Did not want to give up lives they had built ▫ Felt conditions would get better after time Kristallnacht • Anit-Jewish violence that erupted in Germany and Austria on Nov. 9, 1938 ▫ “Night of broken glass” ▫ In response to a Jewish refugee killing a German diplomat in Paris ▫ Hitler organized attacks to look like a public reaction to the news of the murder • Massive destruction ▫ More than 90 Jews dead ▫ 100’s more injured ▫ More than 7,500 Jewish businesses and synagogues destroyed • Gestapo (Nazi secret police) arrested about 30,000 Jewish men in the following days • Took insurance payments owed to Jewish owners of destroyed businesses Jewish Refugees Limits on Immigration • • • • • More than 250,000 Jews escaped Nazi Germany between 1933-1939 Waiting list of more than 100,000 Jewish refugees trying to enter US High unemployment in US made immigration unpopular Anti-Semitism in US No exception for quotas given to refugees or victims of persecution International Response • • • • European countries met in 1938 to discuss issue Most stated regret with their inability to take in refugees Hitler said he would gladly send all German Jews to any country who wanted them Many Jews left ships in 1939 with forged visas; Were denied admittance The St. Louis Affair • • • • • 930 Jewish refugees sailed to Havana, Cuba ; Arrived on May 27, 1939 Were not allowed to come ashore; Documents were not proper Hoped to enter the U.S. eventually Ship circled the coast of Florida but was never granted permission to dock Sailed back to Europe unloading the refugees in France, England, Holland, and Belgium The Final Solution • Wannsee Conference ▫ January 20, 1942 ▫ All Jews in Nazi-controlled Europe would be taken to detention centers Concentration camps Healthy individuals served as slave labor Worked until dead from exhaustion, disease, or malnutrition Extermination Camps Elderly, sick, and young Killed in large gas chambers Concentration Camps • First started in 1933 ▫ Served as jails for political opponents ▫ Built throughout Europe after beginning of WWII • Buchenwald ▫ One of the largest ▫ More than 200,000 prisoners working 12 hour shifts ▫ 100’s of prisoners died every month from exhaustion and horrible living conditions Extermination Camps • Built beginning in late 1941 • Auschwitz ▫ Estimated 1,600,000 killed in Auschwitz gas chambers 1,300,000 Jews ▫ Bodies were burned in giant crematoriums Holocaust Factors (How did it Occur) • • • • Germany’s sense of injury after WWI Severe economic problems Hitler’s control over the German nation Lack of strong tradition of democratic government in Germany • Fear of the Gestapo • Long history of anti-Jewish prejudice and discrimination in Europe Discussion Questions • Why might Jewish people have believed that conditions would improve, not worsen? • Why might the Great Depression have been a deterrent to Jewish immigration from Europe?<|endoftext|>
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A new study questions the use of body mass index (BMI) to predict the effect of obesity on health, saying it ignores important health factors such as how fat is distributed, the ratio of muscle to fat and differences in body composition. The authors call for more accurate and easy-to-use risk assessments.
BMI is a measure of weight in relation to height. According to the American Heart Association, a BMI of less than 25 indicates healthy weight; between 25 and 29.9, overweight; and 30 or higher, obesity. You can determine your BMI using the American Heart Association’s BMI Calculator.
Research has shown higher risk of cardiovascular diseases, Type 2 diabetes and cancer for people with a BMI of 30 or higher. This study discusses the use of BMI in light of the “obesity paradox,” that obesity appears to protect some people against certain diseases.
“BMI is at best a crude biomarker that cannot independently predict risk,” said Robert Eckel, M.D. past president of the American Heart Association. “However, a large majority of people with a BMI of over 30 have at least one risk factor for cardiovascular disease other than age. Even more people with a BMI of over 40 have cardiovascular risk factors.”
Lifestyle factors or genetics can both contribute to metabolic health, even in people who are overweight or obese, but this is the exception rather than rule, he said.
Weight is one of the American Heart Association’s Life’s Simple 7 — seven health factors contributing to heart health and cardiovascular disease risk.<|endoftext|>
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The Makerspaces in Primary School Settings project sought to examine how maker activities using 3D design and 3D printing technology could enhance learning and teaching outcomes. Across the 24 Kindergarten to Year 2 classes that were analysed, students developed a range of 21st century capabilities including creativity, problem-solving, critical thinking, inquiry, design thinking, collaboration, autonomy, literacy, numeracy, scientific understanding, digital literacy, communication, reflective learning capabilities and resilience.
This webinar will explore some of the findings of the study and include a discussion of some practical ways to establish a makerspace or maker culture in primary schools.
About the Presenter:
Dr. Matt Bower is an Associate Professor in the Department of Educational Studies at Macquarie University who focuses on how contemporary technologies can be used to advance learning. He is particularly interested in how innovative technologies such as augmented reality, Web 2.0 tools, virtual worlds, social networking, virtual reality and so on can be most effectively used to support cognitive development and collaborative learning. Matt has over eighty peer-reviewed publications in the area of technology-enhanced learning design, teacher education, and computing education. He has led several funded grant projects and participated in many other research initiatives (total funding exceeds $1M). He has delivered numerous keynote and invited speaker presentations on topics such as augmented reality, blended-synchronous learning and learning design. For more details about Matt’s research and portfolio, see: https://researchers.mq.edu.au/en/persons/matt-bower<|endoftext|>
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The small shark is named for its distinct black-tipped fins. Not to be confused with the blacktip shark, a larger species with similar fin coloration, the blacktip reef shark can be found in shallow inshore waters throughout the Indo-Pacific, including coral reefs, reef flats and near drop offs. It may be seen in mangrove areas and even freshwater environments near to shore, moving in and out with the tide. The blacktip reef shark feeds primarily on fish, including many common reef fishes, but will also consume crustaceans, mollusks, and even snakes!
Based on morphological similarities, Jack Garrick in 1982 grouped this species with the bignose shark (C. altimus) and the sandbar shark (C. plumbeus), while Leonard Compagno in 1988 placed it as the sister species of the grey reef shark (C. amblyrhynchos). A phylogenetic analysis based on allozyme data, published by Gavin Naylor in 1992, indicated that the Caribbean reef shark is the sister taxon to a clade formed by the Galapagos shark (C. galapagensis), dusky shark (C. obscurus), oceanic whitetip shark (C. longimanus), and the blue shark (Prionace glauca). However, more work is required to fully resolve the interrelationships within Carcharhinus.
Although still abundant at Cocos Island and other relatively pristine sites, grey reef sharks are susceptible to localized depletion due to their slow reproductive rate, specific habitat requirements, and tendency to stay within a certain area. The IUCN has assessed the grey reef shark as Near Threatened; this shark is taken by multispecies fisheries in many parts of its range and used for various products such as shark fin soup and fishmeal. Another threat is the continuing degradation of coral reefs from human development. There is evidence of substantial declines in some populations. Anderson et al. (1998) reported, in the Chagos Archipelago, grey reef shark numbers in 1996 had fallen to 14% of 1970s levels. Robbins et al. (2006) found grey reef shark populations in Great Barrier Reef fishing zones had declined by 97% compared to no-entry zones (boats are not allowed). In addition, no-take zones (boats are allowed but fishing is prohibited) had the same levels of depletion as fishing zones, illustrating the severe effect of poaching. Projections suggested the shark population would fall to 0.1% of pre-exploitation levels within 20 years without additional conservation measures. One possible avenue for conservation is ecotourism, as grey reef sharks are suitable for shark-watching ventures, and profitable diving sites now enjoy protection in many countries, such as the Maldives.
Generally a coastal, shallow-water species, grey reef sharks are mostly found in depths of less than 60 m (200 ft). However, they have been known to dive to 1,000 m (3,300 ft). They are found over continental and insular shelves, preferring the leeward (away from the direction of the current) sides of coral reefs with clear water and rugged topography. They are frequently found near the drop-offs at the outer edges of the reef, particularly near reef channels with strong currents, and less commonly within lagoons. On occasion, this shark may venture several kilometers out into the open ocean.
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Off Enewetak, grey reef sharks exhibit different social behaviors on different parts of the reef. Sharks tend to be solitary on shallower reefs and pinnacles. Near reef drop-offs, loose aggregations of five to 20 sharks form in the morning and grow in number throughout the day before dispersing at night. In level areas, sharks form polarized schools (all swimming in the same direction) of around 30 individuals near the sea bottom, arranging themselves parallel to each other or slowly swimming in circles. Most individuals within polarized schools are females, and the formation of these schools has been theorized to relate to mating or pupping.
The coloration is grey above, sometimes with a bronze sheen, and white below. The entire rear margin of the caudal fin has a distinctive, broad, black band. There are dusky to black tips on the pectoral, pelvic, second dorsal, and anal fins. Individuals from the western Indian Ocean have a narrow, white margin at the tip of the first dorsal fin; this trait is usually absent from Pacific populations. Grey reef sharks that spend time in shallow water eventually darken in color, due to tanning. Most grey reef sharks are less than 1.9 m (6.2 ft) long. The maximum reported length is 2.6 m (8.5 ft) and the maximum reported weight is 33.7 kg (74 lb).
Every year, Reef Check trains thousands of citizen scientist divers who volunteer to survey the health of coral reefs around the world, and rocky reef ecosystems along the entire coast of California. The results are used to improve the management of these critically important natural resources. Reef Check programs provide ecologically sound and economically sustainable solutions to save reefs, by creating partnerships among community volunteers, government agencies, businesses, universities and other nonprofits.
$eaworld biodiversity bluefin tuna Climate Change clownfish coral reefs crabs cuttlefishes deep sea dolphins endangered extinction fins fishes frogfishes ghost pipefish global warming Indonesia jellyfish mantas mantis shrimp marine biology Marine Conservation Marine Mammals Marine Protected Areas Marine Science morays nudibranchs octopuses oil spill orca overfishing Papua New Guinea phytoplankton plastics polar bears pollution scuba seafood sea horses sea level rise sea turtles sharks shrimp whales
With more than 60 years of experience in manufacturing and fabricating flexible plastic laminates, Reef Industries has the expertise in helping customers tackle the challenges they face when needing solutions to safeguard and prolong the service life of their investments. From protecting equipment during outdoor storage to ensuring buried utilities are protected from accidental dig-ins, Reef Industries provides customers with dependable and proven high-quality materials for long-term performance.
Founded in 1996, the Reef Check Foundation exists to help preserve the oceans and reefs which are critical to our survival, yet are being destroyed. With headquarters in Los Angeles and volunteer teams in more than 90 countries and territories, Reef Check works to protect tropical coral reefs and California rocky reefs through education, research and conservation.
The Caribbean reef shark feeds on a wide variety of reef-dwelling bony fishes and cephalopods, as well as some elasmobranchs such as eagle rays (Aetobatus narinari) and yellow stingrays (Urobatis jamaicensis). It is attracted to low-frequency sounds, which are indicative of struggling fish. In one observation of a 2 m (6.6 ft) long male Caribbean reef shark hunting a yellowtail snapper (Lutjanus crysurus), the shark languidly circled and made several seemingly "half-hearted" turns towards its prey, before suddenly accelerating and swinging its head sideways to capture the snapper at the corner of its jaws. Young sharks feed on small fishes, shrimps, and crabs. Caribbean reef sharks are capable of everting their stomachs, which likely serves to cleanse indigestible particles, parasites, and mucus from the stomach lining.
Despite its abundance in certain areas, the Caribbean reef shark is one of the least-studied large requiem sharks. They are believed to play a major role in shaping Caribbean reef communities. These sharks are more active at night, with no evidence of seasonal changes in activity or migration. Juveniles tend to remain in a localized area throughout the year, while adults range over a wider area.
Like many sharks, the Caribbean reef shark mainly eats bony fishes. The shark uses six keen senses to locate its prey: olfactory, visual, tactile (including water vibration sensitivity through a lateralis canal system), auditory, gustatory, and electric reception. The Caribbean reef shark is especially adapted to detecting low frequency sounds (indicative of a struggling fish nearby).
A reef is a bar of rock, sand, coral or similar material, lying beneath the surface of water. Many reefs result from abiotic processes (i.e. deposition of sand, wave erosion planing down rock outcrops, and other natural processes), but the best known reefs are the coral reefs of tropical waters developed through biotic processes dominated by corals and coralline algae.<|endoftext|>
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Research in Medecineers from the , hosted at Foto-Cewek, conducted an independent review of the effects of adding a single dose of primaquine (PQ) to malaria treatment to prevent the transmission of the disease.
Mosquitoes become infected with Plasmodium falciparum when they ingest gametocyte-stage parasites from an infected person’s blood. PQ is an antimalarial drug that does not cure malaria illness, but is known to kill the gametocyte stage of the P falciparum parasite, thus potentially limiting transmission to further humans. PQ is known to have potentially serious side effects for those who have the glucose-6-phosphate dehydrogenase (G6PD) enzyme deficiency, which is common in many malaria endemic regions.
In 2012 the World Health Organization (WHO) revised its recommendation for a single dose of PQ alongside treatment for P. falciparum malaria from 0.75 mg/kg down to 0.25mg/kg due to concerns about safety. The three authors looked at a total of 18 trials to examine the evidence of benefits and harms of using PQ in this way by evaluating whether PQ will reduce malaria transmission or the potential that it will reduce transmission based on infectious stages (gametocytes) in the blood. This fresh analysis grouped the studies by dose, to evaluate the effects of the low dose currently recommended by the WHO.
None of the included studies tested whether PQ added to malaria treatment reduced the community transmission intensity of malaria. Two studies showed that PQ at doses of 0.75 mg/kg reduced the number of mosquitoes infected after biting humans. PQ does reduce the proportion of people with gametocyte stages, the numbers of gametocytes and the duration of infectiousness (the period that gametocytes are detected in the blood) at doses of 0.4 mg/kg or above, but two trials at lower doses showed no effect.
Patricia Graves, from James Cook University in Cairns, said: “Evidence shows that at PQ doses of 0.4 mg/kg and above there is reduction in infectiousness of individuals. Whether this translates into an effect at community level in reducing transmission is not clear. Given the lack of safety data in those with G6PD deficiency, the data do not reveal whether the revised dosing recommendation will have a positive impact on health.”<|endoftext|>
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New York - Grade 1 - Math - Operations and Algebraic Thinking - Number Sentences - 1.OA.7 , 1.OA.8
Description
1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2. 1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ – 3, 6 + 6 = _.
• State - New York
• Standard ID - 1.OA.7 , 1.OA.8
• Subjects - Math Common Core
• Grade - 1
Keywords
• Math
• New York grade 1
• Operations and Algebraic Thinking
More New York Topics
Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes.
Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares.
Solve word problems that call for addition of three whole numbers whose sum is less than or equal to 20, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral.
Here is the skill that New York requires you to master
• Grade Level 1
• State Test New York State Assessments
• State Standards New York State P-12 Common Core Learning Standards (CCLS)
• Subject Math
• Topic Name Number Sentences
• Standard ID 1.OA.7 , 1.OA.8
• Description
1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2. 1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ – 3, 6 + 6 = _.<|endoftext|>
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# 2007 iTest Problems/Problem TB2
## Problem
Factor completely over integer coefficients the polynomial $p(x)=x^8+x^5+x^4+x^3+x+1$. Demonstrate that your factorization is complete.
## Solution
Note that $p(x)=(x^5+x+1)+(x^8+x^4+x^3)$. If $x\neq 1$ and $x^3=1$, then $x^5+x+1=x^2+x+1=0$ and $x^8+x^4+x^3=x^2+x+1=0$. Therefore if $x\neq 1$ and $x^3=1$, then $p(x)=0$. Hence $x^2+x+1|p(x)$. Dividing through gives us
$p(x)=(x^2+x+1)(x^6-x^5+2x^3-x^2+1)$
Using the Rational Root Theorem on the second polynomial gives us that $\pm 1$ are possible roots. Only $-1$ is a possible root. Dividing through gives us
$p(x)=(x+1)(x^2+x+1)(x^5-2x^4+2x^3-x+1)$
Note that $x^5-2x^4+2x^3-x+1$ can be factored into the product of a cubic and a quadratic. Let the product be
$x^5-2x^4+2x^3-x+1=(x^2+ax+b)(x^3+cx^2+dx+e)$
We would want the coefficients to be integers, hence we shall only look for integer solutions. The following equations must then be satisfied:
• $be=1$
• $ae+bd=-1$
• $c+ad+e=0$
• $b+ac+d=2$
• $a+c=-2$
Since $b$ and $e$ are integers, $(b,e)$ is either $(1,1)$ or $(-1,-1)$. Testing the first one gives
• $a+d=-1$
• $c+ad=-1$
• $ac+d=1$
• $a+c=-2$
We must have that $(a+c)-(a+d)=c-d=-2--1=-1$ $(ac+d)-(c+ad)=a(c-d)-(c-d)=(a-1)(c-d)=1--1=2$. Therefore $a-1=-2$, or $a=-1$. Solving for $c$ and $d$ gives $(a,b,c,d,e)=(-1,1,-1,0,1)$. We don't need to test the other one.
Hence we have
$p(x)=(x+1)(x^2+x+1)(x^2-x+1)(x^3-x^2+1)$
For any of the factors of degree more than 1 to be factorable in the integers, they must have rational roots, since their degrees are less than 4. None of them have rational roots. Hence $p(x)$ is completely factored.
### Alternate Solution
We write $$p(x)=x^8+x^5+x^4+x^3+x+1=$$ $$x^5(x^3+1)+x^3(x+1)+(x+1)=$$ $$(x+1)(x^5(x^2-x+1) + (x^3+1))=$$ $$(x+1)(x^5(x^2-x+1)+(x+1)(x^2-x+1))=$$ $$(x+1)(x^2-x+1)(x^5+x+1)=$$ $$(x+1)(x^2-x+1)(x^2+x+1)(x^3-x^2+1)$$
The factorization of $x^5+x+1$ is trivial once we look at the exponents modulo $3$; since any root $\omega$ of $x^2+x+1$ satisfies $\omega^3=1$, it follows that $x^2+x+1 | x^5+x+1$ and the cubic factor comes as a result of polynomial division.
To prove that this is a complete factorization, it suffices to note that the factors of degree greater than $1$ have no rational roots (follows from the rational root theorem and some small cases).<|endoftext|>
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## What you need to know
Things to remember:
• Draw a diagram and put the lengths on.
• Calculate the area as you would normally
• To find the area, we just multiply the sides.
Questions don’t always come with diagrams, so you will either have to imagine it in your head or draw it on paper.
What is the area of a rectangle that is 6m long and 2m high?
We can draw the rectangle, and it would look like this:
But we know how to find area of a rectangle, we just need to multiply the two sides.
$$6m \times 3m = 18m^{2}$$
Questions won’t always say which side is which, so you need to decide.
What is the area of a rectangle with side lengths of 5cm and 7cm?
The rectangle could look like one of these:
So, let’s find the area of both and see what happens.
$$5cm \times 7cm = 21cm^{2}$$
$$7cm \times 5cm = 21cm^{2}$$
We get the same answer, so it doesn’t matter! We just need to remember to multiply the sides together.
## Example Questions
#### Question 1: A rectangle has side lengths of 5cm and 3cm, what is the area of the rectangle?
$$5cm \times 3cm = 15cm^{2}$$
$$12cm \times11cm = 131cm^{2}$$<|endoftext|>
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A mass extinction is the eradication of a large number of species within a short period of geological time due to catastrophic factors occur too rapidly for most species to adapt. Today, many scientists think the evidence indicates a sixth mass extinction is under way. The Holocene extinction, also known as the Sixth Extinction or Anthropocene extinction, is ongoing and humans are to blame.
Pollution is one of the primary ways humans have caused severe modifications of wildlife habitat. We have sabotaged the air, water, soil and given little consideration to the ecological consequences of our actions. As a result, wildlife populations are confronted with a bewildering array of pollutants, being suffocated, strangled and eventually killed.
The global temperatures are warming because of greenhouse gases that humans are pumping into the atmosphere. One major consequence is that melting glaciers are raising the sea level. Flooding, increasing temperature and other climate-related consequences make species unable to exist in their original homes.
Hunting is a way for humans to systematically wipe out species very quickly. Animals are poached for cultural medicine, trading, clothing or personal interests.
The more humans convert land to their own purposes, the less habitat left for animals. Natural habitats are being converted for human use at an alarming rate. About half of the earth’s original forests are gone. In fact, we are losing forests at the rate of 20 football fields per minute. If the current rate of deforestation continues, it will take less than 100 years to destroy all the rainforests on Earth.
Recent extinction rates are unprecedented in human history and highly unusual in Earth’s history:
- World Wildlife Fund (WWF) estimates the extinction rate is 1000–10,000 times faster than natural and each year, 200–2000 species go extinct.
- Mother Nature Network (MNN) reports that 38% of all land animals and 81% of freshwater vertebrates went extinct between 1970 to 2012.
- In just the past 40 years, nearly 52 percent of the planet’s wildlife species have been eliminated.
- According to the study published in the journal Science Advances, 75 percent Earth’s species could be lost in the span of two generations.
Humans: The Next Victim
Humans will not be spectators to the phenomenon but rather victims as well. Just before his death in 2010, Professor Frank Fenner left a chilling warning for future generations, saying the end is on the horizon for humanity.
The human race faces a one in 500 chance of extinction in the next year, an expert mathematician has claimed. That is twenty times more likely than dying in a car crash.
So is it all lost? Avoiding a true sixth mass extinction will require us rapid, greatly intensified efforts to conserve already threatened species and minimize the amount of chemical pollutants to the environment. But time is of the essence, the window of opportunity is rapidly closing.<|endoftext|>
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Forest recovering after the B&B; Complex fire in 2003 in the central Oregon Cascade Range. Photo by Garrett Meigs, Oregon State University via Flickr.
Guest blogger Cara Smusiak is a journalist and regular contributor to NaturallySavvy.com's Naturally Green section.
About half of the overstory trees were killed in the B&B; Complex fire in 2003 near Canyon Creek in the central Oregon Cascade Range, yet a high level of tree survival and vegetation rapidly recovered, as can be seen in this photo taken in 2007. This type of burning and re-growth is typical of fires in the Pacific Northwest.
Research has suggested that global warming will change wildfire patterns and cause a leap in the number of forest fires in the Pacific Northwest (though that's hotly debated). But there might just be a silver lining in those billowing clouds of smoke.In an interview with ScienceDaily.com, John Bailey, an associate professor in the Department of Forest Engineering, Resources and Management at Oregon State University, discussed the beneficial role that fire plays in managing and enriching forest ecosystems.
Forest fires are key to long-term forest management, Bailey told Science Daily:
Forests historically had more fire across much of Oregon, and they would love to have more today. Burning is a natural ecosystem process and generally helps restore forest ecosystems. It's ironic that we spend so much money to stop fire, because we should learn to see fire as more of a partner and not always an enemy.
Forest fires are an efficient, natural way for a forest to rid itself of dead or dying plant matter. And the decomposed organic matter enriches the soil with minerals that help new plants sprout up quickly.
Bailey told Science Daily that even the worst case scenario of climate change-related fires may not be as bad for forests as one might assume, since forests have historically seen more fires than they experience today. And the huge cost of trying to prevent fires is unnecessary in many cases, Bailey added:
Right now we're spending billions of dollars to prevent something that is going to happen sooner or later, whether we try to stop it or not, and something that can assist us in sound land management. It may always make sense to put out some fires when they threaten communities, or in other select circumstances. But periodic fire has always been a part of our forests, and we need to accept it as such, sort of like how we plan for and accept a very wet winter that comes along now and then.
Bailey will speak at an upcoming conference at Oregon State University titled, "Forest Health in Oregon: State of the State."<|endoftext|>
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MULTIPLICATION OF A POLYNOMIAL BY A POLYNOMIAL
As with the monomial multiplier, we explainmultiplication of a polynomial by a polynomial by use of an arithmetic example. To multiply (3 + 2)(6 - 4), we could do the operation within the parentheses first and then multiply, as follows:
(3 + 2)(6 - 4) = (5)(2) = 10
However, thinking of the quantity (3 + 2) as oneterm, we can use the method described for a (3 + 2), with the following result:
(3 + 2)(6 - 4) = [(3 + 2) x 6 - (3 + 2) x 4]
Now considering each of the two resultingproducts separately, we note that each is a binomial multiplied by a monomial.
The first is
(3 + 2)6 = (3 x 6) + (2 x 6)
and the second is
-(3 + 2)4 = - [(3 x 4) + (2 x 4)]
Thus we have the following result:
(3 + 2)(6 - 4) = (3 X 6) + (2 X
The complete product is formed by multiplying each term of the multiplicand separately by each term of the multiplier and combining the results with due regard to signs.
Now let us apply this method in two examples involving literal numbers.
1. (a + b)(m + n) = am + an + bm + bn
The rule governing these examples is stated as follows: The product of any two polynomials is found by multiplying each term of one by each term of the other and adding the results algebraically.
It is often convenient, especially when either of the expressions contains more than two terms, to place the polynomial with the fewer terms beneath the other polynomial and multiply term by term beginning at the left. Like 3x2 - 7x - 9 and 2x - 3. The procedure is
Practice problems. In the following problems, multiply and combine like terms:
The products of certain binomials occur frequently. It is convenient to remember the form of these products so that they can be written immediately without performing the complete multiplication process. We present four such special products as follows, and then show how each is derived:
1. Product of the sum and difference of two numbers.
EXAMPLE: (x - y)(x + y) = x2 - y2
2. Square the sum of two numbers.
EXAMPLE: (x+y)2=x2 +2xy+y2
3. Square of the difference of two numbers.
EXAMPLE: (x - Y)2 = x2 -2xy+y2
4. Product of two binomials having a common term.
EXAMPLE: (x + a)(x + b) = x2 + (a + b)x + ab
Product of Sum and Difference
The product of the sum and difference of two numbers is equal to the square of the first number minus the square of the second number. If, for example, x - y is multiplied by x + y, the middle terms cancel one another. The result is the square of x minus the square of y, as shown in the following illustration:
By keeping this rule in mind, the product of the sum and difference of two numbers can be written down immediately by writing the difference of the squares of the numbers. For example, consider the following three problems:<|endoftext|>
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Rd Sharma XII Vol 1 2020 Solutions for Class 12 Science Maths Chapter 6 Adjoint And Inverse Of Matrix are provided here with simple step-by-step explanations. These solutions for Adjoint And Inverse Of Matrix are extremely popular among Class 12 Science students for Maths Adjoint And Inverse Of Matrix Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2020 Book of Class 12 Science Maths Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2020 Solutions. All Rd Sharma XII Vol 1 2020 Solutions for class Class 12 Science Maths are prepared by experts and are 100% accurate.
#### Question 1:
Find the adjoint of each of the following matrices:
(i) $\left[\begin{array}{cc}-3& 5\\ 2& 4\end{array}\right]$
(ii) $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
(iii)
(iv)
Verify that (adj A) A = |A| I = A (adj A) for the above matrices.
Given below are the squares matrices. Here, we will interchange the diagonal elements and change the signs of
the off-diagonal elements.
s.
#### Question 2:
Compute the adjoint of each of the following matrices:
(i) $\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$
(ii) $\left[\begin{array}{ccc}1& 2& 5\\ 2& 3& 1\\ -1& 1& 1\end{array}\right]$
(iii) $\left[\begin{array}{ccc}2& -1& 3\\ 4& 2& 5\\ 0& 4& -1\end{array}\right]$
(iv) $\left[\begin{array}{ccc}2& 0& -1\\ 5& 1& 0\\ 1& 1& 3\end{array}\right]$
Verify that (adj AA = |AI = A (adj A) for the above matrices.
#### Question 3:
For the matrix $A=\left[\begin{array}{ccc}1& -1& 1\\ 2& 3& 0\\ 18& 2& 10\end{array}\right]$, show that A (adj A) = O.
#### Question 4:
If $A=\left[\begin{array}{ccc}-4& -3& -3\\ 1& 0& 1\\ 4& 4& 3\end{array}\right]$, show that adj A = A.
#### Question 5:
If $A=\left[\begin{array}{ccc}-1& -2& -2\\ 2& 1& -2\\ 2& -2& 1\end{array}\right]$, show that adj A = 3AT.
#### Question 6:
Find A (adj A) for the matrix $A=\left[\begin{array}{ccc}1& -2& 3\\ 0& 2& -1\\ -4& 5& 2\end{array}\right].$
#### Question 7:
Find the inverse of each of the following matrices:
(i)
(ii) $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$
(iii) $\left[\begin{array}{cc}a& b\\ c& \frac{1+bc}{a}\end{array}\right]$
(iv) $\left[\begin{array}{cc}2& 5\\ -3& 1\end{array}\right]$
#### Question 8:
Find the inverse of each of the following matrices.
(i) $\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\\ 3& 1& 2\end{array}\right]$
(ii) $\left[\begin{array}{ccc}1& 2& 5\\ 1& -1& -1\\ 2& 3& -1\end{array}\right]$
(iii) $\left[\begin{array}{ccc}2& -1& 1\\ -1& 2& -1\\ 1& -1& 2\end{array}\right]$
(iv) $\left[\begin{array}{ccc}2& 0& -1\\ 5& 1& 0\\ 0& 1& 3\end{array}\right]$
(v) $\left[\begin{array}{ccc}0& 1& -1\\ 4& -3& 4\\ 3& -3& 4\end{array}\right]$
(vi) $\left[\begin{array}{ccc}0& 0& -1\\ 3& 4& 5\\ -2& -4& -7\end{array}\right]$
(vii)
#### Question 9:
Find the inverse of each of the following matrices and verify that .
(i) $\left[\begin{array}{ccc}1& 3& 3\\ 1& 4& 3\\ 1& 3& 4\end{array}\right]$
(ii) $\left[\begin{array}{ccc}2& 3& 1\\ 3& 4& 1\\ 3& 7& 2\end{array}\right]$
#### Question 10:
For the following pairs of matrices verity that
(i)
(ii)
(AB)1=B1 A1 (AB)1=B1 A1(AB)1=B1 A1
Let
#### Question 12:
Given $A=\left[\begin{array}{cc}2& -3\\ -4& 7\end{array}\right]$, compute A−1 and show that $2{A}^{-1}=9I-A.$
#### Question 13:
If $A=\left[\begin{array}{cc}4& 5\\ 2& 1\end{array}\right]$, then show that
#### Question 14:
Find the inverse of the matrix $A=\left[\begin{array}{cc}a& b\\ c& \frac{1+bc}{a}\end{array}\right]$ and show that
#### Question 15:
Given . Compute (AB)−1.
We have,
Let
Show that
(i)
(ii)
(iii) .
​
#### Question 17:
If $A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]$, verify that . Hence, find A−1.
#### Question 18:
Show that $A=\left[\begin{array}{cc}-8& 5\\ 2& 4\end{array}\right]$ satisfies the equation ${A}^{2}+4A-42I=O$. Hence, find A−1.
#### Question 19:
If $A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$, show that ${A}^{2}-5A+7I=O$. Hence, find A−1.
#### Question 20:
If $A=\left[\begin{array}{cc}4& 3\\ 2& 5\end{array}\right]$, find x and y such that ${A}^{2}=xA+yI=O$. Hence, evaluate A−1.
#### Question 21:
If $A=\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]$, find the value of $\lambda$ so that ${A}^{2}=\lambda A-2I$. Hence, find A−1.
#### Question 22:
Show that $A=\left[\begin{array}{cc}5& 3\\ -1& -2\end{array}\right]$ satisfies the equation ${x}^{2}-3x-7=0$. Thus, find A−1.
#### Question 23:
Show that $A=\left[\begin{array}{cc}6& 5\\ 7& 6\end{array}\right]$ satisfies the equation ${x}^{2}-12x+1=O$. Thus, find A−1.
#### Question 24:
For the matrix $A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& -3\\ 2& -1& 3\end{array}\right]$. Show that . Hence, find A−1.
#### Question 25:
Show that the matrix, $A=\left[\begin{array}{ccc}1& 0& -2\\ -2& -1& 2\\ 3& 4& 1\end{array}\right]$ satisfies the equation, ${A}^{3}-{A}^{2}-3A-{I}_{3}=O$. Hence, find A−1.
#### Question 26:
If $A=\left[\begin{array}{ccc}2& -1& 1\\ -1& 2& -1\\ 1& -1& 2\end{array}\right]$. Verify that ${A}^{3}-6{A}^{2}+9A-4I=O$ and hence find A−1.
#### Question 27:
If $A=\frac{1}{9}\left[\begin{array}{ccc}-8& 1& 4\\ 4& 4& 7\\ 1& -8& 4\end{array}\right]$, prove that ${A}^{-1}={A}^{3}$.
#### Question 28:
If $A=\left[\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\right]$, show that ${A}^{-1}={A}^{3}$.
#### Question 29:
If $A=\left[\begin{array}{ccc}-1& 2& 0\\ -1& 1& 1\\ 0& 1& 0\end{array}\right]$, show that ${A}^{2}={A}^{-1}.$
#### Question 30:
Solve the matrix equation $\left[\begin{array}{cc}5& 4\\ 1& 1\end{array}\right]X=\left[\begin{array}{cc}1& -2\\ 1& 3\end{array}\right]$, where X is a 2 × 2 matrix.
#### Question 31:
Find the matrix X satisfying the matrix equation
$X\left[\begin{array}{cc}5& 3\\ -1& -2\end{array}\right]=\left[\begin{array}{cc}14& 7\\ 7& 7\end{array}\right]$.
#### Question 32:
Find the matrix X for which
#### Question 33:
Find the matrix X satisfying the equation
#### Question 34:
If $A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 1& 2\\ 2& 2& 1\end{array}\right]$, find ${A}^{-1}$ and prove that ${A}^{2}-4A-5I=O$
Prove that .
#### Question 36:
We know that (AB)1 = B1 A1.
#### Question 37:
If
We know that (AT)1 = (A1)T.
#### Question 38:
Find the adjoint of the matrix $A=\left[\begin{array}{ccc}-1& -2& -2\\ 2& 1& -2\\ 2& -2& 1\end{array}\right]$ and hence show that .
#### Question 1:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{cc}7& 1\\ 4& -3\end{array}\right]$
#### Question 2:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{cc}5& 2\\ 2& 1\end{array}\right]$
#### Question 3:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{cc}1& 6\\ -3& 5\end{array}\right]$
#### Question 4:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{cc}2& 5\\ 1& 3\end{array}\right]$
#### Question 5:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{cc}3& 10\\ 2& 7\end{array}\right]$
#### Question 6:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{ccc}0& 1& 2\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]$
#### Question 7:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{ccc}2& 0& -1\\ 5& 1& 0\\ 0& 1& 3\end{array}\right]$
#### Question 8:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{ccc}2& 3& 1\\ 2& 4& 1\\ 3& 7& 2\end{array}\right]$
#### Question 9:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\right]$
#### Question 10:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{ccc}1& 2& 0\\ 2& 3& -1\\ 1& -1& 3\end{array}\right]$
#### Question 11:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{ccc}2& -1& 3\\ 1& 2& 4\\ 3& 1& 1\end{array}\right]$
#### Question 12:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{ccc}1& 1& 2\\ 3& 1& 1\\ 2& 3& 1\end{array}\right]$
#### Question 13:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{ccc}2& -1& 4\\ 4& 0& 7\\ 3& -2& 7\end{array}\right]$
#### Question 14:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{ccc}3& 0& -1\\ 2& 3& 0\\ 0& 4& 1\end{array}\right]$
#### Question 15:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{ccc}1& 3& -2\\ -3& 0& -1\\ 2& 1& 0\end{array}\right]$
#### Question 16:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{ccc}-1& 1& 2\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]$
#### Question 17:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{ccc}1& 2& 3\\ 2& 5& 7\\ -2& -4& -5\end{array}\right]$
A = IA
$A=\left[\begin{array}{ccc}1& 2& 3\\ 2& 5& 7\\ -2& -4& -5\end{array}\right]$
$\left[\begin{array}{ccc}1& 2& 3\\ 2& 5& 7\\ -2& -4& -5\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A\phantom{\rule{0ex}{0ex}}{R}_{2}\to {R}_{2}-2{R}_{1}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 1\\ -2& -4& -5\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ 0& 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}{R}_{3}\to {R}_{3}+2{R}_{1}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ 2& 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}{R}_{2}\to {R}_{2}-{R}_{3}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ -4& 1& -1\\ 2& 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}$
${R}_{1}\to {R}_{1}-3{R}_{3}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{ccc}1& 2& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}-5& 0& -3\\ -4& 1& -1\\ 2& 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}{R}_{1}\to {R}_{1}-2{R}_{2}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}3& -2& -1\\ -4& 1& -1\\ 2& 0& 1\end{array}\right]$
Therefore,
${A}^{-1}=\left[\begin{array}{ccc}3& -2& -1\\ -4& 1& -1\\ 2& 0& 1\end{array}\right]$
#### Question 18:
Find the inverse of each of the following matrices by using elementary row transformations:
$\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]$
Let $A=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]$
#### Question 1:
If A is an invertible matrix, then which of the following is not true
(a) ${\left({A}^{2}\right)}^{-1}={\left({A}^{-1}\right)}^{2}$
(b) $\left|{A}^{-1}\right|={\left|A\right|}^{-1}$
(c) ${\left({A}^{T}\right)}^{-1}={\left({A}^{-1}\right)}^{T}$
(d) $\left|A\right|\ne 0$
(a) ${\left({A}^{2}\right)}^{-1}={\left({A}^{-1}\right)}^{2}$
We have, $\left|{A}^{-1}\right|={\left|A\right|}^{-1}$, ${\left({A}^{T}\right)}^{-1}={\left({A}^{-1}\right)}^{T}$ and $\left|A\right|\ne 0$ all are the properties of the inverse of a matrix A
#### Question 2:
If A is an invertible matrix of order 3, then which of the following is not true
(a)
(b) ${\left({A}^{-1}\right)}^{-1}=A$
(c) If , where B and C are square matrices of order 3
(d)
(c) If $BA=CA$, then $B\ne C$ where B and C are square matrices of order 3.
If A is an invertible matrix, then ${A}^{-1}$ exists.
Now,
$BA=CA$
On multiplying both sides by ${A}^{-1}$, we get
$BA{A}^{-1}=CA{A}^{-1}$
Therefore, the statement ​given in (c) is not true.
#### Question 3:
If
(a) is a skew-symmetric matrix
(b) A−1 + B−1
(c) does not exist
(d) none of these
(d) none of these
#### Question 4:
If $S=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, then adj A is
(a) $\left[\begin{array}{cc}-d& -b\\ -c& a\end{array}\right]$
(b) $\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$
(c) $\left[\begin{array}{cc}d& b\\ c& a\end{array}\right]$
(d) $\left[\begin{array}{cc}d& c\\ b& a\end{array}\right]$
(b) $\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$
Adjoint of a square matrix of order 2 is obtained by interchanging the diagonal elements and changing the signs of off-diagonal elements.
Here,
#### Question 5:
If A is a singular matrix, then adj A is
(a) non-singular
(b) singular
(c) symmetric
(d) not defined
(b) singular
#### Question 6:
If A, B are two n × n non-singular matrices, then
(a) AB is non-singular
(b) AB is singular
(c)
(d) (AB)−1 does not exist
(a) AB is non-singular
#### Question 7:
If $A=\left[\begin{array}{ccc}a& 0& 0\\ 0& a& 0\\ 0& 0& a\end{array}\right]$, then the value of |adj A| is
(a) a27
(b) a9
(c) a6
(d) a2
(c) a6
#### Question 8:
If $A=\left[\begin{array}{ccc}1& 2& -1\\ -1& 1& 2\\ 2& -1& 1\end{array}\right]$, then ded (adj (adj A)) is
(a) 144
(b) 143
(c) 142
(d) 14
(a) 144
#### Question 9:
If B is a non-singular matrix and A is a square matrix, then det (B−1 AB) is equal to
(a) Det (A−1)
(b) Det (B−1)
(c) Det (A)
(d) Det (B)
(c) Det (A)
#### Question 10:
For any 2 × 2 matrix, if , then |A| is equal to
(a) 20
(c) 100
(d) 10
(d) 0
#### Question 11:
If A5 = O such that equals
(a) A4
(b) A3
(c) I + A
(d) none of these
#### Question 12:
If A satisfies the equation ${x}^{3}-5{x}^{2}+4x+\lambda =0$ then A−1 exists if
(a) $\lambda \ne 1$
(b) $\lambda \ne 2$
(c) $\lambda \ne -1$
(d) $\lambda \ne 0$
#### Question 13:
If for the matrix A, A3 = I, then A−1 =
(a) A2
(b) A3
(c) A
(d) none of these
#### Question 14:
If A and B are square matrices such that B = − A−1 BA, then (A + B)2 =
(a) O
(b) A2 + B2
(c) A2 + 2AB + B2
(d) A + B
(b) ${A}^{2}+{B}^{2}$
#### Question 15:
If
(a) 5A
(b) 10A
(c) 16A
(d) 32A
(c) 16A
$A=\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒A=2\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒A=2I\phantom{\rule{0ex}{0ex}}⇒{A}^{5}={\left(2I\right)}^{5}\phantom{\rule{0ex}{0ex}}⇒{A}^{5}=16×2I\phantom{\rule{0ex}{0ex}}⇒{A}^{5}=16\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒{A}^{5}=16A$
#### Question 16:
For non-singular square matrix A, B and C of the same order
(a)
(b)
(c) $CB{A}^{-1}$
(d)
Disclaimer: In Quesion, We are to find the inverse of . The inverse is missing in the question.
(d)
We have,
#### Question 17:
The matrix $\left[\begin{array}{ccc}5& 10& 3\\ -2& -4& 6\\ -1& -2& b\end{array}\right]$ is a singular matrix, if the value of b is
(a) − 3
(b) 3
(c) 0
(d) non-existent
(d) non-existent
For any singular matrix, the value of the determinant is 0.
Here,
$A=\left[\begin{array}{ccc}5& 10& 3\\ -2& -4& 6\\ -1& -2& b\end{array}\right]\phantom{\rule{0ex}{0ex}}\left|A\right|=5\left(-4b+12\right)-10\left(-2b+6\right)+3\left(4-4\right)=0\phantom{\rule{0ex}{0ex}}⇒-20b+60+20b-12=0$
Hence, b is non-existent.
#### Question 18:
If d is the determinant of a square matrix A of order n, then the determinant of its adjoint is
(a) dn
(b) dn−1
(c) dn+1
(d) d
(b) dn−1
We know,
$\left|\mathrm{adj}A\right|={\left|A\right|}^{n-1}$
$⇒\left|\mathrm{adj}A\right|={d}^{n-1}$
#### Question 19:
If A is a matrix of order 3 and |A| = 8, then |adj A| =
(a) 1
(b) 2
(c) 23
(d) 26
(d) ${2}^{6}$
#### Question 20:
If ${A}^{2}-A+I=0$, then the inverse of A is
(a) A2
(b) A + I
(c) IA
(d) AI
(c) IA
#### Question 21:
If A and B are invertible matrices, which of the following statement is not correct.
(a)
(b)
(c) ${\left(A+B\right)}^{-1}={A}^{-1}+{B}^{-1}$
(d)
(c) ${\left(A+B\right)}^{-1}={A}^{-1}+{B}^{-1}$
We have, , and all are the properites of inverse of a matrix.
#### Question 22:
If A is a square matrix such that A2 = I, then A1 is equal to
(a) A + I
(b) A
(c) 0
(d) 2A
(b) A
$\mathrm{Given}: {A}^{2}=I\phantom{\rule{0ex}{0ex}}$
On multiplying both sides by ${A}^{-1}$ , we get
${A}^{-1}{A}^{2}={A}^{-1}I\phantom{\rule{0ex}{0ex}}⇒A={A}^{-1}I\phantom{\rule{0ex}{0ex}}⇒A={A}^{-1}$
#### Question 23:
Let and X be a matrix such that A = BX, then X is equal to
(a) $\frac{1}{2}\left[\begin{array}{cc}2& 4\\ 3& -5\end{array}\right]$
(b) $\frac{1}{2}\left[\begin{array}{cc}-2& 4\\ 3& 5\end{array}\right]$
(c) $\left[\begin{array}{cc}2& 4\\ 3& -5\end{array}\right]$
(d) none of these.
(a) $\frac{1}{2}\left[\begin{array}{cc}2& 4\\ 3& -5\end{array}\right]$
#### Question 24:
If $A=\left[\begin{array}{cc}2& 3\\ 5& -2\end{array}\right]$ be such that ${A}^{-1}=kA$, then k equals
(a) 19
(b) 1/19
(c) − 19
(d) − 1/19
(b) 1/19
#### Question 25:
If $A=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 2\\ 2& 1& -2\\ x& 2& y\end{array}\right]$ is orthogonal, then x + y =
(a) 3
(b) 0
(c) − 3
(d) 1
#### Question 26:
If equals
(a) A
(b) − A
(c) ab A
(d) none of these
(d) none of these
#### Question 27:
If , then
(a)
(b)
(c)
(d) none of these
(b)
#### Question 28:
If a matrix A is such that 3 equal to
(a)
(b)
(c)
(d) none of these
(d) none of these
$3{A}^{3}+2{A}^{2}+5A+I=0\phantom{\rule{0ex}{0ex}}⇒3{A}^{3}+2{A}^{2}+5A=-I\phantom{\rule{0ex}{0ex}}⇒{A}^{-1}\left(3{A}^{3}+2{A}^{2}+5A\right)=-I{A}^{-1}\phantom{\rule{0ex}{0ex}}⇒3{A}^{2}+2A+5I=-{A}^{-1}\phantom{\rule{0ex}{0ex}}⇒{A}^{-1}=-3{A}^{2}-2A-5I\phantom{\rule{0ex}{0ex}}$
#### Question 29:
If A is an invertible matrix, then det (A1) is equal to
(a)
(b)
(c) 1
(d) none of these
(b)
We know that for any invertible matrix A, $\left|{A}^{-1}\right|$ = $\frac{1}{\left|A\right|}$.
#### Question 30:
If
(a) $A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, if n is an even natural number
(b) $A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, if n is an odd natural number
(c)
(d) none of these
Disclaimer: In all option, the power of A (i.e. n is missing)
(a) ${A}^{n}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, if n is an even natural number
$A=\left[\begin{array}{cc}2& -1\\ 3& -2\end{array}\right]\phantom{\rule{0ex}{0ex}}{A}^{2}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒A×A=I\phantom{\rule{0ex}{0ex}}⇒{A}^{-1}=A\phantom{\rule{0ex}{0ex}}$
Generally,
#### Question 31:
If x, y, z are non-zero real numbers, then the inverse of the matrix $A=\left[\begin{array}{ccc}x& 0& 0\\ 0& y& 0\\ 0& 0& z\end{array}\right]$, is
(a) $\left[\begin{array}{ccc}{x}^{-1}& 0& 0\\ 0& {y}^{-1}& 0\\ 0& 0& {z}^{-1}\end{array}\right]$
(b)
(c) $\frac{1}{xyz}\left[\begin{array}{ccc}x& 0& 0\\ 0& y& 0\\ 0& 0& z\end{array}\right]$
(d)
(a) $\left[\begin{array}{ccc}{x}^{-1}& 0& 0\\ 0& {y}^{-1}& 0\\ 0& 0& {z}^{-1}\end{array}\right]$
#### Question 32:
If A and B are invertible matrices, then which one of the following is not correct?
(a) adj A$\left|A\right|$ A-1 (b) det (A-1) = [det(A)]-1
(c) (AB)-1 = B-1A-1 (d) (A+B)-1 = B-1 + A-1
(a) adj A = $\left|A\right|$ A1
As we know,
Thus, adj A = $\left|A\right|$ A1 is correct.
(b) det(A1) = [det(A)]1
As we know,
$\left|{A}^{-1}\right|=\frac{1}{\left|A\right|}\phantom{\rule{0ex}{0ex}}⇒\left|{A}^{-1}\right|={\left|A\right|}^{-1}\phantom{\rule{0ex}{0ex}}$
Thus, det(A1) = [det(A)]1 is correct.
(c) (AB)1 = B1A1
As we know,
By reversal law of inverse
(AB)1 = B1A1
Thus, (AB)1 = B1A1 is correct.
(d) (A + B)1 = B1 + A1
Thus, (A + B)1 = B1 + A1 is incorrect.
Hence, the correct option is (d).
#### Question 33:
If A = $\left[\begin{array}{ccc}2& \lambda & -3\\ 0& 2& 5\\ 1& 1& 3\end{array}\right]$, then A-1 exists if
(a) λ=2 (b) λ≠2 (c) λ≠-2 (d) none of these
Given: A = $\left[\begin{array}{ccc}2& \lambda & -3\\ 0& 2& 5\\ 1& 1& 3\end{array}\right]$
A1 exists only if |A| ≠ 0.
$\left|\begin{array}{ccc}2& \lambda & -3\\ 0& 2& 5\\ 1& 1& 3\end{array}\right|\ne 0\phantom{\rule{0ex}{0ex}}⇒2\left(6-5\right)+1\left(5\lambda +6\right)\ne 0\phantom{\rule{0ex}{0ex}}⇒2\left(1\right)+5\lambda +6\ne 0\phantom{\rule{0ex}{0ex}}⇒2+5\lambda +6\ne 0\phantom{\rule{0ex}{0ex}}⇒5\lambda +8\ne 0\phantom{\rule{0ex}{0ex}}⇒5\lambda \ne -8\phantom{\rule{0ex}{0ex}}⇒\lambda \ne \frac{-8}{5}$
Thus, ​A1 exists if λ ∈ $R-\left\{-\frac{8}{5}\right\}$
Hence, the correct option is (a).
#### Question 1:
If A is a unit matrix of order n, then A (adj A) = ___________________.
As we know that, A(adj A) = |A|I.
But it is given that A is a unit matrix of order n
Therefore, A(adj A) = |I|I = (1)I = I
Hence, if A is a unit matrix of order n, then A (adj A) = I.
#### Question 2:
If A is a non-singular square matrix such that A3 = I, then A-1 = _______________.
Given: A3 = I
A
3 = I
Multiplying both sides by A1, we get
⇒ A3A1I A1
⇒ A2(AA1) = I A1
⇒ A2(I) = A1
⇒ A2 = A1
Hence, if A is a non-singular square matrix such that A3 = I, then A1 = A2.
#### Question 3:
If A and B are square matrices of the same order and AB = 3I, then A-1 = __________________.
Given:
A and B are square matrices of the same order
AB = 3I
AB = 3I
Pre-Multiplying both sides by A1, we get
⇒ A1(AB) = A1(3I
(A1A)B = 3(A1I
⇒ (I)B = 3A1
⇒ B = 3A1
⇒ $\frac{1}{3}B$ = A1
Hence, if A and B are square matrices of the same order and AB = 3I, then A= $\overline{)\frac{1}{3}B}$.
#### Question 4:
If the matrix A$\left[\begin{array}{ccc}1& a& 2\\ 1& 2& 5\\ 2& 1& 1\end{array}\right]$ is not invertible, than a = ___________________.
Given: A = $\left[\begin{array}{ccc}1& a& 2\\ 1& 2& 5\\ 2& 1& 1\end{array}\right]$
A is not invertible if |A| = 0.
$\left|\begin{array}{ccc}1& \alpha & 2\\ 1& 2& 5\\ 2& 1& 1\end{array}\right|=0\phantom{\rule{0ex}{0ex}}⇒1\left(2-5\right)-1\left(\alpha -2\right)+2\left(5\alpha -4\right)=0\phantom{\rule{0ex}{0ex}}⇒1\left(-3\right)-1\alpha +2+10\alpha -8=0\phantom{\rule{0ex}{0ex}}⇒-3-\alpha +2+10\alpha -8=0\phantom{\rule{0ex}{0ex}}⇒9\alpha -9=0\phantom{\rule{0ex}{0ex}}⇒9\alpha =9\phantom{\rule{0ex}{0ex}}⇒\alpha =1$
Hence, if the matrix A = $\left[\begin{array}{ccc}1& a& 2\\ 1& 2& 5\\ 2& 1& 1\end{array}\right]$ is not invertible, than a = 1.
#### Question 5:
If A is a singular matrix, then A (adj A) = ____________________.
As we know that, A(adj A) = |A|I.
But it is given that A is a singular matrix
Thus, |A| = 0.
Therefore, A(adj A) = 0I = O, where O is the zero matrix.
Hence, if A is a singular matrix, then A (adj A) = O.
#### Question 6:
Let A be a square matrix of order 3 such that $\left|A\right|$ = 11 and B be the matrix of confactors of elements of A. Then, ${\left|B\right|}^{2}$ = ________________.
Given:
A be a square matrix of order 3
$\left|A\right|$ = 11
B be the matrix of cofactors of elements of A
Since, B be the matrix of cofactors of elements of A
As we know,
Hence, ${\left|B\right|}^{2}$ = 14641.
#### Question 7:
If A is a square matrix of order 2 such that A (adj A) = =______________.
As we know that, A(adj A) = |A|I.
But it is given that A (adj A) = $\left[\begin{array}{cc}10& 0\\ 0& 10\end{array}\right]$
Hence, |A| = 10.
#### Question 8:
If A is an invertible matrix of order 3 and = ___________________.
Given:
A is an invertible matrix of order 3
$\left|A\right|$ = 3
As we know,
#### Question 9:
If A is an invertible matrix of order 3 and = __________________.
Given:
A is an invertible matrix of order 3
$\left|A\right|$ = 5
As we know,
#### Question 10:
If A is an invertible matrix of order 3 and =__________________.
Given:
A is an invertible matrix of order 3
$\left|A\right|$ = 4
As we know,
#### Question 11:
If A = diag (1, 2, 3), then =________________.
Given:
A = diag (1, 2, 3)
⇒ $\left|A\right|$ = 1 × 2 × 3 = 6
As we know,
#### Question 12:
If A is a square matrix of order 3 such that = ________________.
Given:
A is a square matrix of order 3
|A| = $\frac{5}{2}$
As we know,
$\left|{A}^{-1}\right|={\left|A\right|}^{-1}\phantom{\rule{0ex}{0ex}}⇒\left|{A}^{-1}\right|=\frac{1}{\left|A\right|}\phantom{\rule{0ex}{0ex}}⇒\left|{A}^{-1}\right|=\frac{1}{\frac{5}{2}}\phantom{\rule{0ex}{0ex}}⇒\left|{A}^{-1}\right|=\frac{2}{5}\phantom{\rule{0ex}{0ex}}$
Hence, $\left|{A}^{-1}\right|=\overline{)\frac{2}{5}}$.
#### Question 13:
If A is a square matrix such that A (adj A) = 10I, then $\left|A\right|$ = ____________________.
Given:
A is a square matrix
As we know,
Hence, $\left|A\right|=\overline{)10}$.
#### Question 14:
Let A be a square matrix of order 3 and B = _________________.
Given:
A is a square matrix of order 3
= |A|A−1
|A| = −5
Now,
Hence, $\left|B\right|=\overline{)25}$.
#### Question 15:
If k is a scalar and I is a unit matrix of order 3, then adj (kI) = ________________.
Given:
I is a unit matrix of order 3
As we know,
#### Question 16:
If A and A (adj A) = $\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]$, then k = _________________.
Given:
A = $\left[\begin{array}{cc}\mathrm{cos}x& \mathrm{sin}x\\ -\mathrm{sin}x& \mathrm{cos}x\end{array}\right]$
A(adj A) = $\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]$
Now,
$A=\left[\begin{array}{cc}\mathrm{cos}x& \mathrm{sin}x\\ -\mathrm{sin}x& \mathrm{cos}x\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒\left|A\right|=\left|\begin{array}{cc}\mathrm{cos}x& \mathrm{sin}x\\ -\mathrm{sin}x& \mathrm{cos}x\end{array}\right|\phantom{\rule{0ex}{0ex}}⇒\left|A\right|={\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x\phantom{\rule{0ex}{0ex}}⇒\left|A\right|=1$
As we know,
Hence, $k=\overline{)1}$.
#### Question 17:
If A is a non-singular matrix of order 3, then adj (adj A) is equal to ________________.
Given:
A is a non-singular matrix of order 3
As we know,
#### Question 18:
If A = [aij]2×2, where aij = , then A-1 = ____________________.
Given:
A = [aij]2×2, where aij =
Hence, ${A}^{-1}=\overline{)\frac{1}{9}\left[\begin{array}{cc}0& 3\\ 3& 1\end{array}\right]}.$
#### Question 19:
If A$\left[\begin{array}{cc}0& 3\\ 2& 0\end{array}\right]$ and A-1 = λ (adj A), then λ = _____________________.
Given:
A = $\left[\begin{array}{cc}0& 3\\ 2& 0\end{array}\right]$
Hence, $\lambda =\overline{)-\frac{1}{6}}$.
#### Question 20:
If A is a 3×3 non-singular matrix such that AAT = ATA and B = A-1AT, then BBT = ______________.
Given:
A is a 3×3 non-singular matrix
AAT = ATA
B
A−1AT
Hence, BBT = I.
#### Question 21:
If A and B are two square matrices of the same order such that B = -A-1BA, then (A+B)2 = ______________.
Given:
$B=-{A}^{-1}BA\phantom{\rule{0ex}{0ex}}⇒AB=-A{A}^{-1}BA\phantom{\rule{0ex}{0ex}}⇒AB=-IBA\phantom{\rule{0ex}{0ex}}⇒AB=-BA$
Hence, (A + B)2 = A2 + B2.
#### Question 22:
If A is a non-singular matrix of order 3×3, then (A3)-1 = _____________.
ans
#### Question 23:
If A be a square matrix such that , then the order of A is __________________.
Given:
A is a square matrix
As we know,
Hence, the order of is 3.
#### Question 24:
If A = $\left[\begin{array}{ccc}x& 5& 2\\ 2& y& 3\\ 1& 1& z\end{array}\right]$ ,xyz = 80, 3x + 2y + 10z = 20 and A adj A = kI, then k = _________________.
Given:
A = $\left[\begin{array}{ccc}x& 5& 2\\ 2& y& 3\\ 1& 1& z\end{array}\right]$
xyz = 80
3x + 2y + 10z = 20
Now,
As we know,
Hence, k = 79.
#### Question 1:
Write the adjoint of the matrix $A=\left[\begin{array}{cc}-3& 4\\ 7& -2\end{array}\right].$
#### Question 2:
If A is a square matrix such that A (adj A) 5I, where I denotes the identity matrix of the same order. Then, find the value of |A|.
We know
Here,
#### Question 3:
If A is a square matrix of order 3 such that |A| = 5, write the value of |adj A|.
For any square matrix of order n,
#### Question 4:
If A is a square matrix of order 3 such that |adj A| = 64, find |A|.
For any square matrix of order n,
#### Question 5:
If A is a non-singular square matrix such that |A| = 10, find |A−1|.
For any non-singular matrix A,
#### Question 6:
If A, B, C are three non-null square matrices of the same order, write the condition on A such that AB = ACB = C.
Consider $AB=AC$.
On multiplying both sides by ${A}^{-1}$, we get
$A{A}^{-1}B=A{A}^{-1}C$
Therefore, the required condition is A must be invertible or $\left|A\right|\ne 0$.
#### Question 7:
If A is a non-singular square matrix such that ${A}^{-1}=\left[\begin{array}{cc}5& 3\\ -2& -1\end{array}\right]$, then find ${\left({A}^{T}\right)}^{-1}.$
For any invertible matrix A,
$\phantom{\rule{0ex}{0ex}}\left({A}^{T}{\right)}^{-1}=\left({A}^{-1}{\right)}^{T}$
#### Question 8:
Given:
For any two non-singular matrices,
#### Question 9:
If A is symmetric matrix, write whether AT is symmetric or skew-symmetric.
For any symmetric matrix, ${A}^{T}=A$.
Hence, ${A}^{T}$ is also symmetric.
#### Question 10:
If A is a square matrix of order 3 such that |A| = 2, then write the value of adj (adj A).
For any square matrix A, we have
#### Question 11:
If A is a square matrix of order 3 such that |A| = 3, then write the value of adj (adj A).
For any square matrix A, we have
#### Question 12:
If A is a square matrix of order 3 such that adj (2A) = k adj (A), then write the value of k.
#### Question 13:
If A is a square matrix, then write the matrix adj (AT) − (adj A)T.
#### Question 14:
Let A be a 3 × 3 square matrix, such that A (adj A) = 2 I, where I is the identity matrix. Write the value of |adj A|.
#### Question 15:
If A is a non-singular symmetric matrix, write whether A−1 is symmetric or skew-symmetric.
#### Question 16:
If , then find the value of k.
#### Question 17:
If A is an invertible matrix such that |A−1| = 2, find the value of |A|.
#### Question 18:
If A is a square matrix such that , then write the value of |adj A|.
#### Question 19:
If $A=\left[\begin{array}{cc}2& 3\\ 5& -2\end{array}\right]$ be such that then find the value of k.
#### Question 20:
Let A be a square matrix such that ${A}^{2}-A+I=O$, then write ${A}^{-1}$ interms of A.
#### Question 21:
If Cij is the cofactor of the element aij of the matrix $A=\left[\begin{array}{ccc}2& -3& 5\\ 6& 0& 4\\ 1& 5& -7\end{array}\right]$, then write the value of a32C32.
In the given matrix $A=\left[\begin{array}{ccc}2& -3& 5\\ 6& 0& 4\\ 1& 5& -7\end{array}\right]$,
C32 = (−1)3 + 2 (8 − 30) = 22
Therefore, a32C32 = 5 × 22 = 110.
Hence, the value of a32C32 is 110.
#### Question 22:
Find the inverse of the matrix $\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right].$
#### Question 23:
Find the inverse of the matrix .
#### Question 24:
If $A=\left[\begin{array}{cc}1& -3\\ 2& 0\end{array}\right]$, write adj A.
#### Question 26:
If $A=\left[\begin{array}{cc}3& 1\\ 2& -3\end{array}\right]$, then find |adj A|.
#### Question 27:
If $A=\left[\begin{array}{cc}2& 3\\ 5& -2\end{array}\right]$, write ${A}^{-1}$ in terms of A.
$⇒{A}^{-1}=\frac{1}{19}A$
Write
#### Question 29:
Use elementary column operation C2 → C2 + 2C1 in the following matrix equation :
Applying C2 → C2 + 2C1, we get
#### Question 30:
In the following matrix equation use elementary operation R2 → R2 + R1 and the equation thus obtained:
By applying elementary operation R2 → R2 + R1, we get
(Every row operation is equivalent to left-multiplication by an elementary matrix.)
#### Question 31:
If A is a square matrix with $\left|A\right|$ = 4 then find the value of
Given:
A is a square matrix
$\left|A\right|$ = 4
Now,
Hence,
View NCERT Solutions for all chapters of Class 12<|endoftext|>
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Python language has an abstraction of iterator and iterable. This post describes and differentiates between the two.
Iterator is an abstraction that enables walking/looping over a collection. It is a stateful object that represents the state of an iteration over the collection. The state specifies the current location of the iterator in the collection. Iterator also provides the functions to get the value in the collection at the current position and to move to the next element of the collection.
In case of Python, an iterator has to implement
the dunder function
__next__, which returns the next element
of the collection and updates the state to the next position.
Any object in Python, that implements
__next__ is an iterator.
In the context of Python, any object (commonly a collection)
that returns an iterator is called iterable, i.e., one can iterate over it.
Formally, any object that implements the dunder function
which returns a fresh iterator (iterator with position set to the start) is called an iterable.
The returned iterator is used for iterating over the collection.
Note that an iterable can also implement
__getitem__ in case of a sequential index based collection.
Whenever we write a code like
for x in y: ..., then it is expected that
y is an iterable. Python implementation will get a fresh iterator
iter = y.__iter__())
and then iterate over it using (
x = next(iter)).
The following for loop is equivalent to the subsequent while loop.
for x in y: print(x)
iter = y.__iter__() while True: try: print(iter.__next__()) except StopIteration: # part of iter protocol break
Following is an example of iterator and iterable.
RandGen is a class to generate a series of random number given a max count.
RandGen is implemented as an iterable, and it returns an iterator (_RandGenIter) in the function
_RandGenIter is an iterator, it stores the state in the variable pos.
State is initialized to the start (0) and the function
__next__ increments the
in each call.
When the max count is attained it throws the exception
StopIteration as per the protocol
of the iteration mechanism.
class _RandGenIter(): def __init__(self, randgen): self._randgen = randgen self._pos = 0 def __next__(self): if self._pos < self._randgen._max: self._pos += 1 return self._randgen.get_rand() else: raise StopIteration
class RandGen(): def __init__(self, max): self._max = max def __iter__(self): return _RandGenIter(self) def get_rand(self): import random return random.random()
for x in RandGen(4): print(x)
Note: At times, the class supports
It allows to have the same class supporting both the functionalities.
In the above example, we can get rid of _RandGenIter and move its functionality
In this case, the class is both iterator and iterable.
Also, iterators commonly implement the dunder function
__iter__ (returning itself), it allows them
to be used in places where iterables are required. For example, we will be able to use
an iterator in a for loop.
© 2018-19 Manjeet Dahiya<|endoftext|>
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June 20 - 1819 - The SS Savannah Arrives in Liverpool
The SS Savannah was going to inaugurate a new era in ocean going vessels, as it was the first steam powered ship to make a transatlantic crossing. On the other hand, the Savannah was also a sailing ship that had a steam engine and paddle wheel added on in construction, and it only used steam power for a small fraction of its voyage. Steam powered paddle ships had been used on rivers for a decade, but no one had attempted a transatlantic trip. The Savannah's captain, Moses Rogers, convinced the firm of Scarborough & Isaacs in Savannah, Georgia to pay for a steam engine on his ship to make such a voyage. After moving from its shipyard in New York to its home port in Savannah, the ship was so popular that President James Monroe wanted to come aboard and see it. Once it began its historic voyage in May 1819, the ship had a smooth trip. The steam engine was not used much, but it did leave Savannah and enter Liverpool on steam. Once it hit Liverpool, the excitement around the ship was immense, and the intrigue continued as it went to St. Petersburg. Six months after it left home, the Savannah returned to Savannah, but no other steam powered ships would make a voyage for nearly two decades.<|endoftext|>
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Why do the literature review?
Truths about research from your course text,
Truth 1: All research projects are part of a larger conversation and a larger set of evidence (p. 41)
Truth 2: New research almost always has its origins in existing research (p. 42)
Truth 3: Even when it seems there is no literature exactly on point, there will still be literature that discusses the main theories or concepts being used...(p. 42)
Reference: Patten, M. L., & Newhart, M. (2018). Understanding research methods: An overview of the essentials. New York: Routledge.
Step 1: Start combing through the research & collecting sources.
Start with a general topic or idea. What do you want to know more about? Make a list of all the relevant keywords including synonyms. Pull up Google Scholar, Primo (Library Discovery), and maybe a few of the library databases specific to the subject you are researching. Try plugging in your keywords and observe what kind of materials come up. Are they relevant? You might look at what subjects the articles have been given and revise your search based on the categories you are seeing. Not finding much? This would be a great time to contact your librarian!
Step 2: Read & analyze what you've found.
What's been said already? Make a list of your sources so you can keep track of the citations and your notes. Zotero is an awesome free software you can use to store your research. You might also just use a Google Doc. Take notes on each article. What was the author’s research question? What study method did they use? What were the results of their study? Were there any flaws in their research design or methods? What worked? What were their conclusions? Does their study leave me with any questions?
Step 3: Define your research question or hypothesis.
What do you want to know more about? Now that you know what research has already been done, what lingering questions do you have? Are there any studies out there that you can build upon? Are there any gaps in the literature that you can add insight to? It’s your chance to add your own unique voice to the conversation.<|endoftext|>
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# Roulette/Math
Probability
If one understands the basics of probability theory, then in roulette especially it is very easy to test betting systems mathematically. Here is the step by step logic of applying probability in roulette to the possible outcomes.
First, all the mathematics used here is based on a European single 0 wheel since the house edge is half the American version.
We know that the probability of an event happening is the chances of that event compared to all the possible events. For instance, when you flip a coin there are 2 possible outcomes: heads, tails. If you want to know what is the probability that the coin will come up heads, then that would be: heads / (heads + tails) = 1/2 = .5. Likewise when playing an even money bet at roulette, that option covers 18 of the 37 possible outcomes: 18/37=.48648649.
To find out the effect the odds have on a measurable outcome, we can apply that outcome to all possible results. So if we’re playing \$1 on black, then we know that for 18 of 37 outcomes we will net \$1 profit, and for 19 of the 37 possible outcomes we will net a \$1 loss. ((18/37)*1)+((19/37)*-1)=-.02702703. This shows the house advantage on any single spin applied to your bankroll. We know that if you place \$1 on any even number bet on average you will loose almost three cents per spin or \$27 over 100 spins.
This is valuable when looking at more complicated betting within the layout of the table. For instance, if you consider on the thirds position that the return is 2:1. Let’s look at the extremes. If you place a bet on one of the three options, then you are obviously playing against probability: 12/37=.32432432 probability to win. If you place \$1 on all three of the possible options, then for 36 of 37 numbers you will loose \$2, make \$2, and have the bet on the winning third returned to you for a net profit of \$0. This of course makes no sense at all, but you’ll win almost every time if you’re in it to feel like a winner however if your considering a system you’re trying to make money. If we hedge the single bet with the second possible bet and place \$1 on the first two of the thirds, then we cover 24 of 37 numbers 24/37=.64864865. We’re guaranteed to lose one bet, but if the other hits then we make \$2, minus the one lost, plus the winning bet returned makes a net profit of \$1 – and here’s the kicker – our chance of winning on any single bet is greater than 50% (64.86% to be precise).
We know that roulette is an independently random game where the results of one action does not affect the odds of a second action, so presented like this one might see this a winning system of finding a way to shift the odds in your favor. However if we analyze all the possible outcomes we see that the proposition is a losing one. 24 of 37 possible outcomes net us \$1. On 13 of 37 possible outcomes we loose \$2. So we plug in our formula: ((24/37)*1)+((13/37)*-2)=-.05405405. This is even worse than playing even money odds.
Now we come full circle. Almost all systems are based on the premise that the likelihood of an event happening repeatedly gets exponentially smaller the more times in a row one seeks that option. Probability will never rule out a roulette table showing the number 36 100 times in a row, but it will tell us exactly how unlikely it is. The premise is that the probability of an event happening once is multiplied by the likelihood of the second event multiplied by the third event and so on. For instance, for a single number to come up 100 times we multiply (1/37)*(1/37)*(1/37)… for one hundred times. This is a tiny number but we can see how fast it adds up:
(1/37)=.02702703
(1/37)*(1/37)=.00073046
(1/37)*(1/37)*(1/37)=.00001974
(1/37)*(1/37)*(1/37)*(1/37)=.00000053
The likelihood of a number coming up four times in a row is only 0.000053%, but it happens. Just go to Global Player Casino and check out the roulette results for the year. But I digress, the strategies say if you chase a loss long enough it won’t lose any more, and systems like the Martingale set it up so that you realize a profit when that condition fails. However, it’s still a losing system because we can plug in our formula to this just as easily as we can plug it into a single event.
But first let’s examine what it is we’re looking at. If we’re analyzing a system there are only two options we’re interested in: win or loss. Let’s not get too complicated and assume that one loss will exit the system and return the player to the starting state such as the Martingale.
If the first spin loses then we go to a second spin, and if the second spin loses then we go to the third and so on. So we know that for however many levels we examine all the preceding spins will be losses. In other words, if 51.4% of spins will lose, then we are looking at 51.4% of 51.4% will lose twice in a row and 48.6% of 51.4% will win on the second round. Therefore, 51.4% of 51.4% of 51.4% will lose thrice, and 48.6% of 51.4% of 51.4% will win.
For a single level we know that the formula is the probability of a win times the net result and the probability of a loss times the net result.
(((18/37)*1)+((19/37)*-1))= -.02702703
To check the second level, the probability of a loss followed by the probability of a win times the net result is compared to two losses and the net result.
(((19/37)*(18/37))*1)+(((19/37)^2)*-2)= -.27757487
To extrapolate the third, fourth, and fifth level:
((((19/37)^2)*(18/37))*1)+(((19/37)^3)*-4)= -.4133615
((((19/37)^3)*(18/37))*1)+(((19/37)^4)*-8)= -.49040931
We can see no matter how far we go on the Martingale system it’s always more likely a losing proposition than a winning proposition, and the deeper one goes the more likely one is to lose a greater sum of money. Of course this isn’t a surprise since the odds are already against us.
More on other systems and hedge betting later.
Any system can be analyzed like this for any game. If the result is positive, the odds are in the player’s favor. If the result is negative you’re trusting Lady Luck. I haven’t found a formula that results in a positive number. Of course, if I had I'd be in a casino right now.
In practice, most betting systems redistribute the amounts of the wins and losses: an increase in the chance of winning is balanced against a greater loss once it does occur, as it will sooner or later. The oldest and most common betting system is the martingale, or doubling-up, system on even-money bets, in which bets are doubled progressively after each loss until a win occurs. This system probably dates back to the invention of the roulette wheel. [1]
Martingale is the most common betting system in roulette. The popularity of this system is due to its simplicity and availability. When playing on Martingale, it creates the deceptive impression of quick and easy wins. The essence of the Martingale roulette game system is as follows: we bet on an even chance of roulette (red-black, even-odd), for example, on "red": we bet on roulette at \$ 1; if you lose, we double the bet and bet \$ 2. If we lose at roulette, we lose the current bet (\$ 2) and the previous bet (\$ 1) by an amount of \$ 3. If we win, we win \$ 4, eg. wins two bets (1 + 2 = \$ 3) and we have \$ 1 net win from roulette. If you lost at roulette for the second time using the Martingale roulette system, let's double your bet again (it is now \$ 4). If we win, we will win back the previous two bets (1 + 2 = \$ 3) and the current one (\$ 4) from the roulette wheel, and again we win \$ 1 against the casino. [2]
This is a well presented maths explanation of the odds against the player when betting at roulette.But it confuses probability with certainty.Probability Theory deals with uncertainty not certainty. Roulette, like all gambling, is a game of chance so , obviously, chance is involved. This does not mean that only chance is involved. If the roulette wheel is random then no one can predict with certainty that we will win or lose. That certainty belongs to astrology not maths. There is no reason why the wheel should not give the same number continuously for a hundred, a thousand or even a million spins if it is truly random; incidentally, unless we are going to live till eternity then "The Long Run" is irrelevant in real time betting. The writer errs when dealing with betting the First and Second dozens together. Using the 1-18 bet we can lower the odds against us. Placing three chips on 1-18 and one chip on the six-line 19-24 benefits us should zero occur whereas betting the two dozens does not. To my mind, there is an all too casual attitude to discussing roulette and this is exemplified in this article.Also- but not here -there is usually a knee-jerk reaction to anyone who rejects the notion that you are certain to lose. Claims of certainty -to win or lose -are unjustifiable where uncertainty clearly reigns. Gambling is Gambling is Gambling .
1. "Roulette Systems". Britannica.com. `{{cite web}}`: Text "Britannica.com" ignored (help)<|endoftext|>
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Pretigianluca
Tumbling Down Age 7 to 11 Challenge Level: Have a go at this challenging task! Fractions Age 5 to 7 The activities in this feature give you chance to explore fractions. Who first used fractions? Counting Fish Age 14 to 16 Challenge Level:
In this feature we offer rich tasks to build learners’ deep conceptual understanding of fractions. Register for our mailing list. What fractions can you find between the square roots of 65 and 67? How much of the square is coloured blue? A Sudoku with clues as ratios. All of the following tasks provide contexts in which to encourage learners to talk about their ideas and to work towards refining their vocabulary. Investigate the successive areas of light blue in these diagrams.
## 242 Matches for fractions
Fractional Triangles Age 7 to 11 Challenge Level: Doughnut Percents Age 7 to 14 Challenge Level: The large rectangle is divided into a series of smaller quadrilaterals and triangles.
This gives opportunities for different approaches.
Try this matching game and see. How did fractions reach us here? Can you find different ways of showing the same fraction? Egyptian Fractions Age 11 to 14 Challenge Level: All of the following tasks provide contexts in which to encourage learners prkblem talk about their ideas and to work towards refining their vocabulary. Pick two rods of different colours. Folding Fractions Fracctions 14 to 16 Challenge Level: What fractions can you divide the diagonal of a square into by simple folding?
ESSAY AF JESPER WUNG SUNG
## 1213 Matches for fractions problems
Can you find different ways of showing the same fraction? Fractions Age 7 to 11 Explore the idea of fractions using these activities.
Finally, pproblem understand the inverse relation between the denominator and the quantity: So, the tasks in this second group are curriculum-linked but crucially also offer opportunities for learners to develop solcing problem-solving and reasoning skills. The Greedy Algorithm might provide us with an efficient way of doing this. The question should always be, ‘fraction of what? Such tasks also provide valuable opportunities for you to assess where children have got to in their thinking and so support the next steps on their learning journey.
Watch the video to see how to fold a square of paper to create a flower. Counting Fish Age 14 to 16 Challenge Level: Can you see how to build a harmonic triangle?
# Fractions Unpacked :
Stetch it out to match the original length. Were they always written in the same way? Explore how to divide numbers and write fractions in modulus arithemtic.
FREDDIE HIGHMORE Y EMMA ROBERTS HOMEWORK
Can you explain why this happens? Part-whole Concept Age 5 to 14 Written for teachers, this article describes four basic approaches children use in understanding fractions sokving equal parts of a whole.
Can you work out which drink has the stronger flavour? Can you find its length? Linked Chains Age 7 to 11 Challenge Level: An environment which simulates working with Cuisenaire rods. There are three tables in a room with blocks of chocolate on each.
Egyptian Fractions Age 11 to 14 Challenge Level: Fractions – Sequences and Patterns Exploring interesting patterns and sequences generated with fractions.
Take a look at the video and try to find a sequence of moves that will untangle the ropes.<|endoftext|>
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Find solution of equations- Equations given
Chapter 4 Class 12 Determinants
Concept wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Ex 4.5, 11 Solve system of linear equations, using matrix method. 2x + y + z = 1 x – 2y – z = 3/2 3y – 5z = 9 The system of equation is 2x + y + z = 1 x – 2y – z = 3/2 3y – 5z = 9 Writing above equation as AX = B [■8(2&1&[email protected]&−2&−[email protected]&3&−5)][■8(𝑥@𝑦@𝑧)] = [■8([email protected]/[email protected])] Hence A = [■8(2&1&[email protected]&−2&−[email protected]&3&−5)]𝑥= [■8(𝑥@𝑦@𝑧)] & B = [■8([email protected]/[email protected])] Calculating |A| |A| = |■8(2&1&[email protected]&−2&−[email protected]&3&−5)| = 2 |■8(−2&−[email protected]&−5)| – 1 |■8(1&−[email protected]&−5)| + 1 |■8(1&−[email protected]&3)| = 2 (10 + 3 ) – 1(–5 + 0) + 1 (3 – 0) = 2 (13) –1 ( – 5 ) + 1 (3) = 34 Thus, |A| ≠ 0 ∴ The system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 A-1 = 1/(|A|) adj (A) adj (A) = [■8(A_11&A_12&[email protected]_21&A_22&[email protected]_31&A_32&A_33 )]^′ = [■8(A_11&A_21&[email protected]_12&A_22&[email protected]_13&A_32&A_33 )] A = [■8(2&1&[email protected]&−2&−[email protected]&3&−5)] M11 = [■8(−2&−[email protected]&−5)] = 10 + 3 = 13 M12 = [■8(1&−[email protected]&−5)] = –5 + 0 = –5 M13 = [■8(1&−[email protected]&3)] = 3 + 0 = 3 M21 = [■8(1&[email protected]&−5)] = –5 – 3 = –8 M22 = [■8(2&[email protected]&−5)] = –10 + 0 = –10 M23 = [■8(2&[email protected]&3)] = 6 + 0 = 6 M31 = [■8(1&1@−2&−1)] = –1 + 2 = 1 M32 = [■8(2&[email protected]&−1)] = –2 – 1 = –3 M33 = [■8(2&[email protected]&−2)] = –4 – 1 = –5 Now, A11 = (–1)1+1 . M11 = (–1)2 . (13) = 13 A12 = (–1)1+2 . M12 = (–1)3 . (–5) = 5 A13 = (–1)1+3 . M13 = (–1)4 . (3) = 3 A21 = (–1)2+1 . M21 = (–1)3 . (–8) = 8 A22 = (–1)2+2 . M22 = (–1)4 . (–10) = –10 A23 = (–1)2+3 . M23 = (–1)5 . (6) = – 6 A31 = (–1)3+1 . M31 = (–1)4 . (1) = 1 A32 = (–1)3+2 . M32 = (–1)5 . (–3) = 3 A33 = (–1)3+3 . M33 = (–1)6 . (–5) = – 5 Thus, adj (A) =[■8(13&8&[email protected]&−10&[email protected]&−6&−5)] Now, A-1 = 1/(|A|) adj A Putting values = 1/34 [■8(13&8&[email protected]&−10&[email protected]&−6&−5)] Also, X = A-1 B Putting values [█(■8(𝑥@𝑦)@𝑧)] = 1/34 [■8(13&8&[email protected]&−10&[email protected]&−6&−5)][█(■8([email protected]/2)@9)] [█(■8(𝑥@𝑦)@𝑧)] = 1/34 [■8(13(1)+8(3/2)+1(9)@5(1)+(−10)(3/2)+3(9)@3(1)+(−6)(3/2)+(−5)(9) )] [█(■8(𝑥@𝑦)@𝑧)] = 1/34 [■8([email protected][email protected]−9−45)] = 1/34 [█(■8([email protected])@−51)] [█(■8(𝑥@𝑦)@𝑧)] = [█(■8([email protected]/2)@(−3)/2)] Hence x = 1, y = 𝟏/𝟐 & z = (−𝟑)/𝟐<|endoftext|>
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The counterpart of an independent clause. A dependent clause has a subject and verb, but it cannot stand on its own. It is not a complete sentence. A dependent clause begins with a DMW. For example, the following clause is independent:
Jim drove his car over the bridge.
but when a DMW is added to it, it becomes a dependent clause.
When Jim drove his car over the bridge…
The word “when” is a DMW and makes this clause dependent. It is called “dependent” because it must be attached to an independent clause in order for it to function as a complete thought (it is dependent on an independent clause).
Dependent clauses can function either:
When the DMW is a subordinating conjunction, the clause is usually adverbial; when the DMW is a relative pronoun, the clause is usually adjectival. See the presentation below or a video on the same subject.
See here for help diagramming clauses.<|endoftext|>
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# Convert Polar to Rectangular Coordinates and Vice Versa
The rectangular coordinates (x , y) and polar coordinates (R , t) are related as follows. y = R sin t and x = R cos t R 2 = x 2 + y 2 and tan t = y / x To find the polar angle t, you have to take into account the sings of x and y which gives you the quadrant. Angle t is in the range [0 , 2Pi) or [0 , 360 degrees). Problem 1: Convert the polar coordinates (5 , 2.01) and (0.2 , 53 o) to rectangular coordinates to three decimal places. Solution to Problem 1: For the first point (5 , 2.01) R = 5 and t = 2.01 and is in radians. Set your calculator to radians and use the above formulas for x and y in terms of R and t to obtain: x = R cos t = 5 cos 2.01 = -2.126 y = R sin t = 5 sin 2.01 = 4.525 For the second point (0.2 , 53 o) R = 0.2 and t = 53 o and is in degrees. Set your calculator to degrees and use the above formulas for x and y in terms of R and t to obtain: x = R cos t = 0.2 cos 53 = 0.120 y = R sin t = 0.2 sin 53 = 0.160 Problem 2: Convert the rectangular coordinates (1 , 1) and (-2 ,-4) to polar coordinates to three decimal places. Express the polar angle t in degrees and radians. Solution to Problem 2: We first find R using the formula R = sqrt [x 2 + y 2] for the point (1 , 1). R = sqrt [x 2 + y 2] = sqrt [1 + 1] = sqrt ( 2 ) We now find tan t using the formula tan t = y / x. tan t = 1 / 1 Using the arctan function of the calculator, we obtain. t = Pi / 4 or t = 45 o Point (1 , 1) in rectangular coordinates may be written in polar for as follows. ( sqrt ( 2 ) , Pi / 4 ) or ( sqrt ( 2 ) , 45 o ) Let us find find R using for the point (-2 , -4). R = sqrt [x 2 + y 2] = sqrt [4 + 16] = sqrt ( 20 ) = 2 sqrt ( 5 ) We now find tan t. tan t = - 4 / - 2 = 2 Using the arctan function of the calculator, we obtain. t = 1.107 or t = 63.435 o BUT since the rectangular coordinates x and y are both negative, the point is in quadrant III and we need to add Pi or 180 o to the value of t given by the calculator. Hence the polar angle t is given by t = 4.249 or t = 243.435 o Point (-2 , -4) in rectangular coordinates may be written in polar for as follows. ( 2 sqrt ( 5 ) , 4.249 ) or ( 2 sqrt ( 5 ) , 243.435 o ) More references on polar coordinates and trigonometry topics.<|endoftext|>
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# Functions Modeling Change (5th Edition) Edit edition 81% (36 ratings) for this chapter’s solutionsSolutions for Chapter 7.1
We have solutions for your book!
Chapter: Problem:
Step-by-step solution:
Chapter: Problem:
• Step 1 of 8
A function whose values repeat on a regular interval is called a periodic function.
This means that as we move along the x-axis, the shape of the graph must repeat after regular intervals.
So let’s observe carefully each of the given graph and try to find out if the shape of the graph is repeating on regular intervals or not.
• Step 2 of 8
Following is the graph (I):
In this graph we see that the shape is repeating after every interval of length 4.
Start from any point on x-axis and as we move along x-axis, we see that the shape is repeating after every 4 units.
For example if we start from the origin, we see that the shape in the interval is same as the shape in the interval , , , etc.
Hence, this graph might represent a periodic function.
• Step 3 of 8
Following is the graph (II):
In this graph we see that the shape is repeating after a fixed interval.
For example if we start from the point , we see that the shape in the interval is same as the shape in the interval , , ,
Also the lengths of these intervals are same
Hence, this graph might represent a periodic function.
• Step 4 of 8
Following is the graph (III):
In this graph we see that the shape appears to be similar over following intervals:
, , , , ,
But the length of these intervals is not same.
Hence this function is not repeating after a “fixed” interval.
Hence, this graph might not represent a periodic function.
Following is the graph (IV):
• Step 5 of 8
In this graph we see that the shape is repeating after a fixed interval.
For example if we start from the point , we see that the shape in the interval is same as the shape in the interval
Hence, this graph might represent a periodic function.
• Step 6 of 8
Following is the graph (V):
We see that this function is oscillating with increasing amplitude as me move along the
x-axis.
Hence the shape is not repeating on any fixed interval.
Hence, this graph might not represent a periodic function.
• Step 7 of 8
Following is the graph (VI):
We see that the shape is not repeating on any fixed interval.
Hence, this graph might not represent a periodic function.
• Step 8 of 8
Note that in above discussion we used the words “might represent a periodic function” and “might not represent a periodic function”.
The reason for this is that in order to be sure we must either know the complete graph of the function or we must know the formula of the function.
Corresponding Textbook
Functions Modeling Change | 5th Edition
9781118942581ISBN-13: 1118942582ISBN: Authors:
This is an alternate ISBN. View the primary ISBN for: Functions Modeling Change 5th Edition Textbook Solutions<|endoftext|>
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253 |
Course description: A beginning course in a comprehensive welding program. Emphasis on types of welding, kinds of machines, differences in current, and types of electrodes used. Introduction to and basic techniques on, arc welding in the flat position. Basic techniques in arc welding including how to set current and fine settings on power sources. Emphasis is also in striking the arc and motions to be used in running the bead.
Course objectives: The student will select electrodes and amperage settings for various thicknesses of materials and welding positions; define principles of arc welding; and interpret electrode classifications. The student will perform SMAW operations in various positions using selected electrodes and different joint designs.
Lecture: Safety rules and regulations. Safety with welding related tools Oxyacetylene cutting safety. Arc welding safety Maintenance on both oxyacetylene and arc welding equipment Joint design. Demonstrates good arc bead pattern with various electrodes.
Lab: The student will be able to demonstrate safety on arc welding equipment. Be able to demonstrate welding in the flat position; and have the knowledge to select correct electrode. He will have the ability to select correct heat settings on welding machines, and be able to demonstrate good bead patterns on plates.
Course start date: Continuous Enrollment<|endoftext|>
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