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WASHINGTON, DC, March 16, 2010 — A new study involving scientists from 13 different organizations, universities and research institutions states that forest protection offers one of the most effective, practical, and immediate strategies to combat climate change. The study, “Indigenous Lands, Protected Areas, and Slowing Climate Change,” was published in PLoS Biology, a peer-reviewed scientific journal, and makes specific recommendations for incorporating protected areas into overall strategies to reduce emissions of greenhouse gasses from deforestation and degradation (nicknamed REDD).
“Deforestation leads to about 15 percent of the world’s greenhouse gas emissions, more than all the cars, trucks, trains, ships, and planes on earth. If we fail to reduce it, we’ll fail to stabilize our climate,” said Taylor Ricketts, director of World Wildlife Fund’s science program and lead author of the study. “Our paper emphasizes that creating and strengthening indigenous lands and other protected areas can offer an effective means to cut emissions while garnering numerous additional benefits for local people and wildlife.”
The authors highlight analyses showing that since 2002, deforestation in the Brazilian Amazon has been 7 to 11 times lower inside of indigenous lands and other protected areas than elsewhere. Simulation models suggest that protected areas established between 2003 and 2007 could prevent an estimated area of 100,000 square miles of deforestation through 2050. That is roughly the size of the state of Colorado, representing enough carbon to equal 1/3 of the world’s annual CO2 emissions. Within these efforts, location matters; protected areas in regions that face deforestation pressures would be most effective at truly reducing emissions.
“This study reinforces the wisdom behind global investments in protected areas,” says Gustavo A.B. da Fonseca, co-author of the study and Team Leader Natural Resources of the Global Environment Facility (GEF). “In addition to protecting globally important species and ecosystems, the 2,302 protected areas supported by the GEF alone span over 634 million hectares and together store an impressive 30 billion tons of CO2”
International policies for compensating forest nations for REDD are under active negotiation. To access the resulting funds, developing countries will need to develop programs and institutions to reduce forest emissions. “Protected areas represent a valuable component of national REDD programs since they already contain the necessary institutions and infrastructure to handle funds, strengthen protection and generate results,” said Claudio Maretti, Conservation Director, WWF Brazil. “Establishing protected areas usually clarifies land tenure and the associated carbon rights, which has been a sticking point in some negotiations.”
In addition, the study estimates that the cost of creating and better managing protected areas is lower than many other options to reduce emissions from deforestation. Completing and managing a network of protected areas in the developing world might require $4 billion USD annually, which is roughly 1/10 of the capital that could be mobilized by international REDD policies.
According to the study, forest nations can strengthen the role of protected areas in their REDD strategies by:
- Identifying where Indigenous Lands and Protected Areas would most effectively reduce deforestation rates and associated emissions;
- Establishing national monitoring to measure deforestation rates and quantify carbon emissions reductions;
- Establishing insurance mechanisms for illegal logging or forest fires;
- Providing indigenous groups and local communities the information and capacities they need to participate;
- Distributing payments transparently to reward those responsible for reducing emissions.
Note to Editors
The publication “Indigenous Lands, Protected Areas, and Slowing Climate Change” is available at http://www.worldwildlife.org/science/2010pubs/WWFBinaryitem15590.pdf
A MAP showing carbon stocks and potential emissions of selected forest protected areas in the Brazilian Amazon is available at http://www.worldwildlife.org/science/2010pubs/WWFImgFullitem15589.jpg
Potential emissions are estimated by simulating future deforestation through 2050, with and without forest protected areas present. The difference (depicted by orange bars) represents the reductions of CO2 emissions contributed by each forest protected area.
ABOUT WORLD WILDLIFE FUND
WWF is the world’s leading conservation organization, working in 100 countries for nearly half a century. With the support of almost 5 million members worldwide, WWF is dedicated to delivering science-based solutions to preserve the diversity and abundance of life on Earth, halt the degradation of the environment and combat climate change. Visit www.worldwildlife.org to learn more.<|endoftext|>
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We are investigating how cells work together to build a body.
Animals are made up of many different tissues. It is vital that cells within these tissues behave correctly as an organism grows and develops, ensuring that each part of the body is the correct size and shape. If not, this can lead to cancer.
We are using fruit flies and mice as model organisms to understand how cells work together to create sheets of tissue called epithelia. These sheets form the linings that that cover the body and its organs, such as the skin or gut.
To study these tissues, we use powerful microscopes to examine living epithelial cells as they grow and divide. We also search for genes involved in directing the formation of epithelia, and study the consequences of mutations in these genes. And we are making computer simulations to help us understand the underlying processes at work as tissues form.
In order to grow to the correct size, a tissue must control the growth and division of the cells within it. And to make the correct shape, the cells in a tissue must control their individual shapes, orientation, movement and attachments to one another and to the surrounding environment.
We aim to discover the key molecules that control each of these individual cell behaviours, and to discover how the collective behaviour of groups of cells produces tissues of different sizes and shapes.
Our work is revealing how tissues normally develop and maintain themselves in healthy adults, as well as shedding light on what goes wrong in cancer.<|endoftext|>
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1) What is a hotspot?
A volcanic "hotspot" is an area in the upper mantle from which heat rises in a plume from deep in the Earth. High heat and lower pressure at the base of the mantle facilitates melting of the rock. This melt, called magma, rises through cracks to the surface and forms volcanoes. As the tectonic plate moves over the stationary hot spot, the volcanoes are rafted away and new ones form in their place. This results in chains of volcanoes, such as the Hawaiian Islands.
To see history of just one island, visit: http://www.youtube.com/watch?v=t5go-78gCJU
Animation by Jenda Johnson, Earth Sciences Animated
Narrated by Roger Groom, Mt. Taber Middle School
Edited by Anita Grunder and Don Swanson
More animations: http://www.iris.edu/hq/inclass/search/animation<|endoftext|>
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Medical practitioners, such as those at the Centers for Disease Control and Prevention, frequently utilize mathematical models when determining how to best control the spread of mosquito-borne illnesses. Diseases such as chikungunya, dengue fever, malaria, and Zika virus can be life-threatening, and no effective vaccine currently exists. While most mitigation strategies aim to eliminate popular mosquito breeding sites through the use of insecticides, the accompanying costs, logistical difficulties, and resistance evolution make these treatment methods unsustainable.
Figure 1. Female Aedes aegypti mosquito feeding.
is a maternally-transmitted bacteria that occurs naturally in over 60 percent of insect species. Certain strains of Wolbachia
inhibit the transmission of disease-inducing pathogens to humans; this feature gives the microbe potential medical value, and scientists have been studying its effect on mosquitoes for years. Unfortunately, it is not naturally found in Aedes aegypti
mosquitoes, the primary transmitters of mosquito-borne illnesses. If researchers wish to use Wolbachia
to control the spread of these diseases, they must continually reintroduce it to wild mosquito populations. Such repeated introduction is strategically unfeasible.
In an article publishing today in the SIAM Journal on Applied Mathematics, Zhuolin Qu, Ling Xue, and James Mac Hyman use an ordinary differential equation (ODE)-based model to calculate the most effective method of introducing a self-sustaining Wolbachia infection to a wild mosquito population. Their two-sex model accounts for the aquatic life stage, heterosexual transmission, and multiple pregnant states for female mosquitoes, thus capturing the entire transmission cycle. “If a stable Wolbachia infection can be established in wild mosquitoes, this will reduce the spread of mosquito-borne diseases,” Qu said.
Past researchers have used a number of models to study the presence of Wolbachia in wild A. aegypti mosquito populations, though many did not distinguish between mosquitoes’ varied life stages and the fitness costs that accompany Wolbachia. “Most previous models assumed that there is a fixed male/female ratio, and use this assumption to reduce the equations to a single-sex model,” Qu said. “This is a good approximation for isolated wild mosquito populations, but is violated when releasing only infected male mosquitoes into a wild population. Our ODE model for Wolbachia transmission does not assume that there is a fixed ratio of males to females.”
Qu et al. employ a multi-stage system of nine ODEs that integrate both genders, pregnant and non-pregnant females, and all life stages. They group the adult mosquito population into seven compartments based on infection state, pregnancy status, and fertility. Because female mosquitoes typically mate only once during their lifetime and store the sperm for multiple egg clutches, a successful two-sex model differentiates between non-pregnant (unmated) and pregnant (mated) females.
“The Wolbachia infection is transmitted vertically from an infected female mosquito to her offspring,” Qu said. “A female mosquito usually mates successfully once, and if an uninfected female mosquito mates with an infected male mosquito, very few of her offspring survive. Because our compartmental model includes the heterosexual interaction of mosquitoes and multiple pregnant stages for females, it can be used to analyze the threshold condition required to sustain endemic Wolbachia for both perfect and imperfect maternal transmissions.”
Figure 2. Maternal transmission of Wolbachia in the mosquito population.
To reach the threshold at which enough mosquitoes are contaminated for a Wolbachia infection to persist, a model must surmount the shortened lifespan and decreased fecundity (egg-laying rate) that accompanies an outbreak and limits its permanence. “There are three coexisting equilibria for the proposed model: a stable zero-infection equilibrium, an unstable intermediate-infection endemic, and a stable high-infection endemic equilibrium (or complete infection for perfect maternal transmission),” Qu said. “These three equilibria are characterized by a backward bifurcation, with the unstable equilibrium points being the threshold condition for endemic Wolbachia. If the fraction of infected mosquitoes is below this threshold, then the population returns to a zero-infection state. If the fraction is above the threshold, then the infection spreads and eventually almost all of the mosquitoes are infected with Wolbachia.”
After characterizing the threshold condition, the authors simulate and compare real-life mitigation strategies for employment prior to the release of Wolbachia-infected mosquitoes. “Because the threshold condition is characterized by a minimal fraction of mosquitoes that are infected, we could reduce the number of infected mosquitoes that must be released by first reducing the population of uninfected mosquitoes,” Qu said. Tested strategies include indoor residual spraying, since A. aegypti tend to breed in or near houses; larval control, which targets water storage and container removal where mosquitoes often breed; sticky gravid traps/ovitraps, which attract and kill uninfected pregnant females as they prepare to lay eggs; and acoustic attraction, which reduces the number of uninfected males.
“Our simulations indicate that the pre-release mitigations that target pregnant females, such as residual spraying and sticky gravid traps, are more helpful than ones that target only males or the aquatic stage, given that pre-release mitigation stops once the release starts,” Qu said. “Removing uninfected pregnant females greatly slows reproduction of the uninfected offspring, and the gap can be filled up mostly with the infected population.”
Figure 3. Schematic of Wolbachia's complex maternal transmission mating
Ultimately, monitoring both fitness cost and maternal transmission rate are key to establishing and sustaining a Wolbachia outbreak among wild mosquitoes. If researchers introduce enough infected mosquitoes to a natural population to surpass the threshold, the population levels out at a stable endemic Wolbachia-infected equilibrium. As a result, infected mosquitoes will be less likely to spread infectious diseases like dengue, chikungunya, and Zika to the humans they encounter.
Qu et al. recognize that they must further hone their model before their findings can truly guide policy efforts. For example, they assume that all parameters are constant for the sake of simplicity, but temperature, humidity, and other seasonal factors realistically vary. Incorporating seasonality in the model would offer a more accurate projection of multi-season success. “We are currently extending this model to include both spatial heterogeneity and temporal variations using partial differential equations that incorporate the diffusion of mosquitoes and seasonal variations,” Qu said. “Hopefully it could offer more practical insights to help guide mitigation efforts for mosquito-borne diseases.”
Acknowledgments: This research was partially supported by the NSF-MPS/NIH-NIGMS award NSF-1563531.
Source article: Qu, Z., Xue, L., & Hyman, J.M. (2018). Modeling the Transmission of Wolbachia in Mosquitoes for Controlling Mosquito-Borne Diseases. SIAM Journal on Applied Mathematics, 78(2), 826-852.
About the authors: Zhuolin Qu is an applied mathematician researching the modeling of infectious diseases to help public health researchers better understand the spread of disease and optimize mitigation resources. She received her Ph.D. in 2016 and is now a postdoctoral fellow in the Department of Mathematics at Tulane University. Ling Xue is interested in modeling mitigation strategies for infectious diseases, especially vector-borne diseases. Mac Hyman is a past-president of SIAM and an applied mathematician at Tulane University specializing in modeling the spread of infectious diseases.
||Lina Sorg is the associate editor of SIAM News.<|endoftext|>
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# Power Series
Author: John J Weber III, PhD Corresponding Textbook Sections:
• Section 11.8 – Power Series
• Section 11.9 – Representations of Functions as Power Series
## Expected Educational Results
• Objective 24–01: I can use the Geometric Series to rewrite functions as power series.
• Objective 24–02: I can differentiate a power series written in summation notation.
• Objective 24–03: I can explain why the index changes when differentiating a power series written in summation notation.
• Objective 24–04: I can integrate a power series written in summation notation.
• Objective 24–05: I can use the geometric series to find the radius of convergence, R, of a power series.
• Objective 24–06: I can use the geometric series to find the interval of convergence, I, of a power series.
### Bloom’s Taxonomy
A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.
## Power Series
### Algebra
#### Definition: Power Function
A power function is a function with a single term that is the product of a real-valued coefficient, and a variable raised to a fixed real number. In other words, a power function contains a variable base raised to a fixed power.
Example: $f\left(x\right)=3{x}^{4}$$f(x)=3x^4$
$f\left(x\right)$$f(x)$ is a power function because $3{x}^{4}$$3x^4$ is the product of the real value, 3 [the coefficient], and ${x}^{4}$$x^4$ [a variable raised to a single real value].
### Power Series
#### Definition: Power Series
A power series is a sum of power functions.
#### Definition: Power Series Centered at $x=a$$x=a$
$\sum _{n=0}^{\mathrm{\infty }}{c}_{n}\left(x-a{\right)}^{n}$$\sum_{n=0}^{\infty}{c_n(x-a)^n}$
#### Definition: Center of a Power Series
When all variables are subtracted by the same real number, e.g., $\left(x-a{\right)}^{n}$$(x-a)^n$, $a$$a$ horizontally shifts the graph of the function ${x}^{n}$$x^n$
#### Definition: Geometric Series
Recall: $\frac{1}{1-x}=1+x+{x}^{2}+{x}^{3}+\cdots =\sum _{n=0}^{\mathrm{\infty }}{x}^{n}$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots=\sum_{n=0}^{\infty}{x^n}$ for $|\phantom{\rule{0.167em}{0ex}}x\phantom{\rule{0.167em}{0ex}}|<1$$|\,x\,|<1$.
#### Investigation 01
Use the geometric series to rewrite the following functions as a power series:
1. $f\left(x\right)=\frac{1}{1-x}$$\displaystyle f(x)=\frac{1}{1-x}$
2. $f\left(x\right)=\frac{3}{1-x}$$\displaystyle f(x)=\frac{3}{1-x}$
3. $f\left(x\right)=\frac{-2}{1-x}$$\displaystyle f(x)=\frac{-2}{1-x}$
#### Investigation 02
1. $f\left(x\right)=\frac{1}{1-2x}$$\displaystyle f(x)=\frac{1}{1-2x}$
2. $f\left(x\right)=\frac{3}{1-{x}^{2}}$$\displaystyle f(x)=\frac{3}{1-x^2}$
3. $f\left(x\right)=\frac{1}{1-2{x}^{3}}$$\displaystyle f(x)=\frac{1}{1-2x^3}$
#### Investigation 03
Use the geometric series to rewrite the following functions as a power series:
1. $f\left(x\right)=\frac{1}{1+x}$$\displaystyle f(x)=\frac{1}{1+x}$
2. $f\left(x\right)=\frac{4}{1+x}$$\displaystyle f(x)=\frac{4}{1+x}$
3. $f\left(x\right)=\frac{1}{1+{x}^{2}}$$\displaystyle f(x)=\frac{1}{1+x^2}$
#### Investigation 04
Use the geometric series to rewrite the following functions as a power series:
1. $f\left(x\right)=\frac{x}{1-x}$$\displaystyle f(x)=\frac{x}{1-x}$
2. $f\left(x\right)=\frac{{x}^{3}}{1-{x}^{2}}$$\displaystyle f(x)=\frac{x^3}{1-x^2}$
3. $f\left(x\right)=\frac{{x}^{2}}{1+{x}^{2}}$$\displaystyle f(x)=\frac{x^2}{1+x^2}$
#### Investigation 05
Use the geometric series to rewrite the following functions as a power series:
1. $f\left(x\right)=\frac{1}{2-x}$$\displaystyle f(x)=\frac{1}{2-x}$
2. $f\left(x\right)=\frac{x}{3-{x}^{2}}$$\displaystyle f(x)=\frac{x}{3-x^2}$
3. $f\left(x\right)=\frac{{x}^{2}}{{x}^{2}-5}$$\displaystyle f(x)=\frac{x^2}{x^2-5}$<|endoftext|>
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What is Hepatitis C?
Hepatitis C is an acute inflammatory disease of the liver; if it is not resolved, it becomes a chronic infection that can lead to liver cirrhosis in (5-10% of cases) and liver cancer. The World Health Organization estimates that 3% of the world population, approximately 170 million people, suffer from hepatitis C.
Since a patient begins to suffer from chronic hepatitis C there are several stages of evolution from less to more serious. The degree of severity is determined by the presence of fibrosis in the liver, there being five phases: F0 (there is no liver fibrosis), F1 (there is portal fibrosis without bridges), F2 (portal fibrosis with some bridges), F3 (portal fibrosis with many bridges, without cirrhosis) and F4 (liver cirrhosis). The majority of hepatitis C patients are in stages F0-F2 (77%), with F3 being 11% and F4 being 9.6%. The evolution of the disease from F0 to F4 is very slow: they can go from 20 to 47 years or more until the development of liver cirrhosis.
Causes of Hepatitis C
Hepatitis C is caused by infection with hepatitis C virus (HCV), which is an RNA virus of the Hepacivirus family within the genus Flaviviridae.
The VCH has a lipid envelope formed by the E1 and E2 proteins, and a core or Core where the RNA that produces all the structural proteins (Core, E1, E2 / NS1) and non-structural (NS2, NS3, NS4 and NS5) is found ). The NS3 and NS5 proteins are essential for the replication of HCV and therefore are therapeutic targets. There are 6 main types of VCH (called genotypes from 1 to 6, each with several subtypes) that have a different global distribution. Thus, genotypes 1 (subtypes 1b and 1a) and 3 (subtype 3a) predominate in Spain (and Europe), although there are also cases by genotypes 2 and 4 of VCH.
Genotype 5 is found in sub-Saharan Africa and genotype 6 in Asia. Hepatitis C is sometimes associated with infection with hepatitis B virus, which worsens the prognosis. Likewise, it can be associated with infection with human immunodeficiency virus (HIV), which aggravates liver damage.
Most patients do not have symptoms due to Hepatitis C.
Pathways of infection of Hepatitis C
Hepatitis C is transmitted by transfusions, tattoos, acupuncture, intravenous drug abuse, sexually transmitted, vertical route from mother to child. Each year 3 to 4 million people worldwide are infected and around 350,000 die. The probability of transmission of hepatitis C by injection with a contaminated syringe is 1.5-4%. There is no preventive vaccine so in case of exposure to HCV should be taken precautionary measures.
It is important to note that a patient with hepatitis C can perform a normal family, social and work activity, without taking any precautions to avoid the transmission of the disease. Only if at some point you have bleeding you should avoid contact of your blood with other people.
The majority of patients do not have symptoms, so the diagnosis is made by analyzing liver enzymes (transaminases) and detecting antibodies to virus C (anti-HCV) and RNA virus in the blood.
Treatment of Hepatitis C
Among patients with hepatitis C, it is a priority to treat patients with prerhythmic (F3) and cirrhotic (F4). The classic treatment of hepatitis C involves the administration of pegylated interferon alfa (PegIFN) and ribavirin, but due to its limited efficacy (50% response) and its toxicity (flu syndrome, joint pain, etc.) it is no longer used .
Currently there are several new drugs available that have a direct antiviral effect against hepatitis C. Sofosbuvir (Sovaldi @), Simeprevir (Olysio @), Daclatasvir (Daklinza @), Sofosbuvir + Ledipasvir (Harvoni @), Ombitasvir + Paritaprevir + Dasabuvir + Ritonavir (Viekira pak @), etc. These new drugs have great advantages compared to the old ones:
- tolerance is excellent and only 20% of patients notice mild symptoms (headache, tiredness)
- Its effectiveness is much higher than that of previous drugs: they prevent inflammation and destruction and eliminate the virus in about 90% of cases.
The duration of treatment varies between 12 and 24 weeks depending on the characteristics of hepatitis C at the beginning of treatment: patients with cirrhosis, high viral load (concentration of HCV RNA) or who have not responded to previous treatments need to receive the drugs during a longer period.
Currently, new studies are underway to try to reduce the duration of treatment to 6-8 weeks in patients with certain characteristics: low viral load, absence of cirrhosis, etc. There are several combinations of these drugs that can be used: Sofosbuvir + Simeprevir, Sofosbuvir + Daclatasvir, Sofosbuvir + Ledipasvir, etc. Patients with HCV genotypes 1, 2, 4, 5 and 6 respond well to these treatments but those with genotype 3 respond worse. In some cases, ribavirin is added to improve efficacy.
It is important to point out that from the published studies it can be concluded that both the efficacy and the tolerance of the different drugs are similar and no studies have been done comparing the different combinations. Although it is possible to normalize the liver tests and eliminate the virus C from the blood, later it is necessary to follow up the patients until confirming the complete regeneration of the liver. PegIFN can also be used with ribavirin and one of the drugs mentioned above (Sofosbuvir, Simeprevir, Ledipasvir, Daclatasvir, Viekira pak, etc.) although the toxicity is higher.
In patients who do not have access to new drugs or who can wait before being treated, there are measures that can prevent the progression of the disease and that must be applied (phases F0-F2):
- slimming diet if there is overweight;
- do not drink alcohol or smoke;
- metabolic corrections (normalize cholesterol, triglyceride or glucose levels if elevated);
- practicing at least 3 hours of weekly exercise decreases the progression of the disease;
- if the patient has high levels of ferritin (protein that transports iron) the treatment with phlebotomies (blood extractions) and iron-poor diet stops the progression of the disease in 69% of patients;
- there are other drugs that can be useful (ursodeoxycholic acid, vitamin E);
- Intake of 2-5 cups of coffee daily can prevent the progression of the disease. Neither artichokes nor any other type of food are useful.
In patients who do not respond to treatment, hepatitis C disease may require a liver transplant.
In 2004, the Viral Hepatitis Study Foundation discovered a new form of Hepatitis, the hidden C virus, as a result of its research work.
How aggression and evolution can be predicted
To predict the evolution of Hepatitis C patients, the degree of inflammation and the liver fibrosis stage can be determined by performing a liver biopsy. Liver ultrasound and fibroscan are also used.
The determination of PNPLA3 gene polymorphism is very useful to know the future evolution of the disease. For this analysis, a small puncture is made on the patient’s finger (identical to the one done to determine glucose) and blood is deposited on a card that adsorbs human DNA. The analysis can give the following results:
- Homozygous CC: no risk of progression of liver damage.
- Heterozygous CG: low risk of progression of liver damage.
- Homozygous GG: with risk of progression to more severe forms of liver damage (increased liver fibrosis)
Knowing the PNPLA3 gene polymorphism result, the hepatologist can program the frequency of consultations and treatment intensity.
- Carreño V, Castillo I (eds.). Hepatitis víricas: Biología, clínica y tratamiento. 1ª ed. Barcelona: Spriger Verlag Ibérica. Parte IV: Virus C de la hepatitis. 2001; p 241.
- Thein HH, et al. Estimation of stage-specific fibrosis progression rates in chronic hepatitis C virus infection: a meta-analysis and meta-regression. Hepatology 2008;48:418-31.
- Gower E, et al. Global epidemiology and genotype distribution of the hepatitis C virus infection. J Hepatol 2014;61(Suppl.1):S45-57.
- Kabiri M, et al. The changing burden of hepatitis C virus infection in the United States: model-based predictions. Ann Intern Med 2014;161:170-80.
- Liang TJ, Ghany MG. Therapy of hepatitis C-back to the future. N Engl J Med 2014;370:2043-7.
- Carreño V. Review article: management of chronic hepatitis C in patients with contraindications to anti-viral therapy. Aliment Pharmacol Ther 2014;39:148-62.
- Gane E, et al. Strategies to manage hepatitis C virus (HCV) infection disease burden – volume 2. J Viral Hepat 2015;22(Suppl.1):46-73.
Consult our doctor<|endoftext|>
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# Search>Learning Exercise Results> Congruent Triangles / Geometry
## Learning Exercise: Congruent Triangles / Geometry
Title:
Congruent Triangles / Geometry
Material:
Description:
Adjustable lesson for introduction to congruent triangles that includes different complementary technologies.
Course: Mathematics
Submitted by:
Michael Strangio
June 08, 2008
### Exercise
Behavioral Objective(s): SWBAT: -Identify pairs of congruent triangles -Identify corresponding parts of congruent triangles Essential Question (Higher Level Thinking Questions): -How can you tell if two triangles are congruent? Opening (Anticipatory Set): -The teacher will present numerous triangles of different shapes and sizes on the board one at a time. Then ask all of the students to take a long close look at the board. -The teacher will engage the class in a discussion on what they see and write down suggestions on the board. -After all suggestions have been made then the teacher will inform the class that we are going to be working on seeing Congruent Triangles. Modeling/ Guided Practice: -The teacher will say Let me show you congruent triangles. The teacher will take down all of the triangles except for the two triangles that are the same on the board. -Students will be asked what they think congruent triangles are now. -The teacher will explain what congruent means and give examples to the students. Independent Practice: -The teacher will assign the class into pairs; they will have to explore the classroom and find shapes that are congruent. (Preferably triangles if they cant find any another shape will do just fine). Given the writing prompt My shapes are _________ and are congruent because
. They will have to draw and explain what the shape is in their math journals. Students will then have the opportunity to work on the computer to construct congruent triangles at http://nlvm.usu.edu/en/nav/frames_asid_165_g_2_t_3.html?open=instructions&hidepanel=true&from=grade_g_2.html Closure: Students present their findings to the whole group and the teacher will comment on their findings.
### Audience
• Middle School
### Topics
Mathematics Geometry Triangles Congruence
### Requirements
Basic understanding of different types of triangles, isosceles, scalene, equilateral; and that triangles vary.
### Learning Objectives
-Identify pairs of congruent triangles -Identify corresponding parts of congruent triangles<|endoftext|>
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## How do you do double in math?
To get a double of a number, we add the same number to itself. For example, double of 2 is 2 + 2 = 4.
### How do you explain doubles plus one?
Doubles plus 1 is a strategy used to add two consecutive numbers that is, when they are next to each other. We simply add the smaller number twice or double it and then, add 1 to it, to get the final result. Here, for example, consecutive numbers 8 and 9 have been added using the doubles plus one strategy.
What does doubling mean in maths?
two lots of
If you had two cubes, you would have double. Double means two lots of something. For example, double one is two. 1 + 1 = 2.
Is Doubling multiplying by 2?
To multiply by 2. To have 2 of something. Example: Double 4 is 8.
## What is a double lesson?
how do I say we have a double lesson after lunch? a double class is when you have for example 40 minutes of maths followed by another maths class straight after.
### What is the doubling strategy?
Doubling is a strategy that people of all ages frequently use. Young children first learn doubles as an addition of two groups. What multiplication facts can be used by using a doubling strategy? If you said the twos, fours, and eights facts then you are correct! That’s what makes this strategy so powerful.
What is the doubling sequence?
The doubling sequence is my name for the sequence of powers of 2. D = 〈1, 2, 4, 8, 16, 32, 64, 128, . . .〉 Term in the doubling sequence form the basis for the binary. number system. Any natural number can are written as a sum of.
What is the double of 6?
….intersections points double six
each point… 1 8
each line… 10 16
each tritangent plane… 3 36
each double six… 30 36<|endoftext|>
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To decipher maps, several steps should be
followed. Each step seeks pertinent information about the map and its
context in both time and place.
- Content: What is the map's title and/or what is the map about?
- What part of the earth does it represent?
- What is the scale of the map?
- How is the map oriented (that is, what direction is at the top of
- Does the map depict the themes of geography:
- human-environment interactions
- regional identity?
- In what language(s) is the map prepared?
- Authorship: Who made the map (cartographer and/or publisher)? Who
- Sponsorship: Who provided the incentive or funding for the map?
- Date: When was the map made?
- Origin: Where was the map made?
- Creation: By what processes was the map drafted and/or printed?
- Audience: For whom was the map made?
- Function: For what purpose(s) was the map intended, and how was it
- Context: Where does the map fit in the past, what were its sources,
and for what other maps did it serve as a source?
The above nine steps lead to the main historical focus:
Historical Source/Interpretation: What does the map tell us about
its creator and his/her time?<|endoftext|>
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# How do you find the area bounded by x=8+2y-y^2, the y axis, y=-1, and y=3?
May 16, 2017
$\frac{64}{3}$
#### Explanation:
First find the integral of the function:
$\int \left(8 + 2 y - {y}^{2}\right) \mathrm{dy} = - \frac{1}{3} {y}^{3} + {y}^{2} + 8 y$
There are three intervals that you should solve that is $\left[- 3 , - 2\right] , \left[- 2 , 0\right] , \left[0 , 1\right]$.
The problem of plugging $- 3$ directly into the integral would subtract the area the left of the $- 2$ root.
(1)$\int \left[- 2 , 0\right]$
$= 8 \left(- 2\right) + {\left(- 2\right)}^{2} - \frac{1}{3} {\left(- 2\right)}^{3} = \left\mid - \frac{28}{3} \right\mid = \frac{28}{3}$
(2)$\int \left[- 3 , - 2\right] = \int \left[- 3 , 0\right] - \int \left[- 2 , 0\right]$
$= 8 \left(- 3\right) + {\left(- 3\right)}^{2} - \frac{1}{3} {\left(- 3\right)}^{3} - \left(- \frac{28}{3}\right) = - 6 - \left(- \frac{28}{3}\right) = \frac{10}{3}$
(3)$\int \left[0 , 1\right]$
$= 8 \left(1\right) - {\left(1\right)}^{2} - \frac{1}{3} {\left(1\right)}^{3} = \frac{26}{3}$
$\frac{28}{3} + \frac{10}{3} + \frac{26}{3} = \frac{64}{3}$<|endoftext|>
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Temperatures are generally high, including the earth's extremes, but it may be cold at night (daily temperature variation is more extreme in dry climates) and very cold in winter in higher-latitude deserts. The zone is characterized by low precipitation, in different regions varying from highly seasonal to unpredictable to virtually absent; evapotranspiration is always high. Snow is relatively rare, both because of low precipitation and the generally subtropical distribution of the zone. Deserts form where air masses have lost most of their water vapor after traveling overland for long distances (Arabia, North Africa, southwest United States and northern Mexico) or where a hot land mass is adjacent to cold ocean water, moisture from which evaporates quickly over the land (Atacama and Kalahari deserts).
Desert soils are variable in color but often light brown, gray, or yellowish. They are usually calcareous and may be highly saline because of the high evaporation rate and lack of runoff, with a continual buildup of salts (calcium carbonate, gypsum, sodium chloride). Sand is a common substrate, contributing to drought by draining the scarce precipitation very rapidly.
Desert vegetation typically consists of open, well-spaced shrubs with numerous branches from ground level and small, thick leaves. Grasses and forbs may or may not fill in the spaces among the shrubs. Succulents and annuals are well represented, the latter appearing in great diversity and density in occasional favorable years. Shrub and tree growth may be relatively luxuriant along water courses or even in dry washes (temporary stream beds).
Desert species diversity is very much dependent on rainfall and vegetative cover, with the fewest plant and animal species in the driest deserts. The Sonoran Desert is especially rich in species, with a substantial variety of distinct plant associations. The driest deserts (Atacama Desert of Chile/Peru, Sahara Desert of Africa) virtually lack living organisms in some areas. Some plant families are especially diverse in deserts, for example Chenopodiaceae, Crassulaceae, and Cactaceae. No major vertebrate taxa are confined to deserts, but some groups are quite well represented there, and many widespread genera have desert-adapted species. Lizards, snakes, and rodents are well adapted to dry environments, where they are diverse out of proportion to other groups that occur in deserts. Aquatic animals are lacking except where occasional water bodies persist, except for groups such as fairy shrimp that live in ephemeral pools. Amphibians are not very diverse, but some frogs occur even in very dry deserts, as long as rainfall is adequate for breeding at least occasionally.
In typical desert shrubs, the leaves are small and heavily coated with waterproofing materials to prevent excess water loss, gray-green to reflect the sunlight and prevent overheating; the roots are shallow but extensive to take advantage of the shallow organic layer. There is substantial root competition, both within and between species, because this layer is so shallow. There may be chemical competition (allelopathy) between species, with one producing chemicals that inhibit another. Some shrubs associated with water courses have long tap roots to reach the water table. Many species have very thick leaves and/or stems (succulents) that can hold water effectively; some shrink during dry periods and expand during wet periods. As they are very edible to herbivores, most succulents are protected by spines (although this adaptive type is not present in Australia). There is a high proportion of very ephemeral annuals among the forbs, with long-lived, drought-resistant seeds and rapid germination, growth, flowering, and seeding. There are many special adaptations to tolerate high salinity, including salt-excreting glands. One bromeliad, of epiphytic origin, rests on sand in the Atacama Desert, where it is the only plant species, and absorbs fog moisture through its leaves.
Desert animals exhibit many physiological and anatomical adaptations to drought, including the ability to go without free water (their metabolic water is obtained entirely from their diet). Many species are active only at night (or early and late in the day in diurnal species), when the humidity is higher and the temperature lower. Nocturnal activity also leads to reduced predation by visual predators (made easy because of the openness of the environment). Cryptic coloration is at a premium because of the openness, with especially fine substrate color matching. Selective pressures are rigorous enough so there is substantial convergence in morphology and behavior in unrelated organisms in different deserts of the world. Many diurnal ectotherms are pale to reflect sunlight and avoid overheating. Adaptations to lose heat include high surface-to-volume ratios and long appendages (ears, legs). Some aquatic invertebrates persist in the same way as do annual plants, with dormant stages in their life history that are stimulated to develop by occasional sufficient rainfall.
With grazing and farming, much soil is lost to wind erosion on the habitable edges of large desert areas, particularly in Africa, where deserts are expanding markedly ("desertification"). In some areas, especially in developed countries, large areas of desert lands have been lost to irrigation, as desert soils are often very favorable for plant growth if water is provided. Otherwise, this environment is difficult for people to colonize in large numbers, so it is not severely affected by humans in many parts of the world. Without irrigation, agriculture is not possible in this zone, thus very few desert plants are cultivated (prickly pears and agaves only recently). Because of this, many desert peoples are nomadic, and a few large mammals of the desert were domesticated for transportation.<|endoftext|>
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# The 10 times table
## Learning focus
Learn the 10 times table by counting in tens and looking at arrays and number patterns.
This lesson includes:
• one video
• four activities
# Learn
## The 10 times tables
Watch this KS1 Maths video from Supermovers and listen to Webster the Spider sing the 10 times tables.
## Counting in tens
Here is a counting pattern:
10, 20, 30, 40, 50, 60, …
This pattern goes up in tens.
From each number, we add 10 to get the next number.
Can you count on in tens to get to 120?
## Groups of 10
This array has 10 apples in each row.
There are 3 rows, making 3 lots of 10.
10 + 10 + 10
Or 3 lots of 10 is the same as 3 x 10 and 3 x 10 = 30
1 lot of 10 = 102 lots of 10 = 203 lots of 10 = 304 lots of 10 = 40
1 x 10 = 102 x 10 = 203 x 10 = 304 x 10 = 40
This set of number facts are from the 10 times table.
## Example 1:
This number track shows the 10 times table. Some numbers are missing.
• Can you work out which numbers are missing?
• How did you work out the missing numbers?
## Example 2:
A box of pencils contains 10 pencils.
How many boxes of pencils will make 70 pencils?
10 + 10 + 10 + 10 + 10 + 10 + 10 = 70
Or you could do this through multiplication:
7 lots of 10 = 70
so 7 x 10 = 70
## Top tip
Remember that every number in the 10 times table always ends in 0.
# Practise
## Activity 1
Practise counting up and down in tens from 0 to 120.
0102030405060708090100110120
1201101009080706050403020100
Now, have a look at these four different representations showing groups of 10s.
Think about how you would show each one as:
1. an array
3. a multiplication
## Activity 2
Have a go at this 10 times table interactive activity and see if you can get all the questions right.
## Activity 3
Now try this 10 times table interactive activity and see how you get on.
# Play
Test your skills on this topic even further by playing Karate Cats Maths - see if you can collect a new costume for your cat!
# There's more to learn
Have a look at these other resources.<|endoftext|>
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Between 1879 and 1880, six thousand Exodusters left Louisiana and Mississippi for Kansas. Their migration, prompted by the end of racially-integrated Reconstruction governments, by anti-black violence and by sharecropping and tenant farming, brought national attention including a Congressional hearing, and generated a national debate about whether African Americans should migrate from the South to improve their political and economic prospects or remain in the region and continue to demand their rights. Frederick Douglass, the nation’s most prominent African American leader argued against migration because "it leaves the whole question of equal rights on the soil of the South open and still to be settled. " Continued migration "would make freedom and free institutions depend upon migration rather than protection. "
Robert J. Harlan joined Richard T. Greener and other black leaders in opposing Douglass. In a speech on May 8, 1879 he argued that migration was not only a way to escape oppression, it was also a powerful protest against those who would deny African Americans their freedom. Harlan’s speech appears below.
Mr. President, as to the present migration movement of the colored people, let it be understood that we have the lawful right to stay or to go wherever we please. The southern country is ours. Our ancestors settled it, and from the wilderness formed the cultivated plantation, and they and we have cleared, improved, and beautified the land.
Whatever there is of wealth, of plenty, of greatness, and of glory in the South, the colored man has been, and is, the most important factor. The sweat of his brow, his laborer"s toil, his patient endurance under the heat of the semi tropical sun and the chilling blasts of winter, never deterred the laborer from his work.
The blood of the colored man has fertilized the land and has cemented the Union. Aware of these facts, we should be baser than the willing slaves did we consent to the dictation of any men or body of men as to where we may go, when we shall<|endoftext|>
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Last week, I wrote about the big agreement in Rwanda, and I promised to explain HFCs. So as my German relatives once told me, "versprochen ist versprochen!" (a promise is a promise!)
The part of the atmosphere that humans live in is called the troposphere. This layer of the atmosphere is about 78% nitrogen, 21% oxygen, plus water vapor, and many other chemicals including carbon dioxide. The troposphere plays a major role in weather and climate. The next layer up is called the stratosphere. The stratosphere is the part of the atmosphere where we find much less water vapor and we also find the ozone layer that protects us from the sun's UV rays (as opposed to the tropospheric ozone which we commonly call "smog"). We find this ozone layer about 11 to 16 miles above sea level.
Back in the 1970s and 1980s, the global community realized that the ozone layer was thinning, and that humans were a big part of the problem. We focused in on chemicals called CFCs (chloroflourocarbons) which were commonly used in propellants and coolants. Thanks to international agreements (Montreal in 1987 and Copenhagen in 1992), we've phased out or controlled the use of the ozone-depleting chemicals. That's why the person who works on your car air conditioner or your home's heating and AC has to have a special certification for handling freon and other CFCs.
The good news is, we phased out CFCs and are starting to see some recovery of the ozone hole. The bad news is, we replaced many CFCs with HFCs (hydroflourocarbons), which don't damage the ozone layer, but they do affect global warming -- in a BIG way! You've probably heard of heat trapping gases such as methane and CO2. Well, HFCs have over 1000 times the heat trapping potential of CO2! Thus, the Rwanda agreement.
Here are some interesting facts about greenhouse gases that you might not have known:
- Greenhouse gases are necessary to trap heat on the planet, and they're the reason we can live here. The problem is not greenhouse gases alone, it's the amount of EXTRA greenhouse gases humans are adding to the atmosphere.
- Water vapor is the most common greenhouse gas in the atmosphere.
- 95% of CO2 in that atmosphere is naturally occurring. It's that extra 5% that we're adding that's causing the problems.
- Carbon dioxide can last in the atmosphere for 50 to 200 years. Methane breaks down in about 12 years. HFCs are similar to methane with a lifespan of about 14 years. Nitrous oxide lasts over 100 years.
- The atmosphere can store roughly 750 billion tons of carbon without significantly changing Earth's temperature. When we burn fossil fuels, we add about 7 billion additional tons of carbon to the atmosphere each year.
Want to learn more? Just search "climate change" for more posts on this topic!
Be sure to "like" greenmomster on Facebook!<|endoftext|>
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In many cases, the energy to initiate a reaction comes from heat. The rate of most reactions increases regularly with a rise in temperature. The rate roughly doubles for each 18" F. (10° C) rise in temperature. However, other forms of energy can also initiate reactions. Photochemistry is the branch of chemistry concerned with the interaction between light (photons) and molecules. Such reactions are essential to life on earth. The trapping of solar energy in chemical form by plants is the first step in the
Atoms, elements, and molecules: Key chemical reactions 21
down into more than one product is known as a decomposition reaction. For example, heating the compound calcium carbonate produces carbon dioxide gas and quicklime. Quicklime is used in making mortar and glass. If two reactants break down into other products, the reaction is known as a double decomposition reaction. An example of such a reaction is one involving an acid and a base. These compounds react to form a salt and water.
The opposite of decomposition is an addition reaction. The reactants join together instead of breaking down. There is no removal of any part of the reacting molecule. Such a reaction is common in organic chemistry, discussed later in its own section.
If a reaction requires a continuous source of energy (usually heat) to keep it going, it is termed endothermic. In other cases, once a reaction starts, it produces enough energy to sustain itself. There is no need for continued external supplies of energy. Such reactions are termed exothermic.
In some cases, a molecule may break down only partly. It then reforms chemical bonds in a different manner. Such reactions are called rearrangements. They are often significant in synthetic chemistry (organic chemistry) and in the chemistry of living systems (biochemistry). Both organic chemistry and biochemistry are discussed later in this book.
Potassium hydroxide KOH
aiX0 + OO
Elimination reactions are
important in synthetic chemistry. They involve the formation of a small, simple molecule such as water or ammonia /left). The elimination of water (3) accompanies the conversion of an organic bromide (1) into a double-bonded hydrocarbon (2). in the manufacture of nylon(below), the elimination of water accompanies the reaction that creates the synthetic fiber.
food chain for many living species.
Photochemical reactions are also the basis of photography. In black-and-white photography, the key reaction is the conversion of silver salts to finely-divided particles of silver. These particles appear black when light strikes them. Color photography depends upon a range of complex light-sensitive carbon compounds.
In some cases, the effect of a very small amount of light can be dramatic. A mixture of hydrogen and chlorine gases kept in the dark
at room temperature does not react. However, exposure to light starts a series of reactions known as a chain reaction. These reactions are very explosive.
Chemical elementsIn the
following section, all the chemical elements are discussed. There are fact entries at the end of each article. The following abbreviations are used: at. no.—atomic number; at. mass—atomic mass (weight); m.p—melting point (freezing point); b.p.—boiling point.<|endoftext|>
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# If the sum of a positive number and two negative numbers is -12 what are the numbers?
Aug 24, 2017
One possible solution would be $\left\{- 6 , - 7 , + 1\right\}$
#### Explanation:
There are infinitely many possible solutions.
Pick any two negative numbers whose sum is less than $\left(- 12\right)$. [That is the absolute value of their sum must be greater that $12$].
Subtract their sum from $\left(- 12\right)$ to get the required positive number.
Here is another example;
if we pick $- 237$ and $- 106$ their sum is $- 343$ (which is less that $- 12$)
Subtract this from $- 12$ to get the required positive number:
$\textcolor{w h i t e}{\text{XXX}} - 12 - \left(- 343\right) = - 12 + 343 = 331$
$\left(- 237\right) + \left(- 106\right) + \left(+ 331\right) = - 12$<|endoftext|>
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Table of contents
Cluster diagrams (also called cloud diagrams) are a type of non-linear graphic organiser that can help to systematize the generation of ideas based upon a central topic.
Using this type of diagram, the student can more easily brainstorm a theme, associate about an idea, or explore a new subject.
What aspects of learning are being supported?
- Bloom’s Taxonomy: remembering, understanding, applying, analysing, evaluating, creating
- QT – Intellectual Quality: Deep Knowledge, Deep Understanding, Higher-Order Thinking, Metalanguage, Substantive Communication
- QT – Quality Learning Environment: Explicit Quality Criteria, Engagement, High Expectations, Social Support, Students Self-Regulation, Student Direction
Which apps could I use?
- DEC managed collaborative tools (via Portal): Google Docs, Microsoft Office 365, Adobe Connect.
- Related collaborative tools: Evernote, iCloud, BaiBoard, Note Anytime, Web2Access, Cool Tools for Schools, and Exploratree.
- Analogue collaborative tools: Paper, pens and Postit notes.
How do I use it?
- The cluster diagram can be generated electronically or manually, sorting research and organising topics and sub-topics.<|endoftext|>
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# 10 More 10 Less Worksheets
Printable place value worksheets for 1st Grade:
Ten and Ones (eg. 17 = __ ten and __ ones)
Count Tens & Ones (with pictures)
Count Tens & Ones (with place value charts)
Tens and Ones (eg. 73 = __ tens and __ ones)
Add Tens & Ones (with pictures)
Add Tens & Ones (eg. 60 + 4 = )
1 more, 1 less, 10 more, 10 less
### 1 more, 1 less, 10 More, 10 Less Worksheets
Identifying 10 more, 10 less, 1 more, and 1 less than a two-digit number involves understanding the patterns in the base-10 number system.
1. Introduction to Place Value:
Begin by reviewing or introducing the concept of place value. Remind students that each digit in a number has a specific place and value.
2. Visual Representation:
Use visual aids such as place value charts, drawings, or manipulatives to represent two-digit numbers. This helps students visualize the relationship between tens and ones.
3. Identify the Two-Digit Number:
Choose a two-digit number, for example, 45.
4. 10 More:
Explain that to find 10 more than a number, add 10 to the original number.
For 45, 45 + 10 = 55.
Emphasize that only the tens place changes.
5. 10 Less:
Similarly, explain that to find 10 less than a number, subtract 10 from the original number.
For 45, 45 − 10 = 35.
Again, highlight that only the tens place changes.
6. 1 More: For finding 1 more than a number, add 1 to the original number.
For 45, 45 + 1 = 46. Emphasize that this change occurs in the ones place.
7. 1 Less:
For finding 1 less than a number, subtract 1 from the original number.
For 45, 45−1=44.
The ones place is affected in this case.
Understanding the relationships between a two-digit number and its variations (10 more, 10 less, 1 more, and 1 less) helps students develop a deeper grasp of place value and numerical patterns. Regular practice and interactive activities enhance their proficiency in these concepts.
Have a look at this video if you need to review how to find 10 more, 10 less, 1 more, and 1 less than a two-digit number.
Click on the following worksheet to get a printable pdf document.
Scroll down the page for more 10 More 10 Less Worksheets.
### More 1 more, 1 less, 10 More, 10 Less Worksheets
10 More 10 Less Worksheet #1
10 More 10 Less Worksheet #2
10 More 10 Less Worksheet #3 (sprint, practice)
10 More 10 Less Worksheet #4 (sprint, practice)
10 More 10 Less Worksheet #5 (sprint, practice)
10 More 10 Less Worksheet #6 (sprint, practice)
More Printable Worksheets
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.<|endoftext|>
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How do you find a polynomial function that has zeros 2, 4+sqrt5, 4-sqrt5?
Jul 22, 2017
$f \left(x\right) = {x}^{3} - 10 {x}^{2} + 27 x - 22$
Explanation:
If $x = a$ is a zero then $\left(x - a\right)$ is a factor.
So the simplest polynomial function of $x$ that has these zeros is:
$f \left(x\right) = \left(x - 2\right) \left(x - \left(4 + \sqrt{5}\right)\right) \left(x - \left(4 - \sqrt{5}\right)\right)$
$\textcolor{w h i t e}{f \left(x\right)} = \left(x - 2\right) \left(\left(x - 4\right) + \sqrt{5}\right) \left(\left(x - 4\right) - \sqrt{5}\right)$
$\textcolor{w h i t e}{f \left(x\right)} = \left(x - 2\right) \left({\left(x - 4\right)}^{2} - 5\right)$
$\textcolor{w h i t e}{f \left(x\right)} = \left(x - 2\right) \left({x}^{2} - 8 x + 16 - 5\right)$
$\textcolor{w h i t e}{f \left(x\right)} = \left(x - 2\right) \left({x}^{2} - 8 x + 11\right)$
$\textcolor{w h i t e}{f \left(x\right)} = {x}^{3} - 10 {x}^{2} + 27 x - 22$
Any polynomial in $x$ with these zeros will be a multiple (scalar or polynomial) of this $f \left(x\right)$.
$\textcolor{w h i t e}{}$
Footnote
In this example, we were asked for a function with zeros including both $4 + \sqrt{5}$ and its radical conjugate $4 - \sqrt{5}$. If only one had been specified then we would still have had to include the other if we wanted our function to have rational coefficients.<|endoftext|>
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Dear Uncle Colin,
When I express $\frac{x^2}{(x-1)(x-4)^2}$ in partial fractions, why do I need to use three separate fractions? I accept that that’s how it works, I just want to know why!
Pupils Absolutely Raging: “Two Is Adequate, Like!”
Hi, PARTIAL, and thanks for your message!
I’ve got a handwavy explanation and one that goes into a bit more depth.
### Wave your hands in the air like you just don’t care
As a general principle, you need the top of a partial fraction to be one degree lower than the bottom. The $(x-4)^2$ part is quadratic, so it needs to have a linear expression – something like $ax+b$ – on the top.
However, you can rewrite this as $a(x-4) + (b+4a)$ or $a(x-4) + c$, which gives you the three fractions you were expecting.
### Take your hands out of the air and seek a small black square
The top of any (proper) fraction with $(x-1)(x-4)^2$ as a denominator could be any quadratic expression.
An arbitrary quadratic expression has three coefficients, so we will need to solve for three constants.
More specifically, we’re trying to change the basis in our quadratic. At the moment, it’s written as $x^2$, but we’d ideally like to write it in a way that’s more convenient for cancelling down fractions.
The space of quadratic expressions is three-dimensional, so we need three linearly independent expressions, picked as conveniently as possible – by which I mean, so that they will cancel with things on the bottom to leave relatively nice fractions.
Two obvious candidates are $(x-1)$ and $(x-4)^2$, which leave $(x-4)^2$ and $(x-1)$ respectively – but we need a third to span the space.
I stress that we can pick whatever we like here, so long as it’s not a linear combination of the other two, but the most convenient choice is $(x-1)(x-4)$, which just leaves an $x-4$ on the bottom.
So, we can rewrite any quadratic expression as $a(x-1) + b(x-4)^2 + c(x-4)(x-1)$, and anything in that form, when divided through by the original bottom, is relatively simple to integrate, differentiate or apply the binomial expansion to.
Hope that helps!
- Uncle Colin<|endoftext|>
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Mold Health Hazards
Molds can produce a variety of allergenic substances, odorous chemicals, and toxic metabolites. When it multiplies and spreads indoors, high levels of mold can cause a spectrum of health effects.
- People are mainly exposed to mold by inhaling spores and skin/eye contact -- actively-growing mold also releases chemicals to the air which people breathe
- Allergic symptoms are the most common problems (e.g., mucous membrane irritation, rhinitis, and rashes)
- More severe effects (e.g., asthma attacks, hypersensitivity pneumonitis, infections, or toxic reactions) may also occur
Molds have the potential to cause health problems. Molds produce allergens (substances that can cause allergic reactions), irritants, and in some cases potentially toxic substances (mycotocins). Inhaling or touching mold or mold spores may cause allergic reactions in sensitive individuals. Allergic responses include hay fever-type symptoms, such as sneezing, runny nose, red eyes, and skin rash (dermatitis).
Who Does Mold Affect?
- Newborn Children
- Children and the Elderly
- People with Respiratory Problems
- People with Compromised Immune Systems
- Certain individuals, especially children, exhibit severe reactions which include damage to lung tissue, memory loss, and even death
Allergic reactions to mold are common. They can be immediate or delayed. Molds can also cause asthma attacks in people with asthma who are allergic to mold. In addition, mold exposure can irritate the eyes, skin, nose, throat and lungs of both mold-allergic and non-allergic people.<|endoftext|>
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The Texas Department of State Health Services encourages everyone 6 months old and older to get vaccinated now to protect themselves from getting the flu.
“The flu vaccine causes your body to make antibodies to fight influenza, but it takes about two weeks for this to happen. So, it’s important to get the flu vaccine early, before influenza hits your community,” said Dr. Jennifer Shuford, DSHS Infectious Disease Medical Officer. “By getting the flu vaccine every year, you can help protect yourself from influenza, and thereby prevent spreading it to vulnerable people in your family or community.”
It is especially important for people with chronic health conditions, pregnant women, young children, older adults, and the people who live with them to get vaccinated in order to minimize the risk of developing serious complications from the flu. Family members and others around babies should get vaccinated to protect the babies and themselves since infants under 6 months of age cannot get vaccinated.
Influenza is an illness caused by one of a number of related viruses. Symptoms usually start suddenly and include fever, body aches, chills, a dry cough, sore throat, runny nose, headaches and extreme fatigue and can last a week or longer. It is important to note that not all flu sufferers will have a fever.
People can help stop the spread of illness and reduce their chance of catching the flu by getting vaccinated, washing hands frequently, covering coughs and sneezes and staying home if they’re sick. Additional flu information and tips are at TexasFlu.org.
All flu vaccines this season are made to protect against viruses similar to the strains A/Michigan/45/2015 (H1N1), A/Hong Kong/4801/2014 (H3N2) and B/Brisbane/60/2008 (B/Victoria lineage). Some vaccines include an additional vaccine virus strain, B/Phuket/3073/2013 (B/Yamagata lineage).
The CDC advises against the use of the live attenuated influenza vaccine, commonly called the “nasal spray” vaccine and sold under the trade name FluMist. Research from prior flu seasons measured no protective benefit.
People can contact their health care provider, local health department, local pharmacy or use the Vaccine Finder at TexasFlu.org to find out where flu shots are available.
If people are experiencing flu symptoms, health officials encourage them to seek treatment promptly. Antiviral drugs may help shorten the duration or lessen the severity of the flu if started within 48 hours of the onset of symptoms.<|endoftext|>
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South Americans helped colonise Easter Island centuries before Europeans reached it. Clear genetic evidence has, for the first time, given support to elements of this controversial theory showing that while the remote island was mostly colonised from the west, there was also some influx of people from the Americas.
Easter Island is the easternmost island of Polynesia, the scattering of islands that stretches across the Pacific. It is also one of the most remote inhabited islands in the world.
So how did it come to be inhabited in the first place? Genetics, archaeology and linguistics all show that as a whole, Polynesia was colonised from Asia, probably from around Taiwan. The various lines of evidence suggest people began migrating east around 5500 years ago, reached Polynesia 2500 years later, before finally gaining Easter Island after another 1500 years.
But the Norwegian adventurer Thor Heyerdahl thought otherwise. In the mid-20th century, he claimed that the famous Easter Island statues were similar to those at Tiahuanaco at Lake Titicaca in Bolivia, so people from South America must have travelled west across the Pacific to Polynesia. His famous Kon-Tiki expedition, in which he sailed a balsa wood raft from Peru to the Tuamotu islands of French Polynesia, showed that the trip could have been made. But if it was made, no trace remained.
Now Erik Thorsby of the University of Oslo in Norway has found clear evidence to support elements of Heyerdahl’s hypothesis. In 1971 and 2008 he collected blood samples from Easter Islanders whose ancestors had not interbred with Europeans and other visitors to the island.
Thorsby looked at the HLA genes, which vary greatly from person to person. Most of the islanders’ HLA genes were Polynesian, but a few of them also carried HLA genes only previously found in Native American populations.
Because most of Thorsby’s volunteers came from one extended family, he was able to work out when the HLA genes entered their lineage. The most probable first known carrier was a woman named Maria Aquala, born in 1846. Crucially, that was before the slave traders arrived in the 1860s and began interbreeding with the islanders.
But the genes may have been around for longer than that. Thorsby found that in some cases the Polynesian and American HLA genes were shuffled together, the result of a process known “recombination”. This is rare in HLA genes, meaning the American genes would need to be around for a certain amount of time for it to happen. Thorsby can’t put a precise date on it, but says it is likely that Americans reached Easter Island before it was “discovered” by Europeans in 1722.
Thorsby says there may have been a Kon-Tiki-style voyage from South America to Polynesia. Alternatively, Polynesians may have travelled east to South America, and then returned. There is already evidence for that: chicken bones found in Chile turned out to be Polynesian, so we know that the eastward journey did happen at some stage.
However, Thorsby’s findings don’t mean that Heyerdahl’s ideas have been vindicated. The first settlers to Polynesia came from Asia, and they made the biggest contribution to the population. “Heyerdahl was wrong,” Thorsby says, “but not completely.”
The work was presented at a Royal Society discussion meeting on human evolution in London today.
More on these topics:<|endoftext|>
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# Objective Questions (MCQ)
Question:
Objective Questions (MCQ)
The length of a rectangular field exceeds its breadth by 8 m and the area of the field is 240 m2. The breadth of the field is
(a) 20 m (b) 30 m (c) 12 m (d) 16 m
Solution:
Let the breadth of the rectangular field be x m.
∴ Length of the rectangular field = (x + 8) m
Area of the rectangular field = 240 m2
$\therefore(x+8) \times x=240 \quad$ (Area = Length $\times$ Breadth)
$\Rightarrow x^{2}+8 x-240=0$
$\Rightarrow x^{2}+20 x-12 x-240=0$
$\Rightarrow x(x+20)-12(x+20)=0$
$\Rightarrow(x+20)(x-12)=0$
$\Rightarrow x+20=0$ or $x-12=0$
$\Rightarrow x=-20$ or $x=12$
∴ x = 12 (Breadth cannot be negative)
Thus, the breadth of the field is 12 m.
Hence, the correct answer is option C.<|endoftext|>
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# How to expand brackets 2
Last updated: September 28, 2021
## What are brackets and how should they be used?
Brackets are the most commonly used symbols, such as the parentheses in an algebraic expression, to establish groups or explain the order of the operations to be performed.
A factor of a number is an accurate divisor of the given number. Every factor of a number is smaller than or equal to the given number, i.e., it cannot be greater than the given number.
## E2.2: Use brackets and extract the common factors
Expanding brackets means that each item in the brackets is multiplied by the expression outside the brackets.
For example, in the expression $$3(m+7)$$, multiply both $$m$$ and $$7$$ by $$3$$. So: $$3(m+7)=3\times m+3\times 7=3m+21$$
A common factor is a number that can be divisible by two different numbers without a remainder. Numbers can have multiple common factors. You can find the common factors of more than two numbers.
Factorising is also known as the reverse process of expanding brackets. Factoring an algebraic expression means putting it in parentheses and taking the common factors.
The way of factorising is:
• Find the highest common factor of each of the terms in the expression.
• Write the highest common factor in front of the brackets.
• Fill in each term in the brackets by multiplying out.
### Worked examples of using brackets
Example 1: Expand $$(3x+2)(2x+3)$$.
Step 1: Write the given information.
The given expression is $$(3x+2)(2x+3)$$.
Step 2: Multiply both $$2x$$ and $$3$$ by $$(3x+2)$$.
$$(3x+2)(2x+3)=(3x+2)(2x)+(3x+2)(3)$$
Step 3: Expand the equation.
$$(3x+2)(2x+3)=(3x+2)(2x)+(3x+2)(3)$$
It can be written as
$$(3x+2)(2x+3)=3x(2x)+2(2x)+(3x)(3)+2(3)$$
$$6x^{2}+4x+9x+6$$
$$6x^{2}+13x+6$$
Therefore, $$(3x+2)(2x+3)=6x^{2}+13x+6$$.
Example 2: Factorise $$15a^{3}-30a^{2}-360a$$.
Step 1: Write the given information.
The given expression is $$15a^{3}-30a^{2}-360a$$.
Step 2: Write the highest common factor.
The highest common factor is $$15a$$.
Step 3: Write the greatest common factor.
$$15a^{3}-30a^{2}-360a=15a(a^{2}-2a-24)$$
Step 4: Substitute $$-2a=4a-6a$$ in the above equation.
$$15a^{3}-30a^{2}-360a=15a(a^{2}+4a-6a-24)$$
$$15a(a(a+4)-6(a+4))$$
Step 5: Write the common factor.
$$15a^{3}-30a^{2}-360a=15a(a+4)(a-6)$$.
Therefore, the factorisation of $$15a^{3}-30a^{2}-360a$$ is $$15a(a+4)(a-6)$$.
Example 3: A rectangular building is located on a plot that measures $$30\text{m}$$ by $$40\text{m}$$. The building must be placed in such a way that the width of the lawn on all the four sides of the building is the same. Local restrictions state that the building cannot occupy more than $$50%$$. What are the dimensions of the largest building that can be built on the site?
Step 1: Write the given values.
Let $$x$$ represent the width of the lawn.
Let $$40-2x$$ represent the length of the building.
Let $$30-2x$$ represent the width of the building.
The area of the lot is $$1200\text{metre square}$$.
Step 2: Write the maximum area of the building.
The maximum area of the building is $$600\text{metre square}$$.
Step 3: Recall the formula for the area of the rectangle.
$$\text{the area of the rectangle}=\text{length}\times \text{width}$$.
Step 4: Substitute the values in the formula.
$$(40-2x)(30-2x)=600$$.
Step 5: Multiply both $$30$$ and $$-2x$$ by $$(40-2x)$$.
$$(40-2x)(30)+(40-2x)(-2x)=600$$.
Step 6: Expand the brackets.
The equation can be written as follows:
$$40(30)-2x(30)+40(-2x)-2x(-2x)=600$$
$$1200-60x-80x+4x^{2}-600=0$$
$$4x^{2}-140x+600=0$$
Step 7: Factorise the equation $$4x^{2}-140x+600=0$$.
The common factor is $$4$$.
Step 8: Write the common factor in front of the brackets.
$$4(x^{2}-35x+150)=0$$
$$4(x—5)(x-30)=0$$
Now, for $$x-5=0$$:
$$x-5=0$$
$$\therefore x=5$$
Now, for $$x-30=0$$:
$$x-30=0$$
$$\therefore x=30$$
It is not possible because $$x=30$$ and the value of the width is $$30-2(30)=-30$$, i.e., negative.
So, the value of $$x$$ is $$5$$.
Step 9: Write the value of the length.
Substitute $$x=5$$ in $$40-2x$$.
$$\text{Length}=40-2(5)=30\text{m}$$
Step 10: Write the value of the width.
Substitute $$x=5$$ in $$30-2x$$.
$$\text{Width}=30-2(5)=20\text{m}$$
Therefore, the length of the building is $$30\text{m}$$ and the width is $$20\text{m}$$.<|endoftext|>
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In 1854 Vlad the flea and Loxton the rat were enjoying the rotten food and dirt in Scutari Hospital. Their lives were turned upside down by the arrival of Florence Nightingale and her tortoise Jimmy. They know a lot about Florence and her work, but how much do you know?
Find out about Florence Nightingale through puzzles and games and learn how to help people with basic first aid skills.
This 24 page book is printed in black and white so it can be coloured-in. It is packed full of activities to add interesting facts and skills for children interested in Florence Nightingale’s life and legacy.
- What did Florence Nightingale do for us? – complete the sentences.
- First Aid skills – learn what to do if someone is bleeding, choking or has a burn.
- Tortoise shell maze – find your way through the pattern on Jimmy’s shell.
- Sudoku – complete the grid using the letters of the characters in the Vlad books.
- Crutches and Bandages – race to the top of the board, climbing the rungs on the crutches and trying not to slide down the bandages.
- Florence facts with multiple choice answers.
- Word search.
- Label the map – identify the countries that were sending soldiers to the Crimea.
- Make a lantern – follow the instructions to make a lantern like Florence used in the hospital wards.
- Escape! – untangle the lines so that Loxton can get safely to the crate.
There are answers to the quizzes at the end of the book.<|endoftext|>
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Pathology is the examination of tissues and body fluids in order to make a diagnosis.
This involves taking a tissue sample of the tumor when a biopsy or surgery is done to remove the tumor. The sample is then sent to a pathologist for review. Once the diagnostic lab report is completed, it is sent to your doctor who shares it with you.
Ependymoma is less common than other brain or spinal cord tumors. In addition, it is sometimes difficult to distinguish it from other tumor types. This can make it difficult to diagnose or a delay may occur. Seeking a second opinion is always a good option when dealing with rare diseases. Contact us if you would like a second opinion.
What is the classification system for ependymomas?
The most widely used system to classify primary brain tumors is the World Health Organization (WHO) system. The WHO uses morphological features to classify these tumors into various ‘types’ and “grades”. This involves examining tissue under a microscope. The WHO was updated in 2016. In addition to what the tumor looks like under the microscope, molecular changes in the tumor are used to further refine the diagnosis in terms of tumor type and grade.
Unlike other cancers, primary brain and spine tumors generally do not spread (metastasize) outside of the central nervous system (CNS). For this reason, the Tumor-Node-Metastasis (TNM) staging system widely used for most “solid tumor” cancers is not useful for primary brain tumors.
What is classification based on?
The WHO assigns a grade to ependymomas based on the following characteristics:
- Variability in size and shape of the tumor cells (pleomorphism)
- How fast the tumor cells are growing (mitotic count)
- Crowding of tumor cells (cellularity)
- Growth of tumor blood vessels (vascular proliferation)
- Tendency for tumor cells to outgrow their blood supply (tumor necrosis)
- How much the tumor has spread into the surrounding normal tissue (invasion)
- Molecular and genetic features
These criteria apply to both pediatric and adult ependymomas. While a tumor may show characteristics from one or more tumor types and grades, doctors treat patients based on the highest-level tumor grade or what the features are for the majority of the tumor. Visit our treatment section for more details.
What are the types of ependymoma classification?
The WHO classification defines ependymoma and divides them into subtypes.
Subtypes that have been previously recognized by the WHO classification include subependymoma, myxopapillary ependymoma, as well as morphologic variants (for example: cellular ependymoma, papillary ependymoma and tancytic ependymoma).
It is recognized that myxopapillary tumors tend to occur in the spine and subependymomas tend to occur along the ventricles.
Recently, underlying genetic changes have revealed that ependymoma tumor subtypes may be different based on location. There are three locations thought to be important: those that are supratentorial, infratentorial in the posterior fossa, and those located in the spine. Recently, it was discovered that in approximately two-thirds of ependymomas in the supratentorial region, there is a molecular abnormality which is a gene fusion called RELA that results in a unique protein.
This fusion, RELA, results in the uncontrolled activation of a specific tumor cell molecular pathway called the NF-kB pathway. NF-kB contributes to the formation of the tumor and its uncontrolled growth. The 2016 WHO classification recognizes that tumors that are RELA-fusion positive occur in the supratentorial region. If tumors are in the supratentorial region, it is recommend that the tumor is tested for the RELA-fusion protein.
Scientists are beginning to identify characteristic changes in posterior fossa tumors, and specific epigenetic and gene expression profiles have been identified as potential markers to separate posterior fossa ependymomas into clinically distinct groups.
Finally, myxopapillary and subependymoma continue to be classified based on characteristics seen under the microscope.
It’s important to know your tumor subtype, if available, in order to plan treatment and determine prognosis. Ask your physician if your tumor has subtypes and how your tissue can be evaluated.
The WHO classifies ependymomas into three grades:
Grade I – Myxopapillary Ependymoma and Subependymoma
Grade II – Ependymoma (conventional)
Grade III – Anaplastic Ependymoma (which is the more cancerous)
What are the characteristics of each type?
WHO Grade I – Myxopapillary Ependymoma
- Slow growing
- Commonly occurs in young adults in the spinal cord, sometimes in the bottom of the spinal cord, an area referred to as “cauda equina”.
- Tend to have good long-term survival after surgical resection
WHO Grade I – Subependymoma
- Slow growing noninvasive tumor
- Are less cellular masses usually attached to the ventricle wall (cerebrospinal fluid filled cavity in the brain).
- More common in adults and older men
- Associated with long-term survival
- Surgery can be potentially curative
WHO Grade II – Ependymoma (conventional)
- Most common brain tumor in young children
- Most common type of spinal glioma in adults
- Often develop in the ventricles when intracranial
- Several variants exist making diagnosis challenging:
- Cellular ependymoma
- Papillary ependymoma
- Tanycytic ependymoma
- Supratentorial ependymomas can be further subclassified as RELA-positive or RELA-negative
- Can potentially recur as a higher grade tumor even after treatment
WHO Grade III - Anaplastic Ependymoma
- Show evidence of increased tumor cell growth compared to conventional ependymoma
- Show evidence of new blood vessel formation to support active growth
- Exhibit more aggressive behavior than low grade ependymomas
- Often require additional treatment after surgery and can recur
The relevance of conventional grading (grade II versus grade III ependymomas) and the prognostic relevance of this approach is currently a matter of discussion and debate. Researchers are currently looking at how to better define the features that set apart a grade II from a grade III tumor.<|endoftext|>
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# What is the divisibility of 486?
## What is the divisibility of 486?
When we list them out like this it’s easy to see that the numbers which 486 is divisible by are 1, 2, 3, 6, 9, 18, 27, 54, 81, 162, 243, and 486.
## IS 486 can be divided by 2?
486 can be divided by 2 A divisibility rule is a shorthand way of determining whether a given integer is divisible by a fixed divisor without performing the division, usually by examining its digits.
What is the number 450 divisible by?
Factors of 450: 1, 2, 3, 5, 6, 9, 10, 15, 18, 25, 30, 45, 50, 75, 90, 150, 225, and 450.
### What number is 487 divisible by?
The number 487 is divisible only by 1 and the number itself. For a number to be classified as a prime number, it should have exactly two factors. Since 487 has exactly two factors, i.e. 1 and 487, it is a prime number.
### IS 486 a perfect square?
The prime factorization of 486 = 21 × 35. Here, the prime factor 2 is not in the pair. Therefore, 486 is not a perfect square.
What can 345 be divided by?
Factors of 345 are the list of integers that we can split evenly into 345. There are overall 8 factors of 345 i.e. 1, 3, 5, 15, 23, 69, 115, 345 where 345 is the biggest factor. The Pair Factors of 345 are (1, 345), (3, 115), (5, 69), (15, 23) and its Prime Factors are 3 × 5 × 23.
#### What number is divisible by 549?
When we list them out like this it’s easy to see that the numbers which 549 is divisible by are 1, 3, 9, 61, 183, and 549.
#### What Can 1680 be divided by?
The factors of 1680 are 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 16, 20, 21, 24, 28, 30, 35, 40, 42, 48, 56, 60, 70, 80, 84, 105, 112, 120, 140, 168, 210, 240, 280, 336, 420, 560, 840, 1680 and factors of 1319 are 1, 1319.
What is the square of 486?
The square root of 486 is 22.045.
## What is 486 Simplified?
Algebra Examples Rewrite 486 as 92⋅6 9 2 ⋅ 6 .
## What are the factors of 506?
506 and Level 5
• 506 is a composite number.
• Prime factorization: 506 = 2 x 11 x 23.
• The exponents in the prime factorization are 1, 1, and 1.
• Factors of 506: 1, 2, 11, 22, 23, 46, 253, 506.
• Factor pairs: 506 = 1 x 506, 2 x 253, 11 x 46, or 22 x 33.
• 506 has no square factors that allow its square root to be simplified.
Begin typing your search term above and press enter to search. Press ESC to cancel.<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
Inverse Variation Models
Identify and solve y=k/x form equations
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Practice Inverse Variation Models
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Inverse Variation Models
What if you were paid \$500 per week regardless of the number of hours you worked? The more hours you worked in a week (increasing quantity), the less your hourly rate (decreasing quantity) would be. How could you write and solve a function to model this situation? After completing Concept, you'll be able to write inverse variation equations and solve inverse variation applications like this one.
Watch This
Watch this video to see some more variation problems worked out, including problems involving joint variation.
Guidance
Many variables in real-world problems are related to each other by variations. A variation is an equation that relates a variable to one or more other variables by the operations of multiplication and division. There are three different kinds of variation: direct variation, inverse variation and joint variation.
Distinguish Direct and Inverse Variation
In direct variation relationships, the related variables will either increase together or decrease together at a steady rate. For instance, consider a person walking at three miles per hour. As time increases, the distance covered by the person walking also increases, at the rate of three miles each hour. The distance and time are related to each other by a direct variation:
distance=speed×time
Since the speed is a constant 3 miles per hour, we can write: d=3t\begin{align*}d = 3t\end{align*}.
The general equation for a direct variation is y=kx\begin{align*}y = kx\end{align*}, where k\begin{align*}k\end{align*} is called the constant of proportionality.
You can see from the equation that a direct variation is a linear equation with a y\begin{align*}y-\end{align*}intercept of zero. The graph of a direct variation relationship is a straight line passing through the origin whose slope is k\begin{align*}k\end{align*}, the constant of proportionality.
A second type of variation is inverse variation. When two quantities are related to each other inversely, one quantity increases as the other one decreases, and vice versa.
For instance, if we look at the formula distance=speed×time\begin{align*}distance = speed \times time\end{align*} again and solve for time, we obtain:
time=distancespeed
If we keep the distance constant, we see that as the speed of an object increases, then the time it takes to cover that distance decreases. Consider a car traveling a distance of 90 miles, then the formula relating time and speed is: t=90s\begin{align*}t = \frac{90}{s}\end{align*}.
The general equation for inverse variation is y=kx\begin{align*}y=\frac{k}{x}\end{align*}, where k\begin{align*}k\end{align*} is the constant of proportionality.
In this chapter, we’ll investigate how the graphs of these relationships behave.
Another type of variation is a joint variation. In this type of relationship, one variable may vary as a product of two or more variables.
For example, the volume of a cylinder is given by:
V=πR2h
In this example the volume varies directly as the product of the square of the radius of the base and the height of the cylinder. The constant of proportionality here is the number π\begin{align*}\pi\end{align*}.
In many application problems, the relationship between the variables is a combination of variations. For instance Newton’s Law of Gravitation states that the force of attraction between two spherical bodies varies jointly as the masses of the objects and inversely as the square of the distance between them:
F=Gm1m2d2
In this example the constant of proportionality is called the gravitational constant, and its value is given by G=6.673×1011 Nm2/kg2\begin{align*}G = 6.673 \times 10^{-11} \ N \cdot m^2 / kg^2\end{align*}.
Graph Inverse Variation Equations
We saw that the general equation for inverse variation is given by the formula y=kx\begin{align*}y = \frac{k}{x}\end{align*}, where k\begin{align*}k\end{align*} is a constant of proportionality. We will now show how the graphs of such relationships behave. We start by making a table of values. In most applications, x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} are positive, so in our table we’ll choose only positive values of x\begin{align*}x\end{align*}.
Example A
Graph an inverse variation relationship with the proportionality constant k=1\begin{align*}k = 1\end{align*}.
Solution
x\begin{align*}x\end{align*} y=1x\begin{align*}y =\frac {1}{x}\end{align*}
0 y=10=undefined\begin{align*}y=\frac {1}{0} = \text{undefined}\end{align*}
14\begin{align*}\frac {1}{4}\end{align*} y=114=4\begin{align*}y =\frac {1}{\frac{1}{4}}=4\end{align*}
12\begin{align*}\frac {1}{2}\end{align*} y=112=2\begin{align*}y=\frac {1}{\frac{1}{2}}=2\end{align*}
34\begin{align*}\frac {3}{4}\end{align*} y=134=1.33\begin{align*}y =\frac {1}{\frac{3}{4}}=1.33\end{align*}
1 y=11=1\begin{align*}y =\frac {1}{1}=1\end{align*}
32\begin{align*}\frac {3}{2}\end{align*} y=132=0.67\begin{align*}y=\frac {1}{\frac{3}{2}}=0.67\end{align*}
2 y=12=0.5\begin{align*}y =\frac {1}{2}=0.5\end{align*}
3 y=13=0.33\begin{align*}y=\frac {1}{3}=0.33\end{align*}
4 y=14=0.25\begin{align*}y =\frac {1}{4}=0.25\end{align*}
5 y=15=0.2\begin{align*}y =\frac {1}{5}=0.2\end{align*}
10 \begin{align*}y=\frac {1}{10}=0.1\end{align*}
Here is a graph showing these points connected with a smooth curve.
Both the table and the graph demonstrate the relationship between variables in an inverse variation. As one variable increases, the other variable decreases and vice versa.
Notice that when \begin{align*}x = 0\end{align*}, the value of \begin{align*}y\end{align*} is undefined. The graph shows that when the value of \begin{align*}x\end{align*} is very small, the value of \begin{align*}y\end{align*} is very big—so it approaches infinity as \begin{align*}x\end{align*} gets closer and closer to zero.
Similarly, as the value of \begin{align*}x\end{align*} gets very large, the value of \begin{align*}y\end{align*} gets smaller and smaller but never reaches zero. We will investigate this behavior in detail throughout this chapter.
Write Inverse Variation Equations
As we saw, an inverse variation fulfills the equation \begin{align*}y = \frac{k}{x}\end{align*}. In general, we need to know the value of \begin{align*}y\end{align*} at a particular value of \begin{align*}x\end{align*} in order to find the proportionality constant. Once we know the proportionality constant, we can then find the value of \begin{align*}y\end{align*} for any given value of \begin{align*}x\end{align*}.
Example B
If \begin{align*}y\end{align*} is inversely proportional to \begin{align*}x\end{align*}, and if \begin{align*}y = 10\end{align*} when \begin{align*}x = 5\end{align*}, find \begin{align*}y\end{align*} when \begin{align*}x = 2\end{align*}.
Solution
Compare Graphs of Inverse Variation Equations
Inverse variation problems are the simplest example of rational functions. We saw that an inverse variation has the general equation: \begin{align*}y= \frac{k}{x}\end{align*}. In most real-world problems, \begin{align*}x\end{align*} and \begin{align*}y\end{align*} take only positive values. Below, we will show graphs of three inverse variation functions.
Example C
On the same coordinate grid, graph inverse variation relationships with the proportionality constants \begin{align*}k = 1, k = 2,\end{align*} and \begin{align*}k = \frac{1}{2}\end{align*}.
Solution
We’ll skip the table of values for this problem, and just show the graphs of the three functions on the same coordinate axes. Notice that for larger constants of proportionality, the curve decreases at a slower rate than for smaller constants of proportionality. This makes sense because the value of \begin{align*}y\end{align*} is related directly to the proportionality constants, so we should expect larger values of \begin{align*}y\end{align*} for larger values of \begin{align*}k\end{align*}.
Watch this video for help with the Examples above.
Vocabulary
• The general equation for a direct variation is \begin{align*}y = kx\end{align*}, where \begin{align*}k\end{align*} is called the constant of proportionality.
• The general equation for inverse variation is \begin{align*}y= \frac{k}{x}\end{align*}, where \begin{align*}k\end{align*} is the constant of proportionality.
Guided Practice
If \begin{align*}p\end{align*} is inversely proportional to the square of \begin{align*}q\end{align*}, and \begin{align*}p = 64\end{align*} when \begin{align*}q = 3\end{align*}, find \begin{align*}p\end{align*} when \begin{align*}q = 5\end{align*}.
Solution
Practice
For 1-4, graph the following inverse variation relationships.
1. \begin{align*}y= \frac{3}{x}\end{align*}
2. \begin{align*}y= \frac{10}{x}\end{align*}
3. \begin{align*}y= \frac{1}{4x}\end{align*}
4. \begin{align*}y= \frac{5}{6x}\end{align*}
5. If \begin{align*}z\end{align*} is inversely proportional to \begin{align*}w\end{align*} and \begin{align*}z = 81\end{align*} when \begin{align*}w = 9\end{align*}, find \begin{align*}w\end{align*} when \begin{align*}z = 24\end{align*}.
6. If \begin{align*}y\end{align*} is inversely proportional to \begin{align*}x\end{align*} and \begin{align*}y = 2\end{align*} when \begin{align*}x = 8\end{align*}, find \begin{align*}y\end{align*} when \begin{align*}x = 12\end{align*}.
7. If \begin{align*}a\end{align*} is inversely proportional to the square root of \begin{align*}b\end{align*}, and \begin{align*}a = 32\end{align*} when \begin{align*}b = 9\end{align*}, find \begin{align*}b\end{align*} when \begin{align*}a = 6\end{align*}.
8. If \begin{align*}w\end{align*} is inversely proportional to the square of \begin{align*}u\end{align*} and \begin{align*}w = 4\end{align*} when \begin{align*}u = 2\end{align*}, find \begin{align*}w\end{align*} when \begin{align*}u = 8\end{align*}.
9. If \begin{align*}a\end{align*} is proportional to both \begin{align*}b\end{align*} and \begin{align*}c\end{align*} and \begin{align*}a = 7\end{align*} when \begin{align*}b = 2\end{align*} and \begin{align*}c = 6\end{align*}, find \begin{align*}a\end{align*} when \begin{align*}b = 4\end{align*} and \begin{align*}c = 3\end{align*}.
10. If \begin{align*}x\end{align*} is proportional to \begin{align*}y\end{align*} and inversely proportional to \begin{align*}z\end{align*}, and \begin{align*}x = 2\end{align*} when \begin{align*}y = 10\end{align*} and \begin{align*}z = 25\end{align*}, find \begin{align*}x\end{align*} when \begin{align*}y = 8\end{align*} and \begin{align*}z = 35\end{align*}.
11. If \begin{align*}a\end{align*} varies directly with \begin{align*}b\end{align*} and inversely with the square of \begin{align*}c\end{align*}, and \begin{align*}a = 10\end{align*} when \begin{align*}b = 5\end{align*} and \begin{align*}c = 2\end{align*}, find the value of \begin{align*}a\end{align*} when \begin{align*}b = 3\end{align*} and \begin{align*}c = 6\end{align*}.
12. If \begin{align*}x\end{align*} varies directly with \begin{align*}y\end{align*} and \begin{align*}z\end{align*} varies inversely with \begin{align*}x\end{align*}, and \begin{align*}z = 3\end{align*} when \begin{align*}y = 5\end{align*}, find \begin{align*}z\end{align*} when \begin{align*}y = 10\end{align*}.
Vocabulary Language: English
Constant of Proportionality
Constant of Proportionality
The constant of proportionality, commonly represented as $k$ is the constant ratio of two proportional quantities such as $x$ and $y$.
Direct Variation
Direct Variation
When the dependent variable grows large or small as the independent variable does.
Inverse Variation
Inverse Variation
Inverse variation is a relationship between two variables in which the product of the two variables is equal to a constant. As one variable increases the second variable decreases proportionally.
Joint Variation
Joint Variation
Variables exhibit joint variation if one variable varies directly as the product of two or more other variables.<|endoftext|>
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The fat in milk is secreted by specialized cells in the mammary glands of mammals. It is released as tiny fat globules or droplets, which are stabilized by a phospholipid and protein coat derived from the plasma membrane of the secreting cell. Milk fat is composed mainly of triglycerides—three fatty acid chains attached to a single molecule of glycerol. It contains 65 percent saturated, 32 percent monounsaturated, and 3 percent polyunsaturated fatty acids. The fat droplets carry most of the cholesterol and vitamin A. Therefore, skim milk, which has more than 99.5 percent of the milk fat removed, is significantly lower in cholesterol than whole milk (2 milligrams per 100 grams of milk, compared with 14 milligrams for whole milk) and must be fortified with vitamin A.
Milk contains a number of different types of proteins, depending on what is required for sustaining the young of the particular species. These proteins increase the nutritional value of milk and other dairy products and provide certain characteristics utilized for many of the processing methods. A major milk protein is casein, which actually exists as a multisubunit protein complex dispersed throughout the fluid phase of milk. Under certain conditions the casein complexes are disrupted, causing curdling of the milk. Curdling results in the separation of milk proteins into two distinct phases, a solid phase (the curds) and a liquid phase (the whey).
Lactose is the principal carbohydrate found in milk. It is a disaccharide composed of one molecule each of the monosaccharides (simple sugars) glucose and galactose. Lactose is an important food source for several types of fermenting bacteria. The bacteria convert the lactose into lactic acid, and this process is the basis for several types of dairy products.
In the diet lactose is broken down into its component glucose and galactose subunits by the enzyme lactase. The glucose and galactose can then be absorbed from the digestive tract for use by the body. Individuals deficient in lactase cannot metabolize lactose, a condition called lactose intolerance. The unmetabolized lactose cannot be absorbed from the digestive tract and therefore builds up, leading to intestinal distress.
Vitamins and minerals
Milk is a good source of many vitamins. However, its vitamin C (ascorbic acid) content is easily destroyed by heating during pasteurization. Vitamin D is formed naturally in milk fat by ultraviolet irradiation but not in sufficient quantities to meet human nutritional needs. Beverage milk is commonly fortified with the fat-soluble vitamins A and D. In the United States the fortification of skim milk and low-fat milk with vitamin A (in water-soluble emulsified preparations) is required by law.
Milk also provides many of the B vitamins. It is an excellent source of riboflavin (B2) and provides lesser amounts of thiamine (B1) and niacin. Other B vitamins found in trace amounts are pantothenic acid, folic acid, biotin, pyridoxine (B6), and vitamin B12.
Milk is also rich in minerals and is an excellent source of calcium and phosphorus. It also contains trace amounts of potassium, chloride, sodium, magnesium, sulfur, copper, iodine, and iron. A lack of adequate iron is said to keep milk from being a complete food.<|endoftext|>
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Shakespeare is the most famous British playwright in history. People know his name in almost every country in the world. But who exactly was William Shakespeare? That is the question! Shakespeare was born in Stratford-upon-Avon on 23 April, 1564. Families were big in those days. William had seven brothers and sisters. But his parent’s weren’t poor; his father, John Shakespeare, was a successful businessman who bought and sold leather and wool. His mother was the daughter of a rich farmer. When Shakespeare left school, he went to work for his father. But soon after, he met and fell in love with Ann Hathaway, the daughter of a farmer who lived in Stratford. They got married in December 1582, and just five months late, their first daughter, Susanna, was born. William was 18. Ann was 25. What did Shakespeare do for the next ten years?
We don’t know exactly. We don’t know why he gave up a good job in his father’s business and moved to London. We don’t know exactly when or why he became an actor and playwright. All we know is that in 1592 he wrote his first play. After that, his plays became popular very quickly, and he made a lot of money. Four hundred years ago, Shakespeare built a theatre – The Globe – here in the center of London. It was one of London’s first theatres. It was round and had no roof over the center – like the theatres of ancient Rome. OK in Rome – not such a good idea in cold, rainy London! The people of London loves going to the theatre. The globe could hold three thousand people. Some people sat to watch the plays; other stood in the middle, in front of the stage. The audience were usually noisy, often clapping and cheering, and shouting to the actors – and there were only actors, no actresses. Young boys played the parts of women.
It often rained in London then, too. And everyone got very wet. In 1610, after about twenty-five years in London, Shakespeare came back here to Stratford. He was rich, and he had a big house where he enjoyed life with his family and friends. But he didn’t stop writing plays. What kind of plays did William Shakespeare write? Well, he wrote thirty-nine plays. Some of them are comedies, for example, A Midsummer Night’s Dream and Comedy of Errors. They have happy endings. Others are stories from English history, for example, stories about the kings of England. They are very patriotic. Queen Elizabeth 1 often went to see them. And the others are tragedies, such as Hamlet and MacBeth – these are sad, dark stories of murder and revenge. Shakespeare died on his fifty-second birthday in 1616. He is buried in Holy Trinity Church, Stratford. But the characters in his plays are still with us today.
The Renaissance or the revival of learning was the period then european culture was at it’s high. It lasted from the 14’th centure till 17’th centure, and was coursed by complex economic situation and social conditions. The feudal system was been shuttled by the bourgeoisie, thich was getting stronger and stronger. It was more profitable to unite under a single rouler. Absolute monacy came into being. This lead to the forming of nations and the true sense of the world. New social and economic conditions called for the new ideology, because the catholic dogmas didn’t correspond to the new trend of life. For this reason in many european countries the protestant religion sprend up and national churches were established.
Instead of the blind face ordered by the catholic then appeared a new outlook which was called humanism. The time demanded positive recional knowledge and this demand was supplied: in astronomy by Copernicus, in philosophy by Tomas More, in geography by Columbus, Vaska de Gama and others. Leonardo de Vinci was force a new feory of art: “It was the greatest progressive revolution that mankind have so far experience, a time, thich called for “Giants” and produced Giants in power and thought, passion and character in universality and language.”
An example of a typical men of the Renaissance period was: the famous Englishmen sir Walter Raleigh, he was a soldier, sailor, explorer, pirate, coloniser, historian, thilosother and a poet. He was much interested in science and literature. He wrote works of geography and lead expedition to South America. He was an outstanding poet. His poems are full of profound wisdom, written with great elegance and salacity of style. He organised of “academy”. Christother Marlowe the greatest dramatist (before Shakespear). But the most important of most this writer and one of the greatest men of this period was sir Thomas More.
The most brilliant period of English literature was in the second half of the 16’th and begining of 17’th centure.Sometimes it’s called “Elizabethen age” after quen Elizabeth 5. England had become a geat world power. It had established wide commercial contact with countries And rich trading company had been organaized. The english people were now a great nation and the english language inriched was now not unlike the language of Chaucer. Many famous poetical and prose works appeared. Among those who inriched the literary haritage of this period ere sir Philip Sydney, Adnond Spenser and Christother Marlowe. There were fine works of poetry and prose in the Elizabethen age but the greatest hight’s of literature of this period were riached in drama.<|endoftext|>
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# Golden section
The golden section , usually indicated by the letter φ ( phi ) of the Greek alphabet in the honor of Phidias, sculptor and Greek architect of the Parthenon, is the irrational Nombre:
$\ varphi = \ frac \left\{1 + \ sqrt \left\{5\right\}\right\} \left\{2\right\} \ simeq 1,618033988749894848204586834365\dots$.
## Geometrical properties
### Golden section and pentagon
The golden section appears in the proportions of the regular pentagon, like the relationship between the length on the side of the pentagon and that on the side of the Pentagramme registered. This definite purely geometrical report/ratio on a regular polygon was probably the first Greek definition of the golden section.
### Gold rectangle
One calls gold rectangle a rectangle whose relationship between the length and the width is worth the golden section.
The layout of a gold rectangle is done very simply using a compass; it is enough to point the medium on a side of a square, to point one of the two opposed angles, then to fold back the arc of circle on the line passing by the side of the pointed square (to be noted that this construction was a “secrecy” of Compagnonnage to the Moyen-âge).
Here a possible reason of the attraction caused by the right-angled gold : let us consider a rectangle whose sides lengths has and B is in a report/ratio of the golden section:
So of this rectangle, we remove the square on side length B , then the rectangle remaining is again a gold rectangle, since its sides are in a φ report/ratio. Indeed, according to the algebraic properties,
$\ frac \left\{B\right\} \left\{a-b\right\} = \ frac \left\{1\right\} \left\{a/b -1\right\} = \ frac \left\{has\right\} \left\{B\right\} = \ varphi$.
By reiterating this construction, we obtain a succession of increasingly small gold rectangles. This fact is a geometrical interpretation of the development in Fraction continues golden section (see further).
### Gold triangles
The gold triangles are isosceles triangles of which the report/ratio of the east coasts equal to the golden section. There exists about it of two types. Those for which the side ratio/base is worth φ which gives acute triangles called sometimes money triangles and those for which the report/ratio bases/side is worth φ.
In the united figure:
• isosceles triangles BDA and CAB have a common basic angle in A. ABD is thus similar to BCA in a report/ratio of 1/φ.
• Like φ = 1+ 1/φ, cd. = 1 and DBC is isosceles of top D.
• the angle out of B is thus double angle out of C in ABC.
• the sum of the angles of a triangle being worth 180°, one obtains for the angle C the fifth of the flat Angle, either 36° and for the angle B the two fifths of the flat angle, or 72°.
Since it is a question of cutting out a flat angle into 5, he is not surprising to find these gold triangles in the regular Pentagone and the Pentacle.
In an acute gold triangle, one can draw a blunt gold triangle and an acute gold triangle φ time smaller. One finds this same phenomenon in a blunt gold triangle. These facts explain why one finds these two elements in the pavings of Penrose.
### Gold spirals
One can build, starting from a gold rectangle, a Spirale of gold by tracing quadrants in each square. This spiral approaches a Spirale logarithmic curve of center the intersection of the two diagonals of the two rectangles and of polar equation:
$R \left(\ theta\right) = R. \ varphi^ \left\{- \ frac \left\{\ theta\right\} \left\{\ pi/2\right\}\right\}$<|endoftext|>
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# PreTeXt Sample Book: Abstract Algebra (SAMPLE ONLY)
## Section4.4Large Powers of Integers
Computing large powers can be very time-consuming. Just as anyone can compute $$2^2$$ or $$2^8\text{,}$$ everyone knows how to compute
\begin{equation*} 2^{2^{1000000} }\text{.} \end{equation*}
However, such numbers are so large that we do not want to attempt the calculations; moreover, past a certain point the computations would not be feasible even if we had every computer in the world at our disposal. Even writing down the decimal representation of a very large number may not be reasonable. It could be thousands or even millions of digits long. However, if we could compute something like $$2^{37398332 } \pmod{ 46389}\text{,}$$ we could very easily write the result down since it would be a number between 0 and 46,388. If we want to compute powers modulo $$n$$ quickly and efficiently, we will have to be clever.
1
The results in this section are needed only in Chapter 2 (not really).
The first thing to notice is that any number $$a$$ can be written as the sum of distinct powers of 2; that is, we can write
\begin{equation*} a = 2^{k_1} + 2^{k_2} + \cdots + 2^{k_n}\text{,} \end{equation*}
where $$k_1 \lt k_2 \lt \cdots \lt k_n\text{.}$$ This is just the binary representation of $$a\text{.}$$ For example, the binary representation of 57 is 111001, since we can write $$57 = 2^0 + 2^3 + 2^4 + 2^5\text{.}$$
The laws of exponents still work in $${\mathbb Z}_n\text{;}$$ that is, if $$b \equiv a^x \pmod{ n}$$ and $$c \equiv a^y \pmod{ n}\text{,}$$ then $$bc \equiv a^{x+y} \pmod{ n}\text{.}$$ We can compute $$a^{2^k} \pmod{ n}$$ in $$k$$ multiplications by computing
\begin{gather*} a^{2^0} \pmod{ n}\\ a^{2^1} \pmod{ n }\\ \vdots\\ a^{2^k} \pmod{ n}\text{.} \end{gather*}
Each step involves squaring the answer obtained in the previous step, dividing by $$n\text{,}$$ and taking the remainder.
### Example4.4.1.Repeated Squares.
We will compute $$271^{321} \pmod{ 481}\text{.}$$ Notice that
\begin{equation*} 321 = 2^0 +2^6 + 2^8\text{;} \end{equation*}
hence, computing $$271^{321} \pmod{ 481}$$ is the same as computing
\begin{equation*} 271^{2^0 +2^6 + 2^8 } \equiv 271^{2^0} \cdot 271^{2^6} \cdot 271^{2^8} \pmod{481}\text{.} \end{equation*}
So it will suffice to compute $$271^{2^i} \pmod{481}$$ where $$i = 0, 6, 8\text{.}$$ It is very easy to see that
\begin{equation*} 271^{2^1} = \mbox{73,441} \equiv 329 \pmod{481}\text{.} \end{equation*}
We can square this result to obtain a value for $$271^{2^2} \pmod{481}\text{:}$$
\begin{align*} 271^{ 2^2} & \equiv (271^{ 2^1})^2 \pmod{ 481}\\ & \equiv (329)^2 \pmod{ 481}\\ & \equiv \mbox{108,241} \pmod{ 481}\\ & \equiv 16 \pmod{ 481}\text{.} \end{align*}
We are using the fact that $$(a^{2^n})^2 \equiv a^{2 \cdot 2^n} \equiv a^{ 2^{n+1} } \pmod{ n}\text{.}$$ Continuing, we can calculate
\begin{equation*} 271^{ 2^6 } \equiv 419 \pmod{ 481} \end{equation*}
and
\begin{equation*} 271^{ 2^8 } \equiv 16 \pmod{ 481}\text{.} \end{equation*}
Therefore,
\begin{align*} 271^{ 321} & \equiv 271^{ 2^0 +2^6 + 2^8 } \pmod{ 481}\\ & \equiv 271^{ 2^0 } \cdot 271^{ 2^6 } \cdot 271^{ 2^8 } \pmod{ 481}\\ & \equiv 271 \cdot 419 \cdot 16 \pmod{ 481}\\ & \equiv \mbox{1,816,784} \pmod{ 481}\\ & \equiv 47 \pmod{ 481}\text{.} \end{align*}
The method of repeated squares will prove to be a very useful tool when we explore RSA cryptography. To encode and decode messages in a reasonable manner under this scheme, it is necessary to be able to quickly compute large powers of integers mod $$n\text{.}$$
### Remark4.4.2.Sage.
Sage support for cyclic groups is a little spotty — but we can still make effective use of Sage and perhaps this situation could change soon.<|endoftext|>
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Definition - What does Shear Stress mean?
Shear Stress is the shear force per unit area acted on by shear force to sustain a constant rate of fluid movement. It usually occurs during drilling activities when drilling fluids move along solid boundaries.
Shear stress is expressed as:
S = F/A
S represents shear stress
F represents shear force
A represents area on which shear force is acting.
Petropedia explains Shear Stress
Fluids that move along solid boundaries gain shear stress on that given boundary. In fact, "no-slip situation" indicates that the fluid speed relative to the boundary is regarded as zero. However, at some point from that boundary, the speed of flow should be identical to the fluid.
In a flow within a straight container or vessel, fluids do not move at identical velocity at each point within the vessel. In such a case, the fluid flow is slowest near the wall of the vessel and quickest at its focal area. The velocity of fluid assumes a laminar-flow type of profile or "parabolic." This kind of flow is produced by the friction within the wall of the vessel and is associated with the viscosity of the fluid.
The friction produces a tangential force released by the flowing fluid, referred to as the fluid shear stress. The magnitude of this depends on the fluid velocity when moving from and around the vessel.<|endoftext|>
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What is Virtual Reality, and why is it important?
Virtual reality refers to creating illusions of a presence that isn’t really there within an environment, with the use of technology. Virtual reality basically plays a hoax with our brain and senses such as hearing and sight, and fools our brain into having a virtual experience of something that isn’t really tangible or materially present.
Literally, the meaning of virtual reality can be derived from the literal definitions of the words virtual and reality. Virtual refers to something that is real and reality is what we as human beings experience and undergo as first hand observers, hence, when put together, virtual reality means something that is close, near to reality, or something that is an imitation or emulation of reality.
Virtual reality works to create an artificial environment with the help of technological software and technical equipment to completely astound and suspend the viewer and manipulate his/her senses into accepting it as a real environment. Let us know take a detailed look at how virtual reality works to manipulate our senses and brain into accepting a computer generated virtual environment as real.
Our five senses, sight, hearing, smelling, taste, and touch, along with the cognitive perception systems of our brain dictate our interaction with our surroundings and environment. Aside from the basic five senses that our controlled by our basic sense organs, the human bod has several other important senses, for instance, the sense of balance, the sense of intuition etc. that also influence our decisions and interpret whatever information is collected by our brain. Hence, our entire experience with reality is actually what is interpreted by our senses and the brains’ processing mechanisms. Virtual reality works by presenting our sensory organs and brain with a virtual image that is accepted by our brain through a process of interactivity.
Virtual reality works by generating artificial environments by the help of technology and computer software, these experiences primarily involve the receptivity and activity of our two basic senses, sight and hearing. The generation of virtual images involves a complicated and complex technological process that requires the use of advanced computer software and modern technical equipment that are specifically designed to create virtual reality environments.
The simplest and widely used from of virtual reality is a 3-D image that basically allows the content of an image to in any direction or simply zoom in and out. It can be easily created on a personal computer or laptop and does not require much complex equipment or technology. More advanced and complicated virtual reality environments include, rooms augmented with wearable computers, haptics devices that allow viewers to feel the images and wrap-around display screens among others.
With the continuous and rapid evolution of technology, virtual reality has taken over our lives not only in the consumer world for commercial endeavors, but also in the fields of medical science, education, military, aviation, entertainment and much more. Virtual reality has not only revolutionized gaming and films by enhancing the content and presentations of videos, but has also made an impact on enhancing training and learning experiences in countless professions.
Virtual reality has made marked improvements and led to the development of extremely beneficial application that aid the following spheres:
- Psychological treatments and rehabilitation
- Media, films and fashion
Whatever seems to unpractical and unachievable has been made possible and attainable with the advent of virtual reality, all our experimentation have been made successful as by carrying virtual experiments we can identify with the virtual risks, and hence, we can effectively eliminate these risks in reality. From trainee fighters, pilots, surgeons to engineers, architects and businessmen, virtual reality with his productive and efficient application allows us to enhance our knowledge, improve our performance and the productivity of our efforts and resources.<|endoftext|>
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Build your math mind
# What is a Derivative? Derivatives Definition and Meaning
If you’ve got some questions about derivatives — like what they are, why they’re used, and how you can find one — you’ve come to the right place! Derivatives are a fundamental concept on your calculus journey, so don’t hesitate to really spend some time on this.
We don’t want to be overzealous, but we think you’ll derive a lot of value from the details below (sorry, we had to).
Anyway, enough with the jokes! Time to focus on our strengths.
Let’s learn some math, shall we?
## What is a derivative?
A derivative is one of the most important tools in differential calculus. Derivatives give us a flexible way to measure precise rates of change, which is really cool!
You can use derivatives in many different scenarios, including:
### The meaning of derivatives
To put it simply, derivatives show us the instantaneous rate of change at a particular point on the graph of a function. That means we’re able to capture a pretty robust piece of information with relative ease (depending on the level of calculus you’re performing!).
Plus, being able to find derivatives gives us the ability to more accurately model things like velocity, force, acceleration, and more — so they’re not just used to challenge you at homework time. Derivatives are actually crucial to the inner workings of so many things around us.
For example: let’s say you’re at a baseball game, and you want to know the rate of change of the ball’s speed as it leaves the pitcher’s hand. In that case, you’d want to take a derivative!
### Derivative and slope
It’s hard to talk about derivatives without relating them to slope.
Why? Because finding a derivative is actually equivalent to finding the slope of the tangent line at a particular point on a function.
Fun fact: How we calculate a derivative is based on how we calculate slope! It’s rise over run, but with a few interesting twists.
## The definition of derivative
So, knowing the context of derivatives and what they tell us, how are we defining “derivative”? This is our definition:
Derivative: (n) the rate of change of a quantity with respect to a change in a variable; the result of differentiation
Simple enough, right?
## Derivatives in math vs. derivatives in finance
To be clear, we’re here to teach you about derivatives in math, but you may also come across information regarding derivatives in finance or investing.
If that’s what you’re looking for, you’ll want to know that a “derivative” in finance is a contract whose value is derived from the performance of an asset, interest rate, or other “underlying.”
But we’re here to learn math, so let’s get into some equations!
## What will equations with derivatives look like?
We can talk about the “why” of derivatives until we’re blue in the face, but now it’s time to focus on the “how” and take a look at what derivatives will look like on the page.
Depending on whether you’re taking the derivative of a function, integral, or expression, your exact course of action will be different. But your starting problems will probably look something like these examples:
$$f(x)=6x^5+x$$
$$\frac{d}{dx}\int_{\frac{\pi}{4}}^{{x}} \cos(t)dt$$
$$\frac{d}{dx}\int_{-x}^{x} \ln{(2+\sin{t})}dt$$
And you’ll find yourself with solutions that look like this:
$$f^{'}(x)=4x+3$$
$$\frac{dy}{dx}=\frac{98x}{y\times(x^2+49)^2}$$
Note: Those are just examples, NOT the solutions to those problems!
While you’re navigating these equations, don’t forget about the differentiation rules that can help you manipulate calculations more strategically!
As you can tell, there’s quite a variety of derivative equations, so let’s get a little more specific!
### Derivative of a function
Finding derivatives of functions is usually the starting point of your adventure with differentiation, so be sure to master those procedures before moving on to limits and more challenging differentiation tasks.
As with anything in life, it’s helpful to keep your goals in mind while you work! So, what is our goal with derivatives of functions?
When we take the derivative of a function, it’s because we want to find:
• The function’s rate of change
• The slope of the tangent line at a particular point on the function
We’ll go into how to do this in a moment; but first, let’s decode something else you may hear when learning derivatives of functions.
### Derivative with respect to x
Maybe you’ve heard your math teacher talk about a derivative “with respect to” a variable (at the beginning, it’s commonly $$x$$), and you’re not quite sure what that really means.
If we’re working with a function — $$f(x)$$ — and we want to find “the derivative with respect to $$x$$,” that just means we’re looking to find the rate at which $$f$$ changes as $$x$$ changes. Essentially, it means we’re really focusing in on the variable $$x$$ and how it affects our function.
### The process of finding the derivative of a function
If you remember your key vocabulary terms, you’ll remember that the process of finding the derivative is called “differentiation.” (We can help if you need a refresher on calculus vocab!)
Fun fact: the derivative of a function is also a function, and its values are the corresponding derivatives at a point. Essentially, if we determine the derivative of a function $$f$$ at any value $$x$$, then our derivative $$f'(x)$$ represents a function that outputs the value of $$f'(x)$$ for any input $$x$$ (at least, for which this derivative exists).
The actual steps of differentiation will vary based on the individual problem and the complexity — we can explain in detail in the app! — but here’s an overview:
1. Take the derivative on both sides of the equation
2. Use the differentiation rules (also called “derivative formulas”)
3. Find the derivative
4. Simplify the expression, if needed
Need more? Hop on over to dive deeper into derivatives of functions.
Don’t forget: You can also find the derivative of expressions and integrals, which will look a little different.
## What is f’(x)?
$$f’(x)$$ denotes the derivative. So, while you start taking the derivative of a function $$f(x)$$, your solution — the derivative — will be labeled as $$f’(x)$$. You may also see $$f’(x)$$ written as $$\frac{dy}{dx}$$ in certain scenarios.
Just be careful when reading, because it’s easy to miss that little apostrophe. When we see $$f’(x)$$, we’re seeing a derivative, but $$f(x)$$ tells us it’s a function!
## Derivative vs. limit: What’s the difference?
Derivatives and limits aren’t interchangeable terms, but there is some overlap.
The difference depends on whether you’re talking about derivatives and limits of a certain value of a function, or of the function as a whole. Let us explain:
• A derivative of a function at a point is a special type of limit at that point — in other words, every derivative is a limit!
• Numerically, the derivative’s value at a certain point of a function tells us the function’s instantaneous rate of change at that point.
• Geometrically, the derivative’s value at a certain point of a function represents the slope of a line running tangent to the function’s graph at that point.
• The limit of a function at a point represents the behavior of the function around that point. Basically, if the behavior is predictable, then the limit exists.
Does that help you differentiate? (Pun very much intended!)
## Examples of derivatives in math
You know we’re big advocates of practice, so let’s look at some example problems! Using what you’ve learned, you can work through these examples on your own.
1. $$f(x)=e^x+10x$$
2. $$f(x)=\frac{x-3}{\sqrt{8x^2-2}}$$
3. $$\frac{d}{dx}\left(\sqrt{x^2-3} \right)$$
4. $$\frac{d}{dx}\left(\ln{(x)}+7^x\right)$$
5. $$\frac{d}{dx}\int_{x}^{\frac{\pi}4} (t^2-\ln(t))^2dt$$
Here’s how we solve the first practice problem in the app:
Stuck on the other derivative practice problems? Scan the problem with your Photomath app so we can walk you through each step!
FAQ
What are the two definitions of a derivative?
A derivative is described as either the rate of change of a function, or the slope of the tangent line at a particular point on a function.
What is a derivative in simple terms?
A derivative tells us the rate of change with respect to a certain variable.
How are derivatives used in real life?
Derivatives can be used to predict changes like temperature variation in climate change, earthquake magnitude ranges, population census predictions, shifts in momentum, and more.<|endoftext|>
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Fractures are complete or partial bone breaks. Broken bones are common, and occur when the physical force exerted on them is stronger than the bone itself. The main categories of fractures are displaced, non-displaced, open, and closed.
Displaced fractures involve the bone breaking into two or more parts, causing the bone alignment to be off. Non-displaced fractures involve a crack in the bone, without the alignment being thrown off. An open fracture involves the bone breaking through the skin, while a closed fracture involves a bone break without a puncture or open wound in the skin.
What are the treatments for fractures?
Usually x-rays can detect a fracture, but in some cases, other tests (such as MRIs and CT scans) are necessary. This is especially true, depending on the area of the break. If you think you have broken a bone, seek immediate medical help. Make sure you protect the broken area so that you can avoid further damage. Broken arms and legs should be held in place with a splint. If the area is bleeding, be sure to apply pressure to stop the blood before using gauze and a splint. Fractured bones need to be held in place so that they can heal properly over time. Surgery, pins, plates, rods, screws, and glue are all options for holding a fracture in place.<|endoftext|>
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# Golden Ratio
The golden ratio (symbol is the Greek letter "phi" shown at left) is a special number approximately equal to 1.618 It appears many times in geometry, art, architecture and other areas.
## The Idea Behind It
We find the golden ratio when we divide a line into two parts so that: the whole length divided by the long part is also equal to the long part divided by the short part
Have a try yourself (use the slider):
## Beauty
This rectangle has been made using the Golden Ratio, Looks like a typical frame for a painting, doesn't it?
Some artists and architects believe the Golden Ratio makes the most pleasing and beautiful shape.
Do you think it is the "most pleasing rectangle"?
Maybe you do or don't, that is up to you!
Many buildings and artworks have the Golden Ratio in them, such as the Parthenon in Greece, but it is not really known if it was designed that way.
## The Actual Value
The Golden Ratio is equal to:
1.61803398874989484820... (etc.)
The digits just keep on going, with no pattern. In fact the Golden Ratio is known to be an Irrational Number, and I will tell you more about it later.
## Calculating It
You can calculate it yourself by starting with any number and following these steps:
• A) divide 1 by your number (=1/number)
• C) that is your new number, start again at A
With a calculator, just keep pressing "1/x", "+", "1", "=", around and around. I started with 2 and got this:
2 1/2=0.5 0.5+1=1.5
1.5 1/1.5 = 0.666... 0.666... + 1 = 1.666...
1.666... 1/1.666... = 0.6 0.6 + 1 = 1.6
1.6 1/1.6 = 0.625 0.625 + 1 = 1.625
1.625 1/1.625 = 0.6154... 0.6154... + 1 = 1.6154...
1.6154...
It is getting closer and closer!
But it takes a long time to get even close, but there are better ways and it can be calculated to thousands of decimal places quite quickly.
## Drawing It
Here is one way to draw a rectangle with the Golden Ratio:
• Draw a square (of size "1")
• Place a dot half way along one side
• Draw a line from that point to an opposite corner (it is √5/2 in length)
• Turn that line so that it runs along the square's side
Then you can extend the square to be a rectangle with the Golden Ratio.
## The Formula
That rectangle above shows us a simple formula for the Golden Ratio.
When one side is 1, the other side is:
The square root of 5 is approximately 2.236068, so the Golden Ratio is approximately (1+2.236068)/2 = 3.236068/2 = 1.618034. This is an easy way to calculate it when you need it.
Interesting fact: the Golden Ratio is also equal to 2 × sin(54°), get your calculator and check!
## Fibonacci Sequence
There is a special relationship between the Golden Ratio and the Fibonacci Sequence:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
(The next number is found by adding up the two numbers before it.)
And here is a surprise: when we take any two successive (one after the other) Fibonacci Numbers, their ratio is very close to the Golden Ratio.
In fact, the bigger the pair of Fibonacci Numbers, the closer the approximation. Let us try a few:
A
B
B/A
2
3
1.5
3
5
1.666666666...
5
8
1.6
8
13
1.625
...
...
...
144
233
1.618055556...
233
377
1.618025751...
...
...
...
We don't even have to start with 2 and 3, here I chose 192 and 16 (and got the sequence 192, 16, 208, 224, 432, 656, 1088, 1744, 2832, 4576, 7408, 11984, 19392, 31376, ...):
A
B
B / A
192
16
0.08333333...
16
208
13
208
224
1.07692308...
224
432
1.92857143...
...
...
...
7408
11984
1.61771058...
11984
19392
1.61815754...
...
...
...
## The Most Irrational ...
I believe the Golden Ratio is the most irrational number. Here is why ...
One of the special properties of the Golden Ratio is that it can be defined in terms of itself, like this: (In numbers: 1.61803... = 1 + 1/1.61803...) That can be expanded into this fraction that goes on for ever (called a "continued fraction"):
So, it neatly slips in between simple fractions.
But many other irrational numbers are reasonably close to rational numbers (such as Pi = 3.141592654... is pretty close to 22/7 = 3.1428571...)
## Pentagram
No, not witchcraft! The pentagram is more famous as a magical or holy symbol. And it has the Golden Ratio in it:
• a/b = 1.618...
• b/c = 1.618...
• c/d = 1.618...<|endoftext|>
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# Multiplying rational expressions
Learn how to find the product of two rational expressions.
#### What you should be familiar with before taking this lesson
A rational expression is a ratio of two polynomials. The domain of a rational expression includes all real numbers except those that make its denominator equal to zero.
We can simplify rational expressions by canceling common factors in the numerator and the denominator.
If this is not familiar to you, you'll want to check out the following articles first:
#### What you will learn in this lesson
In this lesson, you will learn how to multiply rational expressions.
## Multiplying fractions
To start, let's recall how to multiply numerical fractions.
Consider this example:
$\begin{array}{rl}& \phantom{=}\frac{3}{4}\cdot \frac{10}{9}\\ \\ & =\frac{3}{2\cdot 2}\cdot \frac{2\cdot 5}{3\cdot 3}& & \text{Factor numerators and denominators}\\ \\ & =\frac{\overline{)3}}{\overline{)2}\cdot 2}\cdot \frac{\overline{)2}\cdot 5}{\overline{)3}\cdot 3}& & \text{Cancel common factors}\\ \\ & =\frac{5}{6}& & \text{Multiply across}\end{array}$
In conclusion, to multiply two numerical fractions, we factored, canceled common factors, and multiplied across.
## Example 1: $\frac{3{x}^{2}}{2}\cdot \frac{2}{9x}$
We can multiply rational expressions in much the same way as we multiply numerical fractions.
Recall that the original expression is defined for $x\ne 0$. The simplified product must have the same restictions. Because of this, we must note that $x\ne 0$.
We write the simplified product as follows:
$\frac{x}{3}$ for $x\ne 0$
1) Multiply and simplify the result.
$\frac{4{x}^{6}}{5}\cdot \frac{1}{12{x}^{3}}=$
for $x\ne$
## Example 2: $\frac{{x}^{2}-x-6}{5x+5}\cdot \frac{5}{x-3}$
Once again, we factor, cancel any common factors, and then multiply across. Finally, we make sure to note all restricted values.
The original expression is defined for $x\ne -1,3$. The simplified product must have the same restrictions.
In general, the product of two rational expressions is undefined for any value that makes either of the original rational expressions undefined.
2) Multiply and simplify the result.
$\frac{5{x}^{3}}{5x+10}\cdot \frac{{x}^{2}-4}{{x}^{2}}=$
What are all the restrictions on the domain of the resulting expression?
3) Multiply and simplify the result.
$\frac{{x}^{2}-9}{{x}^{2}-2x-8}\cdot \frac{x-4}{x-3}=$
What are all the restrictions on the domain of the resulting expression?
### What's next?
If you feel good about your multiplication skills, you can move on to dividing rational expressions.
## Want to join the conversation?
• I have absolutely no idea how you figure out which numbers x cannot equal. I know it says it is the numbers that make the original expression undefined but how do you find that out?
• Rational expressions are fractions. Fractions become undefined if the denominator is = 0.
For example: 5/0 = undefined.
Now, if this was 5/x, then it is undefined only when x=0. So x can't = 0.
If you have an expression of: 5/(x-2), then you look at what would make x-2 = 0. If x=2, this fraction would be undefined.
When you have a rational expression that you are simplifying, any time you reduce the fraction, you need to ensure that the restrictions associated with the original fraction are maintained.
For example:
[(x-2)(x-5)] / [(x-2) (x+7)} would have restrictions of x not equal 2 or -7 because these both cause the denominator to become 0. At this point, we don't have to explicitly state the restrictions because they can be derived from the expression. have to explicitly
Once reduced (we cancel out the common factor of (x-2)), then we have (x-5) / (x+7). To ensure we maintain the original restrictions, we must explicitly state that "x not equal 2" because this can no longer be derived from the expression.
Hope this helps.
• on question 2 if you cancel out the whole denominator won't that just make it 0 and make the the answer /no solution ?
• No. If you cancel out the denominator then it would become 1. Thus, the final answer doesn't have a denominator since it is implied that it's denominator is 1.
• For example 2, the answer x +2/x+1 has two Xs. Can't those Xs be cancelled?
• No, because the x is not a factor like say;
x(x+2)/x(x+4) it's a common factor so the x can be canceled out
x+2/x+1, you can't cancel out the x because say x = 1
2+1/1+1 = 3/2
if you were to cancel out the 1 (or x) you'd get 2/1=2 which is not the same as 3/2.
this is because it's not something that you factored out from both numerator and denominator, it's something you're adding/subtracting which could change the value.
(1 vote)
• For question number 3, how does the denominator of the first expression factor into (x−4)(x+2)? Also for the same number why di they included x=/-2 if you can tell that from the simplified expression?
• The denominator is: x^2 - 2x - 8. Find factors of -8 that add to the middle term (-2). The factors are -4 and 2.
That creates the binomial factors of (x-4)(x+2).
The instructions said to select all that apply. x not = -2 applies, as does x not = 4 and 3
• For the example 2:
Why we don't have to care about numerators? if x = -2 for x+2 / x+1, will it be undefined? Why we don't have to specify x != -2 ?
• A 0 in the denominator (in your example: (x+2)/0 ) is undefined, but a numerator in the denominator (in your example: (0/x+1)) isn't undefined. 0/x is 0.
• For Q3, why do we have to specify x does not equal -2 when this is obvious from the simplified form of the expression too? In the videos Sal only gives the values x is undefined for when that value is not clear in the simplified form
(1 vote)
• Because the instructions say to.... Follow the instructions. It asked you to select ALL that apply.
• Find the numerical value for
Start Fraction x minus 9 Over 9 End Fraction
x−9/9 when x=13
• Before simplifying the multiplying of fractions, you can divide out the common factor. true or false
• So im very confused on how to prove whether it's undefined?? What exactly do you do?
• Take just the denominator, then set it equal to zero. Solve that for x. That's where it's undefined. For example:
Y=(bla bla bla)/(x-3)
Is undefined when x-3=0, or in other words when x-3
• On problem # 3 why is X cannot equal -2 when x+2 is part of the remaining expression. I've seen it counted wrong both ways in other examples and I am confused as whether to count it or not.<|endoftext|>
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Whether it is not being able to play with their favourite toy because of a sibling, or witnessing parental conflict in the home, experiences of tension influence how children respond to conflict. As Charles R. Swindoll said, “Each day of your lives, we make deposits in the memory banks of our children.”
We live in a society where conflicts occur at various levels and in various forms. As society evolves, so does the way people interact with each other. Adults and children are often affected by disagreements in their interpersonal relationships. While parents may feel the need to shield their children from conflict, a more effective policy might be to teach children to deal with conflicts from a young age, so they are equipped with the skills and tools to deal with them in a more positive and respectful manner. Basic tools of conflict resolution can help empower children so that they can have the confidence to resolve problems and contribute towards promoting and building healthy civil societies.
As Mawlana Hazar Imam said during a speech at the University of Alberta on 9 June, 2009, “It seems to me to be the responsibility of educators everywhere to help develop ‘ethically literate’ people who can reason morally whenever they analyse and resolve problems, who see the world through the lens of ethics, who can articulate their moral reasoning clearly – even in a world of cultural and religious diversity – and have the courage to make tough choices. And it is clear that the quality of ethical leadership throughout society can in great measure be shaped by our educational institutions.”
There are many valuable lessons that can be learned from conflict resolution training, no matter your age. These include the ability to listen effectively, to be patient and calm, and to be neutral and non-judgmental. Such training also allows children to learn skills and tools in a way that promotes responsibility for their behaviour, and that underlines the importance of expressing their needs in a healthy way and of showing empathy and kindness to others. Some examples of conflict resolution skills that a child can learn include:
- How to cool off when upset.
- How to speak to an adult about their feelings.
- How to speak to each other respectfully.
- How to listen carefully to others.
- How to apologise.
- How to propose solutions to problems.
In 2014, the Aga Khan Conciliation and Arbitration Board for Portugal and ITREB Portugal organised a conflict resolution training program for children between the ages of 7 and 12 entitled the Small Mediators’ Programme. The programme intended to raise awareness among children about the importance of peace, of conflict resolution through mediation, and how the culture of resolving disputes amicably has been rooted in the ethics of our tariqah and history since the beginning of Islam. Children were given tools to resolve conflicts that might occur in certain settings, such as a playground, in order to better develop their listening, problem solving, and peace-keeping skills. They were taught these skills in an experiential game-based environment, which not only made it fun for them but ensured that the learning was properly integrated in a way that allowed them to use the skills in real life, outside the course.
The feedback from the parents demonstrated the immediate impact the program had on their daughters and sons: children appeared to be much calmer, and in the case of one child, his relationship with his older sister noticeably improved. Children shared that they learned several important lessons such as respecting differences and being respectfully tolerant of diversity, listening to opinions, apologising, and appreciating both the importance of mediation and the questions one must ask to mediate correctly. One of the children who attended the program even successfully mediated a minor conflict at home.
Teaching children how to resolve conflicts also has the added advantage of benefiting adult members of the Jamat. In the words of Laozi, “Wise men hear and see as little children do.” Adults learn so much from children every day, and by exposing them to these extra tools, they should be able to use them not only to help other children but also to enable themselves to serve as role models for their peers.
It is often said that children are our future, so let us ensure that we are making positive deposits into their memory banks, so that they can truly make the world a better place in which to live.
Rahim Aly is a Certified Mediation Trainer for the Aga Khan International Conciliation and Arbitration Board. He has more than 15 years of experience in the area of information technology.
Sheila Aly is a Certified Mediation Trainer for the Aga Khan International Conciliation and Arbitration Board. She is also a Barrister with a legal career spanning over 13 years along with serving as a coach and mentor in the United Kingdom.<|endoftext|>
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• Courses
• Data Structure
• ### Number Base Conversion (Binary | Octal | Decimal | Hexadecimal) All in One
Posted on 2021-03-07
Number conversion is the process of converting one number base to another base. In computer science, we have to study about 4 types of number systems, binary, octal, decimal, and hexadecimal number. When we convert one number system to another number system, it will be a total of 12 different number conversion types, which are illustrated below. While converting one number base to another number base it will be better to use the following table.
1. Binary to Decimal
2. Octal to Decimal
4. Decimal to Binary
5. Octal to Binary
7. Binary to Octal
8. Decimal to Octal
1) Binary to Decimal Conversion
i) Convert 11100011101
Convert 11010112 = into equivalent decimal number
Method 1:
```11010112
= 1×26+1×25+0×24+1×23+0×22+1×21+1×20
= 64+32+0+8+0+2+1
= 10710
```
Method 2:
11010112
Place Value 64 32 16 8 4 2 1 Binary 1 1 0 1 0 1 1 Decimal 64 + 32 + 8 + 2 + 1 = 10710
ii) Convert 1111001101112 into an equivalent decimal number
Method 1:
```1111001101112
= 1×211+1×210+1×29+1×28+0×27+0×26+1×25+1×24+0×23+1×22+1×21+1×20
= 2048+1024+512+265+0+0+32+16+0+4+2+1
= 3895
```
Method 2:
Place Value 2048 1024 512 256 128 64 32 16 8 4 2 1 Binary 1 1 1 1 0 0 1 1 0 1 1 1 Decimal 2048+1024+512+265+0+0+32+16+0+4+2+1 = 3895
2) Octal to Decimal Conversion
i) Convert 125638 to a decimal equivalent number
```= 1×84+ 2×83+5×82+6×81+3×80
= 4096 + 1024 + 320 + 48 + 3
= 549110
```
Convert 7A2BD16 into decimal equivalent number
A=10, B=11, C=12, D=13, E=14, F=15
``` = 7×164+10×163+2×162+11×161+13×160
= 458,752 + 40,960+512+176+13
= 50041310
```
4) Decimal to Binary Conversion
Convert 56310 into binary equivalent
Method 1:
2 563 -----1 -----1 -----0 -----0 -----1 -----1 -----0 -----0 -----0 2 281 2 140 2 70 2 35 2 17 2 8 2 4 2 2 1
So, 56310 = 10001100112
Note: (Write binary numbers from bottom to up)
5) Octal to Binary Conversion
Convert 27638 into an equivalent binary number
Octal Number: 2763
3 group binary digits
2 7 6 3 010 111 110 011
:: 27638 = 0101111100112
Convert AD710 into an equivalent binary number
4 group binary digit
A D 7 1010 1101 0111
7) Binary to Octal Conversion
Convert 111000101102 into equivalent octal number
3 group binary digits (Note: group the binary digits from right to left)
011 100 010 110 3 4 2 6
:: 111000101102 = 3468
8) Decimal to Octal Conversion
Convert 8965210 into equivalent octal number
8 89652 -----4 -----6 -----0 -----7 -----5 8 11206 8 1400 8 175 8 21 2
:: 8965210 = 25706410
The process of converting a hexadecimal number into an octal number is that, at first, we convert a hexadecimal number into binary and after that, we convert the obtained binary number into an octal number.
Convert A9D516 into equivalent octal number
Step 1: Converting Hexadecimal number into a binary number
4 group binary digits
A 9 D 5 1010 1001 1101 0101
```:: A9D516 = 10101001110101012
Step 2: Now converting binary to octal number
3 group binary digits
001 010 100 111 010 101
1 2 4 7 2 5
So, A9D516 = 1247258
```
Convert 101110100100012 into equivalent hexadecimal number
``` 4 group binary digits
0010 1110 1001 0001
2 E 9 1
:: 101110100100012 =2E9116```
The process to convert an octal number into hexadecimal first, we will convert an octal number into binary numbers system and after that again we convert the obtained binary number into hexadecimal number by making group of 4 binary digits.
```E.g. Convert 1247258 into hexadecimal number
Step 1: Converting octal to binary
```
1 2 4 7 2 5 001 010 100 111 010 101
:: 1247258 = 0010101001110101012
``` Step 2: Now converting obtained binary number to octal
Binary = 0010101001110101012
Making 4 group binary digits from right
0000 1010 1001 1101 0101
0 A 9 D 5
:: 0010101001110101012 = A9D516
So, 1247258 = A9D516
```
Convert 968510 to equivalent hexadecimal number
16 9685 -----5 -----13 -----5 16 605 16 37 2
:: 968510 = 25D516
1825
#### Recent Guides
Secondary Level Teaching License Examination Sample question paper Secondary Level General Examination Sample question paper-2078 English, Lower Secondary Level Sample question paper-2078 English, Secondary Level Sample Question Paper-2078<|endoftext|>
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# Alternate Interior Angles: Definition, Theorem & Examples
In this lesson, you will learn how to identify alternate interior angles and how to use the theorem to find missing angles and to solve everyday geometry problems.
Definitions
An angle is formed when two rays, a line with one endpoint, meet at one point called a vertex. The angle is formed by the distance between the two rays. Angles in geometry are often referred to using the angle symbol so angle A would be written as angle A.
Angle
A transversal line is a line that crosses or passes through two other lines. Sometimes, the two other lines are parallel, and the transversal passes through both lines at the same angle. The two other lines don’t necessarily have to be parallel in order for a transversal to cross them.
A straight angle, also called a flat angle, is formed by a straight line. The measure of this angle is 180 degrees. A straight angle can also be formed by two or more angles that sum to 180 degrees. Here, angle 1 + angle 2 = 180.
Transversal Line
Parallel lines are two lines on a two-dimensional plane that never meet or cross. When a transversal passes through parallel lines, there are special properties about the angles that are formed that do not occur when the lines are not parallel. Notice the arrows on lines m and n towards the left. These arrows indicate that lines m and n are parallel.
parallel lines
Alternate interior angles are formed when a transversal passes through two lines. The angles that are formed on opposite sides of the transversal and inside the two lines are alternate interior angles. Notice the pairs of blue and pink angles.
Alternate Interior Angles
These pairs are alternate interior angles.
Another way to think about alternate interior angles is by using the z-pattern. Notice that the pair of alternate interior angles makes a Z.
z pattern
In this window pane, angle a and angle b are alternate interior angles because they are on opposite sides of the transversal but inside the parallel lines.
Window Pane
Alternate Interior Angles Theorem
The Alternate Interior Angles theorem states, if two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent.
A theorem is a proven statement or an accepted idea that has been shown to be true. The converse of this theorem, which is basically the opposite, is also a proven statement: if two lines are cut by a transversal and the alternate interior angles are congruent, then the lines are parallel.
These theorems can be used to solve problems in geometry and to find missing information. This diagram shows which pairs of angles are equal and alternate interior. Notice that the lines are parallel.
alternate interior angles
Examples
1. Use the Alternate Interior Angle theorem to find the measure of angle x, angle y, and angle z. Assume the lines are parallel.
alternate interior angles example
First, we need to identify a pair of alternate interior angles. Angle y and 58 are on opposite sides of the transversal and inside the parallel lines, so they must be alternate interior. Since the lines are parallel, angle y = 58.
Next, notice that angle x and 58 form a straight angle. Since a straight angle measures 180 degrees, angle x + 58 = 180 and 180 – 58 = angle x, so angle x = 122.
Finally, angle x and angle z are alternate interior angles, and we know that alternate interior angles are equal. So, angle x = 122 then angle z = 122.
2. Use the Alternate Interior Angles theorem to first find x and y and then to find the measures of the angles. Assume the lines are parallel.
ala example 2
Looking for a Similar Assignment? Order now and Get 10% Discount! Use Coupon Code “Newclient”
The post Alternate Interior Angles: Definition, Theorem & Examples appeared first on Superb Professors.<|endoftext|>
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Let us remember and salute the visionary Nicolaus Copernicus on his birthday.
Born on February 19th, 1473, Copernicus gave our modern world the heliocentric theory of the solar system. He credited the ancient Greek astronomer and mathematician Aristarchus of Samos with originally describing how Earth and her sister planets orbit around the sun and took it upon himself to make the observations and work out the mathematics to prove it. Copernicus reintroduced the heliocentric theory so convincingly that it overcame the dominant earth-centered model preferred by the powerful Christian Church for theological reasons. His rigorous and clear reason simply could not accept the clumsy, assumption-laden model that Claudius Ptolemy had devised in the second century A.D. to explain why the planets did not behave as expected if the earth-centered model was accurate. Copernicus was a religious man, but he did not believe that his faith required him to believe something that his reason and his own eyes demonstrated was untrue.
For emphasizing the primacy of observation-driven reason over theology when it comes to describing and explaining the natural world, Copernicus is widely credited with starting the Scientific Revolution.
Here’s a short list of excellent resources to learn more about the great Nicolaus Copernicus:
Nicolaus Copernicus ~ by Sheila Rabin for The Stanford Encyclopedia of Philosophy
Nicolaus Copernicus ~ by J.J. O’Connor and E.F. Robertson for the MacTutor History of Mathematics Archive, School of Mathematics and Statistics University of St. Andrews, Scotland
Nicolaus Copernicus: Polish Astronomer ~ by Robert S. Westman for Encyclopædia Britannica
*A version of this piece was previously published at Ordinary Philosophy
~ Ordinary Philosophy is a labor of love and ad-free, supported by patrons and readers like you. Any support you can offer will be deeply appreciated!<|endoftext|>
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Teacher Roadmaps - Facilitating Class Discussions and Science Talk
Science Talk Allows Teachers and Mentors to Attend to Students’ Ideas
PlantingScience relies on science talk in the classroom – as a whole class, within a team – before and after online dialogs. Science talk is a powerful tool for the dynamic process of building and refining conceptual understandings. The art of teaching and mentoring for productive science talk is to attend to students’ thinking.
Science talk makes students’ ideas and thinking visible to the student him or herself, to their peers, to you as a teacher, and to the online mentor. In the moment and over the course of the conversation, students self-assess, get feedback, and are coaxed to dig deeper for meaning rather than a rote knowing. In what students say and do, the students’ life experiences become part of the conversation. Even though they may sometimes be contrary to “what science says,” using those experiences and reasoning about them can be a productive way to dig into central concepts.
Together the online and classroom conversations can help students think, reflect, and improve understanding. The more students hear ideas and reasoning of their peers and online mentors, the better they will become at reasoning and recognizing robust evidence and valid arguments. Using evidence to test claims and reasoning through counter-examples to examine alternatives are a significant part of scientific practice. Evaluating evidence is important for all students, not just those who become scientists, as these skills will help them to make informed decisions, personally and as citizens.
- What does that mean?
- How do you know?
- What if…? (present a counter-example)
- Does that mean…? (When conflicting ideas or pieces of evidence are presented)
- Bring the students’ ideas back to the class or team members: “What do you think of that?” followed by “What makes you agree/disagree?”
Tips for Class Discussion Ground Rules:
There are many ways to run productive discussions in classrooms. What works best will vary by the grade level of the students, their background and personalities, and the group dynamics of the class. However, in all contexts it is important to establish classroom norms that emphasize responsibility, respect, and the construction of arguments based on theory and evidence.
Classroom norms for behavior that lead to successful science discussion include:
- Taking and responding to questions
- Asking questions and participating
- Debating ideas, not people
- Respectfully listening to others for understanding
- Speaking loudly enough for the whole group to hear
- Providing agreement or disagreement (with reasoning) in response to other people’s ideas
- Using evidence to weigh competing claims
What You Can Do to Help Students Embrace Talking and Doing Science in Community
Give students the space to put their ideas out onto the floor.
- Take up students’ ideas by asking questions about them and the reasoning behind the ideas.
- Ask students questions that help students reflect, clarify, and go a bit deeper into their thinking.
- Challenge students’ often blind acceptance of a “science fact.”
- Juxtapose the ideas on the floor so that students have to reconcile inconsistencies.
- Establish both successes and failures in the science lab as valuable learning opportunities.
- Encourage students to see themselves as a member of a science community.
Using Juicy Questions to Spawn Science Talk
A juicy question is powerful if:
- it makes you go, “Ah, now that’s an interesting question.”
- you don’t already know the “right” answer.
- you don’t know what will happen once you start investigating it.
- you’ve played around with it in your mind and can’t find a good explanation.
“Juicy questions” are deceptively simple questions that might, on the surface, seem much too easy for your students. For example, “is air matter?” turns out to be a juicy question. By asking students questions that probe beneath the surface of a pat definition, students are forced to make sense of concepts and even at times connect them.
Planting and unpacking juicy questions can be a productive way to help both you and your students. It helps students assess their own understanding – authentic understanding – of how the natural and physical world works. Through artful facilitation by the teacher, the conversation also helps them to connect their observations of these worlds using intuitive sense-making strategies and evidence. Through these conversations, a teacher gains insight into students’ current understanding, helps students rely on innate reasoning skills and validates their life experiences, while building skills of scientific thinking.
Anticipate Some Unease With Uncertainty
For many students, especially the highest-achieving students, the uncertainty of inquiry is unsettling and it may go against most or all of their prior science experience. Students want to know what the “right” answer is so that they can memorize it, whether or not they understand why it is the “right” answer. To help your students become comfortable with the process of science and their active-learning role in it, resist the temptation to immediately give students the answer.
It’s important to monitor the conversations and projects closely to maintain a delicate balance between a productive discomfort with uncertainty that can happen before a mental breakthrough and too much discomfort that then causes students to give up and shut down. A critical hint or reference at the right time can keep the students on a path to building new understandings.
Facilitating Peer Communication and Review
Both scientist and peer mentoring interactions are possible on the PlantingScience platform. If you feel your students are prepared for peer mentoring, you may orchestrate student-to-student dialogs in addition to the scientist mentor matches made by the project staff. Prepare students with examples of helpful, thoughtful peer remarks. Expect some level of off-topic commenting, and be prepared to moderate if needed.
We recommend starting with cross-comments among teams in your class for students to practice giving and responding to constructive feedback with peers. For productive exchanges with students across schools, we recommend you collaborate with a teacher whose students are working on a similar topic and inform mentors. Project staff will gladly assist in setting this up. You may find Peer Communication and Review on the student page helpful.
Here is an example of a helpful student-to-student peer review:
“Hi I’m Lacey from Team Germinators (Southwest High School). Our teacher asked our team to find another high school project similar to ours and review your project. Our project is about which seeds will germinate in cold temperatures, so we thought it would be neat to post on your project about what temperature water is best for radish germination. We think you did a really good job making sure that you controlled your variables and measured that the water was the same temperature every time you watered the seeds. We like your graphs and it’s easy to see that the seeds watered with the room temperature water did best.
Here’s what we think you could improve. You say that room temperature water is best for radish germination. But you only tested three temperatures of water…really hot, room temperature, and really cold. Why did you pick those temperatures? We think gardeners and farmers would be most interested to know if water that’s only a little warmer or a little cooler than room temperature is better for plants. It would be a lot of work to make water for irrigating fields really hot or really cold. But it would be helpful to know if cold water from a hose is worse or better than water that’s been sitting in a bucket in the sun.
We don’t think you can say that room temperature water is BEST for radish germination just from your experiment. We think you might want to change your conclusion to say that room temperature water is BETTER for radish germination than really hot or really cold water. Maybe a little warmer or a little cooler water is even better than room temperature water, but those weren’t included so you don’t know. It could be a future experiment…”
Science Talk Resources and References
The Inquiry Project – Talk Science Professional Development
This site has extensive videos to see talk in the classroom in action and resources about talk moves.
Tools for Ambitious Science Teaching – Discourse Tools
This site has strategies for eliciting ideas and pressing for explanations along with case studies.
Talk Science – contemporary science discussion for the classroom
This site from the United Kingdom has straight-forward tips for setting the stage in the classroom and using particular techniques such as Socratic seminars.
Levin, D.M., T., Gran, and D. Hammer. 2012. Attending and responding to student thinking in science. The American Biology Teacher 74(3):158-162.
Llewellyn, D. and H. Rajesh. 2011. Fostering Argumentation Skills: Doing What Real Scientists Really Do. Science Scope 35(1):22-28.
McNeill, K.L. and J. Krajeik. 2012. Supporting Grade 5-8 Students in Constructing Explanations in Science: The Claim, Evidence, and Reasoning Framework for Talk and Writing. Pearson.
Michaels, S., A.W. Shouse, and H.A. Schweingruber (2008) Ready, Set, SCIENCE: Putting Research to Work in K-8 Science Classrooms. National Academies Press, Washington D.C.
Osborne, J.F. 2010. Arguing to Learn in Science. Science 328: 463-466.
Sampson, V., J. Grooms, and J. Walker. 2009. Argument-Driven Inquiry: A way to promote learning during laboratory activities. The Science Teacher 76(8): 42-47.<|endoftext|>
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Building on the Past
Gould Memorial Library, Bronx Community College, 1901
Lesson Plan by:
Sember Weinman, Education Coordinator
Lehman College Art Gallery
Stanford White (1853-1906) of McKim, Mead, and White
High School Global History and Geography
Unit One: C:3:b: The Roman Republic—Contributions
Students compare the architectural sites of the Gould Memorial Library with its inspiration, the Pantheon in Rome, in order to gain insight into the civilizations in which these structures were built.
Neo-Classical: A style of art that was popular in the 19th Century that was a reaction to Baroque Art. This style was derived from the art and culture of ancient Greece and Rome and imitated this period’s architecture and fascination for order and simplicity.
Beaux Arts: A style of art that borrowed heavily from a range of architectural styles of the past, including Imperial Roman architecture, Italian Renaissance, French and Italian Baroque models, even French late Gothic. Influenced by the Ecole des Beaux-Arts in Paris, American architects of the Beaux-Arts generation often returned to Greek models, which had a strong local history in the American Greek Revival of the early 19th century. Beaux-Arts style attempts an approach to a regenerated spirit within the grand traditions rather than a set of motifs.
Dome: A hemispherical or beehive-shaped vault or ceiling over a circular opening. May be elevated further by placement on a drum. If placed over a square opening, the transition to a round shape is made by use of pendentives in the four corners.
Coffer: A square, rectangular, or polygonal recess in a ceiling to reduce the weight of the structure.
Column: An upright post, bearing the load of the upper part of a building. It consists of a base, a shaft, and a capital. An engaged column is half a column, attached to a wall, and non-weight bearing.
Colonnade: A series of columns, usually supporting lintels or arches.
Corinthial Column: slender fluted column and an elaborate capital decorated with acanthus leaves and scrolls
Capital: The top element of a column, pier or pilaster, usually ornamented with stylized leaves, volutes, animal or human forms. The entablature rests on capitals in Doric, Ionic, and Corinthian orders.
Oculus: A circular opening in a wall or in the top of a dome, from the Latin word for “eye”.
Pantheon: Greek for “all the gods”; therefore, a temple dedicated to all the gods. Specifically, the round-domed building built in Rome in 25A.D.
Pilaster: A rectangular engaged column, sometimes decorative, but sometimes used to buttress a wall.
Portico: A porch with a roof, supported by columns and usually having an entablature and a pediment.
Gould Memorial Library columns
The Gould Memorial Library, originally built for New York University, is an important resource for studying the ideals and values of ancient Rome and the time period in which the Pantheon was built. Since the library is based on the Pantheon in Rome but was actually built at the turn of the 20th century, it serves as a tool to compare the values and ideas of ancient Rome with the interpretations that the architects of the Gilded Age attempted to revive. A premier architect of his time, Stanford White looked to the architecture of the past for elements of classical style, such as the use of perfect symmetry, Roman columns, mosaic tile floors, and a domed ceiling. Although the original Pantheon was built as a temple for the gods and White’s building is a library, he probably intended his creation to conjure classical ideals such as democracy and acceptance of different ideas.
The exterior of the library is yellow brick with limestone pilasters. Inside, the floor plan of the Gould Memorial Library is in the shape of a Greek cross, with a soaring dome overhead. 16 Corinthian columns made of rare green Connemara marble circle the rotunda. Tiffany stained glass windows infuse the interior with a multicolored glow from the area above the mezzanine and in the vestibule. Viewers look up to see statues of the Greek Muses placed around the balcony. Italian mosaic tiles in classical patterns line the floors.
The arcade outside the library houses the Hall of Fame for Great Americans, an open-air colonnade with space for 102 bronze portrait busts of authors, educators, theologians, physicians, lawyers, artists, musicians, actors, scientists, military leaders, social progressives, philosophers, and statesmen. Among those portrayed are Thomas Edison, Booker T. Washington, Harriet Beecher Stowe, Abraham Lincoln, Jane Addams, and Mark Twain. Carved in stone on pediments of The Hall of Fame are the words “By Wealth of thought, or else by mighty deed, They served mankind in noble character. In worldwide good they live forever more.” The Hall of Fame thus connects the achievements of Americans to human accomplishments on a grand historical scale; while commemorating accomplished people in the same way that the Romans celebrated their gods.
The Gould Memorial Library was originally given to New York University by the daughter of Jay Gould, in his memory. He was one of the most well known financiers and speculators of the Gilded Age. New York University turned the library over to Bronx Community College in 1973.
The Architect firm McKim, Mead, and White designed some of the most renowned structures of the time. In addition to the Gould Memorial Library, Stanford White designed the second Madison Square Garden, The North and South Wings of the Metropolitan Museum of Art, The New York Herald Building, and the Washington Square Arch. By using the look of classical architecture for new uses, the architects built into their buildings a grandeur that aimed to infuse chaotic American cities with a sense of order and formality.
The Pantheon is a Roman building that was originally built as a temple to the seven deities of the seven planets in the state religion of Ancient Rome. It has been converted to a Christian church since the 7th century, and has been in continuous use throughout history.
The building is circular with a portico of three ranks of huge granite Corinthian columns (eight in the first rank and two groups of four behind) under a pediment opening into the rotunda, under a coffered, concrete dome, with a central opening called an oculus, or Great Eye, open to the sky. The weight of the dome is concentrated on a ring of voussoirs, which form the oculus. The dome is thin on top and thickens as it meets the walls. A rectangular structure links the portico with the rotunda. The walls of the building are over 20 feet thick so that they can support the tremendous weight of the dome. The height to the oculus and the diameter of the interior circle are the same (144 feet), so the dome would fit perfectly within a cube—or the interior could house a sphere 144 feet in diameter. The Pantheon has the largest surviving dome from antiquity.
Interior of the Pantheon, Rome by Giovanni Paolo Pannini
The interior of the roof was probably intended to symbolize the arched vault of the heavens. The oculus, 27 feet across at the dome’s apex, is the source of all light and is symbolic of the sun. The interior features coffers, which serve to reduce the weight of the dome.
The original Pantheon was built in 27 B.C. to 25 B.C. under the Roman Empire, during the third consulship of Marcus Vipsanius Agrippa, whose name is inscribed on the portico of the building. Agrippa’s Pantheon was destroyed by fire in 80 A.D. It was rebuilt around 125 A.D., during the reign of the Emperor Hadrian. Hadrian was a cosmopolitan emperor who traveled widely in the east and was a great admirer of Greek culture. He may have intended the Pantheon, a temple to all the gods, to be an acknowledgement to varying beliefs and to promote unity among different churches for the subjects of the Roman Empire who did not worship the old gods of Rome—or who worshipped them under different names.
As the best-preserved monument of Roman architecture, the Pantheon has been enormously influential on European and American architects from the Renaissance through to the 19th Century. Included among buildings inspired by the Pantheon is the Gould Memorial Library housed at the Bronx Community College.
Why use architecture? What can it teach us?
Art is part of the complex structure of beliefs and rituals, social and political systems, and the stories of every human society. In architecture, the structure, function, and purpose of a building is fused seamlessly with the aesthetic, social, and political preferences of a particular time and place. The purpose for a building’s use determines its size, its shape, and the materials chosen. The value that a society places on that purpose is reflected in the decoration, the scale, and the amount of time taken to create a building. Architects choose to borrow styles from the past as their interpretation of a particular philosophy or ideal. Buildings embody cultural values in every tile or stone.
This unit enables students to place architecture within a historical context. They will research and create models of the Pantheon in Rome. They will then visit the Gould Memorial library. While they are there, they will use visual literacy techniques to record what they observe. When they return to the classroom, the students will compare and contrast the Gould Memorial Library with its inspiration, the Pantheon in Rome. This will allow the students to sift out social, political, and cultural similarities and differences between ancient Rome and the Gilded Age in New York. Ultimately, the students will be able to bolster their studies of ancient Rome with actual experience, while developing their critical thinking skills by comparing the buildings and considering the cultural atmosphere within which each building was created.
Five class periods plus research that will be conducted as homework.
Students will begin this unit through their study of the social, cultural, and political climate of ancient Rome.
Students will be able to determine why the Pantheon was built, why it looks the way that it does, how it was funded, and what purpose it served in ancient Rome.
Do Now: Students respond to the following question, written on the board prior to class: “What can architecture teach us about the society in which it was built?”
Packet with description and pictures of the Pantheon with interior and exterior views.
Images of Pantheon in cross section and spherical mathematical modeling
© Gert Sperling, “The Quadivium in the Pantheon of Rome”, Fuldatal, Germany
After the students complete their reading on Ancient Rome in class, introduce the vocabulary necessary for them to conduct research on the site by creating a powerpoint presentation that gives examples of the vocabulary terms. See sources for a list of websites that reference the Pantheon. A photocopied set for students to take notes and make sketches can be used as a glossary of terms. They are then given a short reading on the Pantheon in Rome, taken from Historical Precedent above, and work in groups to answer a set of related questions. These may include:
1. What does Pantheon mean? Why do you think Emperor Hadrian had a building
created for this purpose? Who was it intended for?
2. What did the Pantheon borrow from Greek architecture and how is it different than
most Greek buildings?
3. How did Roman culture differ from Greek society? Can you draw any conclusions by
looking at the Pantheon?
4. What are some distinctive features to the structure of the Pantheon—why do you
think it was built this way?
5. Have you seen these aspects of Roman architecture in other buildings? Which ones?
What, if anything, have they come to symbolize?
6. There was a movement of American Architects from the time of Thomas Jefferson
through the turn of the Twentieth Century that wished to revive Greek and Roman
architecture in American buildings. What principles in classical culture and politics do
you think the American architects were interested in bringing to life?
Conduct research to gain structural information about the Pantheon. Find a floor plan of the Pantheon on the internet. Determine the dimensions of the dome, the oculus, and the portico. Try to find out the materials used, how thick the walls are, and what they did to alleviate some of the weight in the dome. Alternatively, this information could be provided to students to read for homework, if they do not have access to the internet.
Students continue to work in groups and begin to problem solve ways to build a model of the Roman Pantheon.
Students will be able to identify the challenges that were faced and accomplishments that were achieved in the building of the Pantheon by building their own scale model.
Students respond to the following questions, written on the board prior to class: “If you had been the architect for the Pantheon, what would you have done differently? Why?
Foam core, Styrofoam, oak tag, scissors, tape, paper towel rolls, cardboard, rulers
Using the materials given, students are asked to work in groups of four to problem solve ways to create scale models of the Pantheon.
One student from each group can study each of the following topics (if students have access to internet they can research online, otherwise information on each topic can be provided):
• Conduct research on the Gould Memorial Library and the Hall of Fame for Great
Americans at the Bronx Community College Campus.
• Look up Stanford White. When did he practice architecture? What architectural
style is he known for? Why was he interested in this style? Identify several buildings
that he has designed.
• What was the Beaux Arts Movement? How did it come about? What did architects
involved with this style borrow from Rome? Why?
• The Gould Memorial Library was built in the Gilded Age. How is this age
characterized? What was happening socially, culturally, and politically?
Samples of finished model, front and top view
Students continue to work in groups on their models of the Pantheon. At the end of class, students share what they had learned from their homework research.
Students will be able to begin to put the Pantheon into a broader context of history through studying the way that architects later in history have borrowed from it.
Students respond to the following questions, written on the board prior to class: “Can you think of a building that you have seen, and or been inside of, that has had an impact on you? What type of building was it? Why do you think it had an effect on you? What does this tell us about the Pantheon? Would it be effective to us?”
Students continue to work in groups on their models of the Pantheon. At the end of class, students share what they had learned from their homework research.
Ask students to make predictions on what they will see when they visit the Gould Memorial Library.
Students visit the Gould Memorial Library at the Bronx Community College Campus.
Students will be able to experience the space of the library and will begin to compare their observations with their study of the Pantheon.
Each student brings a pencil and an unlined notepad or a clipboard with sheets of printer paper.
Activity:Students take a tour of the Gould Memorial Library. While on the tour, they take notes. After the tour, the students are required to create one interior space drawing, one exterior space drawing, and one detail drawing.
Gould Memorial Library interior detail
Students work together in groups to answer a series of questions that are intended to link their experience at the library with their study of the Roman Pantheon.
Objective:Students will be able to compare and contrast the structure and purpose of the Roman Pantheon with the Gould Memorial Library.
Materials:Students will use their own notes from the field trip, their models of the Pantheon, and notes from their research.
Do Now:Students respond to the following question, written on the board prior to class: “If you were to revive one aspect of Roman civilization, what would it be and why?”
Students work in groups to answer the following questions:
1. Compare and contrast the two buildings. What differences and/or similarities are
there in scale? Decoration? Materials? Building technique? Purpose? Impact on
2. Consider the purpose for the buildings. Why would an architect build a library that
is modeled after a temple? Are there guiding principles in common between the two
3. Why would an American Architect be interested in creating a building with Roman
style? What is he saying to his own culture by borrowing from the past?
4. What do you think it would be like to be an immigrant living in the Bronx at the turn
of the century and see the Gould Memorial Library? How would that compare to a
Greek person living within the Roman Empire experiencing the Pantheon for the first
5. Stanford White was an architect working in the Gilded era. Compare the social and
political climate of the Roman Empire with the turn of the twentieth century in
Homework:Take your answer to one of these questions and turn it into an argumentative essay. Use references and cite them.
At the end of the workshop, students turn in their Pantheon models, the answers to the questions from sessions one and five, their notes and drawings, and their final papers. Develop a rubric that accounts for the different aspects of the project.
1. Students can create models of the Gould Memorial Library, paying attention to the
relative scale, dimensions and décor of the two buildings.
2. Students can present oral reports of their findings to the class.
3. Students can design their own library that uses aspects from either of the buildings
studied. This may include floor plans, sketches, and models.
4. Students can study the various ways in which the Pantheon has been used since it
5. Students can explore the cultural biases implicit in both buildings. What class of
people were these buildings intended for—if any in particular? Were these buildings
designed to make their visitors feel empowered, small, or something else?
6. Students can read the New York Times article “Regilding a Stanford White Landmark,”
by Karen Arenson. They can then write a page on whether or not they think it is
important to restore the building and why.
S.S. Standard 1—History of the United States and New York: Students will learn about Roman Society through a visit to the Gould Library in the Bronx. They will compare the historical, social, and intellectual structures in both classical Rome and the Gilded Age in the Bronx, NY at the turn of the 20th Century.
S.S. Standard 2—World History: Students will develop an understanding of the ideas, social and cultural values, beliefs, and history leading to the Roman Republic.
S.S. Standard 3—Geography: Students will study the physical and social environments and human systems of ancient Rome.
S.S. Standard 4—Economics: Students will consider the economic structures and controls in place in both ancient Rome and the Gilded Age.
S.S. Standard 5—Civics, Citizenship, and Government: Students will learn about the reigning political systems of both ancient Rome and the United States at the turn of the 20th Century; and the different assumptions on power, authority, governance, and law held by the people living during both of these times.
Blueprint Strand 1—Arts Making: Students create drawings and models of the Pantheon in Rome, and sketches of the Gould Memorial Library while on site.
Blueprint Strand 2—Literacy In the Arts: Students become familiar with the vocabulary and important considerations implicit in architecture.
Blueprint Strand 3—Making Connections: Students apply their understanding of both Roman and neo-classical architecture to interpret the history within which these structures were created.
Blueprint Strand 4—Community and Cultural Resources: Students take advantage of their access to Bronx Architectural Sites to better understand the purpose and power of architecture both historically and artistically.
Blueprint Strand 5—Careers and Lifelong Learning: Students will obtain exposure to architecture and art history as potential career paths.
RESOURCES FOR RESEARCH:
American Memory Project, U.S. Library of Congress
Bronx Community College, Hall of Fame for Great Americans
Italy Guides, Virtual Panorama and Photo Gallery of Pantheon
Lehman College Art Gallery Public Art in the Bronx
New York City Architecture
Pantheon, Article in Platner’s Topographical Dictionary of Ancient Rome
Pantheon, Virtual Tour
“Regilding a Stanford White Landmark,” by Karen Arenson, The New York Times, 7/30/04, www.gothamgazette.com/community/14/news/666
Metropolitan Museum of Art, Timeline of Art History
Honour, H. and Fleming, J. The Visual Arts: A History, New Jersey: Pearson Education, Inc., 2005.
Salvadori, M., Hooker, S. and Ragus, C. The Art of Construction: Projects and Principals for Beginning Engineers and Architects, Chicago: Chicago Review Press, 1990.
White, N. and Willensky, E. AIA Guide to New York City: The Classic Guide to New York’s Architecture (4th Edition), NY: Three Rivers Press, 2000.<|endoftext|>
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lithosphere lĭth´əsfēr˝ [key], brittle uppermost shell of the earth, broken into a number of tectonic plates. The lithosphere consists of the heavy oceanic and lighter continental crusts, and the uppermost portion of the mantle . The crust and mantle are separated by the Moho or Mohorovicic discontinuity (see earth and seismology ). The thickness of the lithosphere varies from to around 1 mi (1.6 km) at the mid-ocean ridges to approximately 80 mi (130 km) beneath older oceanic crust. The thickness of the continental lithospheric plates is probably around 185 mi (300 km) but is uncertain due to the irregular presence of the Moho discontinuity. The lithosphere rests on a soft layer called the asthenosphere , over which the plates of the lithosphere glide. See plate tectonics .
The Columbia Electronic Encyclopedia, 6th ed. Copyright © 2012, Columbia University Press. All rights reserved.
See more Encyclopedia articles on: Geology and Oceanography<|endoftext|>
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A gigantic cloud of hydrogen evaporating from a planet orbiting a nearby star has been discovered with the Hubble Space Telescope by astronomers.
Dubbed “The Behemoth”, the massive, comet-like object is around 50 times the size of it’s parent star. The hydrogen is venting from a warm, Neptune-sized planet, because extreme radiation from the star.
Study leader, David Ehrenreich of the Observatory of the University of Geneva in Switzerland, explains:
“This cloud is very spectacular, though the evaporation rate does not threaten the planet right now. But we know that in the past, the star, which is a faint red dwarf, was more active. This means that the planet evaporated faster during its first billion years of existence because of the strong radiation from the young star. Overall, we estimate that it may have lost up to 10 percent of its atmosphere over the past several billion years.”
Due to its size and because it is much closer to its star than Neptune is to our sun, the planet, named GJ 436b, is considered to be a “Warm Neptune.”. Even though it is in no danger of having its atmosphere totally evaporated and stripped down to a rocky core, this planet could explain the existence of so-called Hot Super-Earths that are very close to their stars.
Such a phenomenon has never been observed around an exoplanet so small.
It may give clues to how other planets with hydrogen-enveloped atmospheres could have their outer layers evaporated by their parent star, leaving behind solid, rocky cores. Hot, rocky planets such as these that roughly the size of Earth are known as Hot-Super Earths.
Hot Super-Earths could be the remnants of more massive planets that completely lost their thick, gaseous atmospheres to the same type of evaporation. These hot, rocky worlds were revealed by the Convection Rotation and Planetary Transits (CoRoT) and NASA’s Kepler space telescope.
Sice Earth’s atmosphere blocks most ultraviolet light, astronomers needed a space telescope with Hubble’s ultraviolet capability and exquisite precision to find the Behemoth.
“You would have to have Hubble’s eyes,” says Ehrenreich. “You would not see it in visible wavelengths. But when you turn the ultraviolet eye of Hubble onto the system, it’s really kind of a transformation, because the planet turns into a monstrous thing.”
Because the planet’s orbit is tilted nearly edge-on to our view from Earth, the planet can be seen passing in front of its star. Astronomers also saw the star eclipsed by “The Behemoth” hydrogen cloud around the planet.
David Ehrenreich, Vincent Bourrier, Peter J. Wheatley, Alain Lecavelier des Etangs, Guillaume Hébrard, Stéphane Udry, Xavier Bonfils, Xavier Delfosse, Jean-Michel Désert, David K. Sing & Alfred Vidal-Madjar
A giant comet-like cloud of hydrogen escaping the warm Neptune-mass exoplanet GJ 436b
Nature 522, 459–461 (25 June 2015) doi:10.1038/nature14501
Illustration: artist’s concept shows “The Behemoth,” an enormous comet-like cloud of hydrogen bleeding off of a warm, Neptune-sized planet just 30 light-years from Earth. Also depicted is the parent star, which is a faint red dwarf named GJ 436. The hydrogen is evaporating from the planet due to extreme radiation from the star. A phenomenon this large has never before been seen around any exoplanet. Credit: NASA, ESA, and G. Bacon (STScI)<|endoftext|>
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As both were species of hominin, it would have been easier for pathogens to jump populations, researchers said.
Scientists from universities of Cambridge and Oxford Brookes in the UK have reviewed the latest evidence gleaned from pathogen genomes and DNA from ancient bones, and concluded that some infectious diseases are likely to be many thousands of years older than previously believed.
There is evidence that our ancestors interbred with Neanderthals and exchanged genes associated with disease.
There is also evidence that viruses moved into humans from other hominins while still in Africa.
So, researchers argue that it makes sense to assume that humans could, in turn, pass disease to Neanderthals, and that - if we were mating with them - we probably did.
Many of the infections likely to have passed from humans to Neanderthals - such as tapeworm, tuberculosis, stomach ulcers and types of herpes - are chronic diseases that would have weakened the hunter-gathering Neanderthals, making them less fit and able to find food, which could have catalysed extinction of the species, researchers said.
"Humans migrating out of Africa would have been a significant reservoir of tropical diseases," said Charlotte Houldcroft from Cambridge University.
"For the Neanderthal population of Eurasia, adapted to that geographical infectious disease environment, exposure to new pathogens carried out of Africa may have been catastrophic," said Houldcroft.
New techniques developed in the last few years mean researchers can now look into the distant past of modern disease by unravelling its genetic code, as well as extracting DNA from fossils of some of our earliest ancestors to detect traces of disease.
Genetic data shows many infectious diseases have been "co-evolving with humans and our ancestors for tens of thousands to millions of years," researchers said.
The longstanding view of infectious disease is that it exploded with the dawning of agriculture some 8,000 years ago, as increasingly dense and sedentary human populations coexisted with livestock, creating a perfect storm for disease to spread, researchers said.
They say that many diseases traditionally thought to be 'zoonoses', transferred from herd animals into humans, such as tuberculosis, were actually transmitted into the livestock by humans in the first place.
"Hunter-gatherers lived in small foraging groups. Neanderthals lived in groups of between 15-30 members, for example. So disease would have broken out sporadically, but have been unable to spread very far," said Houldcroft.
"Once agriculture came along, these diseases had the perfect conditions to explode, but they were already around," she said.
The findings were published in the American Journal of Physical Anthropology.
Election Results for Lok Sabha Election 2019 will be out on May 23. Get the latest election news and live updates on ndtv.com/elections. Catch all the action on NDTV Live. Like us on Facebook or follow us on Twitter and Instagram for news updates from each of the 543 parliamentary seats for the election 2019<|endoftext|>
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Vail Curious Nature column: Delayed implantation of embryos is natural marvel
The mysteries of nature are numerous and vast, encompassing questions large and small and constantly teasing us to unravel their riddles. One of these mysteries involves the phenomenon of delayed implantation of embryos. If you’ve never heard of this, then you’re probably not alone. In a nutshell, delayed implantation allows a species to delay implantation of the fertilized embryo, sometimes for months, so that the young is born with a better chance of survival.
It sounds pretty amazing, and it is. A fertilized embryo begins its development through cellular division. First, there is one fertilized cell, which divides into two, then four, then eight, and so on, until it forms a hollow ball known as the blastocyst.
The Timing is right
In a typical mammalian pregnancy, the blastocyst implants into the wall of the uterus, where it turns in upon itself and begins differentiating into the many specialized cells that make up the organism.
In the case of species that use delayed implantation, however, the blastocyst simply floats around the uterus for a while, in a state of suspended animation, until cued to implant and continue development by some mysterious, universal calling.
This process wouldn’t be so unexplainable if it only occurred in a few, closely related species, but it actually occurs in a diverse range of species, including mice, seals, weasels, kangaroos, bats, armadillos, red pandas and bears, among others. How did the same exact process occur in such an evolutionarily diverse group of organisms?
Even more confounding is the fact that closely related species have different implantation habits, such as the long-tailed weasel, which delays implantation for months, while its close cousin, the least weasel, experiences no delay. Factors such as these lead researchers to believe that delayed implantation probably developed independently along a variety of evolutionary pathways and in different geologic times.
Locally, black bears are the most famous species that undergo delayed implantation, with mating usually occurring in June, but the embryos don’t implant until sometime in November. Once implanted, the embryos develop in more than two months, and the young are born in the secret confines of their mama’s den sometime in January.
Scientists generally agree that this type of delayed development confers an advantage on the species, allowing the young to be born at a time that is most advantageous for their survival. In the case of bears, being born in the security of the den protects them during their most vulnerable days, allowing them to gain weight and develop before venturing out into the world.
A frustrating Process
Research in the area of delayed implantation has been a somewhat frustrating process, partly because of the diverse range of species involved and partly because it appears that the mechanisms may differ from species to species. Researchers have found certain cues that can enable them to get an entire room full of mice pregnant at the same time, but these tricks become old fast, and scientists seem stuck at determining the specific mechanisms responsible.
But in terms of unraveling the mysteries of the natural world, this phenomenon is thought to have great potential, particularly in the race to fight cancer. Cancer cells are characterized by rapid cell division, and delayed implantation essentially stops, or pauses, cell division. A mechanism that can stop uncontrolled cell division could offer incredible potential if we were able to tap it.
But discoveries such as this one are still a long way off in our never-ending quest to unravel the mysteries of the universe. In the meantime, we can simply appreciate that we are here, right now, able to marvel at this type of mystery.
Jaymee Squires is the director of graduate programs at Walking Mountains Science Center in Avon. Bears are her favorite animals, and she is endlessly fascinated by their unique biology and ecology.
The jury was out just 12 minutes before returning a not-guilty verdict, and another of Artie Loredo’s trials was behind him.<|endoftext|>
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The fossil was originally found on Spitzbergen, the largest island of Svalbard, in 1962. Fifty years later it was rediscovered amongst uncatalogued material in the storage shelves of the University of Oslo’s Natural History Museum.
Palaeontologist Jørn Hurum came across the dusty fossil in 2011 down in the storage shelves of the Natural History Museum in Oslo. It had originally been found on Spitzbergen by the geologist Jenö Nagy in 1962.
Now, five years on, Hurum and a group of scientists have published a study showing this to be a remnant of a small bird or bird-like dinosaur that lived about 113 to 100 million years ago.
The small fossil, from four to five centimetres long, was imbedded in rock and somewhat crushed. It would have been truly difficult to investigate such a fragile artefact five decades ago.
But thanks to computer tomography (CT) scanning technology, it could now be examined.
“We could see that this fossil was a bone, more specifically a femur,” says Aubrey Roberts, a research fellow from the University of Southhampton.
She is one of the scientists behind the new study.
“The bone probably belonged to a bird, or a dinosaur closely related to birds,” explains Roberts.
“We base that conclusion on several factors, but especially because the bone has very thin walls, a bird trait, as they have very hollow bones,” she continues.
In the trees amongst herbivores
The little creature was probably about the size of a common starling.
It would have lived among plant-eating dinosaurs in a temperate environment warmer than Norway’s arctic Svalbard Archipelago as we know it today.
When the creature was alive, in the Lower Cretaceous period, Svalbard was further south, but still at a latitude about 500 km north of today’s Oslo.
The landscape was lush, with conifers and gingko trees in a boggy landscape.
This particular creature appears to have fallen into water when it died. The femur became felicitously protected in the rock beneath a fossil mollusc shell.
This is probably why the fragile bone survived so well.
This is the first fossil vertebrate from this geological period, 113 to 100 million years ago, found in Svalbard or elsewhere in the Arctic. That makes it unique, but by no means a complete surprise.
“Footprints of dinosaurs from this period have been found earlier in Svalbard,” explains Roberts.
The tracks Roberts refers to on Spitzbergen were from herbaceous dinosaurs, ornithopods. These dinosaurs had three-toed feet, like those of birds.
Until now, only these footprints have given evidence of land species that roamed Svalbard in the early Cretaceous. Bones have been lacking.
But here was a fossil, literally right beneath the feet of palaeontologists working in the museum in Oslo.
I am the walrus
Oslo’s Natural History Museum is by far not the only one to have discovered hidden treasures in storage rooms. Scientists at the Norwegian University of Science and Technology recently found a walrus bone from Svalbard that turned out to be 6,600 years old, much older than believed.
Roberts expects there could be other hidden treasures amongst the collections of the museum.
“This could be. Now that the Geology Museum is moving we might discover more new things. This actually happens fairly often, new fossils turn up in museum collections,” says Roberts.
“Museum collections are often gigantic and artefacts can be lying about in storage and untouched a long time,” she explains.
Three million uncatalogued objects
Roberts is right about that. A professor at the Natural History Museum and a co-author of the study, Hans Arne Nakrem, says the museum’s collections contain about six million objects. And there are good reasons for using the qualifier “about”.
“These numbers come and go because an object with us here can be a bag containing from ten to a thousand tiny fossils. So we could count it as one object or one thousand,” he says.
Allowing a wide margin for such uncertainties, Nakrem estimates that about 2.8 million objects are covered by a database. But not everything the museum has in its possession is among these.
“The museums have been chronically understaffed the 30 to 40 years. We haven’t always had people who could catalogue objects properly,” he points out.
In addition, some of the collections are several hundred years old. The collectors of yesteryear were not always astute at describing or labelling what they had found.
“Some crates might be labelled simply ‘North America’, or perhaps, ‘Northern Norway:1910’,” he explains.
Nakrem thinks the museum will eventually get these collections sorted out.
“The quality of the collections is improving. Instead of hanging on to everything that is poorly labelled, the objects are thrown out, at least when they don’t have any scientific relevance,” explains Nakrem.
Back to the spot
But the fossil in question is one exception. Thanks to the geologist Nagy’s meticulous notes, the researchers were able to return to the area where the bone was found.
The scientists hoped to find more fossil bones from the same species.
They did not. But Roberts is still hopeful.
“It’s possible that there are more out there and future expeditions will certainly be looking for birds and dinosaurs. It would be fabulous to find some real huge, terrestrial dinosaurs here in Norway!” she says.<|endoftext|>
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# Cartesian product
The Cartesian product of the two sets and${\ displaystyle A \ times B}$${\ displaystyle A = \ {x, y, z \}}$${\ displaystyle B = \ {1,2,3 \}}$
The Cartesian product or quantity of product is in the set theory a basic structure from given amounts to generate a new set. The ambiguous term “ cross product ” is occasionally used for the Cartesian product . The Cartesian product of two sets is the set of all ordered pairs of elements of the two sets, where the first component is an element of the first set and the second component is an element of the second set. More generally, the Cartesian product of several sets consists of the set of all tuples of elements of the sets, the sequence of the sets and thus of the corresponding elements being fixed. The result set of the Cartesian product is also product quantity , Cross quantity or amount of compound called. The Cartesian product is named after the French mathematician René Descartes , who used it to describe the Cartesian coordinate system and thus established analytic geometry .
## Product of two sets
### definition
The Cartesian product (read "A cross B") of two sets and is defined as the set of all ordered pairs , where one element is out and one element is out. Each element is made with each item of combined. Formally, the Cartesian product is through ${\ displaystyle A \ times B}$ ${\ displaystyle A}$${\ displaystyle B}$ ${\ displaystyle (a, b)}$${\ displaystyle a}$${\ displaystyle A}$${\ displaystyle b}$${\ displaystyle B}$${\ displaystyle A}$${\ displaystyle B}$
${\ displaystyle A \ times B: = \ left \ {(a, b) \ mid a \ in A, b \ in B \ right \}}$
Are defined. In particular, it is also possible to form the Cartesian product of a set with itself and then write
${\ displaystyle A ^ {2} = A \ times A = \ left \ {(a, a ') \ mid a, a' \ in A \ right \}}$.
Occasionally the term “cross product” is also used for the Cartesian product, but this has other meanings, see cross product .
The above definition is any problem (real) on classes and extensible. In particular, pairs are only formed for elements of and ; these cannot be real classes and have no special requirements for pair formation. ${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle A}$${\ displaystyle B}$
### Examples
The Cartesian product of two sets consists of all possible ordered pairs of elements of the sets.
The Cartesian product of the two sets and is ${\ displaystyle A \ times B}$${\ displaystyle A = \ {a, b, c \}}$${\ displaystyle B = \ {x, y \}}$
${\ displaystyle A \ times B = \ left \ {(a, x), (a, y), (b, x), (b, y), (c, x), (c, y) \ right \ }}$.
The Cartesian product , on the other hand, is a different set, namely ${\ displaystyle B \ times A}$
${\ displaystyle B \ times A = \ left \ {(x, a), (x, b), (x, c), (y, a), (y, b), (y, c) \ right \ }}$,
because the order of the elements plays a role in ordered pairs. The Cartesian product of with itself is ${\ displaystyle A}$
${\ displaystyle A \ times A = \ left \ {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c , a), (c, b), (c, c) \ right \}}$.
The real number level arises from the Cartesian product of the real numbers with themselves: ${\ displaystyle \ mathbb {R}}$
${\ displaystyle \ mathbb {R} \ times \ mathbb {R} = \ mathbb {R} ^ {2} = \ {(x, y) \ mid x, y \ in \ mathbb {R} \}}$.
The tuples are also called Cartesian coordinates . The Cartesian product of two real intervals and gives the rectangle${\ displaystyle (x, y)}$ ${\ displaystyle [a, b]}$${\ displaystyle [c, d]}$
${\ displaystyle [a, b] \ times [c, d] = \ {(x, y) \ in \ mathbb {R} ^ {2} \ mid a \ leq x \ leq b, c \ leq y \ leq d \}}$.
### properties
#### Number of elements
a b c d e f G H 8th 8th 7th 7th 6th 6th 5 5 4th 4th 3 3 2 2 1 1 a b c d e f G H
A chessboard has squares identified by a pair of letters from the line and number from the row. ${\ displaystyle 8 ^ {2} = 64}$
If the sets and are finite , then their Cartesian product is a finite set of ordered pairs. The number of pairs corresponds to the product of the numbers of the elements of the initial sets, that is ${\ displaystyle A}$${\ displaystyle B}$ ${\ displaystyle A \ times B}$
${\ displaystyle | A \ times B | = | A | \ cdot | B |}$
In the special case that is, applies ${\ displaystyle A = B}$
${\ displaystyle | A ^ {2} | = | A | ^ {2}}$.
If at least one of the two sets and contains an infinite number of elements and the other is not empty, then its Cartesian product consists of an infinite number of pairs. The Cartesian product of two countably infinite sets is also countable according to Cantor's first diagonal argument . If at least one of the two sets is uncountable , then its product set is also uncountable. ${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle A \ times B}$
#### Empty set
Since no element can be selected from the empty set , the Cartesian product of the empty set with any set results in the empty set. More generally applies
${\ displaystyle A \ times B = \ emptyset ~ \ Longleftrightarrow ~ A = \ emptyset ~ {\ text {or}} ~ B = \ emptyset}$,
that is, the Cartesian product of two sets is empty if and only if at least one of the two sets is empty.
#### Non-commutativity
The Cartesian product is not commutative , that is, for nonempty sets and with is ${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle A \ neq B}$
${\ displaystyle A \ times B \ neq B \ times A}$,
for in the pairs of the set the first element is off and the second is off , while in the pairs of the set the first element is off and the second is off. There is, however, a canonical bijection between the two sets, namely ${\ displaystyle A \ times B}$${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle B \ times A}$${\ displaystyle B}$${\ displaystyle A}$
${\ displaystyle A \ times B \ to B \ times A, \ quad (a, b) \ mapsto (b, a)}$,
with which the quantities can be identified with one another.
#### Non-associativity
The Cartesian product is also non- associative , that is, for non-empty sets , and holds in general ${\ displaystyle A}$${\ displaystyle B}$${\ displaystyle C}$
${\ displaystyle A \ times \ left (B \ times C \ right) \ neq \ left (A \ times B \ right) \ times C}$,
because the set on the left side contains pairs whose first element is off and whose second element is a pair off , whereas the set on the right side contains pairs whose first element is a pair off and whose second element is off. Here, too, there is a canonical bijection between these two sets, namely ${\ displaystyle A}$${\ displaystyle B \ times C}$${\ displaystyle A \ times B}$${\ displaystyle C}$
${\ displaystyle A \ times \ left (B \ times C \ right) \ to \ left (A \ times B \ right) \ times C, \ quad (a, (b, c)) \ mapsto ((a, b ), c)}$.
Some authors identify the pairs and with the ordered triple , which also makes the Cartesian product associative. ${\ displaystyle (a, (b, c))}$${\ displaystyle ((a, b), c)}$${\ displaystyle (a, b, c)}$
#### Distributivity
Illustration of the first distributive law
The following distributive laws regarding union , intersection and subtraction of sets apply to the Cartesian product :
{\ displaystyle {\ begin {aligned} \ left (A \ cup B \ right) \ times C & = \ left (A \ times C \ right) \ cup \ left (B \ times C \ right) \\\ left ( A \ cap B \ right) \ times C & = \ left (A \ times C \ right) \ cap \ left (B \ times C \ right) \\\ left (A \ setminus B \ right) \ times C & = \ left (A \ times C \ right) \ setminus \ left (B \ times C \ right) \\ A \ times \ left (B \ cup C \ right) & = \ left (A \ times B \ right) \ cup \ left (A \ times C \ right) \\ A \ times \ left (B \ cap C \ right) & = \ left (A \ times B \ right) \ cap \ left (A \ times C \ right) \ \ A \ times \ left (B \ setminus C \ right) & = \ left (A \ times B \ right) \ setminus \ left (A \ times C \ right) \ end {aligned}}}
#### Monotony and complement
The Cartesian product behaves monotonically with respect to the formation of subsets , that is, if the sets and are not empty, then the following applies ${\ displaystyle A_ {1}, A_ {2}, B_ {1}}$${\ displaystyle B_ {2}}$
${\ displaystyle (A_ {1} \ times A_ {2}) \ subseteq (B_ {1} \ times B_ {2}) ~ \ Longleftrightarrow ~ A_ {1} \ subseteq B_ {1} ~ {\ text {and} } ~ A_ {2} \ subseteq B_ {2}}$.
In particular, equality applies
${\ displaystyle (A_ {1} \ times A_ {2}) = (B_ {1} \ times B_ {2}) ~ \ Longleftrightarrow ~ A_ {1} = B_ {1} ~ {\ text {and}} ~ A_ {2} = B_ {2}}$.
If we consider the set as the basic set of and the set as the basic set of , then has the complement of in the representation ${\ displaystyle B_ {1}}$${\ displaystyle A_ {1}}$${\ displaystyle B_ {2}}$${\ displaystyle A_ {2}}$${\ displaystyle A_ {1} \ times A_ {2}}$${\ displaystyle B_ {1} \ times B_ {2}}$
${\ displaystyle (A_ {1} \ times A_ {2}) ^ {\ mathsf {C}} = (A_ {1} ^ {\ mathsf {C}} \ times A_ {2} ^ {\ mathsf {C} }) \ cup (A_ {1} ^ {\ mathsf {C}} \ times A_ {2}) \ cup (A_ {1} \ times A_ {2} ^ {\ mathsf {C}})}$.
#### Further calculation rules
Cartesian products of two intervals, their intersections and their unions
It is true
${\ displaystyle \ left (A_ {1} \ cap A_ {2} \ right) \ times \ left (B_ {1} \ cap B_ {2} \ right) = \ left (A_ {1} \ times B_ {1 } \ right) \ cap \ left (A_ {2} \ times B_ {2} \ right)}$,
but in general is
${\ displaystyle \ left (A_ {1} \ cup A_ {2} \ right) \ times \ left (B_ {1} \ cup B_ {2} \ right) \ supseteq \ left (A_ {1} \ times B_ { 1} \ right) \ cup \ left (A_ {2} \ times B_ {2} \ right)}$,
because the set on the left contains pairs and that are not included in the set on the right. ${\ displaystyle A_ {1} \ times B_ {2}}$${\ displaystyle A_ {2} \ times B_ {1}}$
## Product of finitely many quantities
### definition
More generally, the Cartesian product of sets is defined as the set of all - tuples , where for each one element is from the set . Formally, the multiple Cartesian product is through ${\ displaystyle A_ {1} \ times \ dotsb \ times A_ {n}}$${\ displaystyle n}$${\ displaystyle A_ {1}, \ dotsc, A_ {n}}$${\ displaystyle n}$ ${\ displaystyle (a_ {1}, \ dotsc, a_ {n})}$${\ displaystyle a_ {i}}$${\ displaystyle i = 1, \ ldots, n}$${\ displaystyle A_ {i}}$
${\ displaystyle A_ {1} \ times \ dotsb \ times A_ {n}: = \ left \ {(a_ {1}, \ dotsc, a_ {n}) \ mid a_ {i} \ in A_ {i} ~ {\ text {for}} ~ i = 1, \ dotsc, n \ right \}}$
Are defined. With the help of the product symbol , the multiple Cartesian product is also through
${\ displaystyle \ prod _ {i = 1} ^ {n} A_ {i} = A_ {1} \ times \ dotsb \ times A_ {n}}$
written down. The -fold Cartesian product of a set with itself is also written as ${\ displaystyle n}$${\ displaystyle A}$
${\ displaystyle A ^ {n} = \ underbrace {A \ times \ dotsc \ times A} _ {n {\ text {-mal}}} = \ left \ {(a_ {1}, \ dotsc, a_ {n }) \ mid a_ {i} \ in A ~ {\ text {for}} ~ i = 1, \ dotsc, n \ right \}}$.
#### Empty product
The Cartesian product of zero sets is the set that contains the empty tuple as the only element , that is
${\ displaystyle \ prod _ {i = 1} ^ {0} A_ {i} = \ {() \}.}$
In particular, is for any set ${\ displaystyle A}$
${\ displaystyle A ^ {0} = \ {() \}}$.
This is used when constants of a mathematical structure are viewed as zero-digit operations.
With denotes the union of all -fold Cartesian products of a set with itself (for all ), i.e. the set of all tuples with elements of A, including the empty tuple: ${\ displaystyle A ^ {*}}$${\ displaystyle n}$${\ displaystyle A}$${\ displaystyle n \ in \ mathbb {N}}$
${\ displaystyle A ^ {*} = \ bigcup _ {n = 0} ^ {\ infty} A ^ {n}}$.
### Examples
Is then is ${\ displaystyle A = \ left \ {0.1 \ right \}}$
${\ displaystyle A \ times A \ times A = A ^ {3} = \ {(0,0,0), (0,0,1), (0,1,0), (0,1,1) , (1,0,0), (1,0,1), (1,1,0), (1,1,1) \}}$.
In a three-dimensional Cartesian coordinate system, each point is represented as a triplet of coordinates.${\ displaystyle (x, y, z)}$
The Euclidean space consists of three times the Cartesian product of the real numbers : ${\ displaystyle \ mathbb {R} ^ {3}}$${\ displaystyle \ mathbb {R}}$
${\ displaystyle \ mathbb {R} \ times \ mathbb {R} \ times \ mathbb {R} = \ mathbb {R} ^ {3} = \ {(x, y, z) \ mid x, y, z \ in \ mathbb {R} \}}$.
The 3-tuples are the three-dimensional Cartesian coordinates. The Cartesian product of three real intervals , and gives the cuboid${\ displaystyle (x, y, z)}$${\ displaystyle [a, b]}$${\ displaystyle [c, d]}$${\ displaystyle [e, f]}$
${\ displaystyle [a, b] \ times [c, d] \ times [e, f] = \ {(x, y, z) \ in \ mathbb {R} ^ {3} \ mid a \ leq x \ leq b, c \ leq y \ leq d, e \ leq z \ leq f \}}$.
In general, the -fold Cartesian product of the real numbers gives the space and the Cartesian product of real intervals gives a hyper rectangle . ${\ displaystyle n}$${\ displaystyle \ mathbb {R} ^ {n}}$${\ displaystyle n}$
### properties
#### Number of elements
If the sets are all finite, then their Cartesian product is also a finite set, where the number of elements of equals the product of the element numbers of the initial sets , that is ${\ displaystyle A_ {1}, \ dotsc, A_ {n}}$${\ displaystyle A_ {1} \ times \ dotsb \ times A_ {n}}$
${\ displaystyle | A_ {1} \ times \ dotsb \ times A_ {n} | = | A_ {1} | \ cdot \ ldots \ cdot | A_ {n} |}$
or in a different notation
${\ displaystyle \ left | \ prod _ {i = 1} ^ {n} A_ {i} \ right | = \ prod _ {i = 1} ^ {n} | A_ {i} |}$.
In the special case that all sets are equal to one set , the following applies ${\ displaystyle A_ {i}}$${\ displaystyle A}$
${\ displaystyle | A ^ {n} | = | A | ^ {n}}$.
The Cartesian product of a finite number of countably infinite sets is also countable, as can be shown by iterating the argument for the Cartesian product of two sets with the help of Cantor's tuple function .
#### monotony
If the sets and are not empty, then, as with the Cartesian product of two sets, monotony applies ${\ displaystyle A_ {1}, \ ldots, A_ {n}}$${\ displaystyle B_ {1}, \ ldots, B_ {n}}$
${\ displaystyle \ prod _ {i = 1} ^ {n} A_ {i} \ subseteq \ prod _ {i = 1} ^ {n} B_ {i} ~ \ Longleftrightarrow ~ A_ {i} \ subseteq B_ {i } ~ {\ text {for}} ~ i = 1, \ ldots, n}$
and equality
${\ displaystyle \ prod _ {i = 1} ^ {n} A_ {i} = \ prod _ {i = 1} ^ {n} B_ {i} ~ \ Longleftrightarrow ~ A_ {i} = B_ {i} ~ {\ text {for}} ~ i = 1, \ ldots, n}$.
## Product of infinitely many quantities
### definition
It is also possible to define the Cartesian product of an infinite number of sets. If there is an index set and a family of sets for this, then the Cartesian product of the sets is defined by ${\ displaystyle I}$${\ displaystyle (A_ {i}) _ {i \ in I}}$${\ displaystyle A_ {i}}$
${\ displaystyle \ prod _ {i \ in I} A_ {i} = {\ Big \ {} f \ colon I \ to \ bigcup _ {i \ in I} A_ {i} \, {\ Big |} \ , \ forall i \ in I \ colon f (i) \ in A_ {i} {\ Big \}}}$.
This is the set of all images of the union of the sets for which the image in lies. If all are equal to a set , then that is the Cartesian product ${\ displaystyle f}$${\ displaystyle I}$${\ displaystyle A_ {i}}$${\ displaystyle f (i)}$${\ displaystyle A_ {i}}$${\ displaystyle A_ {i}}$${\ displaystyle A}$
${\ displaystyle \ prod _ {i \ in I} A = A ^ {I}}$
the set of all functions from to . If the quantities are different, however, the Cartesian product is much less clear. Even the question of whether any Cartesian product of non-empty sets is non-empty can not be decided with the Zermelo-Fraenkel set theory ZF; the assertion that it is not empty is a formulation of the axiom of choice which is added to ZF in order to obtain the set theory ZFC (“Zermelo-Fraenkel + Choice”). ${\ displaystyle I}$${\ displaystyle A}$${\ displaystyle A_ {i}}$
### Special cases
An important special case of an infinite Cartesian product arises from the choice of natural numbers as the index set. The Cartesian product of a sequence of sets${\ displaystyle \ mathbb {N}}$${\ displaystyle (A_ {i}) _ {i \ in \ mathbb {N}} = (A_ {1}, A_ {2}, \ ldots)}$
${\ displaystyle \ prod _ {i = 1} ^ {\ infty} A_ {i} = \ left \ {(a_ {1}, a_ {2}, \ dotsc) \ mid a_ {i} \ in A_ {i } ~ {\ text {for}} ~ i \ in \ mathbb {N} \ right \}}$
then corresponds to the set of all sequences whose -th sequence member is in the set . For example, if all are , then is ${\ displaystyle i}$${\ displaystyle A_ {i}}$${\ displaystyle A_ {i} = \ mathbb {R}}$
${\ displaystyle \ prod _ {i = 1} ^ {\ infty} \ mathbb {R} = \ mathbb {R} \ times \ mathbb {R} \ times \ dotsb = \ mathbb {R} ^ {\ mathbb {N }} = \ left \ {(a_ {1}, a_ {2}, \ dotsc) \ mid a_ {i} \ in \ mathbb {R} ~ {\ text {for}} ~ i \ in \ mathbb {N } \ right \}}$
the set of all real number sequences. The countable Cartesian product can be mapped bijectively onto the generally defined Cartesian product, because every sequence defines a function with and, conversely, every such function can be written as a sequence . The Cartesian product of finitely many sets can also be understood as a special case of the general definition using finite sequences. ${\ displaystyle (a_ {1}, a_ {2}, \ dotsc)}$${\ displaystyle f}$${\ displaystyle f (1): = a_ {1}, f (2): = a_ {2}, \ dotsc}$${\ displaystyle (f (1), f (2), \ dotsc)}$
### Universal property of the Cartesian product
The family of projections belongs to the Cartesian product . The Cartesian product together with the family has the following property: If there is an arbitrary set and is a family of images, then there is exactly one image with for all . That is, the following diagram is commutative:${\ displaystyle P = \ prod \ limits _ {i \ in I} A_ {i}}$${\ displaystyle \ pi _ {i} \ colon \ prod \ limits _ {i \ in I} \ ni \ alpha \ mapsto \ alpha (i) \ in A_ {i}}$${\ displaystyle \ prod \ limits _ {i \ in I} A_ {i}}$${\ displaystyle (\ pi _ {i} | i \ in I)}$${\ displaystyle X}$${\ displaystyle f_ {i} \ colon X \ to A_ {i}}$${\ displaystyle f \ colon X \ to \ prod \ limits _ {i \ in I} A_ {i}}$${\ displaystyle \ pi _ {i} \ circ f = f_ {i}}$${\ displaystyle i \ in I}$
There is exactly one , so that applies to all :${\ displaystyle f \ colon X \ to P}$${\ displaystyle i \ in I}$${\ displaystyle \ pi _ {i} \ circ f = f_ {i}}$
If this quality is also common with the family , then there is a bijective mapping . ${\ displaystyle Q}$${\ displaystyle p_ {i} \ colon Q \ to A_ {i}}$${\ displaystyle P \ to Q}$
## Derived terms
• A binary relation between two sets is a subset of the Cartesian product of the two sets. In particular, each mapping is therefore a subset of the Cartesian product of the definition and target set of the mapping. More generally, a -digit relation is a subset of the Cartesian product of sets.${\ displaystyle n}$${\ displaystyle n}$
• A projection is a mapping of the Cartesian product of two sets back into one of these sets. More generally, a projection is a mapping from the Cartesian product of a family of sets onto the Cartesian product of a subfamily of these sets that selects elements with certain indices.
• A two-digit link is a mapping of the Cartesian product of two sets into another set. More generally, a -digit link is a mapping of the Cartesian product of sets into another set. Every -digit link can therefore also be understood as a -digit relation.${\ displaystyle n}$${\ displaystyle n}$${\ displaystyle n}$${\ displaystyle (n + 1)}$
• A direct product is a product of algebraic structures , such as groups or vector spaces , which consists of the Cartesian product of the carrier sets and is additionally provided with one or more component-wise links. A direct sum is a subset of the direct product that differs from the direct product only for products of infinitely many quantities; it consists of all tuples that differ from a certain element (usually the neutral element of a link) only in a finite number of places .
• The categorical product corresponds to the Cartesian product in the category of sets and to the direct product in the category of groups and in other categories of algebraic structures.
• In relational databases , the Cartesian product of tables and the join operations based on them are used to link database tables .
## literature
• Oliver Deiser: Introduction to set theory. Georg Cantor's set theory and its axiomatization by Ernst Zermelo. 2nd improved and enlarged edition. Springer, Berlin a. a. 2004, ISBN 3-540-20401-6 .<|endoftext|>
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# Thread: 242.2q.3 find the derivative
1. $\tiny{242.2q.3}$
$\textsf{find the derivative}\\$
\begin{align}
\displaystyle
y&=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}} \\
\ln{y}&=\ln x
+ \frac{1}{2}\ln(x^2+1)
- \frac{2}{3}\ln(x+1)\\
\end{align}
$\textit{thot this would help but what next??}$
2. Use implicit differentiation, knowing that $\d{}{x}\ln\left({x}\right)=\frac{1}{x}$ and $\d{}{x}\ln\left({y}\right)=\frac{1}{y}\d{y}{x}$.
$\tiny{242.2q.3}$
$\textsf{find the derivative}\\$
\begin{align}
\displaystyle
y&=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}} \\
\ln{y}&=\ln x
+ \frac{1}{2}\ln(x^2+1)
- \frac{2}{3}\ln(x+1)\\
\frac{1}{y}\d{y}{x}&=\frac{1}{x}+\dfrac{x}{x^2+1}-\dfrac{2}{3\left(x+1\right)}\\
\d{y}{x}&=\frac{y}{x}+\dfrac{xy}{x^2+1}-\dfrac{2y}{3\left(x+1\right)}
\end{align}
4. $\displaystyle{y=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}}}$
Let $f=x \, \sqrt[]{x^2+1}$ and $g=(x+1)^{2/3}$.
Then $\displaystyle{y'=\frac{f'\cdot g-f\cdot g'}{g^2}} \ \ \ (\star)$
We have the following:
\begin{align*}f' & =(x)' \, \sqrt[]{x^2+1}+x \, (\sqrt[]{x^2+1})'= \sqrt[]{x^2+1}+x \, \frac{1}{2\sqrt[]{x^2+1}}\cdot (x^2+1)' =\sqrt[]{x^2+1}+x \, \frac{2x}{2\sqrt[]{x^2+1}} \\ &=\sqrt[]{x^2+1}+ \, \frac{x^2}{\sqrt[]{x^2+1}} =\frac{\sqrt[]{x^2+1}^2+x^2}{\sqrt[]{x^2+1}}=\frac{x^2+1+x^2}{\sqrt[]{x^2+1}}\\ &=\frac{2x^2+1}{\sqrt[]{x^2+1}}\end{align*}
$$g'=\frac{2}{3}(x+1)^{2/3-1}=\frac{2}{3}(x+1)^{-1/3}$$
$$g^2=(x+1)^{4/3}$$
So, substituting these at the relation $(\star)$ we get:
\begin{align*}y' &=\frac{\frac{2x^2+1}{\sqrt[]{x^2+1}}\cdot (x+1)^{2/3}-x \, \sqrt[]{x^2+1}\cdot \frac{2}{3}(x+1)^{-1/3}}{(x+1)^{4/3}} \\ &=\frac{\sqrt{x^2+1}\left (\frac{2x^2+1}{\sqrt[]{x^2+1}}\cdot (x+1)^{2/3}-x \, \sqrt[]{x^2+1}\cdot \frac{2}{3}(x+1)^{-1/3}\right )}{(x+1)^{4/3}\sqrt{x^2+1}} \\ &=\frac{\left (2x^2+1\right )\cdot (x+1)^{2/3}-x \, (x^2+1)\cdot \frac{2}{3}(x+1)^{-1/3}}{(x+1)^{4/3}\sqrt{x^2+1}} \\ &=\frac{(x+1)^{1/3}\left (\left (2x^2+1\right )\cdot (x+1)^{2/3}-x \, (x^2+1)\cdot \frac{2}{3}(x+1)^{-1/3}\right )}{(x+1)^{1/3}(x+1)^{4/3}\sqrt{x^2+1}} \\ &=\frac{\left (2x^2+1\right )\cdot (x+1)-x \, (x^2+1)\cdot \frac{2}{3}}{(x+1)^{5/3}\sqrt{x^2+1}} \\ & =\frac{3\left (\left (2x^2+1\right )\cdot (x+1)-x \, (x^2+1)\cdot \frac{2}{3}\right )}{3(x+1)^{5/3}\sqrt{x^2+1}} \\ &=\frac{3\left (2x^2+1\right )\cdot (x+1)-2x \, (x^2+1)}{3(x+1)^{5/3}\sqrt{x^2+1}} \\ &=\frac{3\left (2x^3+x+2x^2+1\right )-2 \, (x^3+x)}{3(x+1)^{5/3}\sqrt{x^2+1}} \\ &=\frac{6x^3+3x+6x^2+3-2x^3-2x}{3(x+1)^{5/3}\sqrt{x^2+1}} \\ &=\frac{4x^3+6x^2+x+3}{3(x+1)^{5/3}\sqrt{x^2+1}}\end{align*}
5. Originally Posted by karush
$\tiny{242.2q.3}$
$\textsf{find the derivative}\\$
\begin{align}
\displaystyle
y&=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}} \\
\ln{y}&=\ln x
+ \frac{1}{2}\ln(x^2+1)
- \frac{2}{3}\ln(x+1)\\
\frac{1}{y}\d{y}{x}&=\frac{1}{x}+\dfrac{x}{x^2+1}-\dfrac{2}{3\left(x+1\right)}\\
\d{y}{x}&=\frac{y}{x}+\dfrac{xy}{x^2+1}-\dfrac{2y}{3\left(x+1\right)}
\end{align}
Good job so far, now remember that you already know what y is in terms of x, so you can write the derivative completely in terms of x as well.
Originally Posted by Prove It
Good job so far, now remember that you already know what y is in terms of x, so you can write the derivative completely in terms of x as well.
\begin{align}
\displaystyle
y&=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}} \\
\ln{y}&=\ln x
+ \frac{1}{2}\ln(x^2+1)
- \frac{2}{3}\ln(x+1)\\
\frac{1}{y}\d{y}{x}&=\frac{1}{x}+\dfrac{x}{x^2+1}-\dfrac{2}{3\left(x+1\right)}\\
\d{y}{x}&=\frac{y}{x}+\dfrac{xy}{x^2+1}-\dfrac{2y}{3\left(x+1\right)}
\end{align}
$\displaystyle y=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}} \therefore \frac{y}{x}=\frac{\sqrt[]{x^2+1}}{(x+1)^{2/3}}$
which is the first term
online calculator returned this for the answer so not sure how the 2nd term was derived the third is just 2y in numerator then simplify
$y'=\dfrac{\sqrt{x^2+1}}{\left(x+1\right)^\frac{2}{3}} +\dfrac{x^2}{\left(x+1\right)^\frac{2}{3}\sqrt{x^2+1}} -\dfrac{2x\sqrt{x^2+1}}{3\left(x+1\right)^\frac{5}{3}}$
7. Originally Posted by karush
\begin{align}
\displaystyle
y&=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}} \\
\ln{y}&=\ln x
+ \frac{1}{2}\ln(x^2+1)
- \frac{2}{3}\ln(x+1)\\
\frac{1}{y}\d{y}{x}&=\frac{1}{x}+\dfrac{x}{x^2+1}-\dfrac{2}{3\left(x+1\right)}\\
\d{y}{x}&=\frac{y}{x}+\dfrac{xy}{x^2+1}-\dfrac{2y}{3\left(x+1\right)}
\end{align}
$\displaystyle y=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}} \therefore \frac{y}{x}=\frac{\sqrt[]{x^2+1}}{(x+1)^{2/3}}$
which is the first term
online calculator returned this for the answer so not sure how the 2nd term was derived the third is just 2y in numerator then simplify
$y'=\dfrac{\sqrt{x^2+1}}{\left(x+1\right)^\frac{2}{3}} +\dfrac{x^2}{\left(x+1\right)^\frac{2}{3}\sqrt{x^2+1}} -\dfrac{2x\sqrt{x^2+1}}{3\left(x+1\right)^\frac{5}{3}}$
I suppose you could do it that way, I would have just done...
\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= y \left[ \frac{1}{x} + \frac{x}{x^2 + 1} - \frac{2}{3 \left( x + 1 \right) } \right] \\ &= \frac{x\,\sqrt{ x^2 + 1 }}{\left( x + 1 \right) ^{\frac{2}{3}}} \left[ \frac{1}{x} + \frac{x}{x^2 + 1} - \frac{2}{3\left( x + 1 \right) } \right] \end{align*}<|endoftext|>
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What is Receptive Language Delay?
Receptive language delay is a broad diagnosis that simply means that a child has trouble understanding language. This covers a wide variety of language skills and the child may have trouble with all of those skill, or only one or two. A child with a receptive language delay may also have an expressive language delay. That means that the child would have trouble using language appropriately as well.
What is the Treatment for Receptive Language Delay?
Children with receptive language delays benefit from language therapy from a certified speech-language pathologist. Since this is a broad diagnosis, the therapy is specifically targeted at whatever language skills the child is having the most trouble with. The therapies are specific to the language skill in question. A speech therapist may also be doing language therapy for expressive language skills at the same time to improve overall communication. There are also many things that a parent can do at home to improve receptive language. Keep reading to find activities and resources about the different types of receptive language skills.
One of the best indicators of a child’s receptive language skills is his ability to follow directions. Now, when I say “ability to follow”, I mean that the child understands how to follow directions. Whether or not he chooses to actually do so is an entirely different skill! Here are some activities and resources for helping a child learn to follow directions:
Following Directions: Step-By-Step Guide (Podcast)
Printable Spatial Concepts Game (Follow directions with spatial concepts)
Cooking Activities for Speech and Language (Follow directions on a recipe)
Craft Activities for Speech and Language (Follow directions to complete the craft)
Children with receptive language delays often have difficulty understanding questions that are asked to them. They may confuse the different “wh-” question words. For example, if you ask the child a “who” question, they may respond with a place instead of a person. This means that they are not understanding what that “wh-” question word means.
Children with receptive language delays may also have trouble understanding vocabulary words. This may impact their ability to understand what someone is saying to them as well as impact their ability to learn new concepts. Here is a page all about learning vocabulary:
Children with receptive language delays are likely to have trouble with any activity that involves listening, especially language-heavy tasks. If you are working with a child who is having trouble with listening, start with some simple listening activities that don’t involve language. You can have the child listen to different environmental sounds, like vehicle or animal noises and then guess what those things are called. Then, you can move up to slightly harder listening activities, like repeating back spoken words or sentences. Some other ideas are listening for a repeated word or line in a children’s book (such as listening for the phrase “go away” in Go Away Big Green Monster), answering questions about a story or short paragraph read aloud, or following clues to find something that is hidden when the clues are only given verbally.
Children with language delays often have difficulty understanding and using figurative language such as idioms, similes, and metaphors. Click the link below to find out how to teach these to a child:
Children with language delays often have difficulty making inferences about what’s going on around them or when they are reading. Click the link below to learn more about helping a child make inferences:
Click the link below to join my mailing list so you’ll be updated whenever new receptive language delay resources are posted!<|endoftext|>
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Reproduction in living organisms
Reproduction is the process by which new organisms (offsprings) are
generated. A living organism does not need reproduction to survive, but as a species,
they need that for continuity and to ensure that they are not extinct.
There are two main types of reproduction: these include sexual reproduction
and asexual reproduction.
This involves two individuals of the same species, usually a male and female.
Here the male and female sex cells come together for fertilization to take place. After
this the newly fertilized cell goes on to become a new organism, the offspring. Note
that not all sexual reproduction involve mating.
This form of reproduction occurs without the involvement of another. Asexual
reproduction is very common in single cell organisms and in many plants. There are
many forms of asexual reproduction. Mitosis, fission, budding, fragmentation,
sporulation and vegetative reproduction are all examples of asexual reproduction. In
unicellular organisms, the parent cell just divides to produce two daughter cells. The
term for kind of cell division is Mitosis
Below is an illustration of the process of
a few years and some live for a few days. The term for the length of time an organism
lives is called their ‘Lifespan’. For instance, an adult mayfly lives for only one day, a
mouse lives for 1-2 years and tortoise can live for about 152 years
But can you imagine what will happen to a species if it had no new ones (offspring)
to replace them? They will be extinct. This means reproduction is essential for the
survival of all species. It also ensures that the characteristics of the parents are passed
on to future generations, ensuring continuity.
The Cell Cycle In Living Organisms
The cell cycle is the recurring sequence of events that includes the duplication
of a cell's contents and its subsequent division. This SparkNote will focus on
following the major events of the cell cycle as well as the processes that regulate its
action. In this and the following SparkNotes on cell reproduction, we will see how
the cell cycle is an essential process for all living organisms. In single-cell organisms,
each round of the cell cycle leads to the production of an entirely new organism.
Other organisms require multiple rounds of cell division to create a new individual.
In humans and other higher-order animals, cell death and growth are constant
processes and the cell cycle is necessary for maintaining appropriate cellular
Figure %: The Cell Cycle
reproduction is to create new cells. The cell cycle is the means by which this goal is
accomplished. While its duration and certain specific components vary from species
to species, the cell cycle has a number of universal trends.
DNA packaged into chromosomes must be replicated.
The copied contents of the cell must migrate to opposite ends of the cell.
The cell must physically split into two separate cells.
We will discuss the general organization of the cell cycle by reviewing its two
major phases: M Phase (for mitosis) and interphase. Interphase is generally split into
three distinct phases including one for DNA replication. We will finish with a
discussion of the elements that control a cell's passage through these various stages.
The cell cycle is very highly regulated to prevent constant cell division and only
allows cell that have met certain requirements to engage in cell division.
How long do the different stages of the cell cycle take?
Replication is one of the hallmark features of living matter. The set of
processes known as the cell cycle which are undertaken as one cell becomes two has
been a dominant research theme in the molecular era with applications that extend far
and wide including to the study of diseases such as cancer which is sometimes
characterized as a disease of the cell cycle gone awry. Cell cycles are interesting both
for the ways they are similar from one cell type to the next and for the ways they are
different. To bring the subject in relief, we consider the cell cycles in a variety of
different organisms including a model prokaryote, for mammalian cells in tissue
culture and during embryonic development in the fruit fly. Specifically, we ask what
are the individual steps that are undertaken for one cell to divide into two and how
long do these steps take?
The 150 min cell cycle of Caulobacter is shown, highlighting some of the key
morphological and metabolic events that take place during cell division. M phase is
not indicated because in Caulobacter there is no true mitotic apparatus that gets
assembled as in eukaryotes. Much of chromosome segregation in Caulobacter (and
other bacteria) occurs concomitantly with DNA replication. The final steps of
chromosome segregation and especially decatenation of the two circular
chromosomes occurs during G2 phase.
Arguably the best-characterized prokaryotic cell cycle is that of the model
organism Caulobacter crescentus. One of the appealing features of this bacterium is
that it has an asymmetric cell division that enables researchers to bind one of the two
progeny to a microscope cover slip while the other daughter drifts away enabling
further study without obstructions. This has given rise to careful depictions of the
≈150 minute cell cycle (BNID 104921) as shown in Figure 1. The main components
of the cell cycle are G1 (first Growth phase,
≈30 min, BNID 104922), where at least
some minimal amount of cell size increase needs to take place, S phase (Synthesis,
≈80 min, BNID 104923) where the DNA gets replicated and G2 (second Growth
≈25 min, BNID 104924) where chromosome segregation unfolds leading to
≈15 min). Caulobacter crescentus provides an
interesting example of the way in which certain organisms get promoted to “model
organism’’ status because they have some particular feature that renders them
particularly opportune for the question of interest. In this case, the cell-cycle
progression goes hand in hand with the differentiation process giving readily
visualized identifiable stages making them preferable to cell-cycle biologists over,
say, the model bacterium E. coli.
The behavior of mammalian cells in tissue culture has served as the basis for
much of what we know about the cell cycle in higher eukaryotes. The eukaryotic cell
cycle can be broadly separated into two stages, interphase, that part of the cell cycle
when the materials of the cell are being duplicated and mitosis, the set of physical
processes that attend chromosome segregation and subsequent cell division. The rates
of processes in the cell cycle, are mostly built up from many of the molecular events
such as polymerization of DNA and cytoskeletal filaments whose rates we have
already considered. For the characteristic cell cycle time of 20 hours in a HeLa cell,
almost half is devoted to G1 (BNID 108483) and close to another half is S phase
(BNID 108485) whereas G2 and M are much faster at about 2-3 hours and 1 hour,
respectively (BNID 109225, 109226). The stage most variable in duration is G1. In
less favorable growth conditions when the cell cycle duration increases this is the
stage that is mostly affected, probably due to the time it takes until some regulatory
size checkpoint is reached. Though different types of evidence point to the existence
of such a checkpoint, it is currently very poorly understood. Historically, stages in the
cell cycle have usually been inferred using fixed cells but recently, genetically-
encoded biosensors that change localization at different stages of the cell cycle have
made it possible to get live-cell temporal information on cell cycle progression and
the cell cycle devoted to each of the primary stages of the cell cycle. The area of each
chart is proportional to the overall cell cycle duration. Cell cycle durations reflect
minimal doubling times under ideal conditions. (Adapted from “The Cell Cycle –
Principles of Control” by David Morgan.)
How does the length of the cell cycle compare to the time it takes a cell to
synthesize its new genome? A decoupling between the genome length and the
doubling time exists in eukaryotes due to the usage of multiple DNA replication start
sites. For mammalian cells it has been observed that for many tissues with widely
varying overall cell cycle times, the duration of the S phase where DNA replication
occurs is remarkably constant. For mouse tissues such as those found in the colon or
tongue, the S phase varied in a small range from 6.9 to 7.5 hours (BNID 111491).
Even when comparing several epithelial tissues across human, rat, mouse and
hamster, S phase was between 6 and 8 hours (BNID 107375). These measurements
were carried out in the 1960s by performing a kind of pulse-chase experiment with
the radioactively labeled nucleotide thymidine. During the short pulse, the radioactive
compound was incorporated only into the genome of cells in S phase. By measuring
the duration of appearance and then disappearance of labeled cells in M phase one
can infer how long S phase lasted The fact that the duration of S phase is relatively
constant in such cells is used to this day to estimate the duration of the cell cycle from
a knowledge of only the fraction of cells at a given snapshot in time that are in S
phase. For example, if a third of the cells are seen in S phase which lasts about 7
hours, the cell cycle time is inferred to be about 7 hours/(1/3)
≈20 hours. Today these
kinds of measurements are mostly performed using BrdU as the marker for S phase.
We are not aware of a satisfactory explanation for the origin of this relatively
constant replication time and how it is related to the rate of DNA polymerase and the
density of replication initiation sites along the genome.
The diversity of cell cycles is shown in Figure 2 and depicts several model
organisms and the durations and positioning of the different stages of their cell
cycles. An extreme example occurs in the mesmerizing process of embryonic
development of the fruit fly Drosophila melanogaster. In this case, the situation is
different from conventional cell divisions since rather than synthesizing new
cytoplasmic materials, mass is essentially conserved except for the replication of the
genetic material. This happens in a very synchronous manner for about 10
generations and a replication cycle of the thousands of cells in the embryo, say
between cycle 10 and 11, happens in about 8 minutes as shown in Figure 2 (BNID
103004,103005, 110370). This is faster than the replication times for any bacteria
even though the genome is
≈120 million bp long (BNID 100199). A striking example
of the ability of cells to adapt their temporal dynamics.
Growth And Development
“Development” and “growth” are sometimes used interchangeably in
conversation, but in a botanical sense, they describe separate events in the
organization of the mature plant body.
Development is the progression from earlier to later stages in maturation, e.g. a
fertilized egg develops into a mature tree. It is the process whereby tissues, organs,
and whole plants are produced. It involves: growth, morphogenesis (the acquisition of
form and structure), and differentiation. The interactions of the environment and the
genetic instructions inherited by the cells determine how the plant develops.
Growth is the irreversible change in size of cells and plant organs due to both
cell division and enlargement. Enlargement necessitates a change in the elasticity of
the cell walls together with an increase in the size and water content of the vacuole.
Growth can be determinate—when an organ or part or whole organism reaches a
certain size and then stops growing—or indeterminate—when cells continue to divide
indefinitely. Plants in general have indeterminate growth.
Differentiation is the process in which generalized cells specialize into the
morphologically and physiologically different cells . Since all of the cells produced
by division in the meristems have the same genetic make up, differentiation is a
function of which particular genes are either expressed or repressed. The kind of cell
that ultimately develops also is a result of its location: Root cells don't form in
developing flowers, for example, nor do petals form on roots.
Mature plant cells can be stimulated under certain conditions to divide and
differentiate again, i.e. to dedifferentiate. This happens when tissues are wounded, as
when branches break or leaves are damaged by insects. The plant repairs itself
bydedifferentiating parenchyma cells in the vicinity of the wound, making cells like
those injured or else physiologically similar cells.
Plants differ from animals in their manner of growth. As young animals
mature, all parts of their bodies grow until they reach a genetically determined size
for each species. Plant growth, on the other hand, continues throughout the life span
of the plant and is restricted to certain meristematic tissue regions only. This
continuous growth results in:
Two general groups of tissues, primary and secondary.
Two body types, primary and secondary.
Apical and lateral meristems.
Apical meristems, or zones of cell division, occur in the tips of both roots,
stems of all plants, and are responsible for increases in the length of the primary plant
body as the primary tissues differentiate from the meristems. As the vacuoles of the
primary tissue cells enlarge, the stems and roots increase in girth until a maximum
size (determined by the elasticity of their cell walls) is reached. The plant may
continue to grow in length, but no longer does it grow in girth. Herbaceous plants
with only primary tissues are thus limited to a relatively small size.
Woody plants, on the other hand, can grow to enormous size because of the
strengthening and protective secondary tissues produced by lateral meristems, which
develop around the periphery of their roots and stems. These tissues constitute the
secondary plant body.
Heredity And Variability
Heredity refers to the genetic transmission of traits from parents to offspring.
Heredity helps explain why children tend to resemble their parents, as well as how a
genetic disease runs in a family. Some genetic conditions are caused by mutations in
a single gene. These conditions are usually inherited in one of several straightforward
patterns, including autosomal dominant, autosomal recessive, X-linked dominant, X-
linked recessive, codominant, and mitochondrial inheritance patterns. Complex
disorders and multifactorial disorders are caused by a combination of genetic and
environmental factors. These disorders may cluster in families, but do not have a
clear-cut pattern of inheritance.
Evolution : a process of development in which an organ or organism becomes
more and more complex by the differentiation of its parts; a continuous and
progressive change according to certain laws and by means of resident forces
bathmic or orthogenic evolution : evolution due to something in the organism
itself independent of environment
convergent evolution : the appearance of similar forms and/or functions in two
or more lines not sufficiently related phylogenetically to account for the similarity.
The concept that chance reigns supreme may ring less true when it comes to complex
behaviors. A study of the similarities between the webs of different Tetragnatha
spider species on different Hawaiian Islands provides fresh evidence that behavioral
tendencies can actually evolve rather predictably, even in widely separated places.
The spiders' webs vary significantly, with tissue-like 'sheet webs', disorganized
cobwebs and spiral-shaped 'orb webs' as three of the most common types. Each
species had its own characteristic type of web. But the scientists found that in several
cases, separate species of Tetragnatha spiders on different islands constructed
extremely similar orb webs, right down to the number of spokes, and the lengths and
densities of the sticky spiral that captures bugs. Was this an example of similar
environments producing the same complex behavior, or did the spiders with
corresponding webs share a common ancestor? The tree that linked spiders through
their web-constructing behavior proved highly improbable as it was very
complicated, and contradicted the relationships suggested by their DNA. It is likely
that similar forest types support similar mixes of prey, which could elicit similar web
structures. Previous research has found that physical traits, for example legs or wings,
can arise independently in similar environmental conditions. And various groups
have looked at the evolution of simple behaviors, such as where species locate
themselves within a habitat, like a branch or lake. But the evolution of complex
behaviors is less well understood : predictable evolutionary convergence of behavior
applies far beyond spiders, and happens more often then some believe
- emergent evolution : the assumption that each step in evolution produces
something new and something that could not be predicted from its antecedents.
- organic evolution : the origin and development of species; the theory that
existing organisms are the result of descent with modification from those of past
- parallel evolution : the independent evolution of similar structures in two or
- salutatory evolution : evolution showing sudden changes; mutation or
halmatogenesis / salutatory variation : a sudden alteration of type from
- darwinism / darwinian theory : the theory of evolution by Charles Robert
Darwin according to which higher organisms have developed from lower ones
through the influence of natural selection
adaptive plasticity in response to environmental pressures : snake
that specify a large head size, or head growth may be increased in individual snakes
to meet local demands (adaptive developmental plasticity).
- monogenesis : the theory of evolution according to which the course of
evolution is fixed and predetermined by law, no place being left for chance
- an adaptations programme has dominated evolutionary thought in England
and the United States during the past 40 years. It is based on faith in the power of
natural selection as an optimizing agent. It proceeds by breaking an organism into
unitary 'traits' and proposing an adaptive story for each considered separately. Trade-
offs among competing selective demands exert the only brake upon perfection; non-
optimality is thereby rendered as a result of adaptation as well. Some criticize this
approach and attempt to reassert a competing notion (long popular in continental
Europe) that organisms must be analyzed as integrated wholes, with Bauplane so
constrained by phyletic heritage, pathways of development and general architecture
that the constraints themselves become more interesting and more important in
delimiting pathways of change than the selective force that may mediate change
when it occurs. Some fault the adaptationist programme for its failure to distinguish
current utility from reasons for origin (male tyrannosaurs may have used their
diminutive front legs to titillate female partners, but this will not explain why they got
so small); for its unwillingness to consider alternatives to adaptive stories; for its
reliance upon plausibility alone as a criterion for accepting speculative tales; and for
its failure to consider adequately such competing themes as random fixation of
alleles, production of non-adaptive structures by developmental correlation with
selected features (allometry, pleiotropy, material compensation, mechanically forced
correlation), the separability of adaptation and selection, multiple adaptive peaks, and
current utility as an epiphenomenon of non-adaptive structures. Some support
Darwin's own pluralistic approach to identifying the agents of evolutionary change
- the theory of intelligent design (ID)makes the claim that the existence of
complex systems and phenomena, lacking any justification for their existence that is
known to us, implies that such systems exist as the purposeful result of the activity of
a powerful, conscious being that designed the visible complexity into them. This is
not a scientific explanation, as it posits the existence of something that cannot be
tested or demonstrated by experiment, but must be taken on faith. The contrast
between the theory of intelligent design and the theory of special creation is that the
latter names the designer "God" and declares the story in the biblical book of Exodus
as the whole truth, whereas the former does not name the designer nor does it declare
any particular story of the designer's works and actions to be historical truth.
However, both of these theories are theology, not biology, and while not identical, are
both out of place in a life science journal. Theologians, and even scientists, are
entitled to logically debate questions of faith surrounding the problems of first causes,
complexity, the existence of evil, and so forth, but not in scientific publications.
Albert Einstein is quoted as having said, "Science without religion is lame; religion
without science is blind." Let us be clear, however: science is about knowledge
gained by hypothesis testing, and religion is about faith gained from reason,
inspiration, and introspection. We must keep them properly separated to understand
the difference between that which we can know and that which we must choose, or
choose not, to believe.
- first proposed by W.D. Hamilton in 1964, the theory of kin selection holds
that altruistic cooperative behavior preferentially directed at helping a relative is
favored because it helps that relative do better and reproduce, which indirectly helps
the cooperator to pass on its genes. Generating siderophores is costly to producer
Pseudomonas aeruginosa (cooperators), but others around it can use the siderophores
to their own benefit without paying the price (cheaters). When relatedness is high, the
cooperators spread to fixation and take over; and when relatedness is low, the
cheaters spread to take over, meaning that higher relatedness had a tendency to favor
selection for more altruism or cooperation. Another more subtle effect of kin
selection is the scale of competition—whether competition is local (competition
between close relatives) or global (competition between unrelated bacteria of the
same species). Relatedness increases cooperation, so that over time, a localized group
of highly related organisms emerges. But eventually, these would also become the
closest competitors in the local area, so they were the ones you had to compete with
for spots in the gene pool in the next generation. The experimental effects of
relatedness on the scale of competition explained > 90% of the variation in the
frequency of cooperators versus cheaters at the end of the experiment. The work has
implications for social insects : if individual insects are close relatives but are going
be dispersing to some other area, or maybe foraging in different areas or looking in
different areas for mates, then the scale at which competition might take place is
going to vary quite a bit depending on the ecology of that particular insect.<|endoftext|>
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An ancient art form has taken on new shape at NASA’s Jet Propulsion Laboratory in Pasadena, Calif.
Origami, the Japanese tradition of paper-folding, has inspired a number of unique spacecraft designs here. It’s little wonder that it fascinates NASA engineers: origami can seem deceptively simple, hiding complex math within its creases.
Besides aesthetic beauty, it addresses a persistent problem faced by JPL engineers: how do you pack the greatest amount of spacecraft into the smallest volume possible?
One answer might be found in the Starshade, an immense, folding iris that has been proposed as a way to block light from distant stars. It would unfurl to a diameter of about 85 feet (26 meters) in space, about the size of a standard baseball diamond.
Dampening the brightness of a star’s light would extend the capability of a space telescope to detect orbiting exoplanets. One future project being considered for possible use with Starshade is the Wide Field Infrared Survey Telescope, which will employ a special coronagraph to image larger planets around other stars. If a Starshade is flown, combining it with WFIRST would allow it to detect smaller planets, too.
Something that big is more at risk of micrometeorite strikes; any punctures could mean light getting through and obscuring a telescope’s vision. That’s why JPL turned to an origami-inspired folding pattern, said Manan Arya, a technologist working on Starshade.
“We use multiple layers of material to block starlight, separated by some gaps so that, if we do get hit, there’s a good chance that there won’t be a line-of-sight puncture,” Arya said.
The key was developing algorithms that allow the Starshade to fold smoothly, predictably and repeatedly.
“A huge part of my job is looking at something on paper and asking, ‘Can we fly this?'” Arya said. He could be considered Starshade’s “origamist in chief.” His PhD thesis looked at the use of origami in space superstructures.
A colorful history of space folding inspired him. That includes solar arrays, like those on the International Space Station; experimental wings designed for the space shuttle program in the 1980s; even Echo 1, a 10-story-tall, Earth-orbiting balloon that had to be packed into a 26-inch-diameter (66 centimeters), spherical payload canister before launching.
“Once I realized this is how you fold spacecraft structures, I became interested in origami,” Arya said. “I realized I was good at it and enjoyed it. Now I fold constantly.”
He’s not alone. Robert Salazar, a JPL intern who helped design the Starshade folding pattern, now works on an experimental concept called Transformers for Lunar Extreme Environments. JPL senior research scientist Adrian Stoica leads the project, which would use unfolding, reflective mirrors to bounce the Sun’s rays into deep craters on Earth’s moon. Once deployed, this solar energy could melt water ice or power machinery.
Salazar tests folding designs and materials in a work area littered with scraps, mostly from paper. He also folds Kapton, a tinsel-like material used as spacecraft insulation, and a special polyethylene fabric that doesn’t form permanent creases.
“With most origami, the magic comes from the folding,” Salazar said. “You can’t design purely from geometry. You need to know the qualities of the material to understand how it will fold.”
Salazar has been making origami for 17 years. As a kid, he was inspired by the children’s book “Sadako and the Thousand Paper Cranes.” His own original designs include paper animals. In fact, he folds paper versions of endangered species and donates them to benefit wildlife conservancies.
He said the use of origami in engineering is relatively new and is spurring the publication of technical papers on folding patterns.
“There are so many patterns to still be explored,” Salazar said. “Most designs are for shapes that fold flat. Non-flat structures, like spheres or paraboloids, largely haven’t been done.”
Starshade and the Transformers project are still in their early stages. But Arya points out that we could see space origami very soon. CubeSats are one promising application: these miniaturized satellites are the size of a briefcase, and NASA will launch several key missions using these modular spacecraft in coming years.
Because they require so little space, mass and cost, they’re easier to launch. But CubeSats are limited in what they can do without folding structures, which can pack antennas and other equipment into them.
“That’s an area where I see origami having an increasing role,” Arya said.
Another is robotics. A JPL robot called PUFFER was inspired by origami. Its collapsible body is made from a folding circuit board embedded with fabric. When in use, it pops-up and can climb over rocks or squeeze down under ledges.
In July, NASA placed an open call for origami designs to be used in radiation shielding — another sign that the art form has much to offer the future of space exploration.<|endoftext|>
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Multiples of 11
Here you can find the multiples of 11. The multiples of eleven are the numbers which contain 11 n times; the integer n is called coefficient or multiplier. Also, likewise important, we show you the submultiples of 11. If you like to learn what is a multiple or how to find multiples of a number like 11 then read our article multiples which you can find in the header menu. On this page you can only learn the multiples of the number 11. We start with giving you the definition right below.
A multiple b of 11 is a number such that b = n * 11, n $\epsilon\hspace{3px} \mathbb{Z}$. In other words, a multiple of 11, divided by 11, produces an integer. When you have a certain number b, you can find out if b is multiple of 11 by dividing that number by 11. If the remainder, also known as modulo, is zero, then b is a multiple of 11. If the modulo is different from 0, that is if the division renders decimals, then b is not part of the multiples of 11.
In the same way you can test any number, not just the number eleven.
What are the Multiples of 11
The numbers are 0, 11, 22, 33, … successively. As you can see, getting the list of multiples of 11 is really easy: b = n * 11 with n = 1, 2, 3, … and, although not used very often, 0, by definition, is also a multiple of eleven. Below we have compiled the first 101 numbers:
0, 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132, 143, 154, 165, 176, 187, 198, 209, 220, 231, 242, 253, 264, 275, 286, 297, 308, 319, 330, 341, 352, 363, 374, 385, 396, 407, 418, 429, 440, 451, 462, 473, 484, 495, 506, 517, 528, 539, 550, 561, 572, 583, 594, 605, 616, 627, 638, 649, 660, 671, 682, 693, 704, 715, 726, 737, 748, 759, 770, 781, 792, 803, 814, 825, 836, 847, 858, 869, 880, 891, 902, 913, 924, 935, 946, 957, 968, 979, 990, 1001, 1012, 1023, 1034, 1045, 1056, 1067, 1078, 1089, 1100.
Now you already know what are the multiples of 11. Do you know how many multiples 11 has? The answer is that the number is unlimited. The list of all multiples of 11 is infinite.
BTW: Besides all the multiples of 11, other searched terms on our website include:
The Submultiples of 11
An integer a is a submultiple of 11 only if 11 is a multiple of a. By coming to this page chances are high that you are in fact looking for the submultiples of 11. We will list the numbers right below. Just let us add that you can learn how to get the submultiples of 11 by reading the instructions on our page entitled multiples.
The submultiples of 11 are: 1, 11.
How many submultiples does 11 have? The quantity of submultiples of eleven is 2.
You can obtain the submultiples of any integer using the calculator below. Insert the number, our calculator then produces the result without the need to press a button.
are...
This brings us to the end of our post. We are sure you can now answer questions including, but not limited to, first five multiples of 11?, and name 4 multiples of 11 for example.
If this article about the multiples and submultiples of 11 has been useful to you, then kindly hit the social buttons to let your friends know about our site. If you have any question or suggestion about what are multiples of 11 let us know by filling in the form below or sending us an email.
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# 8.3 Inverse trigonometric functions (Page 4/15)
Page 4 / 15
## Evaluating compositions of the form f ( f−1 ( y )) and f−1 ( f ( x ))
For any trigonometric function, $\text{\hspace{0.17em}}f\left({f}^{-1}\left(y\right)\right)=y\text{\hspace{0.17em}}$ for all $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ was defined to be identical to the domain of $\text{\hspace{0.17em}}{f}^{-1}.\text{\hspace{0.17em}}$ However, we have to be a little more careful with expressions of the form $\text{\hspace{0.17em}}{f}^{-1}\left(f\left(x\right)\right).$
## Compositions of a trigonometric function and its inverse
$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{sin}\left({\mathrm{sin}}^{-1}x\right)=x\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}-1\le x\le 1\hfill \\ \mathrm{cos}\left({\mathrm{cos}}^{-1}x\right)=x\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}-1\le x\le 1\hfill \\ \text{\hspace{0.17em}}\mathrm{tan}\left({\mathrm{tan}}^{-1}x\right)=x\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}-\infty
Is it correct that $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{sin}\text{\hspace{0.17em}}x\right)=x?$
No. This equation is correct if $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ belongs to the restricted domain $\text{\hspace{0.17em}}\left[-\frac{\pi }{2},\frac{\pi }{2}\right],\text{\hspace{0.17em}}$ but sine is defined for all real input values, and for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ outside the restricted interval, the equation is not correct because its inverse always returns a value in $\text{\hspace{0.17em}}\left[-\frac{\pi }{2},\frac{\pi }{2}\right].\text{\hspace{0.17em}}$ The situation is similar for cosine and tangent and their inverses. For example, $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{3\pi }{4}\right)\right)=\frac{\pi }{4}.$
Given an expression of the form f −1 (f(θ)) where evaluate.
1. If $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in the restricted domain of
2. If not, then find an angle $\text{\hspace{0.17em}}\varphi \text{\hspace{0.17em}}$ within the restricted domain of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}f\left(\varphi \right)=f\left(\theta \right).\text{\hspace{0.17em}}$ Then $\text{\hspace{0.17em}}{f}^{-1}\left(f\left(\theta \right)\right)=\varphi .$
## Using inverse trigonometric functions
Evaluate the following:
1. ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{\pi }{3}\right)\right)$
2. ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{2\pi }{3}\right)\right)$
3. ${\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(\frac{2\pi }{3}\right)\right)$
4. ${\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(-\frac{\pi }{3}\right)\right)$
1. so $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{\pi }{3}\right)\right)=\frac{\pi }{3}.$
2. but $\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{2\pi }{3}\right)=\mathrm{sin}\left(\frac{\pi }{3}\right),\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{2\pi }{3}\right)\right)=\frac{\pi }{3}.$
3. so $\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(\frac{2\pi }{3}\right)\right)=\frac{2\pi }{3}.$
4. but $\text{\hspace{0.17em}}\mathrm{cos}\left(-\frac{\pi }{3}\right)=\mathrm{cos}\left(\frac{\pi }{3}\right)\text{\hspace{0.17em}}$ because cosine is an even function.
5. so $\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(\mathrm{cos}\left(-\frac{\pi }{3}\right)\right)=\frac{\pi }{3}.$
Evaluate $\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\left(\mathrm{tan}\left(\frac{\pi }{8}\right)\right)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\left(\mathrm{tan}\left(\frac{11\pi }{9}\right)\right).$
$\frac{\pi }{8};\frac{2\pi }{9}$
## Evaluating compositions of the form f−1 ( g ( x ))
Now that we can compose a trigonometric function with its inverse, we can explore how to evaluate a composition of a trigonometric function and the inverse of another trigonometric function. We will begin with compositions of the form $\text{\hspace{0.17em}}{f}^{-1}\left(g\left(x\right)\right).\text{\hspace{0.17em}}$ For special values of $\text{\hspace{0.17em}}x,$ we can exactly evaluate the inner function and then the outer, inverse function. However, we can find a more general approach by considering the relation between the two acute angles of a right triangle where one is $\text{\hspace{0.17em}}\theta ,\text{\hspace{0.17em}}$ making the other $\text{\hspace{0.17em}}\frac{\pi }{2}-\theta .$ Consider the sine and cosine of each angle of the right triangle in [link] .
Because $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{b}{c}=\mathrm{sin}\left(\frac{\pi }{2}-\theta \right),\text{\hspace{0.17em}}$ we have $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{cos}\text{\hspace{0.17em}}\theta \right)=\frac{\pi }{2}-\theta \text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}0\le \theta \le \pi .\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is not in this domain, then we need to find another angle that has the same cosine as $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and does belong to the restricted domain; we then subtract this angle from $\text{\hspace{0.17em}}\frac{\pi }{2}.$ Similarly, $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{a}{c}=\mathrm{cos}\left(\frac{\pi }{2}-\theta \right),\text{\hspace{0.17em}}$ so $\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(\mathrm{sin}\text{\hspace{0.17em}}\theta \right)=\frac{\pi }{2}-\theta \text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}.\text{\hspace{0.17em}}$ These are just the function-cofunction relationships presented in another way.
Given functions of the form $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{cos}\text{\hspace{0.17em}}x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(\mathrm{sin}\text{\hspace{0.17em}}x\right),\text{\hspace{0.17em}}$ evaluate them.
1. If then $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(\mathrm{cos}\text{\hspace{0.17em}}x\right)=\frac{\pi }{2}-x.$
2. If then find another angle such that $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}y=\mathrm{cos}\text{\hspace{0.17em}}x.$
${\mathrm{sin}}^{-1}\left(\mathrm{cos}\text{\hspace{0.17em}}x\right)=\frac{\pi }{2}-y$
3. If then $\text{\hspace{0.17em}}{\mathrm{cos}}^{-1}\left(\mathrm{sin}\text{\hspace{0.17em}}x\right)=\frac{\pi }{2}-x.$
4. If then find another angle such that $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x.$
${\mathrm{cos}}^{-1}\left(\mathrm{sin}\text{\hspace{0.17em}}x\right)=\frac{\pi }{2}-y$
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# Second Derivative of a Determinant
How does one evaluate the second derivative of the determinant of a square matrix?
Jacobi's formula tells us how to evaluate the first derivative but I can't find anything for the second. This is my attempt:
We can start using the partial derivative formulation of Jacobi's formula, assuming A is invertible: $$\frac{\partial}{\partial \alpha}\det A = (\det A) \text{tr}\left( A^{-1} \frac{\partial}{\partial \alpha} A \right)$$ Taking a second derivative: $$\frac{\partial^2}{\partial \alpha^2}\det A = \frac{\partial}{\partial \alpha}\left[(\det A) \text{tr}\left( A^{-1} \frac{\partial}{\partial \alpha} A \right)\right]$$ Applying product rule: $$= \frac{\partial}{\partial \alpha}(\det A) \cdot \text{tr}\left( A^{-1} \frac{\partial}{\partial \alpha} A \right) + (\det A) \frac{\partial}{\partial \alpha} \text{tr}\left( A^{-1} \frac{\partial}{\partial \alpha} A \right)$$ Replacing $\frac{\partial}{\partial\alpha}\det A$ with Jacobi's formula: $$= (\det A) \text{tr}\left( A^{-1} \frac{\partial}{\partial \alpha} A \right) \cdot \text{tr}\left( A^{-1} \frac{\partial}{\partial \alpha} A \right) +(\det A) \text{tr}\left( \frac{\partial}{\partial \alpha}\left(A^{-1} \frac{\partial}{\partial \alpha} A \right)\right)$$ Factoring out $\det(A)$: $$= (\det A) \left[\text{tr}^2\left( A^{-1} \frac{\partial}{\partial \alpha} A \right) + \text{tr}\left( \frac{\partial}{\partial \alpha}\left(A^{-1} \frac{\partial}{\partial \alpha} A \right)\right)\right]$$ Another product rule: $$= (\det A) \left[\text{tr}^2\left( A^{-1} \frac{\partial}{\partial \alpha} A \right) + \text{tr}\left(\frac{\partial}{\partial \alpha}A^{-1} \frac{\partial}{\partial \alpha} A + A^{-1} \frac{\partial^2}{\partial \alpha^2} A \right)\right]$$ Finally, using $A_{\alpha}$ to denote the partial of A wrt to $\alpha$ we have $$\frac{\partial^2}{\partial \alpha^2}\det A= \det(A) \left[\text{tr}^2\left( A^{-1} A_{\alpha} \right) + \text{tr}\left(A^{-1}_{\alpha} A_{\alpha}\right) + \text{tr} \left( A^{-1} A_{\alpha^2} \right)\right]$$ The second trace actually reduces to N, for an NxN matrix: $$\text{tr}\left(A^{-1}_{\alpha} A_{\alpha}\right)=\text{tr}(I)$$ $$=\sum_{diagonals}1$$ $$=N$$
$$\therefore\frac{\partial^2}{\partial \alpha^2}\det A= \det(A) \left[\text{tr}^2\left( A^{-1} A_{\alpha} \right) + \text{tr} \left( A^{-1} A_{\alpha^2} \right)+N\right]$$
I'm not overly confident with the matrix calculus, but the result is nice enough to seem plausible. Is there another method, or is this proof valid? Thanks!
• Is there any similar results when $A$ is possibly singular? – vulture May 22 '18 at 17:27
## 3 Answers
To keep things simple, assume $A=A(t)$ is a function of a parameter $t$, and we are after the derivatives at zero. Also assume that $A(t)$ is smooth enough in $t$ and that $A(0)$ is invertible. While we are at it, let's replace $A(t)$ by $B(t)=A(0)^{-1}A(t)$ so that $B(0)=I$.
Then $$B(t)=I+tB'(0)+\frac{t^2}2B''(0)+\cdots$$ for some matrices $B_1$ and $B_2$. We now expand $\det B(t)$ by the formula in terms of permutations and their signs. A permutation moving three or more points will lead to a term with a factor of $t^3$ which we can neglect. The identity permutation yields the product of the diagonal elements of $B(t)$ which is \begin{align} &1+t\sum_{i=1}^n B'(0)_{ii} +\frac{t^2}2\sum_{i=1}^n B''(0)_{ii} +t^2\sum_{1\le i<j\le n}B'(0)_{ii}B'(0)_{jj}+\cdots\\ &=1+t\text{ tr}\,B'(0)+\frac{t^2}2\text{ tr}\,B''(0) +t^2\sum_{1\le i<j\le n}B'(0)_{ii}B'(0)_{jj}+\cdots \end{align} The only other permutations counting are the involutions, which together contribute $$-t^2\sum_{1\le i<j\le n}B'(0)_{ij}B'(0)_{ji}+\cdots.$$
The second derivative of $\det B(t)$ at zero is therefore $$\text{tr}\,B''(0)+2\sum_{1\le i<j\le n} (B'(0)_{ii}B'(0)_{jj}-B'(0)_{ij}B'(0)_{ji}).$$ The sum here is the "second trace" of $B'(0)$, the coefficient of $X^{n-2}$ in its characteristic polynomial.
Now if we like we can write $A(t)=A(0)B(t)$ and get a formula for the second derivative of $\det A(t)$.
I think your derivation is good up until you take the derivative of $A^{-1}$. From formula (40) of the Matrix Cookbook the derivative is:
$$\partial \left(X^{-1}\right) = - X^{-1} \left( \partial X \right) X^{-1}$$
so instead of a second term of $\operatorname{tr} \left( A_\alpha^{-1} A_\alpha \right)$ it should instead be $-\operatorname{tr} \left( A^{-1} A_\alpha A^{-1} A_\alpha \right)$. The final result then is:
$$\det \left(A\right) \left( \left(\operatorname{tr}\left(A^{-1} A_\alpha\right)\right)^2 + \operatorname{tr}\left(A^{-1} A_{\alpha\alpha}\right)-\operatorname{tr}\left(A^{-1}A_\alpha A^{-1} A_\alpha\right)\right)$$
It is a simple matter to confirm this formula symbolically for small examples in your CAS of choice (for a Mathematica implementation see my answer to a similar question on MSE).
As pointed out by Carl, your mistake is to permute the inverse and derivative operator:
$$\partial_\alpha(A^{-1}) \neq (\partial_\alpha A)^{-1} \, .$$
You can easily generalize it for whatever two differentiating arguments ($\alpha$ and $\beta$). Starting from Jacobi's formula:
$$\partial_{\alpha}(\mathrm{det}(A)) = \mathrm{det}(A) \, \mathrm{tr}\left(A^{-1} \, \partial_\alpha A\right) \,$$
applying on it the chain rule:
$$\partial_{\alpha\beta}(\mathrm{det}(A)) = \partial_\beta(\mathrm{det}(A)) \, \mathrm{tr}\left(A^{-1} \, \partial_\alpha A\right) + \mathrm{det}(A) \, \mathrm{tr}\left(\partial_\beta \left(A^{-1} \, \partial_\alpha A\right)\right) \, ,$$
using Jacobi's formula on the first $\beta$ derivative and applying the chain rule on the second:
$$\partial_{\alpha\beta}(\mathrm{det}(A)) = \mathrm{det}(A) \, \mathrm{tr}\left(A^{-1} \, \partial_\beta A\right)\,\mathrm{tr}\left(A^{-1} \, \partial_\alpha A\right) + \mathrm{det}(A) \, \mathrm{tr}\left(\partial_\beta \left(A^{-1}\right) \, \partial_\alpha A\right) + \mathrm{det}(A) \, \mathrm{tr}\left(A^{-1} \, \partial_{\alpha\beta} A\right)\, ,$$
and finally, factoring out the determinant and applying the relation for the inverse $\partial_\beta(A^{-1}) = -A^{-1} \, \partial_\beta A \, A^{-1}$:
$$\partial_{\alpha\beta}(\mathrm{det}(A)) = \mathrm{det}(A) \, \left[\mathrm{tr}\left(A^{-1} \, \partial_\beta A\right)\,\mathrm{tr}\left(A^{-1} \, \partial_\alpha A\right) - \mathrm{tr}\left(A^{-1} \, \partial_\beta A \, A^{-1} \, \partial_\alpha A\right) + \mathrm{tr}\left(A^{-1} \, \partial_{\alpha\beta} A\right)\right] \, ,$$
which is the same as what Carl gave for $\beta=\alpha$:
$$\partial_{\alpha\alpha}(\mathrm{det}(A)) = \mathrm{det}(A) \, \left[\left(\mathrm{tr}\left(A^{-1} \, \partial_\alpha A\right)\right)^2 - \mathrm{tr}\left(A^{-1} \, \partial_\alpha A \, A^{-1} \, \partial_\alpha A\right) + \mathrm{tr}\left(A^{-1} \, \partial_{\alpha\alpha} A\right)\right] \, .$$<|endoftext|>
| 4.5 |
1,671 |
Exterior angle are defined as the angles formed in between the next of the polygon and the extended surrounding side that the polygon. The exterior edge theorem says that once a triangle's side is extended, the result exterior angle developed is equal to the sum of the steps of the two opposite inner angles the the triangle. The theorem deserve to be provided to uncover the measure up of one unknown angle in a triangle. To use the theorem, we an initial need to recognize the exterior angle and then the connected two remote internal angles of the triangle.
You are watching: What is the sum of the angle measures in a heptagon?
1 What is Exterior edge Theorem? 2 Proof the Exterior angle Theorem 3 Exterior angle Inequality Theorem 4 FAQs ~ above Exterior angle Theorem
## What is Exterior angle Theorem?
The exterior edge theorem states that the measure up of an exterior edge is equal to the sum of the procedures of the two opposite(remote) internal angles of the triangle. Let united state recall a couple of common properties around the angles of a triangle: A triangle has 3 internal angles which constantly sum up to 180 degrees. It has actually 6 exterior angles and also this to organize gets applied to every of the exterior angles. Keep in mind that one exterior edge is supplementary come its nearby interior angle as they form a straight pair the angles.
We have the right to verify the exterior angle theorem through the known properties that a triangle. Consider a Δ ABC.
The three angles a + b + c = 180 (angle sum property of a triangle) ----- Equation 1
c= 180 - (a+b) ----- Equation 2 (rewriting equation 1)
e = 180 - c----- Equation 3 (linear pair that angles)
Substituting the value of c in equation 3, us get
e = 180 - <180 - (a+b)>
e = 180 - 180 + (a + b)
e = a + b
Hence verified.
## Proof that Exterior edge Theorem
Consider a ΔABC. A, b and also c room the angle formed. Prolong the next BC to D. Now an exterior edge ∠ACD is formed. Attract a heat CE parallel come AB. Currently x and also y are the angles formed, where, ∠ACD = ∠x + ∠y
StatementReason
∠a = ∠xPair of alternating angles. (Since BA is parallel come CE and AC is the transversal).
∠b = ∠yPair of corresponding angles. (Since BA is parallel to CE and also BD is the transversal).
∠a + ∠b = ∠x + ∠yFrom the above statements
∠ACD = ∠x + ∠yFrom the building and construction of CE
∠a + ∠b = ∠ACDFrom the above statements
Hence showed that the exterior edge of a triangle is same to the amount of the 2 opposite interior angles.
## Exterior angle Inequality Theorem
The exterior edge inequality theorem states that the measure up of any kind of exterior angle of a triangle is greater than either of the opposite internal angles. This condition is solve by every the six external angles the a triangle.
### Exterior angle Theorem connected Articles
Check the end a couple of interesting posts related to Exterior edge Theorem.
Important notes
The exterior angle theorem claims that the measure up of one exterior edge is same to the amount of the measures of the 2 remote internal angles of the triangle.The exterior edge inequality theorem states that the measure of any exterior edge of a triangle is higher than one of two people of the opposite inner angles.The exterior angle and also the nearby interior angle are supplementary. All the exterior angles of a triangle amount up to 360º.
## Solved Examples
Example 1: uncover the values of x and also y by using the exterior angle theorem that a triangle.
Solution:
∠x is the exterior angle.
∠x + 92 = 180º (linear pair that angles)
∠x = 180 - 92 = 88º
Applying the exterior angle theorem, us get, ∠y + 41 = 88
∠y = 88 - 41 = 47º
Therefore, the worths of x and y space 88º and 47º respectively.
Example 2: discover BAC and ABC.
Solution:
160º is an exterior angle of the Δ ABC. So, by using the exterior edge theorem, us have, ∠BAC + ∠ABC = 160º
x + 3x = 160º
4x = 160º
x = 40º
Therefore, ∠BAC = x = 40º and ∠ABC = 3xº = 120º
Solution:
Solving the direct pair in ~ vertex D, we acquire ∠ADC + ∠ADE = 180º
∠ADC = 180º - 150º = 30º
Using the angle amount property, because that Δ ACD,
180º - 2 × 30º
∠ACD = 180º - 60º = 120º
∠ACD is the exterior angle of ∠ABC
Using the exterior edge theorem, because that Δ ABC, ∠ACD = ∠ABC + ∠BAC
120º = 60º + ∠BAC
Therefore, ∠BAC = 120º - 60º = 60º.
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## FAQs on Exterior edge Theorem
### What is the Exterior angle Theorem?
The exterior edge theorem states that the measure of one exterior edge is equal to the amount of the actions of the two remote inner angles that the triangle. The remote interior angles are also called opposite inner angles.
### How do you use the Exterior edge Theorem?
To usage the exterior edge theorem in a triangle we an initial need to identify the exterior angle and then the connected two remote internal angles that the triangle. A typical mistake of considering the adjacent interior angle have to be avoided. After identifying the exterior angles and also the related interior angles, we can apply the formula to find the absent angles or to develop a relationship between sides and also angles in a triangle.
### What room Exterior Angles?
An exterior angle of a triangle is developed when any side of a triangle is extended. There are 6 exterior angles of a triangle as each of the 3 sides can be prolonged on both sides and 6 together exterior angles are formed.
### What is the Exterior angle Inequality Theorem?
The measure of an exterior angle of a triangle is constantly greater 보다 the measure of either of the opposite interior angles that the triangle.
### What is the Exterior edge Property?
An exterior angle of a triangle is same to the sum of its 2 opposite non-adjacent interior angles. The amount of the exterior angle and the surrounding interior angle the is not opposite is equal to 180º.
### What is the Exterior edge Theorem Formula?
The sum of the exterior angle = the sum of two non-adjacent inner opposite angles. One exterior edge of a triangle is equal to the sum of its two opposite non-adjacent internal angles.
### Where should We use Exterior edge Theorem?
Exterior edge theorem can be supplied to identify the procedures of the unknown interior and also exterior angles of a triangle.
See more: Is The Sequence Geometric 6 12 24,, Given The Geometric Sequence {6, 12, 24,
### Do every Polygons Exterior Angles add up to 360?
The exterior angles of a polygon are formed when a next of a polygon is extended. Every the exterior angle in every the polygons sum up come 360º.<|endoftext|>
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# Find the cubic polynomial whose graph passes through the points ( -1, -1), (0, 1), (1, 3), (4, -1).
Question
Polynomial graphs
Find the cubic polynomial whose graph passes through the points $$\displaystyle{\left(-{1},-{1}\right)},{\left({0},{1}\right)},{\left({1},{3}\right)},{\left({4},-{1}\right)}.$$
2021-02-15
Step 1
Denote the interpolating cubic polynomial by
$$\displaystyle{p}{\left({x}\right)}={a}_{{{0}}}+{a}_{{{1}}}{x}+{a}_{{{2}}}{x}^{{{2}}}+{a}_{{{3}}}{x}^{{{3}}}$$
Let's substitute the given points in the equation of the polynomial. The points $$\displaystyle{\left(-{1},-{1}\right)},{\left({0},{1}\right)},{\left({1},{3}\right)}{\quad\text{and}\quad}{\left({4},-{1}\right)}$$ thus have to satisfy.
$$\displaystyle-{1}={a}_{{{0}}}-{a}_{{{1}}}+{a}_{{{2}}}-{a}_{{{3}}}$$
$$\displaystyle{1}={a}_{{{0}}}$$
$$\displaystyle{3}={a}_{{{0}}}+{a}_{{{1}}}+{a}_{{{2}}}+{a}_{{{3}}}$$
$$\displaystyle-{1}={a}_{{{0}}}+{4}{a}_{{{1}}}+{16}{a}_{{{2}}}+{64}{a}_{{{3}}}$$
First, let's substitute $$\displaystyle{a}_{{{0}}}={1}$$ into the three remaining equations. That way we reduced the original system to a system of three equations with three unknowns:
$$\displaystyle-{a}_{{{1}}}+{a}_{{{2}}}-{a}_{{{3}}}=-{2}$$
$$\displaystyle{a}_{{{1}}}+{a}_{{{2}}}+{a}_{{{3}}}={2}$$
$$\displaystyle{4}{a}_{{{1}}}+{16}{a}_{{{2}}}+{64}{a}_{{{3}}}=-{2}$$
Let's solve this system using Gauss-Jordan elimination. The augmented matrix of the system is.
$$\begin{bmatrix}-1 & 1 & -1 & -2 \\1 & 1 & 1 & 2\\ 4 & 16 & 64 & -2 \end{bmatrix}$$
Step 2
Multiply the first row by -1.
$$\begin{bmatrix}1 & -1 & 1 & 2 \\1 & 1 & 1 & 2\\ 4 & 16 & 64 & -2 \end{bmatrix}$$
Add -1 times the first row to the second row, Add -4 times the first row to the third row.
$$\begin{bmatrix}1 & -1 & 1 & 2 \\0 & 2 & 0 & 0\\ 0 & 20 & 60 & -10 \end{bmatrix}$$
Multiply the second row by $$\displaystyle\frac{{1}}{{2}}$$
$$\begin{bmatrix}1 & -1 & 1 & 2 \\0 & 1 & 0 & 0\\ 0 & 20 & 60 & -10 \end{bmatrix}$$
Add -20 times the second row to the third row.
$$\begin{bmatrix}1 & -1 & 1 & 2 \\0 & 1 & 0 & 0\\ 0 & 0 & 60 & -10 \end{bmatrix}$$
Multiply the third row by $$\displaystyle\frac{{1}}{{60}}$$.
$$\begin{bmatrix}1 & -1 & 1 & 2 \\0 & 1 & 0 & 0\\ 0 & 0 & 1 & -1/6 \end{bmatrix}$$
Add -1 times the third row to the first row.
$$\begin{bmatrix}1 & -1 & 0 & 13/6 \\0 & 1 & 0 & 0\\ 0 & 0 & 1 & -1/6 \end{bmatrix}$$
Add 1 times the second row to the first row.
$$\begin{bmatrix}1 & 0 & 0 & 13/6 \\0 & 1 & 0 & 0\\ 0 & 0 & 1 & -1/6 \end{bmatrix}$$
The solution is thus
$$=\frac{13}{6}, a_{2}=0,a_{3}=-\frac{1}{6}$$
Along with $$\displaystyle{a}_{{{0}}}={1}$$, the interpolating polynomial is
$$\displaystyle{p}{\left({x}\right)}={1}+{\frac{{{13}}}{{{6}}}}{x}-{\frac{{{1}}}{{{6}}}}{x}^{{{3}}}$$
### Relevant Questions
The quadratic function $$\displaystyle{y}={a}{x}^{2}+{b}{x}+{c}$$ whose graph passes through the points (1, 4), (2, 1) and (3, 4).
Use the appropriate Lagrange interpolating polynomials to find the cubic polynomial whose graph passes through the given points. $$\displaystyle{\left({2},\ {1}\right)},\ {\left({3},\ {1}\right)},\ {\left({4},\ −{3}\right)},\ {\left({5},\ {0}\right)}$$.
Use the appropriate Lagrange interpolating polynomials to find the cubic polynomial whose graph passes through the given points. $$\displaystyle{\left({1},{2}\right)},{\left({2},{1}\right)},{\left({3},{3}\right)},{\left({6},{1}\right)}{\left({1},{2}\right)},{\left({2},{1}\right)},{\left({3},{3}\right)},{\left({6},{1}\right)}$$.
For the following exercises, use the given information about the polynomial graph to write the equation. Double zero at $$\displaystyle{x}=−{3}$$ and triple zero $$\displaystyle{a}{t}{x}={0}$$. Passes through the point (1, 32).
A stunt man whose mass is 70 kg swings from the end ofa 4.0 m long rope along thearc of a vertical circle. Assuming that he starts from rest whenthe rope is horizontal, find the tensions in the rope that are required to make him follow his circular path at each of the following points.
(a) at the beginning of his motion N
(b) at a height of 1.5 m above the bottom of the circular arc N
(c) at the bottom of the arc N
Consider the curves in the first quadrant that have equationsy=Aexp(7x), where A is a positive constant. Different valuesof A give different curves. The curves form a family,F. Let P=(6,6). Let C be the number of the family Fthat goes through P.
A. Let y=f(x) be the equation of C. Find f(x).
B. Find the slope at P of the tangent to C.
C. A curve D is a perpendicular to C at P. What is the slope of thetangent to D at the point P?
D. Give a formula g(y) for the slope at (x,y) of the member of Fthat goes through (x,y). The formula should not involve A orx.
E. A curve which at each of its points is perpendicular to themember of the family F that goes through that point is called anorthogonal trajectory of F. Each orthogonal trajectory to Fsatisfies the differential equation dy/dx = -1/g(y), where g(y) isthe answer to part D.
Find a function of h(y) such that x=h(y) is the equation of theorthogonal trajectory to F that passes through the point P.
For the equation (-1,2), $$y= \frac{1}{2}x - 3$$, write an equation in slope intercept form for the line that passes through the given point and is parallel to the graph of the given equation.
A graph of a linear equation passes through ( -2,0) and (0,-6) is the $$3x-y=6$$, both ordered pairs solutions for the equation
Graph the polynomial in the given viewing rectangle. Find the coordinates of all local extrema. State each answer rounded to two decimal places. State the domain and range. $$\displaystyle{y}={x}^{{{3}}}-{3}{x}^{{{2}}},{\left[-{2},{5}\right]}{b}{y}{\left[-{10},{10}\right]}$$<|endoftext|>
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Listen to today's episode of StarDate on the web the same day it airs in high-quality streaming audio without any extra ads or announcements. Choose a $8 one-month pass, or listen every day for a year for just $30.
You are here
Moon, Jupiter, and Aldebaran
A planet that may have helped give the Moon a pummeling huddles close to the Moon tomorrow morning. Jupiter looks like a brilliant star to the lower left of the Moon. They climb into view by about 2 or 2:30, and are high in the sky at first light. And to spice things up a bit, the bright orange star Aldebaran is about the same distance to the lower right of Jupiter.
Earth, the Moon, and the other planets of the inner solar system took a beating from giant asteroids during the solar system’s first half-billion years or so. But a recent study by researchers at the NASA Lunar Science Institute found that it was especially intense at the end of that period — about four billion years ago.
The researchers studied the contours of craters around an ancient volcanic plain. The observations suggest the craters were formed by asteroids that were moving much faster than those that hit the Moon before or after.
The bombardment may have been triggered by changes in the orbits of Jupiter and Saturn. As their distance from the Sun changed, the gravity of the two planets — particularly Jupiter — kicked a group of large asteroids toward the inner solar system. Some of the asteroids then slammed into the Moon and Earth. Wind, rain, and the motions of the crust have erased any traces of this bombardment on Earth, but it’s preserved in the craters of the Moon — the scars of an ancient pummeling.
More about the Moon and its companions tomorrow.
Script by Damond Benningfield, Copyright 2012
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ON THIS PAGE: You will learn about how doctors describe a cancer’s growth or spread. This is called the stage. Use the menu to see other pages.
Staging is a way of describing where the cancer is located, if or where it has spread, and whether it is affecting other parts of the body.
Doctors use diagnostic tests to find out the cancer's stage, so staging may not be complete until all the tests are finished. Knowing the stage helps the doctor to decide what kind of treatment is best and can help predict a patient's prognosis, which is the chance of recovery. There are different stage descriptions for different types of cancer.
TNM staging system
One tool that doctors use to describe the stage is the TNM system. Doctors use the results from diagnostic tests and scans to answer these questions:
Tumor (T): How large is the primary tumor? Where is it located?
Node (N): Has the tumor spread to the lymph nodes? If so, where and how many?
Metastasis (M): Has the cancer spread to other parts of the body? If so, where and how much?
The results are combined to determine the stage of cancer for each person. There are 5 stages: stage 0 (zero) and stages I through IV (1 through 4), although stage 0 kidney cancer is extremely rare. The stage provides a common way of describing the cancer so doctors can work together to plan the best treatments.
Here are more details on each part of the TNM system for kidney cancer.
Using the TNM system, the "T" plus a letter or number (0 to 4) is used to describe the size and location of the tumor. Tumors are measured in centimeters (cm). One cm is roughly equal to the width of a standard pen or pencil. One inch equals about 2.5 cm.
Stage may also be divided into smaller groups that help describe the tumor in even more detail. This helps the doctor develop the best treatment plan for each patient. If there is more than 1 tumor, the lowercase letter "m" (multiple) is added to the "T" stage category. Specific tumor stage information for kidney cancer is listed below.
TX: The primary tumor cannot be evaluated.
T0 (T plus zero): No evidence of a primary tumor.
T1: The tumor is found only in the kidney and is 7 cm or smaller at its largest area. There has been much discussion among doctors about whether this classification should only include a tumor that is 5 cm or smaller.
T1a: The tumor is found only in the kidney and is 4 cm or smaller at its largest area.
T1b: The tumor is found only in the kidney and is between 4 cm and 7 cm at its largest area.
T2: The tumor is found only in the kidney and is larger than 7 cm at its largest area.
T2a: The tumor is only in the kidney and is more than 7 cm but not more than 10 cm at its largest area.
T2b: The tumor is only in the kidney and is more than 10 cm at its largest area.
T3: The tumor has grown into major veins within the kidney or perinephric tissue, which is the connective, fatty tissue around the kidneys. However, it has not grown into the adrenal gland on the same side of the body as the tumor. The adrenal glands are located on top of each kidney and produce hormones and adrenaline to help control heart rate, blood pressure, and other bodily functions. In addition, the tumor has not spread beyond Gerota's fascia, an envelope of tissue that surrounds the kidney.
T3a: The tumor has spread to the large vein leading out of the kidney, called the renal vein, or the branches of the renal vein; the fat surrounding and/or inside the kidney; or the pelvis and calyces of the kidney, which collect urine before sending it to the bladder. The tumor has not grown beyond Gerota's fascia.
T3b: The tumor has grown into the large vein that drains into the heart, called the inferior vena cava, below the diaphragm. The diaphragm is the muscle under the lungs that helps breathing.
T3c: The tumor has spread to the vena cava above the diaphragm and into the right atrium of the heart or to the walls of the vena cava.
T4: The tumor has spread to areas beyond Gerota's fascia and extends into the adrenal gland on the same side of the body as the tumor.
The “N” in the TNM staging system stands for lymph nodes. These tiny, bean-shaped organs help fight infection. Lymph nodes near the kidneys are called regional lymph nodes. Lymph nodes in other parts of the body are called distant lymph nodes.
NX: The regional lymph nodes cannot be evaluated.
N0 (N plus zero): The cancer has not spread to the regional lymph nodes.
N1: The cancer has spread to regional lymph nodes.
The "M" in the TNM system indicates whether the cancer has spread to other parts of the body, called distant metastasis. Common areas where kidney cancer may spread include the bones, liver, lungs, brain, and distant lymph nodes.
M0 (M plus zero): The disease has not metastasized.
M1: The cancer has spread to other parts of the body beyond the kidney area.
Cancer stage grouping
Doctors assign the stage of the cancer by combining the T, N, and M classifications.
Stage I: The tumor is 7 cm or smaller and is only located in the kidney. It has not spread to the lymph nodes or distant organs (T1, N0, M0).
Stage II: The tumor is larger than 7 cm and is only located in the kidney. It has not spread to the lymph nodes or distant organs (T2, N0, M0).
Stage III: Either of these conditions:
A tumor of any size is located only in the kidney. It has spread to the regional lymph nodes but not to other parts of the body (T1 or T2, N1, M0).
The tumor has grown into major veins or perinephric tissue and may or may not have spread to regional lymph nodes. It has not spread to other parts of the body (T3, any N, M0).
Stage IV: Either of these conditions:
The tumor has spread to areas beyond Gerota's fascia and extends into the adrenal gland on the same side of the body as the tumor, possibly to lymph nodes, but not to other parts of the body (T4, any N, M0).
The tumor has spread to any other organ, such as the lungs, bones, or the brain (any T, any N, M1).
Recurrent: Recurrent cancer is cancer that has come back after treatment. It may be found in the kidney area or in another part of the body. If the cancer does return, there will be another round of tests to learn about the extent of the recurrence. These tests and scans are often similar to those done at the time of the original diagnosis.
Used with permission of the American College of Surgeons, Chicago, Illinois. The original and primary source for this information is the AJCC Cancer Staging Manual, Eighth Edition (2017), published by Springer International Publishing.
It is important for doctors to learn as much as possible about the tumor. This information can help them predict if the cancer will grow and spread or how effective treatment will be. This information includes:
Cell type, such as clear cell or another type
Grade, which describes how similar the cancer cells are to healthy cells
Personal information, such as the person’s activity level and body weight
Presence or absence of fevers, sweats, and other symptoms
Information about the cancer’s stage will help the doctor recommend a specific treatment plan. The next section in this guide is Types of Treatment. Use the menu to choose a different section to read in this guide.<|endoftext|>
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# Exercise 8.1 Class 9 Maths Chapter 8 Quadrilaterals
NCERT Class 9 Maths Solutions PDF: In this post, we have discussed the solution of the Maths Class 9 book Chapter 8. Solutions are given below with proper explanation and utmost care has been taken to ensure that the solutions are correct. The answers provided will not only help in completing all the assignments but also help students in clearing their concepts. Students can download the solutions by printing the chapters by using the command Ctrl+P in Google Chrome and saving it in PDF format. All the best !! Please support us by sharing this website with your school friends.
## Exercise 8.1 Class 9 Maths Chapter 8 Quadrilaterals Solutions
1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
4. Show that the diagonals of a square are equal and bisect each other at right angles.
5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
6. Diagonal AC of a parallelogram ABCD bisects ∠ A (see Fig. 8.19). Show that
(i) it bisects ∠ C also,
(ii) ABCD is a rhombus.
7. ABCD is a rhombus. Show that diagonal AC bisects ∠ A as well as ∠ C and diagonal BD bisects ∠ B as well as ∠ D.
8. ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C. Show that:
(i) ABCD is a square (ii) diagonal BD bisects ∠ B as well as ∠ D.
9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:
(i) ∆ APD ≅ ∆ CQB
(ii) AP = CQ
(iii) ∆ AQB ≅∆ CPD
(iv) AQ = CP
(v) APCQ is a parallelogram
10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that
(i) ∆ APB ≅ ∆ CQD
(ii) AP = CQ
11. In ∆ ABC and ∆ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram<|endoftext|>
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# Cube And Cube Roots
Exercise 1
1. Which of the following numbers are not perfect cubes?
(i) 216
Answer: 216 = 2 x 2 x 2 x 27
= 2 x 2 x 2 x 3 x 3 x 3
As number of 2s and 3s is 3 in the factorization so it is a perfect cube.
(ii) 128
Answer: 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Number of 2s is 7 and 7 is not divisible by three so 128 is not a perfect cube
(iii) 1000
Answer: 1000 = 2 x 2 x 2 x 5 x 5 x 5
Number of 2s and 5s is 3 each so 1000 is a perfect cube.
(iv) 100
Answer: 100 = 2 x 2 x 5 x 5
Number of 2s and 5s is 2 each and not 3 so 100 is not a perfect cube.
(v) 46656
Answer: 46656 = 2 x 2 x 2 x 5832
= 2 x 2 x 2 x 2 x 2 x 2 x 729
= 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
Number of 2s and 3s is 6 each and 6 is divisible by 3 so 46656 is a perfect cube
2. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243
Answer: 243 = 3 x 3 x 3 x 3 x 3
Number of 3s is 5, so we need to another 3 in the factorization to make 243 a perfect cube. 243 multiplied by 3 will be a perfect cube.
(ii) 256
Answer: 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
Number of 2s is 8 so 256 needs to be multiplied by 2 to become a perfect cube.
(iii) 72
Answer: 72 = 2 x 2 x 2 x 3 x 3
Number of 2s is 3 and that of 3s is 2, so 72 needs to be multiplied by 3 to become a perfect cube.
(iv) 675
Answer: 675 = 5 x 5 x 27 = 5 x 5 x 3 x 3 x 3
675 needs to be multiplied by 5 to become a perfect cube.
(v) 100
Answer: 100 = 10 x 10
100 needs to be multiplied by 10 to become a perfect cube.
3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81
Answer: 81 = 3 x 3 x 3 x 3
81 needs to be divided by 3 to become a perfect cube.
(ii) 128
Answer: 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
128 needs to be divided by 2 to become a perfect cube.
(iii) 135
Answer: 135 = 5 x 3 x 3 x 3
135 needs to be divide by 5 to become a perfect cube.
(iv) 192
Answer: 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3
192 needs to be divided by 3 to become a perfect cube.
(v) 704
Answer: 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11
704 needs to be divided by 11 to become a perfect cube.
4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Answer: To find the answer we need to calculate the LCM of 5 and 2. The cube with each sides measuring equal to the LCM will give the dimensions of required cube.
LCM of 5 and 2 = 10
Volume of cuboid = 5 x 5 x 2 cubic cms
Volume of bigger cube=103=1000 cubic cms
Number of cuboids required to make the bigger cube =
Cube And Cube Roots will be available online in PDF book form soon. The solutions are absolutely Free. Soon you will be able to download the solutions.<|endoftext|>
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Cohesion: linking words and phrases. 1.33 Cohesion: linking words and phrases You can use words or short phrases which help to guide your reader through your writing, and to link sentences, paragraphs and sections both forwards and backwards.
Good use will make what you have written easy to follow; bad use might mean your style is disjointed, probably with too many short sentences, and consequently difficult to follow. Your mark could be affected either way. The best way to "get a feel" for these words is through your reading. Most textbooks and articles are well-written and will probably include a lot of these cohesive devices. Don't forget "AND"! There follows a list of words and phrases that can be used. Here are just a few examples of some of the words in action: Desktop computers are cheaper and more reliable than laptops; furthermore, they are more flexible. Prices fell by more than 20% last year. On the whole, his speech was well received, despite some complaints from new members. Top of page Transition word exercise. Online Proofreader: Pre-grade your paper.
Oxford University Press. Academic readiness Q: Skills for Success Second Edition helps students to get ready for academic success.
Enhanced skills support provides four extra pages of reading or listening comprehension in every unit, deepening students’ understanding of the unit topic and better preparing them for the unit assignment. A greater variety of activities encourages students to use critical thinking skills, such as making inferences or synthesizing information from different texts. Video in every unit adds a new dimension to the course, and provides an additional springboard for students to think critically. The documentary-style videos use material from the BBC and CBS, providing authentic, high-interest input related to the unit topic. In the Listening and Speaking strand, the new note-taking skills section provides focused practice on this essential skill in every unit.
Measurable progress The Second Edition has an increased focus on measuring student progress. Blended learning. The Writing for Assignments E-library Project. What is it?
The Writing for Assignments E-Library project is a resource to help people learn about writing at university. It includes examples of student essays and other university assignments from a number of subject areas comments from lecturers about why the writing is good - or how it could be improved How to use it Use the search boxes on the right to find examples of student writing by subject, level or keyword. The speech bubbles are different colours to indicate the different types of comment. What’s in it for me? Learning by example is known to be effective. Tell us what you think! We welcome feedback and comments about how to improve the resource.
Writing for University. Guide to undergraduate dissertations. Writing Tutorials. Welcome to the Purdue University Online Writing Lab (OWL) Academic Writing:Introduction. Just as there are differences in the way we use language for speech and for writing, there are also differences in the way we write for different situations; for example, compare the following written texts: Dear Mark, My accountant friend thinks that phone company shares are a good buy at the moment so I think I'll move some money into them...
Some financial advisors recommend purchasing communication industry shares. Obviously, the second way of phrasing the idea is much more appropriate for your university assignment. Academic Writing Practice - Links. To view each resource click on the links below: Academic writing The links in this section take you to websites providing step by step guidance and/or practice material for different aspects of academic writing.
University of Southampton Academic Skills A useful set of academic skills resources. HKPU: Writing Teacher-produced materials to help with report writing. Available from the Hong Kong Polytechnic University website. Palgrave-Macmillan: Skills4study- essay writing This site provides useful notes guiding you through the different stages of writing an essay. Language and Learning Online A comprehensive set of resources for academic writing. Paraphrase: write it in your own words Practice exercises in paraphrasing with sample answers provided by Purdue University.<|endoftext|>
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An odd extinct mammal that lived in South America during the last ice age had a long neck like a llama's, three-toed feet like a rhino's and what may have been a tapir-like trunk. This peculiar combination of traits fueled a mystery lasting nearly two centuries about how to classify the bizarre beast.
The Macrauchenia genus has puzzled scientists since Charles Darwin discovered limb bones and vertebrae fossils "of some very large animal" in Patagonia and fancied it to be a mastodon, as he wrote in a letter to his mentor, the naturalist John Stevens Henslow, in March 1834. Upon analyzing Darwin's finds, the scientist Sir Richard Owen declared in a species description published in 1838 that the creature resembled a camel, but uncertainty remained about where Macrauchenia fit on the mammal family tree.
The recent discovery of a rare DNA sample from the unusual species provided a crucial missing piece: genetic evidence confirming Macrauchenia lineage and its closest relatives, scientists reported in a new study. [In Images: 'Field Guide' Showcases Bizarre and Magnificent Prehistoric Mammals]
Macrauchenia fossils are fairly plentiful, but paleontologists nonetheless struggled to understand the creature because its combination of features was so unusual, said study co-author Ross MacPhee, curator in the mammalogy department at the American Museum of Natural History in New York.
From these fossils, scientists know that Macrauchenia lived in what is now South America until roughly the end of the Pleistocene epoch (about 1.8 million to 11,700 years ago), and went extinct around 10,000 years ago, MacPhee told Live Science. The long-necked animal was about the size of the average horse, and had a long, narrow skull that was also vaguely horse-like. But its nasal aperture sat squarely between its eyes, leading researchers to speculate that it had either a type of muscular trunk like an elephant's or a fleshy protuberance like a tapir's, MacPhee explained.
Because of these physical features, Macrauchenia was long thought to belong to the branch of the mammalian family tree known as Perissodactyls, which includes tapirs, horses and rhinos. But that group wasn't a perfect fit for Macrauchenia or for other bizarre ice age mammals that were native only to South America, said study co-author Michael Hofreiter, a professor of evolutionary adaptive genomics at the University of Potsdam in Germany.
"These animals are so weird — and their potential relatives are so weird compared to all living mammals," Hofreiter told Live Science. "People went back and forth, and never could put them securely on the tree."
It's not that experts doubted that Macrauchenia was related to Perissodactyls; the trouble was that it looked like it could also be related to a lot of other groups as well, MacPhee said.
Biologists confirm evolutionary relationships of living animals by comparing their DNA. But for paleontologists who are looking at extinct animals, just finding a viable sample of DNA in a fossil can be an enormous challenge (or "a hideous problem," MacPhee said).
"It really depends on the environment," Hofreiter said. Permafrost preserves DNA extremely well, so in those areas, paleontologists can be fairly confident that most fossils will have some viable DNA. But near the equator, where organic matter degrades quickly in the warm, moist environment, hardly any fossils have DNA, he said.
"In between these extremes, it depends on local conditions," Hofreiter said.
And even then, there are limits to DNA preservation; it's unlikely to be preserved for more than a million years, according to MacPhee. That may sound like a staggering amount of time, but in geologic terms, a million years is barely any time at all, MacPhee said.
One out of 17
For the study, researchers looked for DNA in six Macrauchenia fossils and 11 fossils from Toxodon — a genus of South American mammal resembling a hornless rhino and a relative of Macrauchenia. They found one sample of usable mitochondrial DNA, in a Macrauchenia fossil from a cave in Chile. (Mitochondrial DNA resides in energy-making organelles in the body and is passed down only from the mother.)
That sample was about 2 to 3 percent DNA from Macrauchenia, with the rest belonging to assorted microorganisms that had colonized the bone, Hofreiter told Live Science. From that sample, the study authors recovered about 80 percent of Macrauchenia's mitochondrial genome, offering them more precise points of comparison to the Perissodactyl group, to see whether the odd species belonged there.
The researchers learned that Macrauchenia is, in fact, closely related to horses, rhinos and tapirs. However, it is not part of the Perissodactyl group, they found. The odd animal shared a common ancestor with Perissodactyls that dates to roughly 66 million years ago, but around that time, it split off into its own lineage, which died out during the last ice age and left no relatives alive today.
Unlike side-by-side comparisons of physical features in fossils, molecular paleontology can provide definitive answers about genetic relationships, eliminating much of the uncertainty about which animals are related, MacPhee said.
"It gives you 'yes' and 'no' answers instead of lots of 'maybes,'" he said. [What the Heck?! Images of Evolution's Extreme Oddities]
A different branch
A separate study from 2015 found genetic evidence suggesting that Macrauchenia's lineage diverged from Perissodactyls more than 60 million years ago, which the authors discovered by evaluating proteins extracted from collagen in fossils.
But using preserved collagen in this way is still a relatively new process — only a few years old — and the new findings corroborate the 2015 results using more traditional mitochondrial DNA analysis, MacPhee said.
"We were able to show that we got [the] exact same results," MacPhee said. "We placed it [Macrauchenia] next to the modern Perissodactyl group — related to, but not inside modern Perissodactyls," he said.
Resolving where extinct oddballs like Macrauchenia fit on the tree of life answers important questions about ancient evolutionary relationships and biodiversity, and offers insight into how biodiversity millions of years ago came about — and how it could disappear, Hofreiter told Live Science.
"In the Pleistocene, we lost an entire branch of the mammalian family tree — one evolutionary lineage that existed since the age of the dinosaurs," Hofreiter said. "That's quite a substantial part of biodiversity lost at that time, and we wouldn't know this if we didn't have the phylogenetic tree for those species."
The findings were published online today (June 27) in the journal Nature Communications.
Original article on Live Science.<|endoftext|>
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An aural (ear) haematoma is a collection of blood or serum, and sometimes a blood clot within the pinna or ear flap. This blood collects under the skin and causes the ear flap to become thickened. The swelling may involve the entire ear flap or it may involve only a small area.
How does an aural haematoma occur?
Aural haematomas usually occur as a result of local irritation to some part of the ear. When something irritates the ear canal, a dog is likely to respond by scratching or shaking the head. Excessive shaking causes blood vessels to break, resulting in bleeding. An understanding of the ear's anatomy makes the sequence of events more logical.
Understanding ear anatomy
The ear flap is composed of a layer of skin on each side of a layer of cartilage. The cartilage gives the ear flap its shape. Blood vessels go from side-to side by passing through the cartilage. Violent shaking causes the vessels to break as the skin slides across the cartilage.
How are aural haematomas treated?
The first aim of treatment is to drain the haematoma to relieve the pressure and pain associated with the build up of fluid within the ear flap. This is achieved under general anaesthesia where either a single incision or multiple small biopsy holes are made on the inner surface of the ear. The blood is drained and the ear flushed to remove any remaining blood clots. These holes are left open to allow continued drainage of fluid whilst waiting for the ear flap to heal.
Reattachment of the ear cartilage is encouraged with the use of multiple sutures placed through the ear flap (with or without the use of a support to maintain the normal architecture of the ear) and these sutures are left in place for 3 weeks. The specific method used will depend on the size, age and position of the haematoma.
The second major aspect of treatment is to work out why the haematoma formed in the first place. As mentioned, above, any reason that causes the dog to shake its head can result in the formation of an aural haematoma. Some things which can cause this include:
- Grass seed or other foreign body lodged within the ear canal.
- Ear infection.
- Allergies resulting in an itchy ear, scratching and shaking head.
- Fly bites to the tips of the ears.
- Immune mediated disease.
It is essential that the cause of the problem be identified and treated if possible. If a foreign body is found, it is removed. If an ear infection is identified, the ear canal will be thoroughly cleaned during anaesthesia and appropriate medical ointments or medications will be dispensed.
Unfortunately, it is not always possible to identify a cause, or it is difficult to manage the underlying cause (eg allergies). In these cases, another aural haematoma may form in the same ear or in the other ear and management may require long term medications.
Once Your dog is treated will you need to bring him/her back to the vet for further treatment?
The sutures will need to be removed 3 weeks after surgery. At this time, a haematoma is usually healed. If an infection is also being treated your veterinarian will also check to make sure that the infection is gone. It is vitally important that the infection is successfully treated to prevent further head shaking which may result in further haematomas.
What happens if your dog does not have surgery?
If a haematoma is left untreated the blood in the ear flap will separate into serum and a clot and will gradually be absorbed over a period of 10 days to 6 weeks. This is an uncomfortable time for your dog and unfortunately some scarring will take place during this process. It also causes a deformity of the ear flap resulting in a "cauliflower ear" which may cause further problems.<|endoftext|>
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# Volume Formulas
In these lessons, we provide:
1. A table of volume formulas and surface area formulas used to calculate the volume and surface area of three-dimensional geometrical shapes: cube, cuboid, prism, solid cylinder, hollow cylinder, cone, pyramid, sphere and hemisphere.
2. A more detailed explanation (examples and solutions) of each volume formula.
### Table Of Volume Formulas And Surface Area Formulas
The following table gives the volume formulas for solid shapes or three-dimensional shapes. Scroll down the page if you need more explanations about the volume formulas, examples on how to use the formulas and worksheets.
### Volume Of A Cube
A cube is a three-dimensional figure with six matching square sides. The figure below shows a cube with sides s.
If s is the length of one of its sides, then the volume of the cube is s × s × s
Volume of the cube = s3
#### How To Find The Volume Of A Cube?
The formula for the volume of a cube is s × s × s = s3, where s is the length of a side of the cube.
Example:
Find the volume of a cube with sides = 4cm
### Volume Of A Rectangular Solid
A rectangular solid is also called a rectangular prism or a cuboid.
In a rectangular solid, the length, width and height may be of different lengths.
The volume of the above rectangular solid would be the product of the length, width and height that is
Volume of rectangular solid = lwh
#### How To Find The Volume Of A Rectangular Prism Or Cuboid?
The formula for the volume of a cuboid is l × w × h = lwh, where l is the length, w is the width and h is the height of the rectangular prism. This video will give two examples of finding the volume of a rectangular prism.
Examples:
1. Find the volume of a rectangular prism with sides 25 feet, 10 feet and 14 feet.
2. Find the volume of a rectangular prism with sides 5.4 inches, 7.5 inches and 18.3 inches.
### Volume Of A Prism
A prism is a solid that has two parallel faces which are congruent polygons at both ends. These faces form the bases of the prism. The other faces are in the shape of rectangles. They are called lateral faces. A prism is named after the shape of its base.
When we cut a prism parallel to the base, we get a cross section of a prism. The cross section has the same size and shape as the base.
The volume of a right prism is given by the formula:
Volume of prism = Area of base × length
V = Al
where A is the area of the base and l is the length or height of the prism.
#### How To Find The Volume Of A Triangular Prism?
This video shows how to determine which is the base and the height of the triangular prism.
Example:
Find the value of a triangular prism, use the following formula:
Volume = (area of base) × height
### Volume Of A Cylinder
A cylinder is a solid that has two parallel faces which are congruent circles. These faces form the bases of the cylinder. The cylinder has one curved surface. The height of the cylinder is the perpendicular distance between the two bases.
The volume of a cylinder is given by the formula:
Volume = Area of base × height
V = π r2h
where r = radius of cylinder and h is the height or length of cylinder.
How to find the volume of a cylinder?
Example:
Find the volume of a cylinder with radius 9 and height 12
### Volume Of A Hollow Cylinder
Sometimes you may be required to calculate the volume of a hollow cylinder or tube.
Volume of hollow cylinder
where _R_ is the radius of the outer surface and _r_ is the radius of the inner surface.
Volume of hollow containers - cylinder and cone
How you can find the volume of a hollow cylinder and a cone using the formula of volume of prism and pyramid?
Examples:
1. Given a pipe with length = 12cm, outer diameter = 2m and thickness = 40cm. Calculate the amount of concrete used?
2. Given that an ice-cream cone has a diameter of 65mm, height of 15cm and thickness of 2mm. Calculate the volume of wafer in the cone.
### Volume Of A Cone
A cone is a solid with a circular base. It has a curved surface which tapers (i.e. decreases in size) to a vertex at the top. The height of the cone is the perpendicular distance from the base to the vertex.
The volume of a cone is given by the formula:
Volume of cone = 1/3 × Area of base × height
V = 1/3 πr2h
where r is the radius of the base and h is the height of the prism.
How to find the volume of a cone?
Example:
Find the volume of a cone with radius is 12ft and height is 16ft.
### Volume of a Pyramid
A pyramid is a solid with a polygonal base and several triangular lateral faces. The lateral faces meet at a common vertex. The height of the pyramid is the perpendicular distance from the base to the vertex. The pyramid is named after the shape of its base. For example a rectangular pyramid or a triangular pyramid.
The volume of a pyramid is given by the formula:
Volume of pyramid = 1/3 × Area of base × height
V = 1/3 Ah where A
is the area of the base and h is the height of the pyramid.
#### How to find the volume of a pyramid?
Make sure that you use the vertical height to substitute into the formula and not the slant height.
Example:
Find the volume of a pyramid with sides = 9ft, vertical height = 5ft and slant height = 8ft.
### Volume of a Sphere
A sphere is a solid in which all the points on the round surface are equidistant from a fixed point, known as the center of the sphere. The distance from the center to the surface is the radius.
Volume of sphere = 4/3 πr3
where r is the radius.
How to find the volume of a sphere?
Example:
Find the volume of air in the ball with radius = 3cm
### Volume of a hemisphere
A hemisphere is half a sphere, with one flat circular face and one bowl-shaped face.
Volume of hemisphere where r is the radius.
How to find the volume of a hemisphere?
Example:
Find the volume of water in a bowl with diameter = 33cm
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.<|endoftext|>
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|Week #||Q2 W7|
|Unit of Instruction||Evergreen and Deciduous Trees|
Discuss basic classification and characteristics of plants.
|Learning Targets and Learning Criteria|
Students will learn the difference between evergreen (conifers) and deciduous trees. Students will learn how long it takes conifers to grow, particularly some types of firs and spruce, and what a large agricultural commodity these trees are. We’ll also talk about what agricultural commodities are. We will learn about poinsettias, where they originated from, that the colored part is actually a leaf not a flower, and that National Poinsettia Day is December 12th.<|endoftext|>
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7th Grade Ancient Egypt Unit
The Google Street View of Giza, Egypt will be up as the students are coming in and getting settled for the day. The teacher uses questioning techniques to get the students talking about Egypt at the start of class and introduce them to the day's activity.
- Looking at the street view, where do you suspect we are going today?
- What do you think we are going to do?
- What's the name of someone who studies the past? Maybe digs things up? Explores places to learn more about the past?
2 Guided Instruction
Students will investigate the pyramids and toms at Giza and Egyptian objects using The Pyramids app. They will work alone or in pairs, using a guide, to learn more about 3-4 things in The Pyramids app. They can view tombs, objects, and even read about ancient Egyptian times. If students are working alone, they can plug in their earbuds to listen to the guided tour in each of the tombs or pyramids. While the students are using the app, they are encouraged to write one connection, question, or relfection on each item they learn about.
Students will write down one thing they learned and with whom they plan to share that information.
4 Independent Practice
For homework, students will read a Scholastic news article about new discoveries in Egypt.<|endoftext|>
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Scotland Historical Geography
|Scotland Wiki Topics|
|Local Research Resources|
Introduction[edit | edit source]
Learning about the places where your ancestors lived helps you understand the records about them. Local histories and gazetteers contain information about:
- Place names
- Other pertinent information
- Changes in the land and community in which people lived.
Unlike place names in other European countries, many place names in Scotland have not changed for hundreds of years.
Geographically, Scotland is divided into several regions, which include:
- The borderlands with England
- The lowlands
- The highlands
- The islands (to the west and north of Scotland).
These regions were historically divided into thirty-four counties (see below). In addition, there are seven cities--Aberdeen, Dundee, Edinburgh, Glasgow, Inverness, Stirling and Perth--which received their status from the government by letters patent.
Books[edit | edit source]
The following books explain more about Scottish historical geography. You can find these and similar materials at the Family History Library and many other research libraries.
McNeill, Peter, and Ranald Nicholson, eds. 'An Historical Atlas of Scotland, c. 400-c. 1600'. St. Andrews, Scotland: Atlas Committee of the Conference of Scottish Medievalists, 1975. (Family History Library book 941 E3ha.) This book contains many maps to illustrate population movements, settlement patterns, battles, and other important events in Scotland.
Whittington, G., and I.D. Whittington, eds. 'An Historical Geography of Scotland'. London, England: New York: Academic Press, 1983. (Family History Librarybook 941 E3sg.) This is a good overall review of Scotland’s historical geography.
Gilbert, Martin. 'British History Atlas'. New York: Macmillan Co., 1968. (Family History Library book 942 E3bri.)This national historical atlas contains maps to illustrate population movements, railways, battles, plagues, and more.
You can find other sources in the Locality Search of the FamilySearch Catalog under:
SCOTLAND - HISTORICAL GEOGRAPHY
GREAT BRITAIN - HISTORICAL GEOGRAPHY
Maps are also very valuable for determining historical geography. See Scotland Maps for more information.
County Changes Made in 1974[edit | edit source]
In 1974 the British government reorganized the counties of Scotland. Twelve areas, called regions, were created from the original thirty-four counties. See Scotland Counties for the changes.
These changes should not seriously affect genealogical research, but be aware of the following issues:
- Current maps show the new boundaries.
- Current addresses are located in the new counties. The addresses in this outline use the current county structure.
If you are looking for a parish, city, or regional office that houses records, you will need to know the current address and the areas covered by the repository.
You should begin with the pre-1974 county name when you start your genealogical research. Below is a list of the new counties with an indication of the old counties they cover.
|New Region (County)||Old Counties|
|Borders||Berwick,Peebles,Roxburgh,Selkirk, and a small part of Midlothian|
|Central||Clackmannan, parts of Perth and WestLothian, and most of Stirling|
|Dumfries and Galloway||Dumfries, Kirkcudbright, and Wigtown|
|Grampian||Aberdeen, Kincardine, Banff, and most of Moray|
|Highland||Caithness, Nairn, and Sutherland and most of Argyll, Inverness, Moray, and Ross and Cromarty|
|Lothian||East Lothian and most of Midlothian and West Lothian|
|Strathclyde||Bute, Dunbarton, Lanark, Renfrew, Ayr, and parts of Argyll and Stirling|
|Tayside||Angus, Kinross, and part of Perth|
|Island Areas||Old County|
|Shetland||of Zetland (or Shetland)|
|Western Isles||Island areas of Inverness and Ross & Cromarty|<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Instantaneous Rates of Change
## Derivative is the slope of the tangent line at a point.
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Instantaneous Rate of Change
When you first learned about slope you learned the mnemonic device “rise over run” to help you remember that to calculate the slope between two points you use the following formula:
$m=\frac{y_2-y_1}{x_2-x_1}$
In Calculus, you learn that for curved functions, it makes more sense to discuss the slope at one precise point rather than between two points. The slope at one point is called the slope of the tangent line and the slope between two separate points is called a secant line.
Consider a car driving down the highway and think about its speed. You are probably thinking about speed in terms of going a given distance in a given amount of time. The units could be miles per hour or feet per second, but the units always have time in the denominator. What happens when you consider the instantaneous speed of the car at one instant of time? Wouldn’t the denominator be zero?
#### Watch This
http://www.youtube.com/watch?v=7CvLzpzGhJI Brightstorm: Definition of a Derivative
#### Guidance
The slope at a point $P$ (also called the slope of the tangent line) can be approximated by the slope of secant lines as the “run” of each secant line approaches zero.
Because you are interested in the slope as the “run” approaches zero, this is a limit question. One of the main reasons that you study limits in calculus is so that you can determine the slope of a curve at a point (the slope of a tangent line).
Example A
Estimate the slope of the following function at -3, -2, -1, 0, 1, 2, 3. Organize the slopes in a table.
Solution: By mentally drawing a tangent line at the following $x$ values you can estimate the following slopes.
$x$ slope -3 0 -2 0 -1 -1 0 -1 1 2 2 0 3 0
If you graph these points you will produce a graph of what’s known as the derivative of the original function.
Example B
Estimate the slope of the function $f(x)=\sqrt{x}$ at the point $(1,1)$ by calculating 4 successively close secant lines.
Solution: Calculate the slope between $(1,1)$ and 4 other points on the curve:
• The slope of the line between $\left(5,\sqrt{5}\right)$ and $(1,1)$ is: $m_1=\frac{\sqrt{5}-1}{5-1} \approx 0.309$
• The slope of the line between $(4,\ 2)$ and $(1,1)$ is: $m_2=\frac{2-1}{4-1}\approx 0.333$
• The slope of the line between $\left(3,\sqrt{3}\right)$ and $(1,1)$ is: $m_3=\frac{\sqrt{3}-1}{3-1} \approx 0.366$
• The slope of the line between $\left(2,\sqrt{2}\right)$ and $(1,1)$ is: $m_4=\frac{\sqrt{2}-1}{2-1} \approx 0.414$
If you had to guess what the slope was at the point $(1,1)$ what would you guess the slope to be?
Example C
Evaluate the following limit and explain its connection with Example B.
$\lim \limits_{x \to 1} \left(\frac{\sqrt{x}-1}{x-1}\right)$
Solution: Notice that the pattern in the previous problem is leading up to $\frac{\sqrt{1}-1}{1-1}$ . Unfortunately, this cannot be computed directly because there is a zero in the denominator. Luckily, you know how to evaluate using limits.
$m&=\lim \limits_{x \to 1} \left(\frac{\left(\sqrt{x}-1\right)}{\left(x-1\right)} \cdot \frac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)}\right)\\&=\lim \limits_{x \to 1}\left(\frac{\left(x-1\right)}{\left(x-1\right) \left(\sqrt{x}+1\right)}\right)\\&=\lim \limits_{x \to 1} \left(\frac{1}{\left(\sqrt{x}+1\right)}\right)\\&=\frac{1}{\sqrt{1}+1}\\&=\frac{1}{2}\\&=0.5$
The slope of the function $f(x)=\sqrt{x}$ at the point $(1,1)$ is exactly $m=\frac{1}{2}$ .
Concept Problem Revisited
If you write the ratio of distance to time and use limit notation to allow time to go to zero you do seem to get a zero in the denominator.
$\lim \limits_{time \to 0}\left(\frac{distance}{time}\right)$
The great thing about limits is that you have learned techniques for finding a limit even when the denominator goes to zero. Instantaneous speed for a car essentially means the number that the speedometer reads at that precise moment in time. You are no longer restricted to finding slope from two separate points.
#### Vocabulary
A tangent line to a function at a given point is the straight line that just touches the curve at that point. The slope of the tangent line is the same as the slope of the function at that point.
A secant line is a line that passes through two distinct points on a function.
A derivative is a function of the slopes of the original function.
#### Guided Practice
1. Sketch a complete cycle of a sine graph. Estimate the slopes at $0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi$ .
2. Logan travels by bike at 20 mph for 3 hours. Then she gets in a car and drives 60 mph for 2 hours. Sketch both the distance vs. time graph and the rate vs. time graph.
3. Approximate the slope of $y=x^3$ at $(1,1)$ by using secant lines from the left. Will the actual slope be greater or less than the estimates?
1.
$x$ Slope 0 1 $\frac{\pi}{2}$ 0 $\pi$ -1 $\frac{3\pi}{2}$ 0 $2\pi$ 1
You should notice that these are the exact values of cosine evaluated at those points.
2. Distance vs. Time:
Rate vs. Time: (this is the graph of the derivative of the original function shown above)
3.
• The slope of the line between $(0.7,0.7^3)$ and $(1,1)$ is: $m_1=\frac{0.7^3-1}{0.7-1} \approx 2.19$
• The slope of the line between $(0.8,0.8^3 )$ and $(1,1)$ is: $m_2=\frac{0.8^3-1}{0.8-1} \approx 2.44$
• The slope of the line between $(0.9,0.9^3 )$ and $(1,1)$ is: $m_3=\frac{0.9^3-1}{0.9-1} \approx 2.71$
• The slope of the line between $(0.95,0.95^3 )$ and $(1,1)$ is: $m_1=\frac{0.95^3-1}{0.95-1} \approx 2.8525$
• The slope of the line between $(0.975,0.975^3 )$ and $(1,1)$ is: $m_1=\frac{0.975^3-1}{0.975-1} \approx 2.925625$
The slope at $(1,1)$ will be slightly greater than the estimates because of the way the slope curves. The slope at $(1,1)$ appears to be about 3.
#### Practice
1. Approximate the slope of $y=x^2$ at $(1,1)$ by using secant lines from the left. Will the actual slope be greater or less than the estimates?
2. Evaluate the following limit and explain how it confirms your answer to #1.
$\lim \limits_{x \to 1} \left(\frac{x^2-1}{x-1}\right)$
3. Approximate the slope of $y=3x^2+1$ at $(1,4)$ by using secant lines from the left. Will the actual slope be greater or less than the estimates?
4. Evaluate the following limit and explain how it confirms your answer to #3.
$\lim \limits_{x \to 1} \left(\frac{3x^2+1-4}{x-1}\right)$
5. Approximate the slope of $y=x^3-2$ at $(1,-1)$ by using secant lines from the left. Will the actual slope be greater or less than the estimates?
6. Evaluate the following limit and explain how it confirms your answer to #5.
$\lim \limits_{x \to 1}\left(\frac{x^3-2-(-1)}{x-1}\right)$
7. Approximate the slope of $y=2x^3-1$ at $(1,1)$ by using secant lines from the left. Will the actual slope be greater or less than the estimates?
8. What limit could you evaluate to confirm your answer to #7?
9. Sketch a complete cycle of a cosine graph. Estimate the slopes at $0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi$ .
10. How do the slopes found in the previous question relate to the sine function? What function do you think is the derivative of the cosine function?
11. Sketch the line $y=2x+1$ . What is the slope at each point on this line? What is the derivative of this function?
12. Logan travels by bike at 30 mph for 2 hours. Then she gets in a car and drives 65 mph for 3 hours. Sketch both the distance vs. time graph and the rate vs. time graph.
13. Explain what a tangent line is and how it relates to derivatives.
14. Why is finding the slope of a tangent line for a point on a function the same as the instantaneous rate of change at that point?
15. What do limits have to do with finding the slopes of tangent lines?
### Vocabulary Language: English
Average rate of change
Average rate of change
The average rate of change of a function is the change in $y$ coordinates of a function, divided by the change in $x$ coordinates.
instantaneous rate of change
instantaneous rate of change
The instantaneous rate of change of a curve at a given point is the slope of the line tangent to the curve at that point.
limit
limit
A limit is the value that the output of a function approaches as the input of the function approaches a given value.
secant line
secant line
A secant line is a line that joins two points on a curve.
Slope
Slope
Slope is a measure of the steepness of a line. A line can have positive, negative, zero (horizontal), or undefined (vertical) slope. The slope of a line can be found by calculating “rise over run” or “the change in the $y$ over the change in the $x$.” The symbol for slope is $m$
Tangent line
Tangent line
A tangent line is a line that "just touches" a curve at a single point and no others.<|endoftext|>
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# Find all the zeros of the polynomial P(x) 2×4-3×3-5×2+9x-3
Let’s start with the concept of “Find all the zeros of the polynomial (2x4-3x3-5x2+9x-3), it is given that two of its zeros are √3 and -√3.
## Find all the zeros of the polynomial p(x)= (2x4-3x3-5x2+9x-3), it is given that two of its zeros are √3 and -√3.
Solution: √3 and -√3 are zeros of polynomial P(x) = 2x4-3x3-5x2+9x-3.
x = √3 or x = -√3
x -√3 = 0 or x + √3 =0
(x -√3 )(x +√3 ) = 0 x 0
(x)2 – (√3 )2 = 0
x2 – 3 = 0
(x2 – 3) is factor of P(x) = 2x4-3x3-5x2+9x-3.
Now,
(x2 – 3) is completely divisible by P(x) = 2x4-3x3-5x2+9x-3.
When (2x4-3x3-5x2+9x-3) is divided by (x2-3) to get (2x2-3x+1) as a quotient and 0 as a remainder.
Factorise q(x) = (2x2-3x+1)
q(x) = 2x2-3x+1
0 = 2x2-2x-x+1
0 = 2x(x-1) -1(x-1)
0 = (2x-1) (x-1)
Either,
2x-1 = 0 or x-1 = 0
X = ½ or x = 1
Hence, all zeros of polynomial P(x) = 2x4-3x3-5x2+9x-3 are √3 , -√3 , 1 and ½
Some important identity of Polynomial
1. (a+b)2 = a2 + b2 + 2ab
2. (a-b)2 = a2 + b2 – 2ab
3. a2 – b2 = (a+b)(a-b)
4. (a+b+c)2 = a2+b2+c2+2ab+2bc+2ca
5. (a+b)3 = a3+b3+3a2b+3ab2
6. (a-b)3 = a3+b3-3a2b+3ab2
7. a3+b3 = (a+b)(a2+b2-2ab)
8. a3-b3 = (a+b)(a2+b2+2ab)
Polynomials
An expression of the form p(x) = a0+a1x+a2x2+……+anxn, where an is not equal to zero, is called a polynomial in x of degree n.
Degree of polynomials
If P(X) is a polynomial in x, the highest power of x in P(x) is called the degree of the polynomial P(x).
Ex: The degree of polynomial P(X) = 2x3 + 5x2 -7 is 3 because the degree of a polynomial is the highest power of polynomial.
Zero of polynomial
If the value of P(x) at x = K is zero then K is called a zero of the polynomial P(x).
I hope you like this post Find all the zeros of the polynomial P(x) = 2x4-3x3-5x2+9x-3
View more…
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My Name is Mukesh Kumar. I am a Teacher, Blogger, Educational Content Writer, and Founder of CBSE Digital Education.<|endoftext|>
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El'gygytgyn Crater, Russia
About 60 miles north of the Arctic Circle in eastern Russia is Lake El'gygytgyn, which is inside a 3.6-million-year-old meteorite impact crater. The crater is about 15 kilometers (about 9.3 miles) across. "In this false-color image, red indicates vegetation, gray-brown indicates bare land, and deep blue indicates water," NASA writes, noting that the vegetation consists of low-lying plants that make up the Arctic tundra.
Climate researchers find this impact crater particularly valuable, as its lake bed sediments contain a continuous record of past Arctic conditions. They are able to use these sediments to analyze how climate has changed over the past few million years.<|endoftext|>
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Comet “Chury’s” late birth
Comets which consist of two parts, like Chury, can form after a catastrophic collision of larger bodies. Such collisions may have taken place in a later phase of our solar system, which suggests that Chury can be much younger than previously assumed. This is shown through computer simulations by an international research group with the participation of the University of Bern.
In the computer simulations, the research team investigated what happened after two large comet nuclei violently collided together. “The calculations showed that a large part of the material accumulates in many smaller bodies,” explains Martin Jutzi of the Center for Space and Habitability (CSH) at the University of Bern and member of the National Centre of Competence in Research PlanetS. The newly created objects have different sizes and shapes, among them are many elongated bodies, some of which consist of two parts, just like the comet 67P/Churyumov-Gerasimenko, which the University of Bern studied in detail with the Bern mass spectrometer ROSINA on the Rosetta spacecraft.
“We were surprised that in such catastrophic collisions only a small part of the material is considerably compressed and heated,” says Martin Jutzi. Moreover, this material is then ejected and hardly contributes to the formation of the smaller bodies that form a new generation of comet nuclei. On the side of the comet opposite the impact point, volatile substances can withstand even violent collisions. This is why the new generation of comets still has a low density and is rich in volatile substances – properties which have also been found on the comet Chury. Therefore, the duck-shaped comet may well have emerged after a violent, late collision and did not necessarily have to originate from the early formation phase of the solar system, as has been claimed repeatedly. Such collisions could have taken place relatively late in the life of the solar system. This finding has been reported in the journal Nature Astronomy by the research group led by Stephen Schwartz from the University of Côte d’ Azur and the University of Arizona.
Impact with a velocity of several kilometers per second
In previous studies, Martin Jutzi and Willy Benz, professor at the University of Bern and PlanetS director, had already come to the conclusion that Chury did not receive its two-component structure when our solar system was formed 4.5 billion years ago. The researchers showed that the weak point between the two parts of the comet could not have lasted for several billion years and that Chury may have been created by a comparatively gentle impact. “We have now investigated catastrophic collisions involving a lot more energy,” explains Martin Jutzi. The new calculations confirm the previous results and extend the possible formation scenarios.
The research team investigated what happens when different sized bodies collide at different angles at speeds ranging from 20 to 3,000 meters per second. The simulations showed that small fragments merge into many transient aggregates in the hours and days after the collision (see video). The final shape is often the result of two or more large bodies that collide at very low speeds to form a two-component structure.
Possible explanation for “Chury’s” mysterious structures
According to the simulations, during the days and weeks in which the comet received its shape, small aggregates in the vicinity continue to reaccumulate onto it. In reality, this material could be flattened when it hits the surface and thus lead to a layered structure. Moreover, if large blocks accumulate at this stage, cavities may be created which can develop into large pits. Such geological structures were discovered on Chury by the Rosetta mission – these observations were previously considered mysterious. “Our results not only confirm that the comet Chury may be much younger than previously assumed, but also provide a possible explanation for its striking structures,” says Jutzi.
St. Schwartz, P. Michel, M. Jutzi, S. Marchi, Y. Zhang, D. Richardson: Catastrophic disruption as the origin of bilobate comets, Nature Astronomy, 5 March 2018.
Dr. Martin Jutzi
Center for Space and Habitability (CSH) und NCCR PlanetS, Universität Bern
Dr. Stephen Schwartz
Université Côte d’Azur, Nice, France
Dr. Patrick Michel
Université Côte d’Azur, Nice, France<|endoftext|>
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# How do you expand ln ((sqrt(a)(b^2 +c^2))?
Dec 12, 2015
Sticking with Real logarithms, this expands as:
$\ln \left(\sqrt{a} \left({b}^{2} + {c}^{2}\right)\right) = \frac{1}{2} \ln \left(a\right) + \ln \left({b}^{2} + {c}^{2}\right)$
#### Explanation:
If $x , y > 0$ then $\ln \left(x y\right) = \ln \left(x\right) + \ln \left(y\right)$
Assuming we're dealing with Real values here and everything is well defined, we must have $a > 0$ and ${b}^{2} + {c}^{2} > 0$. That is, at least one of $b \ne 0$ or $c \ne 0$, resulting in a strictly positive value for ${b}^{2} + {c}^{2}$.
Also note that if $a > 0$ then $\ln \left(\sqrt{a}\right) = \ln \left({a}^{\frac{1}{2}}\right) = \frac{1}{2} \ln \left(a\right)$
Hence:
$\ln \left(\sqrt{a} \left({b}^{2} + {c}^{2}\right)\right)$
$= \ln \left(\sqrt{a}\right) + \ln \left({b}^{2} + {c}^{2}\right)$
$= \frac{1}{2} \ln \left(a\right) + \ln \left({b}^{2} + {c}^{2}\right)$
If we allow Complex logarithms, then we might try to say something like:
$= \frac{1}{2} \ln \left(a\right) + \ln \left(b + c i\right) + \ln \left(b - c i\right)$
based on the fact that ${b}^{2} + {c}^{2} = \left(b + c i\right) \left(b - c i\right)$, but there are some problems with this.
For example, if $b = - 1$ and $c = 0$ then we find:
$0 = \ln \left(1\right) = \ln \left({b}^{2} + {c}^{2}\right) \ne \ln \left(b + c i\right) + \ln \left(b - c i\right) = \ln \left(- 1\right) + \ln \left(- 1\right) = 2 \pi i$
So this Complex identity does not quite work and is messy to fix.<|endoftext|>
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# Linear Relations and Functions
## Presentation on theme: "Linear Relations and Functions"— Presentation transcript:
Linear Relations and Functions
Objectives: Understand, draw, and determine if a relation is a function. Graph & write linear equations, determine domain and range.
Relation: a set of ordered pairs Domain: the set of x-coordinates
Relations & Functions Relation: a set of ordered pairs Domain: the set of x-coordinates Range: the set of y-coordinates When writing the domain and range, do not repeat values.
Relations and Functions
Given the relation: {(2, -6), (1, 4), (2, 4), (0,0), (1, -6), (3, 0)} State the domain: D: {0,1, 2, 3} State the range: R: {-6, 0, 4}
Relations and Functions
Relations can be written in several ways: ordered pairs, table, graph, or mapping. We have already seen relations represented as ordered pairs.
Table {(3, 4), (7, 2), (0, -1), (-2, 2), (-5, 0), (3, 3)}
Mapping Create two ovals with the domain on the left and the range on the right. Elements are not repeated. Connect elements of the domain with the corresponding elements in the range by drawing an arrow.
Mapping {(2, -6), (1, 4), (2, 4), (0, 0), (1, -6), (3, 0)} 2 1 3 -6 4
Functions A function is a relation in which the members of the domain (x-values) DO NOT repeat. So, for every x-value there is only one y-value that corresponds to it. y-values can be repeated.
Functions Discrete functions consist of points that are not connected. Continuous functions can be graphed with a line or smooth curve and contain an infinite number of points.
Do the ordered pairs represent a function?
{(3, 4), (7, 2), (0, -1), (-2, 2), (-5, 0), (3, 3)} No, 3 is repeated in the domain. {(4, 1), (5, 2), (8, 2), (9, 8)} Yes, no x-coordinate is repeated.
Graphs of a Function Vertical Line Test: If a vertical line is passed over the graph and it intersects the graph in exactly one point, the graph represents a function.
Does the graph represent a function? Name the domain and range.
x y Yes D: all reals R: all reals R: y ≥ -6 x y
Does the graph represent a function? Name the domain and range.
x y No D: x ≥ 1/2 R: all reals D: all reals x y
Does the graph represent a function? Name the domain and range.
x y Yes D: all reals R: y ≥ -6 No D: x = 2 R: all reals x y<|endoftext|>
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• building conceptual understanding
• # Comparing Fractions with Common Denominators & Common Numerators
If students understand what a fraction is and have a good understanding of comparing unit fractions, comparing fractions with a common denominator and numerator will be a confidence booster!
Check out these previous posts:
Fractions Introduction
Unit Fractions
Like I’ve said in my previous post, first, I get students to practice by using fraction bars and then we move onto comparing using number lines. I have found that making students prove their conclusion with a picture and with a sentence has helped their understanding of fractions and helped boost their confidence with fractions! After ample amount of practice, we move away from both.
## Practice using Number Lines: Common Denominators
When comparing fractions that have the same denominator, they are same size pieces. Because they are the same size, we look at the numerator, which tells us how many pieces there are. When comparing fractions with a common denominator, the more pieces it has, the larger the fraction.
Activity Idea: Fraction Memory Match
Directions
1) Lay out the cards, face down, in a 4 x 3 array
2) On your turn, turn two cards face up
3) If the denominators match, compare the fractions, explain your comparison, and take the cards
4) If the denominators don’t match, turn the cards back face down
Sentence Stem
5/8 is larger than 2/8. They have a common denominator (same size pieces), so we need to look at the numerator, (the number of pieces). 5 pieces is more than 2 pieces.”
## Practice using Number Lines: Common Numerators
When comparing fractions that have the same numerator, they have the same number of pieces. Because of this, we look at the denominator, which tells us how big the pieces are or the size of the piece. When comparing fractions with a common numerator, the smaller the denominator, the larger the fraction.
Activity Idea: Fraction War
Directions
1) Players will split up the fraction cards
2) Flip over one fraction card
3) The player with the larger fraction needs to explain why they have the larger fraction. They will take the card if they explain correctly
4) Try to collect the most fraction cards
Sentence Stem: “2/9 is larger than 2/12. They have a common numerator (the same number of pieces), so we need to look at the denominator, (the size of the pieces). Ninths is larger than twelfths.”
Here is another game for comparing fractions with a common numerator. Click here if you’re interested!
Timeline:
1) Introduce Fractions: 1 day
2) Unit Fractions: 1 day
3) Comparing fractions with a common numerator // common denominator: 1 day
4) Mix comparing fractions + introduce making half: 1 day
5) Mix comparing fractions + making half practice + comparing to a half: 1 day
As you can see, I continue to spiral the skills. Spiral practice is very important. One day, I will post my routine for daily math review and daily spiral review. I make time to do both, because both are very important!
This is a part of our fraction series. Check out these other blogposts for more!
How do you teach comparing fractions in a conceptual way?
• building conceptual understanding
• # Pizza Fractions – Unit Fractions
I always start with this activity as a foundation for comparing unit fractions. Every year, this activity is a huge hit with my students! Fraction Pizza Activity: Team Challenge Group students into groups of 4 and put the timer on for 30 minutes. 1) Work with your group to make pizzas 2) Everyone will be assigned […]
• best teaching practices
• # Fractions Introduction
Start off with a quick conversation to activate prior knowledge *tip: have visual pictures available for students to see “Where and where do we use fractions in everyday life?” on the highway, music, shoe size, cooking, pizza, etc “What does ½ mean to you?” there are two total pieces and one is shaded “What does […]
• resources
• # Hedbanz Game
I love the recent hit activity Hedbanz. It is an activity that has been around for a while but it is now finding its way around the classrooms. This activity lends itself for great math conversations and engagement, for whichever concept that is being taught. For the concept of writing and solving equations with an […]
• building conceptual understanding
• # Multiplying Decimals | Why does it get smaller?
We were doing a lesson on multiplying decimals with models when a student asked a question: “Why does our answer get smaller if we’re multiplying?” Great question! This makes us stop and think “How are we teaching the basic understanding of multiplication?” Is it right for us to be saying ‘the number gets larger when we […]
• best teaching practices
• # Decimal Investigation
Building a strong understanding of how our base ten system works is always so important. Taking the time to help students understand this will help with ‘regrouping’ when it comes to adding, subtracting, multiplying, and dividing – it’s worth the time! Ones – Tenths – Hundredths – Thousandths One whole 1.00 One whole is […]
• real talk
• # “I’m going into the teaching field”
What’s the first thought you get when you hear this? We know everyone has different experiences with teaching but we’re sure you can agree with some of these points as a fellow teacher. For reference: Our background = working in a low-income school. After asking the person why they want to become a teacher, our […]<|endoftext|>
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Lewis and Clark’s 1804-1806 Corps of Discovery expedition helped open the gates to the American West. Millions literally followed in their footsteps and many of them created cities which grew and grew and grew.
This geospatial analysis uses the locations of their 1804 campsites over a modern-day map aerial image to determine if the campsites for the journey that helped define America are still around.
Those millions, though, settled in relatively few places along their route. While river towns developed along the Missouri River — particularly the lower Missouri River — relatively few large population centers emerged. That’s Kansas City and St. Louis, really, in Missouri.
We took a look at where the sites were and how they compared with current urban growth. The question: Did the gates to the West flung open by Lewis and Clark and the Corps of Discovery eventually flood population over their campsites?
The answer is not really. Since their campsites were directly adjacent to the river and prone to flooding, there hasn’t been much development around them.
In fact, less than 20 percent of their campsites in Missouri from the 1804 trip are within a half-mile of urban growth. The image below is the St. Louis area from the confluence of the Missouri and the Mississippi to about the Washington, Mo. area (the map reads sequentially from east to west. The furthest east dot is the corps’ winter quarters at the mouth of the Fox River; a sidenote is that the river’s course has shifted significantly since 2004 and the corps’ winter quarters were on the Illinois side).
The third, fourth and fifth campsites are still in the urban growth areas, but none of them have been overrun by growth, as we’ve seen on battlefields such as Gettysburg and Antietam.
The pattern is repeated in the Kansas City area. Only one campsite is deeply in the urban core, the one located at the confluence of the Kansas and Missouri rivers.
Even there, it’s still a preserved area. Our datum point isn’t in an industrial park.
Why is it that these campsites haven’t been overrun by growth when locations of similar historic importance around the country have?
The answer seems to be land value. Camping on flat lands adjacent to the river meant the corps was camping in the alluvial flood plain, where the richest farming land is located. It got so rich and full of nutrients because of regular flooding. And while there has been a spate of building in the Missouri River floodplain around St. Louis, the areas where the campsites were are too close to the river and, consequently, too valuable and too full of risk to build over.<|endoftext|>
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### Measures of central tendency
The first step in dealing with any set of numbers usually involves calculating an average. By average we usually mean mean. However, you should remember from GCSE maths that there are three measures of central tendency or average, (mean, median and mode) and you’ll need to know how to calculate each, and also the good and bad points relating to each.
### Mean
The mean is the sum of average items. To calculate the mean add your numbers up and divide by the number there are.
So, in the case above it would be (3 + 8 + 5)/5 which equals 5. Getting a whole number was pure luck; you are not often likely to end up with a whole number!
The mean is very sensitive because it uses all the data. In the example above it uses all eight numbers.
The mean is most useful when the data approximates to a normal distribution.
### Median
Place your numbers in ascending (or descending) order and pick out the middle one. How much more central could you get. If there’s an even number then find the half way point between the two middle numbers:
19, 21, 23, 24, 32, 33, 36, 45
The middle two numbers are 24 and 32. So our median is 28. If in doubt add together your two middle numbers and divide by two. 24 + 32 = 56. 56/2 = 28. The advantages and disadvantages are the reversal of the mean. One visual example is the following. Imagine a group of children, all similar ages:
• In this example the median is a good measure of central tendency, as it reflects the average age of the group.
• In this example, one child is replaced with an adult. The mean age is still 12.5 years, but fails to take into account the older person. The median is not sensitive to outliers or extreme values.
It isn’t unduly influenced by extreme values. Substituting 32 by 320 19, 21, 23, 24, 33, 36, 45, 320. The median now lies between 24 and 33… 28.5. The median has been adjusted by 0.5.
### Mode
Which number is most common number. In our first set 23 occurs twice so the mode is 23.
• Sometimes: 19, 21, 21, 21, 24, 27, 33, 33, 33, 39 two numbers share the honours, three 21s and three 33sIn this case we have two modes, a bimodal distribution. Often this could be the best way of describing our set of data. For example the males in some species of fish tend to be either big or small with few mediums. A bipolar distribution would best explain these.
Left: a bimodal distribution
Super quick and easy to calculate!
### Measures of Dispersion
Dispersion refers to the extent to which a set of data is spread out, or dispersed from the ‘average’. explanatory context. There are various measures of dispersion, such as range and standard deviation.
### Range
This is the simplest method of calculating dispersion. Basically it is the difference between the largest and the smallest value in the data:
19, 21, 23, 23, 24, 25, 33, 33, 33
In the example above range is 33-19 = 14
Obviously this is easy to calculate but it doesn’t consider all the data and since it is based only on the greatest and smallest values it is very, very prone to outliers!
### Standard deviation
This is the average distance of scores away from the mean. Essentially a mean is calculated and all other pieces of data are then compared to the mean. As a result it uses all of the data. After a little jiggery-pokery involving squaring and square rooting, we end up with a magical number.
To explain its value, we can use the example of IQ. Tests are updated to ensure that the mean IQ is maintained at 100. The standard deviation is 15:
If we drop one standard deviation (SD) below the mean to 85 (100-15) and move one standard deviation up to 115 (100+15), about 68% of our population will fall between these two numbers. Magically this is not just true of IQ and SDs of 15! Given any set of normally distributed data, 68% will always fall within one standard deviation of the mean!!!
Standard deviation therefore doesn’t just tell us the spread; it allows us to quantify that spread.
The smaller the standard deviation the closer our data is to the mean. It isn’t widely spread out, it is consistent. A large standard deviation tells us the opposite. Data is widely spread out around the mean, it is inconsistent.
If I set you a mock (let’s say out of 35) and the average mark was 24 (dreaming J) with a standard deviation of 4, that would tell me that most people (68% in fact) scored between 21 and 28. That’s a tightly packed set of data!
If the SD was 10 that would suggest there was a huge variety of scores, 68% falling between 14 and 34 and the other 32% being even more widely spread than that.<|endoftext|>
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Measuring the financial condition of a company comes in a variety of ways. Net income is the final item on the income statement. Additionally, the net income formula produces that number and shows each line item.
What Is The Net Income Formula?
First of all, net income is a businesses earnings after subtracting expenses such as selling expenses, cost of producing goods, and non-cash items such as depreciation. Most simply, this equation is Net Income = Total Revenues – Total Expenses.
Furthermore, let’s take a look at expense items. The first expense is the cost of goods sold (COGS). Subtracting Revenues from COGS yields gross profit. This shows how much it costs a company to produce sales. Next comes operating expenses. These expenses do not directly relate to production of goods. These include salaries, utilities, and depreciation.
Finally, interest expense and taxes conclude the net income formula. After deducting all the aforementioned expenses, a company finds their net income.
Analyzing The Net Income Formula
As previously mentioned, net income is one of many ways that investors or analysts measure a company’s health. This formula is the root of many ratios in financial analysis. For example, net income shows how well the company returns capital to shareholders through dividends or share repurchases.
Also, net income is not the amount of cash produced by the firm in a given time. How can this be? Non-cash expenses such as depreciation contribute to the net income formula.
This point shows that some consider net income “manipulated”, as companies sometimes take large one-time losses or gains, or choose aggressive depreciation methods. All of these techniques skew net income. Nonetheless, this measure is company specific and provides invaluable information to shareholders and analysts.
In conclusion, accountants use the net income formula as a uniform measure of company profits. The formula is uniform across businesses. This allows ease of analysis for all interested parties.<|endoftext|>
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If you find any mistakes, please make a comment! Thank you.
## There is a unique noncyclic group of order 4
Assume that $G = \{1, a, b, c\}$ is a group of order 4 with identity 1. Assume also that $G$ has no element of order 4. Use the cancellation laws to show that there is a unique (up to a permutation of the rows and columns) group table for $G$. Deduce that $G$ is abelian.
Solution: Let $x,y$ be distinct nonidentity elements of $G$. If $xy = x$, then by left cancellation we have $y = 1$, a contradiction. So $xy$ is either 1 or the third nonidentity element. Also, if $x \neq 1$ we have $x^2 \neq x$ since otherwise $x = 1$. Now we need to find all the possible ways to fill in the following group table under the given constraints.
1 a b c
11abc
aa
bb
cc
Suppose $ab = 1$. Then $ba = 1$, and there are two possibilities for $ac$. If $ac = 1$, then we have $ab = ac$ and so $b = c$, a contradiction. Hence $ac = b$. There are two possibilities for $bc$. If $bc = 1$, then $bc = ba$ and by left cancellation we have $c = a$, a contradiction. Hence $bc = a$. Now since $c$ must have an inverse, we have $c^2 = 1$. Now there are three possibilities for $a^2$. If $a^2 = 1$, we have $a^2 = ab$ and so $a = b$, a contradiction. If $a^2 = b$, then we have $a^2 = ac$ and so $a = c$, a contradiction. Hence $a^2 = c$. But now we have $a^2 = c$, $a^3 = b$, and $a^4 = 1$, so $|a| = 4$, a contradiction. Hence $ab \neq 1$.
Now we have $ab = c$. There are two possibilities for $ba$. If $ba = 1$, then we have $ca = aba = a$ so that $c = 1$, a contradiction. Hence $ba = c$. Now there are three possibilities for $a^2$. If $a^2 = b$, then $a^3 = ab = c$, so that $|a| = 4$, a contradiction. If $a^2 = c$, then $a^2 = ab$ so that $a = b$, a contradiction. Hence $a^2 = 1$. Now there are two possibilities for $ac$. If $ac = 1$, we have $ac = a^2$ so that $a = c$, a contradiction. Hence $ac = b$. Similarly, there are two possibilities for $ca$. If $ca = 1$ we have $ca = a^2$ so that $a = c$, a contradiction. Hence $ca = b$. There are three possibilities for $b^2$. If $b^2 = c$, then we have $b^2 = ab$ so that $a = b$, a contradiction. If $b^2 = a$, then $b^3 = ba = c$ and so $|b| = 4$, a contradiction. Thus $b^2 = 1$. There are two possibilities for $bc$. If $bc = 1$, then $bc = b^2$ so that $b = c$, a contradiction. Hence $bc = a$. Likewise there are two possibilities for $cb$. If $cb = 1$ we have $cb = b^2$ so that $b = c$, a contradiction. Hence $cb = a$. Finally, there are three possibilities for $c^2$. If $c^2 = a$, we have $c^2 = bc$ so that $b = c$, a contradiction. If $c^2 = b$, we have $c^2 = ac$ so that $c = a$, a contradiction. Thus $c^2 = 1$. Thus we have uniquely determined the group table for G, as shown below.
1 a b c
11abc
aa1cb
bbc1a
ccba1
Note that the group table for $G$ is a symmetric matrix. By Exercise 1.1.10, $G$ is abelian.
Another proof to show $G$ is abelian.
Since there is no element of order 4, then the order can only be 1 or 2 or 3 by Exercise 1.1.34.
Suppose there exists an element $x$ such that the order of $x$ is 3. Then $1, x, x^2$ are pairwise distinct. Hence there exists another element $y\in G$ such that $y\ne 1,x, x^2$.
We consider $xy$. It is clear $xy$ can not be $1,x,x^2,y$. So we obtain a new element, a contradiction (more than 4 elements). Thus there are no elements of order 3.
Hence $x^2=1$ for all $x\in G$. Now The group $G$ is abelian can be obtained directly from Exercise 1.1.25.
#### Linearity
This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.<|endoftext|>
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What is a 'launch window'?
The launch window is a term used to describe a time period in which a particular mission must be launched.
If the spacecraft intends to rendezvous with another spacecraft, a planet, or other point in space, the launch must be carefully timed so that the orbits overlap at some point in the future. If the weather is bad or a malfunction occurs during a launch window, the mission must be postponed until the next launch window appropriate for the flight.
On 2 June 2003, ESA launched its Mars Express mission on board a Russian Soyuz-Fregat launcher - almost halfway through the assigned launch window. The Mars Express launch window had opened on 23 May and would last for four weeks.
But why did we need a launch window for Mars Express? If we are going to send a mission to a planet, why not just launch the rocket at any time, find where Mars is in the sky, point the rocket at it and travel there?
Imagine the Solar System as an athletics race track. If you were watching the 400 metres race from the centre of the track and wanted to intercept one of the runners taking part, one way would be to simply chase the runner you wish to stop. If you were fast enough, you might eventually catch up but only after expending a lot of energy and travelling a long way.
A much better way to intercept your athlete is simply to walk across the centre to the other side of the circular track. It is a much shorter distance and you use a lot less energy and time getting there.
You calculate your walk so that you arrive at the other side of the track at the same time as they do. Too early and you are waiting around for them. Too late and you have missed them completely - you'd have to wait one lap until they came around again.
In spaceflight, straight-line paths do not exist for the same reason. All planets move in long, curved paths around the Sun that take the shape of circular and elliptical orbits. In order to reach Mars, Mars Express must leave Earth and then travel in an elliptical orbit around the Sun that will eventually intersect the orbit of Mars.
Calculating the launch window for the ESA Mars Express mission involved timing the launch to allow the spacecraft and Mars to arrive at the same point in space at the same time.
Last update: 28 September 2004<|endoftext|>
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Engineered liver tissue could have a range of important uses, from transplants in patients suffering from the organ's failure to pharmaceutical testing. Now scientists report in ACS' journal Analytical Chemistry the development of such a tissue, which closely mimics the liver's complicated microstructure and function more effectively than existing models.
The liver serves a critical role in digesting food and detoxifying the body. But due to a variety of factors, including viral infections, alcoholism and drug reactions, the organ can develop chronic or acute problems. When it doesn't work well, a person can suffer abdominal pain, swelling, nausea and other symptoms. Complete liver failure can be life-threatening and can require a transplant, a procedure that currently depends on human donors. To curtail this reliance and provide an improved model for predicting drugs' side effects, scientists have been engineering liver tissue in the lab. But so far, they haven't achieved the complex architecture of the real thing. Jinyi Wang and colleagues came up with a new approach.
Wang's team built a microfluidics-based tissue that copies the liver's complex lobules, the organ's tiny structures that resemble wheels with spokes. They did this with human cells from a liver and an aorta, the body's main artery. In the lab, the engineered tissue had a metabolic rate that was closer to real-life levels than other liver models, and it successfully simulated how a real liver would react to various drug combinations. The researchers conclude their approach could lead to the development of functional liver tissue for clinical applications and screening drugs for side effects and potentially harmful interactions.
The authors acknowledge funding from the National Natural Science Foundation of China and the Fundamental Research Funds for the Central Universities.<|endoftext|>
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A source card is an index card with a citation on it to a source, such as a newspaper article, book or website. Source cards can be written in a variety of citation styles such as Modern Language Associate and American Psychological Association formats.
Web Page or Article
When citing a Web page, include the author's name, title of the Web page or article, the date the page or article was accessed and the full URL to the source. When using MLA format, write the citation as: Author Last Name, First Name. "Title of Article/Web Page." Title of Website. Month and Year of Publication. Web. Day Month Year Accessed. The title of the website should be underlined or italicized and all lines after the first should be indented. If the author's name is not given, write "Unknown" in its place.
In APA format, the citation should look like this: Author Last Name, First Name (Date of publication). Title of article or page. Retrieved from http://full article URL. The title of the article or Web page should be italicized. Indent all lines after the first.
Online Journal Article or Encyclopedia
When citing an online journal or encyclopedia, include the author or editor's name, article title, encyclopedia, database or journal title, copyright date, online publisher or sponsoring institution and date accessed. Write the citation in MLA format as follows: Author Last Name, First Name. “Article Title.” Database or Journal Title. Month and Year of Publication. Online Publisher or Institution Name. Web. Day Month Year Accessed. The title of the journal, database or encyclopedia should be underlined or italicized. Indent all lines after the first.
In APA format, write the citation for an online journal like this: Author Last Name, First Name. (Date of publication). Title of article. Title of Journal, volume number, page range. Retrieved from http://full article URL. For an online encyclopedia, write the citation as: Article Name. (n.d.). In Name of Encyclopedia online. Retrieved from http://full article URL. The name of the encyclopedia should be italicized.<|endoftext|>
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Summary: This article talks about the reality of traumatic experiences, known as “Adverse Childhood Experiences” (ACEs) and the negative impact that they can have on learning, health, and wellbeing. The author suggests that a supportive environment based on social-emotional learning can help students reverse the negative effects of this trauma.
Source: Gary G. Abud, Jr., Education Week, July 11, 2017
Traumatic events, such as war, death, or violence can have a serious influence on one’s health, stress, and anxiety; for kids, this is especially true, as they lack the social and emotional skills to deal with the impact of trauma. Trauma can even cause physical pain, including when a traumatic event is non-physical. In recent years, the Center for Disease Control and Prevention has helped to expand what qualifies as trauma to include more social and emotional events, such as poverty, divorce, and food insecurity.
Read the full article here!<|endoftext|>
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Origins Of The Milky Way Galaxy: Newly Discovered Famliy Of Stars May Shed New Insights
Most people are curious about what are the origins of the universe. Thus, more research is being conducted. Currently, an astronomer discovered a family of stars that might lead to the origins of the Milky Way Galaxy.
A new family of stars is recently discovered by an astronomer from LJMU's Astrophysics Research Institute. It was located at the center of the Milky Way Galaxy that provides new light on the early stage of the galaxy's formation. The discovery led them at the root of globular clusters, which is the concentration of typically a million stars developed at the beginning of the history of the Milky Way Galaxy, according to Phys.Org.
LJMU is a member of an international partnership of scientists at numerous institutions called Sloan Digital Sky Survey. One of its projects is the Apache Point Observatory Galactic Evolution Experiment or APOGEE. They collect infrared data for hundreds of thousands of stars in the Milky Way Galaxy.
In a report by Space Daily, it stated that the new family of stars has the possibility to belong to a globular cluster that was ruined during the violent initial formation of the galactic center. In that case, there could have been 10 times more globular clusters found in the Milky Way Galaxy in early life compared today. It simply means that a massive fraction of the old stars located in the inner part of the galaxy today has the possibility that it was formed in globular clusters and was later destroyed.
The project lead researcher Ricardo Schiavon said that "This is a very exciting finding that helps us address fascinating questions such as what is the nature of the stars in the inner regions of the Milky Way, how globular clusters formed and what role they played in the formation of the early Milky Way and by extension the formation of other galaxies."
"From our observations, we could determine the chemical compositions of thousands of stars, among which we spotted a considerable number of stars that differed from the bulk of the stars in the inner regions of the Galaxy, due to their very high abundance of nitrogen."
However, the researchers are not that certain that these stars resulted from globular cluster destruction. They could also be the byproducts of the first episodes of star formation taking place at the beginning of the galaxy's history. They are still conducting further observations to test these hypotheses.<|endoftext|>
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Multiplication And Division: Question Preview (ID: 32772)
Below is a preview of the questions contained within the game titled MULTIPLICATION AND DIVISION: Use This To Study And Practice The Properties Of Multiplication .To play games using this data set, follow the directions below. Good luck and have fun. Enjoy! [print these questions]
5x0=0 a) Commutative Property b) Distributive Property c) Zero Property d) Associative Property 6x7=7x6 a) Commutative Property b) Identity Property c) Distributive Property d) Zero Property The answer to a multiplication problem a) product b) factor c) quotient d) divisor 5x1=5 a) Commutative Property b) Identity Property c) Distributive Property d) Associative Property Numbers we multiply to find the answer to a multiplication problem a) product b) Factor c) Quotient d) Divisor The way in which the factors are grouped does not change the product. a) Associative Property b) Identity Property c) Distributive Property d) Zero Property A problem can be broken apart into simple problems a) Commutative Property b) Identity Property c) Distributive Property d) Zero Property Any factor multiplied by one will equal that factor a) Commutative Property b) Identity Property c) Distributive Property d) Zero Property Any factor multiplied by zero will equal zero a) Commutative Property b) Identity Property c) Distributive Property d) Zero Property Two factors can be multiplied in any order and the product will be the same a) Commutative Property b) Identity Property c) Distributive Property d) Zero Property 6x13=(6x10)+(6x3) a) Identity Property b) Associative Property c) Commutative Property d) Distributive Property (2x3)x4=2x(3x4) a) Associative Property b) Commutative Property c) Identity Property d) Zero Property The answer to a division problem a) factor b) quotient c) divisor d) dividend A number that divides into another number (the smaller number in a division problem) a) factor b) quotient c) divisor d) dividend A number that is to be divided by another number (larger number in a division problem) a) factor b) quotient c) divisor d) dividend
Play Games with the Questions above at ReviewGameZone.com
To play games using the questions from the data set above, visit ReviewGameZone.com and enter game ID number: 32772 in the upper right hand corner at ReviewGameZone.com or simply click on the link above this text.
TEACHERS / EDUCATORS<|endoftext|>
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teaching resource
# Order of Operations – Differentiated Bump Game
• Updated: 18 May 2023
Use this set of Bump games to sharpen your students’ computation skills by using the order of operations to evaluate expressions.
• Non-Editable: PDF
• Pages: 4 Pages
• Differentiated: Yes
Tag #TeachStarter on Instagram for a chance to be featured!
teaching resource
# Order of Operations – Differentiated Bump Game
• Updated: 18 May 2023
Use this set of Bump games to sharpen your students’ computation skills by using the order of operations to evaluate expressions.
• Non-Editable: PDF
• Pages: 4 Pages
• Differentiated: Yes
Use this set of Bump games to sharpen your students’ computation skills by using the order of operations to evaluate expressions.
## Let’s Play an Order of Operations Game!
Understanding how to apply the order of operations can be a fun and exciting step in a student’s math journey. As students are learning about the steps (maybe you’ve taught them PEMDAS or GEMDAS) there are different levels of difficulty that arise when solving problems with multiple operations.
If you are looking for a differentiated resource that will help students of many different abilities conquer their understanding of the order of operations, look no further!
Teach Starter has created a fun Bump game for your students to play as they are learning to evaluate different expressions. This game comes with three different game boards. Each board increases with difficulty.
• Game board 1: Evaluate expressions with two operations and exponents
• Game board 2: Evaluate expressions with two operations, parentheses, and exponents
• Game board 3: Evaluate expressions with three operations, parentheses, and exponents
## How to Play Our Bump! Game!
The first player will roll their dice and add the digits together. The player will find the matching value on the BUMP game board and solve the problem. Find the answer and cover it up with a game piece.
The next player will take the same steps as the first player. If the answer already has a game piece on it, the player can bump the piece off of the board and replace it with his or her own game piece. If the same player places two game pieces on a value, this space is locked and cannot be bumped off.
Through this activity, students will show they use the order of operations to evaluate expressions with parentheses and exponents.
## Tips for Differentiation + Scaffolding
A team of dedicated, experienced educators created this resource to support your math lessons.
In addition to partner work time, use this order of operations game to enhance learning through guided math groups or math centers.
## Easily Prepare This Resource for Your Students
Use the dropdown icon on the Download button to choose between the PDF or editable Google Slides version of this resource.
Print on cardstock for added durability and longevity. Place the game boards in a folder or large envelope for easy access.
This resource was created by Cassandra Friesen, a teacher in Colorado and a Teach Starter Collaborator.
Don’t stop there! We’ve got more activities and resources that cut down on lesson planning time:
### teaching resource
#### Order of Operations – Google Slides Interactive Activity
Simplify expressions by using the order of operations with this Google Slides interactive activity.
### teaching resource
#### Order of Operations Bingo
Review vocabulary and solve expressions with this Order of Operations Bingo game!
### teaching resource
#### Evaluating Expressions With Exponents – Task Cards
Solve expressions with exponents by using the order of operations with this set of 24 task cards.<|endoftext|>
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Comparison of the skeletons of three bipedal mammals:
an Egyptian jerboa, an eastern gray kangaroo and a human.
Anthropology researchers from The University of Texas at Austin have confirmed a direct link between upright two-legged (bipedal) walking and the position of the foramen magnum, a hole in the base of the skull that transmits the spinal cord.
The study, published in a forthcoming issue of the Journal of Human Evolution, confirms a controversial finding made by anatomist Raymond Dart, who discovered the first known two-legged walking (bipedal) human ancestor, Australopithecus africanus. Since Dart's discovery in 1925, physical anthropologists have continued to debate whether this feature of the cranial base can serve as a direct link to bipedal fossil species.
Chris Kirk, associate professor of anthropology and co-author of the study, says the findings validate foramen magnum position as a diagnostic tool for fossil research and sheds further insight into human evolution.Read the rest of this article...<|endoftext|>
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# Common Core: 4th Grade Math : Subtracting multi-digit numbers
## Example Questions
### Example Question #1 : Subtracting Multi Digit Numbers
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
Next, let's look at the numbers in the tens place:
When the top number is smaller than the bottom number, we have to borrow from the number to the left because we can't take away from since is the smaller number. In this case, we are going to look to the . We only ever need to take away from the number to the left. For this problem, that will leave us with a to replace the . So far, your work should look something like this:
Remember, we've borrowed from the hundreds place, so we can put a in front of the number in the tens place. So far, your work should look something like this:
Now, we can subtract the numbers in the tens place:
Next, we can subtract the numbers in the hundreds place:
### Example Question #2 : Subtracting Multi Digit Numbers
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
When the top number is smaller than the bottom number, we have to borrow from the number to the left because we can't take away from since is the smaller number. In this case, we are going to look to the . We only ever need to take away from the number to the left. For this problem, that will leave us with a to replace the . So far, your work should look something like this:
Remember, we've borrowed from the tens place, so we can put a in front of the number in the ones place. So far, your work should look something like this:
Now, let's subtract the numbers in the ones place:
Next, let's look at the numbers in the tens place:
Finally, we can subtract the numbers in the hundreds place:
### Example Question #3 : Subtracting Multi Digit Numbers
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
Next, let's look at the numbers in the tens place:
When the top number is smaller than the bottom number, we have to borrow from the number to the left because we can't take away from since is the smaller number. In this case, we are going to look to the . We only ever need to take away from the number to the left. For this problem, that will leave us with a to replace the . So far, your work should look something like this:
Remember, we've borrowed from the hundreds place, so we can put a in front of the number in the tens place. So far, your work should look something like this:
Now, we can subtract the numbers in the tens place:
Next, we can subtract the numbers in the hundreds place:
### Example Question #4 : Subtracting Multi Digit Numbers
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
When the top number is smaller than the bottom number, we have to borrow from the number to the left because we can't take away from since is the smaller number. In this case, we are going to look to the . We only ever need to take away from the number to the left. For this problem, that will leave us with a to replace the . So far, your work should look something like this:
Remember, we've borrowed from the tens place, so we can put a in front of the number in the ones place. So far, your work should look something like this:
Now, let's subtract the numbers in the ones place:
Next, let's look at the numbers in the tens place:
When the top number is smaller than the bottom number, we have to borrow from the number to the left because we can't take away from since is the smaller number. In this case, we are going to look to the . We only ever need to take away from the number to the left. For this problem, that will leave us with a to replace the . So far, your work should look something like this:
Remember, we've borrowed from the hundreds place, so we can put a in front of the number in the tens place. So far, your work should look something like this:
Now, let's subtract the numbers in the tens place:
Finally, we can subtract the numbers in the hundreds place:
### Example Question #251 : Number & Operations In Base Ten
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
Next, let's look at the numbers in the tens place:
When the top number is smaller than the bottom number, we have to borrow from the number to the left because we can't take away from since is the smaller number. In this case, we are going to look to the . We only ever need to take away from the number to the left. For this problem, that will leave us with a to replace the . So far, your work should look something like this:
Remember, we've borrowed from the hundreds place, so we can put a in front of the number in the tens place. So far, your work should look something like this:
Now, we can subtract the numbers in the tens place:
Next, we can subtract the numbers in the hundreds place:
### Example Question #6 : Subtracting Multi Digit Numbers
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
Next, let's look at the numbers in the tens place:
Finally, we can subtract the numbers in the hundreds place:
### Example Question #7 : Subtracting Multi Digit Numbers
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
Next, let's look at the numbers in the tens place:
Finally, we can subtract the numbers in the hundreds place:
### Example Question #8 : Subtracting Multi Digit Numbers
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
Next, let's look at the numbers in the tens place:
Finally, we can subtract the numbers in the hundreds place:
### Example Question #9 : Subtracting Multi Digit Numbers
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
Next, let's look at the numbers in the tens place:
Finally, we can subtract the numbers in the hundreds place:
### Example Question #1 : Subtracting Multi Digit Numbers
Solve:
Explanation:
When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.
Let's look at the numbers in the ones place:
Next, let's look at the numbers in the tens place:
Finally, we can subtract the numbers in the hundreds place:<|endoftext|>
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Jump to a New ChapterIntroduction to the SAT IIContent and Format of the SAT II Math IICStrategies for SAT II Math IICMath IIC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
4.1 Order of Operations 4.2 Numbers 4.3 Factors 4.4 Multiples 4.5 Fractions 4.6 Decimals
4.7 Percents 4.8 Exponents 4.9 Roots and Radicals 4.10 Logarithms 4.11 Review Questions 4.12 Explanations
Decimals
Decimals are just another way to express fractions. To produce a decimal, divide the numerator of a fraction by the denominator. For example, 1/2 = 12 = .5.
Comparing Decimals
As with fractions, comparing decimals can be a bit deceptive. As a general rule, when comparing two decimals such as .3 with .003, the decimal with more leading zeros is smaller. But if asked to compare .003 with .0009, however, you might overlook the additional zero and, because 9 is the larger integer, choose .0009 as the larger decimal. That, of course, would be wrong. Take care to avoid such mistakes. One way is to line up the decimal points of the two decimals:
• .0009 is smaller than .0030
Similarly,
• .000900 is smaller than .000925
Converting Decimals to Fractions
Knowing how to convert decimals into fractions, and fractions into decimals, is a useful skill. Sometimes you’ll produce a decimal while solving a question and then have to choose from fractions for test choices. Other times, it may be easier to work with fractions. Whatever the case, both conversions can be done easily.
To convert a decimal number to a fraction:
1. Remove the decimal point and use the decimal number as the numerator.
2. The denominator is the number 1 followed by as many zeros as there are decimal places in the decimal number.
3. Reduce the fraction.
Let’s convert .3875 into a fraction. First, we eliminate the decimal point and make 3875 the numerator:
Since .3875 has four digits after the decimal point, we put four zeros in the denominator:
Then, by finding the greatest common factor of 3875 and 10000, 125, we can reduce the fraction:
To convert from fractions back to decimals is a cinch. Simply carry out the necessary division on your calculator, such as for 31/80:
Jump to a New ChapterIntroduction to the SAT IIContent and Format of the SAT II Math IICStrategies for SAT II Math IICMath IIC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
Test Prep Centers
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College Admissions Financial Aid College Life<|endoftext|>
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Labial: Pertaining to the lip, one of the fleshy folds which surround the opening of the mouth or the vagina.
Oral Labia: The upper lip is separated from the nose by the philtrum, the area that lies between the base of the nose and the pigmented edge (called the vermillion border or the carmine margin) of the upper lip. The upper and lower lips meet at the corners (or angles) of the mouth which, in anatomy, are called the oral commissures. The oral commissure normally lies in a vertical line below the pupil.
Small blind pits are sometimes seen at the corners of the mouth; they are known as angular lip pits and are considered normal minor variants.
The lips may be abnormally thin or thick. For example, the upper lip is typically thin and the philtrum flat in children with the fetal alcohol syndrome. If the upper lip is overgrown, the corners of the mouth appear to be downturned.
Labial Sounds: The lips are not only anatomic features of note, they are organs of speech essential to certain articulations. A sound requiring the participation of one or both lips is a labial (labium in Latin means lip) sound or, simply, a labial. All labials are consonants. There are bilabial sounds such as "p" which involve both lips and labiodental sounds such as "v" which involve the upper teeth and lower lip.
Vaginal Labia: There are two pairs of labia (lips) at the entrance to the vagina. They are the labia majora (the larger outside pair) and the labia minora (the smaller inside pair). Together they form part of the vulva (the female external genitalia).
The Word "Labia": The word "lip" can be traced back to the Indo-European "leb" which also produced the Latin "labium" from which came the French "levre." The German "lippe" is just a slip from the English "lip." See also: Labrose.
adj. Of or relating to the lips or labia. Linguistics . Articulated mainly by closing or partly closing the lips, as the sounds (b), (m), or (w). n. Linguistics . A ...
Labial consonants are consonants in which one or both lips are the active articulator. This precludes linguolabials, in which the tip of the tongue reaches for the ...
adjective 1. of, pertaining to, or resembling a labium. 2. of or pertaining to the lips. 3. Phonetics . involving lip articulation, as p, v, m, w, or a rounded vowel ...
Labial (from Latin Labium) may refer to: the lips; the labia (genitalia) In linguistics, a labial consonant; In zoology, the labial scales; See also. All pages beginning with "labial"
Definition of LABIAL. 1: uttered with the participation of one or both lips <the labial sounds \f\, \p\, and \ü\> 2: of, relating to, or situated near the lips or labia<|endoftext|>
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## Overview
In Sections 14.1 - 14.3, we saw:
• How to describe random variables with their probability distributions
• How to combine random variables
In these sections, we will:
• Describe common situations with named distributions
• Discuss properties of these distributions
• Use named distributions to find probabilities
## Discrete Distributions
For now, we will focus on distributions that have discrete outcomes
• Discrete outcomes are outcomes that we can list
• In statistics, the most common of these are distributions which count occurances
• This is usually how we deal with categorical variables
We will see:
• The Bernoulli Distribution
• The Binomial Distribution
• The Poisson Distribution
## The Bernoulli Distribution
A Bernoulli Trial is a random trial where there are only two outcomes
• Yes or No
• Success or Failure
The Bernoulli Distribution describes the behavior of these trials.
• The only thing we need to describe a Bernoulli random variable (RV) is $$p$$, the probability of success
If a random variable follows the Bernoulli Distribution, we write:
• $$X \sim B(p)$$
• "X follows a Bernoulli Distribution with probability of success $$p$$"
## Describing Bernoulli Trials
In Bernoulli Trials, we denote the outcomes as:
• 0 for failure
• 1 for success
Since $$p$$ denotes the probability of success and there are only two outcomes:
• $$P(1) = p$$ is the probability of success
• $$P(0) = 1 - p = q$$ is the probability of failure
Note that "success" and "failure" are completely arbitrary.
## Bernoulli Example: Ebola
Say a doctor tells you that you have a particularly deadly strain of Ebola, which has an 80% mortality rate. Let $$X$$ denote your mortality and $$p = .8$$ be the probability you die.
What's the distribution look like?
• Outcome $$X = x$$ $$P(X = x)$$
Die 1 $$p = 0.8$$
Survive 0 $$q = 0.2$$
## Bernoulli Expectations
What is the expected outcome of a Bernoulli Trial?
• $$E(X) = \sum xP(x) = 1p + 0q = p$$
• The expected value is the probability
• In the long term, this is the proportion of successes
For our Ebola example:
• $$E(X) = p = 0.8$$
• About 80% of people with the disease will die
• You are much more likely to die than survive
## Bernoulli Standard Deviations
What is the variance of a Bernoulli RV?
• $$VAR(X) = \sum \left(x - E(X)\right)^2P(x)$$
• $$VAR(X) = pq$$
• $$SD(X) = \sqrt{pq}$$
Notice that this will be largest when $$p = .5$$, reflecting the increased uncertainty that comes from both outcomes being equally likely.
For our Ebola case:
• $$SD(X) = \sqrt{pq} = \sqrt{0.8\times 0.2} = \sqrt{0.16} = 0.4$$
## The Binomial Distribution: Motivation
Say we want to flip two coins and count the number of heads:
• Outcome $$X = x$$ $$P(X = x)$$
$$HH$$ 2 0.25
$$HT$$ 1 0.25
$$TH$$ 1 0.25
$$TT$$ 0 0.25
• $$P(0) = 0.25 \quad P(1) = 0.5 \quad P(2) = 0.25$$
• What if we flipped three coins?
Outcome $$X = x$$ $$P(X = x)$$
$$HHH$$ 3 $$1/8$$
$$HHT$$ 2 $$1/8$$
$$HTH$$ 2 $$1/8$$
$$HTT$$ 1 $$1/8$$
$$THH$$ 2 $$1/8$$
$$THT$$ 1 $$1/8$$
$$TTH$$ 1 $$1/8$$
$$TTT$$ 0 $$1/8$$
For the three coin flips,
• $P(0) = 1/8 \quad P(1) = 3/8 \quad P(2) = 3/8 \quad P(3) = 1/8$
The Problem
• For two trials, we had 4 rows.
• For 3 trials, we had 8.
• For 4 trials, we'd have 16.
• In general, for $$n$$ trials we have $$2^n$$ unique outcomes
• Tables like this quickly become unmanageable
• The Binomial Distribution is a convenient way to find these probabilities
## The Binomial Distribution
What is the binomial distribution?
• a Binomial Experiment is a series of $$n$$ bernoulli trials
• The Binomial Distribution describes the behavior of these trials
We only need two numbers to describe a binomial distribution:
• $$n$$, the number of trials
• $$p$$, the probability of success in each trial
If $$X$$ follows a binomial distribution:
• $$X$$ counts the number of success in the $$n$$ trials
• $$X$$ has to be a whole number from 0 to $$n$$
## Binomial Conditions
We can use a binomial distribution if:
• Each trial has only two outcomes (i.e., is a Bernoulli trial)
• Each trial is independent of the rest
• The probability of success, $$p$$, is constant for all trials
Notation:
• $$X \sim Binomial(n, p)$$
• $$X$$ follows a binomial distribution with $$n$$ trials and probability of success $$p$$
## Binomial Expectations
If $$X \sim Binomial(n, p)$$, the expected value and standard deviation have simple formulas.
Expectation:
• $$E(X) = \mu = np$$
Standard Deviation:
• $$VAR(X)= \sigma^2 = np(1 - p) = npq$$
• $$SD(X) = \sqrt{VAR(X)} = \sqrt{npq}$$
## Ebola Example
For any one person, survivability is a Bernoulli trial with probability of death $$p = 0.8$$. What if we had twenty patients?
$$X \sim Binomial(n = 20,\; p = .8)$$
Expected Value:
• $$E(X) = np = 20(0.8) = 16$$
• In a group of 20 patients, we would expect 16 to die
Standard Deviation
• $$SD(X) = \sqrt{npq} = \sqrt{np(1-p)} = \sqrt{20(0.8)(0.2)}$$
• $$SD(X) = \sqrt{3.2} \approx 1.79$$
## Binomial Probabilities
For our Ebola patients, what if we wanted to know the probability that fewer than 10 died? More than 18?
• StatCrunch can find them using the Binomial Calculator
• Stat $$\to$$ Calculators $$\to$$ Binomial
• Enter $$n$$ and $$p$$
• Enter the range or value you're interested in
• Hit Compute
## Setting up the Probabilities
The difficult part of finding the probabilities is setting up the problem in the correct notation
• More than 10: $$P(X > 10)$$
• At least 13: $$P(X \ge 13)$$
• Less than 8: $$P(X < 8)$$
• At most 15: $$P(X \le 15)$$
• Equal to 3: $$P(X = 3)$$
Once you figure out exactly what's being asked, solving these problems is a simple matter of entering them in the calculator. Just make sure $$n$$ and $$p$$ are correct.
## Binomial Example
Say a brewery is offering a "free beer" promotion. Under 10% of their bottle caps, you can win a free six pack. Hoping to take advantage of the offer, you buy two cases (48 bottles). Does the number of winning bottles in your cases follow a binomial distribution?
• Each bottle is a winner or a loser
• Each bottle has a 10% chance of being a winner
• There are 48 trials
• One bottle has no effect on the other bottles, assuming the free beer is distributed randomly.
• If $$X$$ is the number of winners: $$X \sim Binomial(n = 48, p = 0.1)$$
## Free Beer: Expectation
$X \sim Binomial(n = 48, p = 0.1)$
What is the expected value?
• $$E(X) = np = 48(0.1) = 4.8$$
What is the standard deviation?
• $$SD(X) = \sqrt{npq} = \sqrt{np(1 - p)} = \sqrt{48(0.1)(0.9)}$$
• $$SD(X) = \sqrt{4.32} \approx 2.08$$
## Free Beer: Probabilities
What are the chances you get exactly 7 free six packs?
• $$P(X = 7)$$
What is the probability that you win fewer than 10?
• $$P(X < 10)$$
What is the probability you get at least two?
• $$P(X \ge 2)$$
What is the probability you get more than 1, but at most 4?
• $$P(1 < X \le 4) = P(2 \le X \le 4)$$
## When the Binomial Model Fails
The binomial distribution is great when we know $$n$$, but the world is often more complicated than that. Say a restaurant owner wants what the 95th percentile is of the customers he will see in a given day so we won't run out of supplies. How do we model this?
• To use the binomial model, we would need to know how many people are potential customers on a given day, and their probability of deciding to eat there
• This is probably unrealistic
• What can we know?
• The average number of customers per day is easier to find
## The Poisson Distribution
The Poisson Distribution models the number of occurances of some event that take place in a unit of space or time
• Customers in a day
• Crimes per square-mile
• Students coming to office hours in an hour
Specifying the model
• The Poisson distribution has only one parameter, $$\lambda$$
• $$\lambda$$ is the average number of occurances per unit of space or time
• If $$X$$ follows the Poisson distribution, we write $$X \sim Poisson(\lambda)$$
## Poisson Expectations
The expected value and standard deviation are particularly straightforward for the Poisson distribution.
Expected Value
• $$E(X) = \lambda$$
Standard Deviation
• $$VAR(X) = \lambda$$
• $$SD(X) = \sqrt{\lambda}$$
## Poisson Example: Restaurant
Say our customer usually sees 100 customers in a given day. Let $$X$$ represent the number of customers in a day, then
• $$X \sim Poisson(\lambda = 100)$$
Expected Value
• $$E(X) = \lambda = 100$$
• The restaurant would expect to see 100 customers on a typical day
Standard Deviation
• $$SD(X) = \sqrt{\lambda} = \sqrt{100} = 10$$
• We would expect any particular day to differ from the mean by 10 customers
## Poisson: Scaling
If we were interested in the number of customers per hour instead of day, how would this change the model?
• If we usually see 100 per day, we would expect $$\frac{100}{24} \approx 4.17$$ per hour
Let $$Y$$ denote the number of customers per hour
• $$Y \sim Poisson(\lambda = 4.17)$$
• $$E(Y) = \lambda = 4.17$$
• $$SD(Y) = \sqrt{\lambda} = \sqrt{4.17} \approx 2.04$$
## Poisson Probabilities
Finding Poisson probabilities can be done the same way as binomial probabilities. The only tricky part is getting the notation right.
What's the probability of seeing more than 20 customers in a day?
• $$P(X > 20)$$
What are the chances of seeing at least twelve customers in a day?
• $$P(X \ge 12)$$
What is the probability of seeing less than 30 customers?
• $$P(X < 30)$$
What is the probability of seeing no more than 15?
• $$P(X \le 15)$$
## Example: Ebola
Say that instead of having 20 cases of ebola, we know that there are an average of 30 new cases per day in a particular city.
What is the Poisson Model?
• $$X \sim Poisson(\lambda = 30)$$
What is the expected number of new cases on a given day?
• $$E(X) = \lambda = 30$$
How much would we expect a given day to vary from the mean?
• $$SD(X) = \sqrt{\lambda} = \sqrt{30} \approx 5.48$$
## Example: Ebola
What's the probability of seeing at least 40 cases on a given day?
• $$P(X \ge 40)$$
What the probability of seeing fewer than 15?
• $$P(X \le 15)$$
What's the probability of seeing more than 50 cases?
• $$P(X > 50)$$
What are the chances of seeing more than 20 but less than 25?
• $$P(20 < X < 25) = P(21 \le X \le 24)$$
## Continuous Distributions
The Bernoulli, Binomial, and Poisson distributions all deal with discrete random variables, which have a distinct set of outcomes. In particular, they are distributions for counting occurances, so they can only take whole numbers as values.
• If we are only dealing with whole numbers, we can count how many values fall in a certain range
• If $$3 \le X \le 5$$, $$X$$ can take the values 3, 4, or 5.
Continuous variables are different
• If $$X$$ is a continous variable, it can take any value in a range.
• If $$3 \le X \le 5$$ if could be 3.01, 3.0000001, 4.9, 4.999999, etc.
• We cannot count how many values there are in a range
## Continuous Probabilities
For discrete variables
• We can find $$P(X = 3)$$, because there are only a certain number of possibilities.
For continunous variables
• We cannot find $$P(X = 3)$$, because there are an infinite number of possibilities
• In fact, the probability that a continuous variable takes any particular value is zero
• We can only find the probability of ranges
## Review: The Normal Distribution
The Normal distribution is a continuous distribution. Recall:
• We define the Normal distribution using its mean $$(\mu)$$ and standard deviation $$(\sigma)$$
• If $$X$$ has a Normal distribution, we write $$X \sim N(\mu, \sigma)$$
Normal Distributions are:
• Bell Shaped
• Unimodal
• Symmetric about $$\mu$$
## Finding Normal Probabilities
In chapter five:
• We found the percent of a population in certain ranges
• This is equivalent to finding the probability and individual is in that range
Using StatCrunch
• Stat $$\to$$ Calculators $$\to$$ Normal
• Enter $$\mu$$ and $$\sigma$$
• Enter the range
• StatCrunch will find the probability
## Normal Distribution Example
Say that the weight of laptops is normally distributed with a mean of 5 lbs. and standard deviation of 1 lb.
Specifying the model:
• $$X \sim N(\mu = 5, \sigma = 1)$$
What is the probability a randomly selected laptop weighs less than 3 lbs?
• $$P(X \le 3)$$
What is the probability a laptop chosen at random weighs at least 4 lbs?
• $$P(X \ge 4)$$?
## Review
• If a random variable has only two outcomes, it can be represented with the Bernoulli distribution
• If a random variable is a series of random trials, we can describe it with a binomial distribution
• If a random variable is the number of occurances per unit of space or time, we can use the Poisson distribution
• The Normal distribution models continuous variables whose shape is unimodal, symmetric, and bell shaped
• For any of these distributions, we can find probabilities using StatCrunch provided we have the necessary parameters to describe them<|endoftext|>
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In an isosceles triangle if the vertex angle is twice the sum of the base angles, calculate the…
In an isosceles triangle if the vertex angle is twice the sum of the base angles, calculate the…
A triangle having two equal sides is known as an isosceles triangle. The two equal sides are called legs, and the third side is known as the base of the isosceles triangle. The angle between the legs is called the vertex angle, and the two angles adjacent to the base are called base angles.
The formula for the area of an isosceles triangle is written as:
Area (A) = $$\frac{1}{2}$$ x b x h
Here, ‘b’ refers to the base and ‘h’ refers to the height of the triangle.
The perimeter of an isosceles triangle is determined by applying the formula:
Perimeter (P) = 2a + b
Here, ‘a’ is the length of the equal sides of an isosceles triangle, and ‘b’ is the length of the base.
The properties of an isosceles triangle are:
Example 1:
The base of an isosceles triangle is 20 centimeters, and the length of each of its legs is 28 centimeters. Calculate its perimeter.
Solution:
AB = AC = a = 28 cm and BC = b = 20 cm.
The formula for finding the perimeter is:
Perimeter (P) = 2a + b
= 2(28) + 20 [Substitute 28 for ‘a’ and 20 for ‘b’]
= 56 + 20 [Simplify]
= 76 cm
Hence, the perimeter of the isosceles triangle is 76 centimeters.
Example 2:
Winnie made a sandwich for her brother. She cut the sandwich into a triangular slice such that the base length of the triangle is 15 centimeters and the height is 20 centimeters. Determine the area of the sandwich.
Solution:
Winnie cut the sandwich into a triangular slice, having a base length (b) of 15 cm and a height (h) of 20 cm.
Area (A) = $$\frac{1}{2}$$ × b × h [Formula for the area of the triangle]
= $$\frac{1}{2}$$ × 15 × 20 [Substitute 15 for ‘b’ and 20 for ‘h’]
= $$\frac{300}{2}$$ [Simplify]
= 150 cm$$^2$$
Therefore, the area of the sandwich is 150 square centimeters.
Example 3:
The area of a triangle is 52 square inches, and the base length is 8 inches. Calculate the height of the triangle.
Solution:
Area (A) = 52 in
base (b) = 8 in
Area (A) = $$\frac{1}{2}$$ × b × h [Area of a triangle formula]
52 = $$\frac{1}{2}$$ × 8 × h [Substitute 52 for A and 8 for b]
52 = 4 × h [Multiply]
$$\frac{52}{4}$$ = h [Simplify]
13 = h
Thus, the height of the given triangle is 13 inches.
If two sides of a triangle are equal, the triangle is said to be an isosceles triangle. Consider a triangle with sides AB, BC, and CA. The triangle is isosceles if it fulfills one of these conditions: AB = BC, or BC = CA, or CA = AB.
Yes, by knowing the two equal angles, we can easily subtract the sum of angles from 180°, as the sum of all angles of a triangle is equal to 180°.
Some of the examples of an isosceles triangle are the roof of a house, a slice of pizza, clothes hangers, traffic signs, and so on.
There are three types of isosceles triangles:
You are watching: Isosceles Triangle (Definition, Properties, Examples). Info created by GBee English Center selection and synthesis along with other related topics.<|endoftext|>
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We are even getting much better optics from graphene. All welcome of course.
Graphene is a continuing revolution and has also led to other elements been similarly mastered as well.
It has made all materials science functionally obsolete and produced astonishing results as we investigate layer by layer fabrications.
Graphene optical lens a billionth of a meter thick breaks the diffraction limit
JANUARY 31, 2016
The graphene lens developed at Swinburne University of Technology is one billionth of a meter thick (Credit: Swinburne University of technology)
With the development of photonic chips and nano-optics, the old ground glass lenses can't keep up in the race toward miniaturization. In the search for a suitable replacement, a team from the Swinburne University of Technology has developed a graphene microlens one billionth of a meter thick that can take sharper images of objects the size of a single bacterium and opens the door to improved mobile phones, nanosatellites, and computers.
One of the key obstacles in advances in optical microscopes is the lens or, more precisely, the diffraction limit, which is the theoretical limit of the resolution of a particular lens. There have been a number of attempts at overcoming the diffraction limit by using such techniques as interferometry, holography, lasers, and electrons, and although scientists have enjoyed some success, it has only been at great cost and complexity.
Another approach has been to explore the use of ultrathin flat lenses that are etched with concentric circles and act like tiny Fresnel lenses. According to the Swinburne team, this has also had some success, but only by crafting the lenses out of gold and other metals that don't lend themselves to mass production.
Swinburne's breakthrough came when Xiaorui Zheng, a PhD student at the Centre for Micro-Photonics, used graphene oxide to form a lens. This material allowed the team to make ultrathin flat lenses that are 300 times thinner than a sheet of paper and weigh a microgram. This was achieved by effectively printing the lens, first spraying a sheet of graphene oxide solution, then molding the circles using a laser beam.
According to the team, the new lens is flexible, can resolve objects as small as 200 nanometers, and can even see into the near infrared. This is possible a it breaks the diffraction limit and allows a focus of less than half the wavelength of light.
Once the technology is mature, the team sees it as having applications beyond microscopy, such as in lighter, thinner mobile phones with thermal imaging capabilities, smaller endoscopes for surgery, as a replacement for conventional lenses in nanosatellites to save a couple of hundred grams of launch weight, and to increase the efficiency of photonic chips in supercomputers and superfast broadband distribution.
The graphene microlens research was published in Nature Communications.<|endoftext|>
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# Question #acf06
Feb 24, 2018
Answer: $5 {\left(x + 2\right)}^{2}$
#### Explanation:
Assuming question is asking for $\sqrt{25 {\left(x + 2\right)}^{4}}$
By definition of square root, we know that:
$\sqrt{{p}^{2}} = p$, where $p$ is any function
Therefore, applying this definition to the original problem we can see that the original problem can be factored into the form ${p}^{2}$:
$\sqrt{25 {\left(x + 2\right)}^{4}}$
$= \sqrt{5 \cdot 5 \cdot {\left(x + 2\right)}^{2} \cdot {\left(x + 2\right)}^{2}}$
$= \sqrt{{5}^{2} \cdot {\left({\left(x + 2\right)}^{2}\right)}^{2}}$
$= \sqrt{{\left(5 {\left(x + 2\right)}^{2}\right)}^{2}}$
Now, we can set our $p$ to be $p = 5 {\left(x + 2\right)}^{2}$, so our answer is
$5 {\left(x + 2\right)}^{2}$<|endoftext|>
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I am going to begin by building anticipation for the book Zero the Hero written and illustrated by Joan Holub and Tom Lichtenheld. I will show this short clip from YouTube, about their character, Zero the Hero, before I read the story. Then I will ask them:
Students, what do you know about the number zero and how we use it in math?
Could we do math without a zero? (Watch the video clip in the resource section for my student's response to this question.)
I will conduct a read aloud with the book Zero the Hero, written and illustrated by Joan Holub and Tom Lichtenheld. I will use this book to discuss the following topics:
There are two videos in the resource section for you to view my students thoughts and understandings of zero. I love how books tap into students prior knowledge and build new connections for them.
When the book discussion is finished I will discuss what happens to numbers when we add one to it. My goal is for my students to develop an understanding that the number just increases to the next number they would say when counting. We will practice several problems aloud together;
Then we will discuss adding 2 to a number and solve several problems aloud together:
CCSS 1.OA.C.6 has first graders developing fluency in addition through the use of multiple strategies. Children who understand why a number remains the same when added to zero, will not only be able to memorize these facts easily, but they will solidify their understanding of the concept of 0, which can be abstract for first graders. I want them to understand the meaning behind the equation so the sum is reached faster. This assists them in developing Math Practice 2 and becoming proficient in reasoning quantitatively. Zero is an important number for students to understand because it actually represents a quantity and makes a major difference in counting, place value understanding, and computation.
Students will practice solving problems adding 0, 1, and 2. It is important for First Graders to develop fluent math facts skills now because it will affect their success in later grades. They need to be able to solve addition and subtraction quickly to move onto more difficult and multi-step problems. Students must be able to reason quantitatively to become math proficient. (MP2). When their work is complete, I will ask them to show me their paper. If everything is correct, they are going to create Zero the Hero, available on Teachers Pay Teachers. You can go here to get the free "Zero the Hero Craftivity." Zero the Hero will be glued to the top of the worksheet. There are several pieces to Zero the Hero and I will develop a central location in the room for students to grab everything when their worksheet is complete. If I were to give them all their supplies up front, it would distract them from the work required on the worksheet. Check the resource section to view pictures and a video of students gathering materials and working.
Here is a great connection between the character "Zero the Hero" and literature. This e-book adds to the children's knowledge of how zero is important in counting. After watching the video, I will have them write their numbers, counting by tens to 120.<|endoftext|>
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The Mammalian Diving Reflex
Did you ever wonder why you instinctively hold your breath when you dive?
It's because of the mammalian diving reflex where your body performs an elaborate response to ensure the survival of your vital organs. The diving reflex is a natural response that every person possess. It is also known as the Diving Response or the Bradycardic Response.
The diving reflex mechanism is a response to immersion of all known air-breathing vertebraes. As vertebraes, mammals have this reflex too, which means, you the reader, have it.
Discovery of the diving reflex
Ever since the scientific world accepted Boyle's law, scientist calculated that diving thirty meters (100ft) were impossible. Under Boyle's Law, the deeper we dive underwater, the higher the pressure. For decades, scientists thought that this high pressure would compress the air in our body and crush our lungs.
But in 1949, two Italian divers proved this theory wrong. Ennio Falco dared Raimondo Bucher (who was also a spearfisher) to dive the limit for 50,000 Lira ($800). Lt. Bucher plunged the thirty meters (100 ft) and won the bet. This feat brought awareness among the scientific community and led them to study more about human physiology.
In 1962, the Swedish physiologist, Scholander discovered the secret of this mechanism. By strapping heart rate monitors to volunteer divers, he found out that these divers had their heart slowed down when they dive.
How does the diving reflex happen?
The human body will conserve the air supply it has when immersed in water. When your body senses that the face is getting wet, it will automatically keep the oxygen inside, and your blood will be moved to your heart and brain.
We have this response even at an early stage of our lives. When you dip a baby in water, it will close its throat to prevent the air from escaping the body and keeping water or anything else out.
Aquatic mammals like the dolphins and whales, even the hippopotamus, are more adept in using their diving reflex. Evolution made their physiology more suited to such conditions.
What triggers the diving reflex?
The diving reflex occurs because of the immersion of the body in water. The need for air made our body more conscious of oxygen consumption. It results in a decrease in heart rate to lessen the oxygen consumption.
Studies found that the diving reflex only works if your face is wet or immersed in water. Scientists have tried simulating different situations with different stimuli but to no avail. Only immersion trigger the mammalian diving response.
Our body responds that way without knowing why it should. It means that the diving reflex is instinctive. But the reflex can become voluntary if you practice on controlling it.
In 1976, the French diver, Jacques Mayol left the world in awe when he dove down to 100 meters underwater (330ft) for the first time. His introduction of yoga to the new sport of freediving gave freedivers the possibility to go deeper.
The changes in the body due to the Reflex
When the diving response kicks in, the body adapts temporarily to being immersed in water. It means that our body will not function the way it usually does.
It starts with a Bradycardia which involves the slowing of the pace of the heart — followed by the Peripheral Vasoconstriction. The second change constricts blood flow and oxygen consumption. Finally, there will be a blood shift. The shift causes blood plasma from different parts of the body to move to the chest cavity. It is to protect the organs in the chest.
These changes allow the human body to stay longer underwater. It also makes sure that no damage is done even with the deprivation of air.
The benefits of enhancing the diving reflex
Practicing to control your diving response can contribute to your health. The diving response makes our body adaptive to conditions where air becomes limited. Your body will function normally even under a similar situation.
Scholander's experiment on divers showed that we could dive deep without the help of any gears. As the divers controlled their breathing, the researchers figured the possibility of treating anxiety.
By practicing apnea, our breathing becomes more controlled, even when our body is lacking in air, we can remain calm.
The Diving Reflex Researches and References
All pictures used for this article are courtesy of Wikimedia Commons and our freediving staff.
If you want to know more about sirens and mermaids, take a look at some of our references below.
- Mammalian Diving Response
- Diving & health
- Butcher Raimondo
- Science of Freediving
- Physiology of Freediver
- An Enigmatic Reflex to Preserve Life?
- Diving Reflex
- Cold Shock Response
Speaking about Mammalian Diving Reflex, do you how deep can a mammal go on a single breath?
Last, but not Least
If you would like to receive interesting content like this in your email Inbox, subscribe to our newsletter.
In addition to our monthly newsletter, we will send you our weekly e-Bulletin with one fascinating topic, like today. There will be no advertising or sales pitch.
We also want to thank YouTube for the amazing videos on this page.
Thanks for reading, and if you wish, see you next week!
The Research and Media Team at Scotty's.
Disclaimer: The views, ideas, and opinions expressed by the writers of Scotty's Media team do not necessarily reflect or represent the views, ideas, and opinions of the company, Scotty's Action Sports Network, Inc.<|endoftext|>
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Small craters point to a cold, dry past for the Red Planet
Ancient Mars may have resembled Antarctica (but without penguins). The frozen Red Planet probably had liquid water on its surface for only relatively short times, according to a new analysis of craters on the surface. The finding that the Martian atmosphere was not dense enough to keep flowing water for more than a few hundred thousand years at a time adds to a growing pile of evidence that Mars probably remained cold and dry throughout most of its history, punctuated by brief periods of relative warmth.
Mars today has a very tenuous atmosphere, not nearly dense enough to keep water from instantly boiling away. But deep canyons and ancient river deposits point to a time when water flowed across the Red Planet. Researchers, however, disagree on whether those features indicate long-lasting temperate climates or brief bursts of warming interspersed throughout a long, dry winter.
To probe the early Martian atmosphere, Edwin Kite, a planetary scientist at Princeton University, and colleagues looked at ancient craters. A crater’s size reflects the size of the rock that hit the surface. In turn, the size of the rocks reflects the density of the air because the atmosphere heats up and breaks apart incoming boulders. Denser atmospheres destroy smaller rocks, while thinner air lets small rocks through.
By looking at crater sizes from Mars’ wetter past, Kite’s team estimated the minimum size of the rocks that made it to the surface. The researchers analyzed high-resolution images taken by the Mars Reconnaissance Orbiter of 316 craters near Gale Crater, where the Mars rover Curiosity has been exploring. These craters, interspersed with ancient river deposits, probably formed during a wetter epoch at least 3.6 billion years ago.
Kite’s team combined the crater sizes with computer simulations of impacts to estimate the atmospheric pressure from that time. The researchers report in the April 13 Nature Geoscience that the pressure was between about 90,000 and 190,000 pascals, which is not enough to sustain liquid water on the surface for an extended period.
Kite says the crater record suggests that the canyons and clays seen on Mars — signs of liquid water — formed during brief warm periods of less than 100,000 years. Such warm periods could be triggered by giant impacts or swings of the planet’s axis.
The early Martian environment, Kite says, was probably similar to certain parts of Antarctica where, despite subzero average temperature and negligible rain or snow, rivers flow and lakes fill during summer glacial melts. Even in this harsh environment microbial life endures, suggesting that life could have gotten a foothold on a mostly cold, dry Mars.
François Forget, a physicist at the Dynamic Meteorology Laboratory in Paris, is not surprised by Kite’s results. The crater data confirm Forget’s own climate simulations suggesting that Mars couldn’t hold on to a thick atmosphere. The Red Planet is so cold, he notes, that in a dense atmosphere any substantial carbon dioxide would condense and form ice caps, reducing the amount of the greenhouse gas in the atmosphere and further chilling the planet. “It’s very difficult,” he says, “to have a wet, warm Mars which looks like Earth does now.”
But some scientists disagree. “They’re opening a big can of worms,” says Ramses Ramirez, a planetary scientist at Pennsylvania State University in University Park. He takes issue with the Kite team’s assumptions about the Martian surface. A hard surface will create smaller craters than a soft one. Kite and colleagues assume the target rock is relatively soft. However, a soft surface may not be realistic, says Ramirez. Unless there was no bedrock or the underlying sediment did not harden much under its own weight, as it does on Earth, then the ancient Martian surface would have been harder than they claim. The sharp cliff faces within the ancient valley networks, great water-carved canyons hundreds of kilometers long and hundreds of meters deep, suggest a hard surface. “If Mars was as soft as they’re saying,” says Ramirez, “these features would have collapsed under their own weight like a sand castle.” But Kite responds that a softer surface is appropriate for the finely grained river deposits where the craters occur, similar to alluvial fans in Earth’s deserts.
Kite plans to analyze craters in other regions of Mars. Studying craters with different ages, he says, should help piece together the history of the atmosphere. Additionally, NASA’s Mars Atmosphere and Volatile Evolution, or MAVEN, mission, scheduled to arrive at Mars in September 2014, will study the present Martian atmosphere, which should help researchers better understand its past as well.
E.S. Kite et al. Low palaeopressure of the martian atmosphere estimated from the size distribution of ancient craters. Nature Geoscience. Published online April 13, 2014. doi: 10.1038/ngeo2137.
M. Rosen. Old rover finds new evidence of water on Mars. Science News. Vol. 185, February 22, 2014, p. 10.
A. Grant. Mars was habitable longer, more recently than thought. Science News Online. December 9, 2013.
A. Grant. Life-friendly environment confirmed on Mars. Science News Online. March 12, 2013.
E. Wayman. Mars clays may have volcanic source. Science News. Vol. 182, October 20, 2012, p. 14.
N. Drake. Mars' history is a fluid situation. Science News. Vol. 180, December 3, 2011, p. 5.<|endoftext|>
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