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1,185	Watch the video or Read the steps below to find out how to use a probability tree: What is a Probability Tree? Probability trees are useful for calculating combined probabilities. It helps you to map out the probabilities of many possibilities graphically, without the use of complicated probability equations. Why Use a probability tree? Sometimes you don’t know whether to multiply or add probabilities. A probability tree makes it easier to figure out when to add and when to multiply. Plus, seeing a graph of your problem, as opposed to a bunch of equations and numbers on a sheet of paper, can help you see the problem more clearly. Parts of a probability tree. A probability tree has two main parts: the branches and the ends. The probability of each branch is generally written on the branches, while the outcome is written on the ends of the branches. Multiplication and Addition Probability Trees make the question of whether to multiply or add probabilities simple: multiply along the branches and add probabilities down the columns. In the following example (courtesy of Yale University), you can see how adding the far right column adds up to 1, which is what we would expect the sum total of all probabilities to be: .9860+ 0.0040 + 0.0001 + 0.0099 = 1 Real Life Uses Probability trees aren’t just a theoretical tool used the in the classroom — they are used by scientists and statisticians in many branches of science, research and by several government bodies. For example, the following tree was used by the Federal government as part of an early warning program to assess the risk of more eruptions on Mount Pinatubo, an active volcano in the Philippines. How to Use a Probability Tree or Decision Tree Sometimes, you’ll be faced with a probability question that just doesn’t have a simple solution. Drawing a probability tree (or tree diagram) is a way for you to visually see all of the possible choices, and to avoid making mathematical errors. This how to will show you the step-by-step process of using a decision tree. How to Use a Probability Tree: Steps Sample question: “An airplane manufacturer has three factories A B and C which produce 50%, 25%, and 25%, respectively, of a particular airplane. Seventy percent of the airplanes produced in factory A are passenger airplanes, 25% of those produced in factory B are passenger airplanes, and 25% of the airplanes produced in factory C are passenger airplanes. If an airplane produced by the manufacturer is selected at random, calculate the probability the airplane will be a passenger plane.” Step 1:Draw lines to represent the first set of options in the question (in our case, 3 factories). Label them (our question list A B and C so that is what we’ll use here). Step 2: Convert the percentages to decimals, and place those on the appropriate branch in the diagram. For our example, 50% = 0.5, and 25% = 0.25. Step 3: Draw the next set of branches. In our case, we were told that 70% of factory A’s output was passenger. Converting to decimals, we have 0.7 P (“P” is just my own shorthand here for “Passenger”) and 0.3 NP (“NP” = “Not Passenger”). Step 4:Repeat step 3 for as many branches as you are given. Step 5: Multiply the probabilities of the first branch that produces the desired result together. In our case, we want to know about the production of passenger places, so we choose the first branch that leads to P. Step 6: Multiply the remaining branches that produce the desired result. In our example there are two more branches that can lead to P. Step 6: Add up all of the probabilities you calculated in steps 5 and 6. In our example, we had: .35 + .0625 + .0625 = .475 Sample Question: If you toss a coin three times, what is the probability of getting 3 heads? The first step is to figure out your probability of getting a heads by tossing the coin once. The probability is 0.5 (you have a 50% probability of tossing a heads and 50% probability of tossing a tails). Those probabilities are represented at the ends of each branch. Next, you add two more branches to each branch to represent the second coin toss. The probability of getting two heads is shown by the red arrow. To get the probability, multiply the branches: 0.5 * 0.5 = 0.25 (25%). This makes sense because your possible results for one head and one tails is HH, HT, TT, or TH (each combination has a 25% probability). In most cases, you will multiply across the branches to get probabilities. However, you may also want to add vertically to get probabilities. For example, if we wanted to find out our probability of getting HHH OR TTT, we would first calculated the probabilities for each (0.125) and then we would add both those probabilities: 0.125 + 0.125 = 0.250. Tip: You can check you drew the tree correctly by adding vertically: all the probabilities vertically should add up to 1.------------------------------------------------------------------------------ Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free! Comments? Need to post a correction? Please post a comment on our Facebook page.<\|endoftext\|>	4.03125
587	# The expression 200 − 32 . ### Single Variable Calculus: Concepts... 4th Edition James Stewart Publisher: Cengage Learning ISBN: 9781337687805 ### Single Variable Calculus: Concepts... 4th Edition James Stewart Publisher: Cengage Learning ISBN: 9781337687805 (a) To determine ## To simplify: The expression 200−32. Expert Solution The expression 20032 is simplified as 62. ### Explanation of Solution Consider the expression 20032. Observe that, 200=2×2×2×5×5=(2)2×(5)2×2=(2×5)2=102 Again observe that, 32=2×2×2×2×2=(2)2×(2)2×2=(2×2)2=42 Substitute 200=102 and 32=42 in 20032 and simplify the above expression as follows, 20032=10242=(104)2=62 Thus, the expression 20032 is simplified as 62. (b) To determine ### To simplify: The expression (3a3b3)(4ab2)2. Expert Solution The expression (3a3b3)(4ab2)2 is simplified as 48a5b7. ### Explanation of Solution Consider the expression (3a3b3)(4ab2)2. Simplify the above expression as follows, (3a3b3)(4ab2)2=(3a3b3)(16a2b4)=3×16a3+2b3+4=48a5b7 Thus, the expression (3a3b3)(4ab2)2 is simplified as 48a5b7. (c) To determine ### To simplify: The expression (3x32y3x2y−12)−2. Expert Solution The expression (3x32y3x2y12)2 is simplified as x9y7. ### Explanation of Solution Consider the expression (3x32y3x2y12)2. Simplify the above expression as follows, (3x32y3x2y12)2=(x2y123x32y3)2(a1=1a)=(x2y12)2(3x32y3)2=(x4y19x3y6)=(x49x3y7) =x9y7 Thus, the expression (3x32y3x2y12)2 is simplified as x9y7. ### Have a homework question? Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!<\|endoftext\|>	4.53125
757	Trying to find out exactly how to convert 18/7 right into a mixed number or fraction? have I obtained the answer for you! In this guide, we"ll to walk you with the step-by-step procedure of converting an improper fraction, in this situation 18/7, to a combined number. Review on! Want to quickly learn or present students how to transform 18/7 to a mixed number? pat this very quick and fun video now! Before us begin, let"s revisit some basic fraction terms so you understand exactly what we"re managing here: Numerator. This is the number over the fraction line. Because that 18/7, the numerator is 18.Denominator. This is the number below the fraction line. For 18/7, the denominator is 7.Improper fraction. This is a portion where the numerator is better than the denominator.Mixed number. This is a means of to express an improper portion by simplifying it to totality units and a smaller overall fraction. It"s an integer (whole number) and also a appropriate fraction. You are watching: 18/7 as a mixed number Now let"s go with the steps needed to convert 18/7 to a mixed number. ## Step 1: discover the totality number We very first want to uncover the entirety number, and also to carry out this we division the numerator by the denominator. Due to the fact that we are just interested in whole numbers, we ignore any kind of numbers come the best of the decimal point. 18/7= 2.5714285714286 = 2 Now that we have our entirety number because that the blended fraction, we need to uncover our new numerator for the fraction part of the mixed number. ## Step 2: acquire the new numerator To work this the end we"ll usage the totality number we calculated in action one (2) and multiply that by the initial denominator (7). The an outcome of the multiplication is then subtracted from the initial numerator: 18 - (7 x 2) = 4 ## Step 3: Our mixed fraction We"ve now simplified 18/7 come a combined number. To watch it, we simply need to put the whole number in addition to our new numerator and also original denominator: 2 4/7 ## Step 4: simple our fraction In this case, our fraction (4/7) have the right to be streamlined down further. In order to do that, we should calculate the GCF (greatest common factor) that those 2 numbers. You deserve to use our handy GCF calculator to work-related this the end yourself if you desire to. We already did that, and the GCF the 4 and also 7 is 1. See more: How Much Does A Casket Weight Of Casket, How Much Does A Casket Usually Weigh We have the right to now division both the brand-new numerator and also the denominator through 1 to simplify this fraction down come its shortest terms. 4/1 = 4 7/1 = 7 When we put that together, we have the right to see that our finish answer is: 2 4/7 Hopefully this tutorial has helped you come understand just how to convert any kind of improper fraction you have into a mixed fraction, complete with a entirety number and also a suitable fraction. You"re complimentary to use our calculator below to job-related out more, however do shot and learn exactly how to execute it yourself. It"s much more fun 보다 it seems, i promise! ## Improper fraction to combined Number Enter one improper portion numerator and denominator<\|endoftext\|>	4.90625
1,360	# Convert revolution to gon Learn how to convert 1 revolution to gon step by step. ## Calculation Breakdown Set up the equation $$1.0\left(revolution\right)={\color{rgb(20,165,174)} x}\left(gon\right)$$ Define the base values of the selected units in relation to the SI unit $$\left(radian\right)$$ $$\text{Left side: 1.0 } \left(revolution\right) = {\color{rgb(89,182,91)} 2.0 \times π\left(radian\right)} = {\color{rgb(89,182,91)} 2.0 \times π\left(rad\right)}$$ $$\text{Right side: 1.0 } \left(gon\right) = {\color{rgb(125,164,120)} \dfrac{π}{2.0 \times 10^{2}}\left(radian\right)} = {\color{rgb(125,164,120)} \dfrac{π}{2.0 \times 10^{2}}\left(rad\right)}$$ Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$ $$1.0\left(revolution\right)={\color{rgb(20,165,174)} x}\left(gon\right)$$ $$\text{Insert known values } =>$$ $$1.0 \times {\color{rgb(89,182,91)} 2.0 \times π} \times {\color{rgb(89,182,91)} \left(radian\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} \dfrac{π}{2.0 \times 10^{2}}}} \times {\color{rgb(125,164,120)} \left(radian\right)}$$ $$\text{Or}$$ $$1.0 \cdot {\color{rgb(89,182,91)} 2.0 \times π} \cdot {\color{rgb(89,182,91)} \left(rad\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} \dfrac{π}{2.0 \times 10^{2}}} \cdot {\color{rgb(125,164,120)} \left(rad\right)}$$ $$\text{Cancel SI units}$$ $$1.0 \times {\color{rgb(89,182,91)} 2.0 \times π} \cdot {\color{rgb(89,182,91)} \cancel{\left(rad\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} \dfrac{π}{2.0 \times 10^{2}}} \times {\color{rgb(125,164,120)} \cancel{\left(rad\right)}}$$ $$\text{Conversion Equation}$$ $$2.0 \times π = {\color{rgb(20,165,174)} x} \times \dfrac{π}{2.0 \times 10^{2}}$$ Cancel factors on both sides $$\text{Cancel factors}$$ $${\color{rgb(255,204,153)} \cancel{π}} \times 2.0 = {\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{π}}}{2.0 \times 10^{2}}$$ $$\text{Simplify}$$ $$2.0 = {\color{rgb(20,165,174)} x} \times \dfrac{1.0}{2.0 \times 10^{2}}$$ Switch sides $${\color{rgb(20,165,174)} x} \times \dfrac{1.0}{2.0 \times 10^{2}} = 2.0$$ Isolate $${\color{rgb(20,165,174)} x}$$ Multiply both sides by $$\left(\dfrac{2.0 \times 10^{2}}{1.0}\right)$$ $${\color{rgb(20,165,174)} x} \times \dfrac{1.0}{2.0 \times 10^{2}} \times \dfrac{2.0 \times 10^{2}}{1.0} = 2.0 \times \dfrac{2.0 \times 10^{2}}{1.0}$$ $$\text{Cancel}$$ $${\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{1.0}} \times {\color{rgb(99,194,222)} \cancel{2.0}} \times {\color{rgb(166,218,227)} \cancel{10^{2}}}}{{\color{rgb(99,194,222)} \cancel{2.0}} \times {\color{rgb(166,218,227)} \cancel{10^{2}}} \times {\color{rgb(255,204,153)} \cancel{1.0}}} = 2.0 \times \dfrac{2.0 \times 10^{2}}{1.0}$$ $$\text{Simplify}$$ $${\color{rgb(20,165,174)} x} = 2.0 \times 2.0 \times 10^{2}$$ Solve $${\color{rgb(20,165,174)} x}$$ $${\color{rgb(20,165,174)} x} = 400 = 4 \times 10^{2}$$ $$\text{Conversion Equation}$$ $$1.0\left(revolution\right) = {\color{rgb(20,165,174)} 4 \times 10^{2}}\left(gon\right)$$<\|endoftext\|>	4.46875
3,205	Russian (pусский язык) belongs to the East Slavic group of the Slavic branch of the Indo-European language family. It is the largest of the Slavic languages. Its closest relatives are Belarusian and Ukrainian. Russian is primarily spoken in the Russian Federation and by older people in the other countries that were once part of the Soviet Union, as well as in Eastern Europe. According to the 2010 census, there were 137 million speakers of Russian in the Russian Federation, and 166 million worldwide (Ethnologue). In addition, Russian is spoken in Canada, China, Finland, Germany, Greece, India, Israel, and the U.S. It is one of the world’s ten most spoken languages. Prior to the 14th century, ancestors of the modern Russians, Belarusians, and Ukrainians spoke varieties of Old East Slavic, a language that was common to all three. Linguists think that it split into what are now Russian, Belarusian, and Ukrainian at the end of the 14th century. Until the end of the 17th century, the official language in Russia was an East Slavic version of Church Slavonic, the liturgical language of the Russian Orthodox Church. The political reforms of Peter the Great in the 18th century included a reform of the Russian alphabet and westernization of the language through numerous borrowings from Western European languages. This resulted in a move away from Church Slavonic norms towards spoken norms. The Revolution of 1917 and the political, social, and economic changes that followed it brought new terminology and greatly increased the number of international words in the Russian vocabulary. The spelling reform of 1918 gave written Russian its modern appearance. Literacy became nearly universal. Accomplishments in military, scientific, technological. and artistic fields as well as space exploration gave modern Russian its world-wide prestige that went along with its superpower image. Until 1917, Russian was the sole official language of the Russian Empire. During the Soviet period, though each of the republics had its own official language, Russian enjoyed a superior status. Following the break-up of the Soviet Union in 1991, several of the newly independent states have promoted the use of their native languages, partly undermining the privileged status of Russian, though its role as the lingua franca of the region has continued. Russian is the official language of the Russian Federation, sharing its official status at the regional level with other languages in various ethnic autonomous regions within the Federation, such as Chuvash, Bashkort,Tatar, and Yakut. Russian is also a co-official language of Belarus, Kazakhstan, and Kyrgyzstan. Russian does not have the status of an official language in Ukraine, but it still functions as a regional and minority language, with the Constitution of Ukraine providing guarantees for its protection and use. Education in Russian, as well as choice of Russian as a second language, are still very popular in many of the former Soviet republics. In the 20th century, Russian was widely taught in the schools of countries that used to be satellites of the USSR, e.g, Poland, Bulgaria, the Czech Republic, Slovakia, Hungary, Albania, former East Germany andCuba. However, today, most young people in these countries know very little or no Russian because it is is no longer mandatory in the school system. Instead of Russian, students in the Eurocentric countries of Eastern Europe prefer to study Western European languages such as English or German. Russian is one of the six official languages of the United Nations, UNESCO, World Bank, World Health Organization, and many other international organizations. Linguists generally divide Russian into three major dialect groups: Northern, Central (transitional), and Southern. There are dozens of smaller variants within each major dialect group. Two features that typically distinguish the Northern from Southern dialects are given in the table below. The Central dialect, spoken around the Moscow area, combines the major features of both dialect groups. \|unstressed vowel /o/\|\|[o]\|\|[a]\|\|[a]\| \|voiced velar fricative /ɣ/\|\|absent\|\|present\|\|absent\| The standard language is based on, but is not identical to the Moscow dialect. The sound system of Russian is quite similar to that of Belarusian and Ukrainian. Its description below is based on the standard language. Russian has 5 vowel phonemes, i.e., sounds that differentiate the meaning of words. The vowels /o/ and /a/ are distinguished only in stressed positions. Unstressed /o/ becomes /a/. The vowels /i/ and /e/ are also distinguished only in stressed position. In unstressed positions /e/ becomes /i/. Each vowel is represented by two letters in the orthography. \|Initially and after unpalatalized consonants\|\|After palatalized consonants and /j/\| The language allows a variety of consonant clusters. These are either all voiced or all voiceless. The last consonant in the cluster determines whether the entire cluster is voiced or voiceless. This rule does not apply to nasals, laterals, or rhotics. All bilabial, labio-dental, and dental consonants have palatalized counterparts pronounced with the blade of the tongue coming in contact with the hard palate. Palatalization is indicated by a small [ʲ] after the consonant in the table below. \|Stops\|\|voiceless\|\|p, pʲ\|\|t, tʲ\|\|k\| \|voiced\|\|b, bʲ\|\|d, dʲ\|\|g\| \|voiced\|\|v, vʲ\|\|z, zʲ\|\|ʒ\| \|Nasals\|\|m, mʲ\|\|n, nʲ\| \|Rhotic (trill, flap)\|\|r, rʲ\| - /x/ = similar to ch in the German pronunciation of Bach - /ʃ/=sh in shape - /ʒ/ = s in measure - /tʃ/ = ch in cheat - /ɾ/ has no equivalent in English; similar to r in Spanish pero - /j/ = y in yet Stress is free and mobile, i.e., it can fall on any syllable of a word and its position can change depending on the form of the word. Stress is not marked in normal orthography but is commonly marked in textbooks and dictionaries. Russian is a richly inflected language with a grammar that is very similar to that of other Slavic languages, especially Belarusian and Ukrainian. Nouns, adjectives, pronouns Russian nouns are marked for gender, number, and case. The three are fused into one ending, as is the case in all Slavic languages. Russian nouns have the following grammatical categories: - gender: masculine, feminine, neuter - four noun and adjective declensions, largely based on gender - number: singular and plural, with a few vestiges of dual - case: nominative, genitive, dative, accusative, instrumental, locative, vocative (only a few remaining vocative forms remain) - animate and inanimate masculine nouns have different endings in the accusative case - There are no indefinite or definite articles. - Adjectives are marked for gender and case only in the singular. In the plural, they are marked only for case. - Adjectives and possessive and demonstrative pronouns precede the nouns they modify and agree with them in gender, number, and case. - There is a distinction in the second person singular between informal (T) and formal (V) forms. - Like all Slavic languages, Russian is a pro-drop language, i.e., personal pronouns can be dropped because the verb ending makes the person clear. Russian verbs agree with their subjects in person and number in the non-past, and in gender and number in the past. They are marked for the following categories: - three persons: first, second, third - four conjugations - two tenses: past, non-past - Present and future tenses have the same endings. - two aspects: imperfective and perfective. - Perfective and imperfective verbs are formed from basic verb roots by adding prefixes and suffixes. Non-past conjugation of perfective verbs indicates future tense, non-past conjugation of imperfective verbs indicates present tense. Imperfective verbs form future tense with the auxiliary verb быть ‘be.’ - three moods: indicative, imperative, conditional - two voices: active, passive - Verbs of motion constitute a special subcategory of verbs. They are characterized by a complex system of directional and aspectual prefixes and suffixes. The neutral word order in Russian is Subject-Verb-Object. However, other orders are possible since inflectional endings take care of clearly marking grammatical relations and roles in the sentence. Word order is principally determined by topic (what the sentence is about, or old information) and focus (new information). Constituents with old information precede constituents with new information, or those that carry the most emphasis. Sasha ljubit Mashu. ‘Sasha loves Masha.’ Neutral word order. No part of the sentence is emphasized. Mashu ljubit Sasha. \|It is Sasha who loves Masha (as opposed to someone else).\| Sasha Mashu ljubit. \|Sasha really loves Masha.\| Russian has a very large vocabulary consisting of a mix of native Slavic and borrowings from other languages. It is difficult to determine what percentage of Russian vocabulary is inherently Slavic and what percentage is borrowed from other languages. By some estimates, about half of Russian vocabulary may consist of words borrowed at one time or another from other languages. This is even more true of scientific, technical, and political vocabulary. Early borrowings into Russian were from Old Church Slavonic, Greek, and Latin, associated with religious sources, and from Altaic languages, associated with the Mongol invasion. Later borrowings came from French, German, Dutch, Italian, and English. Today, the major source of borrowing, particularly in the areas of scientific, political, and technical terminology, is English. However, most of the basic everyday vocabulary is inherently Slavic. Below are a few common words and phrases. \|Good bye\|\|Дo cвидaния\| \|Excuse me, sorry\|\|Извини(тe), пpocти(тe)\| Below are Russian numerals 1-10. The modern Russian alphabet is a variant of the Cyrillic alphabet. It was introduced into Kievan Rus’ at the time of its conversion to Christianity in 988 or earlier, and has undergone significant changes since then, including major reforms in 1708, during the rein of Peter the Great, and in 1918, after the October Revolution. The modern Russian alphabet has the following letters given below in their printed form. The longhand, or cursive, form for some letters is quite different. \|А а\|\|Б б\|\|В в\|\|Г г\|\|Д д\|\|E e\|\|Ë ë\|\|Ж ж\|\|З з\|\|И и\|\|Й й\| \|К к\|\|Л л\|\|М м\|\|Н н\|\|О о\|\|П п\|\|Р р\|\|С с\|\|Т т\|\|У у\|\|Ф ф\| \|Х х\|\|Ц ц\|\|Ч ч\|\|Ш ш\|\|Щ щ\|\|Ы ы\|\|ъ\|\|ь\|\|Э э\|\|Ю ю\|\|Я я\| Take a look at Article 1 of the Universal Declaration of Human Rights in Russian. Вceoбщaя дeклapaция пpaв чeлoвeкa Вce люди poждaютcя cвoбoдными и paвными в cвoeм дocтoинcтвe и прaвax. Oни нaдeлeны paзyмoм и coвecтью и дoлжны пocтyпaть в oтнoшeнии дpyг дpyгa в дyxe бpaтcтвa. Universal Declaration of Human Rights All human beings are born free and equal in dignity and rights. They are endowed with reason and conscience and should act towards one another in a spirit of brotherhood. Russian is the language of major poets and writers whose work has been translated into dozens of the world’s languages. Among them are Pushkin,Lermontov, Gogol, Dostoevsky, Tolstoy, Turgenev, Chekhov, Gorky, Blok, Nabokov, Pasternak, Yevtushenko, and numerous others. English has borrowed a number of words from Russian. Here are some of them: \|balalaika\|\|Russian name for a triangular-shaped stringed instrument\| \|borsch\|\|vegetable and beet soup\| \|cosmonaut\|\|anglicization of Russian kosmonavt\| \|dacha\|\|country or vacation home\| \|glasnost\|\|policy of openness, from glas ‘voice’\| \|gulag\|\|an acronym that means ‘state forced labor camp’\| \|duma\|\|Russian national assembly, from duma ‘think’\| \|matryoshka\|\|wooden dolls stacked inside one another\| \|rouble\|\|unit of currency equivalent to 100 kopecks ‘cents’\| \|samovar\|\|‘hot water urn’ literally ‘auto-boiler’ from sam ‘auto-‘ + var- ‘boil’\| \|sputnik\|\|‘artificial satellite, s- ‘with’ + put ‘trip’ + nik ‘masculine suffix’, literally ‘co-traveller’\| \|steppe\|\|from step’, vast treeless plain of southeastern Eurasia\| \|troika\|\|troika ‘three-horse team abreast, or any group of three’\| \|tundra\|\|tundra ‘Arctic steppe’\| \|vodka\|\|vodka, from vod– ‘water’ + –ka, a diminutive suffix.\|<\|endoftext\|>	3.9375
21,441	1911 Encyclopædia Britannica/Forests and Forestry FORESTS AND FORESTRY. Although most people know what a forest (Lat. foris, “out of doors”) is, a definition of it which suits all cases is by no means easy to give. Manwood, in his treatise of the Lawes of the Forest (1598), defines a forest as “a certain territory of woody grounds, fruitful pastures, privileged for wild beasts and fowls of forest, chase and warren, to rest and abide in, in the safe protection of the king, for his princely delight and pleasure.” This primitive definition has, in modern times, when the economic aspect of forests came more into the foreground, given place to others, so that forest may, in a general way, now be described as “an area which is for the most part set aside for the production of timber and other forest produce, or which is expected to exercise certain climatic effects, or to protect the locality against injurious influences.” As far as conclusions can now be drawn, it is probable that the greater part of the dry land of the earth was, at some time, covered with forest, which consisted of a variety of trees and shrubs grouped according to climate, soil and configuration of the several localities. When the old trees reached their limit of life, they disappeared, and younger trees took their place. The conditions for an uninterrupted regeneration of the forest were favourable, and the result was vigorous production by the creative powers of soil and climate. Then came man, and by degrees interfered, until in most countries of the earth the area under forest has been considerably reduced. The first decided interference was probably due to the establishment of domestic animals; men burnt the forest to obtain pasture for their flocks. Subsequently similar measures on an ever-increasing scale were employed to prepare the land for agricultural purposes. More recently enormous areas of forests were destroyed by reckless cutting and subsequent firing in the extraction of timber for economic purposes. It will readily be understood that the distribution and character of the now remaining forests must differ enormously (see Plants: Distribution). Large portions of the earth are still covered with dense masses of tall trees, while others contain low scrub or grass land, or are desert. As a general rule, natural forests consist of a number of different species intermixed; but in some cases certain species, called gregarious, have succeeded in obtaining the upper hand, thus forming more or less pure forests of one species only. The number of species differs very much. In many tropical forests hundreds of species may be found on a comparatively small area, in other cases the number is limited. Burma has several thousand species of trees and shrubs, Sind has only ten species of trees. Central Europe has about forty species, and the greater part of northern Russia, Sweden and Norway contains forests consisting of about half a dozen species. Elevation above the sea acts similarly to rising latitude, but the effect is much more rapidly produced. Generally speaking, it may be said that the Tropics and adjoining parts of the earth, wherever the climate is not modified by considerable elevation, contain broad-leaved species, palms, bamboos, &c. Here most of the best and hardest timbers are found, such as teak, mahogany and ebony. The northern countries are rich in conifers. Taking a section from Central Africa to North Europe, it will be found that south and north of the equator there is a large belt of dense hardwood forest; then comes the Sahara, then the coast of the Mediterranean with forests of cork oak; then Italy with oak, olive, chestnut, gradually giving place to ash, sycamore, beech, birch and certain species of pine; in Switzerland and Germany silver fir and spruce gain ground. Silver fir disappears in central Germany, and the countries around the Baltic contain forests consisting chiefly of Scotch pine, spruce and birch, to which, in Siberia, larch must be added, while the lower parts of the ground are stocked with hornbeam, willow, alder and poplar. In North America the distribution is as follows: Tropical vegetation is found in south Florida, while in north Florida it changes into a subtropical vegetation consisting of evergreen broad-leaved species with pines on sandy soils. On going north in the Atlantic region, the forest becomes temperate, containing deciduous broad-leaved trees and pines, until Canada is reached, where larches, spruces and firs occupy the ground. Around the great lakes on sandy soils the broad-leaved forest gives way to pines. On proceeding west from the Atlantic region the forest changes into a shrubby vegetation, and this into the prairies. Farther west, towards the Pacific coast, extensive forests are found consisting, according to latitude and elevation above the sea, of pines, larches, fir, Thujas and Tsugas. In Japan a tropical vegetation is found in the south, comprising palms, figs, ebony, mangrove and others. This is followed on proceeding north by subtropical forests containing evergreen oaks, Podocarpus, tree-ferns, and, at higher elevations, Cryptomeria and Chamaecyparis. Then follow deciduous broad-leaved forests, and finally firs, spruces and larches. In India the character of the forests is governed chiefly by rainfall and elevation. Where the former is heavy evergreen forests of Guttiferae, Dipterocarpeae, Leguminosae, Euphorbias, figs, palms, ferns, bamboos and india-rubber trees are found. Under a less copious rainfall deciduous forests appear, containing teak and sal (Shorea robusta) and a great variety of other valuable trees. Under a still smaller rainfall the vegetation becomes sparse, containing acacias, Dalbergia sissoo and Tamarix. Where the rainfall is very light or nil, desert appears. In the Himalayas, subtropical to arctic conditions are found, the forests containing, according to elevation, pines, firs, deodars, oaks, chestnuts, magnolias, laurels, rhododendrons and bamboos. Australia, again, has its own particular flora of eucalypts, of which some two hundred species have been distinguished, as well as wattles. Some of the eucalypts attain an enormous height. Utility of Forests.—In the economy of man and of nature forests are of direct and indirect value, the former chiefly through the produce which they yield, and the latter through the influence which they exercise upon climate, the regulation of moisture, the stability of the soil, the healthiness and beauty of a country and allied subjects. The indirect utility will be dealt with first. A piece of land bare of vegetation is, throughout the year, exposed to the full effect of sun and air currents, and the climatic conditions which are produced by these agencies. If, on the other hand, a piece of land is covered with a growth of plants, and especially with a dense crop of forest vegetation, it enjoys the benefit of certain agencies which modify the effect of sun and wind on the soil and the adjoining layers of air. These modifying agencies are as follows: (1) The crowns of the trees intercept the rays of the sun and the falling rain; they obstruct the movement of air currents, and reduce radiation at night. (2) The leaves, flowers and fruits, augmented by certain plants which grow in the shade of the trees, form a layer of mould, or humus, which protects the soil against rapid changes of temperature, and greatly influences the movement of water in it. (3) The roots of the trees penetrate into the soil in all directions, and bind it together. The effects of these agencies have been observed from ancient times, and widely differing views have been taken of them. Of late years, however, more careful observations have been made at so-called parallel stations, that is to say, one station in the middle of a forest, and another outside at some distance from its edge, but otherwise exposed to the same general conditions. In this way, the following results have been obtained: (1) Forests reduce the temperature of the air and soil to a moderate extent, and render the climate more equable. (2) They increase the relative humidity of the air, and reduce evaporation. (3) They tend to increase the precipitation of moisture. As regards the actual rainfall, their effect in low lands is nil or very small; in hilly countries it is probably greater, but definite results have not yet been obtained owing to the difficulty of separating the effect of forests from that of other factors. (4) They help to regulate the water supply, produce a more sustained feeding of springs, tend to reduce violent floods, and render the flow of water in rivers more continuous. (5) They assist in preventing denudation, erosion, landslips, avalanches, the silting up of rivers and low lands and the formation of sand dunes. (6) They reduce the velocity of air-currents, protect adjoining fields against cold or dry winds, and afford shelter to cattle, game and useful birds. (7) They may, under certain conditions, improve the healthiness of a country, and help in its defence. (8) They increase the beauty of a country, and produce a healthy aesthetic influence upon the people. The direct utility of forests is chiefly due to their produce, the capital which they represent, and the work which they provide. The principal produce of forests consists of timber and firewood. Both are necessaries for the daily life of the people. Apart from a limited number of broad-leaved species, the conifers have become the most important timber trees in the economy of man. They are found in greatest quantities in the countries around the Baltic and in North America. In modern times iron and other materials have, to a considerable extent, replaced timber, while coal, lignite, and peat compete with firewood; nevertheless wood is still indispensable, and likely to remain so. This is borne out by the statistics of the most civilized nations. Whereas the population of Great Britain and Ireland, during the period 1880-1900, increased by about 20%, the imports of timber, during the same period, increased by 45%; in other words, every head of population in 1900 used more timber than twenty years earlier. Germany produced in 1880 about as much timber as she required; in 1899 she imported 4,600,000 tons, valued at £14,000,000, and her imports are rapidly increasing, although the yield capacity of her own forests is much higher now than it was formerly. Wood is now used for many purposes which formerly were not thought of. The manufacture of the wood pulp annually imported into Britain consumes at least 2,000,000 tons of timber. A fabric closely resembling silk is now made of spruce wood. The variety of other, or minor, produce yielded by forests is very great, and much of it is essential for the well-being of the people and for various industries. The yield of fodder is of the utmost importance in countries subject to periodic droughts; in many places field crops could not be grown successfully without the leaf-mould and brushwood taken from the forests. As regards industries, attention need only be drawn to such articles as commercial fibre, tanning materials, dye-stuffs, lac, turpentine, resin, rubber, gutta-percha, &c. Great Britain and Ireland alone import every year such materials to the value of £12,000,000, half of this being represented by rubber. The capital employed in forests consists chiefly of the value of the soil and growing stock of timber. The latter is, ordinarily, of much greater value than the former wherever a sustained annual yield of timber is expected from a forest. In the case of a Scotch pine forest, for instance, the value of the growing stock is, under the above-mentioned condition, from three to five times that of the soil. The rate of interest yielded by capital invested in forests differs, of course, considerably according to circumstances, but on the whole it may, under proper management, be placed equal to that yielded by agricultural land; it is lower than the agricultural rate on the better classes of land, but higher on the inferior classes. Hence the latter are specially indicated for the forest industry, and the former for the production of agricultural crops. Forests require labour in a great variety of ways, such as (1) general administration, formation, tending and harvesting; (2) transport of produce; and (3) industries which depend on forests for their prime material. The labour indicated under the first head differs considerably according to circumstances, but its amount is smaller than that required if the land is used for agriculture. Hence forests provide additional labour only if they are established on surplus lands. Owing to the bulky nature of forest produce its transport forms a business of considerable magnitude, the amount of labour being perhaps equal to half that employed under the first head. The greatest amount of labour is, however, required in the working up of the raw material yielded by forests. In this respect attention may be drawn to the chair industry in and around High Wycombe in Buckinghamshire, where more than 20,000 workmen are employed in converting the beech, grown on the adjoining chalk hills, into chairs and tools of many patterns. Complete statistics for Great Britain are not available under this head, but it may be mentioned that in Germany the people employed in the forests amount to 2.3% of the total population; those employed on transport of forest produce 1.1%; labourers employed on the various wood industries, 8.6%; or a total of 12%. An important feature of the work connected with forests and their produce is that a great part of it can be made to fit in with the requirements of agriculture; that is to say, it can be done at seasons when field crops do not require attention. Thus the rural labourers or small farmers can earn some money at times when they have nothing else to do, and when they would probably sit idle if no forest work were obtainable. Whether, or how far, the utility of forests is brought out in a particular country depends on its special conditions, such as (1) the position of a country, its communications, and the control which it exercises over other countries, such as colonies; (2) the quantity and quality of substitutes for forest produce available in the country; (3) the value of land and labour, and the returns which land yields if used for other purposes; (4) the density of population; (5) the amount of capital available for investment; (6) the climate and configuration, especially the geographical position, whether inland or on the border of the sea, &c. No general rule can be laid down, showing whether forests are required in a country, or, if so, to what extent; that question must be answered according to the special circumstances of each case. The subjoined table shows the forests of various European states:— \|Russia, including Finland\|\|518,000,000\|\|40\|\|61\|\|5.9\| \|Bosnia and Herzegovina\|\|6,400,000\|\|50\|\|78\|\|4.0\| These data exhibit considerable differences, since the percentage of the forest area varies from 3.5 to 50, and the area per head of population from .07 to 9.5 acres. Russia, Sweden and Norway may as yet have more forest than they require for their own population. On the other hand, Great Britain and Ireland, Germany, Denmark, Portugal, Holland, and even Belgium, France and Italy have not a sufficient forest area to meet their own requirements; at the same time, they are all sea-bound countries, and importation is easy, while most of them are under the influence of moist sea winds, which reduces to a subordinate position the importance of forests for climatic reasons. Intimately connected with the area of forests in a country is the state of ownership—whether they belong to the state, corporations or to private persons. Where, apart from the financial aspect and the supply of work, forests are not required for the sake of their indirect effects, and where importation from other countries is easy and assured, the government of the country need not, as a rule, trouble itself to maintain or acquire forests. Where the reverse conditions exist, and especially where the cost of transport over long distances becomes prohibitive, a wise administration will take measures to assure the maintenance of a suitable proportion of the country under forest. This can be done either by maintaining or constituting a suitable area of state forests, or by exercising a certain amount of control over corporation and even private forests. Such measures are more called for in continental countries than in those which are sea-bound, as is proved by the above statistics. Supply of Timber—Imports and Exports.—The following table shows the net imports and exports of European countries (average data, calculated from the returns of recent years). The only timber-exporting countries of Europe are Russia, Sweden, Norway, Austria-Hungary and Rumania; all the others either have only enough for their own consumption, or import timber. Great Britain and Ireland import now upwards of 10,000,000 tons a year, Germany about 4,600,000 tons, and Belgium about 1,300,000 tons. Holland, France, Portugal, Spain and Italy are all importing countries, as also are Asia Minor, Egypt and Algeria. The west coast of Africa exports hardwoods, and imports coniferous timber. The Cape and Natal import considerable quantities of pine and fir wood. Australasia exports hardwoods and some Kauri pine from New Zealand, but imports larger quantities of light pine and fir timber. British India and Siam export teak and small quantities of fancy woods. The West Indies and South America export hardwoods, and import pine and fir wood. The United States of America will not much longer be a genuine exporting country, since they import already almost as much timber from Canada as they export. Canada exports considerable quantities of timber. The Dominion has still a forest area of 1,250,000 sq. m., equal to 38% of the total area, and giving 165 acres of forest for every inhabitant. Although only about one-third of the forest area can be called regular timber land, Canada possesses an enormous forest wealth, with which she might supply permanently nearly all other countries deficient in material, if the governing bodies in the several provinces would only determine to stop the present fearful waste caused by axe and fire, and to introduce a regular system of management. As matters stand, the supplies of the most valuable timber of Canada, the white or Weymouth pine (Pinus strobus), are nearly exhausted, the great stores of spruce in the eastern provinces are being rapidly destroyed, and the forests of Douglas fir in the western provinces have been attacked for export to the United States and to other countries. Net Imports and Exports of European Countries. \|Countries.\|\|Quantities in Tons.\|\|Value in £ Sterling.\| \|United Kingdom\|\|10,004,000\|\|· ·\|\|26,540,000\|\|· ·\| \|Germany\|\|4,600,000\|\|· ·\|\|14,820,000\|\|· ·\| \|Belgium\|\|1,300,000\|\|· ·\|\|5,040,000\|\|· ·\| \|France\|\|1,230,000\|\|· ·\|\|3,950,000\|\|· ·\| \|Italy\|\|620,000\|\|· ·\|\|2,100,000\|\|· ·\| \|Spain\|\|470,000\|\|· ·\|\|1,500,000\|\|· ·\| \|Denmark\|\|470,000\|\|· ·\|\|1,250,000\|\|· ·\| \|Switzerland\|\|204,000\|\|· ·\|\|480,000\|\|· ·\| \|Holland\|\|180,000\|\|· ·\|\|720,000\|\|· ·\| \|Servia\|\|110,000\|\|· ·\|\|160,000\|\|· ·\| \|Portugal\|\|60,000\|\|· ·\|\|200,000\|\|· ·\| \|Greece\|\|35,000\|\|· ·\|\|130,000\|\|· ·\| \|Rumania\|\|· ·\|\|400,000\|\|· ·\|\|840,000\| \|Norway\|\|· ·\|\|1,300,000\|\|· ·\|\|2,200,000\| \|Bosnia and Herzegovina\|\|· ·\|\|3,996,000\|\|· ·\|\|11,400,000\| \|Sweden\|\|· ·\|\|4,460,000\|\|· ·\|\|7,930,000\| \|Russia with Finland\|\|· ·\|\|6,890,000\|\|· ·\|\|10,440,000\| Taking the remaining stocks of the whole earth together, it may be said that a sufficient quantity of hardwoods is available, but the only countries which are able to supply coniferous timber for export on a considerable scale are Russia, Sweden, Norway, Austria and Canada. As these countries have practically to supply the rest of the world, and as the management of their forests is far from satisfactory, the question of supplying light pine and fir timber, which forms the very staff of life of the wood industries, must become a very serious matter before many years have passed. Unmistakable signs of the coming crisis are everywhere visible to all who wish to see, and it is difficult to over-state the gravity of the problem, when it is remembered, for instance, that 87% of all the timber imported into Great Britain consists of light pine and fir, and that most of the other importing countries are similarly situated. In some of these countries little or no room exists for the extension of woodland, but this statement does not apply to Great Britain and Ireland, which contain upwards of 12,000,000 acres of waste land, and 12,500,000 acres of mountain and heath land used for light grazing. One-fourth of that area, if put under forest, would produce all the timber now imported which can be grown in Britain, that is to say, about 95% of the total. The subjoined table shows the movements of timber within the greater part of the British empire:— Net Imports and Exports into and from the British Empire. during the Years during the Years \|United Kingdom\|\|15,000,000\|\|· ·\|\|26,540,000\|\|· ·\| \|Australasia\|\|1,284,000\|\|· ·\|\|568,000\|\|· ·\| \|Africa\|\|72,000\|\|· ·\|\|737,000\|\|· ·\| \|West Indies, Honduras and Guiana\|\|· ·\|\|207,000\|\|· ·\|\|71,000\| \|India, Ceylon and Mauritius\|\|· ·\|\|528,000\|\|· ·\|\|580,000\| \|Dominion of Canada\|\|· ·\|\|4,025,000\|\|· ·\|\|4,789,000\| \|Net Imports\|\|11,596,000\|\|· ·\|\|22,405,000\|\|· ·\| \|Total increase in 16 years\|\|· ·\|\|· ·\|\|10,809,000\|\|· ·\| \|Average annual increase of net imports\|\|· ·\|\|· ·\|\|675,562\|\|· ·\| Forest Management.—In early times there was practically no forest management. As long as the forests occupied considerable areas, their produce was looked upon as the free gift of nature, like air and water; men took it, used it, and even destroyed it without let or hindrance. With the gradual increase of population and the consequent reduction of the forest area, proprietary ideas developed; people claimed the ownership of certain forests, and proceeded to protect them against outsiders. Subsequently the law of the country was called in to help in protection, leading to the promulgation of special forest laws. By degrees it was found that mere protection was not sufficient, and that steps must be taken to enforce a more judicious treatment, as well as to limit the removal of timber to what the forests were capable of producing permanently. The teaching of natural science and of political economy was brought to bear upon the subject, so that now forestry has become a special science. This is recognized in many countries, amongst which Germany stands first, closely followed by France, Austria, Denmark and Belgium. Of non-European countries the palm belongs to British India, and then follow Ceylon, the Malay States, the Cape of Good Hope and Japan. The United States of America have also turned their attention to the subject. Most of the British colonies are, in this respect, as yet in a backward state, and the matter has still to be fought out in Great Britain and Ireland, though many writers have urged the importance of the question upon the public and the government. There can be no doubt that all civilized countries must, sooner or later, adopt a rational and systematic treatment of their forests. For details as to the separate countries, see the articles under the country headings; in this article only some of the more important countries are dealt with, in so far as the history of their forestry is important. A few notes on Germany and France will be given, because in these countries forest management has been brought to highest perfection; Italy is mentioned, because she has allowed her forests to be destroyed; and a short description of forestry in the United Kingdom and in India follows. A separate section is devoted to the United States. Germany is in general well-wooded. The winters being long and severe, an abundant supply of fuel is almost as essential as a sufficient supply of food. This necessity has led, along with a passion for the chase, to the preservation of forests, and to the establishment of an admirable system of forest cultivation, almost as carefully conducted as field tillage. The Black Forest stretches the whole length of the grand-duchy of Baden and part of the kingdom of Württemberg, from the Neckar to Basel and the Lake of Constance. The vegetation resembles that of the Vosges; forests of spruce, silver fir, Scotch pine, and, mingled with birches, beech and oak, are the chief woods met with. Until comparatively recent times large quantities of timber derived from these forests were floated down the Rhine to Holland and also shipped to England. Now the greater part of it is used locally for construction, or it is converted into paper pulp. In the grand-duchy of Hesse the Odenwald range of mountains, stretching between the Main and the Neckar, contains the chief supply of timber. In the province of Nassau there are the large wooded tracts of the Taunus mountain range and the Westerwald. In Rhenish Prussia valuable forests lie partly in the Eifel, on the borders of Belgium, and on the mountains overhanging the Upper Moselle, but they do not furnish such stately trees as the Black Forest and the Odenwald. The Spessart, near Aschaffenburg in Bavaria, is one of the most extensive forests of middle Germany, containing large masses of fine oak and beech, with plantations of coniferous trees, such as spruce, Scotch pine and silver fir. Bavaria possesses other fine forest tracts, such as the Baierischewald on the Bohemian frontier, the Kranzberg near Munich, and the Frankenwald in the north of the kingdom. North Germany has extensive forests on the Harz and Thüringian Mountains, while in East Prussia large tracts of flat ground are covered with Scotch pine, spruce, oak and beech. Every German state has its forest organization. In Prussia the department is presided over by the Oberland Forstmeister at Berlin, while each province, or part of a province, has an Oberforstmeister, under whom a number of Oberförsters administrate the state and communal forests. These, again, are assisted by a lower class of officials called Försters. The Oberförsters throughout Germany are educated at special schools of forestry, of which in 1909 the following nine existed: In Prussia: at Eberswalde and Münden. In Bavaria: at Munich and Aschaffenburg. In Saxony: at Tharand. In Württemberg: at Tübingen. In Baden: at Carlsruhe. In Hesse: at Giessen. In the grand-duchy of Saxony: at Eisenach. The schools at Munich, Tübingen and Giessen form part of the universities at these places; that at Carlsruhe is attached to the technical high school; the others are academies for the study of forestry only, but there is a tendency to transfer them all to the universities. The subordinate staff are trained for their work in so-called silvicultural schools, of which a large number exist. In this way the German forests have been brought to a high degree of productiveness, but the material derived from them falls far short of the requirements, although the forests occupy 26% of the total area of the country; hence the net imports of timber amount already to 4,600,000 tons a year, and they are steadily rising. France.—The principal timber tree of France is the oak. The cork oak is grown extensively in the south and in Corsica. The beech, ash, elm, maple, birch, walnut, chestnut and poplar are all important trees, while the silver fir and spruce form magnificent forests in the Vosges and Jura Mountains, and the Aleppo and maritime pines are cultivated in the south and south-west. About one-seventh of the entire territory is still covered with wood. Forest legislation took its rise in France about the middle of the 16th century, and the great minister Sully urged the enforcement of restrictive forest laws. In 1669 a fixed treatment of state forests was enacted. Duhamel in 1755 published his famous work on forest trees. Reckless destruction of the forests, however, was in progress, and the Revolution of 1789 gave a fresh stimulus to the work of devastation. The usual results have followed in the frequency and destructiveness of floods, which have washed away the soil from the hillsides and valleys of many districts, especially in the south, and the frequent inundations of the last fifty years are no doubt caused by the deforesting of the sources of the Rhone and Saône. Laws were passed in 1860 and 1864, providing for the reforesting, “reboisement,” of the slopes of mountains, and these laws take effect on private as well as state property. Thousands of acres are annually planted in the departments of Hautes and Basses Alpes; and during the summer of 1875, when much injury was done by floods in the south of France, the Durance, formerly the most dangerous in this respect of French rivers, gave little cause for anxiety, as it is round the head waters of this river that the chief plantations have been formed. While tracts formerly covered with wood have been replanted, plantations have been formed on the shifting sands or dunes along the coast of Gascony. A forest of Pinus pinaster, 150 m. in length, now stretches from Bayonne to the mouth of the Gironde, raised by means of sowing steadily continued since 1789; the cultivation of the pine, along with draining, has transformed low marshy grounds into productive soil extending over an area of about two million acres. The forests thus created provide annually some 600,000 tons of pit timber for the Welsh coal mines. The state forest department is administered by the director-general, who has his headquarters at Paris, assisted by a board of administration, charged with the working of the forests, questions of rights and law, finance and plantation works. The department is supplied with officers from the forest school at Nancy. This institution was founded in 1824, when M. Lorentz, who had studied forestry in Germany, was appointed its first director. Italy.—The kingdom of Italy comprises such different climates that within its limits we find the birch and pines of northern Europe, and the olive, fig, manna-ash, and palm of more southern latitudes. By the republic of Venice and the duchy of Genoa forestal legislation was attempted at various periods from the 15th century downwards. These efforts were not successful, as the governments were lax in enforcing the laws. In 1789 Pius VI. issued regulations prohibiting felling without licence, and later orders were published by his successors in the pontifical states. In Lombardy the woods, which in 1830 reached nearly down to Milan, have almost disappeared. The province of Como contains only a remnant of the primitive forests, and the same may also be said of the southern slopes of Tirol. At Ravenna there is still a large forest of stone pine, Pinus pinea, though it has been much reduced. The plains of Tuscany are adorned with planted trees, the olive, mulberry, fig and almond. Sardinia is rich in woods, which cover one-fifth of the area, and contain a large amount of oak, Quercus suber, robur and cerris. In Sicily the forests have long been felled, save the zone at the base of Mount Etna. The destruction of woods has been gradual but persistent; at the end of the 17th century the effects of denudation were first felt in the destructive force given to mountain torrents by the deforesting of the Apennines. The work of devastation continued until a comparatively recent time. In 1867 the monastic property of Vallombrosa, Tuscany, 30 m. from Florence, was purchased by government for the purposes of a forest academy, which was opened in 1869. As only 4% of the total forest area belongs to the state, it is doubtful whether much good can now be done. Great Britain and Ireland.—The British Isles were formerly much more extensively wooded than at present. The rapid increase of population led to the disforesting of woodland; the climate required the maintenance of household fires during a great part of the year, and the increasing demand for arable land and the extension of manufacturing industries combined to cause the diminution of woodland. The proportion of forest is now very small, and yields but a fraction of the required annual supply of timber which is imported with facility from America, northern Europe and the numerous British colonies. Owing to the nature of the climate of the British Islands, with its abundance of atmospheric moisture and freedom from such extremes of heat and cold as are prevalent in continental Europe, a great variety of trees are successfully cultivated. In England and Ireland oak and beech are on the whole the most plentiful trees in the low and fertile parts; in the south of Scotland the beech and ash are perhaps most common, while the Scotch fir and birch are characteristic of the arboreous vegetation in the Highlands. Although few extensive forests now exist, woods of small area, belts of planting, clumps of trees, coppice and hedgerows, are generally distributed over the country, constituting a mass of wood of considerable importance, giving a clothed appearance in many parts, and affording illustrations of skilled arboriculture not to be found in any other country. The principal state forests in England are Windsor Park, 14,000 acres; the New Forest, &c., in Hampshire, 76,000 acres; and the Dean Forest in Gloucestershire, 22,500 acres. The total extent of crown forests is about 125,000 acres. A large proportion of the crown forests, having been formed with the object of supplying timber for the navy, consists of oak. The largest forests in Scotland are in Perthshire, Inverness-shire and Aberdeenshire. Of these the most notable are the earl of Mansfield’s near Scone (8000 acres), the duke of Atholl’s larch plantations near Dunkeld (10,000 acres), and in Strathspey a large extent of Scotch pine, partly native, partly planted, belonging to the earl of Seafield. In the forests of Mar and Invercauld, the native pine attains a great size, and there are also large tracts of indigenous birch in various districts. Ireland was at one time richly clothed with wood; this is proved by the abundant remains of fallen trees in the bogs which occupy a large surface of the island. In addition to the causes above alluded to as tending to disforest England, the long unsettled state of the country also conduced to the diminishing of the woodlands. The forests of Great Britain and Ireland, in spite of the large imports of timber, have not been appreciably extended up to the present time because (1) the rate at which foreign timber has been laid down in Britain is very low, thus keeping down the price of home-grown timber; (2) foreign timber is preferred to home-grown material, because it is in many cases of superior quality, while the latter comes into the market in an irregular and intermittent manner; (3) nearly the whole of the waste lands is private property. As regards prices, it can be shown that the lowest point was reached about the year 1888, in consequence of the remarkable development of means of communication, that prices then remained fairly stationary for some years, and that about 1894 a slow but steady rise set in, showing during the years 1894-1904 an increase of about 20% all round. This was due to the gradual approach of the coming crisis in the supply of coniferous timber to the world. It can be shown that even with present prices the growing of timber can be made to pay, provided it is carried on in a rational and economic manner. Improved silvicultural methods must be applied, so as to produce a better class of timber, and the forests must be managed according to well-arranged working plans, which provide for a regular and sustained out-turn of timber year by year, so as to develop a healthy and steady market for locally-grown material. Unfortunately the private proprietors of the waste lands are in many cases not in a financial position to plant. Starting forests demands a certain outlay in cash, and the proprietor must forgo the income, however small, hitherto derived from the land until the plantations begin to yield a return. In these circumstances the state may well be expected to help in one or all of the following ways: (1) The equipment of forest schools, where economic forestry, as elaborated by research, is taught; (2) the management of the crown forests on economic principles, so as to serve as patterns to private proprietors; (3) advances should be made to landed proprietors who desire to plant land, but are short of funds, just as is done in the case of improvements of agricultural holdings; and (4) the state might acquire surplus lands in certain parts of the country, such as congested districts, and convert them into forests. Action in these directions would soon lead to substantial benefits. The income of landed proprietors would rise, a considerable sum of money now sent abroad would remain in the country, and forest industries would spring up, thus helping to counteract the ever-increasing flow of people from the country into the large towns, where only too many must join the army of the unemployed. Even within a radius of 50 m. of London 700,000 acres of land are unaccounted for in the official agricultural returns. In Ireland more than 3,000,000 acres are waiting to be utilized, and it is well worth the consideration of the Irish Land Commissioners whether the lands remaining on their hands, when buying and breaking up large estates, should not be converted into state forests. Such a measure might become a useful auxiliary in the peaceful settlement of the Irish land question. No doubt success depends upon the probable financial results. There are at present no British statistics to prove such success; hence, by way of illustration, it may be stated what the results have been in the kingdom of Saxony, which, from an industrial point of view, is comparable with England. That country has 432,085 acres of state forests, of which about one-eighth are stocked with broad-leaved species, and seven-eighths with conifers. Some of the forests are situated on low lands, but the bulk of the area is found in the hilly parts of the country up to an elevation of 3000 ft. above the sea. The average price realized of late years per cubic foot of wood amounts to 5d., and yet to such perfection has the management been brought by a well-trained staff, that the mean annual net revenue, after meeting all expenses, comes to 21s. an acre all round. There can be no doubt that, under the more favourable climate of Great Britain, even better results can be obtained, especially if it is remembered that foreign supplies of coniferous timber must fall off, or, at any rate, the price per cubic foot rise considerably. These things have been recognized to some extent, and a movement has been set on foot to improve matters. The Commissioners of Woods and a number of private proprietors had rational working plans prepared for their forests, and instruction in forestry has been developed. There is now a well-equipped school of forestry connected with the university of Oxford, while Cambridge is following on similar lines; instruction in forestry is given at the university of Edinburgh, the Durham College of Science, at Bangor, Cirencester and other places. The Commissioners of Woods have purchased an estate of 12,500 acres in Scotland, which will be converted into a crown forest, so as to serve as an example. The experience thus gained will prove valuable should action ever be taken on the lines suggested by a Royal Commission on Coast Erosion, Reclamation of Tidal Lands and Afforestation, which reported on the last subject in 1909. India.—The history of forest administration in India is exceedingly instructive to all who take an interest in the welfare of the British Empire, because it places before the reader an account of the gradual destruction of the greater part of the natural forests, a process through which most other British colonies are now passing, and then it shows how India emerged triumphantly from the self-inflicted calamity. As far as information goes, India was, in the early times, for the most part covered with forest. Subsequently settlers opened out the country along fertile valleys and streams, while nomadic tribes, moving from pasture to pasture, fired alike hills and plains. This process went on for centuries. With the advent of British rule forest destruction became more rapid than ever, owing to the increase of population, extension of cultivation, the multiplication of herds of cattle, and the universal firing of the forests to produce fresh crops of grass. Then railways came, and with their extension the forests suffered anew, partly on account of the increased demand for timber and firewood, and partly on account of the fresh impetus given to cultivation along their routes. Ultimately, when failure to meet the requirements of public works was brought to notice, it was recognized that a grievous mistake had been made in allowing the forests to be recklessly destroyed. Already in the early part of the 19th century sporadic efforts were made to protect the forests in various parts of the country, and these continued intermittently; but the first organized steps were taken about the year 1855, when Lord Dalhousie was governor-general. At that time conservators of forests existed in Bombay, Madras and Burma. Soon afterwards other appointments followed, and in 1864 an organized state department, presided over by the inspector-general of forests, was established. Since then the Indian Forest Department has steadily grown, so that it has now become of considerable importance for the welfare of the people, as well as for the Indian exchequer. The first duty of the department was to ascertain the position and extent of the remaining forests, and more particularly of that portion which still belonged to the state. Then a special forest law was passed, which was superseded in 1878 by an improved act, providing for the legal formation of permanent state forests; the determination, regulation, and, if necessary, commutation of forest rights; the protection of the forests against unlawful acts and the punishment of forest offences; the protection of forest produce in transit; the constitution of a staff of forest officers, provision to invest them with suitable legal powers, and the determination of their duties and liabilities. The officers who administered the department in its infancy were mostly botanists and military officers. Some of these became excellent foresters. In order to provide a technically trained staff arrangements were made in 1866 by Sir Dietrich Brandis, the first inspector-general of forests, for the training of young Englishmen at the French Forest School at Nancy and at similar institutions in Germany. In 1876 the students were concentrated at Nancy, and in 1885 an English forest school for India was organized in connexion with the Royal Indian Engineering College at Cooper’s Hill. In 1905 the school was transferred to the university of Oxford. The imperial forest staff of India consisted in 1909 of—officers not specially trained before entering the department, 17; officers trained in France and Germany, 23; officers trained at Cooper’s Hill, 143—total 184. In 1878 a forest school was started at Dehra Dun, United Provinces, for the training of natives of India as executive officers on the provincial staff. Since then a similar school, though on a smaller scale, has been established at Tharrawaddy in Burma. About 500 officers of this class have been appointed. In addition, there are about 11,000 subordinates, foresters and forest guards, who form the protective staff. The school at Dehra Dun has lately been converted into the Imperial Forest College. The progress made since 1864 is really astonishing. According to the latest available returns, the areas taken under the management of the department are—reserved state forests, or permanent forest estates, 91,272 sq. m.; other state forests, 141,669 sq. m.; or a total of 232,941 sq. m., equal to 24% of the area over which they are scattered. At present, therefore, the average charge of each member of the controlling staff comprises 1266 sq. m.; that of each executive officer, 446 sq. m.; and that of each protective official, 21 sq. m. It is the intention to increase the executive and protective staff considerably, in the same degree as the management of the forests becomes more detailed. Of the above-mentioned area the Forest Survey Branch, established in 1872, has up to date surveyed and mapped about 65,000 sq. m. From 1864 onwards efforts were made to introduce systematic management into the forests, based upon working plans, but, as the management had been provincialized, there was no central or continuous control. This was remedied in 1884, when a central Working Plans Office, under the inspector-general of forests, was established. This officer has since then controlled the preparation and execution of the plans, a procedure which has led to most beneficial results. Plans referring to about 38,000 sq. m. are now (1909) in operation, and after a reasonable lapse of time there should not be a single forest of importance which is not worked on a well-regulated plan, and on the principle of a sustained yield. While the danger of overworking the forests is thus being gradually eliminated, their yield capacity is increased by suitable silvicultural treatment and by fire protection. Formerly most of the important forests were annually or periodically devastated by jungle fires, sometimes lighted accidentally, in other cases purposely. Now 38,000 sq. m. of forest are actually protected against fire by the efforts of the department, and it is the intention gradually to extend protection to all permanent state forests. Grazing of cattle is of great importance in India; at the same time it is liable to interfere seriously with the reproduction of the forests. To meet both requirements careful and minute arrangements have been made, according to which at present 38,000 sq. m. are closed to grazing; 19,000 sq. m. are closed only against the grazing of goats, sheep and camels; while 176,000 sq. m. are open to the grazing of all kinds of cattle. The areas closed in ordinary years form a reserve of fodder in years of drought and scarcity. During famine years they are either opened to grazing, or grass is cut in them and transported to districts where the cattle are in danger of starvation. The service rendered in this way by a wise forest administration should not be underrated, since one of the most serious calamities of a famine—the want of cattle to cultivate the land—is thus, if not avoided, at any rate considerably reduced. During 1907 the government of India established a Research Institute, with six members engaged in collecting data regarding silviculture, forest botany, forest zoology, forest economics, working plans, and chemistry in connexion with forest produce and production. The institute is likely to lead to further substantial progress in the management of the forests. The financial results of forest administration in India for the years 1865 to 1905 show the progress made: \|Period.\|\| Mean Annual The highest percentage of increase occurred in the period 1880-1885. The revenue since 1886 has been considerably increased by the annexation of Upper Burma. Apart from the net revenue, large quantities of produce are given free of charge, or at reduced rates, to the people of the country. Thus, in 1904-1905, the net revenue amounted to Rs. 11,062,094, while the produce given free or at reduced rates was valued at Rs. 3,500,661, making a total net benefit derived from the state forests during that year of Rs. 14,562,755, or in round figures one million pounds sterling. The out-turn during the same year amounted to 252 million cub. ft. of timber and fuel and 215 million bamboos. The receipts from the sale of other forest produce came to 9 million rupees, out of a total gross revenue of 24 million rupees. These results are highly creditable to the government of India, which has led the way towards the introduction of rational forest management into the British empire, thus setting an example which has been followed more or less by various colonies. Even the movement in the United Kingdom during late years is due to it. Apart from India, substantial progress has been made in Cape Colony, Ceylon, the Straits Settlements and the Federated Malay States. Other British colonies are more backward in this respect. Energetic action is urgently wanted, especially in Canada and Australasia, where an enormous state property is threatened by destruction. Literature.—The following works of special interest may be mentioned: W. Schlich, A Manual of Forestry (London) (vols. i., ii. and iii. by W. Schlich; vols. iv. and v. by W.R. Fisher; 3rd ed. of vol. i., 1906, of vol. ii., 1904, of vol. iii., 1905; 2nd ed. of vol. iv., 1907; 2nd ed. of vol. v., 1908); Baden-Powell, Forest Law (London, 1893); Brown, The Forester (ed. by Nisbet, Edinburgh and London, 1905); Broilliard, Le Traitement des bois (Paris, 1894); Huffel, Économie forestière (Paris, 1904-1907); Lorey, Handbuch der Forstwissenschaft (2nd ed. by Stoetzer, Tübingen, 1903); Rossmässler, Der Wald. - (W. Sch.) The Forest Regions.—The great treeless region east of the Rocky Mountains separates the wooded area of the United States into two grand divisions, which may be called the Eastern and the Western forests. The Eastern forest is characterized by the predominance, on the whole, of broad-leafed trees, the comparative uniformity of its general types over wide areas, and its naturally unbroken distribution. In the Western forest conifers are conspicuously predominant; the individual species often reaches enormous and even unequalled dimensions, the forest is frequently interrupted by treeless areas, and the transitions from one type to another are often exceedingly abrupt. Both divisions are botanically and commercially rich in species. The Eastern forest may conveniently be subdivided into three members: 1. The Northern forest, marked by great density and large volume of standing timber, and a comparative immunity, in its virgin condition, from fire. The characteristic trees are maples, birches and beech (Fagus atropunicea), among the hardwoods and white pine (Pinus strobus), spruce (Picea rubens and Picea mariana) and hemlock (Tsuga canadensis) among conifers. 2. The Southern forest is on the whole less dense than the Northern, and more frequently burned over. Among its characteristic trees are the longleaf (Pinus palustris) and other pines, oaks, gums, bald cypress (Taxodium distichum) and white cedar (Chamaecyparis thyoides). 3. The Central Hardwood forest, which differs comparatively little from adjacent portions of the Northern and Southern forests except in the absence of conifers. Among its trees are the chestnut (Castanea dentata), hickories, ashes and other hardwoods already mentioned. The Western division has two members: 1. The Pacific Coast forest, marked by the great size of its trees and the vast accumulations of merchantable timber. Among its characteristic species are the redwood (Sequoia sempervirens) and the big tree (S. Washingtoniana), the Douglas fir (Pseudotsuga taxifolia), sugar pine (Pinus lambertiana), western hemlock (Tsuga heterophylla), giant arborvitae (Thuja plicata) and Sitka spruce (Picea sitchensis). 2. The Rocky Mountain forest, whose characteristic species are the western yellow pine (Pinus ponderosa), Engelmann spruce (Picea engelmanni) and lodgepole pine (Pinus murrayana). This forest is frequently broken by treeless areas of greater or less extent, especially towards the south, and it suffers greatly from fire. Subarid in character, except to the north and at high elevations, the vast mining interests of the region and its treeless surroundings give this forest an economic value out of proportion to the quantities of timber it contains. This distribution of the various forests is indicated on the first of the two accompanying maps. The second map shows the situation of the national forests hereafter mentioned. The forests of Alaska fall into two main divisions: the commercial though undeveloped forests of the south-east coast, which occur along the streams and on the lower slopes of the mountains and consist chiefly of western hemlock (Tsuga heterophylla), Sitka spruce (Picea sitchensis), yellow cedar (Chamaecyparis nootkatensis) and giant arborvitae (Thuja plicata), usually of large size and uninjured by fire; and the vast interior forests, swept by severe fires, and consisting chiefly of white and black spruces (Picea canadensis and nigra), paper birch (Betula papyrifera) and aspen (Populus tremuloides), all of small size but of great importance in connexion with mining. Northern Alaska and the extreme western coast regions are entirely barren. The National Forest Policy.—The forest policy of the United States may be said to have had its origin in 1799 in the enactment of a law which authorized the purchase of timber suitable for the use of the navy, or of land upon which such timber was growing. It is true that laws were in force under the early governments of Massachusetts, New Jersey and other colonies, providing for the care and protection of forest interests in various ways, but these laws were distinctly survivals of tendencies acquired in Europe, and for the most part of little use. It was not until the apparent approach of a dangerous shortage in certain timber supplies that the first real step in forest policy was taken by the United States. Successive laws passed from 1817 to 1831 strove to give larger effect to the original enactment, but without permanent influence towards the preservation of the live oak (Quercus virginiana Mill.), which was the object in view. A long period of inaction followed these early measures. In 1831 the solicitor of the treasury assumed a partial responsibility for the care and protection of the public timber lands, and in 1855 this duty was transferred to the commissioner of the general land office in the Department of the Interior. The effect of these changes upon forest protection was unimportant. When, however, at the close of the Civil War railway building in the United States took on an unparalleled activity, the destruction of forests by fire and the axe increased in a corresponding ratio, and public sentiment began to take alarm. Action by several of the states slightly preceded that of the Federal government, but in 1876 Congress, acting under the inspiration of a memorial from the American Association for the Advancement of Science, authorized the appointment of an officer (Dr Franklin B. Hough) under the commissioner of agriculture, to collect and distribute information upon forest matters. His office became in 1880 the division of forestry in what is now the United States Department of Agriculture. As the railways advanced into the treeless interior, public interest in tree-planting became keen. In 1873 Congress passed and later amended and repealed the timber culture acts, which granted homesteads on the treeless public lands to settlers who planted one-fourth of their entries with trees. Though these measures were not successful in themselves they directed attention towards forestry. The act which repealed them in 1891 contained a clause which lies at the foundation of the present forest policy of the United States. By it the president was authorized to set aside “any part of the public lands wholly or in part covered with timber or undergrowth, whether of commercial value or not, as public reservations, and the President shall, by public proclamation, declare the establishment of such reservations and the limits thereof.” Some eighteen million acres had been proclaimed as reservations at the time when, in 1896, the National Academy of Sciences was asked by the secretary of the interior to make an investigation and report upon “the inauguration of a rational forest policy for the forest lands of the United States.” Upon the recommendation of a commission named by the Academy, President Cleveland established more than twenty-one million acres of new reserves on the 22nd of February 1897. His action was widely misunderstood and attacked, but it awakened a public interest in forest questions without which the rapid progress of forestry in the United States since that time could never have been made. Within a few months after the proclamation of the Cleveland reserves the present national forest policy took definite shape. Under this policy the national government holds and manages, in the common interest of all users of the forests or its products, such portions of the public lands as have been set aside by presidential proclamation in accordance with the act of 1891. These lands are held against private acquisition under the Homestead Act (except as to agricultural lands as hereafter mentioned), the Timber and Stone Act, and other laws under which the United States disposes of its unappropriated public domain, but not against private acquisition under the Mineral Land Laws. They are selected from lands believed to be more valuable for forest purposes than for agriculture, and are managed with the purpose of securing from them the best and largest possible returns, present and future, whether in the form of water for irrigation or power, of timber, of forage for stock, or of any other beneficial product. The aggregate area of the reserves, or national forests, has been steadily increased until they now include nearly all the timber lands left of the public domain. The general lines of this policy were in part laid down by the commission already mentioned, in its report submitted to the secretary of the interior, May 1, 1897, and by the act of June 4, 1897, which was largely shaped by the work of the commission. Until this act was passed the national forests had been in theory closed against any form of use; nor had the possibility of securing forest preservation by wise use received much thought from those who had favoured their creation. Such a state of affairs could not continue. Before long public opinion would have forced the opening to use of the resources thus arbitrarily locked up, and in the absence of any administrative system providing for conservative use, the national forests would inevitably have been abolished, and the whole policy of government forest holdings would have ceased. The act of June 4, 1897 was therefore of the first importance. This act conferred upon the secretary of the interior general powers for the proper management of the national forests through the general land office of his department. It provided for the designation and sale of dead, mature and large timber; authorized the secretary to permit free use of timber in small quantities by settlers, miners and residents; empowered him to “make such rules and regulations and establish such service as will insure the objects of such reservations, namely, to regulate their occupancy and use and to preserve the forests thereon from destruction”; and made violation of the act or of such rules and regulations a misdemeanour. The statute limited the power to establish forest reservations to the purpose of improving and protecting the forest, securing favourable conditions of water flows, and furnishing a continuous supply of timber for the use and necessities of citizens of the United States. Lands found, upon due examination, to be more valuable for other purposes than for forest uses might be eliminated from any reservation, and all mineral lands within the reservations were left open to private appropriation under the mineral laws. The rights of settlers and claimants were safeguarded, and civil and criminal jurisdiction, except so far as the punishment of offences against the United States in the reservations was concerned, was reserved to the States. While the administration of the national forests was entrusted to the general land office, the same act assigned the surveying and mapping of them to the United States Geological Survey, which has published descriptions and maps of some of the more important. No attempt was made in the general land office to develop a technical forest service. There were, indeed, at the time of passage of the act, less than ten trained foresters in the United States, no means of training more, and very little conception of what forestry actually meant. The purpose of the administration was therefore mainly protection against trespass and fire, particularly the latter. Regulations were made giving effect to the provisions of the act of June 4, set forth above, but in the absence of technical knowledge as to what might safely be done, the tendency was rather to restrict than to extend the use of the forest. Meanwhile, however, there was rapidly developing in another branch of the government service an organization qualified for actual forest management. One year after the passage of the act of June 4, 1897, the division of forestry in the Department of Agriculture ceased to be merely a bureau of information, and became an active agency for introducing the actual practice of forestry among private owners and for conducting the investigations upon which a sound American forest practice could be based. The work awakened great interest among forest owners, and exerted a powerful educational influence upon the country at large. The division extended its work and became (July 1, 1901) the Bureau of Forestry. It drew into its employment for a time nearly all the men who were preparing themselves in increasing numbers (at first abroad, then in the newly-founded schools in the United States) for the profession of forestry, and was soon recognized as qualified to speak authoritatively on technical questions connected with the administration of the national forests. This led to a request from the secretary of the interior for the advice of the bureau on such questions. Working plans were accordingly undertaken for a number of the forests. The general land office, however, was not ready to attempt active forest management. Though some timber was sold and the grazing of stock regulated to some extent, the main object of the land office administration continued to be protection against fire. Many of the regulations which it made could not be enforced. The disadvantages of dispersal of the Federal government forest work among three separate agencies grew more and more apparent, until, on the 1st of February 1905, control of the 63,000,000 acres of forest reserves which up to that time had been set aside was transferred from the general land office to the Bureau of Forestry. In recognition of its new duties the designation of the bureau became the Forest Service. Other provisions of the act which affected the transfer were that forest supervisors and rangers should be selected, so far as possible, from qualified citizens of the state or territory in which each forest was situated, and that all money received from the sale of any products or the use of any land or resources of the national forests should be covered into the treasury and constitute a special fund for their protection, administration, improvement and extension. Five days later a statute gave forest officers the power to arrest trespassers; and on the 3rd of March the lieu land selection law was repealed. This law had opened the way for grave abuses through the exchange of worthless land by private owners within the forests for an equal area of valuable timber lands outside. The law has been modified since by the change of the old name “Forest Reserves” to “National Forests.” The act of June 11, 1906, opened to homestead entry lands within national forests found by examination to be chiefly valuable for agriculture. The administration and improvement of the national forests are now provided for directly by congressional appropriation. The power to create national forests conferred on the president by the act of March 1891 has been repealed for the states of Washington, Oregon, Idaho, Montana, Wyoming and Colorado, but for no others. The Forest Service began in earnest the development of all the resources of the national forests. Mature timber was sold wherever there was a demand for it and the permanent welfare of the forests and protection of the streams permitted, but always so as to prevent waste, guard against fire, protect young growth and ensure reproduction. Regulations were adopted which allowed small sales to be made without formality or delay, secured for the government the full value of timber sold, and eliminated unnecessary routine. Care was taken to safeguard the interests of the government and provide for the maintenance of good technical standards. The conduct of local business was entrusted to local officers. Large transactions with general policies were controlled from Washington, but with careful provision for first-hand knowledge and close touch with the work in the field. Business efficiency and the convenience of the public were carefully studied. In short, an organization was created capable of handling safely, speedily and satisfactorily the complex business of making useful a forest property of vast extent, scattered through sixteen different states of an aggregate area of over 1,500,000 sq. m. and with a population of 9,000,000. The growth since the 1st of July 1897 of the area of the national forests, of the expenditures of the government for forestry, and of the receipts from the national forests, is shown by the statement which follows. Though the act of June 4, 1897, became effective immediately upon its passage, the fiscal year 1899 was the first of actual administration, because the first for which Congress made the appropriation necessary to carry out the law. Area of National Forests, Annual Expenditures of the Federal Government for Forestry and National Forest Administration, and Receipts from National Forests, 1898-1909. at Close of Year \| Division of Forestry (Bureau of Forestry, \| Expenditures upon \| \|1898\|\|40,866,184\|\|20,000.00\|\|· ·\|\|· ·\|\|· ·\|\|· ·\| Until 1906, the sole source of receipts was the sale of timber. In the fiscal year 1907, however, timber sales furnished less than half the receipts. The following statement concerning the timber sales of the fiscal years 1904-1907 will serve to bring out the change that followed the transfer of control to the forest service in the midst of the fiscal year 1905:— \| Receipts from \| These figures show (1) a large excess each year in the amount of timber sold over that cut and paid for; (2) nine times as much timber sold at the end of the four-year period as at the beginning and three times as much cut; and (3) a much higher price obtained per thousand board-feet at the end of the period than at the beginning. Each of these matters calls for comment. The sales are of stumpage only; the government does no logging on its own account. 1. More timber is sold each year than is cut and paid for, because many of the sales extend over several years. With increasing sales the amount sold each year for future removal has exceeded the amount to be removed during that year under sales of earlier years. Large sales covering a term of years are made because the national forests contain much overmature timber, which needs removal, but which is frequently too inaccessible to be saleable in small amounts. To prevent speculation the time allowed for cutting is never more than five years, and cutting must begin at once and be continued steadily. 2. The volume of sales has increased rapidly because much forest is ripe for the axe, the demand is strong, and control by trained men makes it safe to cut more freely. The increase is marked both in small and in large sales, but a score of sales for less than $5000 are made against one for more. The total cut is still far below the annual increment of the forests. As the demand grows restrictions must increase in order to husband the present supply until the next crop matures. 3. The stumpage price would seem on the face of the figures to have risen from about one dollar to more than three dollars per thousand board-feet. The receipts, however, for any one year are not exclusively for the timber cut in that year, since payments are made in advance. In the year 1907 the average price obtained was something less than $2.50 per thousand. It is therefore true that stumpage prices have risen greatly, although conditions new to the American lumbermen are imposed. Full utilization of all merchantable material, care of young growth in felling and logging, and the piling of brush, to be subsequently burned by the forest officers if burning is necessary, are among these conditions. Timber to be cut must first be marked by the forest officers. Sales of more than $100 in value are made only after public advertisement. Only the simplest forms of silviculture have as yet been introduced. The vast area of the national forests, the comparatively sparse population of the West, the rough and broken character of the forests themselves, and the newness of the problems which their management presents, make the general application of intensive methods for the present impracticable. Natural reproduction is secured. The selection system is most used, often under the rough and ready method of an approximate diameter limit, with the reservation of seed trees where needed. The tendency, however, is strongly towards a more flexible and effective application of the selection principle, as a better trained field force is developed and as market conditions improve. One conspicuous achievement was the reduction of loss by fires on the national forests. During the unusually dry season of 1905 there were only eight fires of any importance, and the area burned over amounted only to about .16 of 1% of the total area. In 1900 about .12 of 1% was burned. This was accomplished by efficient patrol, co-operation of the public, and by preventive measures, such as piling and burning the brush on cut-over areas. Since the beginning of 1906 the largest source of income from the national forests was their use for grazing. Stock-raising is one of the most important industries of the West. Formerly cattle and sheep grazed freely on all parts of the public domain. In the early days of the national forests the wisdom of permitting any grazing at all upon them was sharply questioned. Unrestricted grazing had led to friction between individuals, the deterioration of much of the range through overstocking, and serious injury to the forests and stream flow. The forests of the West, however, are largely of open growth and contain many grassy parks, the results of old fires, and many high mountain meadows. Under proper regulations the grass and other forage plants which they produce in great quantity can be used without detriment to the forests themselves, and with great benefit to the stock industry, which often can find summer pasturage nowhere else. Except in southern California grazing is now permitted on all national forests unless the watersheds furnish water for domestic use; but the time of entering and leaving, the number of head to be grazed by each applicant, and the part of the range to be occupied are carefully prescribed. Planted areas and cut-over areas are closed to stock until the young growth is safe from harm, and goats are allowed only in the brushland of the foothills. The results of regulation, in addition to the protection of forest growth and streams, are the prevention of disputes, improved range, better stock, stable conditions in the stock industry, and the best use of the range in the interest of progress and development. The first right to graze stock on the forests is given to residents, small owners and those who have used the range before. Thus the crowding out of the weaker by the stronger and of the settler by the roving outsider has been stopped. In 1906 the forest service began to impose a moderate charge for the use of the national forest range. The following statement shows the amount of stock grazed on the national forests 1904-09, and the receipts for the grazing charge:— Cattle and Horses. Sheep and Goats. A work of enormous magnitude which has now begun is planting on the national forests. At present, with low stumpage prices and incomplete utilization of forest products, clear cutting with subsequent planting is not practicable. There are, however, many million acres of denuded land within the national forests which require planting. Such planting is still confined chiefly to watersheds which supply cities and towns with water. The first planting was done in 1892, in California. Since then similar work has been done on city watersheds in Colorado, Utah, Idaho and New Mexico. Other plantations are in the Black Hills national forest, where large areas of cut-over and burned-over land are entirely without seed trees, and in the sandhill region of Nebraska. Up to 1908 about 2,000,000 seedlings had been planted, on over 2000 acres—a small beginning, but the work was entirely new and presented many hard problems. The nursery operations of the forest service are concentrated at seven stations, located in southern California, Nebraska, Colorado, New Mexico (2), Utah and Idaho, where stock is raised for local planting and for shipment elsewhere. These nurseries are small. Their annual productive capacity is between 8,000,000 and 10,000,000 seedlings. Each nursery is practically an experimental forest-planting station, at which a large variety of species are grown and various methods are tried. The organization of the administrative work of the national forests is by single forests. On the 1st of January 1908 the total number of forests was 165 with a total area of 162,023,190 acres (on April 7, 1909, the numbers were 146 national forests in the U.S. with 167,672,467 acres, besides two in Alaska with 26,761,626 and one in Porto Rico with 65,950 acres). In charge of each forest is a forest supervisor. Under the supervisors are forest rangers and forest guards, whose duties include patrol, marking timber and scaling logs, enforcing the regulations and conducting some of the minor business arising from the use of the forests. Guards are temporary employés; rangers are employed by the year. The supervisors report directly to and receive instructions from the central office at Washington. In this office there are four branches—operation, grazing, silviculture and products—each of which directs that part of the work which belongs to it, dealing directly with the supervisor. For inspection purposes, however, the forests are separated into six districts, in each of which is located a chief inspector with a corps of assistants. The inspectors are without administrative authority, but assist by their counsel the supervisors, and through inspection reports keep the Washington office informed of the condition of all lines of administrative work in progress. Administrative officers alternate frequently between field and office duties. The number of forest officers in the several grades on the 1st of January 1908 were: 6 chief inspectors, 26 inspectors, 106 forest supervisors, 41 deputy forest supervisors, 820 forest rangers and 283 forest guards. The total number of employés of the forest service on the same date, including the clerical force, was 2034. Besides the administration of the national forests, the forest service conducts general investigations, carries on an extensive educational work, and co-operates with private owners who contemplate forest management upon their own tracts. This last work is undertaken because of the need of bringing forestry into practice, the lack of trained foresters outside of the employ of the government, and the lack of information as to how to apply forestry and what returns may be obtained. Co-operation takes the form of advice upon the ground and, on occasion, of the making of working plans. The educational work of the service is performed chiefly through publications, the purpose of which is to spread very widely a knowledge of the importance of forestry to the nation and of the principles upon which its practice rests. The investigations which the service conducts extend from studies of the natural distribution and classification of American forests and of their varied silvicultural problems to statistics of lumber production and laboratory researches which bear upon the economical utilization of forest products. As examples of these researches may be mentioned tests of the strength of timber, studies of the preservative treatment of wood for various uses, wood-pulp investigations and studies in wood chemistry. Forest Instruction.—Most of the men now in the forest service received their training in the United States. There are several professional schools of forestry. The Yale Forest School, which was opened as a department of Yale University in September 1900, offers a two-years’ graduate course with abundant field work, and also conducts a summer school of forestry, especially adapted to the training of forest rangers and special students, at Milford, Pennsylvania. The university of Michigan and Harvard University also offer a two-years’ graduate course in forestry. The Pennsylvania State College has recently established a four-years’ undergraduate course in forestry. The Biltmore Forest School in North Carolina, the oldest of all these schools, offers a one-year course in technical forestry. A large number of the agricultural colleges give instruction in forestry. Among these are Nebraska, Minnesota, Maine, Michigan, Washington and Mississippi agricultural colleges, the university of Georgia and Iowa State College. Berea College, Kentucky, deserves special mention as a college which has done valuable work in teaching forestry without attempting to turn out professional foresters. Forestry among the States.—Among the states forestry has hardly reached the stage of practical application on the ground. New York holds 1,500,000 acres of forest land. It has a commission to care for its forest preserve, and to protect the forest land throughout the state from fire. The constitution of the state, however, prohibits the cutting of timber on state land, and thus confines the work entirely to protection of the forest and to the planting of waste areas. Pennsylvania is at present showing the most efficient activity in working out a forest policy. It has state forests of 820,000 acres, a good fire law more and more satisfactorily enforced, and eight nurseries for growing planting material. In 1905, 160,000 white pine seedlings were set out. It has also a school for forest rangers, to be employed on the state forests, and it has just established a state professional school of forestry. Twenty-six of the states have regularly appointed forest officers, six have carried on studies of forest conditions in co-operation with the forest service, and there is scarcely one which is not actively interested in forestry. Laws, generally good, to prevent damage from forest fires, have been enacted by practically all the states, but their enforcement has unfortunately been lax. Public sentiment, however, is making rapid progress. Among the best laws are those of Maine, New Hampshire, Minnesota, New York, Pennsylvania and Wisconsin. The New York law, for example, provides for the appointment of one or more fire-wardens in each town of the counties in which damage by fire is especially to be feared. In other counties supervisors of towns are ex-officio fire-wardens. A chief fire-warden has general supervision of their work. The wardens, half of the cost of whose services is paid by the state, receive compensation only for the time actually employed in fighting fires. They may command the service of any citizen to assist them. Setting fire to woods or waste lands belonging to the state or to another, if such fire results in loss, is punishable by a fine not exceeding $250 or imprisonment not exceeding one year, or both, and damages are provided for the person injured. Since fire is beyond question the most dangerous enemy of forests in the United States, the measures taken against it are of vital importance. The following table shows the amount of forest land held by the different states, and by the territory of Hawaii:— Area of State Forest Reservations, 1907. Forestry on Private Lands.—The practice of forestry among private owners is of old date. One of the earliest instances was that of Jared Eliot, who, in 1730, began the systematic cutting of timber land to supply charcoal for an iron furnace at Old Salisbury, Connecticut. The successful planting of waste lands with timber trees in Massachusetts dates from about ten years later. But such examples were comparatively rare until recent times. At present the intelligent harvesting of timber with a view to successive crops, which is forestry, is much more common than is usually supposed. Among farmers it is especially frequent. It was begun among lumbermen by the late E.S. Coe, of Bangor, Maine, who made a practice of restricting the cut of spruce from his forests to trees 10, 12 or sometimes even 14 in. in diameter, with the result that much of his land yielded, during his life, a second crop as plentiful as the first. Many owners of spruce lands have followed his example, but until very recently without improving upon it. Systematic forestry on a large scale among lumbermen was begun in the Adirondacks during the summer of 1898 on the lands of Dr W.S. Webb and Hon. W.C. Whitney, of a combined area of over 100,000 acres, under the superintendence of the then Division of Forestry. In these forests spruce, maple, beech and birch predominate, but the spruce alone is at present of the first commercial importance. The treatment is a form of the selection system. Under it a second crop of equal yield would be ripe for the axe in thirty-five years. Spruce and pine are the only trees cut. The work had been executed, at least up to the year 1902, with great satisfaction to the owners and the lumbering contractors, as well as to the decided benefit of the forest. The lumbering is regulated by the following rules, and competent inspectors are employed to see that they are rightly carried out: (1) No trees shall be cut which are not marked. (2) All trees marked shall be cut. (3) No trees shall be left lodged in the woods, and none shall be overlooked by the skidders or haulers. (4) All merchantable logs which are as large as 6 in. in diameter at the small end must be utilized. (5) No stumps shall be cut more than 6 in. higher than the stump is wide. (6) No spruce shall be used for bridges, corduroy, skids, slides, or for any purpose except building camps, dams or booms, unless it is absolutely necessary on account of lack of other timber. (7) All merchantable spruce used for skidways must be cut into logs and hauled out. (8) Contractors must not do any unnecessary damage to young growth in lumbering; and if any is done, they must discharge the men who did it. These two instances of forestry have been most useful and effective among lumbermen and other owners of forest land in the north-east. Among those which have followed their example are the Berlin Mills Paper Company in northern New Hampshire, the Cleveland Cliffs Iron Company in northern Michigan, and the Delaware and Hudson Railroad Company in New York, all of which have employed professional foresters. The most notable instance of forestry in the south is on the estate of George W. Vanderbilt at Biltmore, N.C. This was the first case of systematic forestry under regular working plans in the United States. It was begun in 1891 on about 4000 acres, and has since been extended until it now covers about 100,000 acres. A professional forester with a corps of trained rangers under him is in charge of the work. The Pennsylvania Railroad has recently employed a trained forester and several assistants and has undertaken systematic forestry on a large scale. The effect of the work of the forest service in assisting private owners is evidenced by the fact that down to the year 1908 670 wood lots and timber tracts had been examined by agents of the forest service, of which 250 were tracts over 400 acres in extent, and planting plans had been made for 436 owners covering a total area of 80,000 acres. Expert advice is also given to wood lot owners upon application by many of the state foresters. American Practice.—The conditions under which forestry is practised in Europe and in America differ so widely that rules which are received as axiomatic in the one must often be rejected in the other. Among these conditions in America are the highly developed and specialized methods and machinery of lumbering, the greater facilities for transportation and consequent greater mobility of the lumber trade, the vast number of small holdings of forest land, and the enormous supply of low-grade wood in the timbered regions. High taxes on forest properties, cut-over as well as virgin, notably in the north-western pineries, and the firmly established habits of lumbermen, are factors of great importance. From these and other considerations it follows that such generally accepted essentials of European methods of forestry as a sustained annual yield, a permanent force of forest labourers, a permanent road system and the like, are in most cases utterly inapplicable in the United States at the present day in private forestry. Methods of forest management, to find acceptance, must there conform as closely as possible to existing methods of lumbering. Rules of marked simplicity, the observance of which will yet secure the safety of the forest, must open the way for more refined methods in the future. For the present a periodic or irregular yield, temporary means of transport, constantly changing crews, and an almost total ignorance of the silvics of all but a few of the most important trees—all combine to enforce the simplest silvicultural treatment and the utmost concentration of purpose on the two main objects of forestry, which are the production of a net revenue and the perpetuation of the forest. Such concentration has been followed in practice by complete success. The forests with which the American forester deals are rich in species, usually endowed with abundant powers of reproduction, and, over a large part of their range, greatly dependent for their composition and general character upon the action of forest fires. Of the commercially valuable trees there may be said to be, in round numbers, a hundred out of a total forest flora of about 500 species, but many trees not yet of importance in the lumber trade will become so hereafter, as has already happened in many cases. The attention of the forester must usually be concentrated upon the growth and reproduction of a single species, and never of more than a very few. Thus the silvicultural problems which must be solved in the practice of forestry in America are fortunately less complicated than the presence of so many kinds of trees in forests of such diverse types would naturally seem to indicate. The forest fire problem is one of the most difficult with which the American forester has to deal. It is probable that forest fires have had more to do with the character and distribution of forests in America than any other factor except rainfall. With an annual range over thousands of square miles, in many portions of the United States they occur regularly year after year on the same ground. Trees whose thick bark or abundant seeding gives them peculiar powers of resistance, frequently owe their exclusive possessions of vast areas purely to the action of fire. On the economic side fire is equally influential. The probability, or often the practical certainty, of fire after the first cut, commonly determines lumbermen to leave no merchantable tree standing. Forest fires are thus the most effective barriers to the introduction of forestry. Excessive taxation of timber land is another of almost equal effect. Because of it lumbermen hasten to cut, and afterwards often to abandon, lands which they cannot afford to hold. This evil, which only the progress of public sentiment can control, is especially prevalent in certain portions of the white pine belt. Forest Associations.—Public sentiment in favour of the protection of forests is now widespread and increasingly effective throughout the United States. As the general understanding of the objects and methods of forestry becomes clearer, the tendency, formerly very marked, to confound ornamental tree planting and botanical matters with forestry proper is rapidly growing less. At the same time, the number and activity of associations dealing with forest matters is increasing with notable rapidity. There are now about thirty such associations in the United States. One of these, the Society of American Foresters, is composed exclusively of professional foresters. The American Forestry Association is the oldest and largest. It has been influential in preparing the ground work of popular interest in forestry, and especially in advocating and securing the adoption of the federal forest reservation policy, the most important step yet taken by the national government. It publishes as its organ a monthly magazine called Forestry and Irrigation. The Pennsylvania Forestry Association has been instrumental in placing that state in the forefront of forest progress. Its organ is a bi-monthly publication called Forest Leaves. Other states which have associations or societies of special influence in forest matters are California, Massachusetts, Minnesota, Colorado, New Hampshire, Georgia and Oregon. Arbor Day, instituted in Nebraska in 1872 as a day for shade-tree planting by farmers who had settled on the treeless prairies, has been taken up as a means of interesting school children in the planting of trees, and has spread until it is now observed in every state and territory. It continues to serve an admirable purpose. Lumbering.—According to the census report for 1905 the capital invested in logging operations in the United States was $90,454,596, the number of employés engaged 146,596, and their wages $66,990,000; sawmills represented an invested capital of $381,621,000, and employed 223,674 persons, whose wages were $100,311,000, while planing mills represented a capital of $222,294,000 and employed 132,030 persons whose wages were $66,434,000. \|Product.\|\|Output 1906.\|\| Equivalent \| Total Wood \| \|Wood distillation\|\|108\|\|12\|\|· ·\|\|120\| \|Round mine timbers\|\|165\|\|35\|\|· ·\|\|200\| \|Hewn cross ties\|\|207\|\|503\|\|· ·\|\|710\| All the operations of the lumber trade in the United States are controlled, and to no small degree determined, by the peculiar unit of measure which has been adopted. This unit, the board-foot, is generally defined as a board one foot long, one foot wide and one inch thick, but in reality it is equivalent to 144 cub. in. of manufactured lumber in any form. To purchase logs by this measure one must first know about what each log will yield in one-inch boards. For this purpose a scale or table is used, which gives the contents of logs of various diameters and lengths in board feet. Under such a standard the purchaser pays for nothing but the saleable lumber in each log, the inevitable waste in slabs and sawdust costing him nothing. The table at foot gives the estimated consumption of wood for certain purposes in the United States in 1906. In addition to this amount, an immense quantity of wood is used each year for fuel, posts and other domestic purposes, and the total annual consumption is not less than 20 billion cub. ft. The years 1890 to 1906 were marked by rapid changes in the rank of the important timber trees with reference to the amount of timber cut, and a shifting of the important centres of production. Among coniferous trees, white pine has yielded successively to yellow pine and Douglas fir, while the scene of greatest activity has shifted from the Northern forest to the Southern, and from there is rapidly shifting to the Pacific Coast. The total cut of coniferous lumber has increased steadily, but that of the hardwoods is falling off, and in 1906 it was 15% less than in 1899, while inferior hardwoods are gradually assuming more and more importance, and the scene of greatest activity has passed from the middle west to the south and the Appalachian region. Conifers.—The coniferous supply of the country is derived from four forest regions: (1) The Northern forest; (2) the Southern forest; (3) the Pacific Coast forest; and (4) the Rocky Mountain forest. 1. The Northern forest was long the chief source of the coniferous lumber production in the United States. The principal timber tree of this region is the white pine, usually known in Europe as the Weymouth pine. It has an average height when mature of 110 ft., with a diameter a little less than 3 ft., but the virgin timber is approaching exhaustion. White pine was one of the first trees to be cut extensively in the United States, and Maine, the pine tree state, was at first the centre of production. In 1851 the cut of white pine on the Penobscot river was 144 million ft., that of spruce 14 million and of hemlock 11 million. Thirty years later the pine cut had sunk to 23 million, spruce had risen to 118 million, and hemlock had passed pine by a million feet. Meanwhile, the centre of production had passed from the north woods to the Lake States, and for many years this region was the scene of the most vigorous lumbering activity in the world. The following figures show the cut for the Lake States from 1873 to 1906. It is certain that the remarkable decline in the cut of white pine which these figures show will continue still farther. Second to the white pine among the coniferous lumber trees of the Northern forest is the hemlock (Tsuga canadensis). It is used chiefly for construction purposes and furnishes a comparatively low grade of lumber. The spruce (Picea rubens) is used chiefly for lumber, but it is in large and increasing demand in the manufacture of paper pulp. For the latter purpose hemlock, poplar (Populus tremuloides and P. grandidentata) and several other woods are also employed, but on a smaller scale. The total consumption of wood for paper in the United States for 1906 was 3,660,000 cords, of which 2,500,000 was spruce. Of this, however, 720,000 cords were imported from Canada. 2. The chief product of the Southern forest is the yellow pine. This is the collective term for the longleaf, shortleaf, loblolly and Cuban pines. Of these the longleaf pine (Pinus palustris Mill.), called pitch-pine in Europe, is the most important. Its timber is probably superior in strength and durability to that of any other member of the genus Pinus, and in addition to its value as a timber tree it is the source of naval stores in the United States. The average size of the mature longleaf pine is 90 ft. in height and 20 in. in diameter. Shortleaf (Pinus echinata) and loblolly (P. taeda) are other important members of this group. Their wood very closely resembles that of the longleaf pine and is often difficult to distinguish from it. The trees are also of about the same size and height. Loblolly is, however, of more rapid growth. The total cut of yellow pine in 1906 was 11,661,000,000 board ft.; it has perhaps not yet reached its maximum, but is certainly near it. Another important coniferous tree of the Southern forest is the bald cypress (Taxodium distichum), which grows in the swamps. The cut in 1906 was 839,000,000 board ft., a gain of 69% over 1899. 3. But the great supply of coniferous timber is now on the Pacific Coast. The Douglas fir (Pseudotsuga taxifolia), also known as Douglas spruce, red fir and Oregon pine, is the foremost tree in Oregon and Washington, and the redwood in California. When mature the Douglas fir averages 200 ft. in height and 4 ft. in diameter, and the redwood 225 ft. in height and 8 ft. in diameter. Other important trees of the Pacific Coast are sugar pine (Pinus lambertiana), western red cedar (Thuja plicata), western larch (Larix occidentalis), Sitka spruce (Picea sitchensis), western hemlock (Tsuga heterophylla) and western yellow pine (Pinus ponderosa). These trees will all be of increasing importance. Logging on the Pacific Coast is characterized by the use of powerful machinery and by extreme skill in handling enormous weights. This is especially true in California, where the logs of redwood and of the big tree (Sequoia Washingtoniana) are often more than 10 ft. in diameter. Logging is usually done by wire cables operated by donkey-engines. The journey to the mill is usually by rail. The mills are often of great size, built on piles over tide water and so arranged that their product is delivered directly from the saws and dry kilns to vessels moored alongside. The products of the Pacific Coast forest make their way over land to the markets of the central and eastern states and into foreign markets. Among the lumber-producing states, Washington has in seven years jumped from fifth place to first, and its output has increased from 1,428,000,000 board ft. in 1899 to 4,305,000,000 ft. in 1906. Oregon and California have increased their output from 734,000,000 each in 1899 to 1,605,000,000 and 1,349,000,000 ft. respectively in 1906. Of the total output of these three states (7,259,000,000 ft.) 4,880,000,000 ft. is Douglas fir and 660,000,000 redwood. 4. The important lumber trees of the Rocky Mountain forest are the western yellow pine, the lodgepole pine, the Douglas fir and the Engelmann spruce. The Douglas fir, here extremely variable in size and value, reaches in this region average dimensions of perhaps 80 ft. in height by 2 ft. in diameter, the western yellow pine 90 ft. by 3 ft. and the Engelmann spruce 60 ft. by 2 ft. Mining, railroad and domestic uses chiefly absorb the annual timber product, which is considerable in quantity, and of vast importance to the local population. The lumber output of the Rocky Mountain region is, however, increasing very rapidly both in the north and in the south-west. One of the largest mills in the United States is in Idaho. The following table summarizes the cut of the important coniferous species during the years 1899-1906: \|Kind.\|\|1899.\|\|1904.\|\|1906.\|\| Per Cent Increase \| (+) or Decrease (−) since 1899. \|Yellow Pine\|\|9,659\|\|11,533\|\|11,661\|\|+ 20.7\| \|Douglas Fir\|\|1,737\|\|2,928\|\|4,970\|\|+ 186.2\| \|White Pine\|\|7,742\|\|5,333\|\|4,584\|\|− 40.8\| \|Western Pine\|\|944\|\|1,279\|\|1,387\|\|+ 46.9\| Hardwoods.—The hardwood supply of the country is derived almost entirely from the eastern half of the continent, and comes from each of the three great Eastern forest regions. The following table shows the cut of the important species of hardwoods for 1899 and 1906: or Decrease (−). \|Red gum\|\|285,417\|\|453,678\|\|+ 59.0\| \|All other\|\|208,504\|\|87,637\|\|− 58.0\| Oak, which in 1899 furnished over half the entire output, has fallen off 36.5%. Yellow poplar, which in 1899 was second among the hardwoods, has fallen off 38% and now occupies third place; and elm, the great stand-by in slack cooperage, has fallen 50.8%. On the other hand less valuable species like maple and red gum have advanced 39 and 59% respectively. The decrease is largely due to the fact that the hardwoods grow naturally on the better classes of soil, and in the eastern United States where the population has always been the densest, and as a consequence of this, a large proportion of the original hardwood land has been cleared up and put under cultivation. The hardwood supply of the future must be obtained chiefly from the Appalachian region, where the conditions are less favourable to agriculture. In addition to the lumber cut, enormous quantities of hardwoods are used each year for railroad ties, telephone and other poles, piles, fence posts and fuel, and there is a great, amount of waste in the course of lumbering and manufacture. - These net imports are received from non-European countries. They consist chiefly of valuable hardwoods, like teak, mahogany, eucalypts and others. - The United States fiscal year ends June 30, and receives its designation from the calendar year in which it terminates. Thus, the fiscal year 1898 is the year July 1, 1897-June 30, 1898. - Administration transferred to Bureau of Forestry, February 1, 1905. - Woods waste includes tops, stumps, cull logs and butts, but does not include defective trees left or trees used for road purposes. - Mill waste includes bark, kerf, slabs and edgings. - Not separately reported.<\|endoftext\|>	3.703125
377	Q: # What is 37.5 as a fraction? A: The number 37.5 is expressed as 37 1/2 in simplified terms or 75/2 as a fraction. In simplified form, the 0.5 equals .50 and can be written as 5/10. The fraction 5/10 is simplified further to 1/2, by using 5 as the common factor for both the numerator and denominator. ## Keep Learning Once simplified, 37 1/2 is a mixed number, with a whole number and a fraction. In order to write the answer entirely as a fraction, multiply 37 by the denominator of 2, and then add the numerator of 1. The product of 37 times 2, plus 1, equals 75, which is placed over the denominator of 2, and written as 75/2. Sources: ## Related Questions • A: One way to write the number 7.5 as a fraction is 75/10. This answer is easy to obtain because it only involves moving the decimal one place to the right. Since 7.5 is equal to 7.5 over one, then moving the decimal place to the right gives 75 over 10. Filed Under: • A: To convert a percentage to a ratio, write out the percentage number as a fraction, reduce the fraction to its simplest form and convert the new fraction to a ratio by replacing the slash mark with a colon. Converting a percentage to a ratio takes only a few minutes and requires paper and a pencil. Filed Under: • A: In fractional math, a numerator is the top number in a fraction, and the denominator is the bottom number. In the fraction one-half, for instance, one is the numerator, and two is the denominator.<\|endoftext\|>	4.4375
1,641	The summer and winter of 1862 were hard times for the North. Confederate offensives from Maryland to the Mississippi River sat at the peak of Union setbacks in the summer of 1862. In the end, these offensives were turned back at Corinth, Perryville, and Antietam Creek; the latter also serving to redefine the nation and the war as it now sought to end slavery in the United States due to the Union victory on the banks of Antietam Creek. Abraham Lincoln stated that he issued the proclamation strictly “as a military measure” and as such, only slaves then in territories that “the people whereof shall then be in rebellion against the United States…” would “be then, thenceforward, and forever free….” In order to uphold such “a military measure,” the United States military needed to be winning. Antietam gave Lincoln the opportunity to issue the proclamation, now he needed a victory in late 1862 to put teeth behind it. For this and other reasons, three Union offensives went forward in December 1862 to seek a victory and stop Confederate forces from reinforcing each other in different theaters of war. The first was Ambrose Burnside’s Fredericksburg Campaign; the second was Ulysses Grant’s campaign against Vicksburg; and the last was William Rosecrans’ and the Army of the Cumberland’s offensive against Braxton Bragg’s Army of Tennessee, which culminated in the Battle of Stones River, fought 150 years ago today, and lasting from December 31, 1862 to January 2, 1863. Burnside’s and Grant’s offensives ended in failure, causing Northern morale to reach one of its lowest points in the war. Now all eyes were focused on the campaign begun on December 26 in Tennessee as the Lincoln administration looked to Rosecrans and his army to gain support for its emancipation policy, win a much needed victory for the Union, and end any hopes of British recognition of the Confederacy, which, it was feared, would happen as Britain “might succumb to pressure from English textile concerns and recognize the Confederacy in order to ensure their supply of raw cotton.” The Lincoln administration was facing a similar situation to what it had seen prior to Union victories in Maryland, Kentucky, and Mississippi in the summer of 1862 and was looking to put strength behind the Emancipation Proclamation, one of the most important documents in American history. Rosecrans’ campaign began on December 26 with approximately 46,000 men from the Army of the Cumberland moving southeast towards Murfreesboro, where Braxton Bragg’s Army of Tennessee, 37,000 strong, waited. The Army of the Cumberland reached the outskirts of the central Tennessee city on December 30 and prepared for battle. Both commanders planned offensives but Bragg struck first in the early morning hours of 1862’s last day. In fighting so fierce that soldiers had to plug their ears with cotton, Bragg’s army had pushed Rosecrans’ to the brink of destruction and bent back the Army of the Cumberland’s line into a V-shape. Rosecrans’ army had bent but not broken. That night, as 1862 turned into 1863, Rosecrans held a council of war. Knowing of the dire straits that the Union war effort was in, Rosecrans’ and his subordinates determined to fight it out against Bragg. As George Thomas put, “This army does not retreat,” and indeed it could not retreat and give the North a third defeat in December 1862. Thus, Rosecrans would hold his lines outside of Murfreesboro on the banks of Stones River. 150 years ago today, on New Year’s Day, there was only light skirmishing but the next day witnessed the defeat of another of Bragg’s attacks and ended the Battle of Stones River. On January 3, Braxton Bragg and his army retreated from the field. Of 46,000 men engaged on the Union side, 13,249 were casualties, a rate of 28 percent; 37,000 Confederates were engaged in the battle and lost 10,266 men, or a 27 percent loss. Where the desperate fight to define the nation at Antietam is the bloodiest day in American history, Stones River stands as the Civil War battle with the highest casualty rate, totaling out at 28% and it equally served to save and define the nation in a time of crisis. Though it is often an overlooked battle, the Battle of Stones River had a great impact on the war. Firstly, it served to strengthen Northern spirits after the defeat of Grant at Chickasaw Bluffs and Burnside at Fredericksburg, a defeat which caused Abraham Lincoln to utter his famous words, “If there is a worse place than hell I am in it.” After Rosecrans’ defeat of Bragg, Lincoln wired back “God bless you, and all with you!” When speaking eight months later of the victory, Lincoln told Rosecrans: “I can never forget, whilst I remember anything, that about the end of last year, and beginning of this, you gave us a hard earned victory, which, had there been a defeat instead, the nation could scarcely have lived over.” Not only did Rosecrans’ victory potentially save the war domestically, it also may have done so from a foreign standpoint. The threat that the British might recognize the Confederacy in order to supply themselves with the South’s cotton was a very real one in late 1862. When Rosecrans’ campaign began, Henry Halleck, general-in-chief of all Union armies, told Rosecrans that his campaign “may be, and perhaps is, the very turning point in our foreign relations.” Indeed, it did serve to dissuade the British from recognizing the Confederacy. Last but not least, on the 150th Anniversary of the issuance of the final draft of the Emancipation Proclamation, the Union victory at Stones River provided a much-needed rise in support of the Union war effort and its aims, which was now to free the slaves in territory held by Confederates. Peter Cozzens, author of No Better Place to Die: The Battle of Stones River perhaps put it best when he wrote, “Bragg’s retreat after the battle [of Stones River] gave the North a victory at a time when defeat would have made the Emancipation Proclamation look like the last gasp of a dying war effort….” Just as the victory at Antietam gave Lincoln the opportunity to announce the Emancipation Proclamation, the victory at Stones River garnered support for it and turned it into a main aim of the Northern war effort. The blood spilt on the banks of Antietam Creek and the banks of Stones River had not been in vain and served to reshape not only the course of the Civil War but the nation as a whole. Burlingame, Michael. The Inner World of Abraham Lincoln. United States of America: Board of Trustees of the University of Illinois, 1994. Cozzens, Peter. No Better Place to Die: The Battle of Stones River. Champaign, IL: University of Illinois Press, 1990. Hess, Earl. The Civil War in the West: Victory and Defeat From the Appalachians to the Mississippi. Chapel Hill, NC: The University of North Carolina Press, 2012. Kolakowski, Christopher L. "I Will Die Right Here: The Army of the Cumberland at Stones River." Hallowed Ground 13 (2012): 14-21. Lincoln, Abraham. "The Emancipation Proclamation." Featured Document. Accessed January 1, 2013. http://www.archives.gov/exhibits/featured_documents/emancipation_proclamation/transcript.html.<\|endoftext\|>	3.78125
268	One challenge to improving wheat genetically has been due to the complex genetic mechanisms that go on with wheat DNA. A wheat cell has six copies of its seven chromosomes (42 chromosomes total). Not only in the amount of DNA daunting, but the multiple copies of one gene make genetic manipulation very tricky. CRISPR-Cas9, the DNA editing technology that is changing the face of genetic engineering, can alleviate this issue. How? Because it has what is called multiplex genome editing capacity. This means that many genes can be altered simultaneously and beneficial modifications in multiple genes can be made at the same time. In a new study, mutations in all three copies of the gene TaGW2 resulted in an increase in thousand grain weight, grain area, grain width, and grain length. The group went on to show that these mutations are heritable and were edited by CRISPR- Cas9 in the future generations of wheat by crossing the wheat plants with the gene targeting materials with wheat lines expressing the CRISPR-Cas9 materials. This is an exciting demonstration of gene editing activity in wheat that is passed down from one generation to the next and may provide a useful tool for improving wheat in the future. Editor’s note: Read the full study Read full, original post: Super CRISPR Wheat<\|endoftext\|>	3.828125
504	# How do you solve and graph abs((2t+6)/2)>10? Jun 12, 2017 See a solution process below: #### Explanation: The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent. $- 10 > \frac{2 t + 6}{2} > 10$ First, multiply each segment of the system of inequalities by $\textcolor{red}{2}$ to eliminate the fraction while keeping the system balanced: $\textcolor{red}{2} \times - 10 > \textcolor{red}{2} \times \frac{2 t + 6}{2} > \textcolor{red}{2} \times 10$ $- 20 > \cancel{\textcolor{red}{2}} \times \frac{2 t + 6}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} > 20$ $- 20 > 2 t + 6 > 20$ Next, subtract $\textcolor{red}{6}$ from each segment to isolate the $t$ term while keeping the system balanced: $- 20 - \textcolor{red}{6} > 2 t + 6 - \textcolor{red}{6} > 20 - \textcolor{red}{6}$ $- 26 > 2 t + 0 > 14$ $- 26 > 2 t > 14$ Now, divide each segment by $\textcolor{red}{2}$ to solve for $t$ while keeping the system balanced: $- \frac{26}{\textcolor{red}{2}} > \frac{2 t}{\textcolor{red}{2}} > \frac{14}{\textcolor{red}{2}}$ $- 13 > \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} t}{\cancel{\textcolor{red}{2}}} > 7$ $- 13 > t > 7$ Or $t < - 13$ and $t > 7$ Or, in interval notation: $\left(- \infty , - 13\right)$ and $\left(7 , \infty\right)$<\|endoftext\|>	4.9375
815	# How to Calculate Adjusted R-Squared in R The coefficient of determination, or R^2, provides a measure of how well the predictors in a regression model explain the variance in the dependent variable. However, there is a significant caveat to R^2: it will always increase (or remain the same) as more predictors are added to the model, regardless of whether these predictors are truly meaningful. This can make R^2 somewhat misleading in multiple regression settings. To account for this, statisticians often turn to the adjusted R^2. In this article, we’ll dive deep into the concept of the adjusted R^2, understand its significance, and learn how to calculate it in R. ## Overview: 1. The Problem with R squared 3. Mathematical Background 4. Calculating Adjusted R Squared in R 7. Conclusion ## 1. The Problem with R Squared: While R^2 measures the proportion of variance explained by the predictors in a model, it comes with a caveat. Each time a predictor is added to the model, R^2 will either increase or remain the same, even if the predictor doesn’t significantly explain the dependent variable’s variance. This can give a false sense of a model’s goodness of fit. ## 2. Introducing Adjusted R Squared: Adjusted R^2 accounts for the number of predictors in a model, adjusting the R^2 value based on the necessary predictors. It incorporates the model’s degrees of freedom and can decrease if predictors that don’t enhance the model’s fit are included. ## 3. Mathematical Background: The formula for the adjusted R^2 is: Where: • n is the total number of samples. • p is the number of predictors. As you add more predictors, the denominator becomes smaller, and therefore the whole fraction becomes larger, reducing the value of adjusted R^2 if the new predictors aren’t improving the model’s fit. ## 4. Calculating Adjusted R Squared in R: Thankfully, R simplifies the calculation of the adjusted R^2. Using the lm() function to run a regression, you can retrieve the adjusted R^2 from the model’s summary. # Create some sample data x1 <- c(1,2,3,4,5) x2 <- c(5,3,1,3,2) y <- c(2,4,5,4,5) # Fit a multiple linear regression model model <- lm(y ~ x1 + x2) # Extract the adjusted R-squared value summary(model)\$adj.r.squared This code returns the adjusted R^2 value for the model. ## 5. Interpreting Adjusted R Squared: • An adjusted R^2 closer to 1 indicates that a larger proportion of variance is explained by the model, after accounting for the number of predictors. • If adjusted R^2 is significantly lower than R^2, it may suggest that some predictors are not contributing to the model’s explanatory power and might be redundant. • Conversely, if they’re very close, it suggests that most of the predictors in the model are meaningful. • Offers a more realistic value than R^2 in terms of explanatory power, especially in models with many predictors. • Helps in model selection, by favoring models that only include meaningful predictors. Limitations: • Like R^2, adjusted R2R2 doesn’t indicate whether a regression model is appropriate. • Still doesn’t provide information on the effect size of each predictor. ## 7. Conclusion: While R^2 is a commonly-used statistic to evaluate the fit of a regression model, it has its limitations, especially when dealing with multiple predictors. Adjusted R^2 provides a more nuanced view of the model’s fit by accounting for the number of predictors. In R, this value is readily accessible, making it easy for researchers and data analysts to incorporate it into their model evaluation processes. Posted in RTagged<\|endoftext\|>	4.46875
2,278	# Simultaneous Equations 43 % 57 % Published on July 19, 2008 Author: MoreThanMaths Source: slideshare.net ## Description Presentation explaining what simultaneous equations are, how to use graphs to solve them and the elimination method of sloving. Simultaneous Equations 2x + 3y = 12 x + y = 5 Aims for this topic: You will know what simultaneous equations are You will be able to solve simultaneous equations using graphs You will be able to solve simultaneous equations using an algebraic method You will know what simultaneous equations are You will be able to solve simultaneous equations using graphs You will be able to solve simultaneous equations using an algebraic method What are simultaneous equations? If we have an equation like this with just one letter representing an unknown number, we can solve it. What number does x stand for? Obviously x = 6! x + 2 = 8 If we have an equation like this with just one letter representing an unknown number, we can solve it. What number does x stand for? Obviously x = 6! What are simultaneous equations? If we have two letters there are lots of possible solutions x could be 1 and y could be 4 Or x could be 2 and y could be 3 How many other possibilities can you think of? x + y = 5 If we have two letters there are lots of possible solutions x could be 1 and y could be 4 Or x could be 2 and y could be 3 How many other possibilities can you think of? What are simultaneous equations? Suppose we have two equations We know there are lots of possible pairs of values for x and y that fit the first equation One of these pairs of values also fits the second equation x + y = 5 2x + 3y = 12 Suppose we have two equations We know there are lots of possible pairs of values for x and y that fit the first equation One of these pairs of values also fits the second equation What are simultaneous equations? If x=3 and y=2 both equations are true If you are asked to solve simultaneous equations , you are being asked to find the values for x and y that fit both the equations . x + y = 5 2x + 3y = 12 If x=3 and y=2 both equations are true If you are asked to solve simultaneous equations , you are being asked to find the values for x and y that fit both the equations . How do you solve simultaneous equations? There are two main methods: Using graphs Using algebra There are two main methods: Using graphs Using algebra Solving simultaneous equations using graphs There are just three easy steps: Do a table of values for each equation Draw the two graphs Write down the x and y values where the graphs cross There are just three easy steps: Do a table of values for each equation Draw the two graphs Write down the x and y values where the graphs cross Solving simultaneous equations using graphs Here is an example: Solve simultaneously y = 3 - x x + 2y = 4 Here is an example: Solve simultaneously y = 3 - x x + 2y = 4 1. Do a table of values for each equation: y = 3 -x x + 2y = 4 0.5 1 1.5 2 2.5 3 3.5 y 3 2 1 0 -1 -2 -3 X 0 1 2 3 4 5 6 y 3 2 1 0 -1 -2 -3 X y = 3 -x 2. Draw the two graphs: 3. Write down the x and y values where the graphs cross : The lines cross where x = 2 and y =1 Solving simultaneous equations algebraically There are several methods for solving simultaneous equations using algebra We will use a method called elimination ELIMINATE! There are several methods for solving simultaneous equations using algebra We will use a method called elimination Solving simultaneous equations by elimination There are six steps: O rganise your equations M ake sure two coefficients are equal E liminate a letter S olve the equation S ubstitute … and finally check your answer! There are six steps: M ake sure two coefficients are equal E liminate a letter S olve the equation S ubstitute Solving simultaneous equations by elimination It sounds complicated, but just remember: N O MESS and check your answer … easy… lets try an example… It sounds complicated, but just remember: N O MESS … easy… lets try an example… 1. Organise your equations For example, look at this question: Solve 2a = 16 – 5b a + b = 5 This is much easier if we rearrange the first equation: 2a + 5b = 16 a + b = 5 For example, look at this question: Solve 2a = 16 – 5b a + b = 5 This is much easier if we rearrange the first equation: 2a + 5b = 16 a + b = 5 2. Make Sure Two Coefficients are Equal If we are going to eliminate a letter from our equations, we need to make the number of either a’s or b’s the same in both the equations: 2a + 5b = 16 a + b = 5 If we multiply the second equation by 2, we get: 2a + 5b = 16 2a + 2b = 10 If we are going to eliminate a letter from our equations, we need to make the number of either a’s or b’s the same in both the equations: 2a + 5b = 16 a + b = 5 If we multiply the second equation by 2, we get: 2a + 5b = 16 2a + 2b = 10 3. Eliminate a Letter Now we can get rid of a! Just subtract the second equation from the first: 2a + 5b = 16 2a + 2b = 10 _________ 3b = 6 (Sometimes you will need to add instead of subtracting – a reminder about this will appear later!) Now we can get rid of a! Just subtract the second equation from the first: 2a + 5b = 16 2a + 2b = 10 _________ 3b = 6 4. Solve the equation This is the easy part! We worked out that 3b = 6 So b = 2 This is the easy part! We worked out that 3b = 6 So b = 2 5. Substitute Now we know that b = 2, substitute this into one of our original equations: a + b = 5 a + 2 = 5 a = 3 Now we have solved the equations 2a = 16 – 5b and a + b = 5 , our final solutions are: a = 3 and b = 2 Now we know that b = 2, substitute this into one of our original equations: a + b = 5 a + 2 = 5 a = 3 Now we have solved the equations 2a = 16 – 5b and a + b = 5 , our final solutions are: a = 3 and b = 2 6. Don’t forget to check your answer! All you have to do is substitute your answers into the other equation to make sure it works out correctly If it does, you know your answer is correct If it doesn’t, then you’ve gone wrong somewhere and sadly, you’ll have to do it all again… All you have to do is substitute your answers into the other equation to make sure it works out correctly If it doesn’t, then you’ve gone wrong somewhere and sadly, you’ll have to do it all again… Just checking: Our final solutions were: a = 3 and b = 2 Substitute these into 2 a = 12 – 3 b 2 x 3 = 12 – (3 x 2 ) Both sides equal 6, so it works! Our final solutions were: a = 3 and b = 2 Substitute these into 2 a = 12 – 3 b 2 x 3 = 12 – (3 x 2 ) Both sides equal 6, so it works! Remember the six steps for the elimination method: O rganise your equations M ake two coefficients the same E liminate a letter S ame signs -> S ubtract the equations A lternate signs -> A dd the equations (but ONLY use this rule when deciding whether to add or subtract simultaneous equations!) S olve the equation S ubstitute … and finally check your answer! M ake two coefficients the same E liminate a letter S ame signs -> S ubtract the equations A lternate signs -> A dd the equations (but ONLY use this rule when deciding whether to add or subtract simultaneous equations!) S olve the equation S ubstitute User name: Comment: ## Related pages ### Simultaneous Equations - Mathematics GCSE Revision Simultaneous equations, Maths GCSE revision looking at simultaneous equations and Linear equations. Solving them using elimination and substitution. ### Simultaneous equations - Wikipedia, the free encyclopedia In mathematics, a set of simultaneous equations, also known as a system of equations, is a finite set of equations for which common solutions are sought. ### BBC - GCSE Bitesize: Simultaneous equations Simultaneous equations. Simultaneous equations are two equations with two unknowns. They are called simultaneous because they must both be solved at the ... ### Simultaneous equations - BBC Simultaneous equations. Simultaneous equations are two equations with two unknowns. They are called simultaneous because they must both be solved at the ... ### Simultaneous equations model - Wikipedia, the free ... Simultaneous equation models are a form of statistical model in the form of a set of linear simultaneous equations. They are often used in econometrics. ### Simultaneous linear equations - A complete course in algebra How to solve simultaneous equations. The method of addition. The method of substitution. Cramer's rule: The method of determinants. ### Simultaneous Equations - mathsteacher.com.au Definition of linear simultaneous equations, substitution method and graphical method. ### Solving simultaneous equations - StudyMaths.co.uk Revision notes explaining how to solve linear simultaneous equations. Example questions given with full solutions and an opportunity to practise your skills.<\|endoftext\|>	4.75
216	Q: # Jamie needs to determine the distance across a river. She selects a rock, R, on the opposite river bank. She finds a tree, T, on her side of the river such that is perpendicular to the river. Next, Jamie walks 50 yards down the river from the tree, T, to a spot, C. Using a compass that she made from two sticks, she measures . She then walks from C in a direction that makes the same angle as with the river bank on her side of the river. She keeps walking in that direction until she sees the tree and the rock in a straight line from her position. She marks this point B and then paces from B to T to measure the distance from point B to the tree. If BT turns out to be 75 yards, what is RT, the distance across the river? Accepted Solution A: RT = BT = 75 This is because she used the same angle from point C to mark point B, thus creating a symmetric right triangle. Triangle BCT = Triangle RCT<\|endoftext\|>	4.375
896	Birds of a different feather can flock together when members of one species come to recognize individuals from another, says a pioneering study from the University of Nebraska-Lincoln and University of Chicago. The study, which yielded the first experimental evidence of recognition across non-primate species, took place in an arid region of southern Australia that is home to both variegated and splendid fairy-wrens. Researchers recorded the songs of both species from fairy-wrens residing within the shared territory, just outside that territory or well beyond it: co-residents, neighbors and outsiders, respectively. When the team played those recordings to dominant males of the other species — a variegated song to a splendid fairy-wren, or vice versa — it found that the fairy-wrens tolerated the songs of co-residents but responded with immediate aggression to the songs of neighbors and outsiders. The finding suggests that members of the two species, which do not migrate and may live alongside each other for years, can distinguish long-term companions from passersby or potential trespassers. By contrast, ecologists have generally assumed that even beneficial interactions between species occur without any regard for the individuals taking part in them. “When people are considering or studying these mixed-species associations, it may be important to consider that there is an individual component to the relationships,” said Allison Johnson, the study’s lead author and postdoctoral researcher in biological sciences at Nebraska. “It’s not just: Species A benefits from the presence of Species B, or Species A is competitive with Species B. “There are reasons to recognize (individuals) within species: kinship, reciprocal altruism, competition. It makes sense that recognition would transfer to another species if you have those same sorts of territorial or cooperative interactions.” The researchers proposed that differentiating among members of another species, and forming long-term partnerships with those known members, may have arisen because it makes the exchange of relevant information more efficient and reliable. That, in turn, could facilitate the cooperative behavior observed in many ecosystems — jointly mobbing predators or mutual competitors, for instance. And it could help prevent each species from wasting time and energy competing with a potential ally. Johnson and her colleagues likewise witnessed examples of cooperation between the variegated and splendid fairy-wrens, with members of both species traveling and foraging for food together. The team further discovered that the variegated fairy-wrens, which tend to live in larger groups, spent less time watching for predators and had more success rearing their chicks when familiar splendid fairy-wrens were nearby. “It’s almost as though they’re treating (those) splendids like another group member,” said Johnson, who authored the study with the University of Chicago’s Stephen Pruett-Jones and Christina Masco. Johnson said the findings could have implications for the conservation of species that associate with others in their territories, especially given that there is “plenty of evidence that the presence of one species can facilitate the presence of another.” One question raised by Johnson: If such a species undergoes substantial turnover in its population, could that turnover — and the eventual absence of familiar faces — affect the other species? “From year to year, birds might come to a site that’s perfectly viable,” Johnson said. “But maybe Bob isn’t there, so they think, ‘I don’t know about this,’ and they go someplace else. If there are long-term interactions between specific individuals, that might have an impact on decision-making in terms of where they choose to breed and whether or not they return to the same site. “It would be interesting to see if that’s true for migratory species, as well. Migration tends to re-sort individuals from year to year, so they may be willing to accept associating with whoever is there that year, but we suspect that more cross-species recognition is occurring than has been realized.” The team’s study appeared in the journal Behavioral Ecology. Johnson, Pruett-Jones and Masco received funding in part from the National Science Foundation, the American Ornithological Society and the Norman Wettenhall Foundation.<\|endoftext\|>	4.09375
612	Differentiated instruction allows students with different abilities to advance academically at roughly the same pace. While differentiated instruction requires more preparation on the teacher’s part, if done correctly, it can also ease the challenges of keeping an entire classroom of differently abled students engaged in a single activity. The most basic way of applying differentiated instruction is for a teacher to modify her instruction to appeal to different kinds of learners. For example, a basic lecture can be tailored to appeal to tactile or hands-on learners by incorporating some simple movement or manipulatives as the lecture progresses. The lecturer might toss a soft foam ball around the room to different students who might have questions about the material. The lecturer might also incorporate visual aids such as a computer presentation or video to aid visual learners. And for students that require additional guidance following a lecture, the teacher may distribute his lecture notes, or even a question and answer handout that students can complete as the lecture progresses. When generating differentiation for class activities such as lectures or discussions, teachers include a wide variety of formative assessment strategies to gauge how students are progressing with the material. These activities can range from brief show-of-hand survey questions about the material to more complicated individual reflection assignments such as journaling or one-on-one teacher conferences. Following such brief assessments, teachers adjust their class activities by either progressing forward to more advanced material or taking additional time to review previous material. Occasionally it is possible for a teacher to move some students ahead to more advanced material, while allowing other students to spend additional time reviewing previous material. At the end of a particular unit, a teacher can make minor adjustments to his summative test in an effort to challenge advanced students and provide some additional help for remedial students. Teachers alter some questions by having them engage more advanced cognitive skills for some and less advanced cognitive skills for others. For example, some students might be asked an essay question that requires them to analyze and synthesize two different passages, while other students might be asked an essay question that requires them summarize and analyze only one passage. Additionally, for students with specifically identified learning disabilities or impairments such as dyslexia, teachers can alter the visual dynamics of a test itself to cut down on the visual “clutter” that can sometimes confuse students as they take tests. Following both formative and summative assessments, teachers can elect to group students together for class activities and out-of-class assignments based on students’ abilities. While some teachers elect to form student groups in which each member achieves at roughly the same level, other teachers form student groups in which some members are advanced, while others require additional assistance. The idea behind the former grouping is to allow advanced students to work independently on more complicated activities and assignments, while giving less advanced students an opportunity to work more closely with the teacher. The idea behind the latter groups is to allow advanced students to acquire enrichment by acting as student instructors for the less advanced students requiring remediation. - Digital Vision./Digital Vision/Getty Images<\|endoftext\|>	4.3125
1,857	# Video: Finding the Maximum Error Bound When Approximating a Series by Summing Its First 20 Terms Find the maximum error bound when approximating the series β_(π = 1) ^(β) ((β1)^(π)) β((3π + 7)/(πΒ² + 1) by summing the first 20 terms. Round your answer to 5 decimal places. 05:22 ### Video Transcript Find the maximum error bound when approximating the series the sum from π equals one to β of negative one to the πth power multiplied by the square root of three π plus seven divided by π squared plus one by summing the first 20 terms. Round your answer to five decimal places. The question wants us to find the maximum possible error we could get by approximating this series by summing the first 20 terms. It wants us to round this value to five decimal places. The question wants us to approximate this series by summing the first 20 terms. Thatβs the same as taking the 20th partial sum. And the maximum possible error bound would be a bound on the absolute value of π minus our 20th partial sum, where π would be the value that our series converges to. To help us find this bound, we know that if π π is a positive and decreasing sequence where the limit as π approaches β of π π is equal to zero, then, by the alternating series test, the alternating series the sum from π equals one to β of negative one to the πth power times π π converges β weβll call this value π. Then, we can bound the error between the value of π and our πth partial sum by using the absolute value of π minus the πth partial sum is less than or equal to π π plus one, the absolute value of the first neglected term. By looking at the series given to us in the question, weβll want to set π π equal to the square root of three π plus seven divided by π squared plus one. If we can then show that π π is a positive decreasing sequence whose limit as π approaches β is equal to zero, then, by setting π equal to 20, weβll have the absolute value of π minus the 20th partial sum is less than or equal to π 21. Letβs start by showing that π π is positive. We know that π is greater than or equal to one. So, three π plus seven is positive, and then π squared plus one is positive. And weβre taking the positive square root. So, our sequence π π is positive for all values of π since weβre just taking the positive square root of a positive number. To check that the sequence is decreasing, weβll set π of π₯ equal to the square root of three π₯ plus seven divided by π₯ squared plus one. We know that this sequence will be decreasing if the slope of this function is negative. To help us differentiate this function, weβll start by setting π’ equal to three π₯ plus seven divided by π₯ squared plus one and then using the chain rule. Since π is a function of π’ and π’ is a function of π₯, by the chain rule we have π prime of π₯ is equal to the derivative of the square root of π’ with respect to π’, which we can evaluate by using the power rule for differentiation. This gives us a half multiplied by π’ to the power of negative a half. And then, we need to multiply this by dπ’ by dπ₯. Remember, we only need to calculate whether the slope is negative or positive. Since weβre only interested in the values of π₯ where π₯ is greater than or equal to one, we can see that π’ as positive; itβs the quotient of two positive numbers. So, a half is positive, and π’ to the power of negative a half is also positive. Itβs one divided by the positive square root of a positive number. So, to decide whether the slope is positive or negative, we only need to calculate dπ’ by dπ₯. To find dπ’ by dπ₯, weβll use the quotient rule. Weβll set π£ equal to the numerator, three π₯ plus seven, and π€ equal to the denominator, π₯ squared plus one. This gives us π£ prime is three and π€ prime is two π₯. The quotient rule then tells us dπ’ by dπ₯ is equal to π£ prime π€ minus π£π€ prime all divided by π€ squared. This gives us three times π₯ squared plus one minus three π₯ plus seven times two π₯ all divided by π₯ squared plus one all squared. We can evaluate the numerator to get three π₯ squared plus three minus six π₯ squared plus 14π₯, which we can simplify to give us negative three π₯ squared minus 14π₯ plus three all divided by π₯ squared plus one squared. And we see, for values of π₯ greater than or equal to one, π₯ squared plus one all squared is positive. However, negative three π₯ squared minus 14π₯ plus three is negative. So, dπ’ by the dπ₯ is negative for these values of π₯. This means that our slope, π prime of π₯, is a positive number multiplied by a negative number, which means π prime of π₯ is negative. And if our slope is negative for these values of π₯, our sequence must be decreasing. We now want to check the limit as π approaches β of π π is equal to zero. We see that π π is the square root of a fraction. Weβll divide both the numerator and the denominator of this fraction by the highest power of π which appears in the fraction. Thatβs π squared. Dividing through by π squared gives us the limit as π approaches β of the square root of three over π plus seven over π squared all divided by one plus one over π squared. We can see as π is approaching β, our numerator of three over π plus seven over π squared is approaching zero. And we can see the one over π squared term in our denominator is also approaching zero. However, the term one remains constant. So, this fraction is approaching zero divided by one; itβs approaching zero. And since the limit of a power is equal to the power of the limit, this means that this limit is approaching zero. So, weβve shown as π approaches β, π π approaches zero. This means the maximum error bound when summing the first 20 terms of our series is just π 21. So, if we were to approximate this series by summing the first 20 terms, our error would be at most π 21, which is equal to the square root of three times 21 plus seven divided by 21 squared plus one. And if we calculate this to five decimal places, we get 0.39796.<\|endoftext\|>	4.375
1,200	Area expansion – problems and solutions 1. At 20 oC, the length of a sheet of steel is 50 cm and the width is 30 cm. If the coefficient of linear expansion for steel is 10-5 oC-1, determine the change in area and the final area at 60 oC. Known : The initial temperature (T1) = 20oC The final temperature (T2) = 60oC The change in temperature (ΔT) = 60oC – 20oC = 40oC The initial area (A1) = length x width = 50 cm x 30 cm = 1500 cm2 The coefficient of linear expansion for steel (α) = 10-5 oC-1 The coefficient of area expansion for steel (β) = 2α = 2 x 10-5 oC-1 Wanted : The change in area (ΔA) Solution : The change in area (ΔA) : ΔA = β A1 ΔT ΔA = (2 x 10-5 oC-1)(1500 cm2)(40oC) ΔA = (80 x 10-5)(1500 cm2) ΔA = 120,000 x 10-5 cm2 ΔA = 1.2 x 105 x 10-5 cm2 ΔA = 1.2 cm2 The final area (A2) : A2 = A1 + ΔA A2 = 1500 cm2 + 1.2 cm2 A2 = 1501.2 cm2 Read : Applications of Bernoulli's principle 2. At 30 oC, the area of a sheet of aluminum is 40 cm2 and the coefficient of linear expansion is 24 x 10-6 /oC. Determine the final temperature if the final area is 40.2 cm2. Known : The initial temperature (T1) = 30oC The coefficient of linear expansion (α) = 24 x 10-6 oC-1 The coefficient of area expansion (β) = 2a = 2 x 24 x 10-6 oC-1 = 48 x 10-6 oC-1 The initial area (A1) = 40 cm2 The final area (A2) = 40.2 cm2 The change in area (ΔA) = 40.2 cm2 – 40 cm2 = 0.2 cm2 Wanted : Determine the final temperature (T2) Solution : Formula of the change in area (ΔA) : ΔA = β A1 ΔT The final temperature (T2) : ΔA β A1 (T2 – T1) 0.2 cm2 = (48 x 10-6 oC-1)(40 cm2)(T230oC) 0.2 = (1920 x 10-6)(T230) 0.2 = (1.920 x 10-3)(T2 – 30) 0.2 = (2 x 10-3)(T2 – 30) 0.2 / (2 x 10-3) = T2 – 30 0.1 x 103 = T2 – 30 1 x 102 = T2 – 30 100 = T2 – 30 100 + 30 = T2 T2 = 130 The final temperature = 130oC Read : Determine horizontal displacement of projectile motion 3. The radius of a ring at 20 oC is 20 cm. If the final radius at 100 oC is 20.5 cm, determine the coefficient of area expansion and the coefficient of linear expansion… Known : The initial temperature (T1) = 30oC The final temperature (T2) = 100oC The change in temperature (ΔT) = 100oC – 30oC = 70oC The initial radius (r1) = 20 cm The final radius (r2) = 20.5 cm Wanted : The coefficient of area expansion (β) Solution : The initial area (A1) = π r12 = (3.14)(20 cm)2 = (3.14)(400 cm2) = 1256 cm2 The final area (A2) = π r22 = (3.14)(20.5 cm)2 = (3.14)(420.25 cm2) = 1319.585 cm2 The change in area (ΔA) = 1319.585 cm2 1256 cm2 = 63.585 cm2 Formula of the change in area (ΔA) : ΔA = β A1 ΔT The coefficient of area expansion : ΔA = β A1 ΔT 63.585 cm2 = b (1256 cm2)(70 oC) 63.585 = b (87,920 oC) β = 63.585 / 87,920 oC β = 0.00072 /oC β = 7.2 x 10-4 /oC β = 7.2 x 10-4 oC-1 The coefficient of linear expansion (α) : β = 2 α α = β / 2 α = (7.2 x 10-4) / 2 α = 3.6 x 10-4 oC-1 [wpdm_package id=’698′]<\|endoftext\|>	4.46875
3,163	Week 2-3 - Functions & Limits 1. Functions • Deﬁnition: A function f is a rule that assigns to each element x in a set D exactly one element, called $f(x)$, in a set E. • The set D is called the domain of the function. • The number $f(x)$ is the value of f at x and is read "f of x". • The range of f is the set of all possible values of $f(x)$ as x varies throughout the domain. • A symbol that represents an arbitrary number in the domain of a function f is called an independent variable. A symbol that represents a number in the range of f is called a dependent variable. • Four ways to represent a functions • verbally (by a description in words) • numerically (by a table of values) • visually (by a graph) • algebraically (by an explicit formula) • A catalog of essential functions • linear function • $f(x) = mx+b$ • polynomials • $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{2}x^{2}+a_{1}x^{1}+a_{0}$ • The domain of any polynomial is $\mathbb{R} = (-\infty, + \infty)$. • If the leading coefficient $a_{n}\ne 0$, then the degree of the polynomial is n. • n=2: quadratic function • n=3: cubic function • power function • $f(x) = x^a$ • a = n, where n is a positive integer • a = 1/n, where n is a positive integer. It's a root function. • a = -1: reciprocal function • rational function • A rational function f is a ratio of two polynomials: • $f(x)=\frac{P(x)}{Q(x)}$, $\{Q(x)\ne0\}$ • $f(x)=\frac{2x^{4}-x^{2}+1}{x^{2}-4}$, $\{x\|x \ne \pm 2\}$: • algebraic function • A function f is called an algebraic function if it can be constructed using algebraic operations (such as addition, subtraction, multiplication, division, and taking roots) • trigonometric function • $f(x)=\sin{x}$ • exponential function • $f(x)=b^x$ • law of exponential function: • $b^{x+y}=b^x+b^y$ • $b^{x-y}=\frac{b^x}{b^y}$ • $(b^{x})^{y}=b^{xy}$ • ${ab}^{x}=a^{x}b^{x}$ • logarithmic function • $f(x)=\log_{b}{x}$ 1.1. Combinations of Functions • $(f+g)(x)=f(x)+g(x)$ • $(f-g)(x)=f(x)-g(x)$ • $(fg)(x)=f(x)g(x)$, the domain of fg is $A \cap B$ • $(\frac{f}{g})(x)=\frac{f(x)}{g(x)}$, the domain of f/g is $\{x \in A \cap B\ \|\ g(x) \ne 0\}$. • $(f \circ g)(x) = f(g(x))$ • composition (or composite) of f and g, denoted by $f \circ g$ (“f circle g”). 1.2. Inverse of Functions • Definition: • If a function maps every input to exactly one output, an inverse of that function maps every “output” to exactly one “input.” • denoted by$f^{-1}$ , and read “f inverse”. • to function $N=f(t)$, the inverse function will be $t=f^{-1}(N)$. one-to-one functions • A function is one-to-one if for every value in the range(f(x)), there is exactly one value in the domain(x). • domain of $f^{-1}$ = range of $f$ • range of $f^{-1}$ = domain of $f$ • for example: $f(x)=x^3$ is a one-to-one function, $f(x)=x^2$ is not. 2. Limits • Definition: • $\displaystyle\lim_{x \to a}{f(x)}=L$ • the limit of f(x), as x approaches a, equals L • if we can make the values of f(x) arbitrarily close to L (as close to L as we like) by restricting x to be sufficiently close to a (on either side of a) but NOT equal to a.(This means that in finding the limit of f(x) as x approaches a, we never consider x = a.) 2.1. Limit Laws • Suppose that c is a constant and the limits $\displaystyle\lim_{x \to a}{f(x)}$ and $\displaystyle\lim_{x \to a}{g(x)}$ exist, Then: 1. $\displaystyle\lim_{x \to a}{[f(x)+g(x)]}=\lim_{x \to a}{f(x)}+\lim_{x \to a}{g(x)}$ 2. $\displaystyle\lim_{x \to a}{[f(x)-g(x)]}=\lim_{x \to a}{f(x)}-\lim_{x \to a}{g(x)}$ 3. $\displaystyle\lim_{x \to a}{cf(x)}=c\lim_{x \to a}{f(x)}$ 4. $\displaystyle\lim_{x \to a}{[f(x)g(x)]}=\lim_{x \to a}{f(x)}\cdot \lim_{x \to a}{g(x)}$ 5. $\displaystyle\lim_{x \to a}{\frac{f(x)}{g(x)}}=\frac{\displaystyle\lim_{x \to a}{f(x)}}{\displaystyle\lim_{x \to a}{g(x)}}$, if $\displaystyle\lim_{x \to a}{g(x)} \ne 0$ • These five laws can be stated verbally as follows: 1. Sum Law : The limit of a sum is the sum of the limits. 2. Difference Law : The limit of a difference is the difference of the limits. 3. Constant Multiple Law : The limit of a constant times a function is the constant times the limit of the function. 4. Product Law : The limit of a product is the product of the limits. 5. Quotient Law : The limit of a quotient is the quotient of the limits (provided that the limit of the denominator is not 0). 2.2. Squeeze Theorem • if $g(x) \le f(x) \le h(x)$ and $\displaystyle\lim_{x\to{a}}g(x) = \lim_{x\to{a}}h(x) = L$, then $\displaystyle\lim_{x\to{a}}f(x)=L$ • Sample: to prove $\displaystyle\lim_{x\to{0}}\frac{\sin(x)}{x}=1$ • we know: • $\displaystyle\lim_{x\to{0}}\cos(x)=1=\lim_{x\to{0}}1$ • $\cos(x) \le \frac{\sin(x)}{x} \le 1$ • then: • $\displaystyle\lim_{x\to{0}}\cos(x) \le \lim_{x\to{0}}\frac{\sin(x)}{x} \le 1$ • so: • $\displaystyle\lim_{x\to{0}}\frac{\sin(x)}{x}=1$ 2.3. Continuity • Definition 1: • A function f is continuous at a number a, if $\displaystyle\lim_{x \to a}f(x)=f(a)$ • Notice that this Definition implicitly requires three things if f is continuous at a: 1. f(a) is defined (that is, a is in the domain of f) 2. $\displaystyle\lim_{x \to a}f(x)$ exists 3. $\displaystyle\lim_{x \to a}f(x) = f(a)$ • Definition 2: • A function f is continuous from the right at a number a if $\displaystyle\lim_{x \to a^{+}}f(x) = f(a)$, and f is continuous from the left at a if $\displaystyle\lim_{x \to a^{-}}f(x) = f(a)$ • Definition 3: • A function f is continuous on the interval (a, b), if for all points c so that a < c < b, f(x) is continuous at c. • close intervals: • To say "f(x) is continuous on the interval [a, b]", means: • f(x) is continuous on the interval (a, b) • $\displaystyle\lim_{x \to a^{+}}f(x) = f(a)$ • $\displaystyle\lim_{x \to b^{-}}f(x) = f(b)$ The Intermediate Value Theorem • Suppose that f is continuous on the closed interval [a, b] and let N be any number between f(a) and f(b), where $f(a) \ne f(b)$. Then there exists a number c in [a, b] such that $f(c) = N$. • For example: how to approximate root two? • use function $f(x)=x^2-2$ • when $x=\sqrt{2}$, this function equals 0, • so i just need to look for a positive value, that I can plug into this function to make it equals zero. • We know that f(1)=-1<0 and f(2)=2>0. So base on The Intermediate Value Theorem, there must be a value in domain (1, 2) that exist c such that f(c) = 0. • to continue calculating f(1.5)=0.25>0, we got $c \in (1, 1.5)$, then $c \in (1.4, 1.5)$ ... then we are getting closer and closer to $\sqrt{2}$. Fixed Point • Definition • A fixed point of a function f is a number c in its domain such that f(c) = c. (The function doesn’t move c; it stays fixed.) • if f(x) continuous on [0,1], and $0 \le f(x) \le 1$, then there is an x in domain [0, 1], exist f(x) = x. • To Prove: • Assumption: g(x) = f(x) - x, so g(x) is continuous • g(0) = f(0) - 0 >= 0 • g(1) = f(1) - 1 <= 1 • Base on the IVT(Intermediate Value Theorem), there must be an x such that g(x) = 0, which is f(x) = x. 2.4. Infinity • Definition • $\displaystyle\lim_{x \to a}f(x) = \infty$ means that f(x) is as large as you like provides x is close enough to a. • $\displaystyle\lim_{x \to \pi/2}\tan(x) = \infty$ • $\displaystyle\lim_{x \to \infty}f(x) = L$ means that f(x) is close enough to L provided x is large enough. • $\displaystyle\lim_{x \to \infty}\tan^{-1}(x) = \frac{\pi}{2}$ • if $\displaystyle\lim_{x \to \infty}f(x) = L$ or $\displaystyle\lim_{x \to -\infty}f(x) = L$, then the line y = L is called a horizontal asymptote of the curve y = f(x) : • Potential Infinity vs Actual Infinity (from Wikipedia) • Actual Infinity is the idea that numbers, or some other type of mathematical object, can form an actual, completed totality; • Such as the set of all natural numbers, an infinite sequence of rational numbers. • Potential Infinity is a non-terminating process (such as "add 1 to the previous number") produces an unending "infinite" sequence of results, but each individual result is finite and is achieved in a finite number of steps. • Precise Definitions • $\displaystyle\lim_{x \to a}f(x) = L$ means: • for all $\epsilon > 0$, there is $\delta>0$, • so that if $0 < \|x - a\| < \delta$ ($x \ne a$ and x in within $\delta$ of a), then $\|f(x)-L\| < \epsilon$ (f(x) is within $\epsilon$ of L). • \|x - a\| is the distance from x to a and \|f(x) - L\| is the distance from f(x) to L. • so $\displaystyle\lim_{x \to a}f(x) = L$ means that the distance between f(x) and L can be made arbitrarily small by requiring that the distance from x to a be sufficiently small (but not 0). • For Example: $\displaystyle\lim_{x \to 2}x^2 = 4$ • Let's say $\epsilon = 0.1$, that means $\|f(x)-4\| < 0.1$, 3.9 < f(x) < 4.1, • Base on the definition, there should be a $\delta$, that $2 - \delta < x < 2 + \delta$ to satisfy the demand. • Try $\delta = 0.01$. We got 1.99 < x < 2.01, 3.9601 < x^2 < 4.0401, 3.9 < f(x) < 4.1 which suit the demand. • Another Example: $\displaystyle\lim_{x \to 10}2x = 20$ • Let $\epsilon > 0, \delta = \epsilon / 2$ • if $0<\|x-10\|<\delta$, then, • $0<2\|x-10\|<2\delta=\epsilon$, and so, • $0<\|2x-20\|<\epsilon$<\|endoftext\|>	4.59375
883	Harold Kessler ## Answered question 2021-12-27 Let $f\left(x\right)=\frac{{\left(x+8\right)}^{2}}{{x}^{2}-64}$. Find a) $\underset{x⇒-8}{lim}f\left(x\right)$, b) $\underset{x⇒0}{lim}f\left(x\right)$, and c) $\underset{x⇒8}{lim}f\left(x\right)$. ### Answer & Explanation hysgubwyri3 Beginner2021-12-28Added 43 answers Step 1 a) Plug the given expression for $f\left(x\right)$, then factor the denominator and simplify. $\underset{x⇒-8}{lim}f\left(x\right)$ $=\underset{x⇒-8}{lim}\frac{{\left(x+8\right)}^{2}}{{x}^{2}-64}$ $=\underset{x⇒-8}{lim}\frac{{\left(x+8\right)}^{2}}{{x}^{2}-{8}^{2}}$ $=\underset{x⇒-8}{lim}\frac{{\left(x+8\right)}^{2}}{\left(x+8\right)\left(x-8\right)}$ $=\underset{x⇒-8}{lim}\frac{\left(x+8\right)}{\left(x-8\right)}$ Step 2 Plug $x=-8$ for the limit $=\underset{x⇒-8}{lim}\frac{\left(x+8\right)}{\left(x-8\right)}$ $=\frac{-8+8}{-8-8}$ $=\frac{0}{-16}$ $=0$ eninsala06 Beginner2021-12-29Added 37 answers Step 3 b) Plug expression of $f\left(x\right)$ in the limit then simplify and plug $x=0$ $\underset{x⇒0}{lim}f\left(x\right)$ $=\underset{x⇒0}{lim}\frac{{\left(x+8\right)}^{2}}{{x}^{2}-64}$ $=\underset{x⇒0}{lim}\frac{{\left(x+8\right)}^{2}}{{x}^{2}-{8}^{2}}$ $=\underset{x⇒0}{lim}\frac{{\left(x+8\right)}^{2}}{\left(x+8\right)\left(x-8\right)}$ $=\underset{x⇒0}{lim}\frac{\left(x+8\right)}{\left(x-8\right)}$ $=\frac{0+8}{0-8}$ $=-1$ karton Expert2022-01-04Added 613 answers Step 4 c) We simplify the limit same way and plug x=8 this time. $\begin{array}{}\\ \underset{x⇒8}{lim}f\left(x\right)\\ =\underset{x⇒8}{lim}\frac{\left(x+8{\right)}^{2}}{{x}^{2}-64}\\ =\underset{x⇒8}{lim}\frac{\left(x+8{\right)}^{2}}{{x}^{2}-{8}^{2}}\\ =\underset{x⇒8}{lim}\frac{\left(x+8{\right)}^{2}}{\left(x+8\right)\left(x-8\right)}\\ =\underset{x⇒8}{lim}\frac{\left(x+8\right)}{\left(x-8\right)}\\ =\frac{8+8}{8-8}\\ =\frac{16}{0}\\ =DNE\end{array}$ Do you have a similar question? Recalculate according to your conditions! Ask your question. Get an expert answer. Let our experts help you. Answer in as fast as 15 minutes. Didn't find what you were looking for?<\|endoftext\|>	4.5
1,551	There is a poem, written around 598 AD, which describes hunting a mystery animal called a llewyn. But what was it? Nothing seemed to fit, until 2006, when an animal bone, dating from around the same period, was found in the Kinsey Cave in northern England. Until this discovery, the lynx — a large spotted cat with tasselled ears — was presumed to have died out in Britain at least 6,000 years ago, before the inhabitants of these islands took up farming. But the 2006 find, together with three others in Yorkshire and Scotland, is compelling evidence that the lynx and the mysterious llewyn were in fact one and the same animal. If this is so, it would bring forward the tassel-eared cat's estimated extinction date by roughly 5,000 years. However, this is not quite the last glimpse of the animal in British culture. A 9th-century stone cross from the Isle of Eigg shows, alongside the deer, boar and aurochs pursued by a mounted hunter, a speckled cat with tasselled ears. Were it not for the animal’s backside having worn away with time, we could have been certain, as the lynx's stubby tail is unmistakable. But even without this key feature, it’s hard to see what else the creature could have been. The lynx is now becoming the totemic animal of a movement that is transforming British environmentalism: rewilding. Rewilding means the mass restoration of damaged ecosystems. It involves letting trees return to places that have been denuded, allowing parts of the seabed to recover from trawling and dredging, permitting rivers to flow freely again. Above all, it means bringing back missing species. One of the most striking findings of modern ecology is that ecosystems without large predators behave in completely different ways from those that retain them. Some of them drive dynamic processes that resonate through the whole food chain, creating niches for hundreds of species that might otherwise struggle to survive. The killers turn out to be bringers of life. Such findings present a big challenge to British conservation, which has often selected arbitrary assemblages of plants and animals and sought, at great effort and expense, to prevent them from changing. It has tried to preserve the living world as if it were a jar of pickles, letting nothing in and nothing out, keeping nature in a state of arrested development. But ecosystems are not merely collections of species; they are also the dynamic and ever-shifting relationships between them. And this dynamism often depends on large predators. At sea the potential is even greater: by protecting large areas from commercial fishing, we could once more see what 18th-century literature describes: vast shoals of fish being chased by fin and sperm whales, within sight of the English shore. This policy would also greatly boost catches in the surrounding seas; the fishing industry's insistence on scouring every inch of seabed, leaving no breeding reserves, could not be more damaging to its own interests. Rewilding is a rare example of an environmental movement in which campaigners articulate what they are for rather than only what they are against. One of the reasons why the enthusiasm for rewilding is spreading so quickly in Britain is that it helps to create a more inspiring vision than the green movement's usual promise of ‘Follow us and the world will be slightly less awful than it would otherwise have been'. The lynx presents no threat to human beings: there is no known instance of one preying on people. It is a specialist predator of roe deer, a species that has exploded in Britain in recent decades, holding back, by intensive browsing, attempts to re-establish forests. It will also winkle out sika deer: an exotic species that is almost impossible for human beings to control, as it hides in impenetrable plantations of young trees. The attempt to reintroduce this predator marries well with the aim of bringing forests back to parts of our bare and barren uplands. The lynx requires deep cover, and as such presents little risk to sheep and other livestock, which are supposed, as a condition of farm subsidies, to be kept out of the woods. On a recent trip to the Cairngorm Mountains, I heard several conservationists suggest that the lynx could be reintroduced there within 20 years. If trees return to the bare hills elsewhere in Britain, the big cats could soon follow. There is nothing extraordinary about these proposals, seen from the perspective of anywhere else in Europe. The lynx has now been reintroduced to the Jura Mountains, the Alps, the Vosges in eastern France and the Harz mountains in Germany, and has re-established itself in many more places. The European population has tripled since 1970 to roughly 10,000. As with wolves, bears, beavers, boar, bison, moose and many other species, the lynx has been able to spread as farming has left the hills and people discover that it is more lucrative to protect charismatic Wildlife than to hunt it, as tourists will pay for the chance to see it. Large-scale rewilding is happening almost everywhere — except Britain. Here, attitudes are just beginning to change. Conservationists are starting to accept that the old preservation-jar model is failing, even on its own terms. Already, projects such as Trees for Life in the Highlands provide a hint of what might be coming. An organisation is being set up that will seek to catalyse the rewilding of land and sea across Britain, its aim being to reintroduce that rarest of species to British ecosystem: hope. Write the correct letter, A, B, C or D, in boxes 14 - 18 on your answer sheet. 14.What did the 2006 discovery of the animal bone reveal about the lynx? 15.What point does the writer make about large predators in the third paragraph? 16.What does the writer suggest about British conservation in the fourth paragraph? 17.Protecting large areas of the sea from commercial fishing would result in 18.According to the author, what distinguishes rewilding from other environmental campaigns? Complete the summary using the list of words and phrases A - F below. Write the correct letter, A - F, in boxes 19 - 22 on your answer sheet. Reintroducing the lynx to Britain There would be many advantages to reintroducing the lynx to Britain. While there is no evidence that the lynx has ever put 19 in danger, it would reduce the numbers of certain 20 whose populations have increased enormously in recent decades. It would present only a minimal threat to 21 , provided these were kept away from lynx habitats. Furthermore, the reintroduction programme would also link efficiently with initiatives to return native 22 to certain areas of the country. Do the following statements agree with the claims of the writer in Reading Passage 2? In boxes 23 - 26 on your answer sheet, write TRUE if the statement agrees with the claims of the writer FALSE if the statement contradicts the claims of the writer NOT GIVEN if it is impossible to say what the writer thinks about this 23.Britain could become the first European country to reintroduce the lynx. 24.The large growth in the European lynx population since 1970 has exceeded conservationists’ expectations. 25.Changes in agricultural practices have extended the habitat of the lynx in Europe. 26.It has become apparent that species reintroduction has commercial advantages.<\|endoftext\|>	3.96875
1,752	# Fraction calculator The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression. ## Result: ### 3/5 + 5/6 = 43/30 = 1 13/30 ≅ 1.4333333 Spelled result in words is forty-three thirtieths (or one and thirteen thirtieths). ### How do you solve fractions step by step? 1. Add: 3/5 + 5/6 = 3 · 6/5 · 6 + 5 · 5/6 · 5 = 18/30 + 25/30 = 18 + 25/30 = 43/30 For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(5, 6) = 30. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 5 × 6 = 30. In the next intermediate step, the fraction result cannot be further simplified by canceling. In words - three fifths plus five sixths = forty-three thirtieths. #### Rules for expressions with fractions: Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part. The slash separates the numerator (number above a fraction line) and denominator (number below). Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2. Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3. Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45. The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3. An asterisk * or × is the symbol for multiplication. Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses. The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2 #### Examples: subtracting fractions: 2/3 - 1/2 multiplying fractions: 7/8 * 3/9 dividing Fractions: 1/2 : 3/4 exponentiation of fraction: 3/5^3 fractional exponents: 16 ^ 1/2 adding fractions and mixed numbers: 8/5 + 6 2/7 dividing integer and fraction: 5 ÷ 1/2 complex fractions: 5/8 : 2 2/3 decimal to fraction: 0.625 Fraction to Decimal: 1/4 Fraction to Percent: 1/8 % comparing fractions: 1/4 2/3 multiplying a fraction by a whole number: 6 * 3/4 square root of a fraction: sqrt(1/16) reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22 expression with brackets: 1/3 * (1/2 - 3 3/8) compound fraction: 3/4 of 5/7 fractions multiple: 2/3 of 3/5 divide to find the quotient: 3/5 ÷ 2/3 The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are: PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction. GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction. Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right. ## Fractions in word problems: • Berry Smoothie Rory has 5/8 cup of milk. How much milk does she have left after she doubles the recipe of the smoothie? Berry Smoothie: 2 cups strawberries 1 cup blueberries 1/4 cup milk 1 tbsp (tablespoon) sugar 1/2 tsp (teaspoon) lemon juice 1/8 tsp (teaspoon) vanilla Why does 1 3/4 + 2 9/10 equal 4.65? How do you solve this? We need two tenths kg of a carrot, one tenth of peas and three tenths of of tomatoes to make salad. Express the fraction of the weight of the vegetables to be salad. Convert the result to grams. • Two pizzas Jacobs mom bought two whole pizzas. He ate 2/10 of the pizza and his dad ate 1 1/5. How much is left. • Walk for exercise Anya, Jose, Cali, and Stephan walk for exercise. Anya's route is 2 1/4 kilometers long. Jose's route is 1 1/2 fewer km. Cali's route is 1 1/2 times as long as Jose's route, and 2 fewer km than Stephan's route. What distance (S) is Stephan's route? • Weigh in total I put 3/5 kg of grapes into a box which is 1/4kg in weight. How many kilograms do the grapes and the box weigh in total? • Math homework It took Jose two-thirds of an hour to complete his math homework on Monday, three-fourths of an hour on Tuesday, any two- fifths of an hour on Wednesday. How many hours did it take Jose to complete his homework altogether? • Interior designer To make draperies an interior designer needs 11 1/4 yards of material for the den and 8 1/2 yards for the living room. If material comes only in 20 yard bolts, how much will be left over after completing both sets of draperies? • Tartlets After a special event, a caterer examined the leftovers on the serving table. She saw 10/11 of a tartlet with apples, 2/3 of a tartlet with strawberries, and 10/11 of a tartlet with raspberries. How many leftover tartlets did the caterer have? • Barrels Peter has 42 barrels. One-sixth of them are filled with lemonade, one third is filled with wine and half is empty. Calculate how many barrels are with each filling. • A shopkeeper 3 A shopkeeper sells 8 1/3 kg, 10 1/4 kg and 11 1/5 kg of apples on 3 consecutive days. What is the total weight of apples sold? • Expressions Let k represent an unknown number, express the following expressions: 1. The sum of the number n and two 2. The quotient of the numbers n and nine 3. Twice the number n 4. The difference between nine and the number n 5. Nine less than the number n • Two numbers 11 The sum of two rational numbers is (-2). If one of them is 3/5, find the other.<\|endoftext\|>	4.5625
841	What WebQuests Are (Really) Tom March This section will help your students move through the webquest. It will also help other teachers to see how your lesson flows and make it easier for them to adapt it for their own uses--so the more detail and care, the better. Remember that this whole document is addressed to the student, however, so describe the steps using the second person. For example,... Students create a brochure, diorama and audio guide for a new exhibit on an exotic animal being introduced to a zoo. Although the above activities may involve some reasonable learning, they are not WebQuests because in each case the information can go from the browser to product without altering (even entering?) the learner’s understanding. What is a (real) WebQuest? In the early days The Learning Power of WebQuests Tom March With your team, click into some of the examples of WebQuest (at least 2-3) so that you can become familiar with the process of creating a WEBQUEST and in order to review and discuss the levels of learning (see step 4 for more clarity) for each lesson reviewed.... The process section explains how learners will go about accomplishing the assigned task, including specific clearly stated steps and resources which may include: websites and organizational tools. As part of the process, you may choose to assign student roles for task completion in cooperative Webquests Teaching with Technology Your mission is to create a "WebQuest" that will be a resource for you and your students. WebQuests are meaningful web-based activities designed to challenge the … how to build a car adrian newey book depository A good WebQuest should have students take information in and transform it, using their own judgment and creative problem-solving techniques. If you are concerned about the time involved in Using WebQuests to Teach English etni.org The purposes of this assignment are for you to: Become familiar with some of the resources on the Internet to help you with designing your own Internet lessons. Learn how to create your own, simple WebQuest. Learn how to facilitate student learning using the WebQuest you designed. Having explored how to create multi-level headings word Learners typically complete WebQuests as cooperative groups. Each learner within a group can be given a "role," or specific area to research. WebQuests may take the form of role-playing scenarios, where students take on the personas of various people or roles within the confines of the WebQuest topic. How long can it take? WebQuests Demonstration thirteen.org - What WebQuests Are (Really) Tom March - HOW TO CREATE AN INTERACTIVE LEARNING EXPERIENCE-WEBQUESTS - Webquests Teaching with Technology - Examples of WebQuests for Science Edutopia How To Create A Webquest For Students The best way to learn about WebQuests is to understand the differences between WebQuests that are simply "scavenger hunts" and those which help students understand the material in a deeper and - 21/11/2012 · This feature is not available right now. Please try again later. - It also offers easy-to-use templates for creating your own WebQuests. Getting Started with WebQuests Whether you know a lot about WebQuests or are just learning about these valuable teaching tools, the Web sites described in this article will help you determine the purpose and uses for WebQuests. - For example, if the students have just completed an assemblage of sticks and boxes to make a nonobjective sculpture, you might select something made by a famous sculptor that looks like the sculpture on the Goshen College campus by John Mishler. - 11/11/2018 · A WebQuest is an activity usually created by teachers for their students that leads students to answer inquiries using Web-based resources. There's an emphasis on using the information rather than just gathering it. Therefore, students should have to use analytical and critical-thinking skills to solve a problem or question.<\|endoftext\|>	3.890625
1,177	# What is 51/493 as a decimal? ## Solution and how to convert 51 / 493 into a decimal 51 / 493 = 0.103 The basis of converting 51/493 to a decimal begins understanding why the fraction should be handled as a decimal. Fractions and decimals represent parts of a whole, sometimes representing numbers less than 1. Depending on the situation, decimals can be more clear. We don't say 1 and 1/2 dollar. We use the decimal version of \$1.50. Same goes for fractions. We will say 'the student got 2 of 3 questions correct'. After deciding on which representation is best, let's dive into how we can convert fractions to decimals. ## 51/493 is 51 divided by 493 The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 51 is being divided into 493. Think of this as our directions and now we just need to be able to assemble the project! The numerator is the top number in a fraction. The denominator is the bottom number. This is our equation! We must divide 51 into 493 to find out how many whole parts it will have plus representing the remainder in decimal form. Here's 51/493 as our equation: ### Numerator: 51 • Numerators represent the number of parts being taken from a denominator. Any value greater than fifty will be more difficult to covert to a decimal. The bad news is that it's an odd number which makes it harder to covert in your head. Large numerators make converting fractions more complex. Now let's explore the denominator of the fraction. ### Denominator: 493 • Denominators represent the total number of parts, located below the vinculum or fraction bar. Larger values over fifty like 493 makes conversion to decimals tougher. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Have no fear, large two-digit denominators are all bark no bite. So grab a pen and pencil. Let's convert 51/493 by hand. ## Converting 51/493 to 0.103 ### Step 1: Set your long division bracket: denominator / numerator $$\require{enclose} 493 \enclose{longdiv}{ 51 }$$ We will be using the left-to-right method of calculation. This method allows us to solve for pieces of the equation rather than trying to do it all at once. ### Step 2: Extend your division problem $$\require{enclose} 00. \\ 493 \enclose{longdiv}{ 51.0 }$$ Because 493 into 51 will equal less than one, we can’t divide less than a whole number. Place a decimal point in your answer and add a zero. Now 493 will be able to divide into 510. ### Step 3: Solve for how many whole groups you can divide 493 into 510 $$\require{enclose} 00.1 \\ 493 \enclose{longdiv}{ 51.0 }$$ How many whole groups of 493 can you pull from 510? 493 Multiple this number by our furthest left number, 493, (remember, left-to-right long division) to get our first number to our conversion. ### Step 4: Subtract the remainder $$\require{enclose} 00.1 \\ 493 \enclose{longdiv}{ 51.0 } \\ \underline{ 493 \phantom{00} } \\ 17 \phantom{0}$$ If there is no remainder, you’re done! If you still have numbers left over, continue to the next step. ### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit. Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 51/493 fraction into a decimal is long division just as you learned in school. ### Why should you convert between fractions, decimals, and percentages? Converting between fractions and decimals depend on the life situation you need to represent numbers. They each bring clarity to numbers and values of every day life. This is also true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But they all represent how numbers show us value in the real world. Without them, we’re stuck rounding and guessing. Here are real life examples: ### When you should convert 51/493 into a decimal Investments - Comparing currency, especially on the stock market are great examples of using decimals over fractions. ### When to convert 0.103 to 51/493 as a fraction Carpentry - To build a table, you must have the right measurements. When you stretch the tape measure across the piece of wood, you won't see 10.6 inches. You'll see a tick mark at 10 and 3/5 inches. ### Practice Decimal Conversion with your Classroom • If 51/493 = 0.103 what would it be as a percentage? • What is 1 + 51/493 in decimal form? • What is 1 - 51/493 in decimal form? • If we switched the numerator and denominator, what would be our new fraction? • What is 0.103 + 1/2?<\|endoftext\|>	4.75
916	# Hacktivateed ## Why is converting mixed numbers and improper fractions important? An important part of learning about fractions is becoming comfortable understanding what they mean. Being able to convert between improper fractions and mixed numbers is a great way to be able to understand fractions and recognize how large or small a fraction is. ## What is the efficient way of changing a mixed number into an improper fraction? To convert a mixed number to an improper fraction, follow these steps: Multiply the denominator of the fractional part by the whole number, and add the result to the numerator. Use this result as your numerator, and place it over the denominator you already have. This method works for all mixed numbers. How are improper fractions useful? Improper fractions are easier to do calculations such as addition and subtraction, but mixed fractions are easier to understand. What is the relationship between a mixed number and an improper fraction? A mixed number is made up of a whole number and a fraction. For example: An improper fraction is one that is ‘top-heavy’ so the numerator is bigger than the denominator. ### What is the use of mixed fraction? Kinds of Fractions Types of Fractions Explanation Proper Fraction When the numerator is less than Denominator Improper Fraction When the numerator is greater than the Denominator Mixed Fraction It is an improper function, which is written as a combination of a whole number and a fraction. ### What are the mixed numbers? A mixed number is a whole number, and a proper fraction represented together. It generally represents a number between any two whole numbers. Look at the given image, it represents a fraction that is greater than 1 but less than 2. It is thus, a mixed number. What is 1 and 2/3 as an improper fraction? 5/3 Answer and Explanation: The mixed number 1 2/3 is 5/3 as an improper fraction. What is 3 and 1/3 as an improper fraction? 10/3 The mixed number 3 1/3 can be changed into the improper fraction 10/3. #### What is improper fraction example? An improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number). Fractions such as 65 or 114 are “improper”. #### What is improper fraction in real life? Improper Fraction and Mixed Fraction An improper fraction is a fraction whose numerator is greater than or equal to its denominator. For example, 9/4, 4/3 are improper fractions. Numerically, an improper fraction always equals to or greater than 1. What is 9 4 as a mixed number? 2 1/4 Answer: 9/4 as a mixed number is 2 1/4. What is the example of improper fraction? An improper fraction is a fraction whose numerator is greater than or equal to its denominator. For example, 9/4, 4/3 are improper fractions. ## How do you rewrite a mixed number into an improper fraction? To convert a mixed number into an improper fraction, multiply the whole number by the denominator and add it to the numerator. This becomes the numerator of the improper fraction; the denominator of the new fraction is the same as the original denominator. ## How do you convert a mixed number into an improper? To convert a mixed number to an improper fraction, follow these steps: Multiply the denominator of the fractional part by the whole number, and add the result to the numerator. For example, suppose you want to convert the mixed number. to an improper fraction. First, multiply 3 by 5 and add 2: (3 · 5) + 2 = 17. How do you multiply fractions with mixed numbers? To multiply mixed numbers, first change them to improper fractions. Next, simplify, or reduce, the fractions by removing common factors. Then multiply the numerators, and multiply the denominators. Simplify, or reduce the answer if possible. Finally, if the answer is an improper fraction, you can change it back to a mixed number. How do you convert a mixed number into a fraction? The first step to converting a mixed number to a fraction is to convert the whole number to a fraction. To do this, multiply the whole number by the denominator of the fraction. Then put this over the denominator. For example, let’s convert the whole number 2 in the mixed fraction 2 14 to a fraction.<\|endoftext\|>	4.9375
2,381	<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # Repeating Decimals ## Identify repeating decimals by dividing the numerator of a fraction by the denominator. Estimated4 minsto complete % Progress Practice Repeating Decimals MEMORY METER This indicates how strong in your memory this concept is Progress Estimated4 minsto complete % Repeating Decimals License: CC BY-NC 3.0 Jose has 10 bars of chocolate that he needs to give to 3 of his friends. How many bars of chocolate does each friend receive? In this concept, you will learn to write fractions and mixed numbers as repeating decimals. ### Writing Fractions and Mixed Numbers as Repeating Decimals A terminating decimal is a decimal number that does not go on forever. The word “terminate” means to end. Most of the fractions you have been working with are terminating decimals. Here is a fraction with a terminating decimal. 14\begin{align}\frac{1}{4}\end{align} Divide 1 by 4 to find the decimal value. 4)1.00¯¯¯¯¯¯¯¯¯¯¯¯0.25 8 20200\begin{align}\begin{array}{rcl} && \overset{ \quad 0.25}{4 \overline{ ) {1.00 \;}}}\\ && \ \ \ \underline{-8}\\ && \quad \ \ \ 20 \\ && \quad \underline{ -20} \\ && \qquad 0 \end{array}\end{align} You use zero placeholders, but ultimately, the decimal will divide evenly. A decimal that does not end and repeats the same number or numbers over and over again is called a repeating decimal. When you divide the numerator by the denominator and keep ending up with the same number, you might have a repeating decimal. Convert 23\begin{align}\frac{2}{3}\end{align} to a decimal. First, this does not have a base ten denominator. Divide the numerator by the denominator. 4)2.000¯¯¯¯¯¯¯¯¯¯¯¯¯ 0.66618 20 1820 18 2\begin{align}\begin{array}{rcl} && \overset{ \ \ 0.666}{4 \overline{ ) {2.000 }}}\\ && \underline{ -\; 18}\\ && \quad \ \ 20 \\ && \ \ \ \underline{ -18 }\\ && \qquad 20\\ && \quad \ \underline{ -18 }\\ && \qquad \ \ 2 \end{array}\end{align} The same remainder keeps showing up and the quotient becomes a series of 6’s. It does not matter if you keep adding zero placeholders. A repeating decimal is indicated by adding a line over the last digit or series of digits in the quotient that repeats itself. The decimal value of 23\begin{align}\frac{2}{3}\end{align} is 0.6¯\begin{align}0.\bar{6}\end{align} . ### Examples #### Example 1 Earlier, you were given a problem about Jose and his chocolate bars. Jose wants to give 10 chocolate bars to 3 of his friends. Divide 10 by 3 to find how many chocolate bar each friend receives. Divide 10 by 3. 3)10.000¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 3.333910910910 91¯¯¯¯¯¯¯¯¯\begin{align}& \overset{ \ \ 3.333}{3 \overline{ ) {10.000 \;}}}\\ & \underline{ \; \;\; -9}\\ & \quad 10 \\ & \underline{ \; \; \;-9}\\ & \quad 10 \\ & \underline{ \; \; \;-9}\\ & \quad 10 \\ & \ -9 \\ & \overline{ \; \; \; \; \;1}\end{align} The answer is a repeating decimal 3.3¯\begin{align}3. \bar{3}\end{align}. Jose can give each friend 3.3¯\begin{align}3.\bar{3}\end{align} bars of chocolate. #### Example 2 Is 49\begin{align}\frac{4}{9}\end{align} a repeating decimal or a terminating decimal? Convert the fraction to a decimal. Divide 4 by 9. 9)4.0000¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 0.444436 40 3640 36 40 36 4\begin{align}\begin{array}{rcl} && \overset{ \ \ 0.4444}{9 \overline{ ) {4.0000 \;}}}\\ && \underline{- \; 36}\\ && \quad \ \ 40 \\ && \ \ \ \underline{-36}\\ && \qquad 40\\ && \quad\ \underline{-36} \\ && \qquad \ \ 40\\ && \quad \ \ \ \underline{-36} \\ && \qquad \ \ \ 4 \end{array}\end{align} The same remainder keeps showing up and the quotient will go on and on as a series of 4s. The decimal value of 49\begin{align}\frac{4}{9}\end{align} is a repeating decimal, 0.4¯\begin{align}0.\bar{4}\end{align}. #### Example 3 Determine if the fraction is a repeating or terminating decimal. 13\begin{align}\frac{1}{3}\end{align} Convert the fraction to a decimal. Divide 1 by 3. 3)1.0000¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 0.3333 9 10 910 9 10 91\begin{align}\begin{array}{rcl} && \overset{ \ \ 0.3333}{3 \overline{ ) {1.0000 \;}}}\\ && \ \ \ \underline{ -9}\\ && \quad \ \ 10 \\ && \quad \ \underline{-9} \\ && \qquad 10\\ && \quad \ \ \ \underline{ -9} \\ && \qquad \ \ 10\\ && \qquad \ \underline{ -9} \\ && \qquad \quad 1 \end{array}\end{align} The decimal value of 13\begin{align}\frac{1}{3}\end{align} is a repeating decimal, 0.3¯\begin{align}0. \bar{3}\end{align}. #### Example 4 Determine if the fraction is a repeating or terminating decimal. 18\begin{align}\frac{1}{8}\end{align} Convert the fraction to a decimal. Divide 1 by 8. 8)1.000¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 0.125 8 20 1640 40 0\begin{align}\begin{array}{rcl} && \overset{ \ \ 0.125}{8 \overline{ ) {1.000 \;}}}\\ && \ \ \underline{ \;-8}\\ &&\quad \ \ 20 \\ && \ \ \ \underline{ -16} \\ && \qquad 40\\ && \quad \ \underline{ -40} \\ && \qquad \ \ 0 \end{array}\end{align} The decimal value of 18\begin{align}\frac{1}{8}\end{align} is a terminating decimal, 0.125. #### Example 5 Determine if the fraction is a repeating or terminating decimal. 512\begin{align}5 \frac{1}{2}\end{align} First, convert the fraction part to a decimal. 12=0.5\begin{align}\frac{1}{2} = 0.5\end{align} Then, place the whole number to the left of the decimal point. 512=5.5\begin{align}5 \frac{1}{2}=5.5\end{align} The decimal value of 512\begin{align}5 \frac{1}{2}\end{align} is a terminating decimal, 5.5. ### Review Determine if the fractions are repeating or terminating decimals. 1. 143\begin{align}\frac{14}{3}\end{align} 2. 349\begin{align}\frac{34}{9}\end{align} 3. 233\begin{align}\frac{23}{3}\end{align} 4. 174\begin{align}\frac{17}{4}\end{align} 5. 196\begin{align}\frac{19}{6}\end{align} 6. 125\begin{align}\frac{12}{5}\end{align} 7. 313\begin{align}3 \frac{1}{3}\end{align} 8. 812\begin{align}8 \frac{1}{2}\end{align} 9. 923\begin{align}9 \frac{2}{3}\end{align} 10. 1145\begin{align}11 \frac{4}{5}\end{align} 11. 1614\begin{align}16 \frac{1}{4}\end{align} 12. 443\begin{align}\frac{44}{3}\end{align} 13. 667\begin{align}\frac{66}{7}\end{align} 14. 184\begin{align}\frac{18}{4}\end{align} 15. 747\begin{align}\frac{74}{7}\end{align} ### Review (Answers) To see the Review answers, open this PDF file and look for section 5.21. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Please to create your own Highlights / Notes Show More ### Vocabulary Language: English TermDefinition Repeating Decimal A repeating decimal is a decimal number that ends with a group of digits that repeat indefinitely. 1.666... and 0.9898... are examples of repeating decimals. Terminating Decimal A terminating decimal is a decimal number that ends. The decimal number 0.25 is an example of a terminating decimal. ### Image Attributions 1. [1]^ License: CC BY-NC 3.0 ### Explore More Sign in to explore more, including practice questions and solutions for Repeating Decimals. Please wait... Please wait...<\|endoftext\|>	4.875
962	## MATHEMATICS DIFFERENTIABILITY QUIZ-10 As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background. Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. Q1. • Is continuous at x=0 • Is not continuous at x=0 • Is not continuous at x=0, but can be made continuous x=0 • None of these Solution Q2. • Has no limit • Is discontinuous • Is continuous but not differentiable • Is differentiable Solution Q3. If f(x)=a\|sin⁡x \|+b e\|x\|+c \|x\|3 and if f(x) is differentiable at x=0, then • a=b=c=0 • a=0,b=0;c∈R • b=c=0,a∈R • c=0,a=0,b∈R Solution Q4. • Continuous and differentiable • Continuous and not differentiable • Discontinuous and differentiable • Discontinuous and not differentiable Solution Q5. • Continuous as well as differentiable for all x • Continuous for all x but not differentiable at x=0 • Neither differentiable nor continuous at x=0 • Discontinuous everywhere Solution Q6. If f(x)=x(√x+√(x+1)), then • f(x) is continuous but not differentiable at x=0 • f(x) is differentiable at x=0 • f(x) is not differentiable at x=0 • None of the above Solution > Q7. • c=0,a=2b • a=b,c∈R • a=b ,c=0 • a=b,c≠0 Solution Q8. If f(x)=\|log10⁡x \|, then at x=1 • f(x) is continuous and f'(1+)=log10⁡e,f'(1-)=-log10⁡e • f(x) is continuous and f'(1+)=log10⁡e,f'(1-)=log10⁡e • f(x) is continuous and f'(1-)=log10⁡e,f'(1+)=-log10⁡e • None of these Solution Q9. Let f(x)=[x]+√(x-[x]), where [x] denotes the greatest integer function. Then, • f(x) is continuous on R+ • f(x) is continuous on R • f(x) is continuous on R-Z • None of these Solution Q10. Let f(x) be a function differentiable at x=c. Then, lim(x→c)⁡f(x) equals • f'(c) • f''(c) • 1/(f(c)) • None of these Solution #### Written by: AUTHORNAME AUTHORDESCRIPTION ## Want to know more Please fill in the details below: ## Latest NEET Articles\$type=three\$c=3\$author=hide\$comment=hide\$rm=hide\$date=hide\$snippet=hide Name ltr item BEST NEET COACHING CENTER \| BEST IIT JEE COACHING INSTITUTE \| BEST NEET & IIT JEE COACHING: MATHEMATICS DIFFERENTIABILITY QUIZ-10 MATHEMATICS DIFFERENTIABILITY QUIZ-10<\|endoftext\|>	4.375
869	Edit Article How to Calculate the Product–moment Correlation Coefficient Community Q&A The product-moment correlation coefficient allows you to work out the linear dependence of two variables (referred to as x and y). An example in economics might be that you are the owner of a restaurant. For every 10th customer you record the time he stayed in your restaurant (x, in minutes) and the amount spent (y, in dollars). Is it generally true that the long stayers are also the bigger spenders? This would be a positive correlation. Or is it actually the other way around, e.g., the richer the client the less time he takes for his lunch? This would be a negative correlation. In order to shed some light on this mystery you can calculate the product-moment correlation coefficient, r, sometimes known as Pearson's correlation. Note: The equations are for the linear least squares fit which statistically fits the set of data pairs to a straight line. Steps 1. 1 Remove incomplete pairs. In the next steps, use only the observations where both x and y are known. However do not exclude observations just because one of the values equals zero. 2. 2 Summarize the data into the values needed for the calculation. • n - the number of data pairs. • Σ(x2) - the sum of the squares of the x values. • Σx - the sum of all the x values. • Σ(xy) - the sum of each x value multiplied by its corresponding y value. • Σy - the sum of all the y values. • Σ(y2) - the sum of the squares of the y values. 3. 3 Calculate ssxy, ssxx and ssyy using these values. • ssxy=Σxy-(ΣxΣy÷n)=283-(1293/5)=59.8 • ssxx=Σx2-(ΣxΣx÷n)=40-(1212/5)=11.2 • ssyy=Σy2-(ΣyΣy÷n)=2089-(9393/5)=359.2 4. 4 Insert these values into the equation for r, the product-moment correlation coefficient. The value should be between 1 and -1, inclusive. r=ssxy/(ssxxssyy)0.5=59.8/(11.2359.2)**0.5=0.9428 • A value close to 1 implies strong positive correlation. (The higher the x, the higher the y). • A value close to 0 implies little or no correlation. • A value close to -1 implies strong negative correlation. (The higher the x, the lower the y). Community Q&A • How do you find the product moment correlation coefficient for the given data? • Could you tell me the formula for calculating product-moment correlation coefficient? Tips • Always make a scatter plot. Otherwise you may miss your discovery because the product moment correlation coefficient only takes straight lines into consideration when predicting the value of y from x. • There is often a reason why a lot of questionnaires feature the same questions, making them incredibly boring to answer. The researchers often know a lot about question x and question y, but they don't know yet how they are related or correlated. Warnings • When the correlation is significant you still have not demonstrated causality - that one variable "causes" the other. You have only proven that knowledge of the value of x may help to some degree in predicting the value of y and/or the other way around. • Before you state that two variables are correlated make sure the correlation coefficient is statistically significant. That is to say that the calculated correlation coefficient is unlikely to be a result of pure chance. For example, all your points may lay on the same line, this has a coefficient of +1 or -1, but it would still be inconclusive. (When the coefficient is not significant there is generally no point in reporting its value.) Article Info Featured Article Categories: Featured Articles \| Probability and Statistics In other languages: Thanks to all authors for creating a page that has been read 136,517 times.<\|endoftext\|>	4.53125
923	These following facts about the Black Death would probably give you much information about the disaster. The Black Death was one of the most devastating pandemics in human history, resulting in the deaths of an estimated 75 to 200 million people and peaking in Europe in the years 1346-53. Although there were several competing theories as to the etiology of the Black Death, analysis of DNA from victims in northern and southern Europe published in 2010 and 2011 indicates that the pathogen responsible was the Yersinia pestis bacterium, probably causing several forms of plague. Furthermore, to get to know more about this pandemics, below are some other facts about the Black Death you might like. Facts about the Black Death 1: Death Toll The exact death toll is difficult to measure from medieval sources. The number of deaths varied considerably by area and depending on the source. Current estimates are that between 75 and 200 million people died from the plague Facts about the Black Death 2: Europe Population The Black Death followed a period of population growth in Europe which, combined with two years of cold weather and torrential rains that wiped out grain crops, resulted in a shortage of food for humans and rats. This caused people and animals to crowd in cities, providing an optimal environment for disease. Facts about the Black Death 3: Trading Ships In November 1347, a fleet of Genoese trading ships landed in Messina, Sicily after trading along the coast from the Black Sea to Italy. The ships carried dead and dying sailors, many of whom had strange black growths on their necks, in their armpits, or in their groins. Many coughed blood. Finally, those who were alive died within days. Facts about the Black Death 4: Varieties of Plague Yersinia Pestis causes three varieties of plague: bubonic plague, caused by bites from infected fleas, in which the bacteria moves to lymph nodes and quickly multiplies, forming growths, or buboes; pneumonic plague, a lung infection that causes its victim to cough blood and spread the bacteria from person to person; and septicemic plague, a blood infection that is almost always fatal. Facts about the Black Death 5: Aftermath of the Plague The aftermath of the plague created a series of religious, social, and economic upheavals, which had profound effects on the course of European history. However, it took 150 years for Europe’s population to recover. However, the plague recurred occasionally in Europe until the 19th century. Facts about the Black Death 6: Prevention Other ways purported to prevent or cure the plague were to be happy and avoid bad thoughts, drink good wine, avoid eating fruit. Furthermore, to put fragrant herbs in beverages, avoid lechery, do not abuse the poor, eat and drink in moderation. Finally, maintain a household in accordance with a person’s status, and so on. Facts about the Black Death 7: Bathing Bathing during the plague was discouraged for two reasons. First, along with changing clothes, it was a sign of vanity. This invited the wrath of God and the punishment of sin. Second, bathing was believed to open the pores, making it easier for bad air to enter and exit the body, spreading disease. Ultimately, the latter belief was common throughout Europe well into the the nineteenth century. Facts about the Black Death 8: Third Pandemic A third pandemic began in China and India in the 1890s. And, it eventually reached the United States, with infections being especially dangerous in the San Francisco Bay Area. It was during this pandemic that the real cause (Yersinia Pestis) was discovered, along with a cure. Facts about the Black Death 9: Symptoms Contemporary accounts of the plague are often varied or imprecise. In fact, the most commonly noted symptom was the appearance of buboes (or gavocciolos) in the groin, the neck and armpits. This oozed pus and bled when opened. Facts about the Black Death 10: Great Plague of 1665 After the Black Death, plague epidemics continued to ravage Europe. For example, London was struck by the Great Plague of 1665, with thousands of deaths. This plague was followed almost immediately by the Great Fire, leaving London devastated. Hope you would find those Black Death facts really interesting, useful, and helpful for your additional reading.<\|endoftext\|>	3.703125
802	(a summary by Bill Houchin) Stickler syndrome is a connective tissue disorder, a genetic malfunction in the tissue that connects bones, eyes, and ears. This disorder is associated with problems of vision, hearing, bone & joint, facial and cleft palate. Stickler syndrome received its name from Dr. G. B. Stickler, who first studied and documented the syndrome. The term “syndrome” is derived from the Greek work “syn” meaning “together with” and the Greek word “drome” meaning “to run”. A syndrome is a collection of specific symptoms, all with one cause. Dr. Stickler first studied this syndrome at Mayo Clinic in 1965. His paper titled “Hereditary Progressive Arthro-Ophthalmopathy” associated the severe sight deterioration with joint changes. Other doctors continued the study and have redefined Stickler syndrome to how we know it today. There are several sight problems that may occur to Stickler patients. Common problems include near sightedness, astigmatism, and cataracts, which can be treated with glasses or surgery. More serious problems include the gel which fills the eye deteriorating, the retina deteriorating, eyes moving independent of each other, and glaucoma. Any of these serious problems can lead to blindness. The hearing loss suffered by those who are affected will affect either the middle or inner ear. Deafness can result in the extreme cases. Bone and joint problems consist of arthritis, abnormality to ends of long bones, vertebrae abnormality, curvature of the spine, hunchback, joint pain, knock knee, and double jointed. These will tend to worsen with age. Several facial features are common with Sticklers syndrome. Flat cheeks, flat nasal bridge, small upper jaw, pronounced upper lip groove, small lower jaw, and palate abnormalities are possible, all in varying degrees. 30-40% of patients with Pierre Robin sequence have Stickler syndrome. Patients usually do not have all symptoms that can be attributed to Stickler syndrome. As an example, a patient may only have joint problems, another person will have sight and arthritis problems, and members of the same family may have different symptoms. Some patients may have multiple symptoms, but only one problem is severe enough to be diagnosed. Stickler is believed to be the most common syndrome in the United States and Europe, but one of the rarest to be diagnosed. Most sufferers have such minor symptoms that they do not seek a diagnoses. Those who become patients are generally not correctly diagnosed. One study found a 53% error in original diagnosis of patients found in retrospect to have Stickler. A lot of patients are only diagnosed with one symptom and called, for example, arthritic or near-sighted. It is estimated that 1 in 10,000 people have Stickler Syndrome and only a fraction of them know it. - Arch Opthalmol 1992;.110, 11: 1589-93 - Am J Ophthalmol 1990; 110: 143-48 - Arch Opthalmol 1992; 110, 11: 1589-93 - Am J Hum Genet 1989; 45: 681-88 - J Rheumatology 1992; 19: 1271-5 - J Med Genet 1989; 26: 119-26 - HEARING DEFECTS: - J Am Acad Audiol 1990; 1,1; 37-40 - Smith’s Recog Patrns Hum Malfortn 1988: 242 - MONTEFIORE STUDY: - Birth Defects: Original Article Series; 21, 2: 85-92 - ORIGINAL STUDY: - Mayo Clinic Proceedings 1965; 40, 6: 454-5 - Stickler Syndrome Layperson Information Brochure<\|endoftext\|>	3.9375
308	Unit 1 : Being Financially Responsible What should you think about before making a purchase? Learn to analyze your spending choices and the factors that affect your spending, so you can think before you spend. Imagining Your Financial Future What are your goals for the future? Learn about setting goals and financial success strategies to help you reach those goals. Money habits include the ways you spend, save, and talk about money. Learn the differences between good and bad money habits and how to form good money habits. Supplemental Resource Links Find out how to set goals, determine milestones, and develop good money habits. Use this app, StickK (and similar apps and websites), to set goals, track progress, and get support. Visit this YouTube channel where individuals share about their careless spending and impulse purchases. In this interactive, you will learn about setting SMART savings goals. Check out the worksheet and handout with 10 "Rules for Saving." Try this quiz! After answering 12 questions, you will find out if you are a pathfinder, trooper, nomad, or tenderfoot and learn about financial identities. Download this board game where your challenge will be to live on a “20 Bean Salary.” After entering income and expenses, use this calculator to determine whether a budget is balanced. With this online game, you will serve as the choreographer and financial manager for Angel Alvarez, a competitive dancer trying to dance and save her way to Los Angeles.<\|endoftext\|>	3.71875
716	Today—May 17, 2019—is Endangered Species Day, a day where conservation efforts to protect the United States’ endangered species and their habitats are recognized and observed. On this day, we should also cast our view wider, as scientists at the Smithsonian Institute have done in proposing a “demographic safe space” for Asian elephants and, ultimately, for other large, slow-breeding species. “Critical thresholds in so-called vital rates—such as mortality and fertility among males and females of various ages—can signal an approaching population collapse before numbers drop below a point of no return,” said Dr. Shermin de Silva, president and founder of Trunks & Leaves, an organization dedicated to the conservation of Asian elephants. “We propose that conservation efforts for Asian elephants and other slow-breeding megafauna be aimed at maintaining their ‘demographic safe space’; that is, the combination of key vital rates that support a non-negative growth rate.” These key vital rates governing population growth, the researchers found, are a better indicator of a species’ viability than short-term trends in population size and distribution. “History bears this out,” de Silva said. “Genomic studies of the last mammoths isolated on Wrangel Island—between Russia and Alaska—have shown that although they were able to persist for thousands of years beyond the extinction of mainland populations with just 300 individuals, they had accumulated numerous genetic mutations that may have eventually contributed to their extinction.” In other words, a population of animals can continue to reproduce even though they become “biologically unviable” long before they disappear from the face of the earth. With that in mind, the researchers turned their gaze to mammoths’ descendants, the elephants. Asian elephants are classified as “endangered” under the IUCN Red List because populations are thought to have declined by at least 50 percent in less than a century. Not only that, but Asian elephants breed very slowly; most cows produce just one calf in six years or more. There are fewer than 500,000 of them alive today. While African elephants face extinction due to the horrific practice of killing them just to get their ivory tusks, the vast majority of Asian elephants don’t have tusks. The problem they face is habitat destruction, followed by illegal trade in live animals and parts. “Habitat loss can create something known as ‘extinction debt’ by slowing down birth rates and increasing mortality rates. For slow-breeding, long-lived species, even incremental changes make a big difference, but their longevity can obscure the risk of extinction,” de Silva said. Asian elephants aren’t the only megafauna that could benefit from “demographic safe space.” Giraffes, rhinos, Bactrian camels, and eastern gorillas could also benefit from modeling the interaction between vital rates. Data for these species in the wild is scarce, but, the authors of the study suggest, there is an urgent need to get that data. “Rather than rely on simple population counts or estimates of near-term extinction probability, we urge that conservation resources for slow-breeding megafauna also be invested in identifying demographic tipping points and how to maintain populations within their safe spaces,” de Silva concluded. Photo: Two Asian elephants in a river in Thailand. Credit: Shutterstock<\|endoftext\|>	3.9375
1,440	What is Ocean Wind? Wind is the movement of air and is caused by the difference in atmospheric pressure between high and low pressure systems. Over land, we use instruments called anemometers to measure the surface wind speed and wind direction. These anemometers exist in high-density for many areas and lower-density in less populated regions. But over the oceans, measurement of surface wind characteristics is far more limited, and is primarily obtained from anemometers located at small island weather stations, on ships, and on buoys floating in the ocean. Since the ocean regions are so large, especially the Pacific Ocean, knowledge of the wind characteristics over this vast space is important to weather forecasting, ocean navigation, and climate study. Enter the benefit of satellite wind measurement. One of the first approaches was to use visible images to study cloud motion and indirectly determine wind speed and direction. This method is still in use today. Microwave Measurement of Ocean Wind Two types of microwave instruments measure ocean surface winds, the passive microwave radiometer and the active microwave scatterometer. The radiometer measures ocean surface roughness which we correlate to wind speeds at 10 meters above the water's surface. We can only retrieve wind speeds from radiometers with the exception of the polarimetric radiometer, WindSat, that can measure both wind speed and wind direction. WindSat is the first satellite microwave polarimetric radiometer and was launched in 2003. A typical resolution for radiometer winds is about every 25 km over global oceans. A radiative transfer model and ocean emissivity model is needed to derive the wind speeds from ocean brightness temperatures. We can not obtain winds over land using MW radiometers. The consistent satellite MW radiometer wind speed data record began in 1987 with the launch of the DMSP F08 SSM/I. The scatterometer is an active instrument and sends a signal to the Earth's surface which reflects off the ocean Bragg waves (these are wind generated surface ripples—capillary waves) on the surface of the larger scale ocean waves. The reflected energy measured by the scatterometer is translated using a geophysical model function into a 10 meter neutral wind speed and direction. Scatterometers typically operate at either C-band (~5GHz frequency) or Ku-band (~14 GHz frequency). With special processing techniques, we can obtain wind speeds and directions every 12 km over the oceans. Scatterometers can also be used to measure sea ice and land ice characteristics. The scatterometers that have been in operation longer than a brief period are ERS-2 (C-band), QuikScat (Ku-band), and ASCAT (C-band). RSS scientists are working on merging these long-term wind vectors into a product for climate study. It is surprizing how well these two independent and different measurements agree. In 2003, both a radiometer (AMSR) and scatterometer (SeaWinds) were flown on the Midori-2 satellite. Wind speeds from these two instruments showed excellent agreement in rain-free conditions. RSS Wind Data Products 1-DEGREE, MONTHLY Wind Speed We have merged the wind speed measurements from the many radiometers in operation since 1987, including SSM/I, SSMIS, and WindSat. These data were all processed in a consistent manner using our radiative transfer model and careful instrument intercalibration. The wind speeds from these instruments are used to create a Mean Wind Speed product that is best for use in climate study. This 1-degree, monthly gridded product is further described in the document Merged Monthly 1-degree Wind Speeds. This wind product is available in netCDF format from the RSS ftp server and from the NASA GHRC. We have browse images of mean winds, wind speed anomalies, 12 monthly climatology maps (constructed using 1988 to 2007 data), and a global trend map. The Cross-Calibrated Multi-Platform (CCMP) gridded surface vector winds are produced using RSS V7 radiometer winds, QuikSCAT and ASCAT scattereometer winds, quality-checked moored buoy winds, and the ECWMF ERA-Interim Reanalysis model wind field as a background wind. A 4-dimensional variational analysis (VAM) is used to produce the fully populated nearly-global wind fields from the input data. As such, the CCMP is considered to be a Level-3 ocean vector wind analysis product consisting of four maps daily of a 0.25 degree gridded vector wind field. This product is an update and extension of the original V1.1 CCMP product. RSS has transitioned the CCMP processing code to run using our most up-to-date satellite data observations. All methodology remains the same as that used in the original CCMP product and most of the CCMP processing code is unchanged, with only minor alterations to compensate for the different operating systems and compilers. The CCMP V2.0 data are fully described in the online document and support documentation referenced. Individual Radiometer and Scatterometer Gridded Binary Data Files In addition, wind speed is one of the measurements in the radiometer binary gridded data files and is also available along with wind direction in the scatterometer data files. For detailed product information and data access, see the Missions page for each instrument and quick buttons for accessing the data. \|Instrument\|\|Period of Operation\|\|Version\| \|SSM/I, SSMIS\|\|1987 - present\|\|V7\| \|TMI\|\|1997 - 2015\|\|V7.1\| \|AMSR-2\|\|2012 - present\|\|V7.2\| \|AMSR-E\|\|2002 - 2011\|\|V7\| \|GMI\|\|2014 - present\|\|V8.1\| Wind speed and direction (polarimetric radiometer and scatterometers): \|Instrument\|\|Period of Operation\|\|Version\| \|WindSat\|\|2003 - present\|\|V7.0.1\| \|QuikScat\|\|1999 - 2009\|\|V4\| \|SeaWinds\|\|2003: Apr - Oct\|\|V3a\| \|ASCAT\|\|2006 - present\|\|V1.2\| Related Data Products The RSS data are used to create merged wind products by other researchers. These are a few examples: The original V1.1 Cross-Calibrated Multi-Platform Ocean Winds (CCMP) data set is produced by Atlas et al. The 4x daily data files for 1988 to 2011 are available from the JPL PO.DAAC. The winds are produced using a 4 dimensional variational analysis to derive wind speeds and directions using RSS wind speeds, in situ measurements and ECMWF model data as input. The Blended Sea Winds data set is produced by researchers at NOAA NCDC. This product uses all RSS radiometer wind speeds and an optimal-interpolation processing to merge the winds into a 4x daily product.<\|endoftext\|>	4.25
205	# How do you write an equation in slope intercept form given that the line passes through the point (−6, 4) and has a slope of 4/3? ##### 1 Answer May 30, 2015 The main formula of a line is; $y = a x + b$ We should use one point to find the real equation. If we use (-6,4) point; $y = a x + b \implies 4 = - 6 a + b$; a is the slope of the equation, we found that as $\frac{4}{3}$; $4 = \left(- 6 \cdot \frac{4}{3}\right) + b \implies 4 = - 8 + b \implies b = 12$; So the equation of the line will be; $y = a x + b \implies \underline{y = \frac{4}{3} x + 12}$<\|endoftext\|>	4.40625
3,383	- Avian bornavirus has been shown to be a cause of the disease syndrome formerly known as proventricular dilatation disease or PDD. - Avian bornavirus is a labile virus, susceptible to most disinfectants, detergents, and ultraviolet light. - Birds can harbor asymptomatic avian bornavirus infection. - The avian bornavirus is intermittently shed in feces and urates. - Clinical disease develops secondary to the body’s response to infection, as lymphocytic-plasmacytic infiltrate develops in the brain, spinal cord, and peripheral nerves and is frequently described as neuropathic ganglioneuritis. - Progression of clinical disease, or neuropathic ganglioneuritis, can be slow or quite rapid. - The clinical signs of neuropathic ganglioneuritis typically vary from primarily gastrointestinal, primarily neurological, or both gastrointestinal and neurological. - Definitive diagnosis is challenging antemortem, but typically relies upon a combination of PCR and serology. - Whenever possible, isolation of infected birds is recommended; culling of infected birds is NOT recommended. Proventricular dilatation disease or PDD is one of the most frustrating conditions that avian practitioners, aviculturists, and bird owners encounter today. The recent discovery of avian bornavirus (ABV) as an etiological agent of PDD has not simplified the challenges. The detection of avian bornavirus infection is common in birds with PDD but the virus is detected in so many clinically healthy birds as well as birds with other chronic diseases that it is clear other factors along with ABV may be involved in development of disease. Signalment and distribution Proventricular dilatation disease was first reported in the late 1970s in macaws imported into the United States and Germany. During the early years, this disease was often referred to as “macaw wasting disease” (Payne 2012). Since that time, proventricular dilatation disease has been identified in over 50 different avian species including passerines like the canary (Serinus canaria), honeycreepers, weaver finches, waterfowl, toucans, birds of prey, and psittacine birds (Hoppes 2013, Payne 2012, Smith 2010, Weissenböck 2009). Proventricular dilatation disease has been reported in captive parrots in North America, Latin America, Europe, Australia, Africa, Japan, and the Middle East (Sassa 2013, Last 2012, Doneley 2007). Avian bornavirus* (ABV), a newly discovered member of family Bornaviridae, was identified as a cause of PDD in 2008 (Payne 2012, Gancz 2009, Gray 2010, Gancz 2009, Honkavuori 2008, Kistler 2008). Avian bornavirus is a single-stranded RNA virus. The virus is intermittently shed in the droppings. Transmission is generally fecal-oral (Rinder 2009), however vertical transmission has also been demonstrated (Monaco 2012, Kerski 2012, Lierz 2011). The incubation period for ABV infection can be as brief as days but it has also been hypothesized to be much longer, possibly decades, in some birds. There are ten known ABV genotypes and seven of these genotypes can affect psittacine birds. The virulence of these genotypes can vary, and there is also evidence that an asymptomatic bird harboring one genotype can sometimes become clinically ill after exposure to a second, novel genotype (Rubbenstroth 2013, Nedorost 2012, Lierz 2012, Rinder 2009). Editor’s note: Avian bornavirus in psittacine birds is more accurately described as parrot bornavirus. Proventricular dilatation disease is more accurately described as neuropathic ganglioneuritis, lymphoplasmacytic ganglioneuritis, or avian autoimmune ganglioneuritis. Clinical disease arises when the host mounts an immune response that leads to progressive destruction of brain, spinal cord, and peripheral nerves via lymphocytic-plasmacytic infiltrate. Nerves that supply the proventriculus, ventriculus, and intestines are commonly affected. Progression of clinical disease can be slow or quite rapid. The clinical picture may be primarily gastrointestinal, primarily neurological, or both. (Table 1) (Fig 1, Fig 2). Polyuria, hypotension, and cardiac abnormalities have also been reported. Avian bornavirus infection damages the Purkinje cells, which can potentially cause arrhythmias however this is usually a postmortem finding. \|Table 1. Potential signs of neuropathic ganglioneuritis in the psittacine bird\| \|Gastrointestinal sings\|\|Central or peripheral nervous system signs\| Editor’s Note: Although death would seem to be the most objective of parameters, there are camps of PDD study that disagree vehemently with the statement that symptomatic avian bornavirus disease is invariably fatal and there are reports of birds with chronic disease living long, long lives. Definitive diagnosis of neuropathic ganglioneuritis remains challenging. There are birds at the Schubot Exotic Bird Health Center that have tested positive for ABV via PCR and serology for over 7 years that continue to be clinically healthy. Other individuals consistently tested negative for up to 6 years before becoming acutely ill. These birds only tested positive on PCR and serology at the terminal stages of disease. PCR: Reverse transcription (RT) polymerase chain (PCR) testing of droppings, feathers, and blood has been evaluated in psittacine birds (Table 2) (Hoppes 2013, de Kloet 2011, Dahlhausen 2010). Positive ABV PCR test results do not mean the bird has or ever will develop clinical disease, and most PCR testing does not differentiate between ABV genotypes. Due to intermittent shedding, one negative PCR test result also cannot rule out ABV infection, particularly if the sample is not properly stored. Serology: Serological testing has also been frustratingly inconclusive, with some ABV-infected birds failing to develop a detectable antibody response within weeks of dying of neuropathic ganglioneuritis (Table 2) (Herzog 2010). In other individuals there appears to be a positive correlation between the level of antibodies produced and the development of disease. Tests used to detect ABV titers have included indirect immunofluorescence, ELISA, and Western blot assay (Hoppes 2013). Recommendations based on current practices at Schubot: A combination of cloacal or feather RT-PCR with a serologic assay may offer the best chance of identifying ABV infection. Birds with a high ABV load via PCR combined with high antibody titers appear to have the highest risk of developing clinical disease (Heffels-Redmann 2012), however there is there is no one combination of tests that can be used to reliably predict the progression from ABV exposure and shedding to clinical disease at this time. Multiple tests must be performed before a bird can be considered ABV negative. Ideally three negative PCR test results are needed for a bird to be considered negative. See Questions and Controversies below for additional antemortem diagnosis recommendations. \|Table 2. Laboratories in the United States that perform avian bornavirus diagnostic testing\| \|The Schubot Center\|\|X\| \|Veterinary Molecular Diagnostics\|\|X\|\|X\| The apparently healthy ABV-positive bird There is evidence that ABV is endemic in parrots in the United States, with screening of some homes and aviaries revealing infection rates of 33% to 60% in clinically healthy birds. Considering the difficulty of testing and the intermittent shedding of the virus, these infection rates for exposed birds are likely higher than what is reported. The significance of low or inconsistent antibody levels and low PCR responses remains unclear; however there have been cases of even brief exposure resulting in clinical disease. Current recommendation: Isolate a single ABV-positive bird from other susceptible individuals since the virus is intermittently shed in the feces and urates. Despite this advice, there have been many instances where an infected bird has been housed in direct contact with negative birds for months or even years without apparently transmitting disease. At this stage in our understanding of the disease, it is impossible to predict if individuals housed with positive birds will become positive or clinically ill. A recent study has concluded that horizontal transmission of ABV by direct contact is inefficient in immunocompetent fully-fledged domestic canaries and cockatiels (Nymphicus hollandicus). Birds in this study were followed over a 4-month period (Rubbenstroth 2014). Recommendations based on current practices at Schubot: Determine the genotype in ABV-positive birds and keep birds with different genotypes separate. There is evidence that different genotypes differ in virulence and in the studies performed so far, one genotype is not protective against infection with another ABV genotype. In fact there is evidence that exposure to a novel ABV genotype can be pathogenic in a bird currently harboring ABV without illness. Therefore housing one ABV positive bird does not mean an owner can take in other positive birds without risk (I. Tizard, oral communication, March 2014). There are no recommendations to automatically cull or euthanize infected birds. A clinically healthy positive bird may live its entire life without ever developing clinical disease. Gross necropsy findings in the bird with neuropathic ganglioneuritis may include profound emaciation as well as a dilated proventriculus, ventriculus, and/or intestines (Fig 3). Histopathologically there may be a lymphoplasmacytic ganglioneuritis involving the brain, spinal cord, peripheral nerves, splanchnic nerves, heart, and adrenal gland. In some instances, lesions can also be detected in the lungs or kidneys. Lesions may only be present in the brain and spinal cord in some neurologic cases. Avian bornavirus can consistently be detected by PCR analysis in the tissues of birds that die of neuropathic ganglioneuritis. The virus may be found in all major organs or may be limited to the brain, spinal cord, and peripheral nerves. The vitreous humor can also be a source of ABV, even in apparently healthy carrier birds (Hoppes 2013). Management of disease Given the inflammatory infiltrate that develops, nonsteroidal anti-inflammatory agents like celecoxib (Celebrex, Pfizer) and meloxicam (Metacam, Boehringer Ingelheim) have long been offered to affected birds. Unfortunately neither meloxicam nor the immunosuppressive drug cyclosporine has slowed the progression of disease in birds experimentally infected with ABV at Schubot Exotic Bird Health Center. The Center is currently evaluating cyclosporine and the antiviral agent, ribavirin (Rebetol, Merck) but further study is needed (Hoppes 2013). Visit Questions and Controversies below for additional information. Prevention or control Bornavirus is easily destroyed in the environment, and it is susceptible to most detergents, disinfectants, and ultraviolet light. Although isolation of the ABV-positive bird is recommended, some bird owners have successfully maintained a group of birds that include a positive individual. The risk of disease transmission is reduced when owners offer a good plane of nutrition; provide exemplary husbandry and adequate ventilation, while minimizing stress, employing scrupulous hygiene practices, and basic biosecurity measures. Biosecurity measures that reduce the risk of disease transmission - Wash hands between birds - Use a footbath or disinfect shoes when going from infected to non-infected rooms - Care for ABV-negative birds before handling positive birds Since it has been shown that ABV can be vertically transmitted, incubator hatching does not definitively halt the spread of infection to the chick (Monaco et al 2012, Lierz et al 2011). Questions and controversies - When a bird tests positive for ABV on various antemortem tests does this mean the bird has the disease syndrome called proventricular dilatation disease (PDD)? The detection of ABV infection is common in birds with PDD but ABV infection may occur without clinical evidence of disease or ABV-positive birds can also suffer from other chronic diseases. - Is ABV the only cause of PDD? Although a causal relationship between ABV infection and PDD has been confirmed, ABV has not been confirmed as the one and only cause of this syndrome. - Is ABV a cause of feather destructive behavior? Feather destructive behavior is sometimes seen in birds with ABV infection (Horie 2012). Feather picking and self-mutilation can be observed with any bird with systemic illness, however this feather damaging behavior has also been theorized to occur secondary to peripheral neuropathy. - Is clinical ABV infection invariably fatal? Although death would seem to be the most objective of parameters, there are camps of PDD study that disagree vehemently with the statement that symptomatic ABV disease is invariably fatal. There are reports of birds with chronic PDD disease living long, long lives, however much of this controversy may stem around confirmation of disease. Was ABV infection confirmed in these cases or was PDD diagnosis based on clinical signs and radiographs only? - What other antemortem tests can be performed in the diagnosis of PDD and are there additional considerations for discussion? Since serologic test results for ABV infection are sometimes poorly correlated with clinical disease, additional tests include: - Antiganglioside antibodies: Birds with clinical disease demonstrate elevations in antiganglioside antibodies regardless of PCR test status, and antiganglioside antibodies have been reported to show very good correlation with the presence of histopathologic lesions (B. Dahlhausen, email communication, July 2014). - Histologic evaluation of crop biopsy: Surgical removal of a small, vascular, full thickness segment of the crop wall can be used to identify histologic signs consistent with neuropathic ganglioneuritis (Gregory 1996). Crop biopsy results depend on the presence of segmental disease in the tiny piece of tissue collected. Therefore positive results support a diagnosis of PDD, however negative results have very little meaning (Lierz 2009). Most birds without gastrointestinal signs are almost always negative via crop biopsy. - What can be done for the PDD patient? Although research out of Schubot Exotic Bird Health Center has shown dismal results with the use of meloxicam in cockatiels infected with ABV (Hoppes 2013), anecdotal reports suggest the use of cyclooxygenase (COX)-2 inhibitors may be helpful in some clinical cases (B. Dahlhausen, email communication, July 2014). Positive results have been reported with the pure COX-2 inhibitor, celecoxib. Reported dose ranges vary but typically start at 10-20 mg/kg PO SID to BID. As a COX-2 preferential inhibitor, results with the use of meloxicam (0.5-1 mg/kg PO SID) have been more unpredictable. Most recently positive results have been reported with robenicoxib (Onsior, Novartis) (1 -2 mg/kg IM) paired with immunomodulation therapy in the form a scheduled injection of Mycobacterium bovis inoculum to redirect and ultimately desensitize the bird’s immune system, similar to the management of Guillain–Barré syndrome in humans (Rossi 2013). Although avian bornavirus infection is common in clinically healthy birds, the development of clinical disease is more rare. Symptomatic disease, commonly called neuropathic ganglioneuritis or proventricular dilatation disease, is a progressive, potentially fatal neurological disorder of psittacine birds caused by avian bornavirus. This complex syndrome can present with diverse clinical findings including weight loss, proventricular dilatation, and/or neurological signs. Currently there is no one combination of tests that can be used to reliably diagnose ABV infection and there is no cure, however current recommendations will change as our understanding of this disease continues to grow.<\|endoftext\|>	3.734375
1,229	Open in App Not now # Spherical Cap Volume Formula • Last Updated : 27 Apr, 2022 A spherical cap is a part of a sphere that is obtained by cutting it with a plane. It is the section of a sphere that extends above the sphere’s plane and formed when a plane cuts off a part of a sphere. The base area, height, and sphere radius are all the values that are required to calculate the volume of a spherical cap. Formula V = (1/3)π(3R – h)h2 where, R is the radius of sphere, h is the height of spherical cap, π is a constant with a value of 22/7. Using Pythagoras theorem, we can say that (R – h)2 + a2 = R2. So, the formula can also be written as, V = (1/6)πh(3a2 + h2) Here, a denotes the radius of spherical cap. ### Sample Problems Problem 1. Find the volume of the spherical cap if the radius of the sphere is 7 m and the height of the cap is 10 m. Solution: We have, r = 7 and h = 10. Using the formula we have, V = (1/3)π(3R – h)h2 = (1/3) (22/7) (3(7) – 10) (102) = (1/3) (22/7) (11) (100) = 1152 cu. m Problem 2. Find the volume of the spherical cap if the radius of the sphere is 5 m and the height of the cap is 5 m. Solution: We have, r = 5 and h = 5. Using the formula we have, V = (1/3)π(3R – h)h2 = (1/3) (22/7) (3(5) – 5) (52) = (1/3) (22/7) (10) (25) = 261.8 cu. m Problem 3. Find the volume of the spherical cap if the radius of the sphere is 7.5 m and the height of the cap is 12 m. Solution: We have, r = 7.5 and h = 12. Using the formula we have, V = (1/3)π(3R – h)h2 = (1/3) (22/7) (3(7.5) – 5) (122) = (1/3) (22/7) (17.5) (144) = 1583.4 cu. m Problem 4. Find the radius of the sphere if the height and volume of the cap are 15 m and 2120.6 cu. m respectively. Solution: We have, V = 2120.6 and h = 15. Using the formula we have, V = (1/3)π(3R – h)h2 => 2120.6 = (1/3) (22/7) (3R – 15) (152) => 2120.6 = (1/3) (22/7) (3R – 15) (225) => 3R – 15 = 9 => 3R = 24 => R = 8 m Problem 5. Find the radius of the spherical cap if the height and volume of the cap are 6.5 m and 1305.2 cu. m respectively. Solution: We have, V = 1305.2 and h = 6.5. Using the formula we have, V = (1/3)π(3R – h)h2 => 1305.2 = (1/3) (22/7) (3R – 15) (6.5)2 => 1305.2 = (1/3) (22/7) (3R – 6.5) (42.25) => 3R – 6.5 = 29.5 => 3R = 36 => R = 12 m Now using the formula (R – h)2 + a2 = R2, we have a2 = R2 – (R – h)2 a2 = 122 – (12 – 6.5)2 a2 = 144 – 30.25 a2 = 113.75 a = 10.67 m Problem 6. Find the volume of a spherical cap if its radius is 7 m and height is 14 m. Solution: We have, a = 7 and h = 14. Using the formula we have, V = (1/6)πh(3a2 + h2) = (1/6) (22/7) (14) (3 (7)2 + 142) = (1/6) (22/7) (14) (343) = 2514.3 cu. m Problem 7. Find the volume of a spherical cap if its radius is 4.21 m and height is 9.54 m. Solution: We have, a = 4.21 and h = 9.54. Using the formula we have, V = (1/6)πh(3a2 + h2) = (1/6) (22/7) (9.54) (3 (4.21)2 + 9.542) = (1/6) (22/7) (14) (144.183) = 720.2 cu. m My Personal Notes arrow_drop_up Related Articles<\|endoftext\|>	4.59375
1,490	By John De Gree Fidel Castro, Cuba and the United States of America On November 25th, 2016, one of the world’s brutal dictators, Fidel Castro, died. Fidel Castro led Cuba as its Communist dictator since 1959. During his 57-year rule, Castro was responsible for the murder, torture, and imprisonment of tens of thousands. Castro’s regime did not allow basic civil rights in Cuba such as the freedom of speech, freedom of religion, and freedom of the press. In addition, Castro did not allow Cubans to travel outside of the island and ordered his navy to kill those trying to leave by boat. Under Castro, the Cuban navy sank ships and used fire hoses to drown Cubans in the Caribbean Sea trying to escape island by boat. Cuba, 1492 – 1895 In 1492, Christopher Columbus discovered the New World for Spain, and Spain quickly colonized much of North and South America. For the next four hundred years, Spain ruled Cuba. As Spanish colonies in the New World revolted in the early 1800s and countries such as Mexico gained independence, Cuba remained loyal. In the mid to late 1800s, Cubans wanted independence from Spain and fought for many years. In the Ten Years’ War (1868-78) Cubans struggled to break away, but Spain kept control of the island. The United States of America and Cuba, 1895-1959 During the Second War for Independence (1895-98) the United States entered the war on the side of the Cubans, fighting what Americans call the Spanish-American War. The United States defeated Spain, and at the Treaty of Paris it was decided that Spain would surrender Cuba, Puerto Rico, parts of the West Indies, Guam, and the Philippines to the United States. The Americans gave Cuba its independence in 1901, but the U.S. insisted on the right to have a permanent naval base on Cuba (Guantanamo Bay), and claimed the right to militarily intervene in Cuban affairs should there be unrest. Over the next 60 years, the U.S.A. militarily intervened in Cuba on numerous occasions. Cubans suffered under military dictatorships, and at times, the United States supported Cuban leaders who were undemocratic. Fulgencio Batista (1901-1973) seized power militarily in 1933, was elected Cuban President in 1940, and he made himself rich through his connections as the Cuban President. In 1944, he left office and lived in Florida. While Batista was in Florida, Cuba became unstable and corruption was rampant. The American government supported Batista’s return to power in 1952, and for the next seven years he led Cuba as dictator. Batista did not tolerate anyone going against him, controlled the media, and arrested, tortured and executed those who he believed were Communists. It is not known how many he killed, and the number historians give ranges from 1,000 to 20,000 Cubans. Cuba, 1959 – Present Day In 1959, Communist Fidel Castro and his brother Raul Castro overthrew Batista. The Castro government nationalized (took over) all foreign owned businesses, and eventually took over all businesses owned by Cubans, as well. Communists believe that only the government should own property, and they do not trust businesspeople. Communists are also against religion, and anyone wanting to pray to God in Cuba is punished. In the first few years of the Castro regime, the Cuban government terrorized those who did not obey. Tens of thousands of Cubans were tortured and executed because they did not want to follow the Castro regime. Che Guevara, Castro’s chief enforcer, in response to questions about Castro's firing squads, said, "To send men to the firing squad, judicial proof is unnecessary. These procedures are an archaic bourgeois detail. This is a revolution. And a revolutionary must become a cold killing machine motivated by pure hate." For the next few years, the U.S.A. tried to get Cuba back into its sphere of influence, but failed. In 1961, President Kennedy approved of a plan to support a group of Cubans to invade Cuba and conquer Castro. Called “The Bay of Pigs Invasion”, it failed miserably. In 1962, America realized that the Soviet Union was building a network of nuclear missile launch sites on Cuba. After a U.S. naval blockade, the Soviet weapons were withdrawn, and the U.S. promised to never invade Cuba, again. From 1962 on, Cuba was allied with the Soviet Union, America’s enemy throughout most of the second half of the 1900s. The Soviet Union gave Cuba money, food, and a great amount of support. Cuba after the Fall of the Soviet Union In 1991, the Soviet Union fell apart and could no longer give Cuba subsidies. As a result, Cubans suffer greatly from a lack of food and basic necessities. After 50 years of Communist rule, Cuba has become a poor country run by a government that only cares about itself. The Castro brothers are much older now, but they remain rich, as the average Cuban suffers. Cubans are jailed because they oppose the Castros and the Communists, and some have been executed for their beliefs. Whoever tries to escape Cuba and is caught faces grave danger. President Obama has changed the United States policy towards Cuba from starving the Castro government to recognizing it as legitimate. President Obama thinks that increased American tourism and business will open up Cuba. Other Americans, like Senator Marco Rubio, argue that doing business with Cuba means helping the Castros stay in power. All foreign money spent in Cuba goes straight to the Communist leaders, so more American business means a richer Communist regime. The death of Fidel Castro has been met with diverse reactions from leaders in North America. President Obama wrote, "Fidel Castro altered the course of individual lives, families, and of the Cuban nation. History will record and judge the enormous impact of this singular figure on the people and world around him," in a White House statement. President-Elect Donald Trump wrote, "The world marks the passing of a brutal dictator who oppressed his own people for nearly six decades. Fidel Castro’s legacy is one of firing squads, theft, unimaginable suffering, poverty and the denial of fundamental human rights." Canadian Prime Minister Justin Trudeau said, "Fidel Castro was a larger than life leader who served his people for almost half a century. A legendary revolutionary and orator, Mr. Castro made significant improvements to the education and healthcare of his island nation." It is unclear if the death of Fidel Castro will have a great impact on the lives of Cubans. His brother, 85 year old Raul Castro, is currently the Communist leader of Cuba, and he has so far followed his brother Fidel’s model of repressive leadership. Interesting Questions You Can Discuss With Your Students and Kids: John De Gree John De Gree writes the current events with a look at the history of each topic. Articles are written for the young person, aged 10-18, and Mr. De Gree carefully writes so that all readers can understand the event. The perspective the current events are written in is Judeo-Christian. Receive Articles and Coupons in Your Email<\|endoftext\|>	3.828125
624	"[Lot of 2] Map of the Northern parts of the United States of America... [in set with] Map of the Southern parts of the United States of America…", Bradley, Abraham Jr. Subject: Eastern United States Period: 1802 (circa) Color: Black & White This pair of maps cover the young United States during the tumultuous period following independence when the states and central government fought over control of the western land claims. Seven of the 13 original states had claims to areas in the West (those lands between the Appalachian Mountains and the Mississippi River). These so-called landed states had a great potential advantage over the six landless states. It was assumed that the future sale of western lands would enrich the landed states and possibly allow them to operate without any form of taxation. The landless states feared that they would lose residents and dwindle into insignificance. The matter was further complicated by the fact that the claims often overlapped one another, which pitted one state's interests against another. Both maps are in their second states with "Corrected by the Author" below the title. The map of the northern parts (15.8 x 8.5") shows the new states of Ohio, Indiana, Illinois, Michigan and Wisconsin that were proposed in the Ordinance of 1789 referred to, respectively, as State I, II, III, IV, and V. The northern boundaries of Illinois, Indiana and Ohio are shown at the southern tip of Lake Michigan. The positioning of this line later resulted in a boundary dispute between Michigan and Ohio, with Ohio blocking admission of Michigan to the Union until the question was settled. The mostly blank Wisconsin bears the notation "Large Meadows extending to Lake Michigan." Wisconsin came from the term "Ouisconsin" that is believed to mean "grassy place" in the Chippewa language. The remnants of several bounty land grants are shown including New Connecticut and Seven Ranges. Wayne County is the only county shown in Michigan. Cincinnati is here called Cincinnatus. The boundary line from Gen. Wayne's Treaty of 1795 is shown. The map extends to the Mississippi River with the lands to the west labeled, Re Lands Reserved by the US. The map of the southern parts (15.2 x 7.7") shows much of present-day Mississippi and Alabama in Mississippi Territory, with the coastal region named West Florida. A dotted line running north-south through present-day Georgia demarcates "All West of this line is claimed by the United States & also by Georgia." The overlapping claims were resolved when Spain gave up its claim to West Florida in 1795 and Georgia finally relinquished its claim in 1802 (the last of the landed states to surrender its claim). Another interesting feature of the map is the notation of the number of men in several of the Indian tribes in the western territories. References: Wheat & Brun #158 & 496. Issued folding, now pressed. Several short fold separations expertly repaired. The northern map has a tiny tear in the top right margin.<\|endoftext\|>	3.734375
1,344	## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 14 (Mis Ex) Question 14: Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of −6 – 24i. Let z = (x – iy) (3 + 5i) z = 3x + 5xi – 3yi – 5yi2 = 3x + 5xi – 3yi + 5y = (3x + 5y) + i(5x – 3y) ∴ (3x + 5y) – i(5x – 3y) = −6 – 24i Equating real and imaginary parts, we obtain 3x + 5y = −6 ….(i) 5x – 3y = 24 ….(ii) Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain 9x + 15y = −18 Putting the value of x in equation (i), we obtain 3(3) + 5y = −6 ⇒ 5y = −6 – 9 = −15 ⇒ y = −3 Thus, the values of x and y are 3 and −3 respectively. ## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 13 (Mis Ex) Question 13: Let z = r cos θ + ir sin θ i.e., r cos θ = −1/2 and r sin θ = 1/2 On squaring and adding, we obtain Therefore, the modulus and argument of the given complex number are 1/√2 and 3π/4 respectively. ## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 12 (Mis Ex) Question 12: Let z1 = 2 – i, z2 = −2 + i. Find z1 = 2 – i, z2 = −2 + i (i) z1z2 = (2 – i) (−2 + i) = −4 + 2i + 2i – i2 = −4 + 4i – (−1) = −3 + 4i Question 11: Question 10: ## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 9 (Mis Ex) Question 9 : Solve the equation 21x2 – 28x + 10 = 0 The given quadratic equation is 21x2 – 28x + 10 = 0 On comparing the given equation with ax2 + bx + c = 0, we obtain a = 21, b = −28, and c = 10 Therefore, the discriminant of the given equation is D = b2 – 4ac = (−28)2 – 4 × 21 × 10 = 784 – 840 = −56 Therefore, the required solutions are ## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 8 (Mis Ex) Question 8: Solve the equation 27x2 – 10x + 1 = 0 The given quadratic equation is 27x2 – 10x + 1 = 0 On comparing the given equation with ax2 + bx + c = 0, we obtain a = 27, b = −10, and c = 1 Therefore, the discriminant of the given equation is D = b2 – 4ac =(−10)2 – 4 × 27 × 1 = 100 – 108 = −8 Therefore, the required solutions are ## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 7 (Mis Ex) Question 7: This equation can also be written as 2x3 – 4x + 3 = 0 On comparing this equation with ax2 + bx + c = 0, we obtain a = 2, b = −4, and c = 3 Therefore, the discriminant of the given equation is D = b2 – 4ac = (−4)2 – 4 × 2 × 3 = 16 – 24 = −8 Therefore, the required solutions are ## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 6 (Mis Ex) Question 6: This equation can also be written as 9x2 – 12x + 20 = 0 On comparing this equation with ax2 + bx + c = 0, we obtain a = 9, b = −12, and c = 20 Therefore, the discriminant of the given equation is D = b2 – 4ac = (−12)2 – 4 × 9 × 20 = 144 – 720 = −576 ## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 5 (Mis Ex) Question 5: Convert the following in the polar form : Let r cos θ = −1 and r sin θ = 1 On squaring and adding, we obtain r2(cos2 θ + sin2 θ) = 1 + 1 ⇒ r2(cos2 θ + sin2 θ) = 1 + 1 ⇒ r2(cos2 θ + sin2 θ) = 2 ⇒ r2 = 2 [cos2 θ + sin2 θ = 1] ⇒ r = √2 [Conventionally, r > 0] ∴ √2 cos θ = −1 and √2 sin θ = 1 Let r cos θ = −1 and r sin θ = 1 On squaring and adding, we obtain r2(cos2 θ + sin2 θ)=1 + 1 ⇒ r2(cos2 θ + sin2 θ) = 2 ⇒ r2 = 2 [cos2 θ + sin2 θ = 1] ⇒ r = √2 [Conventionally, r > 0] ∴ √2 cos θ =−1 and √2 sin θ = 1 This is the required polar form.<\|endoftext\|>	4.625
864	### Plants Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this? ### Number Round Up Arrange the numbers 1 to 6 in each set of circles below. The sum of each side of the triangle should equal the number in its centre. ### Weighted Numbers Use the number weights to find different ways of balancing the equaliser. # The Add and Take-away Path ## The Add and Take-away Path Once there was a garden with lots of paths and little patches of grass between. It looked a bit like this. Two children, William and Thea, made up a game. For every patch of grass they passed as they walked to the right on the picture, they added one. For every patch of grass they passed as they walked to the left on the picture, they took away one. If they went North (up on the picture), they added two for every patch of grass, and if they went South (down on the picture), they took away two for each patch of grass. They started where it says S, and ended where it says E on the picture below. They began with ten points at S, and added and took away until they reached E. These are the paths they took: Thea took the yellow path. Remember she started with $10$ points. How many points did she have when she arrived at E? William took the purple path. How many points did he have when he arrived at E? In the picture below, we've taken out the grass and just drawn a grid. You can imagine that the squares are patches of grass and the lines are the paths in the garden. We have marked different starting and end points. What would your score be if you walked along the blue path? And the red path? Make up some more paths between these start and end points. What is your score each time? What do you notice? Can you explain why? You can download a copy of the grid here. ### Why do this problem? This problem will give children a chance to generalise and make predictions. It also provides practice in simple addition and subtraction. It draws out the inverse relationship between these two operations as well as encouraging children to think about the order of operations. In relation to the March 2011 theme, 'actions on objects', we can think of 'going right' as an action performed on the starting number of points (and likewise going in the other directions). We can encourage children to find out what properties these actions have when combined one after the other. You will need copies of this sheet, which could be enlarged if more than two children are going to work using the copy. ### Possible approach You could start by showing the pictures in the first part of the problem to the whole group, making sure that William and Thea's game is understood. A small scale version could be drawn out on the playground or on the hall floor so that the game can be played practically. After this introduction, the group could work in pairs so that they are able to discuss their ideas with a partner, using copies of the sheet. Encourage them to find interesting routes that use subtraction as well as addition. Routes can be drawn using different colours but pairs may well need more than one copy of the sheet. At the end of the lesson the group can come together again and discuss their findings. They could show their most interesting and longest routes. What did they notice about the total each time? Were they surprised by this result? Why do they think this happened? ### Key questions How will you remember which routes you have tried out? How will you remember the total of each route? Can you find a more interesting way to go? Can you check your answers another way? Do you notice anything about the totals? ### Possible extension Learners could try the Stage 2 version of this problem, Journeys in Numberland. ### Possible support Learners might find it useful to have counters to help keep track of the score. Starting with $10$ counters, they could physically add and take away counters, then count them up when they arrive at the end point.<\|endoftext\|>	4.53125
858	# Square In Geometry- Properties, Shape And Formula(Area and Perimeter) ## Square In Geometry Square is a quadrilateral with four equal sides and and four right angles. Square In Geometry- Properties, Shape And Formula(Area and Perimeter) ## Properties of square:- Properties of a square are listed below. • All four sides are equal • All angles are right angle • Opposite sides are parallel to each other and equal in length. • The two diagonals are equal and bisect each other at 90°. • The diagonal bisect vertex angle. ## Shape of square:- A square is two dimensional closed figure with four equal sides and four right angles. Square In Geometry- Properties, Shape And Formula(Area and Perimeter) All four sides are equal. All angles are 90° in measure. The diagonal bisect vertex angle. ## Formulas of Square • Area • Perimeter #### Perimeter: Perimeter of a Plane Figure is the measure of the length of its boundary. The unit of perimeter is same as the unit of length i.e. centimeter(cm),meter(m) ,kilometer(km) etc. Area : Area of a Plane Figure is the measure of the surface enclosed by it. The various units of measuring area are :1 square cm(cm²), 1 square meter(m²),1 hectare etc. 1. Area of a square is = a² =1/2diagonal² where a is the length of the sides of a square. 2. Perimeter of a square = 4a as all four sides of a square are equal. 3. ΔABC=ΔBCD=ΔCDA=ΔDAB= a² /2 (consider above square) 4.ΔAOB=ΔBOC=ΔCOD=ΔDOA= a²/4 5. Diagonal AC=BD=a√2 As we know diagonals are bisect each other at 90° .So, we can apply Pythagoras theorem h²=p²+b² h²=a²+a² h² =2a² h=a√2 h is nothing but the AC=BD=h (diagonal). ### Solved Examples Example 1:-Find the area and perimeter of a square of side 8cm. Answer:-Area of a square = a² . so 8²=64 cm². perimeter of a square= 4a so perimeter of this square=48=32 cm Example 2: Find the area of a square whose diagonal is 2.9 meters long. Sol: Area of a square=1/2diagonal²=1/22.92.9=4.205sq.metres. ### Example 3:-The area of a square whose side is 3cm is _? Answer: – Area of square = a², If a is the side of a square. Here side of the square is given 3cm. Area of square = 3² =9 cm². So Answer of this question is 9 cm². ## FAQs 1.What is square shape in math? Ans: A square is two dimensional closed figure with four equal sides and four right angles. 2.What are the 4 properties of a square? Ans: 1.All four sides are equal 2.All angles are right angle 3.Opposite sides are parallel to each other. 4.The two diagonals are equal and bisect each other at 90°. 3.How can we find area of square? Ans :Area of a square is = a² =1/2diagonal² where a is the length of the sides of a square. 4: Are all square rectangles? Ans: All squares are rectangles but not all rectangles are squares. 5.How many angles are there in Square? Ans: No of angles are available in a square is 4.<\|endoftext\|>	4.53125
1,778	General Astronomy/Kepler's Laws Johannes Kepler was a mathematician who attempted to derive a set of fundamental principles which would explain the motions of the planets. He believed in the heliocentric view of the solar system proposed by Copernicus, and he also possessed a rich set of observations of the planets made by Tycho Brahe. After twenty years of painstaking attempts and various discarded ideas based on geometry, he finally arrived at a mathematical model of orbital motions based on the ellipse. Kepler summarized his findings in the form of three laws of planetary motion, frequently referred to as Kepler's First, Second and Third Laws, respectively: - Kepler's First Law, also known as The Law of Ellipses — The orbits of the planets are ellipses, with the sun at one focus. - Kepler's Second Law, or The Law of Equal Areas in Equal Time — The line between a planet and the sun sweeps out equal areas in the plane of the planet's orbit over equal times. - Kepler's Third Law, or The Law of Harmony — The time required for a planet to orbit the sun, called its period, is proportional to half the long axis of the ellipse raised to the 3/2 power. The constant of proportionality is the same for all the planets. It is often called the Law of Harmony because it shows a harmonic relation between distances and periods. At that time he developed these laws, there was not yet a developed theory of gravity capable of explaining why the planets moved as they were observed to. Later, Isaac Newton, using his universal inverse-square law theory of gravity, was able to show how Kepler's Laws fit into a scientific theory of celestial mechanics. An ellipse is a shape formed by taking a diagonal slice through a cone. It is essentially the shape of a circle viewed at an angle. An ellipse can be drawn by taking a piece of paper, two push-pins, a loop of string, and a pencil. The two pins are pushed through the paper into a suitable surface, providing the two foci for the ellipse. They should be closer together than the loop is long. The loop of string is placed around the base of these pins, leaving some slack. The pencil is now placed so that the pins and the loop form a triangle with a slight tension on the string. Now try to draw a shape by moving the pencil about the pins while keeping the string taunt. The result should be an ellipse. The shape of the ellipse can be varied either by moving the pins closer together or further apart. This shape, according to Kepler, defines the path that a planet takes when it orbits the Sun. \|Kepler's First Law - A planet orbits the Sun on an ellipse with the Sun at one focus.\| Kepler's Second Law says in brief that an object speeds up as it gets closer to the Sun and slows down as it moves further away. Where the distance from the Sun to the orbital path is longer, only a smaller arc needs to be traversed to sweep out an area that requires a wider arc near the Sun. As the planet moves closer to the Sun along its orbit the gravitational force works to increase the velocity. In contrast, as the planet is moving further away, the gravity of the Sun gradually decelerates the body and it is slowed down. \|Kepler's Second Law - A planet in orbit about the Sun sweeps out equal areas in the same time interval .\| A line that divides an ellipse in half and passes through the widest part of the ellipse is called the major axis. A line perpendicular to this axis and dividing the ellipse in half is called the minor axis. Half the length of the major axis is called the semi-major axis, and is represented by . The period required for a planet to complete one full orbit is represented by . The relationship between the period P and the length of the semimajor axis is known as Kepler's Third Law, and can be represented as follows: where the symbol ∝ means "proportional to", and implies that there is a direct mathematical relationship between the period squared and the length of the semi-major axis cubed. The Second and Third Laws provide a basis for calculating the period of any planet orbiting the Sun, as well as determining where the planet will be located along the orbital path. Eccentricity and Orbital Paths The ratio of the distance of a focus from the center of an ellipse to the semi-major axis is called the eccentricity of the orbit. When the two foci of the ellipse are on top of each other, the eccentricity is exactly 0.0 and the shape is a circle. As the eccentricity increases, the orbiting planet moves much further away than at the closest approach. The orbital eccentricities for planets in our Solar system vary from as much as 0.21 for Mercury down to 0.0068 for Venus. The scientific name for the point of closest approach is the periapsis, while the most distance separation is the apoapsis. In the case of planets orbiting the Sun, these are called the perihelion and aphelion, respectively. (The -helion suffix comes from the Greek name for the Sun deity, Helios. This word is also the source of the name for the element Helium.) \|Two elliptical orbits with the same major axis but different eccentricity.\| Perhaps the most counter-intuitive aspect of the Third Law is that for any two identical bodies orbiting the Sun with the same semi-major axis, the orbital period is the same. This is true even if one is orbiting in a perfect circle and the other has an orbit that is highly elliptical (has a relatively high eccentricity). The elliptical shape will fit entirely within the circle except at two points (the ends of the major axis, at which the two curves will be tangent), so it is actually a shorter orbital path. However the aphelion of the ellipse will be located further from the Sun, so the planet will spend more time traversing the distant section of the orbit. The shorter orbit and the slower traverse of the aphelion compensate for each other, resulting in an identical period with the circular orbit. Some Examples Using Kepler's Third Law Originally, Kepler's Third Law was used to describe the motions of the planets around the Sun. As it turns out, it also works very well with regards to other two-body orbital systems like the orbits of moons around Jupiter, or the orbits of binary stars about the center of mass of their system. In all of these cases the period of the orbit squared is proportional to the length of the semi-major axis cubed, with the differences between the orbital systems reflected in the constant of proportionality. In this section we will consider the special case of the planets going around the Sun. If we choose to measure the length of the semi-major axis of an orbit in astronomical units (abbreviated AU, where 1 AU is the distance from the Earth to the Sun) and we measure the orbital period in years (abbreviated as ), then we can express Kepler's Third Law as where is measured in years and is measured in astronomical units. What follows are a few example of how to use this equation. Mars' Orbital Period Repeated measurements of Mars' orbit have determined that the semi-major axis of its orbit has a length of 1.52 AU. How long does it take for Mars to orbit the Sun once? Solution: In this question, we are being asked for the orbital period of Mars. We know from reading the question that the length of the semi-major axis is 1.52 AU. Solving Kepler's Third Law for the period gives us which tells us that it takes 1.87 years for Mars to go around the Sun once. An Unknown Asteroid's Orbital Semi-Major Axis An amateur astronomer spends several months tracking an asteroid and is able to determine that it takes approximately 3/4 of a year for it to orbit once around the Sun. What is the semi-major axis of this asteroid's orbit? Solution: Reading the question tells us that the orbital period of the asteroid is 3/4 of a year, or 0.75 . We are expected to find the length of the semi-major axis , so we need to solve Kepler's Third Law for that. Doing so gives us which means that is has a semi-major axis somewhere between that of Venus' orbit and Earth's orbit.<\|endoftext\|>	4.25
2,421	# Circle. Circle Circle Tangent Theorem 11-1 If a line is tangent to a circle, then the line is perpendicular to the radius drawn to the point of. ## Presentation on theme: "Circle. Circle Circle Tangent Theorem 11-1 If a line is tangent to a circle, then the line is perpendicular to the radius drawn to the point of."— Presentation transcript: Circle Circle Tangent Theorem 11-1 If a line is tangent to a circle, then the line is perpendicular to the radius drawn to the point of tangency AB is tangent to the circle since the segment touches the circle once and intersects with the radius at a 90° angle. Theorem 11-2: If a line in the plane of a circle is perpendicular to a radius at its endpoint on the circle, then the line is tangent to the circle. Theorem 11-2: If a line in the plane of a circle is perpendicular to a radius at its outer endpoint, then the line is tangent to the circle. I is perpendicular to radius OB at B What must be the length of LM for this segment to be tangent line of the circle with center N? The answer is: For segment LM to be a tangent it intersect the radius MN at 90°. Therefore triangle LMN would have to be a right triangle and the Pythagorean theorem provides the necessary length for LM to be a tangent. Theorem 11-3 The two segments tangent to a circle from a point outside the circle are congruent AM = BM CD = CE Theorem 11-3 if two tangents are drawn on a circle and they cross, the lengths of the two tangents (from the point where they touch the circle to the point where they cross) will be the same. Theorem 11-4 Within a circle or in congruent circles Congruent central angles have congruent chords Congruent chords have congruent arcs Congruent arcs have congruent central angles Theorem 11-5 Within a circle or in congruent circles: Chords equidistant from the center are congruent Congruent chords are equidistant from the center Theorem 11-5 When congruent chords are in the same circle, they are equidistant from the center. Theorem 11-6 In a circle, a diameter that is perpendicular to a chord bisects the chord and its arcs Theorem 11-7 In a circle, a diameter that bisects a chord (that is not a diameter) is perpendicular to the chord Theorem 11-8 In a circle, the perpendicular bisector of a chord contains the center of the circle Inscribed Angle = Intercepted Arc Theorem 11-9 The measure of an inscribed angle is half the measure of its intercepted arc. Inscribed Angle = Intercepted Arc Corollaries to the Inscribed Angle Theorem Two inscribed angles that intercept the same arc are congruent An angle inscribed in a semicircle is a right angle. The opposite angles of a quadrilateral inscribed in a circle are supplementary A+C=180º D+B=180º A B 900 140º In any cyclic Quadrilateral opposite corners sum to 1800 C 900 40º D Any 4 points on a circle joined to form a quadrilateral Theorem 11-10 The measure of an angle formed by a tangent and a chord is half the measure of the intercepted arc. Two secants extend from the same point and intersect the circle as shown in the diagram below. What is the value of x? The answer is: x = ½(140-50) x = ½(90) x = 45° Find the measure of angle k, where the two secant segments intersect. The answer is: K = ½(90-30) K = ½(60) K = 30° Theorem 11-11 The measure of an angle formed by two lines that : Theorem 11-11 The measure of an angle formed by two lines that : Intersect inside a circle is half the sum of the measures of the intercepted arcs. Theorem 11-11 2. Intersect outside a circle is half the difference of the measures of the intercepted arcs. Theorem 11-11 Theorem 11-11 Use the theorem for the intersection of a tangent and a secant to find the measure of the angle formed by the intersection of the tangent and the secant. Two Tangents: <ABC is formed by two tangents intersecting outside of circle O. The intercepted arcs are minor arc AC and major arc AC. These two arcs together comprise the entire circle. Theorem 11-12 For a given point circle, the product of the lengths of the two segments from the point to the circle is constant along any line through the point and circle. (DM+ME)DM = (FN+NE)NE (QK+KR)=SR² In the circle below, the chord segments have the following lengths: A= 6, C=3, D=4. Use the theorem for the product of chord segments to find the value of D. Theorem 11-13 An equation of a circle with center (h ,k) and radius r is (x-h)² = r² (standard Form) (x - h)2 + (y - k)2 = r2 (x - h)2 + (y - k)2 = r2 (x - 3)2 + (y - 5)2 = (3sqrt(2))r2 (x - 3)2 + (y - 5)2 = 18 Angles formed by drawing lines from the ends of the diameter of a circle to its circumference form a right angle. So c is a right angle. Definition: A circle is the set of points equidistant from a point C(h,k) called the center. The fixed distance r from the center to any point on the circle is called the radius.The standard equation of a circle with center C(h,k) and radius r is as follows: (x - h)2 + (y - k)2 = r2 Example 1: Find the equation of a circle whose center is at (2, - 4) and radius 5.Solution to Example 1: given (h , k ) = (2 , - 4) and r = 5substitute h, k and r in the standard equation (x - 2)2 + (y - (- 4))2 = 52 (x - 2)2 + (y + 4)2 = 25Go here and set h, k and r parameters into applet and plot the circle. Verify graphically that the equation is that of a circle with the given center and radius.Matched Exercise 1: Find the equation of a circle whose center is at (2 , - 4) and radius 3.Answers. Example 2: Find the equation of a circle that has a diameter with the endpoints given by the points A(-1 , 2) and B(3 , 2).Solution to Example 2: The center of the circle is the midpoint of the line segment making the diameter AB. The midpoint formula is used to find the coordinates of the center C of the circle. x coordinate of C = (-1 + 3) /2 = 1 y coordinate of C = (2 + 2) / 2 = 2The radius is half the distance between A and B. r = (1/2) ([3 - (-1)]2 + [2 - 2]2 )1/2 = (1/2)( )1/2 = 2The coordinate of C and the radius are used in the standard equation of the circle to obtain the equation: (x - 1)2 + (y - 2)2 = 22 (x - 1)2 + (y - 2)2 = 4 Go here and set h, k and r parameters into applet and plot the circle. Verify graphically that the equation is that of a circle with the diameter as given above.Matched Exercise 2: Find the equation of a circle that has a diameter with the endpoints given by A(0 , -2) and B(0 , 2).Answers. Example 3: Find the center and radius of the circle with equation x2 - 4x + y2 - 6y + 9 = 0Solution to Example 3: In order to find the center and the radius of the circle, we first rewrite the given equation into the standard form as given above in the definition. Put all terms with x and x2 together and all terms with y and y2 together using brackets. (x2 - 4x) +( y2 - 6y) + 9 = 0We now complete the square within each bracket.. (x2 - 4x + 4) ( y2 - 6y + 9) = 0 (x - 2)2 + ( y - 3) = 0Simplify and write in standard form (x - 2)2 + ( y - 3)2 = 4 (x - 2)2 + ( y - 3)2 = 22 We now compare this equation and the standard equation to obtain. center at C(h , k) = C(2 , 3) and radius r = 2Matched Exercise 3: Find the center and radius of the circle with equation x2 - 2x + y2 - 8y + 1 = 0Answers. Example 4: Is the point P(3 , 4) inside, outside or on the circle with equation (x + 2)2 + ( y - 3)2 = 9 Solution to Example 4: We first find the distance from the center of the circle to point P. center C at (-2 , 3) radius r = (9)1/2 = 3 distance from C to P = ([3 - (-2)]2 + [4 - 3]2)1/2 = (52 +12)1/2 = (26)1/2Since the distance from C to P is (26)1/2 which approximately equal to 5.1 is greater than the radius r = 3, point P is outside the circle. You can check your answer graphically using this applet.Matched Exercise 4: Is the point P(-1 , -3) inside, outside or on the circle with equation (x - 1)2 + ( y + 3)2 = 4 Alternate Segment Theorem Download ppt "Circle. Circle Circle Tangent Theorem 11-1 If a line is tangent to a circle, then the line is perpendicular to the radius drawn to the point of." Similar presentations<\|endoftext\|>	4.65625
456	Giant Prehistoric Shark Teeth Wash Ashore In North Carolina Recent storms off the coast of North Carolina unearthed fossilized teeth of a Megalodon, the largest species of shark to ever call the ocean home. On average, this prehistoric shark species grew to nearly 60 feet in length, which is about three times the size of modern great whites. "Oh my God, like I said, I felt like I was a lottery winner or something," Denny Bland, one of the lucky fossil finders, told NBC affiliate WITN. "It's like I'm the first one to touch that since it fell out of his mouth back in the day." Megalodons roamed the ancient oceans during the Miocene-Pliocene period, and went extinct roughly 2.6 million years ago, according to a 2014 study published in the journal PLOS ONE by researchers from the University of Florida. Although researchers are unsure why exactly the extremely large marine predators went extinct, many scientists attribute their mass exit to a combination of cooling ocean temperatures, dwindling food supply and new predatory competition. (Scroll to read more...) Megalodons are represented in the fossil record primarily by teeth. The recent teeth discovered in North Topsail Beach and Surf City, North Carolina are roughly the size of an adult's palm, according to WITN. The fossil record indicates that Megalodon's teeth were similar to those of most other sharks, and would have been located in rows that rotated into use as needed. For example, when they were broken, fell out or became to worn down. In total, most sharks have between three and five rows of teeth at all times. The first two rows do the majority of the work and are responsible for catching prey. Researchers believe that Megalodons primarily hunted whales. Since little else is known about this ancient species, the recent finding in North Carolina may unlock more secrets about the ginormous, prehistoric sharks. Saber-Toothed Cat: Ultra-Sharp Teeth Slow to Emerge For more great nature science stories and general news, please visit our sister site, Headlines and Global News (HNGN). -Follow Samantha on Twitter @Sam_Ashley13<\|endoftext\|>	3.8125
1,644	# Search by Topic #### Resources tagged with Making and proving conjectures similar to Logoland - Sequences: Filter by: Content type: Stage: Challenge level: ### There are 37 results Broad Topics > Using, Applying and Reasoning about Mathematics > Making and proving conjectures ### Triangles Within Triangles ##### Stage: 4 Challenge Level: Can you find a rule which connects consecutive triangular numbers? ### Happy Numbers ##### Stage: 3 Challenge Level: Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general. ### Triangles Within Pentagons ##### Stage: 4 Challenge Level: Show that all pentagonal numbers are one third of a triangular number. ### Multiplication Square ##### Stage: 3 Challenge Level: Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice? ### Always a Multiple? ##### Stage: 3 Challenge Level: Think of a two digit number, reverse the digits, and add the numbers together. Something special happens... ### A Little Light Thinking ##### Stage: 4 Challenge Level: Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights? ### Dice, Routes and Pathways ##### Stage: 1, 2 and 3 This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . . ### Loopy ##### Stage: 4 Challenge Level: Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture? ### An Introduction to Magic Squares ##### Stage: 1, 2, 3 and 4 Find out about Magic Squares in this article written for students. Why are they magic?! ### Rotating Triangle ##### Stage: 3 and 4 Challenge Level: What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle? ### Triangles Within Squares ##### Stage: 4 Challenge Level: Can you find a rule which relates triangular numbers to square numbers? ### Problem Solving, Using and Applying and Functional Mathematics ##### Stage: 1, 2, 3, 4 and 5 Challenge Level: Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information. ### To Prove or Not to Prove ##### Stage: 4 and 5 A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples. ### On the Importance of Pedantry ##### Stage: 3, 4 and 5 A introduction to how patterns can be deceiving, and what is and is not a proof. ### Charlie's Mapping ##### Stage: 3 Challenge Level: Charlie has created a mapping. Can you figure out what it does? What questions does it prompt you to ask? ### Alison's Mapping ##### Stage: 4 Challenge Level: Alison has created two mappings. Can you figure out what they do? What questions do they prompt you to ask? ### What's Possible? ##### Stage: 4 Challenge Level: Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make? ### Exploring Quadratic Mappings ##### Stage: 4 Challenge Level: Explore the relationship between quadratic functions and their graphs. ### Close to Triangular ##### Stage: 4 Challenge Level: Drawing a triangle is not always as easy as you might think! ### Multiplication Arithmagons ##### Stage: 4 Challenge Level: Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons? ### Curvy Areas ##### Stage: 4 Challenge Level: Have a go at creating these images based on circles. What do you notice about the areas of the different sections? ### Pericut ##### Stage: 4 and 5 Challenge Level: Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts? ### Center Path ##### Stage: 3 and 4 Challenge Level: Four rods of equal length are hinged at their endpoints to form a rhombus. The diagonals meet at X. One edge is fixed, the opposite edge is allowed to move in the plane. Describe the locus of. . . . ### Few and Far Between? ##### Stage: 4 and 5 Challenge Level: Can you find some Pythagorean Triples where the two smaller numbers differ by 1? ### Polycircles ##### Stage: 4 Challenge Level: Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon? ### Janine's Conjecture ##### Stage: 4 Challenge Level: Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . . ##### Stage: 4 Challenge Level: The points P, Q, R and S are the midpoints of the edges of a non-convex quadrilateral.What do you notice about the quadrilateral PQRS and its area? ### DOTS Division ##### Stage: 4 Challenge Level: Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}. ### Pentagon ##### Stage: 4 Challenge Level: Find the vertices of a pentagon given the midpoints of its sides. ##### Stage: 4 Challenge Level: The points P, Q, R and S are the midpoints of the edges of a convex quadrilateral. What do you notice about the quadrilateral PQRS as the convex quadrilateral changes? ### Tri-split ##### Stage: 4 Challenge Level: A point P is selected anywhere inside an equilateral triangle. What can you say about the sum of the perpendicular distances from P to the sides of the triangle? Can you prove your conjecture? ##### Stage: 4 Challenge Level: Points D, E and F are on the the sides of triangle ABC. Circumcircles are drawn to the triangles ADE, BEF and CFD respectively. What do you notice about these three circumcircles? ### Helen's Conjecture ##### Stage: 3 Challenge Level: Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true? ### How Old Am I? ##### Stage: 4 Challenge Level: In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays? ### Consecutive Negative Numbers ##### Stage: 3 Challenge Level: Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers? ### Exploring Simple Mappings ##### Stage: 3 Challenge Level: Explore the relationship between simple linear functions and their graphs. ### Epidemic Modelling ##### Stage: 4 and 5 Challenge Level: Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths.<\|endoftext\|>	4.46875
1,077	### Is the relation a function? ```Formalizing Relations and Functions Objectives: To determine whether a relation is a function To find domain and range and use function notation Relation: a pairing of numbers in one set with numbers in another set Domain: the set of x-values Range: the set of y-values Problem #1: Identifying Functions Using Mapping Diagram Identify the domain and range of the relation. Represent the relation with a mapping diagram. Is the relation a function? A) {(-2, 0.5), (0, 2.5), (4, 6.5), (5, 2.5)} Problem #1: Identifying Functions Using Mapping Diagram Identify the domain and range of the relation. Represent the relation with a mapping diagram. Is the relation a function? B) {(6,5), (4,3), (6,4), (5,8)} Problem #1 Got It? Identify the domain and range of each relation. Represent the relation with a mapping diagram. Is the relation a function? Vertical Line Test Another method used to determine if a relation is a function. Look at the graph and determine if any vertical line passes through more than one point of the graph, then the relation is not a function. Problem #2: Identifying Functions Using the Vertical Line Test Is the relation a function? Use the Vertical Line Test. A) {(-4, 2), (-3, 1), (0,-2), (-4, -1), (1, 2)} Problem #2: Identifying Functions Using the Vertical Line Test Is the relation a function? Use the Vertical Line Test. B) = − 2 + 3 Problem #2 Got It? Is the relation a function? Use the Vertical Line Test. Function Notation y = -3x + 1 is the same as = −3 + 1 - f(x) replaces y f is the name of the function, not a variable Used to emphasize that the function value f(x) depends on the independent variable x. - Other letters besides f can also be used, such as g and h. Problem #3: Evaluating a Function A) The function w(x) = 250x represents the number of words w(x) you can read in x minutes. How many words can you read in 8 minutes? Problem #3: Evaluating a Function B) The function Y(x) = 1 3 represents the number of yards Y(x) in x feet. How many yards are there in 1 mile? Problem #3 Got It? The function T(x) = 65x represents the number of words T(x) that Rachel can type in x minutes. How many words can she type in 7 minutes? Homework Textbook Page 271; #1 – 3, 5 – 17 All Continued… Objectives: To determine whether a relation is a function To find domain and range and use function notation Problem #4: Finding the Range of a Function A) Multiple Choice The domain of f(x) = 1.5x + 4 is {1, 2, 3, 4}. What is the range? Problem #4: Finding the Range of a Function B) The domain of g(x) = 4x – 12 is {1, 3, 5, 7}. What is the range? Problem #4 Got It? What is the range of f(x) = 3x – 2 with domain {1, 2, 3, 4}? Problem #5: Identifying a Reasonable Domain and Range A) You have 3 qt. of paint to paint the trim in your house. A quart of paint covers 100 2 . The function () = 100 represents the area A(q), in square feet, that q quarts of paint cover. What domain and range are reasonable for the function? What is the graph of the function? Problem #5: Identifying a Reasonable Domain and Range B) Lorena has 4 rolls of ribbon to make party favors. Each roll can be used to make 30 favors. The function F(r) = 30r represents the number of favors F(r) that can be made with r rolls. What are a reasonable domain and range of the function? What is a graph of the function? Problem #5 Got It? A car can travel 32 miles for each gallon of gasoline. The function d(x) = 32x represents the distance d(x), in miles, that the car can travel with x gallons of gasoline. The car’s fuel tank holds 17 gal. Find a reasonable domain and range for each function. Then graph the function. Homework 4 – 6 Worksheet ```<\|endoftext\|>	4.5625
843	Can we make New Zealand pest-free? Lesson 5: Evaluate: So what? Urban ecosanctuary ZEALANDIA, with support from WWF New Zealand, has produced a comprehensive teaching resource supporting schools to explore the pest-free vision with students. This series of lesson plans focuses on students becoming actively involved in contributing to a pest-free New Zealand. This lesson makes connections by mapping biodiversity and its relationship to tracking tunnels and traps, includes a post-unit assessment test and supports embedding the learning and action by reflecting on the process. - What have we discovered? - What is our long-term vision? - Can we track and trap at our homes, at our school, in our community? - Can we formulate a plan to share our knowledge with our community? - What impact would these actions have? Download lesson plan: Lesson 5: Evaluate: So what? Additional supporting resources He tikanga – te reo Māori and English terms – The activities in this ZEALANDIA lesson plan are intended to be used throughout the whole unit, not as a one-off activity. The intent is to help students become familiar with te reo Māori and English terms mentioned throughout the resource. Mathematics and statistics in a real context – The two main elements of this lesson are analysing examples and gathering other sources of data, and collecting and displaying data to analyse to effect real-world outcomes. It can be adapted for higher levels of learning and shows the relevance of mathematical skills in a real-life context. ZEALANDIA lesson plans series - Inspire – overview of the resource - Lesson 1: Investigate: Why do we need to help? - Lesson 2 and 3: Investigate: What is present? - Lesson 4: Instigate: What’s the solution? - Lesson 5: Evaluate: So what? Nature of science Conservation efforts are improved when we understand how living organisms interact and how to effectively target pest species. In the activity Making a tracking tunnel, students monitor the presence of pest species in a neighbouring gully or their school grounds. Careful observation is an important part of science, as outlined in the activity Observation: learning to see. In the activity Mapping the future, students are encouraged to connect and create a sense of belonging by exploring changes that have taken place in their local environment in the last 50–100 years and to plan for the next 50 years. Related Hub's PLD In the recorded online PD session Teachers using the Hub – Bird conservation and literacy, teacher Kim MacPherson talks about the Science Learning Hub’s resources and how she used a literacy approach to engage and explore science issues with year 7 and 8 students. ZEALANDIA has many other educational resources. For advice or assistance in implementing this programme please contact the ZEALANDIA Education team, [email protected]. If your school is in the Wellington region and you would like support to run this programme, access ZEALANDIA’s free Outreach programme by contacting [email protected]. If you want to get involved at an individual or community level, check out Predator Free New Zealand Trust. Read about DOC’s work with Predator Free 2050. The Backyard Sanctuaries website has lots of information on trapping, our wildlife, pests, workshops and more. ZEALANDIA is the world’s first fully fenced urban ecosanctuary. It has an extraordinary 500-year vision to restore a Wellington valley’s forest and freshwater ecosystems as closely as possible to their pre-human state. For 50 years, WWF has been protecting the future of nature as the world’s leading conservation organisation. WWF’s unique way of working combines global reach with a foundation in science, involves action at every level from local to global and ensures the delivery of innovative solutions that meet the needs of both people and nature.<\|endoftext\|>	3.75
663	In AB and BC legs of ABC triangle are points S and R, respectively, such that AS=3/7 of AB and BR=3/8 of BC. Find the area of SBR triangle if the area of ABC is 126 sq.cm 2. Okay, I understand what you mean. (English is not my native language also.) You are saying, In AB and BC (not CD) legs of ABC triangle are points S and R, respectively, such that AS=3/7 of AB and BR=3/8 of BC. Find the area of SBR triangle if the area of ABC is 126 sq.cm. Since only sides are given with values, and the area of triangle ABC, only, and since triangles ABC and SBR have the same angle B, then we will use the formula for the area of any triangle, A = (1/2)absin(theta) where theta is the angle between sides a and b. In triangle ABC, Let x = AB And y = BC So, Area of ABC = (1/2)(x)(y)sinB = 126 xysinB = 252 ------------------------(i) ---------------------------------- In triangle SBR, SB = (4/7) of AB because AS is (3/7) of AB So, SB = (4/7)x BR = (3/8) of BC BR = (3/8)y Area of SBR = (1/2)[(4/7)x][(3/8)y]sinB = [12/(256)](xy)sinB = (3/28)(xy)sinB Substitute into that the xysinB = 252, = (3/28)(252) 3. Hello, blertta! In AB and CD legs of ABC triangle are picked up two points S and R where: .$\displaystyle AS\,=\,\frac{3}{7}(AB),\;BR\,=\,\frac{3}{8}(BC)$ Find the area of $\displaystyle \Delta SBR$ if the area of $\displaystyle \Delta ABC = 126$ cm². Code: C * * * 5/8 * * * * R o * * ** * * * * * 3/8 * * * * * * * * * * * * B * * * * * * o * * * * * * * * * A 4/7 S 3/7 $\displaystyle \Delta SBR$'s height is $\displaystyle \frac{3}{8}$ of $\displaystyle \Delta ABC$'s height. $\displaystyle \Delta SBR$'s base is $\displaystyle \frac{4}{7}$ of $\displaystyle \Delta ABC$'s base. Therefore: .$\displaystyle \Delta SBR \;=\;\left(\frac{3}{8}\right)\left(\frac{4}{7}\rig ht)(126) \;=\;27\text{ cm}^2$<\|endoftext\|>	4.40625
375	A ringwork is a form of fortified defensive structure, usually circular or oval in shape. Ringworks are essentially motte-and-bailey castles minus the motte. Defences were usually earthworks in the form of a ditch and bank surrounding the site. Ringworks originated in Germany in the 10th century as an early form of medieval castle and at first were little more than a fortified manor house. They appeared in England just prior to the Norman conquest and large numbers were built during the late 11th and early 12th centuries. More elaborate versions comprise a ringwork and bailey, the ringwork replacing the more usual motte and the bailey acting as a military stronghold. A survey published in 1969 identified 198 ringwork castles in England and Wales, with a further 50 sites that were considered to possibly be ringworks. \|Wikimedia Commons has media related to Ringworks.\| - Friar, Stephen (2003). A Sutton Companion to Castles. Sutton Publishing. p. 246. ISBN 978-0-7509-3994-2.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles> - Darvill, Timothy (2008). Oxford Concise Dictionary of Archaeology, 2nd ed., Oxford University Press, Oxford and New York, p. 386. ISBN 978-0-19-953404-3. - King, David James Cathcart; Alcock, Leslie (1969). "Ringworks of England and Wales". Château Galliard. 3: 90–127.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles> \|This article relating to archaeology in Europe is a stub. You can help Infogalactic by expanding it.\|<\|endoftext\|>	3.71875
609	Under The Surface: The Story Of Microbes How important are the creatures we can’t see? Microbes (or microorganisms) make up an essential part of our ecosystem. Rachel Spietz, an oceanographer and graduate student at the University of Washington, studies these too-small-to-see (but essential) parts of our ecosystem. She is presenting an interactive exhibit at Life Sciences Research this weekend at Pacific Science Center. While this event often focuses on human health, Spietz will be focusing on smaller critters. Microbes, defined as “microorganisms, especially a pathogenic bacterium,” are often thought of as “bad guys,” especially in the context of disease. “Microbes are essential to developing medicines, which many people know,” says Spietz, because learning about how diseases work helps us fight them. Spietz’s work focuses mostly on the ecological impact of microbes, where they often have a positive role. “One of my favorite things about microbes is their ability to break down landfills,” Spietz explains. “This particularly affects ocean habitats, which I study.” Though garbage often piles up, it cannot do so indefinitely; microbes “digest” the garbage, bringing it back through the carbon cycle naturally. Oceanographers and biologists alike are fascinated by this phenomenon. Microbes keep land, beach, and ocean environments viable. Through this process, they are cycling nutrients back to us. Microbes connect us to our environment. A microbe’s absorption of food can predict things as complex as volcanic eruption. Several years ago, Spietz worked on a project looking at the microbes growing around the Axial Seamount, an underwater volcano,” she says. “I pulled microbes out of the water and tested them; my data helped geologists learn about past and predict future eruptions.” The team collected lava from two kilometers under the sea, and Spietz studied the microbes coming up to the surface. “By seeing patterns in how the microbes use energy, you can start to figure out the history and future of the volcano.” You can watch this process happening live here. While working on these kinds of teams, Spietz got to learn from geologists, physicists, microbiologists, and oceanographers. “Oceanography is very broad and interdisciplinary,” she says. That’s also one of her favorite parts about past Life Sciences Research Weekends. “The sheer number of disciplines represented was impressive.” Even more exciting, perhaps, was the engagement of young guests. “Parents actually had to pull their kids away from these educational displays.” Pacific Science Center is proud to host our 10th annual Life Sciences Research Weekend, November 11-13, 2016. Register here for a chance to see Rachel Spietz’s microbes in action.<\|endoftext\|>	4.03125
5,542	Illusion of control The illusion of control is the tendency for people to overestimate their ability to control events; for example, it occurs when someone feels a sense of control over outcomes that they demonstrably do not influence. The effect was named by psychologist Ellen Langer and has been replicated in many different contexts. It is thought to influence gambling behavior and belief in the paranormal. Along with illusory superiority and optimism bias, the illusion of control is one of the positive illusions. The illusion might arise because people lack direct introspective insight into whether they are in control of events. This has been called the introspection illusion. Instead they may judge their degree of control by a process that is often unreliable. As a result, they see themselves as responsible for events when there is little or no causal link. In one study, college students were in a virtual reality setting to treat a fear of heights using an elevator. Those who were told that they had control, yet had none, felt as though they had as much control as those who actually did have control over the elevator. Those who were led to believe they did not have control said they felt as though they had little control. Psychological theorists have consistently emphasized the importance of perceptions of control over life events. One of the earliest instances of this is when Adler argued that people strive for proficiency in their lives. Heider later proposed that humans have a strong motive to control their environment and Wyatt Mann hypothesized a basic competence motive that people satisfy by exerting control. Wiener, an attribution theorist, modified his original theory of achievement motivation to include a controllability dimension. Kelley then argued that people's failure to detect noncontingencies may result in their attributing uncontrollable outcomes to personal causes. Nearer to the present, Taylor and Brown argued that positive illusions, including the illusion of control, foster mental health. The illusion is more common in familiar situations, and in situations where the person knows the desired outcome. Feedback that emphasizes success rather than failure can increase the effect, while feedback that emphasizes failure can decrease or reverse the effect. The illusion is weaker for depressed individuals and is stronger when individuals have an emotional need to control the outcome. The illusion is strengthened by stressful and competitive situations, including financial trading. Although people are likely to overestimate their control when the situations are heavily chance-determined, they also tend to underestimate their control when they actually have it, which runs contrary to some theories of the illusion and its adaptiveness. People also showed a higher illusion of control when they were allowed to become familiar with a task through practice trials, make their choice before the event happens like with throwing dice, and when they can make their choice rather than have it made for them with the same odds. People are more likely to show control when they have more answers right at the beginning than at the end, even when the people had the same number of correct answers. At times, people attempt to gain control by transferring responsibility to more capable or “luckier” others to act for them. By forfeiting direct control, it is perceived to be a valid way of maximizing outcomes. This illusion of control by proxy is a significant theoretical extension of the traditional illusion of control model. People will of course give up control if another person is thought to have more knowledge or skill in areas such as medicine where actual skill and knowledge are involved. In cases like these it is entirely rational to give up responsibility to people such as doctors. However, when it comes to events of pure chance, allowing another to make decisions (or gamble) on one's behalf, because they are seen as luckier is not rational and would go against people's well-documented desire for control in uncontrollable situations. However, it does seem plausible since people generally believe that they can possess luck and employ it to advantage in games of chance, and it is not a far leap that others may also be seen as lucky and able to control uncontrollable events. In one instance, a lottery pool at a company decides who picks the numbers and buys the tickets based on the wins and losses of each member. The member with the best record becomes the representative until they accumulate a certain number of losses and then a new representative is picked based on wins and losses. Even though no member is truly better than the other and it is all by chance, they still would rather have someone with seemingly more luck to have control over them. In another real-world example, in the 2002 Olympics men's and women's hockey finals, Team Canada beat Team USA but it was later believed that the win was the result of the luck of a Canadian coin that was secretly placed under the ice before the game. The members of Team Canada were the only people who knew the coin had been placed there. The coin was later put in the Hockey Hall of Fame where there was an opening so people could touch it. People believed they could transfer luck from the coin to themselves by touching it, and thereby change their own luck. The illusion of control is demonstrated by three converging lines of evidence: 1) laboratory experiments, 2) observed behavior in familiar games of chance such as lotteries, and 3) self-reports of real-world behavior. One kind of laboratory demonstration involves two lights marked "Score" and "No Score". Subjects have to try to control which one lights up. In one version of this experiment, subjects could press either of two buttons. Another version had one button, which subjects decided on each trial to press or not. Subjects had a variable degree of control over the lights, or none at all, depending on how the buttons were connected. The experimenters made clear that there might be no relation between the subjects' actions and the lights. Subjects estimated how much control they had over the lights. These estimates bore no relation to how much control they actually had, but was related to how often the "Score" light lit up. Even when their choices made no difference at all, subjects confidently reported exerting some control over the lights. Ellen Langer's research demonstrated that people were more likely to behave as if they could exercise control in a chance situation where "skill cues" were present. By skill cues, Langer meant properties of the situation more normally associated with the exercise of skill, in particular the exercise of choice, competition, familiarity with the stimulus and involvement in decisions. One simple form of this effect is found in casinos: when rolling dice in a craps game people tend to throw harder when they need high numbers and softer for low numbers. In another experiment, subjects had to predict the outcome of thirty coin tosses. The feedback was rigged so that each subject was right exactly half the time, but the groups differed in where their "hits" occurred. Some were told that their early guesses were accurate. Others were told that their successes were distributed evenly through the thirty trials. Afterwards, they were surveyed about their performance. Subjects with early "hits" overestimated their total successes and had higher expectations of how they would perform on future guessing games. This result resembles the irrational primacy effect in which people give greater weight to information that occurs earlier in a series. Forty percent of the subjects believed their performance on this chance task would improve with practice, and twenty-five percent said that distraction would impair their performance. Another of Langer's experiments—replicated by other researchers—involves a lottery. Subjects are either given tickets at random or allowed to choose their own. They can then trade their tickets for others with a higher chance of paying out. Subjects who had chosen their own ticket were more reluctant to part with it. Tickets bearing familiar symbols were less likely to be exchanged than others with unfamiliar symbols. Although these lotteries were random, subjects behaved as though their choice of ticket affected the outcome. Participants who chose their own numbers were less likely to trade their ticket even for one in a game with better odds. Another way to investigate perceptions of control is to ask people about hypothetical situations, for example their likelihood of being involved in a motor vehicle accident. On average, drivers regard accidents as much less likely in "high-control" situations, such as when they are driving, than in "low-control" situations, such as when they are in the passenger seat. They also rate a high-control accident, such as driving into the car in front, as much less likely than a low-control accident such as being hit from behind by another driver. Ellen Langer, who first demonstrated the illusion of control, explained her findings in terms of a confusion between skill and chance situations. She proposed that people base their judgments of control on "skill cues". These are features of a situation that are usually associated with games of skill, such as competitiveness, familiarity and individual choice. When more of these skill cues are present, the illusion is stronger. Suzanne Thompson and colleagues argued that Langer's explanation was inadequate to explain all the variations in the effect. As an alternative, they proposed that judgments about control are based on a procedure that they called the "control heuristic". This theory proposes that judgments of control to depend on two conditions; an intention to create the outcome, and a relationship between the action and outcome. In games of chance, these two conditions frequently go together. As well as an intention to win, there is an action, such as throwing a die or pulling a lever on a slot machine, which is immediately followed by an outcome. Even though the outcome is selected randomly, the control heuristic would result in the player feeling a degree of control over the outcome. Self-regulation theory offers another explanation. To the extent that people are driven by internal goals concerned with the exercise of control over their environment, they will seek to reassert control in conditions of chaos, uncertainty or stress. One way of coping with a lack of real control is to falsely attribute oneself control of the situation. The core self-evaluations (CSE) trait is a stable personality trait composed of locus of control, neuroticism, self-efficacy, and self-esteem. While those with high core self-evaluations are likely to believe that they control their own environment (i.e., internal locus of control), very high levels of CSE may lead to the illusion of control. Benefits and costs to the individualEdit Taylor and Brown have argued that positive illusions, including the illusion of control, are adaptive as they motivate people to persist at tasks when they might otherwise give up. This position is supported by Albert Bandura's claim that "optimistic self-appraisals of capability, that are not unduly disparate from what is possible, can be advantageous, whereas veridical judgements can be self-limiting". His argument is essentially concerned with the adaptive effect of optimistic beliefs about control and performance in circumstances where control is possible, rather than perceived control in circumstances where outcomes do not depend on an individual's behavior. Bandura has also suggested that: "In activities where the margins of error are narrow and missteps can produce costly or injurious consequences, personal well-being is best served by highly accurate efficacy appraisal." Taylor and Brown argue that positive illusions are adaptive, since there is evidence that they are more common in normally mentally healthy individuals than in depressed individuals. However, Pacini, Muir and Epstein have shown that this may be because depressed people overcompensate for a tendency toward maladaptive intuitive processing by exercising excessive rational control in trivial situations, and note that the difference with non-depressed people disappears in more consequential circumstances. There is also empirical evidence that high self-efficacy can be maladaptive in some circumstances. In a scenario-based study, Whyte et al. showed that participants in whom they had induced high self-efficacy were significantly more likely to escalate commitment to a failing course of action. Knee and Zuckerman have challenged the definition of mental health used by Taylor and Brown and argue that lack of illusions is associated with a non-defensive personality oriented towards growth and learning and with low ego involvement in outcomes. They present evidence that self-determined individuals are less prone to these illusions. In the late 1970s, Abramson and Alloy demonstrated that depressed individuals held a more accurate view than their non-depressed counterparts in a test which measured illusion of control. This finding held true even when the depression was manipulated experimentally. However, when replicating the findings Msetfi et al. (2005, 2007) found that the overestimation of control in nondepressed people only showed up when the interval was long enough, implying that this is because they take more aspects of a situation into account than their depressed counterparts. Also, Dykman et al. (1989) showed that depressed people believe they have no control in situations where they actually do, so their perception is not more accurate overall. Allan et al. (2007) has proposed that the pessimistic bias of depressives resulted in "depressive realism" when asked about estimation of control, because depressed individuals are more likely to say no even if they have control. A number of studies have found a link between a sense of control and health, especially in older people. Fenton-O'Creevy et al. argue, as do Gollwittzer and Kinney, that while illusory beliefs about control may promote goal striving, they are not conducive to sound decision-making. Illusions of control may cause insensitivity to feedback, impede learning and predispose toward greater objective risk taking (since subjective risk will be reduced by illusion of control). Psychologist Daniel Wegner argues that an illusion of control over external events underlies belief in psychokinesis, a supposed paranormal ability to move objects directly using the mind. As evidence, Wegner cites a series of experiments on magical thinking in which subjects were induced to think they had influenced external events. In one experiment, subjects watched a basketball player taking a series of free throws. When they were instructed to visualise him making his shots, they felt that they had contributed to his success. One study examined traders working in the City of London's investment banks. They each watched a graph being plotted on a computer screen, similar to a real-time graph of a stock price or index. Using three computer keys, they had to raise the value as high as possible. They were warned that the value showed random variations, but that the keys might have some effect. In fact, the fluctuations were not affected by the keys. The traders' ratings of their success measured their susceptibility to the illusion of control. This score was then compared with each trader's performance. Those who were more prone to the illusion scored significantly lower on analysis, risk management and contribution to profits. They also earned significantly less. - Thompson 1999, pp. 187,124 - Plous 1993, p. 171 - Vyse 1997, pp. 129–130 - Hobbs, Christin; Kreiner, Honeycutt; Hinds, Brockman (2010). "The Illusion of Control in a Virtual Reality Setting". North American Journal of Psychology. 12 (3). - Taylor, Shelley E.; Brown, Jonathon D. (1988). "Illusion and well-being: A social psychological perspective on mental health" (PDF). Psychological Bulletin. 103 (2): 193–210. CiteSeerX 10.1.1.385.9509. doi:10.1037/0033-2909.103.2.193. PMID 3283814. Archived from the original on 2010-07-19.CS1 maint: BOT: original-url status unknown (link) - Presson, Paul K.; Benassi, Victor A. (1996). "Illusion of control: A meta-analytic review". Journal of Social Behavior & Personality. 11 (3). - Thompson 1999, p. 187 - Thompson 1999, p. 188 - Fenton-O'Creevy, Mark; Nicholson, Nigel; Soane, Emma; Willman, Paul (2003). "Trading on illusions: Unrealistic perceptions of control and trading performance". Journal of Occupational and Organizational Psychology. 76 (1): 53–68. doi:10.1348/096317903321208880. - Gino, Francesca; Sharek, Zachariah; Moore, Don A. (2011). "Keeping the illusion of control under control: Ceilings, floors, and imperfect calibration". Organizational Behavior and Human Decision Processes. 114 (2): 104–114. doi:10.1016/j.obhdp.2010.10.002. - Enzle, Michael E.; Michael J. A. Wohl (March 2009). "Illusion of control by proxy: Placing one's fate in the hands of another". British Journal of Social Psychology. 48 (1): 183–200. doi:10.1348/014466607x258696. PMID 18034916. - Thompson 2004, p. 116 - Jenkins, Herbert M.; Ward, William C. (1965). "Judgment of contingency between responses and outcomes". Psychological Monographs: General and Applied. 79 (1): SUPPL 1:1–17. doi:10.1037/h0093874. PMID 14300511. - Allan, L.G.; Jenkins, H.M. (1980). "The judgment of contingency and the nature of the response alternatives". Canadian Journal of Psychology. 34: 1–11. doi:10.1037/h0081013. - Langer, Ellen J. (1975). "The Illusion of Control". Journal of Personality and Social Psychology. 32 (2): 311–328. doi:10.1037/0022-35126.96.36.1991. - Langer, Ellen J.; Roth, Jane (1975). "Heads I win, tails it's chance: The illusion of control as a function of the sequence of outcomes in a purely chance task". Journal of Personality and Social Psychology. 32 (6): 951–955. doi:10.1037/0022-35188.8.131.521. - Henslin, J. M. (1967). "Craps and magic". American Journal of Sociology. 73 (3): 316–330. doi:10.1086/224479. - Thompson 2004, p. 115 - McKenna, F. P. (1993). "It won't happen to me: Unrealistic optimism or illusion of control?". British Journal of Psychology. 84 (1): 39–50. doi:10.1111/j.2044-8295.1993.tb02461.x. - Hardman 2009, pp. 101–103 - Thompson 2004, p. 122 - Thompson, Suzanne C.; Armstrong, Wade; Thomas, Craig (1998). "Illusions of Control, Underestimations, and Accuracy: A Control Heuristic Explanation". Psychological Bulletin. 123 (2): 143–161. doi:10.1037/0033-2909.123.2.143. PMID 9522682. - Judge, Timothy A.; Locke, Edwin A.; Durham, Cathy C. (1997). "The dispositional causes of job satisfaction: A core evaluations approach". Research in Organizational Behavior. 19. pp. 151–188. ISBN 978-0762301799. - Judge, Timothy A.; Kammeyer-Mueller, John D. (2011). "Implications of core self-evaluations for a changing organizational context" (PDF). Human Resource Management Review. 21 (4): 331–341. doi:10.1016/j.hrmr.2010.10.003. - Bandura, A. (1989). "Human Agency in Social Cognitive Theory". American Psychologist. 44 (9): 1175–1184. doi:10.1037/0003-066x.44.9.1175. PMID 2782727. - Bandura, A. (1997). Self-efficacy: The exercise of control. New York: W.H. Freeman and Company. - Pacini, Rosemary; Muir, Francisco; Epstein, Seymour (1998). "Depressive realism from the perspective of cognitive-experiential self-theory". Journal of Personality and Social Psychology. 74 (4): 1056–1068. doi:10.1037/0022-35184.108.40.2066. PMID 9569659. - Whyte, Glen; Saks, Alan M.; Hook, Sterling (1997). "When success breeds failure: the role of self-efficacy in escalating commitment to a losing course of action". Journal of Organizational Behavior. 18 (5): 415–432. doi:10.1002/(SICI)1099-1379(199709)18:5<415::AID-JOB813>3.0.CO;2-G. - Knee, C.Raymond; Zuckerman, Miron (1998). "A Nondefensive Personality: Autonomy and Control as Moderators of Defensive Coping and Self-Handicapping" (PDF). Journal of Research in Personality. 32 (2): 115–130. doi:10.1006/jrpe.1997.2207. - Abramson, Lyn Y.; Alloy, Lauren B. (1980). "The judgment of contingency: Errors and their implications.". In Baum, A.; Singer, J. E. (eds.). Advances in Environmental Psychology: Volume 2: Applications of Personal Control. Psychology Press. pp. 111–130. ISBN 978-0898590180. - Msetfi RM, Murphy RA, Simpson J (2007). "Depressive realism and the effect of intertrial interval on judgements of zero, positive, and negative contingencies". The Quarterly Journal of Experimental Psychology. 60 (3): 461–481. doi:10.1080/17470210601002595. PMID 17366312. - Msetfi RM, Murphy RA, Simpson J, Kornbrot DE (2005). "Depressive realism and outcome density bias in contingency judgments: the effect of the context and intertrial interval" (PDF). Journal of Experimental Psychology. General. 134 (1): 10–22. CiteSeerX 10.1.1.510.1590. doi:10.1037/0096-34220.127.116.11. PMID 15702960. Archived from the original on 2011-06-29.CS1 maint: BOT: original-url status unknown (link) - Dykman, B.M., Abramson, L.Y., Alloy, L.B., Hartlage, S. (1989). "Processing of ambiguous and unambiguous feedback by depressed and nondepressed college students: Schematic biases and their implications for depressive realism". Journal of Personality and Social Psychology. 56 (3): 431–445. doi:10.1037/0022-3518.104.22.1681. PMID 2926638.CS1 maint: Multiple names: authors list (link) - Allan LG, Siegel S, Hannah S (2007). "The sad truth about depressive realism" (PDF). The Quarterly Journal of Experimental Psychology. 60 (3): 482–495. doi:10.1080/17470210601002686. PMID 17366313. - Plous 1993, p. 172 - Gollwitzer, P.M.; Kinney, R.F. (1989). "Effects of Deliberative and Implemental Mind-Sets On Illusion of Control". Journal of Personality and Social Psychology. 56 (4): 531–542. CiteSeerX 10.1.1.515.1673. doi:10.1037/0022-3522.214.171.1241. - Wegner, Daniel M. (2008). "Self is Magic" (PDF). In John Baer; James C. Kaufman; Roy F. Baumeister (eds.). Are we free?: psychology and free will. New York: Oxford University Press. ISBN 978-0-19-518963-6. Retrieved 2008-07-02. - Pronin, Emily; Wegner, Daniel M.; McCarthy, Kimberly; Rodriguez, Sylvia (2006). "Everyday magical powers: The role of apparent mental causation in the overestimation of personal influence" (PDF). Journal of Personality and Social Psychology. 91 (2): 218–231. CiteSeerX 10.1.1.405.3118. doi:10.1037/0022-35126.96.36.199. PMID 16881760. Archived from the original on 2011-01-05.CS1 maint: BOT: original-url status unknown (link) - Fenton-O'Creevy, M., Nicholson, N. and Soane, E., Willman, P. (2005) Traders - Risks, Decisions, and Management in Financial Markets ISBN 0-19-926948-3 - Baron, Jonathan (2000), Thinking and deciding (3rd ed.), New York: Cambridge University Press, ISBN 978-0-521-65030-4, OCLC 316403966 - Hardman, David (2009), Judgment and decision making: psychological perspectives, Wiley-Blackwell, ISBN 978-1-4051-2398-3 - Plous, Scott (1993), The Psychology of Judgment and Decision Making, McGraw-Hill, ISBN 978-0-07-050477-6, OCLC 26931106 - Thompson, Suzanne C. (1999), "Illusions of Control: How We Overestimate Our Personal Influence", Current Directions in Psychological Science, 8 (6): 187–190, doi:10.1111/1467-8721.00044, JSTOR 20182602 - Thompson, Suzanne C. (2004), "Illusions of control", in Pohl, Rüdiger F. (ed.), Cognitive Illusions: A Handbook on Fallacies and Biases in Thinking, Judgement and Memory, Hove, UK: Psychology Press, pp. 115–125, ISBN 978-1-84169-351-4, OCLC 55124398 - Vyse, Stuart A. (1997), Believing in Magic: The Psychology of Superstition, Oxford University Press US, ISBN 978-0-19-513634-0<\|endoftext\|>	3.671875
114	Thurgood Marshall was the first African American U.S associate justice of the supreme court. Born in Maryland, he was introduced to the constitution at an early age. Thurgood Marshall was inspired to become a lawyer at an early age. After graduating from the Howard University of Law, Thurgood Marhsall argued civil rights cases before the supreme court. Of his 32 cases, he won 29 before the court. Thurgood Marshall was an important legal figure in securing equality in America Find a school channel on the Fusfoo high school digital network.<\|endoftext\|>	3.84375
1,048	# Sequences and Series ## 1. Sequences A sequence is a set of numbers in a specific order. In years 7 to 10, you will have met these in topics dealing with number patterns. A number pattern is a sequence. Some examples of sequences are: a) 2, 4, 6, 8, 10, 12, … b) 1, 1, 2, 3, 5, 8, 13, …. c) 0, 1, 4, 9, 16, 25. d) 3, -6, 12, -24, 48. A sequence may be finite or infinite. In examples (a) and (b), the sequences go on forever, and so are called infinite sequences. In examples (c) and (d), the sequences terminate at 25 and 48 respectively, and so are called finite sequences. Each number in the sequence is called a term of the sequence. In example (a), the first term is 2, the second term is 4, the third term is 6, and so on. The notation for this is shown below. Tn is the nth term and is called a general term in the sequence • Complete Exercise 1 ## 2. Revision of algebraic skills • Read page 2 of booklet • Complete revision exercises 2.1, 2.4, 3.1, 3.2 ## 3. Arithmetic Sequence Arithmetic Sequence In an arithmetic sequence, the terms increase (or decrease) by the same amount. Each term is a constant amount more (or less) than the previous term. Some examples are: 1. 5, 7, 9, 11, 13, ... 2. -13, -6, 1, 8, 15, … 3. 7, 4, 1, -2, -5, … 4. 20, 19½ , 19, 18½ , 18, … The difference between consecutive terms is the same, and so is called the common difference. In the above examples, the common differences are: 2, 7, -3, -½. Notice that when the sequence decreases, the common difference is negative. • Read p. 3 of booklet • Complete ex. 3 ## 4. The terms of an Arithmetic Sequence • Make notes on and read p. 3-4 of booklet • Complete Ex 4 on p. 5 ## 5. Solving Linear Inequations • Work through the rice distribution problem on p. 5 • Read and make notes on p. 6 • Complete Ex 5 ## 6. Simultaneous Linear Equations • Work through the water tank problem on p. 6 • Work through examples on p. 7 • Complete Ex 6 on p. 8 ## 7. Geometric Sequences • Work through the microbe problem on p. 8 • Work through example 12 and 13 on p. 8-9 • Complete Ex 7 ## 8. Terms of a Geometric Sequence • Read through p. 9-10 and make notes • Complete Ex 8 ## 9. Solving Exponential Equations by Trial and error • Read through The Betting Game problem on p. 11 • Work through example 15 & 16 • Do Ex 3.6 of textbook • Complete Ex 9 on p. 12 • Work through example 15 • Complete Ex 10 on p. 13 ## 10. Solving simultaneous non-linear equations In general, we can use either the substitution method or the elimination method. For the type of equations above, the elimination method is best. • Work through example 18 and 19 • Complete Ex 11 on p. 14 ## 12. Series Series A series is the sum of terms of a sequence. This is a sequence: 8, 4, 2, 1, ... THis is the corresponding series: 8 + 4 + 2 + 1 + ... The symbol Sn is used the represent the sum of "n" terms Sigma Notation Mathematicians often use shorthand notation to simplify the writing of an expression. Sigma notation can be used to simplify the writing of series. Sigma (S) is the Greek letter corresponding to our letter, S, and is used to represent the sum of numbers. • Make notes on p. 14-15 • Complete Ex 12 ## 13. Sum of an Arithmetic Series • Read through the proof on p. 16 and learn the formula on p. 17 • Work through example 22 and 23 • Complete Ex 13 on p. 17 • Work through the pile of logs problem on p. 18<\|endoftext\|>	4.75
158	Year 7 and 8 A modern foreign language gives students the opportunity to appreciate different countries, cultures, communities and people. Learning languages gives students opportunities to develop their listening, speaking, reading and writing skills and to express themselves with increasing confidence, independence and creativity. The skills which are taught in Year 7 and 8 will prepare them for GCSE, where they will extend their linguistic capability in listening, speaking and spoken interaction and they will develop their capacity and independence in reading and writing for a range of purposes and audiences. They will write essays, give presentations on topics which interest them and take part in discussions and role plays. Through modern foreign languages, students will not only learn language skills but they will be encouraged to develop literacy, numeracy, ICT and independent learning skills.<\|endoftext\|>	3.796875
637	Last week was the start to our latest 10-week session of Learn Together, Grow Together. The program is for parents and caregivers and their children ages 3 to 6 years old. While the adults learn about their children’s early learning, and how to support literacy development and success in school, it is also a good opportunity for the parents to brush up on their own literacy skills and connect with other parents of preschool children. We begin each session with stories and singing, followed by gym time. Afterwards, we split the children and parents into separate learning groups, and finally, we bring them back together for fun parent-child learning activities. During parent time for the next few weeks, we will explore “emergent literacy.” Emergent literacy is “the knowledge children have about reading and writing before they can actually read and write.”(http://www.kidsability.ca/en/LiteracyHandouts) We will encourage parents in their role as emergent literacy teachers for their own children. How does a child gain this knowledge about reading and writing? What are some strategies parents can use to foster emergent literacy in their child? Here are a few ideas to try together with your family: - Talk with your child. Your child will learn so much from positive language interactions with you. - Talk about what you see in a picture book, while at the grocery store, at the park, etc. - Explain to your child what you are doing, while you are doing it. For example, if you make cookies, talk about the different ingredients and what steps you have to take, or if you are paying bills, use the time to talk to your child about money and numbers. - Play card and board games together. Turn off the electronic devices and have fun playing a game where there is opportunity to speak with each other. - Sing and rhyme with your child. Sing songs and rhymes together as they provide opportunities to bond with your child as well as expand their vocabulary. You can always make up your own songs and rhymes too – your child will enjoy hearing your voice either way. - Visit your local library and take advantage of their book lending services. - Follow you child’s lead in their interests. For example, if they have an interest in animals, share books about animals, sing songs and rhymes about animals, and play games about animals. If you can, take a trip to a pet store, a farm, or a zoo; take the time to talk about everything you see and experience together. - Model positive literacy behaviours to your child. If your child sees your enjoyment of reading the newspaper, writing a shopping list, talking about the road signs you see, etc., they will think of these literacy activities as positive experiences. Parents have such an important role in cultivating the knowledge of reading and writing in their child, even before they are actually about to read and write. There are so many opportunities to promote emergent literacy in a small child, simply by intentionally interacting with them and involving them in what you are already doing!<\|endoftext\|>	3.6875
565	A processor or micro-processor is the 'brains' of a computer system. It is the processor that controls the working of all of the hardware and software.The processor is sometimes referred to as the Central Processing Unit (CPU).There are many processors available and processor specification is usually one of the first things considered when buying a new personal computer (PC). The type of processor and its speed have the greatest impact on the overall performance of a computer system. Processor performance is related directly to its speed of operation and its architecture Competition among processor manufacturers is fierce and because of this there is a wide and diverse choice of processors in the market place. Processor manufacturers, such as Intel and Advanced Micro Devices (AMD) are continually developing more advanced processors and new models are released within the space of months rather than years. This is in stark contrast to earlier processor developments, such as the 8086, 80286 and 80386 which were released years apart. Types of processor A processor connects directly to the computer's motherboard. Processors can be categorized by the way they are connected to the motherboard.There are two main types of processor connections to motherboard purpose of the processor Our digital devices like Phones, Computers and laptops have grown to become so sophisticated, that they are no longer just devices, they have evolved to become a part of us.A computer processor, or central processing unit (CPU), is the brains of the computer. It runs the operating system and application software installed on the computer to do whatever the user wants to do. The CPU handles most operations on a computer, but some operations are handled by specialized tools such as graphics processing units (GPUs).Processors come in different architectures, you must have come across different types of programs that say they are for 64Bit or 32Bit. What this means is that those programs support that particular processor architecture.If a processor has a 32 Bit Architecture it means that it can process 32 Bits of information in one processing cycle.Similarly, a 64 Bit processor will process 64 Bits in one cycle.Also, The amount of ram you can use also depends on the architecture. The amount of ram you can use depends upon the amount of memory in powers of 2 ^ architecture of the system.For a processor with 16-bit architecture, only 64 KB of ram is accessible. For a 32 Bit processor, the maximum usable ram is 4 GB. And for 64 Bit processor, it is 16 Exa-Byte.Cores are basically processing units in the computer. They receive instructions and act on it. The more Cores you have, the better your processing speed. your purchasing will make be joyfull with respective prices only at paliescart Purchasing High perforamnce processor with amazing affordable prices at paliescart.com<\|endoftext\|>	4.125
592	PDF chapter test Asha wanted to count the numbers between $$-9$$ and $$12$$. She found that totally $$20$$ integers are present, with $$8$$ negative integers, $$zero$$ and $$11$$ positive integers present between $$-9$$ and $$12$$ (excluding $$-9$$ and $$12$$). She also gets to know that there are no integers, present between any two consecutive integers, that is, for example, between $$8$$ and $$9$$ there is no integer. So, between any two consecutive integers the number of integers is $$0$$. She wondered whether the same would happen in case of rational numbers too? She took two rational numbers $\frac{-3}{2}$ and $\frac{-2}{3}$. She converted them to rational numbers with same denominators, for which the denominator of both the numbers should be converted into LCM. She found the LCM of ($$2$$,$$3$$) $$= 6$$. Convert both the numbers with denominator as $$6$$. $\frac{-3×3}{2×3}=\frac{-9}{6}$ and $\frac{-2×2}{3×2}=\frac{-4}{6}$ We have, $\frac{-9}{6}<\frac{-8}{6}<\frac{-7}{6}<\frac{-6}{6}<\frac{-5}{6}<\frac{-4}{6}$ (or) $\frac{-3}{2}<\frac{-8}{6}<\frac{-7}{6}<\frac{-6}{6}<\frac{-5}{6}<\frac{-2}{3}$ She could find rational numbers $\frac{-8}{6},\frac{-7}{6},\frac{-6}{6},\frac{-5}{6}$ between $\frac{-3}{2}$. She doubts that are there only $$4$$ rational numbers between $\frac{-3}{2}$. She gets to know there are more than $$4$$ rational numbers between $\frac{-3}{2}$ because if we find the multiples of denominators, then many more rational numbers can be inserted between $\frac{-3}{2}$. For example, $\frac{-3}{2}=\frac{-3×4}{2×4}=\frac{-12}{8}=\frac{-3×10}{2×10}=\frac{-30}{20}$ and $\frac{-2}{3}=\frac{-2×4}{3×4}=\frac{-8}{12}=\frac{-2×10}{3×10}=\frac{-20}{30}$ Now, between $\frac{-30}{20}$ and $\frac{-20}{30}$, there are $$9$$ rational numbers present.<\|endoftext\|>	4.46875
2,888	Climate Change. Part 32. trash from the "Great Pacific Garbage Patch." About the Garbage Patches: There are massive concentrations of mostly plastic debris floating in the oceans of the world. The most famous concentrations of debris have been collected and trapped within three areas of the North Pacific Gyre (NPG) a.k.a. the North Pacific Subtropical Gyre. A Gyre is a large system of circulating ocean currents. The NPG is one of five major ocean-wide gyres in the world. 1 Two are in the Pacific, two in the Atlantic and one in the Indian Ocean. There are 13 smaller gyres elsewhere in the world's oceans. The NPG is located in a region ranging from about 135°W to 155°W and 35°N to 42° N. 1 Carey Morishige of the federal U.S. National Oceanic and Atmospheric Administration (NOAA) explains that much of the debris found in these areas are microplastics -- small bits of plastic (microplastics): "... that are suspended throughout the water column. A comparison I like to use is that the debris is more like flecks of pepper floating throughout a bowl of soup, rather than a skim of fat that accumulates (or sits) on the surface. 3 There are three areas within the NPG where plastic rash is concentrated, called: - the "Western Garbage Patch," - the "Subtropical Convergence Zone," and - the "Eastern Garbage Patch," a.k.a. the "Great Pacific Garbage Patch." 3 It is about halfway between Hawaii and California. According to the Office of Response and Restoration of NOAA's Marine Debris Program: " 'Patch' is a misleading nickname, causing many to believe that these are islands of trash. Instead, the debris is spread across the surface of the water and from the surface all the way to the ocean floor. The debris ranges in size, from large abandoned fishing nets to tiny micro plastics, which are plastic pieces smaller than 5mm [0.2 inches] in size. This makes it possible to sail through some areas of the Great Pacific Garbage Patch and see very little to no debris." 4 A proposal and an attempt to clean up a garbage patch: During 2012, Dutch student Boyan Slat, aged 18, delivered a TED speech in which he proposed the Ocean Cleanup project to remove plastic from the oceans. Carolyn Gramling, writing for Science News, discussed the project. It involves a series of floating booms about 600 meters (1,970 feet) long that form a "U" shape. They are made made of HDPE plastic. A ship would tow the booms through the "patch," and herd pieces of plastic into a concentrated mess so that they could be more easily picked up by other vessels. The booms drag a 3-meter-long underwater skirt which will collect most of the debris that is located underwater near the surface of the ocean. Laurent Lebreton, Ocean Cleanup’s lead oceanographer, estimated that the Great Pacific Garbage Patch is spread across 1.6 million square kilometers (618,000 square miles) which is about twice the size of the State of Texas. More than 90% of the mass of the debris is made up of pieces that are 5 centimeters (2 inches) or larger in size. These form the debris that the cleanup project is designed to collect. Project leaders believe that once the full fleet od booms is launched, they could remove 50% of the plastic in the Great Pacific Garbage Patch within five years, and 90% by 2040. Environmental scientist Marcus Eriksen at the Los Angeles–based nonprofit group, the 5 Gyres Institute suggests that the Ocean Cleanup project goals are over-optimistic. He suggests that much of the estimated 5 trillion bits of ocean plastic in the oceans is quickly removed from surface waters by natural forces. 5 They now lie out of reach below the booms' skirts. Eriksen notes that 5 Gyres and other conservation organizations believe that the most effective cleanup strategy is to reduce the use of plastics worldwide. Adam Lindquist, the director of the Healthy Harbor Initiative in of Baltimore, MD, said: "Any tech that’s removing plastics from the ocean is a good thing. But the actual worth is getting people to take the problem more seriously, so that we can move toward real solutions -- like source reduction." 2018-DEC: The project ran into a snag: The Ocean Cleanup project's ship with its 2000 foot-long (610 meter) floating boom, was launched on 2018-SEP-08 from Los Angeles, CA. It headed to the Great Pacific Garbage Patch and its estimated 1.8 trillion pieces of trash, weighing a total of about 350 million tons (318 billion kilograms). The group planned to work there for a year. 6 Science News reported that during the Ocean Cleanup shakedown cruise: "Wind and currents have pushed trash into the rig, but the setup hasn't’t kept the trash corralled as planned." 7 Unfortunately, the mission had to be aborted on 2019-JAN-02 because a fracture occurred near one end of the group of booms. An Ocean Cleanup representative said that on: "Saturday, December 29th, unexpectedly, one of the tow heads, together with an 18-metre piece of floater pipe with two stabilizer frames, was found to be drifting away from the rest of Wilson. No prior indications were seen for such a thing to happen, so this came as a surprise for both the offshore crew and the engineering team onshore." "Wilson" is the name given to the "U" shaped boom assembly. The equipment was towed to Hawaii for repairs. They are expected to arrive there in mid January. 8 This topic continues in the next essay, with developments during 2019 The following information sources were used to prepare and update the above essay. The hyperlinks are not necessarily still active today. - "Great Pacific garbage patch," Wikipedia, as on 2019-JAN-07, at: https://en.wikipedia.org/ - Image extracted from the center of a Photoshopped work by "Fangz." It is in the public domain and was downloaded at: https://en.wikipedia.org/ - "How Big Is the 'Great Pacific Garbage Patch'? Science vs. Myth," 2013-FEB-07, at: https://response.restoration.noaa.gov/ - "Garbage Patches," NOAA, 2017-DEC-26, at: https://marinedebris.noaa.gov/ - Carolyn Gramling, "A massive net is being deployed to pick up plastic in the Pacific," Science News, 2018-SEP-07, at: https://www.sciencenews.org/ - "The Ocean Cleanup vessel is on its way to clean up the Great Pacific Garbage Patch," Fast Company, 2018-OCT-07, at: https://www.fastcompany.com/ - Laurel Hamers, "4 ways to tackle ocean trash besides Ocean Cleanup’s broken system," Science News, 2019-JAN-04, at: https://www.sciencenews.org/ - Elanor Gison, "The Ocean Cleanup suspended as device breaks down in Pacific Ocean," De Zeen, 2019-JAN-07, at: https://www.dezeen.com/ The following reports in this section that discuss environment and climate change may be of interest to you: - Part 1: 2013: Quotes. Background. Bay of Bengal incident. Decline in Canada's arctic ice cover. - Part 2: 2013/2014: Canada loses its seasons. 316 North American cities to be under water. - Part 3: 2013/2014: Christians, Republicans' & Democrats' beliefs differ. Volcanoes cause pauses in global warming. Americans' views on climate change. - Part 4: 2014/2015: How Americans viewed climate change. HFC limits. Groups abandoning ALEC. Paris Conference reaches agreement. - Part 5: 2016-FEB to AUG: Climate change concerns: Ocean levels rising. Displacement of animal species. Greenland glacier melting. "Third Pole" glaciers melting. - Part 6: 2016-SEP: Flooding on the U.S. East coast. U.S. and China ratify the 2015 Paris Climate Agreement. Pres. Obama: trends are "terrifying." - Part 7: 2016-SEP: Obama comments on warming (Con't). 31 countries ratify Paris agreement. Trump zigzags on Climate Change. - Part 8: 2016-Summer/Fall: Events showing that climate change is adversely affecting life on Earth - Part 9: 2016-OCT: Did God send Hurricane Matthew as an anti-LGBT message to humanity? Alternate opinions. Hurricane Matthew re-enters Atlantic Ocean. - Part 10: 2016-NOV: U.S. Costal real estate threatened. Rapid temperature rise in the Arctic. Very large poll about climate change. - Part 11: 2016-DEC: President-elect Trump's climate change beliefs. New heads of federal government departments. Extinction of 11 species. - Part 12: 2016-DEC: President-elect Trump's 16 Promised Plans on Energy and the Environment. - Part 13: 2017-JAN: 2016 Was The Hottest Year On Record, Ever. - Part 14: 2017-JAN: Recent Climate Change Measurements! - Part 15:2017-JAN to APR: Climate Change Debate is Changing. What if All the Earth's Glaciers Melted? - Part 16: 2017-JAN to APR: Donald Trump Becomes President. EPA Foe Installed as Head of EPA! Hundreds Die During Flood in Colombia. Carbon Dioxide in Atmosphere Reaches New High. - Part 17: Studies of the Effects of Sea-level Rise in California between Now and the Year 2100. - Part 18: President Trump Decided that the U.S. Will Quit the Paris Agreement. Explosive Reactions to his Plan, nationally and globally. - Part 19: More Reactions to the U.S. Withdraw from the Paris Agreement. Record high temperatures in San Francisco and elsewhere. - Part 20:2017-AUG/SEP : Group of 20 summit meeting, Hurricane Harvey attacks Texas. Is climate change increasing the severity or frequency of hurricanes? - Part 21: 2017-SEP: 2017-SEP: Can government regulations help? U.S. record high temperatures. Did God send Hurricane Harvey to attack TX?. Hurricane Irma attacks Caribbean, FL, SC, GA. Hurricanes Jose & Maria. - Part 22: 2017-SEP/OCT: The most expensive recent hurricanes to attack the United States. Does God use Hurricanes to send a message? Conflicts, climate change cause rise in hunger. - Part 23: 2017-OCT to 2017-DEC: All UN states have signed the Paris Agreement. Government report contradicts the President. Comment by the Union of Concerned Scientists. - Part 24: 2018-JAN to MAY: New temperature records. President signs bill increasing research."Onion's" amusing fake news. Sea levels rising; cities under water. Climate change beliefs as function of age. Proposed Mount Trumpmore project. Volcano in Hawaii. How global warming is affecting California. Animal, Insect, & Bird Ranges Shrink. Nutrients in rice reduced. California Governor's comments. - Part 25: 2018-JUN & JUL: During 2018-JUN: A very few encouraging signs. Hurricanes slowing as winds increase. Changes in Antarctic ice and environment. Norway's growing season increased. A new world highest temperature. Future ocean levels. - Part 26: 2018-JUL: New York City reports to the UN. Impacts of humans burning up all our fossil energy. WWW.CLIMATECENTRAL.ORG's tools. Death Valley, CA, sets new record. Dissapearing wetlands in Louisiana. Climate change affecting Northern lands. Hurricane Florence. - Part 27: 2018-AUG & SEP: Dissapearing wetlands in Louisiana. Climate change affecting Northern countries. Hurricane Florence attacks the Carolinas. Pat Robinson's "Shield of Protection" doesn't work. - Part 28: 2018-SEP & OCT: Trump Admin. predicts: 7ºF temperature rise by the year 2099. Impact of climate change on women. Green Building Week & commitment. New IPCC report predicts a "Strong Risk of as early as 2040. - Part 29: 2018-OCT & NOV: Southern Baptist Convention downplays IPCC Report. Climate changing Fall season. Ozone layer continues to heal. Trump Confuses Weather - Part 30: 2018-NOV: Hope for people who have no access to clean water. Trump Administration releases National Climate Assessment report. Reactions to the report. - Part 31: 2018-DEC: How to lie about climate change. UN head repeats warning. Major report on Arctic climate. Greenhouse gases soar. Can climate scientists be trusted? - Part 33: 2019-JAN-MAR: If Ocean Cleanup fails.... Public opinion on environment. Climate change in early 2019. Sea rise from Antarctic melting. Denmark fights cow flatulence. Disconnect between the EPA and climate scientists. Change is urgent. UN report: One million species could become extinct: Copyright © 2019 by Ontario Consultants on Religious Tolerance Originally posted: 2019-JAN-10 Latest update: 2019-MAR-28 Author: B.A. Robinson<\|endoftext\|>	3.71875
738	Oxygen generators and carbon dioxide scrubbers explained The latest in our series describing how technology works, by Jake Port. Providing breathable air is a challenge in space and under the sea, as is disposing of the carbon dioxide submariners or astronauts exhale. So how do submarines and spacecraft create oxygen and handle the toxic build-up of carbon dioxide? The answer lies in a combination of chemical reactions and electricity. Submarines and the International Space Station produce the majority of the oxygen they need by liberating it from water. A water molecule consists of two hydrogen atoms and one oxygen atom: H2O. Most of the world’s water is found in the oceans – the roaming ground of vessels such as ships and submarines. So it's convenient to use that seawater as an oxygen source. First, it must be distilled to remove impurities such as salts. That is done by heating seawater to create water vapour, leaving the salts behind, and then cooling it to condense in a collection tank. Electricity is passed through the purified water, in a technique known as electrolysis, which separates the water molecule into its constituent parts: hydrogen and oxygen. Free atoms of oxygen and hydrogen happily bind to other atoms of the same element, producing hydrogen and oxygen gas, H2 and O2, at a rate of twice as many hydrogen molecules as oxygen. The positively charged hydrogen ions collect at the negative cathode, while the negatively charged oxygen collects at the positive terminal, or anode. Oxygen is vented into the submarine's air circulation system, and such systems can produce thousands of litres of oxygen each hour. In space, though, water isn’t as plentiful as in the ocean, but it is still the major source of oxygen. To access this, the space station must continuously refill its water tanks and reclaim water vapour produced by the astronauts when they exhale. Electrolysis is powered by the space station's giant solar panels and the oxygen produced hisses into the breathable cabin air system. At the moment, explosive hydrogen is expelled into space, but that may change. Rather than being simply wasted it could be contained for later use, even as a rocket fuel – an idea being considered to fuel future space missions. And water (as a liquid, solid or gas) on another planet or asteroid could provide fuel for the trip home or beyond. But providing the oxygen is only part of the problem solved. While we breathe in oxygen, we also exhale carbon dioxide, or CO2. This gas can be deadly if its concentration in the atmosphere rises above 8%, so controlling it is essential to keep submarine and spacecraft crews alive. The most straightforward way of removing the gas is to vent it. But it must first be captured through a chemical reaction in a carbon dioxide “scrubber”. Traditionally, scrubbers use soda lime (a mixture of chemicals including calcium hydroxide, sodium hydroxide and potassium hydroxide) or amines (a derivative of ammonia) to lock onto CO2 molecules. But in space where every gram counts, lithium hydroxide is used in scrubbers because it has a low molecular weight. When the CO2 reacts with lithium hydroxide it creates create lithium carbonate and water. While this is the primary scrubber type on the space station it's a back-up on a submarine. Scrubbers of all types involve chemical reactions that have the added benefit of producing water, which could be used for drinking or to fuel the oxygen generator.<\|endoftext\|>	3.796875
1,079	Smartick is an online platform for children to master math in only 15 minutes a day Sep08 # Rekenrek Vs. Soroban II: Varied Strategies for the Same Calculation A few months ago we talked in this post about the differences between the Dutch Rekenrek abacus versus the traditional Japanese Soroban and the analogy with hands. Today we will make some comments on the advantages of the Rekenrek compared to the Soroban to encourage varied strategies in children when doing the same calculation. One of the differences between the use of the Soroban and the Rekenrek is that when using the Soroban, each operation can only be done in a single way. Whereas with the Rekenrek students are encouraged to use several strategies for the same calculation. To explain this, we begin by explaining how it is added in the Soroban. Suppose we want to add 6. In the Soroban there are three different ways of adding 6 depending on which number we add to get 6. To illustrate the three forms, we will add from 6 to get 6. We start from position 0 in the Soroban, in which all the beads are far from the horizontal bar that separates the above beads, which are worth 5, from the beads below, which are worth 1. We work on the first rod on the right, the units (the second is the tens, etc.). To add 6, we have to bring the upper and lower beads to the horizontal bar. Adding 6 is adding 5 and 1 (6 = 5 + 1). To add 6 more, there is no problem, since we have the 5 and 1 beads available. The result (of 12 + 6) is 18. To add 6 more, we must do it as we did in the second step (6 = 10 – 5 + 1). Finally, there is a third way of adding 6, which is given in this situation, which is (6 = 10 – 4) and consists of adding 1 ten and removing 4 beads of value 1. Summing up, each sum can be done in three different ways, depending on the availability of abacus beads. The 6 can be added as 5 + 1, 10 – 4, or 10 – 5 + 1, but whenever we make a concrete calculation (4 + 6) we do the same (4 + 6 = 4 + 10 – 4) adding a bead of 10 and removing 4 beads of 1. This is, therefore, a totally algorithmic process that does not allow variations. There are no different strategies for making a sum. However, in rekenrek, as we will see in the following example, different strategies can be used for the same calculation. For example, to do 7 + 5, you can put 7 in the top row and 5 in the bottom row, and count from 7 until you reach the solution. This is a method of counting from a number that is very important to acquire in the first grade of primary school. It can also be immediately recognized (subitization) that the top five red and the bottom red are 10 and that, with the 2 white ones, the result is 12. That is, 7 + 5 = 5 + 5 + 2 = 10 + 2 = 12. This strategy is known as the “use of doubles plus two.” We can remove a bead from above and pass it down, physically or mentally, to recognize that 7 + 5 = 6 + 6 = 12, reducing the calculation again to use of doubles. We can also add two white beads down, recognize that we are doing a double, and then compensate by removing the two we have added. 7 + 5 = 7 + 7 – 2. This strategy is called “use of doubles minus 2.” Finally, we can add the missing beads for 10, breaking down the 5 into 3 (what is missing from the 7 to get 10) and 2 (which remains to be added). We call this strategy “step to 10” and it can be represented as 7 + 5 = 7 + (3 + 2) = (7 + 3) + 2 = 10 + 2 = 12. In summary, with the rekenrek, we can do the same operation (7 + 5) in 5 different ways. We believe that it is of great value that children learn to calculate flexibly, that they know several strategies for the same calculation and that they choose the strategy that seems most appropriate in each case. When we speak of mental calculation, we refer not only to calculations made “by head”, or without external aids, instead we speak of a flexible calculation, which enhances the development of the numerical sense, and not a rigid and algorithmic calculation, which can be developed mechanically. That is why, a few months ago at Smartick, we introduced the rekenrek in the sessions for the youngest students.<\|endoftext\|>	4.875
922	# Comparing Money Raised Alignments to Content Standards: 4.OA.A.2 1. Helen raised \$12 for the food bank last year and she raised 6 times as much money this year. How much money did she raise this year? 2. Sandra raised \$15 for the PTA and Nita raised \$45. How many times as much money did Nita raise as compared to Sandra? 3. Luis raised \$45 for the animal shelter, which was 3 times as much money as Anthony raised. How much money did Anthony raise? ## IM Commentary The purpose of this task is for students to solve three comparisons problems that are related by their context but are structurally different. Multiplicative comparison is purposefully excluded from third grade (see 3.OA.3 and 3.MD.2), making this task appropriate for fourth but not third grade. In these multiplicative comparison problems, one factor and the product are amounts of money and the other factor represents the number of times bigger one amount is than the other. Sometimes this second factor is called a “scale factor.” In part (a), the larger amount (which is the product) is unknown, while in part (b) the scale factor is unknown and in part (c) the smaller amount of money is unknown. Students will study multiplicative comparison problems involving scale factors that are fractions in 5th grade; see 5.NF.B.5. Note that in fourth grade, scale factors must always be bigger than 1, so students often think that “multiplying makes bigger”; however in 5th grade they will see that when the scale factor is less than 1, the product will actually be smaller than the initial quantity. Note that the numbers in parts (b) and (c) are related by the fact family $3\times15=45$. This allows for a classroom discussion about the different interpretations of the factors in a multiplicative comparison context. To see an annotated version of this and other Illustrative Mathematics tasks as well as other Common Core aligned resources, visit Achieve the Core. The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, only one practice connection will be discussed in depth. Possible secondary practice connections may be discussed but not in the same degree of detail. This particular task helps illustrate Mathematical Practice Standard 1, Make sense of problems and persevere in solving them. Problem solving is based upon students engaging in a task in which a solution pathway is not known in advance. As fourth graders approach these three problems, they will analyze the problems to make sense of what each is asking, working to understand the structures and the unknowns. Through this analysis, students will understand that the numbers in parts (b) and (c) are both factors related by the same fact family but each serves a different role in their respective problems. Students need experience with unknowns that play different roles in multiplicative comparison problems in order to support a deeper understanding of this type of problem situation. ## Solutions Solution: Tape diagram 1. She raised six times as much money (as shown in the diagram) so she raised $6 \times 12 = 72$. Helen raised \$72 this year. 2.$? \times 15 = 45$is equivalent to$45 \div 15 = ?$Nita raised 3 times as much as Sandra. 3.$3 \times ? = 45$is equivalent to$45 \div 3 = ?$Anthony raised \$15. Solution: Writing multiplication equations for division problems 1. Helen raised $6 \times \$12$this year, so she raised \$72 this year. 2. This is a “Number of Groups Unknown” problem. We can represent the question as $$? \times 15 = 45$$ or $$45 \div 15 = ?$$ So Nita raised 3 times as much money as Sandra. 3. This is a “Group Size Unknown” problem. We can represent the question as $$3 \times ? = 45$$ or $$45 \div 3 = ?$$ So Anthony raised \\$15.<\|endoftext\|>	4.5625
2,355	# Solutions to Sample Problems 1. ```Solutions to Sample Problems 1. What is the force of friction between a block of ice that weighs 930 N and the ground if Mu = .12? (111.6 N)         W = 930 N Mu = .12 Once again, our formula for Mu is: F = Mu (FN) It's pretty easy to plug-in the stuff we know: F = (.12)(930 N) We then do the math to get 111.6 N. 2. What is the coefficient of static friction if it takes 34 N of force to move a box that weighs 67 N? (.51)    F = 3.0 s      W = 67 N Once again, our formula for Mu is: F = Mu (FN) We know that FN is equal to W so we can plug what we know into the formula and get: 34 N = MuS (67 N) Divide the 34 by 67 and you end up with .507462 which happens to round to the .51. 3. A box takes 350 N to start moving when the coefficient of static friction is .35. What is the weight of the box? (1000 N)         F = 350 Mu = .35 Hey, since FN = W, surprisingly enough we use the formula: F = Mu (FN) which we plug-in the numbers and get 350 = .35 (FN) Divide 350 by .35 to get 1000 N. 4. A car has a mass of 1020 Kg and has a coefficient of friction between the ground and its tires of .85. What force of friction can it exert on the ground? What is the maximum acceleration of this car? In what minimum distance could it stop from 27 m/s? (8500 N, 8.3 m/s/s, 43.8 m)         m = 1020 Kg Mu = .85 Use the formula W = mg with g = 9.8 m/s/s as the force of gravity. We now know the weight to be 9996 N. We can then use the formula F = Mu (FN) with FN equal to the weight. Plugged in it looks like: F = (.85)(9996 N) and equals 8496.6 N but with significant digits it ends up being 8500 N. In order to find the acceleration we need a formula that uses the Force we found and acceleration. Obviously F = ma seems best. We plug-in the numbers and get: 8496.6 N = (1020 Kg) a Solve for a and get 8.33 m/s/s as the acceleration But we're not done yet! We need to solve for the distance. Since this deals with good old velocity, acceleration and distance we need a formula that uses only what we know:            Vo = 37 m/s V = 0 (It comes to a complete stop) a = 8.33 m/s/s Now we can use: x = V2 - VO 2 ----------------2a x = 43.7575 or with sig figs, 43.8 m 5. Clarice moves a 800 gram set of weights by applying a force of 1.2 N. What is the coefficient of friction? (.15)  m = 800 grams or .8 Kg           F = 1.2 N g = 9.8 m/s/s We first need to get the FN or Weight; we must convert the mass to kilograms and then get FN = 7.84 N for the next step. As usual when we look at the formulas we see that: F = Mu (FN) seems to be the only one that will uses Mu We plug-in the numbers to get: 1.2 = Mu (7.84 N) Divide 1.2 by 7.84 to get .15306 or .15 for the coefficient of friction. 6. A car has a coefficient of friction between the ground and its tires of .85. What is the mass of the car if it takes 9620 N of force to make it slide along the ground? (1155 Kg)   Mu = .85       F = 9620 N We have to find the weight first so we use: F = Mu (FN) and that equals 9620 N = (.85)(FN) 9620 divided by .85 = 11317.647 N But we need this in mass so we use: F = mg 11317.647 = (m)(9.8 m/s/s) so the mass equals 1154.8619 or 1155 kg 7. A 5.0 Kg block has a coefficient of friction of .15 on a flat surface. What is its acceleration if you exert a force of 15 N sideways on it when it is at rest? (Find the friction force first) (1.47 m/s/s)            m = 5.0 kg F = 15 N Mu = .15 So we can use the formula: FN = mg or 15 N = (5.0 kg)(9.8 m/s/s) So the FN = 49 N Then we plug that into: F = Mu (FN) that looks like F = (.15)(49 N) And get the actual force to be 7.35 N And then we can plug that back into the formula: F = ma or 7.35 N = (5.0)a Divide 7.35 by 5 and get 1.47 m/s/s 8. A 10 Kg block is at rest on a level surface. It accelerates from rest to 51.2 m/s in 8 seconds when you exert a force of 100 N on it sideways. What is the acceleration of the block? What is the force of friction between the surface and the block, and what is the coefficient of friction? (6.4 m/s/s 36 N, .56)               m = 10 Kg Vo = 0 m/s V = 51.2 m/s t = 8 seconds F = 100 N    Find the acceleration using the our freindly neighborhood aceleration formula once again: V = Vo + at. Plug in a couple of numbers and get: 51.2 m/s = 0 m/s + (a * 8 s) Solve for a, and we get 6.4 m/s/s The force of friction = Force applied ñ The normal force FA = 100 N and FN = 64 N so the FF = 36 N Now we plug that back into the formula: FF = Mu (FN) we use our stuff 36 N = Mu (64 N) Divide 36 by 64 and get .5625 as the coefficient of friction. 9. A 120 Kg log sled accelerates at 1.4 m/s/s when a horse pulls on it. What force must the horse exert if the coefficient of friction between the ground and the sled is .28? (497 N)  m = 120 Kg           Mu = .28 a = 1.4 m/s/s First we need to find FN and we use FN = mg F = (120 Kg)(9.8 m/s/s) and F equals 1176 N We then multiply that by the coefficient of friction (.28)to get the force of friction which is 329.28 N. We then use the formula: F = FA - FF, with the values plugged in we get: 168 = FA - 329.28 solve this little problem and get 497.28 or 497 N. 10. (E.C. - Super Stud Problem) You exert a force of 24 N sideways on an object and it accelerates from 0 - 12 m/s over a distance of 5.2 m. You know that the coefficient of friction between the object and the ground is .58, so what is its mass? (1.23 Kg)          VO = 0 m/s V = 12 m/s Mu = .58         x = 5.2 m FA = 24 N Letís use x = V2 - VO2 ------------2a we plug in what we know and get: 5.2 m = 122 - 2 / 2a solve for acceleration and get a = 13.846 m/s/s we want to find F = ma but weíll need to use: F = FA - FF but we donít know the FF So FF = Mu * FN we plug stuff in and get: Since F = ma and FF = Mu * m * g we can put it into F = FA - FF so it looks like: ma = FA - (Mu * m * g) we can then put in what we know m (13.846) = 24 N ñ (.58 * m * 9.8m/s/s) Wait I donít Know! 13.846 = 24/m ñ 5.684 add 5.684m to both sides and NO That doesnít Work! 6.684m * (13.846) = 24 OHHH MAN! Ohhh dang it this just ends up being 1.23 Kg!!!!! ```<\|endoftext\|>	4.46875
340	Consider the system of two equations in two variables as and a) To solve this system by the method of substitution, we isolate either x or y whichever is easy to be from Suppose that as Substitute this in , it would be just the variable y and constant left. So, bring y to the left side of the equation, the value of y is obtained. Again substitute y in , easily the value of x is obtained. b) To solve the system using elimination, write the equations one above the other. Observe the coefficients of x in both the equations. Suppose the coefficient of x in f(x, y) = a is m and that in g(x, y) = b is n Then multiply f with n and g with m. Subtract one equation from the other. The variable x will be eliminated and the variable y with a constant remains in the equation. On sending the coefficient of y to the side of constant, the value of y is obtained. By substituting this in one of the given equations, the value of x is obtained. c) To solve the system of two equations in two variables by graphing, it can be observed that each function f(x, y) = a and g(x, y) = b are uniquely represented by one curve only. So, the equations have a solution if and only if the respective curves intersect at one or more points. The x- coordinate of the point of intersection on the graph sheet or xy – plane is the value of x and the y - coordinate is the value of y required.<\|endoftext\|>	3.96875
535	Learning outcomes are the measurable products of learning. Assessing these outcomes allows educators to determine if students achieved their goals and met learning objectives. The assessment of learning outcomes aids in accountability among teachers and schools as a whole. Therefore, this weighty results calls for accurate methods of assessment. Educators must develop and implement reliable means of measuring students' performances in learning. Ways to Assess Learning Outcomes Plan your assessments before you begin. When developing a unit or lesson, use your goals and objectives to determine what will be assessed. Make note of what kinds of assessments you'll use and how often you will implement them. Make sure your assessments cover a range of learning outcomes, not just a narrow band. Assess frequently. Make sure you have plenty of checkpoints along the way in order to truly measure student learning. One-time assessments rarely are an accurate measure of learning. A child may know the material but have an "off" day on the day you choose to assess learning. If you assess at multiple points during a unit, you give the student a more reliable way to demonstrate what he or she knows. Frequents assessments also allow you as a teacher to monitor the progress of class as you go, which gives you insight on how you may need to adjust your pacing and instructional methods. Vary your assessments. Your students won't all be strong in the same areas. One student may perform best on written tests, whereas another student may excel in a hands-on project. Be creative in your assessment designs. Traditional tests and worksheets are fine in moderation, but consider other alternatives. Students possibly could demonstrate their knowledge by writing a poem or song, performing a skit, sculpting an artistic design or creating a game that includes concepts recently learned. Record assessments reliably. Make sure you have a way to enter scores in a timely and organized fashion. This may be in a grade book or on your computer. Be specific about certain elements of the assessment: date, time, description, student information and any additional significant notes concerning individual performances. Adjust your instruction according to the outcomes you discover. If you assess frequently during the implementation of the unit, you will begin to see trends in learning outcomes. If the majority of your class is performing well and on-schedule, then you can plan to continue the track you're currently navigating. If a large number of students are producing poor results, you may need to slow down the pace of your instruction or even make sizable changes in the ways you're delivering the material. If you see that only a few students are struggling, you can design and implement interventions that target these specific few.<\|endoftext\|>	4.4375
3,783	\|"Yorkshire, particularly the East Riding, was subjected to very heavy Scandinavian settlement which may partly account for the paucity of Celtic names" (Faull, 1977: 15)\| In the following text, the terms "Celt" and "British" are employed interchangeably. References to "Anglo-Saxon" and "English" are similarly equivalent. Barber (1993) notes that the greatest linguistic influence of the Celts was in terms of place-names. Notwithstanding, Celtic place-names are far exceeded by those of English origin and the tendency is that Celtic survivals are more common in areas of late penetration by the Anglo-Saxons. This assertion is examined in greater detail below. Starting with an overview of the nature of Celtic name-giving, the correlation between the survival of British river-names and the English advance westwards across the country is then examined. Consideration is then given to the subsequent adoption of Celtic place-names by the incoming Anglo-Saxons. Finally, the focus is narrowed to Yorkshire and the Celtic place-names still extant within the county. Faull (1977) suggests that the endurance of British place-names in the Anglo-Saxon period is not necessarily indicative of Celtic settlement. It may simply be that the English adopted the names from nearby British speakers. In cases other than places of particular importance (such as those containing the elements duno- ("fort") and duro- ("walled town") relatively few earlier Celtic place-names are indicative of Celtic settlement. Unlike the Anglo-Saxons who employed precise terms to specifically identify the type of habitation (e.g. homestead, farm etc), British place-names are predominantly topographical rather than habitative, i.e. they named their settlements in relation to neighbouring topographical elements. (e.g. Crayke (NR) "rock, cliff" and Roose (ER) either "heath", "moor" or "promontory" "hill"). Faull proposes that either Celtic settlements were so similar to each other that there was no reason to distinguish them in terms of size in the manner of the Anglo-Saxons (e.g. ham, tun, wic, worð) or that naming settlements after the surrounding topographical features was merely a characteristic of Celtic name-giving. According to Faull (1977) Celtic place-names may be broken down into three categories: The first two groups generally tend to be Old English formations and they are accordingly discussed in the section relating to English place-names. With regard to the final category, Gelling (1988) adds that, in relation to Celtic and pre-Celtic place-names, two general principles can be said to exist: It is notable that, in areas where Celtic place-names are rare, only the larger geographical features bear Celtic names. Conversely, in locations where such names are frequent even minor sites and features bear British designations. Faull (1977) explains this phenomena by drawing attention to the correlation between the progress of the Anglo-Saxons from east to west and the survival of Celtic river-names. She cites Jackson (1953) who has demonstrated that the frequency of such names shows a marked correspondence to the westward movements of the invaders. British River Names As can be seen from the Map 1, Jackson divides England into four areas running from east to west. After Cameron 1996: 46 This area covers everything east of a line running from the Yorkshire Wolds down to Salisbury and the New Forest. It correlates with the territory of primary English settlement around 600 AD and contains very few British names. Those that do exist are generally the large or medium rivers and a high proportion of these are of doubtful etymology, possibly being of pre-Celtic origin. This is the area of intermediate settlement and reflects the English penetration westwards during the during the early 7th century. It runs to the east of Cumberland and Westmorland in the north to the Ribble and thence from Chester to the Bristol Channel along the Dee and Severn valley. Skirting Somerset it passes Selwood and onwards to the Hants-Dorset border. A much larger number of British river names exists in this region added to which their etymologies are more certain. Furthermore, a number of small rivers bear Celtic names. This area correlates to the final stages of the English advance westwards. Located to the west, it includes Cumberland and Westmorland, west Lancashire, the Welsh border counties and south-west England as far as the Tamar. The proportion of river names in this area is high and Celtic etymologies are more certain. There are more woods and hill names than in areas to the east and even stream names are of Celtic origin. This area, which covers the territories of Cornwall, Wales, and the south-west corner of Herefordshire, falls outside the limits of the English territorial settlement. The nomenclature within is almost purely Celtic. The question arises as to why the Anglo-Saxons adopted some Celtic place-names and not others. Adoption of Celtic Place-Names Just as pre-Celtic names would be passed on or taken over by the Britons from the pre-existing Bronze Age culture, so Celtic names were passed on to the Anglo-Saxons. However, this did not happen all at the same time because, as noted above, different parts of England were occupied at different times. The spread of the Anglo-Saxons beginning with the penetration into Kent and coastal areas of the East and Southeast from the middle of the 5th century lasted into the 9th century when advances were made into Cornwall. The very fact that names were adopted by the English presupposes the survival of Celtic speaking people in the districts where the names occur. Cameron (1996) states that, although it is uncertain how these names were communicated, one suggestion is that their adoption by the Anglo-Saxons was the result of bilingualism, specifically among the Britons. In other words, the place-names were passed on rather than borrowed. Notwithstanding, it would appear that, in those territories falling under their immediate control and occupation, the English either failed to adopt Celtic place-names, or changed them to an unrecognizable form. They did however adopt British names for locations on the fringes of these areas and in areas of later settlement. Faull (1977) proposes two possible reasons why the English failed to absorb local British names place-names in areas of early settlement. Firstly, because they were wholly occupied with establishing a foothold in the new territory, they were either disinclined, or had insufficient time, to form any sort of close association with the indigenous population. However, as the English had in fact established some settlements during the period of Roman occupation resulting in the English and the Romano-British living in "close proximity", Faull considers this explanation to be somewhat questionable. She considers that a much more reasonable explanation is that the lack of bilingualism existed within the two groups. As such, the transfer of place-names from one community to the other was impeded. Alternatively, it is possible that the English had already named their settlements and the neighbouring topographical features before the development of bilingualism within the locality. This would explain why British names tend to occur on the fringes of the primary English settlement areas rather than centrally. Celtic place-names in Yorkshire In relation to Yorkshire, the East Riding is situated in Jackson's Area 1 whilst the remainder of the county falls within Area 2. Concerning the first phase of the English incursion into the county, although there is a dearth of British names on the North Yorkshire coast, the Wolds and the Vale of Pickering, there are some on the "coastal strip of good land" and on the fringes of the Wolds (see Map 2). Similarly, British names are rare in Wensleydale, the Howardian Hills and the Vale of York, areas which were penetrated by the Anglo-Saxons in the second phase of their expansion. Those that do exist occur in the margins of these areas. The third and final stage of Anglian push westwards occurred in the early 7th century when they advanced into the West Riding. A large number of British place-names exist around the region of Elmet probably due to the fact that this area maintained its independence until the late 7th century. Further details of the walh, cumbra, brettas etc place-names may be accessed by clicking here. Citing Jackson (1953), Faull claims that, because of the probable changes arising out of phonological substitutions and "folk etymologies", scarcely any pre-English names of pre-1086 vintage exist in their original form in Yorkshire. The following etymologies have been drawn from several sources including Cameron (1996), Gelling (1988), Leith (1996), Strang (1970), Thomson (1964) and Thurlow (1979). The presence of an asterisk before a word (e.g. penno) is indicative of a postulated or constructed form. Any reference to Old English is abbreviated to OE. \|Possible or probable etymology\| \|\|Possibly derived from British Isara "strong river"\| Possibly from a word meaning "violent" and hence "the rough water". This may be of Celtic origin but no acceptable etymology has been suggested. \|\|From British Derva "oak" and thus probably "river lined with oaks".\| \|\|Probably from British dana from the root dan (cf the Danube and the Don in Eastern Europe)\| \|\|From the British dubo-"black" hence the "dark river".\| \|\|Derived from British Isca (cf Romano-British Isca as in Exe. Exe is an Old English metathesis of Esk).\| \|Hodder (WR)\|\|Suggested meaning is "peaceful" although research has shown that Celtic etymology is doubtful. In all likelihood the name is pre-Celtic.\| \|\|Originally regarded as Celtic but etymological reasoning is flimsy and the reconstruction debatable.\| \|Humber (ER)\|\|The suggested Celtic derivation is translated as "good" or "health giving" but this etymology is regarded as doubtful on the same grounds as for the River Hull. It is possible that the name is in fact pre-Celtic.\| \|\|From Celtic word meaning "narrow".\| \|\|From labara "talkative, noisy" and thus"murmuring river".\| \|There are two lines of thought. Either the name is derived from British lano "full" or from an alternative word meaning "good" or "health giving".\| \|\|Is of uncertain etymology but is possibly from a root meaning "brilliant.\| \|Many texts books suggest Celtic origins (possibly from the root udso) but the proposed etymology is unconvincing. The name may therefore be pre-Celtic.\| \|\|Probably means "chief river" from the Celtic prefix ro- and dubro "water".\| \|\|Possibly from a word cognate with Latin rivus "stream" (Also the source of Riccall meaning "calf of" or "little Rye").\| \|Tame (NWR)\|\|From a Celtic adjective meaning "dark"\| \|\|Possibly from a word meaning "surging".\| \|May be based on the river-name Isura linked to the Roman station of Isurium at Aldborough. Following the loss of intervocalic /s / and under the influence of the Old English diphthong io/eo the outcome is Eor. This subsequently becomes Scandinavianised to Jór from whence the modern form of Ure.\| \|\|Suggested meaning is "winding river".\| \|\|Possible or probable etymology\| \|Craike Hill (ER) \|\|The origin is identical to that of Crayke (NR)\| This is a tautology meaning "hill-hill" (from British penno + OE hyll.). Penno as proper noun was taken by the English to be a proper name. \|The Chevin (nr Otley, WR ) \|\|Derived from Old Welsh cefn "ridge".\| Areas / districts \|\|Possible or probable etymology\| \|The more northern of the two Anglian kingdoms in Northumbria, the name is derived from British Briganticia (from the tribal name Brigantes).\| \|Names such as Barwick in Elmet and Sherburn in Elmetare suggest that Elmet was in fact an area rather than a place. Generally regarded as a British kingdom it maintained its independence until the late 7th century AD. The origins of the name are uncertain.\| \|Probably a British enclave similar to Elmet. The name is possibly derived from British kramo- "garlic" although the reason for this is unclear (cf Italian Cremona).\| \|The southern Anglian kingdom of Northumbria, the name probably has its origins in British dobriu the possible meaning of which is "land where there are many rivers".\| Towns /villages / cities \|Name\|\|Possible or probable etymology\| \|Alne (NR)\|\|Certainly pre-English and possibly Celtic, it has been suggested that Alne on the River Kyle was originally as the name of the river itself (compare the rivers Allen, Alne and the Welsh Alun).\| \|Beverley (ER)\|\|The first element may be from British bebro- "beaver".\| \|Cammock (Settle, WR)\|\|Probably originates from a stream-name meaning "bent, twisting".\| \|Catterick (NR)\|\|Probably derived from the old name for the Swale.\| \|Craddock (WR)\|\|Probably an old stream-name representing the Brythonic form of the personal name Caractacus.\| \|Crayke (NR)\|\|From British krakjo "rock" (cf crag)\| \|Dacre (Ripon) (WR)\|\|A stream-name possibly from British dakru "tear", and thus perhaps "trickling".\| \|Dent (WR)\|\|A British name possibly cognate of Old Irish dinn "hill".\| \|Ecclesfield (WR)\|\|All are British / English hybrids incorporating the British egles (the Brythonic form of Latin ecclesia. The Brythonic form was subsequently adopted into Old English as ecles). Note: Exley =Ecclesley\| \|Glaisdale (NR)\|\|Probably Celtic Glas "blue, grey, green".\| \|Ilkley (WR)\|\|Any Celtic etymology is doubtful and the name may in fact be pre-Celtic.\| \|Leeds (WR)\|\|Formerly the name of a district rather than a place. Views on its etymology differ but it may have been either the original name for the River Aire or referred to the local populace whose name contained the element Lat- (cf Romano-British latenses). Smith (1961) supports the notion of it origins as a river name with subsequent progression to a folk name and ultimately to the name of a district.\| \|Leeming (NR)\|\|Probably a river-name turned settlement-name. The etymology may be linked with British lemanio "elm-tree". However, other possibilities exist.\| \|Leven (in Holderness) (ER)\|\|Probably from a river name which could possibly have been Prolemy's Libnios meaning "smooth" or "gliding".\| \|Penistone (WR)\|\|Comprising British Penno "hill" + OE ing + tun\| \|Rossington (WR)\|\|The first element is possibly derived Medieval Welsh ros "moor" (see also Roose (ER) mentioned earlier). However an alternative proposal is the word was not a place-name but a common noun adopted by the English in the same way as they appropriated egles.\| \|York\|\|The name originates from either Celtic Caer Ebruac or Eburacon- based on either personal name Eburos or the noun eburos "yew tree"; The Celtic name was later adopted by the Romans as Eboracum. Thence it became Old English Eorforwic and subsequently Scandinavian Iorvik before assuming its present form in the 13th century.\| \| Caer signifies a fortified place (cf Carlisle)\| Barber, C (1993) The English Language - A Historical Introduction, Cambridge: CUP Cameron, K (1996) English place-names, London: Batsford Faull, M L (1977) British Survival in Anglo-Saxon Northumbria. In Lloyd Laing (Ed)(1977) Studies in Celtic survival, British Archaeological Report 37 Gelling, M (1988) Signposts to the Past, Second Edition, Chichester: Phillimore. Jackson, K. (1953) Language and History in Early Britain, (Edinburgh, 1953) Leith, Dick (1996) The origins of English. In Graddol, D., Leith, D and Swann, J. (1996) English history, diversity and change, New York: Routledge. Smith, A H, (1961) The place-names of the West Riding of Yorkshire, Cambridge, Cambridge University Press, Strang, B M H (1970) A History of English, London: Methuen. Thomson, R L (1964) Celtic Place-names in Yorkshire, In Transactions of the Yorkshire Dialect Society Vol. XI, Part LXIV Thurlow, W. (1979) Yorkshire Place-names, Clapham, North Yorkshire: Dalesman Pubishing.<\|endoftext\|>	3.734375
6,671	# 5.8: Applications of Exponential and Logarithmic Functions $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ $$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$ $$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$ $$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vectorC}[1]{\textbf{#1}}$$ $$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$ $$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$ $$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$ $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left\|#1\right\|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$ ##### Learning Objectives • Apply common exponential models to real-life situations. • Apply common logarithmic models to real-life situations. We have already explored some basic applications of exponential and logarithmic functions. In this section, we explore some important applications in more depth, including radioactive isotopes and Newton’s Law of Cooling. ## Modeling Exponential Growth and Decay In real-world applications, we need to model the behavior with functions. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of population growth, such as a bacteria culture that a biologist is studying for a new medical treatment, we may choose the exponential growth function: $N(t)=N_0e^{rt}\nonumber$ where $$N_0$$ is equal to the value at time zero, $$e$$ is Euler’s constant, and $$k$$ is a positive constant that determines the rate (percentage) of growth. ##### Exponential Growth The model $$N(t)=N_0\cdot e^{kt}$$ represents the population size after a given amount of time $$t$$. $$N_0$$ is the initial population size and $$r$$ is the rate of growth for the population. ##### Example $$\PageIndex{1}$$ A population of bacteria doubles in size every hour. If the culture started with $$10$$ bacteria, find a model that describes the population size after of the bacteria after $$t$$ hours. Use this model to determine how many bacteria there will be after 10 hours. Solution Bacteria growth is exponential. To find $$N_0$$ we use the fact that $$N_0$$ is the amount at time zero, so $$N_0=10$$. To find $$r$$, use the fact that after one hour $$(t=1)$$ the population doubles from $$10$$ to $$20$$. The formula is derived as follows \begin{align} 20&= 10e^{r\cdot 1}\\ 2&= e^r \qquad \text{Divide by 10}\\ \ln2&= r \qquad \text{Convert to exponential form} \end{align} so $$r=\ln(2)$$. Thus the equation that models the bacteria growth is $$N(t)=10e^{(\ln2)t}=10{(e^{\ln2})}^t=10·2^t$$. The graph is shown in Figure $$\PageIndex{2}$$. The population of bacteria after ten hours is $$N(10)=10\cdot2^10=10,240$$. Analysis We could describe this amount is being of the order of magnitude $$10^4$$. The population of bacteria after twenty hours is $$10,485,760$$ which is of the order of magnitude $$10^7$$, so we could say that the population has increased by three orders of magnitude in ten hours. In the previous example, we were provided information about the time it took for the population to double in size. This time is called doubling time. Notice that the model simplified to an exponential function of base 2. We can describe a modified growth model for doubling time as follows: ##### Doubling Time The model $$N(t)=N_0\cdot2^{\frac{t}{d}}$$ represents the population size after a given amount of time $$t$$. $$N_0$$ is the initial population size and $$a$$ is the time it takes for the population to double in size. ##### Example $$\PageIndex{2}$$ Cancer cells sometimes increase exponentially. If a cancerous growth contained 300 cells last month and 360 cells this month, how long will it take for the number of cancer cells to double? Round your answer to the nearest tenth. Solution Defining $$t$$ to be time in months, with $$t = 0$$ corresponding to the initial size of the population, we are given two pieces of data: last month, (0, 300) and this month, (1, 360). From this data, we can find an equation for the growth. Using the form $$N(t)=N_0\cdot2^{\frac{t}{d}}$$, we know immediately $$N_0=300$$, giving $$N(t)=300\cdot2^{t}$$. Substituting in $$(1,360)$$, \begin{align} 360&=300\cdot2^\tfrac{1}{d} \\[4pt] \dfrac{360}{300}&= 2^\tfrac{1}{d}\\[4pt] 1.2&= 2^\tfrac{1}{d} \\[4pt] \ln1.2&=\ln\left(2^{\tfrac{1}{d}}\right) \\[4pt] \ln1.2&=\dfrac{1}{d}\ln2 \\[4pt] d\cdot\ln1.2&=\ln2 \\[4pt] d&=\dfrac{\ln2}{\ln1.2} \\[4pt] d &\approx3.8 \text{ months}\end{align}\nonumber We now turn to exponential decay. One of the common terms associated with exponential decay is half-life, the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay. The model $$M(t)=M_0e^{-\tfrac{ln2}{h}t}$$ represents the amount of material (mass) left after a given amount of time $$t$$. $$M_0$$ is the initial mass size and $$h$$ is the half-life of the material. The formula for radioactive decay is important in radiocarbon dating, which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about $$1\%$$ error for plants or animals that died within the last $$60,000$$ years. Carbon-14 is a radioactive isotope of carbon that has a half-life of $$5,730$$ years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12, which has an atomic weight of $$12$$ and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last $$60,000$$ years, using tree rings and other organic samples of known dates—although the ratio has changed slightly over the centuries. As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated. ##### Example $$\PageIndex{3}$$ The half-life of carbon-14 is $$5,730$$ years. Express the amount of carbon-14 remaining as a function of time, $$t$$. Solution The function that describes this continuous decay is $$M(t)=M_0e^{\left (-\tfrac{\ln(2)}{5730} \right )t}$$. ##### Example $$\PageIndex{4}$$ A bone fragment is found that contains $$20\%$$ of its original carbon-14. To the nearest year, how old is the bone? Solution The percentage of carbon-14 in all living things is $$100\%$$. The percentage of carbon-14 after some amount of time, $$t$$, (which we are trying to find) is 20%. Using decimals for the percent, we can substitute in $$M_0=1 \text{ and } M(t)=0.2$$ into the function we found in the previous example, \begin{align}M(t)&=M_0e^{\left (-\tfrac{\ln(2)}{5730} \right )t}\\[4pt] 0.2&=1\cdot e^{\left (-\tfrac{\ln(2)}{5730} \right )t} \\[4pt] 0.2&=e^{\left (-\tfrac{\ln(2)}{5730} \right )t} \\[4pt] \ln0.2&=\frac{-\ln(2)}{5730}t\\[4pt] -\dfrac{5730}{\ln2}(\ln0.2)&=t\\[4pt] t &\approx13,305 \text{ years}\end{align} The bone fragment is about $$13,305$$ years old. Analysis The instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about $$1\%$$, so this age should be given as $$13,305$$ years $$\pm 1\%$$ or $$13,305$$ years $$\pm 133$$ years. ## Newton’s Law of Cooling Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function. This translation leads to Newton’s Law of Cooling, the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature ##### Newton's Law of Cooling The temperature of an object, $$T$$, in surrounding air with temperature $$T_s$$ will behave according to the formula $T(t)=D_0e^{-kt}+T_s\nonumber$ where • $$t$$ is time • $$D_0$$ is the difference between the initial temperature of the object and the surroundings • $$k$$ is a constant, the continuous rate of cooling of the object ##### Example $$\PageIndex{5}$$ A cheesecake is taken out of the oven with an ideal internal temperature of $$165°F$$, and is placed into a $$35°F$$ refrigerator. After $$10$$ minutes, the cheesecake has cooled to $$150°F$$. If we must wait until the cheesecake has cooled to $$70°F$$ before we eat it, how long will we have to wait? Solution Because the surrounding air temperature in the refrigerator is $$35$$ degrees, the cheesecake’s temperature will decay exponentially toward $$35$$, following the equation $$T(t)=D_0e^{-kt}+35$$ We know the initial temperature was $$165$$, so $$T(0)=165$$. \begin{align} 165&= Ae^{-k0}+35 \qquad \text{Substitute } (0,165)\\ D_0&= 130 \qquad \text{Solve for } D_0 \end{align} We were given another data point, $$T(10)=150$$, which we can use to solve for $$k$$. \begin{align} 150&= 130e^{-k10}+35 \qquad \text{Substitute } (10, 150)\\ 115&= 130e^{-k10} \qquad \text{Subtract 35}\\ \dfrac{115}{130}&= e^{-10k} \qquad \text{Divide by 130}\\ \ln\left (\dfrac{115}{130} \right )&= -10k \qquad \text{Take the natural log of both sides}\\ k &= \dfrac{\ln \left (\dfrac{115}{130} \right )}{-10} \qquad \text{Divide by the coefficient of k} \end{align} This gives us the equation for the cooling of the cheesecake: $$T(t)=130e^{\left(–\tfrac{1}{10} \ln\tfrac{23}{26}t \right)}+35$$. Now we can solve for the time it will take for the temperature to cool to $$70$$ degrees. \begin{align} 70&= 130e^{\left(–\tfrac{1}{10} \ln\tfrac{23}{26} \right)}+35 \qquad \text{Substitute in 70 for } T(t)\\[4pt] 35&= 130e^{\left(–\tfrac{1}{10} \ln\tfrac{23}{26} \right)t} \qquad \text{Subtract 35}\\[4pt] \dfrac{35}{130}&= e^{\left(–\tfrac{1}{10} \ln\tfrac{23}{26} \right)t} \qquad \text{Divide by 130}\\[4pt] \ln \left (\dfrac{35}{130} \right )&= \left(–\tfrac{1}{10} \ln\tfrac{23}{26} \right)t \qquad \text{Take the natural log of both sides}\\[4pt] t&= \dfrac{\ln \left (\dfrac{35}{130} \right )}{\left(–\tfrac{1}{10} \ln\tfrac{23}{26} \right)t}\\[4pt] t&\approx 107 \qquad \text{Divide by the coefficient of t} \end{align} It will take about $$107$$ minutes, or one hour and $$47$$ minutes, for the cheesecake to cool to $$70°F$$. ##### You Try $$\PageIndex{1}$$ A pitcher of water at $$40$$ degrees Fahrenheit is placed into a $$70$$ degree room. One hour later, the temperature has risen to $$45$$ degrees. How long will it take for the temperature to rise to $$60$$ degrees to three decimal places? $$6.026$$ hours ## Modeling with Logarithms For quantities that vary greatly in magnitude, a standard scale of measurement is not always effective, and utilizing logarithms can make the values more manageable. For example, if the average distances from the sun to the major bodies in our solar system are listed, you see they vary greatly. Planet Distance (millions of km) Mercury 58 Venus 108 Earth 150 Mars 228 Jupiter 779 Saturn 1430 Uranus 2880 Neptune 4500 Placed on a linear scale – one with equally spaced values – these values get bunched up, as seen below. 0 500 1000 1500 2000 2500 3000 3500 4000 4500 However, computing the logarithm of each value and plotting these new values on a number line results in a more manageable graph, and makes the relative distances more apparent. Planet Distance (millions of km) log(distance) Mercury 58 1.76 Venus 108 2.03 Earth 150 2.18 Mars 228 2.36 Jupiter 779 2.89 Saturn 1430 3.16 Uranus 2880 3.46 Neptune 4500 3.65 (It is interesting to note the large gap between Mars and Jupiter on the log number line. There is an asteroid belt located there, which some scientists believe is a planet that never formed because of the effects of the gravity of Jupiter.) Logarithms are also used to measure sound intensity levels. Source of Sound/Noise Approximate Sound Pressure in $$\mu$$ Pa (micro Pascals) Launching of the Space Shuttle 2000,000,000 Full Symphony Orchestra 2000,000 Diesel Freight Train at High Speed at 25 m 200,000 Normal Conversation 20,000 Soft Whispering at 2 m in Library 2,000 Unoccupied Broadcast Studio Room 200 Softest Sound a human can hear 20 Logarithms are useful for showing relative changes. For example, comparing the sound of a diesel freight train to a soft whisper, the diesel train is 100 times larger than a soft whisper. $\dfrac{200,000}{2,000} = 100 = 10^2\nonumber$ When one quantity is roughly ten times larger than another, we say it is one order of magnitude larger. The order of magnitude can be found as the common logarithm of the ratio of the quantities. ##### Orders of Magnitude Given two values $$A$$ and $$B$$, to determine how many orders of magnitude $$A$$ is greater than $$B$$, Difference in orders of magnitude = log($$\dfrac{A}{B})$$ One example of a logarithmic scale is the Moment Magnitude Scale (MMS) used for earthquakes. This scale is commonly and mistakenly called the Richter Scale, which was a very similar scale succeeded by the MMS. ##### Moment Magnitude Scale for Earthquakes For an earthquake with seismic moment $$S$$, a measurement of earth movement, the MMS value, or magnitude $$M$$ of an earthquake, is $M = \dfrac{2}{3} log(\dfrac{S}{S_0})\nonumber$ Where $$S_0 = 10^{16}$$ is a baseline measure for the seismic moment. ##### Example $$\PageIndex{6}$$ If one earthquake has a MMS magnitude of 6.0, and another has a magnitude of 8.0, how much more powerful (in terms of earth movement) is the second earthquake? Solution Since the first earthquake has magnitude 6.0, we can find the amount of earth movement for that quake, which we'll denote $$S_1$$. The value of $$S_0$$ is not particularity relevant, so we will not replace it with its value. $6.0 = \dfrac{2}{3} log \left(\dfrac{S_1}{S_0}\right)\nonumber$ $6.0 \left(\dfrac{3}{2}\right) = log \left(\dfrac{S_1}{S_0}\right)\nonumber$ $9 = log\left(\dfrac{S_1}{S_0}\right)\nonumber$ $\dfrac{S_1}{S_0} = 10^9\nonumber$ $S_1 = 10^9 S_0\nonumber$ This tells us the first earthquake has about $$10^9$$ times more earth movement than the baseline measure, $$S_0$$. Doing the same with the second earthquake, $$S_2$$, with a magnitude of 8.0, $8.0 = \dfrac{2}{3} log \left(\dfrac{S_2}{S_0}\right)\nonumber$ $S_2 = 10^{12} S_0\nonumber$ Comparing the earth movement of the second earthquake to the first, $\dfrac{S_2}{S_1} = \dfrac{10^{12} S_0} {10^9 S_0} = 10^3 = 1000\nonumber$ The second value's earth movement is 1000 times as large as the first earthquake. ##### Example $$\PageIndex{7}$$ One earthquake has magnitude of 3.0. If a second earthquake has twice as much earth movement as the first earthquake, find the magnitude of the second quake. Solution Since the first quake has magnitude 3.0, $3.0 = \dfrac{2}{3} log (\dfrac{S}{S_0})\nonumber$ Solving for $$S$$, $3.0 \dfrac{3}{2} = log (\dfrac{S}{S_0})\nonumber$ $4.5 = log (\dfrac{S}{S_0})\nonumber$ $10^{4.5} = \dfrac{S}{S_0}\nonumber$ $S = 10^{4.5} S_0\nonumber$ Since the second earthquake has twice as much earth movement, for the second quake, $S = 2 \cdot 10^{4.5} S_0\nonumber$ Finding the magnitude, $M = \dfrac{2}{3} log (\dfrac{2 \cdot 10^{4.5} S_0}{S_0})\nonumber$ $M = \dfrac{2}{3} log (2 \cdot 10^{4.5}) \approx 3.201\nonumber$ The second earthquake with twice as much earth movement will have a magnitude of about 3.2. As we saw in an earlier section, logarithms are also used to describe $$pH$$ levels. ##### pH Level The pH of a solutions is defined by the following formula, where $$[\ce{H^{+}}]$$ is the concentration of hydrogen ions in the solution \begin{align} {pH}&=−{\log}([\ce{H^{+}}]) \label{eq1} \\[4pt] &={\log}\left(\dfrac{1}{[\ce{H^{+}}]}\right) \label{eq2} \end{align}\nonumber ##### Example $$\PageIndex{8}$$ If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH? Solution Suppose $$C$$ is the original concentration of hydrogen ions, and $$P$$ is the original pH of the liquid. Then $$P=–\log(C)$$. If the concentration is doubled, the new concentration is $$2C$$. Then the pH of the new liquid is $$pH=−\log(2C)$$ Using the product rule of logs $$pH=−\log(2C)=−(\log(2)+\log(C))=−\log(2)−\log(C)$$ Since $$P=–\log(C)$$,the new pH is $$pH=P−\log(2)≈P−0.301$$ This page titled 5.8: Applications of Exponential and Logarithmic Functions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Katherine Skelton.<\|endoftext\|>	4.78125
2,255	What is family violence? Family violence is when someone behaves abusively towards a family member. It is part of a pattern of behaviour that controls or dominates a person and causes them to fear for their own or others’ safety and wellbeing. Violent and abusive behaviour includes physical and sexual violence, and financial, emotional and psychological abuse. Slapping, hitting, rape, verbal threats, harassment, stalking, withholding money, and deliberately isolating someone from their friends and family are some examples of the types of behaviour that occur in family violence. In Victoria, the meaning of family violence is set out in the Family Violence Protection Act 2008. Who does family violence affect? Family violence is most commonly carried out by men against women who are their current or former partners. This is also known as intimate partner violence. Children living in homes where family violence occurs are affected through witnessing the behaviour or themselves being subject to abuse. Family violence is also perpetrated by: - people, including family members, who provide support for people with disabilities - adult children against their elderly parent/s - people in same-sex relationships - family members against others in their extended family1 How common is family violence? In Victoria, data on all family incidents attended by police are collated each year. The latest report for 2015–16 shows: - 78,012 family incidents, an increase of 10% from the previous year - of those affected, 74.8% were female and 24.8% male.4 The incidence of family violence is likely to be much higher than these figures indicate, as many incidents of family violence are never reported. What causes family violence? Family violence is complex. We know from international evidence that the major cause is inequality between women and men – that is, the unequal distribution of power, resources and opportunities. Stereotypical ideas about the roles of women and men in society and the way they should behave fosters an environment for violence against women to occur.5 Gender inequality plays out in society in many different ways, including: - ‘everyday sexism’ such as sexual and verbal harassment of women and girls - demeaning and sexualised portrayals of women and girls in the media - fewer women in leadership roles, giving men more control over decision‐making - the gender pay gap, caused by men being paid more than women for the same or similar work - women’s sport attracting less sponsorship, prize money and media coverage compared to men’s. In individual relationships, this inequality plays out in the belief that a man is entitled to exercise power and control over his partner and children. Individuals – both women and men – are more likely to condone, tolerate or excuse violence against women when they don’t believe women and men are equal, or see them as having firmly set roles or characteristics. For a detailed description of gendered drivers and reinforcing factors, go to Change the Story: A shared framework for the primary prevention of violence against women and their children in Australia. What is the impact of family violence? Family violence by a partner has an enormous impact on the women and children who experience it and on the community as a whole. Abuse and violence can severely limit the activities of women and children and affect their participation in all aspects of life.6 The impact of violence against women and their children costs the Australian economy more than $21.7 billion per year, and this cost is rising.7 Across Australia, police are called out to a family violence incident on average once every two minutes - Every three hours a woman is admitted to hospital as a result of family violence8 - At least one woman a week is killed by a partner or former partner - Intimate partner violence is: - the leading contributor to preventable illness, disability and death for women aged 15 to 44 - the single largest driver of homelessness for women - a common factor in child protection notifications.9 How women experience family violence While every woman’s experience of family violence is unique, for many women experiencing family violence there is a spiral of increasing abuse (this could be physical, emotional, financial, or a combination), rather than a one-off incident. Family violence often starts with an intimate partner’s apparent love transforming into controlling and intimidating behaviour. Over time, the woman is often increasingly isolated from friends and family by her partner. Physical violence may not occur until the relationship is well established, or it may not occur at all. The abusive, violent and controlling behaviours create an environment of fear and constant anxiety in a place where women and children should feel safe and secure. Common experiences for women include: - constant monitoring and regulation of her everyday activities such as phone calls, social interactions and dress - her every move measured against an unpredictable, ever-changing and unknowable ‘rule book’10 - constant put downs by her partner about anything and everything she does - having no control or say over the household finances - criticism of the way she parents her children - disrespectful behaviour towards her in front of their children and others - threats and actual physical violence against her, their children and pets - being blamed for the violence - surveillance using smartphones and other technology.11 Impact on women Abuse and violence have a destructive effect on women over time. It can affect her: - self-esteem, confidence and ability to make decisions - employment, financial independence and long term security - relationships with children and capacity to be a loving, effective parent - social connections with family, friends and community networks - physical health and wellbeing. It can cause: - physical injuries, chronic pain and other medical symptoms and illnesses - anxiety, depression, eating, sleep and panic disorders, suicidal behaviour, traumatic and post-traumatic stress disorders - the adoption of risk behaviours that further affect health such as substance and alcohol abuse, and physical inactivity - unwanted pregnancy, difficulties during pregnancy, foetal abnormalities, stillbirth and low birth weight.13 Impact on children and young people Children often hear or witness family violence, and this has a cumulative impact on them. Effects include:14 - impacts on the brain’s neural pathways, affecting cognitive development and stress response systems15 - low self-esteem and difficulties at school affecting their long term employment and financial security - mental health problems including anxiety, depression, symptoms of trauma, eating disorders and, for some, suicide attempts - increased aggression, anti-social behaviour and likelihood of substance abuse - teenage pregnancy. As well as impacting on health, wellbeing and education, growing up with family violence can affect relationships in later life. Many children who experience family violence do not go on to use violence as adults, but research shows that boys who have been exposed to family violence are more likely to become perpetrators themselves. Meanwhile, girls may be more accepting of intimate partner violence than those who hadn’t experienced family violence as children.16 Why it is hard for women to leave violent relationships17 Leaving a violent relationship is difficult and many women will attempt to leave a number of times before finally separating.18 There are many reasons for this. Increased risk of harm - Violence often escalates when the woman is planning to leave or actually leaves, with an increased risk of assault, stalking and murder. - Many family violence homicides occur during the separation period. Barriers to accessing the system - Women experiencing family violence may not know there are support services that can help them. - Women may not know about the kinds of support available to them; they may feel that services won’t be able to help with their situation. - Women may not have access to money and may not know where financial support is available. - A lack of safe and affordable housing options means women may feel there’s nowhere to go. - Lack of access to money or other resources. - Having to leave her job if she needs to be relocated for safety. Conflicting concerns and priorities - Not wanting to disrupt her children’ lives, education, and links to family and community. - Believing it’s in her children’s best interests to be close to their father. - Continuing to care for her partner and hoping he will change. (Many women don’t want to leave the relationship, they just want the violence to stop.) - For some Aboriginal and Torres Strait Islander women, the fear of putting their connections to extended kinship and family networks and to land at risk. - For some women with disabilities, the fear of losing the person on whom they feel dependent. - For some immigrant and refugee women, the fear of losing their residency entitlements. - Wanting to avoid the stigma associated with being a single parent. Social isolation and its effects - Feelings of shame and guilt about the violence or believing it is her fault. - Fear of being isolated or ostracised by her community. - Difficulty making decisions because she has been cut off from friends and family, is exhausted, and/or lacks confidence in her own judgement. What about men? There is strong evidence that men are overwhelmingly more likely than women to be perpetrators of family violence than victims. But men also experience family violence. Many of the male victims of intimate partner violence are in same-sex relationships. For example in the one of the largest surveys of LGBTI people ever conducted, the Private Lives study, found that of the 33 per cent of participants who had experienced intimate partner abuse, 28 per cent were men19. Many males grow up in homes where family violence is present. That most recent statistics show that over half a million women reported that their children had seen or heard partner violence. And over 400,000 women experienced partner violence during pregnancy20. There is evidence to show that along with the range of detrimental effects (see link to Impact on children and young people section), exposure to family violence in childhood can increase the risk of males becoming perpetrators in adulthood21. Both men and women are three times more likely to assaulted by a man than a woman. For women these assaults are more likely to occur in their home whereas men are more likely to be assaulted in a place of entertainment22. Find out more: Flood, M 2013 He hits, she hits: Assessing debates regarding men’s and women’s experiences of domestic violence. No to Violence, 2014 No to Violence response to the One in Three organisation’s comments about male victims The Line, What about violence towards men? http://www.theline.org.au/what-about-men Stark Evan, 2009, Coercive Control: How men entrap women in personal life, Oxford University Press, Oxford Coercive control: How can you tell whether your partner is emotionally abusive?<\|endoftext\|>	3.671875
1,485	What is labyrinthitis? Labyrinthitis is the inflammation of part of the inner ear called the labyrinth. The eighth cranial nerve (vestibulocochlear nerve) may also be inflamed. The inflammation of these causes a feeling of spinning (vertigo), hearing loss, and other symptoms. In most people, these symptoms go away over time. It is not a common condition. It often only affects one ear. The inner ear has a system of fluid-filled tubes and sacs called the labyrinth. Inside the inner ear, the cochlea gathers information about sound. The vestibular organs gather information about motion and changes in space. These all help to create a sense of balance. The eighth cranial nerve sends all of this information from the inner ear to the brain. When one of the nerves or the labyrinth is infected, it can become inflamed and irritated. This can cause it to not work normally. It may cause hearing loss in one ear. The brain now has to make sense of the information that doesn’t match between the normal nerve and the infected one. This causes vertigo. What causes labyrinthitis? A viral infection of the eighth cranial nerve or labyrinth may cause the condition. The virus may have spread all over your body. Or it may only affect the eighth cranial nerve and labyrinth. In most cases only one nerve is affected. Viruses known to cause labyrinthitis include: - Herpes viruses Bacterial infections of the middle ear are fairly common in children. In rare cases, an infection in the middle part of the ear can spread to the inner ear and cause labyrinthitis. This is more of a risk with middle-ear infections that are long-lasting (chronic) and not treated. In rare cases, bacterial meningitis or a head injury may cause labyrinthitis. In other cases, the cause is not known. Who is at risk for labyrinthitis? Having a viral infection that can cause labyrinthitis increases your risk. Your child’s risk may increase if he or she hasn’t had the recommended vaccines. What are the symptoms of labyrinthitis? Symptoms of labyrinthitis may include: - A feeling of spinning (vertigo) - Lack of balance when walking - Nausea and vomiting - Not able to focus (concentrate) - Periods of uncontrolled, back-and-forth eye movements (nystagmus) - Hearing loss - Ringing in the ears Your symptoms might range from mild to severe. They may come on very quickly. In many people, these symptoms go away over several weeks. Others have symptoms that last longer. A related syndrome is called vestibular neuritis. It causes similar symptoms. But it does not cause hearing problems. It affects only the vestibular part of the labyrinth and nerve. Labyrinthitis does not cause neurological symptoms such as severe headache, speech problems, or loss of arm or leg movement. How is labyrinthitis diagnosed? Your healthcare provider will ask about your health history. You may also have a physical exam. This may include hearing and balance tests. It will also include an exam of your nervous system. Many neurological and other health conditions can cause dizziness and vertigo. Your healthcare provider may need to rule these out. There are no tests for labyrinthitis. But your provider may have you take an imaging test. This can help to rule out other causes of your symptoms, such as stroke. You may have tests such as: - MRI. This is done to rule out stroke. - Electrocardiogram (ECG) or other cardiovascular tests. These can rule out cardiovascular causes. - Electronystagmography (ENG) or videonystagmography (VNG). These record your eye movement. This helps to find the exact area of the problem in your vestibular system and evaluate the cause of your balance disorder. How is labyrinthitis treated? Treatment will depend on your symptoms, age, and general health. It will also depend on how severe the condition is. Immediate treatment for labyrinthitis might include: - Corticosteroid medicines (to help reduce nerve inflammation) - Antiviral medicines - Antibiotics (if there are signs of a bacterial infection) - Medicines to take for a short time that control nausea and dizziness (such as diphenhydramine and lorazepam) If your symptoms go away in a few weeks, you likely won’t need other treatment. If you have symptoms that don’t go away, you may need to do certain exercises. These are known as vestibular rehabilitation exercises. They are a form of physical therapy. These exercises may help your brain learn to adjust to the vestibular imbalance. What are possible complications of labyrinthitis? In most cases, labyrinthitis does not cause any problems. In rare cases, labyrinthitis causes lasting (permanent) damage to the eighth cranial nerve. This can cause lasting problems with balance, and part or total hearing loss. You might need to use a hearing aid. Get treatment right away to help reduce your risk for these complications. When should I call my healthcare provider? Call your healthcare provider if your symptoms get worse or don’t begin to go away after a few days of treatment. Also call your healthcare provider right away if you have new symptoms, such as trouble moving an arm or a leg. Key points about labyrinthitis - Labyrinthitis often results from a viral infection of the eighth cranial nerve or the labyrinth. - Symptoms include vertigo, hearing loss, and dizziness. Symptoms may start suddenly and go away in a few weeks. - Your healthcare provider will need to rule out other more dangerous causes of vertigo, such as stroke. - You might need medicines to treat your symptoms. - If your symptoms don’t go away, you may need rehab exercises to help your brain adjust to the vestibular imbalance. - In rare cases, this condition can cause permanent hearing loss and ongoing vertigo. Tips to help you get the most from a visit to your healthcare provider: - Know the reason for your visit and what you want to happen. - Before your visit, write down questions you want answered. - Bring someone with you to help you ask questions and remember what your provider tells you. - At the visit, write down the name of a new diagnosis, and any new medicines, treatments, or tests. Also write down any new instructions your provider gives you. - Know why a new medicine or treatment is prescribed, and how it will help you. Also know what the side effects are. - Ask if your condition can be treated in other ways. - Know why a test or procedure is recommended and what the results could mean. - Know what to expect if you do not take the medicine or have the test or procedure. - If you have a follow-up appointment, write down the date, time, and purpose for that visit. - Know how you can contact your provider if you have questions.<\|endoftext\|>	3.875
5,924	# Linear differential equation Linear differential equation In mathematics, a linear differential equation is a differential equation of the form : $Ly = f ,$ where the differential operator "L" is a linear operator, "y" is the unknown function, and the right hand side &fnof; is a given function (called the source term). The linearity condition on "L" rules out operations such as taking the square of the derivative of "y"; but permits, for example, taking the second derivative of "y". Therefore a fairly general form of such an equation would be : $a_n\left(x\right) D^n y\left(x\right) + a_\left\{n-1\right\}\left(x\right)D^\left\{n-1\right\} y\left(x\right) + cdots + a_1\left(x\right) D y\left(x\right) + a_0\left(x\right) y\left(x\right) =f\left(x\right)$ where "D" is the differential operator "d/dx" (i.e. "Dy = y' ", "D"2"y = y",... "), and the "ai" are given functions. Such an equation is said to have order "n", the index of the highest derivative of "y" that is involved. (Assuming a possibly existing coefficient "an" of this derivative to be non zero, it is eliminated by dividing through it. In case it can become zero, different cases must be considered separately for the analysis of the equation.) If "y" is assumed to be a function of only one variable, one speaks about an ordinary differential equation, else the derivatives and their coefficients must be understood as (contracted) vectors, matrices or tensors of higher rank, and we have a (linear) partial differential equation. The case where &fnof; = 0 is called a homogeneous equation and its solutions are called complementary functions. It is particularly important to the solution of the general case, since any complementary function can be added to a solution of the inhomogeneous equation to give another solution (by a method traditionally called "particular integral and complementary function"). When the "ai" are numbers, the equation is said to have "constant coefficients". Homogeneous equations with constant coefficients The first method of solving linear ordinary differential equations with constant coefficients is due to Euler, who realized that solutions have the form $e^\left\{z x\right\}$, for possibly-complex values of $z$. The exponential function is one of the few functions that keep its shape even after differentiation. In order for the sum of multiple derivatives of a function to sum up to zero, the derivatives must cancel each other out and the only way for them to do so is for the derivatives to have the same form as the initial function. Thus, to solve :$frac \left\{d^\left\{n\right\}y\right\} \left\{dx^\left\{n + A_\left\{1\right\}frac \left\{d^\left\{n-1\right\}y\right\} \left\{dx^\left\{n-1 + cdots + A_\left\{n\right\}y = 0$ we set $y=e^\left\{z x\right\}$, leading to :$z^n e^\left\{zx\right\} + A_1 z^\left\{n-1\right\} e^\left\{zx\right\} + cdots + A_n e^\left\{zx\right\} = 0.$ Division by "e" "zx" gives the "n"th-order polynomial :$F\left(z\right) = z^\left\{n\right\} + A_\left\{1\right\}z^\left\{n-1\right\} + cdots + A_n = 0.,$ This equation "F"("z") = 0, is the "characteristic" equation considered later by Monge and Cauchy. Formally, the terms :$frac \left\{d^\left\{k\right\}y\right\} \left\{dx^\left\{kquadquad\left(k = 1, 2, dots, n\right).$ of the original differential equation are replaced by "z""k". Solving the polynomial gives "n" values of "z", "z"1, ..., "z""n". Substitution of any of those values for "z" into "e" "zx" gives a solution "e" "z""i""x". Since homogeneous linear differential equations obey the superposition principle, any linear combination of these functions also satisfies the differential equation. When these roots are all distinct, we have "n" distinct solutions to the differential equation. It can be shown that these are linearly independent, by applying the Vandermonde determinant, and together they form a basis of the space of all solutions of the differential equation. ExampleSidebar\|35% :$y$'-2y+2y"-2y'+y=0 , has the characteristic equation : $z^4-2z^3+2z^2-2z+1=0. ,$ This has zeroes, "i", −"i", and 1 (multiplicity 2). The solution basis is then : $e^\left\{ix\right\} ,, e^\left\{-ix\right\} ,, e^x ,, xe^x ,.$ This corresponds to the real-valued solution basis : $cos x ,, sin x ,, e^x ,, xe^x ,.$ The preceding gave a solution for the case when all zeros are distinct, that is, each has multiplicity 1. For the general case, if "z" is a (possibly complex) zero (or root) of "F"("z") having multiplicity "m", then, for $kin\left\{0,1,dots,m-1\right\} ,$, $y=x^ke^\left\{zx\right\} ,$ is a solution of the ODE. Applying this to all roots gives a collection of "n" distinct and linearly independent functions, where "n" is the degree of "F"("z"). As before, these functions make up a basis of the solution space. If the coefficients "Ai" of the differential equation are real, then real-valued solutions are generally preferable. Since non-real roots "z" then come in conjugate pairs, so do their corresponding basis functions nowrap\|"x""k"exp"zx", and the desired result is obtained by replacing each pair with their real-valued linear combinations Re("y") and Im("y"), where "y" is one of the pair. A case that involves complex roots can be solved with the aid of Euler's formula. Examples Given $y"-4y\text{'}+5y=0 ,$. The characteristic equation is $z^2-4z+5=0 ,$ which has zeroes 2+"i" and 2−"i". Thus the solution basis $\left\{y_1,y_2\right\}$ is $\left\{e^\left\{\left(2+i\right)x\right\},e^\left\{\left(2-i\right)x\right\}\right\} ,$. Now "y" is a solution iff $y=c_1y_1+c_2y_2 ,$ for $c_1,c_2inmathbb C$. Because the coefficients are real, we are likely not interested in the complex solutions our basis elements are mutual conjugatesThe linear combinations :$u_1=mbox\left\{Re\right\}\left(y_1\right)=frac\left\{y_1+y_2\right\}\left\{2\right\}=e^\left\{2x\right\}cos\left(x\right) ,$ and :$u_2=mbox\left\{Im\right\}\left(y_1\right)=frac\left\{y_1-y_2\right\}\left\{2i\right\}=e^\left\{2x\right\}sin\left(x\right) ,$ will give us a real basis in $\left\{u_1,u_2\right\}$. Simple harmonic oscillator The second order differential equation :$D^2 y = -k^2 y,$ which represents a simple harmonic oscillator, can be restated as :$\left(D^2 + k^2\right) y = 0.$ The expression in parenthesis can be factored out, yielding :$\left(D + i k\right) \left(D - i k\right) y = 0,$ which has a pair of linearly independent solutions, one for :$\left(D - i k\right) y = 0$ and another for :$\left(D + i k\right) y = 0.$ The solutions are, respectively, :$y_0 = A_0 e^\left\{i k x\right\}$ and :$y_1 = A_1 e^\left\{-i k x\right\}.$ These solutions provide a basis for the two-dimensional "solution space" of the second order differential equation: meaning that linear combinations of these solutions will also be solutions. In particular, the following solutions can be constructed :$y_\left\{0\text{'}\right\} = \left\{A_0 e^\left\{i k x\right\} + A_0 e^\left\{-i k x\right\} over 2\right\} = A_0 cos \left(k x\right)$ and :$y_\left\{1\text{'}\right\} = \left\{A_1 e^\left\{i k x\right\} - A_1 e^\left\{-i k x\right\} over 2 i\right\} = A_1 sin \left(k x\right).$ These last two trigonometric solutions are linearly independent, so they can serve as another basis for the solution space, yielding the following general solution: :$y_H = A_0 cos \left(k x\right) + A_1 sin \left(k x\right).$ Damped harmonic oscillator Given the equation for the damped harmonic oscillator: :$left\left(D^2 + \left\{b over m\right\} D + omega_0^2 ight\right) y = 0,$ the expression in parentheses can be factored out: first obtain the characteristic equation by replacing "D" with λ. This equation must be satisfied for all "y", thus: :$lambda^2 + \left\{b over m\right\} lambda + omega_0^2 = 0.$ Solve using the quadratic formula: :$lambda = \left\{-b/m pm sqrt\left\{b^2 / m^2 - 4 omega_0^2\right\} over 2\right\}.$ Use these data to factor out the original differential equation: :$left\left(D + \left\{b over 2 m\right\} - sqrt$b^2 over 4 m^2} - omega_0^2} ight) left(D + {b over 2m} + sqrtb^2 over 4 m^2} - omega_0^2} ight) y = 0. This implies a pair of solutions, one corresponding to :$left\left(D + \left\{b over 2 m\right\} - sqrt$b^2 over 4 m^2} - omega_0^2} ight) y = 0 and another to :$left\left(D + \left\{b over 2m\right\} + sqrt$b^2 over 4 m^2} - omega_0^2} ight) y = 0 The solutions are, respectively, :$y_0 = A_0 e^\left\{-omega x + sqrt\left\{omega^2 - omega_0^2\right\} x\right\} = A_0 e^\left\{-omega x\right\} e^\left\{sqrt\left\{omega^2 - omega_0^2\right\} x\right\}$ and :$y_1 = A_1 e^\left\{-omega x - sqrt\left\{omega^2 - omega_0^2\right\} x\right\} = A_1 e^\left\{-omega x\right\} e^\left\{-sqrt\left\{omega^2 - omega_0^2\right\} x\right\}$ where ω = "b" / 2"m". From this linearly independent pair of solutions can be constructed another linearly independent pair which thus serve as a basis for the two-dimensional solution space: :$y_H \left(A_0, A_1\right) \left(x\right) = left\left(A_0 sinh sqrt\left\{omega^2 - omega_0^2\right\} x + A_1 cosh sqrt\left\{omega^2 - omega_0^2\right\} x ight\right) e^\left\{-omega x\right\}.$ However, if \|ω\| < \|ω0\| then it is preferable to get rid of the consequential imaginaries, expressing the general solution as :$y_H \left(A_0, A_1\right) \left(x\right) = left\left(A_0 sin sqrt\left\{omega_0^2 - omega^2\right\} x + A_1 cos sqrt\left\{omega_0^2 - omega^2\right\} x ight\right) e^\left\{-omega x\right\}.$ This latter solution corresponds to the underdamped case, whereas the former one corresponds to the overdamped case: the solutions for the underdamped case oscillate whereas the solutions for the overdamped case do not. Nonhomogeneous equation with constant coefficients To obtain the solution to the non-homogeneous equation (sometimes called inhomogeneous equation), find a particular solution "y""P"("x") by either the method of undetermined coefficients or the method of variation of parameters; the general solution to the linear differential equation is the sum of the general solution of the related homogeneous equation and the particular solution. Suppose we face :$frac \left\{d^\left\{n\right\}y\right\} \left\{dx^\left\{n$ + A_{1}frac {d^{n-1}y} {dx^{n-1 + cdots + A_{n}y = f(x). For later convenience, define the characteristic polynomial :$P\left(v\right)=v^n+A_1v^\left\{n-1\right\}+cdots+A_n.$ We find the solution basis $\left\{y_1,y_2,ldots,y_n\right\}$ as in the homogeneous ("f"=0) case. We now seek a particular solution "yp" by the variation of parameters method. Let the coefficients of the linear combination be functions of "x": :$y_p=u_1y_1+u_2y_2+cdots+u_ny_n.$ Using the "operator" notation $D=d/dx$ and a broad-minded use of notation, the ODE in question is $P\left(D\right)y=f$; so :$f=P\left(D\right)y_p=P\left(D\right)\left(u_1y_1\right)+P\left(D\right)\left(u_2y_2\right)+cdots+P\left(D\right)\left(u_ny_n\right).$ With the constraints :$0=u\text{'}_1y_1+u\text{'}_2y_2+cdots+u\text{'}_ny_n$:$0=u\text{'}_1y\text{'}_1+u\text{'}_2y\text{'}_2+cdots+u\text{'}_ny\text{'}_n$:$cdots$:$0=u\text{'}_1y^\left\{\left(n-2\right)\right\}_1+u\text{'}_2y^\left\{\left(n-2\right)\right\}_2+cdots+u\text{'}_ny^\left\{\left(n-2\right)\right\}_n$ the parameters commute out, with a little "dirt": :$f=u_1P\left(D\right)y_1+u_2P\left(D\right)y_2+cdots+u_nP\left(D\right)y_n+u\text{'}_1y^\left\{\left(n-1\right)\right\}_1+u\text{'}_2y^\left\{\left(n-1\right)\right\}_2+cdots+u\text{'}_ny^\left\{\left(n-1\right)\right\}_n.$ But $P\left(D\right)y_j=0$, therefore :$f=u\text{'}_1y^\left\{\left(n-1\right)\right\}_1+u\text{'}_2y^\left\{\left(n-1\right)\right\}_2+cdots+u\text{'}_ny^\left\{\left(n-1\right)\right\}_n.$ This, with the constraints, gives a linear system in the $u\text{'}_j$. This much can always be solved; in fact, combining Cramer's rule with the Wronskian, :$u\text{'}_j=\left(-1\right)^\left\{n+j\right\}frac\left\{W\left(y_1,ldots,y_\left\{j-1\right\},y_\left\{j+1\right\}ldots,y_n\right)_\left\{0 choose f${W(y_1,y_2,ldots,y_n)}. The rest is a matter of integrating $u\text{'}_j.$ The particular solution is not unique; $y_p+c_1y_1+cdots+c_ny_n$ also satisfies the ODE for any set of constants "cj". Example Suppose $y"-4y\text{'}+5y=sin\left(kx\right)$. We take the solution basis found above $\left\{e^\left\{\left(2+i\right)x\right\},e^\left\{\left(2-i\right)x\right\}\right\}$.: Using the list of integrals of exponential functions :(Notice that "u"1 and "u"2 had factors that canceled "y"1 and "y"2; that is typical.) For interest's sake, this ODE has a physical interpretation as a driven damped harmonic oscillator; "yp" represents the steady state, and $c_1y_1+c_2y_2$ is the transient. Equation with variable coefficients A linear ODE of order "n" with variable coefficients has the general form:$p_\left\{n\right\}\left(x\right)y^\left\{\left(n\right)\right\}\left(x\right) + p_\left\{n-1\right\}\left(x\right) y^\left\{\left(n-1\right)\right\}\left(x\right) + cdots + p_0\left(x\right) y\left(x\right) = r\left(x\right).$ Examples A particular simple example is the Cauchy-Euler equation often used in engineering :$x^n y^\left\{\left(n\right)\right\}\left(x\right) + a_\left\{n-1\right\} x^\left\{n-1\right\} y^\left\{\left(n-1\right)\right\}\left(x\right) + cdots + a_0 y\left(x\right) = 0.$ First order equation ExampleSidebar\|35%\|$-3y"+4y\text{'}=5 ,$ with the initial condition : $fleft\left(0 ight\right)=2. ,$ Using the general solution method: : $f=e^\left\{-3x\right\}left\left(int 2 e^\left\{3x\right\}, dx + kappa ight\right). ,$ The integration is done from 0 to x, giving: : $f=e^\left\{-3x\right\}left\left(2/3left\left( e^\left\{3x\right\}-e^0 ight\right) + kappa ight\right). ,$ Then we can reduce to: : $f=2/3left\left(1-e^\left\{-3x\right\} ight\right) + e^\left\{-3x\right\}kappa. ,$ where "κ" is 2 from the initial condition.A linear ODE of order 1 with variable coefficients has the general form :$Dy\left(x\right) + f\left(x\right) y\left(x\right) = g\left(x\right).$ Equations of this form can be solved by multiplying the integrating factor :$e^\left\{int f\left(x\right),dx\right\}$ throughout to obtain :$Dy\left(x\right)e^\left\{int f\left(x\right),dx\right\}+f\left(x\right)y\left(x\right)e^\left\{int f\left(x\right),dx\right\}=g\left(x\right)e^\left\{int f\left(x\right) , dx\right\},$ which simplifies due to the product rule to : $D \left(y\left(x\right)e^\left\{int f\left(x\right),dx\right\}\right)=g\left(x\right)e^\left\{int f\left(x\right),dx\right\}$ which, on integrating both sides, yields : $y\left(x\right)e^\left\{int f\left(x\right),dx\right\}=int g\left(x\right)e^\left\{int f\left(x\right),dx\right\} ,dx+c ~,$ : $y\left(x\right) = \left\{int g\left(x\right)e^\left\{int f\left(x\right),dx\right\} ,dx+c over e^\left\{int f\left(x\right),dx ~.$ In other words: The solution of a first-order linear ODE : $y\text{'}\left(x\right) + f\left(x\right) y\left(x\right) = g\left(x\right),$ with coefficients that may or may not vary with "x", is: :$y=e^\left\{-a\left(x\right)\right\}left\left(int g\left(x\right) e^\left\{a\left(x\right)\right\}, dx + kappa ight\right)$ where "$kappa$" is the constant of integration, and : $a\left(x\right)=int\left\{f\left(x\right),dx\right\}.$ Examples Consider a first order differential equation with constant coefficients: :$frac\left\{dy\right\}\left\{dx\right\} + b y = 1.$ This equation is particularly relevant to first order systems such as RC circuits and mass-damper systems. In this case, "p"("x") = b, "r"("x") = 1. Hence its solution is :$y\left(x\right) = e^\left\{-bx\right\} left\left( e^\left\{bx\right\}/b+ C ight\right) = 1/b + C e^\left\{-bx\right\} .$ ee also * Laplace transform * Fourier transform Wikimedia Foundation. 2010. ### Look at other dictionaries: • Linear differential equation — Linear Lin ear (l[i^]n [ e][ e]r), a. [L. linearis, linearius, fr. linea line: cf. F. lin[ e]aire. See 3d {Line}.] 1. Of or pertaining to a line; consisting of lines; in a straight direction; lineal. [1913 Webster] 2. (Bot.) Like a line;… … The Collaborative International Dictionary of English • linear differential equation — noun : an equation of the first degree only in respect to the dependent variable or variables and their derivatives * * * Math. an equation involving derivatives in which the dependent variables and all derivatives appearing in the equation are… … Useful english dictionary • linear differential equation — Math. an equation involving derivatives in which the dependent variables and all derivatives appearing in the equation are raised to the first power. [1885 90] * * * … Universalium • Differential equation — Not to be confused with Difference equation. Visualization of heat transfer in a pump casing, created by solving the heat equation. Heat is being generated internally in the casing and being cooled at the boundary, providing a steady state… … Wikipedia • Ordinary differential equation — In mathematics, an ordinary differential equation (or ODE) is a relation that contains functions of only one independent variable, and one or more of their derivatives with respect to that variable. A simple example is Newton s second law of… … Wikipedia • Matrix differential equation — A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and of its derivatives of various orders. A matrix differential equation is one containing more… … Wikipedia • differential equation — Math. an equation involving differentials or derivatives. [1755 65] * * * Mathematical statement that contains one or more derivatives. It states a relationship involving the rates of change of continuously changing quantities modeled by… … Universalium • Partial differential equation — A visualisation of a solution to the heat equation on a two dimensional plane In mathematics, partial differential equations (PDE) are a type of differential equation, i.e., a relation involving an unknown function (or functions) of several… … Wikipedia • Hypergeometric differential equation — In mathematics, the hypergeometric differential equation is a second order linear ordinary differential equation (ODE) whose solutions are given by the classical hypergeometric series. Every second order linear ODE with three regular singular… … Wikipedia • Delay differential equation — In mathematics, delay differential equations (DDEs) are a type of differential equation in which the derivative of the unknown function at a certain time is given in terms of the values of the function at previous times. A general form of the… … Wikipedia<\|endoftext\|>	4.40625
1,913	Prior to this date, Basques from Argentina and Chile had migrated to the American West after the discovery of gold in California in 1848 and 1849. Basques from Euskal Herria that migrated directly from their homeland were forced to make the sea voyage across the Atlantic Ocean and around the southern cone of the Americas and up the Pacific coast to North America and the California coast. It was a dangerous, expensive, and what must have seemed like a never-ending voyage. However, once the cross-country railroad was finished, this brought immigrants to the port of New York where they disembarked and then continued on with their direct movement by rail to the west. The majority of those first Basque immigrants to New York had plans to move west, but instead found employment in the ports of New York and New Jersey, and a few had reached the end of their tolerance for traveling and simply refused to move any further. It is often forgotten that "the trip" from the Basque Country to the United States was not only the crossing of the Atlantic, but actually began days earlier with travel by horse or train from a rural town to the cities of Bilbao or San Sebastián-Donostia. The entire next day would include a train trip north to the French ports of Bordeaux or Le Havre. There, emigrants waited additional days completing paperwork and eventually boarded a passenger ship if they were lucky or wealthy, and a cargo ship if they were not. The journey for most Basques at the beginning of the 1900s was lengthy and frightening. Crossing the Atlantic could take anywhere from fourteen to thirty days depending on the itinerary, weather and storms, and the type of ship traveling. Hundreds of interviewed Basque immigrants remember the fear of the voyage and the seasickness they experienced, and advancing the last miles to the entrance of the port of New York was overwhelming for many. It also augmented a new fear of the metropolis, which was far beyond the stature any of them could have ever imagined from their experiences in the rural Basque Country. Between 1855-1890, Basque immigrants arriving to New York were processed at Castle Garden -as one of eight million other new arrivals. Ellis Island officially opened as an immigration processing station in 1892 and remained active until the 1924 National Origins Act was passed by Congress, allowing potential immigrants to undergo their inspections before they left their country of origin. Between 1897 and 1902, there were 636 persons with definite Basque surnames that entered the country through the immigration offices in New York. Eighty-six percent were male and seventy-seven percent were single. There most likely were many other Basques who also entered who were not counted because their surnames were not so obviously recognizable. Basques were only one of hundreds of ethnic groups speaking a myriad of languages that disembarked at Ellis Island. All new applicants were then held at Ellis Island for health inspections, medical inspections, hearings for those detained, literacy tests, and others. Persons with physical deformities, sickness or disease could be refused entry and sent back to their countries of origin. The relief of passing all of the inspections and tests was fantastic. Valentín Aguirre was one of those first Basque pioneers to reach New York City. He arrived from Bizkaia in 1895 and eventually become one of the most significant Basques in the United States. He and his wife, Benita Orbe, had eight children, and together they established the Basque boarding house known as the Santa Lucia Hotel and the Jai-Alai Restaurant. The Santa Lucia was also named the Casa Vizcaína. Though the exact records that Aguirre meticulously kept were unfortunately later thrown away carelessly, it is estimated that several thousand Basque immigrants stayed at the hotel and benefited from the Aguirres' care and assistance in continuing on their journey to Nevada, California, Idaho, and Oregon. Valentín, or one of his sons, would go out to the docks of the city and meet the passenger ships that brought the new immigrants into the city from Ellis Island once they has passed all of their inspections and paperwork. One can certainly imagine the overwhelming relief the Basques felt when from the busy docks of New York City they could hear the shout of "Euskaldunak emen badira?" "Are there any Basques here?" Basques on board shouted back with joy, "Bai, bai! Ni euskalduna naiz!" The Casa Vizcaína served as a travel agency as well, and Valentín Aguirre made arrangements to get Basque immigrants their train tickets to their final destinations inthe west, along with employment information and Basque boarding house along the way information. After staying in New York for a few days to recover from their sea voyages, the majority of Basques continued on to meet the relatives and fellow villagers. Benita Orbe Aguirre made each one a huge basket of food with French bread, tortilla, chorizo, ham and fruit to last them during the first few days of their train journey. Other Basques were thrilled with the energy of the city and decided to stay. Many were from coastal towns in the Basque Country and wanted to remain living in a coastal environment. Others had years of experience working on the docks and in maritime commerce and found jobs in the ports and docks immediately. The original Basque community took root at the foot of the Brooklyn Bridge along the docks of Cherry and Water Streets. Besides the Aguirre hotel there were Basque families that gave room and board to Basque immigrants in their own homes. There were Basque grocery stores and restaurants; Basque delivery businesses; wine and beer distribution businesses. Carmen Moneo sold imported goods from the Basque Country and Spain for more than seventy-five years until the 1980s. Most of the Basques attended Catholic mass at St. Joaquin's Church, St. Joseph's Church, and at the Our Lady of Guadalupe, where there was a Basque priest. Many of the Basque couples were married at the Our Lady of Guadalupe Church. In approximately 1905, Valentín Aguirre, Elias Aguirre, Juan Cruz Aguirre, Escolástico Uriona, and Toribio Altuna gathered together one night to discuss creating a Basque association for Basques in New York. By 1913 they had formed the Central Vasco-Americano Sociedad de Beneficiencia y Recreo, the first Basque Center of the United States, and in 1928 they purchased their first building. Initially this association was a mutual benefit and charity organization dedicated to helping those newly arrived, and to aiding those Basques living in New York that might be in financial difficulties. The Central Vasco-Americano, later renamed Centro Vasco-Americano, began as an all-male member organization although their families participated in all events. The building purchased in 1928 had an indoor fronton, and there was a Basque dancing group for youth. Basques organized their festival picnics at Coney Island and later at various parks. During the late 1940s and early 1950s several of the youth of the Basque organization decided to form their own group called Juventud. At that time, Jon Oñatibia, the organist, txistulari, dance choreographer, and Basque philologist was living in New York and he lead the way to forming the dance troupe Euzkadi, which eventually not only performed in the New York and east coast area, but also conducted a five month tour of Canada, the western United States, and Cuba. Oñatibia directed this group from 1950 to 1963 and later was selected by the Idaho Basque Studies Center and the North American Basque Organizations to teach Basque language, dance and txistu at their annual summer music camps for adolescents. The priest José Mari Larrañaga was another instrumental figure in maintaining Basque identity and collective activity in New York. While at the Church of St. James from 1962 to 1970, he united both generations by helping to organize dinners and dances for the Basque community. He gave masses in euskera. In 1966 the women of the New York Basque community also formed their own group named Andrak. In the 1960s the attendance at Basque picnics, dinners, and dances increased from a usual 150 to over 600 persons. In the decades after World War II, the Centro Vasco-Americano suffered from various problems of circumstances and had to move the seat of the organization various times, renting in different places in Manhattan. In 1973, after many years of renting and moving, the now named Euzko-Etxea of New York, bought their own building in what was a Polish neighborhood in the Brooklyn borough. The building was formerly a two story church and the Basques renovated it to include a large kitchen and bar, a dining room, a small meeting room or classroom for language classes, a small library, and the upstairs is a reception hall for special events that can seat more than 400 persons. A stage and piano complete the second floor reception hall. "The Basques of New York" by GLORIA TOTORICAGÜENA, First Edition 2003<\|endoftext\|>	4
840	# Right Triangles and Trigonometry ## Objective Define angles in standard position and use them to build the first quadrant of the unit circle. ## Common Core Standards ### Core Standards ? • F.TF.A.2 — Explain how the unit circle in the coordinate plane enables the extension of trigonometric functions to all real numbers, interpreted as radian measures of angles traversed counterclockwise around the unit circle. ? • 4.G.A.1 ## Criteria for Success ? 1. Identify an angle in standard position if the vertex is at the origin and one ray is on the positive $x$-axis. 2. Orient the initial side with the ray on the $x$-axis and the terminal side as the other ray. 3. Describe the unit circle as a circle where the radii are one unit and each radius is the hypotenuse of triangles formed inside the unit circle. 4. Describe the importance of the unit circle as it relates to trigonometric ratios. This unit circle will be used throughout higher math. 5. Define reference angles as positive and only in the first quadrant. 6. Know that $(x_\theta,y_\theta)=(\mathrm{cos}\theta,\mathrm{sin}\theta)$ ## Tips for Teachers ? • This lesson and the next lesson cover key concepts in the unit by reinforcing and reviewing G-SRT.6 and G-SRT.7. • The unit circle is used throughout calculus and is built on trigonometric ratios. ## Anchor Problems ? ### Problem 1 Given a blank coordinate plane, draw a 45° angle with a vertex at the origin. ### Problem 2 Suppose that point $P$ is the point on the unit circle obtained by rotating the initial ray 30°. Find ${\mathrm{sin}(30^\circ)}$ and ${\mathrm{cos}(30^\circ)}$ #### References EngageNY Mathematics Algebra II > Module 2 > Topic A > Lesson 4Example 1 Algebra II > Module 2 > Topic A > Lesson 4 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m.. Modified by Fishtank Learning, Inc. ### Problem 3 Each of the points in Quadrant I represent THE point of intersection of a ray that forms a 30° angle, a 45° angle, and a 60° angle with the $x$-axis in the initial position. What are the coordinate points where the rays intersect the circle at 30°, 45°, and 60°? #### References Unit Circle with Reference Triangles by Match Foundation, Inc. is made available by Desmos. Copyright © 2017 Desmos, Inc. Accessed March 13, 2017, 4:27 p.m.. ## Problem Set ? The following resources include problems and activities aligned to the objective of the lesson that can be used to create your own problem set. • Include problems where students are asked to identify and fix the error. For example, give the sine of an angle and a drawing on the unit circle with the angle not in standard position. • Include problems that require students to find the coordinate point based on the angle provided. For example, given an angle, find the coordinate point. Make sure to vary the information given and what is requested for an answer. • Include problems where students need to make a drawing based on the parameters provided. For example, “Draw an angle that has a sine of… Is there only one angle that will give this sine?” Draw an angle in standard position that has an intersection point on the unit circle of ${\left(\frac{1}{2},\frac{\sqrt3}{2}\right)}$. What is the sine of this angle? The cosine?<\|endoftext\|>	4.5
1,225	## Engage NY Eureka Math 1st Grade Module 5 Lesson 2 Answer Key ### Eureka Math Grade 1 Module 5 Lesson 2 Problem Set Answer Key Question 1. Use the key to color the shapes. Write how many of each shape are in the picture. Whisper the name of the shape as you work. a. REDâ€”4-sided shapes: ______7________ b. GREENâ€”3-sided shapes: ____3__________ c. YELLOWâ€”5-sided shapes: ___0___________ d. BLACKâ€”6-sided shapes: _____1_________ e. BLUEâ€”shapes with no corners: _____7________ a. REDâ€”4-sided shapes: 7 b. GREENâ€”3-sided shapes: 3 c. YELLOWâ€”5-sided shapes: 0 d. BLACKâ€”6-sided shapes: 1 e. BLUEâ€”shapes with no corners: 7, Explanation: Used the key to color the shapes. Wrote how many of each shape are in the picture as shown above as a. REDâ€”4-sided shapes: 7 b. GREENâ€”3-sided shapes: 3 c. YELLOWâ€”5-sided shapes: 0 d. BLACKâ€”6-sided shapes: 1 e. BLUEâ€”shapes with no corners: 7 Question 2. Circle the shapes that are rectangles. Explanation: Circled the shapes that are rectangles as shown above. Question 3. Is the shape a rectangle? Explain your thinking. a. _____Yes_______ ________________ Explanation: Yes, its a rectangle with four sides and 4 square corners. b. _______No____________ ___________________________ No, Explanation: No, its not a rectangle because the corners are not squares. ### Eureka Math Grade 1 Module 5 Lesson 2 Exit Ticket Answer Key Write the number of corners and sides that each shape has. Then, match the shape to its name. Remember that some special shapes may have more than one name. Explanation: Wrote the number of corners and sides that each shape has and then, matched the shape to its name as 1. 0 corners, 0 straight sides matched to circle, 2. 3 corners, 3 straight sides matched to triangle, 3. 6 corners, 6 straight sides matched to hexagon, 4. 4 corners, 4 straight sides matched to square respectively. ### Eureka Math Grade 1 Module 5 Lesson 2 Homework Answer Key Question 1. Color the shapes using the key. Write the number of shapes you colored on each line. Key RED 3 straight sides: _____3____ BLUE 4 straight sides: ____4______ GREEN 6 straight sides: ____1______ YELLOW 0 straight sides: ___2_____ RED 3 straight sides: 3 BLUE 4 straight sides: 4 GREEN 6 straight sides: 1 YELLOW 0 straight sides: 2, Explanation: Colored the shapes using the key and Wrote the number of shapes I colored on each line as RED 3 straight sides: 3 BLUE 4 straight sides: 4 GREEN 6 straight sides: 1 YELLOW 0 straight sides: 2. Question 2. a. A triangle has ____3_____ straight sides and ______3___ corners. b. I colored _____3 red_____ triangles. a. A triangle has 3 straight sides and 3 corners, b. I colored 3 red triangles, Explanation: As a triangle has 3 straight sides and 3 corners, So, I colored 3 red triangles as shown above. Question 3. a. A hexagon has _____6____ straight sides and _____6_____ corners. b. I colored ____green______ hexagon. a.Â A hexagon has 6 straight sides and 6Â corners, b. I colored 1 green hexagon, Explanation: As a hexagon has 6 straight sides and 6Â corners, So, I colored 1 green hexagon as shown above. Question 4. a. A circle has ____0___ straight sides and ____0_____ corners. b. I colored ___2 yellow____ circles. a. A circle has 0 straight sides and 0 corners, b. I colored 2 yellow circles. Explanation: As a circle has 0 straight sides and 0 corners, So, I colored 2 yellow circles as shown above. Question 5. a. A rhombus has ___4__ straight sides that are equal in length and ___4__ corners. b. I colored ___1 blue___ rhombus. a. A rhombus has 4 straight sides that are equal in length and 4 corners, b. I colored 1 blue rhombus, Explanation: As a rhombus has 4 straight sides that are equal in length and 4 corners, SoÂ I colored 1 blue rhombus as shown above. Question 6. A rectangle is a closed shape with 4 straight sides and 4 square corners. a. Cross off the shape that is NOT a rectangle. ____________No, 4 straight sides and 4 square corners__, Explanation: Crossed off the shape that is NOT a rectangle as shown above as aÂ rectangle is a closed shape with 4 straight sides and 4 square corners only shape 3 has no 4 straight sides and 4 square corners, so crossed it. Question 7. A rhombus is a closed shape with 4 straight sides of the same length. a. Cross off the shape that is NOT a rhombus.<\|endoftext\|>	4.75
707	Ch 2 Lecture - Vectors.pptx - Chapter 2 Force Vectors(Vector Mathematics Dr Rakhsha Nasseripour PHYS 239 Adding Force Vectors • Scalar quantity Has # Ch 2 Lecture - Vectors.pptx - Chapter 2 Force... • Notes • 59 This preview shows page 1 - 13 out of 59 pages. Chapter 2: Force Vectors (Vector Mathematics) Dr. Rakhsha Nasseripour PHYS 239 Subscribe to view the full document. Scalar quantity Has magnitude, can be 0, positive, or negative number. Vector quantity Has magnitude and direction. Cannot be described by one number. Subscribe to view the full document. Anatomy of a Vector x y q R Vector Addition Triangle Rule (or Tip-to-tail method) 1. Connect tail of B to head of A . 2. Resultant R goes from first tail to last head. A B R Subscribe to view the full document. Vector Addition Parallelogram Law 1. Join tails of vectors. 2. Draw line parallel to A from head of B . 3. Draw line parallel to B from head of A . 4. Diagonal of resulting parallelogram is the resultant vector, R . 5. A and B are also called components of R . A B R Cosine Rule: Subscribe to view the full document. Sine Rule: Example 2-1: The screw eye is subjected to two forces F1 and F2. Determine the magnitude and direction of the resultant force Subscribe to view the full document. Example 2-2: Resolve the horizontal 600-lb force in to components acting along the u and v axes and determine the magnitude of these components. Example 2-3: Find a Missing Component The resultant force of two forces is along the +y axis. Calculate the magnitude of Forces A and B and the angle of Force B such that Force B has a minimum magnitude. F 1 = 200 F 2 30° y x 45 Subscribe to view the full document. Find a Resultant Force (Vector addition of two forces) Find the resultant force on a pin that has two forces applied to it as shown below. • Fall '19 ### What students are saying • As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students. Kiran Temple University Fox School of Business ‘17, Course Hero Intern • I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero. Dana University of Pennsylvania ‘17, Course Hero Intern • The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time. Jill Tulane University ‘16, Course Hero Intern Ask Expert Tutors You can ask 0 bonus questions You can ask 0 questions (0 expire soon) You can ask 0 questions (will expire ) Answers in as fast as 15 minutes<\|endoftext\|>	4.59375
769	# Number of words with length 3 with certain restrictions In today's quiz (about combinatorics) there was a question I was not able to solve mathematically: Find the number of words with length 3 over an alphabet $\Sigma = \{1,2,...,10\}$ where $x \lt y \le z$. ($x$ is the first letter, and so on). How can I solve this in an efficient way? Of course you can just count it like: $9 \times 10 \times 10 \rightarrow$ 1 possiblity $8 \times \{9,10\} \times \{9,10,10\} \rightarrow$ 2 + 1 possibilities .... But how do I solve it with "combinatoric" methods? Think about it like this: If we had to choose $a,b,c$ from $\{1,2,\ldots,10\}$ so that $a < b < c$, we could just choose a subset of $3$ elements from $\{1,2,\ldots,10\}$ and sort them, giving $a < b < c$. The number of such subsets is $\binom{10}{3}$. Now, since we have $x < y \leq z$, we have $x < y < (z + 1)$; we may thus choose $3$ distinct elements from $\{1,2,\ldots,11\}$ (the three elements are $x$, $y$, and $z+1$). We thus have $\binom{11}{3}$ ways to do so, yielding $\binom{11}{3}$ as our answer. In other words, counting $x < y \leq z$ from $\{1,2,\ldots ,10\}$ is the same as counting $x < y < (z + 1)$ from $\{1,2,\ldots,11\}$. • @BrianO, I think there's a misunderstanding. The answer is $\binom{11}{3}$, not the sum of $\binom{11}{3}$ and $\binom{10}{3}$. Commented Oct 20, 2015 at 22:15 • Note that $\binom{11}{3} = \binom{10}{3} + \binom{10}{2}$, so our answers agree. Commented Oct 20, 2015 at 22:18 • Ohhh OK -- doh, sorry, I can plead interruptions :) Yes they're the same (good!). I'll delete my "Hmm, problem". Commented Oct 20, 2015 at 22:40 There are two kinds of length 3 words $w = xyz$ satisfying the conditions that $x < y \le z$ and $x, y, z \in \Sigma = \{1, 2, 3, \dots, 10\}$: 1. Those with $y = z$ These are in 1-1 correspondence with the set of all 2-element subsets of $\Sigma$, so there are $\binom {10} 2$ of them. 2. Those with $y < z$ These are in 1-1 correspondence with the set of all 3-element subsets of $\Sigma$, so there are $\binom {10} 3$ of them. So the number of strings satisfying the conditions is \begin{align} \binom {10} 2 + \binom {10} 3 &= 45 + 120 \\ &= 165 \text{.} \end{align}<\|endoftext\|>	4.4375
848	Science fiction has long dreamed of turning Mars into a second Earth, a place where humans could live without having to put on a space suit. The easiest way to do that would be to use carbon dioxide already on Mars to create a new atmosphere, but now researchers say that is impossible. Terraforming Mars to make its surface habitable for Earth life would involve raising both its temperature and pressure by adding an atmosphere made of heat-trapping greenhouse gases. The only ones present on Mars in any significant amounts are carbon dioxide and water vapour, both of which are currently frozen. What would it be like to live on Mars? Find out at New Scientist Live in London “If there is enough carbon dioxide, we could warm up Mars in 100 years once we start,” says Chris McKay at NASA’s Ames Research Center in California. “We know how to warm up a planet – we’re doing it on Earth. The fundamental question is, is there enough stuff?” No, it turns out. Bruce Jakosky at the University of Colorado, Boulder, and Christopher Edwards at Northern Arizona University used results from several spacecraft to build an inventory of all the carbon dioxide on Mars to figure out whether, if we moved all of it from the ground into the atmosphere, we could create high enough temperatures and pressures for life. Right now Mars has an atmospheric pressure of about six millibars – tiny compared to the one bar at sea level on Earth. “We would need something like a million ice cubes of carbon dioxide ice that are a kilometre across in order to do get to one bar,” says Jakosky. At one bar, the temperature would be just above 0°C, allowing liquid water, and thus life, on the surface. The atmosphere wouldn’t be breathable, but humans could get by with breathing masks, not full space suits, and plants could grow freely, slowly building up oxygen over the course of the next few centuries. But Jakosky and Edwards found that there’s probably only enough carbon dioxide in the Martian polar ice caps, dust and rocks to raise the pressure to 20 millibars at most. So we can’t terraform Mars with existing technology, because there simply isn’t enough carbon dioxide. “It’s not that terraforming itself isn’t possible, it’s just that it’s not as easy as some people are currently saying,” says Jakosky. “We can’t just explode a few nukes over the ice caps.” It ain’t easy There may be hidden reservoirs of carbon deep under the surface that could make the job easier, says Robin Wordsworth at Harvard University. “If you could develop the technology to look for those and extract it, that might get you close to the bar,” he says. “But it’d be kind of a fishing expedition – there’s no guarantee that these things exist.” Without enough carbon, we would have to warm up Mars some other way, perhaps by making chlorofluorocarbons (CFCs) or bombarding the planet with comets or asteroids. That’s going to be difficult, and it will still not be enough to truly make Mars a home. For that, we need nitrogen – and we’re still not sure how much of that Mars has. “If there’s not enough carbon dioxide, terraforming would take thousands of years or more but it’s still possible,” says McKay. “If there’s not enough nitrogen, you need Star Trek. You need warp drive and tractor beams, you need to pull nitrogen from the atmosphere of Jupiter. It becomes science fiction.” Journal reference: Nature Astronomy, DOI: 10.1038/s41550-018-0529-6 Article amended on 13 August 2018 We corrected the amount of CO2 needed to make a Martian atmosphere More on these topics:<\|endoftext\|>	3.671875
2,049	# 9.3: Multiplication of Monomials by Polynomials Difficulty Level: Basic Created by: CK-12 Estimated7 minsto complete % Progress Practice Multiplication of Monomials by Polynomials MEMORY METER This indicates how strong in your memory this concept is Progress Estimated7 minsto complete % Estimated7 minsto complete % MEMORY METER This indicates how strong in your memory this concept is Did you know that the formula for the volume of a pyramid is \begin{align}V= \frac{1}{3}Bh\end{align}, where \begin{align}B\end{align} is the area of the base of the pyramid and \begin{align}h\end{align} is the pyramid's height? What if the area of the base of a pyramid were \begin{align}x^2 + 6x + 8\end{align} and the height were \begin{align}9x\end{align}? What would the volume of the pyramid be? In this Concept, you'll learn how to multiply a polynomial by a monomial so that you can answer questions such as this. ### Guidance When multiplying polynomials together, we must remember the exponent rules we learned in previous Concepts, such as the Product Rule. This rule says that if we multiply expressions that have the same base, we just add the exponents and keep the base unchanged. If the expressions we are multiplying have coefficients and more than one variable, we multiply the coefficients just as we would any numbers. We also apply the product rule on each variable separately. #### Example A Multiply \begin{align}(2x^2 y^3) \times (3x^2 y)\end{align}. Solution: \begin{align}(2x^2 y^3) \times (3x^2 y)=(2\cdot3) \times (x^2 \cdot x^2) \times (y^3 \cdot y)=6x^4 y^4\end{align} Multiplying a Polynomial by a Monomial This is the simplest of polynomial multiplications. Problems are like the one above. #### Example B Multiply the following monomials. (a) \begin{align}(2x^2)(5x^3)\end{align} (c) \begin{align}(3xy^5)(-6x^4y^2)\end{align} (d) \begin{align}(-12a^2b^3c^4)(-3a^2b^2)\end{align} Solutions: (a) \begin{align}(2x^2)(5x^3)=(2 \cdot 5)\cdot (x^2 \cdot x^3) = 10x^{2+3}=10x^5\end{align} (c) \begin{align}(3xy^5)(-6x^4y^2)=-18x^{1+4}y^{5+2}=-18x^5y^7\end{align} (d) \begin{align}(-12a^2b^3c^4)(-3a^2b^2) = 36a^{2+2}b^{3+2}c^4=36a^4b^5c^4\end{align} To multiply monomials, we use the Distributive Property. Distributive Property: For any expressions \begin{align}a, \ b\end{align}, and \begin{align}c\end{align}, \begin{align}a(b+c)=ab+ac\end{align}. This property can be used for numbers as well as variables. This property is best illustrated by an area problem. We can find the area of the big rectangle in two ways. One way is to use the formula for the area of a rectangle. \begin{align}Area \ of \ the \ big \ rectangle & = Length \times Width\\ Length & = a, \ Width = b + c\\ Area & = a \times (b + c)\end{align} The area of the big rectangle can also be found by adding the areas of the two smaller rectangles. \begin{align}Area \ of \ red \ rectangle & = ab\\ Area \ of \ blue \ rectangle & = ac\\ Area \ of \ big \ rectangle & = ab + ac\end{align} This means that \begin{align}a(b+c)=ab+ac\end{align}. In general, if we have a number or variable in front of a parenthesis, this means that each term in the parenthesis is multiplied by the expression in front of the parenthesis. \begin{align}a(b+c+d+e+f+\ldots)=ab+ac+ad+ae+af+ \ldots\end{align} The “...” means “and so on.” #### Example C Multiply \begin{align}2x^3 y(-3x^4 y^2+2x^3 y-10x^2+7x+9)\end{align}. Solution: \begin{align}& 2x^3 y(-3x^4 y^2+2x^3 y-10x^2+7x+9)\\ & = (2x^3 y)(-3x^4 y^2 )+(2x^3 y)(2x^3 y)+(2x^3 y)(-10x^2 )+(2x^3 y)(7x)+(2x^3 y)(9)\\ & = -6x^7 y^3+4x^6 y^2-20x^5 y+14x^4 y+18x^3 y\end{align} ### Guided Practice Multiply \begin{align}-2a^2b^4(3ab^2+7a^3b-9a+3)\end{align}. Solution: Multiply the monomial by each term inside the parenthesis: \begin{align}& -2a^2b^4(3ab^2+7a^3b-9a+3)\\ & = (-2a^2b^4)(3ab^2)+(-2a^2b^4)(7a^3b)+(-2a^2b^4)(-9a)+(-2a^2b^4)(3)\\ & = -6a^3b^6-14a^5b^5+18a^5b^4-6a^2b^4\end{align} ### Practice Multiply the following monomials. 1. \begin{align}(2x)(-7x)\end{align} 2. \begin{align}4(-6a)\end{align} 3. \begin{align}(-5a^2b)(-12a^3b^3)\end{align} 4. \begin{align}(-5x)(5y)\end{align} 5. \begin{align}y(xy^4)\end{align} 6. \begin{align}(3xy^2z^2)(15x^2yz^3)\end{align} Multiply and simplify. 1. \begin{align}x^8 (xy^3+3x)\end{align} 2. \begin{align}2x(4x-5)\end{align} 3. \begin{align}6ab(-10a^2 b^3+c^5)\end{align} 4. \begin{align}9x^3(3x^2-2x+7)\end{align} 5. \begin{align}-3a^2b(9a^2-4b^2)\end{align} Mixed Review 1. Give an example of a fourth degree trinomial in the variable \begin{align}n\end{align}. 2. Find the next four terms of the sequence \begin{align}1,\frac{3}{2},\frac{9}{4},\frac{28}{8}, \ldots\end{align} 3. Reece reads three books per week. 1. Make a table of values for weeks zero through six. 2. Fit a model to this data. 3. When will Reece have read 63 books? 1. Write 0.062% as a decimal. 2. Evaluate \begin{align}ab\left ( a+\frac{b}{4} \right )\end{align} when \begin{align}a=4\end{align} and \begin{align}b=-3\end{align}. 3. Solve for \begin{align}s\end{align}: \begin{align}3s(3+6s)+6(5+3s)=21s\end{align}. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Difficulty Level: Basic Tags: Subjects: ## Concept Nodes: 8 , 9 Date Created: Feb 24, 2012 Aug 16, 2016 Image Detail Sizes: Medium \| Original MAT.ALG.612.L.1<\|endoftext\|>	4.9375
411	Even though the Internet is still a young technology, it's hard to imagine life without it now. Every year, engineers create more devices to integrate with the Internet. This network of networks crisscrosses the globe and even extends into space. But what makes it work? To understand the Internet, it helps to look at it as a system with two main components. The first of those components is hardware. That includes everything from the cables that carry terabits of information every second to the computer sitting in front of you. Other types of hardware that support the Internet include routers, servers, cell phone towers, satellites, radios, smartphones and other devices. All these devices together create the network of networks. The Internet is a malleable system -- it changes in little ways as elements join and leave networks around the world. Some of those elements may stay fairly static and make up the backbone of the Internet. Others are more peripheral. These elements are connections. Some are end points -- the computer, smartphone or other device you're using to read this may count as one. We call those end points clients. Machines that store the information we seek on the Internet are servers. Other elements are nodes which serve as a connecting point along a route of traffic. And then there are the transmission lines which can be physical, as in the case of cables and fiber optics, or they can be wireless signals from satellites, cell phone or 4G towers, or radios. All of this hardware wouldn't create a network without the second component of the Internet: the protocols. Protocols are sets of rules that machines follow to complete tasks. Without a common set of protocols that all machines connected to the Internet must follow, communication between devices couldn't happen. The various machines would be unable to understand one another or even send information in a meaningful way. The protocols provide both the method and a common language for machines to use to transmit data. We'll take a closer look at protocols and how information travels across the Internet on the next page.<\|endoftext\|>	4.3125
2,806	Bubbles large and small signal changes in the Arctic and in Earth’s atmosphere. by Jane Beitler November 12, 2012 Bubbles large and small signal changes in the Arctic and in Earth’s atmosphere. It was nearly winter in Greenland, the tundra patchworked with rumples of earth holding lakes sheathed in smooth ice and snow. Researcher Katey Walter Anthony trudged through the light snow around yet another lake on her survey list, looking for bubbles trapped in the lake ice. “We stumbled across something really weird in a lake right in front of the ice sheet,” she said. “We saw a huge open area in the lake that looked like it was boiling.” Walter Anthony and her team were visiting lakes to measure methane bubbling up. But the roiling seep looked like none other she had seen. “It looked like something deeper and larger, large plumes of bubbles rushing upward,” Walter Anthony said. “So I got curious: where is this gas coming from and what is the mechanism for its release and how widespread is it?” It was a new twist in the problem of lake ice and methane emissions across the changing Arctic. Thawing out the freezer Walter Anthony had been studying methane seeping from Arctic lakes, beginning in northeast Siberia in 2000. Under the lakes, a thick layer of carbon from plants that died hundreds or thousands of years ago stays mostly locked up in permanently frozen ground, like broccoli in the freezer. Today, soils in Siberia and northern Alaska are particularly rich with that organic matter. Now Arctic tundra hovers at a colder temperature that sprouts no trees and only low shrubs and plants, but millions of ponds and lakes. In areas where that permafrost is warming, that organic matter is thawing, rotting, and producing gases that must escape through the lakes. Guido Grosse studies how these lakes, called thermokarst lakes, form and change. “Permafrost keeps the lakes from draining,” Grosse said. “That’s why there are so many lakes.” In recent years, the Arctic has warmed even more strongly than lower latitudes. Now in many areas, the ground is thawing deeper than it used to. “As permafrost degrades, lakes can drain,” said Grosse, at the Permafrost Laboratory at University of Alaska Fairbanks. In other areas, permafrost thaw results in a sinking land surface where new ponds and lakes form, exposing underlying permafrost to even more warming, thawing, and decay. Grosse said, “The lakes are a big emitter of methane in a warmer climate scenario, a warmer Arctic.” Some organic material from vegetation and frozen lake banks normally falls in the lake, thaws, and decays around its edges. This decay stops during the cold season in shallow lakes that freeze to the bottom in harsh Arctic winters. But most lakes deeper than 1.5 meters (5 feet) no longer freeze all the way to the bottom. In these lakes, the organic carbon is beginning to thaw and rot year-round, and the permafrost underneath the lake is beginning to thaw out deeply. Microbes decompose organic carbon in the lake sediments, and in the thawed-out zone under the lake, into methane gas that bubbles to the surface. As the lake surface refreezes in fall, researchers can see the bubbles, trapped in the ice. But they lacked wide-scale measurements of the escaping methane. Bubble, bubble, toil, and trouble In search of methane bubbles, Walter Anthony’s team traveled to lakes by snow machine, helicopter, hiking in, canoe, and bush airplane. “We’ve gone out now on hundreds of lakes and mapped out these methane seeps, in Alaska, Russia, Canada, Finland, Sweden, and Greenland,” she said. It is painstaking work conducted on often dangerously thin first ice in early winter. Melanie Engram, who works with Walter Anthony on the methane studies, explained what it takes to measure emissions at a single lake. Engram said, “There often is snow on top of the ice, so first you shovel a 1-meter wide by 50-meter long [3-foot by 164-foot] transect. Then we drill a hole in the ice on one side, and get a bucket of water and pour it over the transect to remove the last specks of snow so we can see through the ice. Then you can easily see, count, categorize, and measure methane bubbles.” As lakes freeze over in fall, bubbles released from lake sediments get trapped under the freezing surface. The researchers can see stacks of bubbles, separated by thin films of ice, like a time-lapse photograph showing where the bubbles are coming from under the lake. The bubbles and the rate of gas release vary across a lake, and from lake to lake. “If the bubbles are coming up slowly enough, the ice has a chance to grow around them,” Engram said. “Katey has been working to categorize the bubbles. Type A is slow and indicates a small gas flux hardly keeping up with lake ice growth; with type B, some of the bubbles have grouped together by the time the ice forms. Type C has quite large pillows of gas before the ice forms around it. Each of these categories corresponds to a certain rate of gas seepage.” The “boiling” lakes became a fourth type, called “hotspot,” where methane is nearly continuously seeping out at very high rates. The researchers were able to measure seepage rates for each category by installing automated bubble traps, which look like underwater umbrellas, to measure the gas escaping year round. As the permafrost thaws The ultimate goal of the team’s project is Arctic-wide estimates of lake methane emissions. Such estimates are needed for computer climate models, which help test and deepen scientists’ understanding of how Arctic climate responds to change. But with millions of lakes and millions of square miles of Arctic, Engram said, “We can’t go measure every lake. There’s no way of traveling everywhere.” The team thought they could inspect the lakes and compare field observations with satellite images on a larger scale. Then they could apply the bubble cluster classifications and the measurements from their ground studies to estimate how much methane each lake is emitting. This would give them a way to estimate methane emissions from lakes across the entire Arctic. Engram said, “Katey had the idea of looking at Synthetic Aperture Radar (SAR) data.” Other researchers had published studies noting that SAR can detect brighter areas corresponding to tubular bubbles in floating ice. Engram said, “We thought, well, if we see brighter ice where there are tubular bubbles, maybe we can find a SAR wavelength that would be sensitive to the various methane bubble types.” Engram was then working for NASA's Alaska Satellite Facility (ASF) Distributed Active Archive Center (DAAC), which distributes RADARSAT-1 SAR data. A major challenge was to align the data very precisely with the locations of individual lakes, and she thought she knew how to solve it with a new tool from ASF DAAC. “We took SAR data and pushed it through the Convert tool,” Engram said. The tool converted the SAR data into geolocated files that could be used in ArcGIS, a data mapping software. Engram compared the images with their ground observations. The brighter the ice in the SAR imagery, the more bubbles. Early winter SAR images showed the highest correlation with field measurements of methane bubbles. Engram said, “It’s important to know how much methane comes out of northern lakes, because methane is a very potent greenhouse gas that is 25 to 28 times more powerful than carbon dioxide at retaining heat in the atmosphere on a 100-year time scale. If we can do this with SAR remote sensing in a way that’s inexpensive, using NASA’s already available data and tools, we could contribute useful estimates to the Arctic methane budget.” Uncapping the cryosphere But what about the wildly boiling gas plumes? Walter Anthony still wanted to understand what was happening under the ground to cause such a high flow rate. “My husband and I got in little airplanes and started flying around looking for places in the winter where lakes were open because of methane seepage,” she said. “We flew around and looked at about 6,700 lakes in Alaska, but then we needed to ground truth it. So we went to fifty out of the seventy-seven of the sites where we had seen open areas. We found that yes, every one of them does indeed have very large plumes of methane coming up. But the weird thing is, it was only in certain places.” Walter Anthony and her colleagues studied the geology of the areas where they located the big seeps. In the Arctic, frozen ground can keep gas trapped for thousands of years. “Permafrost is a thick cap that seals off deeper geologic layers by blocking pathways through pore spaces with ice,” she said. “There is natural gas underneath some permafrost regions, and that gas cannot escape into the atmosphere because the permafrost is impermeable.” The team did a geospatial analysis, and found that the gas plumes were near places where glaciers and ice sheets are retreating, and where the thickest, most extensive layers of permafrost are now disintegrating from warming and thawing. These methane emissions are strong, but transient. “If you’ve got a pot of water boiling on the stove with a lid on top, you have a bunch of steam that’s building up inside of there, you take the lid off, that steam goes up, poof! But then the air clears,” Walter Anthony said. “And in the same way you pull back this cryosphere cap, it lets the methane out in a poof, over probably a century to thousands of years.” On a human scale, that poof of methane means large amounts of carbon added to an already warming atmosphere. “The lakes are much bigger emitters than we thought before, now that we have come to understand how much methane is actually bubbling out of the lakes,” Walter Anthony said. “In the future we don’t know what will happen. It is a bit of a wild card.” Sorting out all of these contributions helps scientists factor methane emissions into the overall study of Earth’s climate. “Our work is another piece of the puzzle, closely linked to other processes associated to a changing world, and important if you want to know how much methane and carbon dioxide will be emitted in the future,” Grosse said. Walter Anthony, K. M., P. Anthony, G. Grosse, and J. Chanton. 2012. Geologic methane seeps along boundaries of Arctic permafrost thaw and melting glaciers. Nature Geoscience, doi: 10.1038/Ngeo1480. Grosse, G., J. Harden, M. Turetsky, D. A. McGuire, P. Camill, C. Tarnocai, S. Frolking, E. A. G. Schuur, T. Jorgenson, S. S. Marchenko, et al. 2011. Vulnerability of high-latitude soil organic carbon in North America to disturbance. Journal of Geophysical Research - Biogeosciences, 116: G00K06, doi: 10.1029/2010JG001507. Walter, K. M., M. Engram, C. R. Duguay, M. O. Jeffries, and F. S. Chapin III. 2008. The potential use of synthetic aperture radar for estimating methane ebullition from Arctic lakes. Journal of the American Water Resources Association 44(2): 305–315, doi: 10.1111 ⁄j.1752-1688.2007.00163.x. Walter, K. M., S. A. Zimov, J. P. Chanton, D. Verbyla, and F. S. Chapin III. 2006. Methane bubbling from Siberian thaw lakes as a positive feedback to climate warming. Nature 443, doi: 10.1038/nature05040. For more information Alaska Satellite Facility Distributed Active Archive Center (ASF DAAC) \|About the remote sensing data used\| \|Satellite\|\|Canadian Space Agency RADARSAT-1\| \|Sensor\|\|Synthetic Aperture Radar (SAR)\| \|Data sets\|\|Level 1\| \|DAAC\|\|NASA Alaska Satellite Facility Distributed Active Archive Center (ASF DAAC)\| The photograph in the title graphic shows researcher Melanie Engram prodding the snow on the lake surface to check for thin ice, before approaching the snow-free circles that suggest methane seeping from underneath the lake. (Courtesy K. W. Anthony) Last Updated: May 14, 2019 at 8:38 AM EDT<\|endoftext\|>	3.90625
1,335	Physics tries to answer some of the most fundamental questions in the universe. What caused the Big Bang? Why is the observable universe so isotropic? Why is Earth able to nurture life? One important question that has mystified physicists for years is the idea of mass and more importantly its origins. The discovery of the Higgs boson goes some way to explaining this mystery. The most basic explanation of what mass is may be that it is how heavy a physical body is, it is what physical bodies are made up of. Some may even go on to talk about how much matter an object contains. Moreover, you feel mass all around you. When you push on an object, the object has a resistive force which acts in the opposite direction to your push. Part of the resistance you feel is the mass of the object but where does it originate from? Each physical body is made up of its constituent elements and these elements are made up of their subatomic particles: electrons, protons and neutrons. The protons and neutrons are made up of even smaller particles called quarks, but mass is not just a fundamental property of an object like electrical charge. Mass has to be acquired. So how do particles on a nanoscopic scale acquire mass? The theorists: Robert Brout, François Englert and Peter Higgs tried to answer this question in 1964. Their idea was that the universe is a space which is filled with the Higgs field in all directions. A field in physics is a physical quantity that has a value for each point in space and time. The value of the Higgs field in a vacuum is expected to be 246 GeV. In addition the Higgs field has no source whereas a gravitational field is due to the mass of an object. When particles permeate and interact with this Higgs field they experience a resistance which slows them down and makes them heavier. For e.g. the resistance an electron experiences whilst interacting with the surrounding Higgs field is the mass of the electron. However the photon which is a massless particle doesn’t interact with the Higgs field at all. There is a strong relationship between fields and particles which goes back to quantum mechanics. The particles of light, the photons are associated with the electromagnetic field and the gravitational field has a theorised associated particle called the graviton. The particles that are associated with each field are observed as an excitation of that field. Thus when the Higgs field is fluctuated, it must have an associated particle called the Higgs boson, as theorised by Peter Higgs. The existence of the Higgs field can only be confirmed if the Higgs boson is detected. The mass of the Higgs boson was predicted using mathematical modelling but the inaccuracy associated with it was so large that it had a range of over 5 different orders of magnitude. The confirmation of the existence of the Higgs boson, which in turn confirms the existence of the Higgs field came in March 2013 from the Large Hadron Collider (LHC) in Geneva, Switzerland. The LHC is a high energy particle accelerator, which spans 27km (16.5 miles) in circumference, that can recreate the extreme energies of the early universe. Particles such as protons are accelerated to nearly the speed of light in the LHC. The two proton beams travelling in opposite directions have a total energy of 8 TeV. The collision of the two proton beams with one another leaves behind a vast array of other smaller particles. Approximately 0.5 billion proton-proton collisions occur each second. The collision disturbs the space around the particles and this caused a fluctuation in the Higgs field. If done in the right way the fluctuation can be observed as a Higgs boson. Only about 200 out of every 800 trillion collisions produces the Higgs particle. Detection of the Higgs boson in the 2 detectors present at the LHC is another daunting task due to its short mean lifetime of approximately 1.56×10−22 s. Like other heavy particles the Higgs decays into other particles almost immediately, long before it can be observed directly. The two detectors: ATLAS and CMS are arranged in an onion-like structure with various layers of different detectors capable of measuring different decay products. The collision occurs in the centre of the detector and the debris travels radially outwards. The standard model of particle physics can thus be used to predict its decay products and their individual probabilities. Physicists compare the decay paths observed after a particle collision to the decay paths they have simulated using computational methods. They also observe the rate at which the Higgs particle decays to other particles as a function of its mass. When they see a match it is a good sign that the particle that decayed must have come from the Higgs Boson. By studying these decay products at CMS and ATLAS, we now know that the Higgs boson decays into W and Z bosons, 4 electrons and 2 photons. Experimental data also suggests that the Higgs decays into fermions but the probability of this is quite low. The discovery of the Higgs is a new addition to the Standard Model. Until now we had only discovered 2 types of particles: the matter particles which are the quarks and leptons and the force particles which are the photons, gluons and the W and Z bosons. The discovery of the Higgs particle is a new addition to this family of particles. As in much of science, this new discovery has led to a plethora of additional questions. Recent experiments at the LHC are trying to measure the likelihood of each decay mode of the Higgs. Physicists are also trying to calculate the spin of the Higgs by measuring the angular displacement of its decay products. Moreover questions such: ‘Is the Higgs a composite particle or is it made up of something else?’ are being posed. The Higgs field is of immense significance and is the source by which elementary particles acquire mass. It was first theorised in the 1960’s and was confirmed in March 2013. The painstaking 50 year effort involved 5000 physicists and engineers worldwide. The idea of Higgs began as mathematical symbols in an equation on paper. Over 50 years the idea was refined and now mathematics has led the way in the confirmation of the existence of the Higgs boson. The Higgs bridges the gap between the mathematical simplicity as theorised by physicists and the full complexity of the world we live in.<\|endoftext\|>	3.828125
892	NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4. Board CBSE Textbook NCERT Class Class 9 Subject Maths Chapter Chapter 14 Chapter Name Statistics Exercise Ex 14.4 Number of Questions Solved 6 Category NCERT Solutions ## NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 Question 1. The following number of goals were scored by a team in a series of 10 matches 2, 3, 4, 5, 0, 1, 3, 3, 4, 3. Find the mean, median and mode of these scores. Solution: (iii) Mode Arranging the given data in ascending order, we have 0, 1,2, 3, 3, 3, 3, 4, 4, 5. Here, 3 occurs most frequently (4 times) ∴ Mode = 3 Question 2. In a mathematics test given to 15 students, the following marks (out of 100) are recorded 41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60 Find the mean, median and mode of this data. Solution: (i) Mean (ii) Median Arranging the given data in descending order, we have 98, 96, 62, 60, 54, 52, 52, 52, 48, 46, 42, 41,40, 40, 39 Number of observations (n) = 15 which is odd. (iii) Mode Arranging the data in descending order, we have 98, 96, 62, 60, 54, 52, 52, 52, 48, 46, 42, 41,40, 40, 39. Here, 52 occurs most frequently (3 times). ∴ Mode = 52 Question 3. The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x. 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95 Solution: Number of observations (n) = 10which is even. According to question, median = 63 ∴ x + 1 = 63 ⇒ x = 63 – 1 = 62 Hence, the value of x is 62. Question 4. Find the mode of 14, 25,14, 28,18,17,18,14, 23, 22,14 and 18. Solution: The given data is, 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 Arranging the data in ascending order, we have 14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28 Here, 14 occurs most frequently (4 times). ∴ Mode = 14 Question 5. Find the mean salary of 60 workers of a factory from the following table Solution: Hence, the mean salary is ₹ 5083.33. Question 6. Give one example of a situation in which (i) the mean is an appropriate measure of central tendency. (ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency. Solution: (i) Mean marks in a test in mathematics, (ii) Average beauty We hope the NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4, drop a comment below and we will get back to you at the earliest.<\|endoftext\|>	4.96875
229	This nine-minute video shows the phenomenon of prairie dog vocalizations to a variety of threats to their population. The video provides evidence, using sonograms and prairie dog behavioral responses, that each vocalization contains discrete pieces of information that are understood by the local prairie dog population. Teachers may want to prompt students to think more about how the different calls that prairie dogs make elicit different responses. For example: How do you think the prairie dogs respond to different sized predators? Do you think their response is the same no matter what information is encoded in the vocalization? Why do you think the prairie dogs are able to include detailed information in their vocalizations? How does the ability to encode complex information in their vocalizations help a prairie dog population to survive and reproduce? For background information, teachers may access the Scientific American article in which this video was embedded: https://blogs.scientificamerican.com/running-ponies/catch-the-wave-decoding-the-prairie-doge28099s-contagious-jump-yips/<\|endoftext\|>	3.796875
2,033	# INTRODUCTION In statistics, there are three kinds of techniques that are used in the data univariate data analysis. These are univariate analysis, bivariate analysis, and multivariate analysis. How the data analysis technique is selected is based on the variable number and the data type. The statistical inquiry focus is also something to be considered. # UNIVARIATE ANALYSIS Univariate analysis is a basic kind of analysis technique for statistical data. Here the data contains just one variable and does not have to deal with the relationship of a cause and effect. Like for example consider a survey of a classroom. The analysts would want to count the number of boys and girls in the room. The data here simply talks about the number which is a single variable and the variable quantity. The main objective of the univariate analysis is to describe the data in order to find out the patterns in the data. This is done by looking at the mean, mode, median, standard deviation, dispersion, etc. Univariate analysis is basically the simplest form to analyze data. Uni means one and this means that the data has only one kind of variable. The major reason for univariate analysis is to use the data to describe. The analysis will take data, summarise it, and then find some pattern in the data. Univariate analysis is conducted in many ways and most of these ways are of a descriptive nature. These are the Frequency Distribution Tables, Frequency Polygons, Histograms, Bar Charts and Pie Charts Let us get into details here of the kind of analysis that is done to analyze univariate data. • Frequency distribution table Frequency means how often something takes place. The observation frequency tells the number of times for the occurrence of an event. The frequency distribution table may show categorical or qualitative and numeric or quantitative variables. The distribution gives a snapshot of the data and lets you find out the patterns • Bar chart The bar chart is represented in the form of rectangular bars. The graph will compare various categories. The graph could be plotted vertically or these could be plotted horizontally. In maximum cases, the bar will be plotted vertically. The horizontal or the x-axis will represent the category and the vertical y-axis represents the category’s value. The bar graph looks at the data set and makes comparisons. Like for example, it may be used to see what part is taking the maximum budget? • Histogram The histogram is the same as a bar chart which analysis the data counts. The bar graph will count against categories and the histogram displays the categories into bins. The bin is capable of showing the number of data positions, the range, or the interval. • Frequency Polygon The frequency polygon is pretty similar to the histogram. However, these can be used to compare the data sets or in order to display the cumulative frequency distribution. The frequency polygon will be represented as a line graph. • Pie Chart The pie chart displays the data in a circular format. The graph is divided into pieces where each piece is proportional to the fraction of the complete category. So each slice of the pie in the pie chart is relative to categories size. The entire pie is 100 percent and when you add up each of the pie slices then it should also add up to 100. Pie charts are used to understand how a group is broken down into small pieces. The univariate data is the one that consists of just one variable. The analysis of univariate data is the simplest since the information has to deal with a single quantity only and the changes in it. It does have to study the relationship and cause and the analysis is used to describe the data and to find out the pattern that exists in it. Like for example, the height of ten students in a class can be recorded and this is univariate data. There is only one variable which is the height and thus it does not have any relationship and cause attached to it. The description of the pattern that is found in this type of data is made by drawing out conclusions based on dispersion, central measures of tendency, spread, or data, and this is done through the histograms, frequency distribution table, bar charts, etc. Univariate analysis works by examining its effect on a single variable on a given data set. Like for example, the frequency distribution table is a kind of univariate analysis. Here only one variable is involved in the data analysis. There could however be many alternate variables too like height, age, and weight. As soon as a secondary variable gets introduced in the analysis then this is a bivariate analysis. When there is three or more than three variable involved in the data analysis then this is the multivariate analysis. Univariate is a common term that you use in statistics to describe a type of data that contains only one attribute or characteristic. The salaries of people in the industry could be a univariate analysis example. The univariate data could also be used to calculate the mean age of the population in a village. # BIVARIATE ANALYSIS Bivariate analysis means the analysis of the bivariate data. This is a single statistical analysis that is used to find out the relationship that exists between two value sets. The variables that are involved are X and Y. • Univariate analysis is when only one variable is analyzed. • Bivariate data analysis is when exactly two variables are analyzed. • Multivariate analysis is when more than two variables get analyzed. The results that are obtained from the bivariate analysis are stored in a data table that has two columns. Bivariate analysis should not be confused with two sample data analyses where the x and y variables are not related directly. Here is how the bivariate analysis is carried out. • Scatter plots — This gives an idea of the patterns that can be formed using the two variables • Regression Analysis — This uses a wide range of tools to determine how the data post could be related. The post may follow an exponential curve. The regression analysis gives the equation for a line or curve. It also helps to find the correlation coefficient. • Correlation Coefficients –The coefficient lets you know if the data in question are related. When the correlation coefficient is zero then this means that the variables are not related. If the correlation coefficient is a positive or a negative 1 then this means that the variables are perfectly correlated. The kind of bivariate analysis is dependent on the kind of attributes and variables that is used to analyze the data. The variables may be ordinal, categorical, or numeric. The independent variable is categorical like a brand of a pen. In this case, probit regression or logit regression is used. If the dependent and the independent variables are both ordinal which means that they have a ranking or position then the rank correlation coefficient is measured. In case the dependent attribute is ordinal then the ordered probit or the ordered logit is used. It is possible that the dependent attribute could be internal or a ratio like the scale of temperature. This is where regression is measured. Here is how we mention the kinds of bivariate data correlation. • Numerical and Numerical In this kind of variable both the variables of the bivariate data which includes the dependent and the independent variable have a numerical value. • Categorical and Categorical When both the variables in the bivariate data are in the static form then the data is interpreted and statements and predictions are made about it. During the research, the analysis will help to determine the cause and impact to conclude that the given variable is categorical. • Numerical and Categorical This is when one of the variables is numerical and the other is categorical. Bivariate analysis is a kind of statistical analysis when two variables are observed against each other. One of the variables will be dependent and the other is independent. The variables are denoted by X and Y. The changes are analyzed between the two variables to understand to what extent the change has occurred. Bivariate analysis is the analysis of any concurrent relation between either two-variable or attributes. The study will explore the relationship that is there between the two variables as well as the depth of the relationship. It helps to find out if there are any discrepancies between the variable and what the causes of the differences are. The bivariate analysis examples are used is to study the relationship between two variables. Let us understand the example of studying the relationship between systolic blood pressure and age. Here you take a sample of people in a particular age group. Say you take the sample of 10 workers. The first column will have the age of the worker and the second column records their systolic blood pressure. The table then needs to be displayed in a graphical format to make some conclusion from it. The bivariate data is usually displayed through a scatter plot. Here the plots are made on a grid paper y-axis against the x-axis and this helps to find out the relationship between the data sets that are given. A Scatter plot helps to form a relationship between the variables and tries to explain the relationship between the two. Once you apply the age on the y-axis and the systolic blood pressure on the x-axis you will notice possibly a linear relationship between them. How to understand the relationship The graph will show that there is a strong relationship between age and blood pressure and that the relationship is positive. This is because the graph has a positive correlation. So the older is one’s age the higher is the systolic blood pressure. The line of best fit also helps to understand the strength of the correlation. If there is little space between the points then the correlation is strong. The correlation coefficient or R is a numerical value that ranges between -1 to 1. This indicates the strength of the linear relationship between two variables. To describe a linear regression the coefficient is called Pearson’s correlation coefficient. When the correlation coefficient is close to 1 then it highlights a strong positive correlation. When the correlation coefficient is close to -1 then this shows a strong negative correlation. When the correlation coefficient is equal to 0 then this shows no relationship at all. -- -- -- Love podcasts or audiobooks? Learn on the go with our new app.<\|endoftext\|>	4.46875
350	# The length of a postage stamp is 4 1/4 millimeters longer than its width. The perimeter of the stamp is 124 1/2 millimeters. What is the width of the postage stamp? What is the length of the postage stamp? The length and width of the postage stamp are $33 \frac{1}{4} m m \mathmr{and} 29 m m$ respectively. Let the width of the postage stamp be $x$ mm Then, the lengtth of the postage stamp be $\left(x + 4 \frac{1}{4}\right)$mm. Given perimeter is $P = 124 \frac{1}{2}$ We know perimeter of a rectangle is $P = 2 \left(w + l\right)$; where $w$ is the width and $l$ is the length. So $2 \left(x + x + 4 \frac{1}{4}\right) = 124 \frac{1}{2} \mathmr{and} 4 x + 8 \frac{1}{2} = 124 \frac{1}{2} \mathmr{and} 4 x = 124 \frac{1}{2} - 8 \frac{1}{2} \mathmr{and} 4 x = 116 \mathmr{and} x = 29 \therefore x + 4 \frac{1}{4} = 33 \frac{1}{4}$ The length and width of the postage stamp are $33 \frac{1}{4} m m \mathmr{and} 29 m m$ respectively.[Ans]<\|endoftext\|>	4.46875
1,146	## Number Series Questions and Answers Part-8 1. 20, 20, 17, 17, 14, 14, 11, ... choose which pair of numbers comes next? a) 8, 8 b) 11, 11 c) 11, 14 d) 11, 8 Explanation: This is a simple subtraction with repetition series. It begins with 20, which is repeated, then 3 is subtracted, resulting in 17, which is repeated, and so on. 2. 61, 57, 50, 61, 43, 36, 61, ... choose which pair of numbers comes next? a) 29, 61 b) 27, 20 c) 31, 61 d) 29, 22 Explanation: This is an alternating repetition series, in which a random number, 61, is interpolated as every third number into an otherwise simple subtraction series. Starting with the second number, 57, each number (except 61) is 7 less than the previous number. 3. 9, 12, 11, 14, 13, 16, 15, ... choose which pair of numbers comes next? a) 14, 13 b) 18, 21 c) 14, 17 d) 18, 17 Explanation: This is a simple alternating addition and subtraction series. First, 3 is added, then 1 is subtracted, then 3 is added, 1 subtracted, and so on. 4. 4, 8, 22, 12, 16, 22, 20, 24, ... choose which pair of numbers comes next? a) 28, 32 b) 28, 22 c) 22, 28 d) 32, 36 Explanation: This is an alternating repetition series, with a random number, 22, interpolated as every third number into an otherwise simple addition series. In the addition series, 4 is added to each number to arrive at the next number. 5. 40, 40, 31, 31, 22, 22, 13, ... choose which pair of numbers comes next? a) 13, 4 b) 13, 5 c) 4, 13 d) 9, 4 Explanation: This is a subtraction series with repetition. Each number repeats itself and then decreases by 9. 6. What will come at the place of question mark ? 2, 3, 6, 15, ?, 123 a) 47 b) 42 c) 45 d) 50 Explanation: First term → 2 Second term → (2 3 -3) = 3 Third term → (3 3 -3) = 6 Fourth term → (6 3 - 3) = 15 Fifth term → (15 3 - 3) = 42 Sixth term → (42 3 - 3) = 123 7. In the following number series a wrong number is given. Find out the wrong number. 1, 3, 10, 21, 64, 129, 356, 777 a) 129 b) 21 c) 10 d) 356 Explanation: The given pattern is, 2nd term → 12+1 = 3 3rd term → 33+1 = 10 4th term → 102+1 = 21 5th term → 213+1 = 64 6th term → 642+1 = 129 7th term → 1293+1= 388 So, 7th term 356 is wrong and must be replaced by 388. 8. What will come at the place of question mark ? 9, 27, 31, 155, 161, 1127, ? a) 316 b) 1135 c) 1288 d) 2254 Explanation: Given pattern is, 2nd term → 93 = 27 3rd term → 27+4 = 31 4th term → 315 = 155 5th term → 155+6 = 161. 6th term → 1617 = 1127 Missing term → 1127+8 = 1135. 9. What will come at the place of question mark ? 2, 7, 27, 107, 427, ? a) 1262 b) 1707 c) 4027 d) 4207 Explanation: The pattern is, 2nd term → 2 +(5 12) = 7 3rd term → 7 +(5 22) = 27 4th term → 27 +(5 42) = 107 5th term → 107 +(5 82) = 427 Missing term → 427 +(5 *162) = 1707 10. Sunday, Monday, Wednesday, Saturday, Wednesday, Monday, ... a) Sunday, Sunday b) Sunday, Monday c) Sunday, Wednesday d) Sunday, Saturday<\|endoftext\|>	4.53125
2,344	<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 6.8: Absolute Value Equations Difficulty Level: At Grade Created by: CK-12 Estimated10 minsto complete % Progress Practice Absolute Value Equations MEMORY METER This indicates how strong in your memory this concept is Progress Estimated10 minsto complete % Estimated10 minsto complete % MEMORY METER This indicates how strong in your memory this concept is What if you were asked to solve an absolute value equation like \|3x4\|=5\begin{align}\|3x-4\|=5\end{align}? How could you interpret the solution? After completing this Concept, you'll be able to interpret the solutions to absolute value equations like this one by graphing them on a number line. ### Guidance In the previous concept, we saw how to solve simple absolute value equations. In this concept, you will see how to solve more complicated absolute value equations. #### Example A Solve the equation \|x4\|=5\begin{align}\|x-4\|=5\end{align} and interpret the answers. Solution We consider two possibilities: the expression inside the absolute value sign is non-negative or is negative. Then we solve each equation separately. x4=5andx4=5 x=9 x=1\begin{align}& x-4 = 5 \quad \text{and} \quad x-4=-5\\ & \quad \ \ x=9 \qquad \qquad \quad \ x=-1\end{align} x=9\begin{align}x = 9\end{align} and x=1\begin{align}x = -1\end{align} are the solutions. The equation \|x4\|=5\begin{align}\|x-4\|=5\end{align} can be interpreted as “what numbers on the number line are 5 units away from the number 4?” If we draw the number line we see that there are two possibilities: 9 and -1. #### Example B Solve the equation \|x+3\|=2\begin{align}\|x+3\|=2\end{align} and interpret the answers. Solution Solve the two equations: x+3=2and x+3=2 x=1 x=5\begin{align}& x+3 = 2 \quad \text{and} \quad \ \ x+3=-2\\ & \quad \ \ x=-1 \qquad \qquad \quad \ x=-5\end{align} x=5\begin{align}x = -5\end{align} and x=1\begin{align}x = -1\end{align} are the answers. The equation \|x+3\|=2\begin{align}\|x+3\|=2\end{align} can be re-written as: \|x(3)\|=2\begin{align}\|x-(-3)\|=2\end{align}. We can interpret this as “what numbers on the number line are 2 units away from -3?” There are two possibilities: -5 and -1. Solve Real-World Problems Using Absolute Value Equations #### Example C A company packs coffee beans in airtight bags. Each bag should weigh 16 ounces, but it is hard to fill each bag to the exact weight. After being filled, each bag is weighed; if it is more than 0.25 ounces overweight or underweight, it is emptied and repacked. What are the lightest and heaviest acceptable bags? Solution The weight of each bag is allowed to be 0.25 ounces away from 16 ounces; in other words, the difference between the bag’s weight and 16 ounces is allowed to be 0.25 ounces. So if x\begin{align}x\end{align} is the weight of a bag in ounces, then the equation that describes this problem is \|x16\|=0.25\begin{align}\|x-16\|=0.25\end{align}. Now we must consider the positive and negative options and solve each equation separately: x16=0.25andx16=0.25x=16.25 x=15.75\begin{align}& x-16 = 0.25 \qquad \text{and} \quad x-16 =-0.25\\ & \qquad x=16.25 \qquad \qquad \qquad \ \ x=15.75\end{align} The lightest acceptable bag weighs 15.75 ounces and the heaviest weighs 16.25 ounces. We see that 16.2516=0.25 ounces\begin{align}16.25 - 16 = 0.25 \ ounces\end{align} and 1615.75=0.25 ounces\begin{align}16 - 15.75 = 0.25 \ ounces\end{align}. The answers are 0.25 ounces bigger and smaller than 16 ounces respectively. The answer you just found describes the lightest and heaviest acceptable bags of coffee beans. But how do we describe the total possible range of acceptable weights? That’s where inequalities become useful once again. Watch this video for help with the Examples above. ### Vocabulary • The absolute value of a number is its distance from zero on a number line. • \|x\|=x\begin{align}\|x\|=x\end{align} if x\begin{align}x\end{align} is not negative, and \|x\|=x\begin{align}\|x\|=-x\end{align} if x\begin{align}x\end{align} is negative. • An equation or inequality with an absolute value in it splits into two equations, one where the expression inside the absolute value sign is positive and one where it is negative. When the expression within the absolute value is positive, then the absolute value signs do nothing and can be omitted. When the expression within the absolute value is negative, then the expression within the absolute value signs must be negated before removing the signs. • Inequalities of the type \|x\|<a\begin{align}\|x\| can be rewritten as “a<x<a\begin{align}-a < x < a\end{align}.” • Inequalities of the type \|x\|>b\begin{align}\|x\|>b\end{align} can be rewritten as “x<b\begin{align}x < -b\end{align} or x>b\begin{align}x > b\end{align}.” ### Guided Practice Solve the equation \|2x7\|=6\begin{align}\|2x-7\|=6\end{align} and interpret the answers. Solution Solve the two equations: 2x7=62x7=6 2x=13and 2x=1 x=132 x=12\begin{align}& 2x-7 = 6 \qquad \qquad \quad 2x-7=-6\\ & \quad \ \ 2x=13 \qquad \text{and} \qquad \ \ 2x=1\\ & \quad \ \ \ x=\frac{13}{2} \qquad \qquad \qquad \ \ x=\frac{1}{2}\end{align} Answer: x=132\begin{align}x=\frac{13}{2}\end{align} and x=12\begin{align}x=\frac{1}{2}\end{align}. The interpretation of this problem is clearer if the equation \|2x7\|=6\begin{align}\|2x-7\|=6\end{align} is divided by 2 on both sides to get 12\|2x7\|=3\begin{align}\frac{1}{2}\|2x-7\|=3\end{align}. Because 12\begin{align}\frac{1}{2}\end{align} is nonnegative, we can distribute it over the absolute value sign to get x72=3\begin{align}\left \| x-\frac{7}{2} \right \|=3\end{align}. The question then becomes “What numbers on the number line are 3 units away from 72\begin{align}\frac{7}{2}\end{align}?” There are two answers: 132\begin{align}\frac{13}{2}\end{align} and 12\begin{align}\frac{1}{2}\end{align}. ### Practice Solve the absolute value equations and interpret the results by graphing the solutions on the number line. 1. \|x5\|=10\begin{align}\|x-5\|=10\end{align} 2. \|x+2\|=6\begin{align}\|x+2\|=6\end{align} 3. \|5x2\|=3\begin{align}\|5x-2\|=3\end{align} 4. \|x4\|=3\begin{align}\|x-4\|=-3\end{align} 5. 2x12=10\begin{align}\left\|2x-\frac{1}{2}\right\|=10\end{align} 6. \|x+5\|=15\begin{align}\|-x+5\|=\frac{1}{5}\end{align} 7. 12x5=100\begin{align}\left\|\frac{1}{2}x-5\right\|=100\end{align} 8. \|10x5\|=15\begin{align}\|10x-5\|=15\end{align} 9. \|0.1x+3\|=0.015\begin{align}\|0.1x+3\|=0.015\end{align} 10. \|272x\|=3x+2\begin{align}\|27-2x\|=3x+2\end{align*} ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Absolute Value The absolute value of a number is the distance the number is from zero. Absolute values are never negative. linear equation A linear equation is an equation between two variables that produces a straight line when graphed. Show Hide Details Description Difficulty Level: Authors: Tags: Subjects:<\|endoftext\|>	4.9375
1,540	Manipulation of negative square root of a negative term/# • Jun 30th 2012, 05:59 AM daigo Manipulation of negative square root of a negative term/# Suppose I have to solve for y: $\displaystyle x\leq 1$ $\displaystyle (x - 1)^{2} = y$ So I know that (x - 1) will always be 0 or a negative, therefore I must take the negative square root of (x - 1): $\displaystyle -\sqrt{(x - 1)^{2}} = -\sqrt{y}$ Am I to understand that this is the same as: $\displaystyle -1 \cdot \sqrt{(x - 1)^{2}} = -1 \cdot \sqrt{y}$ $\displaystyle =$ $\displaystyle -1 \cdot (x - 1) = -1 \cdot \sqrt{y}$ $\displaystyle =$ $\displaystyle -x + 1 = -1 \cdot \sqrt{y}$ Or how does this work? • Jun 30th 2012, 06:19 AM Plato Re: Manipulation of negative square root of a negative term/# Quote: Originally Posted by daigo Suppose I have to solve for y: $\displaystyle x\leq 1$ $\displaystyle (x - 1)^{2} = y$ So I know that (x - 1) will always be 0 or a negative, therefore I must take the negative square root of (x - 1): $\displaystyle -\sqrt{(x - 1)^{2}} = -\sqrt{y}$ Am I to understand that this is the same as: $\displaystyle -1 \cdot \sqrt{(x - 1)^{2}} = -1 \cdot \sqrt{y}$ $\displaystyle =$ $\displaystyle -1 \cdot (x - 1) = -1 \cdot \sqrt{y}$ $\displaystyle =$ $\displaystyle -x + 1 = -1 \cdot \sqrt{y}$ It is a bit awkward, and you missed a sign. You you could have done this. $\displaystyle (x - 1)^{2} = y \Rightarrow~\|x-1\|=\sqrt{y}$ But $\displaystyle x\le 1 \Rightarrow~1-x=\sqrt{y}$ • Jun 30th 2012, 06:21 AM Soroban Re: Manipulation of negative square root of a negative term/# Hello, daigo! Quote: $\displaystyle \text{Solve for }y\!:\;\;\begin{Bmatrix}x\:\leq\: 1 & [1] \\ (x - 1)^2 \:=\: y & [2]\end{Bmatrix}$ From [1]: .$\displaystyle x\,\le\,1 \quad\Rightarrow\quad x - 1\,\le\,0 \quad\Rightarrow\quad (x-1)^2\,\ge\,0$ Substitute into [2]: .$\displaystyle y \:=\:(x-1)^2 \:\ge\:0$ . . Therefore: .$\displaystyle y \:\ge\:0$ • Jun 30th 2012, 06:32 AM daigo Re: Manipulation of negative square root of a negative term/# I don't really want to focus on solving the problem, I just need some insight on the negative square root operation. I'm trying to understand the way it was done here: $\displaystyle y = (x - 1)^{2}$ $\displaystyle -\sqrt{y} = -\sqrt{(x - 1)^{2}}$ $\displaystyle -\sqrt{y} = x - 1$ $\displaystyle -\sqrt{y} + 1 = x$ Is this correct or incorrect? • Jun 30th 2012, 10:33 AM Wilmer Re: Manipulation of negative square root of a negative term/# Incorrect. • Jun 30th 2012, 04:06 PM thevinh Re: Manipulation of negative square root of a negative term/# Quote: Originally Posted by daigo Suppose I have to solve for y: $\displaystyle x\leq 1$ $\displaystyle (x - 1)^{2} = y$ So I know that (x - 1) will always be 0 or a negative, therefore I must take the negative square root of (x - 1): $\displaystyle -\sqrt{(x - 1)^{2}} = -\sqrt{y}$ Am I to understand that this is the same as: $\displaystyle -1 \cdot \sqrt{(x - 1)^{2}} = -1 \cdot \sqrt{y}$ $\displaystyle =$ $\displaystyle -1 \cdot (x - 1) = -1 \cdot \sqrt{y}$ $\displaystyle =$ $\displaystyle -x + 1 = -1 \cdot \sqrt{y}$ Or how does this work? In this problem, it does not matter what value of x is because the quantity $\displaystyle (x-1)^2$ will always be positive. So $\displaystyle y\geq0$. • Jun 30th 2012, 05:10 PM Plato Re: Manipulation of negative square root of a negative teronfusiom/# Quote: Originally Posted by thevinh In this problem, it does not matter what value of x is because the quantity $\displaystyle (x-1)^2$ will always be positive. So $\displaystyle y\geq0$. That may be a problem. But it is not the problem which is the OP refuses to see that: $\displaystyle \sqrt{(x-1)^2}=\|x-1\|$. That is the source of all this confusion. • Jun 30th 2012, 11:57 PM biffboy Re: Manipulation of negative square root of a negative term/# Is there anything wrong with this. (x-1)= + or -root y So rooty=(x-1) or (1-x) For example if x=-2 y=9 and root y =+3 or -3 • Jul 1st 2012, 01:35 AM BobP Re: Manipulation of negative square root of a negative term/# Can I join in ? What's wrong with this ? $\displaystyle x\leq1,$ so either $\displaystyle x-1<0$ or $\displaystyle x-1=0.$ If $\displaystyle x-1=0$ then $\displaystyle y=0,$ so concentrate on the first case. If $\displaystyle (x-1)^{2}=y,$ then $\displaystyle x-1=\pm\sqrt{y},$ ($\displaystyle y$ is positive), but we know that $\displaystyle x-1<0$ so $\displaystyle x-1=-\sqrt{y}.$<\|endoftext\|>	4.40625
2,616	$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 4.1E: $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ ## Parametric Equations For the following exercises, sketch the curves below by eliminating the parameter t. Give the orientation of the curve. 1) $$\displaystyle x=t^2+2t, y=t+1$$ Solution: orientation: bottom to top 2) $$\displaystyle x=cos(t),y=sin(t),(0,2π]$$ 3) $$\displaystyle x=2t+4,y=t−1$$ Solution: orientation: left to right 4) $$\displaystyle x=3−t,y=2t−3,1.5≤t≤3$$ For the following exercises, eliminate the parameter and sketch the graphs. 5) $$\displaystyle x=2t^2,y=t^4+1$$ Solution: $$\displaystyle y=\frac{x^2}{4}+1$$ For the following exercises, use technology (CAS or calculator) to sketch the parametric equations. 6) [T] $$\displaystyle x=t^2+t,y=t^2−1$$ 7) [T] $$\displaystyle x=e^{−t},y=e^{2t}−1$$ Solution: 8) [T] $$\displaystyle x=3cost,y=4sint$$ 9) [T] $$\displaystyle x=sect,y=cost$$ For the following exercises, sketch the parametric equations by eliminating the parameter. Indicate any asymptotes of the graph. 10) $$\displaystyle x=e^t,y=e^{2t}+1$$ 11) $$\displaystyle x=6sin(2θ),y=4cos(2θ)$$ Solution: 12) $$\displaystyle x=cosθ,y=2sin(2θ)$$ 13) $$\displaystyle x=3−2cosθ,y=−5+3sinθ$$ Solution: 14) $$\displaystyle x=4+2cosθ,y=−1+sinθ$$ 15) $$\displaystyle x=sect,y=tant$$ Solution: Asymptotes are $$\displaystyle y=x$$ and $$\displaystyle y=−x$$ 16) $$\displaystyle x=ln(2t),y=t^2$$ 17) $$\displaystyle x=e^t,y=e^{2t}$$ Solution: 18) $$\displaystyle x=e^{−2t},y=e^{3t}$$ 19) $$\displaystyle x=t^3,y=3lnt$$ Solution: 20) $$\displaystyle x=4secθ,y=3tanθ$$ For the following exercises, convert the parametric equations of a curve into rectangular form. No sketch is necessary. State the domain of the rectangular form. 21) $$\displaystyle x=t^2−1,y=\frac{t}{2}$$ Solution: $$\displaystyle x=4y^2−1;$$ domain: $$\displaystyle x∈[1,∞)$$. 22) $$\displaystyle x=\frac{1}{\sqrt{t+1}},y=\frac{t}{1+t},t>−1$$ 23) $$\displaystyle x=4cosθ,y=3sinθ,t∈(0,2π]$$ Solution: $$\displaystyle \frac{x^2}{16}+\frac{y^2}{9}=1;$$ domain $$\displaystyle x∈[−4,4].$$ 24) $$\displaystyle x=cosht,y=sinht$$ 25) $$\displaystyle x=2t−3,y=6t−7$$ Solution: $$\displaystyle y=3x+2;$$ domain: all real numbers. 26) $$\displaystyle x=t^2,y=t^3$$ 27) $$\displaystyle x=1+cost,y=3−sint$$ Solution: $$\displaystyle (x−1)^2+(y−3)^2=1$$; domain: $$\displaystyle x∈[0,2]$$. 28) $$\displaystyle x=\sqrt{t},y=2t+4$$ 29) $$\displaystyle x=sect,y=tant,π≤t<\frac{3π}{2}$$ Solution: $$\displaystyle y=\sqrt{x^2−1}$$; domain: $$\displaystyle x∈[−1,1]$$. 30) $$\displaystyle x=2cosht,y=4sinht$$ 31) $$\displaystyle x=cos(2t),y=sint$$ Solution: $$\displaystyle y^2=\frac{1−x}{2};$$ domain: $$\displaystyle x∈[2,∞)∪(−∞,−2].$$ 32) $$\displaystyle x=4t+3,y=16t^2−9$$ 33) $$\displaystyle x=t^2,y=2lnt,t≥1$$ Solution: $$\displaystyle y=lnx;$$ domain: $$\displaystyle x∈(0,∞).$$ 34) $$\displaystyle x=t3^,y=3lnt,t≥1$$ 35) $$\displaystyle x=t^n,y=nlnt,t≥1,$$ where n is a natural number Solution: $$\displaystyle y=lnx;$$ domain: $$\displaystyle x∈(0,∞).$$ 36) $$\displaystyle x=ln(5t)$$ $$\displaystyle y=ln(t^2)$$ where $$\displaystyle 1≤t≤e$$ 37) $$\displaystyle x=2sin(8t)$$ $$\displaystyle y=2cos(8t)$$ Solution: $$\displaystyle x^2+y^2=4;$$ domain: $$\displaystyle x∈[−2,2].$$ 38) $$\displaystyle x=tant$$ $$\displaystyle y=sec^2t−1$$ For the following exercises, the pairs of parametric equations represent lines, parabolas, circles, ellipses, or hyperbolas. Name the type of basic curve that each pair of equations represents. 39) $$\displaystyle x=3t+4$$ $$\displaystyle y=5t−2$$ Solutin: line 40) $$\displaystyle x−4=5t$$ $$\displaystyle y+2=t$$ 41) $$\displaystyle x=2t+1$$ $$\displaystyle y=t^2−3$$ Solution: parabola 42) $$\displaystyle x=3cost$$ $$\displaystyle y=3sint$$ 43) $$\displaystyle x=2cos(3t)$$ $$\displaystyle y=2sin(3t)$$ Solution: circle 44) $$\displaystyle x=cosht$$ $$\displaystyle y=sinht$$ 45) $$\displaystyle x=3cost$$ $$\displaystyle y=4sint$$ Solutin: ellipse 46) $$\displaystyle x=2cos(3t)$$ $$\displaystyle y=5sin(3t)$$ 47) $$\displaystyle x=3cosh(4t)$$ $$\displaystyle y=4sinh(4t)$$ Solution: hyperbola 48) $$\displaystyle x=2cosht$$ $$\displaystyle y=2sinht$$ 49) Show that $$\displaystyle x=h+rcosθ$$ $$\displaystyle y=k+rsinθ$$ represents the equation of a circle. 50) Use the equations in the preceding problem to find a set of parametric equations for a circle whose radius is 5 and whose center is $$\displaystyle (−2,3)$$. For the following exercises, use a graphing utility to graph the curve represented by the parametric equations and identify the curve from its equation. 51) [T] $$\displaystyle x=θ+sinθ$$ $$\displaystyle y=1−cosθ$$ Solution: The equations represent a cycloid. 52) [T] $$\displaystyle x=2t−2sint$$ $$\displaystyle y=2−2cost$$ 53) [T] $$\displaystyle x=t−0.5sint$$ $$\displaystyle y=1−1.5cost$$ Solution: 54) An airplane traveling horizontally at 100 m/s over flat ground at an elevation of 4000 meters must drop an emergency package on a target on the ground. The trajectory of the package is given by $$\displaystyle x=100t,y=−4.9t^2+4000,t≥0$$ where the origin is the point on the ground directly beneath the plane at the moment of release. How many horizontal meters before the target should the package be released in order to hit the target? 55) The trajectory of a bullet is given by $$\displaystyle x=v_0(cosα)ty=v_0(sinα)t−\frac{1}{2}gt^2$$ where $$\displaystyle v_0=500m/s, g=9.8=9.8m/s^2$$, and $$\displaystyle α=30degrees.$$ When will the bullet hit the ground? How far from the gun will the bullet hit the ground? Solution: 22,092 meters at approximately 51 seconds. 56) [T] Use technology to sketch the curve represented by $$\displaystyle x=sin(4t),y=sin(3t),0≤t≤2π$$. 57) [T] Use technology to sketch $$\displaystyle x=2tan(t),y=3sec(t),−π<t<π.$$ Solution: 58) Sketch the curve known as an epitrochoid, which gives the path of a point on a circle of radius b as it rolls on the outside of a circle of radius a. The equations are $$\displaystyle x=(a+b)cost−c⋅cos[\frac{(a+b)t}{b}]$$ $$\displaystyle y=(a+b)sint−c⋅sin[\frac{(a+b)t}{b}]$$. Let $$\displaystyle a=1,b=2,c=1. 59) [T] Use technology to sketch the spiral curve given by \(\displaystyle x=tcos(t),y=tsin(t)$$ from $$\displaystyle −2π≤t≤2π.$$ 60) [T] Use technology to graph the curve given by the parametric equations $$\displaystyle x=2cot(t),y=1−cos(2t),−π/2≤t≤π/2.$$ This curve is known as the witch of Agnesi. 61) [T] Sketch the curve given by parametric equations $$\displaystyle x=cosh(t)$$ $$\displaystyle y=sinh(t),$$ where $$\displaystyle −2≤t≤2.$$ Solution:<\|endoftext\|>	4.75
1,087	A new University of Cambridge research project is set to shed light on the history of writing in the ancient world, and explore the longlasting relationship between society and writing that persists today. A new research project at the University of Cambridge is set to shed light on the history of writing, revealing connections to our modern alphabet that cross cultures and go back thousands of years. The project, called Contexts of and Relations between Early Writing Systems (CREWS for short), is to focus on exploring how writing developed during the 2nd and 1st millennia BCE in the ancient Mediterranean and Near East, and will investigate how different writing systems and the cultures that used them were related to each other. The project is led by Dr Philippa Steele of the University’s Faculty of Classics. Described as an “innovative and interdisciplinary approach to the history of writing” the CREWS project aims to enrich our understanding of linguistic, cultural and social aspects of the use, borrowing and development of writing in the ancient world – which can uncover some often surprising links to our modern-day written culture. For instance, today the notion of “alphabetical order” is used to arrange everything from dictionaries to telephone books, but why is the alphabet organised the way it is? Alphabetical order as we would recognise it first appeared over three thousand years ago in Ugaritic, written in a cuneiform script made of wedge-shaped signs impressed on clay tablets. The Ugaritic alphabet was in use in the ancient city of Ugarit, uncovered at Ras Shamra in modern Syria. Some of the surviving tablets discovered by archaeologists are known as “abecedaria”, where the letters of the alphabet are written in order, possibly for teaching or as a training exercise for new scribes. The destruction of Ugarit in around 1200 BCE was not the end for alphabetical order. The Phoenicians, living in what is now modern Syria and Lebanon, used the same order for their own alphabet. While their language was related to Ugaritic, their writing system was not. Instead of cuneiform wedge-shapes, the Phoenicians used linear letters, which were much more similar to those we use in English today. The Phoenician alphabet began with the letters Alep, Bet, Gimel, Dalet, which are strikingly similar to our own A, B, C and D. Dr Steele said: “The links from the ancient past to our alphabet today are no coincidence. The Greeks borrowed the Phoenician writing system and they still kept the same order of signs: Alpha, Beta, Gamma, Delta. They transported the alphabet to Italy, where it was passed on to the Etruscans, and also to the Romans, who still kept the same order: A, B, C, D, which is why our modern alphabet is the way it is today.” That such an apparently simple idea remained so stable and powerful over thousands of years of cultural change and movement is an historic mystery. “The answer cannot be purely linguistic”, Dr Steele said. “There must have been considerable social importance attached to the idea of the alphabet having a particular order. It matters who was doing the writing and what they were using writing for.” The origin of the alphabet is just one of the areas that the CREWS project will explore, along with the social and political context of writing, and drivers of language change, literacy and communication. Because of the high level of interconnectedness in the ancient Mediterranean and Near East, ideas could be spread widely as people moved, traded and interacted with different cultures. “Globalisation is not a purely modern phenomenon”, Dr Steele commented. “We might have better technology to pursue it now, but essentially we are engaging in the same activities as our ancestors.” The CREWS project is the result of a long-term innovative programme of combined and comparative research at the University of Cambridge. It will run for five years and will involve a four-person team working on a variety of ancient cultures and writing systems. The CREWS project has been made possible thanks to the European Research Council, who describe their mission as being “to encourage the highest-quality research in Europe.” Dr Steele, the Principal Investigator on the project and a Senior Research Fellow at Magdalene College, Cambridge, has worked on ancient languages and writing systems for over ten years and previously specialised in the languages of ancient Cyprus. She said: “Cyprus lies right in the middle of an area where ancient people were moving about by land and sea and swapping technologies and ideas. That was one of the inspirations of the CREWS project. By studying how and what ancient people were writing, we will be able to gain more insight into their interactions with each other in ways that have never been fully understood before.” The Contexts of and Relations Between Early Writing Systems (CREWS) project will be based at the University of Cambridge Faculty of Classics, a world-leading centre for the study of the ancient world with a track record for innovative and interdisciplinary research. Running from April 2016, it will continue until 2021. Follow the project blog online at https://crewsproject.wordpress.com/<\|endoftext\|>	3.671875
280	A United States architectural movement better known as the Spanish Colonial Revival architecture commenced in the early 20th century. The movement comprised planning certain cities that were the previous Spanish colonies, which then became American cities, implementing the Spanish architectural style. A principal part of this architectural style can be noticed in California. Santa Barbara used this style as its illustrative line for re-designing the city after an earthquake that took place in1925. Architect George Washington Smith relocated to Montecito and popularized this movement introduced this style. The history of El Pueblo Viejo aesthetic control remains aligned to the Roman and Parisian laws. It tries to keep history intact through the Hispanic architecture. But you may wonder what the Hispanic Architecture is all about. This style is mostly influenced by the architecture of the “white-washed cities” of Andalusia in Southern Spain. In Santa Barbara, local building methods are a product of the natural environment and the materials available in the locale. Kenny Slaught further says that Hispanic architectural types in this area are portrayed by the “minimalism, rural economy, excellence in craftsmanship and direct expression of material”. Designs observed in Santa Barbara demonstrate local handmade quality connected to the sunlight. Besides, colors are also akin to the natural environment, yellow, red, orange and white that remains Santa Barbara’s weather. Real Estate Expert in California<\|endoftext\|>	3.734375
876	## The SBC GMAT Files GMAT Primer: Plane And Coordinate Geometry Basics Haven’t done Geometry since high school? No worries! Here are the basic formulas and definitions you need to know to get started. Create a “Geometry Formulas” cheat-sheet to write down these formulas and refer back to it as needed when you study! An angle is formed by two lines or line segments which intersect at one point. The point of intersection is called the vertex. Angles are measured in either degrees or radians. An acute angle is an angle whose measurement in degrees is between 0 and 90. A right angle is an angle whose measurement in degrees is exactly 90. An obtuse angle is an angle whose degree measure is between 90 and 180. A straight angle is an angle whose degree measure is exactly 180 degrees. All of the angles on one side of a straight line sum to 180 degrees. a + b + c = 180 degrees All of the angles around one point must sum to 360 degrees. a + b + c + d + e = 360 degrees Perpendicular lines are formed when the angle between two lines is 90 degrees. The shortest distance from a point to a line is a line with a length such that the two lines form a 90 degree angle. Two angles are supplementary if they share one line; i.e., if they sum of their angles is 180 degrees. Two angles are complementary if together they make a right angle; i.e., if the sum of their angles is 90 degrees. To bisect an angle means to cut it in half. The two smaller angles will then have the same measurement. If two parallel lines intersect with a third line, the third line is called a transversal. When this happens, all acute angles are equal and all obtuse angles are equal. Each acute angle is supplemental to each obtuse angle. Vertical angles are a pair of opposite angles formed by intersecting lines, and they are equal. There are two main equations for straight lines. One form looks like: For an equation that looks like this the slope is and the y intercept is .For example, in the equation 2x + 3y + 6 = 0, the slope is -2/3 and the y-intercept is -2. The second equation is called slope-intercept form and looks like: . Here m is the slope and b is the y-intercept. Distance Formula Use this to find the distance between two points. Midpoint Formula Use this to find the midpoint between two points (notice how you are essentially finding the average of the x-coordinates and the average of the y-coordinates). Slope = Rise / Run = Change in y / Change in x The standard equation for a parabola is y = ax2 + bx + c. In this equation c represents the y-intercept. A standard equation in which a variable is squared will never make a straight line. The x-intercepts are also called the “roots” or “solutions” of a parabola. On the GRE, parabolas will often be referred to as “functions” interchangeably. The x-intercepts can be found using the quadratic formula: To find the number of x-intercepts a given parabola has, calculate what is called the discriminant: , or the information underneath the radical in the quadatic formula. If the discriminant is positive, the parabola has two intercepts with x-axis; if it is negative there are no intercepts with the x-axis, and if the discriminant is equal to zero there is one intercept with the x-axis. The vertex represents the maximum (or minimum) value of the function. Think of it as the starting point of the function. The vertex of the parabola is located at point for the standard equation. If you are given the standard equation, you can find the vertex and the x-intercepts. The standard equation of a circle is (x – h)2 + (y – k)2 = r2 where (h, k) is the center point of the circle and r is the radius. More Testing Advice from our blog<\|endoftext\|>	4.65625
666	We know that rising carbon dioxide levels in the atmosphere are raising global temperatures and creating unstable and extreme weather patterns that will continue to threaten communities across the globe. And perhaps few people are more familiar with this fact than Kristie Ebi, the University of Washington professor who was the lead author on the Intergovernmental Panel on Climate Change’s (IPCC) blockbuster report last fall on the need to limit global warming to below 1.5 degrees Celsius. But on the stage at TED in Vancouver, Ebi made it clear that climate shifts are far from the only consequence of rising CO2 levels. “Our emission of greenhouse gases from burning fossil fuels is reducing the nutritional quality of our food,” she says. Ebi directly refuted an idea that’s been floating around for a while about the effect of CO2 on food production and global hunger. Technically, plants need CO2 to survive: They bring it in, break it down, and rely on carbon to grow. Some researchers have claimed that more CO2 means that more plants will be able to grow, and higher CO2 levels will then help solve food insecurity. That, according to Ebi, is a hugely, dangerously wrong. Not only will climate change and global warming make agricultural productivity and much more unstable, but when plants take in an excess of CO2, their chemical makeup changes in a way that that’s harmful to the humans and animals that depend on them for nutrition: higher concentrations of CO2, increases the synthesis of carbohydrates like sugars and starches, and decrease the concentrations of proteins and nutrients like zinc, iron, and B-vitamins. “This is very important for how we think about food security going forward,” Ebi says. A lack of iron can lead to outcomes like anemia and stunted development; low levels of zinc contributes to loss of appetite and additional developmental difficulties. B vitamins are crucial for converting our food into energy, and the way our bodies function overall. “It’s not just us,” Ebi says. Animals like cows that rely on grains and plants for their diet produce meat and milk that contain fewer nutrients and vitamins, and people who eat meat take in fewer of those crucial resources. To understand just how profound these effects will be, Ebi’s team studied fields of rice that were saturated with CO2 levels equivalent to those we might experience in 2050. They found that levels of protein in the rice declined by around 10%, iron by around 8%, zinc by 5%, and B vitamins by around 18%. “These don’t sound like major changes, but when you think about the poor in countries who eat mainly starches, this will put them from the edge to over the edge and into nutrient deficiency,” Ebi says. Ebi also calls for more investment from governments and food companies in interventions that could support the nutritional quality of crops. Biofortification, the process of adding vitamins and minerals to seeds, has proven effective in parts of Africa where nutritional deficiency is already a significant problem, and plants can be bred to support greater nutrient diversity. And overall, these findings add more fuel to the argument that we must work to limit CO2 emissions as quickly as possible.<\|endoftext\|>	3.84375
10,813	# University of Lille I PC first year list of exercises n 7. Review Size: px Start display at page: Download "University of Lille I PC first year list of exercises n 7. Review" Transcription 1 University of Lille I PC first year list of exercises n 7 Review Exercise Solve the following systems in 4 different ways (by substitution, by the Gauss method, by inverting the matrix of coefficients of the system, by Cramer s formulas) : { x + y = 3x + 7y = Choose the method that seems the quickest to you and solve, according to the values of a, the following systems : { ax + y = (a + )x + ay = Solution of Exercise : { (a + )x + (a )y = (a )x + (a + )y = (a) By substitution { { { x + y = x + y = 4 3x + 7y = x = 7 3 y 3 y + y = x = 7 3 y { y = 3 x = 7 (b) By the Gauss method { { x + y = x + y = 3x + 7y = y = 3 L L 3L { x = y y = 3 { (c) The inverse of the matrix of coefficients of the system is ( 3 7 ) = ( 7 3 y = 3 x = 7 ) Hence the solution of the system is ( ) x y = ( 7 3 ) ( ) = ( 7 3 ) (d) By Cramer s formulas x = = 7 y = = 3 2 The determinant of the first system is a (a + ) a = a (a) If a / {, }, one can use Cramer s formulas to obtain : a x = { ax + y = (a + )x + ay = y = a (a + ) a a (a + ) a (a + ) a = 4a a = a + a a (b) If a =, the system becomes { { x + y = x + y = x + y = = L L L which is impossible (c) If a =, the system becomes { { x + y = x + y = x y = = 5 L L + L which is also impossible The determinant of the second system is (a + ) (a ) (a ) (a + ) = 4a (a) If a, one can use Cramer s formulas to obtain : { (a + )x + (a )y = (a )x + (a + )y = x = y = (a ) (a + ) 4a (a + ) (a ) (b) If a =, the system becomes { { x y = x y = x + y = = L L + L which is impossible Exercise Solve the following system of 5 equations with 6 unknowns : x + y + z u + 3v w = 3x + y + z 3u + 5v 3w = 4 x + y + z u + 4v 4w = 6 x + y + z u + v w = 3 3x 3u + 3v + 3w = 6 4a = a = a 3 Solution of Exercise : By the Gauss method x + y + z u + 3v w = 3x + y + z 3u + 5v 3w = 4 x + y + z u + 4v 4w = 6 x + y + z u + v w = 3 3x 3u + 3v + 3w = 6 x + y + z u + v w = 3 L L 4 3x + y + z 3u + 5v 3w = 4 x + y + z u + 4v 4w = 6 x + y + z u + 3v w = 3x 3u + 3v + 3w = 6 x + y + z u + v w = 3 y z v + 3w = 5 L L 3L = L 3 L 3 L y z v + 3w = 5 L 4 L 4 L 3y 3z 3v + 9w = 5 L 5 L 5 3L { x + y + z u + v w = 3 y z v + 3w = 5 It follows that the set of solutions is a 4-space in R 6 Let us parametrize the set of solutions by a = z R, b = u R, c = v R, d = w R One obtains x y z u v w = x = y a + b c + d + 3 = b c d y = a c + 3d + 5 z = a u = b v = c w = d 5 + a + b Exercise 3 For each pair (A i, b i ), i 5 of matrices below + c give the nature of the set of solutions of the system A i X = b i ; give a parametric representation of the set of solutions of A i X = b i ; 3 give a basis of the range and a basis of the kernel of A i + d 3 3 4 a) A = c) A 3 = e) A 5 = b = Solution of Exercise 3 : b 3 = b 5 = ; b) A = ; d) A 4 = ; 3 3 b = b 4 = a) Since deta =, the matrix A is invertible hence defines an isomorphism of R 4 The system A X = b has therefore a unique solution given by X = A b = (,,, ) T by a standard computation The range of A is R 4, hence the canonical basis of R 4 is a basis of Im A The kernel of A is { }, hence a basis of kera is b) The rank of A is 4, hence the dimension of the kernel of A is Therefore the set of solutions of A X = b is an affine line in R 5 parallel to kera Denote by (x, y, z, t, u) the coordinates in R 5 Let us parametrize the set of solutions by a = u R The system is equivalent to x + y + t = 3a y + z + t = a z + t = 3a t = a x = 3a + a = a y = a + + a + ( a) = z = 3a + a = a t = a x = 3a y t y = a z t z = 3a t t = a x y z t u = + a ; ;, a R Since A is surjective, the canonical basis of R 4 is a basis of ImA The previous resolution implies that a basis of kera is given by the single vector c) Since the last equation of the system is impossible, the system A 3 X = b 3 admits no solution The rank of A 3 is 4, therefore by the Rank theorem, the dimension of kera 3 is A basis of ImA 3 is given by the 4 columns of A 3 A basis of kera 3 is given by the empty set d) The last equation of A 4 X = b 4 is impossible, hence this system admits no solution The rank of A 4 is 4, hence by the Rank theorem, the dimension of the kernel of A 4 is A basis of ImA 4 is given by the first 4 columns of A 4 A basis of kera 4 is a nontrivial vector X R 5 solution of A 4 X = One finds that generates kera 4 4 5 e) For the basis of ImA 5 and kera 5 see d) The vector b 5 belongs to ImA 5 since the last equation (compatibility condition) is satisfied The kernel of A 5 being a line, the set of solutions of A 5 X = b 5 is an affine line in R 5 parallel to kera 5 Since the vector is a particular solution of the system, one obtains that the set of solutions is parametrized by x y z t = + a, a R u Exercise 4 Compute a basis of the image and a basis of the kernel of the linear application What is the rank of f? f : R 3 R 5 (x, y, z) (x + y, x + y + z, x + y + z, x + y + z, y + z) Solution of Exercise 4 : The matrix of the linear application f is Let us compute a basis of Imf and a basis of kerf One has : Consequently the kernel of f is trivial, and a basis of Imf is given by v = v 3 = The rank of f is the dimension of Imf, that is, 3 Exercise 5 Let A be the matrix 3, v = and 5 6 Consider the matrices B = the matrix A be invertible? and C = Show that AB = AC Can Determine all matrices F of size (3, 3) such that AF = (where denotes the matrix all of whose entries are zero) Solution of Exercise 5 : One has AB = AC = Suppose that the matrix A is invertible Multiply both members of the equation AB = AC on the left by A to get B = C But the matrices B and C are not equal This is a contradiction Hence the matrix A is not invertible Let F be any real matrix (3, 3) F = a b c d e f g h i The equation AF = gives rise to the following system a = b = c = d + g = e + h = f + i = 3a + d + g = 3b + e + h = 3c + f + i = Consequently the set of matrices F such that AF = is the set of matrices of the form F = d e f, d R, e R, f R d e f Exercise 6 For which values of a is the matrix invertible? Compute in this case its inverse A = 4 3 a Solution of Exercise 6 : One has det A = 4 3 a = 4 3 a 3 a + 4 = a (a 3) + = a 7 Hence A is invertible if and only if a 7 In this case, the standard algorithm yields A = a 3 a 4 a a 3 a 7 6 7 Exercise 7 Let a and b be two real numbers, and A be the matrix a b A = Show that rk(a) (where rk denotes the rank) For which values of a and b is the rank of A equal to? Solution of Exercise 7 : Recall that the rank of A is the greatest number of columns of A that are linearly independent Since the second and third columns C, C 3 of A are not proportional, they are linearly independent Therefore the rank of A is at least For the rank of A to be exactly, one has to impose that the first and last columns of A are each a linear combination of C and C 3 (which are fixed) The only linear combination of C and C 3 that has the form (a, 3, 5) T is 3C 3 + C = (, 3, 5) T, hence a = The only linear combination of C and C 3 that has the form (b, 4, ) T is 4C 3 C = (3, 4, ) T, hence b = 3 Consequently the rank of A is if and only if a = and b = 3 Exercise 8 Compute the inverse of the following matrix Solution of Exercise 8 : One obtains A = A = Exercise 9 Let us denote by {e, e,, e n } the canonical basis of R n To a permutation σ S n, one associates the following endomorphism u σ of R n : u σ : R n R n x x n Let τ = (ij) be a transposition Write the matrix of u τ in the canonical basis Show that det(u τ ) = Show that σ, σ S n, u σ u σ = u σ σ Caution! There was a typo in the French original x σ() x σ(n) 3 Show that σ S n, det u σ = ε(σ) where ε denotes the signature Solution of Exercise 9 : 7 8 Let τ be the transposition which exchanges i and j The matrix of u τ in the canonical basis of R n is i j By exchanging the columns i and j of the matrix of u τ one obtains the identity matrix Therefore det u τ = det I =, where I denotes the identity matrix For any σ, σ S n, one has u σ u σ x x n = u σ x σ () x σ (n) = Since the previous equality is satisfied for every x σ (σ()) x σ (σ(n)) x x n = x σ σ() x σ σ(n) = u σ σ x x n in R n, it implies that u σ u σ = u σ σ An alternative proof is to check that u σ sends each e i to e σ (i) (the basis vector whose only nonzero coordinate is the σ (i)-th) : hence, u σ u σ (e i ) = u σ (e σ (i) ) = e σ (σ (i)) = e (σ σ) (i) = u σ σ(i) 3 By, the map which associates u σ to a permutation σ is a group homomorphism from S n into the group of invertible matrices of size (n, n), because u σ u σ = u σ σ = u (σ σ ) Consequently, the map which assigns to a permutation σ the number det u σ is a group homomorphism from S n into {±} Since the transpositions generate the group of permutations S n, two group homomorphisms from S n to {±} which coincide on the set of transpositions coincide on S n By, the group homomorphism from S n into {±} which maps σ onto det u σ coincides with the signature on the set of transpositions, because a transposition is its own inverse Hence σ S n, det u σ = ε(σ) Exercise Compute the eigenvalues and eigenvectors of the following matrix A = 3 4 Compute A n for all n N Solution of Exercise : 8 9 One has det (A λi) = λ λ 3 4 λ = λ λ 3 4 λ 3 4 λ + λ = λ 3 + 3λ λ = λ (λ ) (λ ) Therefore the eigenvalues of A are λ =, λ = and λ 3 = A nontrivial vector in the kernel of A is given by v = Let us find a vector generating the eigenspace associated to λ = One has A λ I I = 3 3 It follows that the vector v = has Consequently the vector v 3 = C C + C C 3 C 3 C C 3 C 3 + C is a basis of the eigenspace associated to λ = Now one A I = 3 3 generates the eigenspace associated to λ 3 = Denote by f the linear application whose matrix in the canonical basis of R 3 is A The vectors v, v and v 3 form a basis of R 3 In this new basis, the matrix of f is D = The relation between A and D is D = P AP where P = P = The inverse of P is Therefore, for n >, we have A n = (P DP )(P DP ) (P DP ) : cancelling all occurrences of P P = I one gets n n n A n = P D n P = n = n + n n n+ + n+ 9 ### Similarity and Diagonalization. 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Introduction Throughout this discussion, n 2. Any cycle in S n is a product of transpositions: the identity (1) is (12)(12), and a k-cycle with k 2 can be written More information ### Notes on Symmetric Matrices CPSC 536N: Randomized Algorithms 2011-12 Term 2 Notes on Symmetric Matrices Prof. Nick Harvey University of British Columbia 1 Symmetric Matrices We review some basic results concerning symmetric matrices. More information ### Solving Linear Systems, Continued and The Inverse of a Matrix , Continued and The of a Matrix Calculus III Summer 2013, Session II Monday, July 15, 2013 Agenda 1. The rank of a matrix 2. 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Question : How do you solve a matrix equation using the matrix inverse? In the previous question, we wrote systems of equations as a matrix equation AX B. In this format, the matrix A contains the coefficients More information ### 1.5 SOLUTION SETS OF LINEAR SYSTEMS 1-2 CHAPTER 1 Linear Equations in Linear Algebra 1.5 SOLUTION SETS OF LINEAR SYSTEMS Many of the concepts and computations in linear algebra involve sets of vectors which are visualized geometrically as More information ### Lecture 1: Systems of Linear Equations MTH Elementary Matrix Algebra Professor Chao Huang Department of Mathematics and Statistics Wright State University Lecture 1 Systems of Linear Equations ² Systems of two linear equations with two variables More information ### ISOMETRIES OF R n KEITH CONRAD ISOMETRIES OF R n KEITH CONRAD 1. Introduction An isometry of R n is a function h: R n R n that preserves the distance between vectors: h(v) h(w) = v w for all v and w in R n, where (x 1,..., x n ) = x More information ### 1.2 Solving a System of Linear Equations 1.. SOLVING A SYSTEM OF LINEAR EQUATIONS 1. Solving a System of Linear Equations 1..1 Simple Systems - Basic De nitions As noticed above, the general form of a linear system of m equations in n variables More information ### Linear Algebra I. Ronald van Luijk, 2012 Linear Algebra I Ronald van Luijk, 2012 With many parts from Linear Algebra I by Michael Stoll, 2007 Contents 1. Vector spaces 3 1.1. Examples 3 1.2. Fields 4 1.3. The field of complex numbers. 6 1.4. More information ### Chapter 20. Vector Spaces and Bases Chapter 20. Vector Spaces and Bases In this course, we have proceeded step-by-step through low-dimensional Linear Algebra. We have looked at lines, planes, hyperplanes, and have seen that there is no limit More information ### 4.5 Linear Dependence and Linear Independence 4.5 Linear Dependence and Linear Independence 267 32. {v 1, v 2 }, where v 1, v 2 are collinear vectors in R 3. 33. Prove that if S and S are subsets of a vector space V such that S is a subset of S, then More information ### Operation Count; Numerical Linear Algebra 10 Operation Count; Numerical Linear Algebra 10.1 Introduction Many computations are limited simply by the sheer number of required additions, multiplications, or function evaluations. If floating-point More information ### is in plane V. However, it may be more convenient to introduce a plane coordinate system in V. .4 COORDINATES EXAMPLE Let V be the plane in R with equation x +2x 2 +x 0, a two-dimensional subspace of R. We can describe a vector in this plane by its spatial (D)coordinates; for example, vector x 5 More information ### 1 Introduction to Matrices 1 Introduction to Matrices In this section, important definitions and results from matrix algebra that are useful in regression analysis are introduced. While all statements below regarding the columns More information ### Solution of Linear Systems Chapter 3 Solution of Linear Systems In this chapter we study algorithms for possibly the most commonly occurring problem in scientific computing, the solution of linear systems of equations. We start More information ### 15.062 Data Mining: Algorithms and Applications Matrix Math Review .6 Data Mining: Algorithms and Applications Matrix Math Review The purpose of this document is to give a brief review of selected linear algebra concepts that will be useful for the course and to develop More information ### Linear Algebra Review. Vectors Linear Algebra Review By Tim K. Marks UCSD Borrows heavily from: Jana Kosecka [email protected] http://cs.gmu.edu/~kosecka/cs682.html Virginia de Sa Cogsci 8F Linear Algebra review UCSD Vectors The length More information ### Lecture 5 Principal Minors and the Hessian Lecture 5 Principal Minors and the Hessian Eivind Eriksen BI Norwegian School of Management Department of Economics October 01, 2010 Eivind Eriksen (BI Dept of Economics) Lecture 5 Principal Minors and More information ### Continued Fractions and the Euclidean Algorithm Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction More information ### 1 0 5 3 3 A = 0 0 0 1 3 0 0 0 0 0 0 0 0 0 0 Solutions: Assignment 4.. Find the redundant column vectors of the given matrix A by inspection. Then find a basis of the image of A and a basis of the kernel of A. 5 A The second and third columns are More information ### Chapter 7. Matrices. Definition. An m n matrix is an array of numbers set out in m rows and n columns. Examples. ( 1 1 5 2 0 6 Chapter 7 Matrices Definition An m n matrix is an array of numbers set out in m rows and n columns Examples (i ( 1 1 5 2 0 6 has 2 rows and 3 columns and so it is a 2 3 matrix (ii 1 0 7 1 2 3 3 1 is a More information ### Lecture Notes 2: Matrices as Systems of Linear Equations 2: Matrices as Systems of Linear Equations 33A Linear Algebra, Puck Rombach Last updated: April 13, 2016 Systems of Linear Equations Systems of linear equations can represent many things You have probably More information ### Section 5.3. Section 5.3. u m ] l jj. = l jj u j + + l mj u m. v j = [ u 1 u j. l mj Section 5. l j v j = [ u u j u m ] l jj = l jj u j + + l mj u m. l mj Section 5. 5.. Not orthogonal, the column vectors fail to be perpendicular to each other. 5..2 his matrix is orthogonal. Check that More information ### 1 Symmetries of regular polyhedra 1230, notes 5 1 Symmetries of regular polyhedra Symmetry groups Recall: Group axioms: Suppose that (G, ) is a group and a, b, c are elements of G. Then (i) a b G (ii) (a b) c = a (b c) (iii) There is an More information ### COMMUTATIVITY DEGREES OF WREATH PRODUCTS OF FINITE ABELIAN GROUPS COMMUTATIVITY DEGREES OF WREATH PRODUCTS OF FINITE ABELIAN GROUPS IGOR V. EROVENKO AND B. SURY ABSTRACT. We compute commutativity degrees of wreath products A B of finite abelian groups A and B. When B More information ### Solution to Homework 2 Solution to Homework 2 Olena Bormashenko September 23, 2011 Section 1.4: 1(a)(b)(i)(k), 4, 5, 14; Section 1.5: 1(a)(b)(c)(d)(e)(n), 2(a)(c), 13, 16, 17, 18, 27 Section 1.4 1. Compute the following, if More information ### Introduction to Matrix Algebra Psychology 7291: Multivariate Statistics (Carey) 8/27/98 Matrix Algebra - 1 Introduction to Matrix Algebra Definitions: A matrix is a collection of numbers ordered by rows and columns. It is customary More information ### Linear Algebra: Determinants, Inverses, Rank D Linear Algebra: Determinants, Inverses, Rank D 1 Appendix D: LINEAR ALGEBRA: DETERMINANTS, INVERSES, RANK TABLE OF CONTENTS Page D.1. Introduction D 3 D.2. Determinants D 3 D.2.1. Some Properties of More information ### Chapter 19. General Matrices. An n m matrix is an array. a 11 a 12 a 1m a 21 a 22 a 2m A = a n1 a n2 a nm. The matrix A has n row vectors Chapter 9. General Matrices An n m matrix is an array a a a m a a a m... = [a ij]. a n a n a nm The matrix A has n row vectors and m column vectors row i (A) = [a i, a i,..., a im ] R m a j a j a nj col More information ### 4.6 Linear Programming duality 4.6 Linear Programming duality To any minimization (maximization) LP we can associate a closely related maximization (minimization) LP. Different spaces and objective functions but in general same optimal More information ### Linear algebra and the geometry of quadratic equations. Similarity transformations and orthogonal matrices MATH 30 Differential Equations Spring 006 Linear algebra and the geometry of quadratic equations Similarity transformations and orthogonal matrices First, some things to recall from linear algebra Two More information ### CITY UNIVERSITY LONDON. BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION No: CITY UNIVERSITY LONDON BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION ENGINEERING MATHEMATICS 2 (resit) EX2005 Date: August More information ### ABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS ABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS John A. Beachy Northern Illinois University 2014 ii J.A.Beachy This is a supplement to Abstract Algebra, Third Edition by John A. Beachy and William D. 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New books will be created during 2013 and 2014) Topic 4 Module 9 Introduction Systems of to Matrices Linear Equations Income = Tickets! More information ### October 3rd, 2012. Linear Algebra & Properties of the Covariance Matrix Linear Algebra & Properties of the Covariance Matrix October 3rd, 2012 Estimation of r and C Let rn 1, rn, t..., rn T be the historical return rates on the n th asset. rn 1 rṇ 2 r n =. r T n n = 1, 2,..., More information ### Eigenvalues, Eigenvectors, Matrix Factoring, and Principal Components Eigenvalues, Eigenvectors, Matrix Factoring, and Principal Components The eigenvalues and eigenvectors of a square matrix play a key role in some important operations in statistics. In particular, they More information<\|endoftext\|>	4.65625
193	The Atomic Clock (or Hydrogen Maser) Radio astronomy and geodesy require the very careful measurement of timing signals in order to extract useful data from observations. Algonquin Radio Observatory has its own atomic clock or hydrogen maser to keep track of time precisely. Accurate to 1 part in 1015, the clock loses less than one second in 30 million years. The clock's timekeeping relies on the measurement of hydrogen atoms as they decay from one electronic state to another. A small viewing window allows visitors touring the facility to see the purple discharge glow that the maser emits during operation. The clock must be kept in operation 365 days a year. The clock is also utilized along with a network of GPS antennas to compute corrections to the orbits of Global Positioning Satellites (GPS) in order to remove orbital precession. Without this vital information, GPS receivers would slowly incur errors in position, and degraded navigation performance would result.<\|endoftext\|>	3.96875
2,210	Stimulus equivalence, as studied by behavior analysts, is a relatively new research area that has been rapidly expanding. Many behavior analysts are excited about the topic, and its possible far reaching implications have not gone unnoticed (Sidman, 1994). The research itself focuses on relations among stimuli. It is based on a mathematical definition, whereby "a relation R is an equivalence relation for set S if the properties of reflexivity, symmetry, and transitivity are satisfied for all members of S" (Hall & Chase, 1991, p. 108). Reflexive properties involve the relation between a stimulus and itself (e.g., A = A). Symmetrical properties involve the relation between a pair of ordered stimuli, such that their order is reversible (e.g., if A = B, then B = A). Transitive properties involve two ordered pairs of stimuli, such that the second element of the first pair is the first element of the second pair, which in turn determines a third ordered pair (e.g., if A = B and B = C, then A = C). When stimuli have these relations to each other, they are said to form an equivalence class. How do behavior analysts study stimulus equivalence, and why do they believe it to be so important? In the standard stimulus equivalence paradigm, relations among stimuli are determined by the choices of a learner. Specifically, the learner is shown a sample stimulus together with some other stimuli that are to be compared to the sample. The learner's task is to select the comparison stimulus that matches the sample stimulus in accordance with a particular relation between the two. Not surprisingly, this paradigm is called match-to-sample. Imagine we set up the following task for a child named Laura. We have three sets of stimuli: the English words cat, dog, horse (Set A); the French words chat, chien, cheval (Set B); and the Spanish words gato, perro, and caballo (Set C). Each word is written on a separate card. We show Laura the cat card from Set A, and beneath it, in random order, are the three French cards from Set B. We instruct Laura to choose the card from the bottom row (comparison stimulus) that goes with the card in the top row (sample stimulus). If she selects the chat card, she is praised; other choices are not praised. Over time, given the cat card in the top row, Laura reliably chooses the chat card from the bottom row. We can summarize her behavior by the rule "If you see cat, then select chat," or more generally, If Stimulus A, then select Stimulus B. Her performance implies that she has learned a relation between the two stimuli, typically abbreviated as AB, with A representing the sample stimulus and B representing the comparison stimulus whose selection is reinforced. The AB relation is illustrated in the top panel of Figure 1. Now we teach Laura a second task. We present chat card from Set B as the sample stimulus. The comparison stimuli are the Spanish words from Set C. In this new task, Laura is reinforced for selecting the gato card from the bottom row, which she eventually learns to do. Consistent with the terminology above, we abbreviate the relation as BC, and it is illustrated in the bottom panel of Figure 1. In the third part of our procedure, we arrange some test trials. In one type of test trial, sometimes we present cat card as the sample stimulus, along with the English words from Set A as the comparisons. Sometimes we present chat card as the sample stimulus, along with the French words from Set B as the comparisons. And sometimes we present gato card as the sample stimulus, along with the Spanish words from Set C as the comparisons. The typical finding is that in test trials like these Laura will select the comparison that is identical to the sample (e.g., If you see cat, select cat), even though she has not been reinforced for doing so. In this untrained performance, called reflexivity, a learner responds to a training stimulus on the basis of its sameness. Reflexivity test trials are illustrated in Figure 2. In another type of test trial, we show Laura the chat card from Set B as the sample stimulus, along with the English words from Set A as the comparisons. Research suggests that Laura will select the cat card, even though she has not been reinforced for doing so. This untrained performance is called symmetry. Because the AB relation was trained, and behavior consistent with a BA relation emerged, we summarize this symmetrical performance by the rule If AB, then BA. We might also present the gato card from Set C as the sample, along with the French words from Set B as the comparisons. Similar to above, Laura will likely select the chat card, and this symmetrical performance is summarized by the rule If BC, then CB. Symmetry test trials are illustrated in Figure 3. On still other test trials, we present the cat card from Set A as the sample stimulus along with the Spanish words from Set C as the comparisons. We are likely to discover that Laura will select the gato card, even though she has not been reinforced for doing so. This untrained performance called, transitivity, is summarized by the rule If AB and BC, then AC. The transitivity test trial is illustrated in the top panel of Figure 4. The untrained relations indicative of reflexivity, symmetry and transitivity are called emergent relations; they are derivatives of other relations that have been trained. They are the defining features of an equivalence class. For Laura, the words cat, chat, and gato form an equivalence class because these stimuli were deemed to be reflexive, symmetrical, and transitive. As Sidman (1986) notes, "A consequence of such class formation is that a subject, when tested, will match any member of a class to any other, even without having encountered the tested relation before" (p. 233). There is an abbreviated way that researchers sometimes use to assess stimulus equivalence. In our example, we would present the gato card as the sample, along with the English words from Set A as the comparisons. No other test trial type would be presented. If Laura is a typical subject, she will likely select the cat card, even though she has not been reinforced for doing so. Her untrained performance can be summarized by the rule If AB and BC, then CA. This abbreviated assessment is called the equivalence test (Sidman, 1994, p. 224). It is considered to be a sufficient test for stimulus equivalence because it simultaneously evaluates both symmetry and transitivity (reflexivity is assumed). Specifically, the combination of the AB and BC relations yields the AC relation via transitivity, and its symmetrical counterpart is the CA relation (but see Pilgram & Galizio, 1995). The equivalence test trial is illustrated in the bottom panel of Figure 4. All trained and tested relations in a standard stimulus equivalence paradigm are summarized in Figure 5. Why are behavior analysts excited about the stimulus equivalence? First, it seems clear that people know more than they have been directly taught. Stimulus equivalence research is discovering specific learning histories that can account for this fact (see also Alessi, 1987). The research is showing that a little bit of training goes a long way. In our example, Laura was only taught only two relations (AB and AC); consequently, she also proved capable of performing in accordance with seven others (AA, BB, CC, BA, CB, AC, CA). And consider this. Suppose we teach Laura only one more relation. Perhaps we show her a picture of a cat, along with the Spanish words from Set C as comparisons, and reinforce her for selecting the gato card (DC). She will then likely prove capable of six more untrained performances: DD, AD, DA, BD, DB, CD (e.g., Sidman & Tailby, 1982). But this is only the beginning. When one member of an equivalence class acquires a new function via a standard behavioral process, other members in that class often take on that same function. This effect has been shown for SDs (Catania, Horne, & Lowe, 1989), reinforcers and punishers (Hayes, Kohlenberg & Hayes, 1991), and elicitors (Dougher, Augustson, Markham, Greenway, & Eldelgard, 1994). In our example, we might reinforce Laura for saying "meow" only in the presence of the cat card, a discrimination training procedure. Given that the printed word cat participates in an equivalence class with the printed words chien and gato, Laura may now be more likely to say "meow" when we show her either chien card or the gato card, even though reinforcement for such a response has never occurred in the presence of either of these stimuli. In other words, the chien and gato cards now act like discriminative stimuli in occasioning behavior without ever having been part of a discrimination training procedure themselves. A second reason why stimulus equivalence has generated excitement is its apparent close link to our everyday terms language and meaning. Consider Laura's CA test trial performance. Laura had never before selected cat given gato; in fact, she had never even seen the two words together. However, she had related the words cat and chat together (AB training trials) and the words chat and gato together (BC training trials). On the CA test trial we might say that Laura responds to Spanish word gato as if it was the French word chat; or, she selects English word cat as if it was the French word chat. The three words appear to be interchangeable. Sidman (1986) summarizes what may be the key contribution of stimulus equivalence research: "By reacting to a word as an equivalent stimulus - the meaning of a word - a person can behave adaptively in an environment without having previously been exposed to it. The emergence of equivalence from conditionality permits Behavior Analysis to account for the establishment of at least simple semantic correspondences without having to postulate a direct reinforcement history for every instance. Instead of appealing to cognitions, representations, and stored correspondences to explain the initial occurrence of appropriate new behavior, one can find a complete explanation in the...units that are the prerequisites for the emergent behavior." (p. 236) Stimulus equivalence is a complicated topic. Sidman (1994) has updated and refined his analysis in considerable detail. Alternative frameworks have also been formulated in recent years by two other research teams to account for the same phenomenon (Hayes & Hayes, 1992; Horne & Lowe, 1996).<\|endoftext\|>	4.125
1,345	Hannibal Barca was born in 247 BCE as a son of the Empire of Carthage, which encompassed all of North Africa and Southern Spain. Hannibal was the son of the great Carthaginian military leader Hamilcar Barca. Hamilcar lead the Carthaginian army in the First Punic War against Rome. Carthage suffered an embarrassing loss to the Romans which included loss of control the city of Sicily. It is widely stated that as a youth Hannibal’s father instilled within him an unrelenting hatred for Rome. When Hannibal was 17 Carthage was able to conquer Hispania which is now the modern day Iberian Peninsula. During the conquest Hamilcar drowned and Hannibal’s brother Hasdrubal the Fair became the commander of the army. This victory was able to expand the Carthaginian Empire which remained a formidable opponent for Rome. Hasdrubal further strengthened the numbers of Carthage by intermarrying the Carthaginians with the conquered Iberians. In 221 BCE Hasdrubal was murdered and Hannibal was elected to assume command of the army at the age of twenty-six. His first campaign was the capturing of the city of Salamanca in 220 BCE. Carthage’s next conquest was of Saguntum which was a close ally of Rome’s. Hannibal’s attack on Saguntum was considered to be a violation of a peace treaty signed between Hasdrubal and Rome. They demanded that Carthage expel Hannibal from the empire. As Hannibal’s fate was being decided, he continued to conquer territories expanding Carthage as far as he could. Hannibal’s brother was appointed as a military commander on the Iberian Peninsula. This move helped the forces of Carthage conquer the peninsula as a whole. Hannibal was determined to bring war to Rome; he remembered what his father told him about the Romans so he launched a military campaign. His conquest of the Iberian Peninsula was an example of his advancement towards conquering the Roman Empire. Because Hannibal conquered the Iberian Peninsula the Roman government declared war against Carthage. This declaration was the beginning of the Second Punic War. Hannibal and his forces invaded Italy in a surprise attack on the Romans, who expected an attack at Sicily. His next move was to cross the Pyrenees Mountains. Before he could cross, his army had to defeat the tribes who dwelled along the foothills of the mountains. His army crossed the Pyrenees and reached the river Rhone. Along the way he managed to pacify the Chiefs of the Gaul’s. This strategic move helped to stall Roman advancement against Carthage. It is said that Hannibal’s army consisted of 50,000 infantry men, 9,000 cavalry and 37 war elephants. Next, Hannibal led his army across the treacherous Alps which took a toll on his army. He lost a number of his solders and some of his war elephants. It is said that he led 38,000 soldiers into the town of Turin, Italy. The Romans became aware of the alliance between the Gaul’s and Carthage, and sent troops of 80,000 to defeat Carthage and the Gaul’s. Their plans were spoiled because Hannibal was able to defeat the forces. This victory gave the Gaul’s confidence in Hannibal so they volunteered the join his army. The Gaul’s were able to add the strength to the army that Hannibal lost crossing the Alps. His army was able to defeat the Romans a second time in a battle at the river Trebbia. In 217 BCE Hannibal and his army crossed the Apennines Mountains and conquered modern day Tuscany. During these battles he lost of one of his eyes but not his hatred for Rome. The Romans retaliated with their own attack against Carthage but was defeated once again at the Trasimene Lake. As Hannibal’s army crossed the Apennines for a second time, Roman forces attacked Iberia and cut off his access to his allies and supplies. In the city of Cannae 80,000 Roman solders attacked Hannibal’s army. Despite being outnumbered, Hannibal’s forces were able to once again gain a victory over Rome. This victory caused Roman allies to pledge their allegiance with Hannibal. The Italian city of Capua became Hannibal’s new military base. In 214 BCE the city of Syracuse became a city of Carthage. 215 BCE King Philip V of Macedonia pledged his allegiance with Carthage. The Romans managed to secure a decisive victory against Carthage. Hannibal was not able to capture the port cities of Cumae and Puteoli. His army was not able to receive reinforcements or supplies. Carthage was losing resources and allies fast. The cities of Syracuse and Capua were regained by the Romans, further weakening Hannibal’s forces. The four year campaign in Italy was taking a toll on his solders. Hannibal sent for his brother located in Iberia to help him fight. Unfortunately Hannibal’s’ brother was defeated crossing the Alps. Rome was able to reclaim Iberia as well as again an ally in the King of Namibia. Aligned with Rome, Namibia attacked Carthage forcing Hannibal to bring his troops home to defend his land. A final battle was fought between Rome and Carthage at Zama in 202 BCE. The Carthaginian forces were weakened from the Punic War and they fell in defeat to Rome. A peace treaty was signed in 201 BCE which forced Carthage to compensate Rome for the damages its forces caused. Also the treaty forced Hannibal to resign as the leader of the Carthaginian army. The mighty Hannibal was able to instill terror into the Roman empire. Rome was the world power at the time and Carthage was a thorn in their side. The size and arrogance of Rome caused them to underestimate the brilliance of Hannibal. He is considered one of the most brilliant military leaders in history. His crossing of the Alps was a feat many may have envisioned but never attempted. Hannibal was determined to keep his promise to his father and annihilate Rome. Though he did not reach his ultimate goal he accomplished more than many military leaders can only dream of. His Empire covered North Africa, Spain and parts of Italy. Hannibal was a true African warrior and a skilled politician. He will never be forgotten within the pages of history. Hannibal was able to show Rome the full potential of his African might. Hannibal Barca the Great, we proudly stand on your shoulders. Click below to view the Hannibal Barca video Click Here to join our mailing list<\|endoftext\|>	3.953125
460	Iron Sulfide FeS2 Pyrite is a compound of iron and sulfur, iron sulfide. Depending upon the conditions under which it forms this mineral can form crystals of different shapes. The crystals are isometric, meaning that they have equal faces. Cubes are common. Octahedral and dodecahedral shapes are also seen frequently. The name comes from the Greek word for fire “pyr”, literally fire rock. This is because it sparks when struck against iron. Because of its high sulfur content pyrite is used in making sulfuric acid and sulfur dioxide. It can be an agent of fossilization creating beautiful fossils, commonly ammonites and brachiopods. Pyrite suns are found in slate seams of coal mines. These "suns" are flat disc shapes with a radial pattern. No one knows exactly how they are formed. There are two prevalent theories; one is that they are pyritized fossils of an unknown animal. Another is that they are pyrite crystals spread out in the thin seams of slate where they are found. In the gold rush era, many people thought they had discovered gold only to realize it was pyrite, which is where it gets its nickname fools gold. The hardness of this mineral is about 6-6.5 on the Mohs scale. It has a metallic luster and normally has a yellow brassy color. Of the sulfide minerals pyrite is the most common. It is usually found in quartz veins with other sulfides and oxides. Mining operations sometimes leave large amounts of pyrite in tailings. This causes environmental damage because it reacts with oxygen and water to form sulfuric acid. Chemical formula: Iron Sulfide FeS2 Streak: greenish black Crystal system: cubic Specific Gravity: 4.8-5 Hardness (Mohs): 6-6.5 Uses: used in making sulfuric acid and sulfur dioxide Location: Rio Tinto, Spain; Quiruvila, Peru; Ontario, Canada; Elba, Italy; Llallagua, Bolivia; In the U.S.A. Utah, Colorado Buy Pyrite Here<\|endoftext\|>	3.75
350	The Civil Rights Act of 1964 (Pub.L. 88–352, 78 Stat. 241, enacted July 2, 1964) is a landmark civil rights and US labor law in the United States that outlaws discrimination based on race, color, religion, sex, or national origin. It prohibited unequal application of voter registration requirements, racial segregation in schools, employment, and public accommodations. Powers given to enforce the act were initially weak, but were supplemented during later years. Congress asserted its authority to legislate under several different parts of the United States Constitution, principally its power to regulate interstate commerce under Article One (section 8), its duty to guarantee all citizens equal protection of the laws under the Fourteenth Amendment and its duty to protect voting rights under the Fifteenth Amendment. The Act was signed into law by President Lyndon B. Johnson on July 2, 1964, at the White House. The bill was called for by President John F. Kennedy in his civil rights speech of June 11, 1963, in which he asked for legislation "giving all Americans the right to be served in facilities which are open to the public—hotels, restaurants, theaters, retail stores, and similar establishments", as well as "greater protection for the right to vote". Kennedy delivered this speech following the immediate aftermath of the Birmingham campaign and the growing number of demonstrations and protests throughout the southern United States. Kennedy was moved to action following the elevated racial tensions and wave of black riots in the spring 1963. Emulating the Civil Rights Act of 1875, Kennedy"s civil rights bill included provisions to ban discrimination in public accommodations, and to enable the U.S. Attorney General to join in lawsuits against state governments which operated segregated school systems, among<\|endoftext\|>	4.28125
730	In the ongoing arms race with humans and their antibiotics on one side, and bacteria with their ability to evolve defenses to antibiotics on the other, humans have enlisted a new ally—other bacteria. Many common antibiotics, including the most famous antibiotic, penicillin, are based around a molecular structure known as a beta-lactam ring. These drugs, aptly named beta-lactam antibiotics, interfere with a bacterium's ability to build its cell wall. As bacteria develop resistance to existing antibiotics, researchers and pharmaceutical companies work to create new ones. That means a lot of work is done creating new kinds of beta-lactams, and that is where Frances Arnold's lab enters the picture. Beta-lactams are made by taking a chainlike molecule and looping it, kind of like taking one end of a string and tying it in a knot to the middle of the string. The paramount challenge is to control precisely where along the molecule the reaction takes place. With traditional synthetic chemistry, chemists have to tack extra pieces onto molecules that they want to turn into beta-lactams. Without those extra pieces, the knots will end up tied in inconsistent spots, resulting in some loops that are large and some that are small. That's undesirable for someone trying to manufacture a consistent batch of antibiotics. But the addition of those extra pieces makes the synthesis more complicated because additional steps are required to add them and still more steps to remove them after the looping is complete. Graduate student Inha Cho and postdoctoral scholar Zhi-Jun Jia, both from Arnold's lab, have developed something simpler by using directed evolution, a technique developed by Arnold, the Linus Pauling Professor of Chemical Engineering, Bioengineering and Biochemistry, and director of the Donna and Benjamin M. Rosen Bioengineering Center. In directed evolution, which Arnold developed in the 1990s and for which she received the 2018 Nobel Prize in Chemistry, enzymes are evolved in a lab until they behave in a desired way. The genetic code of a useful enzyme is transferred into bacteria like Escherichia coli. As the bacteria grow, divide, and go about their lives, they churn out the enzyme. In this case, Cho and Jia took an enzyme known as cytochrome P450, which has been a versatile workhorse in the Arnold lab, and evolved it to produce beta-lactams. Two other versions of enzymes were also created to construct other ring sizes of lactams. One version creates a gamma-lactam, a loop of four carbon atoms and one nitrogen atom. And the other version creates a delta-lactam, a loop of five carbon atoms and one nitrogen atom. "We're developing new enzymes with activity that cannot be found in nature," says Cho. "Lactams can be found in many different drugs, but especially in antibiotics, and we're always needing new ones." Jia points out that the enzymes they have created are also incredibly efficient, with each molecule of enzyme capable of producing up to one million beta-lactam molecules. "They represent the most efficient enzymes created in our lab, and are ready for industrial applications," Jia says. The paper, titled "Site-selective enzymatic C-H amidation for synthesis of diverse lactams" and co-authored by Arnold, appears in the May 10 issue of Science. Support for the research was provided by the National Science Foundation, the Joseph J. Jacobs Institute for Molecular Engineering for Medicine, and Deutsche Forschungsgemeinschaft.<\|endoftext\|>	3.9375
321	Last June astronomers detected the brightest supernova that has ever been observed, known as ASASSN-15lh. The object had been classified as hydrogen-poor superluminous supernova fuelled by a magnetar, altough it is hotter and more luminous than any other hydrogen–poor supernova thus making the classification uncertain. Now scientists try to uncover the nature of this mysterious source using new observational data. Ultraviolet and optical photometry was obtained using the Swift/Ultraviolet Optical Telescope (UVOT), X-ray data from the X-ray Telescope on Swift and spectroscopy with the Hubble Space Telescope (HST). The new observations revealed a remarkable Ultraviolet rebrightening that took place two months after the peak brightness and has not seen in previous superluminous supernova, while Ultraviolet spectra from HST exhibited a lack of broad emission or absorption features also dissimilar to superluminous supernova. This new findings make the classification of ASASSN-15lh even more complex and the development of a model that will successfully describe the source, more challenging. Astronomers estimated that a source like ASASSN-15lh could be detected at redshift beyond 4, when the universe was only about 1.5 billion years old, using the Large Synoptic Survey Telescope (LSST). This would allow scientists to study the properties and conditions of the galaxies that host these sources as well as the interstellar and circumgalactic media at such early times in the history of the universe. Publication: Brown et al. 2016<\|endoftext\|>	3.71875
299	We are coming up on the Christmas season and we all are going to be working on Christmas lessons during the whole month of December. One thing that I like to go over with my class are the different names of Jesus. He is called so many things throughout scripture that really point to who He is. I have created this acrostic poem, using the name of Jesus, and filling it in with other names that He is called. - Click on the preview above to download the printable version - We’ve also uploaded a “blank version” if you would like to change the verses. J – Jesus Name Above Every Name E – Everlasting Father S – Son of God U – Undefiled High Priest S – Savior I have put a Bible verse next to each name of Jesus so that it can also be a great little study to do with the children in your classes. There are verses from the Old and New Testaments to show your children that Jesus is talked about throughout the entire Bible, not just in the Gospels. You could then have the children make their own acrostic about Jesus. You can have them write phrases that start with the letters in the name of Jesus. For example: J – Jerusalem was the city that He traveled to, E – Eternity is secure because of Him, S – Sacrifice, etc. I hope you and your class will enjoy this acrostic about the names of Jesus.<\|endoftext\|>	3.703125
515	# DAT 520 Problem Set 3.docx - DAT 520 Problem Set 3... • Homework Help • 8 • 86% (14) 12 out of 14 people found this document helpful This preview shows page 1 - 3 out of 8 pages. DAT 520 Problem Set 3 Cumulative and Conditional Probability Overview: The last problem set in Module Two exposed you to dependent trials that caused the probability to shift slightly with each choice of sock from the drawer. Kind of sneaky, eh? This problem set gets you a little further into conditional probability with a direct look at Bayes’ theorem. Since this is a course in decision analysis and not a course in probability, we are going to introduce these concepts, use them once or twice by hand, but then let R and Excel do the heavy lifting for us the rest of the time. This problem set, however, requires hand calculations. All you should need is a calculator and your wits. Example Problem: Example 1 : You are flipping a fair coin four times in total, but you are flipping it in groups of two. In order to flip the coin for flips three and four, you have to get heads two times in a row for flips one and two. Problem Notes: Define success: Success for this problem is flipping a head, which is p =50%. Recall the binomial equation: Recall the Law of Total Probability equation: probability(item_1) * (item_1_%contribution) + probability(item_2) * (item_2_ %contribution) + … Recall Bayes’ theorem: Questions for Example Problem: Q1: What is the probability of flipping heads four times in a row, generally ? . Q2: The way this problem is set up, what is the total probability of getting four heads? : We calculate this as 25% chance of flipping two heads on flips one and two. Same thing for flips three and four. Let’s now call the first event p (F1&2) and the second one p (F3&4). Next, build the table . Notice that all the % contributions have to add up to 1. Table of contributions to total probability: P % contrib. p (F1&2) .25 .50 p (F3&4) .25 .50 Using the law of total probability : Total probability for two heads then two heads is (.50 x .25) + (.50 x .25) = 25%.<\|endoftext\|>	4.53125

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