Mathematics Portfolio Sl Essay Example

Mathematics Portfolio Sl Essay Mathematics Standard Level Teacher: Mr. Lazaro Name: Fatema Ismailjee IB 1 2011 Sequence is a set of things (usually numbers) that are in order. e. g. 1, 2, 3, 4, Where 1 is the first term, 2 is the second term and so on. ( in the end means that the sequence goes on forever. Three dots in the middle e. g. 1, 2, 3 7, 8 indicate that the pattern continues until the next number appears. There is finite and infinite sequence, infinite sequence is when the sequence has no end and finite is a set with a function e. g. {1, 3, n} Calculating specific terms leads to an nth  term formula. Before creating a rule of calculation, you need to realize that sequences are functions with the specific domain of the counting numbers {1, 2, 3, 4, 5, }. So the n replaces x  as the input variable and instead of writing  y, we use  an  as the output variable.Arithmetic sequence: the difference between one term and the next is a constant in arithmetic sequence. The general formula is an  = a1à ‚  + (n 1) d Geometric sequence: A geometric sequence is a group of numbers where each term after the first is found by multiplying the previous one by a fixed non zero number called common ratio. The general formula is an = a1 ? rn-1 Series is the sum of terms of a sequence. Sn = x1 + x2 +. xn Arithmetic series: The general formula is Sn  = n/2(a1  + an) Geometric series: a series which has a constant ratio between terms.The general formula is Sn = a1 (1 – rn) 1 r TRIANGULAR NUMBERS Triangular number is the number of dots in an equilateral triangle uniformly filled with dots. This is an investigation task whereby I will try to find number of shapes of geometric figures which form triangular numbers. I will use different sources of information to attain shapes and figures. For the calculations required, different math techniques will be used for the different shape obtained. Aim In this task I will consider geometric shapes which lead to special numbers.The simplest exa mples of these are square numbers, 1, 4, 9, 16, which can be represented by squares of side 1, 2, 3 and 4. The following diagrams show a triangular pattern of evenly spaced dots. The numbers of dots in each diagram are examples of triangular numbers (1, 3, 6, ). .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 3 6 10 15 There is a sequence of the number of dots in the triangular shape above.Complete the triangular sequence with three more terms. . . . . . . . . . . . . . . . . . . . . . 21 dots . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 dots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 dots Find a general statement that represents the nth triangular number in terms of n. In words: The top row has one dot and each successive row under it has one more dot.Using the formula: 1. Find the common difference between the numbers in the sequence. 2. Use the general formula tn = an2 + bn + c. 3. Three equations will be forme d. Using the elimination method find the coefficients i. e. a, b and c. 4. Substitute in the general formula. The general statement can be reached by following the steps above. Common difference: d= U2 – U1 = U3 – U2 1, 3, 6, 10, 15, d= 3-1 = 2 6-3 = 3 10-6 = 4 15-10 = 5 d= 3-2 = 1 4-3 = 1 5-4 = 1 The difference in terms is found in the second stage so the formula will be n2 . 2 Testing: n = 1 , triangular number = 1 22 = 12 1 12 = 12 n = 2, triangular number = 3 n22 = 92 3 92 = 32 32 12 = 12 12 so this will be 12 n therefore, 2 12 n2 = 12n As the common difference in the second stage is 1, it can be deduced that the formula for the nth term is a quadratic equation. I will use the general formula to find the nth term, tn = an2 + bn + c where a and b are the coefficients and c is constant and n is the number of term. 2 n2 = 12 n = n2 + n 2 When n = 1 1 = a (1)2 + b (1) + c 1 = a + b + c . (i) n = 2 3 = a (2)2 + b (2) + c 3 = 4a + 2b + c . (ii) n = 3 6 = a (3)2 + b (3 ) + c 6 = 9a + 3b + c . (iii) Using the elimination method: 3 = 4a + 2b + c 6 = 9a + 3b + c 1 = a + b + c 3 = 4a + 2b + c 2 = 3a + b 3 = 5a + b Now that two equations are obtained: To find the variables i. e. a, b one of them is eliminated.In this case the equations are being subtracted. b will be eliminated first in order to find a. Substitute the values of a in the equation to find the value of b. 3 = 5a + b 3 = 5a + b 1 = a + b + c 2 = 3a + b 3 = 5(1 ) + b 1 = 1 + 1 +c 2 2 2 1 = 2a 3 5 = b 2 2 2 a = 1 b = 1 c = 0 2 Therefore the formula for finding the nth term will be as follows: tn = 1n2 + 1n 2 2 tn = n2 + n 2 Use of technology to find the general statement: Calculator used: CASIO fx-9750 GA PLUS n| 1| 2| 3| 4| 5| 6| 7| y| 1| 3| 6| 10| 15| 21| 28| Let n = x 1. Select STAT. 2. Encode values for x in list 1 and for y in list 2. 3. Select GRPH (by pressing F1). 4. Select GPH1 (by pressing F1 again). 5. Select x^2 (by pressing F3). The display will show: a = 1 2 b = 1 c = 0 y = ax 2 + bx + c 1n2 + 1n y = 2 2 = n2 + n 2 Consider stellar (star) shapes with p vertices, leading to p-stellar numbers. The first four representations for a star with six vertices are shown in the four stages S1 – S4 below. The 6-stellar number at each stage is the total number of dots in the diagram. Find the number of dots (i. e. the stellar number) in each stage up to S6. Stellar numbers are figurate number, based on the number of dots of units that can fit in a centred hexagon or star shapes. S1 – S4 are the numbers of dots in the stars.To find up to S6 find the common difference (d) followed by the addition of numbers of star in the previous star. S1 has 1 dot S2 has 13 dots S3 has 37 dots S4 has 73 dots Find the common difference between the terms. d = S2 – S1 S3 – S2 d = 13 – 1 = 12 37 – 13 = 24 73 – 37 = 36 As the difference is not constant, find the difference within the answers. d = 36 – 24 = 12 13 – 12 = 12 The c ommon difference is 12. S 5 = 36 + 12 = 48 73 + 48 = 121dots S6 = 48 + 12 = 60 121 + 60 = 181dots Find an expression for the 6-stellar number at stage S7. As shown above, the common difference is 12.As it’s a sequence it follows the same trend therefore: To find the next number of dots in the sequence, add it with 12 first and from the second star add it with the multiples of 12, i. e. 24, 36, 48 etc. S6 = 48 + 12 = 60 = S5 + 60 = 121 + 60 = 181 S7 = 60 + 12 = 72 = S6 + 72 = 181 + 72 = 253 S7 = 253 Find a general statement for the 6-stellar number at stage Sn in terms of n. Use the same general formula to obtain the three equations: The general formula: tn = an2 + bn + c When n = 1 1 = a (1)2 + b (1) + c 1 = a + b + c . (i) n = 2 13 = a (2)2 + b (2) + c 13 = 4a + 2b + c . (ii) = 3 37 = a (3)2 + b (3) + c 37 = 9a + 3b + c . (iii) Using the elimination method: 37 = 9a + 3b + c 13 = 4a + 2b + c 13 = 4a + 2b + c 1 = a + b + c 24 = 5a + b 12 = 3a + b After attaining two equations, either of the coefficients should be eliminated. b in this case which will lead us to find a. Substitute value of a in the equation to find b. Hence, substitute values of a and b for c. 24 = 5a + b 24 = 5a + b 1 = a + b + c 2 = 3a + b 24 = 5(6) + b 1 = 6 + (-6) + c 12= 2a 24 – 30 = b 1 – 0 = c 2 2 a = 6 b = -6 c = 1 Substitute a, b and c in the general statement. General statement: tn = 6n2 – 6n + 1 Now repeat the steps above for other values of p Considering stellar (star) shapes when p=7 and when p=8 leading to p-stellar numbers. p = 7 Find the number of dots (i. e. the stellar number) in each stage up to S6. S1 has 1 dot S2 has 15 dots S3 has 43 dots S4 has 85 dots d = 15 – 1 = 14 3 – 15 = 28 85 – 43 = 42 d = 42 – 28 = 14 28 – 14 = 14 e. g. S4 = 43 + 14 = 42 S3 + 42 43 + 42 = 85 dots S 5 = 42 +14 =56 S4 + 56 85 + 56 = 141dots S6 = 56 +14 = 70 S5 + 70 141 + 70 = 211dots Find an expression for the 6-stellar number at stage S7 . As shown above, the common difference is 14. As it’s a sequence it follows the same trend therefore: To find the next number of dots in the sequence, add it with 2 first and from the second star add it with the multiples of 2, i. e. 14, 28, 42 etc. S7 = 70 + 14 = 84 S6 + 84 211 + 84 = 2955dots S7 = 295dots Find a general statement for the 6-stellar number at stage Sn in terms of n.To find the three equations, use the general formula tn = an2 + bn + c. When n = 1 1 = a (1)2 + b (1) + c 1 = a + b + c . (i) n = 2 15 = a (2)2 + b (2) + c 15 = 4a + 2b + c . (ii) n = 3 43 = a (3)2 + b (3) + c 43 = 9a + 3b + c . (iii) Three equations are obtained, to find a, b and c, the equations need to be solved. Elimination method is one of the ways from which we can attain the coefficients and constant. Using elimination method: Firstly, we need to remain with two equations at the end so subtract equations (equation iii – ii and equation ii – i) and two will be remained. 3 = 9a + 3b + c 15 = 4a + 2b + c 15 = 4a + 2b + c 1 = a + b + c 28 = 5a + b 14 = 3a + b Now that there are two equations, find a and b. Subtract the equation to eliminate one variable. After one is found, the other can be easily found by substituting the value of variable attained in the equation. 28 = 5a + b 28 = 5a + b 1 = a + b + c 14 = 3a + b 28 = 5(7) + b 1 = 7 + (-7) + c 4= 2a 28 – 35 = b 1 – 0 = c 2 2 a = 7 b = -7 c = 1 Substitute a, b and c in the general statement. General statement: tn = 7n2 – 7n + 1 p = 8 S1 has 1 dot S2 has 17 dots S3 has 49 dots S4 has 97 dots Find the common difference: d = 17 – 1 = 16 49 – 17 = 32 97 – 49 = 48 As the difference is not constant, subtract the answers to find the common difference. d = 32 – 16 = 16 48 – 32 = 16 To find the following number in the star e. g. S4 = 32 + 16 = 48 S3 + 48 49 + 48 = 97 dots The common difference is 16.If observed carefully the number is found by adding it with mu ltiples of 16 i. e. 32, 48, 64, 80 etc. S 5 = 48 +16 = 64 S4 + 64 97 + 64 = 161dots S6 = 64 + 16 = 80 S5 + 80 161 + 80 = 241dots Find an expression for the 6-stellar number at stage S7. As shown above, the common difference is 16. As it’s a sequence it follows the same trend therefore: To find the next number of dots in the sequence, add it with 16 first and from the second star add it with the multiples of 3, i. e. 32, 48, 64 etc. S7 = 80 + 16 = 96 S6 + 96 241 + 96 = 337dots S7 = 337dots Find a general statement for the 6-stellar number at stage Sn in terms of n.I will use the same general formula to obtain the three equations: The general formula: tn = an2 + bn + c When n = 1 1 = a (1)2 + b (1) + c 1 = a + b + c . (i) n = 2 17 = a (2)2 + b (2) + c 17 = 4a + 2b + c . (ii) n = 3 49 = a (3)2 + b (3) + c 49 = 9a + 3b + c . (iii) Using the elimination method: 49 = 9a + 3b + c 17 = 4a + 2b + c 17 = 4a + 2b + c 1 = a + b + c 32 = 5a + b 16 = 3a + bNow there are two equations, so b has to eliminated by subtracting the two equations to find a. Once, a is obtained one of the equation has to be chosen and substitute the value of a in it. Hence b is obtained. 32 = 5a + b 32 = 5a + b 1 = a + b + c 16 = 3a + b 32 = 5(8) + b 1 = 3 + (-3) + c 16 = 2a 32 – 40 = b 2 2 a = 8 b = -8 c = 1 Substitute a, b and c in the general statement. General statement: tn = 8n2 – 8n + 1Hence, produce the general statement, in terms of p and n,that generates the sequence of p-stellar numbers for any value of p at stage Sn. The general statements produced are: tn = 6n2 – 6n + 1 tn = 7n2 – 7n + 1 tn = 8n2 – 8n + 1 I observed it, and reached to the conclusion that for all the three statements the number of p and number of coefficient that is a and b is the same. Therefore the general statement in terms of p and n that generates the sequence of p-stellar numbers for any value of p at stage Sn is: tn = pn2 – pn + 1 Test the validity of the general st atement. When p = 5 S1 has 1 dot S2 has 11 dots S3 has 31 dotsS4 has 61 dots Common difference: d = 11 – 1 = 10 31 – 11 = 20 61 – 31 = 30 As the difference is not constant, subtract it within the answer obtained: d = 30 – 20 = 10 20 – 10 = 10 As seen, the common difference is 5. As it’s a sequence it follows the same trend therefore: To find the next number in the sequence, add it with multiples of 5 i. e. 10, 15, 20, etc. S 5 = 30 + 10 = 40 61 + 40 = 101dots S6 = 40 + 10 = 50 = S5 + 50 = 101 + 50 = 151dots S7 = 50 + 10 = 60 = S6 + 60 = 151 + 60 = 211 S7 = 211dots The general formula: tn = an2 + bn + c When n = 1 1 = a (1)2 + b (1) + c 1 = a + b + c . (i) n = 2 1 = a (2)2 + b (2) + c 11 = 4a + 2b + c . (ii) n = 3 31 = a (3)2 + b (3) + c 31 = 9a + 3b + c . (iii) Using the elimination method: 31 = 9a + 3b + c 11 = 4a + 2b + c 11 = 4a + 2b + c 1 = a + b + c 20 = 5a + b 10 = 3a + b 20 = 5a + b 20 = 5a + b 1 = a + b + c 10 = 3a + b 20 = 5(5) + b 1 = 6 + (-6) + c 10= 2a 20 – 25 = b 1 – 0 = c 2 a = 5 b = -5 c = 1 Substitute a, b and c in the general statement. General statement: tn = 5n2 – 5n + 1 Below are the values of the formula Sn= 5n2 5n +1 n = Stage number in the 5-stellar shape. | y= total number of dots at stage ‘n’. | 1| 1| 2| 11| 3| 31| 4| 61| 5| 101| 6| 151| 7| 211| If we want to know the number of dots in the 5th term using the formula we replace n with 5 based on the formula Sn= 5n2 – 5n + 1 Sn= 5n2 – 5n + 1 Sn= 5 (5)2 – 5 (5) +1 Sn= 101 Limitation: * The value of p should be greater than or equal to 4 i. e. p ? 4. The value of p cannot be negative. It must be a positive integer. The general statement has some limitations as listed above. It is an arithmetic series as seen. It is derived from the equations generated in the 5, 6, 7-stellar shape. The coefficients in each question are equal to the corresponding stellar number p. References: Sequences.   Math Is Fun Maths Resources. Web. 12 Mar. 2011. Help for a Generic Formula for a Stellar Pattern.? Yahoo! Answers.   Yahoo! Answers Home. Web. 12 Mar. 2011. Triangular Number ENotes. com Reference.   ENotes Literature Study Guides, Lesson Plans, and More. Web. 11 Mar. 2011.

Memo to Bradley Stonefield Essays

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The Ancient Toltec Trade and Economy The Toltec Civilization dominated central Mexico from about 900 - 1150 A.D. from their home city of Tollan (Tula). The Toltecs were mighty warriors who spread the cult of their greatest god, Quetzalcoatl, to the far corners of Mesoamerica. Evidence at Tula suggests that the Toltecs had a trade network and received goods from as far away as the Pacific coast and Central America, either through trade or tribute. The Toltecs and the Postclassic Period The Toltecs were not the first Mesoamerican civilization to have a trade network. The Maya were dedicated merchants whose trade routes reached far from their Yucatan homeland, and even the ancient Olmec - the mother culture of all of Mesoamerica - traded with their neighbors. The mighty Teotihuacan culture, which was pre-eminent in central Mexico from about 200-750 A.D., had an extensive trade network. By the time the Toltec culture reached prominence, military conquest and subjugation of vassal states were on the rise at the expense of trade, but even wars and conquest stimulated cultural exchanges. Tula as a Center of Trade It is difficult to make observations about the ancient Toltec city of Tollan (Tula) because the city was extensively looted, first by the Mexica (Aztecs) before the arrival of the Europeans, and then by the Spanish. Proof of extensive trade networks may have therefore been carried off long ago. For example, although ​jade was one of the most important trade materials in ancient Mesoamerica, only one jade piece has been found at Tula. Nevertheless, archaeologist Richard Diehl has identified pottery from Nicaragua, Costa Rica, Campeche and Guatemala at Tula, and found potsherds traced to the Veracruz region. Shells from the Atlantic and Pacific have also been excavated at Tula. Surprisingly, the Fine Orange pottery associated with the contemporary Totonac culture has not been found at Tula. Quetzalcoatl, God of Merchants As the major deity of the Toltecs, Quetzalcoatl wore many hats. In his aspect of Quetzalcoatl - Ehà ©catl, he was the god of wind, and as Quetzalcoatl - Tlahuizcalpantecuhtli he was the bellicose God of the Morning Star. The Aztecs venerated Quetzalcoatl as (among other things) the god of merchants: the post-conquest Ramirez Codex mentions a feast dedicated to the god by traders. The principal Aztec god of trade, Yacatechutli, has been traced to earlier roots as a manifestation of either Tezcatlipoca or Quetzalcoatl, both of whom were worshiped at Tula. Given the Toltecs fanatical devotion to Quetzalcoatl and that gods later association with the merchant class by the Aztecs (who themselves regarded the Toltecs as the apogee of civilization), it is not unreasonable to surmise that trade played an important role in Toltec society. Trade and Tribute The historical record seems to suggest that Tula did not produce much in the way of trade goods. A great deal of utilitarian Mazapan-style pottery has been found there, suggesting that Tula was, or was not far from, a place that produced it. They also produced stoneware bowls, cotton textiles, and items fashioned from obsidian, such as blades. Bernardino de Sahagà ºn, a colonial era chronicler, claimed that the people of Tollan were skilled metalworkers, but no metal not of later Aztec origin has been found at Tula. It is possible that the Toltecs dealt in more perishable items like food, cloth or woven reeds which would have deteriorated with time. The Toltec did have significant agriculture and possibly exported part of their crops. In addition, they had access to a rare green obsidian found near present-day Pachuca. There is the possibility that the warlike Toltecs produced relatively little themselves, instead relying on conquered vassal states to send them goods as tribute. Tula and the Gulf Coast Traders Toltec scholar Nigel Davies believed that during the Postclassic era trade was dominated by the different cultures of Mexicos Gulf Coast, where mighty civilizations had risen and fallen since the days of the ancient Olmec. During Teotihuacns age of dominance, shortly before the rise of the Toltecs, the gulf coast cultures had been an important force in Mesoamerican commerce, and Davies believes that the combination of Tulas location in the center of Mexico, their low production of trade goods, and their reliance on tribute over commerce placed the Toltecs at the fringes of Mesoamerican trade at the time (Davies, 284). Sources: Charles River Editors. The History and Culture of the Toltec. Lexington: Charles River Editors, 2014. Cobean, Robert H., Elizabeth Jimà ©nez Garcà ­a and Alba Guadalupe Mastache. Tula. Mexico: Fondo de Cultura Economica, 2012. Coe, Michael D and Rex Koontz. 6th Edition. New York: Thames and Hudson, 2008 Davies, Nigel. The Toltecs: Until the Fall of Tula. Norman: the University of Oklahoma Press, 1987.

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Research Paper Tax Questions Problems Example | Topics and Well Written Essays - 750 words

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Why The Berlin Wall Was Assembled Essay

Why The Berlin Wall Was Assembled Essay To fully understand why The Berlin Wall was assembled, one must know of the events that took place. This wall for 28 years separated families, friends and a nation. Perhaps the major reasons it was assembled were for political and economical issues. Politically, the West side was interfering with the Russian Sector (East side). Economically, all of the citizens from East Germany were getting well educated there and moved to West Berlin for work. In this paper, I will explain the events and circumstances that led to the construction of the Berlin Wall. After the World War II in 1945, the Nazi Germany surrendered, the 4 allied countries, the United States, Great Britain, France and Russia signed the Potsdam Agreement treaty which determined the borders for Germany and Berlin. The Potsdam Agreement divided Germany and Berlin into four administrative zones. 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The citizens of East Berlin were no longer allowed to enter West Berlin, including the sixty-thousand workers who worked in West Berlin. However East Berliners still managed to get out through bribery, cigarettes and money. After some people still managed to scale the wall, there was a ban on the sale of rope and twine. On September 20, 1961, to begin construction on the second more permanent concrete wall, the GDR demolished all of the houses near the wall. The Berlin Wall consisted of 67 miles of concrete segment wall which was four meters high, 42 miles of wire mesh fencing, 65 miles of anti-vehicle trenches, 79 miles of signal fence, 302 watchtowers, and 20 bunkers. (Taylor 2007). There behind all of that was a second wall which was called no mans land or death strip. It cut off one hundred-ninety two streets (Taylor 2007). This area made it easy to spot footprints because of the raked gravel; was mined and booby-trapped with tripwires and it offered a clear field of fire to the armed guards who were instructed to shoot on sight. The main crossing point for the American sector of West Berlin was at checkpoint Charley which was six hundred-eighty feet west of the Brandenburg Gate. On October 27, 1961, the United States sent tanks; jeeps and soldiers to Checkpoint to guarantee entrance of US offic ials to West Berlin (Berlin 2002). The wall divided Berlin through the center and the outer part of the city and on the border between West and East Germany, from the Baltic Sea southward through the center of Germany all the way to Hildburghausen. From there it went east toward the border of Czechoslovakia (Taylor 2007). While the wall was being constructed, the United States was opposed to the establishment of the Wall. President John F. Kennedy was crucial to the cause, declaring his commitment with the infamous words: As a free man, I take pride in the words Ich bin ein Berliner (I am a Berliner) (Taylor 2007). At the verge of a nuclear war, the United States and the Soviet Union reached a conclusion, but the Berlin Wall remained but by the mid 80s the relationship between the East and West Germany began to transform. The end of the German Democratic Republic and the Berlin Wall began when Hungary opened its doors to the west. Passage between Communist states was unrestricted; therefore, East Germans could go from East Germany to Hungary and from there to West Germany or any other Western European state. East Germany began to reform. Gunter Schakowsky, the leader of the East Berlin communist party announced on November 9, 1989 that the border to West Berlin would be opened for private trips out of the country. Shortly after his announcement, citizens began hammering and using chisels to knock out pieces of the wall. The Wall had fallen (Taylor 2007). Between November 10, 1989 and later on December 22, 1989 checkpoints were opened for pedestrians at Potsdamer Platz and the Brandenburg Gate. Finally on July 1, 1990 East and West Germany were united and assumed West Germanys old name, The Federal Republic of Germany. All restrictions between East Germany and West Germany were released. The entire wall was taken down (Berlin 2002). In conclusion, the Berlin Wall was erected for political, economical, as a way for the Soviet Union to maintain their communist system and prevent brain drain in East German. These tactics did not improve the situation for East German as the people did not like the communist regime and still found ways to escape. While the erection of The Berlin Wall did not prove to be successful for the Soviet Union; the fall of the wall reunited families, friends and a divided nation back together.

Teacher Interview Report

Actually, this is my second time to do teacher interview. Last time, I went to Shi Pai Junior High School to interview a teacher and asked something about the class management. From the interview, I’ve learned a lot from the teacher and her belief of teaching Chinese. Though to certain degree, this time the report is quite similar to the last report that I’ve done. However, this time, by the demand of teacher Alice and the class objective, my target interviewee must be an â€Å"English† teacher. At first, I was very nervous about how to find an English teacher in junior or senior high school because I’ve got in touch with all my English teachers for several years. it may be very embarrassing to go back and the teacher doesn’t recognize that I used to be her student) Fortunately, one of my best friends in my night school class, knowing my worriment, introduced me an English teacher in Shilin High School of Commerce whom she loves and admires very much. Here, I want to say thanks to my dear friend. Without her, I really cannot finish writing this assignment. Below, there are some basic info of the teacher and the questions that I’ve prepared for the teacher interview,