2D and 3D Sequences.
2D and 3D Sequences Project Plan of Investigation In this experiment I am going to require the following: A calculator A pencil A pen Variety of sources of information Paper Ruler In this investigation I have been asked to find out how many squares would be needed to make up a certain pattern according to its sequence. The pattern is shown on the front page. In this investigation I hope to find a formula which could be used to find out the number of squares needed to build the pattern at any sequencial position. Firstly I will break the problem down into simple steps to begin with and go into more detail to explain my solutions. I will illustrate fully any methods I should use and explain how I applied them to this certain problem. I will firstly carry out this experiment on a 2D pattern and then extend my investigation to 3D. The Number of Squares in Each Sequence I have achieved the following information by drawing out the pattern and extending upon it. Seq. no. 1 2 3 4 5 6 7 8 No. Of cubes 1 5 13 25 41 61 85 113 I am going to use this next method to see if I can work out some sort of pattern: Sequence Calculations Answer =1 1 2 2(1)+3 5 3 2(1+3)+5 13 4 2(1+3+5)+7 25 5 2(1+3+5+7)+9 41 6 2(1+3+5+7+9)+11 61 7 2(1+3+5+7+9+11)+13 85 8 2(1+3+5+7+9+11+13)+15 113 9 2(1+3+5+7+9+11+13+15) +17 145 What I am doing above is shown with the aid of a
A function is a relation between a set of input values
Initial Formative Assessment Exercise A function is a relation between a set of input values (the domain) and the output values (the range). Each input must map to a single output. Let s(x) be the function defined by being equal to the last digit of the value x2 s : x ? (last digit of) x2 x ? Z values I will be investigating the output values (the range) of this function and without a calculator, be able to find the output with any integer of x. Furthermore the function above will be modified i.e. s(kx) where k is between 2 and 10. And the ranges identified. s(7) = 9 To explain why s(7) = 9 firstly we have to understand the function. The function is defined as if you choose any value (of which is a whole number) for x, , you need to square this number, and after doing so, the value of the function is the last digit of this number. 72 = 49 The last digit of 49 = 9 ? s(7) = 9 Further calculations (without a calculator) s(8) = 4 s(10) = 0 s(345) = 5 s(27,560) = 0 s(738,954,683,012) = 4 Using mental arithmetic I can calculate 82 and 102 quite easily. However calculating 738,954,683,0122 is very challenging to say the least. So how can I be confident that my answers are correct? Using the Arabic-Hindu numeral system of hundreds, tens, units etc. and the following simple theory of multiplication I am confident of my answers being correct. Using a three
2D cross shape - Rearranging the Shape
Formulas 2D cross shape Rearranging the Shape I thought by rearranging the shapes individual squares I might be able to come up to a solution for the problem. I may be able to produce a pattern of squares or any other kind of shapes to show a formula for the crosses Firstly I found that the largest sized squares I could make from the cross shape above were 3?3 Then with the squares I had left I could make a rectangle 2?3 But to keep this like the first part I had all ready found I will display it as Then I only had 1 square left Then I looked at each of the shapes I produced in their relation to n or what term the cross shape was in the sequence. The cross shape I made my shapes from was 4th in the sequence the side of the squares were 3?3 and the length of the 2 rectangles were 3 all the measurements I had found were one less than n, all except from the 1 square by itself An equation I could use for one of the squares was +(n-1)² For one of the rectangles was +(n-1) And for the single square by itself was +1 When I put this together I got (n-1)²+(n-)+1 But I had 2 squares and 2 rectangles so then I got 2(n-1)²+2(n-1)+1 Then I simplified this equation: 2(n-1)²+2(n-1)+1 2(n-1)(n-1)+2(n-1)+1 2(n-1)(n-1)+2n-2+1 2(n²-n-n+1)+2n-2+1 2n²-4n+2n+2-2+1 2n²-2n+1 I had simplified the equation as far as I could so now I could apply it to the shapes On the
The Arrhenius Equation and Activation Energy
The Arrhenius Equation and Activation Energy By Jennifer Partenio-Thrasher Course: CM104 Date & Time: Tues. February 4, 2003, 11:20 am- 2:40 pm Lab Partners: Bobbi Weber Tim Hall The Arrhenius Equation and Activation Energy Introduction: In this lab, our task is to determine how different degrees of temperature affect reaction rates. A Swedish chemist, named Svante Arrhenius discovered the relationship between temperature and reaction rate. In finding this relationship a new equation was formed called the Arrhenius Equation: Reaction rate constant k = A e -Ea/RT The factor A represents the frequency of collisions between two molecules in the proper orientation for reactions to occur. The value of A is determined by experiment and will be different for every reaction. The value of the exponential term e -Ea/RT describes the fraction of molecules with the minimum energy required to react, R is the gas constant, 8.314 J/mol-K, T is the temperature in Kelvin and Ea is the activation energy. Activation energy of a reaction is the minimum amount of energy needed to start the reactions. In order to understand and make use of this equation, we must include the Collision Theory. Collision Theory states that in order for a reaction to occur, two molecules must collide in the proper orientation and posses a minimum amount of energy to react. The Arrhenius equation
Mathematics Coursework - Beyond Pythagoras
Mathematics Coursework - Beyond Pythagoras The numbers 3,4 and 5 satisfy the condition 3² + 4² = 5² Because 3² = 3 x 3 = 9 4² = 4 x 4 = 16 5² = 5 x 5 = 25 And so 3² + 4² = 9 + 16 = 25 = 5² This condition also applies to the following sets of numbers (smallest number)² + (middle number)² = (largest number)² The numbers 5, 12 and 13 satisfy the condition 5² + 12² = 13² Because 5² = 5 x 5 = 25 12² = 12 x 12 = 144 13² = 13 x 13 = 169 And so 5² + 12² = 25 + 144 = 169 = 13² This condition also applies to the following sets of numbers (smallest number)² + (middle number)² = (largest number)² The numbers 7, 24 and 25 satisfy the condition 7² + 24² = 25² Because 7² = 7 x 7 = 49 24² = 24 x 24 = 576 25² = 25 x 25 = 625 And so 7² + 24² = 49 + 576 = 625 = 25² The numbers 3, 4 and 5 can be lengths of the sides of a right-angled triangle: The perimeter and area of this triangle are: Perimeter = 3 + 4 + 5 = 12 units Area = 1/2 x 3 x 4 = 6 square units The table below shows the area and perimeter of the (5, 12, 13) and (7, 24, 25) right-angled triangles. Length of shortest side Length of middle side Length of longest side Perimeter Area 3 4 5 2 6 5 2 3 30 30 7 24 25 56 84 To work out the perimeter I added all the sides together, and to get the area I multiplied the shortest length with the middle length then
Concept of Limit in Real Analyses
University of North London School of Education Concept of Limit in Real Analyses by Antonina Sculthorp Course BA/PGCE Secondary Mathematics Year: 1 Module MI 104 Seminar Tutor: M. Karamanou Course tutor: S. Woodage LONDON 2002 Introduction There is probably no other instance in human intellectual history in which so much time and effort was spent merely to reach a satisfactory definition as that for limit. The concept is very closely related with two other fundamental concepts of mathematics and exists alongside both infinity and continuity. The Greek scholars were the first who seriously considered problems of continuity and infinity based on the concept of 'number'. Various attempts were made by them to include the concept of number into geometry. Being able to construct a line segment of any rational m/n length (m,n, the Greeks discovered around 400 B.C. that a diagonal of a unit triangle is an irrational number, which falls out of the number concept. Questions of continuity and infinity seem to have represented a complete mystery to the Greek scholars. The difficulties were clearly indicated by the famous Paradoxes of Zeno, of which it is worth quoting the following: What is the fallacy of the paradox of Achilles and the tortoise? Using concepts established some 2,500 years after Zeno, here is the explanation why Achilles can finally catch up to and pass the
Investigating Hidden Faces.
Investigating Hidden Faces Introduction: My investigation is based on Hidden faces. A hidden face in this investigation is the side of a cube resting on a surface, and also the sides of a cube joining one another. The diagram above has 33 hidden faces, 9 underneath the block (as shown in red), and 24 in between (shown in blue). A seen face in this investigation is a face of a cube which can be seen without disturbing the given block, but you can walk around it. In other words, it is not just what is visible in the diagram, but what is behind as well. Here is a more detailed view of hidden faces. 1. 2. 3. In this investigation I will be looking for patterns between different sized blocks. Using these Patterns I will find many formulae. Using the various formulae, I will try to make formulae to work out any sized block, whether it might be 2x3x4 e.t.c... Planning; Aim: My main aim is to find a formula for working out the number of hidden faces in any XxYxZ cube or cuboid. Method: I will start with simpler orientations of Blocks such as 1x1x1. I will then find the number of hidden faces for that block. From this I will add on an extra cube each time and observe how the pattern emerges. I will then draw a chart, which shows the number of hidden faces. Using the pattern I will make a formula. I will use this information to find an overall formula, which will work
The cubic volume is defined to be 0.478 nm along each edge.
Andy Somody 97300-6222 ENSC 330 Assignment #3 ). The cubic volume is defined to be 0.478 nm along each edge. Since the unit cell has a cubic configuration, the volume of the unit cell can be determined by the following formula: Since each of the sides of the unit cell are defined to be 0.478 nm in length, the volume of the unit cell of CaO can be calculated to be: with significant figures applied The question also states that the cubic volume of CaO contains four Ca2+ ions and four O2- ions. The mass of the unit cell contributed by the ions is given by the following formula: Therefore, the contribution to the mass of the unit cell from each of the Ca2+ and O2- ions is given by modifying the above formula in the following manner. The mass of one mole of Ca2+ ions is equal to the atomic mass of calcium. The atomic mass of calcium is given as 40.08 g/mol. Similarly, the mass of one mole of O2- ions is equal to the atomic mass of oxygen. The atomic mass of oxygen is given as 15.999 g/mol. The number of atoms or ions in one mole of any element (or the number of molecules in one mole of any compound) is defined to be equal to Avogadro's number (Van Vlack, 1989). Avogadro's number states that a mole of any element possesses 6.02 x 1023 atoms or ions of that element (Van Vlack, 1989). Substituting these values into the above equations, we can calculate the contribution to
Investigating the affect that Body Mass Index and Waist-to-hip Ratio has on the Perception of Female Attractiveness.
Investigating the affect that Body Mass Index and Waist-to-hip Ratio has on the Perception of Female Attractiveness. Abstract The perception of the attractiveness of women is influenced by two major contributing factors the Waist-to-Hip ratio (WHR) and the Body Mass Index (BMI). It was concluded that at first, WHR was the more influential of the two, however after further investigation and more accurate regression analysis BMI was the more influential. However, together they only resulted in under 50% of the variance, suggesting that there are many other contributing factors that influence the perception of female attractiveness. Introduction The Body Mass Index (BMI) is a ratio between the height and weight of a person, calculated by the weight of the person (in kilograms) by their height squared (in metres). Waist-to-Hip Ratio (WHR) is considered to be the factor that signals female fertility and health. This is because healthy, premenstrual women deposit fat on their lower body parts resulting in a feminine characteristic, whereas males deposit fat on their upper body parts. When considering the influence of WHR, alone, on attractiveness lower values of the WHR are considered to be more attractive, with values between 0.6-0.7 being maximally attractive. Attractiveness is not only based upon the WHR but also on the BMI. Females with very low BMI values (underweight) and
Pythagoras’ Theorem.
Pythagoras' Theorem Pythagoras Theory is a relatively simple theory used continuously in Standard Grade Mathematics and beyond. It is also used in Physics. It is used not only to simply solve the missing side of a right-angled triangle but also more extensively to solve Reasoning and Application problems and also can be used to solve many higher mathematics problems in trigonometry and in many topics throughout the mathematics syllabus. It is so basic that I'm sure anyone who studied it at school will remember it long after other theorems have been forgotten. Early evidence of the theory can be traced back as far as 2000 BC with the ancient Egyptians. Pythagoras had travelled to Egypt and this may have influenced some of his beliefs. There is some evidence that they used a 3-4-5 triangle to form a perfect right angle. However very little information pre-dates the Greeks so this remains a mystery along with many other ancient Egyptian stories. Hence the Theorem was credited to Pythagoras. What is more likely is that Pythagoras was the first to prove it. Pythagoras then generalised it to all right-angled triangles hence it is Pythagoras' theorem. Pythagoras was born on the island of Samos in around 582 B.C and died in around 500 BC He was a Greek philosopher and mathematician. He had been instructed in the teachings of the early Ionian philosophers Thales, Anaximander and