Height limitations enforced by bone strength

An Analysis of Height Limitations Enforced by Bone Strength Wouldn't it just be amazing if there was no limit to anything? What if we could eat all we want none stop and nothing would happen? Or what if we could write a paragraph and call it an essay? Yes, but unfortunately our world doesn't work like that. There are limits on everything and anything even our growth. There is a certain height limit that we cannot exceed or else our bones won't be able to support our weight and just tumble down. But what is this incredulous height of which we cannot exceed? Well that is one answer I am ready to find out. We know that that no human can exceed 200 Mpa. Mammals like us have a ratio of 3 to five times the safety factors of the working stress to the breaking stress. It would be ridiculous to use 5 times because a human cannot, as I said before, exceed 200 Mpa, and five times the safety factor would exceed it. So I have chosen to work with 3. Since I'm using 3, I have to create a table that shows when our bone stress reaches 200. I am using an average male which would be 189lbs. and 5'9. The bone stress is measured because half of your weight, when walking, goes to one leg and the other half to the other. Multiplication Factor Height (cm) Weight(lbs) Bone Stress Initial X 75 89 50 Double(2X) 350 378 00 Triple (3X) 700 756 200 In this table we see the

  • Word count: 600
  • Level: International Baccalaureate
  • Subject: Maths
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Maths Portfolio Infinite Surds

Maths Portfolio SL Type 1 Infinite Surds In this mathematics portfolio we are instructed to investigate different expression of infinite surds in square root form and then find the exact value and statement for these surds. INFINITE SURDS The following expression is an example of an infinite surd. The first ten terms of the surd can be expressed in the sequence: From the tem terms of the sequence we can observe that the formula for the sequence is displayed as: The results had to be plotted in a graph as shown below: The graph above shows us the relationship between n and. We can observe that as n increases, also increases but each time less than before, suggesting that at a large certain point of n it stops increasing and just follows in a straight line. To find the exact infinite value for this sequence we would use the equation and rearrange in the correct way to find the value. Finding the exact value for the surd The exact value for the surd is 1.618033989 as the second answer does not fit in the problem. Another example of an infinite surd is: The first ten terms of this surd can be expressed in the sequence: From the tem terms of the sequence we can observe that the formula for the sequence is displayed as: The results had to be plotted in a graph as shown below: The graph above also shows us the relationship between n and. We can observe that as n

  • Word count: 598
  • Level: International Baccalaureate
  • Subject: Maths
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IB math project

. My conjecture, for the function bounded by between 'a' and 'b', is, the ratio of area A to area B is equivalent to the ratio of n to 1. Test . n?+ 2. Fraction 3. Negatives 4. Irrational Numbers ) n?+ n=3 n=5 2) Fractions 3) Negatives For all values of n in, no values exist in the region bounded by 0 and 1; therefore no conjecture can be inferred. 4) Irrational Numbers 2. Bounds 0 to 2 Bounds 1 to 2 TEST Bounds 0 to 2 Bounds 0 to 3 Bounds 2 to 4 My conjecture is valid for all functions of , bounded by 0 and 2, and also 1 and 2. For the function , if the area of A is bounded by and , and the area of B is bounded by and , then my conjecture, of n to 1, holds true for all ratios between areas A and B. For Negative Bounds: Bounds -1 to 0 The conjecture of n:1 is the exact opposite for conjecture of the negatively bounded regions in quadrant III. (1:n) If a set of positive bounds were to be changed to negative, then the ratio between A and B in the positive bounds would be equivalent to that of B and A in the negative bounds. (see diagram explanation below) 3. As mentioned in question 2, my conjecture, of n to 1, holds true for all ratios between areas A and B, if a<b; as long the function in question is . Proof: With the function in question being , the area of B was simply calculated by integrating , however the area of A was calculated by

  • Word count: 597
  • Level: International Baccalaureate
  • Subject: Maths
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Math Portfolio - Weather Analysis

Introduction The purpose of this mathematics assignment is to explore the cosine function transformations with the data collected from the average temperature of a city in Ontario. Using www.weatherbase.com, we found average temperatures from the city of our choice. In my case, I selected Bloomfield, Ontario. The data points collected form a cosine function by nature that is already transformed and translated. Cosine Equation y=AcosB(x - C)+D Breakdown In the function, A represents the amplitude. The amplitude is a vertical dilation of either compression or stretch. It directly affects the size of the wavelengths in the function. In this situation, the amplitude is the range of the temperature over a twelve-month period. By calculating subtracting the maximum and minimum temperatures, you will fall upon the middle value. This value is 13.5. Considering the first values of the graph are in the negatives - obviously because of the fact that January and February are winter months - the A value is going to be negative. A period is defined as the time for one full cycle to take place in the graph. The period of this graph is 12 since the data was collected over the span of a year, or 12 months. In this function, the B-value represents the horizontal dilation. The horizontal dilation is the horizontal stretch or compression. This affects how long it takes for the data to be

  • Word count: 593
  • Level: International Baccalaureate
  • Subject: Maths
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SL IA Type 1: Infinite Summation Portfolio

Example Math Portfolio SL Infinite Summation - Type 1: We being this analysis by investigating series of the type , where tn represents the nth term in the series, x and a are two parameters that may be varied and n! = n X (n -1) X (n -2) X ... X 3 X 2 X 1. Initially we will consider the summation, Sn where Sn means to sum the first n terms of the series. We will find the Sn for the first 11 summations of this series, that is those terms that obey . Using Microsoft Excel we can easily obtain these values which are shown in the table below. N Sn 0 .693147 2 .933374 3 .988878 4 .998496 5 .999829 6 .999983 7 .999999 8 2 9 2 0 2 To further clarify how these values were obtained a sample calculation is shown below: , a screenshot highlighting the formula used in Excel to complete the rest is shown in figure 1 below. Figure 1: Showing how Excel was used to calculate Sn If we now take these values and create a graph of n vs. Sn we obtain the result shown in Figure 2 From this graph it is clearly seen that the summation Sn converges on the value 2. It can be said then that as n approaches infinity the summation converges on the value 2. If we complete this same exercise, but this time change the value of parameter a to equal 3 we obtain the following results for Sn shown as a screenshot for Excel. Again if we create a graph of n versus Sn we obtain

  • Word count: 580
  • Level: International Baccalaureate
  • Subject: Maths
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Essay on Russells paradox and Incompleteness theorem

Neil Sanghrajka Essay on Russell’s paradox and Incompleteness theorem For me mathematics is like a tall building, and from time to time mathematicians create new floors of formulas and theorems. A new floor cannot be built without a previous floor, similarly in Mathematics new theorems rely on deductive reasoning using previous knowledge. Mathematics is an axiomatic system; these “truths” build the foundation of the field. However Russell’s paradox and the incompleteness theorem state that the very foundation of mathematics is inconsistent. Russell’s paradox shows an inconsistency in one of the axioms of the set theory. This example shows how mathematics fails the coherent truth test. The coherent truth test states that the premise for deductive reasoning must be logically consistent. Russell’s paradox gives an example of the incompleteness of mathematics. Gödel’s incompleteness theorem says that any axiomatic system cannot be complete and consistent at the same time. Using deductive reasoning, we may reach a point where a mathematical statement can neither be proved nor disproved. This is a simpler interpretation of the incompleteness theorem. To prove or disprove a statement, axioms are used in the proof. However if there is an inconsistency in the axiom itself, the proof can never be definite. An example is the set of all even natural numbers is infinite,

  • Word count: 573
  • Level: International Baccalaureate
  • Subject: Maths
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Graph Theory review notes.

Graph theory review Graph types Simple: no loops or multiple edges Complete: has every pair of vertices adjacent. One move can get you from one vertex to any other vertex. (n vertices--->Kn) Connected: has every vertex accessible from every other vertex, but not necessarily directly. Bipartite: has vertices in two sets and each edge joins one vertex from each set. Vertices within each set are not joined. Complete Bipartite: has every vertex in one set joined to every other vertex in the other set (m vertices in one set, n vertices in the other--->Km,n) Wheel: has every vertex connected to a central hub. (Wn) Two graphs are isomorphic: Equal size + Equal order + Same Degrees + Connectivity is preserved The complement of a graph will have all of the same vertices but its edges will be all of the possible edges that the original graph does NOT have. Handshaking Lemma: For any graph G, the sum of degrees of the vertices in G is twice the size of G. Pigeonhole Principle: n pigeons in m holes, and n>m, then there must be at least one hole containing more than one pigeon. Planar graph: a graph without any edges crossing each other For graph, G, to be planar: If G is a simple graph: If G is a bipartite graph: Any subgraph that contains K5 or K3,3 will not be planar A connected graph has an Eulerian circuit if and only if ALL of its vertices are even. A

  • Word count: 572
  • Level: International Baccalaureate
  • Subject: Maths
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Discover how to solve inverse functions both graphically and algebraically, whilst investigating their relations, properties and patterns.

Year 11 IB Maths HL - Inverse Functions Mathematical Investigation Graphical Determination of the Inverse Fergal Banks Problem Statement: Whilst completing this investigation, I plan to find out three main aspects about inverse functions, * Can we find the inverse of any given function graphically? * What are the properties of the inverses of some common functions? * Do all Functions have an inverse? Whilst using the method, the function f(x) has an inverse f¯¹ (x) if f (f¯¹ (x)) = x. Method: Discover how to solve inverse functions both graphically and algebraically, whilst investigating their relations, properties and patterns. 2. a) The linear function, f(x) = 2x + 4 is reflected. So the points on the graph, (0,4) become (4,0). b) (3x - 1)/(x + 2) is reflected and becomes, (2x + 1)/(3 - x). This results in a mirror image of the original function. c) f(x) = x³ is reflected resulting in the inverse function, g(x) = ³Vx. 3. Using the linear function, f(x) = 4x + 8, it is clear that my results in Q2 are indeed correct, as they are confirmed by the inverse function of the above linear function. It is flipped resulting in g(x) =. I worked out the inverse function by working out the slope and the y-intercept and then plotting those points on the graph. 4. I used the mirror image of

  • Word count: 566
  • Level: International Baccalaureate
  • Subject: Maths
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Matrix Powers

Matrix Powers . Consider the matrix M = a. Calculate Mn for = 2, 3, 4, 5, 10, 20, 50. M2 = M3 = M4 = M5 = M10 = M20 = M50 = b. Describe in words any pattern you observe. Using a TI-83 Plus to calculate each of these power matrices, we are able to find the th matrix. The matrix provided is an identity matrix, which is uniquely defined by the property In M = M thus the value of Mn is equated to two to the power of as well; the value zero stays constant for other values of . The matrix Mn can also be calculated using geometric sequence: Un = U1rn-1;as the common ratio between each matrix is a factor of two. c. Use this pattern to find a general expression for the matrix Mn in terms of . Mn = I × 2n 2. Consider the matrices P = and S = P2 = 2 = = 2 S2 = 2 = = 2 a. Calculate Pn and Sn for other values of and describe any pattern(s) you observe. P3 = = 4 S3 = = 4 P4 = = 8 S4 = = 8 P5 = = 16 S5 = = 16 P7 = = 64 S7 = = 64 P10 = = 512 S10 = = 512 From a standard matrix , we can see that the difference between and , and is two. Therefore we could infer that it follows the pattern , and that in matrix P is two whereas in matrix S is three. However, a coefficient had to be factorised in order to have that difference of two. A pattern for the coefficient was found as 2n-1. 3. Now consider matrices of the form . Steps

  • Word count: 565
  • Level: International Baccalaureate
  • Subject: Maths
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Matrix power. The internal assessment will focused on observing patterns of matrix powers which will be the main key to find the general expression of matrix powers.

Internal Assessment- Matrix Powers The internal assessment will focused on observing patterns of matrix powers which will be the main key to find the general expression of matrix powers. ) Explanation: When matrix M is powered by 2 it gives a result offrom GDC, when M is powered by 3 it gives a result of from GDC and when M is powered by 4 it gives a result of from GDC. The pattern shown is that every time M is powered by a number after its preceding number it is multiply by 2. For instance, 4 shown in matrix, 8 shown in matrix, and 16 shown in matrix . Thus, these results make a general expression that. This formula is proven correct because when from GDC and this formula also works whenfrom GDC. 2) Explanation: When matrix P is powered by 3 it gives a result of , when matrix P is powered by 4 it gives a result of, and when matrix P is powered by 5 it gives a result of . The pattern shown is that results have a common factor of such as 4 shown in matrix, 8 shown in matrix, and then16 shown in matrix. Once the resulting matrix is factor out, the left over numbers inside the matrix has a general pattern of either such as 9 when n = 3, 17 when n = 4 and then 33 when n = 5 or such as 7 when n = 3, 9 when n = 4 and then 31 when n = 5. Thus, these results make a general expression that. This formula is proven correct because when from GDC and this formula also works

  • Word count: 553
  • Level: International Baccalaureate
  • Subject: Maths
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