The Straight Line

The Straight Line Slope-Intercept Form The slope intercept form is probably the most frequently used way to express the equation of a line. The equation can be written in many different ways1, but taken we are in Denmark and are part of a Danish school the equation would be: Where: The slope-intercept form is a type of linear equation. A linear equation is simply an algebraic equation in which each term is either a constant (fixed number) or the product of a constant and (the first power of) a single variable. Y-intercept The Y intercept of a straight line is simply where the line crosses the Y axis, thus it requires no calculation to find. Examples . Find the y-intercept for the following equation. * 2. Find the y-intercept for the following straight line. Slope/Gradient The slope (gradient) ultimately determines the 'steepness' or incline of a line, the higher the slope, the steeper the incline will be. For example, a horizontal line has a slope equal to zero while a line with an angle of 45o has a slope equal to one. The sign (positive or negative) of the slope is very important, as it determines whether the line slopes uphill or downhill. Positive slopes (like the ones above) mean that the line slopes uphill, from left to right. Meanwhile if negative, the line slopes downhill, from right to left. The slope of a line is usually represented as the letter 'a',

  • Word count: 685
  • Level: International Baccalaureate
  • Subject: Maths
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IB Math SL type II Portfolio - Body Mass Index

MATHEMATICS INTERNAL ASSESMENT SL TYPE II Body Mass Index Escaan International School 8/10/2012 ________________ BODY MASS INDEX In this investigation, I will analyse various functions using the data given. A model function will be reached in base with this real-life example: Body mass index (BMI) is a measure of one’s body fat. It is calculated by taking one’s weight (kg) and dividing it by the square of one’s height (m). The table below provides the median BMI for females of varying ages in the US in the year 2000. Age BMI 2 16.40 3 15.70 4 15.30 5 15.20 6 15.21 7 15.40 8 15.80 9 16.30 10 16.80 11 17.50 12 18.18 13 18.70 14 19.36 15 19.88 16 20.40 17 20.85 18 21.22 19 21.60 20 21.65 (Source: http://www.cdc.gov) ________________ When graphed: The independant variable is the age of the women <x>, and the dependant variable is the BMI of these women <y>. Both values have to be >0. The behaviour of this graph can be modelled by the sine function , or by the cosine function as it is periodic and ondulating, which means it repeats a pattern as it goes up and down. Other functions can’t be used in this model as the information given can be reflected in other type of graphs. Finally, we can deduce that it will be a cosine function as the portion that we observe in tha graph doesn’t go through the origin (0,0), as the

  • Word count: 683
  • Level: International Baccalaureate
  • Subject: Maths
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infintite surds

Alice Wang Infinite Surds IB Portfolio An infinite surd is a number that can only be expressed exactly using a root sign. Surds are useful because they can represent irrational numbers that have an infinite number of non-recurring decimals. This expression is known as an infinite surd: It can be considered as a sequence of terms an, where .....etc. It can be inferred from the first ten terms of the sequence that the relation of two consecutive terms is . Figure 1. Relationship between n and an n .414213562 2 .553773974 3 .598053182 4 .611847754 5 .616121207 6 .617442799 7 .617851291 8 .617977531 9 .618016542 0 .618028597 From Figure 1, it can be inferred that the sequence of the two following terms as: . The values of an, are not exact because the values of an, have an infinite number of decimal points since it is a surd. When the value of n increases, the difference between two consecutive values of n are smaller than Graph 1 Graph 1 exemplifies the direct relationship between an and n. As the value of n increases, the value of an increases. In other words, as n gets larger, the value of gets smaller. When n gets larger and gets closer to infinity, an increases at a slower rate (ex:). However, the difference between an and n will always be larger than 0 in terms of and can be represented by the equation an- an+1 > 0. Taking another

  • Word count: 674
  • Level: International Baccalaureate
  • Subject: Maths
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Infinite Surds. As we can see there are ten terms of this sequence where is the general term of the sequence when, is the first term of the sequence.

Math Portfolio Infinite Surds Introduction a surd is an irrational number that can not be written as a fraction of two integers but can only be expressed using the root sign. Bellow an example of an infinite surd: This surd can be turned into a set of particular numbers sequence: 1.414213562 .553773974 .598053182 .611847754 a51.616121207 .617442799 .617851291 .617977531 .618016542 As we can see there are ten terms of this sequence where is the general term of the sequence when, is the first term of the sequence...Etc. A formula has been defined for in terms of: (1) A graph has been plotted to show the relation between and. And it can be oticed that as long as gets larger, gets closer to a fixed value. To investigate more about this fixed value we take this equation into consideration as n gets bigger. n an-an+1 - 0.13956 2 -0.04428 3 -0.01380 4 -0.00427 5 -0.00132 6 -0.00040 7 -0.00013 8 -0.00004 9 -0.00001 We can figure out from the table above that when n gets larger, the term (an+an+1) gets closer to zero but it never reaches it So we can come to the conclusion: When n approaches infinity, lim (an-an+1) ›0 An expression can be obtained in the case of the relation between n and an to get the exact value of the infinite surd: = x If we apply formula (1) to this: › x2=1+x › The equation can be solved using the

  • Word count: 663
  • Level: International Baccalaureate
  • Subject: Maths
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Derivatives of Sine

Mark Andrew Garner Judy Land Math Standard Level April 30, 2008 Derivatives of Sine Functions Method . The following information was my investigation to find the derivative of the function. a) The following is a graph for the function for. TI - 83 Graph Window Xmin = - Xmax = Xscl = Ymin = -5 Ymax = 5 Yscl = b) The table below describes the behavior of the gradient of for. Gradient behavior in given points of x Y Gradient (+, , or 0) 0 + 0 0 0 0 0 + 0 0 0 0 + c) Because the derivative of a function is the gradient at a given point, my conjecture for is as follows: TI - 83 Graph Window Xmin = - Xmax = Xscl = Ymin = -5 Ymax = 5 Yscl = nDeriv(sin(x)) at a given X-value 0 0 0 0 0 Relationship of dy/dx of f(x)=sinx versus Y-value of f(x)=cosx at a given X-value X-Value dy/dx f(x)=sinx f(x)=cosx 0 0 0 0 0 0 0 0 0 TI - 83 Graph Window Xmin = - Xmax = Xscl = Ymin = -5 Ymax = 5 Yscl = The tables on the preceding page show that the derivative of at the following plotted points in increments of starting left to right x =. In the following graph, the points are connected to form , . TI - 83 Graph Window Xmin = - Xmax = Xscl = Ymin = -5 Ymax = 5 Yscl = 2. The following is my investigation of the derivatives of functions in the form. The red dots in the following graphs below represent the

  • Word count: 644
  • Level: International Baccalaureate
  • Subject: Maths
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Investigation Transformations.

Investigation Transformations. Part B: Investigation # 1 These transformations were of sin graphs. When the a value in y = a sin k (x-b) + d is positive the transformation is vertical and the graphs are stretched or compressed by a factor of a. When a is negative the transformation is inverted or reflected and stretched or compressed vertically by a as seen in y = -3sin(x). The period of all these graphs remains the same and all graphs intersect the origin and intersect the x axis at the same points. The maximums of these graphs are the numbers are the positive of the a values. So in y = 2sin(x), the maximum is 2. The minimum is the negative of the a value, so therefore in y = 2sin(x), -2 would be the minimum. Investigation # 2 These transformations were of cos graphs. In this investigation the k value of the equations were changed. As shown when the k value is greater 1 the graph is compressed horizontally by the factor 1/k as seen in y = cos(2x). When the value of k is greater than 0 but less than 1 like in y = cos(1/2) the graph is stretched horizontally by a factor of 1/k. As you can see all the graphs pass through the y-axis at 1 but the graphs do not pass through the x-axis at the same points. The maximums and minimums of these graphs are all the same because the value was the same in all graphs. From these graphs we can determine that the value of

  • Word count: 643
  • Level: International Baccalaureate
  • Subject: Maths
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Investigating Logarithms

Investigating Logarithms log2 + log3 0.7782 log6 0.7782 log3 + log7 .322 log21 .322 log4 + log 20 .903 log80 .903 log0.2 + log11 0.3424 log2.2 0.3424 log0.3 + log 0.4 -0.9208 log0.12 -0.9208 This table to the left clearly shows that the log of 2 numbers added together will equal the log of the number multiplied. The table below clearly shows that log (?) + log (y) will equal log (?y). Let log x = a, let log y = b. Therefore 10a = x and 10b = y, these two equations can then be simplified to 10(a+b) =x*y. it is then possible to convert this back to log (xy) = a + b. log5 + log4 log20 .301 log3 + log2 log6 0.7782 log4 + log8 log32 .505 log6 + log3 log18 .255 log3 + log26 log78 .892 log7 + log4 log28 .447 log12 - log3 0.6021 log4 0.6021 log50 - log2 .398 log25 .398 log7 - log5 0.1461 log1.4 0.1461 log3 - log4 -0.1249 log0.75 -0.1249 log20 - log40 -0.3010 log0.5 -0.3010 This table to the left clearly shows that the log of 2 numbers subtracted from each other will equal the log of the numbers divided by each other.The table below clearly shows the log () - log () will equal log (). Let log x = a, let log y = b. Therefore 10a = x and 10b = y, these two equations can be converted into 10(a - b) = x/y. Finally, this equation can then be converted back into log x - log y = log (x/y). log6 - log2 log3 0.4771 log18 -

  • Word count: 636
  • Level: International Baccalaureate
  • Subject: Maths
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Infinite summation

) Consider the following sequence of terms un , where U0 = U1 = U2 = U3 = ..... This question is about infinite summation, which is a way of expressing an infinite sum. The series consist of (n) terms where (n) effects the sum. The question challenges us to learn about series on ourselves and help us academically in further tasks / exams. (i) Find an expression for un, in terms of n. In this question, we are asked to find an expression for un , so what we have to do is to find a formula which may be applied to all the terms of the sequence and that will help us understand how (n) effect the terms. First of all, we should start by giving the name un to the formula. Un = Then, we should put in the numerator (ln2) since it is constant in all the terms. Un = (ln2) Because (n) is also the power of the numerator, we should place it in the formula. Un = (ln2)n We should place the denominator next. Because it is the product of (n) with values smaller than (n), but greater than 0, we can express it with permutation. Permutation is the product of (n) terms, defined by (n!). Un = (ln2)n Or Un = (ln2)n (ii) Calculate the summation of the first n terms (Sn) of the above sequence accurate to 6 decimal places for . Now, we are asked to put in the n values according to the given domain ( and calculate the summation. We should start with writing down all the

  • Word count: 632
  • Level: International Baccalaureate
  • Subject: Maths
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Maths Portfolio Matrix Binomials

Maths Portfolio SL Type 1 Matrix Binomials In this mathematics portfolio we are instructed to investigate matrix binomials and algebraically find a general statement that combines perfectly with our matrices and equations given. MATRIX BINOMIALS Given that: and We calculated Therefore: And We were then requested to find expressions for by considering the integer powers of X and Y: These expressions were found by observing that the result of was always the matrix to the power of n multiplied by 2 to the power of n-1. The sequence of results gives us: 1, 2, 4 and 8, reaffirming our expressions are correct because . Given that: and where a and b are constants We were asked to find using different values of a and b and then find the expressions for . Therefore: And: Assuming that a = 2 Assuming that a = 3 Assuming that a = 6 Assuming that a = -4 Assuming that b = -1 Assuming that b = 4 Assuming that b = 5 Assuming that b = 7 To find the expression for I observed that the final results achieved were always the value chosen for a multiplied by the matrix X and the product to the power of n. For example (assuming a = 2): As a result we can see that: And since we know Hence: To find the expression for I did the same as for changing the value a to b and X to Y. For example (assuming b = -1): As a result we can see that: And since we know

  • Word count: 625
  • Level: International Baccalaureate
  • Subject: Maths
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Investigating Ratios of Areas and Volumes

Bob Devan Calculus AB/BC 7th & 8th December 6, 2008 Investigating Ratios of Areas and Volumes .Given the function y=, consider the region formed by this function from =0 to =1 and the -axis. Label this area B. Label the region from y=0 to y=1 and the y-axis area A. A. Find the ratio of area A: area B. The area of A for the given function of is. Area A The area of B for the given function of is Area B: Thus the ratio of area A : area B can be given as 2:1. B. Calculate the ratio of the areas for other functions of the type between 0 and . Make a conjecture and test your conjecture for other subsets of the real numbers. Area A: 1- Area B: Conjecture: Given the function the area of A in ratio of the area of B can be given as n:1. Equation Area of A Area of B Ratio 3:1 4:1 5:1 6:1 2. Does your conjecture hold only for areas between x=0 and x=1? Examine for =0 and =2; =1 and =2 etc. Ex 1) The area of A for the function y=x2 [0,2] is 8/3 Area A: The area of B for the function y=x2 [0,2] is 16/3 Area B: Ex 2) The area of for the function is Area A: The area of B for the function is Area B: Ex 3) Area A: Area B: 3. Is your conjecture true for the general case from to such that and for the regions defined below? If so prove it; if not explain why not. Area A: and the y-axis Area B: and the x-axis Area A: = =

  • Word count: 617
  • Level: International Baccalaureate
  • Subject: Maths
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