IA Task I Introduction and purpose of task: The purpose of this task is to investigate the positions of points in intersecting circles and to discover the various relationships between said circles. Circle C1 has center O and radius r. Circle C2 has center P and radius OP. Let A be one of the points of intersection of C1 and C2. Circle C3 has center A and radius r (therefore circles C1 and C3 are the same size). The point P’ (written P prime) is the intersection of C3 with OP. This is shown in
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[pic] A parallelogram is a quadrilateral in which pairs of opposite sides are parallel and are congruent. Opposite sides are parallel and equal in length‚ and opposite angles are equal (angles "a" are the same‚ and angles "b" are the same) NOTE: Squares‚ Rectangles and Rhombuses are all Parallelograms! Name the kind of parallelogram this figure displays? Example 1: [pic] |[pic] |A parallelogram with: | |
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EUROPEAN SCHOOL Mathematics Higher Level Portfolio Type 1 SHADOW FUNCTIONS Candidate Name: Emil Abrahamyan Candidate Number: 006343-021 Supervisor: Avtandil Gagnidze Session Year: 2013 May Candidate Name: Emil Abrahamyan Candidate Number: 006343-021 Mathematics Higher Level Type 1: Shadow Functions SHADOW FUNCTIONS The Aim of the Investigation: The overall aim of this investigation is to investigate different polynomials with different powers and create shadow function
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Sullivan’s Handbags marks up their bags at 45% of the selling price. Pat Sullivan saw a bag at a trade show that she would sell to her customers for $85. What is the most she could pay for the bag and still retain the 45% markup of the selling price? 6. Jeff Jones earns $1‚200 per week. He is married and claims four withholding allowances. The FICA rate is as follows: Social Security rate is 6.2% on $97‚500; Medicare rate is 1.45%. To date his cumulative wages are $6‚000. Each paycheck‚
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1. Solve a. e^.05t = 1600 0.05t = ln(1600) 0.05t = 7.378 t = 7.378/.05 t = 147.56 b. ln(4x)=3 4x = e^3 x = e^3/4 x = 5.02 c. log2(8 – 6x) = 5 8-6x = 2^5 8-6x = 32 6x = 8-32 x = -24/6 x = -4 d. 4 + 5e-x = 0 5e^(-x) = -4 e^(-x) = -4/5 no solution‚ e cannot have a negative answer 2. Describe the transformations on the following graph of f (x) log( x) . State the placement of the vertical asymptote and x-intercept after the transformation. For example‚ vertical shift
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Centre Number For Examiner’s Use Candidate Number Surname Other Names Examiner’s Initials Candidate Signature Pages General Certificate of Secondary Education Higher Tier June 2014 Mark 3 4–5 6–7 Mathematics (Linear) 4365/1H H Paper 1 Monday 9 June 2014 9.00 am to 10.30 am For this paper you must have: 8–9 10 – 11 12 – 13 14 – 15 16 – 17 mathematical instruments. 18 – 19 You must not use a calculator 20 – 21 Time allowed 1 hour 30 minutes 22 – 23 TOTAL Instructions Use black
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Treasure Hunt: Finding the Values of Right Angle Triangles This final weeks course asks us to find a treasure with two pieces of a map. Now this may not be a common use of the Pythagorean Theorem to solve the distances for a right angled triangle but it is a fun exercise to find the values of the right angle triangle. Buried treasure: Ahmed has half of a treasure map‚which indicates that the treasure is buried in the desert 2x + 6 paces from Castle Rock. Vanessa has the other half of the map
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Module 8 Business Decisions Capital Gains Page 705‚ question 30 30A- How much tax will you have saved by waiting? $1‚250 $25‚000 X .10 = $2‚500 $25‚000 X .15 = $3‚750 $3‚750 - $2‚500 = $1‚250 30B- How much would you save in 36% bracket? Between $2‚000 to $4‚400 $25‚000 X .20 = $5‚000 $25‚000 X .28 = $7‚000 to $9‚900 $7‚000 - $5‚000 = $2‚000 $9‚900 - $5‚000 = $4‚400 Interpreting the numbers Page 743‚ Question 20 2‚300 2‚430‚ 2‚018‚ 2‚540‚ 2‚675‚ 4‚800
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Centre Number For Examiner’s Use Candidate Number Surname Other Names Examiner’s Initials Candidate Signature Pages General Certificate of Secondary Education Foundation Tier November 2013 Mark 2–3 4–5 6–7 Mathematics 43601F Unit 1 Wednesday 6 November 2013 9.00 am to 10.00 am For this paper you must have: l mathematical instruments. 10 – 11 12 – 13 14 – 15 16 – 17 a calculator l F 8–9 TOTAL Time allowed l 1 hour Instructions
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1998 9 14 1. 1.1 Markov Property 1.2 Wiener Process 1.3 2. 2.1 2.2 2.3 2.4 2.5 2.6 Taylor Expansion 2.7 3. Stochastic 3.1 3.2 SDE(Stochastic Differential Equation) 4. Stochastic 4.1 Stochastic integration 4.2 Ito Integral 4.3 Ito Integral 4.4 5. Ito’s Lemma 5.1 Stochastic 5.1.1 5.1.2 5.1.3 First Order Term Second Order Term Cross Product Terms “ ” – Ito Integral Riemann (Ordinary Differential Equation) (Chain rule) 5.2 Ito’s Lemma 6. 6.1 6.1.1 6.1.2 Closed-Form Solution Numerical Solution
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