value of r that minimizes this by taking the derivative‚ stetting it equal to 0‚ and solving for r. Use that to find h. You’ll find that the dimensions are different from an actual soda can‚ but I’m sure you can think of why this is the case. THE MATH PROBLEM: The surface area of a cylindrical aluminum can is measure of how much aluminum the can requires. If the can has a radius r and a height h‚ its surface area A and its volume V are given by the equations: A=2(pi)r^2 + 2(pi)rh and V=
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candidates sitting the Year 7 Entrance Tests will automatically be considered for an Academic Scholarship; parents do not need to make a separate application. Year 9 Entry Assessment is made on the basis of three written exam papers in English‚ Maths and Science which are designed to enable candidates to show flair. Each paper lasts one hour. The papers all develop National Curriculum areas which are relevant to the age of entry. Applicants for the Academic Scholarships will come to Bethany
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Write your name here Surname Other names Centre Number Candidate Number Edexcel GCSE Mathematics A Paper 2 (Calculator) Foundation Tier Thursday 8 November 2012 – Afternoon Time: 1 hour 45 minutes Paper Reference 1MA0/2F You must have: Ruler graduated in centimetres and millimetres‚ protractor‚ pair of compasses‚ pen‚ HB pencil‚ eraser‚ calculator. Tracing paper may be used. Total Marks Instructions Use black ink or ball-point pen. Fill in the boxes at the
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Coursework 2 Mathematical Finance Group 27 Q1. Hedging in Complete and Incomplete market Solution: Complete market Suppose we have m states. A complete market A is one with the marketed subspace Span(A.1‚A.2‚ ⋯‚ A.n) includes all possible payoffs over the m states‚ i.e.‚ if it contains all possible m-dimensional vectors. Incomplete market Suppose we have m states. An incomplete market corresponds to a market with fewer linear independent
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Maths Project Class 9 PROJECT WORK: Creative Mathematics Project Ideas General Guidelines: * Each student is required to make a handwritten project report according to the project allotted Please note down your project number according to your Roll Number. Roll Number | Project Number | 1-5 | 1 | 6-10 | 2 | 11-15 | 3 | 16-20 | 4 | 21-25 | 5 | 26-30 | 1 | 31-35 | 2 | 36-40 | 3 | 41-45 | 4 | 46-50 | 5 | * A project has a specific starting date and an end date. *
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Fall 2013 Bldg 2 Room 247 MATH 111 SYLLABUS College Algebra TIME: Mon‚ Wed 12:00 – 2:20 PM Office: CRN#44230 CREDITS: 5 INSTRUCTOR: Jerry Kissick OFFICE HOURS: Mon‚ Wed COURSE TEXT: College Algebra and Trigonometry‚ Custom Edition for Portland Community College‚ Sullivan and Sullivan PREREQUISITES: MATH 95 completed with a C or better and placement into WR 121. 11:30 – 12:00 PM 2:30 – 3:00 PM 3:00 – 4:00 PM 5:30 – 6:00 PM Bldg 2 Room 244C Phone
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1 Asgn 3 Macao Polytechnic Institute School of Business Bachelor of Art in Public Relation Programme st 1 Semester‚ Year 2012/13 Math211521122‚ Assignment III P1103808 Leong Lai Nga‚ Delia Write down your full name and student ID on the first page. Handin your assignment on or before nd the day of 2 midterm. 1. Blood Glucose Level . In the article "Drinking Glucose Improves Listening Span in Students Who Miss Breakfast" ( Educational Research ‚ Vol. 43‚ No. 2‚ pp.201207)
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Item 4B Item 4B Rachel Reiser Maths C Rachel Reiser Maths C Question 1 ab1+f’(x)2 dx y = acosh(xa) If: coshx=12ex+e-x Then: cosh(xa) = 12(exa+e-xa) y = acosh(xa) ∴ y=a(exa+e-xa)2 y=a(exa+e-xa)2 dydx=f’x=ddxa(exa+e-xa)2 dydx=f’x=ddx12aexa+e-xa f’x=12a1aexa+-1ae-xa f’x=exa-e-xa2 f’x2=exa-e-xa22 f’x2=(12exa-12e-xa)(12exa-12e-xa) f’x2=14e2xa-14e0-14e0+14e-2xa f’x2=14e2xa-12+14e-2xa f’x2=14e2xa-2+e-2xa Assuming the catenary is symmetrical‚ the entire length of
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Probability and Statistics Research Project Name: Lakeisha M. Henderson ID: @02181956 Spring 2007 Abstract Table of Contents Principle Component Analysis (PCA) Definition .4 Uses of PCA 5 Illustrative Example of PCA 5 Method to Determine PCA ..6 Basic Analysis of Variance (ANOVA) Purpose and Definition of ANOVA 12 Illustrative Example of ANOVA
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MATH PORTFOLIO NUMBER OF PIECES Kanishk Malhotra 003566-035 (May 2012) In physics and mathematics‚ the ‘DIMENSION’ of a space or object is informally defined as the minimum number of coordinates needed to specify each point within it. Thus a line has a dimension of one because only one coordinate is needed to specify a point on it. A surface such as a plane or the surface of a cylinder or sphere has a dimension of two because two coordinates are needed to specify a point on it (for
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