Properties of Plane Areas Notation: A area x‚ y distances to centroid C Ix‚ Iy moments of inertia with respect to the x and y axes‚ respectively Ixy product of inertia with respect to the x and y axes IP Ix Iy polar moment of inertia with respect to the origin of the x and y axes IBB moment of inertia with respect to axis B-B 1 y x h C b y x Rectangle (Origin of axes at centroid) A Ix bh bh3 12 x Iy b 2 hb3 12 y h 2 Ixy 0 IP bh 2 (h 12 b2) 2 y B Rectangle (Origin of axes at corner)
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quadratic form Q = 6 x 2 + 3 y 2 + 3z 2 − 4 xy − 2 yz + 4 zx into canonical form by an orthogonal transformation. 2 2 2 13. Reduce the quadratic form 8 x1 + 7 x2 + 3x3 − 12 x1 x2 − 8 x2 x3 + 4 x3 x1 to the canonical form by an orthogonal transformation and hence show that it is positive semi-definite. 2 2 2 14. Reduce the quadratic form x1 + 5 x2 + x3 + 2 x1 x2 + 2 x2 x3 + 2 x3 x1 to the canonical form by an orthogonal transformation 15. Reduce the quadratic form x 2 + y 2 + z 2 − 2 xy − 2 yz − 2 zx
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NLSIU-BANGLORE All_India_Gen Y 1 2 130210092 IPSHITA BHUWANIA 28/07/1994 F N GENERAL 4 OTHERS 1 3 NLSIU-BANGLORE All_India_Gen Y 2 3 161210065 SHUBHAM JAIN 24/2/1994 M N GENERAL 4 OTHERS 1 3 NLSIU-BANGLORE All_India_Gen Y 3 4 161010573 RITWIK BHATTACHARYA 2/9/1995 M N GENERAL 4 U.P. 1 3 NLSIU-BANGLORE All_India_Gen Y 4 5 140110072 ABHINAV SINGH 30/9/1995 M N GENERAL 4 OTHERS 1 3 NLSIU-BANGLORE All_India_Gen Y 5 6 140110362 AKSHI RASTOGI
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8-QUEEN CHESS PUZZLE’S SOLUTION IMPLEMENTING BACKTRACKING ALGORITHM I. INTRODUCTION Did you ever think that you can play chess with an eight queen in it? A simple board game‚ that turned into a tricky game. The Queen is the most powerful piece in the game of chess‚ but just how many queens can you fit on a chessboard before they start attacking each other? The answer is eight (8)‚ but positioning so many of these influential ladies on a single board is a tricky challenge. Put them in the
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Product positionMeasuring ValueProposition Description of optionPricing for a 4 week supply of Metabicalat $75Pricing for a 4 week supply of Metabicalat $125Pricing for a 4 week supply of Metabicalat $150OverallAssessment Do not recommend y Competitor basedpricing maybackfire y Metabical doesnot have tocompete at allprice points that aconsumer wants R
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Nature Conservation for Sustainable Societies Professor C Y Jim Department of Geography The University of Hong Kong GEOG1016 Nature Conservation for Sustainable Societies / Professor C Y Jim 1 WATER RESOURCE AND ITS SUSTAINABLE MANAGEMENT INTRODUCTION GLOBAL PATTERN OF SUPPLY AND DEMAND SOURCES AND USES OF WATER HUMAN IMPACTS ON WATERS SUSTAINABLE WATER RESOURCE MANAGEMENT GEOG1016 Nature Conservation for Sustainable Societies / Professor C Y Jim 2 1 INTRODUCTION our water planet (contrary to “Earth”)
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P AS O Y Output o ADC: Shows the combinations of the price level and level of output at which the goods and money markets are simultaneously in equilibrium. o ADC: Downward Sloping ( At higher prices‚ reduction in the Value of money supply‚ demand for output is reduced. P AD O Y Output o Equilibrium level of output and the
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Hooplah‚ Inc. R1. (a) It is acceptable to use the same set of samples when testing both controls‚ as each “transaction” should pass through all of the controls in a business process. It becomes more efficient to test the same set‚ because the audit team could request a comprehensive documents package from the client. The sample size table in Appendix A indicates that a sample size of 58 is acceptable for the first control‚ but the higher estimated population deviation rate for the second control
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Econometrics II Part I The Classical Linear Regression Model (CLRM) and the OLS Estimator 17/02/2010 Notation y1 y y2 yn column vector containing the n sample observations on the dependent variable y. (1) x 1k xk x 2k x nk column vector containing the n sample observations on the independent variable x k ‚ with k 1‚ 2‚ . . . ‚ K. (2) x1 x2 xK X n K data matrix containing the n sample observations on the K independent variables. Usually the
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Let y = x 2 and y 2 = x 4 The above equation becomes: y 2 − 6 y + 8 = 0 y2 − 6 y + 8 = 0 y2 − 4 y − 2 y + 8 = 0 y ( y − 4) − 2( y − 4) = 0 ( y − 2)( y − 4) = 0 y − 2 = 0 and y − 4 = 0 y=2 y=4 2 As‚ y = x x2 = 2 x2 = 4 x = ±2 x=± 2 solution set = { 2‚ − 2‚ 2‚ −2} 2. x −2 − 10 = 3 x−1 Solution: x −2 − 10 = 3 x −1 x −2 − 3 x −1 − 10 = 0 Let y = x −1 and y 2 = x −2 The above equation becomes: y 2 − 3 y − 10 = 0 y 2 − 5 y + 2 y − 10 = 0 y ( y − 5) + 2( y − 5) = 0 ( y + 2)( y − 5)
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