Introduction A traverse consists a series of straight line of known length related one another by known angles between the line. The points defining the ends of the traverse lines are called traverse stations. Traverse survey is a method of establishing control points‚ their position being determined by measuring the distances between the traverse station which serve as control points and the angles subtended at the various stations by their adjacent stations. The angles
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calculations from the Checkpoint Ratio‚ Vertical‚ and Horizontal Analyses‚ providing you show your work. · Horizontal analyses for both companies. You can use your calculations from the Checkpoint Ratio‚ Vertical‚ and Horizontal Analyses‚ providing you show your work. · Ratio analyses for both companies‚ including a test of: o Liquidity. You can use the Current Ratio you compute in the Checkpoint Ratio‚ Vertical‚ and Horizontal Analyses‚ and another ratio. Show your work. o
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-2/3 b= 4 Horizontal and Vertical Lines: - A Horizontal Line has the form y=#. (In an equation of a horizontal line‚ there is no x) - The slope of a horizontal line is 0. Picture: (can walk on the line) - A Vertical Line has the form x=#. (In an equation of a vertical line‚ there is no y.) - The slope of a vertical line is undefined. Picture: (falls off line) - (*y-intercept = none parallel to y-axis unless x=0) 1- If y is the only variable‚ solve for y. 2- Draw a horizontal line that
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TITLE: Trajectory of a Projectiles PROBLEM STATEMENT: A spherical ball was projected from a fixed point ‚O‚ with a speed ‚u‚ angels of elevation : 0‚ 10‚ 20‚ 30‚ 40‚ 50‚ 60‚ 70‚ 80‚ 90 to the horizontal ground surface. Find the value of maximum height travelled by the ball for which the horizontal range‚ R‚ greatest for the same value of u. AIM: To investigate: 1)Range of projected ball. 2)Time of flight for corresponding angel of projection METHODOLOGY: APPARATUS
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MGSM840 Accounting for Management 2012 Term 2 Group Assignment Investment Decision: Cochlear Limited v. Coca-Cola Limited Investment Advice Cochlear v. Coca-Cola Amatil Table of Contents 1 2 3 4 Executive Summary .............................................................................................. 4 Introduction ........................................................................................................... 5 Share Price Analysis ........................
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Time Frame | Objectives | Topics/ Content | Concept/s | Competencies | Teaching Strategy | Values | List of Activities | Materials | Evaluation | References | First Quarter | -Define functions and give examples that depict functions-Differentiate a function and a relation-Express functional relationship in terms of symbols y=f(x)-Evaluate a function using the value of x. | Chapter 1Functions and GraphsFunctions and Function Notations | The equation y=f(x) is commonly used to denote functional relationship
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CHAPTER ONE 1.0 Introduction At the heart of transforming raw ingredients into food for human consumption is the mixing operation. It’s a main task which other food processing steps also share to establish consistency. Whether a food product requires small-scale mixing by hand or high volume blending of multiple ingredients‚ home cooks and food process engineers alike know the importance of proper mixing. Even with the right amount of ingredients and flavors‚ a great recipe will not transform
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To measure the centripetal force for whirling a mass round a horizontal circle and compare the result with the theoretical value given by F = m(2r . Apparatus 12 slotted weights with hanger (0.02kg each) 1 rubber bung with nylon string about 1.5m 1 glass tube about 20cm long 1 triple beam balance 1 meter rule 1 stop watch Several small paper markers Theory When a mass m attached to a string is whirled round a horizontal circle of radius r‚ the centripetal force for maintaining the
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V. Analysis A. Vertical displacemen t= (1/2) x (Vertical Acceleration) x (Time)2 0.92m = (1/2) x (9.8m/s2) x (Time)2 Time = ((2 x 0.92m)/(9.8m/s2))1/2 = 0.43s Horizontal displacement = (Initial horizontal velocity) x (Time) 0.43m = (Initial horizontal velocity) x (0.43s) Initial horizontal velocity = Initial velocity = (0.43m/0.43s) = 1.0m/s Initial Momentum = (Mass) x (Initial Velocity) P0 = (0.008kg) x (1.0m/s) = 0.008kgm/s Time =((2 x Displacement)/(Acceleration))1/2
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with respect to the horizontal. What is the minimum initial speed that would allow the athlete to clear the gap? Neglect air resistance. SOLUTION 1. DEFINE Given: ∆x = 6.0 m q = 35° g = 9.81 m/s vi = ? Unknown: 2. PLAN Diagram: v θ = 35° ∆x = 6.00 m Choose the equation(s) or situation: The horizontal component of the athlete’s velocity‚ vx ‚ is equal to the initial speed multiplied by the cosine of the angle‚ q‚ which is equal to the magnitude of the horizontal displacement‚ ∆x
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