Balancing Equations Balancing equations is a fundamental skill in Chemistry. Solving a system of linear equations is a fundamental skill in Algebra. Remarkably‚ these two field specialties are intrinsically and inherently linked. 2 + O2 ----> H2OA. This is not a difficult task and can easily be accomplished using some basic problem solving skills. In fact‚ what follows is a chemistry text’s explanation of the situation: Taken from: Chemistry Wilberham‚ Staley‚ Simpson‚ Matta Addison Wesley
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Straight Line Equations and Inequalities A: Linear Equations - Straight lines Please remember that when you are drawing graphs you should always label your axes and that y is always shown on the vertical axis. A linear equation between two variables x and y can be represented by y = a + bx where “a” and “b” are any two constants. For example‚ suppose we wish to plot the straight line If x = -2‚ say‚ then y = 3 + 2(-2) = 3 - 4 = -1 If x= -2 -1 -1 1 0 3 1 5 2 7 As you can see‚ we have plotted the
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Prefixes and Equations 1mL = -1x10 3 L = 0.001L 1 kilometers (1km) = 1000m (1x103 m) 1 kilometer (1KL) = 1000L (1x10 3 L) Prefixes Symbol Numerical Value Scientific Notation Equality Kilo | K | 1000 | 103 | 1km = 1x103m 1m = 1x10-3Km | Mega | M | 1 000 000 | 106 | 1Mg = 1x106g1g = 1x10-6Mg | Giga | G | 1 000 000 000 | 109 | 1Gm = 1x109m1m = 1x10-9Gm | Tera | T | 1 000 000 000 000 | 1012 | 1Ts = 1x1012s1s = 1x10-12Ts | Deci | d | 0.1 | 10-1 | 1dL = 1x10-1L1L = 1x101dL (10dL) | *These
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Tak Nga Secondary School 2010-2011 Mid-year Exam Form 4 Mathematics (Paper I) Time allowed: 1 hour 15 minutes Class：________ Name：__________________( ) Marks: ________/ 60 Instructions: 1. Write your name‚ class and class number in the spaces provided on this cover. 2. This paper consists of THREE sections‚ A(1)‚ A(2) and B. Each section carries 20 marks. 3. Attempt ALL questions in this paper. Write your answers in the spaces provided. Supplementary answer sheets will be supplied on request.
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mathematics at a deeper level. Review of homogeneous equations The homogeneous constant coefficient linear equation an y (n) +· · ·+a1 y +a0 y = 0 has the characteristic polynomial an rn +· · ·+a1 r+a0 = 0. From the roots r1 ‚ . . . ‚ rn of the polynomial we can construct the solutions y1 ‚ . . . ‚ yn ‚ such as y1 = er1 x . We can also rewrite the equation in a weird-looking but useful way‚ using the symbol d D = dx . Examples: equation: y − 5y + 6y = 0. polynomial: r2 − 5r + 6 = 0. (factored):
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stream using Equation 1. (Eqn. 1) Where is the flowrate in m3/s and A is the cross-sectional area of the pipe. To find the flowrate‚ we multiply the flowmeter reading by the constant and convert from gallons to cubic meters as follows: The cross sectional area of the 7.75mm pipe is Plugging these values into Equation 1‚ we obtain a bulk velocity . With the bulk velocity value‚ we can find the Reynolds number of the flow using Equation 2. (Eqn. 2) Plugging in known values to
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Diagonally Implicit Block Backward Differentiation Formulas for Solving Ordinary Differential Equations 1.0 Introduction In mathematics‚ if y is a function of x‚ then an equation that involves x‚ y and one or more derivatives of y with respect to x is called an ordinary differential equation (ODE). The ODEs which do not have additive solutions are non-linear‚ and finding the solutions is much more sophisticated because it is rarely possible to represent them by elementary function in close
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ACCOUNTING EQUATIONS 1. Show the accounting equation for the following transaction (i) Ram started business with cash 20000‚ stock 50000‚ building 30000 (ii) Sold goods to Amit for cash 20000 and credit 15000 (iii) Paid rent 500 and rent outstanding 150 (iv) Sold goods costing 12000 for Rs. 15000 (v) Accrued commission 2000 (vi) Furniture purchased from Lalit 12000 and paid 3000 in cash (vii) Received from Amit 13500 in full settlement (viii)
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from now" is six times Miguel’s "age last year" or‚ in math: g + 3 = 6(m – 1) This gives me two equations with two variables: m + g = 68 g + 3 = 6(m – 1) Solving the first equation‚ I get m = 68 – g. (Note: It’s okay to solve for "g = 68 – m"‚ too. The problem will work out a bit differently in the middle‚ but the answer will be the same at the end.) I’ll plug "68 – g" into the second equation in place of "m": g + 3 = 6m – 6 g + 3 = 6(68 – g) – 6 g + 3 = 408 – 6g – 6 g + 3 = 402 – 6g
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Nacua July 20‚ 2013 LEARNING INSIGHTS ON THEORY X / Y / Z AND MASLOW’S HEIRARCHY OF NEEDS McGregor’s Theory X / Y and William Ouichi’s Theory Z I can say that Theory X presents the pessimistic view of employees’ nature and behaviour at work‚ while Theory Y presents the optimistic view. With reference to Maslow’s theory‚ Theory X is based on the assumption that the employees emphasize on the physiological needs and the safety needs; while Theory Y is based on the assumption that the social needs
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