23 limy→6y+6y2-36=limy→6y+6y+6y-6 ⟹limy→61y-6=16-6=10=undefined ∴limit doesn’t exist 26 limx→43-xx2-2x-8=limx→43-xx2-4x+2x-8=limx→4 3-xxx-4+2x-4 =lim x→43-xx-4x+2=3-44-44+2=-10×6=10 =undefined; ∴Limit does not exist. 29 limx→9x-9x-3=limx→9x2-32x-3=limx→9x-3x+3x-3 =limx→9x+3=9+3=3+3=6 30 lim y→44-y2-y=lim y→422-y22-y=lim y→42-y2+y2-y =limy→42+y=2+4=2+2=4 31 fx=x-1‚ &x≤33x-7‚ &x>3 a. limx→3- fx=x-1=3-1=2 b. limx→3+ fx=3x-7=3×3-7=2 c. limx→3 fx=x-1=3-1=2 32
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LINEAR PROGRAMMING FORMULATION PROBLEMS AND SOLUTIONS 7-14 The Electrocomp Corporation manufactures two electrical products: air conditioners and large fans. The assembly process for each is similar in that both require a certain amount of wiring and drilling. Each air conditioner takes 3 hours of wiring and 2 hours of drilling. Each fan must go through 2 hours of wiring and 1 hour of drilling. During the next production period‚ 240 hours of wiring time are available and up to 140 hours of drilling
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JEPPE HIGH SCHOOL FOR BOYS GRADE 9 POLYNOMIAL MATHS LESSON PLAN DATE: Term 2 2012 TIME: 1 HOUR Objective of the lesson Revision of how to: • Use the four basic mathematical operators on various polynomials • Factorise a polynomial depending on its structure • Solve an equation by factorising a polynomial Basic operator use on polynomials Time required: 20 Minutes Method: • Show how each operator works on a polynomial • Show exceptions
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minimization problem the isocost line solution method is used. a) Formulate this as an LP problem to minimize the cost of the dog food. Minimize cost = $0.9X1 + $0.6X2 Subject to 10X1 + 6X2 9 12X1 + 9X2 10 X1‚ X2 0 b) How many pounds of beef and grain should be included in each pound of dog food? Therefore‚ we find that the minimum cost of the dog food happens when X1 = 0.9 and X2 = 0. The minimum
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TAYLMC04_0131961381.QXD 4/14/09 8:33 AM Page 42 Chapter Four: Linear Programming: Modeling Examples PROBLEM SUMMARY 1. “Product mix” example 2. “Diet” example 3. “Investment” example 4. “Marketing” example 5. “Transportation” example 6. “Blend” example 7. Product mix (maximization) 8. Sensitivity analysis (4–7) 9. Diet (minimization) 10. Product mix (minimization) 11. Product mix (maximization) 12. Product mix (maximization) 13. Product mix (maximization) 14. Ingredients mix (minimization)
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The Simplex Method: Learning Team A Mike Smith‚ Todd Jones Math212/Introduction to Finite Mathematics February 1‚ 2011 The Simplex Method: Learning Team A Sam’s Hairbows and Accessories is a small company preparing for the next scheduled craft fair. The owners‚ Sam and Todd‚ both have full-time jobs in addition to owning the company so they are only able to spend a combined total of 80 hours labor to prepare for the fair in four weeks. Sam’s offers five main product lines: basic bows‚ elaborate
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3. Algebra of Polynomials By now‚ you should be familiar with variables and exponents‚ and you may have dealt with expressions like 3x³ or 6x. Polynomials are sums of these "variables and exponents" expressions. Each piece of the polynomial‚ each part that is being added‚ is called a "term". Polynomial terms have variables which are raised to whole-number exponents (or else the terms are just plain numbers); there are no square roots of variables‚ no fractional powers‚ and no variables in the denominator
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Federal Urdu University Karachi. Department of Business Administration. [pic] [pic] Final Assignment [pic] Name: Muhammad Farooq Yaseen Roll#: 49 Class: MBA Section: A | |Index | | |S.No. |Contents |Page No. | |1 |IDENTIFYING
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Part 1. Describe the order of operations and explain how it is used to simplify expressions. Explain specifically the order in which mathematical operations must be performed to correctly simplify an expression. The order of operations is a rule that always applies to mathematical problems. It includes addition‚ subtraction‚ multiplication and division as well as grouping of numbers (such as in parentheses and powers). It gives a definite order of how to do a problem involving multiple operations
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Class Number 13 – 16 IV – St. Hannibal 1. Find m if x3 + 3x2 – mx + 4 leaves a remainder of m when divided by x – 2. 23 + 3(2)2 – m(2) + 4 = m 8 + 12 – 2m +4 = m -3m = -24 -3 -3 m = 8 2. When the expression 6x2 – 2x +3 is divided by x –a‚ the remainder is 3. Find the value of a. 6a2 – 2a + 3 = 3 6a2 – 2a = 0 6a2 = 2a 6a 6a a = 3. The expression x4 + x3 + 2mx2 – 14m4 leaves a remainder of 32 when divided by x + 2m. Find the possible
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