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CHAPTER 6 Polynomials: Factoring (continued) Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 6.1Multiplying and Simplifying Rational Expressions 6.2Division and Reciprocals 6.3Least Common Multiples and Denominators 6.4Adding Rational Expressions 6.5Subtracting Rational Expressions 6.6Solving Rational Equations 6.7Applications Using Rational Equations and Proportions

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CHAPTER 6 Polynomials: Factoring Slide 3Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 6.8Complex Rational Expressions 6.9Direct Variation and Inverse Variation

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OBJECTIVES 6.9 Direct Variation and Inverse Variation Slide 4Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. aFind an equation of direct variation given a pair of values of the variables. bSolve applied problems involving direct variation. cFind an equation of inverse variation given a pair of values of the variables. dSolve applied problems involving inverse variation.

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6.9 Direct Variation and Inverse Variation Direct Variation Slide 5Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. Many problems lead to equations of the form y = kx, where k is a constant. Such equations are called equations of variation. When a situation translates to an equation described by y = kx, with k a positive constant, we say that y varies directly as x. The equation y = kx is called an equation of direct variation.

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Note that for k > 0, any equation of the form y = kx indicates that as x increases, y increases as well. “y varies as x,” “y is directly proportional to x,” and “y is proportional to x” also imply direct variation and are used in many situations. The constant k is called the constant of proportionality or the variation constant. It can be found if one pair of values of x and y is known. Once k is known, other pairs can be determined. 6.9 Direct Variation and Inverse Variation a Find an equation of direct variation given a pair of values of the variables. Slide 6Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 6.9 Direct Variation and Inverse Variation a Find an equation of direct variation given a pair of values of the variables. AIf y varies directly as x and y = 3 when x = 12, find the equation of variation. (continued) Slide 7Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE Solution The words “y varies directly as x” indicate an equation of the form y = kx: y = kx 3 = k · 12 Thus, the equation of variation is y = 0.25x. A visualization of the situation is shown. 6.9 Direct Variation and Inverse Variation a Find an equation of direct variation given a pair of values of the variables. ADirect Variation Slide 8Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 6.9 Direct Variation and Inverse Variation a Find an equation of direct variation given a pair of values of the variables. BFind an equation in which a varies directly as b and a = 15 when b = 25. Then find the value of a when b = 36. (continued) Slide 9Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE Solution a = kb 15 = k 25 Thus, the equation of variation is When b = 36, we have The value of a is when b = 36. 6.9 Direct Variation and Inverse Variation a Find an equation of direct variation given a pair of values of the variables. BDirect Variation Slide 10Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE Solution 1. Familiarize and Translate. The problem states that we have direct variation between B and T. Thus an equation B = kT applies. 6.9 Direct Variation and Inverse Variation b Solve applied problems involving direct variation. CThe number of bolts B that a machine can make varies directly as the time T that it operates. It can make 3288 bolts in 2 hr. How many can it make in 5 hr? (continued) Slide 11Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE Solution 3. Solve. B = kT 3288 = k(2) 1644 = k The equation of variation is B = 1644T. 6.9 Direct Variation and Inverse Variation b Solve applied problems involving direct variation. CThe number of bolts B that a machine can make varies directly as the time T that it operates. It can make 3288 bolts in 2 hr. How many can it make in 5 hr? (continued) Slide 12Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE B = 1644T B = 1644(5) B = 8220 4. Check. The check might be done by repeating the computations. 5. State. The number of bolts a machine can make in 5 hours is 8220. 6.9 Direct Variation and Inverse Variation b Solve applied problems involving direct variation. CDirect Variation Slide 13Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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6.9 Direct Variation and Inverse Variation Inverse Variation Slide 14Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. Note that for k > 0, any equation of the form y = k/x indicates that as x increases, y decreases. When a situation translates to an equation described by y = k/x, with k a positive constant, we say that y varies inversely as x. The equation y = k/x is called an equation of inverse variation.

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EXAMPLE Solution: The words “y varies inversely as x” indicate an equation of the form y = k/x: The equation of variation is 6.9 Direct Variation and Inverse Variation c Find an equation of inverse variation given a pair of values of the variables. DIf y varies inversely as x and y = 30 when x = 20, find the equation of variation. Slide 15Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 6.9 Direct Variation and Inverse Variation d Solve applied problems involving inverse variation. EIt takes 6 hr for 25 people to raise a barn. How long would it take 35 people to complete the job? Assume that all people are working at the same rate. (continued) Slide 16Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE Solution 1. Familiarize. Think about the situation. What kind of variation applies? It seems reasonable that the greater number of people working on a job, the less time it will take. If T = the time to complete the job, in hours, and N = the number of people working, then as N increases, T decreases and inverse variation applies. 6.9 Direct Variation and Inverse Variation d Solve applied problems involving inverse variation. EInverse Variation (continued) Slide 17Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 2. Translate. Since inverse variation applies, we have: 6.9 Direct Variation and Inverse Variation d Solve applied problems involving inverse variation. EInverse Variation (continued) Slide 18Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 3. Solve. We find an equation of variation: The equation of variation is 6.9 Direct Variation and Inverse Variation d Solve applied problems involving inverse variation. EInverse Variation (continued) Slide 19Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE The equation of variation is When N = 35, we have 6.9 Direct Variation and Inverse Variation d Solve applied problems involving inverse variation. EInverse Variation (continued) Slide 20Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 4. Check. A check might be done by repeating the computations or by noting that (4.3)(35) and (6)(25) are both 150. Also, as the number of people increases, the time needed to complete the job decreases. 5. State. It should take 4.3 hours for 35 people to raise the barn. 6.9 Direct Variation and Inverse Variation d Solve applied problems involving inverse variation. EInverse Variation Slide 21Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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