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§5.2 Direct Variation

§5.2 Direct Variation. Warm-ups. Solve each equation for the given variable. 1. nq = m; q 2. d = rt ; r 3. ax + by = 0; y Solve each proportion. 4. = 5. = 6. = . 5 8. x 12. 4 9. n 45. 25 15. y 3. 3. ax + by = 0 ax – ax + by = 0 – ax

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§5.2 Direct Variation

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  1. §5.2 Direct Variation

  2. Warm-ups Solve each equation for the given variable. 1.nq = m; q2.d = rt; r3.ax + by = 0; y Solve each proportion. 4. = 5. = 6. = 5 8 x 12 4 9 n 45 25 15 y 3

  3. 3.ax + by = 0 ax – ax + by = 0 – ax by = –ax = y = – 1.nq = m2.d = rt = = q = = r r = rt t d t m n nq n d t m n –ax b by b d t ax b 4. = 8x = 5(12) 8x = 60 x = 7.5 5 8 x 12 4 9 n 45 25 15 y 3 5. = 6. = 9n = 4(45) 15y = 25(3) 9n = 180 15y = 75 n = 20 y = 5 Warm-ups (solutions)

  4. 1 3 2 3 y = – + xDivide each side by –3. 2 3 y = xDivide each side by –3. The equation has the form y = kx, so the equation is a direct variation. The constant of variation is . 2 3 Example 1: Identifying a Direct Variation Is each equation a direct variation? If it is, find the constant of variation. a. 2x – 3y = 1 –3y = 1 – 2xSubtract 2x from each side. The equation does not have the form y = kx. It is not a direct variation. b. 2x – 3y = 0 –3y = –2xSubtract 2x from each side.

  5. 2 3 – = kDivide each side by –3 to solve for k. 2 3 2 3 y = – xWrite an equation. Substitute – for k in y = kx. 2 3 The equation of the direct variation is y = – x . Example 2: Writing an EquationWhen Given a Point Write an equation for the direct variation that includes the point (–3, 2). y = kxUse the general form of a direct variation. 2 = k(–3) Substitute –3 for x and 2 for y.

  6. Example 3: Graphing a Direct Variation The graph for any direct variation will pass through the origin (0, 0). All direct variations are linear.

  7. xy –1 2 1 2 2 –4 xy –2 1 2 –1 4 –2 y x y x 2 –1 1 –2 = –2 = –0.5 –1 2 2 1 = –0.5 = 2 –2 4 –4 2 = –0.5 = –2 Example 4: Writing a Direct Variationfrom a Table For the data in each table, does y vary directly with x? If it does, write an equation for the direct variation. y x y x Find for each ordered pair. Find for each ordered pair. -2 4 2 -4 -1 2 1 2 1 -2 -1 2 Yes, the constant of variation is –0.5. The equation is y = –0.5x. No, y does not vary directly with x.

  8. Relate: The force of 0.75 lb lifts 48 lb. The force of n lb lifts 210 lb. Define: Let n = the force you need to lift 210 lb. Write: = Use a proportion. Substitute 0.75 for force1, 48 for weight1, and 210 for weight2. 0.75 48 n 210 = 0.75(210) = 48n Use cross products. Solve for n. force2 weight2 force1 weight1 n 3.3 Example ¥: Real-World Problem Solving Suppose a windlass requires 0.75 lb of force to lift an object that weighs 48 lb. How much force would you need to lift 210 lb? You need about 3.3 lb of force to lift 210 lb.

  9. Assignment: pg. 304-305 9-35 Left OMIT #34

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