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72 x. The equation of the inverse variation is xy = 72 or y =. Inverse Variation. Lesson 5-6. Additional Examples. Suppose y varies inversely with x , and a point on the graph of the equation is (8, 9). Write an equation for the inverse variation.
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72 x The equation of the inverse variation is xy = 72 or y = . Inverse Variation Lesson 5-6 Additional Examples Suppose y varies inversely with x, and a point on the graph of the equation is (8, 9). Write an equation for the inverse variation. xy = kUse the general form for an inverse variation. (8)(9) = k Substitute 8 for x and 9 for y. 72 = kMultiply to solve for k. xy = 72 Write an equation. Substitute 72 for k in xy = k.
Inverse Variation Lesson 5-6 Additional Examples The points (5, 6) and (3, y) are two points on the graph of an inverse variation. Find the missing value. x1 • y1 = x2 • y2Use the equation x1 • y1 = x2 • y2 since you know coordinates, but not the constant of variation. 5(6) = 3y2Substitute 5 for x1, 6 for y1, and 3 for x2. 30 = 3y2Simplify. 10 = y2Solve for y2. The missing value is 10. The point (3, 10) is on the graph of the inverse variation that includes the point (5, 6).
Inverse Variation Lesson 5-6 Additional Examples A 120-lb weight is placed 5 ft from a fulcrum. How far from the fulcrum should an 80-lb weight be placed to balance the lever? Relate: A weight of 120 lb is 5 ft from the fulcrum. A weight of 80 lb is x ft from the fulcrum. Weight and distance vary inversely. Define: Let weight1 = 120 lb Let weight2 = 80 lb Let distance1 = 5 ft Let distance2 = x ft
600 80 = xSolve for x. Inverse Variation Lesson 5-6 Additional Examples (continued) Write: weight1 • distance1 = weight2 • distance2 120 • 5 = 80 • xSubstitute. 600 = 80xSimplify. 7.5 = x Simplify. The weight should be placed 7.5 ft from the fulcrum to balance the lever.
a. xy 3 10 5 6 10 3 Check each product xy. xy: 3(10) = 30 5(6) = 30 10(3) = 30 Inverse Variation Lesson 5-6 Additional Examples Decide if each data set represents a direct variation or aninverse variation. Then write an equation to model the data. The values of y seem to vary inversely with the values of x. The product of xy is the same for all pairs of data. So, this is an inverse variation, and k = 30. The equation is xy = 30.
y x b. xy 2 3 4 6 8 12 Check each ratio . 12 8 6 4 = 1.5 y x 3 2 = 1.5 = 1.5 y x The ratio is the same for all pairs of data. Inverse Variation Lesson 5-6 Additional Examples (continued) The values of y seem to vary directly with the values of x. So, this is a direct variation, and k = 1.5. The equation is y = 1.5x.
The cost per souvenir times the number of souvenirs equals the total cost of the souvenirs. this is a direct variation. cost souvenirs Since the ratio is constant at $10 each, Inverse Variation Lesson 5-6 Additional Examples Explain whether each situation represents a direct variation or an inverse variation. a. You buy several souvenirs for $10 each. b. The cost of a $25 birthday present is split among several friends. The cost per person times the number of people equals the total cost of the gift. Since the total cost is a constant product of $25, this is an inverse variation.
