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Mastering Algebra: Direct & Inverse Variations

Review direct and inverse variation functions, solve equations, factor polynomials, and more with examples and practice. Explore various mathematical concepts tied to direct and inverse relationships between variables.

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Mastering Algebra: Direct & Inverse Variations

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  1. Be seated before the bell rings Agenda: Review questions Part 2 ch 7 test (with calculators) Warmup Notes 8.1 DESK Warm-up (in your notes) homework

  2. Notebook 1 Table of content 8.1 Variation Functions Page 19)Add, subtract, multiply polynomials 20) Dividing Polynomials 21) Factor and find roots 22) Fundamental Theorem of Algebra 23)Graph of polynomials 24) Radical Expressions and Rational Exponents 25) Variation Functions 1 HW ; p.573; 5,8,12,16, 20-23

  3. 2.4 2 = x 9 Warm Up 11/26 Solve each equation. 10.8 1. 27.9 2. 1.6x = 1.8(24.8) Determine whether each data set could represent a linear function. 3. no 4. yes

  4. 8.1 Variation Functions A direct variation is a relationship between two variables x and y that can be written in the form y = kx, where k ≠ 0. constant of variation As x’s gets larger, y’s get larger by constant k.

  5. Example 1: Writing and Graphing Direct Variation Given: y varies directly as x, and y = 10 when x = 2.5. Write and graph the direct variation function. y = kx 10 = k(2.5) k = 4 y = 4x y = 4.5x 27 4.5(6) 27 27 

  6. y 6 = 45 30 y 1 = 5 45 Example 2A: Solving Direct Variation Problems The value of y varies directly with x. and y = 6 when x = 30. Find y when x =45. Method 1 Find k. y = kx y = 1/5x 6= k(30) y= 1/5(45) 1/5= k y = 9 Method 2 Use a proportion. 5y= 45 y = 45/5 = 9

  7. 18 75 = 1.5 s You try! Example 2B The perimeter P of a regular dodecagon varies directly as the side length s, and P = 18 in. when s = 1.5 in. Find s when P = 75 in. Method 1 Find k. P = ks P = 12s 18= k(1.5) 75= 12s 12= k 6.25 ≈ s Method 2 Use a proportion. 18s = 112.5 6.25 = s

  8. A joint variation is a relationship among three variables that can be written in the form y = kxz, where k is the constant of variation.

  9. 1 1 1 2 2 2 Example 3: Solving Joint Variation Problems The area A of a triangle varies jointly as the the base B and the height h, and A = 20m2 when B = 5 m and h = 8 m. Find b when A = 60 m2 and h = 6 ft. Step 2 Use the variation function. Step 1 Find k. A = kBh 20= k(5)(8) A = Bh K = 60= B(6) B = 20 The base is 20 meters.

  10. You try ! Example 4 The lateral surface area L of a cone varies jointly as the area of the base radius r and the slant height l, and L = 63 m2 when r = 3.5 mand l = 18 m. Find r to the nearest tenth when L = 8 m2and l = 5 m. Step 1 Find k. Step 2 Use the variation function. L = krl 63= k(3.5)(18) L = rl  = k 8= r(5) 1.6 = r

  11. 12 k k k 6 x x x Example 5: Writing and Graphing Inverse Variation y = Given: y varies inversely as x, and y = 2 when x = 6. Write and graph the inverse variation function. x y -6 y = -4 -3 2= -2 k =12 2 y= 3 4

  12. 1250 1250 k k 1 15 20 v v 3 Example 6A The time t that it takes for a group of volunteers to construct a house varies inversely as the number of volunteers v. If 20 volunteers can build a house in 62.5 working hours, how many working hours would it take 15 volunteers to build a house? Find k. 62.5= t = k =1250 t = t = t ≈83

  13. 26.1954 26.1954 k k 8.82 s s s Example 6b: Sports Application The time t needed to complete a certain race varies inversely as the runner’s average speed s. If a runner with an average speed of 8.82 mi/h completes the race in 2.97 h, what is the average speed of a runner who completes the race in 3.5 h? t = Find k. Substitute. 2.97= Solve for k. k =26.1954 Use 26.1954 for k. t = Substitute 3.5 for t. 3.5= Solve for s. s ≈7.48

  14. 0.05(400) k(300) 0.05T kT 1.5 (1) P P Example 7 combined variation The volume V of a gas varies inversely as the pressure P and directly as the temperature T. A certain gas has a volume of 10 liters (L), a temperature of 300 kelvins (K), and a pressure of 1.5 atmospheres (atm). If the gas is heated to 400K, and has a pressure of 1 atm, what is its volume? Step 1 Find k. Step 2 Use the variation function. V = V = 10= V = 0.05 = k V =20 The new volume will be 20 L.

  15. k(450) kE E E 1 900(0.2) 900m 900 0.1 m You try! Example 8: Chemistry Application The change in temperature of an aluminum wire varies inversely as its mass m and directly as the amount of heat energy E transferred. The temperature of an aluminum wire with a mass of 0.1 kg rises 5°C when 450 joules (J) of heat energy are transferred to it. How much heat energy must be transferred to an aluminum wire with a mass of 0.2 kg raise its temperature 20°C? Step 1 Find k. Step 2 Use the variation function. ΔT = ΔT = 5= 20= = k 3600 = E The amount of heat energy that must be transferred is 3600 joules (J).

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