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

2. Notebook Table of content Page 1 1 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 8.1 Variation Functions HW ; p.573; 5,8,12,16, 20-23

3. Warm Up 11/26 Solve each equation. 1. 2. 1.6x = 1.8(24.8) 10.8 27.9 2.4 = x9 2 Determine whether each data set could represent a linear function. x2468 y12643 x–2–101 y–6–226 3. 4. no 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. Given: y varies directly as x, and y = 10 when x = 2.5. Write and graph the direct variation function. Example 1 : Writing and Graphing Direct Variation y = kx 10 = k(2.5) k = 4 y = 4x y = 4.5x 27 4.5(6) 27

6. The value of y varies directly with x. and y = 6 when x = 30. Find y when x =45. Example 2A: Solving Direct Variation Problems y = kx Method 1 Find k. 6 = k(30) 1/5 = k y = 1/5(45) y = 9 y = 1/5x Method 2 Use a proportion. y = 4530 6 y = 45 5 1 5y = 45 y = 45/5 = 9

7. 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. You try! Example 2B P = ks Method 1 Find k. 18 = k(1.5) 12 = k 75 = 12s 6.25 ≈ s P = 12s Method 2 Use a proportion. 18 = 1.5s 75 18s = 112.56.25 = 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. The area A of a triangle varies jointly as the the base B and the height h, and A = 20m 2 when B = 5 m and h = 8 m. Find b when A = 60 m 2 and h = 6 ft. Example 3: Solving Joint Variation Problems Step 1 Find k. The base is 20 meters. A = kBh 20= k(5)(8) K = 1 2 Step 2 Use the variation function. 60= B(6) B = 20 A = Bh 1 2 1 2

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  m 2 when r = 3.5 m and l = 18 m. Find r to the nearest tenth when L = 8  m 2 and l = 5 m. Step 1 Find k. L = krl 63  = k(3.5)(18)  = k Step 2 Use the variation function. 8  =  r(5) 1.6 = r L =  rl

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

12. 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 = k = 1250 t = k 20 k v t = 1250 v t = 1250 15 t ≈ 83 Example 6A 1 3

13. 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? Find k. Substitute. Solve for k. 2.97 = k = 26.1954 t = k 8.82 k s Use 26.1954 for k. t = 26.1954 s Substitute 3.5 for t. 3.5 = 26.1954 s Solve for s. s ≈ 7.48

14. 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. The new volume will be 20 L. V = 10 = 0.05 = k Step 2 Use the variation function. V = 20 k(300) 1.5 V = 0.05T P V = 0.05(400) (1) kT P

15. 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? You try! Example 8: Chemistry Application Step 1 Find k. The amount of heat energy that must be transferred is 3600 joules (J). ΔT = 5 =5 = = k 1 900 Step 2 Use the variation function. 3600 = E kE m k(450) 0.1 ΔT = E 900m 20 = E 900(0.2)