be seated before the bell rings desk homework warm-up (in your notes) agenda: review questions part...

Be seated before the bell rings

DESK

homeworkWarm-up (in your notes)

Agenda:

Review questions

Part 2 ch 7 test(with calculators)

Warmup

Notes 8.1

NotebookTable of content

Page1

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

Warm Up 11/26Solve each equation.1.

2. 1.6x = 1.8(24.8)

10.8

27.9

2.4 =x 92

Determine whether each data set could represent a linear function.

x 2 4 6 8

y 12 6 4 3

x –2 –1 0 1

y –6 –2 2 6

3.

4.

no

yes

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 variationAs x’s gets larger, y’s get larger by constant k.

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 = kx10 = k(2.5)k = 4y = 4x

y = 4.5x

27 4.5(6)

27 27

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 = kxMethod 1 Find k.

6 = k(30)1/5 = k

y = 1/5(45)

y = 9

y = 1/5x

Method 2 Use a proportion.

y =45 30

6 y =45 5

1 5y = 45

y = 45/5 = 9

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 = ksMethod 1 Find k.

18 = k(1.5)12 = k

75 = 12s6.25 ≈ s

P = 12s

Method 2 Use a proportion. 18 = 1.5 s

7518s = 112.5 6.25 = s

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.

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.

Example 3: Solving Joint Variation Problems

Step 1 Find k.

The base is 20 meters.

A = kBh20= k(5)(8)

K = 12

Step 2 Use the variation function.

60= B(6)

B = 20

A = Bh12

12

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 m and l = 18 m. Find r to the nearest tenth when L = 8 m2 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

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 =

k = 12

y =

k6

kx

y = 12x

y = kx

-6

-4

-3

-2

2

34

x y

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 = 1250t =

k20k

vt = 1250

v

t = 125015

t ≈ 83

Example 6A

13

Example 6b: Sports ApplicationThe 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

ks

Use 26.1954 for k.t = 26.1954s

Substitute 3.5 for t.3.5 = 26.1954s

Solve for s.s ≈ 7.48

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 = 0.05TP

V = 0.05(400)(1)

kTP

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 =

= k1900

Step 2 Use the variation function.

3600 = E

kEmk(450)

0.1

ΔT = E900m

20 = E900(0.2)