Stare at this image…. Do you see the invisible dots?

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Stare at this image…. Do you see the invisible dots?

Transcript of Stare at this image…. Do you see the invisible dots?

Page 1: Stare at this image…. Do you see the invisible dots?

Stare at this image….

Do you see the invisible dots?

Page 2: Stare at this image…. Do you see the invisible dots?

Example #6

Determine the amount of an investment if $1000 is invested at an interest rate of 4% compounded semi-annually for 5 years.

P(1 + )nr

n t

A =

1000(1 + )2.04

2 (5)

A =

A = 1000(1.02)10

A = $1218.99

Page 3: Stare at this image…. Do you see the invisible dots?

Warm-Up

The number of students at West Ottawa High School in 1992 was 1280. Since then, the number has increased 3.2% each year. If this continues, how many will there be in 2012?

y = b(1 + r)x

y = 1280 (1 + ).032 20

y = 1280(1.032)20

y = 2403

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Homework Answers

9. C = 18.9(1 + 0.19)t

10.≈ 8329.24 million computers

11. W = 43.2(1+ 0.06)t

12. ≈ 77 .36 million people

13. ≈ 122,848,204 people

14. ≈ $2097

15.≈ $14,607.78

18. About 17,125,650 visitors

21. ≈ 15.98 %

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Lesson 10-6 B

Objective:

Solve problems involving exponential decay

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Percentage Decay Formula

y =

b = initial amount

r = % decay (as a decimal)

x = time

b (1 - ) xr

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• Words that mean decay (get smaller):

• Depreciates

• Decrease

• Less

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Example # 1

In 2004, the population of Australia was decreasing by 0.8% each year…

(a) Find the growth factor, a.

(b) The 2004 population was 17,800,000. What is the projected population for the year 2010?

y = b ( 1 – r ) x

y = ( 1 – ) 17,800,000 0.008 6

y = ( 0.992 ) 17,800,000 6

(a) (b)0.992 16,962,507

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Example # 2

You bought a car for $28,500 in 2005.

What is the value of the car in 2008 if it depreciates at 13% each year?

y = b ( 1 – r ) x

y = ( 1 – ) 28,500 0.13 3

y = ( 0.87 ) 28,500 3

$ 18, 767

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Half-Life Formula

The half-life of a compound is a measurement of how long it takes for one half of the compound to break down. The

formula for half-life looks like this:

y = b ( .5 ) x

b = Initial Amount of the Compound

x = Number of Half-Life Periods

*** NOT the Half-Life ***

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Example # 3

An isotope of Cesium-137 has a half-life of 30 years. If you start with 20 mg of the substance, how many mg will be left after

90 years? How many after 120 years?

y = b ( .5 ) x

y = ( .5 ) 20 3 y = ( .5 ) 20 4

y = 2.5 mg y = 1.25 mg

= 3 half-lives! = 4 half-lives!

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Example # 4

Radium-226 has a half-life of 1,620 years…

(a) Write an equation for the amount of Radium remaining if you start with 100 mg and x number of half-lives have passed.

(b) If you begin with 4 mg, how much will be left after 3 half-lives?

y = b ( .5 ) x

y = ( .5 ) 4 3

(a) (b)y = 100(.5)x 0.5 mg

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Assignment:

Page 563, problems

1, 3, 16, 17, 19, 20, 23, 24

Quiz Friday 10.5-10.6