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

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Presentation on theme: "Stare at this image…. Do you see the invisible dots?"— Presentation transcript:

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

2 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 + ) n r n t A = 1000 (1 + ) 2.04 2 (5) A = A = 1000(1.02) 10 A = $1218.99

3 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

4 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 %

5 Lesson 10-6 B Objective: Solve problems involving exponential decay

6 Percentage Decay Formula y = b = initial amount r = % decay (as a decimal) x = time b (1 - ) x r

7 Words that mean decay (get smaller): Depreciates Decrease Less

8 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 – )17,800,0000.008 6 y = ( 0.992 )17,800,000 6 (a)(b) 0.99216,962,507

9 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 – )28,5000.13 3 y = ( 0.87 )28,500 3 $ 18, 767

10 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 ***

11 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!

12 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

13 Assignment: Page 563, problems 1, 3, 16, 17, 19, 20, 23, 24 Quiz Friday 10.5-10.6


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