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

Explore exponential functions involving growth, decay, and half-life in investments, populations, and compounds. Practice solving problems and understanding decay percentages. Prepare for a quiz on Friday.

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

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

  2. r n t P (1 + ) A = n Example #6 Determine the amount of an investment if $1000 is invested at an interest rate of 4% compounded semi-annually for 5 years. .04 2 (5) (1 + ) A = 1000 2 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 • C = 18.9(1 + 0.19)t • ≈ 8329.24 million computers • W = 43.2(1+ 0.06)t • ≈ 77 .36 million people • ≈ 122,848,204 people • ≈ $2097 • ≈ $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 x y = b = initial amount r = % decay (as a decimal) x = time b (1 - ) 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 – ) 17,800,000 0.008 6 y = ( 0.992 ) 17,800,000 6 (a) (b) 0.992 16,962,507

  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 – ) 28,500 0.13 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? = 3 half-lives! = 4 half-lives! y = b ( .5 ) x y = ( .5) 20 3 y = ( .5 ) 20 4 y = 1.25 mg y = 2.5 mg

  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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