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Algebra 1 Warm Up 9 April 2012. State the recursive sequence (start?, how is it changing?), then find the next 3 terms. Also find the EQUATION for each. y = a∙b x 1) 12000, 10800,9720, ___, ___, ___ 2 ) 100, 105.25,110.77, ___, ___, ___ Rewrite as a fraction and decimal: - PowerPoint PPT Presentation
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Algebra 1 Warm Up 9 April 2012State the recursive sequence (start?, how is it changing?), then find the next 3 terms.Also find the EQUATION for each. y = a b∙ x
1) 12000, 10800,9720, ___, ___, ___2) 100, 105.25,110.77, ___, ___, ___Rewrite as a fraction and decimal:3) a) 5% b) 50% c) 5.25%
Homework due TuesdayHomework due Tuesday: pg. 345: 1 – 5 ADV: 12: pg. 345: 1 – 5 ADV: 12
OBJECTIVEToday we will explore exponential
growth and decay patterns and write exponential equations.
Today we will take notes, work problems with our groups and present to the class.
Once upon a time, two merchants were trying to work out a deal. For the next month, the 1st merchant was going to give $10,000 to the 2nd merchant, and in return, he would receive 1 cent the first day, 2 cents the second, 4 cents in the third, and so on, each time doubling the amount.
After 1 month, who came out ahead?After 1 month, who came out ahead? THINK- PAIR- SHARETHINK- PAIR- SHARE
Group: Money Doubling?
• You have a $100.00• Your money doubles each year.• How much do you have in 5 years?• Show work. Use a table and/or equation!
Money Doubling
Year 1: $100 · 2 = $200Year 2: $200 · 2 = $400Year 3: $400 · 2 = $800Year 4: $800 · 2 = $1600Year 5: $1600 · 2 = $3200
Earning Interest
• You have $100.00.• Each year you earn 10% interest.• How much $ do you have in 5 years?• Show Work.
• HINT…how much is 10% of $100? HINT…..can you find a constant multiplier?
Earning 10% results
Year 1: $100 + 100·(.10) = $110Year 2: $110 + 110·(.10) = $121Year 3: $121 + 121·(.10) = $133.10Year 4: $133.10 + 133.10·(.10) = $146.41Year 5: $146.41 + 1461.41·(.10) = $161.05Can you find an equation? start at 100, CM = 110/100 = 1.1 Equation?
y = 100(1.1)x y = 100(1.1)5=161.05
Growth Models: Investing
The equation for constant percent growth is
y = A (1+ )x
A = starting value (principal)r = rate of growth (÷100 to put in decimal form)
x = number of time periods elapsedy = final value
100
r
Using the Equation
• $100.00• 10% interest • 5 years• 100(1+ )5 = 100( 1 + 0.10)5
= 100 (1.1)5 = $161.05
10
100
10% as a fraction
Constant multiplier
10% as a decimal
Comparing Investmentswhich is better?
• Choice 1– $10,000– 5.5% interest– 9 years
• Choice 2– $8,000– 6.5% interest– 10 years
Choice 1$10,000, 5.5% interest for 9 years.
Equation: y =$10,000 (1 + )9
=10,000 (1 + 0.055)9
= 10,000(1.055)9
Balance after 9 years: $16,190.94
5.5
100
Choice 2
$8,000 in an account that pays 6.5% interest for 10 years. Equation: y=$8000 (1 + )10
=8,000 (1 + .065)10
=8,000(1 + 0.065)10
Balance after 10 years: $15,071.10
6.5
100
Which Investment?
• The first one yields more money.
– Choice 1: $16,190.94 – Choice 2: $15,071.10
Exponential Decay
Instead of increasing, it is decreasing.
Formula: y = A (1 – )x
A = starting valuer = rate of decrease (÷100 to put in decimal form)
x= number of time periods elapsedy = final value
100
r
Real-life Examples
• What is car depreciation?• Car Value = $20,000• Depreciates 10% a year• Figure out the following values:
– After 2 years– After 5 years– After 8 years– After 10 years
Exponential Decay: Car Depreciation
DepreciationRate
Value after 2 years
Value after 5 years
Value after 8 years
Value after 10 years
10% $16,200 $11,809.80 $8609.34 $6973.57
Assume the car was purchased for $20,000
Formula: y = a (1 – )t
a = initial amountr = percent decrease t = Number of years
100
r
debriefHow does the exponential growth differ from linear growth?
How does the difference show up in the table?
How does the difference show up on the graph?
Worksheet find then towards the end of page http://www.uen.org/Lessonplan/preview.cgi?LPid=24626
http://www.regentsprep.org/regents/math/algebra/AE7/ExpDecayL.htm