algebra 1 warm up 9 april 2012
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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 PresentationTRANSCRIPT
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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
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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.
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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
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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!
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Money Doubling
Year 1: $100 · 2 = $200Year 2: $200 · 2 = $400Year 3: $400 · 2 = $800Year 4: $800 · 2 = $1600Year 5: $1600 · 2 = $3200
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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?
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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
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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
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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
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Comparing Investmentswhich is better?
• Choice 1– $10,000– 5.5% interest– 9 years
• Choice 2– $8,000– 6.5% interest– 10 years
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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
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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
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Which Investment?
• The first one yields more money.
– Choice 1: $16,190.94 – Choice 2: $15,071.10
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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
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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
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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
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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?
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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