# 8.8 – Exponential Growth & Decay

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8.8 – Exponential Growth & Decay. Decay:. Decay: 1. Fixed rate. Decay: 1. Fixed rate: y = a (1 – r ) t. Decay: 1. Fixed rate: y = a (1 – r ) t where a = original amount. Decay: 1. Fixed rate: y = a (1 – r ) t where a = original amount r = rate of decrease. - PowerPoint PPT Presentation

### Text of 8.8 – Exponential Growth & Decay

• 8.8 Exponential Growth & Decay

• Decay:

• Decay:1. Fixed rate

• Decay:1. Fixed rate: y = a(1 r)t

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amount

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body?

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body? 11% indicates that it is fixed-rate decay.

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body? 11% indicates that it is fixed-rate decay. y = a(1 r)t

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body? 11% indicates that it is fixed-rate decay. y = a(1 r)ta = 130

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body? 11% indicates that it is fixed-rate decay. y = a(1 r)ta = 130r = 0.11

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body? 11% indicates that it is fixed-rate decay. y = a(1 r)ta = 130r = 0.11y =

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body? 11% indicates that it is fixed-rate decay. y = a(1 r)ta = 130r = 0.11y = 65

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body? 11% indicates that it is fixed-rate decay. y = a(1 r)ta = 130 r = 0.11y = 65t = ???

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body? 11% indicates that it is fixed-rate decay. y = a(1 r)ta = 130 65 = 130(1 0.11)tr = 0.11y = 65t = ???

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body? 11% indicates that it is fixed-rate decay. y = a(1 r)ta = 130 65 = 130(1 0.11)tr = 0.1165 = 130(0.89)ty = 65t = ???

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body? 11% indicates that it is fixed-rate decay. y = a(1 r)ta = 130 65 = 130(1 0.11)tr = 0.1165 = 130(0.89)ty = 650.5 = (0.89)tt = ???

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body? 11% indicates that it is fixed-rate decay. y = a(1 r)ta = 130 65 = 130(1 0.11)tr = 0.1165 = 130(0.89)ty = 650.5 = (0.89)tt = ??? log(0.5) = log(0.89)t

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body? 11% indicates that it is fixed-rate decay. y = a(1 r)ta = 130 65 = 130(1 0.11)tr = 0.11 65 = 130(0.89)ty = 65 0.5 = (0.89)tt = ??? log(0.5) = log(0.89)t log(0.5) = tlog(0.89) Power Property

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body? 11% indicates that it is fixed-rate decay. y = a(1 r)ta = 130 65 = 130(1 0.11)tr = 0.11 65 = 130(0.89)ty = 65 0.5 = (0.89)tt = ??? log(0.5) = log(0.89)t log(0.5) = tlog(0.89) Power Property log(0.5) = tlog(0.89)

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a persons body? 11% indicates that it is fixed-rate decay. y = a(1 r)ta = 130 65 = 130(1 0.11)tr = 0.11 65 = 130(0.89)ty = 65 0.5 = (0.89)tt = ??? log(0.5) = log(0.89)t log(0.5) = tlog(0.89) Power Property log(0.5) = tlog(0.89) 5.9480 t

• 2. Natural rate:

• 2. Natural rate: y = ae-kt

• 2. Natural rate: y = ae-kta = original amountk = constant of variation t = time y = new amount

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the half-life of Carbon-14 if its constant of variation is 0.00012.

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the half-life of Carbon-14 if its constant of variation is 0.00012.*No rate given so must be Natural.

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the half-life of Carbon-14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = ae-kt

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the half-life of Carbon-14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = ae-kta = 1

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the half-life of Carbon-14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = ae-kta = 1y = 0.5

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the half-life of Carbon-14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = ae-kta = 1y = 0.5k = 0.00012

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the half-life of Carbon-14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = ae-kta = 1y = 0.5k = 0.00012t = ???

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the half-life of Carbon-14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = ae-kta = 10.5 = 1e-0.00012ty = 0.5k = 0.00012t = ???

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the half-life of Carbon-14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = ae-kta = 10.5 = 1e-0.00012ty = 0.50.5 = e-0.00012tk = 0.00012t = ???

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the half-life of Carbon-14 if its constant of variation is 0.00012.*No rate given so m

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