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Mathematics 63 Online
OpenStudy (acxbox22):

this is a question about half-life 1. Cesium-137 has a half life of 30 years. suppose a lab stored a 100 g sample in 1970. a) How many grams remained in the year 2000? b) How many grams will remain in 2090?

OpenStudy (campbell_st):

well you need to find the growth constant.... in 1970 = 100g so 30 years later 2000 = 50g this is the half life... you need to find the value of k, the growth (decay) constant before attempting part (b).

OpenStudy (acxbox22):

@AccessDenied

OpenStudy (campbell_st):

are you using \[P_{t} = P_{0} \times e^{kt}\]

OpenStudy (acxbox22):

i dont know what the variables stand for

OpenStudy (campbell_st):

well in your notes on this question... are they talking about exponential growth... using the exponential e as the base...

OpenStudy (acxbox22):

they dont give any formula all in know is that it is a exponential decay thy taught us the compound interest formula and that is it i dont know anything about half lifes so i am confused :(

OpenStudy (acxbox22):

for example a value of a car goes down 20% each year the car is 3000 dollars find the value of the car after 3 years \[v=3000(0.8)^{3}\] that is my basic knowledge of exponential functions

OpenStudy (campbell_st):

here is how I would do it using exponentials... half life occurs at 30 years so you have 1/2 the original amount... so you need to find the growth(or decay) constant k \[50 = 100 \times e^{k \times 30}\] then \[0.5 = e^{30k}\] take the ln of both sides \[\ln(0.5) = 30k\] solve for k. you should write k to 4 or 5 decimal place.... and in this question it will be negative.... since it is a decay problem. for part (b) you need the constant of decay k, from the above question use \[P_{t} = 100 \times e^{120 \times k}\] the alternate method is every 30 years it gets to half its previous amount 1970 = 100 g 2000 = 50 g (30 years half life) 2030 = 25g (half the previous) 2060 = 12.5g ( half the previous) 2090 = 6.25g ( half the previous life) hope it helps

OpenStudy (campbell_st):

given its a half life question... I'd use the 2nd option

OpenStudy (acxbox22):

thank you so much..i have been on this question for more than an hour

OpenStudy (acxbox22):

you are also the first person who i fanned

OpenStudy (campbell_st):

glad to help

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