Question

if 500 gramsof the substance were present initially and 400 grams are present 50 years later, how many grams will be present after 200 years?

Answer 1

100 grams disappear every 50 years, so in 200 years 400 more grams will disappear, leaving you with no substance after 200 years.

Answer 2

that makes sense :)

Answer 3

haha

Answer 4

:)

Answer 5

mirra you know you can close the question after its been answered so it doesn't show up on the open question list?

Answer 6

Hmm, more explanations pleasee ?

Answer 7

The answer is ?

Answer 8

the answer is 0 grams

Answer 9

okay, it doesn't help at all. the answer is wrong -,-

Answer 10

impossible

Answer 11

@texaschic101 can it be wrong?

Answer 12

oh that is not even close

Answer 13

how

Answer 14

This is radioactive decay... it's not linear.

Answer 15

it decreases as follows;
every 50 hears you have \(\frac{40}{50}=.8\) of the original amount

Answer 16

o.

Answer 17

it must have a calculation. its related to exponential function

Answer 18

I'd assume it's radioactive decay and follows this formula\[\Large A = A_0 e^{-kt}\]
Ao is initial amount, k is rate, t is time

Answer 19

\[50,50\times .8, 50\times .8^2, 50\times .8^3,...\]

Answer 20

every 50 years you have 80% of the amount left,
there are 4 50 year periods in 200 years
compute
\[50\times .8^4\] for your answer (use a calculator)

Answer 21

500 gramsof the substance were present initially and 400 grams are present 50 years later

Ao = 500
when t=50, A=400 \[\Large 400 = 500 e^{-50k}\]solve for k first



Then you need to plug t=200 into the formula \[\Large A = 500 e^{-kt} \]

Answer 22

@agent0smith this is i think too much work, and probably not what the course had in mind (although i could be wrong)
it says "related to exponential function" which my guess is
\[50\times (.8)^t\] where \(t\) is a multiple of 50,
or perhaps if you want times that are not multiples of 50,
\[50\times .8^{\frac{t}{50}}\]

Answer 23

like is said, i could easily be wrong

Answer 24

Fair enough, both methods get the same answer, but yours might be quicker

Answer 25

it would certainly be quicker in this case because \(200=4\times 50\) but your method works as well, although when you truncate the decimal using logs, there will be some error

Answer 26

Yeah yours is faster. The error is pretty negligible though (the amounts are the same to 2 d.p. but i used ~6 decimal places for k)