Question

Cabon-14 has a half life of 5760 years .How old is a fossil that has 1/8 of the substance left?

Answer 1

multiply half-life by 3
3*5760 = 17,280

Answer 2

17,280?

Answer 3

What grade is this for?

Answer 4

Algebra 2

Answer 5

shouldnt it be 1440 yrs?

Answer 6

how?

Answer 7

are you doing y=a*b^kt or y=a*e^kt?

Answer 8

well 1/8 is 1/4 of 1/2 so i divided 5760 by 4...dunno if it should be done that way...is it?

Answer 9

usually for carbon half-life you use y = a*e^kt
1/2 = e^5760k

1/2 ^3 = 1/8

--> 1/8 = e^3(5760k)
thus t = 3*5760

Answer 10

\[y=A_0 e^{kt}\] where A_0 is the initial amount.

when y=0.5A_0, t=5760
\[0.5A_0=A_0 e^{5760k}\]\[0.5=e^{5760k}\]\[ln0.5=lne^{5760k}\]\[-0.693=5760k\]\[k= -0.00012 \]

when y=1/8 A_0
\[A_0/8=A_0 e^{-0.00012t}\]\[1/8=e^{-0.00012t}\]\[ln1/8=lne^{5760k}\]\[-2.079=-0.00012t\]\[t= 17 328 \]

Answer 11

TY