Please help me this question?

Answer 1

Meanlife #= -1/k#

Technically this is a math problem because it involves integration only.

Let

# M= kI#
# I = int_0^oo te^(kt)dt#
Perform integration by parts on I: #u=t, dv= e^(kt)dt rArr v = 1/ke^(kt)# #I= uv-vdu =1/k (te^(kt))_0^oo-1/kint_0^oo e^(kt)dt# #I= uv-vdu =1/k (te^(kt))_0^oo-1/k^2(e^(kt))_0^oo#
#I= 1/k [(te^(-|k|t))_0^oo-1/k(e^(-|k|t))_0^oo]#
Because# lim_(t->oo) te^(-|k|t) ->0 # For a proof, see https://tutor.hix.ai
#and k= -|k| rArr e^(-k|t|) -> 0#, when # t->oo #
#I = 1/k( (0-0)-1/k(0-1))= 1/k^2 #
#M=-kI = -k(1/k^2) = -1/k = 1/|k| = 1/0.000121#

M= 8264 yrs

To verify this answer, compute C-14 half-life:

Half-life = #ln(2)M = 5729 yrs#
and compare it to the standard C-14 half-life = # 5730 +- 40 yrs #
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Answer 2

Of course, I'd be happy to help! What specific question would you like assistance with?

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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