Question

The half-life of a certain radioactive material is 42 days. An initial amount of the material has a mass of 49 kg. Write an exponential function that models the decay of this material. Find how much radioactive material remains after 8 days. Round your answer to the nearest thousandth.

Answer 1

I know this has something to do with this formula I just don't know how to apply it? \[y = ab^{x} + k\]

Answer 2

@Luigi0210

Answer 3

The half-life of a certain radioactive material is 42 days.
An initial amount of the material has a mass of 49 kg.
Write an exponential function that models the decay of this material.
:
A = Ao*2^(-t/h), where:
A = amt after t time
Ao = initial amt (t=0)
t = time of decay


h = half life of substance
:
A = 49*2^(-t/42)
:
:
Find how much radioactive material remains after 8 days.
A = 49*2^(-8/42)
A = 49*.8763
A = 42.9395 ~ 42.940, to the nearest thousandth.

Answer 4

The constant term, "k", has no business in this formulation. It neither grows nor deteriorates. No good.

Answer 5

@jennaknight_03
Could you clarify on how you got all that?
A = 49*2^(-t/42)
How did you times 49 by 2? Where did you get the 2?

Answer 6

im not sure about that 1. Ask @Luigi0210 about it

Answer 7

The half-life of a certain radioactive material is 42 days. An initial amount of the material has a mass of 49 kg. Write an exponential function that models the decay of this material. Find how much radioactive material remains after 8 days. Round your answer to the nearest thousandth.

Given an initial amount, A0, we have: \((1/2)A_{0} = A_{0}e^{k\cdot 42}\). Solving for k, gives:

\((1/2) = e^{k\cdot 42}\)

\(-\ln(2) = k\cdot 42\)

\(k = -\dfrac{\ln(2)}{42}\)

Now, let's test it. Given \(A_{0} = 49\), we should have \(A_{42} = 49/2 = 24.5\)

\(49e^{\left(-\dfrac{\ln(2)}{42}\right)\cdot 42} = 49e^{-\ln(2)} = 49\cdot\dfrac{1}{2} = 24.5\)

Now, let's try it for t = 8.

\(49e^{\left(-\dfrac{\ln(2)}{42}\right)\cdot 8} =\;\;\)??