MATH SOLVE

8 months ago

Q:
# the half life of a certain radioactive material is 68 hours an initial amount of the material has mass of 641 kg. Write an exponential function that models the decay of this material. Find how much radioactive material remains after 4 hours. Round your answer to the nearest thousandth

Accepted Solution

A:

Answer: a) [tex]A=641e^{-0.0101933t}[/tex]b) Radioactive material remain after 4 years = 615.39 kg

Step-by-step explanation:we know that the general equation of exponential decay is:[tex]A=Pe^{kt}[/tex]where,A is amount after time tP is the initial amount =641 kgk is a constantt is time periodThe half-life of a certain radioactive material = 68 hours[tex]\Rightarrow0.5=e^{k\times68}\\\Rightarrow\ \ln(.5) = 68k\\\Rightarrow\ k=\frac{ln(.5)}{68}\\\Rightarrow\ k=\frac{-0.693147181}{68}\\\Rightarrow\ k=-0.010193341[/tex]The required equation will be :[tex]A=641e^{-0.0101933t}[/tex]For t=4 years [tex]A=641e^{-0.0101933\times4}\\\Rightarrow\ A=641e^{-0.0407734}\\\Rightarrow\ A = 641(0.960046687)\\\Rightarrow\ A=615.39\ kg[/tex]Radioactive material remain after 4 years = 615.39 kg

Step-by-step explanation:we know that the general equation of exponential decay is:[tex]A=Pe^{kt}[/tex]where,A is amount after time tP is the initial amount =641 kgk is a constantt is time periodThe half-life of a certain radioactive material = 68 hours[tex]\Rightarrow0.5=e^{k\times68}\\\Rightarrow\ \ln(.5) = 68k\\\Rightarrow\ k=\frac{ln(.5)}{68}\\\Rightarrow\ k=\frac{-0.693147181}{68}\\\Rightarrow\ k=-0.010193341[/tex]The required equation will be :[tex]A=641e^{-0.0101933t}[/tex]For t=4 years [tex]A=641e^{-0.0101933\times4}\\\Rightarrow\ A=641e^{-0.0407734}\\\Rightarrow\ A = 641(0.960046687)\\\Rightarrow\ A=615.39\ kg[/tex]Radioactive material remain after 4 years = 615.39 kg