Respuesta :
Answer:
Step-by-step explanation:
Using the equation A(t) = 400e-.032t
a) replace t with 4 so A(4) = 400e((-.032)(4))
The hardest part about this is making sure to use order of operations. Be certain it works like this:
A(4) = 400e-.128
A(4) = 400(.8799)
A(4) = 351.9 grams
b) A(8) = 400e((-.032)(8)) = 309.7 grams
c) A(20) = 400e((-.032)(20)) = 210.9 grams
Note here that even after 20 years, not quite half of the original amount is gone. So, we can anticipate that in finding the half life, that our answer should be slightly greater than 20 years.
d) 200 = 400e(-.032t)
Divide both sides of the equation by 400.
.5 = e(-.032t)
Change this to logarithmic form.
Ln .5 = -.032t
-.6931≈ -.032t
t ≈ 21.7 years
Hope this helps!
The amount of radioactive lead,
(a).After 3 years is 454.23 grams
(b).After 8 years is 387.07 grams
(c).After 10 years is 363.07 grams.
(d). half life is 21.66 years.
The decay of radioactive lead is given by function,
[tex]A(t)=500e^{-0.032t}[/tex]
The amount of radioactive lead After 3 years is,
[tex]A(3)=500e^{-0.032*3}=0.908*500=454.23g[/tex]
The amount of radioactive lead After 8 years is,
[tex]A(8)=500e^{-0.032*8}=500*0.774=387.07g[/tex]
The amount of radioactive lead After 10 years is,
[tex]A(10)=500e^{-0.032*10}=500*0.726=363.07g[/tex]
Half life is defined as the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay.
So, [tex]250=500e^{-0.032t}[/tex]
[tex]e^{-0.032t}=0.5\\\\-0.032t=ln(0.5)\\\\-0.032t=-0.693\\\\t=0.693/0.032=21.66 years[/tex]
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