SOLUTION: The following table shows the amount of polonium in milligrams remaining after x days from an original sample of 2 milligrams.
x(days) 0 100 200 300
y milligr
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x(days) 0 100 200 300
y milligr
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Question 270807: The following table shows the amount of polonium in milligrams remaining after x days from an original sample of 2 milligrams.
x(days) 0 100 200 300
y milligrams 2 1.22 0.743 0.453
a. is the half life pf te polonium less than 200 days?
b. give a general formula thatr can be used to approximate the amount A of polonium after x days.
c. what is the half-life of the polonium? GRaph A(x) and show th epoint that would show the half-life for x=2 milligrams. Found 2 solutions by ankor@dixie-net.com, stanbon:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! The following table shows the amount of polonium in milligrams remaining after x days from an original sample of 2 milligrams.
x(days) 0 100 200 300
y milligrams 2 1.22 0.743 0.453
:
a. is the half life of the polonium less than 200 days?
Find the half life (h) using values for 100 days and resulting amt of 1.22 mg
2*(2^(-100/h)) = 1.22
:
2^(-100/h) =
;
2^(-100/h) = .61
:
log(2^(-100/h)) = log(.61) log(2) = log(.61) = = -.713
h =
h ~ 140 days is the half life
:
b. give a general formula that can be used to approximate the amount A of polonium after x days.
A = 2*2^(-x/140)
:
:
c. what is the half-life of the polonium?
we found it to be 140 days in (a)
:
GRaph A(x) and show th epoint that would show the half-life for x=2 milligrams.
Graph the equation y = 2*2^(-x/140)
You can see that at 140 days there is about 1 mg of the substance
You can put this solution on YOUR website! The following table shows the amount of polonium in milligrams remaining after x days from an original sample of 2 milligrams.
x(days) 0 100 200 300
y milligrams 2 1.22 0.743 0.453
------------------
Form: y = ab^x ; Find a and b
---
2 = ab^0, so 2 = a
-----------------------
Now y = 2b^x
1.22 = 2b^100
0.743 = 2b^200
----
Divide the 1st into the 2nd to solve for "b"
b^100 = 0.743/1.22 = 0.6090
b = 0.6090^(1/100) = 0.9951
-----
Now y = 2*0.9951^x
----
a. is the half life pf te polonium less than 200 days?
Let y = 1 and solve for "x":
1 = 2*(0.9951)^x
x*log(0.9951) = log(1/2)
x = 141.11 days
Ans to Question: Yes
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b. give a general formula that can be used to approximate the amount A of polonium after x days.
A(t) = 2*(0.9951)^x
------------------------------
c. what is the half-life of the polonium?
Ans: 141.11 days
--------------------------------
Graph A(x) and show the point that would show the half-life for x=2 milligrams.
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Cheers,
Stan H.
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