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Question 162310: CAN SOMEONE PLEASE HELP ME WITH THIS PROBLEM:
Suppose the amount of a radioactive element remaining in a sample of 100 milligrams after x years can be described by A(x) = 100e^-0.01657x. How much is remaining after 257 years? Round to the nearest hundredth of a milligram.
a. 0.01 milli
b. 1.42 milli
c. 425.85 milli
d. 7070.31 milli
: CAN SOMEONE PLEASE HELP ME WITH THIS PROBLEM:
Suppose the amount of a radioactive element remaining in a sample of 100 milligrams after x years can be described by A(x) = 100e^-0.01657x. How much is remaining after 257 years? Round to the nearest hundredth of a milligram.
a. 0.01 milli
b. 1.42 milli
c. 425.85 milli
d. 7070.31 milli

Answer by ankor@dixie-net.com(4668) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose the amount of a radioactive element remaining in a sample of 100
milligrams after x years can be described by A(x) = 100e^-0.01657x.
How much is remaining after 257 years?
Round to the nearest hundredth of a milligram.
:
A(x) = 100*e^(-0.01657x)
:
Substitute 257 for x;
A(x) = 100*e^(-0.01657*257)
:
A(x) = 100*e^-4.25849
:
on a good calc find: e^-4.25849
A(x) = 100 * .01414
:
A(x) = 1.41 milligrams
Question 162310: CAN SOMEONE PLEASE HELP ME WITH THIS PROBLEM:
Suppose the amount of a radioactive element remaining in a sample of 100 milligrams after x years can be described by A(x) = 100e^-0.01657x. How much is remaining after 257 years? Round to the nearest hundredth of a milligram.
a. 0.01 milli
b. 1.42 milli
c. 425.85 milli
d. 7070.31 milli
: CAN SOMEONE PLEASE HELP ME WITH THIS PROBLEM:
Suppose the amount of a radioactive element remaining in a sample of 100 milligrams after x years can be described by A(x) = 100e^-0.01657x. How much is remaining after 257 years? Round to the nearest hundredth of a milligram.
a. 0.01 milli
b. 1.42 milli
c. 425.85 milli
d. 7070.31 milli

Answer by gonzo(568) About Me  (Show Source):
You can put this solution on YOUR website!
formula is given as:
a(x) = 100*e^(-.01657*x)
since x = 257, equation becomes
a(x) = 100*e^(-.01657*257)
which becomes
a(x) = 100*e^(-4.25849)
which becomes
a(x) = 100*.014143643
which becomes
a(x) = 1.41436432
rounded to the nearest hundredth of a milligram equals
1.41
which is close to 1.42 but not quite right on.
since that's the only choice real close, i would assume 1.42 is the answer they are looking for.
answer is b. 1.42