The half-life of a radioactive substance is 12 days. After 11 days a sample of the substance has been reduced to a mass of 7 mg.

What was the initial mass (in mg) of the sample?
Round your answer to 2 decimal places as needed.

Using your answer from above, after how many days will there only be 0.1 mg of substance left?

Step-by-step explanation:

The mass (m) remaining will be a function of the initial mass (m0) and the time in days (t) according to the exponential decay formula ...

  m = m0(1/2)^(t/12)

a) You want to find m0 for m=7 and t=11. Put these numbers in the formula and solve for m0.

  7 = m0(1/2)^(11/12) = 0.529732×m0

  m0 = 7/0.529732 = 13.21 . . . . grams

The initial mass of the sample was 13.21 grams.

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b) We can use the value of m0 we found to determine the number of days until 0.1 mg will remain:

  0.1 = 13.21(1/2)^(t/12)

Dividing by 13.21 and taking logs, we have ...

  log(0.1/13.21) = (t/12)log(1/2)

  t = 12·log(0.1/13.21)/log(1/2) = 84.55

There will be only 0.1 mg left after 84.6 days.

Comment on the calculations

For the latter part of the problem, be sure to use the full-precision value of  m0. Do not round any results until the final answer. Alternatively, you could use 7 mg for m0 and add 11 days to the result you get.

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl