# ``If 18g of a radioactive substance are present initially and 5 yr later only 9 g remain, how much of the substance will be present after 11yr ?

after 11yr there will be ___g of a radioactive substance.
round the final answer to the nearest thousandth as needed. round all intermediate values to the nearest thousandth as needed.''

## half life is 5 years

amount remaining= 18*e^(-.692*11/5)
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18*e^(-.692*11/5)=

## To find out how much of the radioactive substance remains after 11 years, we can use the formula for exponential decay:

A = A0 * (1/2)^(t/h)

Where:
A = the final amount of the substance remaining after time t
A0 = the initial amount of the substance
t = time elapsed
h = half-life of the substance

In this case, the initial amount A0 is 18g, 5 years have passed, and we know that 9g remain. However, we need to determine the half-life of the substance to use in the formula.

To find the half-life, we can use the formula:

A = A0 * (1/2)^(t/h)
9 = 18 * (1/2)^(5/h)

Since we are rounding intermediate values to the nearest thousandth, we can use trial and error or a calculator to solve for h.

By trial and error, let's assume the half-life (h) is 10 years:

9 = 18 * (1/2)^(5/10)
9 = 18 * (1/2)^(1/2)
9 = 18 * (√(1/2))
9 = 18 * 0.70710678118
9 = 12.72831606764

Since 12.72831606764 is not equal to 9, the half-life is not 10 years. We can further refine our calculation by trying a lower half-life value.

Let's try a half-life of 6 years:

9 = 18 * (1/2)^(5/6)
9 = 18 * (0.82287565553)
9 = 14.81176279854

Since 14.81176279854 is not equal to 9, we can conclude that the half-life is greater than 6 years.

Let's continue narrowing down the half-life value:

Trial 1: h = 7 years
9 = 18 * (1/2)^(5/7)
9 = 18 * (0.84589170773)
9 = 15.22605173894

Trial 2: h = 8 years
9 = 18 * (1/2)^(5/8)
9 = 18 * (0.86602540378)
9 = 15.58845977804

By trial and error, we can conclude that the half-life is approximately 8 years (rounded to the nearest whole number).

Now that we know the half-life is 8 years, we can substitute the values into the formula to find the amount of the substance that will be present after 11 years (t = 11):

A = A0 * (1/2)^(t/h)
A = 18 * (1/2)^(11/8)

Calculating this value:

A = 18 * (0.31622776602)
A = 5.69289978835

Rounding the final answer to the nearest thousandth, we find that after 11 years, approximately 5.693 g of the radioactive substance will remain.