A radioactive element decays at a rate of 5% annually. There are 40 grams of the substance present. • How much of the substance remains after 30 years (to the nearest tenth)? • When will the amount of the substance drop to below 20 grams (to the nearest year)?

1) The amount of substance that will remain after 30 years is; A(30) = 8.6 g

2) The time when the amount of the substance drop to below 20 grams is; t < 14 years

A(t) = A_o(e^(-rt))

Where;

A_o is initial amount

r is rate of decay

t is time in years

A(t) is amount left after t years

We are given;

A_o = 40 g

r = 5% = 0.05

1) To find the amount of substance left after 30 years is;

A(30) = 40 × e^(-0.05 × 30)

A(30) = 8.925 g

Approximating to the nearest tenth gives;

A(30) = 8.9 g

2) To find the time when the amount of the substance drop to below 20 grams.

20 > 40 × e^(-0.05 × t)

20/40 > e^(-0.05 × t)

-0.05t < In 0.5

t < (In 0.5)/-0.05

t < 13.86 years

Approximating to the nearest year gives;

t < 14 years

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8.6 grams of the substance remains after 30 years.

The amount of the substance will drop to below 20 grams in 14 years.

Step-by-step explanation:

The amount of the substance after t years is given by the following equation:

In which A(0) is the initial amount and r is the yearly decay rate.

A radioactive element decays at a rate of 5% annually. There are 40 grams of the substance present.

This means, respectively, that

So

How much of the substance remains after 30 years (to the nearest tenth)?

This is A(30).

8.6 grams of the substance remains after 30 years

When will the amount of the substance drop to below 20 grams (to the nearest year)?

This is t when A(t) = 20. So

Rounding to the nearest year

The amount of the substance will drop to below 20 grams in 14 years.