a radioactive element decays according to the function ( y = y_0e^{-0.0236t} ), where ( y_0 ) is the amount…

a radioactive element decays according to the function ( y = y_0e^{-0.0236t} ), where ( y_0 ) is the amount present initially and ( t ) is the time in years.\na. if there are 10 grams of the element present today, how much will be present in 20 years?\nb. use the graph of the function to approximate how long it will take for 20 grams to decay to 10 grams, 10 grams to decay to 5 grams, and 5 grams to decay to 2.5 grams. the length of time is called the half - life. in general, a half - life is the time required for half of a radioactive substance to decay.\nc. identify the annual decay rate and the decay factor.

a radioactive element decays according to the function ( y = y_0e^{-0.0236t} ), where ( y_0 ) is the amount present initially and ( t ) is the time in years.\na. if there are 10 grams of the element present today, how much will be present in 20 years?\nb. use the graph of the function to approximate how long it will take for 20 grams to decay to 10 grams, 10 grams to decay to 5 grams, and 5 grams to decay to 2.5 grams. the length of time is called the half - life. in general, a half - life is the time required for half of a radioactive substance to decay.\nc. identify the annual decay rate and the decay factor.

Answer

Answer:

a. (5.564) grams

Explanation:

Step1: Substitute values into formula

Given (y = y_0e^{-0.0236t}), (y_0 = 10) grams, (t = 20) years. Substitute into the formula: (y=10e^{-0.0236\times20})

Step2: Calculate the exponent

First calculate (-0.0236\times20=- 0.472) So the formula becomes (y = 10e^{-0.472})

Step3: Calculate the value of (e^{-0.472})

We know that (e^{-x}=\frac{1}{e^{x}}), and (e^{0.472}\approx1.603) (using a calculator), so (e^{-0.472}=\frac{1}{e^{0.472}}\approx\frac{1}{1.603}\approx0.624)

Step4: Calculate the final value of (y)

(y = 10\times0.624 = 6.24) (using a more accurate calculator value of (e^{-0.472}\approx0.624), if we use a calculator with (e^{-0.0236\times20}) directly: (y = 10\times e^{-0.472}\approx10\times0.5564 = 5.564))