Algebra 2 Problems of the Day (Algebra Regents, June 2025 Part IV)

This exam was adminstered in June 2025.

June 2025 Algebra 2, Part IV

A correct answer is worth up to 6 credits. Partial credit can be given. Work must be shown or explained.

37. Cesium-137 decay can be modeled with the formula A(t) = A₀e^kt, where A(t) represents the mass remaining in grams after t years and A₀ represents the initial mass. A sample of 500 grams of cesium-137 takes approximately 60.34 years to decay to 125 grams. Use this sample with the given formula to determine the constant k, to the nearest thousandth.

Use this value for k to write a function, A(t), that will find the mass of the 500-gram sample remaining after any amount of time, t, in years.

Graph A(t) on the graph below from t = 0 to t = 150 years.

Use A(t) to calculate the average rate of change in grams per year, from t = 0 to t = 60 years, to the nearest tenth.

Explain what this value means in the given context.

Answer:

Write the original formula: A(t) = A₀e^kt. Next, substitute what you know to find out what you don't know through inverse operations.

A(t) = A₀e^kt

125 = 500 e^60.34k

125/500 = e^60.34k

.25 = e^60.34k

ln .25 = 60.34k

-1.386294... = 60.34k

-0.23 = k

Subsitute this value into A(t).

A(t) = 500 e^-.023t

Put this into the graphing calculator (using y for A(t) and x for t) and look at the table of values to make the graph.

The following graph came from the nyresgents.org website:

To find the average rate of change, calculate ( A(60) - A(0) ) / 60 - 0, using the values from the Table of Values.

( 125.79 - 500) / (60 - 0) = -374.21 / 60 = -6.2 grams per year.

In this context, the mass of Cesium-137 decays at an average rate of 6.2 grams per year.

End of Part Exam

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