All Algebra II Resources
Example Question #1 : Radioactive Decay Equations
Over the past few years, the number of students enrolled at a certain university has been decreasing. Each year there is a 12% decrease in student enrollement. Currently, 14,286 students are enrolled. If this trend continues, how many students will be enrolled in 5 years?
This is an exponential decay problem. The formula for exponential decay is:
= future value
= present value
= rate of decay
= number of periods
This problem requests the number of students five years in the future. The rate of decay is twelve percent. Therefore:
Example Question #2 : Radioactive Decay Equations
Sceintists recently discovered a new type of metal compound. They have roughly 15 grams of this compound, which has a half life of 16 hours. Approximately how much of this substance will the scientists have in 24 hours?
Recall the radioactive decay formula:
The half life formula is:
, where is the half life.
Plug in the given half life:
Plug this value into the radioactive decay formula:
Example Question #3 : Radioactive Decay Equations
The equation for radioactive decay is,
Where is the original amount of a radioactive substance, is the final amount, is the half life of the substance, and is time.
The half life of Carbon-14 is about years. If a fossile contains grams of Carbon-14 at time , how much Carbon-14 remains at time years?
None of the other answers.
Using the equation for radioactive decay, we get:
Example Question #4 : Radioactive Decay Equations
The number of butterflies in an exhibit is decreasing with an exponential decay. The number of butterflies is decreasing by every year. There are butterflies in the exhibit right now. How many butterflies will be in the exhibit in years?
Because the butterflies are decreasing exponentially, we can use this equation
is the final value
is the original value
The decay for this problem is 5% or 0.05
The period of time is 7 years
Using this equation we can solve for
Example Question #5 : Radioactive Decay Equations
The number of fish in an aquarium is decreasing with exponential decay. The population of fish is decreasing by each year. There are fish in the aquarium today. If the decay continues how many fish will be in the aquarium in years?
None of these answers are correct
Every year the population of fish losses 7%. In other words, every year 93% of the fish remain from the previous year. Knowing this, we can use the original number of fish to find the number of fish for the next year. Since we want to know the number of fish 4 years from now, we multiply 1500 by 93% four times.
Example Question #6 : Radioactive Decay Equations
The population of a city is decreasing. The city has a population of , people today, but the population decreases by every year. What will be the population of the city in years if this continues?
Because the population of the city is decreasing every year at 10.5% we can find the population after each year by using
Because this decrease will continue every year for the 6 years, we can continue to multiply the population by the decay for every year.
Example Question #7 : Radioactive Decay Equations
There is water leaking out of a cup. of the water is leaking out every minute. How many kilograms of water will be left in minutes and seconds, if there are kilograms of water ,, in the cup right now?
None of these answers are correct
Because the water is leaking at a continuous rate, we can use the exponential decay equation.
is the decay of the problem, 12% or 0.12. is equal to how many times the water will have a 12% decay. This can be calculated as
To calculate this we must first convert both time to seconds
Our equation is then
Example Question #8 : Radioactive Decay Equations
An animal population is dying out. There is a decrease in this number of animals by every year. In years, there will be of this animal left. What is the current population of this animal today?
This is an exponential decay problem. Therefore, we can use this equation.
is the animal population after the 7 years. is the animal population right now. is the decay of the animal population every year. is the time period of the animal populations decay.
From the problem we know after the 7 years the animal population will be 80, so
The population is decreasing by 8% every year, therefore
= 8% = 0.08
is equal to the number of times a decay of 8% has occurred. Since an 8% decay happens every year, and the population is 7 years from now, the population has decayed 8%, 7 times. In other words,
These values give us,
Rearranging this equation we can solve for
Example Question #9 : Radioactive Decay Equations
A school is losing a certain number of students each year. This year, the school has students. Four years ago the school had students. The yearly rate of the school losing students has been the same for the last four years. What is the school's yearly rate of losing students?
This is an exponential decay problem, meaning that the decay of the school's students can be found using
is the number of students currently at the school, which is 242
is the number of students that were at the school 4 years ago, which is 591
is the number of times the decay has occurred. Since, we are trying to find the yearly decay, the decay that happened to the school from one year to the next, and we have the number of students from 4 years ago, .
is the decay rate of the school that we are trying to find. Because we have every number except for , we can plug the values into the equation to solve for .
Example Question #10 : Radioactive Decay Equations
Cells in a dish have started to decay. The cells are decaying by every minutes. When you left the cells there were cells in the dish. Now there are cells in the dish. Approximately how long did you leave the cells for?
This is an exponential decay problem. That means that after 20 minutes 3% of the cells decay and 97% of the cells in the dish are left. To find the new amount of cells in the dish, we multiple the original number by the 97%.
The number of cells after every 20 minute interval can be calculated this way. Therefore, to find how long the cell were decaying we use,
Which can be rewritten as,
Now we can solve for , which is the amount of time that you were gone
To solve for , we must take the log of both sides to base 10. This will give us,
Remember! is the number of decays that the cells went through. Each decay took 20 minutes to get through, however.
Therefore, the time that you were gone is