Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #319 : Exponents

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?

Possible Answers:

Correct answer:

Explanation:

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 #4 : 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?

Possible Answers:

None of these answers are correct

 kilograms

 kilograms

 kilograms

Correct answer:

 kilograms

Explanation:

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 #1 : 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?

Possible Answers:

Correct answer:

Explanation:

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 #5 : 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?  

Possible Answers:

Correct answer:

Explanation:

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 .

 = 20%

Example Question #6 : 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?

Possible Answers:

Correct answer:

Explanation:

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

Example Question #791 : Mathematical Relationships And Basic Graphs

Suppose 5 milligrams of element X decays to 3.2 milligrams after 48 hours.  What is the decay rate on a day to day basis?

Possible Answers:

Correct answer:

Explanation:

Write the formula for radioactive decay.

Substitute the values in the equation and solve for lambda.

Divide by 3.2 on both sides.

Take the natural log of both sides to eliminate the exponential.

Divide by negative two on both sides.

Convert this to a percentage.

This element's decay rate is approximately:  

Example Question #3451 : Algebra Ii

An element has a half life of 6 days.  What is the approximate amount remaining for a 50 mg sample of this element after 5 days?

Possible Answers:

Correct answer:

Explanation:

Write the formula for half life.

Since the time requested is five out of the six day of the half life, the value of  is:

Substitute all the known values into the equation.

The answer is:  

Example Question #791 : Mathematical Relationships And Basic Graphs

The number of butterflies in an exhibit is decreasing at an exponential rate of 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? 

Possible Answers:

Correct answer:

Explanation:

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 #1 : Other Exponent Applications

A biologist figures that the population  of cane toads in a certain lake he is studying can be modeled by the equation

,

where  is the number of days elapsed in 2015. For example,  represents January 1,   represents January 2, and so forth.

If this model continues, in what month will the population of cane toads in the lagoon reach 5,000?

Possible Answers:

April

June

March

May

February 

Correct answer:

March

Explanation:

Set  and solve for  :

January and February have 59 days total; add March, and this is 90 days. The 76th day is in March.

 

Example Question #2 : Other Exponent Applications

A biologist figures that the population  of cane toads in a certain lake he is studying can be modeled by the equation

,

where  is the number of days elapsed in 2015. For example,  represents January 1,   represents January 2, and so forth.

Assuming that this has been the model for their growth throughout the previous year as well, in what month did the population hit 100 cane toads?

Possible Answers:

August 2014

September 2014

October 2014

December 2014

November 2014

Correct answer:

December 2014

Explanation:

Set  and solve for :

17 days before January 1 was in December of 2014.

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