Call Now to Set Up Tutoring:

(888) 888-0446

Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

Create An Account

All Algebra II Resources

10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2782 : Algebra Ii

Evaluate: $\sum_{k = 0}^{\infty} (-1)^{k} \frac{3}{\pi ^{k}}$

Possible Answers:

$\frac{3 \pi }{ \pi+1 }$

$\frac{ \pi }{3 \pi-1 }$

The series diverges.

$\frac{ \pi }{3 \pi+1 }$

$\frac{3 \pi }{ \pi-1 }$

Correct answer:

$\frac{3 \pi }{ \pi+1 }$

Explanation:

Rewrite the general term so that the sigma notation is as follows:

$\sum_{k = 0}^{\infty} (-1)^{k} \frac{3}{\pi ^{k}}$

$= \sum_{k = 0}^{\infty} (-1)^{k} \cdot 3 \cdot \frac{1}{\pi ^{k}}$

$= \sum_{k = 0}^{\infty} 3 \cdot (-1)^{k} \cdot \left ( \frac{1}{\pi } \right )^{k}$

$= \sum_{k = 0}^{\infty} 3 \left (- \frac{1}{\pi } \right )^{k}$

A series of the form $\sum_{k = 0}^{\infty}a r^{k}$ is an infinite geometric series with common ratio . If |r| < 1 and the initial term is $a_{0}$ , the sum of the series is

$S = \frac{a_{0}}{1-r}$

Here, $r = - \frac{1}{\pi}$ , a = 3 , and

$a_{0}= (-1)^{0} \frac{3}{\pi ^{0}} = 1 \cdot \frac{3}{1} = 3$

Substitute these values in the formula:

$S = \frac{a_{0}}{1-r} = \frac{3}{1- \left ( -\frac{1}{\pi}\right )} = \frac{3}{1+\frac{1}{\pi}}$

Simplifying:

$S = \frac{3}{1+\frac{1}{\pi}}$

$= \frac{3 \cdot \pi }{\left (1+\frac{1}{\pi} \right ) \cdot \pi}$

$= \frac{3 \pi }{1\cdot \pi+\frac{1}{\pi}\cdot \pi }$

$= \frac{3 \pi }{ \pi+1 }$

Report an Error

Example Question #2783 : Algebra Ii

Evaluate: $\sum_{k = 0}^{\infty}7\left ( -\frac{6}{5} \right )^{k}$

Possible Answers:

$3\frac{4}{11}$

$3\frac{9}{11}$

The series diverges.

Correct answer:

The series diverges.

Explanation:

A series of the form $\sum_{k = 0}^{\infty}a r^{k}$ is an infinite geometric series with common ratio . The series converges to a sum if and only if |r|< 1 . However, in the given expression, $r = -\frac{6}{5}$ , so $|r| =\left | -\frac{6}{5} \right | = \frac{6}{5} > 1$ , so the series diverges rather than converges.

Report an Error

Example Question #2784 : Algebra Ii

Evaluate:

$\sum_{k = 0}^{\infty} \frac{4}{e^{k}}$

Possible Answers:

$\frac{e+1}{4 e }$

$\frac{4 e }{e+1}$

$\frac{e-1}{4 e }$

$\frac{4 e }{e-1}$

Cannot be determined

Correct answer:

$\frac{4 e }{e-1}$

Explanation:

Rewrite the general term so that the sigma notation is as follows:

$\sum_{k = 0}^{\infty} \frac{4}{e^{k}} = \sum_{k = 0}^{\infty}\left ( 4 \cdot \frac{1}{e^{k}} \right )= \sum_{k = 0}^{\infty}4 \cdot \left ( \frac{1}{e} \right )^{k}$

A series of the form $\sum_{k = 0}^{\infty}a r^{k}$ is an infinite geometric series with common ratio . If |r| < 1 and the initial term is $a_{0}$ , the sum of the series is

$S = \frac{a_{0}}{1-r}$

Here, $r = \frac{1}{e}$ , a = 4 , and

$a_{0}= \frac{4}{e^{0}} = \frac{4}{1} = 4$

Substitute these values in the formula:

$S = \frac{a_{0}}{1-r}$

$= \frac{4}{1- \frac{1}{e}}$

Simplifying:

$S = \frac{4 \cdot e }{\left (1- \frac{1}{e} \right )\cdot e}$

$S = \frac{4 e }{e-1}$

Report an Error

Example Question #2785 : Algebra Ii

Evaluate:

$\sum_{k = 0}^{\infty} \frac{5}{3^{k}}$

Possible Answers:

$7\frac{1}{2}$

$6\frac{2}{3}$

Cannot be determined

$3\frac{1}{3}$

$3\frac{3}{4}$

Correct answer:

$7\frac{1}{2}$

Explanation:

Rewrite the general term so that the sigma notation is as follows:

$\sum_{k = 0}^{\infty} \frac{5}{3^{k}} = \sum_{k = 0}^{\infty}\left ( 5 \cdot \frac{1}{3^{k}} \right )= \sum_{k = 0}^{\infty}5 \cdot \left ( \frac{1}{3} \right )^{k}$

A series of the form $\sum_{k = 0}^{\infty}a r^{k}$ is an infinite geometric series with common ratio . If |r| < 1 and the initial term is $a_{0}$ , the sum of the series is

$S = \frac{a_{0}}{1-r}$

Here, $r = \frac{1}{3}$ , a =5 , and

$a_{0}= \frac{5}{3^{0}} = \frac{5}{1} = 5$

Substitute these values in the formula:

$S = \frac{a_{0}}{1-r}= \frac{5}{1- \frac{1}{3}} = \frac{5}{ \frac{2}{3}}= 5 \div \frac{2}{3} = \frac{5}{1} \cdot\frac{3}{2} = \frac{15}{2}= 7\frac{1}{2}$ .

Report an Error

Example Question #1 : Other Sequences And Series

Consider the following formula for a recursive sequence:

$a_n=(a_{n-1})^2; a_1=2$

Which answer choice best represents this sequence?

Possible Answers:

2, 4, 16, 32, ...

2, 4, 6, 8, ...

2, 4, 8, 16, ....

2, 4, 16, 256, ...

Correct answer:

2, 4, 16, 256, ...

Explanation:

A recursive formula creates a sequence where each term is defined by the term(s) that precede it. In other words, in order to know term 12, you have to know term 11, etc.

$a_n=(a_{n-1})^2; a_1=2$

The problem already tells us that the first term is 2. Let's find the second term.

$a_2=(a_{2-1})^2=(a_1)^2=2^2=4$

We continue to find the rest of the terms in this way.

$a_3=(a_{3-1})^2=(a_2)^2=4^2=16$

$a_4=(a_{4-1})^2=(a_3)^2=16^2=256$

Report an Error

Example Question #2 : Other Sequences And Series

A sequence is defined recursively as follows:

$a_{1} = 1$

$a_{2} = -1$

$a_{n} = a_{n - 1} a_{n - 2}$ for n = 3, 4, 5, ...

How many of the first twenty terms of the sequence are positive?

Possible Answers:

Correct answer:

Explanation:

Apply the rule to find the first few terms:

$a_{1} = 1$

$a_{2} = -1$

$a_{3} = a_{2} a_{1} = -1 \cdot 1 = -1$

$a_{4} = a_{3} a_{2} = -1 \cdot \left (-1 \right ) = 1$

$a_{5} = a_{4} a_{3} = 1 \cdot \left (-1 \right ) = - 1$

$a_{6} = a_{5} a_{4} = -1 \cdot 1 = - 1$

After the sixth term, it is apparent that this cycle will repeat itself, so the first twenty terms of the sequence will be, in order:

$\left \{ 1, -1, -1, 1, -1, -1,1, -1, -1, 1, -1, -1, 1, -1, -1,1, -1, -1,1, -1\right \}$

Seven of these first twenty terms are positive.

Report an Error

Example Question #3 : Other Sequences And Series

A sequence is defined recursively as follows:

$a_{1} = -5$

$a_{n} = 2 a_{n-1} +6$ for n = 2, 3, 4,...

Which of the following is the first positive term of the sequence?

Possible Answers:

$a_{5}$

$a_{3}$

The sequence has no positive terms.

$a_{4}$

$a_{6}$

Correct answer:

$a_{4}$

Explanation:

Apply the rule to find the first few terms:

$a_{1} = -5$

$a_{2} = 2 a_{ 1} +6 = 2 (-5) + 6 = -10 + 6 = -4$

$a_{3} = 2 a_{ 2} +6 = 2 (-4) + 6 = -8 + 6 = -2$

$a_{4} = 2 a_{ 3} +6 = 2 (-2) + 6 = -4 + 6 = 2$

The first positive term of the sequence is $a_{4}$ .

Report an Error

Example Question #1 : Other Sequences And Series

Which of the following expressions describes the sequence below:

3,8,15,24,35,48,63,...

Possible Answers:

3n-1

2n+2

5n^2-3n+1

n^3+3

n^2+2n

Correct answer:

n^2+2n

Explanation:

In order to determine which expression describes the sequence in the problem, we must determine the relationship each entry has with its position in the sequence. For example, for n=1, we must determine which expression involving n will yield a result of 3, for n=2, we must determine which expression will yield a result of 8, and so on, ensuring that the expression holds true for every n value in the sequence. If we check each of our answers, we can see that only the following expression will give the correct result for each increasing value of n:

n^2+2n

(1)^2+2(1)=3

(2)^2+2(2)=8

(3)^2+2(3)=15

If we continue, we can see that we will obtain the sequence 3,8,15,24,35,48,63,..., so this expression is the correct representation of the sequence given in the problem.

Report an Error

Example Question #3 : Other Sequences And Series

What is the mean of the following quiz scores.

99, 70, 84, 63, 91, 59

Possible Answers:

77.7

78.6

76.7

Correct answer:

77.7

Explanation:

To find the mean of a set of numbers we first must add all the numbers together.

99 + 70 + 84 + 63 + 91 + 59 = 466

Using the formula for mean we get,

$\text{Mean}=\frac{\text{Sum of Numbers}}{\text{Total Number of Entries}}$

Therefore we get,

$\text{Mean}=\frac{446}{6}=77.66=77.7$

Report an Error

Example Question #2 : Other Sequences And Series

What is the median of the following prices.

83, 52, 36, 57

Possible Answers:

54.5

53.5

55.5

Correct answer:

54.5

Explanation:

The median of a set of numbers is the middle value of the set.

To find the middle value of this particular data set put the prices in order of lowest to highest

36, 52, 57, 83

Since we have an even number of entries we will need to find the mean of the two middle numbers and this will become our median.

52 + 57 = 109

109 / 2 = 54.5

Report an Error

View Algebra 2 Tutors

Robert
Certified Tutor

University of Pittsburgh, Bachelor of Science, Physics.

View Algebra 2 Tutors

Jozsef
Certified Tutor

University of Waterloo, Doctor of Philosophy, Applied Mathematics.

View Algebra 2 Tutors

Saarim
Certified Tutor

St Edwards University, Bachelor of Science, Mathematics.

All Algebra II Resources

10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Physics Tutors in San Diego, Statistics Tutors in San Diego, Calculus Tutors in Atlanta, GRE Tutors in Boston, GRE Tutors in San Francisco-Bay Area, Biology Tutors in San Diego, MCAT Tutors in Washington DC, MCAT Tutors in Chicago, SSAT Tutors in Atlanta, SAT Tutors in Los Angeles

Popular Courses & Classes

GMAT Courses & Classes in San Diego, MCAT Courses & Classes in Houston, ISEE Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Seattle, SAT Courses & Classes in Phoenix, SSAT Courses & Classes in Miami, SSAT Courses & Classes in San Diego, SAT Courses & Classes in San Diego, SSAT Courses & Classes in Denver, MCAT Courses & Classes in Atlanta

Popular Test Prep

ISEE Test Prep in Denver, LSAT Test Prep in San Francisco-Bay Area, ISEE Test Prep in San Francisco-Bay Area, SAT Test Prep in Houston, ACT Test Prep in New York City, GMAT Test Prep in San Diego, ISEE Test Prep in Phoenix, SSAT Test Prep in Seattle, GRE Test Prep in Chicago, ISEE Test Prep in Seattle

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #2782 : Algebra Ii

Example Question #2783 : Algebra Ii

Example Question #2784 : Algebra Ii

Example Question #2785 : Algebra Ii

Example Question #1 : Other Sequences And Series

Example Question #2 : Other Sequences And Series

Example Question #3 : Other Sequences And Series

Example Question #1 : Other Sequences And Series

Example Question #3 : Other Sequences And Series

Example Question #2 : Other Sequences And Series

All Algebra II Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: