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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #1 : Other Sequences And Series

Complete the following sequences.

$7, 5, 3, 1, \cdots$

Possible Answers:

Correct answer:

Explanation:

The sequence goes down by 2 so,

7 - 2 = 5 - 2 = 3 - 2 = 1 - 2 = -1 .

Therefore the next number in the sequence is .

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Example Question #2 : Other Sequences And Series

Complete the following sequence

$15, 20, 25, 30, 35, \cdots$

Possible Answers:

Correct answer:

Explanation:

The sequence goes up by 5 so,

15 + 5 = 20 + 5 = 25 + 5 = 30 + 5 = 35 + 5 = 40

Therefore the next term in the sequence will be .

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Example Question #3 : Other Sequences And Series

What is percent of ?

Possible Answers:

100

900

Correct answer:

Explanation:

To find the value related to the specific percentage we need to set up a proportion and solve for x.

$\frac{15}{100}=\frac{x}{60}$

From here we cross multiply and divide to find the value of x.

$15 \times 60 = 900$

$\frac{900}{100}=9$

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Example Question #10 : Other Sequences And Series

What is percent of 135 ?

Possible Answers:

Correct answer:

Explanation:

To find the value for a specific percentage of a number we first need to convert the percentage into a decimal.

50/100 = 0.5

From here we multiply the decimal with the number we are given in the question.

$0.5 \times 135 = 67.5 = 68$

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Example Question #2 : How To Find The Next Term In An Arithmetic Sequence

Find the next term in the sequence:

2, 7, 17, 37, 77,...

Possible Answers:

117

157

Correct answer:

157

Explanation:

The sequence follows the pattern for the equation:

$n_x=2n_{x-1}+3$

$n_{x-1}=77$

Therefore,

n_x=2(77)+3=154+3=157

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Example Question #11 : Other Sequences And Series

If the rule of some particular sequence is written as

x_n=1+3n^2 ,

find the first five terms of this sequence

Possible Answers:

none of these

4,13,28,49,76

4,13,7,9,21

4,17,36,49,33

0,4,13,28,49

Correct answer:

4,13,28,49,76

Explanation:

The first term for the sequence is where n=1 . Thus,

x_1=1+3(1)^2=4

So the first term is 4. Repeat the same thing for the second $\left ( n=2 \right )$ , third $\left ( n=3 \right )$ , fourth $\left ( n=4 \right )$ , and fifth $\left ( n=5 \right )$ terms.

x_2=1+3(2)^2=13

x_3=28

x_4=49

x_5=76

We see that the first five terms in the sequence are

4,13,28,49,76

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Example Question #2791 : Algebra Ii

The harmonic series is $1, \frac{1}{2} , \frac{1}{3} , \frac{1}{4} , ... , \frac{1}{n}$ where the nth term is the reciprocal of n. Which would work as a recursive formula $a _{n} = f(a_{n-1})$ where $a_{n}$ is the nth term?

Possible Answers:

$a_{n} = a_{n-1} - \frac{1}{2}$

$a_{n} = a_{n-1} + 1$

$a_{n} = \frac{1}{a_{n-1}} + 1$

$a_{n} =\frac{1}{\frac{1}{a_{n-1}}+1}$

$a_{n} = \frac{1}{a_{n-1} + 1 }$

Correct answer:

$a_{n} =\frac{1}{\frac{1}{a_{n-1}}+1}$

Explanation:

To go from $a_{n-1}$ to $a_{n}$ , we're adding 1 to the denominator. In words, we're flipping $a_{n-1}$ , adding 1, then flipping it again. For example, to get from $a_{4}=\frac{1}{4}$ to $a_{5 } = \frac{1}{5}$ we would have to flip $\frac{1}{4}$ to be 4, add 1 to get 5, then flip again to get $\frac{1}{5}$ .

The formula that shows this is

$a_{n} =\frac{1}{\frac{1}{a_{n-1}}+1}$

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Example Question #2794 : Algebra Ii

The sum of the first n square numbers can be found using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ . Find the sum of every square number between 1 and 1000.

Possible Answers:

31,248

10,416

338,350

333,833,500

11,440

Correct answer:

10,416

Explanation:

The problem isn't asking us to add the first 1,000 square numbers, but all the square numbers from 1 to 1,000. To figure out this sum, you might need to look at a list of square numbers, or play around with large squares to find the largest one under 1,000. This ends up being 31: 31^2 = 961 while 32^2 = 1024 , which is not between 1 and 1,000. So what we're adding is the first 31 square numbers.

This means we can plug 31 in for n in that formula:

$\frac{31*(31+1)*(31*2 + 1 )}{6} = \frac{31*32*63}{6 } = 10,416$

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Example Question #11 : Other Sequences And Series

The sum of the first n integerss can be found using the formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ .

Find the sum of every number between 17 and 8,043, inclusive.

Possible Answers:

32,348, 946

32, 348, 810

32, 349, 082

136

Correct answer:

32, 348, 810

Explanation:

To find the sum of all the integers in between 17 and 8,043, first we will find the sum of every integer from 8,043, and then we will subtract out the sum of the numbers 1-16, since those aren't between 17 and 8,043.

The sum of the first 8,043 integers is $\frac{(8043)(8044)}{2 } = 32, 348, 946$

The sum of the integers 1-16 is $\frac{(16)(17)}{2 } = 136$

Subtracting gives us 32, 348, 810

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Example Question #14 : Other Sequences And Series

The sum of the first n integers can be found using the formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

Find the sum of all the integers from -2,256 to 4,400.

Possible Answers:

2,299,440

9,682,200

12,228,096

7,136,304

Correct answer:

7,136,304

Explanation:

To calculate this sum, first we will need to find the sum of the positive integers, then the negative interers, then add them together.

To find the sum of the positive integers, use the formula with n = 4,400 :

$\frac{(4400)(4401)}{2 } = 9,682, 200$

To find the sum of the negative integers, we can use the same formula as the positive numbers and then just make that answer negative.

$\frac{(2256)(2257)}{2 } = 2,545, 896$ so the negative numbers add up to -2,545,896 .

The final answer is 9,682, 200 - 2,545, 896 = 7,136, 304

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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Other Sequences And Series

Example Question #2 : Other Sequences And Series

Example Question #3 : Other Sequences And Series

Example Question #10 : Other Sequences And Series

Example Question #2 : How To Find The Next Term In An Arithmetic Sequence

Example Question #11 : Other Sequences And Series

Example Question #2791 : Algebra Ii

Example Question #2794 : Algebra Ii

Example Question #11 : Other Sequences And Series

Example Question #14 : Other Sequences And Series

All Algebra II Resources

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