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1

STANDARDS OF LEARNING

CONTENT REVIEW NOTES

Honors ALGEBRA II

3rd Nine Weeks, 2018-2019

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OVERVIEW

Algebra II Content Review Notes are designed by the High School Mathematics Steering Committee as a

resource for students and parents. Each nine weeks’ Standards of Learning (SOLs) have been identified and a

detailed explanation of the specific SOL is provided. Specific notes have also been included in this document

to assist students in understanding the concepts. Sample problems allow the students to see step-by-step models

for solving various types of problems. A “ ” section has also been developed to provide students with the

opportunity to solve similar problems and check their answers. Supplemental online information can be

accessed by scanning QR codes throughout the document. These will take students to video tutorials and online

resources. In addition, a self-assessment is available at the end of the document to allow students to check their

readiness for the nine-weeks test.

The document is a compilation of information found in the Virginia Department of Education (VDOE)

Curriculum Framework, Enhanced Scope and Sequence, and Released Test items. In addition to VDOE

information, Prentice Hall Textbook Series and resources have been used. Finally, information from various

websites is included. The websites are listed with the information as it appears in the document.

Supplemental online information can be accessed by scanning QR codes throughout the document. These will

take students to video tutorials and online resources. In addition, a self-assessment is available at the end of the

document to allow students to check their readiness for the nine-weeks test.

To access the database of online resources scan this QR code.

The Algebra II Blueprint Summary Table is listed below as a snapshot of the reporting categories, the number

of questions per reporting category, and the corresponding SOLs.

It is the Mathematics Instructors’ desire that students and parents will use this document as a tool toward the

students’ success on the end-of-year assessment.

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Radical Functions

AII/T.1 The student, given rational, radical, or polynomial expressions, will

b) add, subtract, multiply, divide, and simplify radical expressions containing rational numbers and variables, and expressions containing rational exponents;

c) write radical expressions as expressions containing rational exponents and vice versa

Simplifying Radicals

To simplify a radical, you will pull out any perfect square factors (i.e. 4, 9, 16, 25, etc.)

√18 = √9 ∙ 2

The square root of 9 is equal to 3, so you can pull the square root of 9 from underneath

the radical sign to find the simplified answer 3 √2 , which means 3 times the square

root of 2. You can check this simplification in your calculator by verifying that

√18 = 3√2 .

Another way to simplify radicals, if you don’t know the factors of a number, is to create

a factor tree and break the number down to its prime factors. When you have broken

the number down to all of its prime factors you can pull out pairs of factors for square

roots, which will multiply together to make perfect squares.

Example 1: Simplify √128 √2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2

2 ∙ 2 ∙ 2 √2 = 8√2

Example 2: Simplify √32𝑥3𝑦

2 64

8 8

4 2 2 4

2 2 2 2

Scan this QR code to go to a video tutorial on

simplifying radicals.

16 2 x x x y

4 4

2 2 2 2

√2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑦

2 ∙ 2 ∙ 𝑥√2𝑥𝑦 = 4𝑥√2𝑥𝑦

5

To simplify a root of a higher index, you pull out factors that occurs the same number

of times as the index of the radical. As an example, if you are simplifying √64𝑎75 you

would only pull out factors that occurred 5 times, since 5 is the index of the root.

Example 3: Simplify √64𝑎74

√2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎4

2𝑎 √4𝑎34

Radical Operations

In order to multiply or divide radical expressions, they must share the same index. This

means that the power of the root must be the same.

√312𝑥𝑦54 ∙ √125𝑥85

could not be simplified because one is a 4th root and one is a 5th

5𝑦 √27𝑥43÷ 3 √80𝑦43

could be simplified because they are both cube roots.

If √𝑎𝑛

and √𝑏𝑛

are real numbers, then √𝑎𝑛

∙ √𝑏𝑛

= √𝑎𝑏𝑛

If √𝑎𝑛

and √𝑏𝑛

are real numbers, and 𝑏 ≠ 0, then √𝑎

𝑛

√𝑏𝑛 = √

𝑎

𝑏

𝑛

Example 1: Simplify 2𝑥 √16𝑥43∙ √6𝑦3

Because these radicals have the same index, they can be multiplied and simplified.

2𝑥 √96𝑥4𝑦3

2𝑥 √2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 3 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑦3

2𝑥 ∙ 2 ∙ 𝑥 √2 ∙ 2 ∙ 3 ∙ 𝑥 ∙ 𝑦3 = 4𝑥2 √12𝑥𝑦3

Example 2: Simplify 5𝑦 √27𝑦3 ÷ 3 √90𝑥𝑦5

5𝑦

3√

27𝑦3

90𝑥𝑦5 = 5𝑦

3√

3

10𝑥𝑦2 =5𝑦

3√

3

2∙5∙𝑥∙𝑦∙𝑦 =

5𝑦

3𝑦 √

3

10𝑥 =

5

3√

3

10𝑥

Because this is a 4th root, pull out things that occur 4 times.

Because this is a cube root, pull out things that occur 3 times.

6

Radical Functions

1. Simplify √135𝑎𝑏4𝑐63

2. Simplify 3𝑥𝑦 √100𝑥2𝑦5

3. Simplify 5𝑦 √8𝑦3 ∙ 3 √18𝑥𝑦4

4. Simplify 2𝑥 √16𝑥43 ÷ √8𝑥33

Rational Exponents

Properties of Exponents

Property Example

𝑥𝑚 ∙ 𝑥𝑛 = 𝑥𝑚+𝑛 5

12 ∙ 5 2 = 5

12

+2 = 5 52

(𝑥𝑚)𝑛 = 𝑥𝑚 ∙ 𝑛 (8 3)5 = 8 3 ∙ 5 = 8 15

(𝑥𝑦)𝑚 = 𝑥𝑚 ∙ 𝑦𝑚 (10𝑥)2 = 100𝑥2

𝑥−𝑚 = 1

𝑥𝑚 3−3 =

1

33=

1

27

𝑥𝑚

𝑥𝑛= 𝑥 𝑚−𝑛

75

77= 7 5−7 = 7−2 =

1

72=

1

49

( 𝑥

𝑦 )𝑚 =

𝑥𝑚

𝑦𝑚 (

2

3 )3 =

23

33=

8

27

√𝑎𝑛

= 𝑎 1𝑛 √50

3= 50

13

√𝑎𝑚𝑛= 𝑎

𝑚𝑛 √𝑏35

= 𝑏 35

Example 1: Simplify 162 1

4

162 1

4 = √1624

= √2 ∙ 3 ∙ 3 ∙ 3 ∙ 34

3 √24

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Example 2: Write 𝑎−1.5 in radical form.

𝑎−3

2 = 1

𝑎 32

= 1

√𝑎3=

1

√𝑎∙𝑎∙𝑎=

1

𝑎 √𝑎

Example 3: Write in simplest form √𝑛54

√𝑛3

𝑛

54

𝑛 3 2

= 𝑛 5

4 −

3

2 = 𝑛 5

4 −

6

4 = 𝑛− 1

4 = 1

𝑛 14

= 1

√𝑛4

Radical Functions

5. Write 𝑏 7

4 in radical form.

6. Write √𝑥5 in exponential form.

7. Simplify √𝑥6

√𝑥23

Write each radical in exponential form, then use

the properties of exponents to simplify.

Scan this QR code to go to a video tutorial on rational

exponents.

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Radical Equations and Graphs

AII/T.4 The student will solve, algebraically and graphically,

d) equations containing radical expressions.

AII/T.6 The student will recognize the general shape of function (absolute value, square root,

cube root, rational, polynomial, exponential, and logarithmic) families and will convert between graphic and symbolic forms of functions. A transformational approach to graphing will be employed. Graphing calculators will be used as a tool to investigate the shapes and behaviors of these functions.

Radical Equations may require you to square both sides of the equation. This has the potential to introduce extraneous solutions. For this reason, it is very important to always check your solutions to radical equations.

Example 1: Solve for y √5 − 4𝑥3

+ 6 = 5

√5 − 4𝑥3

+ 6 = 5 −6 − 6

√5 − 4𝑥3

= −1

(√5 − 4𝑥3

)3 = (−1)3 5 − 4𝑥 = −1 −5 − 5 −4𝑥 = −6 ÷ (−4) ÷ (−4)

𝑥 = 3

2

√5 − 4(3

2)

3+ 6 = √5 − 6

3+ 6 = √−1

3+ 6 = −1 + 6 = 5

To solve equations with a fractional exponent, raise each side of the equation to the reciprocal power. This is because when you raise a power to a power you multiply the exponents, and multiplying by the reciprocal will make the exponent 1.

Cube both sides of the equation to cancel out the cube root.

Don’t forget to check!


radical equations.

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Example 2: Solve for x: (2𝑥 − 6)3

2 = 8

(2𝑥 − 6)3

2 = 8

[(2𝑥 − 6)32 ]

23 = 8

23

2𝑥 − 6 = 4 +6 + 6 2𝑥 = 10 ÷ 2 ÷ 2 𝑥 = 5

(2(5) − 6) 32 = (10 − 6)

32 = 4

32 = 8

Sometimes it might be helpful to rewrite a radical using a rational exponent to solve.

Example 3: What is the solution of 2 + √(𝑥 − 1)53= 34

2 + √(𝑥 − 1)53= 34

−2 − 2

√(𝑥 − 1)53= 32

(𝑥 − 1) 5

3 = 32

[(𝑥 − 1)53 ]

35 = 32

35

𝑥 − 1 = 8 𝑥 = 9

2 + √(9 − 1)53= 2 + √853

= 2 + √327683

= 2 + 32 = 34 Graphs of Radical Functions

𝑦 = √𝑥 𝑦 = √𝑥3

Only in the 1st quadrant, because square roots are only defined for

positive values of x.

In the 1st and 3rd quadrant because when x is positive y will be positive, and when x is negative y will also be

negative.


Raise both sides to the reciprocal

power. 3

2 ∙

2

3= 1

Isolate the radical expression.

Rewrite in exponential form.

Raise both sides to the reciprocal

power. 5

3 ∙

3

5= 1


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Example 4: The maximum walking speed of a large animal can be modeled by the

equation: 𝑆 = √𝑔𝐿, where S is the speed (ft/sec), g is a constant 32 𝑓𝑡/𝑠𝑒𝑐2,

and L is the length (ft) of the animal’s leg. Use this model to predict how long an animal’s leg would need to be in order to travel 15 ft/sec.

This equation could be solved algebraically or graphically. We will show both examples here.

Algebraically: Plug in the given values for S and g

15 = √32𝐿

(15)2 = (√32𝐿)2 225 = 32𝐿 ÷ 32 ÷ 32 7.03 = 𝐿

Graphically: Graph the equation 𝑆 = √𝑔𝐿 in Y1 as 𝑌1 = √32𝑥.

The y value that we want to consider is 15 ft/sec, so graph 𝑦 = 15 in Y2.

Press 2nd then TRACE to take you to the CALC menu. Scroll down to 5: intersect.

This again shows your answer as X = 7.03 feet.

Radical Equations

1. 4 √2𝑥 + 2 − 6 = 10

2. (4𝑥 + 1)1

4 + 5 = 2

3. 2 √(2𝑥 − 4)23 = 32

4. −3 − √2𝑥5

= −1 5. You can model the population, 𝑃, of St. Louis, Missouri between 1900 and 1940 by

the function 𝑃(𝑥) = 52,000 √𝑥 − 18903

, where 𝑥 is the year. Using this model, in what year was the population of St. Louis 175,000?

Square both sides of the equation to cancel out the square root.

The animal’s leg would need to be just over 7 feet long in order

to travel 15 ft/sec.

You will need to adjust your zoom in order to see the intersection. You can

do this by pressing ZOOM then 0 for ZoomFit.

Your graph should look like this --->

The calculator will ask for first curve?, second curve?

and guess?. Just scroll near the intersection and press ENTER until your intersection is displayed.

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Rational Expressions AII/T.1 The student, given rational, radical, or polynomial expressions, will

a) add, subtract, multiply, divide, and simplify rational algebraic expressions;

A rational expression is the quotient of two polynomials (or a polynomial fraction). To simplify rational expressions you need to factor both the numerator and denominator, and then divide out common factors.

Example 1: Simplify 9𝑥3−12𝑥2

3𝑥2

3𝑥2 (3𝑥−4)

3𝑥2 (1)=

3𝑥−4

1= 3𝑥 − 4 , 𝑥 ≠ 0

Example 2: Simplify 𝑥2−7𝑥+12

(𝑥+3)(𝑥−4)

Factor the numerator! 𝑎 ∙ 𝑐 = 1 ∙ 12 = 12 So, we are looking for factors of 12 that add up to −7 !

To multiply rational expressions, you multiply the numerators and multiply the denominators. You can use factoring to simplify.

Example 3: Multiply and Simplify 9𝑥2−1

3𝑥2 ∙ 3𝑥2

3𝑥2−2𝑥−1

(3𝑥+1)(3𝑥−1)

3𝑥2 ∙ 3𝑥2

(3𝑥+1)(𝑥−1)

3𝑥 − 1

𝑥 − 1 , 𝑥 ≠ 1, 0, −

1

3

𝑥2 −3𝑥

−4𝑥 12

The GCF is 3𝑥2. If you remove this from both the numerator and the denominator, you can divide it out. This is because

3𝑥2

3𝑥2 = 1 where 𝑥 ≠ 0. If x were 0, then

the original problem would be undefined (the denominator would equal 0).

Find the greatest common factor in each row and each column. These will give you your two binomials! Put the now factored trinomial back into the original problem and divide out factors that are in both the numerator and denominator.

𝑥

−4

𝑥 −3

(𝑥−4) (𝑥−3)

(𝑥+3)(𝑥−4)=

𝑥−3

𝑥+3 , 𝑥 ≠ 4, 𝑎𝑛𝑑 𝑥 ≠ −3

Divide out common factors that occur in both the numerator and denominator. Your answer is the product of what remains after common factors are divided out.

Scan this QR code to go to a video tutorial on simplifying rational

expressions.

Don’t forget: anytime the 𝑏 term is negative and the 𝑐 term is positive, your answer will have two minus signs!

12

To divide rational expressions, you can simply multiply the first expression by the reciprocal of the second.

Example 4: Divide and Simplify 12(𝑥−5)

𝑥2−1 ÷

4𝑥−12

𝑥2−2𝑥−3

12(𝑥−5)

𝑥2−1 ∙

𝑥2−2𝑥−3

4𝑥−12

12(𝑥−5)

(𝑥+1)(𝑥−1) ∙

(𝑥+1)(𝑥−3)

4(𝑥−3)

12(𝑥 − 5)

4(𝑥 − 1) =

3(𝑥 − 5)

𝑥 − 1 =

3𝑥 − 15

𝑥 − 1 , 𝑥 ≠ 1, −1, 3

If you want to add or subtract rational expressions, you must first find a common denominator. Once both of your fractions have a common denominator, you can add or subtract the numerators and simplify. You can find the least common denominator (LCD) by finding the least common multiple (LCM) of each denominator. Do this by writing the prime factorization of each expression, and then write the product of the prime factors raised to the highest power in either expression.

Example 5: Find the LCM of 4𝑥 (𝑥2 − 36) 𝑎𝑛𝑑 2𝑥3(2𝑥2 + 15𝑥 + 18) Start by writing the prime factorization of each expression

4𝑥(𝑥2 − 36) = 22𝑥 ( 𝑥 + 6 )( 𝑥 − 6)

2𝑥3(2𝑥2 + 15𝑥 + 18) = 2𝑥3(2𝑥 + 3)(𝑥 + 6) Then, look at each factor individually. If the factor occurs in both expressions, choose the one that has the highest power.

22𝑥3 (𝑥 + 6) (𝑥 − 6) (2𝑥 + 3)

The LCM would be 4𝑥3 (𝑥 + 6) (𝑥 − 6) (2𝑥 + 3)

Multiply by the reciprocal. Factor all polynomials and divide out common factors. Simplify


multiplying and dividing

rational expressions.

Scan this QR code to go to a video tutorial on adding and subtracting rational

expressions.

13

Example 6: Subtract and simplify 𝑥+8

2𝑥+8−

4𝑥−11

𝑥2+11𝑥+28

Start by factoring the denominators and finding the least common denominator (LCD)

2𝑥 + 8 = 2 (𝑥 + 4) 𝑥2 + 11𝑥 + 28 = (𝑥 + 4)(𝑥 + 7) In this case, the LCD is 2(x + 4)(x + 7). This means that you will need to multiply the

first fraction by (x+7)

(x+7), and the second fraction by

2

2. This will give each fraction the same

denominator.

( 𝑥 + 8

2(𝑥 + 4) ) ∙ (

𝑥 + 7

𝑥 + 7 ) − (

4𝑥 − 11

(𝑥 + 4)(𝑥 + 7) ) ∙ (

2

2 )

𝑥2 + 15𝑥 + 56

2(𝑥 + 4)(𝑥 + 7)−

8𝑥 − 22

2(𝑥 + 4)(𝑥 + 7)

𝑥2 + 15𝑥 + 56 − 8𝑥 + 22

2(𝑥 + 4)(𝑥 + 7)

𝑥2 + 7𝑥 + 78

2(𝑥 + 4)(𝑥 + 7) , 𝑥 ≠ −4, −7

A complex fraction is a rational expression that has a fraction in either the numerator, denominator or both (basically this is a fraction in a fraction). There are two methods to simplify a complex fraction. Only one will be shown in these notes. The second method will be explained in a video tutorial.

Example 7: Simplify 5+

1

𝑥𝑦𝑥

𝑥2+2

𝑦

The LCD of all of the rational expressions is 𝑥2𝑦. If we multiply both the numerator and denominator by this LCD, it will help us to simplify.

(5+1

𝑥𝑦 )(𝑥2𝑦)

( 𝑥

𝑥2+2

𝑦 )(𝑥2𝑦)

=5𝑥2𝑦+𝑥

𝑥𝑦+2𝑥2 𝑥(5𝑥𝑦+1)

𝑥(𝑦+2𝑥)=

5𝑥𝑦+1

𝑦+2𝑥

Multiply both fractions to create LCD. Rewrite as one fraction over LCD Simplify numerator and check for common factors between numerator and denominator.

We can now check both the numerator and denominator for common factors. In this

case, we can factor an x out of both. 𝑥 ≠ 0, 𝑦 ≠ 0

14

Rational Expressions

1. 𝑥2−3𝑥−10

𝑥2−10𝑥+25

2. 𝑥2−𝑥−12

2 ∙

𝑥+4

𝑥2−16

3. 3𝑥2−15𝑥

𝑥2+2𝑥+1 ÷

6𝑥2−30𝑥

𝑥+1

4. 2𝑥

𝑥2−𝑥−6+

4

𝑥2+4𝑥+4

5. 𝑥

3𝑥+9−

3

𝑥2+3𝑥

6. 2+

1

𝑦

𝑥2

𝑦2+2

𝑦

Graphing Rational Functions AII/T.6 The student will recognize the general shape of function (absolute value, square

root, cube root, rational, polynomial, exponential, and logarithmic) families and will convert between graphic and symbolic forms of functions. A transformational approach to graphing will be employed. Graphing calculators will be used as a tool to investigate the shapes and behaviors of these functions.

AII/T.7 The student will investigate and analyze functions algebraically and graphically.

a) domain and range, including limited and discontinuous domains and ranges; b) zeros; e) asymptotes; f) end behavior;

AII/T.10 The student will identify, create, and solve real-world problems involving inverse

variation, joint variation, and a combination of direct and inverse variations. To graph a rational function, first check for horizontal and vertical asymptotes. Find x and y-intercepts, look for holes (where the graph is undefined), and then use a few points to determine the general shape of the graph.

Scan this QR code to go to a video tutorial on complex

fractions.

15

Vertical Asymptote

A rational function 𝑓(𝑥) = 𝑃(𝑥)

𝑄(𝑥), has a

vertical asymptote at each real zero of 𝑄(𝑥) if 𝑃(𝑥) and 𝑄(𝑥) have no

common zeros. If they do have a common zero [i.e. (𝑥 − 𝑎)], and the

degree of the term in the denominator is greater, then 𝑓(𝑥) also has a vertical

asymptote at 𝑥 = 𝑎.

𝑓(𝑥) = 𝑥

𝑥 − 1

This graph has a

vertical asymptote at x = 1

Horizontal Asymptote

The horizontal asymptote of a rational function is found by comparing the degree of the numerator (m) and

denominator (n). If m < n, the horizontal asymptote is

𝑦 = 0. If m > n, there is no horizontal

asymptote. If m = n, the horizontal asymptote

is 𝑦 = 𝑎

𝑏, where a is the coefficient of

the highest degree term in the numerator and b is the coefficient of

the highest degree term in the denominator.

𝑓(𝑥) = 𝑥 − 1

2𝑥

This graph has a

horizontal asymptote at 𝑦 = 1/2

The numerator and denominator have

the same degree (1), so the asymptote is

found by dividing the coefficients ½.

The x-intercepts are found by determining where the numerator is equal to zero. The y-intercepts are found by setting x = 0.

Example 1: Graph 𝑓(𝑥) = 𝑥−2

𝑥2−2𝑥−3

Start by factoring the numerator and denominator: 𝑓(𝑥) = 𝑥−2

(𝑥+1)(𝑥−3)

Vertical Asymptote: Because the numerator and the denominator do not share any common zeros, the function will have vertical asymptotes at any value that makes the denominator zero 𝒙 = −𝟏 𝒂𝒏𝒅 𝒙 = 𝟑 Horizontal Asymptote: The degree of the denominator (2) is higher than the degree of the numerator (1) therefore the horizontal asymptote is 𝒚 = 𝟎

X-Intercepts: The x-intercepts are where the numerator is equal to zero. (𝟐, 𝟎)

Y-Intercepts: Set x = 0. Therefore, 0−2

0−2(0)−3=

−2

−3=

2

3 (𝟎,

𝟐

𝟑 )

Now you can sketch your asymptotes, and plot your intercepts.

16

You can solve a rational equation algebraically or graphically. To solve a rational equation graphically, graph both sides of the equation in your calculator and check for intersection points. This procedure will be shown in the video tutorial. Example 2 will show you how to solve algebraically.

Example 2: Solve. 7

𝑥+2+

5

𝑥−2=

10𝑥−2

𝑥2−4 Be sure to check for extraneous solutions!

7

𝑥 + 2+

5

𝑥 − 2=

10𝑥 − 2

(𝑥 + 2)(𝑥 − 2)

( 7

𝑥+2+

5

𝑥−2 ) (𝑥 + 2)(𝑥 − 2 ) =

10𝑥−2

(𝑥+2)(𝑥−2)( 𝑥 + 2)(𝑥 − 2)

7(𝑥 − 2) + 5 (𝑥 + 2) = 10𝑥 − 2

7𝑥 − 14 + 5𝑥 + 10 = 10𝑥 − 2

12𝑥 − 4 = 10𝑥 − 2

2𝑥 = 2

𝑥 = 1

7

1 + 2+

5

1 − 2≟

10(1) − 2

12 − 4

7

3+

5

−1≟

8

−3

−8

3= −

8

3

If you chose to plot a few more points, such as

(4,2

5), 𝑎𝑛𝑑 (−2, −

4

5)

your graph would start to take

shape.

Scan this QR code to go to a video tutorial on graphing

rational functions.

Rewrite with factored denominators so that we can find the LCD. (x+2)(x-2)

Multiply both sides of the equation by the

LCD to clear the denominators.

Solve the equation!

Check for extraneous solutions!

This is not an extraneous solution. Therefore, the

solution to the equation is:

x = 1

Scan this QR code to go to a video tutorial on solving

rational equations.

17

Direct and Inverse Variation

Direct Variation Inverse Variation

𝒚 = 𝒌𝒙 (𝑤ℎ𝑒𝑟𝑒 𝑦

𝑥= 𝑘) 𝒚 =

𝒌

𝒙 (𝑤ℎ𝑒𝑟𝑒 𝑥𝑦 = 𝑘)

Direct variations are always linear and

pass through the origin.

Inverse variations are rational functions and will never pass through the origin.

If the ratio between two variables is a constant, then a direct variation exists. A direct variation can be written in the form 𝑦 = 𝑘𝑥, where 𝑘 is the constant of variation. If the product of two variables is a constant, then an inverse variation exists. An

inverse variation can be written in the form = 𝑘

𝑥 .

Example 3: Determine if each relation is a direct variation, inverse variation, or neither.

x y

1 3

2 4

3 5

x y

-1 2

0 0

1 -2

x y

3 4

1 12

-2 -6

First check the ratios:

Does the ratio 1

3=

2

4 ? NO! Therefore this is NOT a direct variation!

Next check the products: Does 1 ∙ 3 = 2 ∙ 4 ? NO! Therefore this is NOT an inverse variation!


Does the ratio 3

4=

1

12 ? NO! Therefore this is NOT a direct variation!

Next check the products: Does 3 ∙ 4 = 1 ∙ 12? Does this also equal −2 ∙ −6 ? YES! Therefore, this IS an inverse variation!


Does the ratio −1

2=

1

−2 ? YES! Therefore this IS a direct variation!

Notice that we did not use the ordered pair (0,0) to check the ratios. It is impossible to divide by zero. Therefore, we used the other point.

18

To write an equation of a direct variation, use the given point plugged into 𝑦 = 𝑘𝑥, to

solve for 𝑘. For an inverse variation, plug into 𝑦 =𝑘

𝑥 .

Example 4: Suppose 𝑦 varies inversely with 𝑥, and 𝑦 = 10 when 𝑥 = 25. What inverse variation equation relates 𝑥 and 𝑦?

Start with 𝑦 = 𝑘

𝑥.

We are given a value for 𝑥 and 𝑦, so plug those in and solve for 𝑘.

10 =𝑘

25

∙ 25 ∙ 25 250 = 𝑘

𝑦 =250

𝑥 𝒐𝒓 𝑥𝑦 = 250

Once you have a direct or inverse variation equation, you can use that equation to determine other values. Example 5: The distance that you jog, 𝑗, varies directly with the amount of time you jog, 𝑡. If you can jog 4 miles in 0.9 hours, how long will it take you to jog 6.5 miles? Jogging varies directly with time → 𝑗 = 𝑘𝑡

Now we need to solve for 𝑘 in order to write a direct variation equation.

4 = 𝑘(0.9) ÷ 0.9 ÷ 0.9

40

9= 𝑘

𝑗 =40

9𝑡

Now, we can use this equation to solve for the time it takes to jog 6.5 miles. We are given that you jog 6.5 miles. This will be plugged in for 𝑗. Then, solve for 𝑡.

6.5 = 40

9𝑡

÷40

9 ÷

40

9

1.4625 = 𝑡 Therefore, it would take you almost 1.5 hours to jog 6.5 miles.

This is the constant of variation!

This is the inverse variation equation!

This is the constant of variation!

This is the direct variation equation!

Scan this QR code to go to a video tutorial on direct,

inverse, and joint

variations.

19

Sometimes, a quantity varies with respect to more than one quantity. This creates a combined variation. If the quantities vary directly or jointly, those quantities will be in the numerator (with k). If the quantities vary inversely, those quantities will be in the denominator. Example 6: The amount of cake, 𝑐, that it takes to feed a birthday party varies directly with the number of children, 𝑛, and inversely with the age of the children, 𝑎. If a party of sixteen 8 year-old children eat 112 in² of birthday cake, how much cake would a party of ten 6 year-olds require? Start by setting up the formula. 𝑐 varies directly with 𝑛 and inversely with 𝑎.

𝑐 = 𝑘𝑛

𝑎

Substitute for 𝑐, 𝑛, and 𝑎. 𝑐 = 112 in² of cake, 𝑛 = 16 children and 𝑎 = 8 years old.

112 = 16𝑘

8

𝑘 = 56

Write your new combined variation equation with 𝑘 substituted in.

𝑐 = 56𝑛

𝑎

Substitute in your new values for n and a to solve for c.

𝑐 = 56(10)

6

𝑐 = 93.33 𝑖𝑛2 𝑜𝑓 𝑐𝑎𝑘𝑒

Graphing Rational Functions

1. State all asymptotes, x and y-intercepts, and restrictions of 𝑓(𝑥) = 𝑥2−36

2𝑥2+11𝑥−6

2. Sketch the graph of 𝑓(𝑥) = (𝑥+2)(𝑥+5)

(𝑥2−4)

3. Solve for x. 2

𝑥+2−

3

𝑥=

6

𝑥(𝑥+2)

4. y varies inversely with x. If y is 240 when x is 3, what is the constant of variation?

5. The amount of hats knitted by a group is directly proportional to the number of

people in the group. If 5 people can knit 12 hats, how many hats can 12 people

knit?

6. 𝑥 varies jointly with 𝑦 and 𝑧 and inversely with both 𝑤 and 𝑣. Write a formula relating

these variables.

20

Exponential and Logarithmic Functions AII/T.6 The student will recognize the general shape of function (absolute value, square

root, cube root, rational, polynomial, exponential, and logarithmic) families and will convert between graphic and symbolic forms of functions.

AII/T.7 The student will investigate and analyze functions algebraically and graphically.

a) domain and range, including limited and discontinuous domains and ranges; e) asymptotes; f) end behavior;

Rules of Exponents

Multiplying When multiplying two

exponential terms with the same base, you add the exponents.

𝑥4 ∙ 𝑥3 = 𝑥4+3 = 𝑥7

Dividing When dividing two exponential terms with the same base, you

subtract the exponents.

𝑏8

𝑏2= 𝑏8−2 = 𝑏6

Raising a power to a

power

When raising an exponential term to a power, you multiply

the exponents. (𝑛2)5 = 𝑛2∙5 = 𝑛10

Negative Exponents

To simplify negative exponents, write the term as its reciprocal

with a positive exponent. 𝑎−4 =

1

𝑎4

5

𝑥−2=

5𝑥2

1

Example 1: Simplify (−2𝑥³𝑦𝑧²)³(2𝑥³𝑦𝑧²)⁴

(−8𝑥9𝑦3𝑧6) ∙ (16𝑥12𝑦4𝑧8)

−128𝑥21𝑦7𝑧14

Example 2: (𝑎2𝑏𝑐5

𝑎2𝑐2 )3

(𝑎2𝑏𝑐5

𝑎2𝑐2 )

3

= (𝑎2−2𝑏𝑐5−2)3 = (𝑎0𝑏𝑐3)3 = 13𝑏3𝑐9 = 𝑏3𝑐9

Example 3: (2𝑥2𝑦−4

3𝑥𝑦5 )

−3

(2𝑥2𝑦−4

3𝑥𝑦5)

−3

= (3𝑥𝑦5

2𝑥2𝑦−4)

3

= (3𝑥𝑦5𝑦4

2𝑥2 )

3

= (3𝑦9

2𝑥 )

3

= 27𝑦27

8𝑥3

Remember that anything to the zero power equals 1!


rules of exponents.

21

Exponential Growth

𝑦 = 𝑎𝑏𝑥 𝑤ℎ𝑒𝑟𝑒 𝑎 > 0 𝑎𝑛𝑑 𝑏 > 1

The graph crosses the y-axis

at (0, 𝑎). The asymptote is 𝑦 = 0

Exponential Decay

𝑦 = 𝑎𝑏𝑥 𝑤ℎ𝑒𝑟𝑒 𝑎 > 0 𝑎𝑛𝑑

𝑏 𝑖𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 0 𝑎𝑛𝑑 1

The graph crosses the y-axis at (0, 𝑎).

The asymptote is 𝑦 = 0

Logarithmic Functions

𝑦 = log𝑏 𝑥 𝑤ℎ𝑒𝑟𝑒 𝑏 > 0 𝑎𝑛𝑑 𝑏 ≠ 1

A logarithmic function is the inverse of an exponential

function.

You can model exponential growth and decay with the function

𝐴(𝑡) = 𝑎(1 + 𝑟)𝑡 where 𝐴(𝑡) is the final amount, 𝑎 is the initial amount, 𝑟 is the rate of growth or decay,

and 𝑡 is the number of time periods. Example 4: If you invested $5,000 in a savings account today that pays 3.5% interest annually, how much will you have in 10 years?

𝐴(𝑡) = 𝑎(1 + 𝑟)𝑡 𝑎 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 = $5000 𝑟 = 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑟𝑎𝑡𝑒 = 0.035 𝑡 = 𝑡𝑖𝑚𝑒 𝑝𝑒𝑟𝑖𝑜𝑑𝑠 = 10 𝑦𝑒𝑎𝑟𝑠

𝐴(𝑡) = 5000(1 + .035)10

𝐴(𝑡) = 5000(1.035)10 𝐴(𝑡) = 5000(1.41) 𝐴(𝑡) = $7050

Plug in your given values and solve for 𝐴(𝑡).

22

Continuously compounded interest can be calculated using the formula 𝐴(𝑡) = 𝑃 ∙ 𝑒𝑟𝑡 where A(t) is the final amount, P is the initial amount, r is the annual interest rate, and t is the number of years. Example 5: If you invested $2,000 in a savings account today that pays 2.75% interest compounded continuously, in how many years would you have $5,000?

𝐴(𝑡) = 𝑃𝑒𝑟𝑡 𝐴(𝑡) = 𝑓𝑖𝑛𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 = $5000 𝑃 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 = $2000 𝑟 = 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑟𝑎𝑡𝑒 = 0.0275

Exponential and Logarithmic Functions

1. (𝑥4𝑦2

2𝑥𝑦 )

3

2. 𝑛4𝑛2𝑚−5

𝑛6𝑚−2𝑝0

3. A population of 800,000 decreases at a rate of 2.2% annually. After how many

years will the population go below 600,000? (remember that your growth is negative now!)

4. If $1000 is invested in a savings account with 5.1% interest compounded

continuously, how much money will you have after 3 years?

To find the number of years, let’s graph the function and look at the table

of values in our calculator. We will graph 𝑦 = 2000𝑒 .0275𝑥

Once you’ve graphed the

function, press 2nd and GRAPH to show the table

of values.

We want to know where the final

value (Y) equals 5000, so scroll

through the table until you find it.

It would take 34 years for your investment of

$2000 to grow to $5000.

34 years

Scan this QR code to go to a video tutorial on exponential

growth and decay.

23

Properties of Logarithms

log𝑏 𝑥 = 𝑦 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑏𝑦 = 𝑥

𝐹𝑜𝑟 𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑚, 𝑛, 𝑎𝑛𝑑 𝑏 𝑤ℎ𝑒𝑟𝑒 𝑏 ≠ 1, 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑖𝑒𝑠 𝑎𝑝𝑝𝑙𝑦

Product Property

log𝑏 𝑚𝑛 = log𝑏 𝑚 + log𝑏 𝑛 log6 36𝑥 = log6 36 + log6 𝑥

Quotient Property

log𝑏

𝑚

𝑛= log𝑏 𝑚 − log𝑏 𝑛 log2

𝑦

32= log2 𝑦 − log2 32

Power Property

log𝑏 𝑚𝑛 = 𝑛 log𝑏 𝑚 log4 𝑥2 = 2 log4 𝑥

𝐹𝑜𝑟 𝑎𝑛𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑚, 𝑏, 𝑎𝑛𝑑 𝑐 𝑤ℎ𝑒𝑟𝑒 𝑏 ≠ 1 𝑎𝑛𝑑 𝑐 ≠ 1, 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑎𝑝𝑝𝑙𝑖𝑒𝑠

Change of Base Formula log𝑏 𝑚 = log𝑐 𝑚

log𝑐 𝑏 log4 32 =

log2 32

log2 4=

5

2

Example 6: Write log6 12 + log6 18 as a single logarithm. Using the properties of logarithms, this can be condensed to log6(12 ∙ 18)

log6 216 = log6 63 = 3

Example 7: Write log381

𝑥5 in expanded form.

log381

𝑥5= log3 81 − log3 𝑥5

log3 34 − 5 log3 𝑥

4 − 5 log3 𝑥 Example 8: Evaluate log6 49. (Hint: Use Change of Base)

log6 49 = log 49

log 6 ≈ 2.17

Exponential and Logarithmic Functions 5. Write as a single logarithm: log5 600 − log5 24

6. Write as a single logarithm: log8 12 + log8 9

7. Expand: log4 16𝑦2

8. Expand: log2𝑥

64

9. Evaluate: log25 125

10. Evaluate: log9 100

Use a calculator to approximate the answer.

Scan this QR code to go to a video tutorial on properties of

logarithms.

24

Exponential and Logarithmic Equations


c) equations containing rational algebraic expressions;

You can use the properties of exponents and logarithms to help you solve exponential and logarithmic equations. One way to solve an exponential equation is to rewrite terms on both sides of the equation with a common base.

Example 1: Solve 16 = 2 2𝑥−1

24 = 22𝑥−1

4 = 2𝑥 − 1

𝑥 = 5

2

If you cannot find like bases, you can solve by taking the logarithm of both sides. This is only true as long as both bases are positive.

Example 2: Solve 73𝑥 = 45

log 73𝑥 = log 45

3𝑥 log 7 = log 45

𝑥 = log 45

3 log 7

𝑥 ≈ 0.652

You can also solve an exponential or logarithmic equation by graphing both sides of the equation in your calculator and finding where the two graphs intersect.

Exponential and Logarithmic Equations

1. 93𝑥 = 27𝑥−5

2. 124𝑥 = 25 3. log (2𝑥 + 2) = 3 4. log 𝑥 + log(𝑥 + 9) = 1 (Remember that logarithms are only defined for positive numbers)

Rewrite each term with a common base

Set exponents equal to each other (Power Property of Exponents)

Solve

Take the logarithm of both sides

Power Property of Logarithms

Use your calculator to estimate the answer

Scan this QR code to go to a video tutorial on logarithmic and exponential equations.

25

Honors Algebra II

Natural Log and Log Applications


c) equations containing rational algebraic expressions;

AII/T.9 The student will collect and analyze data, determine the equation of the curve of best

fit, make predictions, and solve real-world problems, using mathematical models. Mathematical models will include polynomial, exponential, and logarithmic functions.

The same properties that apply to logarithms also apply to natural logarithms. The natural logarithm is a logarithm with base 𝑒.

𝑥 = ln 𝑦 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑦 = 𝑒𝑥

Example 1: ln(3𝑥 + 4) = 3

3𝑥 + 4 = 𝑒3

𝑥 = 𝑒3 − 4

3

𝑥 ≈ 5.362

Example 2: 4 − 𝑒3𝑥 = 2

−𝑒3𝑥 = −2

𝑒3𝑥 = 2 3𝑥 = ln 2

𝑥 = ln 2

3

𝑥 ≈ 0.231

Natural Log and Log Applications

1. Simplify 3 ln 5 + ln 2

2. ln(𝑥 + 1)2 = 2

3. 2𝑒𝑥

3 + 5 = 37

4. A bacteria population 𝑁 at time 𝑡 (in months) is given by 𝑁 = 𝑁0𝑒2𝑡 where 𝑁0 is the initial population. If the initial population was 100, how long does it take for the population to reach one million?

Rewrite in exponential form. Remember that the natural log uses base e.

Solve for x.


Isolate the 𝑒𝑥 term.

Rewrite in logarithmic form.

Solve for x.


Scan this QR code to go to a video tutorial on natural log

and log applications.

26

Honors Algebra II- Sequences and Series AII/T.2 The student will investigate and apply the properties of arithmetic and geometric

sequences and series to solve real-world problems, including writing the first n terms, finding the nth term, and evaluating summation formulas. Notation will include Ʃ and an.

Arithmetic Sequence Geometric Sequence

A sequence where the difference between consecutive terms is a constant.

(You add or subtract a constant value)

A sequence where the difference between consecutive terms is a common ratio.

(You multiply or divide a constant value)

Examples: 3, 5, 7, 9, 11… (constant is +2) 25, 20, 15, 10… (constant is −5)

Examples: 3, 6, 12, 24, 48… (ratio is 2

1 )

6, 9, 13.5, 20.25, 30.375… (ratio is 3

2 )

Formula

𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑 𝑎 is the starting value, 𝑑 is the common difference and 𝑛 is the number of terms.

Formula

𝑎𝑛 = 𝑎 ∙ 𝑟𝑛−1 𝑎 is the starting value, 𝑟 is the common

ratio and 𝑛 is the number of terms.

Example 1: What is the 35th term of the arithmetic sequence that begins 7, 4… 𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑

𝑎𝑛 = 7 + (35 − 1)(−3)

𝑎𝑛 = −95

Example 2: What is the 20th term of the geometric sequence that begins 1, 2, 4…

𝑎𝑛 = 𝑎 ∙ 𝑟𝑛−1

𝑎𝑛 = 1 ∙ 220−1

𝑎𝑛 = 524,288

Example 3: What is the missing term in this geometric sequence 9, ⎕ , 1 …

𝑎𝑛 = 𝑎 ∙ 𝑟𝑛−1

1 = 9 ∙ 𝑟3−1

1

9= 𝑟2

𝑟 = 1

3

The missing term is 9 ∙ 1

3= 3

Substitute your values (𝑎 = 7, 𝑛 = 35, 𝑑 = −3)

Simplify

Substitute your values (𝑎 = 1, 𝑟 = 2, 𝑛 = 20)

Simplify

Substitute your values (𝑎𝑛 = 1, 𝑎 = 9, 𝑛 = 3)

Simplify

Solve for the common ratio, 𝑟.

27

A series is the sum of a geometric or arithmetic sequence.

Sum of a Finite Arithmetic Series

Sum of a Finite Geometric Series

Sum of an Infinite Geometric Sequence

(Only applicable for |𝒓| < 1)

𝑆𝑛 =𝑛

2 (𝑎1 + 𝑎𝑛)

Where 𝑎1 is the first term, 𝑎𝑛 is

the 𝑛𝑡ℎ term, and n is the number of terms.

𝑆𝑛 =𝑎1(1 − 𝑟𝑛)

1 − 𝑟

Where 𝑎1 is the first term, 𝑟 is the common ratio, and n is the

number of terms.

𝑆𝑛 =𝑎1

1 − 𝑟

. Where 𝑎1 is the first term, and 𝑟 is

the common ratio.

You may see series written in Summation Notation You can write the series 7+9+11+ …+89 as

∑ (2𝑛 + 5

42

𝑛 = 1

)

Example 4: Evaluate Because the explicit formula is linear, this will be an arithmetic series. In order to evaluate an arithmetic series we need to know the first and last term and number of terms.

𝑆𝑛 =𝑛

2 (𝑎1 + 𝑎𝑛)

𝑆𝑛 =42

2 (7 + 89)

𝑆𝑛 = 2016

Sequences and Series 1. What is the 9th term of the geometric sequence that begins 2, 1, … 2. What is the missing term in this arithmetic sequence 12, ⎕ , 25 … 3. Write the series in summation notation 120 + 115 + 110 + 105 + 100 + 95 4. Find the sum of the geometric series

∑ 4𝑛

10

𝑛=1

n = 1 is the lower limit 42 is the upper limit 2n +5 is the explicit formula for each term in the series.

∑(2𝑛 + 5)

42

𝑛=1

Substitute your values

(𝑛 = 42, 𝑎1 = 2(1) + 5 = 7 , 𝑎42 = 2(42) + 5 = 89)

Simplify

Scan this QR code to go to a video tutorial on sequences

and series.

28

Statistics AII/T.12 The student will compute and distinguish between permutations and combinations

and use technology for applications.

You can use multiplication to quickly determine the number of ways a certain event can

happen. You can use Permutations and Combinations to determine the number of

ways. You use Combinations when order is not important and Permutations when the

order is important. These formulas are on your formula sheet, but your formula sheet

does not tell you when to use each formula.

Permutation

𝑃 28

The number of permutations of 8

items of a set arranged 2 items at a time.

𝑃 28 =

8!

(8 − 2)!=

8!

6!=

40320

720= 56

Combination

𝐶 28

The number of

combinations of 8

items taken 2 at a time.

𝐶 28 =

8!

2! (8 − 2)!=

8!

2! (6!)=

40320

1440= 28

Example 1: In a diving meet, the top 8 divers advance to the finals and scores are

cleared. In how many ways can a field of 14 divers qualify to the finals?

Because all of the top 8 advance to the finals, the order that they finish isn’t important

as long as they are in the top 8. This means we will use a combination!

𝐶 814 =

14!

8! (14 − 8)!=

14!

8! (6!)=

8.71782912 × 1010

29030400= 3,003

There are 3,003 possible combinations of divers that can advance to the finals.

29

Example 2: In the finals of the diving meet referenced in Example 1, the top 3

finishers score points for their team. First place receives 10 points, 2nd place receives 8

points, and 3rd place receives 6 points. In how many ways can the 8 finalists finish in

the top 3?

Now, the order is important because 1st place gets more points than 2nd place. We will

use a permutation!

𝑃 38 =

8!

(8 − 3)!=

8!

5!=

40320

120= 336

There are 336 possible ways that the top 8 divers can finish in the top 3.

Statistics 1. A teacher is making a multiple choice quiz. She wants to give each student the

same questions, but have each student's questions appear in a different order. If there are twenty-seven students in the class, what is the least number of questions the quiz must contain?

2. A coach must choose five starters from a team of 12 players. How many different ways can the coach choose the starters?

30

Answers to the problems: Radical Functions

1. 3𝑏𝑐2 √5𝑎𝑏3

2. 30𝑥2𝑦3 √𝑦

3. 180𝑦4 √𝑥𝑦

4. 2𝑥 √2𝑥3

5. √𝑏74

6. 𝑥 5

2

7. 1

𝑥 12

𝑜𝑟 1

√𝑥

Radical Equations 1. 𝑥 = 7 2. 𝑁𝑜 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑟 { } 3. 𝑥 = 34 4. 𝑥 = −16 5. 𝑥 = 1928 Rational Expressions

1. 𝑥+2

𝑥−5 , 𝑥 ≠ 5

2. 𝑥+3

2 , 𝑥 ≠ 4, −4

3. 1

2(𝑥+1) , 𝑥 ≠ −1, 0, 5

4. 2(𝑥2+4𝑥−6)

(𝑥−3)(𝑥+2)2 , 𝑥 ≠ 3, −2

5. 𝑥−3

3𝑥 , 𝑥 ≠ −3, 0

6. 𝑦(2𝑦+1)

𝑥2+2𝑦

Graphing Rational Functions

1. vertical asymptote: 𝑥 = 1

2

horizontal asymptote: 𝑦 = 1

2

x-intercept: (6, 0) y-intercept: (0, 6)

𝑥 ≠ −6,1

2

2.

Graphing Rational Functions 3. 𝑥 = −12 4. 𝑘 = 720 5. ℎ = 28.8 𝑜𝑟 28 ℎ𝑎𝑡𝑠

6. 𝑥 = 𝑘𝑦𝑧

𝑤𝑣

Exponential and Logarithmic Functions

1. 𝑥9𝑦3

8

2. 1

𝑚3

3. 13 𝑦𝑒𝑎𝑟𝑠 4. $1165.32 5. 2

6. log8 108 7. 2 + 2 log4 𝑦 8. log2 𝑥 − 6

9. 1.5 10. ≈ 2.1 Exponential and Logarithmic Equations 1. 𝑥 = −5 2. 𝑥 ≈ 0.324

3. 𝑥 = 499 4. 𝑥 = 1 Natural Log and Log Applications

1. ln 250 ≈ 5.521

2. 𝑥 ≈ −0.757 𝑜𝑟 3.11 3. 𝑥 = 3 ln 16 ≈ 8.318 4. 4.605 𝑚𝑜𝑛𝑡ℎ𝑠 Algebra II Honors

Sequences and Series

1. 1

128

2. 18.5

3. ∑ (−5𝑛 + 125)6𝑛=1

4. 1,398,100

Statistics

1. 5 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛𝑠 𝑤𝑜𝑢𝑙𝑑 𝑔𝑖𝑣𝑒 120 𝑜𝑟𝑑𝑒𝑟 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑖𝑡𝑖𝑒𝑠 2. 792

honors algebra ii - suffolk city public schoolsstar.spsk12.net/math/alg...

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