4
EVALUATING AND
PERFORMING OPERATIONS
ON FUNCTIONS
for General Mathematics
Senior High School (CORE)
Quarter 1 / Week 1
1
FOREWORD
This Self-Learning Kit for General Mathematics is designed
specifically for Grade 11 students in the Senior High School. Thus, a
modest background in grade school mathematics is important, written
in a precise, readable, and conventional manner to facilitate students’
understanding of the subject.
It is aligned with the BEC of the Department of Education
following the prescribed MELCs (Most Essential Learning
Competencies.
It has the following features proven to be valuable aids to learning
Mathematics even at home.
What happened
This section contains pre-activities like review of the prior
knowledge and a pretest on what the learners have learned in their
previous discussions.
What I Need to Know (Discussion)
This section contains definition of terms, different examples of
real-life situations as application of functions. It gives examples and
the corresponding situations that clearly illustrate the applicability of a
mathematical concept.
What I have Learned (Evaluation/Post Test)
The exercises contained in this section are guaranteed to build
mathematical comprehension, skills, and competence. These serve as a
diagnostic tool to identify the learners’ areas of strengths and
difficulties.
2
OBJECTIVES
At the end of the lesson, the learners are expected to:
K: represent real-life situations using functions, including
piecewise;
S: evaluate and perform operations on functions; and
A: appreciate the use of functions to real-life situations.
LESSON 1 REPRESENTING REAL-LIFE
SITUATIONS USING FUNCTIONS
I. WHAT HAPPENED
PRE-TEST:
Instruction: Read each statement carefully then choose the letter of the best answer. Write
your answer in your notebook.
1. What do we call the elements of x or inputs of functions?
A. Domain
B. relation
C. range
D. none of them
2. What do we call the elements of y or outputs of functions?
A. domain
B. relation
C. range
D. none of them
3. In a given relation below, what is the range?
A. {4,5,6}
B. {−2,3}
C. {−2,4,6}
D. none of them
4. Which of the following is not a function?
A. {(0,1), (1,2), (2,3), (3,4)}
C. {(1,3), (4,2), (2,0), (3,4)}
B. {(0,2), (1,3), (4,3), (1,2)}
D. {{1,2}, {2,2}, {3,2}, {4,2}}
5. Among the value of x and y, which is independent variable?
A. x variable
B. y variable
C. x and y variable D. none of them
II. WHAT YOU NEED TO KNOW
3
This image is a visual representation of function because they have an input, process, and
output.
DISCUSSION
What is a function?
Definition of Functions
•
•
•
•
It is a set of ordered pairs of elements x variable and y variable where the first element
x is not repeating.
It is a set of ordered pairs of elements x and y variables where the second element y
is repeating, or it is called many to one function.
Every value of x corresponds to a unique value of y it is called one to one function.
If every value of x corresponds two or more value of y which is called relation not a
function because function has no element of x is repeating.
Note: All functions are a relation but not all relations are functions.
Functions as representations of real-life situations
Identifying an appropriate function will lead to a better understanding of various phenomena.
Example 1: Give a function C that can represent the cost of buying x meals if one meal costs
₱40.
Solution: Since each meal cost ₱40, then the cost of function is 𝐶 (𝑥 ) = 40𝑥
Example 2: One hundred meters of fencing is available to enclose a rectangular area next to a
river (see illustration below). Give a function A that can represent the area that can be enclose,
in terms of x.
Solution:
The area of a
rectangular enclosure is 𝐴 = 𝑥𝑦. Write this as a function of 𝑥. Since only 100 m
of fencing is available, then the steps are as follows.
A. Write the equation 𝑥 + 2𝑦 = 100
B. Find 𝑦?
C. Simplify the equation
𝑥 + 2𝑦 = 100
2𝑦 = 100 − 𝑥; {𝑡𝑟𝑎𝑛𝑠𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑚𝑒𝑡ℎ𝑜𝑑 }
4
𝑦 = 50 − .5𝑥
Since the equation A= xy, substitute the value of 𝑦 = 50 − .5𝑥 to A=xy
𝐴 = 𝑥(50 − .5𝑥 ); {𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦 𝑢𝑠𝑖𝑛𝑔 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦}
𝐴 = 50𝑥 −. 5𝑥 2 ; {𝑖𝑛𝑡𝑒𝑟𝑝𝑟𝑒𝑡 𝐴 𝑎𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝐴(𝑥)}
𝐴 = 50𝑥 −. 5𝑥 2
What is Piecewise Function?
A piece-wise function consists of a set of different intervals of x with the value of y.
Some situations can only be described by more than one formula, depending on the value of
the independent variable.
Example 1: A user is charge₱300 monthly for a mobile plan, which includes 100 free text
messages. Messages more than 100 are charged ₱1 each. Represent the amount of a consumer
pays each month as a function of the number of messages m sent in a month.
Solution: Let 𝑡(𝑚) represent the amount paid by the consumer each month. It can be
expressed by the piece-wise function.
𝑡(𝑚) = {
𝑖𝑓 0 < 𝑚 ≤ 100
𝑖𝑓 > 100
300
300 + 𝑚
Example 2: A jeepney ride cost ₱8 for the first 4km, and each additional integer kilometer
adds ₱1.50 to the fare. Use a piecewise function to represent the jeepney fare in terms of the
distance d in kilometer.
Solution: The input value is distance, and the output is the cost of the jeepney fare. If 𝑓(𝑑 )
represents the far as a function of distance, the function can be represented as follows:
𝑖𝑓 0 < 𝑑 ≤ 4
𝑖𝑓 𝑚 > 4
8
𝐹 (𝑑 ) = {
8 + 1.5𝑑
III. WHAT HAVE I LEARNED
POST TEST
Read and answer the following problems. Answer in your notebook.
1. A videoke machine can be rented for ₱1000 for three days, but for the fourth day onwards
and additional cost of ₱400 per day is added. Represent the cost of renting a videoke machine
as a piecewise function.
2. The fee for hiring a guide to explore a cave is ₱700. A guide can only take care of a maximum
of 4 persons, and additional guides can be hired as needed. Represent the cost of hiring guides
as a function of the number of tourists who wish to explore the cave.
3. The cost of hiring a catering service to serve food for a party is ₱150 per head for 20 persons
or less, ₱130 per head for 21 to 50 persons, and ₱110 per head for 51 to 100 persons. For 100
or more persons, the cost is at ₱100 per head. Represent the total cost as piecewise function of
the number of attendees of the party.
5
4. A courier company charges ₱100 per package weighing 2kg or less. They charge ₱15 for
each additional kilogram or fraction of a kilogram up to 5 kg. Write a piecewise function that
gives the cost C for shipping a package weighing k kilograms.
5. Elaine is paying ₱799 for an internet service with 3GB monthly data and a connection fee
of ₱20 per hour more than the data. Represent her monthly bill as a function of connection
time.
LESSON 2 EVALUATING FUNCTIONS
I. WHAT HAPPENED
Review: PEMDAS (Parenthesis, Exponent, Multiplication, Division, Addition, Subtraction)
Example: 4(2 − 3)2 − 8 ÷ 2𝑥3 + 1
Grouping Symbols
4(−1)2 − 8 ÷ 2 × 3 + 1
Exponents
𝟒 ( 𝟏) − 𝟖 ÷ 𝟐 × 𝟑 + 𝟏
Multiplication and Division {whichever comes first}
𝟒−𝟖÷𝟐×𝟑+𝟏
𝟒−𝟒×𝟑+𝟏
𝟒 − 𝟏𝟐 + 𝟏
Addition and Subtraction {whichever comes first}
−8 + 1
−7
PRE-TEST
Instructions: Evaluate the following functions at 𝑥 = 1.5. Answer in your activity
notebook/sheets.
a.
b.
c.
d.
𝑓(𝑥 ) = 2𝑥 + 1
𝑞(𝑥 ) = 𝑥 2 − 2𝑥 + 2
𝑔(𝑥 ) = √𝑥 + 1
2𝑥+1
𝑟 (𝑥 ) =
𝑥−1
II. WHAT YOU NEED TO KNOW
DISCUSSION:
Evaluating functions means replacing the variable in the function, in this case 𝑥, with a value
from the function’s domain and computing for the result. To denote that we are evaluating of
𝑥 for some 𝑥 in the domain of f, write 𝑓 (𝑥 ).
Example 1.
Evaluate the function 𝑓(𝑥) = 2𝑥 2 − 3𝑥 + 2 when 𝑥 = −1.
6
Solution:
Step 1: Write the original function.
𝑓 (𝑥 ) = 2𝑥 2 − 3𝑥 + 2
Step 2: Substitute the given value to the variable.
𝑓 (−1) = 2(−1)2 − 3(−1) + 2
Step 3: Simplify by applying the order of operations.
𝑓 (−1) = 2(1) + 3 + 2
𝑓 (−1) = 2 + 3 + 2
𝑓 (−1) = 7
Example 2.
Find 𝑔(−4) and 𝑟(1) where 𝒈 and 𝒓 are as defined in 𝑓 (𝑥 ) = 2𝑥 + 1.
Solution: This is not possible because −4 is not the domain of 𝑔(𝑥) and 1 is not in the domain
of 𝑟(𝑥).
Example 3.
Evaluate the following functions, where 𝑓 (𝑥 ) = 2𝑥 + 1 and 𝑞(𝑥 ) = 𝑥 2 − 2𝑥 + 2.
a. 𝑓(3𝑥 − 1)
b. 𝑞(2𝑥 + 3)
Solution:
a. To evaluate 𝑓 (3𝑥 − 1), {𝑟𝑒𝑝𝑙𝑎𝑐𝑒 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑥 𝑖𝑛 𝑓 (𝑥 ) = 2𝑥 + 1 𝑏𝑦 3𝑥 − 1}
𝑓 (3𝑥 − 1) = 2(3𝑥 − 1) + 1
; {𝑎𝑝𝑝𝑙𝑦 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦}
= 6𝑥 − 2 + 1; {𝑎𝑝𝑝𝑙𝑦 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑢𝑛𝑙𝑖𝑘𝑒 𝑠𝑖𝑔𝑛𝑠}
𝑓 (3𝑥 − 1) = 6𝑥 − 1
b. To evaluate 𝑞(2𝑥 + 3); {𝑟𝑒𝑝𝑙𝑎𝑐𝑒 𝑎𝑙𝑙 𝑥 𝑖𝑛 𝑞(𝑥 ) = 𝑥 2 − 2𝑥 + 2 𝑏𝑦 2𝑥 + 3}
𝑞(2𝑥 + 3) = (2𝑥 + 3)2 − 2(2𝑥 + 3) + 2
note: {apply square of binomials and distributive property}
= (4𝑥 2 + 12𝑥 + 9) − 4𝑥 − 6 + 2{𝑐𝑜𝑚𝑏𝑖𝑛𝑒 𝑙𝑖𝑘𝑒 𝑡𝑒𝑟𝑚𝑠}
𝑞(2𝑥 + 3) = 4𝑥 2 + 8𝑥 + 5
III. WHAT HAVE I LEARNED
POSTTEST
A. Instruction: Answer the following given activities in your notebook.
1. Given 𝑓(𝑥) = 𝑥 − 2, find the following values.
𝑎. 𝑓 (0)
b. 𝑓 (3)
c. 𝑓 (−1)
4
2. Given 𝑓(𝑥 ) = 𝑥, find the following values.
a. 𝑓(1)
b. 𝑓 (2)
c. 𝑓(−1)
d. 𝑓(√2)
1
e. 𝑓 (1−𝑥)
7
OPERATIONS ON FUNCTIONS
LESSON 3
I. WHAT HAPPENED
QUICK REVIEW
*Adding, subtracting and multiplying two or more functions together will result in
another function. Dividing two functions together will result in another function if the
denominator or divisor is not the zero function.
PRE-TEST
Perform the indicated operation:
3
6
6
5𝑥
1. 4 + 5 =
4. 𝑥 ÷ 3 =
2.
3.
6
7
5
− =
2
5.
2
2𝑥−1 𝑥 2 −4
𝑥+2
∙
2𝑥−1
=
1
+ 𝑥−3 =
𝑥+7
II. WHAT YOU NEED TO KNOW
DISCUSSION
Function Operations:
Definition.
Let f and g be functions.
(a)
Their sum, denoted by f + g, is the function defined by
(𝒇 + 𝒈)(𝒙) = 𝒇(𝒙) + 𝒈(𝒙)
Let
𝒇(𝒙) = 𝒙 + 𝟑
and
𝒈(𝒙) = 𝒙𝟐 + 𝟐𝒙 – 𝟖
Example 1. Determine the function (𝑓 + 𝑔)(𝑥).
Solution. (𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥)
by definition
2
= 𝑥 + 3 + 𝑥 + 2𝑥 – 8
by substitution
2
= 𝑥 + 3𝑥 – 5
by combining like terms
(b)
Their difference, denoted by f – g, is the function defined by
(𝒇 – 𝒈)(𝒙) = 𝒇(𝒙) – 𝒈(𝒙)
Example 2. Determine the function (𝑓 – 𝑔)(𝑥).
Solution.
(𝑓 – 𝑔)(𝑥) = 𝑓(𝑥) – 𝑔(𝑥)
8
by definition
= 𝑥 + 3 – ( 𝑥 2 + 2𝑥 – 8)
= 𝑥 + 3 – 𝑥 2 – 2𝑥 + 8
= −𝑥 2 − 𝑥 + 11
(c)
by substitution
by distributive property
by combining like terms
Their product, denoted by f ∙ g, is the function defined by
(𝒇 ∙ 𝒈)(𝒙) = 𝒇(𝒙) ∙ 𝒈(𝒙)
Example 3. Determine the function (𝑓 ∙ 𝑔)(𝑥).
Solution. (𝑓 ∙ 𝑔)(𝑥) = 𝑓(𝑥) ∙ 𝑔(𝑥)
= (𝑥 + 3) ∙ ( 𝑥 2 + 2𝑥 – 8)
= 𝑥(𝑥 2 + 2𝑥 – 8) + 3( 𝑥 2 + 2𝑥 – 8)
= 𝑥 3 + 2𝑥 2 – 8𝑥 + 3𝑥2 + 6𝑥 – 24
= 𝑥 3 + 5𝑥 2 – 2𝑥 – 24
(d)
by definition
by substitution
by distributive property
by distributive property
by combining like terms
Their quotient, denoted by f/g, is the function defined by
𝒇
𝒇(𝒙)
( ) (𝒙 ) =
𝑜𝑟 (f/g)(𝒙) = 𝒇(𝒙)/𝒈(𝒙), excluding the values of x
𝒈(𝒙)
𝒈
where g(x) = 0.
𝑓
Example 4. Determine the function (𝑔) (𝑥).
𝑓
( 𝑔 ) (𝑥) =
Solution.
=
=
(e)
𝒇(𝒙)
by definition
𝒈(𝒙)
(𝑥 + 3)
by substitution
( 𝑥 2 + 2𝑥 – 8)
(𝑥 + 3)
by factoring
( 𝑥 +4)(𝑥 – 2)
The composite function, denoted by (𝒇 ◦ 𝒈), is defined by
(𝒇 ◦ 𝒈)(𝒙) = 𝒇(𝒈(𝒙))
Example 5. Find and simplify (𝑓 ◦ 𝑔)(𝑥).
Solution.
(𝑓 ◦ 𝑔)(𝑥) = 𝑓(𝑔(𝑥))
= 𝑔(𝑥) + 3
= 𝑥 2 + 2𝑥 – 8 + 3
= 𝑥 2 + 2𝑥 – 5
by definition
g(x) is the input of f(x)
by substitution
by combining like terms
ILLUSTRATIVE EXAMPLES
Example 1. Consider the following functions below:
𝑓 (𝑥 ) = 𝑥 + 3
𝑝(𝑥 ) = 2𝑥 − 7
Determine the following.
a) (𝑓 + 𝑝)(𝑥)
b) (𝑝/ℎ)(𝑥)
Solution.
𝑔(𝑥 ) = 𝑥 2 + 2𝑥 − 8
𝑥+7
ℎ (𝑥 ) =
2−𝑥
c) (𝑝 – ℎ)(𝑥)
d) (𝑓 ∙ 𝑔)(𝑥)
9
e) (𝑓 ◦ 𝑔)(𝑥)
a) (f + p)(x) = f(x) + p(x)
= x + 3 + 2x – 7
= 3x – 4
by definition
by substitution
combining like terms
b) (p/h)(x) = p(x)/h(x)
2𝑥−7
= 𝑥+7
by definition
by substitution
2−𝑥
= 2𝑥 − 7 ∙
step 1 for the operation division
𝑥+7
4𝑥−2𝑥 2 −14+7𝑥
=
=
2−𝑥
by distributive property
𝑥+7
−2𝑥 2 +11𝑥−14
combining like terms
𝑥+7
c) (𝑝 – ℎ)(𝑥) = 𝑝(𝑥) – ℎ(𝑥)
𝑥+7
= 2𝑥 − 7 − 2−𝑥
by definition
by substitution
2−𝑥
𝑥+7
= (2𝑥 − 7) ∙ 2−𝑥 −
=
=
=
=
4𝑥−2𝑥 2 −14+7𝑥
−
step 2 of the operation
subtraction
2−𝑥
𝑥+7
by distributive property
2−𝑥
2−𝑥
4𝑥−2𝑥 2 −14+7𝑥−(𝑥+7)
subtracting two same denominators
2−𝑥
4𝑥−2𝑥 2 −14+7𝑥− 𝑥−7
by distributive property
2−𝑥
−2𝑥 2 +10𝑥−21
simplifying the numerator
2−𝑥
d) (𝑓 ∙ 𝑔)(𝑥) = 𝑓(𝑥) ∙ 𝑔(𝑥)
= (𝑥 + 3)(𝑥 2 + 2𝑥 − 8)
= 𝑥 3 + 2𝑥 2 − 8𝑥 + 3𝑥 2 + 6𝑥 − 24
= 𝑥 3 + 5𝑥 2 − 2𝑥 − 24
e) (𝑓 ◦ 𝑔)(𝑥) = 𝑓(𝑔(𝑥))
= 𝑔(𝑥) + 3
= 𝑥 2 + 2𝑥 − 8 + 3
= 𝑥 2 + 2𝑥 − 5
by definition
by substitution
by distributive property
combining like terms
by definition
g(x) is the input of f(x)
by substitution
combining like terms
Example 2. Each pair of functions below have a special relationship because
𝑥
2𝑥
(f ◦ g)(x) = x and (g ◦ f)(x) = x. Verify these 𝑓(𝑥 ) = 𝑥−2 ; 𝑔(𝑥 ) = 𝑥−1
Solution.
(f ◦ g)(x) = f(g(x)) =
𝑔(𝑥)
𝑔(𝑥)−2
=
2𝑥
𝑥−1
2𝑥
−2
𝑥−1
10
=
2𝑥
𝑥−1
2𝑥−2(𝑥−1)
𝑥−1
=
2𝑥
𝑥−1
2𝑥−2𝑥+2
𝑥−1
=
2𝑥
𝑥−1
2
𝑥−1
2𝑥
=𝑥−1 ∙
(g ◦ f)(x) = g(f(x)) =
𝑥−1
2
2𝑓(𝑥)
𝑓(𝑥)−1
2𝑥
= 𝑥−2 ∙
𝑥−2
2
=x
=
𝑥
)
𝑥−2
𝑥
−1
𝑥−2
2(
=
2𝑥
𝑥−2
𝑥−1(𝑥−2)
𝑥−2
=
=x
III. WHAT HAVE I LEARNED
POST TEST:
Find and simplify the following functions given f(x) and g(x):
1. f(x) = x + 2 ; g(x) = x2 – 4
(f + g)(x) =
2. 𝑓 (𝑥 ) = √(𝑥 − 2)2 ; g(x) = x2 – 4
(g/f)(x) =
3. 𝑓 (𝑥 ) =
𝑥−2
𝑥+2
; 𝑔 (𝑥 ) =
(f ∙ g)(x) =
4
4. 𝑓 (𝑥 ) = 𝑥+3; 𝑔(𝑥 ) =
(f – g)(x) =
1
𝑥
2
𝑥+3
5. f(x) = x2 – 2x + 1 ; g(x) = x + 2
(f ◦ g)(x) =
11
2𝑥
𝑥−2
𝑥−𝑥+2
𝑥−2
=
2𝑥
𝑥−2
2
𝑥−2
SYNOPSIS
This Self Learning Kit (SLK) discusses on how to represent real-life
situations using functions, evaluate functions and perform operations on
functions. . It also gives examples in solving problems involving functions.
The learners are expected to assess their skills and apply what they have
learned in their everyday life. It also enhances their skills and utilizes what they
have learned.
Bring it on and let us discover jointly in this wonderful journey of learning
how to perform operations on functions including composition of functions.
WRITERS
RONALD G. TOLENTINO is a faculty member of
Negros Oriental High School (NOHS- SHS
Department). He finished his BS Mathematics
degree batch 1993 and earned his 27 units in MATMathematics from Silliman University. He
graduated the CPE at PTC Dumaguete City last
2018. He is currently the SHS DRRR Coordinator of
Negros Oriental High School.
12
ANSWER KEY
LESSON 1
LESSON 3
Pre-Test
39
23
3𝑥+1
1. 20
2. − 14 3. (𝑥+7)(𝑥−3)
Pre-test
1. A 2. C
3. C 4. B
5. A
Posttest
1000
𝑖𝑓 0 ≤ 𝑥 ≤ 3
1. 𝑓 (𝑥 ) = {
1000 + 400(𝑥 − 3) 𝑖𝑓 𝑥 ≥ 4
18
4. 5𝑥 2
Posttest
𝑥
2. 𝑓(𝑥 ) = 700 (4) , 𝑥 ∈ 𝑁
150𝑥 𝑖𝑓 0 ≤ 𝑥 ≤ 20
130𝑥 𝑖𝑓 21 ≤ 𝑥 ≤ 50
3. 𝑓 (𝑥 ) = {
110𝑥 𝑖𝑓 51 ≤ 𝑥 ≤ 100
100𝑥 𝑖𝑓 𝑥 ≥ 100
100 𝑖𝑓 0 < 𝑥 ≤ 2
115 𝑖𝑓 2 < 𝑥 ≤ 3
4. 𝑓 (𝑥 ) = {
130 𝑖𝑓 3 < 𝑥 ≤ 4
145 𝑖𝑓 4 < 𝑥 ≤ 5
5. 𝑓 (𝑥 ) = 799 + 20𝑥
1. 𝑥 2 + 𝑥 − 2
3.
5. 𝑥 2 + 2𝑥 + 1
LESSON 2
Pre-test
a. 𝑓(𝑥 ) = 4
b. 𝑓 (𝑥 ) = 1.25
c. 𝑓(𝑥 ) = √2.5
d. 𝑓 (𝑥 ) = 8
Posttest
1. a. 𝑓 (0) = 2
b. 𝑓 (3) = 1
c. 𝑓 (−1) = −3
2. a. 𝑓 (1) = 4
b. 𝑓 (2) = 2
c. 𝑓 (−1) = −4
d. 𝑓(√2) =
𝑥−2
𝑥(𝑥+2)
4√2
2
1
e. 𝑓 (1−𝑥) = 4 − 4𝑥
13
2. 𝑥 + 2
4.
2
𝑥+3
5. 𝑥 − 2
DEPARTMENT OF EDUCATION
SCHOOLS DIVISION OF NEGROS ORIENTAL
SENEN PRISCILLO P. PAULIN, CESO V
Schools Division Superintendent
JOELYZA M. ARCILLA, EdD
Assistant Schools Division Superintendent
MARCELO K. PALISPIS, EdD
Assistant Schools Division Superintendent
NILITA L. RAGAY, EdD
OIC - Assistant Schools Division Superintendent
CID Chief
ROSELA R. ABIERA
Education Program Supervisor – (LRMS)
ELISA L. BAGUIO, EdD
Division Education Program Supervisor – MATHEMATICS
MARICEL S. RASID
Librarian II (LRMDS)
ELMAR L. CABRERA
PDO II (LRMDS)
NAME OF WRITER
DOMINADOR A. JUMAO-AS
RONALD G. TOLENTINO
NAME OF ILLUSTRATOR/LAY-OUT ARTIST/TYPESETTER
LITTIE BETH S. BERNADEZ
_________________________________
ALPHA QA TEAM
LITTIE BETH S. BERNADEZ
MERCYDITHA D. ENOLPE
RONALD TOLENTINO
DIDITH T. YAP
BETA QA TEAM
ELIZABETH A. ALAP-AP
EPIFANIA Q. CUEVAS
NIDA BARBARA S. SUASIN
VRENDIE P. SYGACO
MELBA S. TUMARONG
HANNAHLY I. UMALI
DISCLAIMER
The information, activities and assessments used in this material are designed to provide
accessible learning modality to the teachers and learners of the Division of Negros Oriental. The
contents of this module are carefully researched, chosen, and evaluated to comply with the set
learning competencies. The writers and evaluator were clearly instructed to give credits to
information and illustrations used to substantiate this material. All content is subject to copyright and
may not be reproduced in any form without expressed written consent from the division.