QUARTER 2
Mathematics
G9
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The Editors
PIVOT 4A CALABARZON Math G9
PIVOT 4A Learner’s Material
Quarter 2
First Edition, 2020
Mathematics
Grade 9
Job S. Zape, Jr.
PIVOT 4A Instructional Design & Development Lead
Angelo D. Uy
Content Creator & Writer
Jhonathan S. Cadavido
Internal Reviewer & Editor
Lhovie A. Cauilan & Jael Faith T. Ledesma
Layout Artist & Illustrator
Jhucel A. del Rosario & Melanie Mae N. Moreno
Graphic Artist & Cover Designer
Ephraim L. Gibas
IT & Logistics
Published by: Department of Education Region IV-A CALABARZON
Regional Director: Wilfredo E. Cabral
Assistant Regional Director: Ruth L. Fuentes
PIVOT 4A CALABARZON Math G9
Guide in Using PIVOT 4A Learner’s Material
For the Parents/Guardians
This module aims to assist you, dear parents, guardians, or siblings
of the learners, to understand how materials and activities are used in the
new normal. It is designed to provide information, activities, and new
learning that learners need to work on.
Activities presented in this module are based on the Most
Essential Learning Competencies (MELCs) in Mathematics as prescribed
by the Department of Education.
Further, this learning resource hopes to engage the learners in
guided and independent learning activities at their own pace. Furthermore,
this also aims to help learners acquire the essential 21st century skills
while taking into consideration their needs and circumstances.
You are expected to assist the children in the tasks and ensure the
learner’s mastery of the subject matter. Be reminded that learners have to
answer all the activities in their own answer sheet.
For the Learners
The module is designed to suit your needs and interests using the
IDEA instructional process. This will help you attain the prescribed
grade-level knowledge, skills, attitude, and values at your own pace outside
the normal classroom setting.
The module is composed of different types of activities that are
arranged according to graduated levels of difficulty—from simple to
complex. You are expected to :
a. answer all activities on separate sheets of paper;
b. accomplish the PIVOT Assessment Card for Learners on page
38 by providing the appropriate symbols that correspond to your
personal assessment of your performance; and
c. submit the outputs to your respective teachers on the time
and date agreed upon.
PIVOT 4A CALABARZON Math G9
Parts of PIVOT 4A Learner’s Material
Development
Introduction
K to 12 Learning
Delivery Process
What I need to know
What is new
What I know
What is in
What is it
Engagement
What is more
What I can do
Assimilation
What else I can do
What I have learned
What I can achieve
Descriptions
This part presents the MELC/s and the desired
learning outcomes for the day or week, purpose of
the lesson, core content and relevant samples.
This maximizes awareness of his/her own
knowledge as regards content and skills required
for the lesson.
This part presents activities, tasks and contents
of value and interest to learner. This exposes
him/her on what he/she knew, what he/she does
not know and what he/she wants to know and
learn. Most of the activities and tasks simply and
directly
revolve
around the concepts of
developing mastery of the target skills or MELC/s.
In this part, the learner engages in various tasks
and opportunities in building his/her knowledge,
skills
and
attitude/values
(KSAVs)
to
meaningfully connect his/her concepts after
doing the tasks in the D part. This also exposes
him/her to real life situations/tasks that shall:
ignite his/ her interests to meet the expectation;
make his/her performance satisfactory; and/or
produce a product or performance which will help
him/her fully understand the target skills and
concepts .
This part brings the learner to a process where
he/she shall demonstrate ideas, interpretation,
mindset or values and create pieces of
information that will form part of his/her
knowledge in reflecting, relating or using them
effectively in any situation or context. Also, this
part encourages him/her in creating conceptual
structures giving him/her the avenue to integrate
new and old learnings.
This module is a guide and a resource of information in understanding the
Most Essential Learning Competencies (MELCs). Understanding the target
contents and skills can be further enriched thru the K to 12 Learning Materials
and other supplementary materials such as Worktexts and Textbooks provided by
schools and/or Schools Division Offices, and thru other learning delivery
modalities, including radio-based instruction (RBI) and TV-based instruction
(TVI).
PIVOT 4A CALABARZON Math G9
WEEKS
1-2
Variation
Lesson
I
There are varied things around you that you should know how things are
related to each other. Like how is time related to the speed of a vehicle or even as
simple as the relationship of the length of your arm span with your height.
Quantities may vary directly that is, as one quantity increases the other
quantity must also increase or they may vary inversely, that is if one quantity increases the other quantity decreases.
When two quantities increase at he same time or decrease at the same
time , it shows direct variation. However if one quantity increases and the other
decreases, it shows inverse variation.
You encounter such situations in your everyday life. For instance, if walk
slowly going to school it takes longer time but walking fast, takes a shorter time
reaching the school. Your speed in walking varies inversely with the time. The
faster you walk the shorter the time it takes you to reach your destination.
Another situation is when your mother asked you to buy rice, which cost P
55.00 a kilo. If one kilo costs P55.00, how much 10 kilos of rice cost?. In this situation the more kilos of rice the more money you will pay. Hence this is an example
of direct variation.
Situations involving 3 or more quantities that varies from one another may
show joint variation or combined variation.
D
Direct Variation
In your previous lesson you were able to represent relationship between
two quantities using graphs, table of values and equations.
Examples:
1. Given:
Graph:
x
1
2
3
4
5
y
2
4
6
8
10
Table of values:
Equation:
y = 2x
You can see from the table of values that as the
value of x increases the value of y also increases. The graph is a line that rises to
the right. This is an illustration of a direct variation. This means that y varies
directly as x. In symbol:
y
 k or y = kx, where k is the constant of variation or constant of
x
proportionality. In this case the constant of variation is 2.
PIVOT 4A CALABARZON Math G9
6
2. You will be celebrating your 14th birthday and you want to have 14 balloons in
your birthday party. The table below shows the number of balloons (x) and the
corresponding cost (y).
As you can see the values of x increase , the values of y also increase. Hence the cost of balloons vary 10 20 30 40 ... 70 ies directly to the number of balloons.
y1 y2 10 20
Since y varies directly as x, find the constant of variation k:



5
x1 x2
2
4
The constant of variation (k) is 5. Thus the equation of the
direct variation is y = 5x.
x
2
4
6
8
... 14
3. y varies directly as x. If y = 12 when x = 4,
A. Find the constant of variation
when y is 36?
b. What is the equation?
C. What is x
Solution:
y 12
A. k 

3
x
4
B . y = 3x
C. using the equation of variation: y = 3x
36
 12
36 = 3x
x=
3
When the value of y increases from 12 to 36, the value of x increases also from
4 to 12.
Direct Square Variation
The relationship between the area and the square of the radius is an example of a direct square variation.
A = πr2 , the constant of variation is π or 3.1416.
Direct square variation states that if y varies directly to the square of x, there is
a non zero constant k such that
y
k
x2
or y = kx2 . The graph of direct square
variation is quadratic in nature it follows a quadratic curve.
Examples.
1. y varies directly as the square of x. If y = 27 when x = 3, find x when y = 81.
Solution: Find the constant of variation:
k 
y
27
27
 2 
3
x2
3
9
Equation y = 3x2
Find x when y = 81
Using the equation y = 3x2 , find x
81 = 3x2
x2 =
81
 27
3
x=
27  3 3
2. If p varies directly as the square of q and p = 256 when q = 8, find:
A. k
B. equation
c. p when q = 2 3
Solution:
A.
k 
p
256

4
q2
82
B. p = 4q 2
C. p = 4q2
7
p = 4( 2 3)2 = 4(12) = 48
PIVOT 4A CALABARZON Math G9
Inverse Variation
In real life there are different situations that shows inverse relation.
Inverse variation states that y varies inversely as x or y is inversely proportional
k
to x if there is non zero constant k, such that y 
or xy = k. The decrease of
x
one quantity results to the increase of the other quantity.
Examples:
1. The table below shows that number of persons working together (x) and the
number of hours (y) the job is completed. Find the constant of variation and
equation. Graph the given table of values.
x
1
2
3
4
5
Constant of variation: xy =
y
2
1
2
3
1
2
2
5
Equation xy = 2 or
y
2
x
2
2
5   3   2(1)  2
5
3
Graph:
2. If y varies inversely as and y = 20 when x = 8, find
A. constant of variation
Solution: A. k = xy
B. variation equation
k = 8(20) = 160
B. xy = 160 or y =
C. y =
C. y when x = 16
160 160

 10
x
16
160
x
As the value of x increases from 8 to 16, the value of y decreses
from 20 to 10.
3. The rate (r ) of the car varies inversely with the time (t). The car travels at a
rate of 60kph for 1.5 hours. Find the rate of the car when it took 1 hour to travel
the same distance (d),
Solution:
In this case, the faster the car travel, the shorter time it will take to cover the
same distance. Hence the constant of variation here is the distance.
Therefore d = rt = 60(1.5) = 90 km. To solve for the rate : r 
d 90

 90 kph.
t
1
The faster the car travels, the shorter the time it takes to travel the same distance.
PIVOT 4A CALABARZON Math G9
8
Learning Task 1:
A. Complete the table if y varies
directly as x or
y
x
30
5
8
B. Complete the table if y varies inversely as x.
x2
k
Equation
y
x
5
6
4
10
18
y = 3x
72
y = 2x2
k
Equation
36
20
y=
12
y=
50
x
24
x
Joint and Combined Variation
Joint and combined variation involves 3 or more quantities that may vary
directly and or inversely to each other.
Joint Variation
If the ratio of one quantity to the product of the other two quantities is cony
 k
stant, the they vary jointly. That is, if y varies jointly as x and z, then
xz
or y = kxz.
Examples:
1. The surface area of a cylinder varies jointly as the radius and height . A = krh,
where A is the surface area , r is the radius and h is the height of the cylinder.
The constant of variation k = 2π. Hence A = 2πrh. Any of the variable r or h increases the area also increases.
2. Translate the following into variation equation:
(a) p varies jointly as q and r.
Equation: p = kqr
(b) area of parallelogram varies jointly as its base and altitude.
Equation: A = bh, k = 1
3. If m varies jointly as p and q and m = 50 when p = 5 and q = 2, find m when
p = 10 and q = 6.
50
y
 k
Solution: Determine the constant of variation:
k = 5(2)  5
xz
m = kpq = 5(10)(6) = 300. As p and q increase m also increases.
Combined Variation
This is a variation where one quantity varies directly to other quantity and
inversely to the other quantity.
The equation
Examples:
v 
kw
means
x
the v varies directly as w and inversely as z.
1. v varies inversely to x and directly to w. If v = 12 when x = 4 and w = 8, find w
when v = 24 and x = 2.
9
PIVOT 4A CALABARZON Math G9
Solution:
Equation: v =
kw
.
x
Find k:
6w
x
Solve for w: v =
12 
24 
6w
3
k (8)
4
12( 4)
6
8
24(3)
w
 12
6
k=
2. If p varies directly as the square of q and inversely as the square root of r, and
p = 20 when q = 2 and r = 64, find p when q = 8 and r = 144.
Solution:
k (2) 2
kq 2
Solve for k: 20 
64
r
2
= 40(8)  40(64)  640
12
3
144
Equation: p =
Solve for p: p
k=
20 64
20(8)

 40
22
4
Learning Task 2
A. Supply the missing value if y varies
B. Supply the missing value if y varies
jointly as x and z
y
x
z
30
3
5
80
k
Equation
4
6
60
directly to v and w and inversely to z
8
Y= 2xz
v
w
z
5
12
14
24
18
4
8
16
4
10
48
y
Y =3xz
12
Y=
20
3
xy
4
12
24
6
6
2
7
15
k
Equation
10
3vw
2z
8vw
y
z
y 
6
E
Learning Task 3
A. Identify if the given equation is a direct, inverse, joint or combined variation
with k as the constant of variation.
1. b = kd
3. m =
2. y = klm
4.
a
 kc
b
kn
p
5. mn = kpq
B. Write the equation of variation and solve for the constant of variation.
1.
2.
p
1
2
3
4
5
q
7
28
63
p
3
6
12
24
48
q
16
8
4
2
1
3.
112 175
PIVOT 4A CALABARZON Math G9
4.
10
p
3
6
12
24
48
q
16
8
4
2
1
p
6
12
18
24
30
q
2
4
6
8
10
A
Learning Task 4
A. Solve the following:
1. The distance (d) from the center of the seesaw varies inversely as the weight (w)
of a person. JB who weighs 50 kg sits 3 feet from the fulcrum. How far from
the fulcrum must JP sit in order to balance with JB if he weighs 35 kg?
2. The number of pages (p) that Ethan reads varies directly as the number of
hours (t) he is reading.
A. write the variation equation
B. If he can read 21 pages in 14 minutes, how may pages can he read in 21
minutes?
3. The pressure of the gas is directly proportional to the temperature and inverse
ly proportional to its volume.
A. Write the variation equation.
B. What happened to the pressure if the volume is reduced to half and the
temperature is doubled?
4. Given the equation y  k
statement is true or false.
pq 2
, where k is the constant of variation, tell whivh
r
A. y and r varies directly.
B. y and q2 are directly proportional
C. y and pq2 varies jointly
D. p and r are inversely proportional
E. y and p varies directly.
5. The volume of a cylinder is given by the formula V = πr2 h. If r is increased by
50% and the height is reduced by 25%, what will happen to the volume? What
is the constant of variation.
B. Give at least three examples of quantities you know or you have or experienced
that show different types of variation.
11
PIVOT 4A CALABARZON Math G9
WEEK
3
Integral and Zero Exponents
I
Lesson
You have learned in previous lessons about laws of exponents. Let us
recall these laws:
(a) Product of a Power: am · an = am+n
(b) Quotient of a Power:
am
 amn
an
(c ) Power of a Power: (am)n = amn
(d) Power of a product: (a · b)m = am bm
m
 
(e )Power of a Quotient:   
a
b
am
bm
In the process of performing the laws of exponents it may result to a
positive, negative, or zero exponents. In this lesson you will learn on how to
simplify
expressions with integral (positive or negative integers) or zero
exponent applying the different laws of exponents.
D
Zero Exponent
Simplify the expression
aa
a2
1
. Write this expression in factored form
aa
a2
Applying the laws of exponent , quotient of a power,
any variable/number a , where a ≠ 0, a0 = 1.
a2
 a22  a0  1
a2
. Thus, for
Examples:
1. 12x0 = 12(1) = 12 , only variable x is raised to zero power
2. (12x)0 = (12)0 (x)0 = 1(1) = 1 applying the law of power of a product, both 12 and
x are raised to zero power
3. (x0 + 3)(x + 3)0 = (1 + 3)( 1) = 4, the expression (x + 3) is raised to zero power
so it is equal to 1.
4. Simplify:
Solution:
45a 4 b 2 , apply laws of exponent in simplifying
5a 4 b
9a4 - 4 b2 - 1 = 9a0 b = 9b
5. Simplify: 2a0 + (2a)0 + 20a
Solution: 2(1) + 1 + 1(a) = 2 + 1 + a = 3 + a
6. Simplify: 4m(n  p ) 0
4m 0
Solution:
4m(1)
m
4(1)
PIVOT 4A CALABARZON Math G9
12
Integral Exponent
An expression has an integral exponent if the exponent is either positive or
negative integers.
The expressions like 3x, 5x-2 + 2, m-8 , are examples or expressions with
integral exponents.
am
 a mn
an
One of the laws of exponent is quotient of a power that is :
Apply this law of exponent in simplifying the following:
1.
2.
x4
 x 42  x 2 . You can use factoring to check the answer:
x2
xxxx
 x  x  x2
xx
x2
 x 2  4  x  2 . The exponent is negative. Use factoring to simplify the expresx4
2
x
x x
1

 2
x x x x
x4
x
sion .
1
x2
. Hence the expression x -2 is equal to
1
The negative sign in the exponent means the reciprocal. Thus a n  a n . Similarly if the expression with negative exponent is in the denominator , get its
reciprocal to express the exponent as positive. Therefore 1 n  a n
a
3. Simplify:
3x3 y 6
x5 y 4
Solution: Observe the exponents of each variable. Let the variable with greater
exponent remain in its position but subtract the smaller exponent from the larger
exponent.
3x3 y 6
3 y64
3y2


5
4
53
x y
x
x2
4. Express all exponent to a positive integers.
(a)
28a 5 b 6 c
4a 0 b 2
4b 2


7a 5b 4 c 3
c2
c2
(b) (5 - 2)5 · 3-3 + (5 + 2)0 = (3)5 · 3-3 + 1 = 32 + 1 = 9 + 1 = 10
(c ) [(8)2]-1 ⋅ (10 - 8)14 ⋅ 4-5
(8)-2 · (2)14 · 4-2
, apply the law of exponent power of a power
(23 )-2 · 214 · (22 )-2 , change to the same base so can apply the law of
2-6 · 214 · 2-4
exponent.
2-6+14+(-4) = 24 = 16, apply the law of exponent product of a power.
(d ) (5 - 2)5 · 3-8 + (5 + 2)0
(3)5 · 3-8 + 1
3-3 + 1
1
+1
33
1
27
28


27 27
27
apply product of a power
get the reciprocal of the number with the negative
exponent to make the exponent positive.
Change to similar fractions before adding.
13
PIVOT 4A CALABARZON Math G9
E
Learning Task 1:
Express the following expression into non zero and non negative exponents. Simplify your answer.
1
8
4
 (ab) 
24
x
y

1. 7-1
6.
11. 
 9 
6 x5 y 4
1
9
 5x0 
2. (14abc)0
7. 
12.

x 2
 y 
3. 10-9
8.
120a 5b 6 c 5
12a 2b 0 c 2
13.
4. 5(xy)0
9.
(4 x) 0 y 5 z 2
(234 xyz ) 0
14.
5. 015
10.
 (5 xy ) o

 10



2
15.
12b 3
a 4
1
a 5 n
1
 
2
A
Learning Task 2
Find the single value of the following with-
1. 2-2 · 3-1
2.
100 + 10
3. 4(60) + 3(3-1 )
4. 5-1 - 2-3
5. (120)- 10
11.
7 2
41
12.
9
21
13. 82
40
14.
3
15.  5 
6. 24 + 2– 2
7. (5 + 5-2)0 (60 + 6)-2
8. 20 • 33
9. 5(-6)0
10. 10-3 ÷ 103
PIVOT 4A CALABARZON Math G9
1
( 4) 3
14
2
out exponent.
5
Rational Exponent and Radical Expressions
I
WEEK
4
Lesson
Your previous lesson is all about integral and zero exponents. This time,
your lesson is on rational exponents and radicals numbers. When simplifying
expressions with rational exponents you should remember the laws of integral
exponents, the rule is the same. However, you need to remember the rules in
adding fractions. In adding fraction, you can only add fractions if they are similar
otherwise you change dissimilar fraction to similar fraction.
Examples:
Simplify the following expressions:
1.
a
1
2
3
2
a
 a
1
3

2
2
, apply the product of a power law of exponent by
adding the exponents of the same base.
4
= a 2  a2
2.
6 x  
1
3
6
, apply the rule in adding fraction then simplify
6
 6 x  3
, apply the power of a power law of exponents by
multiplying the exponents. Use the rule in multi
plying fractions
= (6 x) 2  6 2 x 2 , apply power of a product law of exponent.
= 36x2 , simplify
1
 4 2


3. 
9
=
1
4  2
1
9  2
2 
3 
2
1
2
2
1
2
, apply the quotient of a power law of exponents.
, express 4 and 9 in exponential form
2
=
22
3
4.
2
1
12 3  12 2

2
2
2
, multiply the exponents , then simplify
3
2
 12  3

4
=
1
2
3
126  6
, apply the product of a power law of exponent
7
 12 6
=
12
6
, change the fractions to similar fraction
then add.
 12

1
6
6
1
1
, simplify by
 12 6  12 6  1212 6
applying laws of exponents.
Exponential expressions with rational exponents can be expressed into radical
expressions.
15
PIVOT 4A CALABARZON Math G9
D
Radicals
An exponential number whose exponent is a rational number can be expressed as radicals
Examples:
Exponential
8
Base
Radicals
Exponent
1
3
radical sign
index
3
8
radicand
The denominator of the rational exponent becomes the index of the radical
number.
3
This is read as “cube root of 8”
8
Exponential
Radical
3
122
2
12 3 

12

3
This is read as the
“square root of 12
raised to the 3rd power”
When changing the exponential number with rational exponent whose denominator is 2 to radical number you may or may not write the index 2.
The exponential number
a
m
n ,
when changed to radical is
n
am
read as the nth root of a raised to m.
The nth root of a number is a number that is taken n times as a factor of the
radicand a.
You can express exponential expression to radical expression and vice versa.
Examples:
1. Change
5
1
4
5
=
1
4
to radical expression.
5 (read as the 4th root of 5)
4
3
2. Write 4 2 in radical expression
3
42
4. Write

 4
( read as the square root of 4 cube or 4 raised to the 3rd
power)
3
 8  in exponential form
 8   8 , the index (3) is the denominator of the fractional exponent
3
3
4
4
4
3
and the power (4) is the numerator of the fractional
exponent.
PIVOT 4A CALABARZON Math G9
16
5.
5
Change the expression
32 to exponential form.
 15  , the 5th root of 32 is equal to 32 raised to one fifth.

32  
 32 


5
Laws of Radicals
Laws of radicals are similar to the laws of integral exponents.
Let a, b, m, n are integers:
1.
2.




n
 a   b   a
1
n

n
a

n
 n

a

1
n
n
3.
n
a

b
4.
n m
n
m
a 
1
 b n  ab 
 a n
  
b
 1
  a n

1
n
Examples
, the nth root of a raised to the power of n is equal to a.
1
n
1
1
a  n
1
b  n
a
n
 an  a
n

a
b
n
ab
, the nth root of a times the nth root of b is
equal to the nth root of the product of a and b.
, the nth root of a divided by the nth root of b is
equal to the nth root of the quotient of a and b.
1
1
m
  a mn  mn a , the root of a root is the rth of a, where r = mn


Apply the laws of radicals.
3
1.3 12   12 3   12 3  12

3

1

2.
5
3

4 5 8 
5
( 4)(8) 
5
32  2 ,
multiply if the indices are the same, just multiply
the radicand, copy the common index, then simplify.
The fifth root of 32 is 2, meaning, when you multiply the root 2 by itself 5
times the answer is 32. The index 5 indicates the number of times the root be
multiplied by itself to get the radicand. In this case, 2 5 = 32. Hence 5 32  2
3.
4.
18

2
3
18

2
9 3
, divide the radicands if the indices are the same, then
simplify by extracting the square root of 9 . The 9  3,
since 32 = 9
64  3 8  2 , the square root of 64 is 8 because the 82 = 64 and the cube
root of 8 = 2, since 23 = 8.
5.
3
2
1
2 1

2
12 2 . 12  12 3  12 2  (12) 3
 (12)
43
6
7
 (12) 6  6 12 7  6 12 6  12  6 12 6  6 12  126 12
Apply the laws of rational exponents and radicals to simplify the expression.
Learning Task 1
Use the Laws of rational exponent to simplify:
1.
n 
2.
3b 2  b 2
3
2
4
1
3. 
3

a 




1
2
1
3
4.
17
3
124
1
4 4
5.
a b 
3
1
4 12
PIVOT 4A CALABARZON Math G9
E
Learning Task 2
A. Change exponential expression to radical expression.
2
1. 1253
3
3.
1
2. 5x 2
 15 52 
 5 5 




5.
 12  4
16 




2
xy 5
4.
B. Write each expression in exponential form.
5
3
1. ( 7 )
3.

5

11x
4
4m

2a
8
83 4
5.

3
228
4
A
Learning Task 3
A. State the laws of Radical Exponents and give examples
B. Apply the laws of rational exponents in simplifying the expression.
1.
1
2.
3.
 12 14 
5 x 




a
 6
b



16
y8

3
8
2a  2a 
2
1
2
2
8.
x 
1
10 5
12
3



7.
2
1
6
 4m y

 8 p 20

8
5.
x
 43 3  3
p q 




12
4.
1
 16m 9 n 3


26
6. 




1
4
PIVOT 4A CALABARZON Math G9
9.
10.
 
 x4
5
1

 3q 2 r 4

1
 62

18





2
1
2
1
3
Simplifying Radical Expressions
WEEKS
5-6
I
A radical is in its simplest form if:
1. the radicand does not contain a perfect square, cube or nth power.
Examples:
2
is in its simplest form since 2 is not a perfect square,
meaning the square root of 2 is not exact.
5 Is in its simplest form since 5 is not a perfect cube,
meaning, the cube root of is not exact.
3
2. there is no fraction in the radicand or no radical in the denominator




Example:
3
4




3
3 , the fractions is simplified such that

2
4
there is no radicand in the denominator.
3. the index of the radical id the lowest possible index.
Example:
6
8 is not yet in simplest form since the index 6 can be
expressed as the product of 3 and 2.
3
8  2 since the cube root of 8
To simplify express it as
is 2 and square root of 2 is not exact.
You have to simplify radical expressions with:
1. a perfect nth factor
2. a fraction radicand or denominator
3. reducible index.
Recall that :
m
n
a  a
n
m
a.
n
a m , is a radical expression which indicates the
nth root of am .
b. n is the index which gives the order of radicals; and
c. am is the radicand, the number within radical sign.
D
The properties of radicals which can be useful in simplifying radical expressions
are as follows:
1.
2.
3.
n
ab 
n m
n
p 
p

q
n
nm
n
p
n
q
a  n b , the nth root of a product is equal to the nth root of each factor.
p
, the nth root of the mth root of a number id equal to the nmth root
of the number.
, the nth root of the quotient is equal to the nth root of the numerator and the nth root of the denominator.
19
PIVOT 4A CALABARZON Math G9
Simplifying Radicals by Reducing the Radicand
Reducing radicand is finding a factor of a radicand whose indicated roots
can be found.
Examples:
1. Simplify
50
Solution: 50 is not a perfect square, therefore you have to find a factor of
50 which is a perfect square.
50  25  .2 , the factors of 50 are 25 and 2. 25 is a factor which a
perfect square, meaning you can extract the square
root
25  2 , the square root of the product is equal to the
=
square root of the factors.
= 5 2 , the square root of 25 is 5 , square root of 2 is already
in simplified form.
Therefore
2. Simplify
3
50  5 2 in simplest form
81
Solution: The factor of 81 which is a perfect cube is 27.
3
Therefore
3. Simplify
3
5
81  3 27  3  3 27  3 3  33 3
81  3 3 since 3 3 
3
3
3
3
 13 

3 
 3   27(3)  81


3
32 x 7 y 5 z
Solution: To get the root of variables with integral exponent, the exponent must
be divisible by the index.
5
32 x 7 y 5 z  5 32 x 5 y 5  x 2 z  5 32 x 5 y 5  5 x 2 z  2 xy 5 x 2 z
Take note that when you divide the exponent of x which is 7 by the index 5 there is a remainder of 2 and the exponent of z which is 1 cannot be divided
by 5. Hence the variables tha will remain under the radical sign are x 2 and z.
4 5 6
4
4. Simplify 5 48a b c
Solution: Factor the radicand such that one factor is the 4th power of a certain
number ad the variable such the the exponent is divisible by the index 4.
54 48a 4b10 c 5  54 16a 4b8c 4  3b 2 c  54 16a 4b8c 4  4 3b 2 c  5(2ab 2 )4 3b 2 c  10ab 2 c 4 3b 2 c
The factors of 48 is 16 and 3 where 16 is a perfect 4th power of a number.
The factors of b10 are b8 where the exponent is divisible by 4 and b2 . The factors
of c5 are c4 and c. 5 is the coefficient of the radical expression. Multiply the coefficient 5 with the 4th root of 16a4 b8 c4 .
PIVOT 4A CALABARZON Math G9
20
Simplifying Radicals by Reducing the Order of Radicals
To reduce the order of radicals is to reduce the index to its lowest possible
number.
Examples:
1. Simplify
a12
8
Solution:
8
a 12  8 a 8 .a 4 , find a factor of radicand that is a power of 8
8
=
a 8  8 a 4 , find the 8th root of each factor
4
8
= a a
4
= a  a8
=
aa
= a
2. Simplify
4
, change the radical to exponential form
1
2
, reduce the rational exponent to lowest term
a , change the exponential form to radical form.
25
Solution: 25 is not a perfect power of 4 but a perfect power of 2 . Hence you can
express 25 as 52
25 
4
4
52
2
1
= 5 4  5 2 , change to exponential form and reduce the rational
exponent to lowest term
1
= 52  5
Therefore
3. Simplify
, change the exponential to radical form.
25  5
4
3
, in simplest form
64
Solution: Simplify the innermost radical
3
64 
3
= 2
4. Simplify
, the square root of 64 is 8, since 82 = 64.
8
, the cube root of 6 is 2, since 2 3 = 8
27m 3 n 3
6
Solution:
6
27m 3 n 3  6 33 m 3 n 3  6 (3mn) 3 , express 27 as power of 3: 33 , then
apply the laws of exponent
3
6
1
(3mn) 3  (3mn) 6  (3mn) 2 , change radical to exponential form and re
duce the rational exponent to lowest term
(3mn)
1
2

3mn , change exponential form to radical form
Therefore the simplest form of
5. Simplify
Solution:
8
6
27m 3 n 3
is
3mn
81
1
8
1
1
2
814  
81  818  



4
81 
21
3
, the exponent
1
1 1


8
4 2
PIVOT 4A CALABARZON Math G9
Simplifying Radicals by Rationalizing Denominators
In a radical expression, id there is radical number in the denominator, the
expression is not in simplest form. To eliminate the radical number in the denominator you should apply the rationalization process.
To rationalize denominator, you have to multiply the numerator and denominator by a number such that the resulting radicand in the denominator is a perfect nth power.
Examples:
3
2
1. Simplify
Solution: Apply quotient law of radical
3

2
3

2
3
2
, express as quotient of two radicals
2

2
6
, multiply the radical fraction by a radical which will
4
make the radicand in the denominator a perfect
square.
=
6
, simplify
2
Remember that a radical fraction is in its simplest form if there is no radical
number in the denominator.
2. Simplify
Solution:
3. Simplify:
3
3
3
48
6
48 3
 82
6
32
54
3
3
Solution:
, apply the quotient law of radicals then find the quotient
of the radicand. The quotient 8 is a perfect cube. Thus,
the cube root of 8 is 2.
32
54
3
3
84
Simplify the radicand by finding a factor which is a
perfect cube.
27  2
23 4
= 3
3 2
=
2
3
3
extract the cube root of the factors that are perfect
cube.
2
PIVOT 4A CALABARZON Math G9
simplify by dividing the radicands
22
3. Simplify
64 3
4
2
Solution: Find smallest possible number where you can extract the 4th root.
Since the radicand in the denominator is 2, you need to find a num
ber when multiplied by 2 will give you a number where you can find
the 4th root. In this case the, radicand in the denominator must be
16. Thus multiply both radicand by 8.
64 3  8 64 24 64 24
 4

 34 24
4
2
2 8
16
, simplify by extracting the 4th root
of 16. then divide 6 by 2.
Conjugate of Radical Expression
Examples:
Radical Expression
Conjugate
3
2
3
2
5
2
5
2
The product of the radical expression and its conjugate is an integer.
In finding its product is the same procedure in multiplying the sum and difference
of 2 binomials. Like, (x + y)(x– y) = x2 - y2
Finding the product of radical expression and its conjugate:
A. ( 3 
B. ( 5 
2)( 3 
2) = (3)2 - ( 2 ) 2  9  2  7
2 )( 5 
2  52  3
4. Rationalize the expression
3
2 3
Solution: Find the conjugate of the denominator then multiply the numerator
and the denominator by the conjugate.
3(2  3 )
3(2  3 )

 3(2  3)
43
(2  3 )(2  3 )
23
Simplify the denominator.
PIVOT 4A CALABARZON Math G9
E
Learning Task 1
A. Simplify the radical expressions
1.
63
6.
2.
48
3.
75
4.
5.
3
11.
24
4
625
5
96
7.
3
81
12.
6
8.
3
128
13.
4
243
99
9.
3
40
14.
3
92
10.
3000
3
15.
135
128
B. Simplify radicals by reducing the order or index of radical.
1.
2.
6
16
3
3.
64
10
4.
5.
32
12
12
729
81
A
Learning Task 2
A. Find the product of the radical expression and its conjugate.
Radical Expression
Conjugate
1. 3  5
2.
3 2
3.
6 3
B. Simplify by rationalizing the denominators.
8
2
1.
6.
1
5 2
2.
3
5
4
7.
4
62
3.
4
5
8
8.
2
7 3
14
7
9.
5
4 5
9
6
10.
2
2 1
4.
5.
3
3
PIVOT 4A CALABARZON Math G9
24
Product
Operations of Radicals
WEEK
7
Lesson
I
In this lesson you have to perform operations of radicals such as:
A. adding and subtracting similar and dissimilar radicals.
B. multiply and/or divide radicals
Radical terms are similar if it has the same radicand and index.
Examples:
5 ,3 5 ,
5 . These three radical numbers are similar
since the radicand are the same which is 5 and the index is 2
or all square root.
Radicals are unlike or dissimilar if the radicand are not the same and/or
the index are also different.
Examples:
5 , 3 5 Though the radicals have the same radicand but the
index are different.
5 , 6 The index are the same but the radicand are different.
D
Addition and Subtraction of Radicals
In adding or subtracting radicals, you can only add or subtract similar radicals. To add or subtract similar radicals, add or subtract only the coefficients of
the radical number and copy the common radicals.
Examples:
1.
4 2
2.
5 4  2 4  (5  2) 4  3 4  3( 2)  6. Subtract the numerical coeffi
2  2 2  ( 4  1  2) 2  3 2 . Add and/or subtract the numerical
coefficients. If there is no number before the radical number it is
understood that the coefficient is 1.
cients. Since the radicand is a perfect square, then get the
square root. Multiply the root to the coefficient.
3.
3 3  5 3  2 5  (3  5) 3  2 5  8 3  2 5 . Only add the coeffi
cients of terms that are similar.
4.
2 50 
8  3 2 . In this case we can simplify first the radicand. Since
the lowest radicand is 2 find two factors of the other radicand
in such a way that one factor is 2 and the other is a perfect
square.
2 25  2 
4  2  3 2 . The factors of 50 are 25 and 2, where25 is a perfect
square and the factors of 8 are 4 and 2, where 4 is also a perfect square.
25
PIVOT 4A CALABARZON Math G9
2(5) 2  2 2  3 2 Simplify by getting the square root of numbers
which are perfect square. This time the terms are now similar.
Hence you can already combine them.
10 2  2 2  3 2  (10  2  3) 2  9 2
, simplify.
5. Simplify:
Simplify the radicand by finding a factor
2 x3 8  3x  53 27 x 3  3x
which are perfect cube.
2 x( 2)3 3 x  5(3 x)3 3 x Extract the cube roots
4 x 3 3 x  15 x 3 3 x  19 x 3 3 x Simplify and combine similar terms.
Multiplication of Radicals
There are three cases to be considered in multiplying radicals.
1. Multiplying radicals of the same order or index.
Apply the rule
n
a  n b  n ab
Examples:
A. Find the product of 3 5  2
Solution: 3 5 
4  3(1) 5  2 , multiply the coefficients and the radicands.
= 3 10
B. Find the product of
 43 2  33 4
Solution:  43 2  33 4  (4)(3)3 2  4  123 8 . Multiply both coefficients and both
radicands.
= (-12)(2), extract cube root of 8
= -24
C. Find the product of
32 xy 3  18 x 2
Solution: Simplify first the radicand.
32 xy 3  18 x 2 
16 y 2  2 xy 
= 4y
2 xy  3 x
9x2  2
2  4 y (3 x )
Multiply the coefficients and radicands .
= 12xy
Extract the square root of 4.
= 12xy(2)
Simplify
PIVOT 4A CALABARZON Math G9
= 24xy
26
2 xy ( 2)
4  xy
xy
xy
2. Multiplying Radicals with different indices but same radicands.
If the radicands are the same but different indices, you have to change
them into similar index.
Examples:
33 3
1. Find the product of
1
1
3  3 3  3 2  33 , change to exponential form
Solution:
1 1

3
= 32
= 3
, apply the law of exponent
3
2

6
6
, change the exponential fraction to similar
fractions
5
= 3 6 , add the fractions
=
=
6
2. What is the product of
Solution:
3
2 4 4 
35
, change to radical form
243
, find the power of 35
6
3
3
2 4 4
2 .4 2 2 , rename 4 to make the radicands the same
1
2
23 24 , change to exponential form
1 1

2
= 2 3 2 , reduce
to lowest term and apply the law
=
4
of exponent.
=
2 3
26  6
5
 2 6 , change the exponential fractions to
similar fractions, then add.
=
6
25 , change to radical form
=
6
32 , the power of 25
Binomials can be multiplied using distributive property or applying the rule
of special products of algebraic expressions or binomials.
3. Find the product and simplify: 3 2 4 2 
Solution:
3

2 4
2

3 3

2 4

2 3
2

3
3

, apply distributive property
= 3(4) 2  2  3(1) 2  3 , multiply both coefficients
and both radicands
= 12(2) - 3 6
= 24 - 3 6 , simplify
4. Find the difference of the sum and difference of radical binomials
2

3 3 2 2 3 3 2

Solution: Apply the rule in multiplying sum and difference of binomials.
2

 
3 3 2 2 3 3 2  2 3
  3 2 
2
2
= 4(3) - 9(2) = 12 - 18 = - 6
27
PIVOT 4A CALABARZON Math G9
Dividing Radicals
If the nth root of a number is divided by the nth root of another number,
then the result is the nth root of the quotient.
p

q
n
n
n
p
q
, where q≠0
Dividing radicals with the same indices
Examples:
3
16
2
16

2
1. Find the quotient:
3
Solution:
2. Divide
2
16

2
8 by
Solution:
2
3
3
8 2
, find the quotient of 16 and 2. the quotient
is 8 which is a perfect cube. Hence the
cube root of 8 is 2.
5
8

5
8 5 , rationalize the denominator by a number that will

5 5 make the denominator a perfect square.
40
25
=
, multiply
4  10
25 , simplify
=
2
10
5
3
6
3. Find the quotient 6  3
=
Solution: Change first to same index.
3
6
1
=
=
4. Divide 8 by
2
6 3
1
36
6

3
6

62
(3  2) 2
6
3
3
3 2
3
2
6
6 6
1
36
2
, simplify
 6 3  4  6 12
, find the quotient then simplify
3 3
Solution: To rationalize the denominator, multiply the numerator and
denominator by the conjugate of the denominator.
8

3 3
3 3 3 3

=
8(3  3 )
(3  3)(3  3 )
8(3  3 ) 8(3  3 ) 4

 (3  3 )
93
6
3
Multiply sum and difference of binomial radicals them simplify.
PIVOT 4A CALABARZON Math G9
28
E
Learning Task 1
A. Make each pair of radicals similar by reducing the radicand.
1.
2.
2,
3
3.
4.
5.
3 , 3 24
6.
48 , 12
7.
28 , 63
125
8.
4
32 , 4 162
3 , 5 96
9.
3
40 , 3 135
45 ,
5
8
27 ,
75
10.
8 , 200
B. Referring to each pair in letter A, do the following.
A. Find the sum of each pair of radicals.
B. Subtract the first radical from the second radical.
C. Find the product of each pair of radicals
D. Divide the second radical by the first radical.
A
Learning Task 2.
A. 1. What is/are the rule(s) in adding or subtracting radicals?
2. What is/are the rule(s) in multiplying radicals?
3. How do you divide or rationalize radicals?
4. How do you rationalize if the denominator is a binomial radicals.
B. Perform the indicated operations. Express your answer in simplest radical
form.
1.
11.
9
25
12.
2
4 8
2.
3.
4.
5.
13.
33 2
3
3
6.
7.
5
14. 5  2
8.
9.
15.
3  4 5x 2
4
10.
29
3x 3
PIVOT 4A CALABARZON Math G9
WEEK
8
Radical Equations
I
Lesson
A radical equation is an equation whose unknown quantity is in the radicand. Below are examples of radical equations and not radical equations.
Radical Equations
Not Radical Equations
3 x  5  20
1.
1. x 3  5  20
3
3
2. x  3  2 3
2x  3  x  2
2.
Variable x is not part of radicand.
Some variable x are found
in the radicand.
In this lesson you have to solve radical equations. Hence you have to assume that if two numbers are equal, then the square, cube or n th power are also
n
equal. If x = y, then xn = yn or if n x  y then n x  y n and x = y n .
 
D
Solving Radical Equations
In solving radical equations, you have to follow the following steps:
1. Write the equation such that the radical containing the unknown is on
one side of the equation.
2. Combine similar terms.
3. Raise both sides of the equation to a power same as the index of the
radical. The equation should be free of radical to complete the solution.
4. Check if the value or values obtained will make the original equation
true.
Examples:
1. Solve for the value of x in
x 7
Solution:
 x
2
 72
Since the index is 2, raise both sides of the equation to power of 2.
x = 49
Check:
x 7
49  7
7 = 7 , the square root of 49 is 7.
PIVOT 4A CALABARZON Math G9
30
3
2. Solve:
x6 3
Solution:

x6
3

3
 33 , cube both sides of the equation since the index of the
radical is 3
x - 6 = 27
x = 27 + 6, by APE combine similar terms
= 33
Check: Substitute the value of x to the original equation
33  6  3
3
3
27  3
3 = 3 , the cube root of 27 is 3
3. Solve :
4
x2  x
Solution:
x2  x4

x2

2
, by APE , radical should be in one side of the equation
 x  4
2
, square both sides of the equation since the index
of the radical is 2.
x - 2 = x2 - 8x + 16, square of a binomial
x 2  9 x  18  0 , combine similar terms
x  6( x  3)  0 , solve quadratic equation by factoring.
x - 6 = 0 or x - 3 = 0 , by zero product property
x = 6 or
x=3
Check:
For x = 6
4
For x = 3
x2  x
4 62 6
4
4
x2  x
4
32  3
4
4 6
13
4 1  3
426
6= 6
5 ≠ 3
Only 6 makes the equation true. Hence, 6 is the solution of the equation. 3 is
an extraneous root of the quadratic equation.
31
PIVOT 4A CALABARZON Math G9
4. The square root of the sum a number and 9 is 6. Find the number.
Solution: Let x = the number
x + 9 = the sum of the number and 9
x  9  6 , is the equation

x9

2
2
 6  , square both sides of the equation since the index is 2
x + 9 = 36 ,simplify
x = 36 - 9, by APE
x = 27
Check:
x9  6
27  9  6, replace x by its value
36  6
6= 6
Common problem involving radical is the solution of right triangle applying
the Pythagorean Theorem. This theorem is about the relationships among the
sides of a right triangle. The theorem states the “sum of the squares of the
lengths of the two legs is equal to the square of the length of the hypotenuse.
Thus, if a and b are the legs and c is the hypotenuse of the right triangle, then
c 2  a 2  b 2 or c 
a2  b2
Examples:
1. A man walks 8 meters to the east ,then turn south and walk 6 meters. How
far is he from his starting point?
Solution: Visualize the problem and sketch the figure.
Let A be the starting point , B the ending point and C the turning
point.
The distance of the man from the starting point is
the hypotenuse of the right triangle.
c
a2  b2
c 
6 2  82
c 
36  64
c  100
c = 10
The distance of the man from his starting point is 10 meter.
PIVOT 4A CALABARZON Math G9
32
2. A ladder 12 ft tall leans against the wall of a building. The ladder touches a
point of the wall 8 ft from the ground. How far is the foot of the ladder from the
building?
Solution: Sketch the figure.
Let l = 12, length of the ladder which becomes the hypotenuse of
the right triangle.
h = 6, distance of the top of the ladder from the foot of the build
ing.
d = ? , distance of the foor of the ladder from the foot of the
building.
l 2  h2  d 2
d 2  l 2  h2
d 
by APE
l 2  h2
d  122  62
d  144  36
d 
108
d
36  3  6 3
The distance of the distance of the foot of the ladder from the foot of the
building is 6 3 ft.
3. The area of a square lot is 162 m2 . How long is the side of the square?
Solution:
A = s2 , the area of a square
162 = s2 , substitute the given are to the formula.
s=
s=
162
, solving for s.
81 2
s9 2
, simplify the radicand by finding a factor that is a perfect
square
m, the side of the square
4. The diagonal of the rectangle is 5 x . If the width of the rectangle is x and
the length is twice the width, what is the dimension of the rectangle.
Solution: Using the Pythagorean theorem
5
c2  a 2  b2
5 x 
x
2
 2 x    x 
2
25 x  4 x 2  x 2
2
The width is 5 units, length
is 2(5) = 10 units and the
diagonal is 5 5 units.
25 x  5 x 2
By MPE,
25 x
5x2

5x
5x
x= 5
33
PIVOT 4A CALABARZON Math G9
E
Learning Task 1: Solve the following equations.
x 9
1.
3x  6
12.
3.
4 x 6
13.
4.
7  x  12
5.
4
2.
3
3x  5  8
2 x  2  x  10
11.
x 2  144  0
x
x 2  3  3x
3x  1  4 x
14. 2
15. 2 
x 1  x  5  0
3x  2  5
6.
7.
5
x 5 7
8.
3
x  33 7
9.
84 x  7
33 x  12
10-.
A
Learning Task 2: Solve the following problems.
1. Twice the square root of a number is 12. Find the number.
2. The square root of a number increased by 9 is 27. find the number.
3. The square root of the sum of twice a number and 3 is 6. Find the number.
4. The perimeter of the square is 25 and the side is
x  3 . Find the number.
5. The circumference of the circle( C = 2πr) is 24 cm and the radius is
Find x and its radius.
x2
6. Find the side of the square if the area is 64m 2 .
7. The shorter l6g of a right triangle is one half of the longer leg and the hypote
nuse is 3 cm. How long is each leg of the triangle?
8. Five times the square root of a number is five. Find the number?
9. The cube root of a number decreased by 3 is zero. Find the number.
10. The square root of the cube root of a number is 2. find the number.
PIVOT 4A CALABARZON Math G9
34
5.
5.a b
12
3
4
432
4.
3.
6
n12
2.
3b2
5.
4.
5

5
4
3. 

a
1.
2. 5
A. 1.
Learning Task 1

3
5  5 52
xy

2
16 


2.

B. 1.
2
35
1
6.  1284 


5.
4.
3.
3
x
125
Learning Task 2
1
3
PIVOT 4A CALABARZON Math G9
5.
4.
2a 
1
84 3
3.
1
4
11x 2
8
4m 5
2.
1
7
B. 1.
5
3
m2 y 2
p5 2
2
a
b
2
5
x
q3
p
10
m n 6
5
10. 9q r
6
9. 2x2
8.

6
  a
2a 
7. x6 y3
6.
5
3
m3n
2
Learning Task 3
Week 4
Week 3
Learning Task 1
1.
Learning Task 2
6. 4x3
1
7.
49
4
2.
9.
3.
11.
49
1
4
16
1
5
1.
13.
14.
a5n
7.
3. 5
8. 27
4.
15. 32
11. –64
6.
2. 11
12. 9x 2
8.
4.
10. 100
12
1
b3
12a 4
64
1
9
25
1
ab
9
5
y
9. 6
5. 1
12.
c5
10a 7 b 6
40
3
1000000
1
5.
10.
7
1
13. 18
10 9
1
y5 z 2
1
14.
15.
Week 1 and 2
Learning Task 1
Learning Task 2
A.
A.
y
x
Learning Task 3
z
p
48
q 
q
48
k
10
2.
w
z
x
6
k
k
Equation
3
2
2. a.
B. 21.5
y
x
k
Equation
30
9
b. True
C. True
y = 6x
y = 10x
6
3
Learning Task 4
12
1. 4.3 ft
v
2
3
z
10vw
z
6vw
50
d. False
2
E. True
8
y
68.75%
2. Joint
y
y = 2xz
6
2
80
4. y p= =4xz
3q
2
48
B.
y
Equation
3. a.
B. It will increase 4 times
4. a. False
y
B.
A. 1.Direct
Equation
4. Combined
3. Combined
5. Combined2
B. 1. q = 7p2
3.
192
p
V
kT
2
p 
t
3
P
7z
7
y
5vw
5
5
18
96
x
36
x
y
30
y
2
5
6
5
24
5. k = π, Volume will in
crease by
5
168
Answer Key
Week 5-6
Learning Task 1
Learning Task 2
A.
Week 7
Learning Task 2
Learning Task 1
PIVOT 4A CALABARZON Math G9
36
Week 8
Learning Task 2
Learning Task 1
References
37
PIVOT 4A CALABARZON Math G9
Personal Assessment on Learner’s Level of Performance
Using the symbols below, choose one which best
describes your experience in working on each given task.
Draw it in the column for Level of Performance (LP). Be
guided by the descriptions below.
- I was able to do/perform the task without any difficulty. The task
helped me in understanding the target content/lesson.
- I was able to do/perform the task. It was quite challenging but it still
helped me in understanding the target content/lesson.
- I was not able to do/perform the task. It was extremely difficult. I need
additional enrichment activities to be able to do/perform this task.
Distribution of Learning Tasks Per Week for Quarter 2
Week 1
LP
Learning Task 1
Week 2
LP
Learning Task 1
Week 3
LP
Learning Task 1
Week 4
Learning Task 1
Learning Task 2
Learning Task 2
Learning Task 2
Learning Task 2
Learning Task 3
Learning Task 3
Learning Task 3
Learning Task 3
Learning Task 4
Learning Task 4
Learning Task 4
Learning Task 4
Learning Task 5
Learning Task 5
Learning Task 5
Learning Task 5
Learning Task 6
Learning Task 6
Learning Task 6
Learning Task 6
Learning Task 7
Learning Task 7
Learning Task 7
Learning Task 7
Learning Task 8
Learning Task 8
Learning Task 8
Learning Task 8
Week 5
Learning Task 1
LP
Week 6
Learning Task 1
Week 7
LP
Learning Task 1
LP
LP
Week 8
LP
Learning Task 1
Learning Task 2
Learning Task 2
Learning Task 2
Learning Task 2
Learning Task 3
Learning Task 3
Learning Task 3
Learning Task 3
Learning Task 4
Learning Task 4
Learning Task 4
Learning Task 4
Learning Task 5
Learning Task 5
Learning Task 5
Learning Task 5
Learning Task 6
Learning Task 6
Learning Task 6
Learning Task 6
Learning Task 7
Learning Task 7
Learning Task 7
Learning Task 7
Learning Task 8
Learning Task 8
Learning Task 8
Learning Task 8
Note: If the lesson is designed for two or more weeks as shown in the eartag, just copy your
personal evaluation indicated in the first Level of Performance in the second column up to
the succeeding columns, i.e. if the lesson is designed for weeks 4-6, just copy your personal
evaluation indicated in the LP column for week 4, week 5 and week 6.
PIVOT 4A CALABARZON Math G9
38
PIVOT 4A CALABARZON
For inquiries or feedback, please write or call:
Department of Education Region 4A CALABARZON
Office Address: Gate 2, Karangalan Village, Cainta, Rizal
Landline: 02-8682-5773, locals 420/421
Email Address: lrmd.calabarzon@deped.gov.ph