UC3-Performing estimation and basic calculation

Systeme Internationale (SI) or International System because it is universal.

2. English/British System – originated in England and being used in the
United States.

Prefixes are used in SI units to indicate very, very small to very, very large
numbers and to denote the value for the base units. Large units are used to measure
distances.

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 54 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

Table 1 are the prefixes and their corresponding symbol and meaning (factor).

PREFIX SYMBOL FACTOR PREFIX SYMBOL FACTOR

yotta Y 1024 deci d 10-1

zetta Z 1021 centi c 10-2

exa E 1018 milli m 10-3

peta P 1015 micro µ 10-6

tera T 1012 nano n 10-9

giga G 109 pico p 10-12

mega M 106 femto f 10-15

kilo k 103 atto a 10-18

hecto h 102 zepto ζ 10-21

deka da 101 yocto y 10-24

Table 2 is the Conversion Table of one unit of measure to another unit measure.

LENGTH MASS (WEIGHT) AREA VOLUME

1 km = 1000 m 1 kg = 1000 g 1 acre (a) = 120 sq.yd 1 L = 1.057 qt = 2.1 pt

1 m = 100 cm 1 hg = 100 g 1 hectare (ha) = 2.471 1 L = 1000 mL
acre

1 cm = 10 mm 1 g = 1000 mg 1 ha = 10000 sq.m 1 gal = 3.785 L = 4 qt

1 m = 3.3 ft 1 kg = 2.2 lbs 1 acre = 100 sq.m 1 mL = 1 cu.cm (cc)

1 m = 39.37 in 1 metric ton = 2000 lbs 1 sq. cm = 100 sq.mm 1 cup = 240 mL = 16
tbsp
1 in = 2.54 cm 1 long ton = 2200 lbs 1 sq.km = 100 ha 1 fl. oz = 29.57 mL
1 tbsp = 15 mL = 3 tsp
1 km = 0.62 mi 1 lb = 454 g 1 sq.ft = 144 sq.in 1 tsp = 5 mL
1 cu.yd = 27 cu.ft
1 yd = 3 ft 1 oz = 28 g 1 sq.yd = 9 sq.ft 1 cu.ft = 1728 cu.in
1 cu.m = 1000 cu.dm
1 yd = 36 in 1 lb = 16 oz 1 acre = 4840 sq.ft 1 cu.cm = 1000 cu.mm

1 ft = 12 in 1 cavan of rice = 50 kg 1 sq.mi = 640 acres 1 L = 1000 cu.cm
1 L = 1 cu.dm
1 mi = 5280 ft 1 sq.m = 1.196 sq.yd

TEMPERATURE FORMULA TEMPERATURE

ºC = 5 (ºF – 32)/9 100ºC = 212ºF

ºF = (9ºC/5) + 32 0ºC = 32ºF

CONVERSION OF UNITS

In the metric system, meter (m) is the basic unit of length and gram (g) is also
the basic unit of mass. However, units can be converted from one system to another
or from unit to another unit using the conversion factor (see Table 2). The following
are the rules in the conversion of units:

1. From large to small unit

To convert from a large to a small unit, multiply the given number of the large
unit by the number of the small units contained in one large unit.

Example: 5 km = _____ m
Solution:

Conversion factor: 1 km = 1000 m

5 km x 1000 m (number of small units contained in one large unit – 1 km) 1km

= 5000 m

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 55 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

2. From small to large unit
To convert from a small to a large unit, divide the given number of the small
unit by the number of small units contained in one large unit.

Example: 500 g = ______ kg
Solution:

Conversion factor: 1 kg = 1000 g

500 g = 500 g ÷ 1000 g

1 kg

= 500 g x 1 kg
1000 g

= 0.5 kg

Study the following examples:

Example 1: 3 ha = ________ acres
Solution:

Conversion factor: 1 ha = 2.471 acres

3 ha = 3 ha x 2.471 a
1 ha

= 7.413 acres

Example 2: 2.5 cu.m = _______ liters
Solution:

Conversion factor: 1 cu.m = 1000 cu.dm
1 L = 1 cu. dm

2.5 cu.m = 2.5 cu.m x 1000 cu.dm x 1 L
1 cu.m 1 cu.dm

= 2500 L

Example 3: 110 lbs = _______ kg
Solution:

Conversion factor: 1 kg = 2.2 lbs

110 lbs = 110 lbs x 1 kg
2.2 lbs

= 50 kgs

Example 6: 86ºF = _________ºC
Solution:

ºC = 5 (ºF – 32)/9
= 5 (86ºF – 32)/9
= 5 (54)/9

= 270/9
= 30ºC

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 56 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

SELF-CHECK 3.2-5

I. PRACTICAL EXERCISES
Solve the following problems accurately.

1. A man is 5 ft and 9 inches tall. How tall is he in meters? centimeters?

2. Raul bought 3 hectares of farmland at P100.00 per square meter. How
much will he pay for the land?

3. One bottle of liquid fertilizer can contain 300 cu cm. How many liter/s of
liquid fertilizer will be needed to fill this bottle?

4. One hundred cavans of palay was harvested in a-hectare rice filed. A
kilogram costs P10.50. What is the gross sale?

5. If the temperature of the room is 40ºC, what is its equivalent in ºF?

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 57 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

ANSWER KEY 3.2-5

I. PRACTICAL EXERCISES

1. a. 1.742 m
b. 174.2 cm

2. P3,000,000.00

3. 0.3 L

4. P52,500.00

5. 104ºF

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 58 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

INFORMATION SHEET 3.2-6
PERIMETER AND AREA

Learning Objectives:
After reading this Information Sheet, the trainee must be able to:
1. differentiate perimeter and area;
2. determine the perimeter of different polygons;
3. calculate the area of a given polygon; and
4. solve problems involving perimeter and area.

INTRODUCTION

Perimeter is a measurement of the distance around a shape and area gives us
an idea of how much surface the shape covers. Knowledge of area and perimeter is
applied practically by people on a daily basis, such as architects, engineers, and
graphic designers, and is math that is very much needed by people in general.

PERIMETER

In a plane, simple closed curves formed by three or more line segments are
called polygons. And the simplest polygon is a triangle. It has three sides. The sum of
the three sides is called the perimeter. To find the perimeter (P), simply add the
measure of each side of the polygon. Thus, the triangle below has:

c

a b
P=a+b+c

To determine the perimeter of a rectangle, we must know the length and its
width. Thus, we have:

l

ww P = l + w + l + w = 2l +2w = 2(l + w),
where l is the measure of the length and

w is the width.

MAIS- l Date Developed: Document No. CC-ACP3-03
PERFORM

TESDA QA ESTIMATION AND July 18, 2020 Issued by: Page 59 o 74
SYSTEM BASIC
Developed by: Instructional
CALCULATION Materials Review
Joffrey E. Gacula
Committee
Associate Professor IV
Revision # 00

Below is the list of some common polygon and the formula in determining the
perimeter (P) of each:

1. Parallelogram P = 2(l+ w) where, l = length, w = width 2. Square P
= 4s where, s = side
3. Rhombus P = 4s where, s = side
4. Quadrilateral P = a + b + c + d where, a, b, c, d = sides 5. Equilateral
P = 3s where, s = sides

6. Isosceles P = 2s + a where, s = sides and a = third side

Let us consider the following examples:

Example 1: Find the perimeter of a square whose side is 5 cm.
Solution:

Given: s = 5 cm Find: P

s = 5 cm P = 4s
P = 4 (5) P = 20
cm

Example 2: What is the perimeter of a rectangle whose length is twice the
width of 2.5 feet?

Solution:
Given : l = 2w Find: P
w = 2.5 feet

P = 2 ( 2w + w) P = 2 (l + w)
w = 2.5 feet P = 2 (3w)
P = 6w
L P = 15 feet P – 6 (2.5)

CIRCLE AND ITS CIRCUMFERENCE

A circle is a closed curved whose set of all points are equidistant from a given
point called center. A line segment that joins the center to any point on the circle is

called radius, r. When two points on the circle are joined together passing through
the center, a diameter, d is formed. It is twice the radius. The distance around the
circle is called the circumference of a circle.

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 60 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

A The figure is named Circle O. Its radius (radi), r is
represented by line segments OA and OB whose

r

.O

diameter, d is represented by line segment AB. So,
d = 2r.
r

A

B Circumference, C = πd or where, π = 3.1416
C = 2πr

Example 1: Find the circumference of a circle if its diameter is 5 cm.
Solution:

Given: d = 5 cm Find: C
Formula: C = πd

C = 3.1416 x 5
C = 15.708 cm

Example 2: What is the distance around the circle if the radius is 5 cm?
Solution:

Given: r = 5 cm Find: C
Formula: C = 2πr

C = 2 (3.1416)(5)
C = 31.416 cm

AREAS

Land measure is usually expressed in square unit or square measure. The
concept of finding the area is very vital in this sense. Area is the number of square
units needed to cover a surface. This is shown on the figures below:

a = 1 unit 1 cm

1 cm

b = 1 unit A = 4 sq.cm or 4 cm2

Area = 1 sq. unit

a = 1 unit

A=ab/2A=ab/2
b = 1 unit

A=ab/2

Area = ½ sq.unit So, the area of the shaded triangle A = ½ sq. unit.

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 61 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

Below is the list of formula in finding the area of a polygon:

POLYGON SHAPE FORMULA

Triangle a A = ½ ab where a= altitude or
b height b= base

Rectangle b A = ab where a= length
a b= width

Square s A = s2 where s= side

Parallelogram b A = ab where a= altitude or height
b= base

Trapezoid b2 A = ½ h (b1 + b2) where h= height
b1= lower base b2= upper base

b1

Circle r A = πr2 where π= 3.1461

or r= radius
A = ¼ πd2 d= diameter

Example 1: Find the area of the triangular piece of land as shown in the figure.

Solution: Given: a = 15 cm
Find: A b = 12 cm

12 cm Formula : A = ½ ab
A = ½ (15 cm)(12cm)
A = ½ (180 cm2)

A = 90 cm2

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 62 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

Example 2: The bases of a trapezoid are 10 cm and 15 cm respectively. Its
altitude measures 18 cm. What is the area?

Solution:
Given: b1 = 15 cm Find: A
b2 = 10 cm
h = 18 cm

b2=10cm Formula: A = ½ h (b1 + b2)
A = ½ (18cm)(15cm + 10cm)
A = ½ (18cm)(25cm)
A = ½ (450cm2)

b1=15cm A = 225 cm2

Example 3: Find the area of the circular region shown in the figure.
Solution:

Given: d =10 cm
Find: A
Formula: A = ¼ πd2

A = ¼ (3.1416)(10cm)2
A = ¼ (3.1416)(100cm2)
A = ¼ (314.16 cm2)
A = 78.54 cm2

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 63 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

SELF-CHECK 3.2-6

I. PRACTICAL EXERCISES
Solve the following problems accurately:
1. What is the perimeter of a rectangular garden plot whose length of 8.5m
and width is 175cm.?
2. How long should a barbwire be needed to fence a lot whose shape is a
parallelogram if one side is 43m and the other is 20m?
3. A trapezoidal rice field has an upper base of 25m, a lower base of 35m
and an altitude of 15m. Find the area of the field?
4. Given the figure below, find its perimeter and area.

3 cm

4 cm
8 cm

9 cm

5. How many hectares is a circular cabbage garden with a diameter of
1000 meters?

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 64 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

ANSWER KEY 3.2-6

I. PRACTICAL EXERCISES
1. 20.5 m or 2,050 cm

2. 126 m
3. 450 m2
4. a. P = 34 cm

b. A = 48 cm2
5. 78.54 ha

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 65 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

Title: TASK SHEET 3.2-6

FINDING THE PERIMETER AND AREA

Performance Given the necessary tools, supplies and materials, you are
Objective: required to determine the perimeter and area of the assigned
place or location within 1 hour. Apply your measuring skills
Supplies and correctly and accurately.
Materials
Pen/pencil
Tools and Bond paper
Equipment
Calculator
Procedure Meterstick
Steel tape/Odometer

1. Secure the needed supplies, materials and tools. 2. Get
your partner then proceed to the assigned place or area. 3.
Using a meterstick/steel tape, get the dimensions of the
assigned area and record your data in this table:

Dimension Area
Task

Length
Width
Square
Square

(cm)
(cm)
Meter
Foot

A

B

Perimeter A (in inches)

Perimeter B (in inches)
4. Complete the table.
5. Submit output to trainer for evaluation.

Assessment/ Performance Test with questioning
Evaluation
Method

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 66 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

Performance Criteria Checklist 3.2-6

Performance Standard Yes No N/A

Tool, supplies and materials are complete
Tools are properly and safely used
Data gathered are complete and accurate
Measurements are converted to the desired unit
Table is complete with correct data
Output is submitted on time

Remarks/Recommendations:
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________________________________________
___________________________________________________

_________________________________
Trainer

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 67 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

INFORMATION SHEET 3.2-7
VOLUME OF SOLIDS AND LIQUIDS

Learning Objectives:

After reading this Information Sheet, the trainee must be able to:
1. define volume operationally;
2. calculate the volume of different solids;
3. find the volume of liquids; and
4. solve problems involving volume of solids and liquids.

INTRODUCTION

A solid has three dimensions –
length, width and height. This topic
discusses the geometric figures that are
familiar to us.

Figure 1 is an example of a common
solid composed of unit cubes. Counting
the unit cubes in the solid, we have 30-

unit cubes, so the volume is: 5 units x
3 units x 2 units = 30 cubic units. This

means that there are 30 cubic units
that make up the solid.

Figure 1. Rectangular Solid

VOLUME

The volume of a solid is the measure of how much space an object takes up. It

is measured by the number of unit cubes it takes to fill up the solid. It is the cubic
content of a solid figure. The volume of a regular solid is equal to the product of its
length (l), width (w), and height (h). In symbol,

V = lwh but A = lw
Therefore, V = Ah.

Example 1: Find the volume of the given rectangular solid below.

Solution: Given: l = 15 cm Find: V
w = 5 cm
10 cm h = 10 cm
5 cm Formula: V = lwh
15 cm V = (15 cm)(5 cm)(10 cm) V = 750 cm3

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 68 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

Example 2: Find the volume of a cube whose side is 3 ft.

Solution: Given: s = 3 ft Find: V
Formula: V = Ah

V = s3

V = (3 ft)3

3 ft V = 27 ft3

Below is the list of formula for finding volume of different solids:

GEOMETRIC PICTURE FORMULA
FIGURE

Cube is a s V = s3 where s = side
s
polyhedron whose 6 s
faces are all square.

Rectangular h V = lwh where l = length V =
Parallelepiped is a wl Ah w = width h = height

polyhedron whose six
faces are all
rectangles.

Cylinder is a solid r V = πr2h where r = radius h
h = height
bounded by a closed V = 1/3πr2h where r = radius
cylindrical surface h = height
and two parallel
bases.

Cone is a solid h
r
bounded by a conical
surface whose base
is a circle with a
radius.

Sphere is a solid V = 4πr3 where r = radius 3

bounded by a closed r
surface every point V = 1/3(Ah) where A = area of
of which is the base V = 1/3(lw)h
equidistant from a
fixed point h
called the center.

Pyramid is a

polyhedron of which
one face called base,
is a polygon of any
number of sides
and the other faces
are
triangles which have
a common vertex.

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 69 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

VOLUME OF LIQUIDS

The volume of a liquid, sometimes referred to as capacity, is fairly simple to
measure. The unit liters (L) and milliliters (mL) are used. Remembering that 1000 cc
equals 1L and 1000L equals 1 cm3. Figure 1 are the devices used to measure volume

of liquids that include graduated cylinders, beakers, and Erlenmeyer flasks.

Devices to measure volume of liquids Figure 2. Meniscus Figure 1.

Place the graduated cylinder on a flat surface and view the height of the liquid
in the cylinder with your eyes directly level with the liquid. The liquid will tend to
curve downward. This curve is called the meniscus (see Figure2). Always read the
measurement at the bottom of the meniscus.

Every liquid has a characteristic density (D), which is defined as the ratio of its
mass (m) to its volume (v). Mathematically: D = m/v. If you know what liquid you
have, you can look up its density in a table. Once you know that, all you have to do
to find the mass of the liquid is to measure its volume. Below are the densities of
some common liquids:

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 70 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

Example: A stone weighs 320g in air and 270g when suspended in water.
What is the volume of the stone?

Solution:
Given: Mass of stone in air = 320 g F = V
Mass of stone in water = 270g
Density of water = 1 g/mL

Formula: D = m/V so, V = m/D

Mass of stone = Masson air – Masson water
= 320g – 270g

= 50g

V = 50g = 50 mL
1 g/mL

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 71 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

SELF-CHECK 3.2-7

I. PRACTICAL EXERCISES
Solve The following problems accurately.

1. A truck can carry a load of 30 cu.ft. A piece of land is being excavated and the
soil is being carried away. If the excavation site is 45ft by 42ft by 9ft, how
many full loads of soil must be removed?

2. A farmer builds a warehouse 10m x 20m x 10m. How many cavans of palay
could the farmer store in his warehouse if each cavan/sack of palay measures
0.5m x 0.4m x 1m?

3. How many cubic feet of water are there in a conical-shaped water container, 10
feet high and 14 feet across its base?

4. A cylindrical tank, 11 ft high is used to hold a water volume of 1,050 cu ft for
the supply in the farm. Find the area of the base.

5. A quantity of garden soil dumped upon the ground assumed a conical shape
with a base of 40’6” in diameter and a height of 15’6”. How many cubic yards
of soil are there in the file?

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 72 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

ANSWER KEY 3.2-7

I. PRACTICAL EXERCISES

1. 567 loads

2. 10,000 cavans of palay

3. 513.128 cu.ft

4. 95.45 sq.ft

5. 246.52 cu.yd

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 73 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00

REFERENCES:
Benigno, Gloria D. (2006). Basic Mathematics for College Students, Revised Ed. Rex
Bookstore, Inc. Quezon City
Boyce, John G., et.al (1998). Mathematics for Technical and Vocational Students, 9th
Ed. Prentice Hall International, Inc. Singapore.
Del Rosario, Asuncion C.M., et.al (2003). Applied Mathematics. Academic Publishing
Corp., Mandaluyong City, Philippines.
Kern, Willis F., Bland, James R. (1967). Solid Mensuration, 2nd Ed. John Wileys &
Sons, Inc. Sydney
Navasa, D. C., Valdez, B.J. (1999). You and the Natural World: Physics Textbook 2nd
Ed. Phoenix Publishing House, Inc. Quezon City.
Nocon, F. P, et.al (2001). Essentials of Business Mathematics. Katha Publishing
Co.,Inc. Quezon City.

Workbook in Mathematics for Agriculture. Philippines-Australia Agri-Tech Project.
https://study.com/academy/lesson/how-to-estimate-in-math-definition-lesson
quiz.html
https://www.mathsisfun.com/numbers/estimation.html
https://www.google.com/search?q=introduction+on+perimeter+and+area&rlz=1C1C
HBD_enPH813PH813&oq=introduction+on+perimeter+and+area&aqs=chrome..69i5
7 .11362j0j7&sourceid=chrome&ie=UTF-8
https://www.google.com/search?q=volume+of+liquid+formula&sa=X&ved=2ahUKE

w
iIk4Srp9DqAhXYMd4KHRQOBpMQ1QIoAXoECAwQAg&cshid=1594852320773888&
biw=1366&bih=666

http://www.mykidsadventures.com/how-to-stack-liquids/

MAIS- PERFORM Date Developed: Document No. CC-ACP3-03
TESDA QA ESTIMATION AND July 18, 2020
SYSTEM Issued by: Page 74 o 74
BASIC Developed by:
CALCULATION Instructional
Joffrey E. Gacula Materials Review

Associate Professor IV Committee

Revision # 00