CBI - GROUP 2

Authors Information . . . . . . . . . . . . . . . . . . . . . . . . . .
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Planet 001: General Chemistry . . . . . . . . . . . . . . . . . .

1.1 Arrival 001 . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Quarter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Quarter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Planet 002: Pre – Calculus . . . . . . . . . . . . . . . . . . . . .
2.1 Arrival 002 . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Quarter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Quarter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Planet 003: General Mathematics . . . . . . . . . . . . . . . .
3.1 Arrival 003 . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Quarter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Quarter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Planet 004: Earth Science . . . . . . . . . . . . . . . . . . . . .
4.1 Arrival 003 . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Infographics . . . . . . . . . . . . . . . . . . . . . . . . . . .
Planet 005: EAPP . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1 Arrival 003 . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Best output . . . . . . . . . . . . . . . . . . . . . . . . . . .
Planet 006: Oral Communication . . . . . . . . . . . . . . . .
6.1 Arrival 003 . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Best output . . . . . . . . . . . . . . . . . . . . . . . . . . .
Planet 007: HOPE . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1 Arrival 003 . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Poster . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Planet 008: Filipino . . . . . . . . . . . . . . . . . . . . . . . . . .
8.1 Arrival 003 . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 PPT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Planet 009: 21st Century Literature . . . . . . . . . . . . . .
9.1 Arrival 003 . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2 Best output . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .







INTRODUCTION INTRODUCTION INTRODUCTION INTRODUCTION INTRODUC ORTNI NOITCUDORTNI NOITCUDORTNI NOITCUDORTNI NOITCUDORTNI NOITCUDORTNI NOITCUDORTNI NOITCUDORTNI ORTNI NOITCUDORTNI NOITCUDORTNI NOITCUDORTNI NOITCUDORTNI NOITCUDORTNI NOITCUDORTNI NOITCUDORTNI

Good morning and congratulations! You have been
selected to be an astronaut on our “virtual” Ayres2-11
spacecraft. You are about to embark on a voyage to the
CBI Universe. Only a few astronauts have actually taken
this trip, and it usually lasts 9 days. But through state-of-
the-art virtual technology, we will complete our journey in
just a matter of minutes. We, the space crews, will be

your commander for today.



In this space exploration, we are expecting to see the
wonders that the 9 planets in the CBI Universe possess.
We have various problem-solving with their solutions and

answers for the planets General Chemistry, General
Mathematics, and Pre-Calculus that will surely fuel your

love for math. We also have the best outputs of each
personnel staff for the three English planets namely, Oral
Communication, English for Academic Purposes Program
(EAPP), and 21st Century Literature that you will surely
want to see and explore. As well as we have infographics
where you can learn a lot about the planet Earth Science
and Health Optimizing Physical Education (HOPE). And

last but not the least, we will also explore the amazing
planet of Komunikasyon at Pananaliksik sa Wika at

Kulturang Pilipino which possesses the meaningful history
of the Filipino Language during the different periods in

Philippine History. Overall, this space exploration aims to
review and recall you from all the lessons you have

learned during the first semester. Throughout this trip,
you will enjoy and explore the beauty and breathtaking
things about the 9 planets and learn a lot at the same

time. And with that, we will be heading over to the
General Chemistry Planet in just a moment. So to ensure

a safe flight, please make sure that your spacesuits are
fully functional, all electronic devices are powered off and
your virtual seatbelts are fastened. It looks like mission
control is ready for us to begin boarding. Are you ready to

experience what it would be like to go to the wonderful
place of the CBI Universe? All right, let’s head out to the

launch pad.



Welcome onboard Flight 4B7 with service from planet
Gen Chem to HOPE.

INTRODUCTION INTRODUCTION INTRODUCTION INTRODUCTION INTRODUC



ARRIVAL 001.

Good day ladies and gentlemen,
welcome aboard Ayres2-11 spacecraft
Flight4B7! We are now here at our first
destination; Gen Chem Planet. As we
prepare for our departure, make sure that
you are also already eager and ready to
explore the amazing beauty of our first
planet on this trip.

For this destination, we will be solving
problems divided into two categories
which are the Number of Objects as Molar
Quantity and in a mass of a compound,
and Gas Laws. This will surely be an
exciting trip for everyone! As you will
learn how to calculate the number of
atoms present in the given number of
moles using the formula of what we call
“Avogadro’s Number” and at the same
time, you will know how to analyze and
solve problems using the formula for each
Gas Laws. So what are we guys waiting
for? The whole CBI Universe is already
waiting for all of you! Come and join us as
we go deeper to our first planet
destination; the Gen Chem Planet!

QUARTER 1:

When solving problems dealing with the number of
objects or particles present in a given number of

moles of a substance, Avogadro’s number becomes
a big part of the conversion factor used to relate the

number of objects present to the number of moles
present. The number that 1 mol represent is

expressed as follows: 1 mol = 6.022 x 1023objects.

PROBLEM 1:

There are 0.1125 moles of oxygen in 1.8 grams of
O2 . Calculate the number of atoms present in the
given number of moles of oxygen.

Given: 0.1125 mol of O2

1.8 g of O2

Formula: Avogadro’s Number

Solution: (0.1125)(6.022 × 1023)
= 7.37695 x 1023 atoms

Answer: 7.37695 x 1023 atoms

Explanation: The problem at hand asks for the
number of objects in a given number of moles of
oxygen. In order to determine the number of
objects present in this sample, the given number of
moles is multiplied to the Avogadro's number.
Avogadro's number states that there are 6.022 ×
1023 in one mole of any element. Thus, multiplying
the two yields 7.37695 x 1023, indicating that the
number of atoms present in 0.1225 mole of
oxygen is 7.37695 x 1023 atoms.

PROBLEM 2:

There are 4.721 moles present in
Acetaminophen also known as Paracetamol,
that have 0.65 grams of C8 H9NO2 . Calculate
the number of atoms present in the given
number of moles of Paracetamol.

Given: 4.721 mol of C8H9NO2
1.8 g of O2

Formula: Avogadro’s Number

Solution: (4.721) (6.022 x 1023)
= 2.8429862 x 1027

Answer: 2.8429862 x 1027

Explanation: The question was given as to
how many molecules can be found in a certain
amount of moles of Paracetamol. The
specified number of moles is multiplied by
Avogadro's number to get the number of
particles contained in this sample. According
to Avogadro's number, one mole of any
element contains 6.022 x 1023 particles. Thus,
multiplying the two results given, which give
the result is 2.8429862 x 1027, to sum up
everything, there are 2.8429862 x 1027 atoms
per 4.721 moles present in Paracetamol.

PROBLEM 3:

There are 2.14 moles of chlorine in 150g of
chlorine. Calculate the number of atoms
present in the given number of moles of Cl.

Given: : 2.14 moles of chlorine
150 g of Cl

Formula: Avogadro’s Number

Solution: (2.14) (6.022 x 1023 )
= 1.29 x 1023atoms

Answer: 1.29 x 1023atoms

Explanation: The problem here asks us to
calculate the number of atoms present in the
given number of moles. To calculate the
number of atoms in the given number of
moles. We should first convert the grams of
the given element to moles for us to easily
find the number of atoms. Using Avogadro’s

number that states that there are (6.022 x )1023

in one mole of the given element, we multiply
the number of given moles (2.14moles of Cl)
to Avogadro’s number (6.022 x 1023 ) we get
the answer that is equal to 1.29 x 1023 atoms
of Cl.

PROBLEM 4:

There are 3.33 moles of carbon in a 40
grams of C. Calculate the number of atoms
present in the given number of moles of
carbon.

Given: : 3.33 moles
40 grams of C

Formula: Avogadro’s Number

Solution: (3.33)(6.022 x 1023)
= 2.005326 × 1024

Answer: 2.005326 × 1024

Explanation: The question at hand is how
many items can be found in a certain amount
of moles of carbon. The specified number of
moles is multiplied by Avogadro's number to
get the number of objects contained in this
sample. According to Avogadro's number, one
mole of any element contains 6.022 x 1023

particles. As a result, multiplying the two
gets 2.005326×1024 , suggesting that there
are 2.005326×1024 atoms in a 3.33 mole of
carbon.

QUARTER 2:

Gas laws is a series of laws that predict the behavior
of an ideal gas by describing the relations between the
temperature, volume, and pressure. Boyle's Law tells

us that the volume of gas increases as the pressure
decreases. Charles's Law tells us that the volume of
gas increases as the temperature increases. The ideal
gas law is the combination of the three simple gas

laws.

PROBLEM 1: (BOYLE'S LAW)

A 20.5 mL sample of gas is at 7.500 atm. What
will be the volume if the pressure becomes 2.500
atm, with a fixed amount of gas and temperature?

Given: P1 = 7.500 atm V 1= 20.5 mL
P2 = 2.500 atm V2= ?

Formula: P1 V1 = P2 V2
V2 = P1 V1 / P2

Solution: (7.500 atm)(20.5 mL) / (2.500 atm)
= 61.5 mL

Answer: 61.5 mL

Explanation: The question given is asking the
total number of volume of the gas after its
pressure changes and drops at some point. In
order to calculate it, we are going to use the
Boyle’s Law Formula. Therefore, from the
formula we can derive that in order to get the
volume, we need to multiply the 1st pressure
(7.500 atm) and 1st volume (20.5 mL) then
divided by the 2nd pressure (2.500 atm). As a
result, we will get that the volume of the gas will
be 61.5 after its pressure drops to 2.500 atm
from 7.500 atm.

PROBLEM 2: (CHARLE'S LAW)

A 500 mL sample of nitrogen is heated from
37 °C to 76 °C at constant pressure. What
is the final volume?

Given: : V1 = 500ml
V2 = ?
T1 = 37°C + 273 = 310 K
T2 = 76°C + 273 = 349 K

Formula: Charle's Law
V1 / T1 = V2 / T2
V2 = V1 T2 / T1

Solution: V2 = (500ml) (349 K) / (310 K)
V2 = 562.90ml

Answer: 562.90ml

Explanation: Converting all temperatures to
absolute temperatures should be the initial step in
solving gas law difficulties. To look at it another
way, convert the temperature from Celsius or
Fahrenheit to Kelvin. This can be done by adding
273 to each celsius.For the initial temperature it
will become 37°C + 273 = 310 K and for the
final temperature, 76°C + 273 = 349 K. The
next step is to use Charles' law to find the final
volume which is expressed as V1 / T1 = V2 / T2

where V1 and T1 are the initial volume and
temperature while V2 and T2 are the final
volume and temperature. Since we are finding for
the final volume, we can simplify the equation as
V2 = V1 T2 / T1 We can then enter the values,
V2 = (500ml) (349 K) / (310 K) and the final
answer will be 562.90ml which is the final
volume.

PROBLEM 3: (IDEAL GAS LAW)

A 3.5 L container holds 0.50 moles of N2 gas at
250k, what is the pressure inside of the container?

Given: V = 3.5 L
T = 250 k
R = 0.08206
n = 0.50 mol
P=?

Formula: Ideal Gas Law
Pv = nRT
P = nRT
v

Solution: P = 0.50 X 0.08206 X 250
3.5

Answer: 2.93 atm

Explanation:

First write the following given values; V=

3.5L, n= 0.50 mol, T= 250k, R= 0,08206.

Determine the asked variable which is

pressure or P.

Then use the formula Pv= nRT.

Next, transpose the given to the formula:

P(3.5)=0.50(0.08206) (250)

After you transpose the given formula solve

the problem.

P(3.5)=0.50(0.08206)(250) multiply

0.50+0.08206 x 250 that equals to

10.2575.

Next step:

P(3.5)= 10.2575 next is add3.5 each side to

get P by itself, then cancel to get the P, cancel

the 3.5 P/ cancel 3.5 = 10.2575/3.5 next is

10.2575 divide by 3.5 that is equal to 2.930.

Therefore the pressure inside the containe is

P= 2.930 atm.

PROBLEM 4: (AVOGADRO'S LAW)

If we pump or inflated a lifebuoy to a volume of
56L. At this volume we know that we have 2.5
moles of gas in the lifebuoy. If we keep on adding
air to the lifebuoy and it reaches a volume of
71.68L how many moles of gas would our
lifebuoy contain now?
Given: : V1 = 56 L V2= 71.68 L

n1= 2.5 mol n2 = ?

Formula: Avogadro's Law
V1 / n1= V2 / n2

Solution: n2 = V n / V

n2 = (71.68 L )(2.5 mol) / 56 L

n2 = 3.2 mol

Answer: When the lifebuoy reaches 71.68L of
volume then the total amount of gas that our
lifebuoy contain is 3.2 moles of gas.
Explanation: To solve this problem, first we need
to understand what the problem is all about and
then we should list down our givens to avoid
confusion and to also know what is the unknown
or what the problem is asking. In this problem it
is asking the amount of moles of gas that the
lifebuoy contains when it reaches 71.68L of
volume. Remember when our volume increases
the amount of gas will also increase in that case
our answer in this question should be higher than
2.5 moles of gas.
First, we should identify our given and substitute them
to their corresponding value.
Then we will identify our unknown and use the formula
V1n1 / V2n2
Now that we know that n2 is unknown we will first
transpose our given to the formula and we will have V2
Vm1uwlthipiclihedisb5y2n1tihseenqouuarl to 179.2 then we divide it to the
answer is 3.2 moles of gas. And
as you can see in our solution part, we can also use the
cross multiplication to solve this problem.



ARRIVAL 002.

We hope that you gathered an
information or two from Gen Chem
Planet. Get ready Ayres2-11
spacecraft Flight4B7 because a more
difficult world is up ahead.

Next up!! Planet Pre-Calculus, it
seems that it is a planet full of major
obstacles that we will be
investigating. Conic sections,
specifically Ellipse, Circle, Parabola,
and Hyperbola, are linked to these
problems. However, when we travel
deeper within the planet, we will
encounter a different challenge,
according to previous voyagers. Angle
Measures, Circular Functions, and
Reference Angles are the issues,
according to them.

Every challenge is distinct from
the others, which adds to the intrigue
of this world. As we arrive, we want
you to take notes on how we solve
these said problems.

QUARTER 1:

PROBLEM 1: (ELLIPSE)

The Porac Arc, which is found at the border of the
municipality of Porac and the city of Angeles,
stands with a height of 12 feet and a width of 30
feet. Find the equation for the ellipse, and use that
equation to find the height at a distance of 10 feet
from the center.

Given: a = 152 Solution:
b = 122
x = 10 x2 + y2 = 1
y=? (15)2 (12)2

100 y2 =1
225 + 144

Formula: x2 + y2 x2 + y2 = 1
a2 b2
225 144

Answer: The height of Porac 4 + y2 =1
Arc at a distance of 10 feet 9 144
from the center is 8.94 feet.
y2 = 1 __ 4
144 9

y2 = 5
144 9


y2 = 5
144
144 9
1

y2 = 720
9

√y2 =√80

y = 4√5 ≈ 8.94 ft

Explanation:

The given problem asks to find the equation of the

ellipse and the height at a distance of 10 feet from the

center. To solve this problem, we must first make the

equation of the ellipse. The given ellipse has its major

axis horizontally, which means that its general formula
x2 + y2 , where ‘a’ is bigger than ‘b’.
would be a2 b2

In this case, a = 152 while b = 122. Putting these

values in our general formula would give us the
equation of the ellipse which is x2 + y 2 = 1 .
(15)2 (12)2

We will now use this equation to find the height at a
distance of 10 feet from the center. The first step is to
find the square of the a and b values, which will give us

. x2 + y 2 = 1 . Following that, we would then

225 144

substitute the x2 with the given value of x in the

problem, which is 10 and we would be given the
100 y2 100
equation of 225 + 144 = 1 . The 225 in the

equation would be simplified to 4 and will be
9

transposed to the right side of the equation for us to be

able to find the value of y. The next step is to multiply

the whole equation with 144 to only be left with the

variable ‘y’ on the left side of the equation. This process
720
would then give us y2 = 9 with the fraction

simplified to 80 ( √y2 =√80 ). With this equation,

we can now find the square root of the equation to find

the value of y which results in 8.94 feet.

PROBLEM 2: (CIRCLE)

Architect Vin was tasked to explain how he came
up with the idea of making a round table using the
standard and general form of a circle with a given
center of (-14, 17) and a radius of 25. Now
pretend that you are Architect Vin and find the
standard and general form of a circle using the
center and radius.

Given: center = (-14, 17)
radius = 25

Formula: Standard Formula:
(x - h)2+ (y - k) 2 = r 2

General Formula :
x 2 + y2 -1 = 0

Solution:
Standard Form:

(x - h) 2+ (y - k)2 = r 2
(x + 14)2 + (y – 17)2 = 252
(x + 14)2 + (y – 17)2 = 625
General Form:
(x + 14)2 + (y – 17)2 = 625
x2 + y 2 + Dx + Ey + F = 0
x2 + 28x + 196 + y 2 – 34y + 289 = 625
x2 + 28x + y2 – 34y + 485 = 625
x2 + y 2 + 28x – 34y + 485 – 625 = 0
x2 + y 2 + 28x – 34y -140 = 0
Answer:
Standard Form:
(x + 14)2 + (y – 17)2 = 625
General Form:
x2+ y 2+ 28x – 34y -140 = 0.

Explanation:
The task was to find the general and standard form of a
circle with a given center of (-14, 17) and a radius of 25.
To solve this problem, first we must know the formula of
the standard form, which is (x-h)2 + (y-k)2 = r2 , then
substitute the given into the standard form of the circle by
supplying the center and radius of (x + 14)2 + (y – 17)2
= 252. Tip: we must know the positive and negative
signs that can be found in the value of the center and
radius. For example, the -14 and the -h will have a +
value because of the fact that they are both negative.
After that, we can now calculate the exact value of a
radius, which is the formula that can give us the exact
value of (x + 14)2 + (y – 17)2= 625. After finding the
standard form, we can go on to the general form of a
circle. First, we must know the general formula of a
circle, which is x2+ y 2+ Dx + Ey + F = 0. Then we
supply the given in the standard form by using the
formula of = a2 + 2ab + b2 with that, it will give us the
result of x2+ 28x + 196 + y2– 34y + 289 = 625, After
that, copy the like term, which is 196 + 289 = 485.
Then copy the given based on the general form and
transpose the 625 to the 485. This will come up with a
value of 485–625 =-140 after all, which will give us the
result of the general form: x2+ y 2+ 28x–34y-140 = 0.
To sum up everything that has been said, Architect Vin
needs to use the standard form of a circle, which is = (x
+ 14)2+ (y – 17)2= 625, and the general form that is x2
+ y2 + 28x–34y-140 = 0, to find the exact value of the
round table.

PROBLEM 3: (PARABOLA)

A group of mountaineers discovered a small hill
with a parabolic shape in the forest. It is 40.5ft
high and the base of the hill is8 18ft wide. How far
is the focus from the vertex?

Given: V = (0,0) Solution:
x = 9 , y = 40.5 V = (0,0)

Formula: (x-h)2 =4p (y-k)
(x-h)2 =4p (y-k) x = 9, y = 40.5
92= - 4p (-40.5)
Answer: The distance of 81 = 162
the focus from the vertex 81/162= ½
is ½ (0.5) ft.
therefore p= ½




Explanation: In this problem we are finding the distance
of the focus from the vertex. To solve this problem, we
must first identify the formula we have to use for this
problem. We used the formula (x-h)2 =4p (y-k) but in
this case our vertex is in the origin therefore it is (0,0)
so our formula will be x2 = -4py. Our x is 9 because it is
the endpoint of our base and the y is -40.5 and it is
negative because the parabolic shape of the hill faces
downward when we place it in a cartesian plane. Next
step is substituting the given values that we identify
earlier to find p or the distance of our focus from the

2

vertex, 9 = -4p (-40.5) and solve it we will get 81/162
and simplify the answer so therefore the distance of the
focus from the vertex is ½ ft.

PROBLEM 4: (HYPERBOLA)

Identify the vertices and foci of the hyperbola with
the equation y2 _ x 2 = 1

49 32

Formula: y2 _ x2 = 1
a2 b2

Solution:

y2 _ x2 = 1 y2 = 49

49 32

y2 _ 02 = 1 √y 2 = √49

49 32

y2 = 1 y2 = ±7

49

Therefore, the vertices are located at
(0,±7) or (0,7)(0,-7)

c = √a2 + b2 c = √81
c = √49+32 c = ±9

Therefore, the foci are located at
(0,±9) or (0,9)(0,-9)

Explanation:

The problem presented asks for the vertices and
the foci of the equation y2 _ x2 = 1 .

49 32

Since y is written first before x, we can now infer
that it is a vertical hyperbola. To solve for the
vertices of the equation at hand, we just need to
replace the x into 0. The equation will then be

y2 _ 02 = 1 in which we can just cancel out

49 32

02 . After that we can just multiply 49 on both
32 y2
sides to remove the denominator of 49 which
will look like: (49) y2 = 1 (49) . The result of

49

that will be y2 = 49 , we can just then square
both sides to get y=±7. Since the hyperbola
presented is vertical, we can then say that the
vertices are located at (0,±7) or (0,7)(0,-7).

For the foci on the other hand, we can just use
the equation c = √a2 + b2 . Since a2 and b2 are
given and these are 49 and 32 respectively, we
can just plot those in on the equation. The
equation will then look like: c = √49+32.
Simplify the equation then you will get c = ±81.
And lastly we can finish off the equation by
finding the square root of 81 which is 9. We can
then say that c = ±9. Since the hyperbola did
not change, it means that it is still a vertical
hyperbola that is why to find the foci we will use
(0,±c). Since we already solved for c we can just
plot the answer. The foci of the presented
hyperbola are located at (0,±9) or at (0,9)(0,-9)

QUARTER 2:

PROBLEM 1: (ANGLE MEASURE)

Engr. Villareal wants to know the length of the arc
of the circle in the bridge that they are doing.
Knowing that it has a radius of 25 m that subtends
a central angle of 30 degrees, what do you think is
the length of the arc?

Given: radius = 25 m

subtends a central angle of 30°

Formula: arc length (l)= circumference x fraction

of the circle

Solution: l = 2π(25)(30360)

= 2π(25)(112)

= 50π
12

= 25π
6

= 13.09 m

Answer: The length of the arc is 25π m or
6
13.09 m

Explanation:

In the problem, Engr. Villareal wants to calculate
the length of the arc. In order for us to compute it,
we need to multiply the circumference of the circle
t(eh2qiπsu(a2fto5ior)mn) uitsolamt,haed2efπ.raT(2cht5ieo)r(nef3oo36fr00et,h)t=ehecitr2oc65tleaπl (). And from
=13.09 m

length of the
arc of the circle in the bridge that Engr. Villareal is
working with is 25π m or 13.09 m.
6

PROBLEM 2: (CIRCULAR FUNCTION)

P(Ɵ) = (ᵡ,-3/4 ),
P(Ɵ) is in QIII

Given:
P(θ)=(x, -3/4), P(θ)is in QIII

Solution:

( )x2 + 3 2 = 1 P( )= - (√7 ) , - 3
4
44

x2+ 9 =1 ( ) ( )-√7 2 -3 2
16 4 4
+ =1

x2 = 16 - 9 7 + 9
16 16 16 16

√x2 =√ 7 16 = 1
16
16

= √7 1=1
4

x<0. therefore,

= - (√7 )

4

Answer:
True one circle

Explanation:

In the problem, we have P(θ)=

( )x2 + 3 2 = 1 and P(θ) which is in
4

QIII. In order to solve if it is a true
one circle, equate an equation equal to
one by substituting the value of x and
y from the given P(θ)=(x, -3/4).
With that, the equation x2 + 3 2 = 1

( )4

is made. And then, we will solve for
the value of x through the equation we
made and we’ll get x = - (√7 ) . Now
that we already have the v4 alue of x
((√ )/ ) and y (¾), equate them into
one and if you get 1=1, then it is a
true one circle.

PROBLEM 3: (REFERENCE ANGLE)

In a math class, Lae was asked to find the exact
value of cos(135°) and sin(135°). In order for her
to simplify the calculation of the values of
trigonometric functions, she decided to use the
reference angle. What are the values of cos(135°)
and sin(135°) using a reference angle?

Given:

cos(135°)
sin(135°)

Formula:

180° - given angle = reference angle

sin (45°) = a
a√2

cos (45°) = a
a√2

Solution:

For the reference angle,
135° = second quadrant
180° - 135° = 45°
Therefore, 45° is the reference angle

For cos(45°) and sin(45°),

sin (45°) = a
a√2

sin (45°) = 1
√2

sin (45°) = 1 X √2
√2 √2

Therefore, sin (45°) = √2 (a=1)
2

cos (45°) = a
a√2

cos (45°) = 1
√2

cos (45°) = 1 X √2
√2 √2

Therefore, cos (45°) = √2
2

Answer:

Therefore, using the reference angle (45°) the

sin (45°) = √2 and cos (45°) = √2
2 2

Explanation:

135° is located in the second quadrant. The
angle it makes with the x-axis is 180° −
135° = 45°, so the reference angle is 45°.
This tells us that 135° has the same sine and
cosine values as 45°, except for the sign.
After solving for the reference angle, we will
proceed into finding the values of sin and cos.
We can use the unit circle or the
Pythagorean theorem. This angle can be
connected to the 45 - 45 - 90 degree triangle.
Across the 45° is 1, and across the 90°
angle is √2 . In evaluating the sin (45°), use
any of the 45° in the triangle. Its opposite
which is equals to 1 over the hypotenuse √2
and we need to rationalize it by multiplying
1/√2 to √2/√2 so you will get the answer
√2/2. Therefore, the answer for cos(135°)
and sin(135°) are both √2/2.

PROBLEM 4: (CIRCULAR FUNCTION)

cot(Ɵ) = - 6 , cos(Ɵ) > 0 ; csc(Ɵ)
8

Given: 6
8
cot(Ɵ) = - , cos(Ɵ) > 0

Formula: x cos(Ɵ) = 1
y x
cot(Ɵ) =

csc(Ɵ) = r x2+ y2= r 2
y

Solution: x =- 6
cot(Ɵ) = y 8

Since cos(Ɵ) > 0, 1 x>0
(6,-8)

x 2+ y2 = r 2
(6)2+ (-8) 2= r 2
36 + 64 = r 2
√100 = √r 2

r = 10

csc(Ɵ) = r
y

csc(Ɵ) = 100
-8

csc(Ɵ) = -25
2

Answer:

csc(Ɵ) = -25
2

Explanation:

The problem given states that the cotangent of
angle Ɵ is - 68, if the cosine of angle Ɵ is
greater than 0, what is the cosecant? As we
x
all know, cotangent is y and since the cosine,

which is 1 , is greater than 0, we can now say
x
that the x value should be a positive number.

Since the given fraction is negative and the x

value should be positive, we can then say that

the y value is negative. We can then infer that

(6,-8). Using the standard form of a circle

which is x 2+ y 2= r2 we can just substitute the
values to come up with (6)2 + (-8)2 = r.2 By

simplifying the formula, we can get the value

of the radius which is 10. And since csc(Ɵ) =
crys,c(bƟy)su=bst1-i08t0utinwghtichhe
result will be into
can be simplified

csc(Ɵ) = -25
2



ARRIVAL 003.

Voyagers! Congratulations on arriving
here on Planet 003 safely! Ayres2-11
spacecraft flight4B7, we welcome you to
your next destination, the world of general
mathematics! I'm sure you're already
looking forward to see the wonders of the
world! For your mission, you will join us as
we unlock your knowledge while solving
some problems.

As you explore this world, we advise
you to bring out your pen and paper to take
down some important parts. For the first
part, you will encounter problems revolving
around functions. Come along and solve
some problems involving rational functions,
equations and inequalities, some Real-life
Problems Involving Inverse Functions,
Piecewise functions, and Evaluate
Functions with us! While looking around,
we will enter a portal with another set of
problems! But don’t worry, we are here to
guide you to solve these! The other sets of
issues revolve around the Basic Concepts of
Stocks and Bonds, Simple Interest,
Compound Interest, and General Annuities.
That’s the summary of your tour, voyagers!
I hope you’ll learn a lot more as you delve
deeper into the area of general
mathematics!

QUARTER 1:

PROBLEM 1:

(SOLVING PROBLEMS INVOLVING RATIONAL
FUNCTIONS, EQUATIONS AND INEQUALITIES)

Andreia can finish crocheting a scarf in 5 hours.
Cleah, on the other hand, can finish the same task
in just 4 hours. How long will it take the two to
finish crocheting a scarf if they work together?

Given: Andreia - 5 hours
Cleah - 4 hours

Formula:

No. of Hours Working Together + No. of Hours Working Together = 1

Amount of hours by person A Amount of hours by person b

Solution:

x + x = 1 5x + 4x = 20

5 4

[ ]x+ x = 1 20 9x=20
5
4 9x = 20
99
x 2
10 + x 2
10 =1 2
0
20 2
5 4 x = 9 or 2 9

Answer:
2
2 9 hours

Explanation:

To solve the problem, the formula No. of

hours working together divided by the

amount of hours by person A added to the

No. of hours working together divided by

amount of hours by Person B will be used.

Let x be the number of hours that Andreia

and Cleah can finish crocheting a scarf

together. By the said formula, the equation

x + x =1 is made, the denominator of 5

5 4

being the number of hours by person A and

4 by person B. To solve the rational

equation, every term must be multiplied by a

common denominator in order to cancel the

denominators in the original equation. In

this case, the common denominator of 4 and

5 is 20 which is multiplied to every single

term. This results in a linear equation (5x +

4x = 20) which is solved by combining the

like terms (9x = 20). This will be then

followed by dividing both sides of the

equation by 9 to get the value of x. After

getting the value of x, which is 20/9,

simplify the answer by converting it into a

mixed fraction which yields to 2 2/9 hours.

PROBLEM 2: (SOLVING REAL-LIFE PROBLEMS
INVOLVING INVERSE FUNCTIONS)

The Axie Alert Ph on the Facebook social media
reported the current market value of the Smooth
Love Potion (SLP) on the newest and trending
game online which is the Axie Infinity. This game
on which you can earn has an exchange rate of
one smooth love potion is equivalent to 14 pesos,
which has increased tremendously.

In which Ezekiel gained 1 SLP on Monday, 7
SLP on Tuesday, 14 SLP on Wednesday, 17
SLP Thursday, 19 SLP on Friday, 25 SLP on
Saturday and 30 SLP on Sunday in playing Axie
Infinity. Given the situation, compute his total 1-
week earnings.

Given: SLP = PHP14
Monday = 1
Tuesday = 7
Wednesday = 14
Thursday = 17
Saturday = 25
Sunday = 30

Formula:
amount of SLP X PHP per day

Solution:
Monday: 1 SLP x PHP 14 = PHP 14
Tuesday: 7 SLP x PHP 14 = 98
Wednesday: 17 SLP x PHP 14 = 238
Thursday: 17 SLP x PHP 14 = 238
Friday: 19 SLP x PHP 14 = 266
Saturday: 25 SLP x PHP 14 = 350
Sunday: 30 SLP x PHP 14 = 420

for the total,
14 + 98 + 196 + 238 +266 + 350 +

420
= 1,582

Answer:
Therefore, the total amount of Ezekiel in
playing Axie Infinity will gain in 1 week is
1582 PHP.

Explanation:
To solve this problem, first we must know the
given, which is the amount of Smooth Love
Potion and the current value of the SLP in
pesos per day. Then multiply SLP to PHP per
day (Monday: 1 SLP x PHP 14 = PHP 14,
Tuesday: 7 SLP x PHP 14 = 98,
Wednesday: 17 SLP x PHP 14 = 238,
Thursday: 17 SLP x PHP 14 = 238, Friday:
19 SLP x PHP 14 = 266, Saturday: 25 SLP
x PHP 14 = 350, Sunday: 30 SLP x PHP
14 = 420) and then just add the total gain in
SLP per day (14 + 98 + 196 + 238 +266
+ 350 + 420 = 1,582) in 1 week. To sum
up everything that has been said, the total
amount of Ezekiel in playing Axie Infinity will
gain in 1 week is 1582 PHP.

PROBLEM 3:(EVALUATING FUNCTIONS)

Elijah sells lemonade in the cafeteria of the nearby
school. If one glass of lemonade costs P20 to make and
charges the customers P35 in each glass. If Elijah can
sell 100 glasses of lemonades a day, how much profit
does Elijah make in a week? Let the amount of glasses
be x, f(x)= (35x - 20x)5 shows the function on how to
get for the profit of Elijah in a week without
considering the rent fee.
Given:

f(x)= (35x - 20x)5
x=100

Formula:
f(x)= (35x - 20x)5

Solution:
f(100)= [35(100)-20(100)]5
f(100)= (3500-2000)5
f(100)= (1500)5
f(100)= 7500

Answer:
He makes 7500 a week in profit for selling

lemonade.

Explanation:
To solve the problem at hand, we need to use the
given function f(x)= (35x - 20x)5. Since x is the
amount of lemonade sold and he averages to sell 100 a
day, we can just plot in 100 on every x present in the
function. The function will then be f(100)=
[35(100)-20(100)]5. We can then simplify the
problem and have f(100)= 7500 as an answer. We
can now say that Elijah profits 7500 in a week for
selling lemonade in the cafeteria of a nearby school.

PROBLEM 4: (PIECEWISE FUNCTION)

The Metropolitan Manila Development Authority
(MMDA) officers charge motorists according to the
following function with f(x) representing the total
amount of money to pay and the x represents the
number of violations.
Given:

f(x)= 300 if 0 < x ≤ 1
f(x)= 200x + 100 if 1 < x ≤ 2
f(x)= 850 + 150(x-2) if x > 2

Asked:
1.) If the number of violation is 1, what is the

total amount of money you have to pay?
2.) If the number of violations is 2, what is the

total amount of money you have to pay?
3.) If the number of violations is 3, what is the

total amount of money you have to pay?

Answer:
a.) If the number of violation is 1, what is the total

amount of money you have to pay? 300
b.) If the number of violations is 2, what is the

total amount of money you have to pay? 500
c.) If the number of violations is 3, what is the

total amount of money you have to pay? 1000

Explanation:
In a piecewise function, functions are represented by
a combination of equations, each corresponding to a
part of the domain. In this problem the equation is
given all we have to do is to see carefully what is the
right equation that corresponds to the question, but
we also have to solve for equations that are not exact,
and you have to solve it first after you see the exact
corresponding equation for the certain question, with
that you can easily identify the answers.

QUARTER 2:

PROBLEM 1: (BASIC CONCEPTS OF
STOCKS AND BONDS)

Aling Dory won a good amount of money from a
lottery. In order for her money to grow and be in
good hands, she wants to invest in a company. But
the problem is she cannot choose between Company
A with a current market value of Php 67.00 and a
dividend of Php 9.00 per share for its common
stock, and Company B, with a current market value
of Php 86.00 and a dividend of Php 13.00 per
share. If you are to advice Aling Dory, which
company is a better choice where to invest?

Given: Company B
Dividend per share
Company A = Php 13

Dividend per share

= Php 9

Market Value Market Value
= Php 67 = Php 86

Formula: Dividend per share
market value
Stock Yield Ratio =

Solution: Company B
Php 13 = 0.1512
Company A Php 86
Php 9 = 0.1343
Php 67 Or 15.12%

Or 13.43%

Answer:

Company B is the better choice where to
invest.

Explanation:

The given problem is asking which of the
companies is a better investment. In order
to know the answer, we need to get the
stock yield ratio of the two companies by
dividing the dividend per share by the
market value, then times by 100. From
the formula given, the equation Php
9.00/Php 67.00 = 0.1343 = 13.43%
(Company A) and

Php13.00/Php 86.00 = 0.1512 =
15.12% (Company B) is made. Therefore,
knowing that Company B has a higher
stock yield ratio than Company A means it
is wiser to invest in it. Because if all other
things are okay and equal then investing
in Company B will help you earn more
than investing in Company A.

PROBLEM 2: (SIMPLE INTEREST)

Justin got a 120,000.00 loan for 4 years. He paid
10,000.00 in Interest. What was the interest rate?

Given: i = Php 10,000
P = 120,000
t = 4 years
r=?

Formula: rate = interest
(principal)(time)

Solution: r= 10,000
(120,000)(4)

r= 10,000
480,000

r = 0.02083

r = (0.02083)(100%)

Answer: rate = 2.083%

Explanaition: To solve this problem use the formula

rate(r) equals interest(i) over principal(p) multiplied by
time(t) to find the value of rate. First, you have to
distinguish the value for interest, principal and time.
Justin got a 120,000 loan, that should be the value of
our interest. 120,000 loan for 4 years, 4 years is the
time limit given to Justin, therefore 4 years is the value
of our time. The value of our interest should be 10,000
since Justin paid that much for the interest in his loan.
After completing the given, solve for the value of r by
dividing 10,000 to the product of 120,000 and 4. This
will then be equal to 0.02083 which will be multiplied
to 100 to get the rate which is 2.083%

PROBLEM 3:(COMPOUND INTEREST)

What is the present value of P60,000 due in 5
years if money is worth 10% compounded semi-
annually?

Given: F = Php 60,000
Formula:
m = 2 (semi-annually)

r = 10% or 0.10

t = 5 years

p= F
(1+r/m)(m)(t)

Solution: p= 60,000
(1+0.10/2)(2)(5)

p= 60,000
(1+0.05)(10)

p= 60,000
(1.05)(10)

p = 60,000
(1.628894627)

Answer: p = 36,834.795
or Php 36,834.80

Explanaition: To solve this problem, we will be using

the formula P is equals to F over 1 + r over m raise to
the m times t. First is to substitute the given, P is equal
to F (60,000) over 1 + r (0.10) over m (2) times raise
to m (2) times t (5). Next, divide the r (0.10) to m (2),
and the answer is 0.05, multiply the m (2) and t (5) also
so that the exponent will become 10. Now, our solution
is F (60,000) over 1 + 0.05 raise to 10. Add 1 and
0.05 so that it will become 1.05 raise to 10 and then
simplify the exponent, 1.05 raise to 10 and you will get
1.628894627. Lastly, to get the final answer, divide
the F (60,000) in to 1.628894627 and you will get
36,834.795 or 36,834.80.

PROBLEM 4: (GENEREL ANNUITY)

(What is the future value of an annuity that makes
5 annual payments of Php 11,000, if the interest
rate is 10% per year compounded quarterly?

Given: R = Php 11,000 n= mtrm==05.(042)5= 20
r = 10% or 0.1 j=

t = 5 years F=?

m=4

Formula: F= R[(1+j)n-1]
j

Solution: F= 11,000[(1+0.025)2-01]
0.025

F= 11,000[(1.025)20-1]
0.025

F= 11,000(0.63861644)
0.025

F= 7024.78084
0.025

F = 280,991.2336

Answer:

Php 280,991.23

Explanaition: To solve the problem at hand, we need

to determine the given first. After we determine the
R[(1+j)n-1]
given we need to use the formula F = j then we

substitute the values of the given into the said formula,

it will then be F= 11,000[(1+0.025)2-01] . By simplifying the
0.025
equation, we will come up with the answer

280,991.2336 or Php 280,991.23. This means that the

future value of the annuity is Php 280,991.23



ARRIVAL 004.

We are pleased to see you landed
safely on the fourth planet, Ayres2-
11 spacecraft Flight4B7! Welcome to
Planet 004 Earth Science. We are
tasked to do an Infographic about the
Conservation of Energy on this planet
you have landed. But first, Let's
briefly define it. Energy Conservation
refers to the effort made to reduce
energy consumption by utilizing less
of an energy service. Wait, wait,
wait! Voyagers, are you still with us?
Because it's not only about the
definition of the Conservation of
Energy, there is a lot of information
that is alloted to this Infographic,
like Simple and Easy Ways of
Conserving at Home, Fun Facts, and
the Importance of it that you will
surely enjoy knowing.

I hope you will learn a lot more
when you go much further into Earth
Science. Wishing you a safe flight
into the next planet!