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12/11/2023

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16/08/2020

Hello guys am back for 2020 waec exam to help u guys and it's free dm me to add you in my what's app group

24/06/2019

NECO-MATHEMATICS-ANSWERS
MATHS-OBJ!
1DBBCBDEDAC
11DCDAECDEDC
21CBAAECDCAD
31CCACCBDDCC
41CDDDCABDBA
51BCBDBCAADD
=====================================

(1a)
At the end of year 1
Using; A = P(1 + R/100)N
A = #110,000(1+5/100)
A = #110,000(1.05)
Amount or savings = #115,500.00
At the beginning of year 2,
Principal, p = 115,500 + #50,000 = #165,500.00
At the end of year 2
A = #165,500(1+5/100)¹
A = #165,500 × 1.05
A = #173,775.00
At the beginning of year 3,
Principal, p = #173,775 + #50,000 = #223,775.00
At the end of year 3,
A = #223,775(1+5/100)
A = #223,775 × 1.05
A = 234,963.75

Total savings after 3 years = #234,963.75 + #50,000 = #284,963.75
(1b)
By end of third year
Savings is lesser than #300,000.00 by;
#300,000.00 - 284,963.75
= #15,036.25
= #15,036.25
=====================================

(9ai)
x+1⅔x≤ 2⅓x-1¼
x+5x/3≤ 7x/3 - 5/4
Multiply through with (12) 12x + 20x ≤ 28x - 15
32x ≤ 28x -15
32x - 28x ≤ -15
4x ≤ - 15
x ≤ - 15/4
x ≤ - 3¾.
(9aii)
4x-1/3 - 1+2x/5 ≤ 8+2x
Multiply through with (15) 5(4x-1)-3(1+2x)≤ 15(8+2x)
20x-5-3-6x ≤ 120 + 30x
14x-8 ≤ 120 + 8
-16x ≤ 128
x ≥ 128/-16
x ≥ -8
(9b)
Gradient
m=y²-y¹/x²-x¹
m=9-7/6-3
=⅔
Acute angle θ = Tan-¹ (⅔)
θ=Tan-¹ (0.6667)
θ=33.69degree.
=====================================

(5a)
x² - 5x - 24 = 0
x² - 5x = 24
x² - 5x + (5/2)² = 24 + (5/2)²
(x-5/2)² = 24 + 25/4
96 + 25/4
(x - 5/2)² = 121/4
x - 5/2 = ±√121/4
x - 5/2 = ± 11/2
x = 5/2 ± 11/2
x = 5/2 + 11/2 or 5/2 or 11/2
x = 16/2 or -6/2
x = 8 or -3
(5b)
S0/2 (3x²-4x+2)dx
= 3x²+¹/3 - 4x¹-¹/2 + 2x/1 + C
= 3x³/3 - 4x²/2 +2x/1 + C
= (x³ - 2x² + 2x + C) dx
= y = x³+¹/4 - 2x²+1/3 + 2x¹+¹/2 + C
= y = x⁴/4 - 2x³/3 + 2x²/2 + C
= y = x⁴/4 - 2x³/3 + 2x + C
=====================================
(11)
CLICK HERE FOR THE IMAGE
=====================================

(12a)
Tabulate

Score : 21-30, 31-40, 41-50, 51-60, 61-70, 71-80

Class mark(x) : 25.5, 35.5, 45.5, 55.5, 65.5, 75.5

f : 2,10,12,15,8,3

d(x- x̄): -20, -10, 0, 10, 20, 30

fd : -40, -100, 0, 150, 160, 90

Assumed mean = 45.5
(12b)
Using assumed mean ( x̄) = A.M + Σfd/Σf
x̄ = 45.5+260/50
x̄ = 45.5+5.2 = 50.7
(12c)
Semi inter quartile = Q2-Q7/2

Q3= 3/4 × f
= 3/4 × 50 =150/4 = 37.5

Q1= 1/4 × f
= 1/4 × 50/1 = 12.5

: . Semi inter quartile = 37.5 - 12.5/2 = 25/2
= 12.5
=====================================

(2a)
3^2x-y=1
3^2x-y=3^0
2x-y=0-------------(1)
16^x/4 = 8^3x-y
2^4x/2^2 = 2^3(3x-y)
2^4x-2 = 2^9x-3y
:. 4x-2 = 9x-3y
4x-9x+3y= 2
-5x+3y=2------------(2)
From equation (1):
2x-y=0
y=2x--------(3)
Substitute for y in equation (2)
-5x+3y=2
-5x+3(2x)=2
-5x+6x=2
x=2
Substitute for x in equation (3)
y=2x
y=2(2)=4
:.x=2, y=4
(2b)
x² - 4/3 + x+3/2
2(x² - 4) + 3(x +3)/ 6
2x² - 8 + 3x + 9/6
2x²+3x+1/6
(2x² + 2x)+(x+1)/6
2x(x+1) +1 (x+1)/6.
=====================================

(8ai)
Total surface area
= Total surface of cylinder + Curve surface of hemisphere
= (πr^2+2πrh) + (2πr^2)
= π(r^2 + 2rh) + π(2r^2)
= π[(r^2 + 2rh) + 2r^2]
= π[(7^2 + 2(7)(10) + 2(7)^2]
= π[(49+140) + 98]
= π(287)
= 287πcm^2

Using π=22/7
Total surface area =287×22/7 = 41×22
= 902cm^2
(8aii)
Volume = Volume of cylinder + volume of hemisphere
= πr2h + 2/3πr^3
= π[r^2h + 2/3r^3]
= π[(7^2)(10) + 2/3(7)^3]
= π(490 + 656/3)
= π(2156/3)
= 22/7 × 2156/3
= 22 × 308/3 = 6776/3
= 2258.67cm^3
(8b)
Perimeter of Arc = Φ/360 × 2πr
= 120/360 × 2 × 22/7 × 7
= 1/3 × 44 = 44/3
= 14.67cm
=====================================

(3)
CLICK HERE FOR THE IMAGE
Using SOHCAHTOA
|TM| / |MD| = Tan28°

298.5+1.5/|MD| = 0.5317
|MD| = 300/0.5317 = 564.2m

Similarly,
|TM| / |MC| = Tan34°
300/ |MC| = 0.6745
|MC| = 300/0.6745 = 444.8m

Distance between both , ΔCD
= 564.2 - 444.8
= 119.4m
=====================================

(10ai)
S=t^3 -3t -9t + 1
ds/dt=v
:. 3t^2 - 6t^2 -9

When v=0
3t^2 -6t^2 -9=0
(3t^2 -9t)+(3t-9)
3t(t-3)+3(t-3)=0
(3t+3)(t-3)=0
3t + 3=0
3t= -3
t= -3/3= -1 or t -3=0
t=3seconds

(10aii)
a=dv/dt = 3t^2 -6t -9= 6t -6
a=6t -6
When a=0
6t -6=0
6t=0+6
6t=6
t=6/6
t=1
(10b)
v=3t^2 -6t -9

When t=2seconds
v=3(2)^2 -6(2) -9
= 3*4-9-12-9= -9m/s
v= -9m/s

acceleration, a when t=2seconds

a=6t -6= 6(2) -6= 12-6
a=6m/s^2
(10c)
a=6t -6 = 36-6=30
a=30m/s^2
=====================================

(6a)
For X = a=4a , T6=256
ar^5=256
4ar^5/4=256/4
ar^5=64............(1)

For Y = a=3a, T5=48
ar^4=48
3ar^4/3=48/3
ar^4=16...............(2)

Divide equ (2) by (1):
ar^5/ar^4=64/16
r=4

Substitute for r in equ (2)
ar^4=16
a × 4^4=16
256a/256=16/256
a=1/16

(6ai)
First term of x : a=4a
a=4×1/16=1/4

(6aii)
Sn = a(r^n-1)/r-1
S4 = 1/4(4^4 -1)/4-1
=1/4(256-1)/3
=1/4 × 255/3
=85/4
S4=21.25

(6b)
y(4x+2)^-3

Let u =4x+2, y=u^-3
du/dx=4 , dy/du= -3u^-4

dy/dx=dy/du * du/dx

= -3u^-4 × 4
= -12u^-4
dy/dx= -12(4x+2)^-4

When x =1,
dy/dx= -12(4*1+2)^-4
= -12(4+2)
= -12 * 6^-4
= -12/6^4
= -12/1296
= -1/108
=====================================

(7a)
4x^2 - 9y^2 = 19
2x^2 x^2 - 3^2 y^2=19
(2x-3y)(2x+3y)=19

Substitute for 2x+3y=1
2x-3y=19............(1)
2x+3y=1..............(2)

Subtract equ (2) from (1)
2x-3y-(2x+3y)=19-1
3x-3y-2x-3y=18
-6y/-6=18/-6
y = -3

Substitute for y in equ (1)
2x-3(-3)=19
2x+9=19
2x/2=10/2
x=5

(7b)
√4.033/0.611 × 0.356

Put No and Log In a tabular form

No | log
4.033 | 0.6056 -> | 0.6056
0.611 | 1.7860+ -
0.356 | 1.5514
0.611×0.356|1.3374->|1.3374
4.033/ | --------> | 1.2682
0.611×0.356 ÷2
√4.033/0.611 | -------> | 0.6341
×0.356
Antilog = 4.306
Ans = 4.306
=====================================

(4)
Mark: 1-5, 6-10, 11-15, 16-20, 21-25, 26-30

F : 6,4,5,5,6,4

x : 3,8,13,18,23,28

Fx: 18,32,65,90,138,112

x-x: -12.167, -7.167, -2.167, 2.833, 7.833, 12.833

(x-x)^2: 148.0359, 51.3659, 4.6959, 8.0259, 61.3559, 164.6859

f(x-x)^2: 888.2154, 205.4636, 23.4795, 40.1295, 368.1354, 658.7436

Mean(x) = Σfx/Σf = 455/30 =15.167

Variance = Σf(x-x)^2/Σf = 2184.167/30 = 72.8056
= 72.81(approx.)

Standard deviation = √Variance
= √72.84
= 8.533
= 8.53 (approx.)
=====================================
COMPLETED!
====================================

24/06/2019

*2019 NECO MATHEMATICS ANSWERS*

*===============================================================*
2a

3^(2x-y)=3^0.......eqn1

2^(4x)/2^2=2^3(3x-y)......eqn2

2^(4x-2)=2^(9x-3y).........eqn2

2x-y=0........eqn3

4x-2=9x-3y

5x-3y=-2.........eqn4

From eqn3,y=2x

Put y=2x into equation (4)
5x-3(2x)=-2
5x-6x=-2
-x=-2
X=2
Y=2x=2(2)=4
(X,y)=(2,4)

*take note this sign means ^ raise to power* eg 2^3 means 2³

2b
(X²-4)/3+(x+3)/2
=2(x²-4)+3(x+3))/6
=2x²-8+3x+9)/6
=2x²+3x+1)/6
*===============================================================*
Q5

x²-5x-24=0
x²-5x=24 (completing square method)

Add 1/2 of the coefficient of x on both sides
(-5 X 1/2)² = (-5/2)² = 25/4
x²-5x+(-5/2)²=25/4 +24

(x-5/2)² = 25+96/4

(x-5/2)²=121/4

x-5/2 = ±√121/4

x-5/4 = ±11/4

x=5/4 ±11/4

x=5/4 or 5/4-11/4

x=16/4 or -6/4

x=4 or -3/4
*===============================================================*
No 6

Given (4x + 2)–³

Dy/dx = 4 ( -3) (4x + 2)–³–¹

Dy/dx= - 12 (4x-2)–⁴

Where x = 1

Dy/dx= -12(4(1)-2)–⁴

Dy/dx = - 12(2)–⁴

Dy/dx = -12× 1/2⁴

= -12/16

Dy/dx = -3/4

*take note – means minus*

7bi) see diagram above


Bii) from the diagram above

Teta + teta +70=180
2teta =180-70
Teta=110/2 =55

Also alpha=55+10=65

Thus bearing of Q FROM P will be
180+alpha
=180+65
=245

Wwt typing

8ai) total surface area 2πr
2*22/7*7 =44
8aii)volume =2/3πr^3
=2/3*22/7*7^3
= 2*22*345/21
=15092/21
718.67cm
9ai) x+5/3x≤7/3x-5/4
3x+5x/3≤28x-15/12
12(3x+5x) ≥3(28x-15)
36x+60x≥84x-45
96x-84x≥-45
12x≥-45
X≥-45/

20/06/2019

NECO-PHYSIC-ANSWERS

Note; You Are Advice To Answer Question One (1) And Any Other Five (5) From Section A (Part I) And Four (4) From Section B. Enjoy!

Physic-Obj!
1ECBDDADBDD
11ACBDCBCCDD
21DADAEDDBCD
31DEDEBCECBA
41ADEAAEBEEC
51ADEBABBECB
=====================================

Physic-Theory-Answers

Answers-Six(6) From This Section.

(1a)
(pick 2)
(i) Solar Electric Power Generation
(ii) Solar Cooking
(iii) Solar Thermal Power Production
(iv) Solar-distillation

(1b)
Natural sattelite is a celestial body in space that orbits a larger body, such as Moon Which moves around the Earth.
Artificial satellite is a machine of human creation that is sent to the space or the orbit of the Earth for the collection of data, communication and other ends

(1c)
The applicable relationship is Ns/Np = Vs/Vp meaning the ratio of secondary voltage to primary voltage is equal to the ratio of secondary turns to primary turns
Ns = 200turns , Vs = 300V , Np = ? , Vp= 60V
Ns/Np=Vs/Vp
Np=Ns(Vp/Vs)
Np= 200 × (60/300)
Np = 200 × 0.2
Np= 40 turns
=====================================

(2a)
Force is an agent that changes or tends to change the state of rest or of uniform motion in a straight line of a body

(2b)
Contact force are forces which are in contact with the body to which they are applies while Field force are those forces whose sources do not require contact with the body to which the are applied
=====================================

(3a)
Maximum range (R) : Maximum range is defined as when the angle of the projection (Φ) is equal to 45° ie R = U^2sin90/g and Φ=45°
R= U^2sin90/g and sin90=1
Rmax=U^2/g

(3b)
(i) Stable Equilibrium - A cone resting on its base
(ii) Unstable Equilibrium - A cone resting on its apex or vertex
=====================================

(4a)
It means that the glycerine which is in solid state will require a heat of 13°c to change to liquid state

(4b)
The S.I unit of mass is in kilogram (kg) while that of weight is in Newton (N)
=====================================

(6)
Constructive interference is formed when the wave amplitudes reinforce each other, building a wave of even greater amplitude. Destructive interference is formed when the wave amplitudes oppose each other, resulting in waves of reduced amplitude.
=====================================

(7)
CLICK HERE FOR THE IMAGE
=====================================

(8a)
Faraday's law states that the induced e.m.f. in a conducting circuit is directly proportional to the rate of change of magnetic flux linkage, Φ, with the circuit.

(8b)
(i)Induction motors
(ii)Electric generators
=====================================

(9a)
Electric field acts on charges (Q) while Gravitational field acts on masses (m)

(9b)
Emf =12v
r =0.45Ω
R =5.0Ω
I =?
I =Emf/R+r
I =12/5+0.45=12/5.45
I =2.20A
=====================================

(11)
(Pick Only 1)
-ADVANTAGE-
(i)it is used in generating heat energy required to generate electricity
(ii)it can be operated non-stop with the same nuclear capacity

(Pick Only 2)
-DISADVANTAGE-
(i)the plant set up is capital intensive
(ii)principal raw material (uranuim) is not easily available.
(iii)Careless exposure to raw material affects the cell tissues due to radiation effect.
=====================================

Answers-Four(4) From This Section.

(12a)
Centripetal force and centrifugal force

(12b)
CLICK HERE FOR THE IMAGE

(12ci)
Achimede's principle states that when a body is totally or partially immersed in a fluid, it experience an upthrust which is equal to the weight of the fluid displaced.

(12cii)
eo = 4×10^4 kg/m^3
Mo = 0.4kg
Vo = ?
eL = 1.2×10^2 kg/m^3
Where,
eo = Density of object
Mo = mass of object
Vo = volume of object
eo = Mo/Vo
Vo = Mo/eo = 0.4/4×10^4
Vo = 0.00001m^3
Mo in the liquid = eL × Vo
= 1.2×10^2 × Vo × 3/4
= (1.2×10^2 × 0.00001×3)/4
= 0.0009kg
= 9×10^-4 kg
Reading of the spring = 9×10^-4 ×10 = 9×10^-3 N
=====================================

(13a)
Examples of Renewable energy are :
(i) solar energy
(ii) Wind energy

(13b)
Radius of the wheel = B
Radius of the axle = O
V. R= Distance moved by effort/ Distance moved by load
Effort moves through a distance is equal to the circumference of the wheel 2πB and at the same time interval the load moves through the axle 2πD
V.R=2πB/2πD =B/D

(13ci)
It breaks because ice is less dense than water and water expands when it is frozen

(13cii)
Evaporation takes place at all temperature but boiling takes place at a particular temperature (100°c)

(13d)
m1=200kg
u1 = 16.7m/s
V2 = 0.5m/s
m2 = 200kg
m1u1 + m2V2 = (m1+m2)V
200×16.7 + 200×0.5 = (200+200)V
V = 3340+100/400 = 3440/400
V=8.6m/s
=====================================

(14ai)
A wave is a disturbance which travels through a medium transferring energy from one point to another without causing any permanent displacement of the medium

(14aii)
Tabulate
- Light wave
(i) It is an electromagnetic wave
(ii) It can be polarized

- Sound wave
(i) It is a mechanical wave
(ii) It can can not be polarized

(14b)
Long sight ( hypermetropia ) can be corrected with converging lens

(14c)
F= 440Hz
L1= 18.8cm
L2= 57.8cm

(14ci)
At resonance L+c= λ/4
Where C is the end correction. Assuming the end correction C is negligible,then
L1= λ/4
4L1=
4 × 18.8 = λ
4×18.8/100=0.752m

(14cii)
First position L1=λ
Second position L2 =λ
Hence, L2-L1 = λ/4
V= fλ or f=V/λ
V = 2f(L2-L1)
V= 2×440( 0.5733 - 0.188)
V= 880(0.385)
V=338.8m/s

(14ciii)
Overtime = 5λ/4
5×0.752/4
= 3.76/4
= 0.94m or 94cm
=====================================

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