2019 VERIFIED NECO MATHEMATICS
June 24, 2019
0 comment
*2019 VERIFIED NECO MATHEMATICS ANSWERS*
✔️✔️✔️✔️✔️✔️
π―π―π―π―π―
*MATHS-OBJ*
1-10: DBBCBDEDAC
11-20: DCDAECDEDC
21-30: CBAAECDCAD
31-40: CCACCBDDCC
41-50: CDDDCABDBA
51-60: BCBDBCAADD
=====================================
*===============================================================*
(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.
*===============================================================*
(3)
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 , ΞCD
= 564.2 - 444.8
= 119.4m
SEE THE IMAGE(3)
(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) = Ξ£fx/Ξ£f = 455/30 =15.167
Variance = Ξ£f(x-x)^2/Ξ£f = 2184.167/30 = 72.8056
= 72.81(approximation.)
Standard deviation = √Variance
= √72.84
= 8.533
= 8.53 (approximation.)
SEE IMAGE(4)
*===============================================================*
(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)
Typing
SEE IMAGE(7)
*===============================================================*
(8ai)
Total surface area
= Total surface of cylinder + Curve surface of hemisphere
= (Οr^2+2Οrh) + (2Οr^2)
= Ο(r^2 + 2rh) + Ο(2r^2)
= Ο[(r^2 + 2rh) + 2r^2]
= Ο[(7^2 + 2(7)(10) + 2(7)^2]
= Ο[(49+140) + 98]
= Ο(287)
= 287Οcm^2
Using Ο=22/7
Total surface area =287×22/7 = 41×22
= 902cm^2
(8aii)
Volume = Volume of cylinder + volume of hemisphere
= Οr2h + 2/3Οr^3
= Ο[r^2h + 2/3r^3]
= Ο[(7^2)(10) + 2/3(7)^3]
= Ο(490 + 656/3)
= Ο(2156/3)
= 22/7 × 2156/3
= 22 × 308/3 = 6776/3
= 2258.67cm^3
(8b)
Perimeter of Arc = Ξ¦/360 × 2Οr
= 120/360 × 2 × 22/7 × 7
= 1/3 × 44 = 44/3
= 14.67cm
SEE IMAGE(8)
*===============================================================*
(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
SEE IMAGE(10)
(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 + Ξ£fd/Ξ£f
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
==========
*===============================================================*
COMPLETED!.
*MISSION ACCOMPLISHED*
✔️✔️✔️✔️✔️✔️
π―π―π―π―π―
*MATHS-OBJ*
1-10: DBBCBDEDAC
11-20: DCDAECDEDC
21-30: CBAAECDCAD
31-40: CCACCBDDCC
41-50: CDDDCABDBA
51-60: BCBDBCAADD
=====================================
*===============================================================*
(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.
*===============================================================*
(3)
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 , ΞCD
= 564.2 - 444.8
= 119.4m
SEE THE IMAGE(3)
(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) = Ξ£fx/Ξ£f = 455/30 =15.167
Variance = Ξ£f(x-x)^2/Ξ£f = 2184.167/30 = 72.8056
= 72.81(approximation.)
Standard deviation = √Variance
= √72.84
= 8.533
= 8.53 (approximation.)
SEE IMAGE(4)
*===============================================================*
(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)
Typing
SEE IMAGE(7)
*===============================================================*
(8ai)
Total surface area
= Total surface of cylinder + Curve surface of hemisphere
= (Οr^2+2Οrh) + (2Οr^2)
= Ο(r^2 + 2rh) + Ο(2r^2)
= Ο[(r^2 + 2rh) + 2r^2]
= Ο[(7^2 + 2(7)(10) + 2(7)^2]
= Ο[(49+140) + 98]
= Ο(287)
= 287Οcm^2
Using Ο=22/7
Total surface area =287×22/7 = 41×22
= 902cm^2
(8aii)
Volume = Volume of cylinder + volume of hemisphere
= Οr2h + 2/3Οr^3
= Ο[r^2h + 2/3r^3]
= Ο[(7^2)(10) + 2/3(7)^3]
= Ο(490 + 656/3)
= Ο(2156/3)
= 22/7 × 2156/3
= 22 × 308/3 = 6776/3
= 2258.67cm^3
(8b)
Perimeter of Arc = Ξ¦/360 × 2Οr
= 120/360 × 2 × 22/7 × 7
= 1/3 × 44 = 44/3
= 14.67cm
SEE IMAGE(8)
*===============================================================*
(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
SEE IMAGE(10)
(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 + Ξ£fd/Ξ£f
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
==========
*===============================================================*
COMPLETED!.
Do You Also Need Runs For NECO? Then Pay N800 OR instead pay N4,500 9subjects
while VIP cost N7,000
U CAN ALSO SUBSCRIBE TO GET YOUR OBJ ANSWERS AS TEXT MESSAGE DIRECTLY TO YOUR MOBILE PHONE {SEND UR MTN AIRTIME AS TEXT, FOLLOWED BY UR MOBILE NUMBER TO 09077759576. Or 08170764699.
THEN WAIT PATIENTLY FOR UR ANSWERS
2019 NECO Answers – Legit Runz Expo
Welcome, Are you among those sitting for
2019 NECOEXAM)? Are you among those
searching for 2019 NECO
expo .
expo .
Follow the steps below to get
the live Questions and Answers at least Six
(3-4) hours before your exam
time.
the live Questions and Answers at least Six
(3-4) hours before your exam
time.
SUBSCRIPTION FEE PER SUBJECT
- Text Message/obj: #300 MTN
CARD - Password/Online
Answers: #800 MTN CARD - WhatsApp Message: #600
MTN CARD
OUR DELIVERY PATTERN
- Text Message: In this package,
we send Answers to your phone inbox via
SMS obj only. - WhatsApp Message: In this
package, we add you to an answer broadcast
on WhatsApp were we will post answers to
you. - Password/Online Answers: In this
package we send you the password to unlock
an answer page and access your
answers.
HOW TO SUBSCRIBE OR MAKE PAYMENT
Send
- Your Name,
- Subject Name (e.g
Marketing), - MTN Recharge Card Pin(s)
and - Your Phone Number to
09077759576
Please don’t call, just send us text
message, We don’t attend
to calls because of too many calls. We are
always busy.
No Free Answer will be Posted here at All,
So Subscribe Now and Get Your Answers In The Night Or latest 5hours Before Exam.
Keep your recharge cards safe
until you receive a confirmation
message.
until you receive a confirmation
message.
Make your examination with this help desk without stressing yourself.
YOUR ANSWERS ARE LOADING KEEP ON REFRESHING EVERY 2MINUTES WE ARE ATTENDING TO OUR SUBSCRIBERS BEFORE DROPPING HERE!
I piety those who are waiting for free Expo...
While our subscribers are busy playing with the questions and answers a night before d exams...
0 Response to "2019 VERIFIED NECO MATHEMATICS"
Post a Comment
DROP YOUR COMMENTS HERE