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Neco 2019 Mathematics Theory & Obj Questions and Answers – Free Expo/ Runz

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Neco 2019 Mathematics Theory & Obj Questions and Answers – Free Expo/ Runz by : 5:00 am On June 23, 2019

Hello, welcome to Earboard.com Expo page for the Neco 2019 Mathematics (Maths) questions and answers. Neco 2019 Mathematics Expo Questions and Answers are now available on Earboard.com. For those who have paid for our Expo, kindly click the link below to enter your password. If you have not subscribed, then read further to learn how to subscribe for the Neco GCE 2019 Mathematics Expo Questions and Answers.
NECO 2019 Mathematics/Maths Questions


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Past Neco GCE Mathematics Answers

Mathematics OBJ:
1CDAAEABAEC
11AEDDCDCDCC
21CEBDEDCBBC
31CBEEECBDCC
41DBCBCDDBCA
51BCBDCDCCEC
solving.. please keep refreshing steady

===================================
NECO 2019 Mathematics Answers
NECO 2019 Mathematics Answers

NECO 2019 Mathematics Answers

NECO 2019 Mathematics Answers

NECO 2019 Mathematics Answers

NECO 2019 Mathematics Answers

3)
V = Mg√1 – r²
Square both sides
V² = m²g²(1-r²)
V²/m²g² = 1-r²
r² = 1 – v²/m²g²
r = √1-(v/mg)²
If v = 15, m = 20, and g = 10
r = √1 – (15/20*10)²
r = √1 – (0.075)²
r= √(1.075)(0.925)
r = √0.994375
r = 0.9972

10a)
Obtuse 105 + reflexReflex =255°
Now 2w = reflex2w =255°
W = 255/2 =127.5°
Also 2x = obtuse2x = 105°
X = 105/2 = 52.5°
Now EDF = y(base angles of an isosceles triangle)
BED=X=52.5°(angles in the same segment)
EFD+EDF=BED (sum of interior angles of a triangle equal exterior angle)
Y+y = 52.5°
2y = 52.5°
Y = 52.5°/2

=26.25°

10b)
Draw the diagram
Opp/adj = TanR
|TB|/|BR| = TanR
100/|BR| = Tan60°
|BR| = 100/tan60
|BR| = 100√3
|BR| = 100√3 * √3/√3
=100√3/3m OR 57.7m

==============================
5a)
Mode = mass with highest frequency = 35kg
Median is the 18th mass
= 40kg.
5b)
In a tabular form
Under Masses(x kg)
30,35,40,45,50,55
Under frequency(f)
5,9,7,6,4,4
Ef = 35
Under X-A
-10, -5, 0, 5, 10, 15
Under F(X-A)
-50, -45, 0, 30, 40, 60
Ef(X – A) = 35
Mean = A + (Ef(X – A)/Ef)
= 40 + 35/35
= 40 + 1

= 41kg

11a)
x+y/2 =11
x+y= 11*2
x+y= 22 —(1)
x-y= 4 —-(11)

x+y = 22—-(1)

x-y= 4—-(11)

2y = 18
y= 18/2
y=9
Substitute y=9 in equ 1
x+9=22
x=22-9
x=13
x=13, y=9
x+y= 13+9= 22
Sum of the two number
11b)
(6x + 3) dx
(6x + 3)dx
(6x +3)^6 – (6x + 3)^1
(6 x + 3)^5
(7776x^5 + 243)
38,880x/6 + 243
6480 x^6 + 243x
9(720x^6 + 27x)
11c)
y = x² + 5x – 3 (x = 2)
y = 2² + 5(2) – 3
y = 4 + 10 – 3
y = 14 – 3
y = 11

Gradient of the curve = 1

2a)
(X^2 Y^-3 Z)^3/4/X^-1 Y^4 Z^5
= (X^2)^3/4/X^-1 * (Y^-3)^3/4/Y^4 * Z^3/4/Z^5
= X^3/2/X^-1 * Y^-9/4/Y^4 * Z^3/4/Z^5
=X^3/2+1 * Y^-9/4-4 * Z^3/4-5
=X^5/2 * Y^-25/4 * Z^-17/4
=X^10/4 * Y^-25/4 * Z^-17/4
=(X^10/Y^25 Z^17)^1/4
2b)
√2/k + √2 = 1/k – √2
Multiply both sides by (k+√2)(k-√2)
√2(k-√2) = k+√2
√2k-√2 = k+√2
√2k-k = 2+√2
K(√2 -1) = 2+√2
K = 2+√2/√2-1
K = -(2+√2)/1-√2
Rationalizing
K = -(2+√2) * 1+√2/1-√2
K = -(2+√2)(1+√2)/1 – 2
K = (2+√2)(1+√2)
K = 2+2√2 + √2+2

K = 4+3√2

8)
x=a+by(eqi)
when y=5 and x=19
19=a+5b(eqii)
when y=10 and x=34
34=a+10b(eqiii)
solving eqii and eqiii
a+10b=34
a+5b=19
=>5b=15
b=15/5=3
putting b=3 in eqii
19=a+5(3)
19=a+15
a=19-15
a=4
8i)
Putting a=4 and b=3 in eqi
x=4+3y
This is the relationship between xand y
8ii)
When y=7
x=4+3(7)
x=4+21

x=25

1a)
Log 10(20-10)-log10(+3)=log105
(20-10/+3)=log10 =5
20-10/+3=5
5(+3)=20-10
5+15=20-10
15+10=20-5
25=15*
=25/15
*=5/3=1 2/3
1b)
Discount percent =15%
Discount amount =#600
Actual amount paid on the article =?
Original amount on the article =

15%=#600
15/100
=600
15=600100
15*=60000
*=60000/15
*=#4,000
Therefore actual amount paid on the article
=#4,000-#600
=#3,400

Actual amount paid on the article =#3,400

4i)
length of Arc of the sector
Titter= 72?, r = 14cm
L= titter / 360 x 2 pie r
==> L= 72/360 x 2 x 22/7 x 14
=44352/2520
=17.6cm
4ii) perimeter of the sector
Perimeter = titter/360 x 2 pie r + 2r
=17.6 +(2×14)
=17.6+28
=45.6cm
4iii) Area of the sector
Area = Titter/360 x pie r?
=72/360 x 22/7 x (14)?
=72 x 22 x 196/2520
Area= 310464/2520
=123.2cm?

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