2018 Neco Mathematics Objective and Theory Questions and Answers is available ,Mathematics 2018 Neco Exam Expo, all you need to do is to follow the instructions above, get verified Mathematics Objective and Theory Questions and Answers here, question and answers are now Available. expo for Neco 2018 Mathematics, exam runs for Mathematcs objective and Theory questions and answers Neco maths portal
MATHS OBJ:
1-10: CDAAEABAEC
11-20: ACDDCDCDACu
21-30: CEADEDCABC
31-40: CBEECCBDCC
41-50: DDCBCDDBBA
51-60: BCECCBBCEE
(1a)
Log10(20x - 10) - log10(
x+3) = log10^5
Log10(20x-10/
x+3)= log10^5
20x - 10/
x + 3 = 5
Cross multiplying
20x - 10 = 5(
x + 3)
5(4x - 2) = 5(
x + 3)
4x - 2 =
X + 3
4x -
x = 3+2
3x = 5
X = 5/3 OR 1 whole no 2/3
(1b)
Let actual amount be
#X15% of
#x =
#60015x/100 = 600
X = (100/15)*600
X = 100*40
X = 4,000
Actual amount =
#4,000
(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
(4)
Draw the diagram
(i) Arc length = Tita/360*2πr
= 72/360*2*22/7*14
=1/5*44*2
=88/5
=17.6cm
(ii) Perimeter of Sector = arc length +2r
=17.6+2(14)
=17.6+28
=45.6cm
(iii) Area of sector = Tita/360*πr²
=72/360*22/7*14/1*14/1
=1/5*22*2*14
=616/5
=123.2cm2
(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
(
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
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 = 11
Monday 4th June 2018
OBJ � General Mathematics � 10:00am � 11:45am
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2018 Neco Mathematics Objective and Theory Questions and Answers is available ,all you need to do is to follow the instructions above, get verified Mathematics Objective and Theory Questions and Answers here, question and answers are now Available. expo for Neco 2018 Mathematics, exam runs for Mathematcs objective and Theory questions and answers
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