Neco mathematics obj and theory answers verified by Bmasterz

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Neco mathematics obj and theory answers verified by Bmasterz

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Neco mathematics obj and theory answers verified by Bmasterz
Neco mathematics obj and theory answers verified by Bmasterz

Neco mathematics obj and theory answers | Verified 2018 neco maths expo , 2018 neco maths runz, 2018 neco maths answers,neco runz

NECO MATHS OBJ VERIFIED BY BMASTERZ FOR 2017. THE 2018 Neco mathematics obj and theory answers will be posted here soon.. Keep visiting. Note that if you have paid, you will receive the questions and answers 4 hours before exam.

1-10=CCAADDEECE

11-20=EEBABAADCC

21-30=CBEEDABDDD

31-40=BCCCDDCADE

41-50=BDCDDCDECD

51-60=BCDCAEDCDD

☆BMASTERZ☆

.MATHEMATICS-Answers

Use This No1 pls is more acurate

1)

tabulate

Number-

5932

*6141

num

3679

*3113

den

Log

3.7732

3.7882

7.5612

3.5657

3.4931

7.0588

anti log

7.5612

7.0588/0.5024/3=0.1675

anti log

=1471*10

================================

3a)

P=100I/RT=100*30,000/4*3

P=N250,000

================================

5)

tabulate

Mark- 21-25,26-30,31-35,36-40,41-45,46-50

F- 6,8,12,4,6,4=40

X- 23,28,33,38,43,48

FX- 138,224,396,152,258,192=1360

Class Boundaries- 20.5-20.5, 25.5-30.5,35.5-40.50,40.5-45.5

5i)

MEAN

(X)=EFX/EF=1360/40=34

MODE

L1+[Fm-Fa/2fm-fa-fb]c

=30.5+[12-8/2(12)-8-4]5

=30.5+[4/12]5

=30.5+1.67

=32.17

6i)

7+8+x = 47

15+ 2x =47

2x = 47- 15

2x/2 = 32/2

divide both side by 2

x=16

6ii)

tran only;:

7+8 + 3 + y=30

18 +y =30

y=30 -18 =12

:. Tran only =12 iii) at least two: 7+ 16+ 3 +8 =34

iv)16 + 7 +12 +16 +8 + 3 + Q=95

62+ Q =95

Q= 95 -62

Q=33 :. Car only = 33

(1)

TABULATE

No|Log|

5932|3.7732|

*6141|3.7882|

Num|7.5612|7.5612

3679|3.5657|

*3113|3.4931|

Den|7.0588|7.0588

| |0.5024/3=0.1675

Antilog:-1.471*10^0

=1.471

==================================

(2)

a = 3, b = 20, C = -7

Sum is

(A/1 + 1/B + B/1 + 1/ A) Note A means Alpha and B means beta

A^2 B + B^2 + A + B / AB

= -7/3 (-20/3) + (-20/3) = 140/9 – 20/3 / -7/3

= 80/9 x -3/7 = -80/21

Product:

-7/3 + 2 + 1/-7/3 = -7/3 + 2/1 – 3/7

= -49+42-9/21

= -16/21

x^2 – (sum)x + product= 0

x^2 – 80/21x + (-16/21) = 0

Final Answer : 21x^2 – 80x – 16 = 0

================================

(3a)

P=100I/RT=100*30,000/4*3

P=N250,000

(3b)

#7000 to Francs

#1 = 8 francs

.:.#7000 = ? france

CROSS MULTIPLY

8/1 * 7000 = 56,000 franc

Spending 49,400 franc

56, 000- 49,400 = 6600

Converting 6 ,600 franc to # AT 10 franc TO 1#

10 franc —– 1#

6600 FRANCS —? #

= 6,600/10 X 1

=#660

===================================

(4i )

difference in latitude = 36 degree + 36 degree = 72degree

distance travelled =@/ 360* 2pier

= 72 / 360* 2* 22 / 77* 64

= 20275200 / 2520

= 8045 . 71

= 8050 km ( 3 s.f )

(4ii)

given that speed = 800 km / hr

distance travelled = 804571 km

time = distance travelled / speed

= 8045 . 71/ 800

= 10 .057

= 10 hrs ( to the nearest hour )

===================================

(5)

Tabulate

Mark – 21- 25, 26- 30, 31- 35,36 -40 ,41 -45 ,46- 50

F – 6 ,8, 12,4 ,6, 4= 40

X – 23,28 ,33 ,38, 43, 48

FX – 138, 224, 396, 152, 258, 192= 1360

Class Boundaries- 20 .5 -20 .5, 25 .5 -30 .5, 35. 5- 40. 50,40 .5 -45 .5

(5i)

MEAN

( X )= EFX / EF = 1360/ 40 = 34

MODE

L 1+ [Fm – Fa / 2fm – fa -fb ] c

= 30 .5+ [12 -8/ 2 ( 12 ) -8- 4] 5

= 30 .5+ [4/ 12 ]5

= 30 .5+ 1.67

= 32 .17

===================================

(6)

(i)

7+8+x = 47

15+ 2x =47

2x = 47- 15

2x/2 = 32/2

Divide both side by 2

x=16

(ii)

Train only;:

7+8 + 3 + y=30

18 +y =30

y=30 -18 =12

.:. Train only =12

(iii)

at least two: 7+ 16+ 3 +8 =34

(iv)

16 + 7 +12 +16 +8 + 3 + Q=95

62+ Q =95

Q= 95 -62

Q=33 :. Car only = 33

=======================================

(7a)

a+2d=11—eq1

-(a+8d)=29

-6d/-6=-18/-6

d=3

Sub for d=3 in eq1

a+2(3)=11

a+6=11

a=11-6

a=5

(7b)

L = ar^n-1

729/3 = 3/3 x 3^n-1

243 = 3^n-1

3^5 = 3^-1

5 = n-1

N = 5+1

= 6

Sn = a(r^-1)/r-1 = 3(3^6 – 1)/3-1

Sn = 3( 729 -1)/2

S6 = 3 x 728/2

S6 = 1092

(7c)

3x^3/3 + x | 2,1

(2^3 + 2) – (1^3+1)

(8+2) – (1+1)

= 8‎

=================================

(8ai)

y=(4x+9)^3

Let u=4x+9

y=u^3

du/dx=4,dy/du=3u^2

dy/dx=dy/du*dy/dx

=4*3u^2

=12u^2

Recall u=4x+9

therefore dy/dx=12(4x+9)^2

(8aii)

y=(3x-2)^3(x^2+4)^2

u=(3x-2)^3, v=(x^2+4)^2

du/dx=3.3(3x-2)^3-1

dv/dx=2x.2(x^2+4)^2-1

du/dx=9(3x-2)^2

dv/dx=4x(x^2+4)

dy/dx=Vdu/dx +Udv/dx

=(x^2+4)^2[9(3x-2)^2+(3x-2)^3][4x(x^2+4)]

=(x^2+4)(3x-2)^2(9x^2+36+12x^2-8x)

=(x^2+4)(3x-2)^2(21x-8x+36)

(8b)

y-y1=m(x-x1)

y-6=2(x-2)

y-6=2x-4

y=2x-4+6

y=2x+2

===============================

(9a)

y proportional x proportional 1/x

y=kx/1 +k/x

4=2k/1+k/2

4=(4k+k)/2

8=5k

k=8/5

Relationship:

y=kx/1+k/x

y=8x/5+8/5x

(9b)

4(3^x+1)-3^2x=27

4(3^x *3^1)-362x=3^3

let 3^x=p

4(3p)-p^2=27

12p-p2=27

p^2-12p+27=0

p^2-9p-3p+27=0

p(p-9)-3(p-9)=0

(p-9)(p-3)=0

p=9 or p=3

Recall 3^x=p

when p=9

3^x=3^3

x=3

when p=3

3^x=3^1

x=1

therefore x=2 or 1

==================================

(11a)

Tabulate

X- |-1| |0| |1| |2| |3| |4|

6 |6| |6| |6| |6| |6| |6|

+X |-1| |0| |1| |2| |3| |4|

-X^2 |-1| |0| |-1| |-4| |-9| |16|

Y- |4| |6| |6| |4| |0| |-6|

(11b)

Draw the graph

(11ci)

gradient at x/x

dy/dx =B-A/B-C

=7-4/2-1=3/1=3

(11cii)

roots of equation from the graph

x=-2.2 and x=3

(11ciii)

minimum value of y

=6.4

================

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