*2019-NECO-MATHEMATICS-ANSWERS*

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*MATHS-OBJ*

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11-20: DCDAECDEDC

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(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.

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(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 , ΔCD

= 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) = Σfx/Σf = 455/30 =15.167

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

= 72.81(approximation.)




Standard deviation = √Variance 

= √72.84

= 8.533

= 8.53 (approximation.)

SEE IMAGE(4)

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(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

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(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)

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(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

SEE IMAGE(Cool

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(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)

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Monday 24th June

  • Paper III: Objective – General Mathematics ‹‹›› 10:00am – 11:45am

  • Paper II: Essay – General Mathematics ‹‹›› 12:00noon – 2:30pm



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SEE ALSO : 2018 Nov/Dec Neco Gce Physics Practical Questions and Answers | Exam Runs


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ALSO SEE: Physics Objective and Essay 2018 Neco Gce Questions and Answers


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