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*MATHS OBJ*

1-10: BCBEDDCCDC

11-20: DEBCBBDDEC

21-30: BCBCCBBEBD

31-40: EBABEDCDEB

41-50: DDAECACBAB

51-60: DDBCEACBBE

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

(1)

√((0.0024)×35000)/0.0105

No | Log

0.024 |_3.3802

35000 |4.5441

- |=1.9243

0.0105|_2.0212 -

|3.9031 ÷2

|1.9516

√((0.0024)×35000)/0.0105

Antilog of .9516 =89.45

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

(2a)

Given that the roots of the equation are

X = -2/3 and X = -3/2

3x = -2 and 2x = -3

3x + 2 = 0 and 2x + 3 =0

(3x+2)(2x+3) = 0

6x² + 9x + 4x + 6 = 0

6x² + 13x + 6 = 0

(2b)

R = [3 4 0 ]

[2 0 3 ]

[1 2 2]

(2ci)

2/3R

=2/3[3 4 0]

[2 0 3]

[1 2 2]

= [⅔(3) ⅔(4) ⅔(0)]

[⅔(2) ⅔(0) ⅔(3)]

[⅔(1) ⅔(4) ⅔(2)]

= [2 8/3 0 ]

[4/3 0 2 ]

[2/3 4/3 4/3]

(2cii)

|R|

= |3 4 0|

|2 0 3|

|1 2 2|

=3|0 3| -4|2 3| +0|2 0|

|2 2| |1 2| |1 2|

=3(2×0-3×2)-4(2×2-3×1)

+0(2×2-0×1)

=3(0 - 6)-4(4 - 3) +0(4 - 0)

=3(-6) -4(1) +0(4)

= -18 - 4 + 0 = -22

(2ciii)

The transpose of R

= [3 2 1]

[4 0 2]

[0 3 2]

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

(3a)

Given : R(3,5) and S(-2, -6)

equation to line through them is :

y-5/x-3 = -6-5/-2-3

y-5/x-3 = -11/-5

y-5/x-3 = 11/5

5(y-5)= 11(x-3)

5y-25 = 11x-33

5y-11x = 25-33

5y-11x = -8 or 11x-5y = 8

(3b)

RS= √(X1X2)^2 (y1y2)^2

√(-2-3)^2 + (-6-5)^2

√(-5)^2 + (-11)^2

√25 + 121

√146

= 12.08

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

(4a)

Given: curve; y x² - 3x

gradient ; dy/dx = 2x - 3

At (2-2), gradient = 2(2) - 3

4-3 = 1

(4b)

Given; y= 1+x²/1-x²

dy/dx = (1-x²)(2x) - (1+x²)(-2x)/(1-x²)²

= (1-x²)(2x) - (1+x²)(2x)/(1-x²)²

= 2x(1-x² + 1+x²)/(1-x²)²

= 2x(2)/(1-x²)²

= 4x/(1-x²)²

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

(5)

No of blue balls = 6

No of red balls = 10

(i)

Prob (2 balls of some colour)

= BB or RR

Total no of balls = 6+10 = 16

BB or RR

(6/16 × 5/15) + (10/16 × 9/15)

=30/240 + 90/240

=120/240 = 1/2

(ii)

Prob (2 balls of different colours)

= BR or RB

= (6/16 × 10/15) or (10/16 × 6/15)

= 60/240 + 60/240 = 120/240

=1/2

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

(6a)

27^(2x+1) × 3^-x = 81^(x-2)/9^(x+2)

= 3^3(2x+1) × 3^-x = 3^4(x-2)/3^2(x+2)

=3^6x+3-x = 3^4x-8-2x-4

5x+3 = 2x - 12

5x - 2x = -12-3

3x = -15

3x/3 = -15/3

X = -5

(6b)

X/x+101 = 11/1000

Since all the members are in binary, convert all to denary (base 10)

Xbase2 = Xbase10

101base2 = (1×2^2)+(1×2^0) = 4+1 = 5base10

11base2 = (1×2¹)+(1×2raise to power 0) = 2+1 = 3base10

1000base2 = 1×2^3 = 8base10

X/X+101 = 11/1000 --> X/X+5 = 3/8

3(x+5) = 8(x)

3x+15 = 8x

15 = 8x - 3x

15 = 5x

15/5 = 5/5

X = 3

Convert X=3 to base 10 to base 2

2|3

2|1R1

|0R1

.:. x = 11

(6c)

Given that log5 base 10 = 0.699

and log3 base 10 = 0.477

10

Log75 base 10 = log(3×5×5) base 10

=log(3×5²) base 10

=log3 base 10 + 2log5 base 10

=0.477 + 2(0.699)

= 0.477 + 1.398

= 1.875

log75 base 10 = 1.875

=================÷÷÷

(7a)

Given that Y=2x2+7x-6

To find the gradient of the curve at the point x=3

dy/dx=4x+7 (at x=3)

dy/dx=4(3)+7

dy/dx=19

(7bi)

Ade bought 7kg of maize + 4kg of meat=N4240

Kemi bought 3kg of maize + 5kg meat =N4610

Let x = 1kg of maize and y=1kg of meat

Therefore 7x+4y=4240-----(eq1)

3x+5y=4610---------(eq2)

Substituting simultaneously

Multiply eq1 by 3 and eq2 by 7

21x+12y=12720---eq3

21x+35y=32270---eq4

Substract eq3 from eq4

23y=19550

23y/23=19550/23=850

y=850

Substitute y=850 in1

7x+4y=4240

7x+4(850)=4240

7x=4240-3400

x=840/7

=120

Hence the total cost price per Kg of maize is N120.00 while the total cost per kg of meat is N850.00

(7bii)

Total cost of 10kg of maize and 5kg of meat

=10x+5y

=10(120)+5(850)

1200+4250

=N5450

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

(9)

Draw the diagram

Q(34.5°N, 22.3°W)

P(34.5°N, 38.7°E)

R(35.4°S, 38.7°E)

Distance between P and Q along the parallel of latitude

D = Φ/360 × 2πr

Where r = RcosΦ

Φ = 38.7° + 22.3° = 61°

Q = 34.6°

R = 6400km

D = 61/360 × 22/7 ×6400cos34.6

D =1717760(0.8231)/252

D = 5,610.66768

= 5,610km

(9b)

|PR|=(34.6+35.4)/360 * 2 *22/7*6400

=70/360 * 44/7 *6400

=19712000/2520

=7820km(3 sf)

Therefore the shortest distance between P and R = 7820km(3 sf)

(9c)

Circumference of the circle of latitude through R = 2πr where r = RcosΦ

= 2πRcosΦ

= 2 × 22/7 × 6400cos35.4

=281600/7(0.8151)

= 229,532.16/7

= 32,790.3

= 32,800km (3 s.f)

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

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Author: Mr. Dan

Posted on: 00:04:21 Sat, 30 Nov 2019

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Call: 08064643382 Now..

*MATHS OBJ*

1-10: BCBEDDCCDC

11-20: DEBCBBDDEC

21-30: BCBCCBBEBD

31-40: EBABEDCDEB

41-50: DDAECACBAB

51-60: DDBCEACBBE

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

(1)

√((0.0024)×35000)/0.0105

No | Log

0.024 |_3.3802

35000 |4.5441

- |=1.9243

0.0105|_2.0212 -

|3.9031 ÷2

|1.9516

√((0.0024)×35000)/0.0105

Antilog of .9516 =89.45

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

(2a)

Given that the roots of the equation are

X = -2/3 and X = -3/2

3x = -2 and 2x = -3

3x + 2 = 0 and 2x + 3 =0

(3x+2)(2x+3) = 0

6x² + 9x + 4x + 6 = 0

6x² + 13x + 6 = 0

(2b)

R = [3 4 0 ]

[2 0 3 ]

[1 2 2]

(2ci)

2/3R

=2/3[3 4 0]

[2 0 3]

[1 2 2]

= [⅔(3) ⅔(4) ⅔(0)]

[⅔(2) ⅔(0) ⅔(3)]

[⅔(1) ⅔(4) ⅔(2)]

= [2 8/3 0 ]

[4/3 0 2 ]

[2/3 4/3 4/3]

(2cii)

|R|

= |3 4 0|

|2 0 3|

|1 2 2|

=3|0 3| -4|2 3| +0|2 0|

|2 2| |1 2| |1 2|

=3(2×0-3×2)-4(2×2-3×1)

+0(2×2-0×1)

=3(0 - 6)-4(4 - 3) +0(4 - 0)

=3(-6) -4(1) +0(4)

= -18 - 4 + 0 = -22

(2ciii)

The transpose of R

= [3 2 1]

[4 0 2]

[0 3 2]

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

(3a)

Given : R(3,5) and S(-2, -6)

equation to line through them is :

y-5/x-3 = -6-5/-2-3

y-5/x-3 = -11/-5

y-5/x-3 = 11/5

5(y-5)= 11(x-3)

5y-25 = 11x-33

5y-11x = 25-33

5y-11x = -8 or 11x-5y = 8

(3b)

RS= √(X1X2)^2 (y1y2)^2

√(-2-3)^2 + (-6-5)^2

√(-5)^2 + (-11)^2

√25 + 121

√146

= 12.08

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

(4a)

Given: curve; y x² - 3x

gradient ; dy/dx = 2x - 3

At (2-2), gradient = 2(2) - 3

4-3 = 1

(4b)

Given; y= 1+x²/1-x²

dy/dx = (1-x²)(2x) - (1+x²)(-2x)/(1-x²)²

= (1-x²)(2x) - (1+x²)(2x)/(1-x²)²

= 2x(1-x² + 1+x²)/(1-x²)²

= 2x(2)/(1-x²)²

= 4x/(1-x²)²

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

(5)

No of blue balls = 6

No of red balls = 10

(i)

Prob (2 balls of some colour)

= BB or RR

Total no of balls = 6+10 = 16

BB or RR

(6/16 × 5/15) + (10/16 × 9/15)

=30/240 + 90/240

=120/240 = 1/2

(ii)

Prob (2 balls of different colours)

= BR or RB

= (6/16 × 10/15) or (10/16 × 6/15)

= 60/240 + 60/240 = 120/240

=1/2

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

(6a)

27^(2x+1) × 3^-x = 81^(x-2)/9^(x+2)

= 3^3(2x+1) × 3^-x = 3^4(x-2)/3^2(x+2)

=3^6x+3-x = 3^4x-8-2x-4

5x+3 = 2x - 12

5x - 2x = -12-3

3x = -15

3x/3 = -15/3

X = -5

(6b)

X/x+101 = 11/1000

Since all the members are in binary, convert all to denary (base 10)

Xbase2 = Xbase10

101base2 = (1×2^2)+(1×2^0) = 4+1 = 5base10

11base2 = (1×2¹)+(1×2raise to power 0) = 2+1 = 3base10

1000base2 = 1×2^3 = 8base10

X/X+101 = 11/1000 --> X/X+5 = 3/8

3(x+5) = 8(x)

3x+15 = 8x

15 = 8x - 3x

15 = 5x

15/5 = 5/5

X = 3

Convert X=3 to base 10 to base 2

2|3

2|1R1

|0R1

.:. x = 11

(6c)

Given that log5 base 10 = 0.699

and log3 base 10 = 0.477

10

Log75 base 10 = log(3×5×5) base 10

=log(3×5²) base 10

=log3 base 10 + 2log5 base 10

=0.477 + 2(0.699)

= 0.477 + 1.398

= 1.875

log75 base 10 = 1.875

=================÷÷÷

(7a)

Given that Y=2x2+7x-6

To find the gradient of the curve at the point x=3

dy/dx=4x+7 (at x=3)

dy/dx=4(3)+7

dy/dx=19

(7bi)

Ade bought 7kg of maize + 4kg of meat=N4240

Kemi bought 3kg of maize + 5kg meat =N4610

Let x = 1kg of maize and y=1kg of meat

Therefore 7x+4y=4240-----(eq1)

3x+5y=4610---------(eq2)

Substituting simultaneously

Multiply eq1 by 3 and eq2 by 7

21x+12y=12720---eq3

21x+35y=32270---eq4

Substract eq3 from eq4

23y=19550

23y/23=19550/23=850

y=850

Substitute y=850 in1

7x+4y=4240

7x+4(850)=4240

7x=4240-3400

x=840/7

=120

Hence the total cost price per Kg of maize is N120.00 while the total cost per kg of meat is N850.00

(7bii)

Total cost of 10kg of maize and 5kg of meat

=10x+5y

=10(120)+5(850)

1200+4250

=N5450

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

(9)

Draw the diagram

Q(34.5°N, 22.3°W)

P(34.5°N, 38.7°E)

R(35.4°S, 38.7°E)

Distance between P and Q along the parallel of latitude

D = Φ/360 × 2πr

Where r = RcosΦ

Φ = 38.7° + 22.3° = 61°

Q = 34.6°

R = 6400km

D = 61/360 × 22/7 ×6400cos34.6

D =1717760(0.8231)/252

D = 5,610.66768

= 5,610km

(9b)

|PR|=(34.6+35.4)/360 * 2 *22/7*6400

=70/360 * 44/7 *6400

=19712000/2520

=7820km(3 sf)

Therefore the shortest distance between P and R = 7820km(3 sf)

(9c)

Circumference of the circle of latitude through R = 2πr where r = RcosΦ

= 2πRcosΦ

= 2 × 22/7 × 6400cos35.4

=281600/7(0.8151)

= 229,532.16/7

= 32,790.3

= 32,800km (3 s.f)

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

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