**MATHS OBJ:
1-10: ACBCDDCBAA
11-20: CDCABCCCAC
21-30: DADBABCDAD
31-40: BADADBCACB
41-50: ADDDBDADCA**

1)

On February 28th 2012, value = (100-30/100) * #900,00.00

= 70/100 * #900,00

= #630,000.00

On february 28th 2013, value = (100-22/1000 * #630,00

= 78/100 8 #630,000

= #491,400

On february 28th 2014, value = 78/100 8 #491,400

=383,292

On february 28th 2015, value = 78/100 * #383,292

= #298,967.76

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

NO2)

Given that y = 2pxˆ² – p² x – 14

AT (3, 10)

10 = 2p(3)² – p² (3) – 14

10 = 18p – 3p² – 14

3p² – 18p + 24 = 0

p² – 6p + 8 = 0

using factor method,

p² – 2p -4p + 8 = 0

p(p-2) – 4(p-2) = 0

(p-4)(p-2) = 0

p-4 = 0 or p-4 = 0

p= 4 or p =2

2b) The lines must be solved simultenously

3y – 2x = 21 ——- (1)

4y + 5x = 5 ——-(2)

using elimination method,

(4) 3y – 2x = 21

(30 4y + 5x = 5

12y – 8y = 84 ——— (3)

12y + 15x = 15 ——-(4)

equ (4) minus equ(3)

23x = -69

x = -69/23

x = -3

Put this into equation (1)

3y -2(-3) = 21

3y = 6 = 21

3y = 21 -6

3y = 15

y =15/3

y = 5

coordinates of Q is (-3, 5)

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

Number 3 ( Image )

(3a)

The diagonal = 10.2m and 9.3cm

Using Pythagoras theory

Ac² = 10.2² + 9-3²

Ac² = 104.04 + 86.49

Ac² = 190.53

Ac² = √190.53

Ac² = 13.80

(3b)

DRAW THE DIAGRAM

Using Pythagoras theory

5² = 3² + x²

x² = 5² – 3²

X²= 25 – 9

X² = √16

X= 4cm

CosX = adjacent/hyp

= 4/5

Tan X = opp/adj. = 3/4

5cos x – 4tan x

5(4/5)- 4(3/4)

20/5 – 12/4

4-3= 1

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

Number 4 ( image )

4ai)

sum of angle in a D =180degree

xdegree + 90degree + 180degree – (3x+15)=180degree

xdegree + 90degree + 180degree – 3x+15=180degree

-2x=180degree – 255

+2x/2=+75/2

x=37.5

4aii)

<RsQ =180 – (3x+15)

<RsQ =180-(3*37.5+15)

=180-(112.5 + 15)

=180 – 127.5

<RsQ= 52.5degree

4b)

2N4seven =15Nnine

2*7^2+N*7^1+4*7degree =1*9^2 + 5*9^1+N*9degree

9*49+N*7+4*1=1*81+5*9+N*1

98+7N+4=81+45+N

7N+102=126+N

7N-N=126-102

6N/6 =24/6

N=4

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

Number 5

(5a)

m+n+s+p+q/5=12

m+n+s+p+q=60……(1)

Now;

(m+4)+(n-3)+(5+6)+p-2)+(q+8)/5

=(m+n+s+p+q)+(4-3+3+6-2+8)/5

=60+13/5

=73/5

=14.6

(b)

75% of 500 = 375 people

Number of people above 65 years = 500-375

=125

25% of 500 = 125

Number of people below 15 years = 125

Number between 15 years and 65 years

=500-(125+125)

=500-250

=250 people

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

Number 6 ( Image )

(6a)

Draw the Venn diagram

Let the number of cars with faults in brakes only be x

(6b) Number that passed = 60% × 240 = 144

Number that failed =

240 – 144 = 96

Therefore; 28+2x+x+14+6+6-x+8 = 96

2x + 62 = 96

2x = 96 – 62

2x = 34

X = 34/2

X = 17

(i) faulty brakes cars = 8+6+x+6-x

= 8+6+6

=20

(ii) only one fault = 28+x+2x

=28+3x

=28+3(19)

=28+51

= 79

====================={=

(7a)

(y-y1)/(x-x1)=(y2-y1)/(x2-x1)

(y-5)/(x-2)=(-7-5)/(-4-2)

(y-5)/(x-2)=-12/-6

(y-5)/(x-2)=2

Cross multiply

y-5=2(x-2)

y-5=2x-4

2x-y-4+5=0

2x-y+1=0

(7bi)

DRAW THE DIAGRAM

(7bii)

(I)

p^2=q+r^2-2qrcosP

p^2=8^2+5^2-2*8*5*cos90

p^2=64+25-0

p^2=89

p=sqroot(89)

p=9.4339km

therefore |QR|=9.43km(3 sf)

(II)

q/sinQ=p/sinP

8/sinQ=9.4339/sin90

sinQ=(8*sin90/9.4339

sinq=(8*1)/9.4339 =0.8480

Q=sin^1(0.8480)=57.99 degrees

but Q=30+ A

A=Q-30

=57.99-30

A=27.99 degrees

The bearing of R from Q

=180-A

180-27.99

=155.01

=>152 degrees

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

(8a)

Cost price for Lami= #300.00

Profit made by lami = x%

Ie selling price for lami=(100+x/100)×#300

=#3(100+x)

=#(300+3x)

Bola’s cost price = #3(100+x)

Profit made by bola =x%

Selling price for bola =(100+x/100)×#3(100+x)

=#3/100(100+x)²

James cost price =#3/100(100+x)²=300+(6x+3/4)

expanding;

3/100(10000+200+x²) = 300+3/4+6x

3(10000+200x+x²)=30000+75+600x

30000+600x+3x²=30000+75+600x

3x²=75

X² = 75/3

X² = 25

X = square root 25

X = 5

(8b)

3x-2<10+x<2+5x

3x-2<10+x & 10+x<2+5x

3x-x<10+2 & 10-2<5x-x

2x<12 8<4x

X<12/2 4x>8

X<6 x>8/4

X>2

Also; 3x-2<2+5x

-4<2x 2x > -4

X > -2

Therefore; Range is -2

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

(9a)

Draw the diagram

Angles PTR and PSR are similar

|PT|/|PS| = |TQ|/|SR|

In angle PTR

|TQ|²=|PT|²+|PQ|²-2|PT||PQ|cos30degrees

=4²+6²-2×4×6×cos30

=16+36-48×0.8660

=52-41.568

=10.432

|TQ|=√10.432 =3.22cm

4/10 = 3.22/|SR|

4|SR| = 10×3.22

|SR| = 32.2/4

|SR| = 8.05cn

Approximately 8cm(to the nearest whole number)

(9b)

Atqrs = AΔPSR – AΔPTR

AΔPTR = 1/2×4×6×sin30

=2×6×0.5

=6cm²

AanglePTQ/AanglePSR = |PT|²/|PS|²

6/AanglePSR = 4²/10²

6/AanglePSR = 16/100

16×AanglePSR = 6×100

AanglePSR = 600/16 = 37.5cm2

ATQRS = 37.5 – 6

=31.5cm2

=32cm2

Number 10( image )

(10a) Using Pythagoras theorem from SPQ

|SQ|^2 = 12^2 + 5^2

= 144+25

=169

SQ= sqroot of 169

= 13cm

Sin tita= 5/13 = 0.3846

Tita= Sin^-1(0.3846)

= 22.6degrees

From PRQ

Sin tita= |PR|/12

Sin 22.6 = PR/12

Sin 22.6= PR/12

PR= 12xsin 22.6

PR= 12×0.3843

PR= 4.61cm

(10bii)Let the height at which m touches the wall= y

Cos x^degrees= 8/10= 0.8

x^degrees= Cos^-1(0.8)

= 36.87degrees

Sin x^degrees = y/12

Sin 36.87= y/12

y= 12xsin36.87

y= 12×0.60000

y= 7.2m

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all answer are correct