Maths OBJ:

1ABBCDDBBAA

11ADAACCDCCC

21DADBABCDAD

31BADADBCACB

41ADDDBDADCA

Maths OBJ:

1ABBCDDBBAA

11ADAACCDCCC

21DADBABCDAD

31BADADBCACB

41ADDDBDADCA

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** 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

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

**5a)
Given M,N,S,P and Q J mean = 12
mean = M+n+s+p+q/5=12 (eq1)
m+4+n+3+s+6+p-2+q+8/5=m+n+s+p+q+13/5 (eq2)
from (eq1) m+n+s+p+q=5*12=60
put in (eq2) m+n+s+p+q+13/5
=60+13/5=73/5=14.6**

5b)

total population = 500 people

75th percentile = 75/100*500=373people

their age is 65years

25th percentile = 2

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

MAHTEHMATICS ANSWER AND QUESTION 2018

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