ondo21
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MATHEMATICS ANSWERS
OBJ:

1-10: DBABCBAEDD
11-20: BEBACABCCB
21-30: DACAEDBDEA
31-40: BEEDDDEBBB
41-50: CAADCDDDBC

ESSAY:

Answer All The (5) Question in This Section.

(1B)
X={2,4,6,8}
Y={2,3,7,9}

(XUY) = {2,3,4,6,7,8,9}
Y={2,3,7,9}

Z={4,5,6,7,8}(YUZ) = {2,3,4,5,6,7,8,9}

{XUY} n {YUZ}={2,3,4,6,7,8,9}ANS

…………………………….

(2A)
A=35
X=7
Y=4
A proportion X 1/Y^2
A = XK/Y^2
When K is Constant
35 = 7K/42
35= 7K/16
35*16=7K
560 = 7K
K = 560/7
K=80
.:.
A = XK/Y^2
A=5 * 80/8^2
A=400/64
A=6.25

(2B)
If S = ut + 1/At^2 Make T the subject of formula of the equation.

S = ut + 1/At^2
SAt^2 = ut + 1
SAt^2 – ut =1
SA – u (t^2-t) =1
t^2 – t = 1/SA – u
t = 1/SA-u
(3) Using Ruler And a Pair of Compasses Only,construct triangle ABC with /AB/­=10cm, /BC/=12cm and =120 degree, construct the locus l of ponts equidistant from A and B.
DRAW IT LIKE THE ABOVE IMAGE ON YOUR ANSWER SHEET.

(4A) Using Completing The Square Method Solve The equation. X^2 – X – 12 = 0
X^2-X-12=0
X^2-X/2 – (1/2)^2 = 12/2 + 1/4
(X^2 – 1/2)^2 = 24+1/4 = 25/4
(X-1/2)^2 = √25/4
X-1/2 = ± √25/4
X – 1/2 = ± 5/2
X = 1/2 + 5/2OR1/2 – 5/2
X = 1+5/2OR1-5/2
X = 6/2 OR -4/2X=3OR-2ANS

 

SECTION B: Answer (5) Question in This Section.
9A)AP = Tn + A (n – 1)d
5th term = T5 = a + 4d
14th term = T14 = a + 13d

T5 = a + 4d = 28……………….(equa 1)
T14 = a + 13d = 82 …………..(equa 2)

Solving the simultaneous equationsSubtract equation (1) from equation (2)
a + 4d =28………(1)
a + 13d = 82 …..(2)
9d = 54

:. d = 54/9
d=6
Substitute (d) in equation (1)
a + 4 (6) = 28
a + 24 = 28
a = 28 – 24
a=4
To find the 21st term
T21 = a + 20d
=4 + 20 (6)
= 4 + 120
= 124

12A)
Let the distance be a,c
In ∆ABC ^A=90-205=115°(A FROM glasses emoticon
also ABC=180-205=25°
:.B=60-25=35

:.C=180-(35+115)=180­-150=30°

USING SINE RULE
(i)a/sinA =b/sineB
a=B sine a/sine B= 15 sin 115°/­sin30°=15×0.9404/­0.9880=14.181/0.9880
=14.20a=14.20kmii)si­milarlySin b=C sin B/sin C =15 sin
35°/­sin30°=15×0.4281/­0.9880
=6.4215/0.9880
=6.499km
:.distance of B from A =14.20km
:.bearing of B from A= 6.499 km