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Author Topic: Anambra State University (ANSU) 2021/2022 Post UTME Form, Direct entry is out for sale - Courses, fees Pre-degree, Diploma  (Read 10100 times)

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Offline Ifincy

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  • HomeTown: Okigwe
(1b)
<AEF=<EBH (corresponding angle)
<EBH=3x(degree)
<GHC=<HBE=7x(degree)(same reason)
120(degree)+7x(degree)+3x(degree)=
360(degree) angle at a
point
120+10x=360
10x=360-120
10x=240
x=240/10=24(degree)
(2a)
3sqrt(25*3)-sqrt(4*3)+sqrt(27*4)
15sqrt(3)-2sqrt(3)+2sqrt(27)
15sqrt(3)-2sqrt(3)+2sqrt(9*3)
15sqrt(3)-2sqrt(3)+6sqrt(3)=19sqrt(3)
(2b)
124n=232five
(1*n^2)+(2*n^1)+(4*n^o)=(2*5^2)
+(3*5^1)+(2*5^o)
n^2+2n+4=50+15+2
n^2+2n+4-67=0
n^2+2n-63=0
n(n-7)+9n-63=0
n(n-7)+9(n-7)=0
(n+9)(n-7)=0
n=-9 or n=7:. n=7

Offline Ifincy

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(3a)
Let p=1/x ,q=1/y
p+q=5 ---->equ(1)
q-p=1 ---->equ(2)
p+q=5 ---->equ(1)
p+q=1 ---->equ(2)
Add equ(1) and equ(2)
2q=6 , q=6/2=3
Substitute q=3 into equation(1)
p+q=5
p+3=5
P=5-3
p=2
Recall p=1/x , 2=1/x, x=1/2
q=1/y, 3=1/y , y=1/3
Therefore: x=1/2 , y=1/3

Offline Ifincy

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(3a) Let p=1/x ,q=1/y p+q=5 ---->equ
(1) q-p=1 ---->equ(2) p+q=5 ---->equ(1) p
+q=1
---->equ(2) Add equ(1) and equ(2) 2q=6 ,
q=6/2=3 Substitute q=3 into equation(1) p
+q=5
p+3=5 P=5-3 p=2 Recall p=1/x , 2=1/x,
x=1/2
q=1/y, 3=1/y , y=1/3 Therefore: x=1/2 ,
y=1/3
(3b) Where the road is good 72:x/0.75
x=72*0.75 x=54km Where the road is bad
48:x/0.75 x=48*0.75 x=36km Number of
kilometers of good surface is 54km
(4a) Area(A)= (tita/360*pieR^2) -(1/2R^2Sin
(tita)) 504:((90/360)*22/7*r^2)-
(1/2*r^2*sin90)
504:(0.25*3.142*r^2)-(0.5r2) 504=
0.7855r^2 -
0.5r^2 504= 0.2855r^2 r^2= 504/0.2855
r^2=1765.32 r:sqrt(1765.32) r=42cm

Offline Ifincy

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(5) Pr(x)=0.5, Pr(X^1)=0.5 Pr(Y^1)=0.3 , Pr(y)
=0.7 (a) Pr(both contracts):-pr(x)*Pr(y)
=0.5*0.7=0.35 (5b) Pr(exactly one of the
contracts)= Pr(X)*Pr(Y^1) +Pr(X^1)*Pr(y) :
(0.5*0.3)+(0.5*0.7) =0.15+0.35 =0.5 (5c) Pr
(Neither of the contrasts) = pr(X^1)*Pr(Y^1)
=0.5*0.3 =0.15
(6a) 3/(2p-1/2):(1/3)/(p/4+1/2). 6/(4p-1)
=4/(3p
+6). 6(3p+6)=4(4p-1). 18p+36=16p-4.
2p=-40,
p=-20.
(12a) Given, 5cos(x+8.5)-1=0 C0s(x+8.5)
^o=1/5
cos(x+8.5)^o=0.2 x+8.5^o=Cos^-1(0.2) x
+8.5^o=78.46 x=78.46-8.5 x=69.96^o x=
70(degree)
(12bii) QR^2:(PR^2)+(PQ^2)
-2PR*PCOS(RPQ) P^2=32^2+24^2-2(32)(24)
COS45(DEGREE) P^2=1600-1536*0.7071
P^2=1600-1086.1 P^2=513.88 P:sqrt
(513.88)
p=22.67km Distance QR=22.67km (12biii)
32/sin
(tita)=22.67/sin45(deg), sin(tita)=0.9981
tita:sin^-1(0.9981) tita=86(degree)

Offline Ifincy

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Offline Ifincy

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  • HomeTown: Okigwe

Offline finboy

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  • Location: enugu
  • HomeTown: new haven
(9 ai ) 16 - 2 x + 6 + x + 19 - 3 x + 5 x
+ 8 x + 4 x + 7 x + 4 = 125 45 + 20 x = 125
20 x = 125 - 45 20 x = 80 x = 80 / 20 x = 4 (9 aii )
n (PUQnR )= 16 - 2 x + 5 x + 6 + x = 22 + 4 x = 22 +
4 (4 )= 38 (9 bi) < WYZ =< WXY = 50 (deg )
corresponding angle (9 bii) < W=
50 (degree) Alternate angle YZE =
50 (degree) (< S in the same segment ) YEZ
+ 50 (degree )+ (50 degree )= 180 (de

Offline igbob

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