Monday 4th June 2018
MATHEMATICS OBJ:
110 CDAAEABAEC
1120 AEDDCDCDCC
2130 CEBDEDCBBC
3140 CBEEECBDCC
4150 DBCBCDDBCA
5160 BCBDCDCCEC
110 CDAAEABAEC
1120 AEDDCDCDCC
2130 CEBDEDCBBC
3140 CBEEECBDCC
4150 DBCBCDDBCA
5160 BCBDCDCCEC
Obtuse <BOD + Reflex<BOD = 360degrees (angle at a point)
105 + reflex<BOD = 360degrees
Reflex <BOD= 360  105
=255°
Now 2w = reflex<BOD(angle at centre = twice angle at circumference)
2w =255°
W = 255/2 =127.5°
Also 2x = obtuse<BOD(angle at centre = twice angle at circumference)
2x = 105°
X = 105/2 = 52.5°
Now EDF = y(base angles of an isosceles triangle)
BED=X=52.5°(angles in the same segment)
EFD+EDF=BED (sum of interior angles of a triangle equal exterior angle)
Y+y = 52.5°
2y = 52.5°
Y = 52.5°/2
=26.25°
(10b)
Draw the diagram
Opp/adj = TanR
TB/BR = TanR
100/BR = Tan60°
BR = 100/tan60
BR = 100√3
BR = 100√3 * √3/√3
=100√3/3m OR 57.7m
11a)
x+y/2 =11
x+y= 11*2
x+y= 22 (1)
xy= 4 (11)
x+y = 22(1)

xy= 4(11)
____
2y = 18
y= 18/2
y=9
Substitute y=9 in equ 1
x+9=22
x=229
x=13
x=13, y=9
x+y= 13+9= 22
Sum of the two number
(11b)
(6x + 3) dx
(6x + 3)dx
(6x +3)^6  (6x + 3)^1
(6 x + 3)^5
(7776x^5 + 243)
38,880x/6 + 243
6480 x^6 + 243x
9(720x^6 + 27x)
(11c)
y = x² + 5x  3 (x = 2)
y = 2² + 5(2)  3
y = 4 + 10  3
y = 14  3
y = 11
Gradient of the curve = 11
Log 10(2010)log10(+3)=log105
(2010/+3)=log10 =5
2010/+3=5
5(+3)=2010
5+15=2010
15+10=205
25=15*
=25/15
*=5/3=1 2/3
1b)
Discount percent =15%
Discount amount =#600
Actual amount paid on the article =?
Original amount on the article =
15%=#600
15/100 =600
15=600100
15*=60000
*=60000/15
*=#4,000
Therefore actual amount paid on the article
=#4,000#600
=#3,400
Actual amount paid on the article =#3,400
No 4(i)
length of Arc of the sector
Titter= 72?, r = 14cm
L= titter / 360 x 2 pie r
==> L= 72/360 x 2 x 22/7 x 14
=44352/2520 = 17.6cm
(ii) perimeter of the sector
Perimeter = titter/360 x 2 pie r + 2r = 17.6 +(2×14) =17.6+28= 45.6cm
iii) Area of the sector
Area = Titter/360 x pie r? =
72/360 x 22/7 x (14)? = 72 x 22 x 196/2520
Area= 310464/2520 = 123.2cm?
5a)
Mode = mass with highest frequency = 35kg
Median is the 18th mass
= 40kg.
(5b)
In a tabular form
Under Masses(x kg)
30,35,40,45,50,55
Under frequency(f)
5,9,7,6,4,4
Ef = 35
Under XA
10, 5, 0, 5, 10, 15
Under F(XA)
50, 45, 0, 30, 40, 60
Ef(X  A) = 35
Mean = A + (Ef(X  A)/Ef)
= 40 + 35/35
= 40 + 1
= 41kg
7a)
A) T3=6 & T7 =30
I)common difference using Tn=a+(n1)d
In the 3rd term; n =3
=>T3=a+(31)d=6
=>a+2d=6 equation (1)
In the 7th term ;n =7
=> T7=a+(71)d=30
=>a + 8d=30 equation (1) & (2)
Simultaneous
A+2d =6 equation (1)
A+ 8d=30 equation (2)
0+(16d) =24
=> 6d= 24
=>d = 24/6 =4
II) first term put d=4 into equation
5a)
Mode = mass with highest frequency = 35kg
Median is the 18th mass
= 40kg.
(5b)
In a tabular form
Under Masses(x kg)
30,35,40,45,50,55
Under frequency(f)
5,9,7,6,4,4
Ef = 35
Under XA
10, 5, 0, 5, 10, 15
Under F(XA)
50, 45, 0, 30, 40, 60
Ef(X  A) = 35
Mean = A + (Ef(X  A)/Ef)
= 40 + 35/35
= 40 + 1
= 41kg
11a)
x+y/2 =11
x+y= 11*2
x+y= 22 (1)
xy= 4 (11)
x+y = 22(1)

xy= 4(11)
____________
2y = 18
y= 18/2
y=9
Substitute y=9 in equ 1
x+9=22
x=229
x=13
x=13, y=9
x+y= 13+9= 22
Sum of the two number
(11b)
(6x + 3) dx
(6x + 3)dx
(6x +3)^6  (6x + 3)^1
(6 x + 3)^5
(7776x^5 + 243)
38,880x/6 + 243
6480 x^6 + 243x
9(720x^6 + 27x)
(11c)
y = x² + 5x  3 (x = 2)
y = 2² + 5(2)  3
y = 4 + 10  3
y = 14  3
y = 11
Gradient of the curve = 11
2a)
(X^2 Y^3 Z)^3/4/X^1 Y^4 Z^5
= (X^2)^3/4/X^1 * (Y^3)^3/4/Y^4 * Z^3/4/Z^5
= X^3/2/X^1 * Y^9/4/Y^4 * Z^3/4/Z^5
=X^3/2+1 * Y^9/44 * Z^3/45
=X^5/2 * Y^25/4 * Z^17/4
=X^10/4 * Y^25/4 * Z^17/4
=(X^10/Y^25 Z^17)^1/4
(2b)
√2/k + √2 = 1/k  √2
Multiply both sides by (k+√2)(k√2)
√2(k√2) = k+√2
√2k√2 = k+√2
√2kk = 2+√2
K(√2 1) = 2+√2
K = 2+√2/√21
K = (2+√2)/1√2
Rationalizing
K = (2+√2) * 1+√2/1√2
K = (2+√2)(1+√2)/1  2
K = (2+√2)(1+√2)
K = 2+2√2 + √2+2
K = 4+3√2
(8)
x=a+by(eqi)
when y=5 and x=19
19=a+5b(eqii)
when y=10 and x=34
34=a+10b(eqiii)
solving eqii and eqiii
a+10b=34
a+5b=19
=>5b=15
b=15/5=3
putting b=3 in eqii
19=a+5(3)
19=a+15
a=1915
a=4
(8i)
Putting a=4 and b=3 in eqi
x=4+3y
This is the relationship between xand y
(8ii)
When y=7
x=4+3(7)
x=4+21
x=25
WHATSAPP US
+++++++++++++++++++++++++++++++++++
MATHS ANSWERS TO COME BY 4AM7AM.. NO DOUBT ABOUT THAT..
NO SUBSCRIPTION NO EXPO!
MATHS ANSWERS TO COME BY 4AM7AM.. NO DOUBT ABOUT THAT..
NO SUBSCRIPTION, NO EXPOS.
DON’T EVEN EXPECT ANSWERS ONLINE FOR PUBLIC VIEW
((((((((( HOW TO SUBSCRIBE ))))))))))))))
ONLINE ANSWERS/PASSWORD ==== #600 MTN CARD
MATHS ANSWERS TO COME BY 4AM7AM.. NO DOUBT ABOUT THAT..
NO SUBSCRIPTION NO EXPO!
MATHS ANSWERS TO COME BY 4AM7AM.. NO DOUBT ABOUT THAT..
NO SUBSCRIPTION, NO EXPOS.
DON’T EVEN EXPECT ANSWERS ONLINE FOR PUBLIC VIEW
((((((((( HOW TO SUBSCRIBE ))))))))))))))
(1) Mathematics
DIRECT MOBILE/SMS ==== #1000 MTN CARDONLINE ANSWERS/PASSWORD ==== #600 MTN CARD
Get more stories like on facebook
0 Comments