MATHEMATICS ANSWER
Maths Obj Answer
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Maths Essay Answer
POSTED ANSWER 1,2,3,4,5,6,10,11,12
^ means raise to power
PLS DON'T WRITE (power) on ur paper oooo.
(power) means Raise to power.
Anywhere u see me write (power), just raise d next digit up.
Example: m power 2 means that '2' will be on top of 'm'.
QT(power)2 means that '2' will be on top of QT.
I KNEEL DWN DE BEG U, ABEG READ DIS WELL
(1)
Given, m^2 - 2mn n^2 - 9r^2
m^2 - 9r^2 n^2- 2mn
m^2 - (3r)^2 n(n-2m)
(m-n)^2 - 9r^2
=(m - n - 9r)(m -n 9r)
(1b)
5x - 4y=6 ----eqn1
3^3y -3x=3^-3
3y - 3x = -3
-3x 3y= -3 ----eqn2
5x - 4y =6 ------eqn1
-x y= -1 -------eqn1
5x - 4y =6
-5x 5y =-5
add eqn(1) and (2) together
y=1
substitute y=1 into eqn(2)
-x 1= -1
-x= -1-1, x= -2
x=2, y=1
(2a)
All taxable income is
= 10% 15% 20% 25%
= 70% of 850
= 70/100*850
= #595
(2b)
1: 70/100*120 =#84
2: 70/100*30= #21
3: 70/100*100 =#70
4: 70/1
(3a)
QT(power)2 =QS(power)2 ST(power)2
5(power)2 =4(power)2 ST(power)2
ST(power)2 =25 - 16
ST(power)2 =9
ST=sqrt (9)= 3cm
Cos(tita)=3/5
cos(tita)= 0.6
tita= cos(power)-1(0.6) = 53degree
cos53(degree)=SR/6
0.602 =SR/6
SR= 0.602*6 = 3.6cm
/TR/=3.6/Sin53(degree)=4.5cm
(3b)
Adj=12
Opp=5
Hyp=13
cosx=12/13
Sinx=5/13,
Tanx=5/12
Cosx - 2Sinx/2Tanx = 12/13-2(5/13) / 2(5/12)
= (12/13 - 10/13) / 5/6
= (2/13) / (5/6)
= 2/13*6/5
= 12/65
(4a)
Log10(75/10) - Log10(5/9)(power)2 Log10(100/243)
= Log10(75/10) - Log10(25/81) Log10(100/243)
= Log10(75/10*81/25) Log10(100/243)
= Log10(3*81/10) Log10(100/243)
= Log10(243/10) Log10(100/243)
= Log10( 243/10* 100/243)
=Log10(10)=1
(4b)
X= (10,11,12,13,14,15 )
Y= (2,4,6,8,10,12,14,16 )
U= (10,11,12,13,14,15,16,17,18,19,20 )
(i) XnY = (10,12,14 )
(ii)
X / / /' = (16,17,18,19,20 )
X / / /'nY= (16 )
n(X / / /'nY)= (16 )
(5a)
Considering change in WZX
ZX(power)2 = Zw(power)2 WX(power)2
ZX(power)2 = 6(power)2 8(power)2
= 36 64
ZX(power)2=100
ZX= sqrt(100)
=10cm
ZX= 2ZM
ZM= ZX/2
Zm=10/2 =5cm
Mx=5cm
OM is the height of the pyramid
OX(power)2= OM(power)2 MX(power)2
13(power)2= OM(power)2 5(power)2
169= OM(power)2 25
OM(power)2 = 169 -25
OM(power)2 = 144
OM= sqrt(144)
=12cm
height of pyramid = 12cm
(5b)
COS < 0XZ = 12/13
(5c)
Volume of pyramid = 1/3*Area of base * height.
Area of base = 6*8= 48cm2.
Volume of pyramid = 1/3*48*12.
=192cm3
(6a)
Total stds= 52
No of boys = 1/3*16 = 16/3
No of girls = 1/4*16=4
(6b)
Using Fn= ar(power)n-1
F3=ar(power)3-1= ar
a ar(power)2 = 40 ------eqn(1)
F4/F6 = 1/4 --> ar(power)3/ar(power)5= 1/4
1/r(power)2=1/4
r(power)2= 4
r = sqrt(4)
r=2
i. Common ratio(r)= 2
(ii)
Substitute r = 2 into eqn(1)
a ar(power)2 = 40
a(1 2(power)2) = 40
a(5) = 40
a= 40/5= 8
Fifth term (F5) = ar(power)4
= 8*2(power)4
=8*16
=128
(10ai)
1/2x - 5x/6 - 5/3 -2
(10aii)
----------------------------->
(-2, 0,2)
diagram
(10b)
/PQ/ = 10m
Let KR be the height of pole
tan 45(degree)= KR/PR
KR= tan 45(degree) *(10-x)
KR= (10 -x) --------------eqn(1)
tan 58(degree)= KR/x
KR= tan 58(degree) * X
KR= 1.6 *X
KR= 1.6x ---------------eqn(2)
equation(1) and(2)
10 - x= 1.6x
10= 1.6x x
2.6x = 10
x= 10/2.6
x= 3.85m
(i) Distance fom P to the pole is (10 - x) = (10 - 3.85)= 6.15m
(ii) Height of pole = KR= 1.6x
= 1.6 * 3.85
= 6.16m
(11a)
Pr(Manful pass) = 2/3
Pr(John)= 5/8
Pr(Ernest) = 3/4
Pr(all pass)= 2/3 * 5/8 * 3/4
= 5/16
(11b)
Tabulate
x: 1,2,3,4,5,6
f: 4,2,3,2,6,3
E(f) = 20
f1x1=4 , f2x2=4, f3x3=9 ,
f4x4 =8, f5x5=30, f6x6=18
Efx = 4 4 9 8 30 18 = 73
Cf = 4,6,9,11,17,20
Q1= Ef/4 = 20/4 = 5th
Q3= 3Ef/4 = 3*20/4 =15th
Q1 corresponds to 2
Q3 corresponds to 5
inter-quatile range = Q3 - Q1
= 5 - 2= 3
(12a)
(i)
(ii)
(12bi)
Given
T= sqrt(U/ (1/f 1/g)
square both sides
T(power)2 = (U/ (1/f 1/g)
f g/fg= U/T(power)2
f g = U/T(power)2(fg)
Ufg/T(power)2 - g = f
g(Uf/T(power)2 -1) =f
g= f/(uf/T((power)2 )-1)
(12bii)
T=3, f=4, u=5
g = 4/( 5*4/3(power)2) -1
g= 4*9/11 = 36/11
g= 3 (3/11)
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Maths Obj Answer
Verified MATHS OBJ:
1-10: BAABBCDBCC
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31-40: CACCDCDDDB
41-50: DDDABADCCD
Maths Essay Answer
POSTED ANSWER 1,2,3,4,5,6,10,11,12
^ means raise to power
PLS DON'T WRITE (power) on ur paper oooo.
(power) means Raise to power.
Anywhere u see me write (power), just raise d next digit up.
Example: m power 2 means that '2' will be on top of 'm'.
QT(power)2 means that '2' will be on top of QT.
I KNEEL DWN DE BEG U, ABEG READ DIS WELL
(1)
Given, m^2 - 2mn n^2 - 9r^2
m^2 - 9r^2 n^2- 2mn
m^2 - (3r)^2 n(n-2m)
(m-n)^2 - 9r^2
=(m - n - 9r)(m -n 9r)
(1b)
5x - 4y=6 ----eqn1
3^3y -3x=3^-3
3y - 3x = -3
-3x 3y= -3 ----eqn2
5x - 4y =6 ------eqn1
-x y= -1 -------eqn1
5x - 4y =6
-5x 5y =-5
add eqn(1) and (2) together
y=1
substitute y=1 into eqn(2)
-x 1= -1
-x= -1-1, x= -2
x=2, y=1
(2a)
All taxable income is
= 10% 15% 20% 25%
= 70% of 850
= 70/100*850
= #595
(2b)
1: 70/100*120 =#84
2: 70/100*30= #21
3: 70/100*100 =#70
4: 70/1
(3a)
QT(power)2 =QS(power)2 ST(power)2
5(power)2 =4(power)2 ST(power)2
ST(power)2 =25 - 16
ST(power)2 =9
ST=sqrt (9)= 3cm
Cos(tita)=3/5
cos(tita)= 0.6
tita= cos(power)-1(0.6) = 53degree
cos53(degree)=SR/6
0.602 =SR/6
SR= 0.602*6 = 3.6cm
/TR/=3.6/Sin53(degree)=4.5cm
(3b)
Adj=12
Opp=5
Hyp=13
cosx=12/13
Sinx=5/13,
Tanx=5/12
Cosx - 2Sinx/2Tanx = 12/13-2(5/13) / 2(5/12)
= (12/13 - 10/13) / 5/6
= (2/13) / (5/6)
= 2/13*6/5
= 12/65
(4a)
Log10(75/10) - Log10(5/9)(power)2 Log10(100/243)
= Log10(75/10) - Log10(25/81) Log10(100/243)
= Log10(75/10*81/25) Log10(100/243)
= Log10(3*81/10) Log10(100/243)
= Log10(243/10) Log10(100/243)
= Log10( 243/10* 100/243)
=Log10(10)=1
(4b)
X= (10,11,12,13,14,15 )
Y= (2,4,6,8,10,12,14,16 )
U= (10,11,12,13,14,15,16,17,18,19,20 )
(i) XnY = (10,12,14 )
(ii)
X / / /' = (16,17,18,19,20 )
X / / /'nY= (16 )
n(X / / /'nY)= (16 )
(5a)
Considering change in WZX
ZX(power)2 = Zw(power)2 WX(power)2
ZX(power)2 = 6(power)2 8(power)2
= 36 64
ZX(power)2=100
ZX= sqrt(100)
=10cm
ZX= 2ZM
ZM= ZX/2
Zm=10/2 =5cm
Mx=5cm
OM is the height of the pyramid
OX(power)2= OM(power)2 MX(power)2
13(power)2= OM(power)2 5(power)2
169= OM(power)2 25
OM(power)2 = 169 -25
OM(power)2 = 144
OM= sqrt(144)
=12cm
height of pyramid = 12cm
(5b)
COS < 0XZ = 12/13
(5c)
Volume of pyramid = 1/3*Area of base * height.
Area of base = 6*8= 48cm2.
Volume of pyramid = 1/3*48*12.
=192cm3
(6a)
Total stds= 52
No of boys = 1/3*16 = 16/3
No of girls = 1/4*16=4
(6b)
Using Fn= ar(power)n-1
F3=ar(power)3-1= ar
a ar(power)2 = 40 ------eqn(1)
F4/F6 = 1/4 --> ar(power)3/ar(power)5= 1/4
1/r(power)2=1/4
r(power)2= 4
r = sqrt(4)
r=2
i. Common ratio(r)= 2
(ii)
Substitute r = 2 into eqn(1)
a ar(power)2 = 40
a(1 2(power)2) = 40
a(5) = 40
a= 40/5= 8
Fifth term (F5) = ar(power)4
= 8*2(power)4
=8*16
=128
(10ai)
1/2x - 5x/6 - 5/3 -2
(10aii)
----------------------------->
(-2, 0,2)
diagram
(10b)
/PQ/ = 10m
Let KR be the height of pole
tan 45(degree)= KR/PR
KR= tan 45(degree) *(10-x)
KR= (10 -x) --------------eqn(1)
tan 58(degree)= KR/x
KR= tan 58(degree) * X
KR= 1.6 *X
KR= 1.6x ---------------eqn(2)
equation(1) and(2)
10 - x= 1.6x
10= 1.6x x
2.6x = 10
x= 10/2.6
x= 3.85m
(i) Distance fom P to the pole is (10 - x) = (10 - 3.85)= 6.15m
(ii) Height of pole = KR= 1.6x
= 1.6 * 3.85
= 6.16m
(11a)
Pr(Manful pass) = 2/3
Pr(John)= 5/8
Pr(Ernest) = 3/4
Pr(all pass)= 2/3 * 5/8 * 3/4
= 5/16
(11b)
Tabulate
x: 1,2,3,4,5,6
f: 4,2,3,2,6,3
E(f) = 20
f1x1=4 , f2x2=4, f3x3=9 ,
f4x4 =8, f5x5=30, f6x6=18
Efx = 4 4 9 8 30 18 = 73
Cf = 4,6,9,11,17,20
Q1= Ef/4 = 20/4 = 5th
Q3= 3Ef/4 = 3*20/4 =15th
Q1 corresponds to 2
Q3 corresponds to 5
inter-quatile range = Q3 - Q1
= 5 - 2= 3
(12a)
(i)
(ii)
(12bi)
Given
T= sqrt(U/ (1/f 1/g)
square both sides
T(power)2 = (U/ (1/f 1/g)
f g/fg= U/T(power)2
f g = U/T(power)2(fg)
Ufg/T(power)2 - g = f
g(Uf/T(power)2 -1) =f
g= f/(uf/T((power)2 )-1)
(12bii)
T=3, f=4, u=5
g = 4/( 5*4/3(power)2) -1
g= 4*9/11 = 36/11
g= 3 (3/11)
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