Wednesday, December 14, 2011
Sunday, December 4, 2011
Wednesday, November 30, 2011
Sunday, November 13, 2011
Saturday, November 12, 2011
Trigonmetry Formular O LeveL A Maths That are given
This are the formulars given in Further Trigonmetry
sin2x+cos2x=1
sin(A+B) = sinAcosB+ cosAsinB
sin(A-B) = sinAcosB - cosAsinB
cos(A+B) = cosAcosB-sinAsinB
cos(A-B) = cosAcosB+sinAsinB
tan(A+B) = [tanA+tanB)]/(1-tanAtanB)
tan(A-B) = [tanA-tanB)]/(1+tanAtanB)
sin2A=2sinAcosA
cos2A=cos2A-sin2A
cos2A=2cos2A-1
cos2A=1-2sin2A
tan2A=Tan2A/(1+tan2A)
sinA+sinB=2sin[(A+B)/2]cos[(A-B)/2]
sinA-sinB= 2cos [(A+B)/2] sin [(A-B)/2]
cosA+cosB=2 cos [(A+B)/2]cos[(A-B)/2]
cosA-cosB=-2sin[(A+B)/2] sin [(A-B)/2]
Formulars That are Not given
a cos θ ± b sin θ = R cos (θ ∓ α)
a sin θ ± b cos θ = R sin (θ ± α)
where
and a > 0, b > 0, α is acute
Wednesday, September 28, 2011
Friday, August 5, 2011
Summary for O Level trigonometry
Right angle triangles Use
trigonometric ratio
Tan @ = opposite/adjacent
Cos @ = adjacent/hypo
Sin @ = opposite/hypo
Non right angle use Sin and cosine rule
Sin A/a = SinB/b= SinC/c
trigonometric ratio
Tan @ = opposite/adjacent
Cos @ = adjacent/hypo
Sin @ = opposite/hypo
Non right angle use Sin and cosine rule
Sin A/a = SinB/b= SinC/c
Monday, June 13, 2011
Partial fractions
Partial fractions
Formulars of changing
A) 1/(x-2)(x-3) = A/(x-2) +B/(x-3)
C) 1/(2x2-3)(x-3) = A/(x-3) + Bx+C/2x2-3
D) To solve partial fraction Q(x)/F(x) when the degree of Q(x) is higher than F(x) , use long division
(x3 +2x2-3/ 2x2-3)
A) 1/(x-2)(x-3) = A/(x-2) +B/(x-3)
B) 1/(x-3)2= A/(x-3) + B/(x-3)2
C) 1/(2x2-3)(x-3) = A/(x-3) + Bx+C/2x2-3
D) To solve partial fraction Q(x)/F(x) when the degree of Q(x) is higher than F(x) , use long division
(x3 +2x2-3/ 2x2-3)
Sunday, May 15, 2011
Caculating time line
How to calculate 2004 oct 10 – 2011 may 16
2004 oct 10
2005 oct
2006 oct
2007 oct
2008 oct
2009 oct
2010 oct 10
nov 10
dec 10
jan
feb
march
april
may 10
11
12
13
14
15
16
its 6 years 7 months 6 days
Friday, May 6, 2011
A P6 Challenging question !
John had 993 tables n chairs at first. After he sold 2/5 of the tables n 5/8 of the chairs, he had 459 tables n chairs left. How many tables did he sell?
Wednesday, February 23, 2011
Tuesday, February 22, 2011
Secondary 3 A Maths Logarim
Secondary 3 A Maths
Chapter 2. Logarim
1)Changing indices to
log
ax=
b
x = logab
Example
103=1000
3=log101000
3x=2
x=log3 2
Do IT
Q1)3x=4
Q2)2y=2
2)Changing log to
indices
logab = y
b =
ay
example
log3 x = 2
x = 32
x = 9
Q1) log4y
= 3
Q2) log2x
=2
3) Changing Of Base
Log b
|
|
Logab =
|
-------
|
Log a
|
Log 5
|
|
Log45 =
|
-------
|
Log 4
|
Q1) find log67
Q2) find log89
<Logarithm/> tags
4) Power law of log
alog b = log ba
2log 3 = log 32
Take note
(Log b)2 = (log b)(logb)
not equal to log b2
5) Addition law of
log
Log (ab) = Log a +
logb
Log (12) = Log 3 +
log4
Log (3x) = log 3 +
logx
6) Subtraction law of
log
log (a/b) = log a –
log b
log (x/8) = log x-
log 8
Qn1) Log(4x)
Qn2) log (7/x)
Monday, February 14, 2011
Secondary 3 A Maths Logarim
Secondary 3 A Maths
Chapter 2. Logarim
1)Changing indices to
log
ax=
b
x = logab
Example
103=1000
3=log101000
3x=2
x=log3 2
Qns)
Q1)3x=4
Q2)2y=2
2)Changing log to
indices
logab = y
b =
ay
example
log3 x = 2
x = 32
x = 9
Q1) log4y
= 3
Q2) log2x
=2
3) Changing Of Base
Log b
|
|
Logab =
|
-------
|
Log a
|
Log 5
|
|
Log45 =
|
-------
|
Log 4
|
Q1) find log67
Q2) find log89
<Logarithm/> tags
4) Power law of log
alog b = log ba
2log 3 = log 32
Take note
(Log b)2 = (log b)(logb)
not equal to log b2
5) Addition law of
log
Log (ab) = Log a +
logb
Log (12) = Log 3 +
log4
Log (3x) = log 3 +
logx
6) Subtraction law of
log
log (a/b) = log a –
log b
log (x/8) = log x-
log 8
Qn1) Log(4x)
Qn2) log (7/x)
Tuesday, February 8, 2011
Thursday, January 13, 2011
Facebook message with a high performance student
This is how a great student interact on face book with a tutor :)
Difference between sulfur,sulphide,sulphite and sulfate
8:39pm
Hey
Tuesday, January 4, 2011
Quadratic expansion
(a + b) 2 = a 2 + b2 + 2ab
(a - b) 2 = a 2 + b2 -2ab
(a+b)(a-b) = a2 - b2
Example
(3a-4)(3a+4)
=32a2 -42
=9a2 - 16
(a - b) 2 = a 2 + b2 -2ab
(a+b)(a-b) = a2 - b2
Example
(3a-4)(3a+4)
=32a2 -42
=9a2 - 16
Chapter 1.2 Solving Non linear Equations
Solving Non linear Equations
Example1
2y-3x=1 ------ (1)
y+9x2=8 ------ (2)
from (1) y=(4x+1)/2 ----(3)
sub(1) into (3): (4x+1)/2 + 9x2 - 8 =0
x 2 3x+1+18x2
-16=0
18x2-15+3x=0
(18x-15)(x+1)=0
18x-15=0 or x+1=0
18x=15 x=-1
x=15/18
find y
Example 2
3x/y-2y/x=1 --------- (1)
5x-3y=12 ---------
(2)
(2): x=(12+3y)/5 ----------(3)
multiply (1) by xy
3x2 -2y2=xy ------------(4)
sub (3) into (4)
3(12+3y)
2/5 2 -2y2=y(12+3y)/5
multiply by 25
3(12+3y) 2 -50
y2=5y(12+3y)
3(144+72y+9y2)-50
y2=60y+15y2
432+216y+27y2
-50y2 -60y-15y2 =0
-38y2-60y+432=0
19y2 +30y-216=0
(19y+36)(y-6)=0
y=-36/19 or y=6
<Equations>
tags
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