Basic Trigonometry
Trigonometry from 2 word : Trigono and Metry
5 β α 3
4
sin β =
To solve this problem used trick
Sine
Opposite
Hyphotenuse
Cosine
Adjacent
Hyphotenuse
Tangent
Opposite
Adjacent
Kamis, 15 Januari 2009
The definition, explanation, and the examples of the terms question by Lina Dewi Kurniawati
1. Merupakan : is, are
Mathematics is one of the subject in Senior High School
2. Bilangan bulat : integer number
3. Besaran : value
4. Titik sudut : vertex
The triangle have 3 vertex
5. Bangun ruang : solid figure
Cube is a solid figure
6. Yang sama dan sebangun : congruent
Both of the triangle are congruent
8. huruf vocal : vowel
9. tembereng : curve of a circle
Curve of circle AB have area 4 cm2
10. Lingkaran dalam : subscribe circle
11. lingkaran luar : Inscribe circle
12. lingkaran dalam segitiga : subscribe circle of a triangle
Area of the subscribe circle of triangle is 5 cm2
13. sebidang : coplanar
Both of the points are coplanar
14. tegak lurus : perpendicular
Both of the straight lines are perpendicular.
15. berhimpit : impose
16. hasil kali dalam : inner product
17. himpunan bagian : subset
18. pengayaan : enrichment
Enrichment lesson is not always given
1. Merupakan : is, are
Mathematics is one of the subject in Senior High School
2. Bilangan bulat : integer number
3. Besaran : value
4. Titik sudut : vertex
The triangle have 3 vertex
5. Bangun ruang : solid figure
Cube is a solid figure
6. Yang sama dan sebangun : congruent
Both of the triangle are congruent
8. huruf vocal : vowel
9. tembereng : curve of a circle
Curve of circle AB have area 4 cm2
10. Lingkaran dalam : subscribe circle
11. lingkaran luar : Inscribe circle
12. lingkaran dalam segitiga : subscribe circle of a triangle
Area of the subscribe circle of triangle is 5 cm2
13. sebidang : coplanar
Both of the points are coplanar
14. tegak lurus : perpendicular
Both of the straight lines are perpendicular.
15. berhimpit : impose
16. hasil kali dalam : inner product
17. himpunan bagian : subset
18. pengayaan : enrichment
Enrichment lesson is not always given
I believe in me
In the video, the boy is Dalton. He is very convident. He can stand fearless in front of many people, twenty thousand people. He told that “You believe in me because I believe in me”. If we believe in ourself, we can do anything, we can be anything, become anything, and etc. “Do you believe in me?”. People believe in you, if you believe in yourself. We can do anything if we believe in ourself. Dalton believe in yourself, he need people believe in him, he need people help him.
What ever we do, not only for our generation but also to next generation. Dalton believe that what ever we do if we believe in ourself.
In the video, the boy is Dalton. He is very convident. He can stand fearless in front of many people, twenty thousand people. He told that “You believe in me because I believe in me”. If we believe in ourself, we can do anything, we can be anything, become anything, and etc. “Do you believe in me?”. People believe in you, if you believe in yourself. We can do anything if we believe in ourself. Dalton believe in yourself, he need people believe in him, he need people help him.
What ever we do, not only for our generation but also to next generation. Dalton believe that what ever we do if we believe in ourself.
How difficult to explain mathematics
My name is Isnaini Nurul Noviana. My partner is Esti Nur Kurniawati. We discuss about Differential Equations, the chapter is “The homogenous Linear equation with constant coefficients. In this section we consider the special of the nth-order homogenous linear equation in which all of the coefficients are real constants.
Distinct Real roots
If we have roots which have different, the general solution is y = C1e^M1X +C2e^M2X+…+CNe^MnX, where C1, C2, C3,…,Cn are arbitrary constants.
Repeated real roots
If we have roots which have repeated, the general solution is y = (C1+C2X)e^MX.
This activity was held on Friday, January, 9th 2009 at 01.10pm until 02.00pm, in KOPMA UNY. I started to discuss the differential equation, with the theorem, then exercise. With many exercise, Esti have can do the assignment in the differential equation book, so I am not review my explanation mathematics. This chapter have ever been learn.
My name is Isnaini Nurul Noviana. My partner is Esti Nur Kurniawati. We discuss about Differential Equations, the chapter is “The homogenous Linear equation with constant coefficients. In this section we consider the special of the nth-order homogenous linear equation in which all of the coefficients are real constants.
Distinct Real roots
If we have roots which have different, the general solution is y = C1e^M1X +C2e^M2X+…+CNe^MnX, where C1, C2, C3,…,Cn are arbitrary constants.
Repeated real roots
If we have roots which have repeated, the general solution is y = (C1+C2X)e^MX.
This activity was held on Friday, January, 9th 2009 at 01.10pm until 02.00pm, in KOPMA UNY. I started to discuss the differential equation, with the theorem, then exercise. With many exercise, Esti have can do the assignment in the differential equation book, so I am not review my explanation mathematics. This chapter have ever been learn.
Minggu, 21 Desember 2008
Assignment I
Name : Isnaini Nurul Noviana
NIM : 07305141025
CHAPTER I
INDUCTION OF MATHEMATICS AND BINOMIAL THEOREM
A. Induction of mathematics
Induction of mathematics is one of the proof method from many theorems in theory of number or in other mathematics. Whereas binomial theorem, besides as basic it used to derivation some binomial theorem and solution of mathematics problem. Because of that, the capable of this skill is necessary to who will learn mathematics, because there many treatises in mathematics which use that principal to derivate the theorem or to solve problem. Most of each next treatise use this both principal, either to prove the theorem or to solve the problems.
Induction of mathematics is one of the proof argumentation of a theorem or mathematics statement which whole object is set of integer number or especially set of original number. Pay attention to example mathematics statements bellow.
Example 1.1
1+2+3+…+n= ½ n(n+1) for each original number of n. Is this statement true? For replied this question, we can try to substitute n on this statement into any original number. If n = 1, then that statement become 1= ½ .1(1+1), or 1 = 1, that is got a true statement. If n = 2, then that statement become 1+2= ½ .2(2+1), , or 3 = 3, that is got a true statement. If n = 3, then that statement become 1+2+3= ½ .3(3+1), , or 6 = 6, that is got a true statement too.
The reader continueable to n = 4; 5; or other original number and usually will get true statement. Is with give some example with substitute some of original number into n from origin statement and got true statements can give proof about truth of this statement?
In mathematics, gift some example like that, is not proof of truth of statement which can used in the whole set. The statement in example above, the whole of set is set of all of original number. If we give an example for each original number of n in this statement and get true statement for each number, then that is as truth of proof from that statements.
But this is not efisien and imposible to do, because the element of set of original number unfinite. So, how to prove that statement? One of the way is view the first part from this statement as arithmetics series with first component a = 1, the difference b = 1, final component is Un= n and has n component. So the sum of this series is
Sn= ½ n(a+Un)
= ½ n(1+n)
= ½ n(n+1), that is second part from proved statement.
NIM : 07305141025
CHAPTER I
INDUCTION OF MATHEMATICS AND BINOMIAL THEOREM
A. Induction of mathematics
Induction of mathematics is one of the proof method from many theorems in theory of number or in other mathematics. Whereas binomial theorem, besides as basic it used to derivation some binomial theorem and solution of mathematics problem. Because of that, the capable of this skill is necessary to who will learn mathematics, because there many treatises in mathematics which use that principal to derivate the theorem or to solve problem. Most of each next treatise use this both principal, either to prove the theorem or to solve the problems.
Induction of mathematics is one of the proof argumentation of a theorem or mathematics statement which whole object is set of integer number or especially set of original number. Pay attention to example mathematics statements bellow.
Example 1.1
1+2+3+…+n= ½ n(n+1) for each original number of n. Is this statement true? For replied this question, we can try to substitute n on this statement into any original number. If n = 1, then that statement become 1= ½ .1(1+1), or 1 = 1, that is got a true statement. If n = 2, then that statement become 1+2= ½ .2(2+1), , or 3 = 3, that is got a true statement. If n = 3, then that statement become 1+2+3= ½ .3(3+1), , or 6 = 6, that is got a true statement too.
The reader continueable to n = 4; 5; or other original number and usually will get true statement. Is with give some example with substitute some of original number into n from origin statement and got true statements can give proof about truth of this statement?
In mathematics, gift some example like that, is not proof of truth of statement which can used in the whole set. The statement in example above, the whole of set is set of all of original number. If we give an example for each original number of n in this statement and get true statement for each number, then that is as truth of proof from that statements.
But this is not efisien and imposible to do, because the element of set of original number unfinite. So, how to prove that statement? One of the way is view the first part from this statement as arithmetics series with first component a = 1, the difference b = 1, final component is Un= n and has n component. So the sum of this series is
Sn= ½ n(a+Un)
= ½ n(1+n)
= ½ n(n+1), that is second part from proved statement.
Representing the video of learning mathematics.
Representing the video of learning mathematics.
A. Solving Problem Graph Math
Problem :
13. The figure above shows the graph of . If the function h is defined by , what is the value of ?
Answer :
We are looking for
3
1
2
(substitute , in the function); for look at the graph.
So, the solution of
13. Let the function f be defined by , if . What is the value of ?
Answer :
Looking for , what is f when ?
The information is
Substitute to
So the answer of is 28.
17. In the xy-coordinate plane the graph of intersects line l at (0,p) and (5,t). What is the greatest possible value of the slope of l ?
Answer :
Looking for is greatest m
Line l
Line l :
x y
0 p
5 t
B. Factoring Polynomials
Algebraic long division
For example :
Is a factor of
make along division problem for mathematics school
- bringing down
-
-
0
no remainder
then
is a factor of
Is also a factor of
= ( )( ) can be foctor into
= ( ) , sharing the factor for the equation of
0 = ( )
Thus ever solving all equations , we get
= 0
Or
Or
The roots of are 3, -1, -2.
3 roots for this 3rd degree equation quadratic (2nd degree) equations always have at most 2 roots.
A 4th degree equation would have 4 or fewer roots and so on.
The degree of a polynomial equations always limits the number of roots.
Long division for a 3rd order polynomial =
1 Find a partial quotient of by dividing x into to get .
2 Multiply by the divisor and substract the product from the dividend.
3 Repeat the process until you either “clear it out” or reach a remainder.
Section D
Properties of polynomial graphs
Polynomial have even or odd degrees.
C. Pre-Calculus
Graph of a rational function
Can have discontinuities, because has a polynomial in denominator
For example :
When the function become
imposible
For this function choosing is a bad idea.
break in function graph
Graph
Insert x = 0
(0,-2)
Insert x = 1
that is imposible
x = 1
D.
Discontinuity
BREAK
Rational function don’t always this way
Not all rational functions will give zero in denominator.
never zero, because of the + 1
no break
Don’t forget
Rasional functions denominator can be zero!
For polynomial the graph is a smooth unbroken curve
Rational functions
x zero in thr denominator is apossible situation, there is no value for the functions so break in the graph
break 2 ways
missing point in the graph
for example
the graph like this
A. Solving Problem Graph Math
Problem :
13. The figure above shows the graph of . If the function h is defined by , what is the value of ?
Answer :
We are looking for
3
1
2
(substitute , in the function); for look at the graph.
So, the solution of
13. Let the function f be defined by , if . What is the value of ?
Answer :
Looking for , what is f when ?
The information is
Substitute to
So the answer of is 28.
17. In the xy-coordinate plane the graph of intersects line l at (0,p) and (5,t). What is the greatest possible value of the slope of l ?
Answer :
Looking for is greatest m
Line l
Line l :
x y
0 p
5 t
B. Factoring Polynomials
Algebraic long division
For example :
Is a factor of
make along division problem for mathematics school
- bringing down
-
-
0
no remainder
then
is a factor of
Is also a factor of
= ( )( ) can be foctor into
= ( ) , sharing the factor for the equation of
0 = ( )
Thus ever solving all equations , we get
= 0
Or
Or
The roots of are 3, -1, -2.
3 roots for this 3rd degree equation quadratic (2nd degree) equations always have at most 2 roots.
A 4th degree equation would have 4 or fewer roots and so on.
The degree of a polynomial equations always limits the number of roots.
Long division for a 3rd order polynomial =
1 Find a partial quotient of by dividing x into to get .
2 Multiply by the divisor and substract the product from the dividend.
3 Repeat the process until you either “clear it out” or reach a remainder.
Section D
Properties of polynomial graphs
Polynomial have even or odd degrees.
C. Pre-Calculus
Graph of a rational function
Can have discontinuities, because has a polynomial in denominator
For example :
When the function become
imposible
For this function choosing is a bad idea.
break in function graph
Graph
Insert x = 0
(0,-2)
Insert x = 1
that is imposible
x = 1
D.
Discontinuity
BREAK
Rational function don’t always this way
Not all rational functions will give zero in denominator.
never zero, because of the + 1
no break
Don’t forget
Rasional functions denominator can be zero!
For polynomial the graph is a smooth unbroken curve
Rational functions
x zero in thr denominator is apossible situation, there is no value for the functions so break in the graph
break 2 ways
missing point in the graph
for example
the graph like this
HOW TO EXPRESS MATHEMATICS
A. Empowering : give somebody the power or authority to do something
- Indonesia will be empowering education quality.
B. Pursuing : do something to try to achieve something over a period of time, continue to discuss or involved in something, follow or chase somebody/ something in order to catch them
- Education National Department was pursuing KTSP curriculum.
C. Pengayaan : enrichment, adding material
- Enrichment lesson is not always given.
D. Persamaan garis singgung : tangent equation
- Tangent equation is undefined.
- Indonesia will be empowering education quality.
B. Pursuing : do something to try to achieve something over a period of time, continue to discuss or involved in something, follow or chase somebody/ something in order to catch them
- Education National Department was pursuing KTSP curriculum.
C. Pengayaan : enrichment, adding material
- Enrichment lesson is not always given.
D. Persamaan garis singgung : tangent equation
- Tangent equation is undefined.
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