TUGAS 4
VIDEO 3: ADVERB (KATA KETERANGAN)
How Often
When What for
What use
Adverb, kata untuk mendeskripsikan:
Verb: kata kerja
Adjectives: kata sifat
Kata keterangan lainnya.
Untuk menjawab pertanyaan
Bagaimana? Untuk hal apakah?
Berapa sering? Untuk apa?
Kapan?
Untuk membuat kata slow menjadi kata keterangan dapat dilakukan dengan menambahkan kata –ly dibelakangnya,sehingga menjadi:
Slow + ly = slowly (kata keterangan)
Kapan menggunakan Good and Well
Beberapa triknya, yaitu ketika ada pertanyaan “Bagaimana?” maka dijawab dengan kata keterangan dengan menggunakan Well.
Contoh: Candace can play the accordion very well.
Candace’s playing is good.
VIDEO 4: BASIC TRIGONOMETRY
Welcome to basic trigonometry
Trigonometry is from trigon and metron. Trigonometry is really study of rectangle and the relationship between the side and the angle of rectangle.
To solve them use trig Soh Cah Toa
Soh = Sine is opposite over hypotenuse
Cah = cosine is adjacent over hypotenuse
Toa = Tangent is opposite over adjacent
So, Sin ө= opp/ hyp = 4/5
Cos ө= adj/ hyp = 3/5
Tan ө= opp/ adj = 4/3
If the angle is x, tan x= ¾, the invers of tan ө.
You will know about rectangle of Basic trigonometry II.
VIDEO 5: COMPOUND SENTENCES (KALIMAT MAJEMUK)
Contoh: It’s the end of the world as we know it and I feel fine.
Ada 2 klausa yang dihubungkan dengan 1 kata penghubung yaitu “and”.
Ketika suatu kalimat digunakan sebagai bagian- bagian dalam kalimat yang lebih besar, maka kalimat yang lebih kecil disebut klausa.
Ketika sebuah klausa dapat berdiri sendiri dalam sebuah kalimat, maka klausa tersebut disebut “independent clause”.
Jika kita memiliki 2 klausa dalam kalimat, maka kalimat itu disebut kalimat majemuk.
Untuk menggabungkan 2 independent clause, kita menggunakan:
• Titik dua (:), ketika klausa ke-2 menjelaskan klausa ke-1. contoh: “I love my two sister, they bake me pie”. Untuk menggabungkan kalimat menjadi kalimat majemuk, kita menggunakan titik dua. Contoh: “I love my two sister: They bake me pie”.
• Titik koma (;) untuk menggantikan konjungsi. Contoh: “It’s the end of the world and I feel fine”. Kita dapat mempersingkat kalimat itu menjadi: “It’s the end of the world; I feel fine”.
• Garis penghubung (-)
Kalimatnya terdiri dari beberapa elemen (klausa) sehingga harus dihubungkan dengan menggunakan garis penghubung, karena klausa ke-2 dihubungkan dengan klausa ke-1.
Banyak sekali cara yang menarik untuk menggabungkan kalimat, yaitu:
- Kata penghubung (and)
- Titik dua (:)
- Titik koma (;)
- Garis penghubung (-)
Kalimat Fragmen
Jika kamu mendapat porsi dalam suatu kalimat yang tidak dapat berdiri sendiri sebagai kalimat lengkap. Contoh:
My pet komodo dragon is as gentle as a lamb because he has no teeth
kalimat lengkap kalimat fragmen
Kalimat lengkap terdiri dari Dependent sentences + Independent sentences.
-Klausa dependen (dependent klausa):
Tidak dapat berdiri sendiri
Klausa dependent terdapat pada klausa independent
Bukan berupa kalimat lengkap.
Kalimat lengkap
Contoh: Although Tom sleeps regularly, he is constantly tired.
Dependent clause Independent clause
Kalimat lengkap (kompleks)
VIDEO 6: LIMIT BY INSPECTION
There are two condition:
1. x goes to positive or negative infinity.
2. limit involves polynomial .
For example:
Lim (x^ 3 + 4)/ (x^2+ x+ 1)
x ~
Highest power of x in the numerator is 3.
Highest power of x in the denominator is 2.
This problem is caused of two condition:
polynomial over polynomial
x has to be approaching infinity.
The key to denominator limits by inspection is in looking at power of x in the numerator and the denominator.
To apply this rules:
1. Must be dividing by polynomials.
2. x has to be approaching infinity.
First shortcut rule
If the highest power of x is greatest in numerator, then limits is positive or negative infinity.
Lim x has to be approaching infinity from (x^3+ 4)/ (x^2 + x+1) = + or – infinity. Since all the number are positive and x is going to positive infinity.
If you can’t tell it the answer is positive or negative:
1. Substitute a large number of x.
2. See if you end up with a positive or negative number.
3. Whatever sign you get is the sign of infinity for the limit.
Second shortcut rule
If the highest power of x in the denominator, the limit is zero. Example:
Lim x has to be approaching infinity from (x^2 + 3)/(x^3 + 1)
Highest power of x in the numerator is 2 and highest power of x in the denominator is 3.
Third shortcut rule
Used when: Highest power of x in numerator is same as highest power of x in denominator.
Lim x has to be approaching infinity from (4x^3 + x^2 + 1)/(3x^3 + 4)
Highest power of x in the numerator and denominator is .
According to this rule:
Lim is coefficient of x^3’s is over each other. Example:
Lim x has to be approaching infinity from (4x^3 + x^2+ 1)/(3x^3 + 4)= 4/3
Last shortcut rule
Example: Lim x has to be approaching infinity, the question on the coefficient on the two highest powers.
Remember: Coefficient is the number that goes with a variable.
Example: 2 is the coefficient of 2x^2
75 is the coefficient of 75x^4.
VIDEO 7: PRE-CALCULUS GRAPH
Graph of a rational function which can have discontinuities, because has polynomial in the denominator.
Is possible value x divide by 0 (zero)
Example: f(x) = (x+2)/(x-1)
f (1) the value became (1+2)/(1-1)= 3/0 (bad idea)
Graph f (1) = (1+2)/ 0 (break in function graph)
f (x)= (x+2)/(x-1) (insert 0)
f (0)= (0+2)/(0-1)= -2
Insert 1 became f (1) = (1+2)/ (1-1) = 3/0 (impossible).
Rational function don’t always work this away!
Take graph f(x) = 1/(x^2+1) not all rational function will give zero in denominator. Example: y=(x^2 – x- 6)/(x-3) the graph lose like these.
If x=3, f (3) = (3^2- 3- 6)/ (3-3)= 0/0 that is not possible not allowed. When you see result of 0/0 and also fell you direction be possible. Factor top and bottom of rational function and simplify.
Rational function denominator can be zero if:
• Polynomial have smooth and unbroken curve and for rational function x will be zero in denominator, that impossible situation. Example:
y = (x^2 – x – 5)/ (x – 3)
y = (x – 3) (x + 2)/ (x - 3)
y = x + 2
If x = 3, so that y = x + 2
= 3 + 2 = 5
VIDEO 8: TRIGONOMETRY FUNCTION (TRIG FUNCTION)
To remember you can see:
SOH CAH TOA
SOH = Sine Opposite Hypotenuse
CAH = Cosine Adjacent Hypotenuse
TOA = Tangent Opposite Adjacent
Trig function only need to know values of side to find measure of an angle figure out figure of all part of triangle.
Trig function:
1. Sine
2. Cosine
3. Tangent
4. Cosecant
5. Secant
6. Cotangent
Six basic trigonometry function:
1. Sides of a triangle
2. Angle being measure
Opp: side opposite to theta
Adj: side adjacent to theta
Hyp: side hypotenuse to theta
Sin ө= opp/ hyp Cosec ө= hyp/ opp
Cos ө= adj/ hyp Sec ө= hyp/ adj
Tan ө= opp/ adj Cot ө= adj/ opp
Thursday, January 15, 2009
EXPERIENCE TO EXPLAIN FIRST – ORDER EQUATION WHICH EXACT SOLUTION ARE OBTAINABLE
January 10 th 2009, at 10.00 am in Karang Malang blok B 20A.
I explain about first – order equation which exact solution are obtainable to my friend’s, she is Aida. Aida is just board with me and my classmate. In this first – order equation for which exact solution are obtainable considered certain basic types of first – order equation for with exact solution my be obtained by definite procedures. The purpose of this first – order equation for which exact solution are obtainable is to gain ability to recognize these various types and to apply the corresponding method of solution. Of the types considered here, the so called exact equation in a sense the most basic. First, I explain standard form of first –order differential equation. I given example of standard form of first order differential equation is dy/dx = f (x,y) or the differential form M (x,y) dx + N (x,y) dy = 0. An equation in one of form these may readily be written in the other form. For example the equation dy/dx = (x^2/y^2)/(x – y) is the form of (1) it may be written (x^2 + y^2) dx + (y – 1) dy = 0.Of the form (2), may be written in the form (1) as dy/dx = - (sin x + y)/(x + 3 y).
In the form (1) it is clear from the notation it self that y is regarded as the dependent variable and x as the independent one, but in the form (2) we may actually regard either variable as the dependent one and the other as the independent. However, in this text, all different differential equation of the form (2) in x and y, we shall regard y as dependent and x as independent, unless the contrary is specifically stated. For example above, she understands.
Then, I that given know definition of exact differential equation is let F be a function of two real variables such that F has continuous first partial derivatives in a domain D. the total differential dF of the function F is defined by the formula dF (x,y) = dF (x,y)/dx dx + dF (x,y)/dy dy for all (x,y) E D.
I am give example to Aida, and she can to finish true. I am happy given explanation to her. She can to finish problem mathematic with easy because she is very like mathematic. She is also university student mathematic in UNY. She is know about definition of M (x,y) dx + N (x,y) dy is called an exact differential in a domain D if there exist a function F of two variables such that is expression equal the total differential dF(x,y) for all (x,y) element D. example of the differential equation y^2 dx + 2 xy dy = 0 is an exact differential equation, since the expression y^2 dx + 2 xy dy is an differential. Indeed, it is the total differential of the function F defined for all (x,y) by F(x,y) = xy^2, since the coefficient of dx is dF(x,y)/dx = y^2 and that of dF(x,y)/dy = 2xy. On the other hand, the more simple appearing equation y dx + 2x dy = 0, obtained from y^2 dx + 2 xy = 0 by dividing through by is not y, is not exact.
I also explain about theorem of the differential equation M(x,y) dx + N(x,y) dy = 0. where M and N have continuous first partial derivative at all point (x,y) in a rectangular
1. if the differential equation M(x,y) dx + N(x,y) dy = 0 is exact in D, then dM(x,y)/dy = dN(x,y)/dx for all (x,y) element D.
2. conversely, if dM(x,y)/dy = dN(x,y)/dx
With that theorem she become very understanding manner working differential equation. Manner the solution of exact differential equation, let proceed to solve exact differential equation. If the equation M(x,y) dx + N(x,y) dy = 0 is exact in a rectangular domain D, then there exist a function F such that dF(x,y)/dx = M(x,y) and dF(x,y)/dy = N(x,y) for all (x,y) element D.
Then the equation may be written (dF(x,y)/dx) dx + (dF(x,y)/dy) dy = 0 or simply dF(x,y) = 0. the relation F(x,y) = c is obviously of this, where c is an arbitrary constant. To very know in differential equation I explain theorem again, the theorem is suppose the differential equation M(x,y) dx + N(x,y) dy = 0 satisfies the differential ability requirement of theorem above and is exact in a rectangular domain D. Then, a one parameter family of solution of this differential equation is given by F(x,y) = c, where is a function such that dF(x,y)/dx = M(x,y) and dF(x,y)/dy = N(x,y) for (x,y) element D. with theorem she more know, and I also given example to her in order to he is know. Example’s (3 x^2 + 4 xy) dx + (2 x^2 + 2 y) dy = 0, first duty is to determine whether or not the equation is exact. Here M(x,y) = 3 x^2 + 4 xy, N(x,y) = 2 x^2 + 2 y
dM(x,y)/dy = 4 x, dN(x,y)/dx = 4 x
for all real (x,y), and so the equation is exact in every rectangular domain D.
I finish explain to Aida at 11.00 am, and she is look happy with my explain.
January 10 th 2009, at 10.00 am in Karang Malang blok B 20A.
I explain about first – order equation which exact solution are obtainable to my friend’s, she is Aida. Aida is just board with me and my classmate. In this first – order equation for which exact solution are obtainable considered certain basic types of first – order equation for with exact solution my be obtained by definite procedures. The purpose of this first – order equation for which exact solution are obtainable is to gain ability to recognize these various types and to apply the corresponding method of solution. Of the types considered here, the so called exact equation in a sense the most basic. First, I explain standard form of first –order differential equation. I given example of standard form of first order differential equation is dy/dx = f (x,y) or the differential form M (x,y) dx + N (x,y) dy = 0. An equation in one of form these may readily be written in the other form. For example the equation dy/dx = (x^2/y^2)/(x – y) is the form of (1) it may be written (x^2 + y^2) dx + (y – 1) dy = 0.Of the form (2), may be written in the form (1) as dy/dx = - (sin x + y)/(x + 3 y).
In the form (1) it is clear from the notation it self that y is regarded as the dependent variable and x as the independent one, but in the form (2) we may actually regard either variable as the dependent one and the other as the independent. However, in this text, all different differential equation of the form (2) in x and y, we shall regard y as dependent and x as independent, unless the contrary is specifically stated. For example above, she understands.
Then, I that given know definition of exact differential equation is let F be a function of two real variables such that F has continuous first partial derivatives in a domain D. the total differential dF of the function F is defined by the formula dF (x,y) = dF (x,y)/dx dx + dF (x,y)/dy dy for all (x,y) E D.
I am give example to Aida, and she can to finish true. I am happy given explanation to her. She can to finish problem mathematic with easy because she is very like mathematic. She is also university student mathematic in UNY. She is know about definition of M (x,y) dx + N (x,y) dy is called an exact differential in a domain D if there exist a function F of two variables such that is expression equal the total differential dF(x,y) for all (x,y) element D. example of the differential equation y^2 dx + 2 xy dy = 0 is an exact differential equation, since the expression y^2 dx + 2 xy dy is an differential. Indeed, it is the total differential of the function F defined for all (x,y) by F(x,y) = xy^2, since the coefficient of dx is dF(x,y)/dx = y^2 and that of dF(x,y)/dy = 2xy. On the other hand, the more simple appearing equation y dx + 2x dy = 0, obtained from y^2 dx + 2 xy = 0 by dividing through by is not y, is not exact.
I also explain about theorem of the differential equation M(x,y) dx + N(x,y) dy = 0. where M and N have continuous first partial derivative at all point (x,y) in a rectangular
1. if the differential equation M(x,y) dx + N(x,y) dy = 0 is exact in D, then dM(x,y)/dy = dN(x,y)/dx for all (x,y) element D.
2. conversely, if dM(x,y)/dy = dN(x,y)/dx
With that theorem she become very understanding manner working differential equation. Manner the solution of exact differential equation, let proceed to solve exact differential equation. If the equation M(x,y) dx + N(x,y) dy = 0 is exact in a rectangular domain D, then there exist a function F such that dF(x,y)/dx = M(x,y) and dF(x,y)/dy = N(x,y) for all (x,y) element D.
Then the equation may be written (dF(x,y)/dx) dx + (dF(x,y)/dy) dy = 0 or simply dF(x,y) = 0. the relation F(x,y) = c is obviously of this, where c is an arbitrary constant. To very know in differential equation I explain theorem again, the theorem is suppose the differential equation M(x,y) dx + N(x,y) dy = 0 satisfies the differential ability requirement of theorem above and is exact in a rectangular domain D. Then, a one parameter family of solution of this differential equation is given by F(x,y) = c, where is a function such that dF(x,y)/dx = M(x,y) and dF(x,y)/dy = N(x,y) for (x,y) element D. with theorem she more know, and I also given example to her in order to he is know. Example’s (3 x^2 + 4 xy) dx + (2 x^2 + 2 y) dy = 0, first duty is to determine whether or not the equation is exact. Here M(x,y) = 3 x^2 + 4 xy, N(x,y) = 2 x^2 + 2 y
dM(x,y)/dy = 4 x, dN(x,y)/dx = 4 x
for all real (x,y), and so the equation is exact in every rectangular domain D.
I finish explain to Aida at 11.00 am, and she is look happy with my explain.
Terjemahan buku SMP kelas 2
Pengarang ASYONO
Penerbit Bumi Aksara
TUGAS 5
LINEAR EQUATION SYSTEM TWO VARIABLES (SPLDV)
- study porpuse
after studying this chapter, student can hoped :
1. the form explain linear equation system two variables (SPLDV)
2. finish SPLDV
- this chapter in subject matter
1. linear equation with two variables (SPLDV)
2. linear equation system two variables (SPLDV)
Before studying this chapter, to work problem this following!
1. to determine solution the set equation following!
a. 2x = 5
b. 2x – 5 =4 – x
c. 1/(4-2x) = 2
d. (x + 2)/5 = 1
e. ½ = 2/(x – 3)
2. To make graph of y = 2x + 1 !
A. Linear equation with two variable (PLDV)
Before discuss linear equation with two variable, please more fast we to came back remembers form following equation.
1. 2x + 5 = 4
2. 3y + 2 = y + 1
3. x^2 + 5 = 2
you try watch form equation above.
Furthermore, do you still remember finish the way linear equation with one variable (PLSV)? To attention example following!
Example :
To determine solution of equation (2/3)x + 4 = 6!
Solution :
(2/3) + 4 = 6
(2/3) x + 4 – 4 = 6 -4 (twice space between joint minus with 4)
(2/3) x = 2
2/3 (2/3) x = 3/2 x 2 (space between joints match with 3/2)
x = 3
so, the solution of equation’s is (3)
1. discuss linear equation with two variable (PLDV)
can you make equation ? if you answer with true so can written the equation as following.
2 x + 3 y = 6000
At the equation above measles beside variable’s hold on an important one highest also consist of two variable. So the equation it is the equation with two variables.
Example :
1. 2 x + y = 4 is it the linear equation with two variable, because has two variable the highest rank 1 is x and y.
2. x^2 + y = 2 the linear no equation with two variable, because the there rank variable than one is x^2
exercise 4.1
1. mention which of following equation the form linear equation with two variable !
a. 2 x – 3 = x + 2
b. (x/y) = 4
c. 2 x – y = 0
d. x + y^2 = 0
e. (y/x) + 1 = 4
f. 2 x = y + 3
2. To obvious true of false pronouncement following !
a. 2 x + 1 is the linear equation with two variable, because it has the rank variable 1 highest
b. 2 x + y^2 = 5 is the linear equation with two variable, because it has the two variables
c. (x/y) + 1 = 3 the linear no equation with two variable, because x divide y
3. Explain one variable to another one (PLDV)
In PLDV, a certain variable can obvious become the besides form variable. For more, to attention example and explanation this following!
Let, PLDV of 3 x + y = 6
This equation has two variable x and y, the exhausted at same math, is math left. These form can change another to form as the form obvious in variable other variable or the form variable one with besides variable exhausted at different the math. It so happen method’s as following :
X = 2 – (1/3) y
X = -(1/3) y + 2
3 x + y = 6
Y = 6 – 3 x (to move x and coefficient to right math)
Y = -3 x + 6
So, form 3 x + y = 6 can obvious in form
Y = -3 x + 6 or x = - (1/3) y +2
So, a certain PLDV can changed to form the variable one obvious to besides variable
Exercise 4.2
1. To fill points at following obvious until x become true !
a. 3 x + 6 = 9
3 x = 9 - …
3 x = - 6 y + …
X =…+ …
b. 2 x + 4 y – 20 = 0
2 x + 4 y =…
2 x = … - 4 y
2 x =…. + 20
X = … + …
2. To obvious to for form the explain variable y to x following!
a. 3 y = 6 x + 9
b. 2 x – 3 y = 6
c. 4 x + 2 y = 8
d. – 3 x + 4 y = 12
e. 3 x – 2 y + 6 = 0
f. – 3 x – 4 x – 12 = 0
B. Linear equation two variable (SPLDV)
1. Discuss mean linear equation with two variables
To attention example following!
Purchase 2 a books and 5 pencils with total price Rp 4.500,00. Form mentioned the problem if price a book is x and a pencil is y so obtained two the linear equation two variables; it is 3 x + 4 y = 5.000,00 and 2 x + 5 y = 4.500,00.
If there two linear equation with be found two the variable in a problem and each the variable that represent a thing so second this the equation mentioned as the linear equation two variable system. The answer it while form this system is value fastened form each variable fulfill the second equation. Remember, if at equation system this isn’t pair which fulfill that the second equation, so is called the equation system has the answer or the answer association is the empty set.
The kind of the linear equation system, among others :
a. Linear equation system with two variable
Example : 2 x + y = 4
3 x + y = 5
b. Linear equation system with three variable
Example : x + y + z = 6
2 x + y + 3 z = 13
X + 2 y + z = 8
Further more, the discuss this only limited at the linear equation system with two variable.
To attention the linear equation system two variable, it is
2 x + y = 4
3 x + y = 5
From mentioned SPLDF, the first equation consist of the variable x with 2 coefficient, the variable y with 1 coefficient, and 4 constant, the second equation consist of the variable y with 3 coefficients, the variable y with 1 coefficient and 5 constant.
Exercise 4.3
From following equation, which was form the linear equation with two variable ?
1. x + 3 v = 4
2. x^2 + 2 v = 3
3. v
4. x + v = 3
2. the root and the not two variables linear equation system
to attention this the linear equation !
x + y = 3
x – y = 1
if we watch closely, the equation system this consist of two variables.
For x = 1, y = 1, so x + y = 3
1 + 1 = 3 (sentence is false)
x – y = 1
1 – 1 = 1 (sentence is false)
For x = 2, y = 1, so x + y = 3
2 + 1 = 3 (sentence is true)
x – y = 1
2 – 1 = 1 (sentence is true)
For x = 2, y = 1, so x + y = 3
1 + 2 = 3 (sentence is true)
x – y = 1
1 – 2 = 1 (sentence is false)
From mentioned explanation, it appears if x substituted y substituted 1 so obtained at SPLDV the equation became the sentences. The substituted fastened is call the solution SPLDV or the root SPLDV. The values besides 2 and wouldn’t cause at SPLDV the equation become sentences which true. These value from the solution SPLDV>
So, the solution SPLDV or the roots SPLDV is the substituted variable’s until the second equation at SPLDV become be sentences with true. Further more, process of writing SPLDV with use the word “and” as SPLDV : x + y = 3 and x – 1 =1
Can substituted with use sign and
SPLDV : x + y = 3
x – y = 1
remember :
the word “and” at the linear equation system two variable has mean is bound or the relation another the equation.
So, the solution of equation system this must fulfill which is the second the equation.
Exercise 4.4
1.to obvious each this pronouncement true of false !
If SPLDV is x – y = 2 and x+ y = 5 so, the solution of SPLDV is (1,3)
2. try with understand the value x and y, determine solution SPLDV of
a. 2 x + 3 y = 7
3 x – 2 y = 4
b. x + 5 y = 8
2 x + y = 7
3. price 3 apples and 2 manggo is Rp 8.000,00. price 2 apples and 3 manggo is Rp. 7.000,00. how price are 1 apple and mango ? to working base on your way !
3. determining the solution SPLDV
there some the way for finish an the linear equation system with two variables, it is as the following!
a. finishing equation with the method substitute
finish equation with other variable. For more clearly, to attention example following!
Example :
1.to finish the equation system following with the method substitute!
2 x + y = 5 and substituted to y = 3 x
Solution :
Y = 3 x, equation 2 x + y = 5
Result’s can to obtained as following
2 x + y = 5
2 x + 3 y = 5
5 x = 5
X = 1
Because x = 1 so value 1 can substituted at y = 3 x. result’s :
Y = 3 x
Y = 3 (1)
Y = 3
So, root solution’s = (1,3)
2.to determine solution set the equation follow with the method substitute !
X – y - 1 = 0 and 3 x + 2 y = 8
Solution :
For this problem we takes form :
x – y -1 = 0
x – 1 = y or y = x – 1
further more, to substitute y = x -1 at 3 x + 2 y = 8, be obtained :
3 x + 2 y = 8
3 x + 2 (x – 1 ) = 8
3 x + 2 x – 2 = 8
5 x – 2 = 8
5 x = 10
X = 2
If x = 2 substituted at y = x – 1, so be obtained :
Y = x – 1
y = 2 – 1
y = 1
so, root solution’s (2,1)
exercise 4.5
to determine solution the equation system with the method substitute following !
1. x = 2 y, 2 x – y = 3
2. y = x – 1, x + 2 y = 7
3. 3 x – y = 1, 2 x – y = 0
4. x – 2 y = 2, x + 2 y – 6 = 0
5. y = 2 x + 1, 4 x – y = 1
6. x + 2 y = 4, 2 x – 3 y -1 = 0
7. 3 x + 2 y = 12, 2 x – 3 y – 1 = 0
8 . (1/3) x + (1/2) y = 3, (2/3) x – (1/4) y = 1
b. finish the equation with the method substitute
eliminate means’s the cause. Finish the equation system with the method substitute eliminate is the cause one fault obtain value variable.
Further more, for understand solution the equation system with the method eliminate to attention example following !
Example :
1. with the method eliminate, to finish the equation system following !
x + y = 5 and x – y = 1
solution :
the second equation arranged vertical, letter to loss one fault variable’s with count up or subtract. Until be obtained :
with add
x + y = 5
for x = 3 so, x + y = 5
3 + y =5
Y = 2
So, root solution’s (3,2)
With subtract
x + y = 5
for y = 2, so x + 5
x + 2 = 5
x = 3
so root solution’s (3,2)
2. to finish equation with the method eliminate following !
2 x – 3 y = 3 time 1 2 x – 3 y = 3
3 x + y = 10 time 3 9 x + 3 y = 30
So x = 3
For x = 3 so be obtained :
2 x – 3 y = 3
2 (3) – 3 y = 3
6 – 3 y = 3
- 3 y = - 3
Y = 1
So, root solution’s (3.1)
Exercise 4.6
To determine solution the equation system with the method eliminate following !
1. x + y = 4 and 2 x – y = 5
2. 3 x + 2 y = 13 and 2 x + y = 8
3. x + 2 y = - 1 and 2 x – y = 3
4. 3 x + 4 y + 2 = 0 and x + 3 y – 1 = 0
5. 5 x – 2 y = 4 and 3 x = 4 y – 6
6. 4 x – 3 y – 11 = 0 and 3 x – 2 y – 8 = 0
7. 2 x + 91/3) y = 5 and (1/2) x – y = - 2
8. (x – 2 )/3 + y = 2 and (y + 1)/2 + x = 6
c. finish equation system with the method graph
Finish equation system with the method graph is drawing graph the second equation at one draw
At coordinate plane, and the point coordinate graph the second equation this form solution’s.
For than explain, to attention example following !
Solution
The way with as has explain in the lesson drawing graph it is with take some example the fulfill equation x + y = 3 and x – y = 1.
Remember !
Determine solution with the method graph rather difficult if point piece not round number. From picture 4.1 measles the second graph equation mentioned have the looks of in point (2.1). so, root solution (2.1)
The draft from this third method can be found when you study the linear equation system more of two variable, matriks, and the quadrate equation.
Exercise 4.7
To determine solution of the equation system with the method graph following.
1. x = 2, y = 3
2. y = x, y = 2
3. y = x + 1, y = 2 x
4. x + y = 3, x + 2 y = 4
5. x + y = 0, 2 x + y = 1
6. x + 2 y = 8, 2 x – y = 3
7. 3 x + 2 y = 8, 2 x – y = 3
8. 2 x – y = 5, x – 2 y = 1
To try !
To make graph of SPLDV
2 x + y = 6
x – y = 1
according to graph, to determine root solution of SPLDV
4.finish story the problem interrelation with system.
Example :
Price two a pencil and a book is Rp. 1.000,00. if price a book Rp. 4000,00 more expensive of price a pencil, how price is a pencil a book ?
Solution
a. understand problem
be cognizant of price two pencil and one book is Rp 1.000,00. price book Rp 4.000,00 more expensive of price a pencil
question : price one pencil and price one book.
b. Arranging planning
The draft used is solution linear equation system two variable using the method substitute.
c.perform planning
let, price a pencil is x and a book is y, be obtained until linear equation system with two variable as following. 2 x + y = 1000 and y = x + 400. for the finish this equation system, to use the method substitute. y = x + 400 substitute at 2 x + y =1000 be obtained :
2 x + y = 1000
2 x (x + 400) =1000
2 x + x + 400 =1000
3 x + 400 = 1000
3 x = 600
x = 200
for x = 200 so, y = 200 + 400
y = 600
so, price a pencil Rp 200,00 and a book Rp 600,00
d. examination
for x = 200 and y = 600, so
2 x + y = 1.000,00
2 (200) + 600 = 1000 (proven)
y = x + 400
600 = 200 + 400
600 = 600 (proven)
Exercise 4.8
1. if add two number is and different is 1, to determine
2. two a corner interspace straight. If different the second corner is 20 degree, to determine each corner’s
3. a certain point (2,3) and (4,5) exhaust at y = m x + c with two variable m and c, to determine m and c.
4. the cost wash two dress and three wastern-style shirt Rp 6.000,00. if the cost wash a dress and wastern – style shirt Rp 2.500,00, to determine each the cost to wash
5. there two corner interspace right. If one fault corner two times of other’s, to determine each the corner following
5.finish SP nonlinear two variable from to used SPLDV
In more discuss the mathematic, ordernary could nonlinear equation system. This non linear equation system more found in counting matriks and vector. For finish nonlinear equation system, change more ago to for from linear equation system. For more explain to attention this example following.
Example :
To determine solution of 6/x + 2/y = 3
3/x + 8/y = 5
Solution :
6/x + 2/y = 3 time xy 6 y + 2 x = 3 xy…..(1)
3/x + 8/y = 5 time xy 3 y + 8 x = 5 xy….. (2)
So obtained new SPLDV
- 14 x = - 7 xy
Substitute y = z to for (1) , be obtained :
6 y + 2 x = 3 xy
6(2) + 2 x = 3 x (2)
12 + 2 x = 6 x
12 = 6 x – 2 x
12 = 4 x
x = 12/4
x = 3
so root solution is (3,2)
exercise 4.9
to determine solution of SPLDV following
1. 2/x + 3/y = 2
6/x – 3/y = 2
2. 9/2x + 4/y = 5
3/x + 8/y = 6
3. 3/x – 2/y = 2
4/x + 2/y = 5
4. 5/x + 3/2x = 6
3/x + 6/y = 7
SUMMARY
1. linear equation with two variable is equation with two variable the rank one highest.
2. solution SPLDV or root SPLDV is change variable until equation at SPLDV following become the sentence true.
3. the way determine root solution SPLDV, it is :
a. substitute
b. eliminate
c. graph
Reflection
From discuss SPLDV at these chapter, which are you more understand of the third method the root determine SPLDV or solution or solution SPLDV ?
Which aren’t you understand ?
What is problem ?
Can you do inlaid work this the method
Pengarang ASYONO
Penerbit Bumi Aksara
TUGAS 5
LINEAR EQUATION SYSTEM TWO VARIABLES (SPLDV)
- study porpuse
after studying this chapter, student can hoped :
1. the form explain linear equation system two variables (SPLDV)
2. finish SPLDV
- this chapter in subject matter
1. linear equation with two variables (SPLDV)
2. linear equation system two variables (SPLDV)
Before studying this chapter, to work problem this following!
1. to determine solution the set equation following!
a. 2x = 5
b. 2x – 5 =4 – x
c. 1/(4-2x) = 2
d. (x + 2)/5 = 1
e. ½ = 2/(x – 3)
2. To make graph of y = 2x + 1 !
A. Linear equation with two variable (PLDV)
Before discuss linear equation with two variable, please more fast we to came back remembers form following equation.
1. 2x + 5 = 4
2. 3y + 2 = y + 1
3. x^2 + 5 = 2
you try watch form equation above.
Furthermore, do you still remember finish the way linear equation with one variable (PLSV)? To attention example following!
Example :
To determine solution of equation (2/3)x + 4 = 6!
Solution :
(2/3) + 4 = 6
(2/3) x + 4 – 4 = 6 -4 (twice space between joint minus with 4)
(2/3) x = 2
2/3 (2/3) x = 3/2 x 2 (space between joints match with 3/2)
x = 3
so, the solution of equation’s is (3)
1. discuss linear equation with two variable (PLDV)
can you make equation ? if you answer with true so can written the equation as following.
2 x + 3 y = 6000
At the equation above measles beside variable’s hold on an important one highest also consist of two variable. So the equation it is the equation with two variables.
Example :
1. 2 x + y = 4 is it the linear equation with two variable, because has two variable the highest rank 1 is x and y.
2. x^2 + y = 2 the linear no equation with two variable, because the there rank variable than one is x^2
exercise 4.1
1. mention which of following equation the form linear equation with two variable !
a. 2 x – 3 = x + 2
b. (x/y) = 4
c. 2 x – y = 0
d. x + y^2 = 0
e. (y/x) + 1 = 4
f. 2 x = y + 3
2. To obvious true of false pronouncement following !
a. 2 x + 1 is the linear equation with two variable, because it has the rank variable 1 highest
b. 2 x + y^2 = 5 is the linear equation with two variable, because it has the two variables
c. (x/y) + 1 = 3 the linear no equation with two variable, because x divide y
3. Explain one variable to another one (PLDV)
In PLDV, a certain variable can obvious become the besides form variable. For more, to attention example and explanation this following!
Let, PLDV of 3 x + y = 6
This equation has two variable x and y, the exhausted at same math, is math left. These form can change another to form as the form obvious in variable other variable or the form variable one with besides variable exhausted at different the math. It so happen method’s as following :
X = 2 – (1/3) y
X = -(1/3) y + 2
3 x + y = 6
Y = 6 – 3 x (to move x and coefficient to right math)
Y = -3 x + 6
So, form 3 x + y = 6 can obvious in form
Y = -3 x + 6 or x = - (1/3) y +2
So, a certain PLDV can changed to form the variable one obvious to besides variable
Exercise 4.2
1. To fill points at following obvious until x become true !
a. 3 x + 6 = 9
3 x = 9 - …
3 x = - 6 y + …
X =…+ …
b. 2 x + 4 y – 20 = 0
2 x + 4 y =…
2 x = … - 4 y
2 x =…. + 20
X = … + …
2. To obvious to for form the explain variable y to x following!
a. 3 y = 6 x + 9
b. 2 x – 3 y = 6
c. 4 x + 2 y = 8
d. – 3 x + 4 y = 12
e. 3 x – 2 y + 6 = 0
f. – 3 x – 4 x – 12 = 0
B. Linear equation two variable (SPLDV)
1. Discuss mean linear equation with two variables
To attention example following!
Purchase 2 a books and 5 pencils with total price Rp 4.500,00. Form mentioned the problem if price a book is x and a pencil is y so obtained two the linear equation two variables; it is 3 x + 4 y = 5.000,00 and 2 x + 5 y = 4.500,00.
If there two linear equation with be found two the variable in a problem and each the variable that represent a thing so second this the equation mentioned as the linear equation two variable system. The answer it while form this system is value fastened form each variable fulfill the second equation. Remember, if at equation system this isn’t pair which fulfill that the second equation, so is called the equation system has the answer or the answer association is the empty set.
The kind of the linear equation system, among others :
a. Linear equation system with two variable
Example : 2 x + y = 4
3 x + y = 5
b. Linear equation system with three variable
Example : x + y + z = 6
2 x + y + 3 z = 13
X + 2 y + z = 8
Further more, the discuss this only limited at the linear equation system with two variable.
To attention the linear equation system two variable, it is
2 x + y = 4
3 x + y = 5
From mentioned SPLDF, the first equation consist of the variable x with 2 coefficient, the variable y with 1 coefficient, and 4 constant, the second equation consist of the variable y with 3 coefficients, the variable y with 1 coefficient and 5 constant.
Exercise 4.3
From following equation, which was form the linear equation with two variable ?
1. x + 3 v = 4
2. x^2 + 2 v = 3
3. v
4. x + v = 3
2. the root and the not two variables linear equation system
to attention this the linear equation !
x + y = 3
x – y = 1
if we watch closely, the equation system this consist of two variables.
For x = 1, y = 1, so x + y = 3
1 + 1 = 3 (sentence is false)
x – y = 1
1 – 1 = 1 (sentence is false)
For x = 2, y = 1, so x + y = 3
2 + 1 = 3 (sentence is true)
x – y = 1
2 – 1 = 1 (sentence is true)
For x = 2, y = 1, so x + y = 3
1 + 2 = 3 (sentence is true)
x – y = 1
1 – 2 = 1 (sentence is false)
From mentioned explanation, it appears if x substituted y substituted 1 so obtained at SPLDV the equation became the sentences. The substituted fastened is call the solution SPLDV or the root SPLDV. The values besides 2 and wouldn’t cause at SPLDV the equation become sentences which true. These value from the solution SPLDV>
So, the solution SPLDV or the roots SPLDV is the substituted variable’s until the second equation at SPLDV become be sentences with true. Further more, process of writing SPLDV with use the word “and” as SPLDV : x + y = 3 and x – 1 =1
Can substituted with use sign and
SPLDV : x + y = 3
x – y = 1
remember :
the word “and” at the linear equation system two variable has mean is bound or the relation another the equation.
So, the solution of equation system this must fulfill which is the second the equation.
Exercise 4.4
1.to obvious each this pronouncement true of false !
If SPLDV is x – y = 2 and x+ y = 5 so, the solution of SPLDV is (1,3)
2. try with understand the value x and y, determine solution SPLDV of
a. 2 x + 3 y = 7
3 x – 2 y = 4
b. x + 5 y = 8
2 x + y = 7
3. price 3 apples and 2 manggo is Rp 8.000,00. price 2 apples and 3 manggo is Rp. 7.000,00. how price are 1 apple and mango ? to working base on your way !
3. determining the solution SPLDV
there some the way for finish an the linear equation system with two variables, it is as the following!
a. finishing equation with the method substitute
finish equation with other variable. For more clearly, to attention example following!
Example :
1.to finish the equation system following with the method substitute!
2 x + y = 5 and substituted to y = 3 x
Solution :
Y = 3 x, equation 2 x + y = 5
Result’s can to obtained as following
2 x + y = 5
2 x + 3 y = 5
5 x = 5
X = 1
Because x = 1 so value 1 can substituted at y = 3 x. result’s :
Y = 3 x
Y = 3 (1)
Y = 3
So, root solution’s = (1,3)
2.to determine solution set the equation follow with the method substitute !
X – y - 1 = 0 and 3 x + 2 y = 8
Solution :
For this problem we takes form :
x – y -1 = 0
x – 1 = y or y = x – 1
further more, to substitute y = x -1 at 3 x + 2 y = 8, be obtained :
3 x + 2 y = 8
3 x + 2 (x – 1 ) = 8
3 x + 2 x – 2 = 8
5 x – 2 = 8
5 x = 10
X = 2
If x = 2 substituted at y = x – 1, so be obtained :
Y = x – 1
y = 2 – 1
y = 1
so, root solution’s (2,1)
exercise 4.5
to determine solution the equation system with the method substitute following !
1. x = 2 y, 2 x – y = 3
2. y = x – 1, x + 2 y = 7
3. 3 x – y = 1, 2 x – y = 0
4. x – 2 y = 2, x + 2 y – 6 = 0
5. y = 2 x + 1, 4 x – y = 1
6. x + 2 y = 4, 2 x – 3 y -1 = 0
7. 3 x + 2 y = 12, 2 x – 3 y – 1 = 0
8 . (1/3) x + (1/2) y = 3, (2/3) x – (1/4) y = 1
b. finish the equation with the method substitute
eliminate means’s the cause. Finish the equation system with the method substitute eliminate is the cause one fault obtain value variable.
Further more, for understand solution the equation system with the method eliminate to attention example following !
Example :
1. with the method eliminate, to finish the equation system following !
x + y = 5 and x – y = 1
solution :
the second equation arranged vertical, letter to loss one fault variable’s with count up or subtract. Until be obtained :
with add
x + y = 5
for x = 3 so, x + y = 5
3 + y =5
Y = 2
So, root solution’s (3,2)
With subtract
x + y = 5
for y = 2, so x + 5
x + 2 = 5
x = 3
so root solution’s (3,2)
2. to finish equation with the method eliminate following !
2 x – 3 y = 3 time 1 2 x – 3 y = 3
3 x + y = 10 time 3 9 x + 3 y = 30
So x = 3
For x = 3 so be obtained :
2 x – 3 y = 3
2 (3) – 3 y = 3
6 – 3 y = 3
- 3 y = - 3
Y = 1
So, root solution’s (3.1)
Exercise 4.6
To determine solution the equation system with the method eliminate following !
1. x + y = 4 and 2 x – y = 5
2. 3 x + 2 y = 13 and 2 x + y = 8
3. x + 2 y = - 1 and 2 x – y = 3
4. 3 x + 4 y + 2 = 0 and x + 3 y – 1 = 0
5. 5 x – 2 y = 4 and 3 x = 4 y – 6
6. 4 x – 3 y – 11 = 0 and 3 x – 2 y – 8 = 0
7. 2 x + 91/3) y = 5 and (1/2) x – y = - 2
8. (x – 2 )/3 + y = 2 and (y + 1)/2 + x = 6
c. finish equation system with the method graph
Finish equation system with the method graph is drawing graph the second equation at one draw
At coordinate plane, and the point coordinate graph the second equation this form solution’s.
For than explain, to attention example following !
Solution
The way with as has explain in the lesson drawing graph it is with take some example the fulfill equation x + y = 3 and x – y = 1.
Remember !
Determine solution with the method graph rather difficult if point piece not round number. From picture 4.1 measles the second graph equation mentioned have the looks of in point (2.1). so, root solution (2.1)
The draft from this third method can be found when you study the linear equation system more of two variable, matriks, and the quadrate equation.
Exercise 4.7
To determine solution of the equation system with the method graph following.
1. x = 2, y = 3
2. y = x, y = 2
3. y = x + 1, y = 2 x
4. x + y = 3, x + 2 y = 4
5. x + y = 0, 2 x + y = 1
6. x + 2 y = 8, 2 x – y = 3
7. 3 x + 2 y = 8, 2 x – y = 3
8. 2 x – y = 5, x – 2 y = 1
To try !
To make graph of SPLDV
2 x + y = 6
x – y = 1
according to graph, to determine root solution of SPLDV
4.finish story the problem interrelation with system.
Example :
Price two a pencil and a book is Rp. 1.000,00. if price a book Rp. 4000,00 more expensive of price a pencil, how price is a pencil a book ?
Solution
a. understand problem
be cognizant of price two pencil and one book is Rp 1.000,00. price book Rp 4.000,00 more expensive of price a pencil
question : price one pencil and price one book.
b. Arranging planning
The draft used is solution linear equation system two variable using the method substitute.
c.perform planning
let, price a pencil is x and a book is y, be obtained until linear equation system with two variable as following. 2 x + y = 1000 and y = x + 400. for the finish this equation system, to use the method substitute. y = x + 400 substitute at 2 x + y =1000 be obtained :
2 x + y = 1000
2 x (x + 400) =1000
2 x + x + 400 =1000
3 x + 400 = 1000
3 x = 600
x = 200
for x = 200 so, y = 200 + 400
y = 600
so, price a pencil Rp 200,00 and a book Rp 600,00
d. examination
for x = 200 and y = 600, so
2 x + y = 1.000,00
2 (200) + 600 = 1000 (proven)
y = x + 400
600 = 200 + 400
600 = 600 (proven)
Exercise 4.8
1. if add two number is and different is 1, to determine
2. two a corner interspace straight. If different the second corner is 20 degree, to determine each corner’s
3. a certain point (2,3) and (4,5) exhaust at y = m x + c with two variable m and c, to determine m and c.
4. the cost wash two dress and three wastern-style shirt Rp 6.000,00. if the cost wash a dress and wastern – style shirt Rp 2.500,00, to determine each the cost to wash
5. there two corner interspace right. If one fault corner two times of other’s, to determine each the corner following
5.finish SP nonlinear two variable from to used SPLDV
In more discuss the mathematic, ordernary could nonlinear equation system. This non linear equation system more found in counting matriks and vector. For finish nonlinear equation system, change more ago to for from linear equation system. For more explain to attention this example following.
Example :
To determine solution of 6/x + 2/y = 3
3/x + 8/y = 5
Solution :
6/x + 2/y = 3 time xy 6 y + 2 x = 3 xy…..(1)
3/x + 8/y = 5 time xy 3 y + 8 x = 5 xy….. (2)
So obtained new SPLDV
- 14 x = - 7 xy
Substitute y = z to for (1) , be obtained :
6 y + 2 x = 3 xy
6(2) + 2 x = 3 x (2)
12 + 2 x = 6 x
12 = 6 x – 2 x
12 = 4 x
x = 12/4
x = 3
so root solution is (3,2)
exercise 4.9
to determine solution of SPLDV following
1. 2/x + 3/y = 2
6/x – 3/y = 2
2. 9/2x + 4/y = 5
3/x + 8/y = 6
3. 3/x – 2/y = 2
4/x + 2/y = 5
4. 5/x + 3/2x = 6
3/x + 6/y = 7
SUMMARY
1. linear equation with two variable is equation with two variable the rank one highest.
2. solution SPLDV or root SPLDV is change variable until equation at SPLDV following become the sentence true.
3. the way determine root solution SPLDV, it is :
a. substitute
b. eliminate
c. graph
Reflection
From discuss SPLDV at these chapter, which are you more understand of the third method the root determine SPLDV or solution or solution SPLDV ?
Which aren’t you understand ?
What is problem ?
Can you do inlaid work this the method
Wednesday, January 7, 2009
TUGAS 1 B
MATEMATIKA YUNANI SETELAH EUCLIDES
1. Latar Belakang Sejarah
Selama pemerintahan dinasti Ptolomeus, yang berlangsung hamper 300 tahun, kota Iskandaria bebas dari pertentangan-pertentangan didalam maupun diluar. Ini diakhiri dengan suatu masa pertentangan yang singkat sewaktu Mesir menjadi sebagian kerajaan Romawi, dan setelah itu Pax Romana singgah didalam itu. Tidaklah mengherankan bahwa Iskadaria menjadi surga bagi para cendikiawan dan hamper selama seribu tahun banyak hasil ilmiah lahir di kota itu.
Masa akhir pada zaman itu dikuasai oleh Roma, pada 212 SM. Syracusa tunduk pada pengepungan Romawi; pada 146 SM, Carthago jatuh dibawah kekuasaan kerajaan Romawi dan pada tahun yang sama, yang terakhir dari kota Yunani, yaitu Corintha jatuh pula. Yunani menjadi bagian kerajaan Romawi. Mesopotamia tidak ditahlukkan sampai pada 65 SM dan Mesir tetap diperintah keluarga Ptolomeus sampai pada 30 SM. Kebudayaaan Yunani berkembang dalam kehidupan Romawi dan agama Nasrani mulai berkembang, terutama dikalangan budak-budak belian dan kaum miskin. Penguasa-penguasa Romawi memungut pajak-pajak yang beray, tetapi selebihnya tidak mencampuri organisasi ekonomi yang mendasari dari pajajah-penjajah Timur.
Konstatin yang Agung adalah kaisar Romawi pertama yang memeluk agama Kristen dan sebagai agama resmi. Pada tahunh 330, konstatin memindahkan ibu kota dari Roma ke Byzantium, yang diberi nama Konstantinopel. Pada tahun 395, kerajaan Romawi dibagi menjadi dua yaitu kerajaan Timur dan Barat.
Struktur ekonomi dari kedua kerajaan itu pada dasrnya adalah pertanian dengan menggunakan tenaga budak belian. Keadaan serupa menimbulkan kebekuan pada kerja ilmiah yang asli dan pemikiran yang produktif berangsur-angsur menurun terutama dibagian Barat, karena di sana perbudakan dilakukan secara besar-besaran. Pasaran budak belian mulai menurun sampai tingkat biasa. Madzab Iskandaria lenyap bersama hancurnya masyarakat kuno. Pemikiran kreatif diganti penyusunan(komplikasi) dan ulasan. Pada tahun 641, Iskandaria jatuh ketangan Arab.
2. Archimedes
Salah satu ahli matematika yang terbesar sepanjang zaman dan terbesar di zaman kuno. Archimedes adalah penduduk Syracusa. Ia dilahirkan pada 287 SM dan meninggal ketika terjadi perampokan di kota Syracusa oleh orang Romawi pada 212 SM. Ia putra seorang ahli astronomi dan ia merupakan orang terpandang di mata Raja Heron dari Syracusa. Ada riwayat yang menyatakan, bahwa ia singgah beberapa waktu di Mesir, mungkin sekali di Universitas Iskadaria, karena diantara teman-teman yang ia sebut Canon, Dositheus, dan Eratosshenes. Dua orang yang pertama dan terakhir pengganti Euklides adalah seorang pustakawan Universitas. Banyak penemuan matematika Archimedes telah diberitahukan pada orang-orang tersebut.
Ahli sejarah bangsa Romawi meneruskan kisah-kisah yang menarik dari Archimedes. Yang sangat terkenal adalah uraian-uraian tentang alat-alat canggih yang direncanakan untuk mempertahankan Syracusa terhadap pengepungan yang dipimpin jendral Romawi Marcellus. Ada ketapel yang berjarak lempar yang dapat diatur, tiang-tiang yang menonjol yang dapat digerakkan untuk menjatuhkan benda-benda berat di atas kapal-kapal musuh yang dating terlalu dekat pada dinding-dinding kota, dank ran penekan yang besar mengangkat kapal-kapal musuh. Terdapat kisah bahwa ia menggunakan kaca-kaca besar untuk membakar kapal-kapal. Adapun kisah tentang cara ia membuat orang percaya pada pernyataan “Berilah aku tempat untuk berdiri dan aku akan menggerakkan dunia”, dengan memindahkan sebuah kapalyang dibebani berat, tanpa susah payah dan seorang diri dengan menggunakan katrol berganda. Padahal kapal itu dengan susah payah telah ditarik ke atas oleh pekerja-pekerja.
Jelas bahwa Archimedes memiliki kemampuan yang besar untuk memusatkan pikirannya dan ada kisah tentang bagaimana ia melupakan keadaan sekitarnya kalau sedang tenggelam dalam suatu persoalannya, contoh yanh khas ialah cerita yang sering diulangi tentang makota Raja Heron pandai emas yang dicurigai. Agaknya Raja Hero mempunyai sebuah mahkota emas yanh dikhawatirkan mengandung perak yang disembunyikan dan ia menyampaikan masalahnya pada Arcimedes dan Archimedes mandapatkan jawabannya ketika sedang mandi dengan menemukan hokum hydrostatika 1. ia lupa be4rpakaian, keluar dari kamar mandinya dan lari pulang sambil berteriak-teriak “Eureka, Eureka!” disepanjang jalan.
Archimedes banyak mengerjakan geometri dengan menggunakan lukusan-lukisan yang dibuatnyapada minyak-minyak sehabis mandi yang diusapkan pada badanya. Karya-karya Archimedes merupakan karya-karya ulung Ia juga mengarang buku-buku lain yang berjudul Data, Phaenomena, Optika, Unsur Musik, Pembagian Bentuk, Porisme (3 jilid), Pseudoria, Katoptrika, Irisan Kerucut (4 jilid). Tapi buku-buku tersebut sudah musnah.Kita tahu dari laporan orang lain, misalnya laporan Proclus, ahli filsafat Yuanai, yang menulis tentang euclides kira-kira 700 tahun sesudah Euclides meninggal. Kapan dan dimana Euclides lahir, tak ada orang yang tahu. Kapan dan dimana Euclides meningal, juga tak ada orang yang tahu. Meskipun demikian, di bidang geometri Euclides adalah orang yang paling berpengaruh di dunia. Maka tak mengherankan kalau ia sering mendapat julukan geometri. Di bidang matematika, penemuannya terhadap nilai phi lebih mendekati dari ilmuan sebelumnya, yaitu 223/71 dan 220/70Archimedes adalah orang yang mendasarkan penemuannya dengan eksperiman. Sehingga, ia dijuluki Bapak IPA Eksperimental.
3. ERATOSTHENES
Eratosthenes dilahirkan di Cyrene (Libya saat ini), tetapi bekerja dan meninggal di Alexandria. Dia tidak pernah menikah dan dikenal sebagai seorang yang sombong.Eratosthenes belajar di Alexandria dan untuk beberapa tahun di Athena. Pada 236 SM ia ditunjuk oleh Ptolemy III Euergetes I sebagai pustakawan Perpustakaan Alexandria, menggantikan pustakawan pertama, Zenodotos. Dia membuat beberapa sumbangan penting pada matematika dan sains, dan merupakan teman baik Archimedes. Sekitar 255 SM ia menciptakan bola armilar, yang digunakan secara luas hingga diciptakannya oreri pada abad 18. Pada 195 SM ia menjadi buta dan setahun kemudian diduga membiarkan dirinya kelaparan hingga meninggal dunia.Ia dicatat oleh Cleomedes dalam On the Circular Motions of the Celestial Bodies sebagai orang yang telah menghitung keliling Bumi pada sekitar tahun 240 SM, menggunakan metode trigonometri dan pengetahuan mengenai sudut kemiringan Matahari saat tengah hari di Alexandria dan Eratosthenes mengetahui bahwa pada saat titik musim panas pada siang lokal di kota Syene yang terletak di Tropic of Cancer, Matahari akan tampak di zenit, tepat di atas kepala. Ia juga mengetahui dari pengukuran bahwa di kampung halamannya, Alexandria, sudut kemiringan Matahari pada saat yang sama adalah 7,2° di selatan zenit. Dengan asumsi bahwa Alexandria berada di utara Syene ia menyimpulkan bahwa jarak dari Alexandria ke Syene adalah 7,2/360 dari total keliling Bumi. Jarak antara kedua kota tersebut diketahui dari para pedagang/pengelana sekitar 5000 stadia: sekitar 800 km. Dia mendapatkan angka akhir 700 stadia per derajat, yang berarti keliling Bumi adalah 252.000 stadia. Ukuran pasti dari stadion yang dia gunakan saat ini tidak lagi diketahui dengan pasti (ukuran stadion Attic sekitar 185 m), tetapi umumnya dipercaya bahwa keliling Bumi yang dihitung Eratosthenes adalah sekitar 39.690 km.Meskipun metode Eratosthenes cukup baik, akurasi perhitungannya masih terbatas. Akurasi pengukuran Eratosthenes terkurangi oleh fakta bahwa Syene tidaklah tepat berada di Tropic of Cancer, tidak juga tepat berada di selatan Alexandria, dan Matahari sebetulnya adalah sebuah piringan yang berada pada suatu jarak tertentu dari Bumi dan bukan sebuah "sumber titik" pada jarak yang tak hingga. Sumber lain dari galat pengukurannya adalah: ketelitian tertinggi pengukuran sudut pada jaman itu hanyalah seperempat derajat, dan pengukuran jarak melalui perjalanan darat masih diragukan. Maka akurasi dari perhitungan Eratosthenes adalah mengejutkan, sebab keliling Bumi yang diukur melewati kutub-kutubnya saat ini diketahui berharga 40.008 km.Percobaan Eratosthenes pada saat itu sangat dipandang, dan perkiraannya tentang ukuran Bumi diterima hingga ratusan tahun sesudahnya. Metodenya digunakan oleh Posidonius sekitar 150 tahun kemudian.
4. BILANGAN-BILANGAN PRIMA
Suatu kali ada seseorang yang datang kepada saya dengan maksud ingin memulai belajar Java, ia banyak bertanya kepada saya masalah Java dan perkembangannya. Salah satu pertanyaan yang muncul dari salah satu cerita adalah bagaimana membuat sebuah program Java yang mampu menghasilkan deretan bilangan prima?, yang jadi masalah adalah, saya sendiri saat itu sudah lupa apa perngertian bilangan prima, sehingga kita kesulitan untuk membuat programnya. he..he..he..(alasan...!!!).
Kemarin saya teringat kejadian itu, dan segera saya googling mencari tahu pengertian bilangan prima. Dari wikipedia versi indonesia saya mendapatkan pengertian bilangan prima secara baku, kemudian saya coba memahami arti sebenarnya pengertian bilangan prima. Bilangan prima adalah Suatu bilangan asli yang dimulai dari 2, dimana bilangan ini adalah bilangan yang tidak habis dibagi dengan bilangan apapun dibawahnya kecuali dibagi dengan satu dan bilangan itu sendiri. Begitu pemahaman yang saya ambil dari pengertian bilangan prima. Mari kita coba ambil sebuah angka dan coba memeriksanya dengan pemahaman yang saya dapat, misal kita ambil angka 15, apakah angka 15 adalah bilangan prima ? Pemeriksaan kita, akan kita urutkan seperti ini :
1. Apakah 15 lebih besar sama dengan 2 (15 >= 2 ) ?, sebab salah syarat pertama bilangan prima adalah bilangan asli yang dimulai dari 2.
2. Apakah 15 akan habis dibagi dengan bilangan-bilangan yang dimulai dari 2 sampai dengan 14 ?, sebab syarat bilangan prima yang lain adalah bilangan yang tidak habis dibagi dengan bilangan apapun dibawahnya kecuali satu dan dirinya sendiri.
Ok kita mulai pemeriksaan kita satu demi satu, untuk pemeriksaan pada tahap yang pertama 15 lolos, artinya 15 memang lebih besar dari pada 2, kemudian kita lanjutkan dengan pemeriksaan tahao ke dua. Apakah 15 habis jika di bagi dengan 2, hasilnya 15 tidak habis di bagi dengan 2, karena 15/2= 7 dan sisa 1. Pemeriksaan kita lanjutkan dengan membagi lagi 15 dengan 3, pada pembagian yang ini 15 jelas bukanlah bilangan prima, sebab 15 dibagi dengan 3, habis, 15/3 = 5, habis tidak ada sisa. Jelas 15 bukanlah bilangan prima sebab 15 habis di bagi dengan 3.
Sekarang kita akan ambil satu bilangan lagi yaitu 17, dan kita akan coba periksa lagi dengan metode yang sama dengan ketika kita mencari bilangan prima untuk angka 15. Untuk pemeriksaan pertama bilangan 17 pasti lolos sebab 17 lebih besar dari pada 2, dan untuk pemeriksaan ke dua, kita lihat seperti gambar di bawah ini :
17 / 2 = 8 sisa 1
17 / 3 = 5 sisa 2
17 / 4 = 4 sisa 1
17 / 5 = 3 sisa 1
17 / 6 = 2 sisa 5
17 / 7 = 2 sisa 3
17 / 8 = 2 sisa 1
17 / 9 = 1 sisa 8
17 / 10 = 1 sisa 7
17 / 11 = 1 sisa 6
17 / 12 = 1 sisa 5
17 / 13 = 1 sisa 4
17 / 14 = 1 sisa 3
17 / 15 = 1 sisa 2
17 / 16 = 1 sisa 1
Pada gambar diatas, kita bisa melihat bahwa 17 itu selalu sisa jika dibagi dengan bilangan apapun dibawah bilangan 17 kecuali 1 dan 17 itu sendiri. Dari situ kita bisa simpulkan bahwa 17 adalah bilangan prima.
Terjemahan dari buku : AN INTRODUCTION TO THE HISTORY OF MATHEMATICS
Penulis :Howard Eves
Penerbit : the united states of America
MATEMATIKA YUNANI SETELAH EUCLIDES
1. Latar Belakang Sejarah
Selama pemerintahan dinasti Ptolomeus, yang berlangsung hamper 300 tahun, kota Iskandaria bebas dari pertentangan-pertentangan didalam maupun diluar. Ini diakhiri dengan suatu masa pertentangan yang singkat sewaktu Mesir menjadi sebagian kerajaan Romawi, dan setelah itu Pax Romana singgah didalam itu. Tidaklah mengherankan bahwa Iskadaria menjadi surga bagi para cendikiawan dan hamper selama seribu tahun banyak hasil ilmiah lahir di kota itu.
Masa akhir pada zaman itu dikuasai oleh Roma, pada 212 SM. Syracusa tunduk pada pengepungan Romawi; pada 146 SM, Carthago jatuh dibawah kekuasaan kerajaan Romawi dan pada tahun yang sama, yang terakhir dari kota Yunani, yaitu Corintha jatuh pula. Yunani menjadi bagian kerajaan Romawi. Mesopotamia tidak ditahlukkan sampai pada 65 SM dan Mesir tetap diperintah keluarga Ptolomeus sampai pada 30 SM. Kebudayaaan Yunani berkembang dalam kehidupan Romawi dan agama Nasrani mulai berkembang, terutama dikalangan budak-budak belian dan kaum miskin. Penguasa-penguasa Romawi memungut pajak-pajak yang beray, tetapi selebihnya tidak mencampuri organisasi ekonomi yang mendasari dari pajajah-penjajah Timur.
Konstatin yang Agung adalah kaisar Romawi pertama yang memeluk agama Kristen dan sebagai agama resmi. Pada tahunh 330, konstatin memindahkan ibu kota dari Roma ke Byzantium, yang diberi nama Konstantinopel. Pada tahun 395, kerajaan Romawi dibagi menjadi dua yaitu kerajaan Timur dan Barat.
Struktur ekonomi dari kedua kerajaan itu pada dasrnya adalah pertanian dengan menggunakan tenaga budak belian. Keadaan serupa menimbulkan kebekuan pada kerja ilmiah yang asli dan pemikiran yang produktif berangsur-angsur menurun terutama dibagian Barat, karena di sana perbudakan dilakukan secara besar-besaran. Pasaran budak belian mulai menurun sampai tingkat biasa. Madzab Iskandaria lenyap bersama hancurnya masyarakat kuno. Pemikiran kreatif diganti penyusunan(komplikasi) dan ulasan. Pada tahun 641, Iskandaria jatuh ketangan Arab.
2. Archimedes
Salah satu ahli matematika yang terbesar sepanjang zaman dan terbesar di zaman kuno. Archimedes adalah penduduk Syracusa. Ia dilahirkan pada 287 SM dan meninggal ketika terjadi perampokan di kota Syracusa oleh orang Romawi pada 212 SM. Ia putra seorang ahli astronomi dan ia merupakan orang terpandang di mata Raja Heron dari Syracusa. Ada riwayat yang menyatakan, bahwa ia singgah beberapa waktu di Mesir, mungkin sekali di Universitas Iskadaria, karena diantara teman-teman yang ia sebut Canon, Dositheus, dan Eratosshenes. Dua orang yang pertama dan terakhir pengganti Euklides adalah seorang pustakawan Universitas. Banyak penemuan matematika Archimedes telah diberitahukan pada orang-orang tersebut.
Ahli sejarah bangsa Romawi meneruskan kisah-kisah yang menarik dari Archimedes. Yang sangat terkenal adalah uraian-uraian tentang alat-alat canggih yang direncanakan untuk mempertahankan Syracusa terhadap pengepungan yang dipimpin jendral Romawi Marcellus. Ada ketapel yang berjarak lempar yang dapat diatur, tiang-tiang yang menonjol yang dapat digerakkan untuk menjatuhkan benda-benda berat di atas kapal-kapal musuh yang dating terlalu dekat pada dinding-dinding kota, dank ran penekan yang besar mengangkat kapal-kapal musuh. Terdapat kisah bahwa ia menggunakan kaca-kaca besar untuk membakar kapal-kapal. Adapun kisah tentang cara ia membuat orang percaya pada pernyataan “Berilah aku tempat untuk berdiri dan aku akan menggerakkan dunia”, dengan memindahkan sebuah kapalyang dibebani berat, tanpa susah payah dan seorang diri dengan menggunakan katrol berganda. Padahal kapal itu dengan susah payah telah ditarik ke atas oleh pekerja-pekerja.
Jelas bahwa Archimedes memiliki kemampuan yang besar untuk memusatkan pikirannya dan ada kisah tentang bagaimana ia melupakan keadaan sekitarnya kalau sedang tenggelam dalam suatu persoalannya, contoh yanh khas ialah cerita yang sering diulangi tentang makota Raja Heron pandai emas yang dicurigai. Agaknya Raja Hero mempunyai sebuah mahkota emas yanh dikhawatirkan mengandung perak yang disembunyikan dan ia menyampaikan masalahnya pada Arcimedes dan Archimedes mandapatkan jawabannya ketika sedang mandi dengan menemukan hokum hydrostatika 1. ia lupa be4rpakaian, keluar dari kamar mandinya dan lari pulang sambil berteriak-teriak “Eureka, Eureka!” disepanjang jalan.
Archimedes banyak mengerjakan geometri dengan menggunakan lukusan-lukisan yang dibuatnyapada minyak-minyak sehabis mandi yang diusapkan pada badanya. Karya-karya Archimedes merupakan karya-karya ulung Ia juga mengarang buku-buku lain yang berjudul Data, Phaenomena, Optika, Unsur Musik, Pembagian Bentuk, Porisme (3 jilid), Pseudoria, Katoptrika, Irisan Kerucut (4 jilid). Tapi buku-buku tersebut sudah musnah.Kita tahu dari laporan orang lain, misalnya laporan Proclus, ahli filsafat Yuanai, yang menulis tentang euclides kira-kira 700 tahun sesudah Euclides meninggal. Kapan dan dimana Euclides lahir, tak ada orang yang tahu. Kapan dan dimana Euclides meningal, juga tak ada orang yang tahu. Meskipun demikian, di bidang geometri Euclides adalah orang yang paling berpengaruh di dunia. Maka tak mengherankan kalau ia sering mendapat julukan geometri. Di bidang matematika, penemuannya terhadap nilai phi lebih mendekati dari ilmuan sebelumnya, yaitu 223/71 dan 220/70Archimedes adalah orang yang mendasarkan penemuannya dengan eksperiman. Sehingga, ia dijuluki Bapak IPA Eksperimental.
3. ERATOSTHENES
Eratosthenes dilahirkan di Cyrene (Libya saat ini), tetapi bekerja dan meninggal di Alexandria. Dia tidak pernah menikah dan dikenal sebagai seorang yang sombong.Eratosthenes belajar di Alexandria dan untuk beberapa tahun di Athena. Pada 236 SM ia ditunjuk oleh Ptolemy III Euergetes I sebagai pustakawan Perpustakaan Alexandria, menggantikan pustakawan pertama, Zenodotos. Dia membuat beberapa sumbangan penting pada matematika dan sains, dan merupakan teman baik Archimedes. Sekitar 255 SM ia menciptakan bola armilar, yang digunakan secara luas hingga diciptakannya oreri pada abad 18. Pada 195 SM ia menjadi buta dan setahun kemudian diduga membiarkan dirinya kelaparan hingga meninggal dunia.Ia dicatat oleh Cleomedes dalam On the Circular Motions of the Celestial Bodies sebagai orang yang telah menghitung keliling Bumi pada sekitar tahun 240 SM, menggunakan metode trigonometri dan pengetahuan mengenai sudut kemiringan Matahari saat tengah hari di Alexandria dan Eratosthenes mengetahui bahwa pada saat titik musim panas pada siang lokal di kota Syene yang terletak di Tropic of Cancer, Matahari akan tampak di zenit, tepat di atas kepala. Ia juga mengetahui dari pengukuran bahwa di kampung halamannya, Alexandria, sudut kemiringan Matahari pada saat yang sama adalah 7,2° di selatan zenit. Dengan asumsi bahwa Alexandria berada di utara Syene ia menyimpulkan bahwa jarak dari Alexandria ke Syene adalah 7,2/360 dari total keliling Bumi. Jarak antara kedua kota tersebut diketahui dari para pedagang/pengelana sekitar 5000 stadia: sekitar 800 km. Dia mendapatkan angka akhir 700 stadia per derajat, yang berarti keliling Bumi adalah 252.000 stadia. Ukuran pasti dari stadion yang dia gunakan saat ini tidak lagi diketahui dengan pasti (ukuran stadion Attic sekitar 185 m), tetapi umumnya dipercaya bahwa keliling Bumi yang dihitung Eratosthenes adalah sekitar 39.690 km.Meskipun metode Eratosthenes cukup baik, akurasi perhitungannya masih terbatas. Akurasi pengukuran Eratosthenes terkurangi oleh fakta bahwa Syene tidaklah tepat berada di Tropic of Cancer, tidak juga tepat berada di selatan Alexandria, dan Matahari sebetulnya adalah sebuah piringan yang berada pada suatu jarak tertentu dari Bumi dan bukan sebuah "sumber titik" pada jarak yang tak hingga. Sumber lain dari galat pengukurannya adalah: ketelitian tertinggi pengukuran sudut pada jaman itu hanyalah seperempat derajat, dan pengukuran jarak melalui perjalanan darat masih diragukan. Maka akurasi dari perhitungan Eratosthenes adalah mengejutkan, sebab keliling Bumi yang diukur melewati kutub-kutubnya saat ini diketahui berharga 40.008 km.Percobaan Eratosthenes pada saat itu sangat dipandang, dan perkiraannya tentang ukuran Bumi diterima hingga ratusan tahun sesudahnya. Metodenya digunakan oleh Posidonius sekitar 150 tahun kemudian.
4. BILANGAN-BILANGAN PRIMA
Suatu kali ada seseorang yang datang kepada saya dengan maksud ingin memulai belajar Java, ia banyak bertanya kepada saya masalah Java dan perkembangannya. Salah satu pertanyaan yang muncul dari salah satu cerita adalah bagaimana membuat sebuah program Java yang mampu menghasilkan deretan bilangan prima?, yang jadi masalah adalah, saya sendiri saat itu sudah lupa apa perngertian bilangan prima, sehingga kita kesulitan untuk membuat programnya. he..he..he..(alasan...!!!).
Kemarin saya teringat kejadian itu, dan segera saya googling mencari tahu pengertian bilangan prima. Dari wikipedia versi indonesia saya mendapatkan pengertian bilangan prima secara baku, kemudian saya coba memahami arti sebenarnya pengertian bilangan prima. Bilangan prima adalah Suatu bilangan asli yang dimulai dari 2, dimana bilangan ini adalah bilangan yang tidak habis dibagi dengan bilangan apapun dibawahnya kecuali dibagi dengan satu dan bilangan itu sendiri. Begitu pemahaman yang saya ambil dari pengertian bilangan prima. Mari kita coba ambil sebuah angka dan coba memeriksanya dengan pemahaman yang saya dapat, misal kita ambil angka 15, apakah angka 15 adalah bilangan prima ? Pemeriksaan kita, akan kita urutkan seperti ini :
1. Apakah 15 lebih besar sama dengan 2 (15 >= 2 ) ?, sebab salah syarat pertama bilangan prima adalah bilangan asli yang dimulai dari 2.
2. Apakah 15 akan habis dibagi dengan bilangan-bilangan yang dimulai dari 2 sampai dengan 14 ?, sebab syarat bilangan prima yang lain adalah bilangan yang tidak habis dibagi dengan bilangan apapun dibawahnya kecuali satu dan dirinya sendiri.
Ok kita mulai pemeriksaan kita satu demi satu, untuk pemeriksaan pada tahap yang pertama 15 lolos, artinya 15 memang lebih besar dari pada 2, kemudian kita lanjutkan dengan pemeriksaan tahao ke dua. Apakah 15 habis jika di bagi dengan 2, hasilnya 15 tidak habis di bagi dengan 2, karena 15/2= 7 dan sisa 1. Pemeriksaan kita lanjutkan dengan membagi lagi 15 dengan 3, pada pembagian yang ini 15 jelas bukanlah bilangan prima, sebab 15 dibagi dengan 3, habis, 15/3 = 5, habis tidak ada sisa. Jelas 15 bukanlah bilangan prima sebab 15 habis di bagi dengan 3.
Sekarang kita akan ambil satu bilangan lagi yaitu 17, dan kita akan coba periksa lagi dengan metode yang sama dengan ketika kita mencari bilangan prima untuk angka 15. Untuk pemeriksaan pertama bilangan 17 pasti lolos sebab 17 lebih besar dari pada 2, dan untuk pemeriksaan ke dua, kita lihat seperti gambar di bawah ini :
17 / 2 = 8 sisa 1
17 / 3 = 5 sisa 2
17 / 4 = 4 sisa 1
17 / 5 = 3 sisa 1
17 / 6 = 2 sisa 5
17 / 7 = 2 sisa 3
17 / 8 = 2 sisa 1
17 / 9 = 1 sisa 8
17 / 10 = 1 sisa 7
17 / 11 = 1 sisa 6
17 / 12 = 1 sisa 5
17 / 13 = 1 sisa 4
17 / 14 = 1 sisa 3
17 / 15 = 1 sisa 2
17 / 16 = 1 sisa 1
Pada gambar diatas, kita bisa melihat bahwa 17 itu selalu sisa jika dibagi dengan bilangan apapun dibawah bilangan 17 kecuali 1 dan 17 itu sendiri. Dari situ kita bisa simpulkan bahwa 17 adalah bilangan prima.
Terjemahan dari buku : AN INTRODUCTION TO THE HISTORY OF MATHEMATICS
Penulis :Howard Eves
Penerbit : the united states of America
Tugas 2
IF YOU BELIEVE IN YOURSELF
So here you sit, by this time tomorrow I will have ground you way through an hour workout on the bike trainer, slung yourself around the corners of a track at a six minute mile pace. If you believe in your self so other people also will believe in you. All something beginning and growing from yourself. Anywhere, anytime, we able to produce something if you believe in yourself. We don’t worry and weekness because if you believe in your ability so your strengt for produce something will be grow.
IF YOU BELIEVE IN YOURSELF
So here you sit, by this time tomorrow I will have ground you way through an hour workout on the bike trainer, slung yourself around the corners of a track at a six minute mile pace. If you believe in your self so other people also will believe in you. All something beginning and growing from yourself. Anywhere, anytime, we able to produce something if you believe in yourself. We don’t worry and weekness because if you believe in your ability so your strengt for produce something will be grow.
Friday, January 2, 2009
TUGAS 4
A. grammar (tata bahasa)
adalah bagaimana menggunakan bahasa yang tepat untuk membuat kalimat. Tata bahasa pada kalimat adalah ilmu kalimat (syntak). Tipe dasar dari kalimat :
simple sentence (kalimat sederhana)
yaitu semua elemen yang paling dasar/elemen dasar dari kalimat yang terdiri dari subjek/predikat. Subjek adalah yang melakukan pekerjaan.
Contoh :
Anak kecil yang bergembira menendang boneka sampai pagar. (simple sentence)
Anak kecil yang bergembira = subjek; menendang = kata kerja
Predikat dalam simple sentence :
Main verb (kata kerja pokok/inti + sesuatu yang dilakukan. (complete predicate)
Contoh :
1. anak kecil yang bergembira menendang boneka sampai pagar.
Menendang = simple predicate; sampai pagar = complete predicate
2. anak kecil yang bergembira menendang boneka sampai pagar.
Anak kecil yang bergembira = simple subject complete
menendang boneka sampai pagar = complete predicate
simple sentence (kalimat sederhana), tanpa subjek atau predikat
command (perintah) : kalimat langsung untuk orang kedua
• kick that gnome over the fence (tendang boneka itu sampai pagar)
hey, kamu, tendang bola itu sampai pagar
implied predicate (predikat yang dapat disimpulkan)
contoh :
• siapa orang yang menendang boneka itu sampai pagar ? cindy(implied predicate)
• cindy, menendang boneka itu sampai pagar
B.KATA KERJA
Verb, tindakan yang ditunjukan untuk mendiskripsikan keadaan suatu kata benda dan pengganti nama benda.
Contoh : dave berlari. Berlari = kata kerja
Tindakannya, apa yang dilakukan Dave ? dia berlari
Dalam tata bahasa inggris kata kerja diganti bentuknyauntuk menunjukkan siapa yang melakukan tindakkan.
• Saya melakukan……….(I do)
• Kamu melakukan………(you do)
• Dia melakukan…………(he does)
• Kita melakukan………...(we do)
• Mereka melakukan……..(they do)
Subjek kata kerja adalah saya, kamu, mereka, dia perempuan, itu, dia laki-laki.
Dengan menggunakan kata kerja berlari, subjek itu menjadi :
• I run (saya berlari)
• You run (kamu berlari)
• We run (kami berlari)
• She run (dia berlari)
Mengambil kata kerja dan melihat bagaimana itu diganti dengan subjek yang berbeda disebut kata penghubung.
Kata kerja (to be)
I am, you are, he is, she is, we are, they are
I, you, she, he, it (singular subject/subjek tunggal)
We. They (plural subject/subjek majemuk)
Kata benda
Tunggal/majemuk
Contoh : ibu Midori (subjek tunggal)
Yodel bernyanyi (kata kerja)
Tata bahasa penunjuk
Kata benda tunggal
Kata benda majemuk
I yodel (saya bernyanyi dengan suara yodel)
You yodel (kamu bernyanyi dengan suara yodel)
We yodel (kami bernyanyi dengan suara yodel)
They yodel (mereka bernyanyi dengan suara yodel)
She yodel (dia(perempuan) bernyanyi dengan suara yodel)
He yodel (dia(laki-laki) bernyanyi dengan suara yodel)
It yodel
Kata kerja yang diganti bentuk menggantungkan pada apa tititk yang pada waktu bertindak yang dilakukan.
A. grammar (tata bahasa)
adalah bagaimana menggunakan bahasa yang tepat untuk membuat kalimat. Tata bahasa pada kalimat adalah ilmu kalimat (syntak). Tipe dasar dari kalimat :
simple sentence (kalimat sederhana)
yaitu semua elemen yang paling dasar/elemen dasar dari kalimat yang terdiri dari subjek/predikat. Subjek adalah yang melakukan pekerjaan.
Contoh :
Anak kecil yang bergembira menendang boneka sampai pagar. (simple sentence)
Anak kecil yang bergembira = subjek; menendang = kata kerja
Predikat dalam simple sentence :
Main verb (kata kerja pokok/inti + sesuatu yang dilakukan. (complete predicate)
Contoh :
1. anak kecil yang bergembira menendang boneka sampai pagar.
Menendang = simple predicate; sampai pagar = complete predicate
2. anak kecil yang bergembira menendang boneka sampai pagar.
Anak kecil yang bergembira = simple subject complete
menendang boneka sampai pagar = complete predicate
simple sentence (kalimat sederhana), tanpa subjek atau predikat
command (perintah) : kalimat langsung untuk orang kedua
• kick that gnome over the fence (tendang boneka itu sampai pagar)
hey, kamu, tendang bola itu sampai pagar
implied predicate (predikat yang dapat disimpulkan)
contoh :
• siapa orang yang menendang boneka itu sampai pagar ? cindy(implied predicate)
• cindy, menendang boneka itu sampai pagar
B.KATA KERJA
Verb, tindakan yang ditunjukan untuk mendiskripsikan keadaan suatu kata benda dan pengganti nama benda.
Contoh : dave berlari. Berlari = kata kerja
Tindakannya, apa yang dilakukan Dave ? dia berlari
Dalam tata bahasa inggris kata kerja diganti bentuknyauntuk menunjukkan siapa yang melakukan tindakkan.
• Saya melakukan……….(I do)
• Kamu melakukan………(you do)
• Dia melakukan…………(he does)
• Kita melakukan………...(we do)
• Mereka melakukan……..(they do)
Subjek kata kerja adalah saya, kamu, mereka, dia perempuan, itu, dia laki-laki.
Dengan menggunakan kata kerja berlari, subjek itu menjadi :
• I run (saya berlari)
• You run (kamu berlari)
• We run (kami berlari)
• She run (dia berlari)
Mengambil kata kerja dan melihat bagaimana itu diganti dengan subjek yang berbeda disebut kata penghubung.
Kata kerja (to be)
I am, you are, he is, she is, we are, they are
I, you, she, he, it (singular subject/subjek tunggal)
We. They (plural subject/subjek majemuk)
Kata benda
Tunggal/majemuk
Contoh : ibu Midori (subjek tunggal)
Yodel bernyanyi (kata kerja)
Tata bahasa penunjuk
Kata benda tunggal
Kata benda majemuk
I yodel (saya bernyanyi dengan suara yodel)
You yodel (kamu bernyanyi dengan suara yodel)
We yodel (kami bernyanyi dengan suara yodel)
They yodel (mereka bernyanyi dengan suara yodel)
She yodel (dia(perempuan) bernyanyi dengan suara yodel)
He yodel (dia(laki-laki) bernyanyi dengan suara yodel)
It yodel
Kata kerja yang diganti bentuk menggantungkan pada apa tititk yang pada waktu bertindak yang dilakukan.
TUGAS I A
Pythagorean Theorem
For a right triangle with legs a and b and hypotenuse c ,
a^2 + b^2 = c^2 (1)
Many different proofs exist for this most fundamental of all geometric theorems. The theorem can also be generalized from a plane triangle to a trirectangular tetrahedron, in which case it is known as de Gua's theorem. The various proofs of the Pythagorean theorem all seem to require application of some version or consequence of the parallel postulate: proofs by dissection rely on the complementarity of the acute angles of the right triangle, proofs by shearing rely on explicit constructions of parallelograms, proofs by similarity require the existence of non-congruent similar triangles, and so on (S. Brodie). Based on this observation, S. Brodie has shown that the parallel postulate is equivalent to the Pythagorean theorem.
After receiving his brains from the wizard in the 1939 film The Wizard of Oz, the Scarecrow recites the following mangled (and incorrect) form of the Pythagorean theorem, "The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side." In the fifth season of the television program The Simpsons, Homer J. Simpson repeats the Scarecrow's line (Pickover 2002, p. 341). In the Season 2 episode "Obsession" (2006) of the television crime drama NUMB3RS, Charlie's equations while discussing a basketball hoop include the formula for the Pythagorean theorem.
A clever proof by dissection which reassembles two small squares into one larger one was given by the Arabian mathematician Thabit ibn Kurrah (Ogilvy 1994, Frederickson 1997).
Another proof by dissection is due to Perigal (left figure; Pergial 1873; Dudeney 1958; Madachy 1979; Steinhaus 1999, pp. 4-5; Ball and Coxeter 1987). A related proof is accomplished using the above figure at right, in which the area of the large square is four times the area of one of the triangles plus the area of the interior square. From the figure, d = b-a , so
4(1/2 ab) + d^2 (2)
2ab + (b-a)^2 (3)
2ab + b^2 - 2ab + a^2 (4)
a^2 + b^2 (5)
c^2 (6)
The Indian mathematician Bhaskara constructed a proof using the above figure, and another beautiful dissection proof is shown below (Gardner 1984, p. 154).
c^2 + 4(1/2 ab) = (a + b)^2 (7)
c^2 + 2ab = a^2 + 2ab + b^2 (8)
c^2 = a^2 + b^2 (9)
Several beautiful and intuitive proofs by shearing exist (Gardner 1984, pp. 155-156; Project Mathematics!).
Perhaps the most famous proof of all times is Euclid's geometric proof (Tropfke 1921ab; Tietze 1965, p. 19), although it is neither the simplest nor the most obvious. Euclid's proof used the figure below, which is sometimes known variously as the bride's chair, peacock tail, or windmill. The philosopher Schopenhauer has described this proof as a "brilliant piece of perversity" (Schopenhauer 1977; Gardner 1984, p. 153).
Let triangle ABC be a right triangle,squere CAFG,squere CBKH, squere ABED and CL parallel BE be squares,triangle FAB and triangle CAD. The triangles and are equivalent except for rotation, so
2 triangle FAB = 2 triangle CAD (10)
Shearing these triangles gives two more equivalent triangles
2 triangle CAD = ADLM (11)
Therefore,
squere ACGF = ADLM (12)
Similarly,
Squere BC = 2 triagle ABK (13)
so
a^2 + b^2 = cx + c(c-x) = c^2 (14)
Heron proved that AK, CL and BF intersect in a point (Dunham 1990, pp. 48-53).
Heron's formula for the area of the triangle, contains the Pythagorean theorem implicitly. Using the form
K = 1/4 root of 2a^2b^2 + 2a^2c^2 + 2b^2c^2-(a^4 + b^4 + c^4) (15)
and equating to the area
K = 1/2 AB (16)
gives
1/4 a^2b^2 = 1/16 [2a^2b^2 + 2a^2c^2 + 2b^2c^2-(a^4 + b^4 + c^4) (17)
Rearranging and simplifying gives
a^2 + b^2 = c^2 (18)
the Pythagorean theorem, where is the area of a triangle with sides a, b,c and (Dunham 1990, pp. 128-129).
A novel proof using a trapezoid was discovered by James Garfield (1876), later president of the United States, while serving in the House of Representatives (Gardner 1984, pp. 155 and 161; Pappas 1989, pp. 200-201; Bogomolny). 1/2 [bases * altitude] (19)
1/2 (a + b) (a + b) (20)
1/2 ab + 1/2 ab + 1/2 c^2 (21)
Rearranging,
1/2 (a^2 + 2ab + b^2) = ab + 1/2 c^2 (22)
a^2 + 2ab + b^2 = 2ab + C^2 (23)
a^2 + b^2 = c^2 (24)
An algebraic proof (which would not have been accepted by the Greeks) uses the Euler formula. Let the sides of a triangle be a , b and c , and the perpendicular legs of right triangle be aligned along the real and imaginary axes. Then
a^2 + bi = ce^-18 (25)
Taking the complex conjugate gives
a^2 - bi = ce^-18 (26)
Multiplying (25) by (26) gives
a^2 + b^2 = c^2 (27)
(Machover 1996).
Another algebraic proof proceeds by similarity. It is a property of right triangles, such as the one shown in the above left figure, that the right triangle with sides x , a ,d and (small triangle in the left figure; reproduced in the right figure) is similar to the right triangle with sides d , b ,y and (large triangle in the left figure; reproduced in the middle figure), giving
x/a = a/c y/b = b/c (28)
x = a^2/c y = b^2/c (29)
so
c = x + y = a^2/c + b^2/c = (a^2 + b^2)/c (30)
c^2 = a^2 + b^2 (31)
(Gardner 1984, p. 155 and 157). Because this proof depends on proportions of potentially irrational numbers and cannot be translated directly into a geometric construction, it was not considered valid by Euclid.
http://mathworld.wolfram.com/PythagoreanTheorem.html
Pythagorean Theorem
For a right triangle with legs a and b and hypotenuse c ,
a^2 + b^2 = c^2 (1)
Many different proofs exist for this most fundamental of all geometric theorems. The theorem can also be generalized from a plane triangle to a trirectangular tetrahedron, in which case it is known as de Gua's theorem. The various proofs of the Pythagorean theorem all seem to require application of some version or consequence of the parallel postulate: proofs by dissection rely on the complementarity of the acute angles of the right triangle, proofs by shearing rely on explicit constructions of parallelograms, proofs by similarity require the existence of non-congruent similar triangles, and so on (S. Brodie). Based on this observation, S. Brodie has shown that the parallel postulate is equivalent to the Pythagorean theorem.
After receiving his brains from the wizard in the 1939 film The Wizard of Oz, the Scarecrow recites the following mangled (and incorrect) form of the Pythagorean theorem, "The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side." In the fifth season of the television program The Simpsons, Homer J. Simpson repeats the Scarecrow's line (Pickover 2002, p. 341). In the Season 2 episode "Obsession" (2006) of the television crime drama NUMB3RS, Charlie's equations while discussing a basketball hoop include the formula for the Pythagorean theorem.
A clever proof by dissection which reassembles two small squares into one larger one was given by the Arabian mathematician Thabit ibn Kurrah (Ogilvy 1994, Frederickson 1997).
Another proof by dissection is due to Perigal (left figure; Pergial 1873; Dudeney 1958; Madachy 1979; Steinhaus 1999, pp. 4-5; Ball and Coxeter 1987). A related proof is accomplished using the above figure at right, in which the area of the large square is four times the area of one of the triangles plus the area of the interior square. From the figure, d = b-a , so
4(1/2 ab) + d^2 (2)
2ab + (b-a)^2 (3)
2ab + b^2 - 2ab + a^2 (4)
a^2 + b^2 (5)
c^2 (6)
The Indian mathematician Bhaskara constructed a proof using the above figure, and another beautiful dissection proof is shown below (Gardner 1984, p. 154).
c^2 + 4(1/2 ab) = (a + b)^2 (7)
c^2 + 2ab = a^2 + 2ab + b^2 (8)
c^2 = a^2 + b^2 (9)
Several beautiful and intuitive proofs by shearing exist (Gardner 1984, pp. 155-156; Project Mathematics!).
Perhaps the most famous proof of all times is Euclid's geometric proof (Tropfke 1921ab; Tietze 1965, p. 19), although it is neither the simplest nor the most obvious. Euclid's proof used the figure below, which is sometimes known variously as the bride's chair, peacock tail, or windmill. The philosopher Schopenhauer has described this proof as a "brilliant piece of perversity" (Schopenhauer 1977; Gardner 1984, p. 153).
Let triangle ABC be a right triangle,squere CAFG,squere CBKH, squere ABED and CL parallel BE be squares,triangle FAB and triangle CAD. The triangles and are equivalent except for rotation, so
2 triangle FAB = 2 triangle CAD (10)
Shearing these triangles gives two more equivalent triangles
2 triangle CAD = ADLM (11)
Therefore,
squere ACGF = ADLM (12)
Similarly,
Squere BC = 2 triagle ABK (13)
so
a^2 + b^2 = cx + c(c-x) = c^2 (14)
Heron proved that AK, CL and BF intersect in a point (Dunham 1990, pp. 48-53).
Heron's formula for the area of the triangle, contains the Pythagorean theorem implicitly. Using the form
K = 1/4 root of 2a^2b^2 + 2a^2c^2 + 2b^2c^2-(a^4 + b^4 + c^4) (15)
and equating to the area
K = 1/2 AB (16)
gives
1/4 a^2b^2 = 1/16 [2a^2b^2 + 2a^2c^2 + 2b^2c^2-(a^4 + b^4 + c^4) (17)
Rearranging and simplifying gives
a^2 + b^2 = c^2 (18)
the Pythagorean theorem, where is the area of a triangle with sides a, b,c and (Dunham 1990, pp. 128-129).
A novel proof using a trapezoid was discovered by James Garfield (1876), later president of the United States, while serving in the House of Representatives (Gardner 1984, pp. 155 and 161; Pappas 1989, pp. 200-201; Bogomolny). 1/2 [bases * altitude] (19)
1/2 (a + b) (a + b) (20)
1/2 ab + 1/2 ab + 1/2 c^2 (21)
Rearranging,
1/2 (a^2 + 2ab + b^2) = ab + 1/2 c^2 (22)
a^2 + 2ab + b^2 = 2ab + C^2 (23)
a^2 + b^2 = c^2 (24)
An algebraic proof (which would not have been accepted by the Greeks) uses the Euler formula. Let the sides of a triangle be a , b and c , and the perpendicular legs of right triangle be aligned along the real and imaginary axes. Then
a^2 + bi = ce^-18 (25)
Taking the complex conjugate gives
a^2 - bi = ce^-18 (26)
Multiplying (25) by (26) gives
a^2 + b^2 = c^2 (27)
(Machover 1996).
Another algebraic proof proceeds by similarity. It is a property of right triangles, such as the one shown in the above left figure, that the right triangle with sides x , a ,d and (small triangle in the left figure; reproduced in the right figure) is similar to the right triangle with sides d , b ,y and (large triangle in the left figure; reproduced in the middle figure), giving
x/a = a/c y/b = b/c (28)
x = a^2/c y = b^2/c (29)
so
c = x + y = a^2/c + b^2/c = (a^2 + b^2)/c (30)
c^2 = a^2 + b^2 (31)
(Gardner 1984, p. 155 and 157). Because this proof depends on proportions of potentially irrational numbers and cannot be translated directly into a geometric construction, it was not considered valid by Euclid.
http://mathworld.wolfram.com/PythagoreanTheorem.html
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