Transformasi Geometri Matematika Translasi Refleksi Rotasi Dilatasi

Pengertian Transformasi

Transformasi adalah pemetaan suatu titik A pada suatu bidang ke titik A' dan titik A' tersebut adalah bayangan dari titik A.

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Jenis-Jenis Transformasi Geometri

Translasi (Pergeseran)

Suatu transformasi yang memindahkan setiap titik pada bidang dengan jarak dan arah tertentu.

Rumus Translasi (Pergeseran)

A(x, y) --> A' (x + x', y + y')

Tentukan bayangan titik A(2, 3) B(-1, 2) C(2, -3) oleh translasi [ 2 , 3 ]

A(2, 3) --> = A'(2 + 2, 3 + 3) = A'(4, 6)

B(-1, 2) --> = B'(-1 + 2, 2 + 3) = B'(1, 5)

C(2, -3) --> = C'(2 + 2, -3 + 3) = C'(4, 0)

Rumus Induksi Elektromagnetik Fisika SMA dan Pembahasan Soal

Mencari Fluks, Medan Magnetik, GGL Induksi

1. Sebuah medan magnetik yang besarnya 0,8 T berarah ke sumbu X. sebuah kumparan sekunder dengan panjang sisi 5 cm memiliki satu lilitan dan membuat sudut tetha terhadap sumbu. tentukan fluks magnetik melalui kumparan ketika :

 Jawab

Pembahasan Soal Bunga Majemuk dan Anuitas Matematika SMA

Bunga Majemuk

Sebuah modal sebesar $M_o$ (modal pokok/awal), dibungakan selama n periode dengan sistem bunga majemuk sebesar b = i% per periode

Modal tersebut setelah n periode ditentukan oleh :

$M_t$ = $M_o$ ${(1 + b)}^n$

Anuitas

Dalam teori keuangan adalah suatu rangkaian penerimaan atau pembayaran tetap yang dilakukan secara berkala pada jangka waktu tertentu.

karena tetap, nilai anuitas tiap pembayaran sama besar. yang berbeda proporsi angsuran dan bunganya.

Anuitas = angsuran + bunga

     

Rumus Bunga Tunggal Biasa dan Eksak Matematika SMA dan Pembahasan Soal

Maksud Bunga Tunggal Biasa & Bunga Tunggal Eksak

Bunga tunggal biasa, 1 bulan dianggap 30 hari. Sehingga dalam satu tahun ada 12 x 30 = 360 hari. 

Bunga tunggal eksak, berdasarkan jumlah hari sesungguhnya, ada dua macam :

-Kabisat = 365 hari (tahun tidak bisa dibagi 4 contoh : 2015, 2017, 2018)

PHYSICS – STATIC ELECTRIC – ELECTRIC FIELD

PHYSICS – STATIC ELECTRIC – ELECTRIC FIELD

Electric field is defined as the electr force per unit charge. The electric field radially outward from positive charge and radially toward a negative point charge.

 


Define the electrc field in the center of the square if :

a.   The 4 charges are negative

b.   4 charges  in sequences negative-positive-negative-positive

c.   4 charges  in sequences positive-positive-negative-negative

d.   One of the charge is negative and the others are positive

a. The 4 charges are negative :

there are the same amount electric field towards each charges but opposite directions, so, all electric fields get eliminated. And the center have no electric field left. Zero electric field

b. The 4 charges in sequences negative-positive-negative-positive:

A and C charges have same amount electric field (because same amount of charges), but opposite directions, so do C-B charges.

Rumus Listrik Statis Fisika SMA dan Pembahasan Soal

1. Diagram di samping menunjukkan tiga bola kecil bermuatan listrik. Besar gaya yang bekerja pada bola B adalah … 

(k = 9 x 109 N m2/c2)

Pembahasan

Contoh Soal Rumus Listrik Statis Fisika 1

1. Diagram di samping menunjukkan tiga bola kecil bermuatan listrik. Besar gaya yang bekerja pada bola B adalah … 

(k = 9 x 109 N m2/c2)

Jawab

Pembahasan Soal Fungsi Matematika

1.      G(x) = x² – x + 3

      Fog)(x) = 3x² – 3x + 4

      F(x-2) = ....

    

Function Composition

WELCOME EVERYBODY! ^^ TODAY I'LL POST ABOUT FUNCTION COMPOSITION, HIGHSCHOOL GRADE. THIS MATERIAL IS SO CHALLENGING. BUT YOU CAN GET USED TO IT! ^^
 

The Basic Characteristic Of Function Compositions

       I.          Composition Function is :

( f o g ) (x) =  f( g(x) )  = f composition g for x

Example ^^ :

1.  f(x) = 5x – 4 and g(x) = 2x + 8

( f o g )? = f( g(x) ) 

f( 2x + 8 )  ----->  5( 2x + 8 ) – 4 ---> 10x + 40 – 4 ----> 10x + 36

as you can see above ...

Basically change the X value of F(x) into g(x) so become f( g(x) ) = f( 2x + 8 )   and because the F(x) = 5x – 4 logically F( 2x + 8) = 5( 2x + 8) – 4     ^^   ----->   10x + 8

How about ( g o f )?

Value still the same : f(x) = 5x – 4 and g(x) = 2x + 8

( g o f ) =  g( f(x) )

g(5x -4) ------>  2(5x – 4) + 8   ----->  10x – 8 + 8 = 10x    ^^

The answer is different so this is mean ( f o g )  IS NOT THE SAME AS ( g o f )

2.      F(x) = x² + 2x – 3 and g(x) = 5x -1

( f o g )? = f( g(x) ) 

f(5x – 1) ------>  (5x – 1)² + 2(5x – 1) – 3

               ------->  25x – 10x + 1 + 10x – 2 - 3

               ------->  (25x – 4) the answer!

How about ( g o f )(-1) ?

Value still the same :  F(x) = x² + 2x – 3 and g(x) = 5x - 1

( g o f ) =  g( f(x) ) ---> ( g o f )(-1) = g( f(-1) )

5(x² + 2x – 3) – 1 ----->  What was asked is ( g o f )(-1) . so change the x from ( f(x) )into value -1 FIRST like this  -----> 5((-1)² + 2(-1) – 3) – 1 

---> 5(1 -2 – 3) – 1

---> 5(-4) – 1 = -21 the answer!

Here some other Characteristic ^^   :

1.  Function Compositions is NOT commutative 

           ( f o g )  IS NOT THE SAME AS ( g o f )

     2.  Function Compositions IS Associative

            ( ( f o g ) o h)(x)  =  ( f o (g o h ) )(x)

     3.  There is Identity Function I(x) = x

           so ---> ( f o I )(x) = ( I o f )(x) = f(x)

How To Determine Circle Equations

II.        LOOK FOR CIRCLE EQUATIONS ^^

         Now the reverse, how to find the equation of the circle if known its center and radius.

     Determine the equation of the circle below :

1.     Center (0,0) r = 3

2.     Center (-3, -1) r = 6
3.   Point (6, 8) is on a Circle A the center is (0.0). The circle equations?
            Answer 

  1. For the center (0,0) The circle equation  : x² + y² = r²     

o   So, the equation circle = x² + y² = 3

---

2. For the center not (0,0) r The circle equation  : (x-a)² + (y-b)² =r²

o   So, the equation circle = (  x-(-3) )² + ( y – (-8) )² = 6²

o   The equation circle = (x+3)² + (y+8)² = 6²

---

3. For the Center (0,0) and point (6.8) are in the circle, 
because passed on the circle. (6.8) as (x, y)
• Equation circle when center (0,0) = x² + y² = r²   (x,y) = (6,8)
Because radius unknown we seek it first:
x² + y² = r²
6² + 8² = r²
r² = 100
then the equation circle = x² + y² = 10²

---

So now, how to determine Circle Equations if the circle passed on some 3 points  that located in the circle itself ^^

Example :

 A Circle x² + y² + Ax + By + C = 0 passed on 3 points (2,0) , (0, -2) , (4, -2) The Circle Equations is?

 

Insert the points to the x² + y² + Ax + By + C = 0 as (x,y)

(2,0) ---> 2² + 0² + A(2) + B.0 + C = 0

                 4 + 2A + 0 + C = 0 ........................ (1)

(0, -2) --> 0² + (-2)² + A.0 + B(-2) + C = 0

                             4 – 0 - 2B + C = 0 .......................... (2)

(4, -2) ---> 4² + (-2)² + A(4) + B(-2) + C = 0

                  16 + 4 + 4A - 2B +  C = 0

                  20 + 4A - 2B +  C = 0 ....................(3)

 
there are 3 quations and let's solve the equations!

Eliminate (1) dan (2)

4 + 2A + 0 + C = 0 .... (1) 

4 – 0 - 2B + C = 0 ..... (2)        

--------------------------------   -

0 + 2A + 2B + 0 = 0 .... (4)  

 

Eliminate (3) dan (2)

20 + 4A - 2B +  C = 0 ...(3)

4 – 0 - 2B + C = 0 ..... (2)    

------------------------------    -

 16 + 4A = 0

 

A = -4 -------> Substitute A as -4 to .... (4)

2A + 2B  = 0 ---> -8 + 2B = 0 --> B =4 

Substitute A and B as its value to equations number (1) / (2) / (3) 

4 + 2A + 0 + C = 0 .... (1) 
 4 – 8 + 0 + C = 0 

C = 4

after we found all the value A, B, C insert them to : x² + y² + Ax + By + C = 0 

So the Circle Equations  x² + y² - 4x + 4y + 4 = 0 ^^