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TIAR AYU K.
11313244013
Pend. Matematika Internasional 2011
English Assignment 1
Mathematics for Junior High School Grade 2

CHAPTER 1
ALGEBRA AND ITS APPLICATION
 page 34 , no. 8
    The simplified form of   (12x+6)/(4x^2- 1)  is
solution :
The nominator and dominator of fraction (12x+6)/(4x^2- 1) is 12x + 6 and 4x^2 – 1. The nominator and dominator have similar fraction, which ( 2x + 1 ). Therefore
(12x+6)/(4x^2- 1) =  (6 ( 2x+1))/(2x+1)(2x-1)  = (6 )/(2x-1) .

CHAPTER  3
EQUATIONS OF A STRAIGHT  LINE
 page 103 , no. 14
The equation of line s through point (-2,5) and has a gradient   2/3 is
solution :
we have point p (-2,5) has values x_1 = -2 and y_1 = 5 . where the gradient of the line is   2/3 . Therefore m =   2/3 , and then to find the line equation through point (x_1,y_1 ) and has gradient m, we use formula y-y_(1 ) = m ( x - x_1 ).
therefore the equation line is :
            y-y_1 = m ( x - x_1 )
            y – 5 = 2/3 ( x + 2 )
            y – 5 = 2/3 x + 4/3
                  y = 2/3 x + 4/3 + 5
                  y = 2/3 x + 61/3
thus, the equation of the line through p (-2,5) and has a gradient of 2/3 is y = = 2/3 x + 6 1/3

CHAPTER 7
POLYHEDRAL
 page 276, no. 13
    Given the lenght side of the base of a square based pyramid is 4cm. If the height of the lateral face is 10 cm, then the surface area of the pyramid is ...
Solution :
Given AB = 4cm and TE = 10cm
    The area of ABCD      = side x side
                                       = 4 X 4 = 16
    so, the area of ABCD is 16cm^2
The area of triangle BCT = 1/2 x area of the base x height
                                       = 1/2 x 4 x 10
                                       = 20
    so, the area of triangle BCT is 20cm^2
The surface area of pyramid T.ABCD = area of ABCD + ( 4 x area of triangle BCT )
                                                          = 16 + ( 4 x 20  )
                                                          = 96
    so, the surface area of pyramid T. ABCD is 96cm^2.

CHAPTER 4
SYSTEM OF LINEAR EQUATIONS IN TWO VARIABLES
page 128, no. 11
    Mono is seven year older than Edi. While cahyo is four year older then Mono. If the sum of Mono's age and Edi's age is equal to Cahyo's age, then Mono is ... year old.
solution :
Let  A is age cahyo , B is age of mono , and C is age of edi. where the age is older than mono 7 age edi, then it is analogika b = (c +7), while the older age Cahyo 4 age of mono, then it is analogika a = (b +4), and the last one is that if the old mono plus age Cahyo edi same age, then his analogika a = (b + c).
 Therefore , a=(b+4) -----> b=(c+7) and we insert it into the value b ...,  a=(c+7+4) => a=(c+11). And then the input value a = (c +11) and the value of b = (c +7) into a = (b + c) to be c +7 +11 = c + c which produces  c +7 +11 = 2c . Now we solve these equations, c we moved to the left and 7 right to change its sign to negative, then we get the result c is c = 4 . Last step is to input the results of c = 4 in analogika b = (c +7), so that a b = (4 +7) and we get the result b = 11.
so, age is 11 years old mono.

CHAPTER 5
THE PYTHAGOREAN THEOREM
page 154, no. 3
    A rectangle has length and width are 12 cm and 9cm length respectively. The length of the diagonal is
Solution :
Given A rectangle has length and width are 12cm and 9cm and then we must find the length of the diagonal. we can use formula of The Pythagorean Theorem for the solution of the problem as follows.
 c^2 = a^2 + b^2
 c^2 = 12^2 +  9^2
 c^2 = 144 + 81
 c^2 = 225
 c     = √225
 c     = 15
Thus, the length of the diagonal of a rectangle is 15cm