Number 7

T : v -> v by definition :

If P (x,y) then :

i)                     T (P) = ( x+1,y) for x ≥ 0

ii)                    T (P) = ( x -1,y) for x < 0

Answer :

Is T injective ?

If A prapeta for point of A (x,y)

X + 1,y = x,y = y

X = x,y – 1,y

Y = y

For example A (x,y1) and B (x2, y2)

T (A) = (x1+1,y1) and T(B) = (x2-1, y2)

So x1 + 1 = x2 – x1 and y1 = y2, such that x1 = x2 and y1 = y2 so A = B

Is T transformation ?

If T injective and T is bijective domain. So that T is transformation from V to V.

Number 3

Ascertainable : three point of A, R, S aren’t collinear. They have domain T that definition :

T (A) = A. T (P)= P’ such that P as point center of AP

The Question :

      a.       Draw R’ = T(R)

      b.       Draw Z such that T (Z) = S

      c.        Is T transformation ?

 

The explained of c :

Is that function v to v ?...

Yes, because every element has onto functions.

      Ø  Is that surjektif ? yes, because they have P’ = T (P),  A  = T(A). the definition at question

      Ø  Is that Injektif ? yes, because every domain has different relation.

Those are proved transformation

 

Number 2

Number 4

Ascertainable : P = {0,0}  C1 = {(x,y) | x^2 + y^2 =1}

                                            C1 = {(x,y) | x^2 + y^2 =25}

T  : C1 -> C2 are domain that definition :

If x members of  C1 then T (x) = x^1 = p segment of x slice of C2

Question :

·         If A = (0,1). Determine of T(A)

·         Determine range of B (4,3)

·         If z arbitrary of point at range of T.

Determine the distance of ZZ’ with Z’ = T (Z)

·         If E and F are two point at range of T.  are they can called that  E’F’ ?

Answer:

      a.A = (0,1)

    T (A) = (0,5)

      b.Through  point of (0,0) and (4,3)

    y-y1/y2-y1 = x-x1/x2-x1

    y-0/3-0 = x-0/4-0

    x=4/3y

   16y^2 + y1/9 =1

   16 y^2 + 9y^2/9 =1

   9 = 25 y^2

   Y^2 = 9/25

   Y = 3/5

     c. Zz’ = Domain C1

    Z’ (T(x)) = Range E

    Then G-C1 = J-T = 4

      d. E’F’/EF = ¼ of circle circumference/ ¼ of circle circumference

                  = r2 /r1 = 5/1

    E’F’= 5 EF