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