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  1. J

    Mathematics: Post your doubts here!

    Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!! much thanks
  2. J

    Mathematics: Post your doubts here!

    Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!! yes
  3. J

    Mathematics: Post your doubts here!

    Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!! Here's another one: f-1 (inverse f - how do i write that?) = px + q g-1 (inverse g) = 3-2x. Find p and q if gf9x) = 5/2 - x So I worked that f(x) = (x-q)/p and that g(x) = (3-x)/2 ..... correct? So that makes gf(x) = (3-((x-q)/p))/2...
  4. J

    Mathematics: Post your doubts here!

    Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!! oh, you mean we can just compare the left and right? no need further workings? thanks! (sorry about the later edit ....)
  5. J

    Mathematics: Post your doubts here!

    Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!! Please help. Question: f:x -> (x+2)/a; a not 0. g:x -> 2x^2+1. If gf(x) = 2x^2 - bx +9, find the values of a and b. The answer is a=1, b= -8 and a= -1, b= -8 So here is what I worked out so far: gf(x) = 2 ( (x+2)/a )^2 + 1 = 2x^2...
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