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Math Help - All igcse questions

  1. #1
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    Question All igcse questions

    hey everybody
    i couldnt figure out this question from a past paper about quadratic formula
    it was pretty straight foward-
    the total area of rectangle R and Q is 64cm^2
    calculate the value of x correct to one decimal place
    5x^2+30x+24
    i got 0.95 (for subtraction) and 5.05

    i am jus confused on what i am suppose to choose so that the answer is 64
    plus forgot what it meant by correct to one decimal place...was it rounding up? ...n also 3 significant figures....actually i m mixed up in both of them -.-

    thanks
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  2. #2
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    Quote Originally Posted by darkpinkish View Post
    hey everybody
    i couldnt figure out this question from a past paper about quadratic formula
    it was pretty straight foward-
    the total area of rectangle R and Q is 64cm^2
    calculate the value of x correct to one decimal place
    5x^2+30x+24
    i got 0.95 (for subtraction) and 5.05

    i am jus confused on what i am suppose to choose so that the answer is 64
    plus forgot what it meant by correct to one decimal place...was it rounding up? ...n also 3 significant figures....actually i m mixed up in both of them -.-

    thanks
    From what I can gather 64 is what the quadratic function is equal to.

    5x^2-30x+24=64 which is equal to (x-2)(x-4)=0

    I have no idea how you got the quadratic
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  3. #3
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    same here...but the asnwer is 1.1,
    from the first part of the question i've arrived to 5x^2+30x+24
    and the second part said find x correct 1 decimal place and the total area equals 64cm^2
    i got the answer as 0.95 and 5.05
    i was wondering if 0.95 corrected to one decimal place is 1.1, since i m confused between corrct to 1 decimal place and correct to 1 significant figure
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  4. #4
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    Quote Originally Posted by darkpinkish View Post
    hey everybody
    i couldnt figure out this question from a past paper about quadratic formula
    it was pretty straight foward-
    the total area of rectangle R and Q is 64cm^2
    calculate the value of x correct to one decimal place
    5x^2+30x+24
    Surely the "past paper" you got this from must have some information about "Rectangle R and Q" and what they have to do with 5x^2+ 30x+ 24!

    As stated they don't appear to have anything to do with one another and there is no way to find "x" with just a polynomial and not an equation.

    i got 0.95 (for subtraction) and 5.05

    i am jus confused on what i am suppose to choose so that the answer is 64
    plus forgot what it meant by correct to one decimal place...was it rounding up? ...n also 3 significant figures....actually i m mixed up in both of them -.-

    thanks
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  5. #5
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    Arrow Functions

    Find ff^-1(8)

    and i even want to know how is it different from f^-1(x) , the inverse

    =)
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  6. #6
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by darkpinkish View Post
    Find ff^-1(8)

    and i even want to know how is it different from f^-1(x) , the inverse

    =)
    ff^{-1}(8) = 8. This is because a function and it's inverse cancel each other out.

    ff^{-1}(x) means substitute the inverse of f(x) into the original f(x). However, because these two are "opposite" functions the answer is always x.

    See spoiler for examples:

    Spoiler:
    Example 1:

    f(x) = \frac{a}{x} \, , \: a,x \neq 0 then f^{-1}(x) = \frac{a}{x}.

    Therefore ff^{-1}(x) = \frac{a}{\frac{a}{x}} = x. This follows because f(x) = f^{-1}(x)

    Example 2:

    f(x) = x^2 \, , \: x \geq 0

    y = x^2 \: \rightarrow \: x = \sqrt{y}

    \therefore f^{-1}(x) = \sqrt{x} \: , \: x \geq 0

    ff^{-1}(x) = (\sqrt{x})^2 = x



    If you have any more questions on functions and their inverses I suggest you start a new thread
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