Recent content by Badsiii

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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon $State the range and domain for$ $i)$ \ \ \ \sqrt{4-x^2}
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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon 0 lol (x-x)
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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Ln(xroot3)^2=ln(6-7x) 3x^2=6-7x 3x^2+7x-6=0 X=2/3 or x=/=-3 Therefore x=2/3
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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Shut it
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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Correction: 8x^9sinxcosx
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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon It's correct now
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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Sorry I read it as twelve not twenty T10= 39 t20=79 S= 11/2(39+79)=649
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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon I) t17= 3+16*4= 67 ii) 129
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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Yeh that's correct
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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Tan3x + 3xsec^2(3x)
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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Yes true
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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Because the gradient isn't 6 it's 2.. Plus you can use that way.. Just with the gradient of 2 not 6
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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Use that only if you have the points and gradient but not the equation. For example finding tangent to the line or the normal equation. Here they give you the derivative of the curve. So just integrate find C by subbing the points and that's your equation; y=2x^3-6x-5
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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Differentiate: 2yln(y) __y
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    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Yeh that's what I got... Mines just not simplified
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