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Thread: Can someone help with part I and II

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    Can someone help with part I and II

    https://imgur.com/gallery/R9sWS

    My solutions are attached. Can you help with part I and how to find the area of the circle and square

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    617 pages fan96's Avatar
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    Re: Can someone help with part I and II

    The square has length . Its perimeter, which is the amount of wire used to make the square, must therefore be .

    The remaining amount of wire, cm (not ), is used to make the circumference of the circle ().

    You now know the radius of the circle and the side length of the square, so you can find the area of both shapes now.



    When you find , notice that the required expression is , so the hint is that you should leave the expression as is without expanding the term in the square or combining fractions.



    You can take out the constant in the second term's denominator and differentiate using the chain rule, which will give you the required answer.
    Last edited by fan96; 16 Mar 2018 at 7:29 PM.
    HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]

    1(3√3) t2 dt cos(3π/9) = log(3√e) | Integral t2 dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.

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    Re: Can someone help with part I and II



    THis should be right if i didnt make any silly mistake

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    Re: Can someone help with part I and II

    For part iv I got 42.21cm^2. Don’t you substitute x= 40/Pi+4 into the total area formula and use your calculator to get the answer

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    Re: Can someone help with part I and II

    HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]

    1(3√3) t2 dt cos(3π/9) = log(3√e) | Integral t2 dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.

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    Re: Can someone help with part I and II

    no it should be 56.01 (2dp) fan96 is right

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