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    Hi guys

    Is anybody able to tell me what the happens in part b in the last step of the model answer?

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    Thanks 😄
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    Common denominator in the square root.
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    (Original post by jasminetwine)
    Hi guys

    Is anybody able to tell me what the happens in part b in the last step of the model answer?

    Name:  image.png
Views: 161
Size:  356.9 KB

    Thanks 😄
    Specifically: 1 - \frac{x^2}{16} = \frac{16 - x^2}{16} so \sqrt{1 - \frac{x^2}{16}} = \sqrt{\frac{16 - x^2}{16}} = \frac{\sqrt{16-x^2}}{\sqrt{16}} = \frac{\sqrt{16-x^2}}{4}.

    Hence: 8\sqrt{1 - \frac{x^2}{16}} = 8 \times \frac{\sqrt{16-x^2}}{4} = 2\sqrt{16-x^2}

    GLORIOUS LATEX.
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    (Original post by Zacken)
    Specifically: 1 - \frac{x^2}{16} = \frac{16 - x^2}{16} so \sqrt{1 - \frac{x^2}{16}} = \sqrt{\frac{16 - x^2}{16}} = \frac{\sqrt{16-x^2}}{\sqrt{16}} = \frac{\sqrt{16-x^2}}{4}.

    Hence: 8\sqrt{1 - \frac{x^2}{16}} = 8 \times \frac{\sqrt{16-x^2}}{4} = 2\sqrt{16-x^2}

    GLORIOUS LATEX.
    Thank you very much Zacken

    I'm not sure I would have ever spotted that!

    Do you think a student would still be able to achieve full marks if the answer was left unsimplified as dy/dx=+- 1/(8(1-1/16x^2)) (i.e. The penultimate line)?
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    (Original post by jasminetwine)
    Thank you very much Zacken

    I'm not sure I would have ever spotted that!

    Do you think a student would still be able to achieve full marks if the answer was left unsimplified as dy/dx=+- 1/(8(1-1/16x^2)) (i.e. The penultimate line)?
    I'm not entirely sure, but unless asked for a specific form, I don't think they should lose any marks.
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    (Original post by NotNotBatman)
    Common denominator in the square root.
    That is right! you have razor-sharp eyes for understanding mathematical steps. :congrats:

    (Original post by jasminetwine)
    Hi guys

    Is anybody able to tell me what the happens in part b in the last step of the model answer?

    Name:  image.png
Views: 161
Size:  356.9 KB

    Thanks 😄
    What about the steps before? did you understand them? well.
 
 
 
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