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Thread: Inequality help?

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    Inequality help?

    I'm having trouble answering these questions, especially b) and c). Any help/answers would be greatly appreciated

    maths.bmp

    a and b.bmp
    Last edited by atamiabwv; 8 Apr 2018 at 12:16 AM.

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    Re: Inequality help?

    b) R.T.P.

    Squaring both sides,





    Multiplying both sides by 16,





    Can you finish it from here?

    c) seems to be missing some context... what are and ?

    Also this looks like it would belong in the Mathematics subforum (i.e. the 2U one).
    Last edited by fan96; 7 Apr 2018 at 11:31 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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    Senior Member sida1049's Avatar
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    Re: Inequality help?

    For part (b), literally follow the hint: after squaring both sides and simplifying, you should get
    32 + 16√2 < 38 + 12√2,
    which simplifies further to
    √2 < 3/2,
    which is the same inequality in part (a).

    I can't help you with part (c), since I don't know what a and b are.

    Bachelor of Science (Advanced Mathematics)/Bachelor of Arts III, USYD

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    Re: Inequality help?

    Quote Originally Posted by fan96 View Post
    b) R.T.P.

    Squaring both sides,





    Multiplying both sides by 16,





    Can you finish it from here?

    c) seems to be missing some context... what are and ?

    Also this looks like it would belong in the Mathematics subforum (i.e. the 2U one).
    Yes sorry this is in the wrong forum! But I'm a bit confused at how 6+√2/4 squared does not give 36+2/16? I've attached a picture above now showing the values for a and b. Sorry, I probably seem like a maths noob

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    Re: Inequality help?

    Quote Originally Posted by sida1049 View Post
    For part (b), literally follow the hint: after squaring both sides and simplifying, you should get
    32 + 16√2 < 38 + 12√2,
    which simplifies further to
    √2 < 3/2,
    which is the same inequality in part (a).

    I can't help you with part (c), since I don't know what a and b are.
    Thanks for your help!! I've now attached values for a and b

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    Re: Inequality help?

    Quote Originally Posted by atamiabwv View Post
    how 6+√2/4 squared does not give 36+2/16?


    With , :

    Last edited by fan96; 8 Apr 2018 at 1:36 AM.
    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: Inequality help?

    Quote Originally Posted by fan96 View Post


    With , :

    That makes more sense now, thank you heaps

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    Re: Inequality help?

    Quote Originally Posted by fan96 View Post
    b) R.T.P.

    Squaring both sides,





    Multiplying both sides by 16,





    Can you finish it from here?

    c) seems to be missing some context... what are and ?

    Also this looks like it would belong in the Mathematics subforum (i.e. the 2U one).
    Do you perhaps have any idea how to do part (c) now, after the attachment?

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    Re: Inequality help?

    Quote Originally Posted by atamiabwv View Post
    Do you perhaps have any idea how to do part (c) now, after the attachment?
    Given

    R.T.P.

    Square both sides





    Divide both sides by 64:





    This is starting to look like the result of part b). Can you take it from here?
    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: Inequality help?

    Quote Originally Posted by fan96 View Post
    Given

    R.T.P.

    Square both sides





    Divide both sides by 64:





    This is starting to look like the result of part b). Can you take it from here?
    That's really helpful, thank you very much! I've taken it a bit further for the LHS, but I'm not sure what do to for the RHS. I've attached my working: IMG_6433.bmp

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    Re: Inequality help?

    Quote Originally Posted by atamiabwv View Post
    That's really helpful, thank you very much! I've taken it a bit further for the LHS, but I'm not sure what do to for the RHS. I've attached my working: IMG_6433.bmp
    Don't square it - you want to get it to the result in part b).

    You'll be able to see the answer by doing just one subtraction.
    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: Inequality help?

    Quote Originally Posted by fan96 View Post
    Don't square it - you want to get it to the result in part b).

    You'll be able to see the answer by doing just one subtraction.
    Hmm I’ve tried subtracting -2 from both sides, other than that I don’t know what else I could really subtract. I’m still not getting the answer, and my classmates are confused too. Sorry for all the hassle

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    Re: Inequality help?





    Subtract 2:





    Multiply by -1 (which flips the inequality sign)





    Which is the result in b)
    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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