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Thread: VCE Maths questions help

  1. #451
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    Re: VCE Maths questions help

    Quote Originally Posted by InteGrand View Post
    Answer to your last question: yes.

    Thanks. Also, do you know how to find the general solutions to the equations I posted above?
    Thanks
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    Thanks. Also, do you know how to find the general solutions to the equations I posted above?
    Thanks
    Yes, I do.

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    Re: VCE Maths questions help

    Quote Originally Posted by InteGrand View Post
    Yes, I do.
    Could you please explain how to do it as read my textbook but got confused with it's explanation
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    Also, can someone please explain how to find the general solution to the following equations.
    Cos 2(x+pi/3) = 1/2

    2 tan 2(x+pi/4) = 2√(3)

    2 sin(x+pi/3) = -1

    Thanks

    Silly question, but is
    sin 2(x+pi/3) the same as sin(2x+2pi/3)
    Quote Originally Posted by boredsatan View Post
    Could you please explain how to do it as read my textbook but got confused with it's explanation





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    Re: VCE Maths questions help

    If (x-1)^3 = 8
    then would x-1 = +- cube root of 8, or just + cube root of 8 and why?

    Also, does anyone know how you would work out sin(11pi/4), which is sin(2.75pi), and is in quadrant 2, and the rule for quadrant 2 is pi - base, but since it's 2.75, would you do (2pi) + (pi-base)?
    And how would you work out the value of the base?

    How would you find the basic angle for a negative value?
    For instance cos(x) = -1/2

    How would you work out (x+1)^3 = -2 without a calculator?
    Thanks
    Last edited by boredsatan; 19 Jan 2018 at 9:30 AM.
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    If (x-1)^3 = 8
    then would x-1 = +- cube root of 8, or just + cube root of 8 and why?

    Also, does anyone know how you would work out sin(11pi/4), which is sin(2.75pi), and is in quadrant 2, and the rule for quadrant 2 is pi - base, but since it's 2.75, would you do (2pi) + (pi-base)?
    And how would you work out the value of the base?

    How would you find the basic angle for a negative value?
    For instance cos(x) = -1/2

    How would you work out (x+1)^3 = -2 without a calculator?
    Thanks
    Anyone?
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    Re: VCE Maths questions help

    Bump!
    Any one know how to do the above questions?

    Also, does anyone know how to solve the equation sin x = 1/2 for [-pi, to 2pi]
    I'm confused with the negative domain
    Thanks
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    Re: VCE Maths questions help

    Bump
    Can somebody PLEASE ANSWER all my questions above?
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    Also, does anyone know how to solve the equation sin x = 1/2 for [-pi, to 2pi]
    I'm confused with the negative domain
    Thanks
    The sine wave has symmetry across the origin, so it's an odd function, which means that .

    Basically that implies that the sine of a negative angle is the same as reversing the sign of the sine of the positive angle, i.e.

    .

    When we go in the positive direction, the first angle for which is .

    So that means that when we go in the negative direction, the first angle for which is .

    But is outside the range of , so it's not a solution to the equation.
    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: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    Also, does anyone know how you would work out sin(11pi/4)
    The period of is . So,



    Also, . So,



    Quote Originally Posted by boredsatan View Post
    How would you work out (x+1)^3 = -2 without a calculator?
    Thanks
    If you want the answer in exact form, then





    Last edited by fan96; 21 Jan 2018 at 8:01 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: VCE Maths questions help

    The equation of the image of the graph of y = sin x under a transformation of a dilation of factor 1/2 from the y-axis followed by a translation of pi/4 units in the positive direction of the x axis is:
    I'm narrowed it down two 2 options, but don't know which one is correct. Can someone please explain which one is correct and why
    y = sin(2x - pi/4) or y = sin2(x-pi/4)
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    Re: VCE Maths questions help

    let f: [0, pi/2] → R, where f(x) = cos(3x) -2. The graph of f is transformed by a reflection in the x-axis followed by a dilation of factor 3 from the y-axis. How would you work out the resulting graph, super confusing so wold greatly appreciate if someone helped out
    Last edited by boredsatan; 23 Jan 2018 at 8:57 PM.
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    The equation of the image of the graph of y = sin x under a transformation of a dilation of factor 1/2 from the y-axis followed by a translation of pi/4 units in the positive direction of the x axis is:
    I'm narrowed it down two 2 options, but don't know which one is correct. Can someone please explain which one is correct and why
    y = sin(2x - pi/4) or y = sin2(x-pi/4)
    I'm not familiar with that specific terminology, but if "dilation of factor 1/2 from the y-axis" means compressing the function horizontally, then you replace with and to shift it right you would replace with , in that order.



    Dilate:



    Shift graph:



    Quote Originally Posted by boredsatan View Post
    let f: [0, pi/2] → R, where f(x) = cos(3x) -2. The graph of f is transformed by a reflection in the x-axis followed by a dilation of factor 3 from the y-axis. How would you work out the resulting graph, super confusing so wold greatly appreciate if someone helped out


    Reflect across x axis:



    Dilate:



    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: VCE Maths questions help

    An open tank is to be made from a sheet of metal 84 cm by 40 cm by cutting congruent squares of side length x cm from each of the corners.
    I've found the volume to be = (84-2x)(40-2x)(x)

    State the maximal domain for V when it is considered as a function of x. Confused how to do this part
    Any help would be appreciated
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    An open tank is to be made from a sheet of metal 84 cm by 40 cm by cutting congruent squares of side length x cm from each of the corners.
    I've found the volume to be = (84-2x)(40-2x)(x)

    State the maximal domain for V when it is considered as a function of x. Confused how to do this part
    Any help would be appreciated
    bump
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    Re: VCE Maths questions help

    A rectangle is defined by vertices N and P(x,y) on the curve with equation y = 16-x^2 and vertices M and Q on the x axis
    a.i. Find the area, A of the rectangle in terms of x
    ii. state the implied domain for the function defined by the rule given in part i

    Can someone please help me with this?
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    Re: VCE Maths questions help

    A metal worker is required to cut a circular cylinder from a solid sphere of radius 5 cm. Express r in terms of h, where r cm is the radius of the cylinder and h cm is the height of the cylinder. Hence show that the volume, V cm^3, of the cylinder is given by V = (1/4)(pi)(h)(100-h^2)
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    Re: VCE Maths questions help

    n a tidal river, the time between high tides and low tides is 12 hours. The average depth of water at a point in the river is 5m. At high tide the depth is 8 m. Assume that the depth of the water, h(t) m, at this point is given by
    h(t) = A sin(nt+e)+b, where t is the number of hours after noon. At noon there is a high tide.

    a. find the values of A, n,b, and e

    b. at what times is the depth of the water 6 m?
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    Re: VCE Maths questions help

    Bump
    Can somebody PLEASE ANSWER all my questions above?
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