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Thread: VCE Maths questions help

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    Re: VCE Maths questions help

    Quote Originally Posted by InteGrand View Post
    4
    How do you work it out?
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    How do you work it out?
    It's the slope of the tangent at the given point (and we're given that tangent's equation (we're given that the line is tangent)).

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    Re: VCE Maths questions help

    Quote Originally Posted by InteGrand View Post
    It's the slope of the tangent at the given point (and we're given that tangent's equation (we're given that the line is tangent)).
    so the gradient of 4x-4 which is 4
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    so the gradient of 4x-4 which is 4
    Yep
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    Re: VCE Maths questions help

    Quote Originally Posted by Squar3root View Post
    Yep
    thanks
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    Re: VCE Maths questions help

    how would you find f'(1) for f(x) = 5x
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    how would you find f'(1) for f(x) = 5x
    Note f'(x) = 5 for all x, so f'(1) = 5.

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    Re: VCE Maths questions help

    What is this? boredsatan's do my maths homework and sanity check thread?

    Most of the time, you don't show appreciation for help you receive... and even dare to ignore some solutions.

    This thread is more harm than good; it's an obstacle that lessens the opportunities for you to exercise your own judgement.

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    Re: VCE Maths questions help

    He is not the only one. Many on BOS posed questions and mostly made no acknowledgement nor words of appreciation. I suppose that is the norm of the current generation.
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    Re: VCE Maths questions help

    Quote Originally Posted by He-Mann View Post
    What is this? boredsatan's do my maths homework and sanity check thread?

    Most of the time, you don't show appreciation for help you receive... and even dare to ignore some solutions.

    This thread is more harm than good; it's an obstacle that lessens the opportunities for you to exercise your own judgement.
    Is there a problem with not being that good at maths?
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    Is there a problem with not being that good at maths?
    No

    Quote Originally Posted by He-Mann View Post
    Most of the time, you don't show appreciation for help you receive... and even dare to ignore some solutions.

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    Re: VCE Maths questions help

    Quote Originally Posted by He-Mann View Post
    What is this? boredsatan's do my maths homework and sanity check thread?

    Most of the time, you don't show appreciation for help you receive... and even dare to ignore some solutions.


    This thread is more harm than good; it's an obstacle that lessens the opportunities for you to exercise your own judgement.
    If I say thank you every time, won't it count as shitposting?
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    If I say thank you every time, won't it count as shitposting?
    Why would it be considered as 'shitposting'?

    There are other ways to express gratitude other than words. Imagine you and a group of kind helpers spare some time out of multiple days to help a struggling homeless person and that person just takes your gifts without making eye contact, rarely expresses gratitude, begs for more... How would you feel?


    Also, I'm not making the comments based on my views on your ability to do maths but it's because I've noticed you've become dependent on this thread to do some maths. How will you handle yourself when there is no help?
    Last edited by He-Mann; 25 Jun 2017 at 4:29 PM.

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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    If I say thank you every time, won't it count as shitposting?
    I think another thing He-Mann said in the last line was to try questions yourself before posting, e.g. derivative of 5x
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    Re: VCE Maths questions help

    Alot of the time I do try the questions myself but still find it hard
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    Alot of the time I do try the questions myself but still find it hard
    Hello there, just wondering, how is finding the derivative of 5x hard?

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    Re: VCE Maths questions help

    Quote Originally Posted by -s- View Post
    Hello there, just wondering, how is finding the derivative of 5x hard?
    i didn't ask to find the derivative of 5x, i asked this
    "how would you find f'(1) for f(x) = 5x"
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    i didn't ask to find the derivative of 5x, i asked this
    "how would you find f'(1) for f(x) = 5x"
    f'(x) is a derivative so it's pretty much differentiating 5x
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    Re: VCE Maths questions help

    Quote Originally Posted by pikachu975 View Post
    f'(x) is a derivative so it's pretty much differentiating 5x
    if you say so
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    Re: VCE Maths questions help

    Quote Originally Posted by boredsatan View Post
    Alot of the time I do try the questions myself but still find it hard
    Quote Originally Posted by boredsatan View Post
    how would you find f'(1) for f(x) = 5x
    even though i am 99.9% sure you're a troll but for that 0.1% of doubt, im gonna be honest; if you find these basic questions "hard" you should really consider dropping math or dropping down to general maths. It won't do anything good for your atar if you get 50 in 2U versus getting 65+ in general
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    Re: VCE Maths questions help

    Are my answers right?
    r(x) = -200x^2 + 4800x
    find r(x+h)
    -200x^2 - 400xh -200h^2 + 4800x + 4800h
    use differenetiation by first principles to find the equation of the rate of change in revenue respect to the month of the year
    -400x + 4800
    find the instantaneous rate of change in revenue at B (6,21600)
    -400(6) + 4800 = 2400
    find the equation of the tangent to the revenue at C (10,28000)
    -400(10) + 4800 = -4000 + 4800 = 800
    y = 800x + c
    28000 = 800(10) + c
    y = 800x + 20,000

    2. p(x) = x^3 - 4x + 2
    find p'(x)
    3x^2 - 4
    find p'(3)
    3(3)^2 - 4 = 27-4 = 23
    find the equation of the perpendicular line to p(x) at x = 3
    p(3) = (3)^3 - 4(3) + 2 = 17
    (3,17)
    p'(3) = 3(3)^2 - 4 = 27-4 = 23
    m1*m2 = -1
    m1*23 = -1
    m1 = -1/23
    17 = -1/23(3) + c
    c = 394/23
    y = -1/23x + 394/23
    use newton's methods to calculate the root of the equation x^3 - 4x + 2 = 0 that lies near x = 3. Express the answer correct to 3 decimal places

    3. c(x) = (x^3 - 9x^2 + 26x - 24)/(x-3)
    is the function continuous?
    Yes, the function is continuous as it is drawn without lifting the pen of paper
    calculate lim x (3) for c(x)
    (x^3 - 9x^2 + 26x - 24)/(x-3) = (x-2)(x-3)(x-4)/(x-3)
    = (x-2)(x-4)
    = (3-2)(3-4)
    = 1 * -1
    = -1

    4. The rate of change is expressed by s'(t) = 3x^2 - 36x + 72, x [0,12]. Find an expression for the total inventory, s(t) if the company had 200 items initially
    6x - 36 + 200
    b. Determine the area bounded by s'(t) and the x-axis between 2.54 ≤ t ≤ 9.46, correct to 2 decimal places
    using calculus, 166.28
    using the graph of the gradient, draw a sketch of a possible curve of s(t), label turning points, end points and intercepts
    not sure about this question
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    Re: VCE Maths questions help

    Any help will be greatly appreciated. Have a test this friday
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    Re: VCE Maths questions help

    bump
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    Re: VCE Maths questions help

    bump
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    Re: VCE Maths questions help

    bump
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