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Secondary Maths questions, give me your maths questions

Discussion in 'Education and Employment' started by Nasma, Aug 29, 2009.

Put it out there
  1. Speck

    Speck SWAG

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    magic_johnson what chapter are you up to in Specialist?

    Seems strange for you to be doing that SAC when we are only a week from finishing the course (halfway through dynamics).

    Had my Kinematics/Vector Calculus SAC today which was very tough.

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  2. Speck

    Speck SWAG

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    Don't you just use a calculator program to find that?
  3. magic_johnson!

    magic_johnson! All Australian

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    Yeah, i kinda learnt how to do it by hand now, i wanted to know how to do it both ways. We've finished the book, just had the 40 mark sac on everything though so had to have a bit of a refresher. All went well :thumbsu: So far i've got c, c+ and c and am sitting about middle of quite a smart class, but i think i'm pushing a b+ maybe even an a this time. Hopefully...
  4. RobbieGray17

    RobbieGray17 All Australian

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    Ok I know this will probably be easy for you guys but its doing my head in. I'm studying algebra for the first time since school (15 years) and its so frustrating finding the answer in the book, but not knowing if you got there correctly, secondly not having an answer and having no bloody idea if its right. The joys of self study:(

    I have : - 2z^2 - x / 2x - y^2

    The first - is over the whole equation

    z = -4, x = 3 and and y = -2

    I have no answer in the book and can't punch it into the online calculator.

    But I get -29/2 as an answer is that right or am I doing something wrong? I got every other question on the test right but just this one has screwed me over.

    Oh and one more

    2/y+1 = 12/y+1 - 5

    Again its even number question so the stupid book doesnt give me the answer.

    So I get 6y -11. am I right or wrong here.
  5. jessCFC

    jessCFC Rookie

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    Firstly make sure you use brackets:

    Substitution into what I think you meant by your equation gives:

    -2(-4)^2-3/(2x3)-(-2)^2 = -2(16)-1/2-4 = -73/2

    Secondly do you mean:

    2/(y+1) or 2/y +1 because these are completely different. Let's assume the first, then

    2/(y+1)-12/(y+1)=-5

    10/(y+1)=-5

    10/(-5)=y+1

    -2=y+1

    y=-3
  6. RobbieGray17

    RobbieGray17 All Australian

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    TX heaps for the reply.
  7. jessCFC

    jessCFC Rookie

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    In the second one, you can't have brackets matching up with an equals sign in between. Also if you have terms with a plus or minus in between and you multiply one term then you also have to multiply the other term so the minus five would also need to be multiplied. However, remember that you always do the opposite of what is there, plus to minus etc. But the first thing you should do is collect the like terms to one side and constants to the other.
  8. jessCFC

    jessCFC Rookie

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    In the first one, everything is correct up until you add your fractions... check that part.
  9. RobbieGray17

    RobbieGray17 All Australian

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    I got 2(16) Step 1
    32-3 = 29 step 2

    6 + -4 = 2

    I still get 29/2 following steps.
  10. Speck

    Speck SWAG

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    Do you mean:

    [2(16)-3]/[6-4]
    [32-3]/[2]
    29/2

    or

    2(16)-3/6-4
    32-0.5-4
    27.5 or 55/2

    Brackets change things massively.
  11. jessCFC

    jessCFC Rookie

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    You cannot add fractions with different denominators. So for example:

    -32-1/2 -4 = 36-1/2 = -72/2 - 1/2 = -73/2

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  12. liam_13

    liam_13 All Australian

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    Can anyone here give me some help with a collision question?

    A body of mass 400g is moving on a smooth surface at a velocity of 1.25m/s towards the East. It strikes a body of mass 600g, initially at rest, and then the 400g body moves at a velocity of 1m/s in a direction of 36.9 degrees North of east

    a) What is the easterly component of the total momentum of the system before and after the collision?

    b) What is the northerly component of the total momentum of the system before and after the collision?

    c) Determine the final velocity of the 600g body

    I've drawn a diagram, and calculated the total momentum (kg x v), which initially is .4 x 1.25 + 0 x .6, which is .4 x 1.25 = .5 kg/ms-1

    Is the initial easterly component the 1.25 x .4 = .5, and the northerly 0 due to no movement up before the collision? Then using the principle of conservation of momentum, the momentum after the collision is .5, but this where I get stuck, calculating the easterly and northerly components of the system.

    If anyone finds that easy enough, would they also be happy to answer one other as well? Thanks.
  13. jessCFC

    jessCFC Rookie

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    It's been a while since i did this but aren't you just resolving forces like resolving vectors?
  14. liam_13

    liam_13 All Australian

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    Yes it is to do with vectors. It's the application of them I'm not good at.
  15. SchumiUCD

    SchumiUCD Rookie

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    Correct.

    Each component of the momentum is conserved so the easterly momentum afterwards is 0.5 and northerly is 0.

    The question gives you the speed and direction of the 400g body after the collision so you can get its momentum in each direction. The remainder of the total then belongs to the 600g one and you can work out its velocity from that.

    Fire ahead.
  16. Starlin Castro

    Starlin Castro Team Captain

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    Using the formula, cos(2x)=1-2sin^2(x) and calculus, find the coordinates where the tangent to the curve of sin(2x)+2cos(x) is parallel to the x axis.
  17. treefingers

    treefingers the mighty hawks

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    y=sin(2x)+2cos(x)

    y'=2cos(2x) -2sin(x)
    =2(1-2sin^2(x))-2sin(x)
    =2 - 4.sin^2(x) - 2sin(x)

    Tangent is parallel to x axis when derivative = 0

    so y'=0
    2 - 4.sin^2(x) - 2sin(x) = 0

    let A = sin(x)

    so 2-4A^2-2A=0

    4A^2+2A-2=0
    4a^2+4A-2A-2=0
    4A(A+1) -2(A+1) = 0

    (4A-2).(A+1)=0

    so 4A-2=0
    A=1/2

    and A+1=0
    A=-1

    Therefore sin(x)=1/2, x=pi/6 and sin(x)=-1, x = 3pi/2

    Coordinates, just sub answers into y
  18. DESTRUCTOR

    DESTRUCTOR Rookie

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    Where should I start with relearning maths from a year 7-ish level (That's the main question)

    I was epic at maths in primary school (it helps when all it was was arithmetic and no homework) and I sucked hard at maths in high school.

    But for some stupid reason I seem to find things here and there in maths to be interesting and I would like progress (however slowly) past knowing how to count.

    So back to the original question, with what should I start? (I'm also extremely lazy so I'm not going to go back to school, buy any books or go through a whole course from the internet (yes, I know I suck)) I've also not understood anything in this thread.
  19. jessCFC

    jessCFC Rookie

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    check out khan academy . org the guy has a video on most general maths topics. Start with the basics in year seven: order of operations, fractions, ratios, basic algebra for rearranging equations, types of numbers.
  20. DESTRUCTOR

    DESTRUCTOR Rookie

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    Thanks, I'll have a look.
  21. treefingers

    treefingers the mighty hawks

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    let x £ R+ and t £ T, where T={x£R: 0<x<1} £=element of

    Prove by contradiction that...

    If x≤x^t then x≤1

    HELP
  22. jinny1

    jinny1 Team Captain

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    Oh I forgot everything about math methods probability section already Lol :(
  23. SchumiUCD

    SchumiUCD Rookie

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    To prove something by contradiction, take the opposite of what you have to prove (x ≤ 1) and your starting assumption (x ≤ x^t), work though those and either contradict one of those or get something obviously untrue (0 > 1 or 1 = 2 or something).

    So here, you start with (A) x > 1 and (B) x ≤ x^t

    I'd start by getting the log of both sides of (B). This won't change the > sign because log is monotonically increasing (i.e. for larger numbers, their log keeps getting bigger).
    You then have log(x) t.log(x).
    Then [1-t]log(x) ≤ 0
    As 0<t<1, [1-t] > 0 and as x > 1, log(x) > 0.
    Therefore, [1-t]log(x) > 0 (two positive numbers multiplied give a positive number).
    The two bold inequalities contradict each other so (A) and (B) can't both be true so (B) being true must imply that (A) isn't so x ≤ 1.
    QED.
    Bob Vance likes this.
  24. Pie Legend

    Pie Legend Team Captain

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    Laplace Transform..anyone have any knowledge of this? Absolutely annoying the **** out of me!
  25. treefingers

    treefingers the mighty hawks

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    Exam revision...
    Screen Shot 2013-06-07 at 12.03.51 PM.png