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    Quote:
    Originally Posted by nollaig
    I think people who who write posts in text gibberish, trying to mock people who enjoy doing maths problems, are the real sad ones!


    here, here.......that adds you to the list
    Is that the best you can do?
    Not only can you not write, you can't read properly either!
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    (Original post by Pencil Queen)
    so you do want to get banned again?
    banned for what?
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    what about my answers no one acknowledged them...
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    (Original post by Economics)
    banned for what?
    spamming
    abusing members for posting maths question on a maths subforum on a forum called UK-Learning
    encouraging people to take up smoking on a website frequented by people under 16 (ie encouraging illegal behavior)
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    (Original post by Pencil Queen)
    spamming
    abusing members for posting maths question on a maths subforum on a forum called UK-Learning
    encouraging people to take up smoking on a website frequented by people under 16 (ie encouraging illegal behavior)
    i aint encouraging,,,,,
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    (Original post by Economics)
    i aint encouraging,,,,,
    (Original post by Economics)
    u should try one there really very nice
    http://www.uk-learning.net/showpost....8&postcount=11
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    oh give it a rest.............
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    (Original post by Economics)
    oh give it a rest.............
    when you stop spamming
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    (Original post by Economics)
    do u get a kick out of being so annoying?
    Do you not realise that spamming is annoying?

    I have no problem with annoying spammers
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    (Original post by Economics)
    do u guys realise just how sad u are sat at ur computer at 1:24 in the afternoon doing maths!
    I did Further Maths this morning, we were sketching graphs of rational functions. It was good.
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    (Original post by Invisible)
    I did Further Maths this morning, we were sketching graphs of rational functions. It was good.

    Ah, curve sketching. A favourite passtime of mine!
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    (Original post by Ralfskini)
    Ah, curve sketching. A favourite passtime of mine!
    Oblique Asymptotes! Wow! Asymptotes don't just have to be horizontal or vertical! They can be straight line graphs to!
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    (Original post by integral_neo)
    what about my answers no one acknowledged them...

    That's because this thread was hijacked. You'd think they'd at least leave the academic subforum alone! :mad:
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    (Original post by Invisible)
    Oblique Asymptotes! Wow! Asymptotes don't just have to be horizontal or vertical! They can be straight line graphs to!

    Which module is this?
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    (Original post by Ralfskini)
    Which module is this?
    P4 OCR.

    We're also doing Complex Numbers! So all that time when the teacher taught us that if the discriminant is -ve you have no solutions to the quadratic equation, it was false! Liars!
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    (Original post by Invisible)
    P4 OCR.

    We're also doing Complex Numbers! So all that time when the teacher taught us that if the discriminant is -ve you have no solutions to the quadratic equation, it was false! Liars!

    LOL

    I got stuck in year 10 trying to solve a quadratic equation using the formula and then i had to take the square root of a negative numbers... I didnt know about complex numbers so I asked my teacher..he didnt lie to me and instead gave me P4 book hienamann and told me to look at the complex numbers chapters..i thought aha so thats the trick
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    (Original post by Ralfskini)
    That's because this thread was hijacked. You'd think they'd at least leave the academic subforum alone! :mad:
    are my answers correct?
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    Answers! Well done everyone

    1. How many factors, which themselves are multiples of 3, does the number 241920 have? (not including 3 itself)

    divide 241920 by 3:

    3*80640 = 241920

    Now express 80640 as a product of primes:

    80640 = (2^8)(3^2)(5)(7)

    This means there are 9*3*2*2 factors, or 108. But one of them is 3, so 107 factors.

    I think this is right? lol someone check plz

    2. The difference and product of two consecutive numbers which are prime are added together. Is the number produced a prime itself?

    As one of two consecutive numbers is even, the only two consecutive numbers which are both prime are 2 and 3. The result is (2)(3) + (3-2) = 7, which is prime.

    3. Evaluate the series 3 + 7 + 11 + ... + 199 + 203.

    Sum = (average term)*(number of terms)

    average term can be found by finding the average of the first and last, as the series is .. what's the word.. uniformly distributed between 3 and 203. Average is 103. Taking the series, the nth term is 4n-1, so if:

    4n - 1 = 203,
    4n = 204
    n = 51 terms.

    103 * 51 = 5253.

    4. Find the value of (6! + 5!)/5!

    Taking out the common factors 5! from the numerator,

    5!(6 + 1)/5! = (6+1) = 7

    5. Simplify: [(n+1)!/(n-1)!]!

    Some simple algebra here.

    [(n+1)!/(n-1)!]! = [(n)(n+1)]!
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    (Original post by mik1a)
    Answers! Well done everyone

    1. How many factors, which themselves are multiples of 3, does the number 241920 have? (not including 3 itself)

    divide 241920 by 3:

    3*80640 = 241920

    Now express 80640 as a product of primes:

    80640 = (2^8)(3^2)(5)(7)

    This means there are 9*3*2*2 factors, or 108. But one of them is 3, so 107 factors.

    I think this is right? lol someone check plz

    2. The difference and product of two consecutive numbers which are prime are added together. Is the number produced a prime itself?

    As one of two consecutive numbers is even, the only two consecutive numbers which are both prime are 2 and 3. The result is (2)(3) + (3-2) = 7, which is prime.

    3. Evaluate the series 3 + 7 + 11 + ... + 199 + 203.

    Sum = (average term)*(number of terms)

    average term can be found by finding the average of the first and last, as the series is .. what's the word.. uniformly distributed between 3 and 203. Average is 103. Taking the series, the nth term is 4n-1, so if:

    4n - 1 = 203,
    4n = 204
    n = 51 terms.

    103 * 51 = 5253.

    4. Find the value of (6! + 5!)/5!

    Taking out the common factors 5! from the numerator,

    5!(6 + 1)/5! = (6+1) = 7

    5. Simplify: [(n+1)!/(n-1)!]!

    Some simple algebra here.

    [(n+1)!/(n-1)!]! = [(n)(n+1)]!
    i solved 3 of them
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    (Original post by integral_neo)
    i solved 3 of them
    60% success rate.

    This would get you a C in a Maths exam.
 
 
 
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