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    (Original post by ByronicHero)
    Is this correct? I skipped some steps as it isn't an exam.
    Yep, that's fine - what I'd have done though would be to use the difference of two squares: a^2 - b^2 = (a-b)(a+b), to get:

    (x+7)^2 - (x-3)^3 = (x+7 - x + 3)(x+7 + x - 3) = 10(2x + 4) = 20(x +2) but your method is entirely correct and is likely the intended method at GCSE, well done!
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    (Original post by Zacken)
    Yep, that's fine - what I'd have done though would be to use the difference of two squares: a^2 - b^2 = (a-b)(a+b), to get:

    (x+7)^2 - (x-3)^3 = (x+7 - x + 3)(x+7 + x - 3) = 10(2x + 4) = 20(x +2) but your method is entirely correct and is likely the intended method at GCSE, well done!
    Your way is better.

    I covered that today actually so I am pretty annoyed not to have realised the shortcut.

    At least I was correct though :lol:

    Thanks
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There are n sweets in a bag.

6 of the sweets are orange.

The rest of the sweets are yellow.



Hannah takes a random sweet from the bag. 

She eats the sweet. 



Hannah then takes at random another sweet from the bag. 

She eats the sweet. 



The probability that Hannah eats two orange sweets is 1/3. 

Show that n^2 – n – 90 = 0.
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    (Original post by OGGUS)
    This
    



  (x+7)^2-(x-3)^2

=(x^2+14x+49)-(x^2-6x+9)

=20x+40

=20(x+2)
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    (Original post by jakepds)
     

There are n sweets in a bag.

6 of the sweets are orange.

The rest of the sweets are yellow.



Hannah takes a random sweet from the bag. 

She eats the sweet. 



Hannah then takes at random another sweet from the bag. 

She eats the sweet. 



The probability that Hannah eats two orange sweets is 1/3. 

Show that n² – n – 90 = 0.
    lol that's supposed to be an N² also i hate this question i still don't know how to do this... it was on my GCSE ;/
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    (Original post by OGGUS)
    This calculator paper question
    Spoiler:
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    I assume I am missing something as this seems far too simple and I don't recall covering this a few days ago when I did the GCSE material, but would it not just be 64 + 16 = 80 (sqr root of) = 8.944... = 8.94?
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    (Original post by ByronicHero)
    Spoiler:
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    I assume I am missing something as this seems far too simple and I don't recall covering this a few days ago when I did the GCSE material, but would it not just be 64 + 16 = 80 (sqr root of) = 8.944... = 8.94?
    Yep. All correct.
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    (Original post by Zacken)
    Yep. All correct.
    Thanks again!
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    (Original post by thefatone)
    lol that's supposed to be an N² also i hate this question i still don't know how to do this... it was on my GCSE ;/
    I don't really know how to use latex so apologies for the mess...

    6/n * 5/n-1 = 1/3

    30/(n^2-n) = 1/3

    30 = (n^2-n)/3

    90 = n^2 - n

    n^2 - n - 90 = 0
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    (Original post by jakepds)
     

There are n sweets in a bag.

6 of the sweets are orange.

The rest of the sweets are yellow.



Hannah takes a random sweet from the bag. 

She eats the sweet. 



Hannah then takes at random another sweet from the bag. 

She eats the sweet. 



The probability that Hannah eats two orange sweets is 1/3. 

Show that n^2 – n – 90 = 0.
    (Original post by thefatone)
    lol that's supposed to be an N² also i hate this question i still don't know how to do this... it was on my GCSE ;/
    Spoiler:
    Show

    The answer came into my head very quickly but I had to think a little about how to represent it.

    Well, 6/n x 5/n-1 = 1/3 (first sweet x second sweet)

    so multiplied = 30/(n^2 - n) = 1/3

    n^2 - n thus is 100 so N = 10

    Thus 10^2 - 10 - 90 = 0

    I don't think I have put that very elegantly.
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    (Original post by ByronicHero)
    Spoiler:
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    The answer came into my head very quickly but I had to think a little about how to represent it.

    Well, 6/n x 5/n-1 = 1/3 (first sweet x second sweet)

    so multiplied = 30/(n^2 - n) = 1/3

    n^2 - n thus is 100 so N = 10

    Thus 10^2 - 10 - 90 = 0

    I don't think I have put that very elegantly.

    I don't get where you got the 10 from, have a look at mine, the number 100 or 10 doesn't come up but the answer is correct?
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    (Original post by jakepds)
    I don't get where you got the 10 from, have a look at mine, the number 100 or 10 doesn't come up but the answer is correct?
    I just replaced N with the value of N in my calculations. It looks like we did more of less the same thing. 100 should have read 90, however.

    I have written it down correctly on paper but messed it up here :facepalm2:
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    (Original post by jakepds)
    I don't really know how to use latex so apologies for the mess...

    6/n * 5/n-1 = 1/3

    30/(n^2-n) = 1/3

    30 = (n^2-n)/3

    90 = n^2 - n

    n^2 - n - 90 = 0
    (Original post by ByronicHero)
    Spoiler:
    Show

    The answer came into my head very quickly but I had to think a little about how to represent it.

    Well, 6/n x 5/n-1 = 1/3 (first sweet x second sweet)

    so multiplied = 30/(n^2 - n) = 1/3

    n^2 - n thus is 100 so N = 10

    Thus 10^2 - 10 - 90 = 0

    I don't think I have put that very elegantly.
    it was that easy?...

    FK this shiz, i drew a bloody huge tree diagram for this bull and i didn't get **** from it >.>
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    (Original post by thefatone)
    it was that easy?...

    FK this shiz, i drew a bloody huge tree diagram for this bull and i didn't get **** from it >.>
    You could do the tree but would only need to branch off for orange x orange. Knowing that you could trial and error a few simple cases and very swiftly arrive at N = 10. No idea how many marks that would get you but I would assume at least one method mark and possibly the answer mark :dontknow:

    When I was taking GCSE papers a few days ago I think the probability questions were easier than this one as a tree yielded a reasonably simple answer, and that is the method I assume they teach in school (I never did GCSEs in school).

    Oh well. No point dwelling on it now!
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    (Original post by ByronicHero)
    You could do the tree but would only need to branch off for orange x orange. Knowing that you could trial and error a few simple cases and very swiftly arrive at N = 10. No idea how many marks that would get you but I would assume at least one method mark and possibly the answer mark :dontknow:

    When I was taking GCSE papers a few days ago I think the probability questions were easier than this one as a tree yielded a reasonably simple answer, and that is the method I assume they teach in school (I never did GCSEs in school).

    Oh well. No point dwelling on it now!
    stupid tree o.o cost me a few marks, i always did them my way and they always worked and i decided to experiment in the actual thing >.>
    at least i still came out with an A*
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    (Original post by thefatone)
    stupid tree o.o cost me a few marks, i always did them my way and they always worked and i decided to experiment in the actual thing >.>
    at least i still came out with an A*
    Well it's all good then
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    (Original post by ByronicHero)
    Well it's all good then
    i'm gonna slap me in the face, i filled the whole page with stuff and didn't get an answer -.-
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    (Original post by thefatone)
    i'm gonna slap me in the face, i filled the whole page with stuff and didn't get an answer -.-
    I'm curious. Can you post your working?
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    A function, defined on the set of positive integers, is such that f(xy)=f(x)+f(y) for all x and y. It is known that f(10)=14 and f(40)=20. What is the value of f(500)?
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    (Original post by Kvothe the arcane)
    I'm curious. Can you post your working?
    i'm afraid my page of work about a year ago during my GCSE has probably been binned
 
 
 
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