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    (Original post by mariam687)
    my laptop is so ****, it dont download nothing lol
    getting a new one after gcse's lol

    i'll try n download it tho, see if it works?
    yeah, go for it skype shouldnt be a problem, not that much of a hard application to run or download. lemme know your name on it when your done
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    (Original post by junayd1998)
    yeah, go for it skype shouldnt be a problem, not that much of a hard application to run or download. lemme know your name on it when your done
    alright imma let u kno

    but i doubt it'll work, its not skypes problem, its the laptop tbh

    but yh imma let u kno dw
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    (Original post by mariam687)
    alright imma let u kno

    but i doubt it'll work, its not skypes problem, its the laptop tbh

    but yh imma let u kno dw
    Okay sure!
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    (Original post by junayd1998)
    Yeah you got skype we doing in a group so we can all help each other
    Whats ur skype? ill add you
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    Does anyone know where the predicted 2016 edexcel paper is?
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    (Original post by SmallTown)
    Does anyone know where the predicted 2016 edexcel paper is?
    Yes i do!
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    (Original post by _Xenon_)
    Yes i do!
    tell me / send me the link plz
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    (Original post by ShahbazKHAN)
    tell me / send me the link plz
    I wouldn't rely on them unless you've revised absolutely everything?
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    (Original post by _Xenon_)
    I wouldn't rely on them unless you've revised absolutely everything?
    yh i have
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    I can join, I'm doing A level though
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    (Original post by ShahbazKHAN)
    tell me / send me the link plz
    Hey, I found this, it may be useful

    http://www.mathsgenie.co.uk/resources/june2016p1.pdf

    But obvs some questions are easy, but if you get these questions right, you know what kinda questions are like
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    (Original post by ShahbazKHAN)
    yh i have
    OK good luck.

    1H QUESTION PAPER: http://www.mathsgenie.co.uk/resources/june2016p1.pdf
    1H SOLUTIONS: http://www.mathsgenie.co.uk/resources/june2016p1ans.pdf

    Once you've done these try the ones by OnMaths which can be done ONLINE.
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    add me Guys
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    (Original post by kandaman)
    add me Guys
    Drop your skype name
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    I need help with this Name:  image.jpg
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Size:  87.8 KB
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    (Original post by Rajive)
    I need help with this Name:  image.jpg
Views: 59
Size:  87.8 KB
    Well we were told that the perimeter of rectangle is 28cm. So if we denote the length of the rectangle as y, then the perimeter would be the whole distance around the rectangle. Which would be 2x + 2y.

    We now that the perimeter is 2x +2y, therefore we can equate this to 28 as we were told that the perimeter of the rectangle is 28cm.

    Equation 1:

    2x + 2y =28

    Dividing through by two. Therefore x + y = 14. Then we can rearrange the equation making y the subject. Which gives us, y = 14 - x


    Another fact the we now is that the width is xcm and the length of the diagonal is 12 cm. As we have let the length of the rectangle equal to y, we can now you Pythagoras which gives us x^2 + y^2 = (12)^2. Which simplifies to x^2 +y^2 =144

    We now have two equations.

    Equation 1: x + y = 14 and Equation 2: x^2 + y^2 =144. I would now like you to solve these simultaneous equations. Hope this helped!
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    (Original post by phat-chewbacca)
    Well we were told that the perimeter of rectangle is 28cm. So if we denote the length of the rectangle as y, then the perimeter would be the whole distance around the rectangle. Which would be 2x + 2y.

    We now that the perimeter is 2x +2y, therefore we can equate this to 28 as we were told that the perimeter of the rectangle is 28cm.

    Equation 1:

    2x + 2y =28

    Dividing through by two. Therefore x + y = 14. Then we can rearrange the equation making y the subject. Which gives us, y = 14 - x


    Another fact the we now is that the width is xcm and the length of the diagonal is 12 cm. As we have let the length of the rectangle equal to y, we can now you Pythagoras which gives us x^2 + y^2 = (12)^2. Which simplifies to x^2 +y^2 =144

    We now have two equations.

    Equation 1: x + y = 14 and Equation 2: x^2 + y^2 =144. I would now like you to solve these simultaneous equations. Hope this helped!
    Wooow that makes a lot of sense.

    I feel scared because I wouldn't know to do this in the exam...

    Thanks though
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    (Original post by Rajive)
    Wooow that makes a lot of sense.

    I feel scared because I wouldn't know to do this in the exam...

    Thanks though
    I suggest you do it again in a few days then!
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    (Original post by phat-chewbacca)
    I suggest you do it again in a few days then!
    I got 7+_ root 23?
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    Anyone got the aqa Linear B predicted paper?
 
 
 
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