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Edexcel: 11th and 14th June 2013 (Linear) 1MA0/1H + 2H (Official Thread) Watch

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    (Original post by ameliaamelia)
    I think that you factorise both equations, on the top and bottom to get

    (3x+5)(x-7) and on the bottom (3x+5)(3x-5).

    Expand them out to check that they are the original equations.

    The bottom is difference of two squares, so noticing that makes it quicker to simplify. Then you cross out the (3x+5) and (3x+5) because they cancel out, this leaves x-7 on the top and 3x-5 on the bottom.

    Not sure if that helped, but it's basically factorising then crossing out which brackets are the same.

    I should of put this before. I understand everything you have done :| I know you factorise the bottom bit because it is the difference of two squares and I know how to cancel.

    I just don't understand how they got (3x+5)(x-7) at the top. How do you get that? Can you factorise and simplify it in steps to show me?
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    (Original post by TomR10)
    For the area of a triangle what does ½ab sin(c) mean?

    Don't understand what sin(c) is ? What do you do? I'm confused about this. Someone enlighten me ?
    The Sine and Cosine rule will be given to you at the front of the paper and you use this to find the missing value (length or angle)
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    Posted from TSR Mobile[/QUOTE]

    (Original post by ameliaamelia)
    I think that you factorise both equations, on the top and bottom to get

    (3x+5)(x-7) and on the bottom (3x+5)(3x-5).

    Expand them out to check that they are the original equations.

    The bottom is difference of two squares, so noticing that makes it quicker to simplify. Then you cross out the (3x+5) and (3x+5) because they cancel out, this leaves x-7 on the top and 3x-5 on the bottom.

    Not sure if that helped, but it's basically factorising then crossing out which brackets are the same.
    I got x-7/ 3x- 5 as well. If you want me to explain how I done it, I will more than likely confuse you more
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    Oh my god. The school has messed up and told me I have this exam tomorrow. Can anyone help?

    Is there anyway I can get a decent grade tomorrow if I do some quick revision. Past papers, exam notes - anything??
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    If you only have time for one more past paper, having already done June 2012 Non calculator which one is best to try?
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    http://tinypic.com/view.php?pic=dfbpcx&s=5

    On this, I have got to the point where I've worked out all of the values around the outside. Being 25m on the hypotenuse, 24m along the bottom and 7m on the remaining side. I am now confused as to how I substitute these numbers in order to find the area of the triangle. Can any one explain?
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    (Original post by cmorga1)
    If you only have time for one more past paper, having already done June 2012 Non calculator which one is best to try?
    Oh no. My printer doesn't even work and I need to print a paper!
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    Hi everyone, I was wondering if anyone could help me out with a question:

    It's not a specific question but it's about that table you do where it gives you an equation for Y and you have to fll it in using the different values of x and then make a line on a graph.

    I get all that but what I don't get is when it askes you to use the graph to find the solutions of the equation and the paper has some thing like:
    X=........
    X=........
    I have no idea what to do...

    (for example question 14 in the linerer mathematics paper 3 on june 2010)
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    (Original post by w0tSuhail)
    I should of put this before. I understand everything you have done :| I know you factorise the bottom bit because it is the difference of two squares and I know how to cancel.

    I just don't understand how they got (3x+5)(x-7) at the top. How do you get that? Can you factorise and simplify it in steps to show me?
    Ok, I will do it here.

    3x^2-16x-35. 3X35=105 What times to give 105 and adds to give -16? -21 AND +5

    From there you get the equation 3x^2 -21x+5x-35=0 You factorise the first bit to get 3x(x-7) and then the second to get 5(x-7). You find you have a common factor of x-7!

    That leaves you with 3x +5. Therefore you get (x-7) + (3x +5).

    Factorise the bottom and you will find that the two (3x+5) cancel out!

    Leavin x-7/3x -5
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    I got x-7/ 3x +5 as well. If you want me to explain how I done it, I will more than likely confuse you more [/QUOTE]

    Yes please
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    who knows how to do this question-
    400 letters were sent
    ratio sent was 5:3
    first class-46p second class-36p- work out the total cost of sending all of them
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    (Original post by w0tSuhail)
    I got x-7/ 3x +5 as well. If you want me to explain how I done it, I will more than likely confuse you more
    Yes please[/QUOTE]
    Done it above ^^
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    (Original post by lolface32)
    Ok, I will do it here.

    3x^2-16x-35. 3X35=105 What times to give 105 and adds to give -16? -21 AND +5

    From there you get the equation 3x^2 -21x+5x-35=0 You factorise the first bit to get 3x(x-7) and then the second to get 5(x-7). You find you have a common factor of x-7!

    That leaves you with 3x +5. Therefore you get (x-7) + (3x +5).

    Factorise the bottom and you will find that the two (3x+5) cancel out!

    Leavin x-7/3x -5

    thank you - I just didnt find what multiplied to get 105 which was 21 * 5 :|
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    If you only have time for one more past paper, having already done June 2012 Non calculator which one is best to try?
    Have you done June 2012 non calculator?
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    best of luck everyone for tomorrow
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    (Original post by TomR10)
    http://tinypic.com/view.php?pic=dfbpcx&s=5

    On this, I have got to the point where I've worked out all of the values around the outside. Being 25m on the hypotenuse, 24m along the bottom and 7m on the remaining side. I am now confused as to how I substitute these numbers in order to find the area of the triangle. Can any one explain?
    First find the value of X: (3x) + (3x +1) + (x-1) = 56

    simplify: 7x = 56
    x = 8

    Now substitute x=8 and find the area.

    Area = 1/2(width*length)

    Area = 1/2((x-1)*(3x))

    Area = 1/2 ((8-1)*(3*8))

    Area = 1/2 (7)*(24)

    Area = 1/2 (168)

    Area = 84


    I hope I am correct.
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    (Original post by student.wizz)
    Sorry about it being messy but you get the idea

    Some other guy explained it too. Thanks anyway
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    (Original post by MichelleQ)
    who knows how to do this question-
    400 letters were sent
    ratio sent was 5:3
    first class-46p second class-36p- work out the total cost of sending all of them
    You do 5 + 3 which is 8 and then you do 400/8 which gives you 50.

    Then you do 50 x 5 and 50 x 3 to get the ratio 250:150

    Then you do 250 x 46 to get 11500 and 150 x 36 which is 5400

    Then you add them up to get the answer 16900p (£169)

    I'm prety sure this is the answer but i'd be great if someone else can confirm.
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    (Original post by w0tSuhail)
    First find the value of X: (3x) + (3x +1) + (x-1) = 56

    simplify: 7x = 56
    x = 8

    Now substitute x=8 and find the area.

    Area = 1/2(width*length)

    Area = 1/2((x-1)*(3x))

    Area = 1/2 ((8-1)*(3*8))

    Area = 1/2 (7)*(24)

    Area = 1/2 (168)

    Area = 84


    I hope I am correct.
    Yeah, that's correct on the mark scheme. Well done. I got up to
    x=8 and then got confused! Could you go over the section following x=8. I'm still a bit unsure, cheers mate!
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    yeah, checked the book, its the answer, thank you
 
 
 
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