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    Any predictions then?
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    Please could anyone explain question 22 on here http://madasmaths.com/archive/maths_...ons_part_a.pdf, how do we know that set (b^-1)= 1/3? Thanks
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    (Original post by thesmallman)
    jan 2016 (IAL) is difficult af- is everything there in our spec??
    it wasn't that hard - got 62/65 (not 75 due to the roots of polynomial qn)

    Things to take away from that paper:

    - You may have an ugly expression for x or y when doing conics - use the formula rather than hoping to factorising
    - Be mindful when carrying out summations and proofs (eg one k+1 power and one k-1)


    rest of it was alright
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    Which is the hardest paper in your guys opinion? Or rather the topic that will most likely tear us a new one come friday mid-morning?
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    (Original post by tazza ma razza)
    Which is the hardest paper in your guys opinion? Or rather the topic that will most likely tear us a new one come friday mid-morning?
    Jan 2016 was relatively harder
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    (Original post by LewisClothier)
    So, you have the geometrical transformation of the matrix A, should be 135 degree clockwise rotation, then you need to find a multiple of 135 that equals a multiple of 360.
    In other words, how many times would you need to apply this transformation to I to get back to I. the answer cannot be 0 because it needs to be a positive integer, although that would work... so you would need to apply the transformation 8 times to get back to I.
    To get 8, would you keep dividing multiples of 360 by 135 until you get a whole number?
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    (Original post by techfan42)
    To get 8, would you keep dividing multiples of 360 by 135 until you get a whole number?
    I think It's easier to imagine the rotation, and count on your fingers how many times it takes you to get back to I xD but that's just me.

    I'm joking but what that is essentially doing is multiplying by the matrix each time, hence multiplying the rotation (135 degrees) each time. Keep doing this until you get to a multiple of 360 and then you'll be back to where you started.
    Spoiler:
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    totes not joking...
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    (Original post by economicss)
    Please could anyone explain question 22 on here http://madasmaths.com/archive/maths_...ons_part_a.pdf, how do we know that set (b^-1)= 1/3? Thanks
    The determinant of the product of two square and equal dimension matrices is equal to the product of both matrix's determinants. Also, the determinant of the inverse of a matrix is equal to the inverse of the determinant of the original matrix.

    So if the determinant of the inverse is a third, what does that tell you about \det(\mathbf{A})? Then can you find \det(\mathbf{B}) after that?
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    (Original post by tazza ma razza)
    it wasn't that hard - got 62/65 (not 75 due to the roots of polynomial qn)

    Things to take away from that paper:

    - You may have an ugly expression for x or y when doing conics - use the formula rather than hoping to factorising
    - Be mindful when carrying out summations and proofs (eg one k+1 power and one k-1)


    rest of it was alright
    Yeah I got the same mark as you but yh just realised the roots of polynomial qs wasn't in the spec ahaha
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    Divisibilty proof will definitely come up, there were none last year
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    The IAL F1 papers have this one question on every paper where it's to do with roots of some equation and you have to add the roots or cube the roots or something. And I swear it's not in the textbook for FP1

    Is this where the specs differ or is it meant to be like a harder question or something?
    Should I bother learning these for the FP1 on Friday.
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    (Original post by Siddhart1998)
    The IAL F1 papers have this one question on every paper where it's to do with roots of some equation and you have to add the roots or cube the roots or something. And I swear it's not in the textbook for FP1
    It's not on the UK spec.
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    Name:  Screenshot 2016-05-19 at 00.52.48.png
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    Guys, I've been doing the F1 Jan 2016 paper and found it very good so far, and then I got to this question. I got to the 2 triangles, correct coordinates etc, and then I proceeded to attempt work out the area by splitting the shapes which is obviously not needed as I know you can do 1/2bh.
    However, and this has bugged me for quite long in a level maths, I really don't see how either of those are a right angle triangle? I thought 1/2bh was only for right angle triangles. I don't get how either of those are right angle triangles to be able to do 1/2bh really.

    It's probably really simple, but it really annoys me so someone help me please
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    (Original post by yesyesyesno)
    It's probably really simple, but it really annoys me so someone help me please
    The formula 1/2 * base * perpendicular height applies to all triangles as long as what you've found is the perpendicular height from the base.

    In this case, the perpendicular height is easy to find because it's just the length of the blue line which is the distance from the x-axis to the point B which is just the y coordinate.

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    (Original post by Zacken)
    The formula 1/2 * base * perpendicular height applies to all triangles as long as what you've found is the perpendicular height from the base.

    In this case, the perpendicular height is easy to find because it's just the length of the blue line which is the distance from the x-axis to the point B which is just the y coordinate.

    But then doesn't this include an extra area between the red and blue line, where it has the capital A from 'Area'
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    (Original post by yesyesyesno)
    But then doesn't this include an extra area between the red and blue line, where it has the capital A from 'Area'
    No. I'm not talking about the length of the red line at all. I am only talking about the length of the blue line. The blue line is the perpendicular distance from the base to the third vertex.

    The fact that I use the base as being the length OS means I am finding the area of the triangle OBS.

    The area of the triangle is the 1/2 * base * perpendicular height.

    So, in this example the perpendicular height is the dotted line. The base is the base. The area is 1/2 * 4 * 6. Even if the triangle is not right angled.



    Our example makes it look as though we are building some other random triangle, but we are not.

    I only drew the red line as an extension to the base to show what the perpendicular height from the base is.

    Look at this:



    In the first one (a) we can find the perpendicular height. In the second example, we need to extend the base to find the perpendicular height, but the area of the triangle is the triangle with the solid lines, the dotted line is just an extension that allows us to find the perpendicular height.
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    (Original post by Zacken)
    No. I'm not talking about the length of the red line at all. I am only talking about the length of the blue line. The blue line is the perpendicular distance from the base to the third vertex.

    The fact that I use the base as being the length OS means I am finding the area of the triangle OBS.

    The area of the triangle is the 1/2 * base * perpendicular height.

    So, in this example the perpendicular height is the dotted line. The base is the base. The area is 1/2 * 4 * 6. Even if the triangle is not right angled.



    Our example makes it look as though we are building some other random triangle, but we are not.

    I only drew the red line as an extension to the base to show what the perpendicular height from the base is.

    Look at this:



    In the first one (a) we can find the perpendicular height. In the second example, we need to extend the base to find the perpendicular height, but the area of the triangle is the triangle with the solid lines, the dotted line is just an extension that allows us to find the perpendicular height.
    Thank you so much !
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    (Original post by yesyesyesno)
    Thank you so much !
    Makes sense now?
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    (Original post by Zacken)
    Makes sense now?
    Yes! this always bugged me because it never made sense to me so I would always start making right angled triangles and squares out of the triangle however it was virtually impractical for that question.
    What I gather is that if it's not necessarily a right angled triangle but you can find a perpendicular height, whether it be a dotted line extended from your actual base or just somewhere along your base you apply the normal formula, with your actual base (not dotted) and your perpendicular height.
    I really can't thank you enough, I'm pretty sure any question I've asked you've answered it with excellent clarity.
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    (Original post by yesyesyesno)
    Yes! this always bugged me because it never made sense to me so I would always start making right angled triangles and squares out of the triangle however it was virtually impractical for that question.
    What I gather is that if it's not necessarily a right angled triangle but you can find a perpendicular height, whether it be a dotted line extended from your actual base or just somewhere along your base you apply the normal formula, with your actual base (not dotted) and your perpendicular height.
    I really can't thank you enough, I'm pretty sure any question I've asked you've answered it with excellent clarity.
    Awesome. And yep, that's precisely right. (it's actually what the sine area rule thingy says in disguise, if you've seen that before.)

    Thanks, much appreciated.
 
 
 
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