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    • Thread Starter
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    If anyone has more questions I won't be able to reply from now as I won't be online. Best of luck in the exam to everyone! 😊
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    (Original post by Habiba12)
    go on https://www.youtube.com/watch?v=-z4SUypJZxo its a bit annoying but trust me itll stay in your head i heard this once in yr 7 and i still remember it helps with circles and pi
    LOOL its funny! & Ty
    • Welcome Squad
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    Welcome Squad
    (Original post by TJ2001)
    Proof is hard to explain over typing, this short video may help.. http://m.youtube.com/watch?v=DxODoNGetVY
    With vectors the basic rule to remember is if the are multiples they're parallel, and if they're multiples and share a common point it's a straight line. Sometimes it can be very hit or miss with the complicated problems but providing you remember those rules you can give it a good go.
    Surds again can be very hit or miss, but this short video should outline the basic principle - http://m.youtube.com/watch?v=Aq8FavvjSII - and a video is probably easier to understand that if I was to try and type out an explanation.
    Congruent triangles, make sure you remeber the four rules of what needs to be equal to be congruent.
    SSS (three side)
    AAS (two angles and corresponding side)
    SAS (two sides and included angle)
    RHS (right angle, hypotenuse, side)
    For similar shapes know these rules and you'll be fine...
    (K being the scale factor)
    Length - times by K
    Surface area - times by K^2
    Volume - time by K^3

    Hope this has helped. Good Luck in the exam 😊
    Thanks
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    I have a test tomorrow I'm in college doing level 2 do you think these questions will be on the test
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    (Original post by Ranadosa)
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    Here. Sorry can't explain it - don't have time, going to bed now.
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    (Original post by Ranadosa)
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    The size of each angle in an n-sided regular polygon is \frac{180(n-2)}{n} You can use this to find the size of each angle in the octagon.

    Use 180(n-2) to find the sum of angles in the hexagon.

    And the size of the unknown angles in the hexagon will be equal, so you can work that out by dividing (the sum of angles in a hexagon - 280) by 4.
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    (Original post by Ranadosa)
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    In regular Octagon
    Ex.Angle = 360/8 = 45 degrees
    Int.Angle- 180-45 = 135 degrees

    JK is line of symmetry so draw line of symmetr to divide hexagon into two trapeziums. Now the angle 140 degrees has been split into two so now angle JKF and angle KJF are each equal to 70 degrees.

    In a trapezium, there a two paralell lines therefore interior angles can add up to 180 degrees. 180-70= 110 degrees

    Now take angle inside trapezium away from interior angle of octagon:
    135- 110 = 25 degrees
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    how do you work out the surface area of a solid hemisphere?
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    (Original post by Yasminxb)
    how do you work out the surface area of a solid hemisphere?
    Half of the surface area of a sphere + area of a circle

    A = 2 \pi r^2 + 2 \pi r
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    (Original post by Yasminxb)
    how do you work out the surface area of a solid hemisphere?
    Surface area of a whole sphere is  4\pi r^2 . So half of that outer surface would be  2\pi r^2 but you have to remember because you 'cut' the sphere in half there is a circle face at the bottom of the hemisphere so we have to add on  \pi r^2 .
    So as the other user said the total surface area of a hemisphere is  2\pi r^2 + \pi r^2 = 3\pi r^2 .
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    (Original post by TheOtherSide.)
    Half of the surface area of a sphere + area of a circle

    A = 2 \pi r^2 + 2 \pi r
    (Original post by B_9710)
    Surface area of a whole sphere is  4\pi r^2 . So half of that outer surface would be  2\pi r^2 but you have to remember because you 'cut' the sphere in half there is a circle face at the bottom of the hemisphere so we have to add on  \pi r^2 .
    So as the other user said the total surface area of a hemisphere is  2\pi r^2 + \pi r^2 = 3\pi r^2 .
    thank youu :heart:
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    Right its hard to explain but you need to find the midpoint of the current line first. Then this will give you an x and y co-ordinate which you use later on. You then have to find the gradient from the current line, it will either be in an equation you are given or you will have to work it out from the diagram given (height÷length). This gradient will probably be in a fraction form, you flip this fraction to find the gradient of the perpendicular line. Then tou have to substitute everything you have found out into the equation y=mx+c...so... you would take the x and y from the midpoint and the gradient (m). Then you just need to make c the subject of your equation.then finally substitue the gradient and c into your equation y=mx+c to get your new equation for the perpendicular line.... sooo recap.
    -find the midpoint xy
    -find the gradient of current line (ab)
    -find the gradient of new line (ab flipped)
    -substitue the gradient and x and y into y=mx+c
    -then find c and substitue one last time to get you answer

    I hope this made some sense... good luck!


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    (Original post by Jaff_)
    Yo! Can someone explain the alternate segment therom and proof about consecutive numbers. Thanks, much appreciated!
    Always use n and n+1 to show consecutive numbers, there was one a few years back about the difference of two squares of consecutive numbers and it was a case of multiplying n squared and n+1 squared and collecting like terms, sorry if it's vague
    Good luck!
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    Yo guys I did my IGCSE Maths a year early in 2015 June, got 97% lol. Anyways, reply to this or PM me if u got any questions, free and happy to help (if I can) lol!
 
 
 
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