Maths year 11

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    (Original post by RDKGames)
    How would you approach this?
    Well I would count all the rectangles inside that shape and work out the perimeter as usual.

    But the triangles... I would subtract the 8 -2 to find one side... and the other side are 8?

    Maybe? Idk

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    (Original post by z_o_e)
    Well I would count all the rectangles inside that shape and work out the perimeter as usual.

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    Yes you can do that; just remember that dont need the perimeters of these rectangles, you only need to consider one side of each rectangle for the perimeter of the 8 sided shape.

    (Original post by z_o_e)
    But the triangles... I would subtract the 8 -2 to find one side... and the other side are 8?

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    That's exactly right. Since these are right-angled triangles, you can use those sides to work out the hypotenuse which is one of the edges for the 8 sided shape.
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    (Original post by RDKGames)
    Yes you can do that; just remember that dont need the perimeters of these rectangles, you only need to consider one side of each rectangle for the perimeter of the 8 sided shape.



    That's exactly right. Since these are right-angled triangles, you can use those sides to work out the hypotenuse which is one of the edges for the 8 sided shape.
    Omgg I think I did it right. It was a guess tbh


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    (Original post by z_o_e)
    Omgg I think I did it right. It was a guess tbh


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    Perfect.

    It makes though, just think about it rather than leaving it as a guess.
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    (Original post by RDKGames)
    Perfect.

    It makes though, just think about it rather than leaving it as a guess.


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    When I was doing my GCSE's I found it useful to create an extra column or two where I label one 'midpoints' and the other 'frequency x midpoints'. Then to find the mean you simply add up the values from 'freq x midpoints' and divide by the total number of objects; in this case 40 runners.
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    (Original post by RDKGames)
    When I was doing my GCSE's I found it useful to create an extra column or two where I label one 'midpoints' and the other 'frequency x midpoints'. Then to find the mean you simply add up the values from 'freq x midpoints' and divide by the total number of objects; in this case 40 runners.


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    I believe that is correct.

    I like the way you draw your \sum (capital sigma, meaning 'sum') symbol as a 3 other way around ;D
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    (Original post by RDKGames)
    I believe that is correct.

    I like the way you draw your \sum (capital sigma, meaning 'sum') symbol as a 3 other way around ;D
    Hah tyy



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    (Original post by RDKGames)
    I believe that is correct.

    I like the way you draw your \sum (capital sigma, meaning 'sum') symbol as a 3 other way around ;D


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    Bingo.
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    (Original post by RDKGames)
    Bingo.
    I'm getting the hang of these



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    (Original post by z_o_e)
    I'm getting the hang of these

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    I used to separate these into parts, so a hemisphere and a cone, and mark on any relevant information on both that can help me plug in values into their volume formulae. In this case you have to work backwards to get the radius using the cone volume formula.
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    (Original post by RDKGames)
    I used to separate these into parts, so a hemisphere and a cone, and mark on any relevant information on both that can help me plug in values into their volume formulae. In this case you have to work backwards to get the radius using the cone volume formula.


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    Write down the volume formulae for both; the sphere and the cone, then see what variables you know within those equations, such as the volume of the cone as it's given in the question.
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    (Original post by RDKGames)
    Write down the volume formulae for both; the sphere and the cone, then see what variables you know within those equations, such as the volume of the cone as it's given in the question.
    I don't get this question

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    (Original post by z_o_e)
    I don't get this question

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    Ok, let's first get our formulae for the volumes. These are given in the formula booklet, I believe.

    Cone: V_c=\frac{1}{3}\pi hr^2
    Sphere: V_S=\frac{4}{3}\pi r^3 therefore half a sphere would be V_s=\frac{2}{3}\pi r^3

    We want to find the volume of the entire shape. We are told the volume of the cone therefore we only need to work out the volume of the hemisphere.

    What information do we need to work out the hemisphere? Well, by the looks of the equation we only need to find the radius of the hemishere.

    How do we find the radius? For this we can use the volume and height of the cone so help us. We are told that the volume is 270\picm3, and that the height is 10cm.

    Since we know both of these, we can plug them into the equation of the cone giving us:
     270\pi = \frac{1}{3}\pi \cdot 10 \cdot r^2

    We can see that pi cancels and we can solve for the radius. Once we know the radius, it's only the matter of referring back to what we said before and plugging it into the hemisphere volume formula to find that.

    Once we have the volume of the hemisphere, we can add that onto the volume of the cone and we get the overall volume of the solid shape.

    Important note: The question asks you to give your answers in terms of \pi. This means that rather writing out something like 2\cdot 3\pi as 18.849... we instead leave our answer as 6\pi. No need to convert into decimals.
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    (Original post by RDKGames)
    Ok, let's first get our formulae for the volumes. These are given in the formula booklet, I believe.

    Cone: V_c=\frac{1}{3}\pi hr^2
    Sphere: V_S=\frac{4}{3}\pi r^3 therefore half a sphere would be V_s=\frac{2}{3}\pi r^3

    We want to find the volume of the entire shape. We are told the volume of the cone therefore we only need to work out the volume of the hemisphere.

    What information do we need to work out the hemisphere? Well, by the looks of the equation we only need to find the radius of the hemishere.

    How do we find the radius? For this we can use the volume and height of the cone so help us. We are told that the volume is 270\picm3, and that the height is 10cm.

    Since we know both of these, we can plug them into the equation of the cone giving us:
     270\pi = \frac{1}{3}\pi \cdot 10 \cdot r^2

    We can see that pi cancels and we can solve for the radius. Once we know the radius, it's only the matter of referring back to what we said before and plugging it into the hemisphere volume formula to find that.

    Once we have the volume of the hemisphere, we can add that onto the volume of the cone and we get the overall volume of the solid shape.

    Important note: The question asks you to give your answers in terms of \pi. This means that rather writing out something like 2\cdot 3\pi as 18.849... we instead leave our answer as 6\pi. No need to convert into decimals.
    Got it.

    So I could do 270/10 which gives me the radius.

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    (Original post by z_o_e)
    Got it.

    So I could do 270/10 which gives me the radius.

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    After you cancel pi, \frac{270}{10} would give you \frac{1}{3}r^2. I'm sure you know what to do from there to get the radius.
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    (Original post by RDKGames)
    After you cancel pi, \frac{270}{10} would give you \frac{1}{3}r^2. I'm sure you know what to do from there to get the radius.
    Nooo:*(

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