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Maths year 11

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Reply 660
Original post by RDKGames
How would you approach this? :smile:


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.
Reply 662
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. :smile:

It makes though, just think about it rather than leaving it as a guess.
Reply 664
Original post by RDKGames
Perfect. :smile:

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.
Reply 666
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. :smile:

I like the way you draw your \sum (capital sigma, meaning 'sum') symbol as a 3 other way around ;D
Reply 668
Original post by RDKGames
I believe that is correct. :smile:

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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Reply 669
Original post by RDKGames
I believe that is correct. :smile:

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




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Reply 671
Original post by RDKGames
Bingo.


I'm getting the hang of these :biggrin:



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Original post by z_o_e
I'm getting the hang of these :biggrin:

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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.
(edited 7 years ago)
Reply 673
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.
Reply 675
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 :frown:

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Original post by z_o_e
I don't get this question :frown:

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

Cone: Vc=13πhr2V_c=\frac{1}{3}\pi hr^2
Sphere: VS=43πr3V_S=\frac{4}{3}\pi r^3 therefore half a sphere would be Vs=23πr3V_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 270π270\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:
270π=13π10r2 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 π\pi. This means that rather writing out something like 23π2\cdot 3\pi as 18.849... we instead leave our answer as 6π6\pi. No need to convert into decimals.
(edited 7 years ago)
Reply 677
Original post by RDKGames
Ok, let's first get our formulae for the volumes. These are given in the formula booklet, I believe.

Cone: Vc=13πhr2V_c=\frac{1}{3}\pi hr^2
Sphere: VS=43πr3V_S=\frac{4}{3}\pi r^3 therefore half a sphere would be Vs=23πr3V_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 270π270\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:
270π=13π10r2 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 π\pi. This means that rather writing out something like 23π2\cdot 3\pi as 18.849... we instead leave our answer as 6π6\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, 27010\frac{270}{10} would give you 13r2\frac{1}{3}r^2. I'm sure you know what to do from there to get the radius.
Reply 679
Original post by RDKGames
After you cancel pi, 27010\frac{270}{10} would give you 13r2\frac{1}{3}r^2. I'm sure you know what to do from there to get the radius.


Nooo:*(

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