Maths year 11

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    Multiply both sides by 3, then square root both sides.
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    (Original post by RDKGames)
    Multiply both sides by 3, then square root both sides.


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    That would indeed be the radius, but you should get used to working like this in the exams:

    27=\frac{1}{3}r^2 (27 comes from 270 over 10 as you might expect)
    \rightarrow 81=r^2
    \rightarrow 9=r
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    (Original post by RDKGames)
    That would indeed be the radius, but you should get used to working like this in the exams:

    27=\frac{1}{3}r^2 (27 comes from 270 over 10 as you might expect)
    \rightarrow 81=r^2
    \rightarrow 9=r
    Yepp

    Then

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    (Original post by z_o_e)
    Yepp

    Then

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    Yes, but my layout would be clearer for the examiner/whoever is marking it as it shows logical progression to get to r and convinces them that its value is 9. You can move on nonetheless
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    (Original post by RDKGames)
    Yes, but my layout would be better for the examiner/whoever is marking it. You can move on nonetheless
    Yepp

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    (Original post by B_9710)
     \displaystyle \int \frac{1}{\cos x } \ dx.
    sorry im at work so quick typing this.

    ∫ 1/Cosx dx. This is a standard integral but you can use the quotient rule to prove it. This is just workings but if you dont get any part of it then ask.

    ∫ 1/Cosx dx = ((Cosx)(0) - (1)(-Sinx)) / (Cosx)^2 +C

    = Sinx / Cos^2x +C

    = (Sinx /Cosx)*(1/Cosx) +C

    =SecxTanx +C
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    (Original post by z_o_e)
    Yepp

    Then

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    The question asks you for the total volume though, not just finding the radius.
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    (Original post by TSRPAV)
    sorry im at work so quick typing this.

    ∫ 1/Cosx dx. This is a standard integral but you can use the quotient rule to prove it. This is just workings but if you dont get any part of it then ask.

    ∫ 1/Cosx dx = ((Cosx)(0) - (1)(-Sinx)) / (Cosx)^2 +C

    = Sinx / Cos^2x +C

    = (Sinx /Cosx)*(1/Cosx) +C

    =SecxTanx +C
    Integrate, not differentiate.
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    (Original post by TSRPAV)
    sorry im at work so quick typing this.

    ∫ 1/Cosx dx. This is a standard integral but you can use the quotient rule to prove it. This is just workings but if you dont get any part of it then ask.

    ∫ 1/Cosx dx = ((Cosx)(0) - (1)(-Sinx)) / (Cosx)^2 +C

    = Sinx / Cos^2x +C

    = (Sinx /Cosx)*(1/Cosx) +C

    =SecxTanx +C
    Wrong.
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    (Original post by RDKGames)
    Wrong.
    Can you do it?
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    (Original post by B_9710)
    Can you do it?
    Indeed, but only because I remember what to multiply the integral by. If I didn't, I would probably struggle without help.
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    (Original post by RDKGames)
    Indeed, but only because I remember what to multiply the integral by. If I didn't, I would probably struggle without help.
    There's a very neat substitution but I won't say it and give it away (it may be quite obvious).
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    (Original post by B_9710)
    Integrate, not differentiate.
    Sorry I'm hangover and at work so my brain no work ahah yeah but all you do is 1/cosx = secx and then use data sheet as its on there,the proof is difficultName:  image.png
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    (Original post by TSRPAV)
    Sorry I'm hangover and at work so my brain no work ahah yeah but all you do is 1/cosx = secx and then use data sheet as its on there,the proof is difficultName:  image.png
Views: 27
Size:  49.5 KB
    That's what I was asking for, the proof.
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    (Original post by RDKGames)
    Indeed, but only because I remember what to multiply the integral by. If I didn't, I would probably struggle without help.
    So the radius is 9?

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    (Original post by z_o_e)
    So the radius is 9?

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    yes
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    (Original post by RDKGames)
    yes


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    Why are you working out the volume of the cone if you're given it in the question? Also that wouldn't give you the right volume anyway because 9^2 \not= 18 and not to mention you're missing \pi
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    (Original post by RDKGames)
    Why are you working out the volume of the cone if you're given it in the question? Also that wouldn't give you the right volume anyway because 9^2 \not= 18 and not to mention you're missing \pi


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