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    (Original post by frozo123)
    Have you got it now? Sub a back into any of the equations
    then to understand what you've done
    Solve

    sin(ax-b)=0 0<x<360 ( the inequality sign meant to be or equal to )

    so let (ax-b)= theta

    so values are 0, pi, 2pi
    then add b, divide a
    each of your values
    you should get what the coordinates are in the question
    Ahh, that makes sense! So it'll be 0+b/2 which must equal pi/10..which is why b is pi/5? I don't think I'll be able to do this in an exam though :/

    Thanks
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    (Original post by userxx990)
    So guys, ive summed up my very dumb maths doubts, very simple and quick to answer like weird equations etc, can any one show me a paper work solution of these, if not, one of these? Its hell important, as my basics realllyyy sucks! Please help!
    I'm not sure whether because the sun is blinding my view or your pdf file is upside down
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    (Original post by ToLiveInADream)
    Ahh, that makes sense! So it'll be 0+b/2 which must equal pi/10..which is why b is pi/5? I don't think I'll be able to do this in an exam though :/

    Thanks
    I can't remember whether a is 2 but if it was then yeah that's it
    just work backwards
    Idk if you need any help just ask
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    (Original post by frozo123)
    I'm not sure whether because the sun is blinding my view or your pdf file is upside down
    Sorrry! is it finee nooww?
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  1. File Type: pdf 4444444.pdf (373.6 KB, 112 views)
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    (Original post by userxx990)
    So guys, ive summed up my very dumb maths doubts, very simple and quick to answer like weird equations etc, can any one show me a paper work solution of these, if not, one of these? Its hell important, as my basics realllyyy sucks! Please help!
    okay first one, I don't know where the 25 comes from
    but the square root of y
    because log5y= b
    1/2log5y=1/2b
    and 1/2log5y is the same as log5y^1/2
    ask someone else about the 25.. sorry!

    second one you could put x1/3 as the denominator and then multiply everything by x1/3 to get rid of the negative power
    so you'll get -x^1/3=-2
    then solve for x

    third one well cubing it is (2root2)(2root2) which is 8 so then another 2root2 is 16root2 then you can subtract as the roots are the same
    fourth one it's an arithmetic sequence so use the formula Sn= blah blah
    there will be 21 terms if you think about it because the 10 and 30 are inclusive
    so Sn=10.5(54+(21-1)2)

    3tan^2theta -1=0
    you put the 1 on the other side then divde by 3 and solve it how you would
    if you don't know how to solve it let me know

    next one you put the -1/2 as the denominator
    then multiply every term by 1/2
    so you'll get x^2-8=0
    then solve for x
    next one turn it into a top heavy fraction so (49/9)^-1/2
    flip it because the power is negative so (9/49)^1/2 then square root each term

    final one, you multiply every term by x to get rid of the denominator then multiply by 2 to get rid of the 2
    although I don't get how that equals what you've written down
    ask someone else

    hope that helps!

    are you an international student or something? cus half of that is c1
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    Innnn on this thread
    Not looking forward to this :lol:
    Hope everyone's well though
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    (Original post by frozo123)
    okay first one, I don't know where the 25 comes from
    but the square root of y
    because log5y= b
    1/2log5y=1/2b
    and 1/2log5y is the same as log5y^1/2
    ask someone else about the 25.. sorry!

    second one you could put x1/3 as the denominator and then multiply everything by x1/3 to get rid of the negative power
    so you'll get -x^1/3=-2
    then solve for x

    third one well cubing it is (2root2)(2root2) which is 8 so then another 2root2 is 16root2 then you can subtract as the roots are the same
    fourth one it's an arithmetic sequence so use the formula Sn= blah blah
    there will be 21 terms if you think about it because the 10 and 30 are inclusive
    so Sn=10.5(54+(21-1)2)

    3tan^2theta -1=0
    you put the 1 on the other side then divde by 3 and solve it how you would
    if you don't know how to solve it let me know

    next one you put the -1/2 as the denominator
    then multiply every term by 1/2
    so you'll get x^2-8=0
    then solve for x
    next one turn it into a top heavy fraction so (49/9)^-1/2
    flip it because the power is negative so (9/49)^1/2 then square root each term

    final one, you multiply every term by x to get rid of the denominator then multiply by 2 to get rid of the 2
    although I don't get how that equals what you've written down
    ask someone else

    hope that helps!

    are you an international student or something? cus half of that is c1
    bro, firstly thank you so much for such a great help!

    Just the 3tan^2 one, i solved it properly but they are asking to give the answer in terms of Pie, now how do you express this number in terms of pie?

    Next, the one you told for - x^1/3 = -2 how to solve this equation? i really dont know :P

    and the 2 under root 2 one if you cube it out how will you end up with 8?

    And yes you guessed right, im international have C12 on wednesday..
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    (Original post by userxx990)
    bro, firstly thank you so much for such a great help!

    Just the 3tan^2 one, i solved it properly but they are asking to give the answer in terms of Pie, now how do you express this number in terms of pie?

    Next, the one you told for - x^1/3 = -2 how to solve this equation? i really dont know :P

    and the 2 under root 2 one if you cube it out how will you end up with 8?

    And yes you guessed right, im international have C12 on wednesday..
    Cube both sides


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    (Original post by userxx990)
    bro, firstly thank you so much for such a great help!

    Just the 3tan^2 one, i solved it properly but they are asking to give the answer in terms of Pie, now how do you express this number in terms of pie?

    Next, the one you told for - x^1/3 = -2 how to solve this equation? i really dont know :P

    and the 2 under root 2 one if you cube it out how will you end up with 8?

    And yes you guessed right, im international have C12 on wednesday..
    was it tan^2theta=1/3 ?
    square root both sides
    so you get tantheta= +-root3/3
    put your calculator in radians mode
    it should work use a graph of tan to work it out
    or CAST but graph is better
    and its aite
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    (Original post by frozo123)
    was it tan^2theta=1/3 ?
    square root both sides
    so you get tantheta= +-root3/3
    put your calculator in radians mode
    it should work use a graph of tan to work it out
    or CAST but graph is better
    and its aite
    ohh so radians mode was needed, after that ill simply put the answer and thats in terms of pie right?
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    (Original post by physicsmaths)
    Cube both sides


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    yeep thankss!
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    1 more question i had forgot, how to factorise, 2x^3 - 16 ?
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    Why is arn used instead of arn-1 in question 6 January 2010?

    Question Paper
    Mark Scheme
    Examiner's Report
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    (Original post by userxx990)
    1 more question i had forgot, how to factorise, 2x^3 - 16 ?
    You can't
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    (Original post by Cosmi)
    Why is arn used instead of arn-1 in question 6 January 2010?

    Question Paper
    Mark Scheme
    Examiner's Report
    hard to explain but it says 3 years after it was purchased, one year after it was purchased would be x0.8, 2 years after would be x0.8^2 etc.
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    [QUOTE=userxx990;55935917]1 more question i had forgot, how to factorise, 2x^3 - 16 ?[/QUOTE

    Hi you can try factor theorem and see if anything can factor in to it . Then factorise x-2 is a solution and then you can divide by it hope that helps
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    (Original post by Cosmi)
    You can't
    Yes you can.


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    (Original post by userxx990)
    1 more question i had forgot, how to factorise, 2x^3 - 16 ?
    X=2 is asolution so divide by x-2


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    (Original post by physicsmaths)
    X=2 is asolution so divide by x-2


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    how did you get x-2 from?
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    where can you get the info for 100UMS on each year
 
 
 
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