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    (Original post by NiamhM1801)
    Neither...its come up in multiple past papers that I've done. It's basically Pythagoras and trig in 3 dimensions, so it might be under that in the spec?
    Oh I see. I thought you meant intersection between say the line
     \displaystyle \frac{x-2}{3} = y+5=\frac{z-1}{2} and the plane

     \displaystyle x-2y+z=2 .
    I know what you're talking about now.
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    (Original post by B_9710)
    You said factorise  \displaystyle x^{-4}-25y^2 . Expand yours out and see what it comes out as. You posted in the question a power of x^-2.
    Unless that was a mistake in your question?
    Ah, I've just realised it is indeed an error in my question. The power of is supposed to be positive. Sorry about that.
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    (Original post by Wolfram Alpha)
    Ah, I've just realised it is indeed an error in my question. The power of is supposed to be positive. Sorry about that.
    The answer you posted is correct then.
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    (Original post by zara_ruby)
    Just though I would start a thread about the further maths exams and how everyone is revising.
    Thanks for making this thread OP! I moved this into the Maths Exams forum and also put it on the GCSE Exam Directory Thread (which is here) so other students can find it easily
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    Hello everyone. I have a question.
    1/2x2a^2xsin30 = 18.
    If I wanted to get rid of the 1/2, I multiply both sides by two. Does this mean the new equation is as follows:
    2a^2xsin30 = 18.
    I wasn't very sure since in the first equation, not everything is being halved. Thanks in advance.
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    (Original post by Wolfram Alpha)
    Hello everyone. I have a question.
    1/2x2a^2xsin30 = 18.
    If I wanted to get rid of the 1/2, I multiply both sides by two. Does this mean the new equation is as follows:
    2a^2xsin30 = 18.
    I wasn't very sure since in the first equation, not everything is being halved. Thanks in advance.
    No, as you haven't multiplied the rhs by 2. It should be (I think):
    2a²×sin30 = 36
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    (Original post by Wolfram Alpha)
    Hello everyone. I have a question.
    1/2x2a^2xsin30 = 18.
    If I wanted to get rid of the 1/2, I multiply both sides by two. Does this mean the new equation is as follows:
    2a^2xsin30 = 18.
    I wasn't very sure since in the first equation, not everything is being halved. Thanks in advance.
    Assuming it's 2(a^2) and not (2a)^2 the easiest way is just to go straight to a^2 sin30=18
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    (Original post by Wolfram Alpha)
    Hello everyone. I have a question.
    1/2x2a^2xsin30 = 18.
    If I wanted to get rid of the 1/2, I multiply both sides by two. Does this mean the new equation is as follows:
    2a^2xsin30 = 18.
    I wasn't very sure since in the first equation, not everything is being halved. Thanks in advance.
    Is the question
     \displaystyle \frac{1}{2}\times 2a^2 \sin 30^{\circ} = 18 ?

    If so the 1/2 and the 2 on the left hand side of the equation cancel right away.
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    (Original post by B_9710)
    Is the question
     \displaystyle \frac{1}{2}\times 2a^2 \sin 30^{\circ} = 18 ?

    If so the 1/2 and the 2 on the left hand side of the equation cancel right away.
    Ah, I see. Thanks very much. Since I post quite a few maths questions, may I just ask how I can post the proper...equation so it looks like yours do. It's much less confusing.
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    (Original post by Wolfram Alpha)
    Ah, I see. Thanks very much. Since I post quite a few maths questions, may I just ask how I can post the proper...equation so it looks like yours do. It's much less confusing.
    http://www.thestudentroom.co.uk/wiki/LaTex
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    (Original post by NiamhM1801)
    No, as you haven't multiplied the rhs by 2. It should be (I think):
    2a²×sin30 = 36
    Oh yes, I just realised. Thanks for your help Niamh.
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    Thanks very much.
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    (Original post by Wolfram Alpha)
    Thanks very much.
    No problem, just to warn you it isn't the most intuitive thing to use
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    (Original post by Wolfram Alpha)
    Oh yes, I just realised. Thanks for your help Niamh.
    No problem, although the other persons method of just cancelling the ½ and the 2 is easier (I didn't see it at first)
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    Hi, I need help on a particularly awkward question.
    The question is :

    Simplify fully the expression

     \displaystyle \frac{3x^3}{x^2} \times \sqrt{\frac{9}{x^2}} .
    I get an answer but it's just cant be right.
    Could someone quickly just post an answer so I can see if I'm right?
    Thank you!
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    (Original post by Ano123)
    Hi, I need help on a particularly awkward question.
    The question is :

    Simplify fully the expression

     \displaystyle \frac{3x^3}{x^2} \times \sqrt{\frac{9}{x^2}} .
    I get an answer but it's just cant be right.
    Could someone quickly just post an answer so I can see if I'm right?
    Thank you!

    I got 9*
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    (Original post by NiamhM1801)
    I got 9*
    So you're saying for any x value, the expression will always be equal to 9. That's interesting.
    Try x=-1 into the original expression and see what answer you get. Post the answer you get when x=-1.
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    (Original post by Ano123)
    So you're saying for any x value, the expression will always be equal to 9. That's interesting.
    Try x=-1 into the original expression and see what answer you get. Post the answer you get when x=-1.
    Ah, now I think you should have 2 solutions? 9 and -9? As √9 has 2 answers?
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    (Original post by NiamhM1801)
    Ah, now I think you should have 2 solutions? 9 and -9? As √9 has 2 answers?
    Not quite.  \sqrt 9 = 3 .
     \sqrt 9 \neq -3 .
    What's actually happening here is you made a mistake when square rooting [/tex] x^2 [/tex]. If you put any negative number as x into the expression, say -1 you get  \sqrt {(-1)^2} = \sqrt 1 = 1 . So as you can see if x is a negative number then it returns a positive number.
    So actually  \sqrt {x^2} = | x | .
    The vertical lines mean the absolute value. Meaning we just take the magnitude of the number. So |-2| = 2.
    |-100|=100 etc.
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    (Original post by Ano123)
    Not quite.  \sqrt 9 = 3 .
     \sqrt 9 \neq -3 .
    What's actually happening here is you made a mistake when square rooting [/tex] x^2 [/tex]. If you put any negative number as x into the expression, say -1 you get  \sqrt {(-1)^2} = \sqrt 1 = 1 . So as you can see if x is a negative number then it returns a positive number.
    So actually  \sqrt {x^2} = | x | .
    The vertical lines mean the absolute value. Meaning we just take the magnitude of the number. So |-2| = 2.
    |-100|=100 etc.
    Woah there's no way this could be in further maths right? And I thought that √9 could be 3 or -3? As (-3)² is 9?
    What did you get as your answer then? This has really confused me!
 
 
 
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