Maths year 11

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    (Original post by RDKGames)
    ba+2a=2a^2b ??? You're not multiplying them here.
    Then how do I add them?

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    (Original post by z_o_e)
    Then how do I add them?

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    Well certainly not by multiplying them, that's a whole other operation.

    It's just that, though you can factor out the a which is helpful to simplify the overall fraction further.
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    (Original post by RDKGames)
    Well certainly not by multiplying them, that's a whole other operation.

    It's just that, though you can factor out the a which is helpful to simplify the overall fraction further.
    3ab?

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    No. ab+2a=a(b+2)

    \rightarrow \frac{a(b+2)}{2a} = \frac{b+2}{2}
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    (Original post by RDKGames)
    No. ab+2a=a(b+2)

    \rightarrow \frac{a(b+2)}{2a} = \frac{b+2}{2}
    Is that the final answer?

    So do I do the rest just like the same way?

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    (Original post by z_o_e)
    Is that the final answer?

    So do I do the rest just like the same way?

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    Yes. That would be the final simplified answer.
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    (Original post by RDKGames)
    Yes. That would be the final simplified answer.
    Okay can you do a quick example on root 500 and simplifying it. On paper please x

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    (Original post by RDKGames)
    Yes. That would be the final simplified answer.
    I can't really factories this and the numerators have no like terms



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    (Original post by RDKGames)
    Yes. That would be the final simplified answer.
    I can't really factories this and the numerators have no like terms...



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    (Original post by z_o_e)
    Okay can you do a quick example on root 500 and simplifying it. On paper please x

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    Lol sure, root 500 it is then.
    Attachment 574822

    Also that fraction with a, b, c and d cannot be factorised any further. Just make it a single fraction and you're done


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    (Original post by z_o_e)
    I can't really factories this and the numerators have no like terms...



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    Hmmm confusing.


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    (Original post by z_o_e)
    Hmmm confusing.


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    What's confusing about it? I'm splitting 500 into 2 and 250 as they make up the 500. Obviously I cannot decompose the 2 any further so I split the 250 into 2 and 125, and finally I split 125 into 5 and 25. There are all surds multiplied together so I just simplify the one's I can.
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    (Original post by RDKGames)
    What's confusing about it? I'm splitting 500 into 2 and 250 as they make up the 500. Obviously I cannot decompose the 2 any further so I split the 250 into 2 and 125, and finally I split 125 into 5 and 25. There are all surds multiplied together so I just simplify the one's I can.
    Doesn't 25 decompose into 5 and 5

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    (Original post by z_o_e)
    Doesn't 25 decompose into 5 and 5

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    Yeah. But for one, I know that 25 is a square number so when it is square rooted, I would just get 5.

    Otherwise, I could've written it as root 5 times root 5 but I would've gotten a 5 out of these two multiplied. I just skipped this step.
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    (Original post by RDKGames)
    Yeah. But for one, I know that 25 is a square number so when it is square rooted, I would just get 5.

    Otherwise, I could've written it as root 5 times root 5 but I would've gotten a 5 out of these two multiplied. I just skipped this step.
    So using these circled numbers how would I find simplified of 500?


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    yusss I got it!



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    (Original post by z_o_e)
    So using these circled numbers how would I find simplified of 500?


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    It would be \sqrt{500}=\sqrt2 \cdot \sqrt2 \cdot \sqrt5 \cdot \sqrt5 \cdot \sqrt5 then just simplify whatever you can.
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    (Original post by RDKGames)
    It would be \sqrt{500}=\sqrt2 \cdot \sqrt2 \cdot \sqrt5 \cdot \sqrt5 \cdot \sqrt5 then just simplify whatever you can.
    What about this?


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    (Original post by z_o_e)
    What about this?


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    Correct method except you must've gotten carried away because at the very bottom, 5 and 5 do not make up 10 with multiplication.
 
 
 
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