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musicalrose

  • 3 years ago

( (1/7)x^2 - (2/5)) * ( (1/2)x + (1/2) )

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  1. apoorvk
    • 3 years ago
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    multiply each term of the anyone binomial into the other. Like this: (a+b)(c+d) = a(c+d) + b(c+d)

  2. musicalrose
    • 3 years ago
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    so..[ (1/7)x^2 * ((1/2)x^2) + (1/2)) ] + [ ((2/5) * ((1/2)x^2) + (1/2)) ] ?

  3. math456
    • 3 years ago
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    yup thts right..

  4. apoorvk
    • 3 years ago
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    good except that the (2/5) has a 'minus' sign with it.

  5. math456
    • 3 years ago
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    \[\left(\begin{matrix}1 \\ 14\end{matrix}\right)x ^{3}+\left(\begin{matrix}1 \\ 14\end{matrix}\right)x ^{2}-\left(\begin{matrix}2 \\ 10\end{matrix}\right)x-\left(\begin{matrix}2 \\ 10\end{matrix}\right)\]

  6. apoorvk
    • 3 years ago
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    @math456 oop - that notation is actually used for 'combination' or 'C' :P. Other wise it's good.

  7. musicalrose
    • 3 years ago
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    crap im so lost.

  8. math456
    • 3 years ago
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    ohh yea i know i couldnt find the division sign thts y i use this..

  9. apoorvk
    • 3 years ago
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    forget that comment, @musicalrose , you're almost there, just keep in mind that it's a "-2/5" over there, so you include the 'minus' sign as well while multiplying.

  10. math456
    • 3 years ago
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    @musicalrose i will make it clear.. u use the same concept..

  11. math456
    • 3 years ago
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    [ (1/7)x^2 *((1/2)x+(1/2) ] - (2/5*((1/2)x^2) + (1/2)) ] ?

  12. musicalrose
    • 3 years ago
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    yes itd be written like that

  13. musicalrose
    • 3 years ago
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    .07x^3 + 1/2 for the first part?

  14. math456
    • 3 years ago
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    (1/14)x3+(1/14)x2−(2/10)x−(2/10) <-- like this

  15. musicalrose
    • 3 years ago
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    oooh because you made like denominators?

  16. math456
    • 3 years ago
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    yup... u'll get 0.07 for 1/14

  17. musicalrose
    • 3 years ago
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    so shouldnt the 1/14x2 actually be 7/14 since it was 1/2x2? or am I not getting it lol

  18. math456
    • 3 years ago
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    |dw:1339011637725:dw|

  19. musicalrose
    • 3 years ago
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    I think I get it now..

  20. math456
    • 3 years ago
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    kool..

  21. musicalrose
    • 3 years ago
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    thank you tons for spending so much time helping me :)

  22. math456
    • 3 years ago
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    no problem..!

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