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imqwerty

  • one year ago

math for KIDS :P

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  1. imqwerty
    • one year ago
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  2. pooja195
    • one year ago
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    Much more like math for adults.

  3. Jadeishere
    • one year ago
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    I don't know... gah

  4. Afrodiddle
    • one year ago
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    Looks like I'm still a toddler.

  5. freckles
    • one year ago
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    *

  6. anonymous
    • one year ago
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    Why many of us still feel like a kid?

  7. anonymous
    • one year ago
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    whoa thats a long problem

  8. ganeshie8
    • one year ago
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    my first guess is to play with \[17\mid a ~~\text{and}~~17\mid b \implies 17\mid(ma\pm nb)\]

  9. Empty
    • one year ago
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    \[\left[ \begin{array}c 1 & -3/2\\-3/2 & 2\\\end{array} \right]\] and \[\left[ \begin{array}c 0 & 1/2\\1/2 & 0\\\end{array} \right]\] have the same determinant

  10. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} x^2-3xy+2y^2+x-y=17m\hspace{.33em}\\~\\ (x-2y+1)(x-y)=17m\hspace{.33em}\\~\\ \end{align}}\) we can consider here 4 cases \(\large \color{black}{\begin{align} x-2y+1=17,\ x-y=m \hspace{.33em}\\~\\ x-2y+1=m,\ x-y=17 \hspace{.33em}\\~\\ x-2y+1=17m,\ x-y=1 \hspace{.33em}\\~\\ x-2y+1=1,\ x-y=17m \hspace{.33em}\\~\\ \end{align}}\)

  11. Empty
    • one year ago
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    @mathmath333 How did you factor that, I wanna learn how you figured that out.

  12. mathmath333
    • one year ago
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    do u agree m should be an integer

  13. ganeshie8
    • one year ago
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    Ahh nice, factoring the quadratic \(x^2-3xy+2y^2\) is easy using factor by grouping : \[x^2-3xy+2y^2 = x^2-xy-2xy+2y^2 = x(x-y)-2y(x-y)\]

  14. ganeshie8
    • one year ago
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    a bit enlightening to notice that any quadratic equation of form \(ax^2+bxy+cy^2=0\) represents a pair of straight lines passing thru origin and thus can be factored as \((y+m_1x)(y+m_2x)=0\)

  15. Empty
    • one year ago
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    Ahhh ok thanks @ganeshie8 I worked through a few but I like the fact that all equations of that form in n variables seems to imply that it's really n lines through the origin, interesting.

  16. imqwerty
    • one year ago
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    well done guys. heres the solution -

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  17. ikram002p
    • one year ago
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    i wonder if i played with the linear combination, how it would end nicely

  18. ikram002p
    • one year ago
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    that is also neat

  19. Empty
    • one year ago
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    I tried to see if the final one was a linear combination of the two but it doesn't appear to be, try subtracting one from the other to get rid of the x^2 but you'll still have the y^2 that you can't get rid of @ikram002p

  20. ikram002p
    • one year ago
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    well yeah (there is no linear combination ) see why :- m x^2-2mxy+my^2-5mx+7my nx^2-3nxy+2ny^2+nx-ny ------------------------- xy -12x+15y set of equations:- m=n 2n=m -2m-3n=1 -5m+n=-12 7m-n=15

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