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imqwerty
 one year ago
math for KIDS :P
imqwerty
 one year ago
math for KIDS :P

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pooja195
 one year ago
Best ResponseYou've already chosen the best response.0Much more like math for adults.

Jadeishere
 one year ago
Best ResponseYou've already chosen the best response.0I don't know... gah

Afrodiddle
 one year ago
Best ResponseYou've already chosen the best response.0Looks like I'm still a toddler.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why many of us still feel like a kid?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0whoa thats a long problem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2my first guess is to play with \[17\mid a ~~\text{and}~~17\mid b \implies 17\mid(ma\pm nb)\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.1\[\left[ \begin{array}c 1 & 3/2\\3/2 & 2\\\end{array} \right]\] and \[\left[ \begin{array}c 0 & 1/2\\1/2 & 0\\\end{array} \right]\] have the same determinant

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2\(\large \color{black}{\begin{align} x^23xy+2y^2+xy=17m\hspace{.33em}\\~\\ (x2y+1)(xy)=17m\hspace{.33em}\\~\\ \end{align}}\) we can consider here 4 cases \(\large \color{black}{\begin{align} x2y+1=17,\ xy=m \hspace{.33em}\\~\\ x2y+1=m,\ xy=17 \hspace{.33em}\\~\\ x2y+1=17m,\ xy=1 \hspace{.33em}\\~\\ x2y+1=1,\ xy=17m \hspace{.33em}\\~\\ \end{align}}\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.1@mathmath333 How did you factor that, I wanna learn how you figured that out.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2do u agree m should be an integer

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Ahh nice, factoring the quadratic \(x^23xy+2y^2\) is easy using factor by grouping : \[x^23xy+2y^2 = x^2xy2xy+2y^2 = x(xy)2y(xy)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2a 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 \((y+m_1x)(y+m_2x)=0\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Ahhh 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.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.3well done guys. heres the solution 

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1i wonder if i played with the linear combination, how it would end nicely

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I 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

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1well yeah (there is no linear combination ) see why : m x^22mxy+my^25mx+7my nx^23nxy+2ny^2+nxny  xy 12x+15y set of equations: m=n 2n=m 2m3n=1 5m+n=12 7mn=15
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