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Deaths: 136 Mom kills: 0 Gonna quit today haha (my best was basement five)

U know that this is a parallelogram because opposite sides are parallel and opposite sides are equal in length Area of parallelogram = base x perpendicular height...

Binding of Issac is too addictive

Re: MX2 Integration Marathon nahh its just the length, good effort man! in the end its all about answering the question

Re: MX2 Integration Marathon $let u^2 = 1-x^2, du = -xdx \\ \\ \therefore I = $\int_0^1 (1-u^2)u^2du = \left ( \frac{1}{3}- \frac{1}{5} \right ) = \frac{2}{15}

lol!

LOL! To show u x\rightarrow b

Mg(NO_{3})_{2(aq)} \rightarrow Mg^{2+}_{(aq)}+ 2NO3^-_{(aq)}

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Therefore lactus rectum = 4a

no thats 3U

Avogadro's constant

beaten by pappa

yeh v(max) is which acceleeration is zero or u can find the vertex

'pie'?

There is a reason why they wrote the equation of PQ to be like that instead of y = \frac{1}{2}(p+q)x - apq

i dont believe u, last time i saw u, u didnt look like a 192

there was some confusion with the gradient thing lol m(norm. P) = -1/t .: m(norm. Q) = t .: m(tang. Q) = -1/t dy/dx = x/2a For the point of contact of tangents with gradient -1/t -1/t = x/2a x = -2a/t y = 2a/t^2 Q[-2a/t, 2a/t^2] (in my previous try, i subbed in dy/dx as gradient of...

yeh got that too