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but what does it actually "mean" to differentiate a complex number? how do we justify this? (just raising some q's for discussion)

in my previous answer i made the mistake of not multiplying i(5pi/6) by 9. this amended answer gives the correct solution: \\f(t)=\frac{\sin t}{e^{\sqrt{3}{t}}}=\Im \left(e^{(i-\sqrt3)t}\right)\\f^{(9)}(t)\\=\Im \left((i-\sqrt{3})^9e^{(i-\sqrt3)t}\right)\\=\Im...

\\8abc<(a+b)(b+c)(c+a)\\$Since, $x+y-z>0, y+z-x>0 $ and $ z+x-y>0\\$We can let $a=x+y-z, b=y+z-x $ and $c=z+x-y\\8(x+y-z)(y+z-x)(z+x-y)\\<((x+y-z)+(y+z-x))((y+z-x)+(z+x-y))((z+x-y)+(x+y-z))\\=2y\times 2x \times 2z\\=8xyz\\$Hence, $(x+y-z)(y+z-x)(z+x-y)< xyz

\\$What you have done:$\\\int \cos^3x dx\\=\int (1-\sin^2x)\cos xdx\\=\int (1-\sin^2x)d(\sin x)\\=\sin x -\frac{1}{3}\sin^3 x+C_1\\\\$What they have done:$\\\cos3x = 4\cos^3 x - 3\cos x\Rightarrow \cos^3 x=\frac{1}{4}(\cos 3x +3\cos x)\\\therefore \int \cos^3 x dx\\=\frac{1}{4}\int (\cos 3x...

\\f(t)=\frac{\sin t}{e^{\sqrt{3}{t}}}=\Im \left(e^{(i-\sqrt3)t}\right)\\f^{(9)}(t)\\=\Im \left((i-\sqrt{3})^9e^{(i-\sqrt3)t}\right)\\=\Im \left(2^9e^{i\frac{5\pi}{6}}e^{(i-\sqrt3)t}\right)\\=512\times \Im \left(e^{i\frac{5\pi}{6}}e^{(i-\sqrt3)t}\right)\\=\frac{512}{e^{\sqrt3t}}\times \Im...

hahaha happy gilmore?

pretty good, become much bulkier. though, i am drinking >2L of milk a day (well, on my good days)

3x5 back squats @ 105, 115, 127.5 3x5 flat bench @ 55, 60, 62.5 1x5 deadlift @ 135 cbf for ancillary exercises today

shame on you trebla always has the best mathematical posts

they are defined to be perpendicular. the x axis and the y axis arent magical lines that are by nature perpendicular

back squat 3x5 @ 105, 115, 125 overhead press 3x5 @ 37.5, 40, 42.5 bent over rows 3x5 @ 55, 65, 75 3x12 weighted situps 3 sets to failure dips

but how can physical testing generalise that F_f = kN in all instances, on all surfaces

why is friction directly proportional to the normal force? whos to say there isn't an exponential, sinusoidal etc relationship?

\\OB=CO=1\\$arc$(AB)=\theta \times OB=\theta\\\sin\theta=\frac{CA}{CO}\Rightarrow CA=\sin\theta\\\tan\theta=\frac{DB}{OB}\Rightarrow DB=tan\theta\\$Area$(OCA)\leq $ Area$(OCB)\leq $ Area$(ODB)\\\frac{1}{2}\times OA\times CA\leq \pi(1)^2\times\frac{\theta}{2\pi}\leq \frac{1}{2}\times OB\times...

meh, they're everywhere. my gym is full of uneducated curlers & 300kg partial squatters

absolutely love the gym progression makes me so happy :o

i actually like it hated first week with a passion, but it's not that bad :D

green agree with meilz, green's the only normal capsicum :confused:

\\\sin\left(2\sin^{-1}\frac{3}{5} \right )\\=2\sin \left(\sin^{-1}\frac{3}{5} \right )\cos\left(\sin^{-1}\frac{3}{5} \right )\\=2\sin \left(\sin^{-1}\frac{3}{5} \right )\sqrt{1-\left(\sin\left(\sin^{-1}\frac{3}{5} \right...

i came into uni dreading programming. its pretty fun! : D