Harmonize wrote:Anoohilator wrote:Why not just use normal right-angled triangle trigonometry? SOH CAH TOA and all that jazz..
...Cause its half 11 and I'm too tired to use my brain lol
Haha that's fair enough
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guineadan wrote:Whilst flicking through my homework planner, I see that I have maths due in on tuesday
I'm stuck on the last question (I thought it was going a bit too well xD). So, here it is ^^
Find the maximum and minimum points (if any) of the following graphs:
e) y= 3x^4 - 8x^3 + 6x^2 + 1
Therefore:
dy/dx = 12x^3 - 24x^2 + 12x , which simplifies to:
dy/dx = x^3 - 2x^2 + x
Is it even possible to factorise this to get values of x?
I've tried making the -2x^2 + x part into brackets:
(-2x-1)(x-1), then adding the x to make it cubic:
x(-2x-1)(x-1) - but that doesn't work because there's a silly little 2 in front of that x^2 :p To factorise the quadratic bit, you need a 2 there though =S I know how to use the (d^2y)/(dx^2) thing - I just need help on this first bit
Suggestions, anyone?





Anoohilator wrote:What is the overall capacitance of two capacitors in parallel, C1 and C2?

Skynetmain wrote:Anoohilator wrote:What is the overall capacitance of two capacitors in parallel, C1 and C2?
1/Ctot = 1/C1 + 1/C2
Ctot = 1/(1/C1 + 1/C2) = C1*C2/(C1 + C2)
Or something like that. Electronics wasn't my strong suit, but I do know the E-field of a capacitor is q/(epsilon naught)