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so the particle doesnt slide down when: F < μN F/N < μ tan θ < μ rather then going through everything else

iii) to stop slipping down plane the centripetal force and friction has to balance or be greater then the weight component acting down the plane mg sin@ < (mv^2/r)cos@ + uN mg sin@ - (mv^2/r)cos@ < uN [mg sin@ - (mv^2/r)cos@] / N < u since (mv^2/r)cos@ cant be negative, the LHS...

its right, try to simplify cos(tan^-1x) to get the answer in the book.

by parts? I (sinx.cosx) dx = sinx.sinx - I (sinx.cosx) dx 2 I (sinx.cosx) dx = (sinx)^2 I (sinx.cosx) dx = (sinx)^2 / 2 + C

f(x) = x^3 + 3x + 2i f(-i) = 0 & f'(-i) = 0 .'. A = -i x^3 + 3x + 2i = (x + i)^2(x - B) sub x = 0, B = 2i

they are the same

its spastic

thank god for maths. change your sig.

inverse functions 1 to 1

its not really worth studying that much anyway, you'll get a high enough uai without having to get 100% in every school assessment. you'd also get a higher uai if you didnt go to class for any of your prelim subjects, and read hsc excel summaries at home. in the end youre just gonna study your...

if you dont need a high uai to do what you want at uni its hard to study for hsc.

cos (@/2) = sqrt[0.5(cos@ + 1)] -----(1) sin (@/2) = sin@ / [2 sqrt(0.5cos@ + 0.5] -----(2) (1) / (2): tan (@/2) = sin@ / (cos@ + 1) sub, @ = cos^-1(x) tan[0.5 cos^-1(x)] = sin[cos^-1(x)] / (x + 1) = sqrt(1 - x^2) / (x + 1) = sqrt[(1 - x) / (1 + x)] tan^-1 [sqrt[(1 - x) / (1 + x)]] = 0.5...

you need to take into consideration the payment each month along with the interest. let, A = amount owing C = original cost of car on the day: A = C - 850 in 1 month: A = (C - 850)1.0085 - 850 in 2 months: A = [(C - 850)1.0085 - 850]1.0085 - 850 ... in 36 months: A = C(1.0085)^36 -...

cool

what are the best math topics you learn at an undergraduate level?

wow, what a terrible survey

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