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that's basically right but it could be improved a little bit: in a normal exam, you probably don't need to write down the first few lines of explanation, but it's not wrong either for the line before LHS=dP/dt, "Now dP/dt=kp(1-rp)" , some ambiguity could occur as this is not shown yet...

http://seoulbeats.com/2011/07/k-pop-extravaganza-in-australia-this-november/

b anything/b education is like all below 80

wat mark?

probably i didn't clarify it enough, but that's meant to help you to find dP/dt

you don't need to solve the differential equation yourself , in exams you're given the solution and you're expected to verify it(which is much easier than solving it) i.e. find its derivative , and substitute back

alright step 1: find dP/dt for LHS, i think that's what u meant by "substitute the expression for P into dP/dt" step 2: sub P into kP(1-RP) for RHS step 3: tidy up your answers for LHS and RHS to make the equality apparent

if u have difficulty in differentiating try this easier example first d/dx (5+3x)^3 for which you'll get 3*(5+3x)^2*3, where the 3 is the derivative of whatever is in side the brackets, note the derivative involves an exponential term in ur question ...

LHS=dP/dt , so differentiate the expression for P with respect to t, which means you'll need the chain rule for the RHS , just sub in the expression for P u may need to change ur expression a bit so that both sides look exactly the same

b science = b science(adv science) =b science(adv maths), in terms of the stuff you can potentially choose to learn the only difference is what gets printed on your transcript

e.g after 9 seconds, the sides are 1 , and the diagonal is 10sqrt(2)- 9*sqrt(2)=sqrt(2) which means it's still a square

the diagonal and the side are of the same dimension, it makes sense that they're both decreasing at constant rates to keep the shape as a square

-sqrt(2)= ??

they publish their errata like every 1000 years?

so to use ur method correctly, we need the product rule for V=A*r/3 dV/dA= r/3 + A/3*dr/dA, where A=4pi*r^2 so dA/dr=8pi*r, dr/dA= 1/8pi*r, sub this back into dV/dA equation dV/dA= r/3 + A/3*dr/dA = r/3 + (4pi*r^2 /3) * (1/8pi*r) = r/3 + r/6 =r/2

V=A*r/3 here both A and r are variables

something doesn't seem right here, hm...

+1 also don't forget stuff like (a-b)^2=(a+b)^2-4ab

let's call the displacment x, velocity v, acceleration a displacement is like distance , except for it's a vector. i.e. it's got direction for example, if u travel due east for 20m , then west 20m your displacement would be 0 cos you're back to where you've started but the actual distance...

^also when they ask you to "explain" something , you actually have something to talk about whereas in english, i doubt that even the authors themselves could answer some of the hsc questions about their own texts well enough