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Yes, but I posted it on a thread to get a direct answer...Well?

I sat my HSC last year but looking at this question would this be valid: let angle QBA = x therfore angle BAP = x (alternate angles in parallel lines...) Similarly, let angle BQP = y therfore angle APQ = y (alternate angles in parallel lines...) Now, angle QTA = x + y ( the...

I sat my HSC last year but looking at this question would this be valid: let angle QBA = x therfore angle BAP = x (alternate angles in parallel lines...) Similarly, let angle BQP = y therfore angle APQ = y (alternate angles in parallel lines...) Now, angle QTA = x + y ( the exterior...

You guys talk about recent implementation of macroeconomic policy? Stimulus question so I thought it was necessary...

How DID operations influences affect the business from achieving its operations objectives??

Did you have to change the tension from Newtons?

with CAD/BOP what did it ask, for anyone who has it now and online

Hey guys, just wondering what is an appropriate length for the module essays for the trials?

What is the difference between civil engineering and civil engineering (construction management) at usyd. thanks

HI, has any past students been accepted by sydney or nsw uni through eas for youth allowance, thank you:awesome:

Hi, has anyone successfully been eligible for the Broadway Scheme for Youth Allowance at Sydney Uni, thank you

looking at 60-61 couldnt get the binomial or parametrics :(

B

8 (iii) use the velocity found in part (ii) i.e. v = 7/(t+4)^2 and stick that in v for dv/dt. so you get d/dt [7/(t+4)^2] and do the derivative of that equation.

6) b(iii) so it's like finding an asymptote you have to see what it approaches as e^-0.05t approaches infinity so if you start sticking big numbers, 100, 1000, 1000000 etc you'll see it = 0 therefore p = 150 + 300(0) = 150 7) a(ii) need to do sum of the roots so alpha + beta and from part (i)...

At the point of inflexion you can draw a tangent to that point which is horizontal hence horizontal point of inflexion

yes it actually is possible

You're welcome :)

all you need to do is find the gradient of the line after the point 30000. so you would take the point (30000,1000) and (40000,3000) and do rise over run to get your gradient i.e: the rise is 2000 (3000-1000) and the run 10000 (40000-30000) therefore 2000/10000 = 0.2 or 20cents so c is the answer

Nah you don't really have to and you wouldn't lose any marks