Search results

Just ask the head teacher or whoever if you can atleast try it until the first test, usually they will let you.

Lol, Don't worry it's not as hard as it looks :)

1. y=x^2/8\\\therefore dy/dx=x/4\\\text{ when: }x=4,dy/dx=-1\\\text{ When: } x=-4,dy/dx=-1\\\text{ The general equation of a line is : } y-y_{1}=m[x-x_{1}]\\ \therefore \text{tangent at : } (4,2)\text{is } y-2=1(x-4) =>y=x-2\\\text{similarly,second tangent is : } y=-2-x\\\text{ Now, when }x=o ...

x^2+y^2=8x\\x^2-8x+16+y^2=16\\(x-4)^2+y^2=16\\ \text{ draw the thing if you like}\\ \text{a Typical slice: } \Delta v= 2xy/3\\\therefore v=\lim_{\Delta x->0}\sum_{x=0}^{8}2xy/3.\Delta x\\=\lim_{\Delta x->0}\sum_{x=0}^{8}=2x/3\sqrt{16-(x-4)^2}.\Delta x \text{,By rearranging the equation of the...

1)conjugate of z= conjugate of {10i-10/-8-4i]=conjugate of (10i-10)/conjugate of [-8-4i]- this is a rule therefore, conjugate of z=-10-10i/-8+4i =(-10-10i)(-8-4i)/(-8+4i)(-8-4i) =40+120i/80 =1/2+3/2i 2) a. (x+iy)^2=-3-4i \text{ ,where x and y are real}\\x^2+2xyi-y^2=-3-4i\\x^2-y^2=-3...

a) arg(z)=0 b) arg(z)=-pi c) arg(z)=pi/2 d) 0 < arg(z) < pi/2 e) -pi < arg(z) < -pi/2

d=sqrt[(x2-x1)2+(y2-y1)2)]-distance formula. 4=sqrt{(x+3)2+(y-2)2} 16=(x+3)2+(y-2)2

Hey guys, Currently I'm enrolled in : MATH1141 PHYS1131 COMP1911 ENGG1000 Does anyone know where I can buy the text books for these courses? Do past first years sell them for cheaper prices? As I heard they can be quite expensive. Thank you,

Hey, For some reason when I attempt to enrol into MATHS1141, there's only 2 tutorials I can choose from and since 2 tuts are required , I have to choose both. However, one of them is a talented students tutorial and it says you need combined marks greater then 189 for mx1 and mx2, however my...

Ψ=sin(nx/L) dΨ/dx=(n/L)cos(nx/L), assuming n and L are constants- if not then this is prob wrong lol d2Ψ/dx2=-(n2x)/L2)sin(nx/L) therefore, E.sin(nx/L)=(- ħ²/2m) .-(n2x)/L2)sin(nx/L) E=(- ħ²/2m) .-(n2x)/L2) =ħ²n2x/2mL2

P(-4,4), Q(1,1/4), R(-3/2,-1) RTP:MPRxMRQ=-1 MPR=[4+1]/[-4+(3/2)]=5/[-5/2]=-2 MQR=[(1/4)+1]/[1+3/2]=1/2 LHS=MPRxMRQ=2x(-1/2)=-1= RHS The other one is done similarly.

You probably forgot to change the speed into ms-1 y''(let the' indicate the . on top)=-g y'=-gt+40sin7 y=-5gt+40tsin7+2.4 x''=0 x'=40cos7 x'=40tcos7

For n=3, LHS= (1-2/3)=1/3 RHS=2/(3(2))=1/3=LHS Assume true for n=k i.e. assume: (1-2/3)(1-2/4)(1-2/5)...(1-2/k)=2/(k(k-1)) Prove True for n=k+1, i.e. prove (1-2/3)(1-2/4)(1-2/5)...(1-2/k)(1-2/(k+1))=2/((k+1)k) LHS=(1-2/3)(1-2/4)(1-2/5)...(1-2/k)(1-2/(K+1)) =2/(k(k-1))x(1-2/(k+1)), by...

=cosecx, as they cancel out.

Most probably the integral has a double angle in it. E.g. Cos2x- so just pretend the 2x = a, and then you can let t=tan(a/2) ,if you like.

I don't think you can buy them in stores, rather you have to buy them off his website. HSC Coaching – Shop

Re: If a person you know quite well still hasn't told you or other friends his/her UA About 91.

Most probably, well bad for his standards- could be 99.95 for all you know :P

You're meant to know both methods, because in the HSC they generally have 2 questions each asking for a different method anyway...

Maths Ext 2 people.