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2t + 2 + h = 12 \implies h = 10 - 2t Also V = \left(\frac{4\pi h}{3}\right)\left(t^2 + t + 1\right) Substituting h into the equation above \therefore V = \left(\frac{4\pi \left(10 - 2t\right)}{3}\right)\left(t^2 + t + 1\right) Let u = \frac{4\pi \left(10 - 2t\right)}{3} and v = t^2 + t +...

Here is a website that has video tutorials about limits and calculus in general. It will really help with your question =) Tutorials for the Calculus Phobe

Do you have the answer? Have you tried letting the roots be \alpha and \alpha + 1 for the first one and for the second letting the roots be 2\alpha and 3\alpha?

I completely agree... but the Fitzpatrick text is a good text to use while in year 11 to get a feel for 4 unit maths and get to know the basic concepts. But i would completely suggest using a much harder text when in year 12.

(note: for these problems drawing a diagram is absolutely essential!) Lets find the equation of the tangent at x = 2 y = e^x y' = e^x $When$\,\,x =2, y = e^2\,\,$and$\,\,y' = e^2 So the gradient of the tangent is m = e^2 and the tangent goes through the point \left(2, e^2\right) So by the...

New Senior Mathematics (ie Fitzpatrick) is probably the easiest of the 4 unit texts. You could just start going through that now (ie in year 11) and then when you start year 12 just move onto Cambridge or Patel for 4 unit.

That is true, but the program run by ruse is a school based thing. I'm fairly sure no other school in the state has the same kind of program run at their schools for the NQE.

Your lucky as your school actually provides proper training for the comp. Are you participating in the NQE training at your school?

Do you think that marks would be lost if you were to go straight into y' for a quotient rule in the HSC exam?

Just to clarify, i am just doing the 2 unit HSC Mathematics exam this year, and 3/4 unit HSC exam next year. But i would have been doing the 3 unit HSC maths exam this year but unfortunately the times table at my school has English on during some of the 3 unit lessions and so i couldnt do it...

That too =)

I personally think that establishing u' and v' is a more "safer" method both as it rules out fewer mistakes which is very important for example if they through in a large quotient rule or product rule its easier to have this step by step reference. Secondly, it lets the marker know that you...

I also hope you go well in your HSC exams this year =)

y = \frac{3x + 1}{\sqrt{x+1}} $Let$\,\,u = 3x + 1\,\,$and$\,\,v = \sqrt{x+1} = \left(x+1\right)^{\frac{1}{2}} u' = 3 $if$\,\,y = f\left(x\right)^{n},\,\,$then$\,\,y' = n \cdot f'\left(x\right) \cdot f\left(x\right)^{n-1} \therefore v' = \frac{1}{2} \cdot 1 \cdot \left(x+1\right)^{-...

If there are any problems from this exam that you cannot do then just post them here.

a) Pythagoras - 8 m (b)There is the way you mentioned (ie name one angle whateva), but the more mathematically appealing "proof" is as follows. \angle AFD = \angle ACB = 90^{\circ} \therefore DF//BC \,\,$(corres. angles equal)$ \therefore \angle ADF = \angle DBE \,\,$(DF//BC, corres. angles...

I am going to assume that DF and DE are perpendicular to AC and BC respectively (does the question mention this?) In which case proving the two triangles is easy. The "test" you have to use is AAA, that is, all angles are equal (angle-angle-angle) You could let \angle DAF = \theta, and then...

In part (b) we will consider the areas in order to solve the problem The general formula for the area of a trapezium is: A = \frac{h}{2}\left(a+b\right) Now, at the point P drop a line such that it is perpendicular to the interval SR and name the point where it touches the line SR "C". Let...

AB//SR \implies AB//PQ \,\,\left(PQ//SR\right) $Also$\,\, BC//PQ \therefore AC//PQ \implies ABC\,\, $is a straight line$

Regarding your general question, yes, it is a well known theorem. 1(a) FD//BC (line joining the midpoint of two sides is parallel to the third) In triangles AFG and ACE, both share angle CAE and because FD//BC both other corresponding angles are equal. \therefore \triangle AFG ||| \triangle...