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The curves are arcs of circles whose centers are vertices of the square, side 6cm. Find the area of the shaded region.

No worries :D It's good that your getting help, but try as hard as you can to find the problem before asking. Look through notes and worked example to see how it's done and if after that you can't see your mistake, then post.

You just did the reciprocal, not the negative reciprocal. m should = -1/(-3/2)=2/3 not -2/3. You can't just do 1/gradient, must be -1/gradient.

http://www.mediafire.com/?ssdiret1h9sdl7g ^2U papers, not all trial though. about 70 there.

The perpendicular to any gradient is the negative reciprocal, there is no formula, its just that. You can test by observation from a graphing site ( www.fooplot.com ) so you understand it more. e.g if a gradient of a line is -3, then the gradient of the line perpendicular to it is 1/3. See...

Find the gradient of the line x-2y=9. Now because it has to be perpendicular, the negative reciprocal of that will be the gradient you use for the straight line. i.e You found the gradient to be 1/2, but to be perpendicular, you take the negative reciprocal which is -2. That's the gradient for...

Wow, one is in year 9...

It depends what you want to do. Obviously if you want to be a brick layer, university isn't the best option. And if you want to be a doctor, TAFE might not be the best option.

3(d)\\\\&LHS=\frac{1}{cos^2{x}}+\frac{sin^{2}x}{cos^{2}x}\\\\&=\frac{1-sin^{2}x}{cos^{2}x}\\\\&=\frac{cos^2{x}}{cos^2{x}}\\\\&=1\\\\&RHS=\frac{1}{sin^2{x}}+\frac{cos^{2}x}{sin^{2}x}\\\\&=\frac{1-cos^{2}x}{sin^{2}x}\\\\&=\frac{sin^2{x}}{sin^2{x}}\\\\&=1\\\\&=LHS EDIT: Beaten again D:

\begin{align*}&3(c)\\&LHS=3+\frac{3sin^{2}x}{cos^{2}x}\\\\&=\frac{3cos^{2}x+3sin^{2}x}{cos^{2}x}\\\\&=\frac{3(sin^2{x}+cos^2{x})}{1-sin^2{x}}\\\\&=\frac{3}{1-sin^2{x}}\\\\&=RHS\end{align*}

By the way buddy, Dux College have one that uses 2011 results, but thanks for the info. And the thing you said about school principals is BULLSHIT for most schools.

What is it your asking?

\begin{align*}&Let~\log_{A}B=\log_{B}A=x\\&A^{x}=B~~and~~B^{x}=A\\&\therefore AB= A^{x} \cdot B^{x}=(AB)^{x}\\&(AB)^{x-1}=1\\&AB=1~~or~~x=1\\&but~it~is~given~A\neq B ,~so~x\neq 1\\\therefore &AB=1\end{align*} ^I tried to do it without log laws, so that's how I did it.

haha yeh, but you have to show how to get the answer. :D I guessed that the first time aswell :P

This thread is practically dead, so thought I might post a question. But unlike many of the others posted so far, a 2U student can do it. \begin{align*}&A~and~B~are~positive~real~numbers~for~which~\log_{A}B=\log_{B}A.\\&If~neither~A~or~B~is~1~and~A\neq{B},~find~the~value~of~AB.\end{align*}

^^ 12 months from when you get you L's.

http://exampapers.web.fc2.com/ep/yr11.html#Maths - 2U ^Heaps here. Plus check nerdasdasd thread's where he has uploaded a ton of stuff

+1 He has them completely mixed up. I don't know about other humanities but business studies is all rote learned. (except for structure of responses etc.)

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Re: 98.5+ plan I saw you said somewhere else that people in your cohort hated 4U Maths yet still did well. And now you are saying that this "work hard and get whatever ATAR you want" strategy has worked for everyone in your cohort and students. All these people it has worked for have been...