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Odlwmn: why have you had this procedure at such a young age? I've already had 5 colonoscopies. Sick of it. Not doing it anymore; preparing for the procedure is quite a bother. But you are right. Have one every 5 years or so after you are 45?? unless you have related symptoms that need checking...

This is how I would do it. \text {Expression = } \left(x+\frac {1}{x} \right)^7 \times \left(x-\frac{1}{x} \right)^3 \\ \\ = \sum _{i=0} ^7 \binom 7 i x^{7-i} x^{-i} \times (x^3 - 3x +3x^{-1} - x^{-3})\\ \\ = (..x^7 + .. x^5 + .. x^3 + .. x + .. x^{-1} + .. x^{-3} \cdots) \times...

Q3(c) 1+\omega + {\omega}^2 + {\omega}^3 + \cdots {\omega}^8 = 0 \implies \omega + \omega^2 + \omega ^3 + \cdots + \omega ^8 = -1\\ \\ z = cos \theta + i sin\theta \implies \bar z = cos \theta - isin\theta \implies z + \bar z = 2cos \theta \\ \\ $Now $ \omega $ and $ \omega ^8 ,\omega ^2 $ and...

Easy. Area of smaller pentagon : area of larger pentagon = 4:25 = 2^2:5^2 .: side length of smaller pentagon : side length of larger pentagon = 2 : 5 [the regular pentagons being similar figures] .: side length of smaller pentagon = (2/5)*22.5 cm = 9 cm Note: the principle being simply: if...

y = cos x means y'= -sin x = cos(x+pi/2) For n=1 cos(x+\frac{n\pi}{2} = cos(x + \frac {\pi}{2}) .: true for n=1 Suppose formula is true for n=k>=1 $i.e. $y^k = cos(x + \frac{k\pi}{2})\\ \\ \therefore y^{k+1} = \frac{d}{dx}cos(x+\frac{k\pi}{2}) \\ \\ =-sin(x+\frac{k\pi}{2}) =...

= \left(px^3-qx^{-2} \right)^9 = \sum^9 _{i = 0} \binom 9 i (px^3)^{9-i} (qx^{-2})^i = \sum ^9 _{i=0}\binom 9 i p^{9-i} q^i x^{27-3i-2i}\\ \\ x^{17} $ term $ \implies 27 - 5i = 17 \implies i = 2 $ \\ \\ and $ x^{27} $ term $ \implies 27 - 5i = 27 \implies i = 0\\ \\ \therefore \binom 9 0...

You can get them from Muji stores in Chatswood or The Galleries(?), near Town Hall Station.

With 1 or 2 years, I think it's possible. I've been thinking of starting a class for 5 or 6 year olds, in the hunch that they need to be taught how to write while they're still young; this includes how to hold (grip) the pen. From my past experience, trying to teach some Yr 11 or 12 students how...

Maybe problem is not the pen, but the way you write. I've no problem with my Muji pens.

If you want to do 4U Maths, you cannot rely on MIF text alone. In fact, to do 4U, you need higher order capabilities. Whilst your school may adopt MIF, you must supplement yourself with textbooks with more solid treatment and more challenging exercises, like Cambridge. If you are getting...

4pi/3 ?? -pi/2 < theta < pi/2

Hi Forenza I'd be happy to tutor you in Maths & Maths Ext 1. Can you tell me: - how are you in your Maths currently? so so, good or very good? - what do you hope to achieve in yr maths? - any area, if any, you currently find difficult? Yr school? Yr suburb? Drongoski

I have in my possession a complete? set of old Prior Education HSC Physics course notes (including regular Tests)- maybe around 15 - 20 year-old. They come in 8 folders. It'd be a pity to throw them away. Whatever the age, most of the contents are still current. The physics of 20 years ago is...

(a+b+c+d)^2 = a^2 + b^2 + c^2 + d^2 + 2\times(ab +ac+ad+bc+bd+cd)\\ \\ a^2+b^2 \geq 2ab $ -------- $ c^2+d^2 \geq 2cd\\ \\ \therefore (a+b+c+d)^2 \geq 2ab + 2cd + 2(ab + bc+cd+ad) + 2ac + 2 bd\\ \\ = 4ab + 2bc + 4cd + 2ad + 2ac + 2bd\\ \\ = ? ? ? Wonder if this can be of any help .

be good in your Integration by Parts

LHS =(sin(A+B) + sin(A-B)) \times (sin(A+B) - sin(A-B))\\ \\ =[(sinAcosB + cosAsinB) + (sinAcosB - cosAsinB)] \times [(sinAcosB + cosAsinB) - (sinAcosB - cosAsinB)] \\ \\ = 2sinAcosB \times 2sinBcosA \\ \\ = 2sinAcosA \times 2sinBcosB \\ \\ sin2A \times sin2B

Earlier I did this in a similar way, using my crazy method: \int sin\sqrt{2x}dx = \frac {1}{2}\int sin\sqrt{2x}\cdot d(\sqrt{2x})^2 \\ \\ = \frac {1}{2}\int sin \sqrt{2x} \cdot 2\sqrt {2x} \cdot d\sqrt{2x}= -\int \sqrt{2x} \cdot dcos\sqrt{2x} \\ \\ =-\sqrt{2x}cos\sqrt{2x} + \int cos\sqrt{2x}...

Someone asked me how to do: \int sin\sqrt{2x}dx

possibly.

a) \therefore (a^2+b^2) + (b^2+c^2) + (a^2+c^2) \geq 2ab + 2bc + 2ac \\ \\ \therefore 2(a^2 + b^2 + c^2) \geq 2(ab + bc + ac ) b) a^4 + b^4 = (a^2)^2 + (b^2)^2 \geq 2a^2b^2\\ \\ \therefore c^4 + d^4 \geq 2c^2d^2 \\ \\ \therefore a^4 + b^4 + c^4 + d^4 \geq 2((ab)^2 + (cd)^2) \geq 2 \times...