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Re: New MX2 Game woops, thanks for the correction. 2nd attempt: (x_1+x_2+...+x_k)+x_{k+1}>\sqrt{(x_1+x_2+...+x_k)x_{k+1}}\\\ln(x_1+x_2+...+x_k+x_{k+1})\\>\frac{1}{2}(\ln(x_1+x_2+...+x_k)+\ln(x_{k+1}))\\>\frac{1}{2}(\frac{1}{2k-1}(\ln x_1+\ln x_2+...+\ln x_k+(2k-1)\ln...

Re: New MX2 Game New Q: Show that \sin nt+\sin(nt+\alpha)+\sin(nt+2\alpha)+...+\sin(nt+(k-1)\alpha)=0\,\,where\,\,\alpha =\frac{2\pi}{k} Sums to products might be useful here.

Re: New MX2 Game (i) (\sqrt{x_1}-\sqrt{x_2})^2\geq 0\\x_1+x_2\geq 2\sqrt{x_1x_2}\geq \sqrt{x_1x_2} (ii) $for n=2,$\\\ln(x_1+x_2)>\ln(\sqrt{x_1x_2})\\=\frac{1}{2}\ln(x_1x_2)\\>\frac{1}{2(2)-1}\ln(x_1x_2)\\\\$assume n=k is true,\\for...

is it? why

Let major and minor axes of elliptical cross sections be a and b respectively. $Since axes are proportional to given ellipse$,b=\frac{3}{5}a\\9=25(1-e^2)\\e=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\\ae=y\\a(\frac{4}{5})=y\\a=\frac{5}{4}y\\\delta V=\pi ab\delta...

x_A=Ut\\y_A=-\frac{g}{2}t^2+4h\\x_B=V(t-10)\\y_B=-\frac{g}{2}(t-10)^2+h\\when\,\,y_A=0,T_A=\sqrt{\frac{8h}{g}}\\when\,\,y_B=0,T_B=10+\sqrt{\frac{2h}{g}}\\T_A=T_B\\\sqrt{\frac{8h}{g}}=10+\sqrt{\frac{2h}{g}}\\\sqrt{\frac{2h}{g}}\left ( 2-1 \right...

oh ok then, thanks

yeh i was thinking that too, but if you consider it using the locus definition of hyperbola, you get a different answer. SR = ST = b (equal radii) b=e(x- (a/e)) (SR=ST=ePM) x=(a+b)/e = 2a/e = asqrt(2) S(ae,0)=>(asqrt(2),0) How come the original method didnt work out the same way?

2x^2 - 2aex + (a^2e^2 - 2a^2) = 0 subbing in e=√2, i get 2x^2 - 2a√2x + (2a^2 - 2a^2) = 0 x^2 - a√2x = 0 x(x-a√2) = 0 x=0, a√2 so midpoint would be a√2/2 :S

The hyperbola x^2/a^2 - y^2/b^2 =1 has one focus S on the positive x axis. The circle with centre S and radius b cuts the hyperbola at points R and T. If b=a, show that RT is the diameter of the circle. when i attempted this question i got: (x-ae)^2 + y^2 = a^2 x^2 - y^2 = a^2 Solving...

upload them onto some file hosting site like MEGAUPLOAD - The leading online storage and file delivery service or RapidShare: Easy Filehosting it would be greatly appreciated

oh right, i didnt read the part where it says the foci lie on the given ellipse. no wonder...

A solid is formed about the ellipse x^2/25 + y^2/9 =1 in such a way that each plane section of the solid perpendicular to the x-axis is an ellipse whose foci are on the given ellipse. The major and minor axes of each section are proportional to those of the given ellipse. Determine the volume of...

You should have used absolute values for ST and ST' as they are both distances (ie. >0).

you probably missed the d(e^x) part after it.

For the first one, i guess you could use t-substitution. cant think of a better method at the moment. let\,\,t=\tan x\\dt=\sec^2xdx=(1+t^2)dx\\dx=\frac{dt}{1+t^2}\\\\\int \frac{1+\sin^2x}{1+\cos^2x}dx\\=\int \left ( \frac{1+\frac{t^2}{1+t^2}}{1+\frac{1}{1+t^2}} \right ).\frac{dt}{1+t^2}\\=\int...

let\,\,u^2=e^x\\2udu=e^xdx\\dx=\frac{2udu}{e^x}=\frac{2udu}{u^2}=\frac{2du}{u}\\\\I=\int \frac{1}{\sqrt{1-e^x}}dx\\=2\int \frac{du}{u\sqrt{1-u^2}}\\\\let u=\sin\theta\\du=\cos\theta d\theta\\\\I=2\int \frac{\cos\theta d\theta}{\sin\theta \cos\theta}\\=2\int \csc\theta d\theta\\=2\int...

y_1^2=4ax\,\,and\,\,y_2=\frac{2}{3}x\\\delta V=\frac{\pi}{2}(y_1^2-y_2^2)\delta x\\=\frac{\pi}{2}(4ax-\frac{4}{9}x^2)\delta x\\V=\frac{\pi}{2}\int_{0}^{9a}(4ax-\frac{4}{9}x^2)dx\\=2\pi\int_{0}^{9a}(ax-\frac{x^2}{9})dx\\=27\pi a^3 i might have misunderstood the question.

Hey, are the questions from the Coroneos supplement book any good for topics like conics and volumes?

New Question: $Show that$\,\,1+(1+x)+(1+x)^2+...+(1+x)^n=\frac{(1+x)^{n+1}-1}{x}\\$Hence show that$\,\,\binom{n}{r}+\binom{n-1}{r}+\binom{n-2}{r}+...+\binom{r}{r}=\binom{n+1}{r+1}