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again, just do it by parts d/dx ln(x+sqrt[x^2-1]) = 1/sqrt[x^2-1] then just a basic application of ibp

do it by parts \int \sqrt{9+x^2}dx \\ =x\sqrt{9+x^2} - \int \frac{x^2}{\sqrt{9+x^2}}dx\\ =x\sqrt{9+x^2} - \int \frac{x^2+9}{\sqrt{9+x^2}}dx + 9\int\frac{dx}{\sqrt{9+x^2}}dx\\ = x\sqrt{9+x^2} - \int \sqrt{9+x^2}dx + 9ln(x+\sqrt{9+x^2})\\ 2\int \sqrt{9+x^2}dx = x\sqrt{9+x^2} +...

needs some conics knowledge an ellipse w/ foci (-3,0), (3,0) since for an ellipse, the sum of the distances from a point on the ellipse to the two foci is a constant and 2a = 8, a=4 so b = rt(7) x2/16 + y2/7 = 1

Re: 2009 3U Maths Marathon well i see no error in gurmies' working, and time =/= displacement

new line is \\

well we have our equations, velocity v at 7 seconds: vcos@ = Vcos@ = V/2 vsin@ = Vsin@ + gt = rt3 V/2 - 70 v2 = vcos2@ + vsin2@ = V2 + 702 -70 rt3 V = (3/4 V)2 = 9/16 V2 so 7/16 V2 - 70 rt3 V + 702 = 0 V2 - 160rt3 V + 70*160 = 0 solve as a quadratic V = 160rt3 /2 +- rt(1602*3 - 4*70*160)/2 =...

write down/derive the equations, do some algebra,

Re: 2009 3U Maths Marathon ummmm..

work = change in energy = force*distance energy gain of watermelon is 20*10J = 200J F= ma = mv/t v = Ft/m = 20*0.025/0.025 m/s = 20m/s

you can put \ then a space, at the end of the word

do you just solve the equations simultaneously? we don't do linear algebra in the hsc i.e. 1) x - 7y + z = 0 2) -w + 6x + 9y = 0 3) x + 6y + z = 0 3) - 1): 13y = 0 y = 0 so x = -z = -t -w -6z + 0 = 0 w = -6z = -6t

if the velocity is too small, then it would start to move down the slope, so it needs to be kept up by the inner rail (which is pretty much what i said before so i don't know if it answers your question). the fact that it needs to have the rail apply an upwards force implies that it is moving...

when the train travels at 10m/s, the velocity is too small to keep it moving in a circle, so it has to be kept up by the inner rail (ie. the inner rail exerts a force up the slope) when the train travels at 20m/s, the velocity is too large to keep it moving in a circle, so it has to be kept...

arcsinh is sinh-1 which is inverse hyperbolic sine. you put 9+x2 instead of minus are you sure the power is written as -1/2 rather than just 1/2? http://integrals.wolfram.com/index.jsp?expr=(9-x^2)^(1%2F2)&random=false gives the books answer

i did a google search and found these: 1+1 - Google Search 1+0 - Google Search 1+2 - Google Search not sure if its right tho??? but it agrees w/ rocktheyard and i tried the finger method and got the same answers where did these q's come from??

\int \frac{sin^2x + cos^2x}{sinxcosx}dx\\ \int \frac{sinx}{cosx} + \frac{cosx}{sinx}dx\\ -ln|cosx| + ln|sinx|\\ = ln|tanx| + C or \frac{1}{sinxcosx} = \frac{sec^2x}{tanx} which gets you it in one step, but hard to spot

..huh?

^ ah yes, i missed the 1+xyz pair

expand, you get 1 + x + y + z + xy + yz + zx + xyz = 8 using the fact (for numbers >= 0) a + b >= 2sqrt(ab) for each pair of eg. x and yz we get: 6sqrt(xyz) + xyz =< 7 let sqrt(xyz) = a, then we have a2 + 6a - 7 <= 0 (a+7)(a-1) <= 0 as a is positive, a =< 1 for this to be true so sqrt(xyz) =< 1...

\int \frac{1}{sin^2x + 2cos^2x}dx\\ \int \frac{sec^2x}{tan^2x + 2}dx which is then a standard integral