LJP07 Posted November 20, 2006 Report Share Posted November 20, 2006 I can't figure out these questions, so I need help as to how to figure them out:1. SQUARE ROOT of X + Y = 1/2 X - Y The square root is over both the x + y and x - y, and the question is to express y as the subject of the formula, by the way the X-Y is a fraction under X+Y, it won't submit that in the thread though , the answer at the back of the book is -3x 5 2.Another one I'm stuck on is : T = 2Pi Square Root of L gTo express g in terms of the other variables, the answer at the back of the book is g = 4 pi squared l T squared. 3.The surface area S, of a cylinder is given by S = 2 PI r squared + 2 Pi r hIt's volume V is given by V = Pi R squared hExpress V in terms of S and r only? Quote Link to comment Share on other sites More sharing options...

moo Posted November 20, 2006 Report Share Posted November 20, 2006 Just a posting tip: You can use the "code" tags to line things up visually. Sometimes it helps to use an editor with a fixed width font to view it first though, or you can just keep previewing the post until it's correct.SQUARE ROOT of [u]X + Y[/u] = [u]1/2[/u] X - Y The tags I used are visible by hitting the "quote" button. :warped: moo Quote Link to comment Share on other sites More sharing options...

LJP07 Posted November 20, 2006 Author Report Share Posted November 20, 2006 Ok, thanks, pity you couldn't be as helpful for solving the problems lol. :warped: Quote Link to comment Share on other sites More sharing options...

moo Posted November 20, 2006 Report Share Posted November 20, 2006 Ok, thanks, pity you couldn't be as helpful for solving the problems lol.You're welcome. Pity you didn't use the info I offered to clarify the problems lol. :warped: moo Quote Link to comment Share on other sites More sharing options...

CraigD Posted November 20, 2006 Report Share Posted November 20, 2006 Taking your first question first,1.SQUARE ROOT of X + Y = 1/2 X - YFirst, a note on writing math:“Function of …” isn’t a very good way to do it. You could better write the problemSqr((X+Y)/(X-Y)) = 1/2,or((X+Y)/(X-Y))^.5 = 1/2,Or, if you feel like learning a little LaTeX (click the “quote” button to see how this simple one is done, or see the 6576 for an introduction),[math]\sqrt{\frac{X+Y}{X-Y}}=\frac12[/math] Now, a hint on how to solve it((X+Y)/(X-Y))^.5 = 1/2 is hard to manipulate, because of the ^.5 (square root) exponent, so begin by raising each side to the 2nd power (squaring) it(((X+Y)/(X-Y))^.5)^2 = (1/2 )^2(X+Y)/(X-Y) = 1/4 You should be able to take it from there, and solve for X in terms of Y, or Y in terms of X.A hint: the solution will look like Y = (?/?)X - no exponents, just a simple equation of a line. Post your progress, or if you get stuck again. Quote Link to comment Share on other sites More sharing options...

cwes99_03 Posted November 20, 2006 Report Share Posted November 20, 2006 Well done, Craig. Hope we see an answer soon from Prolu Quote Link to comment Share on other sites More sharing options...

LJP07 Posted November 22, 2006 Author Report Share Posted November 22, 2006 That's as far as I got CraigD, and got: x + y = 1/4(x-y)x + y = 1/4x - 1/4yy + 1/4y = -3/4xy( 1 + 1/4 ) = -3/4xy = -3/4x 1 + 1/4 y = -3/4x 5/4 y = -3/5x wrong as the answer is -3x Quote Link to comment Share on other sites More sharing options...

CraigD Posted November 22, 2006 Report Share Posted November 22, 2006 y = -3/5x wrong as the answer is -3x[math]y = -\left(\frac{3}5\right) x[/math] is correct. :thumbs_up Reread your first post. You appear to have missed the underlining and the 5 due to it not aligning properly on the page. Quote Link to comment Share on other sites More sharing options...

sebbysteiny Posted November 23, 2006 Report Share Posted November 23, 2006 Hi prolu, Craig D removed my reply from this post. I think it was quite reasonable to do so because my post went into it at a higher level than what you need to get full marks. But I would like to say, even if you don't need it for that question, you should always be very careful when you squre or take the square root of both sides of an equation. This is because sqr (x^2) = + or - x. When you square both sides you should gain a solution. eg, y = x. Square both sides. Y^2 = x^2. But one solution here is y=1, x = -1. This is not a solution of the original equation. It is a solution of the corresponding equation y = -x. Similarly, solutions are lost when you take the (positive) square root of both sides. I wouldn't worry about this in too much detail. Just be aware that taking the squares and square roots of both sides is a bit dodgy and await with glea the time when you will be shown how to do such opperations correctly. If you want to get a taster into how this works, we are discussing exactly this in the physics and Mathematics forum. Chacmool 1 Quote Link to comment Share on other sites More sharing options...

sebbysteiny Posted November 24, 2006 Report Share Posted November 24, 2006 Glad my question posed some debate, but as regards my other thread, can ye solve problem two? 2.Another one I'm stuck on is : T = 2Pi Square Root of LgTo express g in terms of the other variables, the answer at the back of the book is g = 4 pi squared lT squared. 3.The surface area S, of a cylinder is given by S = 2 PI r squared + 2 Pi r hIt's volume V is given by V = Pi R squared hExpress V in terms of S and r only? Yes I can solve problem 2. The question is, can you? It's not difficult. It's simply a matter of rearranging the equation to make g the subject of the equation. Short of actually doing the question for you, there is not much more I can say. If you need any futher help, ask, but I'm sure you can do it on your own. There are not even any complexities about -ve roots here. 3. This is simply substitution. Make R the subject of the Surface Area equation; sub R for S into the volume equation. You would be adviced to know 3 backwards and in your sleep. Solving simultaneous equations by substitution is one of the most useful parts of mathematics you learn. If you want to get top grades, you could try using elimination here as well. To do very well, you need to be fluent in both methods. Quote Link to comment Share on other sites More sharing options...

LJP07 Posted November 24, 2006 Author Report Share Posted November 24, 2006 Ok thanks, eventually got it out!, although I would like you to tell me the method of solving the following equation: Express T in terms of R, S and V? Where V = 1/2(3Pi/2)Squared and R= Square of x-y/x+y V cubed = (rtSquared - vt)Squared/ V/ VSquared T - R Cubed. Sorry if that looks messy, but I can't seem to finish it. Quote Link to comment Share on other sites More sharing options...

sebbysteiny Posted November 24, 2006 Report Share Posted November 24, 2006 Ok thanks, eventually got it out!, although I would like you to tell me the method of solving the following equation: Express T in terms of R, S and V? Where V = 1/2(3Pi/2)Squared and R= Square of x-y/x+y V cubed = (rtSquared - vt)Squared/ V/ VSquared T - R Cubed. Sorry if that looks messy, but I can't seem to finish it. would you now. hmmm. Well, if you WANT it, then I must obey, surely. After all, I wouldn't want you to not get what you want, would I??? I'm only kidding. I'm trying to replace all my "can I have"s and "I need"s, with "I want" or "I would like" myself. It's much more assertive and most people will accept the above frame without thinking about it. Arn't we social manipulators!!! However, I still will not give you what you want since I can't understand it :confused: . You need to add brackets to make it clear exactly what is dividing what. You also need to be more precise with what you mean by T and t. Are they the same. Also, define x, y, R, S, V and T, and all their little cousins, and try to do it on one post. Otherwise, happy to help. Quote Link to comment Share on other sites More sharing options...

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