Guest Stefkov Posted March 11, 2007 Posted March 11, 2007 I thought I could do thisbut its annoying me. Continuing on from (3) Hence show that (big S with 4 at top and 1 at bottom)(x^5/2+1)dx = (65/12) .....................................................x^2 I'm guessing you get rid of the 65/12, that means its equal to so you have to show it equals that. You have to do the square brackets. Also I remember being told you have to use the x^(1/2) + x^(-2), from the other part.
DCK Posted March 11, 2007 Posted March 11, 2007 I thought I could do thisbut its annoying me. Continuing on from (3) Hence show that (big S with 4 at top and 1 at bottom)(x^5/2+1)dx = (65/12) .....................................................x^2 I'm guessing you get rid of the 65/12, that means its equal to so you have to show it equals that. You have to do the square brackets. Also I remember being told you have to use the x^(1/2) + x^(-2), from the other part. [1->4]S(x^2.5+1)/x^2 dx) = [1->4]S(x^.5+1/x^2 dx) = [1->4][2/3 * x^1.5 - 1/x] = (2/3 * 4^1.5 - 1/4) - (2/3 * 1^1.5 - 1/1) = (16/3 - 1/4) - (2/3 - 1) = (64/12 - 3/12) + 1/3 = 64/12 - 3/12 + 4/12 = 65/12 Hope it's readable Edit: here's a more legible version:
Supergrunch Posted March 11, 2007 Posted March 11, 2007 Yep, DCK's done it right. By the way, if it ever says "hence", then you're supposed to use the previous answer.
Guest Stefkov Posted March 11, 2007 Posted March 11, 2007 Cheers guys, but just the one more, I've just cant figure out who to do this. I've got to findthe co-ordinates of A on the picture, you cant see. But its 3x^2 - 10x + 3 When you gotta find co-ordinates you gotta make dy=0 dx so 3x^2 - 10x + 3 = 0 (3x - ?)(x - ?) = 0 And thanks for your help.
Fresh Posted March 11, 2007 Posted March 11, 2007 Cheers guys, but just the one more, I've just cant figure out who to do this.I've got to findthe co-ordinates of A on the picture, you cant see. But its 3x^2 - 10x + 3 When you gotta find co-ordinates you gotta make dy=0 dx so 3x^2 - 10x + 3 = 0 (3x - ?)(x - ?) = 0 And thanks for your help. I don't understand what your asking since nothing will go in those brackets.
Guest Stefkov Posted March 11, 2007 Posted March 11, 2007 We were told when you have to find a point dy/dx = 0. So the equation 3x^2 - 10x + 3 = 0. Now I just have to factorise (that the right word?) it. I'm pretty sure up to now its (3x - ?)(x - ?) cos its 3x^2 it has to have 3x in one bracket and x in the other to make the 3x^2. Its just hte numbers that confuse me.
Supergrunch Posted March 11, 2007 Posted March 11, 2007 We were told when you have to find a point dy/dx = 0. So the equation 3x^2 - 10x + 3 = 0. Now I just have to factorise (that the right word?) it. I'm pretty sure up to now its (3x - ?)(x - ?) cos its 3x^2 it has to have 3x in one bracket and x in the other to make the 3x^2. Its just hte numbers that confuse me. It's (3x - 1)(x - 3).
Guest Stefkov Posted March 11, 2007 Posted March 11, 2007 Thanks Supergrunch. Now I will do at least 1 question without any help
The fish Posted March 11, 2007 Posted March 11, 2007 It's (3x - 1)(x - 3). So he can do, as he says, one question without any help, surely some one should explain how you get that... What you need to do is find two numbers that add to make -10, and also multiply to make 3. They are -1 and -3 as: 3 (as it's 3x) times -3=-9 -9 + -1=-10 -3 times -1=3
Guest Stefkov Posted March 11, 2007 Posted March 11, 2007 I kinda know how to do it, its just remembering how to do it and when theres a 3x like that question it just baffles me sometimes. Thanks for clearing it up Fish.
Supergrunch Posted March 11, 2007 Posted March 11, 2007 So he can do, as he says, one question without any help, surely some one should explain how you get that... What you need to do is find two numbers that add to make -10, and also multiply to make 3. They are -1 and -3 as: 3 (as it's 3x) times -3=-9 -9 + -1=-10 -3 times -1=3 I've already explained how to solve quadratics enough times in this thread...
Ginger_Chris Posted March 12, 2007 Posted March 12, 2007 The quadratic formula? btw, use www.s-cool.co.uk lots of good stuff there for maths. I used to use it alot for basic trig identities and such. Edit: @blender depends on notation: 100/3 exactly (best form really, as its exact); 33 and 1/3 (also exact but clumsy); 33.3333 (recurring) (written as 33.3 with a dot over the last 3)(not exact); or 33.33 (2 decimal points)(not exact); or 33.333 (5 significant figures)(not exact).
Atomic Boo Posted March 12, 2007 Author Posted March 12, 2007 translate: tu viens chez moi? Je passe chez toi l'apres-midi avec la casette de roch voisine. edit: dont worry about it now.
Twozzok Posted March 15, 2007 Posted March 15, 2007 Curve of x^2, the normal to the curve is at A(1,1), The normal cuts the curve to the left at B. Use differentiation to find the equation of the normal at A. Verify the point where the normal cuts the curve again has coordinates (-3/2 , 9/4). I do this by differentiating X^2 ( which is 2x), so the equation of the normal is y= -1/2x ? Then all I need to do is prove (-3/2 , 9/4) satisfies y=-1/2x? (which i don't think it does !_!) EDIT: Also, how do I find where: y= cuberootX - X^2 crosses the x-axis?
Zell Posted March 15, 2007 Posted March 15, 2007 Differentiating gets you the gradient function, not the equation of the tangent (because the gradient varies along the curve). Substitute (1,1) into the gradient function to find the gradient of the tangent, which is 2. So the gradient of the normal is -1/2. Now we need to find out the value of +c. So we substitute the known co-ordinate into "y=mx+c". 1 = -1/2(1) + c Therefore c = 3/2 So equation of normal: y = -1/2x + 3/2 To find out the co-ordinates of B, we solve the simultaneous equations (I'll let you do that )
Goron_3 Posted March 15, 2007 Posted March 15, 2007 i need to know a 'left wing literary critic' by 2moro...someone help! It's VERY important as my coursework needs to be finished within an hour
Daft Posted March 19, 2007 Posted March 19, 2007 I REALLY NEED HELP!! I've got a 2500 word essay in tomorrow on why the Tokugawa bakufu collapsed! The problem is the essay isn't supposed to be more than 2500 words!!! ANY tips on how to cut down!! I would really appreciate it!!
|Laguna| Posted March 19, 2007 Posted March 19, 2007 Didn't know whether to post this in the Politics thread, or this one. But I think it is more suited for here. So/ Hey, any Scottish people out there, who know a wee thing about politics, do the Scottish Parliamentry elections get much antetnion? It's for an essay I'm doing contrasting General elections with Scottish and European elections, and I know that not much attention is given to European elections, (29% turnout!) but I haven't the foggiest on how big a deal Scottish Parliamentry Elections are. Any help would be greatly appreciated.
Supergrunch Posted March 19, 2007 Posted March 19, 2007 I REALLY NEED HELP!! I've got a 2500 word essay in tomorrow on why the Tokugawa bakufu collapsed! The problem is the essay isn't supposed to be more than 2500 words!!! ANY tips on how to cut down!! I would really appreciate it!! Write it in Japanese- 1 word per sentence.
Rummy Posted March 19, 2007 Posted March 19, 2007 Who knows about statistical analyses? Namely, mutliple regressions in spss and how to interpret what you get out. I have 3 IVs(extraversion, neuroticism and gender) and 1 DV(luckiness score). They're all continuous, except gender of course, because...welll, because.
McMad Posted March 20, 2007 Posted March 20, 2007 I need some help from you Chemistry wizzes out there, I need examples of Metallic Bonds, Covalent Bonds and Ionic Bonds due to me being useless at the subject and can't figure them out for myself. P.S. The more, the better.
MoogleViper Posted March 20, 2007 Posted March 20, 2007 Covalent bonds are non metals only, metallic bonds are metals only and ionic bonds are metals and non metals.
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