Welcome, Guest: Register On Nairaland / LOGIN! / Trending / Recent / NewStats: 3,195,586 members, 7,958,815 topics. Date: Thursday, 26 September 2024 at 03:04 AM |
Nairaland Forum / Nairaland / General / Education / Nairaland Mathematics Clinic (495467 Views)
Mathematics Clinic / 2015 Cowbell Mathematics Champion, Akinkuowo Honoured By School. / Lead City University Clinic Welcomes First Ever Baby Since 10 Years Of Opening (2) (3) (4)
(1) (2) (3) ... (9) (10) (11) (12) (13) (14) (15) ... (284) (Reply) (Go Down)
Re: Nairaland Mathematics Clinic by cigie(m): 8:58pm On Jan 04, 2013 |
biolabee: truman u are a demon(a+b+c) ^2 = 64 this move has made your effort futile a^2 + b^2 + c^2 = 64 is not equal to (a+b+c) ^2 = 64 |
Re: Nairaland Mathematics Clinic by Nobody: 9:04pm On Jan 04, 2013 |
coolexy2: ](1)what is the real application of mathematics in our live? Question 2 (a): y = cos ^3 x Since cos^3 x = (cos x)^3 Lets assume that: cos x = t The equation becomes y = (t)^3 Applying chain rule: dy/dt= 3 [t] ^2 . dt/dx Substituting (t) back = 3 (cos ^ 2 x) . ( - sin x ) = - 3 cos ^ 2 x sin x So first derivative: dy/dx = - 3 cos ^ 2 x sin x Solving the derivative of -3cos^2 x sin x which is also the 2nd derivative of cos^3 x. Lets assume that : -3cos^2 x = U and sin x = V Applying product rule to -3cos^2 x sin x d^2y/d x^2 = U.dv/dx + V.du/dx = -3cos^2 x . cos x + sin x . -(6cos x) . (-sin x) = -3cos^3 x + 6sin^2 x cos x = 6sin^2 x cos x - 3cos^3 x = 3cos x (2sin^2 x - cos^2 x). Thus the second derivative of cos^3 x d^2y/d x^2 = 3cos x (2sin^2 x - cos^2 x) Simplifying further, 3cos x (2sin^2 x - cos^2 x) Since, sin^2 x = 1 - cos^2 x replacing will give us : = 3cos x ( 2[1 - cos^2 x ] - cos^2 x) = 3cos x (2 - 2cos^2 x - cos^2 x) Final Answer : d^2y/dx^2 = 3cos x (2 - 3cos^2 x) 2 Likes |
Re: Nairaland Mathematics Clinic by ositadima1(m): 9:19pm On Jan 04, 2013 |
^^^ cool 1 Like |
Re: Nairaland Mathematics Clinic by calculusx(m): 10:01pm On Jan 04, 2013 |
ositadima1: ^^^ cool Bro. I want to confirm if you are the one that purchased 99 Camry Xle from me some years ago? I was doing my NYSC then if you can remember. I misplaced your contact number |
Re: Nairaland Mathematics Clinic by ositadima1(m): 10:19pm On Jan 04, 2013 |
calculusx: No be me, probably a namesake. |
Re: Nairaland Mathematics Clinic by calculusx(m): 10:41pm On Jan 04, 2013 |
ositadima1: OK. Thx |
Re: Nairaland Mathematics Clinic by Obinoscopy(m): 11:21pm On Jan 04, 2013 |
ositadima1: I didn't fabricate anything but i made a little mistake in the number, will correct it though. Just admit it however that u still can't solve it I'll repeat the qxn once more: x^x+y^y=46912, x+y=10 Find x and y. |
Re: Nairaland Mathematics Clinic by Nobody: 12:04am On Jan 05, 2013 |
Boring thread |
Re: Nairaland Mathematics Clinic by Nobody: 1:21am On Jan 05, 2013 |
Mathano: At cryptic,It is finitely generated.A group is said to be finitely generated when there exist a^n=e. a is d generator and n is d order of d group.for a polynomial ring generated by a group dere will b a point where a^n will be equal to the identity element.Guy, i have seen you really like Abstract Algebra especially Groups and Rings. I'm a Part 3 Maths student at OAU and i have to be honest; i do not enjoy this area. But i 've got exam to do on this course next wk monday and i have to read it to have a chance at doing well. The note is really voluminous and our lecturer 'll make sure he left no stone unturn. I'm convinced you love this aspect of pure maths. Can you tell me how you 've come to understand it ? 1 Like |
Re: Nairaland Mathematics Clinic by biolabee(m): 1:36am On Jan 05, 2013 |
cigie a + b +c = 8 square both sides (a+b+c)^2 = 8^2 = 64 |
Re: Nairaland Mathematics Clinic by ositadima1(m): 8:01am On Jan 05, 2013 |
Obinoscopy: , the answers are x=6, y=4. but u still fabricated it though |
Re: Nairaland Mathematics Clinic by Calebsworld(m): 8:17am On Jan 05, 2013 |
Prove by the principle of induction.1/1.3 +1/3.5 +1/5.7+...+1/(2n-1)(2n+1)=n/2n+1.thanks |
Re: Nairaland Mathematics Clinic by Obinoscopy(m): 9:24am On Jan 05, 2013 |
ositadima1: Show ur workingd dude |
Re: Nairaland Mathematics Clinic by Obinoscopy(m): 9:27am On Jan 05, 2013 |
ositadima1: Show ur workings dude |
Re: Nairaland Mathematics Clinic by ositadima1(m): 10:22am On Jan 05, 2013 |
Obinoscopy: Solving it algebraically will take days of toil, I don't think I got the brains. Quick method is to assume x and y are real, positive whole numbers. So, since x=10-y is relatively simple, I went on trying all possible combinations, trial and error style. Another approach would be to plot x+y=10 and x^x+y^y=46912, and get the values for x and y at there intersections. I used a plotting cal. for this. 3 Likes |
Re: Nairaland Mathematics Clinic by Nobody: 11:43am On Jan 05, 2013 |
* following * |
Re: Nairaland Mathematics Clinic by Calebsworld(m): 4:38pm On Jan 05, 2013 |
Calebsworld: Prove by the principle of induction.1/1.3 +1/3.5 +1/5.7+...+1/(2n-1)(2n+1)=n/2n+1.thankshello Richiez am waiting. |
Re: Nairaland Mathematics Clinic by WAM1(f): 5:42pm On Jan 05, 2013 |
<3<3<3 loving this |
Re: Nairaland Mathematics Clinic by marcangelo(m): 6:24pm On Jan 05, 2013 |
math-diva: |
Re: Nairaland Mathematics Clinic by Nobody: 7:36pm On Jan 05, 2013 |
@calculusx nd 2good I disagree wit u guyz wen u say anything raise to power zero is one, I can prove to u dat d definition is erroneous and ambiguous!and even any number raise to power zero is not one! So pls kindly edit ur definition! |
Re: Nairaland Mathematics Clinic by 2good(m): 7:49pm On Jan 05, 2013 |
masperano: @calculusx nd 2good I disagree wit u guyz wen u say anything raise to power zero is one, I can prove to u dat d definition is erroneous and ambiguous!and even any number raise to power zero is not one! So pls kindly edit ur definition!proof me wrong then. |
Re: Nairaland Mathematics Clinic by j2012(m): 8:29pm On Jan 05, 2013 |
Pls prove the following trig identities 1) Sin(t)/1+cos(t)=1/cosec(t)+cot(t) 2)2tan(t)/1-tan square(t)=2sin(t)cos(t)/(cos square(t)-sin square (t) 3)1+cot square(t)sin(t)=cosec(t) |
Re: Nairaland Mathematics Clinic by Nobody: 10:00pm On Jan 05, 2013 |
2good: proof me wrong then. |
Re: Nairaland Mathematics Clinic by Nobody: 10:45pm On Jan 05, 2013 |
j2012: Pls prove the following trig identities Question 1. Prove that sin t/1+ cos t = 1/cosec t + cot t. Taking from the RHS, Since cosec t = 1/sin t and cot t = 1/tan t or cos t/sin t, 1/cosec t + cot t becomes : = 1/[1/sin t + cos t/sin t] =1/(1+cos t)/sin t, inverting: = sin t/(1+ cos t) Hence: sin t/(1+ cos t) = 1/cosec t + cot t. Question 2. Prove that: 2tan t/(1-tan^2 t) = 2cos t sin t/(cos^2 t - sin^2 t) From the LHS, Since tan t = sin t/cos t 2tan t/(1-tan^2 t) becomes: =2sin t/cos t /[1/1 - (sin^2 t)/cos^2 t], inverting: =2sin t/ cos t * cos^2 t/(cos^2 t - sin^2 t) =2sin t cos^2 t / cos t [cos^2 t - sin^2 t] =2sin t cos t/(cos^2 t - sin^2 t) Hence 2tan t/(1-tan^2 t) = 2cos t sin t/(cos^2 t - sin^2 t) Question 3. Prove that 1+cot^2(t) sin (t) = cosec (t). This question is prolly wrong because from the LHS, If you substitute cot t = cos t/sin t 1 + cot^2 t sin t =1/1 + cos^2 t * sin t/sin^2 t sin^2 t + cos^2 t * sin t/ sin^2 t Since the above can only be multiplied before addition is initiated according to BODMAS, there is no way the result can cancel out to give cosec t If we add before multiplying sin^2 t + cos^2 t would be 1 to give us sin t/sin^2 t = 1/sin t or cosec t, but it's mathematically wrong to add before multiplying, meaning the question is logically incorrect! Thus : 1+cot^2(t) sin (t) is NOT equal to cosec (t) 1 Like |
Re: Nairaland Mathematics Clinic by Nobody: 11:56pm On Jan 05, 2013 |
At johnpaul,D first thing is dat most students cram in groups and rings,If u can try to understand it and knw d basis of hw to proof it wil b easy to play with,I wil give u an example to proove dat d inverse of any group is unique,u will first make an assumption dat there are two inverses,since we are trying to proove dat it is unique d two must be equal.pf suppose dere are two inverses b and b1.let a be an element of d group ab=e,ab1=e,since b and b1 are inverses,there multiplication must give d identity element,b=eb(identity multiplied by any number will give dat number)but e=ab1,we can nw subst dat,b=ab1b,b=(ab)b1(Just reordering)but ab=e,there4 b=eb1,b=b1(since identity multiplied by any no will give d no)b=b1.Therefore d inverse is unique.Dere are loads of textbook onlyn 4 abstract algebra,u can also use kuku's textbuk,A prof in UI. |
Re: Nairaland Mathematics Clinic by Nobody: 8:17am On Jan 06, 2013 |
Mathano: At johnpaul,D first thing is dat most students cram in groups and rings,If u can try to understand it and knw d basis of hw to proof it wil b easy to play with,I wil give u an example to proove dat d inverse of any group is unique,u will first make an assumption dat there are two inverses,since we are trying to proove dat it is unique d two must be equal.pf suppose dere are two inverses b and b1.let a be an element of d group ab=e,ab1=e,since b and b1 are inverses,there multiplication must give d identity element,b=eb(identity multiplied by any number will give dat number)but e=ab1,we can nw subst dat,b=ab1b,b=(ab)b1(Just reordering)but ab=e,there4 b=eb1,b=b1(since identity multiplied by any no will give d no)b=b1.Therefore d inverse is unique.Dere are loads of textbook onlyn 4 abstract algebra,u can also use kuku's textbuk,A prof in UI.Are you a student in UI ? We use this Kuku's textbook but it's not our major text. Like you said, some student cram the course - and me too. I understand little and i cannot wait to do this course one last time. I do not expect to come back to it again. Your example is on point. |
Re: Nairaland Mathematics Clinic by biolabee(m): 10:32am On Jan 06, 2013 |
Superb Doubledx For no 3 it came to sint + cos2t = 1 There is a missing sint I think the original Q shd have been sin t + cot2tsint = cosec t 1 Like |
Re: Nairaland Mathematics Clinic by Nobody: 11:10am On Jan 06, 2013 |
Pls who can proove dat continuity is a necessary condition but not a sufficient one for deriviability |
Re: Nairaland Mathematics Clinic by kasbeats(m): 4:24pm On Jan 06, 2013 |
At richiez,aw cn U̶̲̥̅̊ be so sure dat y=3...was dat a guess? |
Re: Nairaland Mathematics Clinic by j2012(m): 4:34pm On Jan 06, 2013 |
doubleDx: Thanks man, I appreciate. |
Re: Nairaland Mathematics Clinic by Nobody: 6:23pm On Jan 06, 2013 |
biolabee: Superb Doubledx @biolabee, Good observation bruv! He made a mistake in the process of typing the question. It should be, prove that : sin t + cot^2 t sin t = cosec t or (1 + cot^2t) sin t = cosec t. He omitted the brackets, which changed the entire question. Thanks for the observation. j2012: @ j2012 Question 3. I figured you omitted the brackets. Note that 1+cot^2 t sin t and (1+cot^2 t) sin t are not the same. Next time you are coping a question make sure you don't omit anything, any omission will change the question. So, question 3 should be: Prove that : ( 1+cot^2 t) sin t = cosec t From the LHS, if you substitute cot t = cos t/sin t (1 + cot^2 t) sin t will become: = {1/1 + cos^2 t / sin^2 t} *sin t (sin^2 t + cos^2 t) * sin t/ sin^2 t Since sin^2 + cos^2 t = 1 Substituting will give us: = (1) sin t/sin^2 t = 1/sin t = cosec t :. ( 1+cot^2 t) sin t = cosec t 1 Like |
(1) (2) (3) ... (9) (10) (11) (12) (13) (14) (15) ... (284) (Reply)
Mastercard Foundation Scholarship, Enter Here / Jamb Result Checker 2012/2013 – UTME Result Checker / DIRECT ENTRY Admission.
(Go Up)
Sections: politics (1) business autos (1) jobs (1) career education (1) romance computers phones travel sports fashion health religion celebs tv-movies music-radio literature webmasters programming techmarket Links: (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) Nairaland - Copyright © 2005 - 2024 Oluwaseun Osewa. All rights reserved. See How To Advertise. 65 |