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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)
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Re: Nairaland Mathematics Clinic by Zainey338: 10:06pm On Sep 14, 2016 |
Hello every one, how are you? ![]() |
Re: Nairaland Mathematics Clinic by Trone: 4:17pm On Sep 16, 2016 |
Convert 0.75 in base 10 to a number in base 5. Kindly help out! ASAP |
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 7:49pm On Sep 16, 2016 |
MhizVeronica: What level and deptmt? Am Mathematics set 2014 |
Re: Nairaland Mathematics Clinic by Aksimple47(m): 11:01pm On Sep 16, 2016 |
Pls d Gurus, kindly state d best mathematics textbook with both pure/further maths to buy because I need one urgent for secondary school student. |
Re: Nairaland Mathematics Clinic by MhizVeronica(f): 4:16pm On Sep 17, 2016 |
ladokuntlad: 100l mathematics too |
Re: Nairaland Mathematics Clinic by 2nioshine(m): 8:03pm On Sep 17, 2016 |
Trone: Convert 0.75 (base 10) to base 5 : Algorithm : (1) Multiply number in base 10 by 5 : 0.75 * 5 = 3.75 (2) Separate integer from fraction : 3.75 = 3 + 0.75 (3) Integer is the first place in the decimal of the base 5 number, i.e., decimal so far is 0.3. (4) Go back to (1) and multiply decimal just obtained by 5 again. In this case, we'll just be repeating the same thing (0.75 *5 = 3.75) over and over. Thus, the decimal expansion is 0.333 NB: This is a rare case were the decimal part gives a repeated fraction. Most of this type of questions comes much in handy... See goggle Summary The key idea of converting a fraction (i.e., a value with a finite number of digits right of the radix point) is to multiply repeatedly by K, and determine if the result is bigger than 1 or not. If so, store the integer part in an array. If not, store 0 in the array. Then, subtract off the integer part. Repeat until the number is 0. If the number is 0, it may simply be a repeated fraction. 2 Likes |
Re: Nairaland Mathematics Clinic by Trone: 6:26am On Sep 18, 2016 |
2nioshine:Thank you! 2 Likes |
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 9:58pm On Sep 18, 2016 |
MhizVeronica: Stil av a long way to go. Mathematics is fun. Get mentally prepared for abstract reasoning 1 Like |
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 10:01pm On Sep 18, 2016 |
@moderator and d likes. Why was my solution hidden? Trone: I gave a solution to this particular problem about 2-3days now. And on checking tonight, it's nowhere to be found |
Re: Nairaland Mathematics Clinic by Umartins1(m): 11:17pm On Sep 18, 2016 |
Aksimple47: Project 1 Scholastic Series 2 Project 3 Engineering Maths by K.A Stroud of for very advanced level, engineering maths by H.K. Dass. |
Re: Nairaland Mathematics Clinic by megabeatsonline: 5:43am On Sep 19, 2016 |
As students are we utilising past questions to their full? Correct usage can actually boost our chances of passing. Jobbers practice GMAT past questions. Its all about Full UTILISATION. check http://genesisonlinetutorials..co.ke/2016/09/exam-tips-past-questions-benefits-of-hind-sight.html |
Re: Nairaland Mathematics Clinic by MhizVeronica(f): 6:14pm On Sep 20, 2016 |
ladokuntlad: drop ur whatsapp or fb id pls |
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 10:30pm On Sep 20, 2016 |
MhizVeronica: My WhatsApp number is on my signature while my fb id is ladokun olatunji |
Re: Nairaland Mathematics Clinic by Ayomide002: 4:24pm On Sep 21, 2016 |
smiles. .........
nairaland maths clinic.....
long time old friends....2nioshine,ladokuntlad and others.....
thanx for keeping the thread alive
@calculusf (x) 2 Likes |
Re: Nairaland Mathematics Clinic by kingmayak: 1:32pm On Sep 28, 2016 |
Please help with this. 1) find an equation of the tangent line to the curve y=tan^2 x at the point ((pi÷3),3) 2) find integral of cos ^6 x DX 3) if xy + y^2 =1 find (d^2 y ÷ dx^2) at (0,1) Thanks |
Re: Nairaland Mathematics Clinic by Umartins1(m): 6:20pm On Sep 28, 2016 |
D |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 8:42am On Sep 29, 2016 |
kingmayak:here 1 Like 1 Share
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Re: Nairaland Mathematics Clinic by kingmayak: 2:56pm On Sep 29, 2016 |
Thanks a lot buh (3) didn't see it clearly. |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 5:51pm On Sep 29, 2016 |
kingmayak: ok leme type it out Given that. xy+y² =1 getting y" at (0,1) first differentiating implicitly => y+xy' +2yy' =0 .......(*) at (0,1) . => 1 +0(y') +2y'=0 y'= -1/2 again , from (*) y'+ y' +xy"+2(y')² +2yy" =0 at (0,1) we have 2(-1/2) +0(y" ) + 2(-1/2)² +2(1)y"=0 thus y"= 1/4 thats it hope you get ? 1 Like 1 Share |
Re: Nairaland Mathematics Clinic by kingmayak: 10:18pm On Sep 29, 2016 |
Thanks alot |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 10:23pm On Sep 29, 2016 |
kingmayak: uwc buddy... |
Re: Nairaland Mathematics Clinic by Soneh(m): 10:38am On Oct 01, 2016 |
Pls guys help on these questions 1. Find an equation of the tangent line to the curve y=tan^2 x. At the point (pi/3,3) (2).find the integral of cos^6 xdx (3).if xy+y^2=1.find d^2y/dx^2 at (0,1). |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 11:39am On Oct 01, 2016 |
Soneh: scroll up . its solved bro. |
Re: Nairaland Mathematics Clinic by Cortana2: 7:51pm On Oct 03, 2016 |
Please I need help with this question. I just can't get around it. The Circle Geometry theorems I know don't seem to work. The question is, Find x, i.e. AB.
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Re: Nairaland Mathematics Clinic by oluwasammy(m): 9:09pm On Oct 03, 2016 |
@ agentofchange is dat nt numerical analysis qustion frm HK DASS |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:15pm On Oct 03, 2016 |
Re: Nairaland Mathematics Clinic by profmathsland(m): 12:12pm On Oct 04, 2016 |
Good day to you all. It's been quite a while. I have been on a huge research concerning the order of operations in Mathematics but I am back and better now. How have you guys been faring? |
Re: Nairaland Mathematics Clinic by Cortana2: 10:50am On Oct 06, 2016 |
Cortana2: Please I still need help with this question. |
Re: Nairaland Mathematics Clinic by 2nioshine(m): 3:22pm On Oct 06, 2016 |
Cortana2: Maybe you should try this theorem If a chord and a tangent intersect externally, then the product of the length of segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection Thus AB*BC =(CT) ^2 6x=8^2 And this gives x=64/6=10.67 Try and do a cross check it's been long i worked on circle geometry |
Re: Nairaland Mathematics Clinic by Joe82834(m): 4:30am On Oct 08, 2016 |
Cortana2: The correct theorem states that,"If a secant and a tangent of a circle are drawn from a point outside the circle, then the product of the lengths of the secant and its external segment equals the square of the length of the tangent segment. Interpreted as AC × BC = CT² (6+x)6 = 8² = 64 6 + x = 64 ÷ 6 = 10.67 x = 10.67 - 6 x = 4.67 |
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