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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 2nioshine(m): 7:05am On Oct 08, 2016 |
Joe82834: Check the bolded Bro.... Modified Logically, considering the length of the line this seems more like it. seems like i omitted something in my approach |
Re: Nairaland Mathematics Clinic by oluwasammy(m): 10:40am On Oct 08, 2016 |
i mean d question u gave solutn to |
Re: Nairaland Mathematics Clinic by Cortana2: 9:54am On Oct 09, 2016 |
Joe82834: Thank you very much. The little problem is that this theorem was not given in the text, so I believe there must be a simpler way of doing this using one of the eight theorems found in New General Maths for SS2. |
Re: Nairaland Mathematics Clinic by Cortana2: 9:54am On Oct 09, 2016 |
Joe82834: Thank you very much. The little problem is that this theorem was not given in the text, so I believe there must be a simpler way of doing this using one of the eight ( ![]() |
Re: Nairaland Mathematics Clinic by Cortana2: 9:55am On Oct 09, 2016 |
2nioshine: Thank you very much. The little problem is that this theorem was not given in the text, so I believe there must be a simpler way of doing this using one of the eight found in New General Maths for SS2. 1 Like |
Re: Nairaland Mathematics Clinic by ayokunlei(m): 5:43pm On Oct 18, 2016 |
. |
Re: Nairaland Mathematics Clinic by busuyem: 3:58pm On Oct 19, 2016 |
Cortana2: You can actually solve it without the theorem but it will take a longer process. join A and T with a line that passes through the centre (diameter), then join B and T as well and u get a chord. U will notice u get a right angle at B and a right angle at T from the known theorems. then u can get X by equating results got for AT from the two right-angled triangles formed. i.e CBT and ATB. |
Re: Nairaland Mathematics Clinic by Cortana2: 10:25am On Oct 27, 2016 |
busuyem: Thank you for replying late. First, there is no way to convincingly prove that a line joining A and T passes through the centre of the circle. Secondly, I got a right angle at T but not at B, since B is not a point of tangency. I would nevertheless appreciate if you can attach a picture of what you're trying to convey. Thank you. |
Re: Nairaland Mathematics Clinic by busuyem: 10:48am On Oct 27, 2016 |
Cortana2: Since A and T are on the circumference, that means u can draw two radii from them to meet at the centre which will form diametre. And if you get diameter from AT, that means angle ABT is right-angled and that settles it. |
Re: Nairaland Mathematics Clinic by Joe82834(m): 2:13pm On Oct 28, 2016 |
busuyem:A and B being on the circumference doesn't mean a straight line joining them must be a diameter,it could be a chord and that spoils d angle abt being a right angle theory.you are simple trying to use d eyes to assume dat line AT passes through d centre which isn't satisfactory.remember,every diameter is a chord but not every chord is a diameter |
Re: Nairaland Mathematics Clinic by busuyem: 2:46pm On Oct 28, 2016 |
Joe82834: Ok. That was what I did then I got the answer. And I'm very sure that questn has solution without using that suggested theorem. Besides, if CT is tangent to the circle, it is certain that the tangent is perpendicular to the circle radius. If that is possible, can u extend the radius from T to A which must form a diameter since the extension passes through the center of the circle? Anyway, since u don't accept this, I think ur opinion counts. Bye. |
Re: Nairaland Mathematics Clinic by Joe82834(m): 3:04pm On Oct 28, 2016 |
busuyem:Hope u no de vex,cos ppl argue to learn. Definitely d tangent is perpendicular to d radius but nothing (except with d eyes) tells us that a straight line from d tangent to d centre, if extended passes through A. |
Re: Nairaland Mathematics Clinic by busuyem: 3:51pm On Oct 28, 2016 |
Joe82834: Lolz...I no dey vex but I no fit write online like that. Are u trying to say u can't get a triangle through points ATB and if u can, do u mean it can't be possible for T to join A through the centre? Better still, have u used what I ealier said to solve the question and get it? It's like u didn't bother to use at all because if u did, u would get better explanation urself than written here. I only responded to your questn because u mentioned General Maths as the texbk u used which I also used while teaching Maths in senior classes. Anyway, I might not be right afterall because this is Maths which has no mr know-it-all. Even the theorem used to solve that questn came out of this because it's never one of the fundamental theorems of cirgle theorms. #edited |
Re: Nairaland Mathematics Clinic by Nobody: 1:48am On Oct 30, 2016 |
Please, I need someone to solve this maths: 1. A store determines the retail price of a DVD Player by marking up its wholesale price by 60 percent. After a 25 percent discount off the DVD Player's retail price, the DVD Player cost $162. What is the wholesale price of the DVD Player? 2. Evaluate 6P3-6C3 |
Re: Nairaland Mathematics Clinic by fawaz050(m): 1:47pm On Oct 31, 2016 |
Mrphantom99:.1 #135 |
Re: Nairaland Mathematics Clinic by bayo1991: 4:49pm On Oct 31, 2016 |
I love dis Forum, pls direct some question to me, if u have any problem with math. |
Re: Nairaland Mathematics Clinic by bayo1991: 4:51pm On Oct 31, 2016 |
Dis is my watapp no 08133550169 |
Re: Nairaland Mathematics Clinic by Mathemagician1(m): 7:13am On Nov 01, 2016 |
Mrphantom99: 1. (x + 60% of x) - 25% of x = 162 x + 3/5x - 1/4x = 162 1.35x = 162 x = 162/1.35 x = N120 Wholesale price of the DVD is N120 2. 6P3 = 6! / (6-3)! 6! / 3! 6*5*4 = 120 6C3 = 6!/3!3! 6*5*4 / 3*2*1 = 20 6P3-6C3 120 - 20 = 100 |
Re: Nairaland Mathematics Clinic by Obinoscopy(m): 7:23am On Nov 01, 2016 |
Mrphantom99:Mathemagician1 has solved question 1. However your question 2 isn't clear. Do you mean simplify or evaluate? |
Re: Nairaland Mathematics Clinic by Mathemagician1(m): 7:27am On Nov 01, 2016 |
Obinoscopy: I've solved both. I was working on the second question before your msg. Cheers |
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 10:39am On Nov 01, 2016 |
Mathemagician1: 1. (x+0.6x)*0.75=162 1.6x*0.75=162 1.2x=162 x=162/1.2 x=135 @ fawaz050 is right |
Re: Nairaland Mathematics Clinic by fawaz050(m): 10:44am On Nov 01, 2016 |
ladokuntlad:boss longtym fa..u just ignore ur students |
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 11:03am On Nov 01, 2016 |
fawaz050: Life for here dey roughly smooth oo ![]() ![]() ![]() I no ignore una ooo. U know there is real life after school which is what man is facing now |
Re: Nairaland Mathematics Clinic by fawaz050(m): 2:35pm On Nov 01, 2016 |
ladokuntlad:E go work out sir |
Re: Nairaland Mathematics Clinic by Agnesgrace(f): 8:25pm On Nov 01, 2016 |
Richiez:An integer P is a ............iff (p-1)! is congruence to -1(modp) fill in the missing word and provide the name of the Mathematician that gave that definition........ Happy Sunday to you. |
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 9:26pm On Nov 02, 2016 |
Agnesgrace: congruence modulo by Euler |
Re: Nairaland Mathematics Clinic by Agnesgrace(f): 8:53am On Nov 04, 2016 |
ladokuntlad:No. Euler definition for congruence is different from that........ Good morning and have a wonderful day. |
Re: Nairaland Mathematics Clinic by Cortana2: 9:55am On Nov 10, 2016 |
busuyem: The one at T will be form a right angle with the radius but the one at A will not. Remember the theorem says: A tangent to a circle is perpendicular to the radius drawn to its point. |
Re: Nairaland Mathematics Clinic by thankyouJesus(m): 8:01am On Nov 13, 2016 |
Long time here |
Re: Nairaland Mathematics Clinic by timmykaydude: 9:02am On Nov 19, 2016 |
I love this thread,I have been searching for this kind of thing on facebook,the one I came in contact with they just tell u the answer,so glad I found this thread. |
Re: Nairaland Mathematics Clinic by Hadampson(m): 3:49pm On Nov 19, 2016 |
Gurus in the house, please help me with diz question on sum of geometric progression... IN AN EXPONENTIAL SERIES OF INCREASING POSITIVE TERMS, THE SUM OF THE FIRST SIX TERMS IS NINE TIMES THE SUM OF THE FIRST THREE TERMS. THE SUM OF THE FOURTH AND THE FIFTH TERM IS 120. FIND THE (1) CONSTANT RATIO (2) FIRST TERM (3) SUM OF THE FIRST TWENTY TERMS. |
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 10:35pm On Nov 20, 2016 |
Hadampson:.
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