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Re: Nairaland Mathematics Clinic by seno408(m): 10:36pm On Mar 31, 2013 |
Just new here. So what are the questions? |
Re: Nairaland Mathematics Clinic by echibuzor: 11:52pm On Mar 31, 2013 |
seno408: Just new here. So what are the questions? Its not a contest o, Just capacity building... Bring your.problems and pple will help u solve them... 1 Like |
Re: Nairaland Mathematics Clinic by seno408(m): 9:26am On Apr 01, 2013 |
echibuzor:I know its not a competition. |
Re: Nairaland Mathematics Clinic by Mbahchiboy(m): 3:24pm On Apr 02, 2013 |
Plz factorise x^6+8 |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 4:43pm On Apr 02, 2013 |
plz help me with this: if 2 bread and 4 butter cost 20naira. If 3 bread and 6 butter cost 60naira. Find the cost of 1 bread and 1 butter. |
Re: Nairaland Mathematics Clinic by Richiez(m): 5:56pm On Apr 02, 2013 |
johnpaul1101: plz help me with this: if 2 bread and 4 butter cost 20naira. If 3 bread and 6 butter cost 60naira. Find the cost of 1 bread and 1 butter.This is one very interesting application of mathematics, but check the question very well and ensure that you have provided us with the right values... i think the values are inconsistent. |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 6:10pm On Apr 02, 2013 |
Richiez: This is one very interesting application of mathematics, but check the question very well and ensure that you have provided us with the right values... i think the values are inconsistent.tnx richiez, she says d question is ok(i.e there is no mistakes in it) i tried solving it, but i got to point in which the variable i'm luking for canceled each other giving me 0. How can i find a variabke wen the variable itself is zero |
Re: Nairaland Mathematics Clinic by Richiez(m): 6:10pm On Apr 02, 2013 |
Mbahchiboy: Plz factorise x^6+8SOLUTION x^6 + 8 this is same as; (x^2)^3 + 2^3 from our knowledge of addition of two cubes, a^3 + b^3 = (a+b)(a^2 -ab + b^2) therefore x^6 + 8 =(x^2)^3 + 2^3 =(x^2 + 2)(x^4 -2x^2+ 4) 1 Like |
Re: Nairaland Mathematics Clinic by Richiez(m): 6:26pm On Apr 02, 2013 |
johnpaul1101: tnx richiez, she says d question is ok(i.e there is no mistakes in it) i tried solving it, but i got to point in which the variable i'm luking for canceled each other giving me 0. How can i find a variabke wen the variable itself is zerosame problem here, if it is an assignment given to her, tell her to ask her friends for the correct values... 1 Like |
Re: Nairaland Mathematics Clinic by echibuzor: 9:09pm On Apr 02, 2013 |
johnpaul1101: The Question has a serious typo.... Someone should try plotting a graph for the equations, I am pretty sure there will be no intersection.... |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 9:27am On Apr 03, 2013 |
goodday every1, plz i would like someone to help me with these questions: (1) a student is told to draw the graph of y=x^2+4x-6. He is then told to draw a linear graph on the same axis such that the intersection of the two graphs will give the solution to the equation x^2 + 4x - 7=0.what is the equation of the linear graph he needs to draw? (2)find the product xy if x,3/2,6/7,y, are in a GP |
Re: Nairaland Mathematics Clinic by Nobody: 10:30am On Apr 03, 2013 |
I'm really impressed...like I said before "I'm following every equation solved" |
Re: Nairaland Mathematics Clinic by Adol16(m): 11:49am On Apr 03, 2013 |
Larrysamuel: Good day my maths guru, pls can u help me with this simultaneous equation (one quadratic and one linear) : x^2-y^2= 56....equation 1, xy=20....equation 2. 10xu must b an oau student cos dats 1 of d assignment we were given last week in ssc105 |
Re: Nairaland Mathematics Clinic by Richiez(m): 11:57am On Apr 04, 2013 |
johnpaul1101: goodday every1, plz i would like someone to help me with these questions: (1) a student is told to draw the graph of y=x^2+4x-6. He is then told to draw a linear graph on the same axis such that the intersection of the two graphs will give the solution to the equation x^2 + 4x - 7=0.what is the equation of the linear graph he needs to draw? (2)find the product xy if x,3/2,6/7,y, are in a GPjohnpaul, there is no tool to enable one plot a graph here... |
Re: Nairaland Mathematics Clinic by echibuzor: 3:28pm On Apr 04, 2013 |
johnpaul1101: goodday every1, plz i would like someone to help me with these questions: Richiez: @johnpaul..... Equate y=x^2+4x-6 and x^2 + 4x - 7=0 that should give you the Equation of the linear graph...After doin it myself, I got y = -1.. Dont know but i think that means the linear graph is parallel to the x-axis at y = -1... And for the Second one.... (3/2)/x = (6/7) / (3/2) = y/(6/7) (common ratio of the GP) 3/2x = 4/7 = 7y/6 Hence : 3/2x = 7y/6 14xy = 18 xy = 18/14... |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 4:14pm On Apr 04, 2013 |
echibuzor:tnx chibuzor, i think i get it now |
Re: Nairaland Mathematics Clinic by echibuzor: 5:08pm On Apr 04, 2013 |
johnpaul1101: My Pleasure... |
Re: Nairaland Mathematics Clinic by Mbahchiboy(m): 9:22am On Apr 05, 2013 |
I am new hear,can any one plz tell me how to reply someone's question |
Re: Nairaland Mathematics Clinic by Mbahchiboy(m): 9:26am On Apr 05, 2013 |
Dats right |
Re: Nairaland Mathematics Clinic by Esiri111(m): 10:14am On Apr 05, 2013 |
Lets start |
Re: Nairaland Mathematics Clinic by Ibbaba(m): 11:41am On Apr 05, 2013 |
A |
Re: Nairaland Mathematics Clinic by Adol16(m): 9:14pm On Apr 05, 2013 |
This is it solve for a 5loga(base 4) + 48log4(base a)= a/8. |
Re: Nairaland Mathematics Clinic by Ayowumie(m): 1:07am On Apr 06, 2013 |
echibuzor: I'm still awaiting your response to the email that I sent to you. Thanks |
Re: Nairaland Mathematics Clinic by Nobody: 5:52pm On Apr 06, 2013 |
I hail all the gurus in the house. Nice works! 3 Likes |
Re: Nairaland Mathematics Clinic by teadrake(m): 6:41am On Apr 07, 2013 |
Much gratitude to the 'Gurus'.Nice job u all doing 1 Like |
Re: Nairaland Mathematics Clinic by echibuzor: 9:18am On Apr 08, 2013 |
Ayowumie: Replied.. Sorry I was out network coverage...... |
Re: Nairaland Mathematics Clinic by Richiez(m): 2:26pm On Apr 10, 2013 |
Adol16: This is it solve for a 5loga(base 4) + 48log4(base a)= a/8. SOLUTION loading.... |
Re: Nairaland Mathematics Clinic by adstick: 3:50pm On Apr 10, 2013 |
LOVE U GUYS U ARE TOO MUCH BUT HOW I WISH I CAN SOME WHO CAN BROUGHT ME UP MORE THAN THIS HERE OOOOOOOOOOOO |
Re: Nairaland Mathematics Clinic by Adex097: 2:37am On Apr 11, 2013 |
Ibbaba: Hello nairaland mathematicians,am an undergraduate whose course deals with math(computer science),I wasn't that good in math while in secondary school as a result of my foundation and also my math secondary school teacher but I have so much fasion for math,I really would love to be tutoring others in math like I kinder see lovely student in my school do(they love calculations to anything else)go get engineering mathematics..after using that book, your "maths" life will never remain thesame again.... My sis gave hers to me as a gift some years back |
Re: Nairaland Mathematics Clinic by Richiez(m): 6:50pm On Apr 11, 2013 |
Adol16: This is it solve for a 5loga(base 4) + 48log4(base a)= a/8. SOLUTION 5log4 a + 48loga 4 = a/8 multiply thru by 8 40log4 a + 384loga 4 = a next we convert log(base a) to log(base 4) 40log4 a + 384log4 4/log4 a = a but log4 4 = 1 40log4 a + 384/log4 a = a 40(log4 a)2 +384 =alog4 a 40(log4 a)2 - alog4 a + 384 = 0 let log4 a = x...........(1) 40x2 -ax + 384 =0 ............(2) solving simultaneously, x=4 and a=256 hence a=256 |
Re: Nairaland Mathematics Clinic by Onuels(m): 7:46pm On Apr 11, 2013 |
Hey maths gurus wch of u guys lives in abeokuta, I wud like to get in contact wit you. Here is ma contact 08184135053 |
Re: Nairaland Mathematics Clinic by Richiez(m): 12:00am On Apr 12, 2013 |
Ghettoguru: I hail all the gurus in the house. Nice works!bro where av u been? |
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