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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 MrCalculus(m): 7:24am On Jun 07, 2013 |
Calculusf(x):OK SIR...I think i get u nw |
Re: Nairaland Mathematics Clinic by MrCalculus(m): 8:38am On Jun 07, 2013 |
Mr Calculus: plz guyz need ur help:::Convert 1111.100110 base 2 to basePlz guys need ur help hear |
Re: Nairaland Mathematics Clinic by MrCalculus(m): 8:43am On Jun 07, 2013 |
Mr Calculus: I really need ur help hear guys.Plz guruz help out |
Re: Nairaland Mathematics Clinic by Edis4christ(m): 12:22pm On Jun 07, 2013 |
Mr Calculus: I really need ur help hear guys.I ave tried half way, let me post dis 1 first, still trying harder 4rm laws of indices x * x=x^2 So x^(a^2 b^-1 c^-1)+(a^-1 b^2 c^-1)+(a^-1 b^-1 c^2) Bring all d ^-1 nd ^2 2geda x^(a^2(bc)^-1)+((ab)^-1 b^2)+((ab)^-1 c^2) x^(a^2+b^2+c^2)(ab+bc+ac)^-1 Am still solving 4 d final ans |
Re: Nairaland Mathematics Clinic by Edis4christ(m): 12:22pm On Jun 07, 2013 |
Mr Calculus: I really need ur help hear guys. |
Re: Nairaland Mathematics Clinic by Tuljaking(m): 12:23pm On Jun 07, 2013 |
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Re: Nairaland Mathematics Clinic by algorithm(m): 11:33pm On Jun 07, 2013 |
estilo: I greet all the maths gurus in d house ooooo. Please have come again I want you guyz to assist me to solve some statistics questins.am a Mass comm student and I dnt knw anytin abt calcultn(olodo gurl)pls help ur sis oooo.HEYA..... God will see you through.. AMEN |
Re: Nairaland Mathematics Clinic by papindinho(m): 9:13am On Jun 09, 2013 |
ladokuntlad: pls help me find d value of x if x^2=4x. show workings oooo. hint x=4this way is also cool. X^2-4x=0 factorise x(x-4)=0 so, x=0 or x-4=0 x=0 or x=4 |
Re: Nairaland Mathematics Clinic by papindinho(m): 9:21am On Jun 09, 2013 |
pls maths gurus help with this o. The speed of 30km per min expressed in cm per sec = ? Note= km means kilometre min means minute, cm =centimetre and sec=second |
Re: Nairaland Mathematics Clinic by mathefaro(m): 10:43am On Jun 09, 2013 |
papindinho: pls maths gurus help with this o.30km/min means; 30km = 1 min..........(i) then convert the km to cm and min to sec (1km = 100000cm, 1 min = 60 sec) therefore, (i) becomes: 3000000cm = 60sec; divide both sides by 60; 50000cm = 1 sec; therefore 30km/min = 50000cm/sec. |
Re: Nairaland Mathematics Clinic by rashywire: 3:48pm On Jun 09, 2013 |
Hello my fellow aspirants, I'm new on dis thread, pls carry me along. M going 4 elect/elect in lag |
Re: Nairaland Mathematics Clinic by rashywire: 3:51pm On Jun 09, 2013 |
Mr Calculus: Plz guruz help out. Hi pls i also wnt 2 be ok in ma maths, hw cn i? |
Re: Nairaland Mathematics Clinic by rashywire: 3:53pm On Jun 09, 2013 |
I'm new on dis thread, i hail all maths gurus oooo |
Re: Nairaland Mathematics Clinic by estilo(f): 10:09am On Jun 10, 2013 |
Ur sis hail ooo. @ortarico, me appreciate all ur help,tenks so much cockroach will nt eat ur brain oo lol. Tenk sir A grateful heart 2 Likes |
Re: Nairaland Mathematics Clinic by BRAINEE123: 11:45am On Jun 10, 2013 |
If a body moves with a final velocity of 30km/hr and accelerates at 12km/hr^2 in 2hrs. What is the distance covered in 2.5hr? |
Re: Nairaland Mathematics Clinic by BRAINEE123: 11:56am On Jun 10, 2013 |
Plz pals , solve this for me: If a body moves with a final velocity of 30km/hr and accelerates at 12km/hr^2 in 2hrs. What is the distance covered in 2.5hrs? |
Re: Nairaland Mathematics Clinic by echibuzor: 12:55pm On Jun 10, 2013 |
BRAINEE123: Plz pals , solve this for me: If a body moves with a final velocity of 30km/hr and accelerates at 12km/hr^2 in 2hrs. What is the distance covered in 2.5hrs?. . This is a problem under the Laws of motion.. The first part of the question..(a body moves with a final velocity of 30km/hr and accelerates at 12km/hr^2 in 2hrs)... V = u + at, where V = 30, a = 12, t = 2, by substitution, u = 6..... . The second part of the problem (What is the distance covered in 2.5hrs?)... S = ut + 1/2 at2.... where u = 6, a = 12, t = 2.5.. by substitution, S = 36km... |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 6:09pm On Jun 10, 2013 |
goodday richiez and all respectable mathematicians, plz i urgently need help with this: If log x=bar 2.3675 and log y=0.9750, find x+y. Thanks |
Re: Nairaland Mathematics Clinic by labodinho: 6:42pm On Jun 10, 2013 |
Plz help me out.Hw do i solve logs with diff base like this No 1 Log9 base 3-Log3 base 27+ Log9 base sqaure root3.........No 2.Solve the equatn LogX base 2-Log(X-1) base 2=2.........This one is on Inequalities.No 3. (2x-1)(x+4) divided by x-5 both numerator n denominator Greater than(sign) 0 ....Tanx. |
Re: Nairaland Mathematics Clinic by echibuzor: 7:05pm On Jun 10, 2013 |
labodinho: Plz help me out.Hw do i solve logs with diff base like this No 1 Log9 base 3-Log3 base 27+ Log9 base sqaure root3.........No 2.Solve the equatn LogX base 2-Log(X-1) base 2=2.........This one is on Inequalities.No 3. (2x-1)(x+4) divided by x-5 both numerator n denominator Greater than(sign) 0 ....Tanx.. . . I couldnt type it but I hope the picture of my workbook is clear enough...
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Re: Nairaland Mathematics Clinic by labodinho: 7:21pm On Jun 10, 2013 |
echibuzor:Oh,tanx.can't really see them clearly.Can u mail me on labodihoo@gmail.com or exercise patient by writing it.Plz,jst @ yur conveniency.... |
Re: Nairaland Mathematics Clinic by MrCalculus(m): 7:29pm On Jun 10, 2013 |
johnpaul1101: goodday richiez and all respectable mathematicians, plz i urgently need help with this:x=antilog of bar 2.3675 y=antilog of 0.9750.. anytin u get add it together |
Re: Nairaland Mathematics Clinic by echibuzor: 7:37pm On Jun 10, 2013 |
labodinho: Oh,tanx.can't really see them clearly.Can u mail me on labodihoo@gmail.com or exercise patient by writing it.Plz,jst @ yur conveniency.... Alrite man.. I ll type it on Word and attach it to the mail I am gonna send gonna send to u.. |
Re: Nairaland Mathematics Clinic by labodinho: 7:39pm On Jun 10, 2013 |
echibuzor:Tanx bro. |
Re: Nairaland Mathematics Clinic by echibuzor: 8:22pm On Jun 10, 2013 |
labodinho: Tanx bro. Check your Email box... |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 8:43pm On Jun 10, 2013 |
Mr Calculus:thanks very much |
Re: Nairaland Mathematics Clinic by oladoya(m): 8:44pm On Jun 10, 2013 |
labodinho: Plz help me out.Hw do i solve logs with diff base like this No 1 Log9 base 3-Log3 base 27+ Log9 base sqaure root3.........No 2.Solve the equatn LogX base 2-Log(X-1) base 2=2.........This one is on Inequalities.No 3. (2x-1)(x+4) divided by x-5 both numerator n denominator Greater than(sign) 0 ....Tanx.here comes the answer... (1) log9-log3+log9 since they are in diff base just represent it with letter so we have A-B-C so lets solve A... Log9base3=a, then let the base have a as its power then we have 9=3^a, 3^2=3^a equate the base we have a=2.... Then lets take B... Log3base27=b, 3=27^b, 3^1=3^3b equ base, we have b=1/3.... Now C..... Log9basert3=c, 9= 3^c/2 we have 3^2=3^c/3 equ base we have 2=c/2 cross mult. We c=4 then remenber A-B+C = 2-1/3+4=17/3.... 2nd question (logx)base2-(logx-1)base2=2... According to the rule take a log(x/(x-1))=2, x/(x-1)=2^2, x/(x-1)=4 cross multiply you have x=4x-4, x-4x=4, -3x=-4, x=4/3...... No3 solving it quadratically we have x>1/2 or x>-4 |
Re: Nairaland Mathematics Clinic by echibuzor: 9:00pm On Jun 10, 2013 |
oladoya: here comes the answer... (1) log9-log3+log9 since they are in diff base just represent it with letter so we have A-B-C so lets solve A... Log9base3=a, then let the base have a as its power then we have 9=3^a, 3^2=3^a equate the base we have a=2.... Then lets take B... Log3base27=b, 3=27^b, 3^1=3^3b equ base, we have b=1/3.... Now C..... Log9basert3=c, 9= 3^c/2 we have 3^2=3^c/3 equ base we have 2=c/2 cross mult. We c=4 then remenber A-B+C = 2-1/3+4=17/3.... 2nd question (logx)base2-(logx-1)base2=2... According to the rule take a log(x/(x-1))=2, x/(x-1)=2^2, x/(x-1)=4 cross multiply you have x=4x-4, x-4x=4, -3x=-4, x=4/3...... No3 solving it quadratically we have x>1/2 or x>-4. . Good man. but the problem here i that if I dont a good knowledge of the subject, it will look like jibberish to me.... I think Seun has to upgrade the Text Editor to one that can handle complex typefaces... |
Re: Nairaland Mathematics Clinic by rashywire: 9:04pm On Jun 10, 2013 |
Log(x sq + 7x - 2) base 2 equal to Log(x sq + 3x -6) base 2 + Log8 base 4 |
Re: Nairaland Mathematics Clinic by rashywire: 9:09pm On Jun 10, 2013 |
rashywire: Log(x sq + 7x - 2) base 2 equal to Log(x sq + 3x -6) base 2 + Log8 base 4pls urgent answer |
Re: Nairaland Mathematics Clinic by labodinho: 9:15pm On Jun 10, 2013 |
echibuzor:sir,i got the mail.Tanx for the help.The answers were not in d options buh if datz d procedure for working ,there's no probs.No 1 optionz (A)6(1/3)(B)5(1/2)(C)9 (D)5(2/3).....No. 2 (A)2 (B)1(1/3)(C)1(1/99)(D)No solution....No. 3....No options |
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