Nairaland Mathematics Clinic - Education (204) - Nairaland
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| Re: Nairaland Mathematics Clinic by Nobody: 6:19pm On Nov 12, 2015 |
ladokuntlad:You didn't solve for n factorial n! --> n(n-1) |
| Re: Nairaland Mathematics Clinic by ladokuntlad(m): 6:24pm On Nov 12, 2015 |
all4naija:u said when n=8 so y should i boda myself expanding n again? |
| Re: Nairaland Mathematics Clinic by Nobody: 6:28pm On Nov 12, 2015 |
ladokuntlad:Yes, you are right. But, n is 8 not the 'n factorial' in this case. Look at it from that logical perspective. |
| Re: Nairaland Mathematics Clinic by bolkay47(m): 6:29pm On Nov 12, 2015 |
ladokuntlad:He's question is rather ambiguous. I made a mistake at first. |
| Re: Nairaland Mathematics Clinic by Nobody: 6:32pm On Nov 12, 2015 |
| Re: Nairaland Mathematics Clinic by Madmathecian(m): 6:40pm On Nov 12, 2015 |
Half of 5 = 3, based on that proportion, what is one third of 10? |
| Re: Nairaland Mathematics Clinic by ladokuntlad(m): 6:46pm On Nov 12, 2015 |
all4naija:You mean the expansion n(n-1)(n-2).... is to (n-8 ) baa? well dat still makes it 8! Or beta stil explain wat u meant by n=8 |
| Re: Nairaland Mathematics Clinic by ladokuntlad(m): 6:50pm On Nov 12, 2015 |
| Re: Nairaland Mathematics Clinic by Nobody: 6:59pm On Nov 12, 2015 |
ladokuntlad:No, it is not. Just like you have the 0! is 1 n in this case is just 8 then look for the n factorial (n!). That means you would have to apply the recursive formula at that point in time. |
| Re: Nairaland Mathematics Clinic by ladokuntlad(m): 7:22pm On Nov 12, 2015 |
all4naija:bt that is still d final ansa using recursive formular 8*(8-1)!=8*7! still giv the same result (9!/3!-8*7!)*1-3! (12*7!-8*7!)-3! 4*7!-3!(is this wat u looking for) ansa still givs 20,154 except u looking for a different approach to d question. i believe we stil av the same result |
| Re: Nairaland Mathematics Clinic by Nobody: 7:43pm On Nov 12, 2015 |
ladokuntlad:I don't think so. Look at the formula and plug in 8 and solve for it. It can't be the same thing. It will look like this n(n-1) --> 8(8-1) --> 8*7 = 56. That means there is no need for 56 factorial or 56! but just 56 because that is now the n!. |
| Re: Nairaland Mathematics Clinic by ladokuntlad(m): 7:46pm On Nov 12, 2015 |
all4naija:pheew sighs out |
| Re: Nairaland Mathematics Clinic by Madmathecian(m): 10:14am On Nov 13, 2015 |
Integrate x/x^2 + 3x + 2 dx |
| Re: Nairaland Mathematics Clinic by killsmith(f): 11:01am On Nov 13, 2015 |
Is this thread limited to calculus o.d.e and the likes?..... ![]() |
| Re: Nairaland Mathematics Clinic by MathsChic(f): 11:12am On Nov 13, 2015 |
Madmathecian:Not clear x/(x^2 + 3x + 2)? |
| Re: Nairaland Mathematics Clinic by inkon: 11:24am On Nov 13, 2015 |
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| Re: Nairaland Mathematics Clinic by Madmathecian(m): 11:58am On Nov 13, 2015 |
MathsChic:Yup |
| Re: Nairaland Mathematics Clinic by ladokuntlad(m): 12:09pm On Nov 13, 2015 |
Madmathecian:using integration by partial fraction gives 2ln(x+2)-ln(x+1)+constant |
| Re: Nairaland Mathematics Clinic by ladokuntlad(m): 12:15pm On Nov 13, 2015 |
my desert for today
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| Re: Nairaland Mathematics Clinic by Madmathecian(m): 9:10pm On Nov 13, 2015 |
Problem 7 is a word problem. |
| Re: Nairaland Mathematics Clinic by Nobody: 9:25pm On Nov 13, 2015 |
killsmith:Not really. |
| Re: Nairaland Mathematics Clinic by Laplacian(m): 10:10pm On Nov 13, 2015 |
ladokuntlad:too tiny |
| Re: Nairaland Mathematics Clinic by Madmathecian(m): 11:04pm On Nov 13, 2015 |
| Re: Nairaland Mathematics Clinic by Madmathecian(m): 2:40am On Nov 14, 2015 |
Choose wisely .
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| Re: Nairaland Mathematics Clinic by Ayomide002: 7:33am On Nov 14, 2015 |
Madmathecian:....x/(x^2+3x+2)..x/(x+1)(x+2)....find the partial fraction...x/(x+1)(X+2)=2/(x+2) - 1/(x+1)......integrate now to get...2ln(x+2)-ln(x+1)+c..... Ln[(x+2)^2/(x+1)]+c. |
| Re: Nairaland Mathematics Clinic by ladokuntlad(m): 10:24am On Nov 14, 2015 |
Laplacian:try use chrome or any oda browser apart from opera mini dats if u one mobile. then pinch to zoom |
| Re: Nairaland Mathematics Clinic by agentofchange1(m): 3:51pm On Nov 14, 2015 |
Hey guys, no solution to my interal question yet ? ![]() |
| Re: Nairaland Mathematics Clinic by agentofchange1(m): 4:00pm On Nov 14, 2015 |
This man is sure on point Nice piece sir, planning/thinking of doing mathematical statistics or statistical modelling or statistical simulations in either Staford , Princeton or Cambridge, any requites to note before embarking on any ? currently and undergrad, with statistics major ( loves anything that involves calculus or D.Es, in solving real problems. tnx in advance masperano: |
| Re: Nairaland Mathematics Clinic by Soneh(m): 5:52pm On Nov 14, 2015 |
please gurus in the house help me 1.EVALUATE the following limits (1)limit as x approaches 0 of 3x-2x/x (2)limit as x approaches 0+ of xsinx (3)limit as x approaches 0 of 1-cos22x/x2 2.show that if aline L has the equation Ax+By=C, then a line M perpendicular to L has an equation of the form -Bx+Ay=E 3.IF x2+y2=a2, find the value of dy/dx (4)Evaluate the following integrals (1)∫x3+2/x4-1 dx (2)∫π/4 0 dx/1+cot2x |
| Re: Nairaland Mathematics Clinic by Nobody: 6:20pm On Nov 14, 2015 |
Soneh:You'll have to use L'Hopital's rule here: Differentiate the numerator then differentiate the denominator (separately, not using the quotient rule) Doing that you'll get: limx->0/[sub] (3xlog(3) - 2xlog(2) )/ 1 Substituting 0 for x gives you log(3/2) (2)limit as x approaches 0+ of xsinxEvaluating directly gives you the indeterminate form 00 so you'll have to apply L'Hopital's Rule. But there's no derivative for xf(x), so you'll have to write it in a form that has a derivative. Use the identity eln(x) = x. Rewrite the question as elnx^sin(x) which is equivalent to eln(x)*sin(x) L'Hopital's rule only works for quotients so you'll have to rewrite it as eln(x)/1/sin)x), or simply as eln(x)/cosec(x) The exponential function has a derivative so we can proceed. However when you substitute 0 for x, ln(0) is undefined. Applying L'Hopital's rule you get: lim[sub]×->0 e-sin[sup]2x / x*cos(x)[/sup] Evaluating directly gives you 0/0 but you can rewrite it as e{-sin^2 ×/x*1/cosx which leaves you with e-sin^2×/x Applying L'Hopital's rule again you get limx->0 e2cos(x)*sin(x) Which gives e0 = 1. |
| Re: Nairaland Mathematics Clinic by agentofchange1(m): 6:22pm On Nov 14, 2015*. Modified: 9:48pm On Nov 14, 2015 |
Soneh:
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| Re: Nairaland Mathematics Clinic by agentofchange1(m): 6:25pm On Nov 14, 2015 |
Q4
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