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Re: Nairaland Mathematics Clinic by Oxide65: 1:07pm On Aug 14, 2014 |
Arithmetic: Question: |
Re: Nairaland Mathematics Clinic by Oxide65: 1:09pm On Aug 14, 2014 |
Arithmetic: Question: I'm trying to send d solution But It's not appearing |
Re: Nairaland Mathematics Clinic by Oxide65: 1:13pm On Aug 14, 2014 |
Arithmetic: Question:
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Re: Nairaland Mathematics Clinic by Nobody: 1:13pm On Aug 14, 2014 |
benbuks: Reasons why some questions on this forum might not be /are not solved includes For me, helping out with solution to questions on this thread became problematic the moment I changed to android. Unlike my former qwerty phone, typing out solution to problems with this one takes too much time. |
Re: Nairaland Mathematics Clinic by 1Book(m): 2:46pm On Aug 14, 2014 |
[quote author=Oxide65][/quote] ...Just upload the final answer where y=? so that it will be clear. |
Re: Nairaland Mathematics Clinic by Oxide65: 3:17pm On Aug 14, 2014 |
1Book: y=cosx (1+cosx ) |
Re: Nairaland Mathematics Clinic by Oxide65: 3:17pm On Aug 14, 2014 |
1Book: y=cosx (1+cosx ) That's what I found After solving it. |
Re: Nairaland Mathematics Clinic by Oxide65: 3:27pm On Aug 14, 2014 |
1Book: y=cosx (1+Acosx ) |
Re: Nairaland Mathematics Clinic by Nobody: 11:49am On Aug 15, 2014 |
hmmmm. |
Re: Nairaland Mathematics Clinic by Nobody: 11:50am On Aug 15, 2014 |
Tolzeal: Help Gurus At Home: hmm seems...like just four numbers can do d job . .leme try sha... |
Re: Nairaland Mathematics Clinic by Nobody: 11:54am On Aug 15, 2014 |
Arithmetic: x is a very small value which ranges from 0 to infinity. Hence x approaches infinity. x=9/4 prove by solution if u can bro... i think newton-raphson cud b d guy 2 call .. what thinketh thou .? |
Re: Nairaland Mathematics Clinic by Nobody: 12:09pm On Aug 15, 2014 |
smurfy:. ok bro u can also try when chanced ...m also using Android ... greetings sir.. |
Re: Nairaland Mathematics Clinic by Arithmetic(m): 3:38pm On Aug 15, 2014 |
benbuks:Oh!, I'm very sorry sir, I made a silly mistake along the way. :::SOLUTION::: 18¥x=729¥x-1. Let a=¥x, we have; NB:¥ rep. square root. 18a=729a-1, =>32.2a=36(a-1). =>2a=3{6(a-1)-2}. =>2a=3(6a-6-2). =>3{6a-8}-2a=0. Using Binomial Expansion, we have; (1+2)6a-8-2a=0. =>1+2{6a-8}+22.{6a-8}(6a-9)/2+...-2a=0. =>1+12a-16+2(36a2-102a+72)-2a=0. =>1+12a-16+72a2-204a+144-2a=0. On simplifying, we have; =>72a2-194a+129=0. Solving the affected quadratic eqn. =>a=1.5 or 1.194444. =>a=3/2 or 43/36. We take the value of a=3/2, which satisfies the equation. Recall a=¥x, =>x=a2. ...x=(3/2)2=9/4. Newton Raphon's iterative method will also do.#stillyourboy |
Re: Nairaland Mathematics Clinic by Drniyi4u(m): 4:03pm On Aug 15, 2014 |
Tolzeal: Help Gurus At Home:I guess u mean we should take 9 numbers(some may be repeated) from d set of numbers that can sum up to 30, m I right? |
Re: Nairaland Mathematics Clinic by dejt4u(m): 4:14pm On Aug 15, 2014 |
Tolzeal: Help Gurus At Home:after spending minutes on this question, i concluded within myself dat it is not possible.. There is no way you will add odd numbers together in odd numbers of way and you will get even number..it is jst nt possible QED |
Re: Nairaland Mathematics Clinic by Soneh(m): 6:05pm On Aug 16, 2014 |
if|x y z | | a b c | =5 | 1 2 3 | use properties of determinants to find |1 2 9. | |a b 3c | |x y 3z | pls I need the solution with explanations |
Re: Nairaland Mathematics Clinic by efficiencie(m): 9:13pm On Aug 16, 2014 |
Soneh: if|x y z | Let the matrix A with element a(ij) be defined as: A=[row(1)(a(11) a(12) a(13) ...a(1m), row(2)(a(21) a(22) a(23) ...a(2m), ... row(n)(a(n1) a(n2) a(n3) ...a(nm)] Hence if d determinant: |row(1)(x y z), row(2)(a b c), row(3)(1 2 3)|=5 Then |row(1)(1 2 9), row(2)(a b 3c), row(3)(x y 3z)| Multiple of a column =3. |row(1)(1 2 3), row(2)(a b c), row(3)(x y z)| Interchange of rows =-3. |row(1)(x y z), row(2)(a b c), row(3)(1 2 3)| =-3(5) =-15 hence |row(1)(1 2 9), row(2)(a b 3c), row(3)(x y 3z)|=-15 I hope i'm ryt!!! |
Re: Nairaland Mathematics Clinic by Soneh(m): 12:14pm On Aug 17, 2014 |
efficiencie:pls explain better, i'm preparing for an exam |
Re: Nairaland Mathematics Clinic by efficiencie(m): 12:28pm On Aug 17, 2014 |
Soneh: the determinant of a matrix A with row elements r(.j)=k×a(.j) or column elements c(i.)=k×a(i.) equals the product 'k' and the matrix with elements a(.j) or a(i.) respectively. hence |1 2 9| |a b 3c| |x y 3z| equals 3 multiplied by d determinant |1 2 3| |a b c| |x y z| it's lik factoring 3 out of d 3rd column The second move involves interchanging d 1st n 3rd rows and d@ renders the determinant negative. |
Re: Nairaland Mathematics Clinic by Nobody: 11:00am On Aug 18, 2014 |
dejt4u: yea bro.....this very enigma had made me do some crazy things with numbers & algebra . @ a glance of it , one could conclude that , its mathematically "impossible " to sum odd integers , odd number of times to give an even number & vice versa . however , for some infinitesimal or infinitude reasons , i'm kinda reluctant to also accept the notion . (maybe am not crazy enough yet ..lolz. ) .. its well. |
Re: Nairaland Mathematics Clinic by Nobody: 3:39pm On Aug 18, 2014 |
Two coins are biased in such a way so that when they are both tossed once: i) the probability of getting two heads is the same as the probability of getting two tails ii) the probability of getting one head and one tail is 5/8 For each coin, what is the probability of getting a head? |
Re: Nairaland Mathematics Clinic by Soneh(m): 9:49pm On Aug 19, 2014 |
pls gurus in the house should help with this: write in standard form, (a)3i^105+5i^73-8i^50-2i^119 (b) (3i-5)^4 -(2-I)^4 PLS I will appreciate if the solutions are accompanied with explanation |
Re: Nairaland Mathematics Clinic by Oxide65: 12:48pm On Aug 20, 2014 |
Guys Please I'm trying to prove that lim(SQROOT(n^2 +n)-n)=1/2 But the estimates I'm making aren't moving well. Please can someone help?.Even After derationalizing I still got stuck. I need help Please.
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Re: Nairaland Mathematics Clinic by Nobody: 1:28pm On Aug 20, 2014 |
Oxide65: Guys Please I'm trying to prove that lim(SQROOT(n^2 +n)-n)=1/2 But the estimates I'm making aren't moving well. Please can someone help?.Even After derationalizing I still got stuck. I need help Please. as n --> ?... |
Re: Nairaland Mathematics Clinic by Nobody: 2:08pm On Aug 20, 2014 |
Soneh: pls gurus in the house should help with this: --------SOLUTION-------------- (a) = 3(i104 )i + 5(i72 )i -8(i2 )25 -2(i118 ) i =>3i(1) +5i(1) - 8(-1) -2i(-1) = 10i +8 (b) by difference of two squares. we have [(3i-5)2 ]2 - [(2-i)2 ]2 =>(9i2 -30i+25)2 -(4-4i+i2)2 =>(16-30i)2 - (3-4i)2 => (256-960i +900i2 )-(9-24i+16i2 ) =>-544-960i +7 +24i hence we have , -537 -936i |
Re: Nairaland Mathematics Clinic by Oxide65: 2:50pm On Aug 20, 2014 |
benbuks: as n--->.infinity. Thanks for your time. |
Re: Nairaland Mathematics Clinic by Soneh(m): 3:01pm On Aug 20, 2014 |
benbuks:pls I don't get the (a) part, what happened to the powers. thanks |
Re: Nairaland Mathematics Clinic by Nobody: 3:57pm On Aug 20, 2014 |
Oxide65: ok """""""""SoLuTiOn""""""""""""""" take product by the conjugate expression √(n2+ n ) + n at the numerator & denominator => Lim n-->infinity =[[ √(n2 +n) - n ] [[√(n2 + n) + n ] ] / [√(n2 + n ) + n ] => Lim n-->infinity ( n2 +n - n2 ) / [ √[n 2(1 + 1/n) ] + n ] => simplifying & dividing throughout by 'n' we have 1/√( 1 + 1/n ) + 1 since n>0 ,as n--> infinity we thus have 1/[ √(1+0) +1 ] =1/2 ......... (proved ) |
Re: Nairaland Mathematics Clinic by Oxide65: 5:14pm On Aug 20, 2014 |
benbuks: Very well. Thanks alot. Im very sorry i could have been more specific. Could you try using the €-N argument? I'm sure you must be familiar with it because from what I've seen, I must confess all my mathematical problems died when I found this link. |
Re: Nairaland Mathematics Clinic by Oxide65: 5:16pm On Aug 20, 2014 |
benbuks:Very well. Thanks alot. Im very sorry i could have been more specific. Could you try using the €-N argument? I'm sure you must be familiar with it because from what I've seen, I must confess all my mathematical problems died when I found this link. |
Re: Nairaland Mathematics Clinic by Oxide65: 5:25pm On Aug 20, 2014 |
Soneh: Use the concept of indices.
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Re: Nairaland Mathematics Clinic by Nobody: 10:40pm On Aug 20, 2014 |
Soneh: ok...you"ll have to go back to your basics of complex numbers rationally , negative roots are invalid e.g ± √-1 , ±√-2 ,±√-3 e.t c. solving x2 + 1 =0 ..........(*) produces complex roots , however mathematicians have worked severally on such irrational roots , which later discovered & agreed that solution to say (*) =±√-1 we now use ' i ' or ' j ' = ±√-1 ..........(**) hence solution of (*) ,x = ± i having believed that x=± i =>i2 =-1 .......( from (**) ) i3 = i(i2) =i(-1) =-i i4 = (i2 ) (i2 ) = (-1)(-1)=1 i5 = (i4 ) i = (1)i = i we could observe that powers of ' i ' in . 2 ,6 , 10 ,14 . . . =-1 3 ,7 , 11 ,15 . . . =-i 4 , 8 , 12 , 16 . . . = 1 5 , 9, 13 , 17 , . . . =i & their multiples in that order , gives same result . . |
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