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Re: Nairaland Mathematics Clinic by donfourier(m): 9:08am On Jan 14, 2015 |
a matrix of 5x5 hmmmmmmmmmmmm...... |
Re: Nairaland Mathematics Clinic by donfourier(m): 9:24am On Jan 14, 2015 |
a positive integer is called tico if it is the product of three different prime number that add up to 74 . verify that 2014 is tico. which year will be the next tico year? which one will be the last tico year in the history? |
Re: Nairaland Mathematics Clinic by donfourier(m): 11:47am On Jan 14, 2015 |
Laplacian: |
Re: Nairaland Mathematics Clinic by Nelsonesq(m): 2:47pm On Jan 14, 2015 |
Mathematicians in the house, Pls solve this The transformation equations from ashy to cartesian coordinates are X=e^u.secvcoshw Y=e^u.tanvcoshw Z=e^usinhw. If f=f(x,y,z) is a differentaible function and H=y.df/dx+x.df/dy+df/dz, derive an appropriate expression for H in ashy coordinates. |
Re: Nairaland Mathematics Clinic by jackpot(f): 7:04am On Jan 15, 2015 |
Laplacian:In an hour, Mr A must have done (1/t_1) times the work Mr B - (1/t_2) times the work so, in an hour, both of them working together must have done (1/t_1+1/t_2) times the work so, working together, they completes the work in 1 / (1/t_1 + 1/t_2) hours. This is the same as t1t2/(t1+t2) hours. @Laplacian and others, please feel free to criticize. |
Re: Nairaland Mathematics Clinic by Nobody: 7:37am On Jan 15, 2015 |
Laplacian: Let the farm work done be 1 farmwork. Hence the rate at which Mr. A works is 1/t1 (Work/hrs). Similarly Mr. B's rate since the quantity of work is similar (1 farm work) is 1/t2. Hence both of them working together would accomplish 1/t3 rate. = 1/t1 + 1/t2 = 1/t3 I'm sure you can take it from there. |
Re: Nairaland Mathematics Clinic by Nelsonesq(m): 9:23am On Jan 15, 2015 |
Richiez:Here is one for you to solve..... Abeg no fall my hand o The transformation equations from ashy to cartesian coordinates are X=e^u.secvcoshw Y=e^u.tanvcoshw Z=e^usinhw. If f=f(x,y,z) is a differentaible function and H=y.df/dx+x.df/dy+df/dz, derive an appropriate expression for H in ashy coordinates. |
Re: Nairaland Mathematics Clinic by Laplacian(m): 11:50am On Jan 15, 2015 |
jackpot:that's pretty!! (i didn't think of dis approach) Scrypt:hahahaha...u pay me back in my coin? (nice work!!) |
Re: Nairaland Mathematics Clinic by donfourier(m): 2:18pm On Jan 15, 2015 |
by using identity ie trigonometric identity |
Re: Nairaland Mathematics Clinic by Kentnickole(m): 3:40pm On Jan 15, 2015 |
Laplacian: Mr A's farm work per hr = 1/t1 Mr B's farm work per hr = 1/t2 A and B working together = 1/t1+ 1/t2 = t1t2/(t2 + t1) Hope this helps. |
Re: Nairaland Mathematics Clinic by Laplacian(m): 8:29pm On Jan 15, 2015 |
Show that; If 2n+1 is a prime, then, n is a power of 2 |
Re: Nairaland Mathematics Clinic by bigwig97(m): 9:03pm On Jan 15, 2015 |
can someone help to solve these questions in d pix..thankx
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Re: Nairaland Mathematics Clinic by naturalwaves: 3:30pm On Jan 16, 2015 |
agentofchange1: Let T, F, O represent the 10kobo,5kobo and 1kobo notes respectively. The probability of picking anyone of them is. 1/3 and pr of not picking anyone is 2/3. Pr ( they are not of the same colour) Implies pr ( TF) or pr (TO)or pr(FO) = (1/3 *1/3) + (1/3*1/3) + (1/3*1/3) = 1/9 + 1/9 +1/9 =3/9 = 1/3. Pr( same value) Implies pr (TT) or pr (FF) or pr (OO) = (1/3*1/3) + (1/3*1/3) + (1/3*1/3) =1/9+1/9+1/9 =3/9 =1/3. |
Re: Nairaland Mathematics Clinic by naturalwaves: 3:44pm On Jan 16, 2015 |
bigwig97:Hahahaha, go and study the bisection method very well and apply it to the question, you don't expect someone to start drawing all those lines and iterations for you. |
Re: Nairaland Mathematics Clinic by Temieasy(m): 6:41pm On Jan 16, 2015 |
bigwig97:xup, guy I solved the bisection method, but when I got to the 10th iteration. but did not include the error tolerance in ur question. I think the answer should b 0.373046875 as d midpoint of the interval of [0, 1] I used. shikena.. would try and upload d pic |
Re: Nairaland Mathematics Clinic by bigwig97(m): 7:47pm On Jan 16, 2015 |
Temieasy:thanks man! dnt bother 2 upload pix av gotten t here also..bt the C++ part 4rm 2nd pix z nt stil clear |
Re: Nairaland Mathematics Clinic by Nobody: 9:20pm On Jan 16, 2015 |
The solution to the question: x+y=5 --------------(1) x^x+y^y=31 --------------(2) therefore by breaking down equation 2 we have x^x+y^y=2^2+3^3. By equating, we have x^x=2^2 and y^y=3^3. So x=2(twice) and y=3(twice) |
Re: Nairaland Mathematics Clinic by Nobody: 9:25pm On Jan 16, 2015 |
Solve this: x^2=2^x, find the value of x |
Re: Nairaland Mathematics Clinic by Temieasy(m): 11:23pm On Jan 16, 2015 |
bigwig97: did u get the same answer I got? as for the c++,am not too good at it just learning |
Re: Nairaland Mathematics Clinic by ameer2: 6:56am On Jan 17, 2015 |
gurus in d house.pls help mee solve dis questions. Find the absolute and relative errors in Y given that the function y=2.0Sin x + 3 ln x to be evaluated for X=1.26. |
Re: Nairaland Mathematics Clinic by bolkay47(m): 7:00am On Jan 17, 2015 |
jgenius:use iteration method:: F(x)=x^2-2^x F“(x)=2x-2^xln2 use x0=0 as starting point. You will get 2.. |
Re: Nairaland Mathematics Clinic by bigwig97(m): 7:14am On Jan 17, 2015 |
Temieasy:i stopped at 4th iteration.. and my ans at dt point z 0.375...xo definately u would b correct. 4 d c++, i wil find how 2 do t sha.. thanks |
Re: Nairaland Mathematics Clinic by mdee1(m): 8:23am On Jan 17, 2015 |
Pls guys help with this assignment . Find the equation of the line whose perpendicular from the origin is 3-units and is at an angle of 40 degree to the + y-axis . . And the second picture, just show that the first answer PQ(up) equals is equal to the second answer PQ(down) equals. thanks in advance
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Re: Nairaland Mathematics Clinic by Nobody: 10:01pm On Jan 17, 2015 |
Adol16:the simplest way of solving it is by breaking down eqn 2.i.e x^x+y^y=2^2+3^3,now by equating we have x^x=2^2 and y^y=3^3.Therefore x=2 and y=3. |
Re: Nairaland Mathematics Clinic by 2nioshine(m): 4:06pm On Jan 18, 2015 |
badmus45: check this out Q1. Hint: Obtain the auxiliary equation and solve the L.H.S ie m2 + 2m + 5=0 m=-1+2i or -1-2i thus the first soln becomes Y(t)=e^-t (Asin2t +B cos 2t) since an expression in sine already appears in the soln the R.H.S is solved as thus. Y=Ctsint +Dtsin # Differentiate the above to the second power and order # substitute values to the actual equation given in the Question #compare coeficient and obtain a general equation in C and D # combine the general solution of the LHS and RHS #Input the given boundry condition Hope it helps................... |
Re: Nairaland Mathematics Clinic by efficiencie(m): 12:58am On Jan 21, 2015 |
efficiencie: the partial derivative of (δx/δy) with respect to 'x' is equal to what? |
Re: Nairaland Mathematics Clinic by rashywire: 6:43am On Jan 21, 2015 |
pls help me with dis question, if (x-y)^3 = A(x+y), prove d@ (2x+y)dy/dx = x+2y source: KA STROUD FURTHER PROBLEM 7 |
Re: Nairaland Mathematics Clinic by efficiencie(m): 9:51am On Jan 21, 2015 |
rashywire: given that: (x-y)^3 = A(x+y), the derivative wrt x is: 3(x-y)^2 (1-dy/dx)=A(1+dy/dx) And since A=(x-y)^3/(x+y) then 3(x-y)^2 (1-dy/dx)=((x-y)^3/(x+y))(1+dy/dx) 3(x+y)(1-dy/dx)=(x-y)(1+dy/dx) 3(x+y)-3(x+y)(dy/dx)=(x-y)+(x-y)(dy/dx) ((y-x)-3(x+y))(dy/dx)=(x-y)-3(x+y) 2(-y-2x)(dy/dx)=2(-x-2y) (2x+y)(dy/dx)=x+2y which is the required relation. 1 Like |
Re: Nairaland Mathematics Clinic by jaryeh(m): 3:51am On Jan 22, 2015 |
Chai! Long time........ I need help on this pls.. tag: jackpot, Alphamaximus, dejt4u, doubledx, Richie and other generals. I don forget una names finish o.
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Re: Nairaland Mathematics Clinic by efficiencie(m): 2:10pm On Jan 22, 2015 |
jaryeh: Given that: arccosa+arccosb+arccosc=π arccosa+arccosb=π-arccosc Take sin of both sides sin(arccosa+arccosb)=sin(π-arccosc) sin(arccosa)cos(arccosb)+ cos(arccosa)sin(arccosb) =sinπ.cos(arccosc)- cosπ.sin(arccosc) √(1-cos^2(arccosa)) cos(arccosb)+ cos(arccosa) √(1-cos^2(arccosb)) =sinπ.cos(arccosc)- cosπ. √(1-cos^2(arccosc)) b√(1-a^2)+a√(1-b^2)=√(1-c^2) Similarly since we can rewrite the equation as: arccosa+arccosc=π-arccosb We have that: c√(1-a^2)+a√(1-c^2)=√(1-b^2) c√(1-a^2)-√(1-b^2)=-a√(1-c^2) on solving for √(1-a^2) in the simultaneous equations below (using Crammer's rule): b√(1-a^2)+a√(1-b^2)=√(1-c^2) ...1 c√(1-a^2)-√(1-b^2)=-a√(1-c^2) ...2 We have that: √(1-a^2)=(-√(1-c^2)+a^2.√(1-c^2))/(-b-ac) √(1-a^2)=(√(1-c^2)-a^2.√(1-c^2)/(b+ac) (b+ac)√(1-a^2)=(1-a^2)√(1-c^2) On squaring both sides: (b+ac)^2.(1-a^2)=(1-a^2)^2.(1-c^2) And dividing thru by (1-a^2) (b+ac)^2=(1-a^2)(1-c^2) b^2+2abc+a^2.c^2=1-a^2-c^2+a^2.c^2 b^2+2abc=1-a^2-c^2 a^2+b^2+c^2 +2abc=1 QED |
Re: Nairaland Mathematics Clinic by bolkay47(m): 2:20pm On Jan 22, 2015 |
With each roll of a cutlass,a man cut 0.1% of grass in a field. if there were 20 square feet of grass to be cut. how much grass will remain after 60th stroke? Please help me out...thanks |
Re: Nairaland Mathematics Clinic by AlphaMaximus(m): 6:34pm On Jan 22, 2015 |
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