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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 Nobody: 1:30pm On Nov 14, 2014 |
I really need help with this question. Two circles touch externally at T. A chord of the first circle XY is produced and touches the other at Z. The chord ZT of the second circle, when produced, cuts the first circle at W. Prove that angle XTW = angle YTZ. Thanks in advance. 1 Like |
Re: Nairaland Mathematics Clinic by benny1620: 3:08pm On Nov 15, 2014 |
HAY guys.....greet thee all its BENBUKS .... THIS's NOW MA NEW ACCOUNT ... NICE SOLVING.. -SHALOM-... |
Re: Nairaland Mathematics Clinic by benny1620: 3:13pm On Nov 15, 2014 |
jackpot: hay "miss" jackport , honey i'm here . anything for the boys . ? |
Re: Nairaland Mathematics Clinic by benny1620: 3:20pm On Nov 15, 2014 |
Calculusfx: hay prof. its "benbuks" greet thee and other generals.. 1luv.. |
Re: Nairaland Mathematics Clinic by Nobody: 4:05pm On Nov 15, 2014 |
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Re: Nairaland Mathematics Clinic by meelorlah(f): 7:47pm On Nov 15, 2014 |
This is d Lord's doing and it is marvelous in my sight....People Are Goooooooodddddd.I throway salute oo |
Re: Nairaland Mathematics Clinic by Faitty20(f): 5:58pm On Nov 16, 2014 |
hello guys...which of this courses would you advice someone to study..pure and applied maths or statistics..pls state some reasons..thanks |
Re: Nairaland Mathematics Clinic by sundaylinus: 11:59pm On Nov 17, 2014 |
pls help with this... Out of 500 families with 5 children each, how many would you expect to have: (a)exactly 2 boys (b)more than 1 boy (c)less than 2 boys roundup ur answer tothe nearest whole number. |
Re: Nairaland Mathematics Clinic by benny1620: 7:21am On Nov 18, 2014 |
sundaylinus: Lets use E(x) = m =Np known that 1= q+p P(boy) =P(girl) =1/2 1/2 =p =q N=500 n=5 by binomial distribution P(X=r) = nCr*q(n-r) * pr a) P(X=r=2) =5C2*(1/2)5 = 10/32 we would expect (500)(10/32) =5000/32 … finish up b) P(r>1) =p(2) +p(3)+p(4) +p(5) = .... finish up then multiply your answer by 500 c) p(r <2) = p(0) +p(1) ..... similar as former hope you get... 2 Likes 1 Share |
Re: Nairaland Mathematics Clinic by sundaylinus: 12:10pm On Nov 18, 2014 |
benny1620:pls. Benny put more light, this topic was introduced to us by this assignment and we are to submit it on 20th pls. |
Re: Nairaland Mathematics Clinic by benny1620: 8:26am On Nov 19, 2014 |
sundaylinus: ok bro , didn't have much time , was in the lecture hall . But go get a probability /statistics text book for details . Now on the question , considering 500 families ( maybe in Lagos ) we're told that each of them have 5 children . could be all boys, or all girls or mixed .Using probability techniques , we want to predict / say on average the number of boys the family could consist of . we use binomial distribution , p= probability of a boy in a family q= probability of NOT a boy ( i.e a girl) since we know that most usually , you're either a boy or a girl ( ignore special cases ) , just like tossing a coin , we either have 'heads' or 'tails' ( i.e only two possible outcomes ) , thus in that case , we have equal probabilities we should know that p+q=1 ...........( from axioms) p=1/2 ........(probability of success ) q= 1 -p = 1- 1/2= 1/2 ....(probability of failure ) now , we see that p=q=1/2 from here go to my initial solution above . follow it carefully and finish up , you don't expect me to do everything for you do you.? or you tell me where you're confused . gracia... 1 Like |
Re: Nairaland Mathematics Clinic by dRector(m): 12:44pm On Nov 22, 2014 |
Good afternoon Gurus in the house . please help with this trigonometry and analytical geometry questions . 1. cosx + cos3x+cos5x+cos7x=0 2. cos2x+cosx+1=0 3.sin30+sin2tita =0 Thanks in advance |
Re: Nairaland Mathematics Clinic by edaloba(m): 7:20am On Nov 29, 2014 |
Good day to u all, it has been a while since I visited this great thread, I am the former 'boladearo' a student of all the maths generals on this thread. |
Re: Nairaland Mathematics Clinic by edaloba(m): 7:42am On Nov 29, 2014 |
Find the quadratic equation which has the difference of its roots to be 2, and difference of the square of its roots to be 5. Soln. X - y = 2 ...............1 X^2 - y ^2 = 5 .....2 X = 2 + y .............3 Subt d value of x into eqtn 2 (2+y)^2 - y^2 = 5 4 + 4y + y^2 - y^2 = 5 4y = 5 - 4 Y = 1/4 . X =9/4 X+y =5/2. xy =9/16 D eqtn is then x^2 - 5/2x + 9\16 16x^2 - 40x + 9 Soln 2 X-y = 2 X^2 - y^2 = 5 (X-y)^2 + 2xy = 5 2^2 + 2xy = 5 2xy = 5 - 4 Xy = 1/2 (X-y)(x+y) = 5 2(x+y) = 5 X + y = 5/2 So equtn is x^2 - 5/2x + 1/2 =2x^2 -5x +1 Pls ma generals in the house why is ma ans not the same. I ought to get the same ans since I followed maths rule in all ma solving . |
Re: Nairaland Mathematics Clinic by mathefaro(m): 8:17am On Nov 29, 2014 |
edaloba:the bold expression is where you got the second solution wrong. x^2 - y^2 =/= (x - y) ^2 + 2xy(not equal) (x - y) ^2 + 2xy = x^2 + y^2 |
Re: Nairaland Mathematics Clinic by edaloba(m): 10:04am On Nov 29, 2014 |
mathefaro:thank you very much sir. Hav seen ma mistake 1 Like |
Re: Nairaland Mathematics Clinic by Nobody: 1:47pm On Nov 29, 2014 |
There's a new smartphone app for solving maths. you can download it here http://www.figureout.com.ng/photomath-download-phone-app-can-solve-mathematics/ |
Re: Nairaland Mathematics Clinic by Soneh(m): 7:03pm On Nov 30, 2014 |
pls the gurus in the house I need yur assistance on this assigment and please I will appreciate it if the solutions are accompanied with explanation : 1.determine the radius and center coordinate of the circle whose equation is 4x^2+4y^2-16x+32y-116=0 2.find the gradient of the tangent to the curve. 3x^2.y^3-5x^3.y+8x=11y^2-3x^3. at(1,1)and (0,-2). 3. evaluate lim. (e^-×_1/x) x--0 4.given f(x)={x-5. X<a .......................{7/x+1. x>=a find 'a' such that f(x) is continuous at. x=a 5.evaluate . (a) ∫e^-x sin3xdx. (b)∫^2. x^-2in(2x)dx ....5 (pls note that the 5 is an underscore on the integration sign) now over to my bosses.benbuks,jackpot,rhydex,calculus fx,efficiency,laplacian........ |
Re: Nairaland Mathematics Clinic by rhydex247(m): 2:00pm On Dec 01, 2014 |
5.evaluate . (a) ∫e^-x sin3xdx. solution. Let I=∫e^-x sin3xdx. using integration by part. let u=sin3x and dv=e^-x du=3cos3xdx and v=-e^-x UV-$VDU {-e^-x}{sin3x} -$ {-e^-x}{3cos3x}dx. -e^-xsin3x + 3${e^-x}{cos3x}dx similarly, let u=cos3x and dv=e^-x du=-3sin3xdx and v= -e^-x -e^-xsin3x + 3{ uv-$vdu} -e^-xsin3x + 3{(cos3x)( -e^-x) -$ (-e^-x)(-3sin3x)dx} -e^-xsin3x + 3{-e^-xcos3x- 3$(e^-x)(sin3x)dx} recall dat I=∫e^-x sin3xdx. I=-e^-xsin3x + 3{-e^-xcos3x- 3(I)} I=-e^-xsin3x- 3e^-xcos3x-9I 10I=-e^-xsin3x- 3e^-xcos3x I=1/10 (-e^-xsin3x- 3e^-xcos3x) I=-1/10(e^-xsin3x+3e^-xcos3x) |
Re: Nairaland Mathematics Clinic by Soneh(m): 4:14am On Dec 02, 2014 |
@rhydex, thanks for that I will appreciate it if u'll assist me in solving the rest of the questions. still calling on other gurus in the house to cum to my rescue @..bosses.benbuks,jackpot,calculus fx,efficiency,laplacian. |
Re: Nairaland Mathematics Clinic by rhydex247(m): 1:20pm On Dec 03, 2014 |
1.determine the radius and center coordinate of the circle whose equation is 4x^2+4y^2-16x +32y-116=0 Solutn Recall dat d eqn of a circle is given by x^2+y^2+2gx+2fy+c=0........ eqn * 4x^2+4y^2-16x+32y-116=0 (dividing thru by 4) yield x^2+y^2-4x+8y-29=0......... eqn ** by comparing eqn * &** we av g=-2 f=4 c=-29 Now, for centre = (-g,-f)=(2,-4). also radius= √(g^2+f^2-c) =√(-2)^2 +4^2+29= √49=7 clearly, centre= (2,-4) and radius=7. |
Re: Nairaland Mathematics Clinic by rhydex247(m): 1:31pm On Dec 03, 2014 |
3. evaluate lim. (e^-×_1/x) x--0 (b)∫^2. x^-2in(2x)dx ....5 (pls note that the 5 is an underscore on the integration sign) 2.find the gradient of the tangent to the curve. 3x^2.y^3-5x^3.y +8x=11y^2-3x^3. at(1,1)and (0,-2). please i will really appreciate if u can re- write these three questions in an appropriate way. |
Re: Nairaland Mathematics Clinic by 2muchopoTBdope(m): 6:20pm On Dec 03, 2014 |
Soneh:#2. Solving by implicit differentiation >>> 3(x^2 . 3y^2 dy/dx + y^3 . 2x) - 5(x^3 dy/dx + 3x^2 . y) + 8 = 22y dy/dx - 9x^2 Opening the Bracket >>>> 9x^2 . y^2 dy/dx + 6xy^3 - 5x^3dy/dx - 15x^2 . y + 8 = 22y dy/dx - 9x^2 Collect Like terms,then factorize dy/dx out >>>> dy/dx (9x^2 . y^2 - 5x^3 - 22y) = 15x^2 . y - 6xy^3 -8 - 9x^2 >>>> dy/dx = (15yx^2 - 6xy^3 - 8 - 9x^2)/(9x^2 . y^2 - 5x^3 - 22y) >>>> For the first cordinate (1,1) ,Substituting x = 1 and y =1 into the above equation dy/dx = 2/9. >>>> For the 2nd coordinate (0, -2), substituting x = 0 and y = -2 into the same equation dy/dx = -2/11. |
Re: Nairaland Mathematics Clinic by 2muchopoTBdope(m): 6:41pm On Dec 03, 2014 |
rhydex247:Another way, so u ve a choice The equation of a circle cud be written as (x - a)^2 + (y -b)^2 = r^2 The idea is trying to reduce the equation u were given to look like the general equation of a circle,so u cud relate them,then find the centre and radius. 4x^2 + 4y^2 - 16x + 32y = 116 Divide through by 4 x^2 + y^2 - 4x + 8y = 29 x^2 - 4x + y^2 + 8y = 29 Using completing the square method to simplify (x^2 -4x + 4) + (y^2 + 8y + 16) = 29 +4 + 16 (x - 2)^2 + (y + 4)^2 = 49 Relating the above equation to the equation of a circle a = 2 b = -4 r^2 = 49 r = √49 r = 7 Centre (a,b) = (2 , -4) Radius (r) = 7. |
Re: Nairaland Mathematics Clinic by FuckYou: 12:36am On Dec 06, 2014 |
Please help!! Angle of elevation: -an engineer wishes to bridge a river.he observe that the angle of elevation of the top of a tree on the far bank is 27degree from a point on the near bank and 14degree from a point 30m further back.. How wide was the river Thanks (2)from point P on level ground the angle of elevation of the top of a tower is 26degree 51'(min),from a point 25m closer to the tower and on the same line with p and the base of the tower of the angle of elevation of the top is 53degree 30min.... Calculate to three significant figures the height of the tower.. Please help me drawing(photo) will be highly appreciated. Thanks |
Re: Nairaland Mathematics Clinic by Soneh(m): 7:15am On Dec 06, 2014 |
rhydex247:Sorry for replying late, hope the modifications gives you a clearer understanding of the questions |
Re: Nairaland Mathematics Clinic by jackpot(f): 9:28am On Dec 06, 2014 |
FuckYou:Hello, Mr FuckYou! |
Re: Nairaland Mathematics Clinic by jackpot(f): 9:48am On Dec 06, 2014 |
Soneh:baby question. Already answered. 2.find the gradient of the tangent to the curve. 3x^2.y^3-5x^3.y+8x=11y^2-3x^3. at(1,1)and (0,-2).wrongly posed question since the points (1, 1) and (0, -2) doesnt even lie on the given curve. 3. evaluatenot clear. 4.given f(x)={x-5. X<asolve x-5= 7/x + 1 to get the value(s) of a. 5.evaluate . (a) ∫e^-x sin3xdx.a is baby. For b, express x^-2 ln(2x) as (ln 2)/(x^2) + (ln x)/(x^2) and go from there. |
Re: Nairaland Mathematics Clinic by tuwayz(m): 11:07am On Dec 06, 2014 |
Pls help me out on this, showing out clear workings |
Re: Nairaland Mathematics Clinic by tuwayz(m): 11:10am On Dec 06, 2014 |
Pls help me out on this, showing out clear workings (1)Calculate the mid point of the line of segment. Y - 4X + 3 = 0 which lies between the X-axis and Y-axis (2)Find the equation of the straight line through (-2,3) and perpendicular to 4X + 3Y - 5 = 0 |
Re: Nairaland Mathematics Clinic by Miscellaneous(m): 8:22pm On Dec 06, 2014 |
jackpot: this girl... I just don't know wat to say |
Re: Nairaland Mathematics Clinic by Miscellaneous(m): 8:41pm On Dec 06, 2014 |
tuwayz: 1. Given: Y-4x+3=0 when y=0, 0-4x+3=0 -4x=-3 =>>> x=3/4 the point is thus =(3/4,0) when x=0, y-4(0)+3=0 y =-3 =>>> y=-3 thus the point is =(0, -3) but, midpoint=(x1+x2)/2, (y1+y2)/2 for x, (3/4+0)/2=3/8 for y , (-3+0)/2 =-3/2 >>> therefore, midpoint is (x,y) = (3/8,-3/2) 2. Given 4X + 3Y - 5 = 0 we proceed by putting it in gradient intercept form i.e , Y= -4/3X +5/3 for 2 lines to be perpendicular, m1 * m2 = -1 thus, -4/3 * 3/4 will = -1 so new gradient for perpendicular will be 3/4 from formula for gradient, (y-y1)/(x-x1) = m2 so, 4(y-3) = 3(x+2) 4y -12= 3x+6 =>>>>>> 4y=3x+18 |
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