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Nairaland Forum / Nairaland / General / Education / Nairaland Mathematics Clinic (493980 Views)
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 badmus45: 5:21am On Feb 09, 2015 |
those smileys are -8 respectivelyi.e (18,- ![]() ![]() |
Re: Nairaland Mathematics Clinic by makasimatics(m): 8:49am On Feb 09, 2015 |
Youngsage: Explain further on how you applied the elimination method |
Re: Nairaland Mathematics Clinic by makasimatics(m): 9:08am On Feb 09, 2015 |
Richiez: Yours in more complex for leaners |
Re: Nairaland Mathematics Clinic by makasimatics(m): 9:15am On Feb 09, 2015 |
Who can prove the almighty formula using this magic formula? F'(x)=±sqrt(f'(x)^2 -4f(x)) This formula derives the quadratic formula. Do you know? @ Richez |
Re: Nairaland Mathematics Clinic by Profmaojo: 6:30pm On Feb 11, 2015 |
Limit n tend to infinity square root of n^2 +n _n =1/2 note _n is not in d square root |
Re: Nairaland Mathematics Clinic by Profmaojo: 6:31pm On Feb 11, 2015 |
Find d supremum and infimum of n-1/2n |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:11pm On Feb 11, 2015 |
Profmaojo: snap & post ..not clear |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:12pm On Feb 11, 2015 |
makasimatics: derive it let's see asap.... |
Re: Nairaland Mathematics Clinic by rhydex247(m): 2:29am On Feb 12, 2015 |
Which textbook can it be of help with this courses 1. Fluid mechanics 2. Mathematical modelling. |
Re: Nairaland Mathematics Clinic by efficiencie(m): 12:26am On Feb 14, 2015 |
Arithmetic: ∫(1/lnt)dt =∫(1/lnt)(t.dt/t) =∫(t/lnt)dlnt =∫(e^lnt/lnt)dlnt Change of variable! Put lnt=x =∫(e^x/x)dx =∫(1/x+1+x/2!+x^2/3!+x^3/4!...) dx =lnx+x+x^2/2.2!+x^3/3.3!+x^4/4.4!+...+x^n/n.n!+... now put x=lnt =ln(lnt)+lnt+(lnt)^2/2.2!+(lnt)^3/3.3!+(lnt)^4/4.4!+...+(lnt)^n/n.n!+... I dey laff o! Shey i nevr jonezz |
Re: Nairaland Mathematics Clinic by efficiencie(m): 12:27am On Feb 14, 2015 |
Arithmetic: ∫(e^x/x)dx =∫(1/x+1+x/2!+x^2/3!+x^3/4!...) dx =lnx+x+x^2/2.2!+x^3/3.3!+x^4/4.4!+...+x^n/ n.n!+... |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:00am On Feb 14, 2015 |
efficiencie: seems you're on track. series's cool. |
Re: Nairaland Mathematics Clinic by Laplacian(m): 6:25pm On Feb 15, 2015 |
Find d area of d region between d curves; y=x^4 & y=2x-x^2. First we find their point of intersection and ensure that none of d two curves crosses d x-axis bewteen dis pionts of intersection. x^4=2x-x^2 or (x^3+x-2)x=0. Hence, x=0, x=1 are d only real solutions. Equatin each of d given eqns to zero, we obtain; x^4=0 i.e x=0. Then, 2x-x^2=0, x=0 or 2. Hence, none of d two equations crosses d x-axis between x=0 and x=1. So, area bounded by y=x^4 is; A1=§x^4dx=x^5/5 limit is from 0 to 1. So, A1=1/5. Area bounded by y=2x-x^2 is, A2=§(2x-x^2)dx=x^2-x^3/3 limit is from 0 to 1. Hence, A2=2/3. So area between d two curves; A=A2-A1=2/3-1/5=7/15 units |
Re: Nairaland Mathematics Clinic by Laplacian(m): 6:45pm On Feb 15, 2015 |
Find d eqn of d cicle which passes thru d point (1,1), half a radius of sqr(10)/2 and whose centre lies on d line y=3x-7. If d centre passes thru d given line, then d co-ordinate of d centre must satisfy d line. Let d coordinate of d centre b (a,b), then; b=3a-7, the radius r=sqr(10) and d eqn of d circle is; (a-1)^2+(b-1)^2=10 or substitutin for b gives (a-1)^2+(3a - 8 ) ^2=10 put a-1=z, then, z^2+(3z-5)^2=10 or 10z^2-30z+15=0 or 2z^2-6z+3=0 z=[3(+ or -)sqr(3)]/2. From z=a-1, get a & from b=3a-7, get b. Substitute for a and b in d eqn of d circle above |
Re: Nairaland Mathematics Clinic by Laplacian(m): 7:05pm On Feb 15, 2015 |
What must be d value of k in order that; 2x+y-3=0, kx+3y+1=0, x+y+7=0 may meet in a point. Discuss d cases when k=3 and k=6 d point of intersection of d first and last lines are obtained first. Subtractin their eqns give; x=10 and y=-17. For the second line to satisfy this point, we must have; 10k-3(17)+1=0, or k=52/10. (i)caseI; if k=3, we have, 3x+3y+1=0 or by dividing thru by 3; x+y+1/3=0. Hence the line is parallel to the third line (since their slope is the same) and is higher the third line (and also the point of intersection) since the intercept is greater. (ii) caseII; if k=6, then, 6x+3y+1=0 or by dividing by 3, 2x+y+1/3=0. Again, this line is parallel to the first line and is lower, below it. |
Re: Nairaland Mathematics Clinic by Laplacian(m): 7:26pm On Feb 15, 2015 |
Find the eqns of d sides of d triangle ABC where A, B, C are d points (5,7), (3,3), (7,1) respectively. Hence, show that the triangles ABC has angles of 90, 45 and 45. Verify this results by finding the lengths of d triangle. Slope of line AB is (3-7)/(3-5)=2 eqn of line AB is y=2x-3 slope of line BC is (1-3)/(7-3)=-1/2 eqn of line BC is y=-x/2+9/2 slope of line AC is (1-7)/(7-5)=-3 eqn of line AC is y=-3x+22 now, two lines are perpendicular if the product of their slope is -1. Multiplyin slope of AB and BC we get; 2*-1/2=-1. Now, d angle between two lines is found from; tan@=(m1-m2)/(1+m1*m2), where m1 and m2 are the slope of d lines. I dont have calculator, complete the rest. |
Re: Nairaland Mathematics Clinic by Laplacian(m): 8:00pm On Feb 15, 2015 |
D length of d tangent from d point (3,2) to d circle x^2 +y^2-2x-3y+k=0 is 9 units, find k. Centre of d circle, C=(1,3/2). Length of of given point from center L^2=(3-1)^2+(2-3/2)^2=17/4. Since d the system forms a pythagorean triangle, the radius is found from; r^2=17/4-9^2. This shows that the circle does not have a REAL radius, and so, k cannot be real |
Re: Nairaland Mathematics Clinic by Laplacian(m): 8:09pm On Feb 15, 2015 |
Find the equation and radius of d second circle of the triangle formed by d three lines; 2y-9x+26=0, 9y+2x+32=0, 11y-7x-27=0. I dont understand this questio. Is it the circumcircle we are lookin 4? or the inscribed circle? |
Re: Nairaland Mathematics Clinic by emmyeuler1: 9:42pm On Feb 15, 2015 |
Laplacian:it is not second circle oooo......it is circumcircle i meant |
Re: Nairaland Mathematics Clinic by Laplacian(m): 9:57pm On Feb 15, 2015 |
emmyeuler1:k. Solve three equations choosing two at a time and u'll get the points of intersections. Substitute each points of intersection in the general equation of the circle; x^2+y^2+2ax+2by+c=0, and solve the three simultaneous equations for a, b and c.....its too bulky 2 solve without calculator... |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 7:54am On Feb 16, 2015 |
hmmmm see mathematicians a beg.... @sir laplacian & prof. emmyeuler. I greet thee in CAPITAL letters... |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 7:59am On Feb 16, 2015 |
here's a lit2 problem that is becoming a problem to the problem solvers ... try out 1. Find the value of x,y and z if x^2-yz=1, y^2-xz=4, and z^2-xy=9. anticipating. |
Re: Nairaland Mathematics Clinic by yemstok(m): 8:06am On Feb 16, 2015 |
agentofchange1: Here 1. Integral 7x-4/2x^2-3x-2 dx 2. Integral x^3/x+1 dx 3. Integral cosx/7+sinx dx = In (7+sinx)+C 4. Integrate with respect to X A. Sinxcosx B. Sin6x C. X^2 e^3 5. Y= sin^-1 (2x-1) b. Y=tan^-1 (1+x/1-x) |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:08am On Feb 16, 2015 |
yemstok: am not with pen/paper now. Q3 put. u=7+sinx du=cosxdx dx= du/sinx thus => $cosx/u *du/cosx = $du/u = ln (u) +C hence we have ln(7+sinx) +C . ...(as expected) Q4. A $sinxcosxdx set u= sinx dx=du/cosx => $u*sinx*du/sinx = >u^2 /2 + C = 0.5sin^2 x + C . B. similarly let t= 6x dt= 6dx dx= dt/6 =>1/6 $sint .dt = -1/6 cost + C = -1/6 cos6x + C . C) guess the question is like this x^2 e^(3x) let's use integration by parts (I.B.P) put u=x^2 ; du =2xdx dv=e^(3x) dx ; v=e^(3x) /3 by $udv =u.v -$vdu. , we have x^2 e^(3x) /3 -2/3 $ x e^(3x) dx..............(*) again let's compute $xe^(3x)dx from. (*) above set p= x. ; dp = dx dq=e^(3x)dx ; q=e^(3x) /3 by $pdq = p.q -$qdp , we have xe^(3x) /3 - 1/3 $ e^(3x) dx =>xe^(3x) /3. -e^(3x) /9 ..........(**) putting (**) in (*) to get the final ans.. =>x^2 e^(3x) /3 -2/3 [ xe^(3x) /3 -e^(3x) /9 ]. + C simply. further.. rest solution coming later... m not with pen/paper ...will use partial fraction for 1 &2 . check Q5. are we to integrate or differentiate .? |
Re: Nairaland Mathematics Clinic by bolkay47(m): 4:03pm On Feb 16, 2015 |
yemstok:no 1::: 3ln(2x+1)+2ln(x-2)+K |
Re: Nairaland Mathematics Clinic by yemstok(m): 4:39pm On Feb 16, 2015 |
agentofchange1: Thanks bro Q5 is integration |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 5:21pm On Feb 16, 2015 |
bolkay47: yea man you try just a li2 correction (the bolded its 3/2. not 3 ) nice 1 try post full solutions next time 1luv ..happy solving. |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 5:24pm On Feb 16, 2015 |
yemstok: OK bro... will post solutions asap.. but wait oo..how much you go pay me for. burning my MB....? lolz. ( just kidding ..never mind ) |
Re: Nairaland Mathematics Clinic by yemstok(m): 5:57pm On Feb 16, 2015 |
agentofchange1: Oya mail me your number make I give you small change. |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 6:45pm On Feb 16, 2015 |
yemstok: Q5 A) ...let p=sin^-1 (2x-1) Sinp=(2x-1) Cosp dp=2dx dx=(cosp dp)/2 ∫sin^-1 (2x-1)dx= ∫p(cosp dp)/2= 1/2∫pcosp dp Using integration by parts U=p du=dp dv=cosp v=sinp 1/2(p sinp- ∫sinp dp) 1/2(p sinp+cosp)+k But p=sin^-1 (2x-1), sinp=(2x-1), cosp= √(1-sin^2p)= √(1-(2x-1)^2)= 2√(x-x^2) 1/2((2x-1)sin^-1 (2x-1)+2√(x-x^2)) 1/2(2x-1)sin^-1 (2x-1)+√(x-x^2)+k B) ans = xarctan[(1+x)/(1-x)] - 1/2 ln | 1+x2 | + C
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Re: Nairaland Mathematics Clinic by tobillionaire(m): 7:45pm On Feb 16, 2015 |
Given that CosZ =L, where Z is an acute angle, find an expression for (CotZ - CosecZ )/(SecZ +TanZ ) 2.Find there numbers in G.P whose sum is 28 and whose product is 512. 3. Find the derivative of Y with respect to X If Y=√{(1+ x )/(1- x ) Where √=square root |
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