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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 johnpaul1101(m): 6:46pm On Jan 16, 2013 |
Richiez:ok richiez, i'm reliefed now, but i will try 2 reduce d question, but i hav one problem;logical reasoning.....wen it comes 2 logical questions such as WORD PROBLEM LEADING TO........it wud b like my brain is locked, i.e.....d incapability of representing an equation from a word question(written question) 1 Like |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 6:48pm On Jan 16, 2013 |
Ghettoguru:thanks |
Re: Nairaland Mathematics Clinic by biolabee(m): 8:00pm On Jan 16, 2013 |
sorry johnpaul seened u were trying to show urself pls ask away dont reduce o johnpaul1101: ok richiez, i'm reliefed now, but i will try 2 reduce d question, but i hav one problem;logical reasoning.....wen it comes 2 logical questions such as WORD PROBLEM LEADING TO........it wud b like my brain is locked, i.e.....d incapability of representing an equation from a word question(written question) Chelcy10: Make x the subject of the formula; STX^3-2WX=10X^2T Chelcy10: Make x the subject of the formula; STX^3-2WX=10X^2Tthis q looks iffy rewrite it again with brckets |
Re: Nairaland Mathematics Clinic by Nobody: 8:35pm On Jan 16, 2013 |
Chelcy10: Make x the subject of the formula; STX^3-2WX=10X^2T STX^3 - 2WX = 10X^2T stx^3 - 2wx - 10x^2t = 0 Lets divide through by x stx^2 - 10tx - 2w = 0 We can now use quadratic formula to solve for x a = st, b = -10t and c = -2w x = [-b±√(b^2 - 4ac)]/2a x = -(-10t) ±√[(-10t)^2 - 4(st)(-2w)]/2(st) x = [10t±√(100t^2 + 8stw)]/2st x = [10t±√4(25t^2 + 2stw)]/2st x = [10t± 2√(25t^2 + 2stw)]/2st x = [5t± √(25t^2 + 2stw)]/st X = [5T ± √(25T^2 + 2STW)]/ST 2 Likes |
Re: Nairaland Mathematics Clinic by Richiez(m): 10:07pm On Jan 16, 2013 |
Johncasey: Evaluate d cube root of (10+6*sqrt of 3) 3√(10+ 6√3) the above can be re-written as; (10+ 6√3)^1/3 = 10^1/3[1+ 0.6√3]^1/3 for fractional index, the binomial series is given by; (1+x)^1/n = 1 + x/n + 1/n(1/n -1)x^2/2! + 1/n(1/n -1)(1/n -2)x^3/3!+.......... Note that more accuracy is attained when more terms in the series are used in our case, 1/n=1/3 and x=0.6√3 2.154[ 1 + (1/3)(0.6√3) + 1/3(1/3 -1)(0.6√3)^2/2! + 1/3(1/3 -1)(1/3 -2)(0.6√3)^3/3!] 2.154[ 1 + 0.346 - 0.12 + 0.069 ] = 2.789 (approx) remember due to approximations, and the fact that we used only the first three terms, this result will only be approximate to the calculator value...but this is the best method to use in the absence of a calculator thanks 2 Likes |
Re: Nairaland Mathematics Clinic by Johncasey(m): 12:11am On Jan 17, 2013 |
Leav d ansa in a surd form |
Re: Nairaland Mathematics Clinic by Fetus(m): 1:35am On Jan 17, 2013 |
Richiez:....Yepa! I doff my hat 4 u oooo.....tankz so much man... |
Re: Nairaland Mathematics Clinic by Fetus(m): 1:41am On Jan 17, 2013 |
Richiez:....kindly help with dis...simplify tan^2(1/4¥ - 1/2€)....where ¥ and € are pie and teetha respectively....tanks |
Re: Nairaland Mathematics Clinic by Nobody: 6:58am On Jan 17, 2013 |
Fetus: ....kindly help with dis...simplify tan2(1/4¥ - 1/2€)....where ¥ and € are pie and teetha respectively....tanks tan2(π/4 - Ө/2) => Lets put A = (π/4) and B = (Ө/2) Since tan (A - B ) = (tan A + tan B )/(1 + tan A tan B ) => tan trig identity for difference of two angles. tan2(A - B ) = [(tan A + tan B )/(1 + tan A tan B )]×[(tan A + tan B )/(1 + tan A tan B ) = (tan A + tanB )2/ (1 + tanAtanB )2 Remember that : A = (π/4) and B = (Ө/2), substituting back yields => = [tan(π/4) + tan(Ө/2)]2/[1 + tan(π/4)tan(Ө/2)]2 Now, since we know that tan(π/4) = sin(π/4)/cos(π/4) = 1/√2 ÷ 1/√2 = 1, substituting back yields => = [1 + tan(Ө/2)]2/[1 + (1)tan(Ө/2)]2 = = 1 :. tan2(π/4 - Ө/2) => 1 2 Likes |
Re: Nairaland Mathematics Clinic by Nobody: 7:15am On Jan 17, 2013 |
Please for those of you posting advanced maths questions, I believe you can handle the simple ones when we are not online. So, please do not ignore questions you know you will be able to solve if they are posted and no one is online to solve them. At least that way, others who are not as advanced are you are can learn from you too! Lets keep the thread alive people. Good morning all! 2 Likes |
Re: Nairaland Mathematics Clinic by Nobody: 7:48am On Jan 17, 2013 |
TOSIN ACCA: I have tried solving it yet no solution.I stopped at this point: a-ac+or -sqrt(4a^2c^2-8a^2c+8rc). I guess there is no solution to it. Even the past questions booklet said so thanks for your help. I'll be back with a handful of questions tomorrow.God loves you. The bolded part is not entirely correct. Biolabee did a good job with your question. Maybe you did not solve and simplify the quadratic equation properly. However, if the answer isn't among the options then they prolly made some typo-errors or intentionally wanted to waste people's time on that question! To save time, it's advisable to move on to the next question if you are sure of your workings and there is no option for it or you can't handle it as you are expected to answer each question in 45secs. Jamb tricks for you! 2 Likes |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 10:20am On Jan 17, 2013 |
plz help me with this: in a two-digit number, the sum of the digits is 8, the difference between this number and the number with the digits reversed is 54. What is this number? |
Re: Nairaland Mathematics Clinic by ositadima1(m): 11:08am On Jan 17, 2013 |
johnpaul1101: plz help me with this: in a two-digit number, the sum of the digits is 8, the difference between this number and the number with the digits reversed is 54. What is this number? 1) in a two-digit number... "x" and "y" 2) the sum of the digits is 8... x+y=8 3) the difference between this number and the number with the digits reversed is 54... [10y+x]-[10x+y]=54 Now, 10y+x-10x-y=54 9y-9x=54 y-x=6 x+y=8 -x+y=6 add these two, 2y=14 y=7 x=1 What is this number?... 71 2 Likes |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 11:17am On Jan 17, 2013 |
ositadima1: |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 11:18am On Jan 17, 2013 |
[quote author=johnpaul1101][/quote]thanks a lot |
Re: Nairaland Mathematics Clinic by Nobody: 11:20am On Jan 17, 2013 |
Find the value of the lettered angles in the following diagram: Hint: /MN/ and /SK/ are straight lines. O is the centre of the circle. Find X and T
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Re: Nairaland Mathematics Clinic by Freiburger(m): 11:21am On Jan 17, 2013 |
Prove that Tanx = Sinx/Cosx. |
Re: Nairaland Mathematics Clinic by Nobody: 11:41am On Jan 17, 2013 |
Freiburger: Prove that Tanx = Sinx/Cosx. Supposing (x) is an angle in a right angled triangle, then: sin (x) = opp/hyp .....(1) cos (x) = adj/hyp ......(2) tan(x) = opposite/adj ....(3) Equation (1) can be rewritten as opp = (hyp).sin (x) and equation (2) as adj = (hyp).cos(x) Substitute the value of opp and adj in equation (3) tan(x) = (hyp.sin x)/(hyp.cos x) thus tan(x) = sin(x)/cos(x) 2 Likes |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 11:49am On Jan 17, 2013 |
a woman's age and her son's add to 45years. Five years ago, the woman was 6times as old as her son, how old was the woman when the son was born? After solving it, i got 45years which is wrong, plz help |
Re: Nairaland Mathematics Clinic by Freiburger(m): 12:11pm On Jan 17, 2013 |
johnpaul1101: a woman's age and her son's add to 45years. Five years ago, the woman was 6times as old as her son, how old was the woman when the son was born?Answer should be 25 Lets assume the woman is presently 35 and her son is 10 that adds up to be 45 as stated in the question. 5 years ago the woman was 30, since she is now 35 and her son was 5, since he is now 10. so the woman was 6 times older so, 5years before the son was born the woman should be 25 @ johnpau1101 how you are O.k with that explanation. 1 Like |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 12:17pm On Jan 17, 2013 |
Freiburger:i'm very ok with it, but cant it be represented with equations and solved? 1 Like |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 12:19pm On Jan 17, 2013 |
this is another similar question:bisi's and fibie's ages add up to 29, 7years ago, bisi was twice as old as fibie, find their present ages. |
Re: Nairaland Mathematics Clinic by biolabee(m): 12:45pm On Jan 17, 2013 |
johnpaul1101: this is another similar question:bisi's and fibie's ages add up to 29, 7years ago, bisi was twice as old as fibie, find their present ages. b + f = 29 ---------(1) 7 years go bisi was b - 7 years old fibie was f - 7 However she was twice as old as fibie thus b - 7 = 2(f - 7) expand b - 7 = 2f - 14 b = 2f - 7 ---------(2) substitute in equation 1 (2f-7) + f = 29 3f - 7 = 29 3f = 36 f =12 bisi is thus 29 - 12 = 17 years 4 Likes |
Re: Nairaland Mathematics Clinic by Richiez(m): 1:31pm On Jan 17, 2013 |
biolabee: b + f = 29 ---------(1) 7 years go bisi was b - 7 years old fibie was f - 7 However she was twice as old as fibie thus b - 7 = 2(f - 7) expand b - 7 = 2f - 14 b = 2f - 7 ---------(2) substitute in equation 1 (2f-7) + f = 29 3f - 7 = 29 3f = 36 f =12 bisi is thus 29 - 12 = 17 years nice work biolabee @johnpaul, with this you can solve the exercise u pasted earlier 1 Like |
Re: Nairaland Mathematics Clinic by Johncasey(m): 2:47pm On Jan 17, 2013 |
Richiez:tanks hav goten it.. Its 1+root3 |
Re: Nairaland Mathematics Clinic by biolabee(m): 4:27pm On Jan 17, 2013 |
^^ Richiez thanks! |
Re: Nairaland Mathematics Clinic by Nobody: 4:43pm On Jan 17, 2013 |
@Johnpaul1101, I'm happy that my name sake is here. Congrats for coming here and improving yourself. |
Re: Nairaland Mathematics Clinic by Nobody: 7:28pm On Jan 17, 2013 |
Help with these ones guys. 1. Integrate : (a) ∫(x^3 + 3x^2 + 5x +1)e^x dx (b) ∫ln(x^2 + 2x) dx 2. Find the antiderivatives of the following: (a) x sec^2(x^2)tan^2(x^2)dx (b) x^4dx/(4 + x^2) |
Re: Nairaland Mathematics Clinic by Nobody: 8:57pm On Jan 17, 2013 |
Ghettoguru: Help with these ones guys. Question 1a ∫(x3 + 3x2 + 5x +1)ex dx To solve this question, we have to use integration by parts: Lets put u = x3 + 3x2 + 5x +1, differentiating, du = (3x2 + 6x + 5)dx and lets put => dv = ex dx, integrate, then v = ex The formula for integration by parts is : ∫udv = uv - ∫vdu substituting yields => ∫(x3 + 3x2 + 5x +1)ex dx => = (x3 + 3x2 + 5x +1)ex - ∫(3x2 + 6x +5)ex dx The bolded part will have to be integrated by parts too => Lets put u = 3x2 + 6x + 5, differentiating, du = (6x + 6)dx. Let dv = ex dx, integrating, v = ex. Substituting back in the formula for integration by parts => ∫udv = uv - ∫vdu ∫(x3 + 3x2 + 5x +1)ex dx => = (x3 + 3x2 + 5x +1)ex - (3x2 + 6x +5)ex - ∫(6x + 6)ex dx Looking at the bolded part, we cannot integrate it directly, so we have to apply integration by parts here again. Lets put u = 6x + 6, differentiating, du = 6dx. Lets put dv = exdx, integrating, v = ex Substituting back in the formula for integration by parts => ∫udv = uv - ∫vdu ∫(x3 + 3x2 + 5x +1)ex dx => = (x3 + 3x2 + 5x +1)ex - (3x2 + 6x +5)ex + (6x + 6)ex - ∫6ex dx Now looking at the bolded, it can now be integrated directly => i.e ∫6ex dx = 6ex + C :. ∫(x3 + 3x2 + 5x +1)ex dx => = (x3 + 3x2 + 5x +1)ex - (3x2 + 6x +5)ex + (6x + 6)ex - 6ex + C = ex[x3 + 3x2 + 5x +1 - 3x2 - 6x - 5 + 6x + 6 - 6] + C = ex[ x3 + 5x - 4] + C I hope this helps! I'll check the rest tomorrow morning. 1.(b) ∫ln(x2 + 2x) dx Using integration by parts here again, Let u = ln(x2 + 2x), then du = (2x + 2)/(x2 + 2x) Let dv = dx, the v = x ∫udv = uv - ∫vdu ∫ln(x2 + 2x) dx => xln(x2 + 2x - ∫x(2x+ 2)dx/(x2 + 2x) = xln(x2 + 2x - ∫x(2x+ 2)dx/x(x + 2) Divide (2x + 2)/(x + 1) and we have 2 & -2/(x + 1) = xln(x2 + 2x) - ∫2dx + ∫2dx/(x + 2) :. ∫ln(x2 + 2x) dx => = xln(x2 + 2x) - 2x + 2ln|x + 2| + C = xlnx(x +2) - 2(x - ln|x + 2|) + C = 2(ln|x + 2| - x) + xlnx(x +2) + C 1 Like |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 6:44am On Jan 18, 2013 |
Johnpaul88: @Johnpaul1101,dat's nice, i only wish 2 improve my mathematical knowledge, and believe me, this thread has helped a lot |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 6:45am On Jan 18, 2013 |
back to school |
Re: Nairaland Mathematics Clinic by Richiez(m): 7:18am On Jan 18, 2013 |
Excellent work! General DoubleDx 3 Likes |
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