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Re: Nairaland Mathematics Clinic by Nobody: 1:59pm On Nov 18, 2013 |
rhydex 247:,..9c 1 log _e (u) = ln u |
Re: Nairaland Mathematics Clinic by Mbahchiboy(m): 4:11pm On Nov 18, 2013 |
echibuzor:@ d first 1 i did something like: (1+3)^x=8x. and i applied (1+x)^n method but my answers deviated thoroughly. d second 1 hmm.....i no even know hw to start am |
Re: Nairaland Mathematics Clinic by Laplacian(m): 6:02pm On Nov 18, 2013 |
Mbahchiboy: plz u guys should help solve dis using "d citizen" approach.i tried doing it but i dont know my mistake.2^2x=8x, or 2^(2x-3)=x or 2^(2x-3)/4^(2x-3)=x/4^(2x-3) (1/2)^(2x-3)=x*(1/4)^(2x-3) (0.5)^(2x-3)=x*(0.25)^(2x-3) (1-0.5)^(2x-3)=x*(1-0.75)^(2x-3) by binomial approx. 1-(2x-3)*0.5...= x*[1-(2x-3)*0.75+...] re-arrangin gives 1-x+3/2=x-3x/4*(2x-3) or 4-4x+6=4x-6x^2+9x or 6x^2-17x+6=0 for d second question; (4/x)^x=2x or 4^x=2x^(x+1) or 2^(2x-1)=x^(x+1)...i hope u can go on from there? 1 Like |
Re: Nairaland Mathematics Clinic by Mbahchiboy(m): 8:41pm On Nov 18, 2013 |
Laplacian:THANKU VERY MUCH SIR 4 D HIGHLIGHT.....BUT PLZ COMPLETE D NO 2 I DON PAUSE |
Re: Nairaland Mathematics Clinic by Laplacian(m): 10:52pm On Nov 18, 2013 |
Mbahchiboy: THANKU VERY MUCH SIR 4 D HIGHLIGHT.....BUT PLZ COMPLETE D NO 2 I DON PAUSE2^(2x-1)=x^(x+1).....(eqn1) let y=x^(x+1).........(eqn2) substitut (2) in (1) to get 2^(2x-1)=y........(eqn3) from (3), 2^(2x-1)/4^(2x-1)=y/4^(2x-1) or (0.5)^(2x-1)=y*(0.25)^(2x-1) or (1-0.5)^(2x-1)=y*(1-0.75)^(2x-1) or 1-0.5(2x-1)=y[1-0.75(2x-1)] .......(eqn4) from (2), y=x^(x+1) or y*[1/(4x)]^(x+1)=(x/4x)^(x+1) or y*[1+1/(4x)-1]^(x+1)=(0.25)^(x+1) or y[1+(x+1){1/(4x)-1}+...]= (1-0.75)^(x+1) or y[1+(x+1)(1-4x)/4x]=1+0.75(x+1)....(eqn5) eliminat y from eqn4 and eqn5 and solve d cubic to obtain x...both equtaion can b easier solved using logarithmic approximation...but obviously dis thread is not a convenient place to use logarithmic operations... |
Re: Nairaland Mathematics Clinic by rhydex247(m): 9:09am On Nov 19, 2013 |
Question 1. Show that the four points whose position vectors are 3i-2j+4k, 6i+3j+k, 5i+7j+3k and 2i+2j+6k are coplanar. Question 2. If a ^ r=b+¥a and a.r=3 where a=2i+j-k and b=-i-2j+k. Then find r and ¥. Question 3. Prove the identity. (a ^ b).(c ^d)=(a.c)(b.d)-(a.d)(b.c). Where ^ means cap or cross |
Re: Nairaland Mathematics Clinic by Nobody: 10:04am On Nov 19, 2013 |
An error of P% is made in measuring the radius of a circle. Calculate the approximate percentage error in the measurement of the circumference. |
Re: Nairaland Mathematics Clinic by echibuzor: 10:21am On Nov 19, 2013 |
smurfy: An error of P% is made in measuring the radius of a circle. Calculate the approximate percentage error in the measurement of the circumference.(P2 - 2rP) / ^r2 x 100 |
Re: Nairaland Mathematics Clinic by factorial1(m): 10:29am On Nov 19, 2013 |
How many odd numbers with three digits can be formed from the digits 1, 2, 3, 4, 5 if you can use any of the digits more than one? |
Re: Nairaland Mathematics Clinic by factorial1(m): 10:30am On Nov 19, 2013 |
explain hw you arrived @that bro...? echibuzor: |
Re: Nairaland Mathematics Clinic by echibuzor: 10:51am On Nov 19, 2013 |
factorial1: explain hw you arrived @that bro...?I sort of read the question wrong, my previous answer is for the percentage error in area. The circumference is p/r x 100 |
Re: Nairaland Mathematics Clinic by Nobody: 12:00pm On Nov 19, 2013 |
factorial1: How many odd numbers with three digits can be formed from the digits 1, 2, 3, 4, 5 if you can use any of the digits more than one? Here goes... The last digit can be arranged in 3 ways, since it must end with 1, 3 or 5. Each of the first two digits can be arranged in 5 ways, since each digit can be used more than once. The answer therefore is 5 * 5 * 3 = 75 ways. |
Re: Nairaland Mathematics Clinic by Nobody: 12:05pm On Nov 19, 2013 |
echibuzor: If the above is your answer, then it's wrong. The options are A. (1/2)P% B. P% C. (3/2)P% D. 2P% Note: Clear working'll be appreciated. |
Re: Nairaland Mathematics Clinic by echibuzor: 12:42pm On Nov 19, 2013 |
smurfy:Correct Measurement gives you 2^r (note that '^' means pi) measurement with the error in it gives 2^(r+p) == 2^r + 2^p Error in measurement is 2^p (in whatever unit it is m, cm, km) percentage is (2^p / 2^r) x 100 So you see how I arrived at my answer... I used Linear Approximation and got the same thing.. So it's either I am hungry (which I am BTW) or the Option is just not there or there is something I am not seeing (This one is most likely) |
Re: Nairaland Mathematics Clinic by factorial1(m): 12:57pm On Nov 19, 2013 |
Correct! smurfy: |
Re: Nairaland Mathematics Clinic by factorial1(m): 12:59pm On Nov 19, 2013 |
Find the equation of the tangent(s) where x = 2 on the curve x^2 + y^2 - 2x + y = 6. |
Re: Nairaland Mathematics Clinic by Nobody: 1:16pm On Nov 19, 2013 |
smurfy: An error of P% is made in measuring the radius of a circle. Calculate the approximate percentage error in the measurement of the circumference. The only way I could solve this is by using numbers. Here goes... Let True Radius = 20cm Let Error Radius = 18cm. PE = (TR - ER)/TR * 100% This gives 10% (My P is therefore 10. Now let's find PE for the circumference. Whatever I get is directly compared with the value of P above.) Now, C = 2 * pi * r True C = 2 * pi * 20 = 40 * pi cm Error C = 2 * pi * 18 = 36 * pi cm PE = (TC - EC)/TC * 100%. This gives (4pi/40pi) * 100% = 10%. This shows that if there's a 10% error in the measurement of the radius of a circle, there'll be a 10% corresponding error in the measurement of its circumference. Hence, the correct answer is P%. The issue is whether this can be solved neatly, algebraically without using this primary school method. That, I'd like to know. |
Re: Nairaland Mathematics Clinic by echibuzor: 1:24pm On Nov 19, 2013 |
smurfy:This is literally what I showed in my workings. My answer is right, it just lacked the conclusion. P% in the question is gotten by error p / actual r x 100. which is repeated in the final answer.. |
Re: Nairaland Mathematics Clinic by Nobody: 1:25pm On Nov 19, 2013 |
factorial1: Find the equation of the tangent(s) where x = 2 on the curve x^2 + y^2 - 2x + y = 6. Differentiate the equation of the circle implicitly. Find y when x =2. Make dy/dx the subject, slot in x = 2, y = ... to find the gradient. Then use y - y1 = m(x - x1) to find the required equation. Note: x1 = 2, y1 = -3 or 2 |
Re: Nairaland Mathematics Clinic by rhydex247(m): 1:33pm On Nov 19, 2013 |
rhydex 247: Question 1. |
Re: Nairaland Mathematics Clinic by ameer2: 2:27pm On Nov 19, 2013 |
Prove dat cosecX-1/cosecX+1=[1-tan(x/2)/1+tan(x/2)]^2. . .any1 dat cn prove dis will have a special reward.I promise. |
Re: Nairaland Mathematics Clinic by Laplacian(m): 3:43pm On Nov 19, 2013 |
smurfy: An error of P% is made in measuring the radius of a circle. Calculate the approximate percentage error in the measurement of the circumference.let C denote d circumference and let # denot pi then, C=2#r, now differentiat C wit respet to r to get dC/dr=2# or the error in the circumference; dC=2#*dr, now divide thru by C=2#r to get dC/C=2#*dr/(2#r) or dC/C=dr/r=p% showin dat d percentag error in measurin d circumference must be d same as dat made in measurin d radius |
Re: Nairaland Mathematics Clinic by Laplacian(m): 3:52pm On Nov 19, 2013 |
ameer!:...i suggest u use brackets apprioprately to for ambiguity (for those who 'll solv d question)... |
Re: Nairaland Mathematics Clinic by Mbahchiboy(m): 3:55pm On Nov 19, 2013 |
factorial1: Find the equation of the tangent(s) where x = 2 on the curve x^2 + y^2 - 2x + y = 6.BEFORE I START PLZ cross check d ques i think u omitted d value of y |
Re: Nairaland Mathematics Clinic by Nobody: 6:01pm On Nov 19, 2013 |
Mbahchiboy: BEFORE I START PLZ cross check d ques i think u omitted d value of y Yeah. I think you're right. I got two values for y, thereby giving two different values for dy/dx, -2/5 and 2/5. Equation of tangent is 2x + 5y - 14 = 0 or -2x + 5y + 19 = 0 I don't think we're meant to get two answers. He probably missed something. |
Re: Nairaland Mathematics Clinic by Mbahchiboy(m): 6:55pm On Nov 19, 2013 |
[quote author=Laplacian] 2^(2x-1)=x^(x+1).....(eqn1) let y=x^(x+1)...(eqn2) substitut (2) in (1) to get 2^(2x-1)=y(eqn3 from (3), 2^(2x-1)/4^(2x-1)=y/4^(2x-1) or (0.5)^(2x-1)=y*(0.25)^(2x-1) or (1-0.5)^(2x-1)=y*(1-0.75)^(2x-1) or 1-0.5(2x-1)=y[1-0.75(2x-1)] from (2), y=x^(x+1) or y*[1/(4x)]^(x+1)=(x/4x)^(x+1) or y*[1+1/(4x)-1]^(x+1)=(0.25)^(x+1) or y[1+(x+1){1/(4x)-1}+...]= (1-0.75)^(x+1) or y[1+(x+1)(1-4x)/4x]=1+0.75(x+1[\quote/]thanks sir i think i'l try it 4 better understandin |
Re: Nairaland Mathematics Clinic by Nobody: 7:47pm On Nov 19, 2013 |
Laplacian: Thanks Laplacian. Thanks for this method. Tell me, assuming you got something like dC/C = 1/2(dr/r), what would you have said about P%? |
Re: Nairaland Mathematics Clinic by factorial1(m): 9:58pm On Nov 19, 2013 |
I omitted nothing der, the qt is correct n' accurate! Mbahchiboy: BEFORE I START PLZ cross check d ques i think u omitted d value of y |
Re: Nairaland Mathematics Clinic by Calculusfx: 10:13pm On Nov 19, 2013 |
Hmmmm |
Re: Nairaland Mathematics Clinic by Calculusfx: 10:21pm On Nov 19, 2013 |
I have not been to the thread since morning...kudos to master laplacian,master echibuzor,master factorial(who used logic to abuse me on the quiz thread),my main oga rhydex,master chiboy and master smurfy...for not allowing the clinic to be silent as mortuary...i respect you all. 1 Like |
Re: Nairaland Mathematics Clinic by Calculusfx: 10:24pm On Nov 19, 2013 |
ameer!:...hmmm.from rhs {(1-tan(1/2)x)/(1+tan(1/2)x)}^2 since tan(1/2)x=sin(1/2)x/cos(1/2)x...then {[1-sin(1/2)x/cos(1/2)x]/{1+sin(1/2)x/cos(1/2)x}^2...{LET'S DEAL WITH THE ONE IN THE BRACKET FIRST)multiply the numerator and denominator by cos(1/2)x which gives {(cos(1/2)x-sin(1/2)x)/(cos(1/2)x+sin(1/2)x)}^2...then square to give (sin^2(1/2)x-2sin(1/2)xcos(1/2)x+cos^2(1/2)x)/(sin^2(1/2)x+2sin(1/2)xcos(1/2)x+cos^2(1/2)x)...don't forget from trig. that cos^2(1/2)x+sin^2(1/2)x=1 and 2sin(1/2)xcos(1/2)x=sinx...substitute those to give...(1-sinx)/(1+sinx)...multiply the numerator and the denominator by 1/sinx...to give {1/sinx(1-sinx)/1/sinx(1+sinx)=(1/sinx-1)/(1/sinx+1)...don't forget 1/sinx=cosecx...substitute that to give cosecx-1/cosecx+1 QED |
Re: Nairaland Mathematics Clinic by statusnet(m): 12:23am On Nov 20, 2013 |
Integrate (tan^2) x / tan2x dx |
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