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Re: Nairaland Mathematics Clinic by yungryce: 8:36am On Feb 07, 2013 |
@doubledx & richiez. Keep d gud work going jor 2 Likes |
Re: Nairaland Mathematics Clinic by Adol16(m): 12:51pm On Feb 07, 2013 |
nice.one: ans x=4 or x=2 |
Re: Nairaland Mathematics Clinic by Bolaji16(m): 6:47pm On Feb 07, 2013 |
solve for x in root x - root x-2 = 1. Sum1 shud plz help me solve this. God Bless |
Re: Nairaland Mathematics Clinic by Richiez(m): 4:30am On Feb 08, 2013 |
Bolaji 16: solve for x in root x - root x-2 = 1. Sum1 shud plz help me solve this. God Blessthe question is not clearly interpretable pls retype in a way that we can understand so we can help u out quickly |
Re: Nairaland Mathematics Clinic by Nobody: 9:01am On Feb 08, 2013 |
Bolaji 16: solve for x in root x - root x-2 = 1. Sum1 shud plz help me solve this. God Bless Is this what you meant √(x ) - √(x - 1) = 1? If it is then here is the solution> √(x ) - √(x - 1) = 1 rearranging> √(x - 1) = √(x) - 1 Squaring both sides to get rid of the roots> [√(x - 1)]^2 = (√(x) - 1)^2 x - 1 = x - 2√(x) + 1 -1 -1 = - 2√(x) 1 = √(x) Squaring both sides again yields [√(x)]^2 = 1^2 x = 1 1 Like |
Re: Nairaland Mathematics Clinic by Nobody: 9:14am On Feb 08, 2013 |
^Nice one |
Re: Nairaland Mathematics Clinic by Nobody: 9:18am On Feb 08, 2013 |
yungryce: @doubledx & richiez. Keep d gud work going jor Thanks bruv, I have solved the remaining questions. I'll take out time to type and post them later 2day or 2mrrw. 1luv |
Re: Nairaland Mathematics Clinic by Bolaji16(m): 12:04pm On Feb 08, 2013 |
Ghettoguru:thank u very much bro.. |
Re: Nairaland Mathematics Clinic by Bolaji16(m): 12:37pm On Feb 08, 2013 |
Help me check dis out
|
Re: Nairaland Mathematics Clinic by Nobody: 2:00pm On Feb 08, 2013 |
^Solution 2√(x + 4) - x = 1 2√(x + 4) = x + 1 Square both sides [2√(x + 4)]^2 = (x + 1)^2 4(x + 4) = x^2 + 2x + 1 4x + 16 = x^2 + 2x + 1 Rearrange and collect like terms => x^2 - 2x - 15 = 0 factorizing => x^2 - 5x + 3x - 15 = 0 x(x - 5) + 3(x - 5) = 0 (x - 5)(x + 3) = 0 :. x - 5 = 0 or x + 3 = 0 x = 5 or -3 1 Like |
Re: Nairaland Mathematics Clinic by Richiez(m): 3:08pm On Feb 08, 2013 |
Great job Ghetoguru and double dy/dx 2 Likes |
Re: Nairaland Mathematics Clinic by biolabee(m): 5:34pm On Feb 08, 2013 |
doubleDx: ^Solution Sorry my esteemed gurus but doing a check with -3 does not solve the equation Please check |
Re: Nairaland Mathematics Clinic by Nobody: 9:08pm On Feb 08, 2013 |
biolabee: No probs bruv, check this out => 2√(x + 4) - x = 1 2√(x + 4) = x + 1 For x = -3 on the LHS 2√(x + 4) = 2 . [±√(-3 + 4)] = 2. [±√(1)] = 2 .(+ 1 ) or 2. (- 1) = 2 or -2 For x = -3 on the RHS = x + 1 = -3 + 1 = - 2 It's true for x = -3 that LHS = RHS with a value of (-2) Note=> (.) is for multiplication! 1 Like |
Re: Nairaland Mathematics Clinic by biolabee(m): 2:01am On Feb 09, 2013 |
nice one we learn everyday 1 Like |
Re: Nairaland Mathematics Clinic by Fetus(m): 4:38am On Feb 09, 2013 |
doubleDx:..u are nt solving my questions ni?...kindly asist me na 1 Like |
Re: Nairaland Mathematics Clinic by Nobody: 9:10am On Feb 09, 2013 |
Fetus: Kindly help with dis questions....1) write down d first four terms of d bionomial series expansion in ascending power of x of d functions (1-x)^-1/2 and hence show dat root 10= 3(1+1/20+1*3/20*40+1*3*5/20*40*60+........) No(2) find d common ratio of d geometric sequence sin2€, -sin€cos2€, sin2€cos^2(2€),.....prove dat for @<€<₹/2...the series sin2€-sin2€cos2€ + sin2€sin^2(2€) +....has a sum to infinity and show dat d sum to infinity is tan€... Where €,₹ and @ are alpha, pie and teetha respectively.... Tanks Sorry bruv, we miss this one => Solution to Question 1a (1 - x)^(-1/2) Since => (1 + r)^k = 1 + kr + k(k - 1)(r)^2/2! + k(k - 1)(k - 2)(r)^3/3! + .... r = (-x) and k = -1/2 in this case. Substituting yields => = 1 + (-1/2)(-x) + (-1/2)( -3/2)(-x)^2/2! + (-1/2)(-3/2)(-5/2)(-x)^3/3! + .... = 1 + x/2 + 3x^2/8 + 15x^3/8(6) + ... Hence, the first 4 terms of the expression (1 - x)^(-1/2) => = 1 + x/2 + 3x^2/8 + 5x^3/16 I'm a little busy now; will check question 2 later. 1luv man! 1 Like |
Re: Nairaland Mathematics Clinic by ayoope(f): 6:26pm On Feb 09, 2013 |
Weldone Richiez, Biolabee, DoubleDx and other maths gurus. Much respect! 2 Likes |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 7:09pm On Feb 09, 2013 |
happy weekend to all my maths gurus......it's been quite a long time....please i would like to ask this: from a given set of data, is there any mathematical way of calculating the range apart from HIGHEST DIGIT-LOWEST DIGIT........I'm asking because i have never been taught about range.....and whenever i tell my frnds that these is the only way of finding the range from a given data, they will just laugh and say; it doesnt sound like a mathematical method...so plz my mathematicians...help me 1 Like |
Re: Nairaland Mathematics Clinic by Nobody: 11:21am On Feb 10, 2013 |
ayoope: Weldone Richiez, Biolabee, DoubleDx and other maths gurus. Much respect! Thanks. Do you have any math problem for us? |
Re: Nairaland Mathematics Clinic by Nobody: 11:27am On Feb 10, 2013 |
Fetus: Kindly help with dis questions....1) write down d first four terms of d bionomial series expansion in ascending power of x of d functions (1-x)^-1/2 and hence show dat root 10= 3(1+1/20+1*3/20*40+1*3*5/20*40*60+........) No(2) find d common ratio of d geometric sequence sin2€, -sin€cos2€, sin2€cos^2(2€),.....prove dat for @<€<₹/2...the series sin2€-sin2€cos2€ + sin2€sin^2(2€) +....has a sum to infinity and show dat d sum to infinity is tan€... Where €,₹ and @ are alpha, pie and teetha respectively.... Tanks Solution to Question 1b To prove that √10 = 3[1 + 1/20 + 1(3)/[(20)(40)] + [(1)(3)(5)/[(20)40(60) + ....] Evaluating the RHS yields => = 3[1 + 1/20 + 1(3)/[(20)(40)] + [(1)(3)(5)/[(20)40(60) + ....] = 3 (1.0541) = 3.162 approx. √10 can be rewritten as => = √[9(1 + 1/9)] = 3√(1 + 1/9) = 3 (1 + 1/9)^(1/2) Expanding the bolded part yields => = 3[1 + (1/2)(1/9) + (1/2)(-1/2)(1/9)^2/2! + (1/2)(-1/2)(-3/2)(1/9)^3/3! + ....] Simplifying further yields => = 3[ 1.0541 approx] = 3.162 approx Thus, √10 = 3(1 + 1/20 + 1(3)/[(20)(40)] + [(1)(3)(5)/[(20)40(60) + ...] = 3[1 + (1/2)(1/9) + (1/2)(-1/2)(1/9)^2/2! + (1/2)(-1/2)(-3/2)(1/9)^3/3! + .... = 3.162 approx 1 Like |
Re: Nairaland Mathematics Clinic by biolabee(m): 6:07pm On Feb 10, 2013 |
ayoope: Weldone Richiez, Biolabee, DoubleDx and other maths gurus. Much respect! thanks ma you are far too kind |
Re: Nairaland Mathematics Clinic by ayoope(f): 12:50am On Feb 11, 2013 |
@doubleDx, I don't have any. Been following this thread since the very first day. Just amazing the way you all provide solutions to the questions asked. 2 Likes |
Re: Nairaland Mathematics Clinic by babsomotde(m): 2:05am On Feb 11, 2013 |
Maths is all about accuracy and memory. One you have those two, it will be easy for you to pound on any equation. www.edubabs.com |
Re: Nairaland Mathematics Clinic by Fetus(m): 2:27pm On Feb 11, 2013 |
doubleDx:...God bless u man.....tanx alot..... 2 Likes |
Re: Nairaland Mathematics Clinic by Nobody: 5:14pm On Feb 12, 2013 |
Fetus: ...God bless u man.....tanx alot..... Thanks and you're welcome anytime. ayoope: @doubleDx, I don't have any. Thanks a lot ma'am. Your compliment is appreciated! Fetus: Kindly help with dis questions....1) write down d first four terms of d bionomial series expansion in ascending power of x of d functions (1-x)^-1/2 and hence show dat root 10= 3(1+1/20+1*3/20*40+1*3*5/20*40*60+........) No(2) find d common ratio of d geometric sequence sin2€, -sin€cos2€, sin2€cos^2(2€),.....prove dat for @<€<₹/2...the series sin2€-sin2€cos2€ + sin2€sin^2(2€) +....has a sum to infinity and show dat d sum to infinity is tan€... Where €,₹ and @ are alpha, pie and teetha respectively.... Tanks Your second question has some issues. I doubt if it's a geometric series! Check this out => The geometric sequence is given as => sin2α, -sinαcos2α, sin2αcos^2(2α),.. For Ө<α<π₹/2 The common ratio (r) => a2/a1 or a3/a2 where a1, a2 and a3 are first, second and third term of the series respectively. Using the 1st and 2nd term, r = a2/a1 = [-sin(α)cos(2α)]/sin(2α) => Since cot(2α) = cos(2α)/sin(2α), then=> r1 = -cot(2α) sin(α) Using the 2nd and 3rd term, r = a3/a2 = [sin 2α cos^2(2α)]/[-sin α cos 2α] Since tan(2α) = sin(2α)/cos(2α), then=> r2 = -tan (2α)cos^2^(2α) If the series is geometric, r1 should be equal to r2; but in this case r1≠ r2, thus I doubt if the question is correct! Re-check it bruv. Sum to infinity for r1 => S∞ = a1/(1 - r1) If we substitute r1, it yields=> = sin(2α)/[1 - (-cot(2α) sin(α))] = sin(2α)/[1 + cot(2α) sin(α)] or Sum to infinity for r2=> S∞ = a1/(1 - r2) If we substitute r2, it yields=> = sin(2α)/[-tan (2α)cos^2^(2α)] Which is not equal to the result for r1. Recheck the question. If possible write it out on a paper and scan it. Take care. |
Re: Nairaland Mathematics Clinic by Nobody: 8:07pm On Feb 12, 2013 |
Youngsage: Spoiler Alert: Hello bro, I don't want to cause spoiler-alert but you got your solution wrong. You might be lucky on your answer. The properties applied on the shaded equation is invalid: log(ab) = log(a) + log(b, which implies that log(a+b) NotEqual log(a)+log(b). I'm glad you've been corrected by other people. This is a nonlinear equation which can only be solved numerically or by graphical method. |
Re: Nairaland Mathematics Clinic by Nobody: 5:07am On Feb 13, 2013 |
chikis: Let initial cost of living = C Let the new increase in cost of living by 15% = 1.15C He then decreased the quantity of food bought by 10%. In order word, he now purchases 90% of the new cost = 90% * 1.15C = 0.9*1.15C Thus the fractional increase w.r.t the initial cost (C) = (New Spending - Initial Cost)/(Initial Cost) = (0.9*1.15C - C)/C = (0.9*1.15 - 1)= 0.035 Thus, Fractional Increase = 0.035 or 3.5% An illiterate, I believe, can follow my follow solutions. 1 Like |
Re: Nairaland Mathematics Clinic by Nobody: 5:21am On Feb 13, 2013 |
Ehinafe: Richiez: Spoiler Alert: You actually didn't solve the simultaneous equation. You're meant to reach the final solution analytically and not by estimation or shooting. |
Re: Nairaland Mathematics Clinic by Adol16(m): 10:17am On Feb 13, 2013 |
Solve 1. 1/X + 1/y + 1/z = 1 find x,y and z 2. 5Loga base 4 + 48log4 base a = a/8. |
Re: Nairaland Mathematics Clinic by Richiez(m): 10:17am On Feb 13, 2013 |
donedy: Spoiler Alert: Hello bro, I don't want to cause spoiler-alert but you got your solution wrong. You might be lucky on your answer. The properties applied on the shaded equation is invalid: log(ab) = log(a) + log(b, which implies that log(a+b) NotEqual log(a)+log(b). I'm glad you've been corrected by other people. This is a nonlinear equation which can only be solved numerically or by graphical method.he had already been corrected severally, anyways thanks for the observation and welcome to the thread |
Re: Nairaland Mathematics Clinic by Richiez(m): 10:20am On Feb 13, 2013 |
@double dy/dx, i dey see ur handwork, u'r a great man, i salute you! 2 Likes |
Re: Nairaland Mathematics Clinic by ositadima1(m): 10:23am On Feb 13, 2013 |
donedy: Mr Spoiler alert, it would enlighten us some if you gave us the correct answer. 2 Likes |
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