Stats: 3,243,097 members, 8,123,474 topics. Date: Thursday, 03 April 2025 at 06:36 AM |
Nairaland Forum / Calculusfx's Profile / Calculusfx's Posts
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Royver: It depends. Is he bigger physically? Are you into violence? Or would you rather let things go? How sure are you that he took it? Do you believe in negative prayers?.thank you very much my brother,i understand everything you said,i decided to let it go,but some people who know the value of what the memory card contains are ready to fight for it with their lives coz the guy is a cultist....i'm confused and the only thing he can use on it is songs,even not all of it coz i have about 650 songs on it.....i love your advise bro |
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My memory card which comprises many vital things was stolen by a guy....though i didn't see him when he took it,but i knew he was the one who took it...what can i do? |
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Hmmmmm Mbahchiboy: still pending.....plz i need answers to dis...s=a(r^n-1)/(r-1)...s(r-1)=a(r^n-1)...sr-s=ar^n-a...sr-ar^n=s-a...r{s-ar^(n-1)}=s-a....but don't forget that ar^(n-1)=nth term of a gp...r(s-nth)=s-a...then r=(s-a)/(s-nth)....i guess that's the solution to the problem...though the last term wasn't specified in the question..... |
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I nominate sir tunechi...he will make it to a standard level..believe me |
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koliks: Gud day 2 d guruus in d auz,can any1 find d integral of dx/e^2x-2e^x...£ represents my integral sign...then £dx/(e^2x-2e^x)...let u be e^x....du=e^xdx...du=udx...dx=du/u...substitute those to get £du/u(u^2-2u)...£du/u^2(u-2)...then let's solve using integration by partial fraction...1/u^2(u-2)=A/u+B/u^2+C/(u-2)...multiply through by u^2(u-2) to get 1=A.u(u-2)+B(u-2)+Cu^2...expand the rhs to get 1=u^2A-2uA+Bu-2B+Cu^2...then 1=(A+C)u^2+(B-2A)u-2B...equate them to give,(A+C)=0,(B-2A)=0 2B=1...then,B=1/2...then 1/2-2A=0...A=1/4...1/4+C=0,then C=-1/4...substitute that,then £1/4u.du+£1/2u^2.du-£1/4(u-2).du...then 1/4lnu-1/2u+1/4ln(u-2)....then lnu^(1/4)+ln(u-2)^(1/4)-1/2u...then ln{u(u-2)}^(1/4)-1/2u...then 1/4ln(e^2x-2e^x)-1/2e^x....i'm rushing in typing,i may be wrong,but i just want you to get the steps...any criticism is allowed |
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wisemania:...even people have been intimidating me on physics111 before i wrote my postutme...but i never get scared,i'm a mathematician by passion,i shouldn't get afraid of physics...i love this bro...more power and power again...i pray we all succeed...i'm just bothered on td...coz i don't know anything in it,but i believe that if i'm being taught,i will surely comprehend...but i need help of people like you to put me through,i don't know much on mathematics ooo,i just pray i do better...pls,mail me at omoyeleolalekanjacob@gmail.com sir... |
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wisemania:...hmmm,i love this my brother,but it's not good enough,how would you recommend h.k.dass for a fresher(though i have never seen the textbook before but i know it's advanced oaths textbook),what i can just ask the person to get engineering maths by k.a stroud...but he should not plan of buying pure mathematics by bunday for now...coz bunday wrote his own as if schaum outline series write theirs...the ways you can become a maths guru are 1.get a fundamental math textbooks for now before progressing to higher ones 2.move very close to those who know it(though nobody knows it,it just depends on the aspect one is) 3.get exposed to questions...but the above of all is ur determination and interest on it... |
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wisemania: |
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wisemania:...my brother,i'm uniben mechanical engineer...fresher,and with ur texts.you are an engineer studying in uniben...can you pls help me on some stuffs...like guiding? |
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Kendzyma: tnx man...iz td also compulsory der cous I dnt do td in second scul o and I dnt tink I can learn it all by myself at om...my question iz...wen I get der...will I b abl to cope or catch up wit td...we are almost the same bro...i never knew what furthermaths was when i was in sec.school...likewise td...i represent mechanical engineering...UNIBEN...fresher... |
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Hmmmm |
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benbuks: Wow.....my oga at the top...i respect |
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Ortarico: Where have my masters been? Pls, save the life of this great thread!...my masters,it's been a while....how maths life Olarewajub: Solve 2^2y + 2 = 33 X 2^y - 8...i think the question is 2^(2y+2)=33*2y-8.... |
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d citizen:...i can see we are the same bro...i'm a mechanical engineer to be,by profession and a mathematician by passion...pls,if you don't mind being my friend,mail me at omoyeleolalekanjacob@gmail.com |
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Ortarico:...the polynomial is advance,try and read lodovico ferarri's method for quartic equation...don't ask me,if you don't get it ooo coz the formula is very advance...someone said those who study masters in algebra can be taught the formula...BUT YOU CAN USE IT FOR ANY POLYNOMIAL |
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X^4+2x^3+4x^2+8x+16=0 |
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busuyem:...let the number be xy...from the question,we were told that the difference is 1,and the unit digit is greater than the ten digit,which means y-x=1...then y=1+x...EQUATION 1....the question states again that the original question is one more than 5 times the sum of the digits...from xy...the digits are x and y but the original one is 10x+y...assuming 23 which will equal to 10*2+3...then 10x+y=5(x+y)+1...10x+y=5x+5y+1....5x-4y=1...don't forget that y=1+x from equation 1,substitute that to give 5x-4(1+x)=1...5x-4x-4=1...therefore x=5...from equation 1...y=x+1=5+1 to give 6...since the number is xy...then the number is 56... |
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Ortarico:...great master but i guess this is newton-raphson iterative method |
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benbuks: ....u try bro..bt it seems. U didnt get d question....integrate sqrt[(1 +sqrt(x)]/x..dats d question...re-try......nice 1 dia "calculus(fx)"....y=∫√(1+√x)/x.dx...let √x=a²...x=a^4...dx=4a^3.da...then,∫√(1+a²)/a^4.4a^3da...∫√(1+a²)/a..da...if u²=1+a²...udu=ada...da=udu/a...then ∫u/a*udu/a...∫u²/a².du but u²=1+a²...a²=u²-1...then ∫u²/(u²-1).du...do the division to get...∫1+1/(u²-1).du...from here...let's solve 1/u²-1...resolve to partial fraction,1/u²-1=(A/u-1)+(B/u+1)....1=A(u+1)+B(u-1)...if u=1...A=1/2...and B=-1/2....which gives,1/2(u-1)-1/2(u+1)...then,∫1+1/2(u-1)-1/2(u+1).du...then,{u+1/2ln(u-1)-1/2ln(u+1}...{u+1/2ln(u-1)/(u+1)}+c...don't forget that u²=1+a²...u=√1+a²...don't forget that a²=√x...u=√1+√x...substitute that to the answer to give...{√(1+√x)+(1/2)ln(√1+√x-1)/(√1+√x+1)+c...{√(1+√x)+(1/2)ln(x^1/4)/(√2+√x)+c...i guess.....waiting for better solutions from my generals |
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emmyeuler1: guy u said u got admission to read mechanical engr......but how come u know all this.....did u do any A-LEVEL program?...i'm still your boy master...i haven't done any A-level work ooo.....i pray God helps |
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For the benbuks question....y=∫√1+√x.dx.....let √x be a²...then x=a^4...dx=4a^3da...substitute to give ∫4a^3√1+a².da....let u be 1+a²...du/2a=da...then ∫4a²√u...4∫a²√u but u-1=a²...then,4∫(u-1)√u...4∫(u^3/2)-(u^1/2)...4{[2u^(5/2)]/5-[2u^(3/2)]/3}...8/5.u^(5/2)-8/3.u^(3/2)...but u=a²+1 and a^4=x...substitute those |
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Odehfamily: 5/12 of a number is subtracted from 3/4 of the number. The positive difference is 7 less than 5/6 of the number. Find the number. plz help me...let the number be x...(3x/4)-(5x/12)=(5x/6)-7...(9x-5x)/12=(5x-42)/6...4x=10x-84...x=14 |
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busuyem: Help solve dis o-the numerator of a fraction is 5 less than its denominator.if 6 is added to the numerator and 4 to the denominator,the fraction is doubled.what is the fraction?Pls, help me out with this:...let the denominator be x...since the numerator is 5 less than the denominator...then the original fraction is (x-5)/x...but the question continues that 6 is added to the numerator and 4 to the denominator...then (x-5+6)/(x+4)...which is equal to 2 times the original equation{(x-5)/x}...then (x+1)/(x+4)=2(x-5)/x...multiply through by x(x+4){which is the same as cross multiplication} to give x(x+1)=2(x-5)(x+4)...x^2+x=2x^2-2x-40...x^2-3x-40....x=8 or -5...since the original equation is (x-5)/x...substitute 8 which gives 3/8 OR substitute -5 which gives 2 |
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d citizen:...y=x^x^x^x^x to infinity...which gives y={x^(x^4)}^infinity....but x^4*infinity=infinity...then,y=x^infinity...dy/dx=infinity.x^infinity-1...dy/dx=infinity.x^infinity...since infinity-1=infinity...then dy/dx=infinity... |
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Fetus: X+y =5...it's obvious that the answer is (3,2) or (3,2) as (x,y)...but if it's of no simple root...use newton raphson method...that x'=x-f(x)/f'(x)...i also heard that the great newton was unable to solve the question |
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2004^2004 |
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rhydex 247: 1. Solve the equation. (x-1)^4+(x+3)^4=82....(3)...x²+xy+y²=84...(i) and x+√xy+y=14...(ii)from (ii).x+y-14=√xy...square both sides...(x+y-14)²=xy...then x²+2xy+y²-28x-28y+196=xy...then x²+2xy-xy+y²-28x-28y+196=0...x²+xy+y²-28x-28y+196...don't forget that from (i)x²+xy+y²=84...substitute that to give...84-28x-28y+196=0...28x+28y=280...divide through by 28...x+y=10...{*}...don't forget that x²+y²=(x+y)²-2xy...substitute that to (i)...(x+y)²-2xy+xy=84...(x+y)²-xy=84 from (*)...x+y=10...i.e...100-xy=84...xy=16...(*')...taking * and *' x+y=10 and xy=16...solve simultaneously to give x=2,y=8 OR x=8,y=2...PLS MY GURUS,IF YOU DON'T UNDERSTAND MY EXPLANATIONS,KINDLY TELL ME TO MAKE MY EXPLANATIONS CLEAR...in my own respective,i think this thread is not for teens... |
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rhydex 247: 1. Solve the equation. (x-1)^4+(x+3)^4=82....(1)...(x-1)^4+(x+3)^4=82...let {a be x-1}...then x+3=a+4...then a^4+(a+4)^4=82...expand to give...2a^4+16a^3+96a^2+256=82...divide through by 2...a^4+8a^3+48a^2+128a+87=0......solve the polynomial to give (a+1)(a+3)(a^2+4a+29)=0...a=-1 and -3...not that a^2+4a+29=0 gives imaginary number.....don't forget that x-1=a....when a=-1 then x=-1+1=0...when a=-3...x-1=-3...x=-3+1=-2....therefore x=0 or -2 |
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jackpot: Transform "7^sin x" to "e^(sin x ln 7)"....try to show humility sister...i could remember a question benbuks posted,and you asked him not to post simple question again...why can't you solve this,you are just using approximate method...Try to know this that:THERE'S NO MATHS GURU...you can just know up to the level you are and the aspect you are studying......WELL SHA,you tried... 1 Like |
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The are some numbers that there factors expect themselves are added up to them...e.g...6 with factors 1,2,3 and 6...without 6,the others are added to 6.....then 28 with factors 1,2,4,7,14and28,without 28...the others are added to 28...find the fifth number that can exhibit this |
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Sir-Tunechi:...got you bro...thank you very much...that's arithmetric mean |
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Kudos to everybody here |
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