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Re: How To Calculate Quickly And Correctly In Mathematics by Nobody: 7:47pm On Jun 18, 2013 |
Ortarico: A little error there Chief Ortarico => log25^{x}_{5} - log1/25_{5} => log(5)^{^2(x)}_{5} - log(25)^{(-1)}_{5} => (2x)log5_{5} - log(5)^{2(-1)}_{5} => (2x)log5_{5} - (-2)log(5)_{5} Since log5_{5} = 1 => (2x)log5_{5} - (-2)log(5)_{5} => 2x .(1) + 2 => 2x + 2 => 2(x + 1) So the final answer should be => 2(x + 1) which is option B as solved by Omosivie 1 Like |
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 7:51pm On Jun 18, 2013 |
doubleDx: Ok thanks general doubleDx 1 Like |
Re: How To Calculate Quickly And Correctly In Mathematics by Calculusfx: 8:52am On Jun 19, 2013 |
To find the sum of positive integers...1+2+3+...+n=n(n+1)/2... |
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 2:32pm On Jun 19, 2013 |
Pls help me solve this: 1/root 2 -1 minus 1/root 2 +1. Thanks |
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 4:56pm On Jun 19, 2013 |
Omosivie,: NB: _/ is for the root 1/(_/2 - 1) - 1/(_/2 + 1) The LCM is (_/2 - 1)(_/2 + 1): => (_/2 + 1) - (_/2 - 1)/(_/2 - 1)(_/2 + 1) Clearing the bracket we'll get: => _/2 + 1 - _/2 + 1/_/4 + _/2 - _/2 - 1 => 1 + 1/2 - 1 => 2. . . . . . . .ans 1 Like |
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 6:36pm On Jun 19, 2013 |
Ortarico: |
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 1:28pm On Jun 21, 2013 |
Pls help me out with this two simultaneous equations: 1, xy=-3; x+y=2 2, x+2y=0; x^2+y^2=20 |
Re: How To Calculate Quickly And Correctly In Mathematics by Nobody: 6:54pm On Jun 21, 2013 |
Omosivie,:wat r d opti0ns? |
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 8:06pm On Jun 21, 2013 |
Omosivie,: Let's go. . . . 1. xy = -3. . . . . .eqn(i) x + y= 2. . . . . . .eqn(ii) from eqn(ii): x= 2 - y. . . . . .eqn(iii) substitute eqn(iii) in2 eqn(i): xy= -3 (2 - y) y= -3 2y - y^{2}= -3 Re-arrangng gives: y^{2} - 2y - 3= 0 (y - 3)(y + 1) y= 3 or -1 :. x= 1 or -3 To check: xy= -3; (-1*3)= -3 x + y= 2; (-1 + 3)= 2 2. x + 2y= 0. . . . . .eqn(i) x^{2} + y^{2}= 20. . . . .eqn(ii) from eqn(i): x + 2y= 0 x= -2y. . . . . . .eqn(iii) substitute eqn(iii) in2 eqn(ii): x^{2} + y^{2}= 20 (-2y)^{2} + y^{2}= 20 4y^{2} + y^{2}= 20 5y^{2}= 20 y^{2}= 4 y= +/- 2 substitute y= 2 in2 eqn(i): x + 2y= 0 x + 2(2)= 0 x + 4= 0 x= -4 To check: x + 2y= 0; -4 + 2(2)= 0 x^{2} + y^{2}= 20; (-4)^{2} + (2)^{2}= 20 1 Like |
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 8:42pm On Jun 21, 2013 |
Ortarico:Thanks, but the answer is not the same with my textbook's answer. According to my textbook, the answer for number one is -3 or 1 and the answer for number two is (4,2) or (-4,2). I don't understand dis textbook and i'm totally fed up. |
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 8:45pm On Jun 21, 2013 |
Leebliss13: wat r d opti0ns?No options |
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 8:50pm On Jun 21, 2013 |
Omosivie,: But that's it now. At no 1 x is (-3 , 1) then y is (-1 , 3) and at no 2 x is -4, then y is 2. To check that you saw was just to confirm. That's what I got sister 1 Like |
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 9:17pm On Jun 21, 2013 |
Ortarico:Thanks, I definately need glasses |
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 9:42pm On Jun 21, 2013 |
Omosivie,: Thanks too, it happens to everyone. |
Re: How To Calculate Quickly And Correctly In Mathematics by Calculusfx: 4:25pm On Jun 22, 2013 |
Remember this bro,the square root of a natural number is + or -...so,from where you got y^2=4...therefore y=+ or - 2...therefore y=+2 or -2...from the first equation x+2y=0...substitute y=-2...therefore x+2(-2)=0...x-4=0...x=4...substitute y=2...x+2y=0...x+2(2)=0...x+4=0...x=-4...therefore,when y=-2,x=4 and when y=2,x=-4 |
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 4:51pm On Jun 22, 2013 |
Calculusf(x): All right, but if you look at it well, you'll see that I typed +/- 2, I know the rule boss. 1 Like |
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 1:46pm On Jun 24, 2013 |
I hope i'm not bothering u guys with these "easy" questions but can u please help me out with them 1, m^3 + mn^2-nm^2-n^3/m^4-n^4 2, -12a^2x/-8a^2b^2 |
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 2:39pm On Jun 24, 2013 |
Omosivie,: 1. m^{3} + mn^{2} - nm^{2} - n^{3}/m^{4} - n^{4} m^{3} + n^{3} + mn^{2} - nm^{2}/m^{4} - n^{4} (m - n)^{3} + mn(n - m)/(m - n)^{4} mn(n - m)/(m - n) . . . . .ans |
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 2:40pm On Jun 24, 2013 |
2. Please re-type, there's a mistake or I don't get you well. |
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 3:09pm On Jun 24, 2013 |
Ortarico: 2. Please re-type, there's a mistake or I don't get you well.-12a^2 x / -8a^2 b^2 |
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 3:17pm On Jun 24, 2013 |
Ortarico:The answer @ the back of my textbook is 1/m+n |
Re: How To Calculate Quickly And Correctly In Mathematics by Nobody: 6:47pm On Jun 24, 2013 |
Omosivie,: {m^{3} + mn^{2} - nm^{2} - n^{3}}/(m^{4} - n^{4}) Simplifying m^{4} - n^{4} yields => (m^{2}^2 - n^{2}^2 => (m^{2} - n^{2})(m^{2} + n^{2})...difference of 2 squares. Rearranging the numerator & factorising yields => => mn^{2} - nm^{2} + m^{3} - n^{3} => mn(n - m) - (n^{3} - m^{3}) Now, (n^{3} - m^{3}) can be rewritten as => (n - m)^{3} + 3mn(n - m) - difference of two cubes, which makes the numerator => mn(n - m) - {(n - m)^{3} + 3mn(n - m)} Putting the numerator & denominator together yields: =>> [ mn((n - m) - {(n - m})^{3} + 3mn((n - m))]/(m^{2} - n^{2})(m^{2} + n^{2}) Factorising (n - m) from the numerator above yields=> =>> [(n - m)][mn - {(n - m)^{2} + 3mn}]/(m^{2} - n^{2})(m^{2} + n^{2}) =>> [n - m][mn - {(n^{2} - 2mn + m^{2} + 3mn)]/(m^{2} - n^{2})(m^{2} + n^{2}) =>> [n - m][mn - {(n^{2} + m^{2} + mn)]/(m^{2} - n^{2})(m^{2} + n^{2}) =>> (n - m)(-n^{2} - m^{2})/(m^{2} - n^{2})(m^{2} + n^{2}) =>> (n - m)(-1)(n^{2} + m^{2})/(m^{2} - n^{2})(m^{2} + n^{2}) =>> (m - n) =>> (m - n)/(m^{2} - n^{2}) =>> =>> 1/(m + n) Hence, {m^{3} + mn^{2} - nm^{2} - n^{3}}/(m^{4} - n^{4}) = 1/(m + n) 1 Like |
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 7:12pm On Jun 24, 2013 |
doubleDx:Wow!!!! This is so complex(one of the reasons why I hate maths). Could this be further maths?? I just pray I'm not given this question in my upcoming exam cos it's too long, difficult and complex. Thanks for solving it. |
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 7:14pm On Jun 24, 2013 |
@ Ortarico, Calucus(fx) and DoubleDx. Pls help me out with the number 2 question |
Re: How To Calculate Quickly And Correctly In Mathematics by Nobody: 7:33pm On Jun 24, 2013 |
Omosivie,: It's not Further Maths lol, the solution is not as long as it appears, though my explanation is, so you could understand better. It's just a matter of understanding two expressions and using them at the right time => 1. That A^{2} - B^{2} = (A - B)(A + B) - difference of two squares 2. That A^{3} - B^{3} = (A - B)^{3} - 3AB(A - B) - difference of two cubes So that A^{4} - B^{4} = (P^{2} - Q^{2}) => (P - Q)(P + Q), such that =>> P = A^{2} and Q = B^{2}. Understanding these expressions and knowing when to apply/fix them in when you factorize makes the question weak and easy to solve. I hope you catch my drift... so, you see? It's easy, don't be discouraged god sister! 1 Like |
Re: How To Calculate Quickly And Correctly In Mathematics by Nobody: 7:47pm On Jun 24, 2013 |
Omosivie,: Solution -12a^{2} x /-8a^{2} b^{2} => This is easy seeing as -4a^{2} is a factor of the numerator and denomitor, cancelling out & reducing the expression to => For the numerator => -12a^{2}/-4a^{2} = 3x For the denomitor => -8a^{2} b^{2}/-4a^{2} = 2b^{2} Putting the simplified numerator and denomitor together yields => => 3x/2b^{2} |
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 8:01pm On Jun 24, 2013 |
doubleDx: Respect to you @ general doubleDx. I supposed to have used the law of cubic and quadratic expression but for my hurry I missed it. Please help explain more at the five starred phase of your work, thanks general! 1 Like |
Re: How To Calculate Quickly And Correctly In Mathematics by Nobody: 9:12pm On Jun 24, 2013 |
Ortarico: [ mn (n - m) - {(n - m)^{3} + 3mn (n - m)}]/(m^{2} - n^{2})(m^{2} + n^{2}) If you look at the numerator of the above expression carefully, you will realize that (n - m) is a common factor/coefficient @ mn (n - m) , (n - m)^{3} and 3mn (n - m) So that if (n - m) is replaced with say Q, the numerator would be => [ mn Q - {(Q)^{3} + 3mn(Q)}] which reduces the numerator to => mnQ - {Q^{3} + 3mnQ} mnQ - Q^{3} - 3mnQ Now, if Q is factorized from the above, it becomes => Q(mn - Q^{2} - 3mn) => Q(-2mn - Q^{2}) => -Q(2mn + Q^{2}) Remember that Q = (n - m), so that -Q = (m - n), putting back the values of Q & -Q into the numerator's expression, yields => (m - n)(2mn + (n - m)^{2}) (m - n)(2mn + n^{2}+ m^{2} - 2mn) (m - n)(n^{2} + m^{2}) Putting the simplified numerator back into the expression yields => =>> (m - n) =>> (m - n)/(m^{2} - n^{2}) =>> =>> 1/(m + n) I hope this is clearer... 2 Likes |
Re: How To Calculate Quickly And Correctly In Mathematics by Ortarico(m): 9:41pm On Jun 24, 2013 |
doubleDx: Oh yes, thanks boss, one love! 1 Like |
Re: How To Calculate Quickly And Correctly In Mathematics by Calculusfx: 9:45pm On Jun 24, 2013 |
You have really done a great job@general doubledx... |
Re: How To Calculate Quickly And Correctly In Mathematics by Nobody: 10:56pm On Jun 24, 2013 |
^ Thanks bruv! |
Re: How To Calculate Quickly And Correctly In Mathematics by Omosivie1(f): 9:31am On Jun 25, 2013 |
doubleDx:Thanks, I'll cram these algebraic identities/expressions. |
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