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Re: Algebra Cafe: Solve Math Problems Here by JUNIT1(m): 6:37am On Apr 07, 2008 |
LET ME JUST SHW TO THE PEOPLE THAT DON'T KNOW ALREDY A BASIC MATHS CALCULATION. ANY NUMBER TIMES 11 WLL HAVE THE SAME NUMBER TWICE. E.G 9 X 11= 99 ANY DOUBLE NUMBER TIMES 11 WILLL ADD THE TWO NUMBERS AND THE SUM WILL GO IN THE MIDDLE E.G 24 X 11=2[b]6[/b]4 (2+4+6) SPECIAL CASE IF THE TWO NAMBER ADDED GIVE YOU A DOUBLE NUMBER JUST FOLLOW THIS PROCEDURE. E.G 39X 11= 429( 3+9= 12 SO YOU JUST ADD 1 TO THE THREE AND PUT THE 2 IN THE MIDDLE) CHECK YOUR CALCULATOR FOR EASY REFERENCE. AND ASK ME ANY QUESTIONS IF YOU DON/T UNDERSTAND. |
Re: Algebra Cafe: Solve Math Problems Here by JUNIT1(m): 6:38am On Apr 07, 2008 |
LET ME JUST SHW TO THE PEOPLE THAT DON'T KNOW ALREDY A BASIC MATHS CALCULATION. ANY NUMBER TIMES 11 WLL HAVE THE SAME NUMBER TWICE. E.G 9 X 11= 99 ANY DOUBLE NUMBER TIMES 11 WILLL ADD THE TWO NUMBERS AND THE SUM WILL GO IN THE MIDDLE E.G 24 X 11=2[b]6[/b]4 (2+4=6) SPECIAL CASE IF THE TWO NUMBER ADDED GIVE YOU A DOUBLE NUMBER JUST FOLLOW THIS PROCEDURE. E.G 39X 11= 429( 3+9= 12 SO YOU JUST ADD 1 TO THE THREE AND PUT THE 2 IN THE MIDDLE) CHECK YOUR CALCULATOR FOR EASY REFERENCE. AND ASK ME ANY QUESTIONS IF YOU DON/T UNDERSTAND. |
Re: Algebra Cafe: Solve Math Problems Here by JUNIT1(m): 6:40am On Apr 07, 2008 |
engineered if you have a problem with this thread and mathematics. ddon't come on this thread go to a science one or something, everyone has different opinions |
Re: Algebra Cafe: Solve Math Problems Here by Nobody: 7:01am On Apr 07, 2008 |
well, i dont have a problem with this thread, cause i do love mathematics also, but i think i was just trying to make a point. well maybe i can start my own thread, for people who want to think outside the box. GOD BLESS NAIJA. |
Re: Algebra Cafe: Solve Math Problems Here by RichyBlacK(m): 9:40am On Apr 07, 2008 |
engineerd: @engineerd, You are a fool! For addressing people you don't know in that manner makes you a complete and utter idiot! Do you think we are all "armchair engineers" with only theoretical knowledge? It's small boys like you with profound emptiness that make Nigeria look really bad. How can you come up here and insult us all like that? You are a senseless slowpoke, an empty barrel, a nonentity, and a charlatan. Rubbish! |
Re: Algebra Cafe: Solve Math Problems Here by PLC1: 10:21am On Apr 07, 2008 |
But where are the ladies? They can't come here and solve maths with guys. |
Re: Algebra Cafe: Solve Math Problems Here by Olumide7(m): 10:35am On Apr 07, 2008 |
For which values of r is the following sequence convergent? an=nrn |
Re: Algebra Cafe: Solve Math Problems Here by donchichi: 3:32pm On Apr 07, 2008 |
The questions here have now gone beyond algebra. |
Re: Algebra Cafe: Solve Math Problems Here by Olumide7(m): 4:06pm On Apr 07, 2008 |
If it's not algebra, then what is it? Geometry? Trigonometry? Someone had to put something challenging up here. Anyway I need people to try before I show you how it's done, that's if no one gets it right. The question is: For which values of r is the following sequence convergent? an=nrn |
Re: Algebra Cafe: Solve Math Problems Here by bawomolo(m): 4:15pm On Apr 07, 2008 |
well, i don't have a problem with this thread, cause i do love mathematics also, but i think i was just trying to make a point. well maybe i can start my own thread, for people who want to think outside the box. GOD BLESS NAIJA. so what problems have u solved and what designs have u proposed mr.nerd. it's mugu's like u that probably hated on theoreticians like albert einstein. theoreticians and experimentalists should work together rather than abuse each other. |
Re: Algebra Cafe: Solve Math Problems Here by bawomolo(m): 4:18pm On Apr 07, 2008 |
an=nrn this series never converges since it doesn't approach zero. oh wait let me guess. can't it convert for -1<r<1 ? |
Re: Algebra Cafe: Solve Math Problems Here by Olumide7(m): 5:00pm On Apr 07, 2008 |
bawomolo: Wrong
close, you need to be more precise and you also have to show your workings. |
Re: Algebra Cafe: Solve Math Problems Here by JUNIT1(m): 6:06pm On Apr 07, 2008 |
.The question is: For which values of r is the following sequence convergent? an=nrn [quote][/quote] [b]OLUMIDE THIS QUESTION IS AN INSULT TO MY INTELLIGENCE, DON'T BRING HIGH SCHOOL QUESTIONS HERE, (WELL HIGH SCHOOL MATHS IN S.A)[/b][/color][color=#990000][/color][color=#990000][/color][color=#990000]ANYONE HAVE SOMETHING CHALLENGING FOR ME. |
Re: Algebra Cafe: Solve Math Problems Here by bawomolo(m): 6:18pm On Apr 07, 2008 |
i don't see why we need workings when it can be gotten through visual examinations. for a sequence to converge, the sum of the the series must not be infinity, it has to be converge at a finite number i believe. 0<r<1 rn all approach infinity beyond this range. numbers within the range converge to zero. it oscillates for numbers between -1<r<0 |
Re: Algebra Cafe: Solve Math Problems Here by Olumide7(m): 6:38pm On Apr 07, 2008 |
@ J-UNIT Reverse psychology at its best. @ Bawomolo You are 40% right. if 0<r<1, then 1/r >1 and we apply l'hospital to the function f(x) = x/(1/r)x. Hence lim x tends to infinity of f(x) = 0. Hence, if 0<r<1 then the limit as n tends to infinity of an = the limit as n tends to infinity of of nrn which is all equal to 0, therefore an is convergent. If -1<r<0 then by the first case |an| tends to 0 and hence an tends to 0. if r=0 then an=0 and an converges. if r>1 then an=nrn >n and so an diverges if r<-1 then an=n(-1)n diverges Hence, the sequence converges whenever |r|<1. |
Re: Algebra Cafe: Solve Math Problems Here by RichyBlacK(m): 6:39pm On Apr 07, 2008 |
Olumide7: Using the ratio test: lim |an+1/an| = p; p < 1 <=> series converges; p > 1 <=> diverges; p=1, inconclusive Given an = nrn So, an+1 = (n+1)r(n+1) an+1/an = r + r/n lim |r + r/n| = |r| Hence, sequence converges if |r| < 1. |
Re: Algebra Cafe: Solve Math Problems Here by Olumide7(m): 6:43pm On Apr 07, 2008 |
@RichyBlack I see u used the ratio test, How did u arrive at that answer? Did u skip any steps? and I don't think the ratio test can be used in this case. Well, your solution is not convincing enough. and for the record this is a sequence not a series. Ratio tests can only be used for series |
Re: Algebra Cafe: Solve Math Problems Here by RichyBlacK(m): 6:48pm On Apr 07, 2008 |
Olumide7: The d'Alembert ratio test can be used to solve the problem. I skipped the absolute condition initially, but later corrected it. |
Re: Algebra Cafe: Solve Math Problems Here by bawomolo(m): 6:55pm On Apr 07, 2008 |
my answer is correct using the absolute value theorem. i just used visual examination. obviously this isn't my forte. |
Re: Algebra Cafe: Solve Math Problems Here by RichyBlacK(m): 7:15pm On Apr 07, 2008 |
Olumide7: @Olumide7 That statement is incorrect! You fail to see the link between series and sequences. Theorems governing series are used to determine the behavior of sequences. Every series has its corresponding sequence - the series is just a function of the sequences (addition). A convergent series implies that the corresponding sequence also converges. Though, the reverse is not necessarily the case. Also, if the series is divergent, no credible statements can be made about its corresponding sequence. Illustration: an = 1/n The series Σan diverges but the sequence an converges. an = 1/2n The series converges, and the sequence MUST converge. So, showing that a series converges is sufficient to show that the corresponding sequence converges. |
Re: Algebra Cafe: Solve Math Problems Here by bawomolo(m): 12:44am On Apr 08, 2008 |
solve the simultaneous equation |
Re: Algebra Cafe: Solve Math Problems Here by Cadet(f): 12:47am On Apr 08, 2008 |
bawomolo:Using elimination or substitution? |
Re: Algebra Cafe: Solve Math Problems Here by MrPataki1: 1:32am On Apr 08, 2008 |
bawomolo: [center]x - y + z = 10. . . . . . . . . . . .(1) 3x + y + 2z = 34. . . . . . . . . . (2) -5x + 2y -z = -14. . . . . . . . . .(3)[/center] (from equation 1) :- [center]z = 10 - x + y. . . . . . . . . . .(4)[/center] Substitute equation (4) into equation (2) :- [center]3x + y + 2(10 - x + y) = 34[/center] [center]x + 3y = 14. . . . . . . . . . . . . (5) [/center] Substitute equation (4) into equation (3) :- [center]-5x + 2y - (10 -x + y) = -14 4x - y = 4. . . . . . . . . . . . .(6)[/center] From equation (5) :- [center]x = 14 - 3y. . . . . . . . . . .(7)[/center] Substitute equation (7) into equation (6) [center]4(14 - 3y) - y = 4 56 - 4 = 12y + y 52 = 13y[/center] :- [center]y = 4. . . . . . . . . (8 )[/center] Substitute equation (8 ) into equation (7) :- [center]x = 14 - 3y = 14 - 3(4) = 2[/center] Substitute both (x) and (y) into equation (1) :- [center]2- 4 + z = 10 z = 12.[/center] x =2, y = 4, z = 12 |
Re: Algebra Cafe: Solve Math Problems Here by bawomolo(m): 5:11am On Apr 08, 2008 |
Using elimination or substitution? any method, there's more if u want to try your hands. x + y + z = 9 x + 2y + 3z = 23 x + 5y - 3z = -7 it's more comfortable solving it with linear algebra by the way. |
Re: Algebra Cafe: Solve Math Problems Here by RichyBlacK(m): 11:41pm On Apr 08, 2008 |
bawomolo: Using Crammer's Rule (Linear Algebra) Aw = b where A = 1 1 1 1 2 3 1 5 -3 w = x y z b = 9 23 -7 Determinant of A , or det(A) = -12 Replacing each column of A with b to get A1, A2, and A3 respectively: A1 = 9 1 1 23 2 3 -7 5 -3 A2 = 1 9 1 1 23 3 1 -7 -3 A3 = 1 1 9 1 2 23 1 5 -7 Get the determinants of these matrices: det(A1) = -12 det(A2) = -24 det(A3) = -72 Get solution: x = det(A1)/det(A) = 1; y= det(A2)/det(A) = 2; z = det(A3)/det(A) = 6 |
Re: Algebra Cafe: Solve Math Problems Here by Scarlett(f): 12:33pm On Apr 09, 2008 |
Someone once said "As long as there is Algebra, there will always be prayer in Schools" If x > -2, simplify the expression 2| x + 2 | - 3x - (-2 - x) + | 6 - 9 | Easy. |
Re: Algebra Cafe: Solve Math Problems Here by bawomolo(m): 3:40pm On Apr 09, 2008 |
2| x + 2 | - 3x - (-2 - x) + | 6 - 9 | 2x+4-2X+2+3 4+2+3 =9?? |
Re: Algebra Cafe: Solve Math Problems Here by bawomolo(m): 3:57pm On Apr 09, 2008 |
[img]http://www.phx.devry.edu/academics/test_center/concentric.jpg[/img] In the figure above, both circles have the same center, and the radius of the larger circle is R. If the radius of the smaller circle is 3 units less than R, what represents the area of the shaded region? |
Re: Algebra Cafe: Solve Math Problems Here by Scarlett(f): 4:50pm On Apr 09, 2008 |
Large Circle Rad= R Small Circle rad= R-3 Area of a Circle= TT R^2 Area of Shaded Circle = Area of Large Circle-Area of Small Circle Where; Area of Large Circle= TT R^2 Area of Small Circle = TT(R-3)^2 Therefore, Area of Shaded protion= TT R^2 - TT (R-3)^2 ?? |
Re: Algebra Cafe: Solve Math Problems Here by bawomolo(m): 12:12am On Apr 10, 2008 |
you are right, if u simplify it further, it's pi(6R-9) Perform the function operations given the following functions: F(x) = 3x^2 + 1 and g(x) = square root of x a. f(x+h) [img]http://www.mc.maricopa.edu/dept/d25/math/AssetReview/COLST/Image84.gif[/img] f(x)/g(x) f(g(x)) f(g(4)) |
Re: Algebra Cafe: Solve Math Problems Here by bawomolo(m): 4:38pm On Apr 11, 2008 |
Solve the following equations: [img]http://www.mc.maricopa.edu/dept/d25/math/AssetReview/COLST/Image102.gif[/img] [img]http://www.mc.maricopa.edu/dept/d25/math/AssetReview/COLST/Image104.gif [/img] [img]http://www.mc.maricopa.edu/dept/d25/math/AssetReview/COLST/Image103.gif[/img] |
Re: Algebra Cafe: Solve Math Problems Here by FOD(m): 5:04pm On Apr 11, 2008 |
bawomolo:2x + 1 ------- = 3 x2 2x + 1 = 3x2 Rearranging, 3x2 - 2x - 1 =0 (3x+1) (x-1) = 0 x= -1/3 or 1 bawomolo:3/(x2 +2) = 4 - 1 3/(x2 + 2) = 3 Cube both sides, (3/x2 + 2)3 = 33 x2 + 2 = 27 x2 = 27 - 2 x2 = 25 x = +5 or -5 |
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