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Education › Re: Nairaland Mathematics Clinic by jackpot(f): 12:08pm On Dec 06, 2013

benbuks: ..wow thanks man, you just safed my job.

....*madam jackpot , a'v cracked it ma..,oya give me my allowance, else i 'll go on strike.*.....lolz.

lolz.

I've got a better offer na. What would a N250.00k allowance do for you? undecided

I am thinking about raising your retirement age from 65 to 80 years. cheesy

how about that? wink

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 10:51pm On Dec 03, 2013

Laplacian: ?

BOUNDARY(P n Q) IS A SUBSET OF [BOUNDARY(P) u BOUNDARY(Q)]z....u didnt back up d above assertion in ur proof and its a vital part of d proof...a more logical consequence of ur explanation should b;
BOUNDARY(P n Q) IS A SUBSET OF [(P) u (Q)]...bcuz since boundary(P) and boundary(Q) lies outside P n Q, it follows directly dat boundary PnQ lies in P u Q

again; (BUT NOT BOTH AT THE SAME TIME)...OBVIOUSLY does not follow from d fact dat P & Q are open: as an illustration, supposin Q lies wholly in P, then x€ boundary of both P and Q....which contradicts ur assertion...

somehow, I perceived maybe you are mixing up the term "boundary" of a set and upper/lower bounds of a set. These are very different concepts.

To prove that your argument is a basket, lets come down to IR.

Let P=(2,7) and Q=(4,5). Clearly, P and Q are open sets and Q lies wholly in P.
Now, the boundary of P is the 2-point set {2,7};
boundary of Q is the 2-point set {4,5}.
Therefore,
boundary(P) u boundary(Q)={2,7}u{4,5}={2,4,5,7}

Now,
P n Q = (2,7) n (4,5) = (4,5).
Therefore,
boundary(PnQ)= {4,5}.

Is {4,5} not a subset of {2,4,5,7} ?
That is, in this example,
is boundary(P n Q) not a subset of [boundary(P) u boundary(Q)] ?

I will withdraw the statement that "if x€ boundary(PnQ), then either x€P or x€Q" and replace it with "Let x€boundary(PnQ), then if x€P, then x is not in Q, and if x€Q, then x is not in P.

As for the proof of the "assertion", draw pictures of two overlapping sets like circles with dotted edges (the way you draw venn diagram). Label the two sets P and Q

boundary of the set P are those points lying on the edges, but since it is dotted lines, it is not inside the set.

Only pictures will clarify things more, bro Lap! wink

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 7:31am On Dec 02, 2013

@Laplacian and Humphrey,

I said criticize, not to type question mark and to reply "GOOD". grin

Well, if you can draw pictorially two open sets(remember to use dotted lines for the edges), then you will understand this proof. wink

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 3:08pm On Dec 01, 2013

Laplacian: @Humphry, i suggest u & Benbuks should always provide d incorrect steps in any wrong solutions like jackpot alwaz does and myselt too, not 2 simply dismis it & leave d solver in doubt&confusion...even if my solution does not cover d general case, it has @ least settled d case pointed out by jackpot abov....and dat should worth some apprreciation...
u 'r modest lady jackpot & i like ur styl of criticism....i 've not really wrked on provin d general case though, but i 'll giv it a second thought now that u hav underlined it...but i know it will hav 2 incoperate d NORM and follow a similar chain of reasonin as abov....by d way, wat's preventin u from supplying a proof? Or 're u just watchin 4rm d sidelines just 2 handpick my errors?....
@ALL, am not here 2 enter into competition wit anybdy, 4 those dat copy questions (witout havin challenges wit them) from textbooks & paste here.....i'll only attempt questions only by individuals who hav difficulties wit them & are genuinely desperate 4 their solution...

I will provide a lazy solution to the problem sha.

A set is open if none of its points are boundary points.

Now, let P and Q be open sets. Then boundary(P) lies outside P and boundary(Q) lies outside Q.

Now, P n Q is a subset of P. P n Q is also a subset of Q. This means that boundary(P) and boundary(Q) lies outside P n Q.

Boundary(P n Q) is a subset of [boundary(P) u boundary(Q)]. Infact if x € boundary(P n Q), then either x € P or x € Q (but not in both at the same time) since P and Q are open sets.
Thus, boundary(PnQ) lies outside PnQ.

Hence, PnQ is open.

Feel free to critize this solution, folks.

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 1:25am On Dec 01, 2013

Humphrey77: sin2x+sinx is equal to 1 we know that sin2x is equal to 2sinxcosx. clearly; we have 2sinxcosx +sinx equal to 1 . so let sinx equal to y; by subtitution into

sin2x + sinx equal to 1 we obtain the equation: 4y⁴-3y²-2y+1 equal to zero ; by solving for y we obtain y equal to 1; clearly recall that sinx is equal to y so we have x equal to arcsin(1) we have x equal to 90 so x is equal to 90

Sweetheart, I am sorry to burst your bubbles but your answer is regrettably very myopic and careless. For you to see what I meant,

4y⁴-3y²-2y+1=(y-1)(4y³+4y²+y-1)
This is a quartic equation with four roots: 2 real and 2 complex conjugate roots.

The real roots are y=1 and y=0.347810385. . .
Now, for y=1,
sin x=1 --> x= sin^-1(1)=90 + 360k, where k is an integer.
i.e., x={. . .-630,-270,90,450,810,. . .}.
Clearly this is an infinite set of solutions for x.

Next, for y=0.347810385. . .,
x=sin^-1(0.347810385. . .)=20.35344681+360k, where k is an integer.
This also yields an infinite set of solutions.

For the other 2 complex roots of y, complex solutions of x exists, but i will stop here, since the person that posed the question was probably expecting only real solutions.

So, dear Humphrey, if I should score your solution,

As an SS1 student, I'll score you 75%.

As an SS3 student, I'll score you 50% because you failed to get the other real solution of y.

As a year one student, I'll score you 40%. #LetMyPeopleGoThings# tongue

As a final year student, na 0% for sure, since the number of solutions for x (in order to get the full marks) is infinite, and you merely provided one.
See how I arrived at your score:
1/infinity * 100%=0% tongue

1 Like

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 12:25am On Dec 01, 2013

Humphrey77: good work! BUT U NEED TO GO AND STUDY YR NUMBER THEORY VERY WELL

This guy, e be like say u snuff a little alcohol, bah? Why are u quoting people and replying nonsense? I drafted a solution that will convince even non-mathematicians on how the answer is gotten. I am not a stereotyped mathematician.

A mere 5 seconds look at the question, I pictured the answer. Integral powers of 3 ends with the cycle of digits {3, 9, 7, 1}.

1998Mod4=2

and 3 raised to power of the residue class of 2 yields 9.

and 9Mod5=4.

Maybe thats how you want the solution to look like?

SMH.

1 Like

Jokes Etc › Re: Offtopic "Bar Joint" For Jokers (season II) by jackpot(op): 9:19pm On Nov 30, 2013

sorry to burst your bubbles, Sir.

ayam not a Lesbo!! tongue

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 9:15pm On Nov 30, 2013

^you are assuming that all your sets are subsets of IR, the set of real numbers and you are also assuming that elements can be compared by your use of less than sign. ( < )

my question is, what if the sets are open in R²? Or what if they are domains in the Complex Plane?

Can your proof work for an arbitrary open set?

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 7:17pm On Nov 30, 2013

Laplacian: guy am stil an amatur ooo, (not yet a topologist)..
Let a<m<A and b<n<B denot d two open sets...
suppos, on d contrary, dat their intersection, p, is a closed set...
Then we can find x and y such dat
x<=p<=y, where a, b<x and y<A, B
now suppose, without loss of generality, dat A<B....then we can always find d numbers z such dat
y<z<A, since z<A<B, z should b in their intersection, but y<z a contradiction, showing dat z is not in their intersection...hence our supposition dat p is a closed set is untenable....and d result follows

Hi Sir Laplacian

how can a set be less than or equal to a point x?

Jokes Etc › Re: Offtopic "Bar Joint" For Jokers (season II) by jackpot(op): 3:33pm On Nov 30, 2013

Efemena_xy:

Good morning Jackie!

morning Efe_seXY! wink

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 3:02pm On Nov 30, 2013

Humphrey77: if 3 raise to power 1998 is divided by 5 what is the remainder

Laplacian: 4

If you're not cleared as to how the answer above is gotten, note that for any natural number n,

3⁴ⁿ=(3⁴)ⁿ = 81ⁿ = 10M+1, for some integer M.

Similarly,
3⁴ⁿ⁺¹=3(3⁴ⁿ)=30M+3
3⁴ⁿ⁺²=3²(3⁴ⁿ)=90M+9
3⁴ⁿ⁺³=3³(3⁴ⁿ)=270M+27.

Observe that 3¹⁹⁹⁸ can be written as 3⁴ⁿ⁺², where n=499.

Now, for some integer M,

3¹⁹⁹⁸=3⁴ⁿ⁺²=90M+9=18(5)M+(5+4)=5(18M+1)+4.

From, the last equation, it is obvious why 3 raised to power 1998 leaves a remainder of 4 when divided by 5.

Gracias.

Politics › Re: Anambra Supplementary Election: Live Updates by jackpot(f): 8:39am On Nov 30, 2013

[size=16pt]If you are so eager(just like me wink

) waiting for this supplementary election to be concluded and our dear APGA to be declared winner and waiting for the imminent political demise of Dr Chris Olaniyi Ngige, hit LIKE!!! [/size] :-D cool

11 Likes

Jokes Etc › Re: Offtopic "Bar Joint" For Jokers (season II) by jackpot(op): 12:23am On Nov 30, 2013

Efemena_xy: Na only for Jackie's bar, pessin go dey advertise for business...

If for say I arrest dem, come kick them enta police cell, dem go say na dem be bread-winner and i no go hear word for human rights people. Dusts wey Oshiomhole raise recently neva settle. wink

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 7:34am On Nov 29, 2013

^lolz. . .okay dear

so, here we go! cool

express that number inside the limit sign as that same over 1.

Now, to make things look nice, multiply numerator and denominator with the square root of x.

If we substitute now, you will get (1-1)/0=0/0

so, we can now apply L'Hospital's rule to get the limit as tends to zero of (2sqrt(x))/(sqrt(1+2x))

substitute 0 now to get 0/1=0

Kia kia cheesy

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 12:12am On Nov 29, 2013

benbuks: ..here

limit equals 0.

Politics › Re: Chief Willie Obiano Declared Winner of Anambra Governorship Election by jackpot(f): 6:33am On Nov 18, 2013

up APGA

1 Like

Politics › Re: Anambra Decides: Results Coming In by jackpot(f): 8:20pm On Nov 16, 2013

If your ear is like mine itching for APGA to be declared winner tomorrow morning, hit LIKE!!! wink

8 Likes

Politics › Re: Anambra Decides: Results Coming In by jackpot(f): 8:20pm On Nov 16, 2013

If your ear is like mine itching for APGA to be declared winner tomorrow morning, hit LIKE!!! wink

12 Likes

Politics › Re: Liveupdate On Anambra Election From The 21 Local Government-pls No Comments by jackpot(f): 4:12pm On Nov 16, 2013

Update:

Obiano winning landslide in Ihialla, Agulu, Obeledu, Aguluzoigbo, Onitsha South, Adazi. Stay tuned for more.

1 Like

Politics › Re: Anambra Election: Inec Citizens Contact Centre: Reporting Live by jackpot(f): 12:36pm On Nov 16, 2013

First time I'm seeing a Police Helicopter with Siren monitoring the elections.

Thank God for the credible elections we have witnessed so far.

Up APGA!!!

6 Likes

Islam › Re: Is It Permissible To Close Your Eyes During Solat? by jackpot(f): 8:58pm On Nov 15, 2013

LONGFELLOW02: [size=30pt]BOKO HARAM PPLE[/size]

Islam › Re: Is It Permissible To Close Your Eyes During Solat? by jackpot(f): 5:37pm On Nov 15, 2013

LONGFELLOW02: [size=30pt]BOKO HARAM PPLE[/size]

Lolz

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 5:20pm On Nov 15, 2013

yeah, my post is to you.

Yeah, sure. Your's is shorter and sexier. I agree.

I commend your novel approach.

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 12:46pm On Nov 15, 2013

^ ^absolutely brilliant and novel!!!

But that doesn't mean that your solution is very brief though, since one has to compute P(5<=x<=10). See as you just quote am as 27/36 as if you won't draw sample space first to get am. cheesy

Once again, I commend your novel approach. It's very sexy!!! wink

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 10:23pm On Nov 14, 2013

smurfy: Really? I thought my method was watery and was expecting some rebukes from thee followed by a grandiose way of solving it.

Is there another method?

well, you could have used permutations anyway.

Like arrangement of 6,5, 1 is 3P3=6
then, do the same for all and add.

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 7:50pm On Nov 14, 2013

smurfy: Here goes...

(a) Sample space = 6^3 = 216

Set of combinations which gives a sum of 11 include:
164, 146, 461, 416, 641, 614, 155, 515, 551, 245, 254, 425, 452, 524, 542, 236, 263, 326, 362, 623, 632, 344, 434, 443, 335, 353, 533.

That's 27 ways of getting the sum 11. So answer to question (a) is 27/216 = 3/24

(b) Set of combination which gives the sum 12 include:
156, 165, 516, 561, 615, 651, 246, 264, 426, 462, 624, 642, 255, 525, 552, 336, 363, 633, 345, 354, 435, 453, 534, 543, 444.

That's 25 ways of getting the sum of 12. That gives 25/216.

Nice!!!

Education › Re: Nairaland Math Quiz Winner l::::::JARYEH::::::l by jackpot(f): 4:56pm On Nov 14, 2013

doubleDx: Yeah, you did great! keep it up! Wishing you and others luck in the next round!

General jackpot...howdy?

I'm fine, Super-General doubleDx

never mind my non-participation sha. I dont wanna start what I may not finish cos the quiz is strictly timed. My schedule is not very flexible.

I don think am before. Refreshing mails, collecting mails, resending correct candidates to Richiez, responding to Richiez's mail, posting questions, solutions and replying people's post; all these in a record time.

You guys are doing a great job of doing all these.

Kudos!

2 Likes

Education › Re: Nairaland Mathematics Clinic by jackpot(f): 4:00pm On Nov 14, 2013

Hi all, please solve this.

If three fair dice are rolled simultaneously ONCE, what is the probability of getting:
(i) a sum of 11?
(ii) a sum of 12?

Show FULL working cheesy

Education › Re: Nairaland Math Quiz Winner l::::::JARYEH::::::l by jackpot(f): 3:56pm On Nov 14, 2013

smurfy: 8.35 and 10. Where's the approximation?

Hehehe. I understand. The question didn't say the denomination was naira and that coins are not accepted.

For example, if na for Yankee, you can still pay 8$ 33cents cheesy

#TeamPureMathematiciansDontApproximateOften# cheesy

2 Likes

Education › Re: Nairaland Math Quiz Winner l::::::JARYEH::::::l by jackpot(f): 3:49pm On Nov 14, 2013

sorry, I was fast to type it so that some may find it useful since the time allowed to solve it was short. Hence d lil mistake. My bad!

smurfy: You mean p + q = 1 formula?

I split the probability into three for each three odd and like wise even numbers.
i meant
use 2/9 for number 2,
2/9 for 4
2/9 for 6
all in all 2/9+2/9+2/9 = 2/3 (total prob. for even)

1/9 for number 1
1/9 for 3
1/9 for 5
all in all 1/9 +1/9 + 1/9 = 1/3 (total prob for odd)

Education › Re: Nairaland Math Quiz Winner l::::::JARYEH::::::l by jackpot(f): 3:18pm On Nov 14, 2013

Expected Value Problem, I see.

Hint: use 1/9 for probability(even){e.g. Numbers 2,4,6} and 1/18 for probability(odd){Numbers 1, 3, 5}

then apply the formula

only geniuses can solve this without aid cheesy

Education › Re: Nairaland Math Quiz Winner l::::::JARYEH::::::l by jackpot(f): 1:59pm On Nov 14, 2013

Hint to Richiez question! : -X

find the sum of the AP with a=101 as the first term, common difference d=3 and the number of terms n=300 cheesy

**just passing ooo** wink

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