₦airaland Forum

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shame shame shame

[quote author=WHξξ∟s]I thought about that earlier but i think one can understand the question more if he/she thoroughly examine the nature of the question by visiting the sister-thread . Anyway i will post the question as you suggested and if one needs further digging he shall just view the sister-threads content and how other people try to deal with the question b4.

It goest this way:
3x+2y-7=19 ------(1)
4x-y+27=4 --------(2)
2x+4y-52=32 -------(3)

in this case collecting like terms wont work cos the author of this question want us to find the unknown variables preferably using the elimination method.. I know it looks peculiar[cos 3 variables need to be there not 2] but the author manage to provide solution out of this obfuscating equation.. Hence you knw of the question, now i will stick 2 dis thread for a while and see how our maths gurus will get d numbers of the unknown[3] variables[z included]..

Check the siser thread here: www.nairaland.com/1199966/maths-homework-please[/quote]Hey, what's this?

Sapphire Ȋ̊§ one of those..., no no i must not talk to her, i must not give in to passion and thirst.

Hi, I am back from slumber.I know... some would see my user name and pass, I have acquired a wondrous reputation, the cross eyed dimwit who's known to open whack threads. Honestly, I don't give!

This is about what I feel when I see "agreeable ladies" and its driving be bunkers. I feel a burst of desire so strong I have to leave such location, I crave for sex more than ever before, the more I indulge the more the craving. I need to grip my self before I do something stupid and rash. As hard as I try to pull away I end up seeing them wherever..., scantly dressed chocolate damsels, with bodies that would make angels envious. I need help...

I see them on nl too, I want I want I want...

People like her are granted second chances, she is probably somewhere happy with her new family. Life is a circle, and some worthy souls get to be recycled...

WackyJ1: they have been trying since but the guy did not hear, it was time to use fire

Lol

fazbuk: (1)The power developed in a resistor R by a battery of emf E and internal resistance r is given by P=E^2R/(R r)^2.Differentiate P wit respect to R and show dat the power is a maximum when R=r. (2)An electrical voltage E is given by E=(15sin50pie 't' 40cos50pie 't')volts.where 't' is the time in seconds.Determine d Maximum value of voltage.

1) This is a max/min problem under differentiation.

P=E^2R/(R r)^2
P=E^2R(R r)^(-2) differentiating by product rule
dP/dR=-2(E^2)R(R r)^(-3) (E^2)(R r)^(-2)

For max/min point dP/dR=0

-2(E^2)R(R r)^(-3) (E^2)(R r)^(-2)=0

2(E^2)R(R r)^(-3)=(E^2)(R r)^(-2)
2R=(R r)
R=r

If you wanted to know whether it was a maximum point, take a second derivative and slot in value for R if d^2P/(dR)^2<0 then you are ok.

for question 2

E=15sin(50 πt)+40cos(50 πt) volts

once again we find dE/dt and equate to zero to find max/min value of E

dE/dt=15(50 π)cos(50 πt)-40(50 π)sin(50 πt)

15(50 π)cos(50 πt)-40(50 π)sin(50 πt)=0
15(50 π)cos(50 πt)=40(50 π)sin(50 πt)
15/40=sin(50 πt)/cos(50 πt)
15/40=tan(50 πt)

now we would have gone on to find time t and substitue inside E to get its maximum value, men... no time, me like shortcut.If you can remember sohcahtoa , we use this to manipulate

15/40=tan(50 πt) so, opp=15, adj=40 hyp=(15^2+40^2)^1/2

sin(50 πt)=15/(15^2+40^2)^1/2
cos(50 πt)=40/(15^2+40^2)^1/2

E=15sin(50 πt)+40cos(50 πt) volts, substituting our values

E=15[15/(15^2+40^2)^1/2]+40[40/(15^2+40^2)^1/2] volts
E=15^2/(15^2+40^2)^1/2+40^2/(15^2+40^2)^1/2 volts
E=(15^2+40^2)/(15^2+40^2)^1/2 volts
E=(15^2+40^2)^1/2 volts
E=42.72 volts

[quote author=mr.drizzy]fbk group-0:3.whatsapp group[/quote]you are funny

OCTAVO: So you mean this all about you? It's only your own selfish interest you are protecting? What about the interest of others? So you mean it's you first before others? I see the kind of leader you can be. All you did was to simply outrule the suggestion of whatsapp? Which in every sense is better and safer than facebook. Mehn I detest your personal interest in this. And fyi there's nothing bad in running the two together! I expected you to come up with that sensible idea in order to balance the situation. But your personal interest in facebook won't let you do that. And fyi, either you like it or not, we are going to create the whatsapp group and run it alongside with fb simultaneously.

what's wrong with you man, how old are you sef? You just displayed your immaturity

colombiana: Since its stinking stop kissing and start givin h.ead cuz at times dat place and ur mouth might smell the same and she wud love u more

I'd rather have bad breath than be a fool

luluosas: Men and brenthren, all hope is not lost yet in using this Glo BB plans on non BB device. This is what Glo twitted at me today, though I am yet to try it, but will do so now. Update will be rolling in shortly.

This response is vague.

TOSIN ACCA: Kindly help me solve this question:
Quantitative Ability
Fifteen men take 10 days to complete the job working 12 hours a day. How many hours a day should 10men work to complete the job in 20 days?
(A) 5
(B) 9
© 6
(D) 8
(E) 7

15men take 10days.....12h/day to do a job
15men take (10x12)h to do the job..............[10x(3/2)men]
10men take (3/2)x(10x12)h to do the job........[less men more hours, by the same ratio]
10men take (180)h to do the job
10men take (20)x(9)h to do the job
10men take 20days.....9h/d to do the job

Answer is 9

luluosas: Register now now by visiting www.twitter.com

whats your twitter handle?

xzit: I haven't yet checked their rates in other countries. But this is their resident country and we are not just customers but citizens. That has to count for something. And if the saying that ''the customer is always right'' is anything to go by then we have a standing right to demand how we want to be treated. Least we get is a reasonable explanation and the best we get is what we demand. At the end of the day we would have made them take us seriously.

I am with you sir, don't have a twitter handle yet, but I think its time to get one. It's up to us to stand up and ask for a better service.

donedy: This was a question I asked myself back in SS2 when I was so into circular geometry, fortunately my physics teacher asked that same question during my third year, which I answered arrogantly. This is the answer:

A straight line is the arc of an infinite circle. In other words, as the radius of a circle tends to infinity, the arc takes on the form of a straight Line.

The principle is useful somewhere in Quantum Theory.

I salute all the yet-to-be geniuses that tried the problem. It gives me joy that Nigeria still has a bright future ahead.

I thought about that, but I shrug off the idea simply because as diameter increases so does d circle or arc dimensions. I would have said as the arc length (s) gets smaller and smaller to an infinitesimal small distanse (ds) the arc tends to a straight line, in short all curves are made of infinitesimal straight lines joined together back to back. I thought you wanted an analytical answer. lol, @ yet-to-be geniuses. I was busy playing ps1 and sega in ss2, contra hard cops was my favorite at the time...

For question 2)

I don't really grasp the question sha, let me give it a try though

Equation of an arc(circle) is x^2+y^2=r^2
To find slope dy/dx
dy/dx=[r^2-x^2]^1/2

so a line tangent to the arc at any point would be y=(dy/dx)x+d where d is a constant.

so, y=(-x/[r^2-x^2]^1/2)x+d at any point (x,y) on the arc

I don't know it joo, lol

donedy: You got the answer right, but I'm confused on why you implied that y^² x^²=(2a)^² because that will expand the rectangle away from the circle. However, you only know the logic behind your algorithm. The money is still on: Solve the remaining.

I would have done it this way:

we that x² y² = a²

Area (A) = x*y (Which is one quarter of the area)

Thus A(x) = 4xy (Total Area, since each has to be equal in that the center of the circle is at the origin)

==> A(x) = 4x(a² - x²)^1/2

Then find the min/max point as you did.

Ok, you have a square inside a circle, and for the square to be the biggest size posible its verteces most touch the circle at four ponts. Now, two ponts of contact happen to define diameter of the circle, as well as diagonal of the square

We have diagonal of square=diameter of circle

Diagonal of square=2a
Remeber, sides of square(rectangle) are x and y

Square is made of two right angle triangles, so pythagoras applies. Therefore x^2+y^2=(2a)^2

donedy: Hello Richiez,questions 2 and 3 still need a deeper thinking. However, you tried though. Let me rephrase them.

2) Define the shape a Straight Line in terms of an arc and it's radius. (Hint: limit)

3) Calculate the area of the biggest rectangle that be inscribed in a circle of form x² + y² = a², where a is a constant. (Hint: zero slope)

*** The price is still on. Two attempts remaining.

let me start with no 3)
Assuming the rectangle to be a square (very logical), lets see...

Let ABCD be the rectangle inscribed in the circle of radius a
Let x be the length, y be the width of the rectangle
AC is the diagonal of ABCD and is the diameter (2a) of the circle
so y^2+x^2=(2a)^2 using pythagoras

Now,
y^2=(2a)^2-x^2
area of rectangle is A=xy

A=xy
A=x[(2a)^2-x^2]^1/2

dA/dx=0 at max/min point

dA/dx=x(1/2){[(2a)^2-x^2]^(-1/2)}(-2x)+[(2a)^2-x^2]^1/2
0=x(1/2){[(2a)^2-x^2]^(-1/2)}(-2x)+[(2a)^2-x^2]^1/2
x^2[(2a)^2-x^2]^(-1/2)=[(2a)^2-x^2]^1/2
x^2=(2a)^2-x^2
2(x^2)=(2a)^2
x=a(2)^1/2

to test whether it is actualy maximum we find second derivative which is negative and therefore less than 0, so we have a maximum here.

y^2=y^2=(2a)^2-x^2 substituting for x gives y=a(2)^1/2
A=[a(2)^1/2][a(2)^1/2]
A=2a^2

solving...

[quote author=Ukwu_Nnu]Mmehie ala

Eradicating mmehie for real is my call to service. [/quote]So you dey tie ukwu and amu for a living, lol

A tuo otu nga, eke amu eriri, lol

Ukwu_nu, explain?

Hotdiamond: wow! Thats's nice. I neva knew you were glo's New PRO. MtcheewWWWW SMH

Hey mr, if you are not interested in glo, why don't you f..uck uff

Good news, airtel is so slow

na wao, I no fit shout.

dmcdad: You are right... Tall peeps tend to weigh more. I was weighing 80kgs some two years ago and now I weigh just 65kgs. Same me oooo. It varies anywayz

Dude, were you ill? 15kg is a lot oooo.

Ok, just saw ur pix, you look good. You still got a big neck.

wonderful! I am tall and slimmmmmm, and lean with packs that pop out when I press my tommy . But I noticed tall people tend to be heavier, my friend who weighs same as me, this dude is shorter, bulky and round, was I flabbergasted ?

Richiez: nice work, u'r on ur way to earning the title 'GENERAL'

Bros don't forget to invite me at the inaugural ceremony, long live the generals, long live all maths luvers, you all rock!

[quote author=Ileke-IdI]

I'm sorry but this ish is hilarious. The fact that someone wold harm himself for a cause that Romanians don't care about is very very foolish[/quote]Romanians, hahaha. You obviously don't know shit about geography. lol

doubleDx: ^ Write out the question properly, if possible write it out on a paper, then scan and attach. That way, we'll understand your question properly!

whats up man, was looking for ur pix on the other thread.

obadiah777: I DONT THINK HE GOT WHIPPED BECASE OF $20. ITS A REPUTATION THING. HE IS A DRUG DEALER. HIS REPUTATION IS ON THE LINE. IF EVERY CRACKHEAD IN TOWN KNOW THAT THEY CAN GET AWAY WITH NOT PAYING FOR THEIR DRUGS THEN THE GUY IS OUT OF BUSINESS. SO YOU HAVE TO MAKE AN EXAMPLE OUT OF SOMEONE SO THE OTHER CRACKHEADS WILL BE SCARED. IN THIS CASE THE CRACKHEADS SON JUST HAPPENED TO BE THE EXAMPLE. NOW IF THE CRACKHEAD HEARS ABOUT THIS HE WILL QUICKLY COME UP WITH THE $20. ITS ALL BUSINESS. NEVER PERSONAL. YALL CANT BE MESSING WITH MY MANS REP. HE GOT BILLS TO PAY. HIS REP IS ALL HE GOT. HIS REP IS IMPT IN THIS LINE OF BUSINESS

this is sad but true, it's business

[quote author=Ileke-IdI]I'm still the same old me you dislike. But if you think I care about your feelings for one second, you've got another thing come.
Lets just enjoy the thread and do away with trifling sentiments.[/quote]I heard you.