Do my Penguin for me

[Paj!]

I actually hate English AS level.

 

I've literally been staring at my first draft of "In what ways does J.D Salinger present Holden Caulfield as an innocent?" for 7 hours 11 minutes. I deleted one sentence and corrected one spelling mistake.

[MoogleViper]

I actually hate English AS level.

 

I've literally been staring at my first draft of "In what ways does J.D Salinger present Holden Caulfield as an innocent?" for 7 hours 11 minutes. I deleted one sentence and corrected one spelling mistake.

 

That book is amazingly boring. I can't stand that piece of crap.

[Twozzok]

I think so but it's ages since I've done electicity.

 

Apparently either our school has faulty equipment or taking the Ammeter out affects the voltage significantly :/

 

Variable Resistor time ¬_¬

[Shino]

Just a pointer, to use the ammeter you need to close the circuit the ammeter itself.

[chairdriver]

I hate my core 2 maths teacher, since he doesn't actually teach us the course, just gives us shite hand-outs.

 

 

The first three terms of a geometric progression are 4, 2, 1.

 

Find the twentieth term, expressing your answer as a power of 2.

 

Find also the sum to infinity of this progression.

 

 

And

 

 

Simplify loga(a^2) - 4loga(1/a)

 

 

Thanks.

[Twozzok]

Just a pointer, to use the ammeter you need to close the circuit the ammeter itself.

 

Well, I don't understand what you mean but I ended up using this circuit.

 

 

EDIT: Excuse the bmp filebeing massive !_!

[Supergrunch]

I hate my core 2 maths teacher, since he doesn't actually teach us the course, just gives us shite hand-outs.

 

 

The first three terms of a geometric progression are 4, 2, 1.

 

Find the twentieth term, expressing your answer as a power of 2.

 

Find also the sum to infinity of this progression.

 

 

And

 

 

Simplify loga(a^2) - 4loga(1/a)

 

 

Thanks.

Yeah, I just got hand outs for stats, 'twas annoying.

 

4, 2, 1,... is a geometric series with first term, a, of 4 and a common difference, r, or 1/2, or 2^(-1).

 

The nth term is given by a*r^(n-1), making the 20th term:

 

4*(2^(-1))^19 = (2^2)*(2^(-19)) = 2^(-17)

 

And the sum to infinity is given by a/(1-r) = 4/(1-(1/2)) = 8.

 

For the second:

 

loga(a^2) - 4loga(1/a)

 

By loga you mean log to the base a right? If so, this simplifies to:

 

2 - 4(-1) = 6

[chairdriver]

Thanks. I got 2^-17, but by a weird way.

 

For the second:

 

loga(a^2) - 4loga(1/a)

 

By loga you mean log to the base a right? If so, this simplifies to:

 

2 - 4(-1) = 6

 

Yeah I mean log to base a.

 

Can you go through the steps, because I'm really confused about log rules?

 

I mean, can you just add together two log functions as you would other stuff?

 

 

Thanks.

[Supergrunch]

Yeah I mean log to base a.

 

Can you go through the steps, because I'm really confused about log rules?

 

I mean, can you just add together two log functions as you would other stuff?

 

 

Thanks.

Well, you need to understand the definition of a logarithm, which is the following:

 

If a^x = y, then loga(y) = x.

 

This takes a bit of getting used to, but it helps to think of logging as the opposite of raising to a power, much as division is the opposite of multiplication. You can also express loga(y) in words, it means "what power do you raise a to to get y?"

 

This definition has a few important consequences, whatever the base (but providing all logs are of the same base), which are:

 

log(x^y) = y*log(x)

log(x) + log(y) = log(xy)

log(x) - log(y) = log(x/y)

 

Also, implicit in the definition is:

 

logx(x) = 1

 

Logarithms of zero and negative numbers are undefined.

 

I'm not sure what you mean by "add together two log functions as you would other stuff", but the rules you'll need are above.

 

Anyway, the question was:

 

loga(a^2) - 4loga(1/a)

= 2loga(a) + 4loga(a)

= 2 + 4

= 6

[chairdriver]

Cool.

 

Thanks alot.

[Ellmeister]

Given that x is sufficiently small find the first three terms of the binomial expansions of (1+x)^-2 and (1-4x)^-1

 

Hence find the first three terms of the expansion of 3+2x^2/ (1+x)^2(1-4x)

 

 

I can do first bit easy.

 

I got, 1-2x+3x^2. And 1+4x+16x.

 

For the second part, I'm stuck on what to do. I tried to expand the top part into root3(1+2x/3)^2 and then got weird numbers: root 3 + 4root3x/3 + 4root3x^2/9

 

I'm just a little stuck on what to do, can someone like give me a clue or just go through it clearly please? thanks.

[jayseven]

I actually hate English AS level.

 

I've literally been staring at my first draft of "In what ways does J.D Salinger present Holden Caulfield as an innocent?" for 7 hours 11 minutes. I deleted one sentence and corrected one spelling mistake.

I've no idea if you're still doing this essay, but can I ask what aspects you're looking at? His interactions with other characters? His speech? Nis naivity? If you want to send me your first draft I can give you a few pointers, hints, tips, clues, hazards, chances, bungo bongos.

[Supergrunch]

Given that x is sufficiently small find the first three terms of the binomial expansions of (1+x)^-2 and (1-4x)^-1

 

Hence find the first three terms of the expansion of 3+2x^2/ (1+x)^2(1-4x)

 

 

I can do first bit easy.

 

I got, 1-2x+3x^2. And 1+4x+16x.

 

For the second part, I'm stuck on what to do. I tried to expand the top part into root3(1+2x/3)^2 and then got weird numbers: root 3 + 4root3x/3 + 4root3x^2/9

 

I'm just a little stuck on what to do, can someone like give me a clue or just go through it clearly please? thanks.

Do you mean (3+2x)^2/ (1+x)^2(1-4x)? If so, you need only multiply out the bracket, divide by each of the two terms you calculated earlier, then simplify. Remember that binomial expansions are finite for powers that are postive integers.

[Ellmeister]

No its actually 3 + 2x^2 on top. Which is what has confused me. That and what you do when putting it all together :S

[Supergrunch]

No its actually 3 + 2x^2 on top. Which is what has confused me. That and what you do when putting it all together :S

Then it's even easier, as you don't need to multiply out anything at all.

 

You have A = (3+2x^2)(1+x)^(-2)(1-4x)^(-1)

(I arbitrarily called it A)

 

We know that: (1+x)^(-2) = 1 - 2x + 3x^2 + ... and (1-4x)^(-1) = 1 + 4x + 16x^2 + ...

 

So A = (3+2x^2)(1)(1) + (3+2x^2)(-2x)(4x) + (3+2x^2)(3x^2)(16x^2) + ...

 

= 3 + 2x^2 - 24x^2 - 16x^4 + 144x^4 + 96x^6 + ...

 

= 3 - 22x^2 + 128x^4 + ... (the x^6 term is the 4th, so I removed it)

 

Anyway, that's the approach. The calculations might be a bit wrong, I did it quickly.

 

Also: For 3+2x^2, root3(1+2x/3)^2 is incorrect. it is only correct for (3+2x)^2.

[Ellmeister]

So basically you do,

 

(3+2x^2) x (1-2x+3x^2) x (1+4x+16x^2)?

 

Or did I go wrong again? I ended up getting 3 + 6x +35x^2 from what I did (after reading maths book thingy)

[Supergrunch]

So basically you do,

 

(3+2x^2) x (1-2x+3x^2) x (1+4x+16x^2)?

 

Or did I go wrong again? I ended up getting 3 + 6x +35x^2 from what I did (after reading maths book thingy)

Erm, actually I think that's right, and mine's wrong.

[Ellmeister]

Thanks for the help! I followed what you said and then checked the last little part and I'm guessing you just multiplied a bit wrong due to you doing it so quick.

 

Thanks for all the help, again!

[Paj!]

I've no idea if you're still doing this essay, but can I ask what aspects you're looking at? His interactions with other characters? His speech? Nis naivity? If you want to send me your first draft I can give you a few pointers, hints, tips, clues, hazards, chances, bungo bongos.

 

Hi, thanks for replying, and I MIGHT actually take you up on that offer, but for Part 2, which I have to redraft seriously (got it totally wrong first time, but apparently what I said was right, just muddled it up).

 

The essay I posted about just got an "Excellent work!", so i'm happy. (For now... )

[jayseven]

Well that's good news Sure, any part is fine.

[The Bard]

I love CITR, great book. I'll write your essays for you :p.

[nightwolf]

Ok sorry to be dragging this thread up but I really could do with the help:

 

Specify a web server, and associated resources, suitable for the website and understand the security issues and copyright issues associated with using a web server.

 

 

Basically I need a web server to whack the website I made (the website isn't real etc) and I need specs like firewalls and stuff and then talk about security and things like that.

 

I'm stuck because I really don't understand anything about using one what the ''issues'' are or where I can find one that tells me everything i need to know.

[Paj!]

Catcher In The Rye IS amazing...

[system_error]

Ok sorry to be dragging this thread up but I really could do with the help:

 

Specify a web server, and associated resources, suitable for the website and understand the security issues and copyright issues associated with using a web server.

 

 

Basically I need a web server to whack the website I made (the website isn't real etc) and I need specs like firewalls and stuff and then talk about security and things like that.

 

I'm stuck because I really don't understand anything about using one what the ''issues'' are or where I can find one that tells me everything i need to know.

 

Dunno if I fully understand what you want but why not start simple and "make" your own webserver. You need a simple computer, Apache or IIS and you are done. I know you don't actually have to do that but when you try it out it might be easier to see where the possible loopholes are.

 

Forgot to change the default password, allow remote access, every directory is accessible, outdated software, spyware, DoS attacks - there are plenty of mistakes you can make and this is just the server itself. Websites can also be dangerous with all that ActiveX stuff, scripts, faulty browsers. Addons like php, asp and similar can increase security problems if they are not really maintained that good.

[Strider]

I need some help. Im revising for stats and i'm crap at permutations. Anyway will someone explain to me how i'd do this:

 

Each of the 7 letters in the word DIVIDED is printed on a seperate card. The cards are arranged in a row:

 

In how many arrangements are all three D's together?

 

I'm thinking it's something dead easy and it's me being thick.

 

Also, how many arrangements start with I?