Pink Poogle Toy Forum

Have you been given scenes to compare? If not, choose main scenes, perhaps the climatic endings, and another, I haven't studied this play, so, sorry IU can't be more specific.

Then look at film effects. What has the director chosen to do with the camera, what lighting is there. Look at what the charecters are doing.

Then, for each of these effects, write a paragraph, about half a page, comparing how each of these differ between the three films. Remember to include at the end of each, a little explanation of why each film did it like this, which is effective, to be honest, WHY?

I'd suggest also the death of Mercutio & Tybalt scenes, the balcony scenes, and the opening scene. They're all portrayed differently, so you should be able to do a good comparison.

I'm sorry this is something that's bugging me, just in case anyone uses it for reference: You divide both sides by 4, which equals 6. It was obviously a typo because it comes out fine in the end, but I just needed to say that. Thank you.

Edit: I hope someone replies to this soon. Anyway ...
In maths we were given a ten-by-ten grid. We would take a three-by-three sample of it, multiplied the diagonally opposite corners, and subtracted the smaller from the larger. We experimented with different sized grids as well. I've written formulae for why they always end up the same for some of them (I know this is confusing, but its not that important yet) and then I wrote the totals down, and wrote a formula for the difference between them. Then we got an eight-by-ten grid, and experimented with that. I found a link between them both. The ten-by-ten grid had a formula of (N^2)10 and the eight by ten grid was (N^2)8. I would like help with a formula for this too. If someone's willing to help, but needs more detail, just say so. I'd really appreciate this.

Edit: Deleted

Tuna, of course I'll help, I like this kinda thing, but, what numbers were put into a grid? Is it like a number grid counting from 1 to x?

Also, in your formula, what does N stand for?

Thank you! Hang on while I condense my posts into one ... there! By the way, you're the first to call me by my nickname, and I didn't even have to tell anyone what it was, yay! So it was rom one to a hundred on the first case, but because the second grid was smaller it was one to eighty. I'll try and make everything clearer by giving some examples.

3-by-3 section of a 10-by-10 grid:

3, 4, 5
13, 14, 15
23, 24, 25

3*25=75
5*23=115

115-75=40

If we call the first number (In the example 3) X, then the equation is
(X+2)(X+20) - (X)(X+22)
or
X^2+22X+40 - X^2+22X

So it must always equal 40, in a two-by two this would be 10 in a four-by-four 90. So the sequence:

N 1 2 3
T 10 40 90
So the formula is (N^2)10

For 8-by-10s the formula ends up as,
(N^2)8. So obviously, if we take the width of the square as X, then the sequence of totals is,
(N^2)X

I would like the formula for why, if you can't do it then never mind. I can feel the answer sort of, but I can't quite grasp it.

Okay, I think I get this.

Let's take the frid again:

3 4 5
13 14 15
23 24 25

However, if you call the first number n. you can write it like this:

n n+1 n+2
n+10 n+11 n+12
n+20 n+21 n+22

In which case, the formula would be:

(n+20)(n+2) - n(n+22)

Which simplifies:

n[sup]2[/sup] + 22n + 40 - n[sup]2[/sup] - 22n

Which simplifies to 40

Now lets take it with width 8:

n n+1 n+2
n+8 n+9 n+10
n+16 n+17 n+18

In this case, it equals:

(n+16)(n+2) - n(n+18)

Which simplifies:

n[sup]2[/sup] + 18n + 32 - n[sup]2[/sup] - 18n

Which simplifies to 32

Now, what happens if we take it width x?


n n+1 n+2
n+x n+x+1 n+x+2
n+2x n+2x+1 n+2x+2


You should be able to do it from there

Is that what you wanted?

That should do nicely! I'd already got all of the the first two bits down, but yes, that was the grid I was missing. Thank you, that's perfect! I think you have to do an open and close html tag for html to work.

Cool. If you need any help, please don't hesitate to reask.

I think it may be the DoHtml tags...

**tests

[html]
<table>
<tr><td>Hi!</td></tr>
</table>
[/html][/code]

Does anyone know how to say "You all are stupid" in Spanish?

I think it's "Ustedes son estupidos" but can somebody check that?

mm... i beileve thats fine... but you can also use "tonto/a" or "bobo/a" to get the point across as well.

You can also say "Vosotros sois estupidos" if you're familiar with the crowd. Ustedes is formal, vosotros is not.

http://e.1asphost.com/channy27/scan.jpg

The thing is, I haven't been taught about improper integrals yet, so I'm totally confused about how to approach this problem. Can anyone help? By the way, I've only been taught about u-substitution, so, if possible, use an integral integrable (say that ten times fast) by u-sub. Thanks.

Edit: Nevermind, we just covered improper integrals today.

Maths problem!
Integration:
∫ cosx e^x with upper limit 0, lower limit 2
My brain is too tired to think the answer...

You have to use Integration by Parts to solve this one. Let's solve it first, and then plug in the limits of integration after. Remember the formula:
uv - ∫udv.

∫e^x*cosx dx

Let v = cosx, dv = -sinx.
Let du = e^xdx, u = e^x.

Plug in, and this yields: ∫e^x*cosx dx = e^x*cosx - ∫e^x*(-sinx) dx

Factor out the negative from sinx to get: ∫e^x*cosx dx = e^x*cosx + ∫e^x*sinx dx.

Integrate ∫e^x*sinx dx by parts...
Let v = sinx, dv = cosx.
Let du = e^xdx, u = e^x.

Plug in, yielding: ∫e^x*cosx dx = e^x*cosx + e^x*sinx - ∫e^xcosx dx.

Notice ∫e^x*cosx dx on both sides of the equation. Combine them to get:

2∫e^x*cosx dx = e^x*cosx + e^x*sinx

Divide each side by 2.

∫e^x*cosx dx = (e^x*cosx + e^x*sinx)/2
Because we are in reality computing a definite integral, we do not need the constant of integration.

Now evaluate it from [0,2].

You should get 1.321959.