I need some help with my homework. Does anyone know what the first known record of a witch was? I've looked everywhere, but I couldn't find it. Also, I need to know what the proper name for the place where witches were put to death. Any help would be greatly appreciated. :mrgreen:

 

"Witches" were around long before written record. The first recorded one? Idk. But! The death penalty for witchcraft was first issued by Emperor Charlemagne of Frankia in the 8th century.

 

Proper name? Europe or the Americas. Or if you want to get technical and go with the stereotypical "burning" of witches, that would be stake. eg. Burned at the stake.

 

Generally they just hanged them, though.

 

 

 

Don't cite me as your source.

 

 

 

Dusty668, www.tip.it/forum, 10th September 2009, "Burning Witches Comment", First, Online, 10th September 2009, http://forum.tip.it/viewtopic.php?f=10&t=808746

 

 

 

If he wants to ... this is the prescribed Harvard System of Referencing. :thumbup:

 

Heh, just put that into my Works Cited page. Had to change it a little so it wold be MLA.

does anyone know how to (*if its possible*) to add password protections to folders in vista, without having to download any extra programs? im sick of my siblings looking through my files :@ thank god they havent got the code for my current "folder maze", but its overcomplicated to do, even for me...

I don't believe Vista has a feature that allows you to lock a folder so you can only open it with a password.

 

 

 

Hunter

http://windowshelp.microsoft.com/Windows/en-US/help/93e008e4-929c-4904-ab54-65f5a40fe7fa1033.mspx

 

I don't know what else to tell you :/

I've been studying for my maths exam (which is in .. 2 hours.. eek.) and I came across a question which I didn't know how to answer. Could anyone please explain it to me?

 

 

 

 

 

 

Generally when I'm trying to make a quadratic equation I use this 'model quadratic equation':

 

 

 

ax² + bx + c

 

 

 

But it didn't work for me this time :/

 

 

 

Edit: I ended up forming the correct equation (x² + 3x + 1) and I ended up with the correct answer, 20.

 

:D

wongtong what the hell did you do ?

 

 

 

The [general] correct equation for the number of tiles in any one set is

 

 

 

n² + (n-1)

 

 

 

Where n is the number of tiles arranged vertically.

 

 

 

So, with 461 such tiles. Solve for n.

 

 

 

So, the answer would be 21 [for n]

 

 

 

EDIT: Explained

 

 

 

21² + (21-1) = 441 + 20 = 461

 

 

 

But you already knew that. :P

 

 

 

To solve n² + (n-1)

 

 

 

I usually just draw a graph. Too long to solve it algebraically.

Dear god. I can't do high school maths anymore.

Dear god. I can't do high school maths anymore.

 

 

 

:lol:

 

 

 

I was never good at maths either.

Dear god. I can't do high school maths anymore.

 

 

 

Hell, I'm in high school and I can't even figure it out. :?

I'm a sophomore in Honors Pre-Cal and I couldn't get it. At least I'm not alone. :wall:

You have three points: (1,5) (2,11) (3,19)

 

5= a+b+c

 

11=4a+2b+c

 

19=9a+3b+c

 

 

 

You can use elimination to solve for a,b, and c.

Wongtong, your solution was right.

 

Adrenal, n² + (n-1) gives 1 tile for pattern 1, and 5 tiles for pattern 2. You're actually off by 1 pattern, so the answer is 20.

 

dave's approach is correct and will lead you to the right soution, though it's a bit too complicated for what you need.

Here is my explanation of quadratic patterns.

 

 

 

[hide=]The model equation: ax² + bx + c

 

The model equation translated to the problem we're working with:

 

T = an² + bn + c

 

 

 

 

You can determine that a pattern is a quadratic pattern because the 'DIFFERENCE OF DIFFERENCE' is the same. (2)

 

Now to get a (for the model equation), it is half of the difference of difference.

 

So half of 2 is ... 1!

 

So a = 1.

 

 

 

T = 1n² + bn + c

 

 

 

Let's substitute a number for n. (n is the pattern number)

 

Let's say n = 1.

 

 

 

(1 x 1²) + (b x 1) + c = 5 (I got the 5 from the table)

 

1 + b + c = 5

 

b + c = 5 - 1

 

b + c = 4

 

 

 

Substitute another number for n.

 

Let's say n = 3.

 

 

 

(1 x 3²) + (b x 3) + c = 19

 

9 + 3b + c = 19

 

3b + c = 19 - 9

 

3b + c = 10

 

 

 

Now take the underlined equations and make a simulatneous equation.

 

 

 

3b + c = 10

 

b + c = 4

 

2b = 6

 

b = 3

 

 

 

Substitute b back into an equation such as b + c = 4

 

c = 4 - 3

 

c = 1

 

 

 

The model equation is now

 

T = n² + 3n + 1

 

 

 

Substitute T = 461

 

461 = n² + 3n + 1

 

460 = n² + 3n

 

n² + 3n - 460 = 0

 

I type this into my graphics calculator to get

 

(n-20)(n+23) = 0

 

n is therefore 20 and -23.

 

It can't be negative so it's 20.[/hide]

 

Adrenal, I can't graph things because it's algebra, not graphs. It asks for an algebraic equation :P

All you people SUCK ! :P

 

 

 

Adrenal, n² + (n-1) gives 1 tile for pattern 1, and 5 tiles for pattern 2

 

 

 

Lolwut ?

 

 

 

Not the sequence of the pattern but the number of tiles.

 

 

 

So... Pattern 1 has two vertical tiles.

 

 

 

2² + (2-1)

 

 

 

= 5

 

 

 

What's wrong ?

 

 

 

I think I read the actual question wrong. :wall:

 

 

 

In that case, for the pattern number - the equation is - n² + 3n + 1

 

 

 

wongtong, your method is, long, drawn-out and unnecessarily complicated.

What is wrong with you people? The answer is clear if you just look at the pictures of the tiles. The two consistent elements are the squares and the individual tiles to the right. The squares' sides are one tile larger than the pattern number (n+1) so the number of tiles in that square is (n+1)^2 since it's a freaking square. Then, to the right of each square, they added the same number of tiles as the pattern number, or +n.

 

 

 

So we get (n+1)^2 +n = #tiles

 

n^2 +3n +1 =#tiles

 

 

 

After you plug in the 461 just use that crazy formula that solves quadratic equations in the form ax^2 +bx +c that they teach you in highschool

i thought the forums weren't for homework help?

It wasn't for homework, I was studying.

 

And it's not long and drawn out, it only looks long because I take ages to explain it.

 

I can figure it out in 3 minutes. (or less) and it works in basically every situation so :P

still sounds like homework help to me. it's schoolwork, it's almost the same. the only difference is it wasn't specifically assigned by a teacher or something.

still sounds like homework help to me. it's schoolwork, it's almost the same. the only difference is it wasn't specifically assigned by a teacher or something.

 

What's wrong with that? :(

still sounds like homework help to me. it's schoolwork, it's almost the same. the only difference is it wasn't specifically assigned by a teacher or something.

 

What's wrong with that? :(

 

Not your personal study group.

So we get (n+1)^2 +n = #tiles

 

n^2 +3n +1 =#tiles

 

 

 

fastortoise, that's what I did.

 

 

 

You have to solve it relative to the pattern number, rather than number of tiles.

still sounds like homework help to me. it's schoolwork, it's almost the same. the only difference is it wasn't specifically assigned by a teacher or something.

 

What's wrong with that? :(

 

Not your personal study group.

 

To be honest I don't see how it's different from any other question. "I'm not your psychiatrist."

 

Anyways, it's not like other people mind helping.

So we get (n+1)^2 +n = #tiles

 

n^2 +3n +1 =#tiles

 

 

 

fastortoise, that's what I did.

 

 

 

You have to solve it relative to the pattern number, rather than number of tiles.

 

I don't think it's about helping anymore.

Not your personal study group.

 

Yeah, but we all learned something new about our math professionals here.

 

 

 

And how Lateralus is not one of those.

I know this is really vague, but I need to get a new cell phone. Any ideas?