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[Ronike]

I'm your friend Xiivi :D

ANyone here have Castlevania Harmony of Despair?

 

[Tom]

xiivi has no friends

im willing to host a newbie game full of non newbies

i will then paw it off to xiivi as i shirk my responsibilities and take cat naps

 

[~ Gheb ~]

ANyone here have Castlevania Harmony of Despair?

I only have Aria of Sorrow [used to have Symphony of the Night, Order of Ecclesia and Portrait of Ruin]. I never even heard of Harmony of Despair tbh. What system is it?

Edit: Don't you mean Harmony of dissonance?

 

[Ronike]

No I mean harmony of despair. Its on the XBOX, and its multiplayer.

 

[~ Gheb ~]

.joel might have it. I know he mentioned it before.

 

[Tom]

so much work.... it will be so worth it....

 

[Tom]

also

^^ xiivi, after a successful day of scum hunting

 

[Tom]

 

[M.K]

Xiivi is a shiny Zangoose, not a Buizel.

Does not compute :[

 

[Tom]

he is a buizel

>:C

 

[adumbrodeus]

I'm mad cause I didn't get invited to FF7 mafia. *pouts* Obviously I'm the best player here and should therefore be invited to all games, right?


...


Guys?


*crickets chirp*

 

[~ Gheb ~]

Can't write adumbrodeus without "dumb"...

 

[#HBC | FrozeηFlame]

OH SNAP!

You can't get at Gheb lol

 

[adumbrodeus]

But, but, I'm the best evar, how can you not realize this gheb?


*gets pimpslapped*

 

[#HBC | Mac]

lezz bring back aimmafia

omni, tom, marshy, kevinm
lez goo

 

[Xiivi]

everyone should know my fav pokemon is buizel

 

[KevinM]

All I know is I feel attraction to you.

 

[X1-12]

"In the event of a modkill, the Day will typically end if the player being killed is town-aligned and continue otherwise. Deadline extensions and exceptions to this rule may be made based on the state of the game and my personal judgment."

why is this rule in place in most mafia games?

 

[Kirbyoshi]

If a player does something worthy of a modkill, and they are town, it gives scum the immediate chance to NK. If the player is scum, it gives town a continued chance to lynch scum. It's basically there to punish the offending player's entire faction, not just the single player. Also, people could just go "ok, prove to us you're town by doing something to get modkilled so we don't waste a lynch on you."

 

[Tom]

SHERLOCK, the 3-episode series by the BBC, is the greatest television show ever.

The first episode holds a special interest to each and every one of you in Decisive Games. I would go download it, immediately, if I were you.

Seriously.

 

[KevinM]

I miss most of the people I talked to in Dgames all the time.

:\

 

[Pythag]

Are you stating that you miss the people you used to talk to,

or that you are always missing the people that talk in Dgames?

 

[KevinM]

Both Py-baby :\

 

[M.K]

Calculus problem.
I wouldn't normally post here, but god, I hate my teacher. She gives us a sheet of **** we've never done before, tells us that she's gonna take NO questions about it, but it counts as a quiz grade....

lim
x-->0

of 1-cos(x)/2sin^2(x)

 

[adumbrodeus]

I miss most of the people I talked to in Dgames all the time.

:\

Which is why you should join bioware mafia.

 

[#HBC | Mac]

SHERLOCK, the 3-episode series by the BBC, is the greatest television show ever.

The first episode holds a special interest to each and every one of you in Decisive Games. I would go download it, immediately, if I were you.

Seriously.

if it has to do with sherlock holmes or something related then...

IM me a dl link please?

 

[Ronike]

if it has to do with sherlock holmes or something related then...

IM me a dl link please?

Same here please?

 

[vanderzant]

Calculus problem.
I wouldn't normally post here, but god, I hate my teacher. She gives us a sheet of **** we've never done before, tells us that she's gonna take NO questions about it, but it counts as a quiz grade....

Basically, you've got an equation of the form 0/0. I can't see any way to really simplify I it, so use L'hopital's rule which basically means that if you have a function of the form 0/0, its limit will be equal to the limit if you differentiate both top and bottom. So do just that

1-cos(x)/2sin^2(x)
= sin(x)/(4 sin(x) cos(x))
= 1/4cos(x)

So, as x -> 0; cos(x) approaches 1, therefore the limit is
= 1/4

 

[Clownbot]

I wouldn't necessarily expect to enter but is another PNS gonna happen anytime soon? I'd love reading more and there was some interest in creating one not too far back- hell, I think enough people showed interest to fill up a roster.

 

[Nicholas1024]

Basically, you've got an equation of the form 0/0. I can't see any way to really simplify I it, so use L'hopital's rule which basically means that if you have a function of the form 0/0, its limit will be equal to the limit if you differentiate both top and bottom. So do just that

1-cos(x)/2sin^2(x)
= sin(x)/(4 sin(x) cos(x))
= 1/4cos(x)

So, as x -> 0; cos(x) approaches 1, therefore the limit is
= 1/4

Or, if you'd rather play around with the algebra...

(1-cos(x))(1+cos(x))/(2sin^2(x)(1+cos(x)))
= (1-cos^2(x))/(2sin^2(x)(1+cos(x)))
= sin^2(x)/(2sin^2(x)(1+cos(x)))
=1/(2(1+cos(x)))
= 1/4 as x goes to zero.

So yeah, it's 1/4.

 

[M.K]

Or, if you'd rather play around with the algebra...

(1-cos(x))(1+cos(x))/(2sin^2(x)(1+cos(x)))
= (1-cos^2(x))/(2sin^2(x)(1+cos(x)))
= sin^2(x)/(2sin^2(x)(1+cos(x)))
=1/(2(1+cos(x)))
= 1/4 as x goes to zero.

So yeah, it's 1/4.

Basically, you've got an equation of the form 0/0. I can't see any way to really simplify I it, so use L'hopital's rule which basically means that if you have a function of the form 0/0, its limit will be equal to the limit if you differentiate both top and bottom. So do just that

1-cos(x)/2sin^2(x)
= sin(x)/(4 sin(x) cos(x))
= 1/4cos(x)

So, as x -> 0; cos(x) approaches 1, therefore the limit is
= 1/4

Oh. My. ****ing. God.
I love you both SO much. You guys are wayyy too helpful to be free.

So, uh, this problem set is literally making me cry, so I'm just gonna dump some of the harder problems here and if you have the time, PLEASE PLEASE PLEASE help me solve them ;_;

If lim x-->3 f(x) = 7, which of the following must be true?

1) f is continuous at x = 3
2) f is differentiable at x = 3
3) f (3) = 7

a) None
b) 2 only
c) 3 only
d) 1 and 3 only
e) 1, 2, and 3

4cos(x+pi/3) =

a) 2√ 3 cos (x) - 2 sin (x)
b) 2√ 3 cos (x) + 2 sin (x)
c) 2 cos x - 2 √ 3 sin (x)
d) 4 cos (x) +2
e) 2 cos (x) +2√ 3 sin (x)

At x = 3, the function given by f(x) =

x^2 , x < 3
6x -9, x ≥ 3

is:

a) undefined
b) continuous but not differentiable
c) differentiable but not continuous
d) neither continuous nor differentiable
e) both continuous and differentiable

This is due Tuesday, but PLEASE help me ;_; I will love you all FOREVER (not that I already don't, but more so now <3)

btw:

√ = Square root

 

[Overswarm]

Calculus problem.
I wouldn't normally post here, but god, I hate my teacher. She gives us a sheet of **** we've never done before, tells us that she's gonna take NO questions about it, but it counts as a quiz grade....

Judging by the time of year, it appears your class is being tested. With your correct answers, you're prolly gonna look like the star pupil since this is how she's gauging your class' performance levels. Good luck!

 

[M.K]

Judging by the time of year, it appears your class is being tested. With your correct answers, you're prolly gonna look like the star pupil since this is how she's gauging your class' performance levels. Good luck!

Pshhh.
This is how my class is set up. We have ~25 kids.

Probably 15 of us are Seniors, the rest are Juniors.

I have an 89.3 in the class 4 weeks into the year.
My quiz grades have been 96, 67, and 76. We've had 1 test and I got a 91. I've gotten 100s on all my homework questions.

Almost EVERY SINGLE JUNIOR has an A, and around 5 of the Seniors have As. I have a freaking B, and I've NEVER gotten a B in my entire high school career (Final Grade that is, I've ended mid terms with Bs before).
It's just so disheartening, because she'll pass out a quiz, say "Alright now, I'm not trying to trick you or trip you up here", and yet, SHE IS because the problems are twice as hard as the hardest problems on the homework.

-_- In a world where I'm usually the star pupil, I'm not, and it's so....depressing

 

[#HBC | Mac]

and i assumed she scrubbed your floor going by the state of her knees

so far so good, tho i dont much like the actor who plays sherlock

 

[vanderzant]

First question:

If lim x-->3 f(x) = 7, which of the following must be true?

1) f is continuous at x = 3
2) f is differentiable at x = 3
3) f (3) = 7

Not entirely sure, but I'm inclined to say that none are true as f(x) could be ANYTHING. A limit is when the function gets very close to something, but doesn't actually equal it. So I think it is possible to have a function that is approaching 7 at x = 3 (3) but doesn't ever equal it. Think discontinuous functions, which implies that (1) is untrue, which therefore means you can not differentiate at x=3 (2). tl;dr: it's either all or none (I think!).

Second question answer is (c). You need to use the trig identity:
1) cos(x+y) = cos(x)cos(y) - sin(x)sin(y)

As well as know that:
cos(pi/3) = 1/2
sin(pi/3) = √3/2

Apply all these and simplify, and you'll get (c)

Edit: Last question:

Lower limit:
As x -> 3-;
x^2 approaches 9

Upper limit
As x -> 3+;
6x -9 approaches 9

And since f(3) = 9, the function is continuous at x = 9.

Not sure whether you can differentiate it (I've never done that before) though. Instincts tell me you can, but I'd check with someone else

 

[Tom]

and i assumed she scrubbed your floor going by the state of her knees

so far so good, tho i dont much like the actor who plays sherlock

cumberbatch is the best

you should like him

 

[Nicholas1024]

All right then...

@MK

First question is definitely a). Just consider the function f(x) = 7 when x =/= 3 and f(3) = 5. It's not continuous, not differentiable, and f(3) is definitely not 7, even though it satisfies the requirements of the question.

Second question...

4cos(x+pi/3) = 4(cos(x)cos(pi/3) - sin(x)sin(pi/3))
= 4(cos(x)/2 - sqrt(3)sin(x)/2)
= 2(cos(x) - sqrt(3)sin(x))
= 2cos(x) - 2sqrt(3)sin(x)

So yeah, it is c.

Third question:
First off, you can eliminate answers a) and c) right off the bat, as it's obviously defined, and differentiable -> continuity.

Second off, it's definitely continuous, as evaluating both at c,

3^2 = 9 = 6*3 - 9.

And taking the derivative of both sides,

d/dx(x^2) = 2x, d/dx(6x - 9) = 6
So, the derivative is 6 at both ends.

Therefore, it's both continuous and differentiable.

 

[Nicholas1024]

By the way MK, what did you find so hard about the homework? Maybe I can help you with the concepts...

Also, are you allowed to get the answers from the internet/other people? (I assumed yes because you are coming to us here for help, but I just want to make sure.)

 

[M.K]

By the way MK, what did you find so hard about the homework? Maybe I can help you with the concepts...

Also, are you allowed to get the answers from the internet/other people? (I assumed yes because you are coming to us here for help, but I just want to make sure.)

Well, I would say that my "basic skills" are really lacking. My Pre-Calculus teacher literally dumped us as a class, she HATED us. It all stemmed from the fact that she wasn't even in school for the first 2 weeks, and the sub handed us our very first "trig" stuff, and we were all like "O__O".
It also didn't help that I realized my calculator was in the wrong mode >_>;;.

Yeah, the only person we can't ask for help is the actual teacher. i mean, she helped us A LITTLE today, but I've got almost no self-confidence.
I guess I'm use to being good at this stuff. I BREEZED through Geometry, Algebra II, Pre-Calc to an EXTENT, but here I'm struggling, and that's a foreign feeling to me.
:/

EDIT

ONE more question, that's IT and then I'm DONE :

If f(x)= 2 + |x-3| for all x, then the value of the derivative f'(x) at x=3 is:

A) -1
B) 0
C) 1
D) 2
E) nonexistent.

That's it! then I'm finished!!

 

[#HBC | Mac]

wtf tommm!?

wtf was tht ending