Digital Paint Discussion Board
Digital Paint Community => Other Stuff => Topic started by: y00tz on August 24, 2007, 01:56:33 AM
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I just thought I'd let you know that if you need help with your homework, regardless of grade or subject, this forum is cluttered with intelligent people that could help you, so don't be afraid to post here and ask for help.
I'm sure between the intelligence of the forum users, our Googling power, and the random references cluttering our bookcases, we could probably help you out.
-y00tz
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Well then anybody want to do my online Algebra class for me? ;D
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WHY DONT U DO ALGEBRA URSELF!!!! ITS THE EASIEST PART OF MATHS ALL TOGETHER. I CAN MABYE HELP OUT A BIT BUT IT DEPENDS WAT GRADE UR IN
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WHY DONT U DO ALGEBRA URSELF!!!! ITS THE EASIEST PART OF MATHS ALL TOGETHER. I CAN MABYE HELP OUT A BIT BUT IT DEPENDS WAT GRADE UR IN
KnacKster Translation:
" Golly Gee Wilikers KiLo I'd LOVE to do your online algebra, but I have to get permission from mommy to go online first. "
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Yeah well I hate that class because it's all online and I am too lazy to actually do it. I mean seriously I am paying a $500 a credit hour and the professor gets paid for giving me a password to do a test. >:(
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Buy a textbook and start reading. It's much cheaper (and you can go at your own pace ;)).
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geometry sucks.. i have had to make 3 powerpoints already.. and its only the 2nd week of school....
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Haha, help me with my College algrebra & trigonometry class when i start to get more involved in it
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i dont know about you guys but do you work for credits or watever?
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i dont know about you guys but do you work for credits or watever?
I work for money... Which goes from my bank account, to Lexi's purse, then to Bloomingdales...
/me just balanced his checkbook
.. *sigh*
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y00tz you should make an IRC channel for people to get homework help, so they can play even MORE DP.
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Which goes from my bank account, to Lexi's purse, then to Bloomingdales...
/me just balanced his checkbook
.. *sigh*
Be like listen here future wifey you don't need to go shopping at the most expensive place in town just go down to K-Mart and buy your clothes. ;D :P
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hahahah wal-mart FTW
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Be like listen here future wifey you don't need to go shopping at the most expensive place in town just go down to K-Mart and buy your clothes. ;D :P
I wouldn't' mind if she shopped at the most expensive store IN TOWN, but there isn't even a Bloomingdales in Missouri, meaning my Shabbat is often spent driving to Illinois.
Totally worth the time spent with her though.
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Haha, where do you live? I mean i live in illinois and there's nothing here besides a few things in chicago :P
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Springfield, Missouri.
Old Orchard, Illinois I know for sure has a few 'bigger' store, such as Bloomingdales and Macy's.
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shoot the biggest thing here in chattanooga is maybe a SEARS or Belk...or maybe even pac-sun
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I need a phone number, a address with a zip code, and a contact email for a police academy in Michigan.
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Here you go!! (http://www.google.com/search?q=police%2Bacademy%2Bmichigan&sourceid=navclient-ff&ie=UTF-8&rlz=1B3GGGL_en___US219)
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Thanks Knack, I am working on an Advance Placement U.S. history timeline, so i dont have time to do it myself.
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;D
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Ah…AP US History. I currently have a D in that class: the first and only D I have ever gotten since I came to the America.
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The only D's I ever got in school were in programming, IB Computer Science, IB HL Math, and lastly IB Physics... ironically because I spent the class-time building trebuchets and trebuchet similulators.. :(
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and you only built those to sling ur nasty underware across the room to the hamper?
;D
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Totally worth the time spent with her though.
Oo, so she reads the forums also?
:P
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Ok, I need helps.
Greek philosopher is trapped in jail. There are 2 doors in the jail and 2 guards guarding the doors. 1 door leads to freedom, other to hell. The guard guarding the door to heaven never lies. The guard guarding the door to hell always lies. The philosophist can go out from either of those doors and he can ask 1 question from those guards. He asks "Are you guarding the door to heaven?". The guards answers either "yes" or "no". Now, how can the philospher be sure which guard is guarding which door and get out safely?
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Is he allowed to ask each guard, or can he only ask one of the guards?
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Wait, so that's the only question he can ask?
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yes thats the only question he can ask. He asks both guards.
Well basically i need to know the logic behind this
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Post removed
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(http://imgs.xkcd.com/comics/labyrinth_puzzle.png)
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+1 if I could Eiii.
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Ok, I need helps.
Greek philosopher is trapped in jail. There are 2 doors in the jail and 2 guards guarding the doors. 1 door leads to freedom, other to hell. The guard guarding the door to heaven never lies. The guard guarding the door to hell always lies. The philosophist can go out from either of those doors and he can ask 1 question from those guards. He asks "Are you guarding the door to heaven?". The guards answers either "yes" or "no". Now, how can the philospher be sure which guard is guarding which door and get out safely?
"What would the other guard say is the door to Hell?" Then take that one.
Eiii: Haha!
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How can you answer yes/no to that question zorch? Labyrinth is a smelly film anyway.
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It was ok but mostly sucked in my view.
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Ah…AP US History. I currently have a D in that class: the first and only D I have ever gotten since I came to the America.
Yeah I just switched into regular U.S. history today. ;D
It sucks though because my teacher seriously has elephantitus of the nuts.
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Just checked, now B-. ;D
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Being the fantasticly smart girl I am (jaykay.), its embarassing to admitt that I am oblivous to everything my French teacher says. Does anyone have experience in French? I'm in French III and on the verge of a C. :|
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Any ! wanna help me in my math?almost have a F.Well heres one problem.Choose the correct answer.
if 2x over three =5,2y-2 over 4 =3,and z over 2 +z over 3=5 which of the following is true
A. x>y
B.y<z
C.x=z
D.z>x
Thanks every 1
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Oo, so she reads the forums also?
:P
Haha, that would have been smooth of me...
As for Sarah, if it were Hebrew you'd have better luck, but French is just Delphi to me...
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yootz look at my last post on thread and help please.
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d.
z>x
x=5/2
y=5/2
z=3
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if 2x over three =5,2y-2 over 4 =3,and z over 2 +z over 3=5 which of the following is true
2X/3 = 5
2X = 5(3)
2x = 15
X = 15/2
(2Y-2)/4 = 3
2Y-2 = 3(4)
2Y-2 = 12
2Y = 14
Y=7
Z/2 + Z/3 = 5
3Z/6 + 2Z/6 = 5
5Z = 5(6)
Z = 6
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A. X>Y
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omg I see the error of my ways
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Thank you for the help.As for you knack I think you should go back to school..LOL by
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Ok I just need to learn calc and trig I would be straight.
I need to know this for what im goin into
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Is there some kind of higher level trigonometery because I learnt it a year ago...
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It can get more complex (see pre-calc with limits)
But I think I learned trig as a 7th grader here in The States.
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xy – a = ax + y How do you solve that for x? Would it be x = (ax + y) + a) / y?
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xy -a = ax + y
x = (a + ax + y) / y is what I get. But I'm probably wrong.
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Well, it depends. Do you divide the y or add the a first? I can't remember.
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add the a 1st. trust me, I'm Asian.
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Simplify:
1-csc x
csc x
sin²u x cot²u + sin²u
sin(x-(pie/6))
tan7u + tan5u
1-tan7u x tan5x
Any type of showing steps shown would be wonderful.
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Given:
Cos x = (2/3) and 3pie/2<x<2pie, find cos(x/2)
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Only going to do the first one since I'm reading a book.
Just change csc(x) to 1/sin(x), and then you've got a fraction over a fraction. The rest should be algebra -- no more identities needed.
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Thanks.
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Okay, second one:
sin^2(u)(cot^2(u)+1) -- distributive property
Then, change cot^2(u) to cos^2(u)/sin^2(u), use some algebra to un/re-distribute it, sin^2(u) should cancel on the first term (the fraction), and then use the identity cos^2(u)+sin^2(u) = 1
Third one, I guess you could use the sum/difference identities, but I don't know if that really counts as simplifying. (sin(u-v)=sin(u)cos(v)-cos(u)sin(v))
Fourth looks like a royal bastard and I don't see a way to do it easily off my head.
And fifth, it's been too long since I've done trig and it's too late for me to try too hard. :P
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K I need some french help. I am a standard French level two student and I need help with some easy sentences.
let me know if you can help and dont use a translator i really need help.
ill post the sentences when i find one of you who can help. prefrebly someone who is fluent in francais.
-erad
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Most of the people who are fluent in french are far away from fluent in english so...=\
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y00tz you should make an IRC channel for people to get homework help, so they can play even MORE DP.
I like this idea too. I am pretty good in alg too.
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O Gawd im so glad im done with trig. Them identities made me want to slit my wrists.
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That's why i dropped trig. It just didn't click in my mind so i took something more real life applicable.
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Ok, circle theorems: anyone got any good mnemonics for remembering these?
i said mnemonics
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Hello old thread.
The function f(x) = ax^3 - 7x^2 - bx - 20 is divisible by x^2 - 3x - 10. Find the values of a and b.
I've spent more than the past hour working on this question with a friend, and we just can't get anything out of it. The only place I could find any other question like this was in my textbook, but it had no help and description of how to do it, just the answer to it only.
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whats 2 + 2 .... last time someone said it was 22.
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Whoa, think I may be a little to late but still.
@Cam
x^2 - 3x - 10 = (x - 5)(x + 2), let's call it w(x). w(x) zeroes for x = 5 and x = -2, and so does f(x). Put 5 and -2 on place of x in f(x) and it equals zero, like:
{ a*5^3 - 7*5^2 - b*5 - 20 = 0
{ a*(-2)^3 - 7*(-2)^2 - b*(-2) - 20 = 0
This leads to an answer that a=3 and b=36, unless I made a mistake.
Hope that solves the problem, if it still exists.
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Post removed
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Whoa, think I may be a little to late but still.
@Cam
x^2 - 3x - 10 = (x - 5)(x + 2), let's call it w(x). w(x) zeroes for x = 5 and x = -2, and so does f(x). Put 5 and -2 on place of x in f(x) and it equals zero, like:
{ a*5^3 - 7*5^2 - b*5 - 20 = 0
{ a*(-2)^3 - 7*(-2)^2 - b*(-2) - 20 = 0
This leads to an answer that a=3 and b=36, unless I made a mistake.
Hope that solves the problem, if it still exists.
Ahh it leaves 2 equations which can be solved simultaneously. I didn't think of doing it like that. Thanks for that, gets that out of my head :P.
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Permutation and Combinations
1. How many 13 card hands are there having exactly nine cards from any suit? Answer: 235 237 860
2. How many ways are there of picking 2 cards, one after the other, from a deck of 52 cards if the
A) First card is replaced? Answer: 2704
B) First card is not replaced? Answer: 2652
3. In how many ways can a 5 question test be answered if each question is to be answered true or false? Answer:32
I need help on how to do it, please. I'm betting i could get these questions if i took my time, but i don't feel like doing it right now.
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Haha, probability.
1) Don't understand the question. Is it how many 13 card hands can be made up of 9 cards of the same suit?
2a) nCr(52,1) x nCr(52,1) = 2704. Basically thats 52 ways of getting the first card AND (in probability X is always and, + is always or) 52 ways of getting the next card.
b) nCr(52,1) x nCr(51,1) = 2652, quicker still is nPr(52,2), use that when its not being replaced :P.
3) 2^5 = 32. Which is essentially nCr(5,0) + nCr(5,1) + nCr(5,2) + nCr(5,3) + nCr(5,4) + nCr(5,5). Meaning there is 1 way of giving no answers, 5 ways of giving only one answer, 10 ways of giving 2 answers, etc. Add em together because its not giving 1 answer AND 2 answers, its 1 answer OR 2 answers.
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Oh okay, thanks :) I had a test on it today and i think i failed it, woot
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Haha, I aced probability last year, hoping to do the same this year :)
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I got the test back today and i got 31 out of 44. Wow, i guess i do kinda know what i'm doing.