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story645
Post #342824 - Reply to (#342780) by daisukidesuyo
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6:27 am, Dec 18 2009
Posts: 506


Quote from daisukidesuyo
Will anyone here be willing to help me with Calculus? It's a few topics because I have a chapter quiz tomorrow.

Most of the people who can help with physics, plus some others. Again, give us the questions and someone will go for it. (And hopefully try to avoid the drama a simple physics question caused.)

silencer
Post #342858 - Reply to (#342772) by BoxBox
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The last Blood Elf
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1:18 pm, Dec 18 2009
Posts: 200


Quote from BoxBox
Dude, let's try to be civilized around here. It's not 4chan where you rant and flame people.

Hey, that's not what i did !!
O_O i tried to keep this from afftecting your ego, well, atleast i tried
These are what you said, right ?
Quote from BoxBox
SO Let me make it absolutely simple and clear: Velocity is speed with direction. Or Speed is the magnitude of velocity.
I said that this is half correct ( one of my effort to make it sound less harsh,erm, my friends woudn't use that word if they see this) Even from the most basic context, there are cases that the magnitude of a velocity vector is not speed, at all. The best terms to conceptualize the relationship between them are "distance traveled","displacement", and "instantaneous".
Quote from BoxBox
Besides....when objects zig zags as you said, there's no way it can have constant speed

This is wrong, i tried to give out the hint, but i was disappointed. And i was serious when i said you should do research on this if you want to discuss. Why would i try to flame you, lol But look at how this has turned out. O_O
Quote from BoxBox
Besides, a particle travelling in a zig zag motion will definitely change in acceleration. There needs a change in it's X and Y component to change direction. I'm sure YOU of all people should agree.
In a zigzag trajectory, the object must have acceleration, i never disagree with that -_- And there's nothing wrong if an object with such motion has constant speed. When the acceleration vector changes the velocity vector, its speed can be constant.
A simple example, a ball zigzags when it bounces back and forth elastically between 2 parallel walls. Its velocity does change (due to elastic collision), its speed doesn't. And you tell me what is the acc ? how does it affect the ball's motion ? (just asking for fun)
Quote from BoxBox
And the average speed of such a particle cannot be found just by looking at the start and end point of the path. It's a bit more complicated.
The average speed = distance traveled / time taken -_-
And the distance traveled is different from the displacement (the change in position)
Quote from BoxBox
p.s. please do not say that your country's education is superior to the U.S's education system....it's racist and mean....I hate it when people bad mouth the U.S. when they haven't even been to the U.S. before
O_O you said it all, read what i post again if you like.
You are being over sensitive...
I'm just not sure of the system in US, and i haven't even mention about my country's. Plus i don't follow the general educational system of my country, well, not that i care my curriculum is a bit different.
Quote from BoxBox

To mod: I'm sorry to rant, but it's just when I see people giving other people false information, intentionally or not, I like to correct them. I mean this is a place where people get their homework helped....and I try to maintain the quality of the answer....

O_O



And i was trying to optimize the answer laugh Come to think of it, if there's no argument, there's no optimization, and i don't feel sorry, not one bit.
Perhaps i m a self righteous troll eek


PS. Uncivilized, lol

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story645
Post #342896
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5:35 pm, Dec 18 2009
Posts: 506


Silencer and Box Box-take it to pm.

You guys are getting into far more detail than MasamiAkane needs-seriously one half right answer would have been fine 'cause in physics 101 it doesn't matter so much which option you choose. The solutions usually are such that as long as she uses the formula in her book for speed, she'll be fine. I also think you scared the poor girl off from asking any more questions.

Dr. Love
Post #342897
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5:41 pm, Dec 18 2009
Posts: 8862


Next post directed at each other will be immediately deleted.

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Dragonfiremule
Post #346313
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12:10 am, Jan 5 2010
Posts: 2098


ANYWAY......

So, I was doing some review problems on a test, and I got them wrong. Only, I'm pretty sure I did it right, so it might just because my teacher had his 10 year old kid grade the paper AGAIN. (I hate my teacher)

(Basically, I just want you guys to check if these are right, so I can go complain to him about it.)

So like, a problem was x^2 + 5x =0. I put the answer as -5.

Also, another was was x^2 + x = 56. I put the answer x=7.

Or one was factoring, as was like,
8x^2 + 8x + 2, and I put the factor as (4x+2)(2x+1)

Another one was 4x^2 = 81, and I put the answer as 9/2

and x^2 + 18x +81 = 0 (Factoring), I put (-18,-81) as the answer.

Last edited by Dragonfiremule at 12:21 am, Jan 5

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BoxBox
Post #346315
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12:33 am, Jan 5 2010
Posts: 1668

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Quote
Another one was 4x^2 = 81, and I put the answer as 9/2


The complete answer is 9/2 and -9/2

Notice this is a positive/negative answer combination.

Quote
x^2 + 18x +81 = 0 (Factoring), I put (-18,-81) as the answer.


(x+9)^2=0

So x+9=0

x=-9


Quote
Also, another was was x^2 + x = 56. I put the answer x=7.


Answer is x=7, x=-8


Quote
So like, a problem was x^2 + 5x =0. I put the answer as -5.

Answer x=0, x=-5

Quote
Or one was factoring, as was like,
8x^2 + 8x + 2, and I put the factor as (4x+2)(2x+1)


deduce down to 4x^2+4x+1=0 so the final factor form is (2x+1)^2

Last edited by BoxBox at 12:40 am, Jan 5

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Gay book discussion thread
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this is not about cannibalism...please get back on topic

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Dragonfiremule
Post #346316 - Reply to (#346315) by BoxBox
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12:37 am, Jan 5 2010
Posts: 2098


Quote from BoxBox
Quote
Another one was 4x^2 = 81, and I put the answer as 9/2


The complete answer is 9/2 and -9/2

Notice this is a positive/negative answer combination.

Quote
x^2 + 18x +81 = 0 (Factoring), I put (-18,-81) as the answer.


(x+9)^2=0

So x+9=0

x=-9


Damn. I was hoping it was the teachers fault.
What about the rest?


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Does the walker choose the path, or the path the walker?
ssakura3
Post #346317
Member

12:39 am, Jan 5 2010
Posts: 1


Dragonfiremule: I haven't done algebra for a long time, but the following is what I thought the answer should be.

x^2 + 5x =0 --> x(x+5)=0 --->therefore, x =0 and x=-5

x^2 + x = 56 --> x^2 + x -56 = 0 -->(x-7)(x+8)=0 --> therefore x=7 and x=-8

8x^2 + 8x + 2 --> 2(4x^2 +4x +1) --> 2(2x+1)(2x+1) or 2(2x+1)^2



BoxBox
Post #346318 - Reply to (#346317) by ssakura3
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12:41 am, Jan 5 2010
Posts: 1668

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Quote from ssakura3
Dragonfiremule: I haven't done algebra for a long time, but the following is what I thought the answer should be.

x^2 + 5x =0 --> x(x+5)=0 --->therefore, x =0 and x=-5

x^2 + x = 56 --> x^2 + x -56 = 0 -->(x-7)(x+8)=0 --> therefore x=7 and x=-8

8x^2 + 8x + 2 --> 2(4x^2 +4x +1) --> 2(2x+1)(2x+1) or 2(2x+1)^2




Yep, exactly. It's strange though, I would have thought the teacher might accept his/her 8x^2 + 8x + 2 answer. I guess that teacher is particularly b-tchy....

I would still go to the teacher to complain about that problem though.

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Gay book discussion thread
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this is not about cannibalism...please get back on topic

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I think it is exactly the topic. I see nothing wrong.
reid1
Post #346320 - Reply to (#346313) by Dragonfiremule
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12:44 am, Jan 5 2010
Posts: 2021


Quote from Dragonfiremule
ANYWAY......

So, I was doing some review problems on a test, and I got them wrong. Only, I'm pretty sure I did it right, so it might just because my teacher had his 10 year old kid grade the paper AGAIN. (I hate my teacher)

So like, a problem was x^2 + 5x =0. I put the answer as -5.

Also, another was was x^2 + x = 56. I put the answer x=7.

Or one was factoring, as was like,
8x^2 + 8x + 2, and I put the factor as (4x+2)(2x+1)

Another one was 4x^2 = 81, and I put the answer as 9/2


First one, answers are 0 and -5.

Second one, -8 and 7.

Third one, 8(x+1/2)^2.
You have to isolate x^2 here, so:
8x^2+8x+2=
8(x^2+x+1/4)=
8(x+1/2)(x+1/2)=
8(x+1/2)^2
(I think, last time I did this stuff was ages ago bigrazz ).

Fourth one, 9/2 and -9/2.


Remember that second grade equations always have two solutions when b^2-4ac > 0.


Edit: Well, someone already answered but still I'm leaving here the post smile


Last edited by reid1 at 12:51 am, Jan 5

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Dragonfiremule
Post #346508
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7:41 pm, Jan 5 2010
Posts: 2098


Thanks guys. I get why he marked \them wrong. I forgot to put the second part of the answer or didn't simplify all the way. Thanks for the help. (Boxbox, I'm a she.) smile wink grin


Oh, I have one last question.

Its factoring.

(x-4)^2=9 so I did

(x-4)(x-4)=9

x^2 - 8x - 16 = 9
x^2-8x-25=0

So to solve it, I would have to factor that out more, but I can't seem to do so, so I think it may be prime. Is this correct? And would that be the answer?


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NightSwan
Post #346511
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Mad With a Hat
Member

7:53 pm, Jan 5 2010
Posts: 4592


If I still remember something...

(x-4)^2=9

x^2-8x+16=9
x^2-8x+7=0
X1=7
X2=-1

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Dragonfiremule
Post #346512 - Reply to (#346511) by NightSwan
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7:54 pm, Jan 5 2010
Posts: 2098


Quote from NightSwan
If I still remember something...

(x-4)^2=9

x^2-8x+16=9
x^2-8x+7=0
X1=7
X2=-1

OH! Thanks. I didn't realize I had not turned the 16 into a positive 16. Thanks.

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Does the walker choose the path, or the path the walker?
NightSwan
Post #346514
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Mad With a Hat
Member

7:57 pm, Jan 5 2010
Posts: 4592


NP.
I like the algebra I can solve.
I think I screwed up there though.
It's -7, -1
^^;;;;

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you would be my rainbow in shades of grey.


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Dragonfiremule
Post #346515 - Reply to (#346514) by NightSwan
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8:06 pm, Jan 5 2010
Posts: 2098


Quote from NightSwan
NP.
I like the algebra I can solve.
I think I screwed up there though.
It's -7, -1
^^;;;;

The 7 is correct, but I don't know about the -1.

-1^2 - 8(-1) +7 = 0

1 + 8 + 7 =/= 0

A regular 1 works though.

1 + -8 + 7 = 0 biggrin

Unless you do the opposite? I can't remember...

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