Homework Help

15 years ago
Posts: 510
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.)

15 years ago
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 🤣 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 😲
PS. Uncivilized, lol

15 years ago
Posts: 510
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.
Next post directed at each other will be immediately deleted.
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15 years ago
Posts: 2126
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.
Does the walker choose the path, or the path the walker?

15 years ago
Posts: 1668
Warn: Banned
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.
x^2 + 18x +81 = 0 (Factoring), I put (-18,-81) as the answer.
(x+9)^2=0
So x+9=0
x=-9
Also, another was was x^2 + x = 56. I put the answer x=7.
Answer is x=7, x=-8
So like, a problem was x^2 + 5x =0. I put the answer as -5.
Answer x=0, x=-5
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
Quote from you_no_see_me_
this is not about cannibalism...please get back on topic
Quote from Toto
I think it is exactly the topic. I see nothing wrong.

15 years ago
Posts: 2126
Quote from BoxBox
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.
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?
Does the walker choose the path, or the path the walker?
15 years ago
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

15 years ago
Posts: 1668
Warn: Banned
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.
Quote from you_no_see_me_
this is not about cannibalism...please get back on topic
Quote from Toto
I think it is exactly the topic. I see nothing wrong.
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 😛 ).
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 🙂
What I like in Trivial Pursuit style. Pick your category:[img]http://i379.photobucket.com/albums/oo233/Reid4891/Manga/trivia-1.jpg[/img]

15 years ago
Posts: 2126
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.) 😁
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?
Does the walker choose the path, or the path the walker?

15 years ago
Posts: 4764
If I still remember something...
(x-4)^2=9
x^2-8x+16=9
x^2-8x+7=0
X1=7
X2=-1
Hrodulf and Bjornolfr, you will not be forgotten.
[img]http://i411.photobucket.com/albums/pp199/SyberAngelGabrielle/couplesig.jpg[/img]
[color=black]And if the world were black and white,
you would be my rainbow in shades of grey.[/color]
If I had a fantasy self, it'd be a tentacle monster.

15 years ago
Posts: 2126
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.
Does the walker choose the path, or the path the walker?

15 years ago
Posts: 4764
NP.
I like the algebra I can solve.
I think I screwed up there though.
It's -7, -1
^^;;;;
Hrodulf and Bjornolfr, you will not be forgotten.
[img]http://i411.photobucket.com/albums/pp199/SyberAngelGabrielle/couplesig.jpg[/img]
[color=black]And if the world were black and white,
you would be my rainbow in shades of grey.[/color]
If I had a fantasy self, it'd be a tentacle monster.

15 years ago
Posts: 2126
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 😀
Unless you do the opposite? I can't remember...
Does the walker choose the path, or the path the walker?