Sat Apr 22, 2006 1:15 am
Out of curiosity, October Revolution= Bolshevik revolution?
Sat Apr 22, 2006 3:06 am
Yes, although there were Mensheviks involved as well.
Sun Apr 23, 2006 11:28 pm
Does anybody know how many newtons 1Kg will give me? i cant decide weather its just 10 or i must times it by 9.81 (force of gravity). i'm trying to find young modulus.
Mon Apr 24, 2006 1:39 pm
I would use 9.81 if I were you.
(nitpick: it's Young's modulus, rather than young modulus. The second sounds like a character in Star Trek).
EDIT: At any rate, Newtons are a measure of force, and are calculated by using F=ma, so in your case, F=(1kg)(9.81 m/s^2) = 9.81 kg*m/s^2, or 9.81 N
(nitpick: it's Young's modulus, rather than young modulus. The second sounds like a character in Star Trek).
EDIT: At any rate, Newtons are a measure of force, and are calculated by using F=ma, so in your case, F=(1kg)(9.81 m/s^2) = 9.81 kg*m/s^2, or 9.81 N
Tue Apr 25, 2006 7:58 pm
shapu wrote:I would use 9.81 if I were you.
(nitpick: it's Young's modulus, rather than young modulus. The second sounds like a character in Star Trek).
EDIT: At any rate, Newtons are a measure of force, and are calculated by using F=ma, so in your case, F=(1kg)(9.81 m/s^2) = 9.81 kg*m/s^2, or 9.81 N
thank shapu, you're a lifesaver.
and the young/young's thing was just a typo as i was in a rush. i agree however that it sounds like a character from star trek.
Fri May 05, 2006 8:25 pm
Does anyone know how Oliver Cromwell was killed? I was asked to do a history essay on it, and I can't find a place that actully explain his death! 
Fri May 05, 2006 8:28 pm
Twitchy wrote:Does anyone know how Oliver Cromwell was killed? I was asked to do a history essay on it, and I can't find a place that actully explain his death!
Here you go,
http://en.wikipedia.org/wiki/Oliver_Cro ... _execution
We actually studied him in History this semester, although, our teacher just said he died at this time. Not how. Or if he did, I wasn't listening.
Fri May 05, 2006 8:38 pm
Thankyou!!!! Thats a life saver!
My history teacher is a bit.....
and the essay was due in last week! But thats been a big help! Thankyou!
My history teacher is a bit.....
and the essay was due in last week! But thats been a big help! Thankyou!
Tue May 16, 2006 9:11 pm
I have a poetry assignment. I have to analyze five poems I can pick from these six poets: Carl Sandburg, Robert Frost, Langston Hughes, Lewis Carroll, Edgar Allen Poe, and Henry W. Longfellow. I can't do one we've gone over, so that kills the road less traveled. So far, I have a boundless moment, annabel lee, and the arrow and the song. Any suggestions?
P.S.- It has to be at least three stanzas.
P.S.- It has to be at least three stanzas.
Tue May 16, 2006 9:27 pm
Oh, Annabel Lee is a classic. Stay with that one.
For Lewis Carroll, you could always do "Jabberwocky." It's always fun.
"Dream Deferred" is a classic hughes poem, and inspired the play "raisin in the sun." You might also try "Cultural exchange" which is longer and less well-known.
For Lewis Carroll, you could always do "Jabberwocky." It's always fun.
"Dream Deferred" is a classic hughes poem, and inspired the play "raisin in the sun." You might also try "Cultural exchange" which is longer and less well-known.
Tue May 16, 2006 9:32 pm
THanks, Shapu! Question: About the Jabberwocky, does that have any literary devices? Like similes, metaphors, etc?
Wed May 17, 2006 1:45 am
This is a bit of plug for my hometown, but you should do Carl Sandburg's "Chicago."
Wed May 17, 2006 10:41 am
Thanks! I'll keep that in mind in case she rejects my poems...
Wed May 17, 2006 7:31 pm
Hey guys, I come asking for some serious help.
Okay, so here's the deal. I just got a 120 problem Algebra II review packet, right? Normally if I don't understand something I'll go to my notes, and then the textbook if I have to. However, I have been absent the past two days and I have to get this done by tomorrow, and I have no notes on the topic and my teacher already collected our textbook.
Like I said, there are about 120 problems on this page, and there's a couple I don't understand. I'll put them here - if you can help, even with one - I'd greatly appreciate it. (All x'es are variables, not multiplication. I know most people know, but just to bring it up . . .)
I've tried my best to answer the problems, but don't know if they are right. After the ========== are my answers, just so you guys don't think I'm trying to bum off of you. The problem with my answers is I've only Greatest Common Factor'ed most of them, and that's not what he wants, I don't think. I think my teacher wants them in the UNFOIL/UNFACE (x+3)(x+2) method or something. Not sure.
Factor as a Binomial Square
40x^2-25y^2 ====== (7x-5y)(7x+5y)
9x^2+16 ====== ? (3x+4)(?)
Factor as a Binomial Cube
x^3-27========== (x-3)(x^2+3x+9)
8x^3+125 ========== (2x+5)(4x^2-10x+25)
Factor as a Trinomial Square
16x^2-40x+25 ====== (4x+5)(4x-5)
9x^2+12x+4 ======= (3x+2)(3x+2)
Factor
8x^2-14x-15 ====== (2x-50)(4x+3)
10x^2-73x+21 ====== (10x-3)(x-7)
Solve
9x^2-25=0 ====== prime?
12x^2=38x-6
x^4+36=13x^2
x^2+90=21x
x^2=36 === x=6
Factor
5s(c-7)-3(7-c) ======== (5a-3)(c-7)
3x^4+24x ======= 3x(x^3+8)
50x^2 - 18 =========== 25(x^2-9)
12x^2y+18xy^2 ===== (6xy+6xy)(2x+3y) OR 6xy(2x)+6xy(3y)
*confused*
Okay, so here's the deal. I just got a 120 problem Algebra II review packet, right? Normally if I don't understand something I'll go to my notes, and then the textbook if I have to. However, I have been absent the past two days and I have to get this done by tomorrow, and I have no notes on the topic and my teacher already collected our textbook.
Like I said, there are about 120 problems on this page, and there's a couple I don't understand. I'll put them here - if you can help, even with one - I'd greatly appreciate it. (All x'es are variables, not multiplication. I know most people know, but just to bring it up . . .)
I've tried my best to answer the problems, but don't know if they are right. After the ========== are my answers, just so you guys don't think I'm trying to bum off of you. The problem with my answers is I've only Greatest Common Factor'ed most of them, and that's not what he wants, I don't think. I think my teacher wants them in the UNFOIL/UNFACE (x+3)(x+2) method or something. Not sure.
Factor as a Binomial Square
40x^2-25y^2 ====== (7x-5y)(7x+5y)
9x^2+16 ====== ? (3x+4)(?)
Factor as a Binomial Cube
x^3-27========== (x-3)(x^2+3x+9)
8x^3+125 ========== (2x+5)(4x^2-10x+25)
Factor as a Trinomial Square
16x^2-40x+25 ====== (4x+5)(4x-5)
9x^2+12x+4 ======= (3x+2)(3x+2)
Factor
8x^2-14x-15 ====== (2x-50)(4x+3)
10x^2-73x+21 ====== (10x-3)(x-7)
Solve
9x^2-25=0 ====== prime?
12x^2=38x-6
x^4+36=13x^2
x^2+90=21x
x^2=36 === x=6
Factor
5s(c-7)-3(7-c) ======== (5a-3)(c-7)
3x^4+24x ======= 3x(x^3+8)
50x^2 - 18 =========== 25(x^2-9)
12x^2y+18xy^2 ===== (6xy+6xy)(2x+3y) OR 6xy(2x)+6xy(3y)
*confused*
Thu May 18, 2006 11:07 pm
Even though I think that I might be too late.. (which I hope I'm not), I'm going to try to help, because it might be better for the long term anyway.
I skipped certain sections because all of the problems in them were correct.
Factor as a Binomial Square
You can only factor as a binomial square when the roots are a difference of each other. Otherwise, such as in the case of the second problem, you have to use the quadratic formula, because the roots are imaginary.
You did the first problem correctly, except that 7 is not the square root of 40. (I'm assuming that you meant 49)
Factor as a Trinomial Square
When you factor as a Trinomial square, you're roots have to be the same. The second one is correct, the first one is not. (All you have to do for the first one is change one of your signs.)
Solve
9x^2-25=0
I'm going to go through this one. You're objective at first is to get x^2 by itself, so you can put it under the radical and get rid of the square.
1. Subtract 25
2. Divide 25 by 9
3. x^2=25/9 <-- put both sides under the radical. 25 and 9 are both
perfect squares, so you get 5/3.
In problems such as 2-4, you need to set them equal to zero and factor. Then you set your roots equal to zero and solve.
Simple ex. x^2+12=-7x
x^2+7x+12=0 (I automatically put them in decreasing power order, it makes it easier)
(x+3)(x+4)=0
x+3=0 x+4=0
x=-3 x=-4
The last problem is almost correct, but remember, when you find the radical of anything, its both is positive and negative value.
Factor
For the first problem, you have the right idea. But, the objects in the parentheis (c-7),(7-c), need to be the same. To do this, you need to factor out a -1 in one of them:
5a(c-7)-3(7-c)
5a(c-7)-3(-1)(c-7)
5a(c-7)+3(c-7)
(5a+3)(c-7)
The second and third problems are right so far, but since (x^3+8 ) and (x^2-9) are still factorable, you just need to factor them too.
For the last problem, you are almost right. When you distribute, you only need to factor out one value for your binomial. (i.e. if you have 2x+4, and you chose to factor out a 2, your answer will be 2(x+2) rather than (2+2)(x+2) ).
I really hope that this helps, feel free to PM me if anything that I said is confusing and I'll try to explain it more.
I'm also super sorry if I'm too late.
I skipped certain sections because all of the problems in them were correct.
Factor as a Binomial Square
You can only factor as a binomial square when the roots are a difference of each other. Otherwise, such as in the case of the second problem, you have to use the quadratic formula, because the roots are imaginary.
You did the first problem correctly, except that 7 is not the square root of 40. (I'm assuming that you meant 49)
Factor as a Trinomial Square
When you factor as a Trinomial square, you're roots have to be the same. The second one is correct, the first one is not. (All you have to do for the first one is change one of your signs.)
Solve
9x^2-25=0
I'm going to go through this one. You're objective at first is to get x^2 by itself, so you can put it under the radical and get rid of the square.
1. Subtract 25
2. Divide 25 by 9
3. x^2=25/9 <-- put both sides under the radical. 25 and 9 are both
perfect squares, so you get 5/3.
In problems such as 2-4, you need to set them equal to zero and factor. Then you set your roots equal to zero and solve.
Simple ex. x^2+12=-7x
x^2+7x+12=0 (I automatically put them in decreasing power order, it makes it easier)
(x+3)(x+4)=0
x+3=0 x+4=0
x=-3 x=-4
The last problem is almost correct, but remember, when you find the radical of anything, its both is positive and negative value.
Factor
For the first problem, you have the right idea. But, the objects in the parentheis (c-7),(7-c), need to be the same. To do this, you need to factor out a -1 in one of them:
5a(c-7)-3(7-c)
5a(c-7)-3(-1)(c-7)
5a(c-7)+3(c-7)
(5a+3)(c-7)
The second and third problems are right so far, but since (x^3+8 ) and (x^2-9) are still factorable, you just need to factor them too.
For the last problem, you are almost right. When you distribute, you only need to factor out one value for your binomial. (i.e. if you have 2x+4, and you chose to factor out a 2, your answer will be 2(x+2) rather than (2+2)(x+2) ).
I really hope that this helps, feel free to PM me if anything that I said is confusing and I'll try to explain it more.
I'm also super sorry if I'm too late.