studies support thread

[kryss]

[ a ] Ba(OH)2 + [ b ]HNO3 --> [ c ] Ba(NO3)2 + [ d ] H2O
Ba: 1a = 1c
H: 2a + 1b = 2d
O: 2a + 3b = 6c + 1d
N: 1b = 2c

i think is what you meant.

[Hikaru]

guh I didn't exctract it right -.-;
calculating using only 3 elementric functions - multiply an equetion by a constant(=/= 0), adding an equation * constant to another equation, and changing the order of equations.

>.< I though you said something else and then I solved it, and heck it's errorious with a keyboard and the forums font and all >.<
so...

the =0 is not needed because 0 is neutral to subsraction and stays the same with multiplication
(1,0,-1,0)  ---> (1,0,-1,0)  ----> (1,0,-1,0)   ---> (2,0,0,-1)
(2,1,0,-2)  ---> (0,1,2,-2)  ----> (0,1,2,-2)   ---> (0,1,0,-1)
(2,3,-6,-1) ---> (0,3,-4,-1) ----> (0,0,-10,5) ---> (0,0,2,-1)
(0,1,-2,0)  ---> (0,1,-2,0)  ----> (0,0,-4,2)   ---> (0,0,0,0)

and back to eqatuions:
a = 1/2d
b = d
c = 1/2d

that's an infinite number of solutions, like say, (a,b,c,d)=(1,2,1,2)

[ 1 ] Ba(OH)2 + [ 2 ]HNO3 --> [ 1 ] Ba(NO3)2 + [ 2 ] H2O

[kryss]

i did equation-balancing when i was doing A-level chemistry.
we didn't use formulae, we just looked at it and wrote down what worked.
your method is the best, but more time-consuming.

[Hikaru]

yea I did that too, theoretically it's the best if you have huge formulas, but you don't really get any of those.
you can see that a = c because of Ba, and then you can comlete the others.

[kryss]

show me the equation and i'll tell you.

[Hikaru]

so... after doing the SAME INTEGRAL, x^3/(2-x^2)^(1/2)dx ALL DAY LONG and not finding the right solution, I found that I was right all allong and had absolutly no errors, that's if you don't include the fact that I entered x^3*(2-x^2)^(1/2)

=..=
the error repeated itself because of Firefox's saved filled forms. my suspicion aroused when I typed x^3/ and the saved form suddently disappeared.





damn you Red, it's all your fault.

[Hikaru]

after refreshing my memory of trigo substitution, I finnaly solved (x-x^2)^(-5/3)!
you need to complete it to a square, then mark the square as u, subsitute for a*sin(teta) and you should get integral of sec(teta)^4

you should get somewhere, but where I got to was twice as big as it should be =/
where the hell is that missing ½?

OK! solution!

[Hikaru]

*scratches head*
well you do need to change it back to an x.

[kryss]

[HClO4] -> [H]+ + [ClO4]-

or am i being remarkably dull? oh, and i'd never heard of perchloric acid before, love wikipedia...

[Zanibas]

Well, this confuses me, and I just want to stay in Algebra I.

God, if I have to learn that this year, I will most likely run!

Well, let's see who's smart!

((5((9(x+2-8)^2)+16y)/2(97/sq.rt. 49))/7=193675.571

Solve for x & y

[Zanibas]

Yeah, I just made that on the top of my head, so I have no idea...

Find x if y=0, and y when x=0, like x&y intercepts. Forgot to finish

[Zanibas]

Not meaning to offend anyone...

[Zanibas]

Yep, was reading a wiki on Propaganda, and I found tons of these posters.

Some of them crack me up instead of scaring me .

[NarutakiTwinsRule]

Aw shit, I coulda used this for homework.

Damnit!

I'mma need help later since I'm a noob and everyone's ahead of me [i'm in algebra 1, leave me alone].

[NarutakiTwinsRule]

Uff, I'll post the problem tommorrow. It's bed time, like 2 hours ago.