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StarlightDreams
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Post #325950 |
Member
10:08 pm, Oct 6 2009
Posts: 3870
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@_@
Okay.
SOO. My teacher told me the answer.
He said it was a redox reaction.
@_@
And the answer...I wrote it down somewhere...5Ag...something...
:<
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TimeManInJail
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Post #328370 |
Member
1:37 am, Oct 18 2009
Posts: 536
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oh um.. i forgot my dam text book and uh well my essay is due on monday >.> so was wondering if anyone can give me pros on the protestants reformation.. martin luther john hus.. whycliff.. examples would be darn useful please and thank yoU!
mad typist
lost on the research
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silencer
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Post #328440 - Reply to ( #325748) by StarlightDreams |
The last Blood Elf
Member
7:31 am, Oct 18 2009
Posts: 200
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Quote from StarlightDreams <.<
*had to search for this thread*
Ag (s) + HNO−3 (aq) -> H−2 (g) + NO + AgNO−3 (aq)
− = Subscript.
Someone please balance it for me. D:
I notice this thread too late. One doesn't have to know about the change of oxidation state ( happens in the so called redox reactions) to solve this. Pure algebra with a small hint that we are in aqueous solution are what you need :
5Ag(s)+6HNO-3(aq) -> H-2(g)+NO+5AgNO-3(aq)+2 H-2O
( the will always be collision with water molecules )
post this in case someone still cares.
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Calíbre
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Post #328443 - Reply to ( #328440) by silencer |
Madman
 Moderator
7:59 am, Oct 18 2009
Posts: 3341
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Quote from silencer Quote from StarlightDreams <.<
*had to search for this thread*
Ag (s) + HNO−3 (aq) -> H−2 (g) + NO + AgNO−3 (aq)
− = Subscript.
Someone please balance it for me. D: I notice this thread too late. One doesn't have to know about the change of oxidation state ( happens in the so called redox reactions) to solve this. Pure algebra with a small hint that we are in aqueous solution are what you need :
Yea...
Quote from dictionary.com stoichiometry [stoi-kee-om-i-tree]
–noun
the calculation of the quantities of chemical elements or compounds involved in chemical reactions.
Hurray Math
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StarlightDreams
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Post #328739 |
Member
4:43 am, Oct 19 2009
Posts: 3870
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;_;
I didn't know!
I didn't get the extra credit either. Evil teacher. D:
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mike0dude
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Post #328744 |
Member
4:54 am, Oct 19 2009
Posts: 256
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i have 2 pratical work
and 2 exams tomorow an i didnt do shit this weekend!!!
im so screwed... 
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Vildur
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Post #330898 |
Dungeon Master
Member
1:36 am, Oct 28 2009
Posts: 38
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v = [x,y,z]
[2,-3,1] x v = [1,0,2]
Solve for x, y and z. (This is the cross product of 2 vectors)
I don't think it works but I'd appreciate it if someone could confirm that for me or tell me how to solve it.
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silencer
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Post #331164 |
The last Blood Elf
Member
4:15 am, Oct 29 2009
Posts: 200
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you know what is cross product and what is scalar product right ?
u = [2,-3,1]
v = [x,y,z]
w = [1,0,2]
u x v = w means u and w must be perpendicular to each other so that v exists, while, clearly, they aren't,
you can check by scalar product and can find the angle between them if you want to.
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Althaea
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Post #331398 |
Tea Leaf
Member
2:05 am, Oct 30 2009
Posts: 191
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wow I didn't know this existed! I guess I have one.
Given r(x)= (8x^10) - (2x^5)
rewrite r(x) as a composition of 3 functions.
Find:
f(x)
g(x)
h(x)
such that (f) o (g) o (h) = r(x)
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story645
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Post #331404 - Reply to ( #331398) by Althaea |
forum bunny
Member
2:21 am, Oct 30 2009
Posts: 504
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Quote from Althaea Given r(x)= (8x^10) - (2x^5)
assuming that r(x) = f(g(h(x)))
8x^10 factors out to 2*4x^10 -> 2*2x^5*2x^5
some factoring rules:
x^a = x^b+c, a=b+c
j*x =k*l*x, j=k*l
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Althaea
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Post #331408 |
Tea Leaf
Member
2:40 am, Oct 30 2009
Posts: 191
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Thanks! makes much more sense now and I didn't learn those factoring rules yet.
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silencer
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Post #331436 |
The last Blood Elf
Member
4:18 am, Oct 30 2009
Posts: 200
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Yeah, though there are infinitely many ways to find f(x), g(x) and h(x) in that case, most of them require math at high school level though.
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Noobsrus
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Post #331611 - Reply to ( #330898) by Vildur |
Wall-o-text
Member
11:01 pm, Oct 30 2009
Posts: 367
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Quote from Vildur v = [x,y,z]
[2,-3,1] x v = [1,0,2]
Solve for x, y and z. (This is the cross product of 2 vectors)
I don't think it works but I'd appreciate it if someone could confirm that for me or tell me how to solve it.
Well you try the determinant of the equation so you have the matrix
( i j k )
( 2 -3 -1 )
( x y z )
If you find the determinants you end up with
i(-3z - y) - j(2z - x) + k(2y + 3x)
you know that the value for i = 1 and j = 0 and k = 2
so you have 3 equations
-3z -y = 1
-2z + x = 0
2y + 3x = 2
Then you solve using an augmented matrix
( 0 -1 -3 ] 1 )
( 1 0 -2 ] 0 )
( 3 2 0 ] 2 ) Let Row 3 be equation to (Row 3) - (Row 2)*3
( 0 -1 -3 ] 1 )
( 1 0 -2 ] 0 )
( 0 2 6 ] 2 ) Let Row 3 be equation to (Row 3) + (Row 1)*2
( 0 -1 -3 ] 1 )
( 1 0 -2 ] 0 )
( 0 0 0 ] 4 )
There is an infinite number of solution so I'm really confused about the question now... lol
I'm doing the working in my head so check out the working out, I haven't done this for about 2years now and a bit rusty lol
Last edited by Noobsrus at 11:12 pm, Oct 30
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Althaea
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Post #332393 |
Tea Leaf
Member
6:23 am, Nov 4 2009
Posts: 191
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ive got another problem i need help on and its on functions again!
f(x)=ax+b
inverse of f(x)=bx+a
with a,b real, what is the value of a+b?
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qwert123
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Post #332443 |
Member
2:29 pm, Nov 4 2009
Posts: 157
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f(x)=ax+b
to find f^-1(x):
y = ax+b
x = ay+b
y = (x-b)/a
f^-1(x) = (x-b)/a
but : f^-1(x) = bx+a
so
x/a = bx and -b/a = a
b = 1/a and b = -a²
1/a = -a²
a(1/a) = a(-a²)
1 = -a³
a = -1
b = 1/-1 = -1
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a + b = -2
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