Dead Men Walking

dMw Chit Chat => The Beer Bar => It's my Birthday! => Topic started by: Jewelz^ on September 14, 2008, 07:25:16 PM

Title: Help
Post by: Jewelz^ on September 14, 2008, 07:25:16 PM
This math question from my homework has baffled me and mates..
if y = 3 - 3x, express x in terms of y
 
means nothing to me!
Help please :)
Title: Help
Post by: Anonymous on September 14, 2008, 07:29:27 PM
Quote from: Jewelz^;244725This math question from my homework has baffled me and mates..
if y = 3 - 3x, express x in terms of y
 
means nothing to me!
Help please :)
y=3-3x
y-3 = -3x (take 3 from each side)
(y-3)/-3 = x (divide both sides by -3)
(3-y)/3 =x (rearrange to tidy up the minus signs)

so x = (3-y)/3
Title: Help
Post by: Jewelz^ on September 14, 2008, 07:34:10 PM
another one for you's (i dont like algebra it confuses me very much)
 
x(sqaured) - 7x = 0                          x = ???
Title: Help
Post by: Anonymous on September 14, 2008, 07:45:58 PM
Quote from: Jewelz^;244727another one for you's (i dont like algebra it confuses me very much)
 
x(sqaured) - 7x = 0                          x = ???
xsquared =7x (add 7x to each side)
x = 7 (divide both sides by x)
Title: Help
Post by: Jewelz^ on September 14, 2008, 07:49:16 PM
Thanks blueball you Genius!
i get stuck on the A*/A grade stuff and abit confused sometimes with the B grade stuff..
Jewelz hates algebra :ranting2:
Title: Help
Post by: Anonymous on September 14, 2008, 07:52:07 PM
I love Algebra - always have done.
Title: Help
Post by: Penfold on September 14, 2008, 08:13:27 PM
Quote from: BlueBall;244730I love Algebra - always have done.

I bet you say that to all the girls :flirty:

PEN
Title: Help
Post by: Lexander on September 14, 2008, 08:14:45 PM
Quote from: BlueBall;244728xsquared =7x (add 7x to each side)
x = 7 (divide both sides by x)

Am I missing something but I think thats not right :g:
   As x = 7 doesnt make the equation work.  (squared)7 is not 7 times 7.
Title: Help
Post by: Anonymous on September 14, 2008, 08:21:09 PM
any number squared is that number multiplied by itself. 7 squared = 7x7
Title: Help
Post by: Lexander on September 14, 2008, 08:25:00 PM
Quote from: BlueBall;244735any number squared is that number multiplied by itself. 7 squared = 7x7

yeah rite, sorry mixed square root and ^2 now I got it.
Though squarred = squareroot, my bad :getmecoat:
Title: Help
Post by: Soon Jung on September 14, 2008, 08:34:50 PM
Quote from: Penfold;244732I bet you say that to all the girls :flirty:

PEN

Would've worked on me, Oth said something similar once :narnar:


I see its a second degree equation, was just about to say the answer is 7 and 0, but I see now that since we are missing the final part of one, it can be solved without :D
Nvm me!
Title: Help
Post by: Jewelz^ on September 14, 2008, 08:40:20 PM
i should be getting more tough cookies on other mathmatic subjects everyweek!
i will post up the ones i am not sure about :)
see if you guys are up to GCSE standards :flirty:
Title: Help
Post by: Aquilifer on September 14, 2008, 10:35:36 PM
Quote from: Jewelz^;244727another one for you's (i dont like algebra it confuses me very much)
 
x(sqaured) - 7x = 0                          x = ???

x^2-7x = 0

I suggest write it open like this:
x^2-7x = 0
x(x-7)=0
=> either x=0 or x-7=0 => x=0,x=7
(because if we have A*B=0, then either A or B (or both) have to be 0)

You can work it out also by dividing with x...
x^2-7x = 0      || :x   (x not 0)
x-7=0  => x=7

..but be careful. Remember that you cannot divide by 0 and you have to check that separately...

if x=0 => 0^2-7*0=0  => 0=0   => x=0 is valid

=> x=0, x=7

There is also a general formula to give the roots of a polynome of the 2nd degree (i.e ax^2+bx+c=0), but it makes no sense using it as it is trivial when c=0.
Title: Help
Post by: Doorman on September 14, 2008, 10:57:45 PM
So that's how you get A levels! :rolleyes: Beats learning it eh? :lmfao:
Title: Help
Post by: Browne on September 14, 2008, 11:51:39 PM
For God's sake, there should be a warning on this post!

Sobriety Required
Title: Help
Post by: OldBloke on September 15, 2008, 01:17:44 AM
Dead Men Walking homework service. How can we help? :doh:

Jewelz - a bit of advice. You got the answers you wanted but, at the very least, please make an effort to understand the solution. If you can't see how the answer was derived then speak to your teacher and explain that you are struggling and need help. If you just hand in correct answers then your teacher will think (wrongly) that you are up to speed. Don't cheat yourself or the teacher.
Title: Help
Post by: T-Bag on September 15, 2008, 02:16:07 AM
Tips to solving Algebra if you ever need to use it and your internet is unavailable (ie for an exam).

Simplify as much as possible:
2x+ 3 -x +2 = 5
is just the same as saying
x =0
add everything with the same power together. x^2s xs and numbers can be grouped together.

Remember you can always change both sides in the same way without breaking the equation. Add 2 to each side, multiply by 8 etc. This is most useful when you "Move something" from one side to the other.

eg:

x + 3 = 0
You add -3 to both side (some would call it subtracting but there's not real difference between positive and negative numbers in this case)
x + 3 - 3 = -3
Simplify
x = -3.

Next thing to remember is factorising:
2x^2 + x + 3= 0
Moving things around won't tell you much as you've got two instances of x to different powers so you can't combine them.
If you remember one formula from maths let it be this one:
x = {-b +/- SQRT[b^2 - 4ac]}/2a
a = number of x^2
b = number of x
c = number of x^0 (lone numbers)
That will give you the roots of any quadratic even if they are negative, complex, shared.

If you don't give up on maths you might get to be cool like me and study physics. :narnar:

P.S if I've made any mistakes go easy on me, it's really late.
Title: Help
Post by: Aquilifer on September 15, 2008, 01:38:48 PM
Quote from: T-Bag;244754...
Next thing to remember is factorising:
2x^2 + x + 3= 0
Moving things around won't tell you much as you've got two instances of x to different powers so you can't combine them.
If you remember one formula from maths let it be this one:
x = {-b +/- SQRT[b^2 - 4ac]}/2a
a = number of x^2
b = number of x
c = number of x^0 (lone numbers)
That will give you the roots of any quadratic even if they are negative, complex, shared.

Yes, that is exactly the general solution to the 2nd order polynome. I didn't write that because... a) It looks like Jewelz hasn't been taught that yet (and at least here if you use some method which hasn't been taught and the result is wrong, you get 0 points) b) it looks stupid if you input a,b,c into the formula and you have c=0.

Some more general stuff:
Generally polynome of n'th order has n roots. Some of them can be imaginary or multiples though. That means you can write it (x-r1)(x-r2)*...(x-rn)=0 where the r1,...rn are the roots. This means if you know one or more roots, you could reduce it to something simpler. E.g. If you have a 3rd order polynome and you know one root is 3, then you know it is (x-3)*f2(x)=0, where the f2 is 2nd order. And 2nd order polynome can be solved always.
Title: Help
Post by: Doorman on September 15, 2008, 01:50:18 PM
Quote from: Aquilifer;244789Yes, that is exactly the general solution to the 2nd order polynome. I didn't write that because... a) It looks like Jewelz hasn't been taught that yet (and at least here if you use some method which hasn't been taught and the result is wrong, you get 0 points) b) it looks stupid if you input a,b,c into the formula and you have c=0.

Some more general stuff:
Generally polynome of n'th order has n roots. Some of them can be imaginary or multiples though. That means you can write it (x-r1)(x-r2)*...(x-rn)=0 where the r1,...rn are the roots. This means if you know one or more roots, you could reduce it to something simpler. E.g. If you have a 3rd order polynome and you know one root is 3, then you know it is (x-3)*f2(x)=0, where the f2 is 2nd order. And 2nd order polynome can be solved always.
You're strange Marcie. :blink:
Title: Help
Post by: OldBloke on September 15, 2008, 05:00:43 PM
I'm sure Jewelz is avidly following this and can't wait for the next installment :D
Title: Help
Post by: Anonymous on September 15, 2008, 05:11:58 PM
Quote from: OldBloke;244819I'm sure Jewelz is avidly following this and can't wait for the next installment :D
He might even want to start more advanced maths early :)
Title: Help
Post by: Jewelz^ on September 15, 2008, 06:32:57 PM
Quote from: OldBloke;244753Dead Men Walking homework service. How can we help? :doh:
 
Jewelz - a bit of advice. You got the answers you wanted but, at the very least, please make an effort to understand the solution. If you can't see how the answer was derived then speak to your teacher and explain that you are struggling and need help. If you just hand in correct answers then your teacher will think (wrongly) that you are up to speed. Don't cheat yourself or the teacher.

There was 18 other questions. but on these two i just got abit confused and sometimes it can help to be taught in others ways.
I have learnt this before but cant remember how to do them :)
 
Edit: and i understood the answers
Edit2: Thanks for help
Title: Help
Post by: Doorman on September 15, 2008, 06:36:20 PM
Next, English language. :norty:
Title: Help
Post by: Jewelz^ on September 15, 2008, 06:53:24 PM
I actually have some redrafts to do :narnar:
got to get from a D to a C on
Frankenstein
An Inspector Calls
booring :sad:
Title: Help
Post by: Browne on September 15, 2008, 06:58:24 PM
Quote from: Doorman;244835Next, English language. :norty:

Couldn't resist, now could you?
Title: Help
Post by: Jabbs on September 15, 2008, 09:41:43 PM
Quote from: BlueBall;244730I love Algebra - always have done.

BB and others, please rearrange these letters into a well known word :norty:

D S A

Quote from: Browne;244840Couldn't resist, now could you?

Ah, Doorman was being subtle there for a moment, almost restrained!  :dribble:
Title: Help
Post by: Anonymous on September 15, 2008, 09:43:59 PM
Quote from: Jabbs;244860D S A
Ads, its an abbreviation of advertisement :lmfao: