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i am solving a set of simlutaneous equations. they are as follows:
[SIZE=14] x[SIZE=1] 2[SIZE=14] +4y[SIZE=1] 2[SIZE=14] =16
[SIZE=14] xy=4
the small numbers indicate exponents. i have two of these for my math hw, but for now, if anyone knows how to solve this one, pleasse help, fast
edit- this isnt coming out right, so in words-
x squared + 4y squared= 16
xy=4
help please
That does look quite complicated...what does S stand for? :blink:Quote:
Originally posted by Legolas@15 March 2004 - 01:07
i am solving a set of simlutaneous equations. they are as follows:
[SIZE=14] x[SIZE=1] 2[SIZE=14] +4y[SIZE=1] 2[SIZE=14] =16
[SIZE=14] xy=4
the small numbers indicate exponents. i have two of these for my math hw, but for now, if anyone knows how to solve this one, pleasse help, fast
x^2 +4y^2 = 16
xy = 4
is that what you meant to do?
looks like there will be more then one answer to that question try taking the square root of the top equation and it should be alot more simple.
2 2
x+ 4y = 16
xy=4
4play, yes that is what i meant. it will have 2 answers for x and 2 answers for y, according to the exponent number
FFS if xy=4 then one of them is 1 and one is 4.
Work the rest out.
my math teacher doesnt like logic, so does anyone know how to solve using substitution? (solve for a variable in one equation and substitute the value into the second equation)
They can't both be 2, that would be ridiculous. Variables can't be the same number.Quote:
Originally posted by Legolas@15 March 2004 - 03:13
2 2
x+ 4y = 16
xy=4
no no, those are sposed to exponents
Learn English, the maths will follow.Quote:
Originally posted by Legolas@15 March 2004 - 03:16
no no, those are sposed to exponents
Are you Neotheone at all.
no, and i know english lol- im only in an 8th grade advanced class and my teacher expects us to know things perfectly an hour after he teaches them, not even thorougly enough
xy=4
therefore
x = 4/y
therefore
(4/y)^2 + 4y^2 = 16
which is
16/(y^2) + 4y^2 = 16
which is
16 + 4y^4 = 16y^2
which is
4y^4 - 16y^2 + 16 = 0
which is
y^4 - 4y^2 + 4 = 0
which is
(y^2 - 2)^2 = 0
which is
y^2 - 2 = 0
which is y = root 2
therefore when y = root 2, x = (4/root 2)
so the answer is x = (4/root 2) and y = root 2.
They're indices, not exponents.Quote:
Originally posted by Legolas@15 March 2004 - 01:16
no no, those are sposed to exponents
thanks a lot lamsey- are u a college professor or something? u know your stuff
:oQuote:
Originally posted by Lamsey@14 March 2004 - 20:24
xy=4
therefore
x = 4/y
therefore
(4/y)^2 + 4y^2 = 16
which is
16/(y^2) + 4y^2 = 16
which is
16 + 4y^4 = 16y^2
which is
4y^4 - 16y^2 + 16 = 0
which is
y^4 - 4y^2 + 4 = 0
which is
(y^2 - 2)^2 = 0
which is
y^2 - 2 = 0
which is y = root 2
therefore when y = root 2, x = (4/root 2)
so the answer is x = (4/root 2) and y = root 2.
m4d skillz!
First year university student. Oh, and it's half one in the morning.
somebody has a late start monday then.
i thought you were doing english at uni :huh:
MEng Computer Science :geek:Quote:
Originally posted by 4play@15 March 2004 - 01:34
somebody has a late start monday then.
i thought you were doing english at uni :huh:
a masters hey, /me adds your email to my list of people i can bug for help. :)
:oQuote:
Originally posted by 4play@15 March 2004 - 01:40
a masters hey, /me adds your email to my list of people i can bug for help. :)
No, I'm doing... um... a PhD in History of Turnips, that's it! :unsure:
uno mas problemo, senor profesor
x+2y= 4
y^2- xy= 7
do this one yourself or you'll never learn and fail all your exams and have to take a job working voluntary for a sex education scheme for the council.
:oQuote:
Originally posted by Lamsey+15 March 2004 - 03:44--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (Lamsey @ 15 March 2004 - 03:44)</td></tr><tr><td id='QUOTE'> <!--QuoteBegin-4play@15 March 2004 - 01:40
a masters hey, /me adds your email to my list of people i can bug for help. :)
No, I'm doing... um... a PhD in History of Turnips, that's it! :unsure: [/b][/quote]
At a Poly :lol:
Hey Legolas that's some basic stuff, just follow what Lamsey did, if you keep asking for the solutions you won't learn anything.
Gawd damn it, I was just about to go to bed <_<Quote:
Originally posted by Legolas@15 March 2004 - 01:51
uno mas problemo, senor profesor
x+2y= 4
y^2- xy= 7
Ok...
x + 2y = 4
so
x = 4 - 2y
so
y^2 - xy = 7
is
y^2 - y(4 - 2y) = 7
which is
y^2 - 4y + 2y^2 = 7
which is
3y^2 - 4y - 7 = 0
which is
(3y - 7)(y + 1) = 0
so
y = 7/3 or y = -1
so
when y = 7/3, x = (4 - 14/3) = -2/3
when y = -1, x = (4 - -2) = 6
so answers are (-2/3, 7/3) and (6, -1)
No more, I'm off to bed. Do the rest yourself :P
Give a man a fish...
Excellent point.Quote:
Originally posted by Nimdock@15 March 2004 - 04:00
Give a man a fish...
he's gonna want a boat.... <_<Quote:
Originally posted by Nimdock@14 March 2004 - 18:00
Give a man a fish...
lol thanks for all the help lamsey and others. no more. thanks again
Can't sleep :angry:
Ok, so now that this is homework corner, let me ask a question (physics, not maths):
"1 litre of water is heated from 20 degrees celsius until it has boiled and completely evaporated. The steam then condenses onto a second surface at 20 degrees celsius. How much energy has been transferred from the first surface to the second?"
Now, the way I'm thinking, it's just gonna be the specific heat needed to cool the water from 100 degrees celsius down to 20 degrees (i.e. 83600J), plus the latent heat of vapourisation of water (i.e. 2260000J). Any thoughts though people? :)
sorry i cant help 4thgen, but im only in 8th grade and we havent learned a bit of physics
would you like me to add the size too?Quote:
Originally posted by Legolas@15 March 2004 - 02:07
i am solving a set of simlutaneous equations. they are as follows:
[SIZE=14] x[SIZE=1] 2[SIZE=14] +4y[SIZE=1] 2[SIZE=14] =16
[SIZE=14] xy=4
the small numbers indicate exponents. i have two of these for my math hw, but for now, if anyone knows how to solve this one, pleasse help, fast
edit- this isnt coming out right, so in words-
x squared + 4y squared= 16
xy=4
help please
hang on let me think.... :huh:
Nope, it's a simple problem requiring only water's specific heat relating to condensation and vaporization.
But how is that physics?
its actually not that complicated, we learned that stuff in algebra 1. i see lamsey beat me to it though :P
Do you mean latent heat?Quote:
Originally posted by haxor41789@15 March 2004 - 04:55
Nope, it's a simple problem requiring only water's specific heat relating to condensation and vaporization.
But how is that physics?
It's more physics than anything else...(except possibly chemistry)
would you like me to add the size too?[/b][/quote]Quote:
Originally posted by RapFan+15 March 2004 - 04:19--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (RapFan @ 15 March 2004 - 04:19)</td></tr><tr><td id='QUOTE'><!--QuoteBegin-Legolas@15 March 2004 - 02:07
i am solving a set of simlutaneous equations. they are as follows:
[SIZE=4] x[SIZE=1] 2[SIZE=14] +4y[SIZE=1] 2[SIZE=14] =16
xy=4
the small numbers indicate exponents. i have two of these for my math hw, but for now, if anyone knows how to solve this one, pleasse help, fast
edit- this isnt coming out right, so in words-
x squared + 4y squared= 16
xy=4
help please
I think you should just close the tags :lol:
@nigel :lol: :lol:
Edit: feeling too much in a holiday mood to read and answer all through :lol:
This is just stupid.Quote:
Originally posted by 4th gen@15 March 2004 - 02:26
Can't sleep :angry:
Ok, so now that this is homework corner, let me ask a question (physics, not maths):
"1 litre of water is heated from 20 degrees celsius until it has boiled and completely evaporated. The steam then condenses onto a second surface at 20 degrees celsius. How much energy has been transferred from the first surface to the second?"
Now, the way I'm thinking, it's just gonna be the specific heat needed to cool the water from 100 degrees celsius down to 20 degrees (i.e. 83600J), plus the latent heat of vapourisation of water (i.e. 2260000J). Any thoughts though people? :)
It would be a lot easier to carry it in a bucket.