This post makes my head hurt...
You will get another one from me.. just have to wait some hours for itjimmanycricket wrote:
nice to know we helped.SteikeTa wrote:
Ok, got it now and the test it self! Thanks all of you!!!
now wheres that karma.
x=.6 Ti-83+ FTW!
Ha!FrankieSpankie3388 wrote:
x=.6 Ti-83+ FTW!
<--- Ti-84 + Silver Edition FTW!
factor out the x-3 then reduce and foil out the 4, lol i have no idea what im talking about
Last edited by Masterstyle (2006-09-24 16:55:57)
I am not going to read all the responses.. but i did this problem in a second or two.. and heres how
Theres more than one way to skin a cat... but this is how i did it.
I first set them equal...
(4x-2)/5 = (3x-1)/10
I hope you realize why i can do this. I suppose you could say the 0 is still there on the right side.. but no need to write it as it isnt needed..
Now cross multiply... all that is .. is taking numerator times denominator of other side... and vice versa... just look at this next step for more clairification...
(4x-2)*10 = (3x-1)*5
now you dont have any fractions to deal with.
you could again do a few different things here... what i did is just multiply each side out... by that i mean multiply the left side by 10.... and right by 5.. producing
40x-20 = 15x-5
you now get the x's on one side.. and numbers on other..
40x-15x = -5 + 20
25x = 15
x = 3/5
feel free to send me a message if you have any other questions... these really take no time once you understand it all.
edit: rofl apparently im gay because i know math... hehe thanks for the +1 anyway... but yes i guess i <3 math.. i am going to be a computer enginner someday.. hopefully ><
Theres more than one way to skin a cat... but this is how i did it.
I first set them equal...
(4x-2)/5 = (3x-1)/10
I hope you realize why i can do this. I suppose you could say the 0 is still there on the right side.. but no need to write it as it isnt needed..
Now cross multiply... all that is .. is taking numerator times denominator of other side... and vice versa... just look at this next step for more clairification...
(4x-2)*10 = (3x-1)*5
now you dont have any fractions to deal with.
you could again do a few different things here... what i did is just multiply each side out... by that i mean multiply the left side by 10.... and right by 5.. producing
40x-20 = 15x-5
you now get the x's on one side.. and numbers on other..
40x-15x = -5 + 20
25x = 15
x = 3/5
feel free to send me a message if you have any other questions... these really take no time once you understand it all.
edit: rofl apparently im gay because i know math... hehe thanks for the +1 anyway... but yes i guess i <3 math.. i am going to be a computer enginner someday.. hopefully ><
Last edited by slicknic (2006-09-24 18:31:46)
"I'm an excellent driver." "I'm an excellent driver." "Wapner is on at 4."
Quoted from Rainman for you youngins...
Quoted from Rainman for you youngins...
Sorry about that I didn't see the sqrt, it was really late. But if you look closely, sqrt 200 is the same thing as 10x(sqrt2).DrunkFace wrote:
You get squareroot (102 + 102) = squareroot of 200... which i put. Its a right angle triangle. You talked about imaginary numbers before yet your getting simple year 10 maths wrong, whats wrong with you.. getting old and forgetful?Vub wrote:
You don't get 20m/s NE though, you get 10x(squareroot of 2) m/s NE.DrunkFace wrote:
No adding 2 vectors will give you a third vector of different size and direction. His example with the triangle explains it the best.
If you have 2 engines pushing you in differnt directions (vectors).
1. 10m/s north
2. 10m/s east
and you add them, you get sqrt 200 m/s North East your new vector.
Just complete the triangle and that third line is your new vector with a new size and direction.
EDIT: just found a diagram (has nothing to do with my example tho)
http://www.physics.uoguelph.ca/tutorial … 25anew.gif
By adding vectors A, B and C you get D.
ABCD is a cyclic quadrilateralNessie09 wrote:
anyone?Nessie09 wrote:
Ok, sinds where doing maths:
http://img1.putfile.com/thumb/9/26613451733.jpg(click to enlarge)
prove that ED is parallel to CF
been working on this all day, but can't figure it out
Therefore <FCA = 180 - <FBA
FE is a straight line
Therefore <ABE = 180 - <FBA = <FCA
But <ABE = <ADE (equal chords subtend equal angles at circumference)
Therefore <ADE = <ABE = <FCA
Therefore CF parallel to DE (Alternate angles equal)
I'm just wondering, this position vector should have an x and y component (I assume that they're i and r respectively), so how can i and r be part of the calculations? You'll end up with an integer, how does one integer refer to a position vector?bennisboy wrote:
try this one:
r is a position vector of a particle given by:
r=t2i+0.5t3j
t is time and i and j are unit vectors in the horizontal and vertical respectively.
give an equation for the velocity v of the particle
The only way for differentiation to give you the correct velocity vector is if i and r are constants. Are they meant to be constants?
haven't we already been through this. I take it you mean i and j? they are unit vectors in the horizontol and vertical. therefore the are constants. maybe i should have put in brackets, but when in bold, that means it is a vector. you dont actuall differentiate the unit vector, just the numbers that give you the quantity of them. therefore you end up with integer quantities for both unit vectors.Vub wrote:
I'm just wondering, this position vector should have an x and y component (I assume that they're i and r respectively), so how can i and r be part of the calculations? You'll end up with an integer, how does one integer refer to a position vector?bennisboy wrote:
try this one:
r is a position vector of a particle given by:
r=t2i+0.5t3j
t is time and i and j are unit vectors in the horizontal and vertical respectively.
give an equation for the velocity v of the particle
The only way for differentiation to give you the correct velocity vector is if i and r are constants. Are they meant to be constants?
Last edited by bennisboy (2006-09-25 07:25:01)
hey search the problems on google 90% of the time someone posted the same thing about the problem on some other site and there should be answers. Hope that helps
Last edited by blademaster (2006-09-25 07:30:41)
Well then why not represent it in parametric form?bennisboy wrote:
haven't we already been through this. I take it you mean i and j? they are unit vectors in the horizontol and vertical. therefore the are constants. maybe i should have put in brackets, but when in bold, that means it is a vector. you dont actuall differentiate the unit vector, just the numbers that give you the quantity of them. therefore you end up with integer quantities for both unit vectors.Vub wrote:
I'm just wondering, this position vector should have an x and y component (I assume that they're i and r respectively), so how can i and r be part of the calculations? You'll end up with an integer, how does one integer refer to a position vector?bennisboy wrote:
try this one:
r is a position vector of a particle given by:
r=t2i+0.5t3j
t is time and i and j are unit vectors in the horizontal and vertical respectively.
give an equation for the velocity v of the particle
The only way for differentiation to give you the correct velocity vector is if i and r are constants. Are they meant to be constants?
The answer = Pineapple
I have no idea, its jus not the done thingVub wrote:
Well then why not represent it in parametric form?bennisboy wrote:
haven't we already been through this. I take it you mean i and j? they are unit vectors in the horizontol and vertical. therefore the are constants. maybe i should have put in brackets, but when in bold, that means it is a vector. you dont actuall differentiate the unit vector, just the numbers that give you the quantity of them. therefore you end up with integer quantities for both unit vectors.Vub wrote:
I'm just wondering, this position vector should have an x and y component (I assume that they're i and r respectively), so how can i and r be part of the calculations? You'll end up with an integer, how does one integer refer to a position vector?
The only way for differentiation to give you the correct velocity vector is if i and r are constants. Are they meant to be constants?
omg!!!!
i seriously dotn even kno how to do circumfrence, or perimiters, area, so on.. i will have no chance
my ansewr is 0, CUZ IT SAYS SO!!
i seriously dotn even kno how to do circumfrence, or perimiters, area, so on.. i will have no chance
my ansewr is 0, CUZ IT SAYS SO!!
The dude answered his own question in his first post. X=1 3/5-3/5= 0.... but i hope he got that a long time ago....
wow, disregard my last post and just concentrate on my sig....
I am not doing your math homework. . .
I took some time and didn't get anywhere, but these pages should help:Nessie09 wrote:
anyone?Nessie09 wrote:
Ok, sinds where doing maths:
http://img1.putfile.com/thumb/9/26613451733.jpg(click to enlarge)
prove that ED is parallel to CF
been working on this all day, but can't figure it out
http://www.algebralab.org/lessons/lesso … Chords.xml
http://www.algebralab.org/Word/Word.asp … Chords.xml