Math problem

[leone]

Originally posted by a_iver+Jan 7 2008, 12:13 AM--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (a_iver @ Jan 7 2008, 12:13 AM)</td></tr><tr><td id='QUOTE'>

Originally posted by leone@Jan 7 2008, 01:11 AM
<!--QuoteBegin-a_iver

@Jan 7 2008, 12:08 AM
oh no, wait... the coordinates would be equal to (0,0) and (a, 'B').... i'm going in circles i swear..

Dude hang on, what are the givens of this question?

i dunno anymore ut: [/b][/quote]
I hate you. My head hurts now. :lol: :lol:

 

[a_iver]

i used 0,0 for g,h but that might've been bad?

 

[leone]

*twitch* What does the problem say?

 

[neo979]

Originally posted by leone+Jan 6 2008, 10:08 PM--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (leone @ Jan 6 2008, 10:08 PM)</td></tr><tr><td id='QUOTE'>

Originally posted by givemfitz@Jan 6 2008, 11:55 PM
<!--QuoteBegin-a_iver

@Jan 6 2008, 09:41 PM


i need to make the equation without using 'b' (i don't have access to it), but from the two sets of coordinates that make the hypotenuse

Well now I'm confused. How can you have the Pythagorean theorem without b? Unless you mean b is always what you're solving for........yeah confused :lol:

I think he means that there are two unknowns. In which case you have to make an equation using a and b

ie: a+b=c
a=c-b

then you substitue your "known" a into the original equation.

(c-B)+b=c

The thing is in this case it keeps turning into zero. So I'm either out of practice or the numbers are off. h34r: [/b][/quote]
This is close to being the correct strategy for handling two unknowns, but the problem is that you can't substitute one equation back into itself. If you have two unknowns, you need to have two equations. In fact, for N unknowns, you need N different equations. Otherwise, you do not have enough information to determine concrete values for the variables.

 

[leone]

I'm outta practice. :woo:

Where's Ravi?

 

[givemfitz]

Originally posted by neo979@Jan 6 2008, 10:16 PM

This is close to being the correct strategy for handling two unknowns, but the problem is that you can't substitute one equation back into itself. If you have two unknowns, you need to have at least two equations. In fact, for N unknowns, you need at least N different equations. Otherwise, you do not have enough information to determine concrete values for the variables.

I think that the only real answer would be to plot all possible values on a graph but only on the positive side as stated earlier.

EDIT - all possible values for A and B that is

 

[a_iver]

is there another value i can use? like the angle of the hypotenuse if x is 90° and y is 0°

 

[a_iver]

Originally posted by givemfitz+Jan 7 2008, 01:24 AM--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (givemfitz @ Jan 7 2008, 01:24 AM)</td></tr><tr><td id='QUOTE'> <!--QuoteBegin-neo979@Jan 6 2008, 10:16 PM

This is close to being the correct strategy for handling two unknowns, but the problem is that you can't substitute one equation back into itself. If you have two unknowns, you need to have at least two equations. In fact, for N unknowns, you need at least N different equations. Otherwise, you do not have enough information to determine concrete values for the variables.

I think that the only real answer would be to plot all possible values on a graph but only on the positive side as stated earlier. [/b][/quote]
but the values are always changing for x and y.

 

[leone]

If they're always changing than how can they even have values?

 

[givemfitz]

Well anthony...............is that because X and Y are actually A and B or B and A?
(depending on which arrangement keeps the values on the positive side of the graph)


Cos ya knowwww. *shoots Anthony* <_<

 

[a_iver]

their values are always changing. this isn't one triangle shape, it's always shifting values for 'a' and b, but the length of 'c' always stays the same

 

[a_iver]

i'm sorry fitz! i'm horribly lost. i can't use a graph, i need to give the computer an equation... A and B are segments. x is the length of A, and B is the height of y. but.... let me see if i can draw this...

 

[givemfitz]

Well.........yes. But isn't your (0,0) point going to always be +3 and not the actual (0,0) on the graph for your purposes? or am I just getting further and further from what you mean?

You're making my head hurt and I've really earned my mindless TV time. :lol: :lol:

 

[givemfitz]

Originally posted by a_iver@Jan 6 2008, 10:44 PM
i'm sorry fitz! i'm horribly lost. i can't use a graph, i need to give the computer an equation... A and B are segments. x is the length of A, and B is the height of y. but.... let me see if i can draw this...

Well if you can't use a graph then I'm pretty sure the answer can only be expressed as an equation or more correctly multitudes of equations given one known variable.

Pretty much what neo already told you..............now go away :lol: :lol: :lol: :lol: :lol:

 

[Scythe]

When life gets complex, multiply it by it's complex conjugate

 

[leone]

When life gets complex, grab a beer. k:

 

[neo979]

a_iver, if you still need help with this, please just post the whole problem verbatim (if there are pictures, you could scan them, describe them, or re-draw them yourself) so we don't go off on any more tangents (pun intended) - or just ask your teacher.

 

[iamkgb]

Originally posted by a_iver+Jan 6 2008, 10:21 PM--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (a_iver @ Jan 6 2008, 10:21 PM)</td></tr><tr><td id='QUOTE'> an equation to plug the numbers in, not so much trying to solve the problem, because the values will be constantly changing. So I have this right triangle. a^2 + b^2 = c^2, right? Well c is always equal to 3. But 'a' and 'b' are always changing. [/b]

If it's an equation you're after then it's "simply":

3 = SqRt(a^2 + b^2) (as already stated by neo979)

However, given that you've also stated

Originally posted by a_iver@
two points connected make up the hypotenuse 'c'. their coordinates are (g,h) and (i,j)

AND

<!--QuoteBegin-a_iver
A and B are segments. x is the length of A, and B is the height of y[/quote]

I think the only way for us to help you is for you to put the WHOLE question on here...b/c the answer cannot "simply" be the equation above.

Ahhhh trig....

"SIN" ing off now

 

[givemfitz]

Just a guess.......but I don't think that what he was doing was assigned. It may have been inspired by an assigned problem but I think he had a late night, sleep deprived brain storm and decided to create a program that could solve for two unknown variables. Or something along those lines.

Just guessing.

 

[leone]

I want some special water.