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9x - 7i > 3(3x - 7u)



i = square root of -1




There is a surprise for the first person that solves the problem.

ef1
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I don't think that was the answer.

It's my process, don't censor me.

haermm

I'm actually very aware i is basic college maths but we did that last fall semester and I don't remember how to do it because I was an art major and dumb as all hell

I think it involved flipping fractions or dividing by -1 or something

Therefore I'm justified in spamming more EF.

ef1
ef1
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you're such a ****

bork pls

where is dudathong with our prize?

it's probably him clogging up the OTF again by obnoxiously replying to threads

I just mark forum read.

too much work

Lazy shitbag

duh

still no dadudeisamom

dadudelookslikealady

Heh heh yeah

Originally posted by dadudemon
9x - 7i > 3(3x - 7u)



i = square root of -1




There is a surprise for the first person that solves the problem. -7i > -21u

-i > -3u

i < 3u

-1 < 9u(sq)

1 > -9u(sq)

1/9 > -u(sq)

1/3 > -u

-1/3 < u

I think.

I was never good at sums so I'm gonna go with

8=> - 7175

parenthesis


You're close but not quite right.

i/3 < u is the correct answer but there is another way to express that same exact inequality that is the answer I am looking for.


Jaden, you're way off.

ef1

Originally posted by dadudemon
parenthesis


You're close but not quite right.

i/3 < u is the correct answer but there is another way to express that same exact inequality that is the answer I am looking for.


Jaden, you're way off.

You're so lame....haermm

Originally posted by dadudemon
9x - 7i > 3(3x - 7u)

i = square root of -1

There is a surprise for the first person that solves the problem.
First and foremost, the imaginary unit i isn't the square root of -1. As the space of real numbers is a subspace to complex numbers, still making the square root of negative numbers nonexistent. Also, see proof.

http://img135.imageshack.us/img135/2558/99049670.gif

Secondly you can't compare two complex numbers. As such a comparison wouldn't have the basic properties required of the subspace such as symmetry and reflexivity, and would thus be nonsensical.

You tell'em!!!

Oh damn, if it has a solution, it's gonna be tough to figure it out

Originally posted by dadudemon
9x - 7i > 3(3x - 7u)



i = square root of -1




There is a surprise for the first person that solves the problem.
Whats the surprise though

i <3 u

Originally posted by silver_tears
You're so lame....haermm

Originally posted by silver_tears
i <3 u





You won the prize!

I love you, Irene! hug





Paranthesis, you were so very close to getting the prize, man. SO CLOSE!

First time I've seen someone winning by being conventionally wrong.

If i wasn't the imaginary unit, and you'd reformed the relation then this could've worked out.

Originally posted by Astner
First and foremost, the imaginary unit i isn't the square root of -1. As the space of real numbers is a subspace to complex numbers, still making the square root of negative numbers nonexistent. Also, see proof.

http://img135.imageshack.us/img135/2558/99049670.gif

Secondly you can't compare two complex numbers. As such a comparison wouldn't have the basic properties required of the subspace such as symmetry and reflexivity, and would thus be nonsensical.

Originally posted by Astner
First time I've seen someone winning by being conventionally wrong.

If i wasn't the imaginary unit, and you'd reformed the relation then this could've worked out.

Astner, I had no idea you failed College Algebra 201.

Originally posted by Astner
First time I've seen someone winning by being conventionally wrong.

If i wasn't the imaginary unit, and you'd reformed the relation then this could've worked out. I agree, the additional statement about i made me first think it said "it wasn't the root of minus 1", then when I read it again I thought the problem was to find the value of u as you would any algebraic problem. To that, I only have one thing left to say.

Originally posted by Barker
ef1

Originally posted by dadudemon



You won the prize!

I love you, Irene! hug


blush2

Originally posted by Astner
First time I've seen someone winning by being conventionally wrong.
You're not an American voter, are you.

he's like from estonia or something.

i'm going to drop toe hold all of you

Stalky the Clown answered this on my forum right after you posted it, and, parenthesis actually answered it as well:

Originally posted by parenthesis
-7i > -21u

-i > -3u

i < 3u

-1 < 9u(sq)

1 > -9u(sq)

1/9 > -u(sq)

1/3 > -u

-1/3 < u

I think.

Originally posted by Ax3l
You're not an American voter, are you.

Ohohoho, classic political comedy, epic!

Quadruple post FTW

Breaking combos and asking for attention, don't mind me. ef1

Mega burn

Originally posted by dadudemon
So you really did fail Algebra 201?

Too bad. Well, you tried.
Can't you read? I never took a course called algebra 201 and thus I couldn't have failed it.

Either way, courses with names such as calculus 101 and algebra 201 are of substandard quality. They cover less content and details over more time. It's the layman's math courses.

In general, I'm better at math than you.

Authority aside, you ignored my proofs.

Originally posted by Astner
In general, I'm better at math than you.


Bow down, DDM, his maths is stronger than yours!!!

Originally posted by dadudemon


Jaden, you're way off.

Did I only hit the belly and not the ****?

Originally posted by Astner
Can't you read? I never took a course called algebra 201 and thus I couldn't have failed it.

Either way, courses with names such as calculus 101 and algebra 201 are of substandard quality. They cover less content and details over more time. It's the layman's math courses.

In general, I'm better at math than you.

Authority aside, you ignored my proofs.

Originally posted by Astner
Either way, courses with names such as calculus 101 and algebra 201 are of substandard quality.

That's...my point.



Originally posted by Astner
They cover less content and details over more time. It's the layman's math courses.


Exactly.

Originally posted by Astner
In general, I'm better at math than you.

I agree.

Originally posted by Astner
Authority aside, you ignored my proofs.

So you really did fail Algebra 201?

Bump

Solve for

76x-8y+3z^2=7
3x+3y+8z=6
y+89z=-3
6x=76y+z

Use distribution as opposed to elimination.

Personally, I wouldn't even want to touch this problem.

Oops, if you're confused find X, Y, Z.

The answer will by like (X,Y,Z); think a three dimensional graph.

Originally posted by Dolos
That's why they never went ever three rows in the problems I had.

For four rows I guess you'd need w, x, y, z. Four variables, and a four dimensional graph???

Maybe I'm just being pseudo-intellectual right now. But would that make a 4d graph? If you had 4 variables and 4 rows????
If you have as many unique equations as you do variables you'll get one solution, this will manifest itself as a dot.

If you have one less unique equation than you do variables, you'll get a solution in terms of a single variable function that will manifest the answers as a curve.

If you have two less unique equations than you do variables, you'll get a solution in terms of a two-variable function that will manifest the answers as a surface.

The next step is will represent the solutions as a three-dimensional body, a four-dimensional body, etc.

Originally posted by Astner
If you have as many unique equations as you do variables you'll get one solution, this will manifest itself as a dot.

If you have one less unique equation than you do variables, you'll get a solution in terms of a single variable function that will manifest the answers as a curve.

If you have two less unique equations than you do variables, you'll get a solution in terms of a two-variable function that will manifest the answers as a surface.

The next step is will represent the solutions as a three-dimensional body, a four-dimensional body, etc.

Curve as in parabola?

What do you know about the fourth dimension in terms of math? Do they ever go over that in any of your courses?

Originally posted by Dolos
Curve as in parabola?
Curve as in a line that can bend in any direction.

Let's examine the following basic equation.

x + 2y = 5

Where x and y are real numbers.

Two variables, one equation. This is called an underdetermined system.

Evidently there are infinitely many solutions to the equation above each of which can be expressed as x = f(y) = 2y - 5.

This function would describe x as a function of y as a straight line, with each point of the line being a solution to the equation.

So if y = 0 then x = -5. This is a solution to the equation. If y = 0.1 then x = -4.8, and so on.

So (x, y) = (-5, 0) and (-4.8, 0.1) are two of the infinitely many solutions.

Originally posted by Dolos
What do you know about the fourth dimension in terms of math? Do they ever go over that in any of your courses?
The fourth dimension as in the mathematically Euclidean sense (or alternative geometries), or the fourth dimension in the sense of physics as expressed in Quantum Field Theory?

Because the answer to both questions are yes. In fact we worked with curve- and surface integrals than bended and twisted in n dimensions in one of the first courses we took.

Originally posted by Astner
Curve as in a line that can bend in any direction.

Let's examine the following basic equation.

x + 2y = 5

Where x and y are real numbers.

Two variables, one equation. This is called an underdetermined system.

Evidently there are infinitely many solutions to the equation above each of which can be expressed as x = f(y) = 2y - 5.

This function would describe x as a function of y as a straight line, with each point of the line being a solution to the equation.

So if y = 0 then x = -5. This is a solution to the equation. If y = 0.1 then x = -4.8, and so on.

So (x, y) = (-5, 0) and (-4.8, 0.1) are two of the infinitely many solutions.

Yes, we just finished up that part of College Algebra. Now we're moving onto exponents. Haven't look at the homework yet.



Nice.

Originally posted by Dolos
Yes, we just finished up that part of College Algebra. Now we're moving onto exponents. Haven't look at the homework yet.
College?

Originally posted by Dolos
Nice.
In Euclidean geometry the transition from three to four dimensions is the same as that from two to three dimensions, you just add another axis orthogonal to the rest of them.

Originally posted by Astner
College?

Like I said it's a Junior College, but not all courses are remedial. I tested out of all my Remedial Courses, so I'm still taking college level courses. Also I should earn about 13,000 dollars this year as a Mover Helper, which will help with more courses.



Therein lies the fundamental difference between the extra-dimensional concept.

Originally posted by Dolos
Like I said it's a Junior College, but not all courses are remedial. I tested out of all my Remedial Courses, so I'm still taking college level courses. Also I should earn about 13,000 dollars this year as a Mover Helper, which will help with more courses.
Oh that's right, American colleges and universities aren't tax funded so you have to pay for your education straight out of your own pocket.

Originally posted by Dolos
Therein lies the fundamental difference between the extra-dimensional concept.
What?

Originally posted by Astner
Oh that's right, American colleges and universities aren't tax funded so you have to pay for your education straight out of your own pocket.








A plane of existence that has one extra dimension. Hence, extra-dimensional.

Some of the things described about the 4thD seem unreal, I think it's pretty awesome.

N0WjV6MmCyM

The notion of learning how to work with this mathematically does intrigue me.

How to work with dimensional projections? That's basic linear algebra.

You ignorant people get impressed by the simplest of things. http://streamonster.com/wp-content/uploads/2012/05/Kappa.png

Originally posted by Astner
How to work with dimensional projections? That's basic linear algebra.

You ignorant people get impressed by the simplest of things. http://streamonster.com/wp-content/uploads/2012/05/Kappa.png

I'm not ignorant, I'm just no where near linear algebra. Linear algebra is MAT 201, I'm way back in college algebra at MAT 121.

Originally posted by Dolos
I'm not ignorant, I'm just no where near linear algebra. Linear algebra is MAT 201, I'm way back in college algebra at MAT 121.
My three first courses were: Single Variable Analysis Linear Algebra and Geometry The Physics Around Us
These courses were held simultaneously over the first period (seven weeks) in our program.

So we jumped straight from high school maths to Linear Algebra. Which leads me to wonder whether or not you were even taught multiplication in high school.

I almost dropped out of high school. The minimum requirements was rudimentary algebra. As in below MAT 096.

However, I have a friend who's graduating this year, a senior in high school, who's exactly where you were as far as math, suffice to say he has a perfect GPA, he also won the State Wrestling Tournament and placed 2nd or 3rd in a Nationally Ranked Tournament when not going ballistic on his education. He has scholarship offerings into so many universities.

I just decided not to go to school and instead to **** around. I'm not the best one for you to compare your education with.

Originally posted by Dolos
I almost dropped out of high school. The minimum requirements was rudimentary algebra. As in below MAT 096.

However, I have a friend who's graduating this year, a senior in high school, who's exactly where you were as far as math, suffice to say he has a perfect GPA, he also won the State Wrestling Tournament and placed 2nd or 3rd in a Nationally Ranked Tournament when not going ballistic on his education. He has scholarship offerings into so many universities.

I just decided not to go to school and instead to **** around. I'm not the best one for you to compare your education with.
Makes sense.

So why the change of heart? Why not continue to **** around?

Originally posted by Astner
Makes sense.

So why the change of heart? Why not continue to **** around?

Because some things never change....but some things do.

http://www.expoweekly.com/site/wp-content/uploads/2010/03/Morpheus-Red-or-Blue-Pill-the-matrix-1957140-500-568.jpg

Idk, the notion of being mediocre no longer appeals to me. My eyes have been opened to the wonders of science.

Only the brightest will have a place in the future.

Well if you ever need help with either math, physics, or programming, sent me a personal message.

Originally posted by Astner
Well if you ever need help with either math, physics, or programming, sent me a personal message.

I wouldn't give you the satisfaction of seeing how far behind you I am now.

Originally posted by Astner
My three first courses were: Single Variable Analysis Linear Algebra and Geometry The Physics Around Us
These courses were held simultaneously over the first period (seven weeks) in our program.

So we jumped straight from high school maths to Linear Algebra. Which leads me to wonder whether or not you were even taught multiplication in high school.


I tested out of all algebra and geometry related classes. I didn't have to take those in college.* My Freshman year, we did cryptography for our "math". We not only had to select all values and encrypt and decrypt a set using the Merkle-Hellman knapsack cryptosystem, we had to write a program that would do the same....all of it manually (by hand...and on the programming, no shortcuts: we had to write the entire algorithms). Certainly, it was an NP-Complete, but it certainly felt hard. ho ho ho!



*That was my way of saying that some ignorant people are burdened with the simplist of things when they first go to college...because you said this "You ignorant people get impressed by the simplest of things."

Cryptography falls under Linear Algebra too. Nice try though.

Originally posted by Astner
Cryptography falls under Linear Algebra too. Nice try though.

Uhhh....very little of it does.


Here is some of the math used in it (note: none of it listed is "Linear Algebra":

Abstract Algebra
Combinatorics
Algebraic Geometry
Discrete Math
Calculus
Statistics


Nice try, though...but you have quite a bit to learn if you think Cryptography falls under "linear algebra." That would be like saying all colored pencils are blue.

Originally posted by dadudemon
Uhhh....very little of it does.

Here is some of the math used in it (note: none of it listed is "Linear Algebra":

Abstract Algebra
Combinatorics
Algebraic Geometry
Discrete Math
Calculus
Statistics

Nice try, though...but you have quite a bit to learn if you think Cryptography falls under "linear algebra." That would be like saying all colored pencils are blue.
The underlined are part of Linear Algebra as well.

Originally posted by Astner
The underlined are part of Linear Algebra as well.

You have that backwards.


Linear Algebra is part of those groups. All of them.

Originally posted by dadudemon
Nice try, though...but you have quite a bit to learn if you think Cryptography falls under "linear algebra." That would be like saying all colored pencils are blue.