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what if it was %100 going to hit earth in 3 weeksThen I would have read the part about how it is 270 meters in diameter and realized that it probably wouldn't do much after it got through the atmosphere, and that they could probably blow it up.
what would you do
i never even said it will make your computer faster, stupid, it will cut your electricity bill by a few dollars and make your PC more energy efficient.It will cut your bill by a few pennies a year. Not worth the cost of gold over copper.
He obviously died right before his team was going to push down the nexus, so he sold his items and got new ones.
son, what the forget are you doing
You can undock or redock as many apps as you want.
wow, both of them are full of stuff i'll never use. it's almost like the default apps on my phone.
If you trace your finger around side 1 you will eventually arrive at side 2, meaning they are the same side.
???
0s do not matter in physics.Let me rephrase this.
[10 , 0°] + [5 , 180°] = [5 , 0°] = 5
[10, 0° , 0°] + [5 , 180° , 0°] = [5 , 0° , 0°] = [5 , 0°] = 5
[10, 0° , 0° , 0°] + [5 , 180° , 0° , 0°] = [5 , 0° , 0° , 0°] = [5 , 0° , 0°] = [5 , 0°] = 5
Squares having no depth has nothing to do with physics breaking down. Take any two-dimensional planar slice of reality.Except you cannot divide the depth on anything into a quantity of planes because planes have a depth of 0. So you are wrong because in order for you to be correct you must divide by 0.
YES I KNOW YOU CAN'T TAKE A PLANAR SLICE OF AN ATOM forget YOU JUST TAKE THE INTERSECTED PORTION OF IT ASDF
That planar slice of reality is in three dimensions and it has no depth. Therefore, since reality is made up of an infinite number of objects with no depth whatsoever, reality does not and cannot exist. Well done, you just proved reality doesn't exist. You are a nonexistant Ipquarx.
Is it possible to have a 3-d square?Vectors are not objects. They are forces in directions. There is no limit to the amount of dimensions they can exist in. It is possible to do the math without 0s, but I didn't pay attention enough to make up the formula to convert 4 dimensional hyperspherical polar coordinates into 4 dimensional hyperrectantgular coordinates to add them.
No, because everything 3-d has to have physical depth, if it doesn't have depth then by definition it is 2d.
What I'm saying is, adding a zero onto the end of a vector does not add a dimension. An object can exist in only one dimension.
See? Because of this, yes our 3-d physics equations may be right, but they cannot function properly in the next dimension without having some sort of breaking point.0s do not matter in physics.
If it's mathematically impossible and hasn't been created in reality then why are people saying it exists?It is mathematically possible. That is the whole point.
The Klein bottle is not four dimensional. If you don't believe me look at its Wikipedia pageFrom your beloved wikipedia:
Note that this is an "abstract" gluing in the sense that trying to realize this in three dimensions results in a self-intersecting Klein bottle. The Klein bottle, proper, does not self-intersect. Nonetheless, there is a way to visualize the Klein bottle as being contained in four dimensions.
By adding a fourth dimension to the three dimensional space, the self-intersection can be eliminated. Gently push a piece of the tube containing the intersection along the fourth dimension, out of the original three dimensional space. A useful brown townogy is to consider a self-intersecting curve on the plane; self-intersections can be eliminated by lifting one strand off the plane.tl;dr A true Klein bottle must exist in the fourth dimension.
Blockland forums: Certified theoretical physicistsIt is theoretical math, which is not complex at all because theoretical math (for the most part) is defined by its consistency and patterns.
Actually, the equations for Volume and area are v=L x W x H and a=L x W.I use length^dimension because we don't have a name for fourth and beyond dimensional measurements, so it keeps it consistent an easier to understand. Substituting L^3 for LWH or L^4 for LWH(IV) doesn't change my point.
