Add Single Cells to a Table

I was much too slow but just in order to show siunitxWith booktabs:

1. Change Partition Table with GParted

partition length of 5854484482 sectors exceeds the msdos-partition-table-imposed maximum of 4294967295 This signals clearly that the (logical) disk is over the 2 TB limit.To work around it - you have to use the GPT partition tabel. To change it do:If you change the partition table from msdos to gpt, you will lose all your data! Ensure that you have a backup to restore the data.If you decide to reorganize your disk, consider to use the whole raid array as lvm physical volume. It is no problem with a current ubuntu to put the /boot partition inside the lvm. This has also the benefit that you can easily resize this partition.

2. Placement of table in IEEE paper

You should consider not changing any of the layout when submitting a paper. If tables should float to the top based on the conference paper class, then so-be-it.However, you can force the floating behaviour of the table environment to stop based on the following two approaches:.

3. What is the name of this table "tool"?

Maybe it's a desk organizer. Try using Desk Trays to place papers in, or hang utility trays from the sides to place journals or other regularly used books when not in use. Could not bring up your link, so I hope these ideas help anyway. Your English, was just fine.

4. i need a multiplication table?

If you have a Spreadsheet program, such as Quattro or Excel, make one yourself. Its quite simple

5. How to improvise a table for a Thanksgiving dinner?

you could go to the hardware store and buy a set of saw horses and a sheet of plywood. set the sawhorses up place the plywood on top and place a sheet on top. thats what we do. we have carpenters that we borrow the saw horses from.

6. Missing number in the table [closed]

The answer is 1because 2 6 8 plus 3 7 3 is exactly 6 4 1 (not sure yet)

7. Writing descriptions of columns of table

If you need colors, you may use colortbl or table from xcolor and more preferably tikz. But this is a simple way with ordinary tabular:

8. Why was the medal table manipulated ?

Um. how do you know that the first one is ranked by gold medals? o.O USA has both most gold and most medals overall in the first one so it could be most medals.

9. Excel Pivot Table help?

sophisticated factor. browse at google or bing. that can assist!

10. Formatting database like table

If word length exceeds field width wrap still should do something sensible. As coded I expect the hell breaking loose (size_t is unsigned).Consider renaming max_linebreaks to box_height and factoring out a get_row_height function. Along the same line, it is not quite obvious what is the task of the inner loop. Consider to factor it out into a function, just for sake of giving it a descriptive name

11. Source formatting of Markdown Table

I would simplify the formating of the rows using list-comprehensions/generator expressions rather than building a list to feed into join:

12. Character Table From Presentation

You could also proceed like this, once you have (correctly) established that $G$ has four irreducible characters of degree 1 (linear characters). Note that $x^2$ is in $G^prime$, the derived group of $G$, since $x^-1y^-1xy = x^2.$ Hence $x^2$ is in the kernel of each linear character of $G$, so $x^4$ is certainly in all these kernels. But there must be some irreducible character $chi$ of $G$ whose kernel does not contain $x^4.$ Let $sigma$ be a representation affording $chi$, so you have established already that $sigma$ must be two-dimensional. Now $sigma(x)$ must have order $8$, since $sigma(x^4)

eq I_2 times 2$, by the way we chose $chi$. What can the eigenvalues of $sigma(x)$ be? At least one, say $omega$, is a primitive $8$-th root of unity. But then $omega^3$ is also an eigenvalue of $sigma(x)$, since if $sigma(x).v = omega v$, then $sigma(y^-1xy).sigma(y^-1).v = omega^3sigma(y^-1)v$. This tells us that $chi(x) = omega omega^3$ (and, more generally, tells us that $chi(x^r) = omega^r omega^3r$ for $ 0 leq r leq 7).$ You could finish by noting that $sum_j=0^7 |chi(x^j)|^2 = 16,$ so that $chi$ must vanish outside $langle x

angle$, but there are several other ways to use the orthogonality relations to deduce this. Note that this really gives us two different irreducible characters, since we could have used $bar omega$ instead of $omega$. However, this procedure is rather specific to this particular group.Here are a couple of well-known useful general facts. If $G$ is a finite non-Abelian $p$-group, then the number of distinct linear characters of $G$ is divisible by $p^2.$ This is because $p^2$ divides $|G|$, and divides $chi(1)^2$ for each non-linear irreducible character of $G$. Also, if $G$ is a finite $p$-group, and $chi$ is an irreducible character of $G$, then $chi(1)^2$ divides $[G:Z(G)].$ (Since both of these are powers of $p$, it suffices to prove that $chi(1)^2 leq [G:Z(G)].$ But $$|G| = sum_g in G|chi(g)^2| geq sum_z in Z(G) |chi(z)|^2 = |Z(G)|chi(1)^2).$$