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2021-01-27
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Aniqa O'Neill

Answered 2021-01-28
Author has **100** answers

32 You can add 10 to both sides to get $\frac{c}{2}$ = 16 then multiply both sides by 2 to get c = 32.

asked 2020-10-31

asked 2022-06-29

Let $RLS(k)$ be the problem where we try to determine whether a Linear System of $n$ unknown variables and $m$ equations with Rational coefficients, has a $k$-solution which is basically a vector $x=({x}_{1},{x}_{2},\dots ,{x}_{n})$ such that at most $k$ of the ${x}_{i}$'s are non-zero.

Prove that $RLS(k)$ is NP-hard

Prove that $RLS(k)$ is NP-hard

asked 2021-09-11

The reduced row echelon form of the augmented matrix of a system of three linear equations in three variables must be of the form

asked 2022-04-01

For which value of b function

asked 2022-05-20

In the euclidean space, the points $(x,y,z)$ belonging to a regular octahedron are those that satisfy the inequalities

$\pm x\pm y\pm z\le a$

where $a\ge 0$. These eight inequalities can be divided into two groups of four according to the number (even or odd) of negative signs they contain. For example, the inequalities

$\begin{array}{r}x-y+z\le a\\ -x+y+z\le a\\ x+y-z\le a\\ -x-y-z\le a\end{array}$

all have one or three negative signs and the points satisfying these form a tetrahedron. The other four inequalities correspond to the dual tetrahedron of the first, which shows that the intersection of two regular dual tetrahedra form a regular octahedron. Moreover, the vertices of the two tetrahedra can be seen as the eight vertices of a cube.

I am wondering if there exists a similar relationship between regular polytopes in four dimensions. As it is another case of a regular cross-polytope, the hexadecachoron (or 16-cell) is defined by the sixteen inequalities

$\pm x\pm y\pm z\pm w\le a.$

If one were to take the eight inequalities containing an odd number of negative signs, say

$\begin{array}{r}x-y-z-w\le a\\ -x+y-z-w\le a\\ -x-y+z-w\le a\\ -x-y-z+w\le a\\ x+y+z-w\le a\\ x+y-z+w\le a\\ x-y+z+w\le a\\ -x+y+z+w\le a\end{array}$

which 4-polytope would be obtained ? I doubt it would be a regular 5-cell, since (obviously) the number of cells and the number of hyperplanes don't add up. Besides, the intersection of the two 4-polytopes corresponding to the two sets of eight inequalities should technically correspond to the 16-cell.

The tesseract, having eight cells, could be a candidate, but I have been unable to show that these eight inequalities define one (or any other 4-polytope). Any ideas?

$\pm x\pm y\pm z\le a$

where $a\ge 0$. These eight inequalities can be divided into two groups of four according to the number (even or odd) of negative signs they contain. For example, the inequalities

$\begin{array}{r}x-y+z\le a\\ -x+y+z\le a\\ x+y-z\le a\\ -x-y-z\le a\end{array}$

all have one or three negative signs and the points satisfying these form a tetrahedron. The other four inequalities correspond to the dual tetrahedron of the first, which shows that the intersection of two regular dual tetrahedra form a regular octahedron. Moreover, the vertices of the two tetrahedra can be seen as the eight vertices of a cube.

I am wondering if there exists a similar relationship between regular polytopes in four dimensions. As it is another case of a regular cross-polytope, the hexadecachoron (or 16-cell) is defined by the sixteen inequalities

$\pm x\pm y\pm z\pm w\le a.$

If one were to take the eight inequalities containing an odd number of negative signs, say

$\begin{array}{r}x-y-z-w\le a\\ -x+y-z-w\le a\\ -x-y+z-w\le a\\ -x-y-z+w\le a\\ x+y+z-w\le a\\ x+y-z+w\le a\\ x-y+z+w\le a\\ -x+y+z+w\le a\end{array}$

which 4-polytope would be obtained ? I doubt it would be a regular 5-cell, since (obviously) the number of cells and the number of hyperplanes don't add up. Besides, the intersection of the two 4-polytopes corresponding to the two sets of eight inequalities should technically correspond to the 16-cell.

The tesseract, having eight cells, could be a candidate, but I have been unable to show that these eight inequalities define one (or any other 4-polytope). Any ideas?

asked 2022-04-20

If $\left|a\right|>|a+b+c|$ prove that there is complex root such that $\left|z\right|<2$ .

Let$a{x}^{2}+bx+c=0$ be a quadratic equation where $\left|a\right|>|a+b+c|$ , a, b, $c\in \mathbb{R}$ . Prove that this equation has at least one solution $z\in \mathbb{C}$ such that $\left|z\right|<2$ .

Let

asked 2022-05-07

What is $2\sqrt{2}$ plus $5\sqrt{2}$?