Results for query "  ab  x c   "

Boolean algebra: (A'+B)(A+C)

I have a little problem with this expression:

x = (A'+B)(A+C)


I know it can be simplified to:

A'C+AB


since ive used some software to simplify it, but i simply can't see how it is done.This is what i've done so far:

(A'+B)(A+C) =>
A'A + AB + A'C + BC =>
0 + AB + A'C + BC =>
AB + A'C + BC


I just fail to see how i can do this differently and get to the correct result.

Why are ring actions much harder to find than group actions?

And of course, a group $G$ acting on $X$ is just a group homomorphism $G \to \text{Aut }X$.In fact, any ordinary category $C$ has a free $\text{Ab}$-enriched category $\mathbb{Z}[C]$ equipped with the universal functor from $C$ to an $\text{Ab}$-enriched category.

Invertible matrices satisfying $[x,y,y]=x$.

Let $x,y$ be invertible matrices (say, over $\mathbb C$) and $[x,y,y]=x$ where $[a,b]=a^{-1}b^{-1}ab$, $[a,b,c]=[[a,b],c]$.Consider the one-relator group $\langle x,y \mid [x,y,y]=x\rangle$.

Do these conditions on a semigroup define a group?

For all $a,b,c\in S$, if $ab=ac$ then $b=c$.The example is the Baer-Levi semigroup: the semigroup of all one-to-one mappings $\alpha$ of a countable set $X$ into itself such that $X\setminus \alpha(X)$ is not finite.

Calculate coordinates of third point on a triangle

I have two points coordinates, A (x,y) and B (x,y) and a distance L, that distance is AC and BC, I don't have the distance for AB, and I have to find the coordinates for the point C


1) From $A(x_A, y_A)$ and $B(x_B, y_B)$, find $x_M$ and $y_M$, where M is the midpoint of AB.2) Find $m_{AB}$, the slope of AB.

Can you apply XOR to the following expression?

So I know that

$$\bar{x}y + x\bar{y} = x \oplus y$$

Can this be applied to something like

$$ \bar{a}\bar{b}cd + ab\bar{c}\bar{d} $$

to get
$$ab \oplus cd$$


(Assuming I'm interpreting your notation correctly: $\overline{x}$ is "not $x$", etc.)

No, because the negation of $ab$ isn't $\overline{a}\overline{b}$, it's $\overline{a}+\overline{b}$.

Prove this inequality $\sum_{cyc}\frac{a}{ab+1}\geq \frac{3}{2}$

For $a>0$,$b>0$ and $c>0$ such tat $a+b+c=3$ prove that: $$\frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}\geq \frac{3}{2}$$




By AM-GM
$$\frac{a}{ab+1}=a-\frac{a^2b}{ab+1}\ge a-\frac{a^2b}{2\sqrt{ab}}=a-\frac{a\sqrt{ab}}{2}$$

$$\Rightarrow L.S\ge a+b+c-\frac{a\sqrt{ab}+b\sqrt{bc}+c\sqrt{ca}}{2}$$

$$\ge 3-\frac{a^2+b^2+c^2+ab+bc+ca}{4}$$

Help me to continue


By Vasc's inequality $(x^2+y^2+z^2)^2\geq3(x^3y+y^3z+z^3x)$ and by AM-GM we obtain:
$$\sum_{cyc}\frac{a}{ab+1}=a+b+c+\sum_{cyc}\left(\frac{a}{ab+1}-a\right)=$$
$$=3-\sum_{cyc}\frac{a^2b}{ab+1}\geq3-\sum_{cyc}\frac{a^2b}{2\sqrt{ab}}=3-\frac{1}{2}\sum_{cyc}\sqrt{a^3b}\geq3-\frac{3}{2}=\frac{3}{2}.