# 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.

# Negative coefficient in an almost cyclotomic polynomial

We show that the coefficient of $X^c$ or of $X^{b+c-1}$ of $P(X)$ is negative.Note that $ac>b+c$ and so on, hence

P(X)\equiv\frac{(1-X^a)(1-X^b)(1-X^c)(1-X^d)}{(1-X)^2(1-X^{ab})}\pmod{X^{b+c}}.

# Existence of state on a C*-algebra satisfying $|\tau(ab)|=\|ab\|$

Is it the case there is a state $\tau$ on $A$ such that $|\tau(ab)|=\|ab\|$?I'm particularly interested in the case when $a,b$ are projections and $\|ab\|=1$.

# 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.

# Groups with triple system of self-normalizing subgroups

$A = N_G(A)$, $B = N_G(B)$, $C = N_G(C)$;
$AB = BC = CA = G$;
$A \cap B = B \cap C = C \cap A = 1$.One is $\Delta(X) = \{ (x,x): x \in X \}$.

# 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.

# Properties of a matrix product - if AB=C and XY=C and BA=D, does YX=D?

I'm trying to figure out whether a property for a matrix product exist.Let's say we have 2 sets of matrices: $A$, $B$ and $X$, $Y$ ($A \neq X$) and let's say that their product is the same $AB=XY=C$.

# What is $a: b: c$ if $a: b$ is $\frac ab$?

If $a:b$ means $\displaystyle\frac ab$, then what does $a:b:c$ mean?Suppose
$$a:b:c=x:y:z$$
Then ,
$$\frac{a}{b}=\frac{x}{y}$$
And
$$\frac{b}{c}=\frac{y}{z}$$

A ratio is not exactly a fraction, but somewhat related.

# 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 $\overline{a}+\overline{b}$.

# Simplify X = A(AB)' + A'B'C + ABC

Why is this way of simplifying the Boolean expression wrong?Can someone explain to me why they are not the same?

# Find polynomial : $P(x)=x^3+ax^2+bx+c$

Find all polynomials in $\mathbb{Q}[x]$ is of the form $P(x)=x^3+ax^2+bx+c$ which has $a, b, c$ as its roots.By Vieta's formulas,

$$a+b+c=-a,\\ab+bc+ca=b,\\abc=-c. # Maximization with inequality constraints I have to solve this optimization problem$$\underset{x}{\max} \left(AB-\frac{xC}{D}(E+F)\right)$$subject to$$
\frac{AB}{x}-\frac{C}{D}(E+F) \leq G

# Find $AB=c, BC=a$ and $CA=b$ where $a,b,c \in \mathbb Z$

Find $AB=c, BC=a$ and $CA=b$ where $a,b,c \in \mathbb Z$ and $MA, MB, MC$.(Without multiplying by $k$)

My work so far:

Let $C(0;0); A(0;b); B(a;0)$ and $M(x;y)$.

# Proof of an equation

If $$\sqrt{a-x} + \sqrt{b-x} + \sqrt{c-x} = 0$$ then prove that (a+b+c+3x)(a+b+c-x) = 4(ab+bc+ca).So $a=x$, $b=x$ and $c=x$.

# If $ab+bc+ca=0$, find the value of $x^{a-1}.x^{b-1}.x^{c-1}$. [closed]

If $ab+bc+ca=0$, find the value of $x^{a-1}.$a\$.