text:
\section{Introduction}

Throughout this article, the rings are associative but not necessarily assumed to have unity unless indicated otherwise. Also an ideal means a two-sided ideal. Let $G$ be a group with identity $e$ and $R$ be a ring. Then $R$ is called $G$-graded if $R=\displaystyle\bigoplus_{g\in G} R_{g}$ with $R_{g}R_{h}\subseteq R_{gh}$ for all $g, h\in G$, where $R_{g}$ is an additive subgroup of $R$ for all $g\in G$. The elements of $R_{g}$ are called homogeneous of degree $g$. If $a\in R$, then $a$ can be written uniquely as $\displaystyle\sum_{g\in G}a_{g}$, where $a_{g}$ is the component of $a$ in $R_{g}$. The component $R_{e}$ is a subring of $R$ and if $R$ has a unity $1$, then $1\in R_{e}$. The set of all homogeneous elements of $R$ is $h(R)=\displaystyle\bigcup_{g\in G}R_{g}$. Let $P$ be an ideal of a graded ring $R$. Then $P$ is called a graded ideal if $P=\displaystyle\bigoplus_{g\in G}(P\cap R_{g})$, i.e., for $a\in P$, $a=\displaystyle\sum_{g\in G}a_{g}$ where $a_{g}\in P$ for all $g\in G$. It is not necessary that every ideal of a graded ring is a graded ideal (\cite{Alshehry Dawwas}, Example 1.1). For more details and terminology, see \cite{Hazart, Nastasescue}.

For commutative graded rings, graded $2$-absorbing ideals, which are a generalization of graded prime ideals, were introduced and investigated in \cite{Zoubi Dawwas Ceken}. Recall from \cite{Atani} that a proper graded ideal $P$ of a commutative graded ring $R$ is said to be a graded weakly prime ideal of $R$ if whenever $x, y\in h(R)$ and $0\neq xy\in P$, then either $x\in P$ or $y\in P$. Also from \cite{Zoubi Dawwas Ceken} a proper graded ideal $P$ of a commutative
graded ring $R$ is called a graded $2$-absorbing ideal of $R$ if whenever $x, y, z\in h(R)$ and $xyz\in P$, then either $xy\in P$ or $xz\in P$ or $yz\in P$. The concept of a graded weakly $2$-absorbing ideal of a commutative graded ring $R$ was introduced in \cite{Zoubi Dawwas Ceken}. A proper graded ideal $P$ of a commutative graded ring $R$ is called a graded weakly $2$-absorbing ideal of $R$ if whenever $x, y, z\in h(R)$ and $0\neq xyz\in P$, then either $xy\in P$ or $xz\in P$ or $yz\in P$.

Graded prime ideals over non-commutative graded rings have been introduced and examined by Abu-Dawwas, Bataineh and Al-Muanger in \cite{Dawwas Bataineh Muanger}. A proper graded ideal $P$ of $R$ is said to be graded prime if whenever $I$ and $J$ are graded ideals of $R$ such that $IJ\subseteq P$, then $I\subseteq P$ or $J\subseteq P$. As a generalization of graded prime ideals over non-commutative graded rings, the concept of graded $2$-absorbing ideals over non-commutative graded rings has been introduced and investigated by Abu-Dawwas, Shashan and Dagher in \cite{Dawwas Shashan Dagher}. A proper graded ideal $P$ of $R$ is said to be graded $2$-absorbing if whenever $x, y, z\in h(R)$ such that $xRyRz\subseteq P$, then $xy\in P$ or $yz\in P$ or $xz\in P$. Recently, graded weakly prime ideals over non-commutative graded rings have been introduced and studied by Alshehry and Abu-Dawwas in \cite{Alshehry Dawwas}. A proper graded ideal $P$ of $R$ is said to be graded weakly prime if whenever $I$ and $J$ are graded ideals of $R$ such that $0\neq IJ\subseteq P$, then $I\subseteq P$ or $J\subseteq P$. In this article, we are following \cite{Groenewald} to introduce and study the concept of graded weakly $2$-absorbing ideals as a generalization of graded weakly prime ideals in a non-commutative graded ring, and show that many of the results in commutative graded rings also hold in non-commutative graded rings.

\section{Graded Weakly $2$-Absorbing Ideals}

In this section, we introduce and study the concept of graded weakly $2$-absorbing ideals. In what follows a ring $R$ is a non-commutative ring with unity unless indicated otherwise.

\begin{defn} Let $P$ be a proper graded ideal of a graded ring $R$. Then $P$ is said to be a graded weakly $2$-absorbing ideal of $R$ if $0\neq xRyRz\subseteq P$ implies $xy\in P$ or $yz\in P$ or $xz\in P$ for all $x, y, z\in h(R)$. If $0\neq xyz\in P$ implies $xy\in P$ or $yz\in P$ or $xz\in P$ for all $x, y, z\in h(R)$, we say that $P$ is graded completely weakly $2$-absorbing.
\end{defn}

Apparently, if $R$ is a commutative graded ring with unity, then the two concepts of graded weakly $2$-absorbing and graded completely weakly $2$-absorbing
coincide. The next example shows that this will not be the case for non-commutative graded rings:

\begin{exa}\label{Remark 4.6} Consider $R=M_{2}(\mathbb{Z})$ (the ring of all $2\times 2$ matrices with integer entries) and $G=\mathbb{Z}_{4}$. Then $R$ is $G$-graded by $R_{0}=\left(\begin{array}{cc}
                             \mathbb{Z} & 0 \\
                             0 & \mathbb{Z}
                           \end{array}\right)$, $R_{2}=\left(\begin{array}{cc}
                             0 & \mathbb{Z} \\
                             \mathbb{Z} & 0
                           \end{array}\right)$ and $R_{1}=R_{3}=0$. Consider the graded ideal $P=M_{2}(2\mathbb{Z})$ of $R$. Clearly, $P$ is a graded prime ideal of $R$ and hence also a graded weakly $2$-absorbing ideal of $R$. On the other hand, $P$ is not a graded completely weakly $2$-absorbing ideal of $R$ since $A=\left(\begin{array}{cc}
                             3 & 0 \\
                             0 & 2
                           \end{array}\right)$, $B=\left(\begin{array}{cc}
                             0 & 3 \\
                             5 & 0
                           \end{array}\right)$ and $C=\left(\begin{array}{cc}
                             7 & 0 \\
                             0 & 4
                           \end{array}\right)\in h(R)$ with $0\neq ABC=\left(\begin{array}{cc}
                             0 & 36 \\
                             70 & 0
                           \end{array}\right)\in P$, $AB=\left(\begin{array}{cc}
                             0 & 9 \\
                             10 & 0
                           \end{array}\right)\notin P$, $AC=\left(\begin{array}{cc}
                             21 & 0 \\
                             0 & 8
                           \end{array}\right)\notin P$ and $BC=\left(\begin{array}{cc}
                             0 & 12 \\
                             35 & 0
                           \end{array}\right)\notin P$.
\end{exa}

Undoubtedly, every graded $2$-absorbing ideal of a graded ring $R$ is a graded weakly $2$-absorbing ideal of $R$. If $R$ is any graded ring, then $P=\{0\}$ is a graded weakly $2$-absorbing ideal of $R$ by definition. On the other hand, $P=\{0\}$ is not necessary to be a graded $2$-absorbing ideal, check the following example:

\begin{exa} Consider $R=\mathbb{Z}_{8}[i]$ and $G=\mathbb{Z}_{2}$. Then $R$ is $G$-graded by $R_{0}=\mathbb{Z}_{8}$ and $R_{1}=i\mathbb{Z}_{8}$. Undeniably, $P=\{0\}$ is not a graded $2$-absorbing ideal of $R$ since $2\in h(R)$ with $2R2R2\subseteq P$ but $2.2\notin P$.
\end{exa}

\begin{lem}\label{1} Let $P$ be a graded ideal in a graded ring $R$. Assume that $P$ is a graded weakly prime ideal of $R$. If $I$ and $J$ are graded right (left) ideals of $R$ such that $0\neq IJ\subseteq P$, then $I\subseteq P$ or $J\subseteq P$.
\end{lem}

\begin{proof} Suppose that $I$ and $J$ are graded right (left) ideals of $R$ such that $0\neq IJ\subseteq P$. Let $(I)$ and $(J)$ be the graded ideals generated by $I$ and $J$ respectively. Then $0\neq(I)(J)\subseteq P$, whence $I\subseteq(I)\subseteq P$ or $J\subseteq(J)\subseteq P$.
\end{proof}

\begin{lem}\label{Lemma 1.2} Let $P$ be a graded ideal in a graded ring $R$. If $P$ is a graded weakly prime ideal of $R$ and $0\neq xRyRz\subseteq P$ for $x, y, z\in h(R)$, then $x\in P$ or $y\in P$ or $z\in P$.
\end{lem}

\begin{proof} Suppose that $x, y, z\in h(R)$ such that $0\neq xRyRz\subseteq P$. Then $0\neq (Rx)RyRz\subseteq P$ and it follows from Lemma \ref{1} that $x\in Rx\subseteq P$ or $0\neq RyRz\subseteq P$. By reiterating this, the result follows.
\end{proof}

\begin{prop}\label{Proposition 4.2} Let R be a graded ring and $P$ be a proper graded ideal of $R$. If $P$ is a graded weakly prime ideal of $R$,
then $P$ is a graded weakly $2$-absorbing ideal of $R$.
\end{prop}

\begin{proof} Let $x, y, z\in h(R)$ such that $0\neq xRyRz\subseteq P$. By Lemma \ref{Lemma 1.2}, we have $x\in P$ or $y\in P$ or $z\in P$. Accordingly, $xy\in P$ or $yz\in P$ or $xz\in P$, and the result holds.
\end{proof}

\begin{prop}\label{Corollary 4.4} If $P$ and $K$ are two distinct graded weakly prime ideals of a graded ring $R$, then $P\bigcap K$ is a graded weakly $2$-absorbing ideal of $R$.
\end{prop}

\begin{proof} If $P\bigcap K=\{0\}$, then it is clear that $P\bigcap K$ is a graded weakly $2$-absorbing ideal of $R$. Let $x_{1}, x_{2}, x_{3}\in h(R)$ such that $0\neq x_{1}Rx_{2}Rx_{3}\subseteq P\bigcap K$. Then $0\neq x_{1}Rx_{2}Rx_{3}\subseteq P$ and $0\neq x_{1}Rx_{2}Rx_{3}\subseteq K$. It now follows from Lemma \ref{Lemma 1.2} that $x_{i}\in P$ and $x_{j}\in K$ for some $i, j$, and then $x_{i}x_{j}\in P\bigcap K$. Thereupon, $P\bigcap K$ is a graded weakly $2$-absorbing ideal of $R$.
\end{proof}

If $R$ and $T$ are two $G$-graded rings, then $R\times T$ is $G$-graded by $(R\times T)_{g}=R_{g}\times T_{g}$ for all $g\in G$. By (\cite{Bataineh Dawwas}, Lemma 4), $P\times K$ is a graded ideal of $R\times T$ if and only if $P$ is a graded ideal of $R$ and $K$ is a graded ideal of $T$. The next example shows that there are graded weakly $2$-absorbing ideals which are not graded weakly prime:

\begin{exa}\label{Remark 4.5} Consider $R=\mathbb{Z}_{2}[i]$, $T=\mathbb{Z}_{4}[i]$ and $G=\mathbb{Z}_{2}$. Then $R$ is $G$-graded by $R_{0}=\mathbb{Z}_{2}$ and $R_{1}=i\mathbb{Z}_{2}$. Also, $T$ is $G$-graded by $T_{0}=\mathbb{Z}_{4}$ and $T_{1}=i\mathbb{Z}_{4}$. So, $R\times T$ is $G$-graded by $(R\times T)_{j}=R_{j}\times T_{j}$ for all $j=0, 1$. Now, $\{0\}$ is a graded ideal of $R$ and $2T$ is a graded ideal of $T$ as $2\in h(T)$, so $P=\{0\}\times 2T$ is a graded ideal of $R\times T$ by (\cite{Bataineh Dawwas}, Lemma 4). Indeed, $P$ is not a graded weakly prime ideal of $R\times T$ since $x=(0, 1), y=(1, 2)\in h(R\times T)$ with $(0, 0)\neq xy=(0, 2)\in P$, $x\notin P$ and $y\notin P$. On the other hand, $P$ is graded $2$-absorbing and hence a graded weakly $2$-absorbing ideal of $R\times T$.
\end{exa}

\begin{thm}\label{Theorem 4.7} Let $P$ be a proper graded ideal of a graded ring $R$. Suppose that for graded left ideals $A, B$ and $C$ of $R$ such that $0\neq ABC\subseteq P$, we have $AC\subseteq P$ or $BC\subseteq P$ or $AB\subseteq P$. Then $P$ is a graded weakly $2$-absorbing ideal of $R$.
\end{thm}

\begin{proof} Let $x, y, z\in h(R)$ such that $0\neq xRyRz\subseteq P$. Then $RxRyRzR\subseteq P$, and as a consequence, since $R$ has a unity, $0\neq xRyRz=1.xR.1.yR.1.z.1\subseteq(RxR)(RyR)(RzR)\subseteq P$. By assumption, we have $xy\in(RxR)(RyR)\subseteq P$ or $yz\in(RyR)(RzR)\subseteq P$ or $xz\in(RxR)(RzR)\subseteq P$. Thus, $P$ is a graded weakly $2$-absorbing ideal of $R$.
\end{proof}

If $R$ is a $G$-graded ring and $K$ is a graded ideal of $R$, then $R/K$ is $G$-graded by $(R/K)_{g}=(R_{g}+K)/K$ for all $g\in G$. By (\cite{Saber}, Lemma 3.2), if $K$ is a graded ideal of $R$ and $P$ is an ideal of $R$ such that $K\subseteq P$, then $P$ is a graded ideal of $R$ if and only if $P/K$ is a graded ideal of $R/K$.

\begin{prop}\label{Proposition 4.8} Let $R$ be a graded ring and $P$ be a graded weakly $2$-absorbing ideal of $R$. If $K$ is a graded ideal of $R$ with $K\subseteq P$, then $P/K$ is a graded weakly $2$-absorbing ideal of $R/K$.
\end{prop}

\begin{proof} Let $x+K, y+K, z+K\in h(R/K)$ such that $0+K\neq(x+K)(R/K)(y+K)(R/K)(z+K)\subseteq P/K$. Then $x, y, z\in h(R)$ such that $0\neq xRyRz\subseteq P$. As $P$ is a graded weakly $2$-absorbing ideal of $R$, we have $xy\in P$ or $yz\in P$ or $xz\in P$, and hence $(x+K)(y+K)\in P/K$ or $(y+K)(z+K)\in P/K$ or $(x+K)(z+K)\in P/K$. So, $P/K$ is a graded weakly $2$-absorbing ideal of $R/K$.
\end{proof}

\begin{prop}\label{Proposition 4.9} Let $K\subseteq P$ be proper graded ideals of a graded ring $R$. If $K$ is a graded weakly $2$-absorbing ideal of $R$ and $P/K$ is a graded weakly $2$-absorbing ideal of $R/K$, then $P$ is a graded weakly $2$-absorbing ideal of $R$.
\end{prop}

\begin{proof} Let $x, y, z\in h(R)$ such that $0\neq xRyRz\subseteq P$. Then $x+K, y+K, z+K\in h(R/K)$ such that $(x+K)(R/K)(y+K)(R/K)(z+K)\subseteq P/K$. If $0\neq xRyRz\subseteq K$, then $xy\in K\subseteq P$ or $yz\in K\subseteq P$ or $xz\in K\subseteq P$ since $K$ is a graded weakly $2$-absorbing ideal of $R$. If $xRyRz\nsubseteq K$, then $0+K\neq(x+K)(R/K)(y+K)(R/K)(z+K)\subseteq P/K$. Since $P/K$ is a graded weakly $2$-absorbing ideal of $R/K$, $(x+K)(y+K)\in P/K$ or $(y+K)(z+K)\in P/K$ or $(x+K)(z+K)\in P/K$, which implies that $xy\in P$ or $yz\in P$ or $xz\in P$. Therefore, $P$ is a graded weakly $2$-absorbing ideal of $R$.
\end{proof}

Let $R$ and $T$ be two $G$-graded rings. In \cite{Nastasescue}, a ring homomorphism $f:R\rightarrow T$ is said to be a graded homomorphism if $f(R_{g})\subseteq T_{g}$ for all $g\in G$.

\begin{prop} Let $R$ and $T$ be two $G$-rings and $f:R\rightarrow T$ be a graded homomorphism. Then $Ker(f)$ is a graded ideal of $R$.
\end{prop}

\begin{proof} Clearly, $Ker(f)$ is an ideal of $R$. Let $x\in Ker(f)$. Then $x\in R$ such that $f(x)=0$. Now, $x=\displaystyle\sum_{g\in G}x_{g}$, where $x_{g}\in R_{g}$ for all $g\in G$, which implies that $f(x_{g})\in f(R_{g})\subseteq T_{g}$ for all $g\in G$. So, for $g\in G$, $f(x_{g})\in h(T)$ with $0=f(x)=f\left(\displaystyle\sum_{g\in G}x_{g}\right)=\displaystyle\sum_{g\in G}f(x_{g})$, which yields that $f(x_{g})=0$ for all $g\in G$ as $\{0\}$ is a graded ideal. Therefore, $x_{g}\in Ker(f)$ for all $g\in G$, and hence $Ker(f)$ is a graded ideal of $R$.
\end{proof}

\begin{thm}\label{Theorem 4.10} Let $R$ and $T$ be two $G$-rings and $f:R\rightarrow T$ be a surjective graded homomorphism.
\begin{enumerate}
\item If $P$ is a graded weakly $2$-absorbing ideal of $R$ and $Ker(f)\subseteq P$, then $f(P)$ is a graded weakly $2$-absorbing ideal of $T$.

\item If $I$ is a graded weakly $2$-absorbing ideal of $T$ and $Ker(f)$ is a graded weakly $2$-absorbing ideal of $R$, then $f^{-1}(I)$ is a graded weakly $2$-absorbing ideal of $R$.
\end{enumerate}
\end{thm}

\begin{proof}
\begin{enumerate}
\item By (\cite{Refai Dawwas}, Lemma 3.11 (2)), $f(P)$ is a graded ideal of $T$. Since $P$ is a graded weakly $2$-absorbing ideal of $R$ and $Ker(f)\subseteq P$, we conclude that $P/Ker(f)$ is a graded weakly $2$-absorbing ideal of $R/Ker(f)$ by Proposition \ref{Proposition 4.8}. Since $R/Ker(f)$ is isomorphic to $T$, the result holds.

\item By (\cite{Refai Dawwas}, Lemma 3.11 (1)), $f^{-1}(I)$ is a graded ideal of $R$. Let $K=f^{-1}(I)$. Then $Ker(f)\subseteq K$. Since $R/Ker(f)$ is isomorphic to $T$,we conclude that $K/Ker(f)$ is a graded weakly $2$-absorbing ideal of $R/Ker(f)$. Since $Ker(f)$ is a graded weakly $2$-absorbing ideal of $R$ and $K/Ker(f)$ is a graded weakly $2$-absorbing ideal of $R/Ker(f)$, we conclude that $K=f^{-1}(I)$ is a graded weakly $2$-absorbing ideal of $R$ by Proposition \ref{Proposition 4.9}.
\end{enumerate}
\end{proof}

Motivated by Theorem \ref{Theorem 4.7}, we have the following question:

\begin{question} Suppose that $P$ is a graded weakly $2$-absorbing ideal of $R$ that is not a graded $2$-absorbing ideal and $0\neq ABK\subseteq P$ for some graded ideals $A, B$ and $K$ of $R$. Does it imply that $AB\subseteq P$ or $AK\subseteq P$ or $BK\subseteq P$?
\end{question}

We will give a partial answer through the coming discussions. Motivated by (\cite{Zoubi Dawwas Ceken}, Definition 3.3), we introduce the following:

\begin{defn} Let $R$ be a $G$-graded ring, $g\in G$ and $P$ be a graded ideal of $R$ with $P_{g}\neq R_{g}$.
\begin{enumerate}
\item $P$ is said to be a $g$-$2$-absorbing ideal of $R$ if whenever $x, y, z\in R_{g}$ such that $xR_{e}yR_{e}z\subseteq P$, then $xy\in P$ or $yz\in P$ or $xz\in P$.

\item $P$ is said to be a $g$-weakly $2$-absorbing ideal of $R$ if whenever $x, y, z\in R_{g}$ such that $0\neq xR_{e}yR_{e}z\subseteq P$, then $xy\in P$ or $yz\in P$ or $xz\in P$.

\item Let $P$ be a $g$-weakly $2$-absorbing ideal of $R$ and $x, y, z\in R_{g}$. We say that $(x, y, z)$ is a $g$-triple-zero of $P$ if $xR_{e}yR_{e}z=0$, $xy\notin P$, $yz\notin P$ and $xz\notin P$.
\end{enumerate}
\end{defn}

Note that if $P$ is a $g$-weakly 2-absorbing ideal of $R$ that is not a $g$-$2$-absorbing ideal, then $P$ has a $g$-triple-zero $(x, y, z)$ for some $x, y, z\in R_{g}$.

\begin{prop}\label{Proposition 4.12} Let $P$ be a $g$-weakly $2$-absorbing ideal of $R$ and suppose that $xR_{e}yK_{g}\subseteq P$ for some $x, y\in R_{g}$ and some graded left ideal $K$ of $R$. Assume that $(x, y, z)$ is not a $g$-triple-zero of $P$ for every $z\in K_{g}$. If $xy\notin P$, then $xK_{g}\subseteq P$ or $yK_{g}\subseteq P$.
\end{prop}

\begin{proof} Suppose that $xK_{g}\nsubseteq P$ and $yK_{g}\nsubseteq P$. Then there exist $r, s\in K_{g}$ such that $xr\notin P$ and $ys\notin P$. Since $xR_{e}yR_{e}r\subseteq xR_{e}yK_{g}\subseteq P$ and since $(x, y, r)$ is not a $g$-triple-zero of $P$ and $xy\notin P$, $xr\notin P$, we obtain that $yr\in P$. Also, since $xR_{e}yR_{e}s\subseteq xR_{e}yK_{g}\subseteq P$ and since $(x, y, s)$ is not a $g$-triple-zero of $P$ and $xy\notin P$, $ys\notin P$, we obtain that $xs\in P$. Now, since $xR_{e}yR_{e}(r+s)\subseteq xR_{e}yK_{g}\subseteq P$ and since $(x, y, r+s)$ is not a $g$-triple-zero of $P$ and $xy\notin P$, we get that $x(r+s)\in P$ or $y(r+s)\in P$. If $x(r+s)\in P$, then since $xs\in P$, $xr\in P$, a contradiction. If $y(r+s)\in P$, then since $yr\in P$, $ys\in P$, a contradiction. Hence, $xK_{g}\subseteq P$ or $yK_{g}\subseteq P$.
\end{proof}

\begin{defn} Let $R$ be a $G$-graded ring, $g\in G$ and $P$ be a $g$-weakly $2$-absorbing ideal of $R$ and suppose that $A_{g}B_{g}K_{g}\subseteq P$ for some graded ideals $A, B$ and $K$ of $R$. We say that $P$ is free $g$-triple-zero with respect to $ABK$ if $(x, y, z)$ is not a $g$-triple-zero of $P$ for every $x\in A_{g}, y\in B_{g}$ and $z\in K_{g}$.
\end{defn}

The next proposition is clear.

\begin{prop}\label{Remark 4.14} Let $P$ be a $g$-weakly $2$-absorbing ideal of $R$ and suppose that $A_{g}B_{g}K_{g}\subseteq P$ for some graded ideals $A, B$ and $K$ of $R$ such that $P$ is free $g$-triple-zero with respect to $ABK$. If $x\in A_{g}, y\in B_{g}$ and $z\in K_{g}$, then $xy\in P$ or $xz\in P$ or $yz\in P$.
\end{prop}

\begin{thm}\label{Theorem 4.15} Let $P$ be a $g$-weakly $2$-absorbing ideal of $R$ and suppose that $0\neq A_{g}B_{g}K_{g}\subseteq P$ for some graded ideals $A, B$ and $K$ of $R$ such that $P$ is free $g$-triple-zero with respect to $ABK$. Then $A_{g}K_{g}\subseteq P$ or $B_{g}K_{g}\subseteq P$ or $A_{g}B_{g}\subseteq P$.
\end{thm}

\begin{proof} Suppose that $A_{g}K_{g}\nsubseteq P$, $B_{g}K_{g}\nsubseteq P$ and $A_{g}B_{g}\nsubseteq P$. Then there exist $x\in A_{g}$ and $y\in B_{g}$ such that $xK_{g}\nsubseteq P$ and $yK_{g}\nsubseteq P$. Now, $xR_{e}yK_{g}\subseteq A_{g}B_{g}K_{g}\subseteq P$. Since $xK_{g}\nsubseteq P$ and $yK_{g}\nsubseteq P$, it follows from Proposition \ref{Proposition 4.12} that $xy\in P$. Since $A_{g}B_{g}\nsubseteq P$, there exist $a\in A_{g}$ and $b\in B_{g}$ such that $ab\notin P$. Since $aR_{e}bK_{g}\subseteq A_{g}B_{g}K_{g}\subseteq P$ and $ab\notin P$, it follows from Proposition \ref{Proposition 4.12} that $aK_{g}\subseteq P$ or $bK_{g}\subseteq P$.

\underline{Case (1):} $aK_{g}\subseteq P$ and $bK_{g}\nsubseteq P$. Since $xR_{e}bK_{g}\subseteq A_{g}B_{g}K_{g}\subseteq P$ and $xK_{g}\nsubseteq P$ and $bK_{g}\nsubseteq P$, it follows from Proposition \ref{Proposition 4.12} that $xb\in P$. Since $aK_{g}\subseteq P$ and $xK_{g}\nsubseteq P$, we obtain that $(x+a)K_{g}\nsubseteq P$. On the other hand, since $(x+a)R_{e}bK_{g}\subseteq P$ and neither $(x+a)K_{g}\subseteq P$ nor $bK_{g}\subseteq P$, we have that $(x+a)b\in P$ by Proposition \ref{Proposition 4.12}, and hence $ab\in P$, a contradiction.

\underline{Case (2):} $bK_{g}\subseteq P$ and $aK_{g}\nsubseteq P$. Using an analogous altercation to case (1), we will have a contradiction.

\underline{Case (3):} $aK_{g}\subseteq P$ and $bK_{g}\subseteq P$. Since $bK_{g}\subseteq P$ and $yK_{g}\nsubseteq P$, $(y+b)K_{g}\nsubseteq P$. But $xR_{e}(y+b)K_{g}\subseteq P$ and neither $xK_{g}\subseteq P$ nor $(y+b)K_{g}\subseteq P$, and hence $x(y+b)\subseteq P$ by Proposition \ref{Proposition 4.12}. Since $xy\in P$ and $(xy+xb)\in P$, we have that $xb\in P$. Since $(x+a)R_{e}yK_{g}\subseteq P$ and neither $yK_{g}\subseteq P$ nor $(x+a)K_{g}\subseteq P$, we conclude that $(x+a)y\in P$ by Proposition \ref{Proposition 4.12}, and hence $ax\in P$. Since $(x+a)R_{e}(y+b)K_{g}\subseteq P$ and neither $(x+a)K_{g}\subseteq P$ nor $(y+b)K_{g}\subseteq P$, we have $(x+a)(y+b)\in P$ by Proposition \ref{Proposition 4.12}. But $xy, xb, ay\in P$, so $ab\in P$, a contradiction. Consequently, $A_{g}K_{g}\subseteq P$ or $B_{g}K_{g}\subseteq P$ or $A_{g}B_{g}\subseteq P$.
\end{proof}

\begin{lem}\label{Lemma 4.16 (1)} Let $P$ be a $g$-weakly $2$-absorbing ideal of $R$ and suppose that $(x, y, z)$ is a $g$-triple-zero of $P$ for some $x, y, z\in R_{g}$. Then $xR_{e}yP_{g}=\{0\}$.
\end{lem}

\begin{proof} Suppose that $xR_{e}yP_{g}\neq\{0\}$. Then there exist $r\in R_{e}$ and $p\in P_{g}$ such that $0\neq xryp$. Now, $xry(p+z)=xryp+xryz=xryp\neq0$. Hence, $0\neq xR_{e}yR_{e}(p+z)\subseteq P$. Since $P$ is $g$-weakly $2$-absorbing, we have that $x(p+z)\in P$ or $y(p+z)\in P$. Thus $xz\in P$ or $yz\in P$, a contradiction.
\end{proof}

\begin{lem}\label{Lemma 4.16 (2)} Let $P$ be a $g$-weakly $2$-absorbing ideal of $R$ and suppose that $(x, y, z)$ is a $g$-triple-zero of $P$ for some $x, y, z\in R_{g}$. Then $P_{g}yR_{e}z=\{0\}$.
\end{lem}

\begin{proof} Suppose that $P_{g}yR_{e}z\neq\{0\}$. Then there exist $r\in R_{e}$ and $p\in P_{g}$ such that $0\neq pyrz$. Now, $(x+p)yrz=xyrz+pyrz=pyrz\neq0$. Hence, $0\neq(x+p)R_{e}yR_{e}z\subseteq P$. Since $P$ is $g$-weakly $2$-absorbing, we have that $(x+p)y\in P$ or $(x+p)z\in P$. Hence, $xy\in P$ or $xz\in P$, a contradiction.
\end{proof}

\begin{lem}\label{Lemma 4.16 (3)} Let $P$ be a $g$-weakly $2$-absorbing ideal of $R$ and suppose that $(x, y, z)$ is a $g$-triple-zero of $P$ for some $x, y, z\in R_{g}$. Then $xP_{g}z=\{0\}$.
\end{lem}

\begin{proof} Suppose that $xP_{g}z\neq\{0\}$. Then there exists $p\in P_{g}$ such that $0\neq xpz$. Now, $x(y+p)z=xyz+xpz=xpz\neq0$. Hence, $0\neq xR_{e}(y+p)R_{e}z\subseteq P$. Since $P$ is $g$-weakly $2$-absorbing, we have that $x(y+p)\in P$ or $(y+p)z\in P$. Hence, $xy\in P$ or $yz\in P$, a contradiction.
\end{proof}

\begin{lem}\label{Lemma 4.16 (4)} Let $P$ be a $g$-weakly $2$-absorbing ideal of $R$ and suppose that $(x, y, z)$ is a $g$-triple-zero of $P$ for some $x, y, z\in R_{g}$. Then $P_{g}^{2}z=\{0\}$.
\end{lem}

\begin{proof} Suppose that $P_{g}^{2}z\neq\{0\}$. Then there exist $p, q\in P_{g}$ such that $0\neq pqz$ .Now, $(x+p)(y+q)z=xyz+xqz+pyz+pqz=pqz\neq0$ by Lemma \ref{Lemma 4.16 (2)} and Lemma \ref{Lemma 4.16 (3)}. Hence, $0\neq(x+p)R_{e}(y+q)R_{e}z\subseteq P$. Since $P$ is $g$-weakly $2$-absorbing, we have that $(x+p)z\in P$ or $(y+q)z\in P$ or $(x+p)(y+q)\in P$. Hence, $xz\in P$ or $yz\in P$ or $xy\in P$, a contradiction.
\end{proof}

\begin{lem}\label{Lemma 4.16 (5)} Let $P$ be a $g$-weakly $2$-absorbing ideal of $R$ and suppose that $(x, y, z)$ is a $g$-triple-zero of $P$ for some $x, y, z\in R_{g}$. Then $xP_{g}^{2}=\{0\}$.
\end{lem}

\begin{proof} Suppose that  $xP_{g}^{2}\neq\{0\}$. Then there exist $p, q\in P_{g}$ such that $0\neq xpq$. Now, $x(y+p)(z+q)=xyz+xyq+xpz+xpq=xpq\neq0$ by Lemma \ref{Lemma 4.16 (1)} and Lemma \ref{Lemma 4.16 (3)}. Hence, $0\neq xR_{e}(y+p)R_{e}(z+q)\subseteq P$. Since $P$ is $g$-weakly $2$-absorbing, we have that $x(y+p)\in P$ or $x(z+q)\in P$ or $(y+p)(z+q)\in P$. Hence, $xy\in P$ or $xz\in P$ or $yz\in P$, a contradiction.
\end{proof}

\begin{lem}\label{Lemma 4.16 (6)} Let $P$ be a $g$-weakly $2$-absorbing ideal of $R$ and suppose that $(x, y, z)$ is a $g$-triple-zero of $P$ for some $x, y, z\in R_{g}$. Then $P_{g}yP_{g}=\{0\}$.
\end{lem}

\begin{proof} Suppose that $P_{g}yP_{g}\neq\{0\}$. Then there exist $p, q\in P_{g}$ such that $0\neq pyq$. Now, $(x+p)y(z+q)=xyz+xyq+pyz+pyq=pyq\neq0$ Lemma \ref{Lemma 4.16 (1)} and Lemma \ref{Lemma 4.16 (2)}. Hence, $0\neq(x+p)R_{e}yR_{e}(z+q)\subseteq P$. Since $P$ is $g$-weakly $2$-absorbing, we have that $(x+p)y\in P$ or $y(z+q)\in P$ or $(x+p)(z+q)\in P$. Hence, $xy\in P$ or $yz\in P$ or $xz\in P$, a contradiction.
\end{proof}

The next theorem is a consequence result from Lemma \ref{Lemma 4.16 (1)} - Lemma \ref{Lemma 4.16 (6)}.

\begin{thm}\label{Theorem 4.17} Let $R$ be a $G$-graded ring, $g\in G$ and $P$ be a ideal of $R$ such that $P_{g}^{3}\neq\{0\}$. Then $P$ is a $g$-weakly $2$-absorbing ideal of $R$ if and only if $P$ is a $g$-2-absorbing ideal of $R$.
\end{thm}

\begin{proof} Suppose that $P$ is a $g$-weakly $2$-absorbing ideal which is not a $g$-$2$-absorbing ideal of $R$. Then $P$ has a $g$-triple-zero, say $(x, y, z)$ for some $x, y, z\in R_{g}$. Since $P_{g}^{3}\neq\{0\}$, there exist $p, q, r\in P_{g}$ such that $pqr\neq0$, and then $(x+p)(y+q)(z+r)=pqr\neq0$. Hence, $0\neq(x+p)R_{e}(y+q)R_{e}(z+r)\subseteq P$. Since $P$ is $g$-weakly $2$-absorbing, we have either $(x+p)(y+q)\in P$ or $(x+p)(z+r)\in P$ or $(y+q)(z+r)\in P$, and thus either $xy\in P$ or $xz\in P$ or $yz\in P$, a contradiction. Hence, $P$ is a $g$-$2$-absorbing ideal of $R$. The converse is obvious.
\end{proof}

\begin{cor}\label{Corollary 4.18} If $P$ is a $g$-weakly $2$-absorbing ideal of $R$ that is not a $g$-$2$-absorbing ideal, then $P_{g}^{3}=\{0\}$.
\end{cor}

Let $R\ $be a $G$-graded ring. Then an $R$-module $M$ is said to be
$G$\textit{-graded} if $M=\bigoplus_{g\in G}M_{g}$ with $R_{g}M_{h}\subseteq
M_{gh}$ for all $g,h\in G$, where $M_{g}$ is an additive subgroup of $M$ for
all $g\in G$. The elements of $M_{g}$ are called homogeneous of degree $g$. It
is clear that $M_{g}$ is an $R_{e}$-submodule of $M$ for all $g\in G$. We
assume that $h(M)=\bigcup_{g\in G}M_{g}$. Let $N$ be an $R$-submodule of a
graded $R$-module $M$. Then $N$ is said to be graded $R$-submodule if
$N=\bigoplus_{g\in G}(N\cap M_{g})$, or equivalently, $x=\sum_{g\in G}x_{g}\in
N\ $implies that $x_{g}\in N$ for all $g\in G$. It is known that an
$R$-submodule of a graded $R$-module need not be graded. For more terminology
see \cite{Hazart, Nastasescue}.

Let $M$ be an $R$-bi-module. The idealization $R\ltimes M=\{(r,m):r\in R,m\in
M\}$ of $M$ is a ring with componentwise addition and
multiplication: $(x,m_{1})+(y,m_{2})=(x+y,m_{1}+m_{2})$ and $(x,m_{1}%
)(y,m_{2})=(xy,xm_{2}+m_{1}y)$ for each $x,y\in R$ and $m_{1},m_{2}\in M$. Let
$G$ be an Abelian group and $M$ be a $G$-graded $R$-module. Then $X=R\ltimes
M$ is $G$-graded by $X_{g}=R_{g}\bigoplus M_{g}$ for all $g\in G$ \cite{RaTeShKo}.

\begin{thm}\label{Theorem 5.1} Let $R$ be a graded ring with unity, $M$ be a graded $R$-bi-module and $P$ be a proper graded ideal of $R$. Then $P\ltimes M$ is a graded $2$-absorbing ideal of $R\ltimes M$ if and only if $P$ is a graded 2-absorbing ideal of $R$.
\end{thm}

\begin{proof} Assume that $P\ltimes M$ is a graded $2$-absorbing ideal of $R\ltimes M$ and $xRyRz\subseteq P$ for some $x, y, z\in h(R)$. Then $(x, 0), (y, 0), (z, 0)\in h(R\ltimes M)$ with $(x, 0)R\ltimes M(y, 0)R\ltimes M(z, 0)\subseteq P\ltimes M$, and then $(x, 0)(y, 0)=(xy, 0)\subseteq P\ltimes M$ or $(x, 0)(z, 0)=(xz, 0)\subseteq P\ltimes M$ or $(y, 0)(z, 0)=(yz, 0)\subseteq P\ltimes M$. Hence, $xy\in P$ or $xz\in P$ or $yz\in P$, as required. Conversely, let $(x,m)R\ltimes M(y, n)R\ltimes M(z, p)\subseteq P\ltimes M$ for some $(x,m), (y, n), (z, p)\in h(R\ltimes M)$. Hence, $x, y, z\in h(R)$ with $xRyRz\subseteq P$, and then we have $xy\in P$ or $xz\in P$ or $yz\in P$. If $xy\in P$, then $(x,m)(y, n)=(xy, xn+ym)\subseteq P\ltimes M$. Similarly, if $xz\in P$, then $(x, m)(z, p)\in P\ltimes M$, and if $yz\in P$, then $(y, n)(z, p)\in P\ltimes M$, and this completes the proof.
\end{proof}

\begin{thm}\label{Theorem 5.2 (1)} Let $R$ be a graded ring with unity, $M$ be a graded $R$-bi-module and $P$ be a proper graded ideal of $R$. If $P\ltimes M$ is a graded weakly $2$-absorbing ideal of $R\ltimes M$, then $P$ is a graded weakly 2-absorbing ideal of $R$.
\end{thm}

\begin{proof} Let $0\neq xRyRz\subseteq P$, where $x, y, z\in h(R)$. Then $(0, 0)\neq(x, 0)R\ltimes M(y, 0)R\ltimes M(z, 0)\subseteq P\ltimes M$, and then $(xy, 0)\in P\ltimes M$ or $(xz, 0)\in P\ltimes M$ or $(yz, 0)\in P\ltimes M$. Hence, $xy\in P$ or $xz\in P$ or $yz\in P$. So, $P$ is graded weakly $2$-absorbing.
\end{proof}

\begin{thm}\label{Theorem 5.2 (2)} Let $R$ be a $G$-graded ring with unity, $M$ be a graded $R$-bi-module, $g\in G$ and $P$ be a graded ideal of $R$ with $P_{g}\neq R_{g}$. Then $P\ltimes M$ is a $g$-weakly $2$-absorbing ideal of $R\ltimes M$ if and only if $P$ is a $g$-weakly 2-absorbing ideal of $R$ and for any $g$-triple zero $(x, y, z)$ of $P$ we have $xR_{e}yR_{e}M_{g}=M_{g}R_{e}yR_{e}z=xM_{g}z=0$.
\end{thm}

\begin{proof} Suppose that $P\ltimes M$ is a $g$-weakly $2$-absorbing ideal of $R\ltimes M$. Let $0\neq xR_{e}yR_{e}z\subseteq P$, where $x, y, z\in R_{g}$. Then $(0, 0)\neq(x, 0)R_{e}\ltimes M_{e}(y, 0)R_{e}\ltimes M_{e}(z, 0)\subseteq P\ltimes M$, and then $(xy, 0)\in P\ltimes M$ or $(xz, 0)\in P\ltimes M$ or $(yz, 0)\in P\ltimes M$. Hence, $xy\in P$ or $xz\in P$ or $yz\in P$. So, $P$ is $g$-weakly $2$-absorbing. Suppose that $(x, y, z)$ is a $g$-triple zero of $P$. Assume that $xR_{e}yR_{e}M_{g}\neq0$. Then there exist $r, s\in R_{e}$ and $m\in M_{g}$ such that $xrysm\neq0$, and then $(0, 0)\neq(xrysz, xrysm)=(x, 0)(r, 0)(y, 0)(s, 0)(z,m)\in (x, 0)R_{e}\ltimes M_{e}(y, 0)R_{e}\ltimes M_{e}(z,m)\subseteq xR_{e}yR_{e}z\ltimes M_{g}=0\ltimes M_{g}\subseteq P\ltimes M$. But $(x, 0)(y, 0)\notin P\ltimes M$ and $(x, 0)(z,m)\notin P\ltimes M$ and $(y, 0)(z,m)\notin P\ltimes M$, which contradicts the fact that $P\ltimes M$ is a $g$-weakly $2$-absorbing ideal. If $M_{g}R_{e}yR_{e}z\neq0$, then there exist $n\in M_{g}$ and $r, s\in R_{e}$ such that $nrysz\neq0$. As above, we have $(0, 0)\neq(xrysz, nrysz)=(x, n)(r, 0)(y, 0)(s, 0)(z, 0)\in(x, n)R_{e}\ltimes M_{e}(y, 0)R_{e}\ltimes M_{e}(z, 0)\subseteq xR_{e}yR_{e}z\ltimes M_{g}=0\ltimes M_{g}\subseteq P\ltimes M$. But $(x, n)(y, 0)\notin P\ltimes M$ and $(x, n)(z, 0)\notin P\ltimes M$ and $(y, 0)(z, 0)\notin P\ltimes M$ and again contradicting the fact that $P\ltimes M$ is a $g$-weakly $2$-absorbing ideal. If $xM_{g}z\neq0$, then there exists $t\in M_{g}$ such that $xtz\neq0$. Now, $(0, 0)\neq(xyz, xtz)=(x, 0)(1, 0)(y, t)(1, 0)(z, 0)\in(x, 0)R_{e}\ltimes M_{e}(y, t)R_{e}\ltimes M_{e}(z, 0)\subseteq xR_{e}yR_{e}z\ltimes M_{g}=0\ltimes M_{g}\subseteq P\ltimes M$. But $(x, 0)(y, t)\notin P\ltimes M$ and $(x, 0)(z, 0)\notin P\ltimes M$ and $(y, t)(z, 0)\notin P\ltimes M$ so contradicting the fact that $P\ltimes M$ is a $g$-weakly $2$-absorbing ideal. Conversely, suppose that $(0, 0)\neq(x, n)R_{e}\ltimes M_{e}(y,m)R_{e}\ltimes M_{e}(z, t)\subseteq P\ltimes M$ for $(x, n), (y,m), (z, t)\in R_{g}\ltimes  M_{g}$. Then $x, y, z\in R_{g}$ with $xR_{e}yR_{e}z\subseteq P$.

\underline{Case (1):} $xR_{e}yR_{e}z\neq0$. Since $P$ is $g$-weakly $2$-absorbing, $xy\in P$ or $xz\in P$ or $yz\in P$. Hence, $(x, n)(y,m)\in P\ltimes M$ or $(x, n)(z, t)\in P\ltimes M$ or $(y,m)(z, t)\in P\ltimes M$, as desired.

\underline{Case (2):} $xR_{e}yR_{e}z\neq0$. If $xy\notin P$ and $xz\notin P$ and $yz\notin P$, then $(x, y, z)$ is a $g$-triple zero of $P$ and by assumption $xR_{e}yR_{e}M_{g}=M_{g}R_{e}yR_{e}z=xM_{g}z=0$. Now, $(x, n)R_{e}\ltimes M_{e}(y, m)R_{e}\ltimes M_{e}(z, t)\subseteq\left(xR_{e}yR_{e}z, M_{g}R_{e}yR_{e}z+xM_{g}z+xR_{e}yR_{e}M_{g}\right)=(0, 0)$, a contradiction.
\end{proof}

\begin{defn} Let $R$ be a graded ring and $P$ be a proper graded ideal of $R$. Then P is said to be a graded strongly weakly $2$-absorbing ideal of $R$ if whenever $A, B$ and $C$ are graded ideals of $R$ such that $0\neq ABC\subseteq P$, then $AC\subseteq P$ or $BC\subseteq P$ or $AB\subseteq P$.
\end{defn}

\begin{prop}\label{Proposition 6.9} Let $P$ be a proper graded ideal of $R$. Then $P$ is a graded strongly weakly $2$-absorbing ideal of $R$ if and only if for any graded ideals $A, B$ and $C$ of $R$ such that $P\subseteq A$, $0\neq ABC\subseteq P$ implies that $AB\subseteq P$ or $AC\subseteq P$ or $BC\subseteq P$.
\end{prop}

\begin{proof} If $P$ is a graded strongly weakly $2$-absorbing ideal of $R$, then the result holds by definition. Conversely, let $K, B$ and $C$ be graded ideals of $R$ such that $0\neq KBC\subseteq P$. Then $A=K+P$ is a graded ideal of $R$ such that $0\neq ABC\subseteq P$, and then by assumption, $AB\subseteq P$ or $AC\subseteq P$ or $BC\subseteq P$. Hence $KB\subseteq P$ or $KC\subseteq P$ or $BC\subseteq P$. So, $P$ is a graded strongly weakly $2$-absorbing ideal of $R$.
\end{proof}

\begin{prop}\label{Proposition 6.10} Let $R$ be a graded ring. Then every graded ideal of $R$ is graded strongly weakly $2$-absorbing if and only
if for any graded ideals $I, J$ and $K$ of $R$, $IJ=IJK$ or $IK=IJK$ or $JK=IJK$ or $IJK=0$.
\end{prop}

\begin{proof} Suppose that every graded ideal of $R$ is graded strongly weakly $2$-absorbing. Let $I, J$ and $K$ be graded ideals of $R$. If $IJK\neq R$, then $IJK$ is graded strongly weakly $2$-absorbing. Suppose that $IJK\neq0$. Then $0\neq IJK\subseteq IJK$ and $IJ\subseteq IJK$ or $IK\subseteq IJK$ or $JK\subseteq IJK$ and hence $IJ=IJK$ or $IK=IJK$ or $JK=IJK$. If $IJK=R$, then $I=J=K=R$. Therefore, $R=R^{3}$. Conversely, let $P$ be a proper graded ideal of $R$, $0\neq IJK\subseteq P$ for some graded ideals $I, J$ and $K$ of $R$. Then $IJ=IJK\subseteq P$ or $IK=IJK\subseteq P$ or $JK=IJK\subseteq P$. Hence, $P$ is a
graded strongly weakly $2$-absorbing ideal of $R$.
\end{proof}

\begin{cor}\label{Corollary 6.11} Let $R$ be a graded ring such that every graded ideal of $R$ is graded strongly weakly $2$-absorbing. Then for any graded ideal $I$ of $R$, $I^{3}=I^{2}$ or $I^{3}=0$.
\end{cor}

\begin{question} As a proposal for future work, we think it will be worthy to study non-commutative graded rings in which every graded ideal is graded weakly $2$-absorbing or graded strongly weakly $2$-absorbing. What the kind of the results that will be achieved?
\end{question}

redpajama_set_name: RedPajamaArXiv


of North Haven, since 2018. After earning his bachelor's degree in Mechanical Engineering from Western New England University, Greg began working for SERVPRO in 2005. He found that working in the restoration industry gave him the opportunity to help others during difficult times. Impressed with the possibilities that franchise ownership presented, Greg purchased his first franchise, SERVPRO of Milford-Orange-Stratford, in 2008.
and Lead and Abatement. From the beginning, he has worked to establish a strong core team of employees who have also been professionally trained both through certification classes and on the job experience.
with the purchase of SERVPRO of Southbury/Torrington in 2013 and, most recently, SERVPRO of Newtown and Southern Litchfield County in 2018.
Ted White has worked for and owned SERVPRO franchises for over 30 years. Ted worked for SERVPRO when he was in high school and continued to work over breaks while he attended college and eventually owned his own franchise. Although Ted recently decided to sell this SERVPRO franchise to Greg, he continues to work for SERVPRO as a production manager.
Ted is IICRC certified in Water, Fire, OSHA, Mold and Advanced Structural Drying. He is also trained in Subrogation. In his spare time, he volunteers his time as the Chairman of the Board of Trustees for The Institute for American Indian Studies in Washington, CT.
Chuck is our Construction and Repair Manager here at SERVPRO of Newtown and Southern Litchfield County. With over 20 years in the mitigation, restoration and construction industry, and 25 in the construction industry, Chuck performs full service construction and general contracting needs to many of our clients.
Chuck is proactive with his clientele and works hard to ensure that our customers receive the very best workmanship and value on every project. He is meticulous in his work and has an eye for detail that our customers appreciate.
Keri Perna works at SERVPRO of Newtown and Southern Litchfield County as an Administrative Assistant. Keri is most likely the person that you will speak to when you call our office. She has worked with us since 2013.
Roseanne works as a business development associate at SERVPRO of Newtown and Southern Litchfield County. In this role, Roseann spends her time in the community and helps us ascertain how we can best serve it. Prior to coming to SERVPRO, Roseann worked in the medical and legal fields and for the past 15 years, owned an auto glass company. She is a familiar face to many in the Milford community due to her volunteer work with various local charities.
Matt is a production manager for SERVPRO of Newtown and Southern Litchfield County. When we do work in your home or business, Matt will be available to help guide you through the restoration process. He has been with our franchise for five years and comes from the construction and mortgage industries. Matt's commitment to our customers is evident in the way that he always goes above and beyond in his work.
Kathy is our office manager and has been with SERVPRO for over 5 years.
Davi DeOliveira is a Crew Chief who has 13 years of experience at SERVPRO of Newtown and Southern Litchfield County. Davi knows all of the ins and outs of our business and is invaluable to our team!
Danielle is the Director of Business Development here at SERVPRO of Newtown and Southern Litchfield County. In this role, she is the eyes and ears for our business - spending time out in the area, meeting people, learning how we can best serve the community and spreading the word about the services we offer. Danielle has been with us for 4 years, having come to our company with 20 years of marketing, account management and business development experience. She is a graduate of the University of Connecticut with a degree in Business Administration.
Amado "Junior" Delgado is a Production Manager who has been with SERVPRO for 4 years. He is SERVPRO ECTP and Water Certified and has a NYS Mold Abatement Worker Certification. His co-worker's and customer's appreciate the perpetual smile, easy going nature and positive attitude he brings wherever he goes.
Wantuir "Von" DeOliveira is a crew chief for SERVPRO of Newtown and Southern Litchfield County production team. He has been with us for 8 years and is a wealth of experience and information.
Theny works in our office as an administrative assistant. Theny was born and raised in the Philippines and completed her education there. Traveling and reading are her favorite pastimes. Theny is glad to have the opportunity to work at SERVPRO on a team that is of a caring environment. What she loves most about working at SERVPRO is being able to value work-life balance.
Alexis works in our office as an administrative assistant. She's been in this industry for 4 years and she is certified in Water Damage Restoration. She is a New Haven, CT native. Alexis' favorite hobbies include collecting, doing make-up and cheering on the Golden State Warriors. Her favorite food is almost everything from her culture, Laos and Thai. Last but not least, her favorite person in the entire world? Her 5 year old daughter Sophia!
Mario is one of our production managers. He has over 18 years of experience in construction and mitigation services. Mario holds certifications in Water Damage Restoration, Applied Structural Drying, AMRI-Mold and Carpet Cleaning. Mario's objective when he is out on the job is to provide the best customer service and outcome for his customers.
Kirkland Hart is one of our restoration technicians. What stands out about Kirk is his can-do attitude and willingness to help our customers as well as others on the crew.
Vinny is a Business Developer for SERVPRO of Milford-Orange-Stratford.
Marco is a Crew Chief.
Erick is a crew chief for our production team.
Marc is our warehouse manager.
Christian is a production technician.
Italo is one of our production technicians.
We are happy to welcome Aaron Hayden as a new addition to our team. He has joined our team as a restoration technician.
We recently welcomed a new restoration technician, Cody Borowski, to our team.
Erickson Bernado works on our crew as a restoration technician.
Evan is a restoration technician at SERVPRO of Newtown and Southern Litchfield County.
Farith is new to our team as a restoration technician.
Why Should You Have Your Ducts Cleaned?


Today's weather is turning out to be moderate or heavy rain shower. The visibility is going to be around 15 km i.e. 9 miles and an atmospheric pressure of 1010 mb . The daytime temperature is going to reach 32 °c and the temperature is going to dip to 22 °c at night. We expect around 8.6 mm of precipitation to fall and cloud covering 58% of the sky, the humidity will be around 83%.
Tomorrow weather is forecasted to be light rain shower. The visibility is going to be around 14 km i.e. 8 miles and an atmospheric pressure of 1011 mb. The daytime temperature is going to reach 33 °c and the temperature is going to dip to 23 °c at night. We expect around 18.1 mm of precipitation to fall and cloud covering 77% of the sky, the humidity will be around 86%.
On Thursday weather will be moderate or heavy rain shower with daytime temperature reaching 32 °c. Night time temperature are expected to be 23 °c.We expect around 20.2 mm of precipitation to fall. The visibility is going to be around 11 km i.e. 6 miles and an atmospheric pressure of 1012 mb. We expect around 20.2 mm of precipitation to fall and cloud covering 63% of the sky, the humidity will be around 91%.
Friday seems to be moderate or heavy rain shower. Yurupar, Colombia visibility is going to be around 12 km i.e. 7 miles and an atmospheric pressure of 1012 mb. The daytime temperature is going to reach 30 °c and the temperature is going to dip to 23 °c at night. We expect around 22.9 mm of precipitation to fall and cloud covering 78% of the sky, the humidity will be around 91%.
Light rain shower will be the weather pattern for the Saturday. The visibility is going to be around 8 km i.e. 5 miles and an atmospheric pressure of 1012 mb. The daytime temperature is going to reach 26 °c and the temperature is going to dip to 23 °c at night. We expect around 24.8 mm of precipitation to fall and cloud covering 73% of the sky, the humidity will be around 97%.
Looking at the weather in Yurupar, Colombia over the next 7 days, the maximum temperature will be 33℃ (or 91℉) on Wednesday 24th April at around 1 pm. In the same week the minimum temperature will be 21℃ (or 70℉) on Sunday 28th April at around 4 am.
The national weather service for Yurupar, Colombia is reporting Friday 26th April to be the wettest day in the coming week with around 11.30mm (or 0.4 inches) of rainfall. Make sure to carry an umbrella if you are out and about in Yurupar, Colombia.
The windiest of all days will be Tuesday 23rd April as wind will reach 3mph (or 5kmph) at around 10 am.


text:
Fan Engagement, Team Performance
Austria Returns: A Reminder of the Current State of Play in the Austrian Bundesliga
With top-flight football in Austria all set to resume next week, Stats Perform applies data to provide a quick refresher of what is shaping up to be a four-way fight for the Bundesliga title.
By: Philipp Ertl
Following intensive negotiations between the Austrian Bundesliga, the federation and the government, an agreement was struck to resume the league on 2nd June. Prior to that, title challengers Red Bull Salzburg will play second tier side SC Austria Lustenau in the UNIQA ÖFB Cup Final on 29th May, the first game of competitive football in the country since early March.
To help get you back up-to-speed with how the competition was shaping up prior to the enforced hiatus, Stats Perform is on hand to provide a data-driven reminder of the Bundesliga's key talking points.
Points Deductions and a Contrast of Styles in the Title Race
The stage is set for a titanic battle for the Bundesliga title, with Salzburg's domination of domestic football under serious threat from LASK. For the past two seasons, the league has adopted a format similar to that used in other European competitions where after each team has played each other twice, they are split into two groups resulting in the top six entering a championship round, with the remaining six participating in a Europa League play-off/relegation round.
LASK finished the regular season six points clear at the top, which under normal circumstances would have meant they would enter the championship phase with a three point lead after every team's points had been halved and rounded-up. However the team from Linz were hit with a six point deduction for the violation of rules introduced during the Covid-19 outbreak. The penalty comes after footage emerged of the club allegedly holding prohibited training sessions during the hiatus. The club have made it clear they intend to appeal the penalty, having previously accused unnamed third-parties of 'industrial espionage'.
The controversy adds another dimension to a closely-fought title race, especially when you take into account LASK's outstanding form before the league paused. They were the only team to win all four of their league games following the winter break and made Bundesliga history by winning their first 11 away games – a league record.
In contrast Salzburg, who entered 2019/20 having won a record six consecutive league titles, ended a four game winless run with a 2-0 victory over Sturm Graz in their final game before the break. It was their longest run without a win since the 2011/12 season under Ricardo Moniz.
Both team's successes have been built on different strengths. If the championship was awared based on the American Football adage "offense wins games, defence wins championships", then LASK would be in the driving seat. Whilst Salzburg have set a new league record of 74 goals scored in just 22 games, Linz have been resolute at the back, conceding just 20 goals so far.
When we compare both team's approaches using the Stats Perform Playing Styles Framework, we can see that both teams play at a high tempo, with LASK focusing particularly on high pressing. Salzburg also press high up the pitch, but when they are in possession they spend a higher proportion of time engaged in the build-up phase, which is defined as enjoying periods of long, controlled possession in the opposition's half, between the halfway line and the opposition box.
Salzburg's use of controlled possession is also reflected by the Sequence Framework. Jesse Marsch's side have completed 164 sequences comprising of 10 or more passes during the season, significantly more than their title rivals. Their expected goals output has also been substantially higher, demonstrating their ability to generate a high volume of high quality chances.
Rapid Not To Be Discounted
LASK's points deduction, coupled with the halving of regular season points, mean that Rapid now sit just four points off Salzburg at the top. The side from Vienna were another team who were in excellent form when the season stopped, and are unbeaten in their last nine Bundesliga games (5 wins, 4 draws). It was their best run of form since August-November 2017, when they went 10 games unbeaten.
Dietmar Kühbauer's men have scored seven of their goals directly from the opposition's loss of possession, the most of any team in the league. According to the Playing Styles Framework, they are engaged in a higher proportion of counter-attacking possessions than any other team in the league – and 32% above the league average. Two of their key counter attacking players are Stefan Schwab and Thomas Murg. Schwab contributes 19% of Rapid's total distance gained as a result of passing, whilst no other player generates as much territory through dribbling during counter attacks as Murg.
Rapid also possess a striker in top form. Since joining from St. Pölten in the summer, Taxiarchis Fountas has scored 14 goals his first 19 games, a record which is similar to that of Robert Beric during 2014/15, who is considered to be the last prolific striker to wear Rapid's colours. Fountas has exceeded his expected goals output by over five goals, a differential surpassed only by WAC's Shon Weissman.
Weissman outperforms Haaland and Messi
Fourth placed WAC are another team who could have a say in the title race, given that they possess two players, Shon Weissman and Michael Liendl, who are leading the league's goalscoring and assist rankings by some margin.
In his debut Bundesliga season, Weissman is five goals clear of his nearest rival, Patson Daka, in the goalscoring charts, with 22 goals from his first 21 games. He was also one goal clear of Erling Haaland before the Norwegian moved to Dortmund. But had the 19-year-old stayed in Salzburg, who would have come out on top in the golden boot race at the end of the season?
It would undoubtedly have been a close race between them, but one area where Weissman has particularly excelled has been in his Expected Goals on Target (xGOT) output.
We can use xGOT values as an indicator, over a given time period, for how well a player is shooting. For example, a player whose xGOT is exceeding their xG is executing better quality shots, given the quality of chances he has attempted shots from.
So far this season, Weissman's xGOT has exceeded his xG output by nearly five, demonstrating his ability to increase the probably of a shot resulting in a goal based on the end location of the shot, compared to the probability of the shot location. This differential is highest of any player participating in the Bundesliga and across the big-five European leagues – pipping Lionel Messi to top spot.
Liendl and Weissman Providing a Lethal Combination
Veteran Michael Liendl has assisted nine of Shon Weissman's 22 goals this season. In total he has contributed 15 assists during the campaign, ten of which have come from open play.
Ranking second in the league for assists is set piece specialist Peter Michorl, who has contributed ten during 19/20. His dead ball prowess is reflected by the fact that only three of his assists have come from open play, and two of those were short lay-offs for teammates to shoot from distance. Compared to Liendl, he has contributed very few passes with an expected assists value above 0.1 into dangerous locations inside the box.
Liendl's involvement in generating chances also comes through in the sequence framework. In total, Liendl is involved in 54 of every 100 shot-ending open play sequences created by WAC, a ratio which is bettered only by Hartberg's Rejko Rep. They are the only two players to be involved in more than half of their team's sequences resulting in a shot.
Little to Separate Out Winners and Losers in the Fight for Survival
At the other end of the league, six teams will now be fighting it out for a single place in the Bundesliga's end-of-season Europa League play-offs, as well as looking to secure their top flight status for 2020/21.
To highlight how tight the race could end up being, we have generated a final league table based on the points each team will have once the season resumes, and added points from the remaining fixtures based on if the results were to replicate the same outcomes from the match-ups in the regular season.
If that scenario were to happen, we would end up with three teams tied on 26 points at the top, with the other teams all tied on 20 points at the bottom. In this scenario Mattersburg would be relegated, despite having a superior goal difference, because their rivals had their regular season points rounded down (i.e they had half a point chalked off from their average regular season points).
Projected League Table: Qualification Round – Bundesliga 2019/20
1 FK Austria Wien* 32 8 15 9 50 49 1 26
2 FC Flyeralarm Admira* 32 8 12 12 39 54 -15 26
3 SCR Altach 32 11 5 16 53 58 -5 26
4 Swarovski Tirol* 32 8 6 18 41 73 -32 20
5 St. Pölten* 32 5 14 13 33 70 -37 20
6 SV Mattersburg 32 8 5 19 37 66 -29 20
* Regular season points were rounded down when they were halved prior to the season split (meaning they are ranked above teams whose points were not rounded down)
With potential match winners spread across all the teams involved in a four-horse title race, together with a tightly congested fight for survival, it would appear that all the elements are there for the 2019/20 season to deliver a classic finish.
Shape Analysis: Automatically Detecting Formations
Jonny Whitmore
Artificial Intelligence, Fan Engagement, Team Performance
Webinar: AI for Soccer
Andy Brown
'Can We Put Numbers on That?': An Interview with Baseball Analytics Expert Eno Sarris
Kevin Chroust
Women in Sports: Rushton Helps Build Atlanta United into a Winner
Amy West and Taylor Bechtold

redpajama_set_name: RedPajamaCommonCrawl


19. Compare the falls of Mayan, Aztec, and Incan civilizations. (A.) How were their falls similar? (B.) How were they different? (C.) Give an example of how an aspect of each civilization survives today.
A. Similar – They all were located in Latin America and they all lost their power quickly when compared to their rise to power.
B. Different - The Mayans were much earlier in history, around 250 A.D. and they were not conquered. They seem to have just disappeared as a civilization, but moved into the surrounding area. No one knows how or why they disbanded their empire. The Incans and Aztecs were conquered by the Spanish (Pizzaro/Incan, Cortes/Aztec).
C. Today's Civilization – Descendants of Mayan still live in Latin America, today. Aztec festivals still exist in Mexico, in their dance, music and religious ceremonies. The Incan language Quechua is still spoken in South America.
20. A. Explain what motivated explorers to travel to the New World? Give three (3) examples.
B. Explain what motivated settlers to travel to the New World? Give three (3) examples.
21. How did the decimation of Native Americans lead to the slave trade? Give three (3) examples.
- Once the number of Native Americans were gone, the Spanish needed workers for their mines and plantations, both to work the fields and as servants in the home.
22. List and explain at least three reasons why a relatively small band of Spanish soldiers was able to conquer much of Latin America and the people who lived there.
23. How might the cultures of Latin America be different if the Spanish and Portuguese had never arrived? What kind of culture do you think would dominate the region today?
24. Explain the similarities and differences between how the Spanish and French dealt with the indigenous peoples in North America.
- French were the most civil. They wanted to introduce religion, but not by force. They wanted to trade with the Native Americans, but not by force. The French believed in treaties, and then, they would honor the treaties that were signed with the Native Americans.
- Spanish were out for riches, gold, and land. They would use any force and method to attain their goals. They even used force to convert Native Americans to Christianity.
25. Discuss the origins and the positive/negative effects of the Colombian Exchange by using examples of goods, animals or diseases that were part of the exchange?
Latin America's disadvantages were the loss of lives to diseases and war. They lost their land. They lost their identity/culture/way of life.


We have six bowl games on Monday (New Year's Eve), and five more Tuesday to bring in the New Year. Let's see how sharps (professional wagerers) have been betting the final stage of this season's bowl bonanza.
PITTSBURGH VS. STANFORD: A lot of dog interest here. An opener of Stanford -6 has come down to -4, probably because the Pac 12 is having another miserable bowl campaign so far. Arizona State and California looked awful. Washington State was dominated in game stats but eked out a close win over Iowa State. There's some interest in the favorite on the four. That might set up a tug-of-war on game day if the ill-informed public reflexively takes all the favorites.
MICHIGAN STATE VS. OREGON: Big time matchup for a bowl with such a weird name. Oregon found out its star quarterback was staying for another year. That created initial buzz for the Ducks…which was then erased when the Pac 12 opened so poorly. An opener of Oregon -1.5 is back where it started after a temporary surge. Sharps seem to like Michigan State at +2 or more. Oregon would have soared to the key number of three if sharps liked them.
MISSOURI VS. OKLAHOMA STATE: Missouri opened at -9.5. That struck some old-timers as steep. But, Okie State's defense was awful this season…and the Big 12 as a whole isn't setting the world on fire this postseason. Missouri's from the SEC, that has been getting good results. As I put together this report, we're seeing a variety of lines in that -9 to -9.5 area. Sharps would definitely take this dog if they see a +10 available.
UTAH VS. NORTHWESTERN: Not much betting interest here. An opener of Utah -6.5 was nudged up to the key number of -7 early on. Sharps haven't been interested in Northwestern as much as they should have been this season. Utah rose despite the recent Pac 12 bias. It is a shorter trip for the Utes. Northwestern has been a feisty dog this season. Maybe that will create a tug-of-war before kickoff.
TEXAS A&M VS. NC STATE: A lot of love for the Aggies since lines went up. An opener of Texas A&M -4.5 has been bet up to -7, with some stores testing -7.5. Sharps obviously loved A&M at anything below a TD. We're seeing some buy back on the dog with the hook. We'll have to see what the public does. This is the last game of the day, which gives squares several hours to lay the points. That would probably set up a tug-of-war between A&M -7 and NC State +7.5. Though, that could happen a half-point higher depending on what happens in earlier games.
MISSISSIPPI STATE VS. IOWA: 2019 begins with the traditional SEC vs. Big 10 bowls. Here, the SEC entry has been getting a lot of support. Mississippi State opened at -6, and was driven up to -7.5 after Auburn and Florida crushed Purdue and Michigan. Iowa does historically get respect as a "defensive dog" under this head coach. We'll see if that causes a tug-of-war, or just freezes the number at 7.5. Sharps like the SEC favorite at -7 or less (especially below the seven).
PENN STATE VS. KENTUCKY : Penn State is up from -5.5 to -6.5. Tricky one to read because the public loves betting favorites on New Year's Day, and is much more likely to bet a traditional power like Penn State rather than a traditional doormat like Kentucky. That said, Kentucky is the SEC team facing a possibly overrated Big 10 team. Sharps would definitely take this dog at +7. Oddsmakers may not give them that opportunity unless shops are just flooded with Nittany Lions money. Could happen.
LSU VS. CENTRAL FLORIDA: An opener of LSU -8 has been bet down to -7. SEC respect shows up on the key number, which is why some shops have posted the hook. Probably a tug-of-war between LSU -7 and UCF +7.5. Though, if the SEC keeps getting good results, that money battle could happen a half-point higher. This is a great situational spot for UCF, who exploited it to upset Auburn last season. Here they'd have to do it with a talented but relatively inexperienced backup quarterback. Sharps definitely like that backup QB at +8…and some kept betting at +7.5.
OHIO STATE VS. WASHINGTON: Ohio State is up from -5 to -6.5, with some stores now testing the key number of seven. Again, we have the Pac 12 issue in play. Maybe the conference will get better results Monday than they had in prior days. For now, sharps like the "Urban Meyer" factor with the head coach retiring after the game for the favorite. Sharps have laid -6 or less with OSU. We'll see if dog lovers come in hard at +7, or soft at +7 but harder at +7.5.
GEORGIA VS. TEXAS: Strong sharp sentiment early for Georgia, as an opener of -10.5 went up to -12.5. A few spots have tested -13, but Texas money does come in at that price. Most sharps believe Georgia was a final four caliber team that was much better than Notre Dame and at least a field goal better than Oklahoma. The public had no qualms laying double digits with Clemson and Alabama this past Saturday. I think we'll see a more solid -13 by kickoff unless some rich Texans pull out their wallets.
You can purchase my daily BEST BETS right here at the website with your credit card. If you have any questions call the VSM office at 1-888-777-4155. Be sure to ask about combination packages that include college basketball. That's my favorite sport to handicap and bet. Conference play really gets rolling Wednesday. So, it's time for you to start thinking about the road to March Madness!
Thanks for reading. I won't be back until Friday to cover sharp betting for NFL's Wildcard Weekend. Let me be among the first to wish you a Happy New Year! Hope the games go your way. See you Friday.


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Young Londoners
Young Londoners & their families
Young Londoners & their peers
Housing & communities
VOICE OF LONDON
We support young Londoners, their families, peers and communities affected by violence and exploitation.
Young Londoners are full of ambition and aspiration. We want to ensure that continues into their adult lives and that they can thrive without the challenges of exploitation and violence.
This is not yet the reality for enough young Londoners. Many are facing challenges from multiple avenues; be it personal experience, or simply in the places where they live, learn and have fun.
We believe that all young Londoners can have a positive and productive future. Whatever their history, when a child or young person comes to Safer London, they are just that – a child or a young person. We work alongside them to help unlock their potential, so they can move forward positively.
We put young Londoners at the centre of their support. Our focus is on them and their needs. By building strong, trusting relationships we gain an insight into their world, begin to understand what they need and together find the best ways to keep them safe. We work not only with them and their families, but also their peer networks and directly within the community and the places where they live and spend their time.
We're committed to learning, innovating, and delivering high quality, evidence-based support. With a footprint in every borough in London, we build trusting, professional partnerships with statutory, charitable and community partners to embed our approaches and work towards achieving our Vision.
Our vision is a world where all young Londoners can thrive, free from exposure to violence and exploitation.
Our values define what we stand for and believe in. They are the principles, ideals and characteristics that define our culture, standards and aspirations.
We are change-makers: This isn't just a job for us. We get up every morning to create lasting change for young people and all of London.
We are inclusive: All that we do recognises and values the diverse experiences of young people; our services are for any young person who needs us.
We are trusted: People trust us to build lasting relationships, to be transparent, and to deliver results.
We are courageous: We are brave, unafraid to challenge, and committed to empowering young people to stay safe and happy.
We are collaborative: We stand with individuals and organisations who share our vision; our people and our partners are central to everything we do
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The tenth subject of Mimisbrunnr.info's Six Question series is Norwegian singer and musician Lindy-Fay Hella. Best known for her work as female vocalist for the popular musical project Wardruna, Hella has appeared throughout the group's discography and has performed, for example, on the Norwegian government-owned NRK1 television network and in front to the Gokstad ship at the Oslo Viking Ship Museum (Norwegian Vikingskipshuset på Bygdøy). Additionally, her voice can be heard throughout the extremely popular television show Vikings (2013-ongoing), which prominently employs tracks from the Wardruna discography.


Improve an individual's ability to process, understand, and respond to language.
Having difficulty nursing or bottle feeding? We can help.
Gain a better understanding of your child's or your own learning style.
Fluency-enhancing device designed for people who stutter.
Diagnostic and therapeutic services to Spanish speaking individuals and their families.
Develop skills that enable you to interact positively with your child.
Providing speech, language, learning, occupational and behavioral services.
Looking for Sound or Language Development therapy in San Antonio? Greater Learning LP provides speech, language, learning, occupational and behavioral services for all ages. Contact us today to learn more!
Greater Learning LP is dedicated to helping people reach their full potential. Services are based on the premise that individuals have unique learning abilities, styles and needs. Clients are seen individually to maximize the benefits of intervention. Intervention programs are designed to meet an individual's unique and specific learning needs.
Individualized programs and time enhances the potential for success. We know that all learning does not occur best in a group or classroom situation, the individual is our goal.


The university of music (Hochschule für Musik) was founded by Carl Muellerhartung, a pupil of composer Franz Liszt. Liszt spent several years in Weimar, encouraging the founding of a school for the education of musicians in orchestral instruments.
The main building Bauhaus University was designed by Henry van de Velde to house the studios of the sculptors at the Grand Ducal Saxon Art School. The edifice is part of the UNESCO World Heritage Site 'Bauhaus and its sites in Weimar and Dessau'.
In Germany many universities had one or more dentention rooms for punishing students. Their German name was Karzer. The Karzer of the university of Jena is located behind this cross-barred window depicted above.


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2012 NBA Playoffs, Celtics Vs. Heat: LeBron's First Half Was Close To Perfect
By Ryan O'Hanlon Jun 8, 2012, 8:36am EDT
Share All sharing options for: 2012 NBA Playoffs, Celtics Vs. Heat: LeBron's First Half Was Close To Perfect
Last night in Boston, the Miami Heat forced a game seven with a 98-79 victory over the Celtics. LeBron James was the catalyst, producing one of the greatest playoff performances in NBA history. James had 45 poins on 19-for-26 shooting with 15 rebounds and five assists.
He dropped 30 points in the first half, tying an NBA playoff record. As Diego Quezada of Peninsula Is Mightier wrote:
Coming into this game with a tremendous amount of pressure, LeBron James responded with a magnificent game to lead the Heat in a 98-89 win. He a nearly impeccable first half, scoring 30 points and making 12 of his first 13 shots. He made a series of contested jumpers, worked well out of the post and even had a nasty tip-dunk in the second quarter. On the other side of the court, James guarded Paul Pierce, Rajon Rondo and Kevin Garnett at different times. To put it simply, he had a half for the ages. James missed a few shots in the third, but still made a number of big-time baskets, always responding when the Celtics threatened with a mini-run.
Over at Celtics Blog, Evans Clinchy says that LeBron's epic performance covered up some not-so-epic turns by his teammates:
The Heat's three-time MVP was so dominant that his lackluster teammates went unnoticed. It was an afterthought that Dwyane Wade missed 10 of his first 12 shots, that Chris Bosh still wasn't his normal self, that Mario Chalmers' play was scatterbrained on both ends, that Mike Miller was a complete no-show. Wade in particular got a lucky bailout -- it was his fourth consecutive game missing the majority of his shots, and several of them were embarrassing ones. Layups. Open jumpers with LeBron double-teamed. Looks he should have made -- but no one noticed or cared, because LeBron covered up every Miami miscue.
Stick with this StoryStream for news and updates throughout the Eastern Conference Finals. For more analysis and discussion on the Heat, visit Peninsula Is Mightier. Visit CelticsBlog and SB Nation Boston for the opponent's perspective during the series. For all your Conference Finals analysis, visit SB Nation's NBA hub.
NBA Playoffs 2012, Heat Vs. Celtics: Miami Advances To NBA Finals For Second Straight Year
NBA Playoffs 2012, Heat Vs. Celtics: Game 7 Reaction
LeBron James Is Not A Championship Player


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Conference Insider: Big Sur Children's Writing Workshop
Need a boot camp for kidlit authors? Look no further than this bicoastal conference.
Years ago, author Jeff Stone attended the Big Sur Children's Writing Workshop with the beginning of what would become his The Five Ancestors series for middle-school readers. At the workshop, he met literary agent Laura Rennert as well as an editor at Random House.
Big Sur Writers Workshop
Then, he landed a six-figure book deal.
"He's our biggest success story," says Andrea Brown, executive director of the workshop and president of the Andrea Brown Literary Agency. "But we have an enormous number of success stories."
Participants are asked to bring polished drafts of their manuscripts to the three-day event, which is geared toward intermediate – rather than beginning – children's writers. Those interested in attending are asked to email three sample pages of their work for faculty review before registering.
The workshop, which takes place once a year in both Cape Cod and Northern California, caps at 100 participants who spend a long weekend in groups of no more than five writers at a time, guided in critique and revision by a professional author, agent, or editor. "There's a warm, congenial feeling all weekend because a lot of the same writers and faculty attend each year," Brown says. "We've created a writing community."
In between critique group sessions, participants listen to panels made up of literary agents and – separately – editors, who offer insights into writing, publishing, and marketing children's literature. "Editors talk about what they're looking for and how acquisitions work at their house," Brown explains. "One of our regular faculty, Eric J. Adams, does a wonderful workshop on structure, and sometimes we teach about query letters."
However, the focus of the weekend is critique and revision. Brown says this rigor and focus leads to little "miracles" by the time the weekend is over.
"Some people come in on Friday thinking they're picture book writers and leave Sunday thinking they're middle-grade writers," Brown says. "Some have been writing in first person, and they realize their story is much better in third person. They experiment during the weekend, and then go home and concentrate on what they have to do in order to make their work more publishable."
Featured Presenters
The Cape Cod workshop, held in May 2020, features New York Times bestselling author David Elliott, series author Debbi Michiko Florence, young adult novelist Sarah Darer Littman, and author/illustrator Anne Sibley O'Brien, among others.
Editor Nikki Garcia from Little, Brown Books for Young Readers and editor Jennifer Ung from Simon Pulse/Simon & Schuster will attend, as will Charlesbridge editorial director Yolanda Scott and Page Street Publishing YA editor Ashley Hearn.
Agents Jennifer Laughran, Jennifer Mattson, Kelly Sonnack, and Jennifer March Soloway will be on hand, along with Brown. Each agent represents picture and chapter books, middle grade, and young adult fiction.
In December 2020, the workshop takes place in Monterey rather than in Big Sur. Brown suggests that writers leave time before and after the weekend to take advantage of the Monterey Bay Aquarium and beaches and shopping. "We want writers to come and hole up for the weekend and focus on their writing," she says. "Most people take the revising time really seriously. In the evenings, we end by 8:30 so that they can write at night, and then there's time to revise on Saturday afternoon. That's when we see the miracles," she says.
Brown urges first-timers to show up with an open mind and to focus on craft and revision rather than on finding an agent. "We don't want you to be there to pitch," she explains. "This is boot camp for writers. Here, you find out what you still have to do to make your manuscript publishable."
She cautions writers to avoid being defensive during critique sessions. "We're not here to pat you on the back but to tell you what your weak spots are in your manuscript," she says. "Then, you can spend the weekend working on a few pages, or a particular character, or dialogue."
Brown asks attendees to bring a laptop and a thumb drive with works in progress, as well as three picture book manuscripts if available, or four to six chapters of a novel. "I like everyone to start Friday's two-hour critique group with just the first two to four pages of a manuscript," she says. "That first page or two is so important; that's what hooks an agent or editor. If you don't knock our socks off on page one, we'll never get to see how great page four is."
Contributing editor Melissa Hart is the author of Better with Books: 500 Diverse Books to Ignite Empathy and Encourage Self-Acceptance in Tweens and Teens (Sasquatch, 2019). Twitter/Instagram @WildMelissaHart.
Originally Published November 14, 2019
By Melissa Hart


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Toc show (de la expresión talk show) es un programa de televisión chileno de conversación, conducido por Juan Carlos Valdivia. Fue estrenado a través de UCV Televisión (actual TV+) el 4 de marzo de 2013.

Historia
Tras la abrupta salida de Juan Carlos Valdivia y su esposa Claudia Conserva del canal La Red en septiembre de 2011, el matrimonio se embarcó en una búsqueda de proyectos en los cuales ambos pudiesen participar. A principio de 2012 comenzaron las conversaciones entre UCV Televisión y la pareja televisiva para que se integraran al canal, pero a fin de cuentas únicamente Valdivia optó por este, ya que Conserva fichó por TVN. La integración fue confirmada por el director ejecutivo de UCV Televisión, Enrique Aimone, en noviembre de 2012. Según informó Juan Carlos Valdivia al diario La Tercera, este nuevo proyecto sería "un programa de conversación sobre la actualidad, con grandes dosis de humor. Cuenta con panelistas estables e invitados".

Una semana antes del estreno de Toc show, una especie de «marcha blanca» se dio en UCV Televisión. Entre el 25 de febrero de 2013 y el 1 de marzo del mismo año se emitió «Eo, eo, eo, que siga el veraneo», espacio que permitió a la producción experimentar con las transmisiones en alta definición. Entre los invitados a este espacio estuvieron Sebastián «Lindorfo» Jiménez, Daniel Fuenzalida, Constanza Michelson, entre otros.

El programa contaba originalmente con la participación de Constanza Michelson, Mariana Derderián, Soledad Bacarreza y Gonzalo Egas, pero a medida que este fue avanzando la totalidad de los panelistas fue cambiando. Contaba con la participación estable de Pedro Bastidas, Pauli Bolatti, Francisca Undurraga y Verónica Vieyra.

El programa fue reemplazado desde el 29 de febrero de 2016 por Nessun dorma: nadie duerma. El programa volvió  emitirse a reemplazar a Nessun dorma: nadie duerma, dos años después, el 12 de marzo de 2018.

En junio de 2022, y luego de que Claudia Conserva, la esposa del conductor Juan Carlos Valdivia, anunciara que padece cáncer, el conductor se ausentó durante dos semanas del programa, y anunció que su participación en este sería intermitente para acompañar a su esposa. La conducción la asume Felipe Vidal.

Equipo

Conductor
  Juan Carlos Valdivia (2013-actualidad)
Panelistas estables
  Felipe Vidal (2020)
  Manuel Rodríguez (2020)

Expanelistas
  Bernardita Ruffinelli (2019)
  Álvaro García (2019)
  Nathalie Nicloux (2019)
  Ruth Gamarra (2019)
  Marcela Brane (2019)
  Carolina Brethauer (2019)
  Mariela Sotomayor (2019)
  Pablo Zúñiga (2018-2019)
  Mery Salas (2018-2019)
  Javiera Acevedo (2018-2019)
  Eugenio Figueroa (2018-2019)
  Daniela Palavecino (2018)
  Yunet Guerra (2018)
  Verónica Vieyra (2016)
  Francisca Ayala (2015)
  Daniela Urrizola (2015)
  Lagarto Murdock (2015)
  Constanza Solar (2015)
  Dominique Lattimore (2015)

  Nayara Vit (2015)
  Pauli Bolatti (2015)
  Lorena Gálvez (2014)
  Kika Silva (2014)
  Pedro Bastidas (2014-2016)
  Francisca Undurraga (2013-2014; 2016)
  Flavia Fucenecco (2013; 2014-2015)
  Katherine Bodis (2013-2014)
  Constanza Michelson (2013-2014)
  Gonzalo Egas (2013-2014)
  Laura Prieto (2013)
  Glenny Rosado (2013)
  Mey Santamaría (2013)
  Soledad Bacarreza (2013)
  Paola Camaggi (2013)
  Paloma Cifuentes (2013)
  Mariana Derderián (2013)

Línea de Tiempo Panelistas (1ª etapa)

Línea de Tiempo Panelistas (2ª etapa)

Invitados
Es costumbre que, por la ausencia momentánea de uno o más panelistas estables, distintos personajes del mundo del modelaje, la televisión y el espectáculo participaron del programa, ya sea a modo de reemplazo por uno o más capítulos, durante un período de transición por la búsqueda de un nuevo panelista estable, a presentar secciones especiales, o simplemente en calidad de invitados. Además, quien ocupaba el puesto de conductor del programa cuando Juan Carlos Valdivia se ausentaba, generalmente era Pedro Bastidas o, en su defecto, Giancarlo Petaccia. Entre quienes pasaron por el espacio, destacaron:

  Javiera Acevedo
  Gabriela Zambrano
  María José Maldonado
  Daniella Chávez
  Sabrina Sosa
  Melina Figueroa
  Kathy Contreras
  Sofía Clerici

  Natalia Navarro
  Carla Ochoa
  Kim Gutiérrez
  Yazmín Vásquez
  Jessica Alonso
  Romina Ansaldo
  Camila Grove
  Catalina Vera

  Andrea Iglesias
  Tamara Sepúlveda Harnisch
  Julia Prado
  Mariana Missaglia
  Giselle Gómez Rolón
  Regina Fernández
  Katty Pérez

Producción
La producción se lleva a cabo por la propia productora de Juan Carlos Valdivia, GOA Films, la cual ocupa todo un piso de un edificio. Además, está altamente equipada con cámaras de alta definición, salas de redacción y switch. Al tener una productora propia, Valdivia tiene control total de Toc show.

Episodios
Toc show fue estrenado el 4 de marzo de 2013. El primer episodio tuvo de invitados al comentarista deportivo Claudio Palma y al vocero de gobierno del primer mandato de Michelle Bachelet, Francisco Vidal. Pero además, durante los primeros episodios destacaron como invitados Andrés Allamand, Paola Camaggi, Claudia Conserva, Cristian «Chico» Pérez, Mey Santamaría, José Miguel Viñuela, y Nicolás Copano, entre otros. A medida que los episodios fueron avanzando, las invitaciones a personalidades públicas comenzaron a cesar y destacaron únicamente las participaciones de los panelistas estables.

Originalmente, el programa contaba con 220 episodios.

Véase también
 Así somos

Referencias

Enlaces externos
 Sitio oficial de Toc show

Programas de televisión de Chile
Programas de televisión de TV+
Televisión de Chile en los años 2010
Televisión de Chile en 2013
Televisión de Chile en 2014
Televisión de Chile en 2015
Televisión de Chile en 2016
Televisión de Chile en 2017
Televisión de Chile en 2018
Televisión de Chile en 2019
Televisión de Chile en 2020
Televisión de Chile en 2021
Televisión de Chile en 2022
Televisión de Chile en 2023

redpajama_set_name: RedPajamaWikipedia


text:
About Media Relations
Dreamers communities stand to lose in DACA bargain
Industrial and Labor Relations
Rebecca Valli
rv234@cornell.edu
As Congress nears a deadline next week to approve a new budget or budget extension, the fate of the Deferred Action for Childhood Arrivals (DACA) program hangs in the balance. In September, the Trump administration discontinued the program, which has granted residence and work permits to about 690,000 undocumented immigrants who arrived in the United States as children.
Shannon Gleeson
Professor of Labor Relations, Law & History
Shannon Gleeson, a Cornell University professor of labor relations and an expert on immigrant labor, has been working on a three-year long collaborative project to study how the DACA policy is implemented at the local level in Houston, New York City, San Francisco and San Jose. She says that on the ground, DACA protection empowered undocumented immigrants with an education, means to provide for their families and protection from labor abuses.
Gleeson is available for interviews in Spanish.
Gleeson says:
"The DACA program provided many undocumented immigrants with hope for their educational and professional futures. It also gave beneficiaries the critical work authorization that they needed to provide for their families. For these workers, DACA was a crucial reprieve from deportation that also helped empower them to confront abusive employer behavior.
"With the rescission of DACA, the President and Attorney General have needlessly stripped these immigrants of this critical support, putting the well-being of these immigrants, their families, and their communities in danger. The same is true for the Temporary Protected Status program that Trump is dismantling piece by piece. The proposed border militarization measures instead being proposed are an irresponsible use of resources that will further terrorize immigrant communities.
"This punitive approach also ignores the central role that the United States has played in creating those conditions of violence and economic instability that propel migrants to this country in the first place. In contrast, a broad-based legalization program for all undocumented immigrants is a sensible and humane solution, which research shows was successful three decades ago with the 1986 Immigration Reform and Control Act.
"We would do well to follow this example, and that of other countries that shoulder a far heavier portion of the world's burden of refugees, and that regularize their undocumented populations far more frequently."
Maria Lorena Cook
Professor of International and Comparative Labor
Maria Lorena Cook, a professor of comparative labor at Cornell University and author of several books on labor movements in Latin America and unauthorized migration, argues that in the current political environment, protection of the Dreamers may come at the expense of further harm to their communities.
Cook is available for interviews in Spanish.
Cook says:
"Unless Congress acts, hundreds of thousands of DACA recipients stand to lose their deportation relief and work authorization on March 5, 2018, the date that DACA is set to expire. The future of DACA beneficiaries lies in the hands of the Republican-dominated Congress and the Trump administration. Assuming, in the best of cases, that Democrats are able to wrest a commitment to protect Dreamers, the outcome is likely to be unpalatable.
"Dreamers may face a Hobson's choice: either their own protection from deportation in exchange for even more restrictive immigration enforcement – and the greater likelihood that their family members and communities will suffer its effects – or nothing at all."
Cornell University has television, ISDN and dedicated Skype/Google+ Hangout studios available for media interviews.
312 College Ave. · Cornell University
mediarelations@cornell.edu
@CornellMedia


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To add the lean muscle tissue in the body is most commonly referred to as the bulking. Actual weight can be very deceiving so many people get trapped in the weight gaining games. Adding quality weight is all you need in bulking. If the weight is increasing rapidly you're not adding a quality weight to your body. If the scale goes up and your strength increases too you think you have added the lean tissue but when you add some fats you have added water gains in your body. There is no key to add weight to your body; not even the anabolic androgenic steroids can work if your food intake is inadequate.


Welcome to our 5-part blog series on why to Bin the Pin. In this fourth post, Henrik Knudtzon, Chief Financial Officer at IDEX Biometrics explains The Power of Your Fingertip.
Don't you think a minuscule four-digit PIN is too vulnerable to be used in this hacker-active age? Now you might consider elongated alternatives such as passwords but most of the time it becomes more of a hassle because you can simply can't remember all the different passwords you might have. Did you know the power to fight payment fraudulence is in your hands? Literally!
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Our previous post in Bin the Pin looked at whether fingerprint identification is the future of payment security?.


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text:
AMD Athlon 5050e Dual Core
AMD Athlon 4450B Dual Core
Buy Athlon 5050e Dual Core
Buy Athlon 4450B Dual Core
Athlon 5050e Dual Core
Athlon 4450B Dual Core
In terms of overall gaming performance, the AMD Athlon 5050e Dual Core is marginally better than the AMD Athlon 4450B Dual Core when it comes to running the latest games. This also means it will be less likely to bottleneck more powerful GPUs, allowing them to achieve more of their gaming performance potential.
The Athlon 5050e Dual was released less than a year after the Athlon 4450B Dual, and so they are likely to have similar levels of support, and similarly optimized performance when running the latest games.
The Athlon 5050e Dual and the Athlon 4450B Dual both have 2 cores, and so are quite likely to struggle with the latest games, or at least bottleneck high-end graphics cards when running them. With a decent accompanying GPU, the Athlon 5050e Dual and the Athlon 4450B Dual may still be able to run slightly older games fairly effectively.
The Athlon 5050e Dual and the Athlon 4450B Dual are from the same family of CPUs, and thus their clock speeds are directly comparable. With this in mind, it is safe to say that with a 0.3 GHz faster base clock rate, the Athlon 5050e Dual manages to provide slightly better performance than the Athlon 4450B Dual.
The Athlon 5050e Dual and the Athlon 4450B Dual have the same L2 cache size, and neither CPU appears to have an L3 cache. They even have the same L1 cache size, so are identical in terms of cache size.
Both the Athlon 5050e Dual and the Athlon 4450B Dual have the same TDP of 45 Watts, and were created with the same manufacturing size of 65 nm, which means they will affect your yearly electricity bill about equally.
Brisbane Brisbane
Socket AM2 Socket AM2
21 Oct 2008 28 Apr 2008
On April 21, 2005, less than a week after the release of Venice and San Diego, AMD announced its next addition to the Athlon 64 line, the Athlon 64 X2. Released on May 31, 2005, it also initially had two different core revisions available to the public, Manchester and Toledo, the only appreciable difference between them being the amount of L2 cache. Both were released only for Socket 939. The Athlon 64 X2 was received very well by reviewers and the general public, with a general consensus emerging that AMD's implementation of multi-core was superior to that of the competing Pentium D. Some felt initially that the X2 would cause market confusion with regard to price points since the new processor was targeted at the same enthusiast, US$350 and above market already occupied by AMD's existing socket 939 Athlon 64s. AMD's official breakdown of the chips placed the Athlon X2 aimed at a segment they called the prosumer, along with digital media fans. The Athlon 64 was targeted at the mainstream consumer, and the Athlon FX at gamers. On April 21, 2005, less than a week after the release of Venice and San Diego, AMD announced its next addition to the Athlon 64 line, the Athlon 64 X2. Released on May 31, 2005, it also initially had two different core revisions available to the public, Manchester and Toledo, the only appreciable difference between them being the amount of L2 cache. Both were released only for Socket 939. The Athlon 64 X2 was received very well by reviewers and the general public, with a general consensus emerging that AMD's implementation of multi-core was superior to that of the competing Pentium D. Some felt initially that the X2 would cause market confusion with regard to price points since the new processor was targeted at the same enthusiast, US$350 and above market already occupied by AMD's existing socket 939 Athlon 64s. AMD's official breakdown of the chips placed the Athlon X2 aimed at a segment they called the prosumer, along with digital media fans. The Athlon 64 was targeted at the mainstream consumer, and the Athlon FX at gamers.


text:
package com.mossle.user.persistence.manager;

import com.mossle.core.hibernate.HibernateEntityDao;

import com.mossle.user.persistence.domain.UserRepo;

import org.springframework.stereotype.Service;

@Service
public class UserRepoManager extends HibernateEntityDao<UserRepo> {
}

redpajama_set_name: RedPajamaGithub


"Visual Silence" by M.C. Chambers – Grand Prize winner of the 2008 National Space Society fiction contest. The Return to Luna short story contest, jointly run by the National Space Society and Hadley Rille Books, had open submissions from February 28, 2008, through June 15, 2008. We received many fine stories and we thank all the authors who entered.
It's Midsummer in early Renaissance Scotland. A troubled clan chief charges a mysterious courier called Silk with a message that will determine the course of war in the Highlands. On the road to Edinburgh, Silk meets handsome musician "Velvet" Hugh MacToro. Will he prove distraction or salvation? Will she bring him luck – or ruin? The air is full of music and secrets…and more magic than you know.


text:
<?php

namespace Cola\Database\DataTypes;

/**
 * DateTimeDataType
 *
 * @version 1.0.0
 * @since 1.0.0
 * @author dazarobbo <dazarobbo@live.com>
 */
abstract class DateTimeDataType extends DataType {
}