From AI influencers to fake personas and a gold phone that never arrived, these online ‘scams’ are exposing hard truths MAGA men were never The post This Online ‘Scam’ Is Forcing MAGA Men to Confront Hard Truths appeared first on Mystical Raven.

In its tournament tuneup, Bay City Western lost three straight games.

MUSKEGON COUNTY, MI - Three wastewater leaks were discovered this week in Muskegon and Ottawa counties and were reported to the state.

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MORGANTOWN, W.Va. – Ticket requests for WVU football season ticket holders and Mountaineer Athletic Club (MAC) members are now available for the Duke's Mayo Classic, WVU vs. Virginia, on Sept. 19, at Bank of America Stadium, in Charlotte, North Carolina.

American-point-guard-Tyler-Harris-was-the-top-scorer-of-the-day-in-the-world-yesterday-with-the-34-points-he-had-in-the-Danish-league-Harris-graduated-from-University-of-South-Florida-three-years-ago-Despite-his-strong-performance-his-team-Naestved-lost-to-the-Bakken-Bears-81-to-87-He-is-the-5th-best-scorer-in-the-league-averaging-almost-18-points-and-2-rebounds-per-game-Among-many-awards-he

Senatorial candidate James Talarico is taking the world by storm thanks to his no-nonsense approach to compromising on Christianity. Now, with the new James Talarico Bible translation, you can wield the Word of God with just as much inaccuracy.

Why does this $$h_z = \begin{cases} -z(L+x) \; , & x\in \left[-L,-z\right]\\ (L-z)x \; , & x\in \left[-z,z\right]\\ z(L-x) \; , & x\in \left[z,L\right] \end{cases}$$ define a $2L$-periodic sawtooth like function with $z \in (0,L)$? In the picture, $L$ is marked as $l$. In the interval $[-z,z]$, for example, I get $\tan(\alpha) = \frac{H}{z}$ with $H$ being the height of the triangle between $0$ and $L$, and $\alpha$ being the angle from the $x$-axis at point $0$ and the triangle. But why is this also equal to $L-z$?

I am reading about circular arrangements in this pdf on page 103. This is an image taken from that pdf 6.2. CIRCULAR PERMUTATIONS Proposition 6.2.15. Let $P_1, \ldots, P_n$ be all the arrangements of an m-multiset. Then, $$ [R_{0}+\cdots+R_{m-1}](P_{1}+\cdots+P_{n})=m(P_{1}+\cdots+P_{n}). $$ Let $P$ be an arrangement of an $m$-multiset with orbit size $k$. Then, by Proposition 6.2.10 $k$ divides $m$. Now, from the understanding obtained from the above example, we note that $[R_0 + \cdots + R_{m-1}](P)$ accounts for $\frac{m}{k}$ counts of the same circular arrangement, where $\frac{m}{k}$ is nothing but ‘the number of rotations fixing $P'$. Also, by Proposition 6.2.11, we know that two orbits are either disjoint or the same and hence the next two results are immediate. But I have trouble understanding the sentence "$(R_0 + \cdots + R_{m−1})(P)$ accounts for $\frac mk$ counts of the same circular arrangement" If we make all $m$ rotations of $P$ into circular arrangement don't we get $m$ counts of the same circular arrangement? Why only $\frac mk$? I understand there are $\frac mk$ counts of each distinct arrangement that gives the same circular arrangement. Example: If I have the arrangement $ABCABCABC$, On $R_1$ ,I get $BCABCABCA$ On $R_2$, I get $CABCABCAB$ On $R_3$, I get $ABCABCABC$ So if apply the rotations from $R_0$ to $R_8$, I get 3 of each. But all of them are referring to same circular arrangement, right? Or is my understanding incorrect?

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NOVI, MI - Sports stars galore have descended upon Metro Detroit at the largest sports card convention in the state.

Photo by George Walker/Icon Sportswire via Getty ImagesRyan Gerard sits just one stroke adrift of the leader, Eric Cole, heading into the final round of the Charles Schwab Challenge.Gerard turned what would have been a decent round into a good one on day three of the Charles Schwab Challenge, after he birdied his final two holes.On what ended up being a really challenging third round for everyone in Fort Worth, Texas, the 26-year-old American proved that he has the strength of mind required to achieve huge success in the game.Ryan Gerard’s late flourish will undoubtedly have put him in great spirits heading into the final round of the Charles Schwab Challenge.Photo by Sam Hodde/Getty ImagesThe firmness of the greens at the Charles Schwab Challenge has surprised players so far this week.And Gerard’s impressive ball-striking should stand him in good stead in that regard heading into Sunday’s final round.However, he did highlight one big mistake that he made on Saturday at Colonial Country Club.Ryan Gerard admits what he did at the Charles Schwab was ‘not acceptable’The 2025 Barracuda Championship winner addressed the media in Fort Worth after posting a third-round 68.Gerard responded when asked by reporters to explain what caused him to make a really sloppy bogey on the first hole at Colonial on Saturday.Photo by Sam Hodde/Getty Images“Yeah, we made a really, really bad bogey on 1,” he said.“I hit it, I hit the wedge too far. I thought the green was going to be softer, it took a big bounce, sat straight down.“I had absolutely nothing. Just kind of made a mess of the first hole. That’s not acceptable.“Ryan Gerard’s elite mentality enabled him to bounce backAfter his bogey on the first hole on Saturday, Gerard made birdies on two and three to steady the ship immediately.“You don’t knock yourself out of it early on in the day on Saturday by making one kind of bad decision,“ the 26-year-old insisted.“That’s kind of something that we talked about was, you don’t need to press it, you don’t need to force it, you don’t need to just push it to the edge right away, there’s a lot of golf left to be played and a lot of opportunities.“This course isn’t just going to present them to you, you’re going to have to go out and execute good golf shots and take advantage of them when they show up.“But it’s not one of those things where that was a day ruiner, that wasn’t going to be a deal breaker, that wasn’t going to be the end of the week.“Ryan Gerard will be hoping to make a better start on Sunday at the Charles Schwab Challenge, as he goes in search of his second PGA Tour win.

Go behind the scenes with Walt Disney Imagineering as a Show Quality Art Director reveals what it takes to keep the Country Bears show-ready every day

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KASHYYYK — A therapist working with famed Star Wars icon Chewbacca told his colleagues this week that he was trying to teach the Wookiee how to deal with his anger issues in more constructive ways that don't involve ripping people's arms off. Dr. Aris Thorne, a licensed galactic psychotherapist, confirmed on Friday that he had begun intensive cognitive behavioral therapy with Chewbacca to address the Wookiee's well-known anger issues. Thorne admitted that the therapy intensified after a routine game of holochess resulted in an Imperial IT technician losing both arms.

Will Shortz notes:Jeffrey Martinovic and Evan Park are fellow students at Western University in London, Ontario. They became friends over cards (bridge, the Korean game Mighty) and intramural sports (ultimate Frisbee, water polo, softball, and others). This puzzle started when Evan observed that the two shaded words in the grid could intersect. Together, they brainstormed theme answers related to that crossing, then filled the grid and composed clues over a month of video calls.Jeffrey Martinovic and Evan Park notes:We are good friends from school, bonding through activities like frisbee, water polo, cards, outdoor adventures, and of course, crosswords. We are both graduating now. Jeffrey will be starting work soon and Evan will be traveling the world for a year, volunteering along the way. What a nice graduation gift it is to have a puzzle together in the Times as we go our separate ways for a while. This puzzle arose when Evan had an idea that included a literal representation of a bow and arrow. Jeffrey thought to link it to William Tell because every morning his high school would play the William Tell Overture in the halls right before class. The concept developed organically from there, and the puzzle came together over many iterations, spanning months, all over video call. We hope you enjoyed solving!

The View-Master in the Buzz Lightyear Space Ranger Spin queue once had a clever detail that went unnoticed by most guests. Now it's gone. The post Sly View-Master Trick Quietly Vanishes from Buzz Lightyear Queue appeared first on Parkeology.

Halfway through stroke play, Arizona is in good position to not just make it to Monday but get into match play for the first time. And leading the charge is one of its seniors, who has himself atop the leaderboard.Filip Jakubcik shot a 6-under par 66 on Saturday during the second round of the NCAA Men’s Golf Championships at Omni La Costa Resort and Spa in Carlsbad, Cali. Jakubcik is at -10 for the tournament, two strokes ahead of Alabama’s William Jennings and Louisiana’s Malan Potgieter for the individual lead.As a team, the UA shot 3-under on Saturday and is at -4 through 36 holes, which is good for a tie for sixth place out of 30 teams. The top 15 remain in contention after Sunday’s third round, and the Wildcats are seven shots above that current cut line, while the top eight after the fourth round advance to match play.Senior Zach Pollo and freshman William Wistrand both shot even par in the second round, but the paths they took to get there were very different. While Wistrand parred his first nine holes and 14 of 18, Pollo only had seven pars to go with five birdies, an eagle, three bogeys and two double bogeyes.Pollo had an eagle on his first hole, the par-5 10th, then was plus-4 on the next four holes only to birdie the next two and then finish the back nine by going bogey-double bogey. He rebounded on the front nine with three birdies to finish even for the day.

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