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[razgueado]

Morning, muchachos.

morning RazMan.

That's classic. Morning, Deano.

[FloridaDean]

shot the banter all to hell.

[FloridaDean]

think I'll join the Redfish.

[A Friend of Charlie]

think I'll join the Redfish.

You all creating a retired person's banter?

[FloridaDean]

think I'll join the Redfish.

You all creating a retired person's banter?

we do text back and forth. a little livelier. 😂

[A Friend of Charlie]

think I'll join the Redfish.

You all creating a retired person's banter?

we do text back and forth. a little livelier.

That's how it starts.

[FloridaDean]

think I'll join the Redfish.

You all creating a retired person's banter?

we do text back and forth. a little livelier.

That's how it starts.

they're short texts - we're old.

[FloridaDean]

grilling a couple of blu cheese burgers in the rain.

[A Friend of Charlie]

grilling a couple of blu cheese burgers in the rain.

Sunny day here. Rain is due back though.

[FloridaDean]

grilling a couple of blu cheese burgers in the rain.

Sunny day here. Rain is due back though.

73° and light steady rain. not RA friendly.

[bluecollar]

Sampled the Dias de Gloria yesterday Rick.  Enjoyed it, nice flavors, a bit one note but not boring.  Expected AJ construction and burn.

I had the 6x58 and it was good but it missed the WoW factor with me. I had high expectations .......

TWSS

Haha Bean!!!

[FloridaDean]

understand wemen is like:

\displaystyle Substitute\; x=cos\alp, \alp\in[0: ;\: \pi ]Substitutex=cos\alp,\alp∈[0:;π]
\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+
1−x 2∣= 2(2x 2 −1)\Leftright∣cos\alp+sin\alp∣=
2(2cos
2\alp−1)
\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N
2cos(\alp−
4π )∣=N 2cos(2\alp)\Right\alp∈[0;
4π]∪[ 43π;π]1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;
4π]\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dotscos(\alp−
4π )=cos(2\alp)…
2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[
43π ;π]\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−
4π
 )=cos(2\alp)…
1

[A Friend of Charlie]

understand wemen is like:

displaystyle Substitute; x=cosalp, alpin[0: ;: pi ]Substitutex=cosalp,alp∈[0:;π]
displaystyle Then; |x+sqrt{1-x^2}|=sqrt{2}(2x^2-1)Leftright |cosalp +sinalp |=sqrt{2}(2cos^2alp -1)Then∣x+
1−x 2∣= 2(2x 2 −1)Leftright∣cosalp+sinalp∣=
2(2cos
2alp−1)
displaystyle |N {sqrt{2}}cos(alp-frac{pi}{4})|=N {sqrt{2}}cos(2alp )Right alpin[0: ;: frac{pi}{4}]cup [frac{3pi}{4}: ;: pi]∣N
2cos(alp−
4π )∣=N 2cos(2alp)Rightalp∈[0;
4π]∪[ 43π;π]1) displaystyle alp in [0: ;: frac{pi}{4}]alp∈[0;
4π]displaystyle cos(alp -frac{pi}{4})=cos(2alp )dotscos(alp−
4π )=cos(2alp)…
2. displaystyle alpin [frac{3pi}{4}: ;: pi]alp∈[
43π ;π]displaystyle -cos(alp -frac{pi}{4})=cos(2alp )dots−cos(alp−
4π
 )=cos(2alp)…
1

That makes perfect sense.

[Threebean]

understand wemen is like:

displaystyle Substitute; x=cosalp, alpin[0: ;: pi ]Substitutex=cosalp,alp∈[0:;π]
displaystyle Then; |x+sqrt{1-x^2}|=sqrt{2}(2x^2-1)Leftright |cosalp +sinalp |=sqrt{2}(2cos^2alp -1)Then∣x+
1−x 2∣= 2(2x 2 −1)Leftright∣cosalp+sinalp∣=
2(2cos
2alp−1)
displaystyle |N {sqrt{2}}cos(alp-frac{pi}{4})|=N {sqrt{2}}cos(2alp )Right alpin[0: ;: frac{pi}{4}]cup [frac{3pi}{4}: ;: pi]∣N
2cos(alp−
4π )∣=N 2cos(2alp)Rightalp∈[0;
4π]∪[ 43π;π]1) displaystyle alp in [0: ;: frac{pi}{4}]alp∈[0;
4π]displaystyle cos(alp -frac{pi}{4})=cos(2alp )dotscos(alp−
4π )=cos(2alp)…
2. displaystyle alpin [frac{3pi}{4}: ;: pi]alp∈[
43π ;π]displaystyle -cos(alp -frac{pi}{4})=cos(2alp )dots−cos(alp−
4π
 )=cos(2alp)…
1

That makes perfect sense.

For a moment there I thought we had another troll visit.

[A Friend of Charlie]

understand wemen is like:

displaystyle Substitute; x=cosalp, alpin[0: ;: pi ]Substitutex=cosalp,alp∈[0:;π]
displaystyle Then; |x+sqrt{1-x^2}|=sqrt{2}(2x^2-1)Leftright |cosalp +sinalp |=sqrt{2}(2cos^2alp -1)Then∣x+
1−x 2∣= 2(2x 2 −1)Leftright∣cosalp+sinalp∣=
2(2cos
2alp−1)
displaystyle |N {sqrt{2}}cos(alp-frac{pi}{4})|=N {sqrt{2}}cos(2alp )Right alpin[0: ;: frac{pi}{4}]cup [frac{3pi}{4}: ;: pi]∣N
2cos(alp−
4π )∣=N 2cos(2alp)Rightalp∈[0;
4π]∪[ 43π;π]1) displaystyle alp in [0: ;: frac{pi}{4}]alp∈[0;
4π]displaystyle cos(alp -frac{pi}{4})=cos(2alp )dotscos(alp−
4π )=cos(2alp)…
2. displaystyle alpin [frac{3pi}{4}: ;: pi]alp∈[
43π ;π]displaystyle -cos(alp -frac{pi}{4})=cos(2alp )dots−cos(alp−
4π
 )=cos(2alp)…
1

That makes perfect sense.

For a moment there I thought we had another troll visit.

Maybe we did.