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  • JCSR

    NO STAGE PLAN
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    May 11, 2017
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    Santa Claus
    I know you support the Second Amendment, but what are your thoughts on the First Amendment?
    From what I see the First is being breached by special interest groups in the US at a rate I never thought I'd live to see. People losing jobs over something they said as a teenager etc. I could go on but I don't want to rant!
     

    JCSR

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    May 11, 2017
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    Santa Claus
    What is the airspeed velocity of an unladen swallow?

    Pretty simple really....

    [FONT=&quot][FONT=KaTeX_Main][FONT=KaTeX_Math]
    S
    [/FONT][FONT=KaTeX_Math]u[/FONT][FONT=KaTeX_Math]b[/FONT][FONT=KaTeX_Math]s[/FONT][FONT=KaTeX_Math]t[/FONT][FONT=KaTeX_Math]i[/FONT][FONT=KaTeX_Math]t[/FONT][FONT=KaTeX_Math]u[/FONT][FONT=KaTeX_Math]t[/FONT][FONT=KaTeX_Math]e[/FONT][FONT=KaTeX_Math]x[/FONT]=[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT]\alp,\alp∈[0:;[FONT=KaTeX_Math]π[/FONT]][/FONT][/FONT]

    [FONT=&quot][FONT=KaTeX_Main]\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)[FONT=KaTeX_Math]T[/FONT][FONT=KaTeX_Math]h[/FONT][FONT=KaTeX_Math]e[/FONT][FONT=KaTeX_Math]n[/FONT]∣[FONT=KaTeX_Math]x[/FONT]+1∣=2(2[FONT=KaTeX_Math]x[/FONT]2−1)\Leftright∣[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT]\alp+[FONT=KaTeX_Math]s[/FONT][FONT=KaTeX_Math]i[/FONT][FONT=KaTeX_Math]n[/FONT]\alp∣=2(2[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT]2\alp−1)[/FONT][/FONT]
    [FONT=&quot][FONT=KaTeX_Main]\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]∣[FONT=KaTeX_AMS]N[/FONT]2[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT](\alp
    4[FONT=KaTeX_Math]π[/FONT]​​
    )∣=[FONT=KaTeX_AMS]N[/FONT]2[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT](2\alp)\Right\alp∈[0;
    4[FONT=KaTeX_Math]π[/FONT]​​
    ]∪[
    43[FONT=KaTeX_Math]π[/FONT]​​
    ;[FONT=KaTeX_Math]π[/FONT]][/FONT][/FONT]

    [FONT=&quot]1) [/FONT][FONT=&quot][FONT=KaTeX_Main]\displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;
    4[FONT=KaTeX_Math]π[/FONT]​​
    ][/FONT][/FONT]

    [FONT=&quot][FONT=KaTeX_Main]\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT](\alp
    4[FONT=KaTeX_Math]π[/FONT]​​
    )=[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT](2\alp)…[/FONT][/FONT]

    [FONT=&quot]2. [/FONT][FONT=&quot][FONT=KaTeX_Main]\displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[
    43[FONT=KaTeX_Math]π[/FONT]​​
    ;[FONT=KaTeX_Math]π[/FONT]][/FONT][/FONT]

    [FONT=&quot][FONT=KaTeX_Main]\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT](\alp
    4[FONT=KaTeX_Math]π[/FONT]​​
    )=[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT](2\alp)…
    [/FONT][/FONT]
     

    Cameramonkey

    www.thechosen.tv
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    35   0   0
    May 12, 2013
    31,970
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    Camby area
    Pretty simple really....

    [FONT=&amp][FONT=KaTeX_Main][FONT=KaTeX_Math]
    S
    [/FONT][FONT=KaTeX_Math]u[/FONT][FONT=KaTeX_Math]b[/FONT][FONT=KaTeX_Math]s[/FONT][FONT=KaTeX_Math]t[/FONT][FONT=KaTeX_Math]i[/FONT][FONT=KaTeX_Math]t[/FONT][FONT=KaTeX_Math]u[/FONT][FONT=KaTeX_Math]t[/FONT][FONT=KaTeX_Math]e[/FONT][FONT=KaTeX_Math]x[/FONT]=[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT]\alp,\alp∈[0:;[FONT=KaTeX_Math]π[/FONT]][/FONT][/FONT]

    [FONT=&amp][FONT=KaTeX_Main]\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)[FONT=KaTeX_Math]T[/FONT][FONT=KaTeX_Math]h[/FONT][FONT=KaTeX_Math]e[/FONT][FONT=KaTeX_Math]n[/FONT]∣[FONT=KaTeX_Math]x[/FONT]+1∣=2(2[FONT=KaTeX_Math]x[/FONT]2−1)\Leftright∣[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT]\alp+[FONT=KaTeX_Math]s[/FONT][FONT=KaTeX_Math]i[/FONT][FONT=KaTeX_Math]n[/FONT]\alp∣=2(2[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT]2\alp−1)[/FONT][/FONT]
    [FONT=&amp][FONT=KaTeX_Main]\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]∣[FONT=KaTeX_AMS]N[/FONT]2[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT](\alp
    4[FONT=KaTeX_Math]π[/FONT]​​
    )∣=[FONT=KaTeX_AMS]N[/FONT]2[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT](2\alp)\Right\alp∈[0;
    4[FONT=KaTeX_Math]π[/FONT]​​
    ]∪[
    43[FONT=KaTeX_Math]π[/FONT]​​
    ;[FONT=KaTeX_Math]π[/FONT]][/FONT][/FONT]

    [FONT=&amp]1) [/FONT][FONT=&amp][FONT=KaTeX_Main]\displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;
    4[FONT=KaTeX_Math]π[/FONT]​​
    ][/FONT][/FONT]

    [FONT=&amp][FONT=KaTeX_Main]\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT](\alp
    4[FONT=KaTeX_Math]π[/FONT]​​
    )=[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT](2\alp)…[/FONT][/FONT]

    [FONT=&amp]2. [/FONT][FONT=&amp][FONT=KaTeX_Main]\displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[
    43[FONT=KaTeX_Math]π[/FONT]​​
    ;[FONT=KaTeX_Math]π[/FONT]][/FONT][/FONT]

    [FONT=&amp][FONT=KaTeX_Main]\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT](\alp
    4[FONT=KaTeX_Math]π[/FONT]​​
    )=[FONT=KaTeX_Math]c[/FONT][FONT=KaTeX_Math]o[/FONT][FONT=KaTeX_Math]s[/FONT](2\alp)…
    [/FONT][/FONT]
    FAIL.

    The correct answer is "African or European swallow? "

    [video=youtube;y2R3FvS4xr4]http://www.youtube.com/watch?v=y2R3FvS4xr4[/video]
     

    actaeon277

    Grandmaster
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    Rating - 100%
    4   0   0
    Nov 20, 2011
    93,284
    113
    Merrillville
    Better be careful. You have vastly increased your rate of posting. You're going to break the next 100 in 3 weeks.

    And mouse isn't the one to worry about. He gives warnings first. If you heed the warning... no worry.
     

    JCSR

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    May 11, 2017
    9,033
    133
    Santa Claus
    Better be careful. You have vastly increased your rate of posting. You're going to break the next 100 in 3 weeks.

    And mouse isn't the one to worry about. He gives warnings first. If you heed the warning... no worry.

    :yesway: Had no idea I would get swamped with all these questions. :ingo:
     
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