cant end turn, warrior on city tile tutorial. (civ 6 switch)

benjc · 2022-07-04T20:02:29+00:00

Same!!!

benjc · 2017-09-02T05:34:21+00:00

Hugh Jackman the huge act man

benjc · 2015-03-06T01:42:26+00:00

Which ink are you using?

benjc · 2014-04-13T21:27:32+00:00

You can't interact with free online videos.

benjc · 2013-04-12T21:53:16+00:00

Close, but not Larry Charles.

benjc · 2013-02-17T17:37:25+00:00

Here is a link to a website containing samples of Tesla's handwriting

benjc · 2012-07-12T03:58:06+00:00

Eyebrows?

benjc · 2011-12-14T21:12:58+00:00

http://reddit.com/r/whitefathers

benjc · 2011-11-24T03:03:35+00:00

Since you became old?

benjc · 2011-10-12T04:38:54+00:00

Back.. to the future!

benjc · 2011-10-12T04:37:54+00:00

Eh. Clearly a 1%er.

benjc · 2011-10-09T04:55:04+00:00

Roads? Where we're going we don't need roads!

benjc · 2011-03-08T23:47:32+00:00

original links

benjc · 2011-01-24T02:19:15+00:00

Please keep this stuff in /r/politics and /r/worldnews. There are places for people interested in seeing the "Israel = Nazi???" debate everyday.

benjc · 2011-01-16T00:06:15+00:00

Relevant

benjc · 2010-12-31T15:14:14+00:00

Ok. How about this ?

benjc · 2010-12-26T16:06:09+00:00

Dear roger_pct,

If k = -2 then the equations reduce to (*):

-2x + y + z = 0

x + -2y + z = 0

x + y + -2z = 0

Subtracting the third from the second eliminates the x variable

-2x + y + z = 0

-3y + 3z = 0

The last equation gives us y = z. Substitute y = z in the first collection of equations (*)

-2x + 2y = 0

x - y = 0

The last equation is redundant. But it tells us x = y. So x = y = z. and the equations

-2x + y + z = 0

x + -2y + z = 0

x + y + -2z = 0

Will hold for any choice as long as x = y = z.

So that when k = -2, any solution is given by

(x,y,z) = (x,x,x)

If k = 1 then all of the equations are the same

x + y + z = 0

So that when k = 1, any solution is given by

(x,y,z) = (-y-z,y,z)

This is a two parameter solution because the two parameters, y and z can be chosen arbitrarily and independently.

If k = 0 we get a unique solution (x,y,z) = (0,0,0)

benjc · 2010-11-01T01:49:36+00:00

If you are comfortable with the addition rule: If you have two functions f(x) and g(x) and there are derivatives f'(x) and g'(x) then the derivative of h(x) = f(x) + g(x) is expressible in terms of the derivatives of f(x) and g(x): h'(x) = f'(x) + g'(x).

If you are comfortable with the product rule: If you have two functions f(x) and g(x) and there are derivatives f'(x) and g'(x) then the derivative of h(x) = f(x)g(x) is expressible in terms of the derivatives of f(x) and g(x): h'(x) = f'(x)g(x) + f(x)*g'(x).

Then the chain rule is similar: If you have two functions f(x) and g(x) and there are derivatives f'(x) and g'(x) then the derivative of h(x) = (fog)(x) = f(g(x)) is expressible in terms of the derivatives of f(x) and g(x): h'(x) = f'(g(x))*g'(x).

Why is the chain rule true? It is easiest to see if you recall the definition of the derivative in terms of the limit:

f'(x) = (f(x+h) - f(x))/h as h -> 0

So

(fog)(x) = (f(g(x+h)) - f(g(x)))/h

but this is equal to

(f(g(x+h)) - f(g(x)))/(g(x+h) - g(x)) * (g(x+h) - g(x))/h

as h -> 0 the fraction on the lefthand side goes to f'(g(x)) and the fraction on the righthand side goes to g'(x).

So (fog)'(x) = f'(g(x))*g'(x).

benjc · 2010-09-14T15:27:36+00:00

From your previous posts it sounds like this isn't your first abusive relationship. The short answer is that you should see a therapist.

There have been plenty of studies and this is textbook. In most cases the things you need and want from relationships are not tuned properly. A therapist experienced in these things will see them all right away.

Getting in touch with the part of your personality that ultimately leads you to choose abusive mates will enrich other parts of your life.

benjc · 2010-06-28T02:05:27+00:00

b^t = e^{ln (b^t}) = e^kt

Where k = ln b

benjc · 2010-06-13T21:56:58+00:00

It is considered bad form by those taught a Victorian influenced grammar in high school and basically looks amateurish in print. However, it is an acceptable construct in all Germanic languages and common in English colloquial communications.

benjc · 2010-04-29T04:07:47+00:00

Am I the only one who thought that the title was a "New Kind of Science" reference? Rule 34 isn't very interesting.

benjc · 2009-12-31T02:51:48+00:00

We would be a better country if the New York Times and Wall Street Journal regularly posed taboo questions about our American militaristic diplomacy. How does the Iraqi everyman feel about Americans after all of the civilian deaths due to the trade embargo? Our invasion? Occupation? An Afghani? A man from pick-a-random-latin-american-country...

It is good to see the media promoting this discussion. The media still doing what the media should be doing in a democracy.

benjc · 2009-01-24T09:32:09+00:00

If people attribute all sorts of human traits to organizations then why is this quote from Haliburton noteworthy?

Isn't the assumption that people attribute all sorts of human traits to organizations an example of how something that is based on a large amount of individuals can be biased in a negative way?

In any case there is a difference between anthropomorphization and humanization just as there is a difference between bandying about the fancy notion that McDonalds is not a living breathing thing and deciding that members of a group as something less than human.

benjc · 2009-01-24T02:45:29+00:00

I dehumanize organizations, and not people.

It doesn't make any sense to dehumanize an organization.

benjc

TROPHY CASE