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Wouldn't you know it, there's a wikipedia article for that. I personally have used 7digital and bandcamp, but qobuz has been mentioned several times in other comments and hdtracks seems like it might work after you create an account.

They might be the most common because they're the easiest, but there are also still plenty of people actually paying for the games. I'll never be convinced that piracy is an actual threat to making money. Piracy has never been easier, just see /c/piracy@lemmy.dbzer0.com for proof, and yet pretty much all forms of entertainmment are as profitable as ever.

(C )Any digital good that is advertised or offered to a person that the seller cannot revoke access to after the transaction, which includes making the digital good available at the time of purchase for permanent offline download to an external storage source to be used without a connection to the internet.

It shouldn't be that hard, gog.com manages to do it

4chan anon who made significant contribution to a math problem in order to find the best way to watch an anime

Looks like this one https://en.wikipedia.org/wiki/Superpermutation based on the content of the wikipedia article

Then go away and live your life. Let the people who are angry and energized yell at companies to try and get better conditions for consumers.

https://www.iso-ne.com/ Looking at my own region of New England, renewables are only at about 8% right now. And that includes burning wood, refuse, and landfill gas as renewable sources.

Given r=f(θ), we are generally not concerned with r′=f′(θ); that describes how fast r changes with respect to θ

You're using the derivative of a polar equation as the basis for what a tangent line is. But as the textbook explains, that doesn't give you a tangent line or describe the slope at that point. I never bothered defining what "tangent" means, but since this seems so important to you why don't you try coming up with a reasonable definition?

I think we fundamentally don't agree on what "tangent" means. You can use

x=f(θ)cosθ, y=f(θ)sinθ to compute dydx

as taken from the textbook, giving you a tangent line in the terms used in polar coordinates. I think your line of reasoning would lead to r=1 in polar coordinates being a line, even though it's a circle with radius 1.

Given r=f(θ), we are generally not concerned with r′=f′(θ); that describes how fast r changes with respect to θ

I think this part from the textbook describes what you're talking about

Instead, we will use x=f(θ)cosθ, y=f(θ)sinθ to compute dydx.

And this would give you the actual tangent line, or at least the slope of that line.

Polar Functions and dydx

We are interested in the lines tangent a given graph, regardless of whether that graph is produced by rectangular, parametric, or polar equations. In each of these contexts, the slope of the tangent line is dydx. Given r=f(θ), we are generally not concerned with r′=f′(θ); that describes how fast r changes with respect to θ. Instead, we will use x=f(θ)cosθ, y=f(θ)sinθ to compute dydx.

From the link above. I really don't understand why you seem to think a tangent line in polar coordinates would be a circle.

A straight line in polar coordinates with the same tangent would be a circle.

I'm not sure that's true. In non-euclidean geometry it might be, but aren't polar coordinates just an alternative way of expressing cartesian?

Looking at a libre textbook, it seems to be showing that a tangent line in polar coordinates is still a straight line, not a circle.

The tangent of all points along the line equal that line