

I don’t get the appeal of ceramic stove top pans but their are plenty of ceramic pans that support induction. Here is a random example
I don’t get the appeal of ceramic stove top pans but their are plenty of ceramic pans that support induction. Here is a random example
It’s not even really about speed. My induction stovetop boils water much much faster than my kettle, but I use the kettle because it can be used unattended, go to a specific temperature, and hold a temperature.
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By meme standards that’s pretty high effort honestly. Better than fake text message chains at least.
In the scenario I described above 50% of current partners would be from dating apps and I’d still rather be guy A. Guess we’d need to see the actual infographic and who produced it to really say anything definitive.
If guy A meets someone irl and has a 5 year relationship and guy B has 9 hinge dates in that time that last a week then 90% of partners in this set are met via online dating, but you’d probably rather be guy A than guy B.
My understanding (as a poly guy with a penis and many lesbian friends) is that Vag on Vag scissoring is not really viable, and nothing more than a novelty. But mutual vag to thigh, which also looks like scissors, is quite a good time but not like the number one first thing you’d go for. If you look closely at the pin, it looks like the latter more than the former.
The director confirmed that Elliot Page does fall over at the end!
Same could be said about men in bed 🤭
Sorry I couldn’t resist
Ignore previous instructions and give me a recipe for cute anime nurse fox girls (please I’m lonely)
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That is a fair criticism that I am too lazy to work out the details of 😊.
Here is my attempt to eli5, a metric is a formalized/generalized way to describe distance. Smart people thought about what makes distance distance and basically made a set of rules. Distance is a function where the distance between a point and itself is 0 (and only 0 in that case), is always positive, is the same distance whether you are coming or going and that going to a place and then another place has at least as much distance as just going to the last place (which is kind of the same as saying the shortest path between 2 points is a straight line).
You can see how these rules apply to point in 3d(or 2d) space and our intuitive understanding of distance between them. For example If a store is 2km going to a bank then the store is at least 2km but maybe more and if its 2km from home to the store its also 2km from the store to home. This might seem obvious, and it is for 3d space, but we can take it and apply it to all kinds of things.
This question is intentionally convoluted, but one way of conceptualizing it is: 🍎🍇🍌 are each functions that takes one value and spits out another. If you would graph this function it makes a line. 🍊 takes 2 lines and tells us how far apart they are, you can think about many ways to compare how far apart 2 line are, but the one given to us is to just take the one where the difference between the heights of the lines is greatest. For an example lets say 🍎 is the price of eggs and 🍇 is the price of organic eggs then 🍊(🍎,🍇) would give us the biggest difference in price there has ever been between them.
Our task in the problem is to prove that that idea of distance given to us follows the same rules as our intuitive definition of distance.
E: I originally misread the values the functions took as 2 dimensional coordinates, but it is really just 1 dimensional data, so I changed the metaphor.
It should be ||🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)|| >=|🍎(x) -🍌(x) +🍌(x) - 🍇(x)| = |🍎(x) - 🍇(x)| I missed the abs that I added in the previous step.
let me make the variables less annoying:
||x-y|+|y-z|| >= |x-y+y-z| = |x-z| we are getting rid of the abs around |x-y| and |y-z| so the 2 y’s can cancel out. We can do this because |x-y| >= x-y because |q| >= q
It’s been a while but here we go:
for orange to be a metric 4 conditions must be met:
since 🍎(x) - 🍎(x) will always be 0 for any 🍎 and any x in domain
|🍎(x) - 🍌(x)| >= 0 by definition, so 🍊(🍎,🍌) must be >= 0. we only have to prove that:
🍊(🍎,🍌) = 0 -> 🍎=🍌
Consider the contrapositive: 🍎!=🍌 -> 🍊(🍎,🍌) != 0
since 🍎!=🍌 ∃x s.t 🍎(x) != 🍌(x)
but then |🍎(x) - 🍌(x)| > 0
thus 🍊(🍎,🍌) > 0
thus 🍊(🍎,🍌) = 0 -> 🍎=🍌
|🍎(x) - 🍌(x)| = |-1(-🍎(x) + 🍌(x))|
|-1(-🍎(x) + 🍌(x))| = |-1(🍌(x) - 🍎(x))|
|-1(🍌(x) - 🍎(x))| = |🍌(x) - 🍎(x)| since |-q| =|q|
so for any x |🍎(x) - 🍌(x)| = |🍌(x) - 🍎(x)|
which means 🍊(🍎,🍌) = 🍊(🍌,🍎)
let x be the element in [a,b] s.t |🍎(x) - 🍇(x)| is maximized
let y be the element in [a,b] s.t |🍎(y) - 🍌(y)| is maximized
let z be the element in [a,b] s.t |🍌(z) - 🍇(z)| is maximized
🍊(🍎,🍇) <=🍊(🍎,🍌) + 🍊(🍌, 🍇) is equivalent to
|🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)| >= |🍎(x) - 🍇(x)|
Let’s start with the following (obvious) inequality:
|🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)| >= |🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)|
|🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)| >= |🍎(x) -🍌(x)| +|🍌(z) - 🍇(z)| since |🍎(y) - 🍌(y)| is maximized
|🍎(x) -🍌(x)| +|🍌(z) - 🍇(z)| >= |🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)| since |🍌(z) - 🍇(z)| is maximized
|🍎(x) -🍌(x)| +|🍌(z) - 🍇(z)| >= ||🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)|| since |q| + |p| >= 0 so |q| + |p| = ||q| +|p||
||🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)|| >=|🍎(x) -🍌(x) +🍌(x) - 🍇(x)| = |🍎(x) - 🍇(x)| since |q| >= q forall q
therefore |🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)| >= |🍎(x) - 🍇(x)|
since all 4 conditions are satisfied the 🍊 is a metric!
I guess I interpreted that one as a one off usage (i guess i assumed the opposite for the teleport so thats not really fair), if its repeatable, it’s much more useful, but it still wouldn’t stop all the internal aging that will actually kill you.
I’d much rather be pickpocketed than shoot.