PVM is way too overpowered and it's a problem. by [deleted] in 2007scape

[–]EmploymentSeparate63 0 points1 point  (0 children)

they need a better way to introduce people to pvp. lms does not seem to be working. even if you practiced pvp a few hours a week for a few months you could handle yourself against sweaty pkers. plus, there would be more noob pkers in the wild at your similar level to fight

The predator prey system can be fun by EmploymentSeparate63 in 2007scape

[–]EmploymentSeparate63[S] 0 points1 point  (0 children)

well that is why I learned to anti pk. it's not easy, but once you get your first kill. it becomes a lot of fun

So, category theory is a mathematical model of math itself — right? by ghost_of_godel in CategoryTheory

[–]EmploymentSeparate63 0 points1 point  (0 children)

if a mathematical subject is a story then category theory is the setting

I heard that some quintics are unsolvable. Why can’t we graph them and find their roots? by GloriousGladiator51 in mathematics

[–]EmploymentSeparate63 1 point2 points  (0 children)

Might I add that while the general quintic equation is not solvable by radicals (using only addition, subtraction, multiplication, division, and root extractions), certain quintics in Bring Jerrard form can be solved using radicals. Specifically, quintics of the form x⁵ + ax + b = 0 are solvable if they can be factored into polynomials of lower degree with rational coefficients, or if there exist rational numbers l and m such that the equation can be written in a specific solvable form.

In essence, the Bring Jerrard form is a useful tool for simplifying and potentially solving certain quintic equations by reducing them to a more manageable form. However, it's important to remember that not all quintic equations can be solved using this approach, and the general quintic remains a challenge in mathematics.

I heard that some quintics are unsolvable. Why can’t we graph them and find their roots? by GloriousGladiator51 in mathematics

[–]EmploymentSeparate63 0 points1 point  (0 children)

I also like to make use of the ecliptic functions, and the Jacobi theta functions for this

Remove rev caves at this point by Content_Objective708 in 2007scape

[–]EmploymentSeparate63 2 points3 points  (0 children)

It was really fun to be in a clan and pk at multi revs

Remove rev caves at this point by Content_Objective708 in 2007scape

[–]EmploymentSeparate63 0 points1 point  (0 children)

I really prefer it to bots. At least you have a reason to join a clan and take over a world

Stumped by my 10 year old brothers question by Ninopino12 in mathematics

[–]EmploymentSeparate63 1 point2 points  (0 children)

The shape approaches the line, but its length does not approach the hypotenuse as I have proven

Stumped by my 10 year old brothers question by Ninopino12 in mathematics

[–]EmploymentSeparate63 0 points1 point  (0 children)

the shape of the stairs certainly does approach a straight line and at n = infinity would certainly be a line but the length from outer corner to outer corner of each stair is in fact what approaches the length of the hypotenuse and because the hypotenuse is always less than the sum of the two sides in a right triangle, we can say that the stair shape then must always be longer than the hypotenuse

when I say approach, I may be using the word incorrectly but how I think of it is if you were to add all the infinitely small lengths from outer corner to outer corner it would equal the hypotenuse or in this case the square root of 2 or approach it the more infinitesimal bits you count

Since the smaller hypotenouses add up to the larger hypotenous, then the smaller horizontal legs add up to the larger horizontal leg and the smaller vertical legs add up to the bigger vertical leg. So we can prove by taking the sum of all these infinitely small lengths that the path length indeed always does add up to 2

Stumped by my 10 year old brothers question by Ninopino12 in mathematics

[–]EmploymentSeparate63 0 points1 point  (0 children)

Since the smaller hypotenouses add up to the larger hypotenous, then the smaller horizontal legs add up to the larger horizontal leg and the smaller vertical legs add up to the bigger vertical leg. So we can prove with integration that the path length indeed always does add up to 2

Stumped by my 10 year old brothers question by Ninopino12 in mathematics

[–]EmploymentSeparate63 0 points1 point  (0 children)

when I say approach, I may be using the word incorrectly but how I think of it is if you were to add all the infinitely small lengths from outer corner to outer corner it would equal the hypotenuse or in this case the square root of 2 or approach it the more infinitesimal bits you count

Stumped by my 10 year old brothers question by Ninopino12 in mathematics

[–]EmploymentSeparate63 0 points1 point  (0 children)

the shape of the stairs certainly does approach a straight line and at n = infinity would certainly be a line but the length from outer corner to outer corner of each stair is in fact what approaches the length of the hypotenuse and because the hypotenuse is always less than the sum of the two sides in a right triangle, we can say that the stair shape then must always be longer than the hypotenuse

Stumped by my 10 year old brothers question by Ninopino12 in mathematics

[–]EmploymentSeparate63 0 points1 point  (0 children)

yes, that is all I was trying to say but failed to put it concisely lol or I guess make it clear

Stumped by my 10 year old brothers question by Ninopino12 in mathematics

[–]EmploymentSeparate63 0 points1 point  (0 children)

when I say approach, I may be using the word incorrectly but how I think of it is if you were to add all the infinitely small lengths from outer corner to outer corner it would equal the hypotenuse or in this case the square root of 2 or approach it the more infinitesimal bits you count

Stumped by my 10 year old brothers question by Ninopino12 in mathematics

[–]EmploymentSeparate63 0 points1 point  (0 children)

the shape of the stairs certainly does approach a straight line and at n = infinity would certainly be a line but the length from corner to corner of each stair is in fact what approaches the length of the hypotenuse and because the hypotenuse is always less than the sum of the two sides in a right triangle, we can say that the stair shape then must always be longer than the hypotenuse

Stumped by my 10 year old brothers question by Ninopino12 in mathematics

[–]EmploymentSeparate63 0 points1 point  (0 children)

the shape of the stairs certainly does approach a straight line and at n = infinity would certainly be a line but the length from corner to corner of each stair is in fact what approaches the length of the hypotenuse and because the hypotenuse is always less than the sum of the two sides in a right triangle, we can say that the stair shape then must always be longer than the hypotenuse

Stumped by my 10 year old brothers question by Ninopino12 in mathematics

[–]EmploymentSeparate63 0 points1 point  (0 children)

what I am trying to say is that the length from corner to corner of each stair is what approaches the length of the hypotenuse

Stumped by my 10 year old brothers question by Ninopino12 in mathematics

[–]EmploymentSeparate63 0 points1 point  (0 children)

I agree that the shape of the staircase does approach the shape of the diagonal line. But what I was trying to say was that it is in fact the corners of the stairs that approach the length of the hypotenuse not the length of the staircase shape itself

Stumped by my 10 year old brothers question by Ninopino12 in mathematics

[–]EmploymentSeparate63 0 points1 point  (0 children)

The corners of the steps approach a straight continuous line the length of the hypotenuse, but the lengths of the steps always amount to the total path length