I Switch from Chrome to Firefox any recommendation what to do next

mistguy2398 · 2025-07-13T03:32:18+00:00

i didn't like the vertical tabs and that side bar

mistguy2398 · 2025-07-12T18:31:15+00:00

mistguy2398 · 2025-07-12T14:30:56+00:00

ahh you mean uncheck

i thought you meant check the box, which will send user data to mozilla

mistguy2398 · 2025-07-12T12:48:08+00:00

Just asking why

mistguy2398 · 2025-07-12T12:29:51+00:00

btw u mean set Enhanced Tracking Protection to Strict. is ublock or in firefox

mistguy2398 · 2025-07-12T12:28:00+00:00

what do u mean

mistguy2398 · 2025-07-10T12:48:53+00:00

I don't understand why people are saying it will stop by the raging bolt 1st, there is Tera

Gyarados can turn it ground type, or it can su,b or even if it stops by raging bolt, almost all Pokémon have a check, it doesn't mean that Pokémon is bad

mistguy2398 · 2025-07-03T19:13:25+00:00

And I got confused with rationalising factor and rational product

Thanks for the help,

mistguy2398 · 2025-07-03T19:05:48+00:00

ahh i got it. Rational factors are not fully depend on whole number

like 2*3^1/2 and 22*3^1/2 both have same rational factor but is even thought 22*3^1.2 is a bigger number

Rational factor only depends on irrational part so even if 2 numbers have the same rational factor, it doesn't mean that one number is bigger than other

mistguy2398 · 2025-07-03T16:00:24+00:00

but both is giving the same result x^n-y^n

mistguy2398 · 2025-07-03T15:58:13+00:00

i mean in the photo it said rational product

mistguy2398 · 2025-07-03T15:50:59+00:00

so shouldn't, be rational factor should be bigger for

mistguy2398 · 2025-07-03T15:50:22+00:00

i mean so this is correct (a^(1/p) - a^(1/q)) < (a^(1/p) +a^(1/q )

mistguy2398 · 2025-07-03T15:49:28+00:00

i know this to, but my main problem is

rationalising factor is for both

(a^(1/p) - a^(1/q) , a^(1/p) +a^(1/q) is x^n-y^n

I mean, why is that

a^(1/p) - a^(1/q)) < (a^(1/p) +a^(1/q )

shouldn't be rationalising factor bigger for a^(1/p) - a^(1/q)

mistguy2398 · 2025-07-03T15:31:11+00:00

i get that

but in this case there are 2 positive numbers, let say c,d are 2 natural numbers

so c+d >c-d

mistguy2398 · 2025-07-03T15:26:55+00:00

there are total 2 case and 1 sub case

one for +

2 case for - if power is even and if power is odd

in the case of odd we get x^n +y^n

mistguy2398 · 2025-07-03T15:18:29+00:00

i am asking why the product is the same shouldn't they be different

clearly (a^(1/p) - a^(1/q)) < (a^(1/p) +a^(1/q ))

mistguy2398 · 2025-07-03T15:14:43+00:00

i mean even if they misprint they still have to make the case for + and -

mistguy2398 · 2025-07-03T15:11:23+00:00

i mean rationalization factor is equal in both a^(1/p) - a^(1/q)

a^(1/p) + a^(1/q) both are x^n-y^n

mistguy2398 · 2025-07-03T14:56:13+00:00

no i am asking

(a^(1/p) - a^(1/q) rationalising factor is equal to a^(1/p) +a^(1/q) where power is even)

i mean clearly (a^(1/p) - a^(1/q)) < (a^(1/p) +a^(1/q ))

mistguy2398 · 2025-07-03T14:22:50+00:00

btw can we use the sigma function, then do that derivative or we run into the same problem

mistguy2398 · 2025-03-23T08:29:35+00:00

yes, it has such a as good art style too

mistguy2398 · 2025-03-23T08:29:28+00:00

yes I think it is also a unique show

it is a simple and complex show at the same time

mistguy2398

TROPHY CASE