help-17

[river kettle]

that was simpler than i thought

what about this one

how do i know what probability rules to apply

@paper depot any ideas

[paper depot]

this is the kind of problem where a tree diagram works great

the first level of the tree should branch into Rain vs No Rain, and the second should branch into Fire vs No Fire on each

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[pearl crest]

hey guys im abit confused
the roots of the characteristic poly of a triangular matrix,which every matrix can be converted to,are supposed to be the eigenvalues but then triangular matrices just show that some subspaces are invariant and not the eigenvectors which we only get from the characteristic poly of the diagonal matrix of eigenvalues
could someone elaborate

[flat whale]

Elaborate what

Pick an example of a simple triangular matrix from your book

[pearl crest]

i mean should the characteristic poly of the triangular matrices give the roots ? @flat whale

because i thought that would happen only in the case of a diagonal matrix,where the diagonal entries are the eigenvalues themselves

[flat whale]

Nope

What are the eigenvalues of an upper triangular matrix

[pearl crest]

but the expression for both are like prod(t-diag entries)

the roots of its characteristic opolynomial right?

[pearl crest]

which would be this

the product of the diagonal entries

$$\prod_{i} (t-a_{ii})$$

[twin meteor]

Staasi

[pearl crest]

@flat whale
am i in deeep shit

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[final sierra]

What happens when I get this? In previous problems I’d factor a common coefficient out but that isn’t the case here : (

[dreamy viper]

put it in terms of sin or cos

cos^2 = 1-sin^2

sin^2=1-cos^2

[final sierra]

and then distribute from there?

[dreamy viper]

yeah

[final sierra]

youre the best, thank you so much!

.close

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[undone aurora]

What should I do for this question

[hushed pewter]

What does "unique solution" mean to you?

[undone aurora]

if b^2 - 4ac = 0 then the equation has a unique solution

[hushed pewter]

Correct

So what would you do?

[paper depot]

https://tenor.com/view/just-do-it-shia-la-beouf-do-it-gif-4531935

[brisk moss]

🥇 is an emoji that will forever correspond to swr for me

[undone aurora]

$(2k - 2)^2 - 4(3 - 3k) = 0 \$
$4k^2 - 8k + 4 - 12 + 12k = 0 \$
$4k^2 + 4k - 8 = 0 \$
$k^2 + k - 2 = 0 \$
$(k + 2)(k - 1) = 0 \$
$k = -2$ or $k = 1 \$
$k = 1$ is the only solution in $\mathbb{N}$

[twin meteor]

Calc II Victim

[hushed pewter]

Looks perfect

[undone aurora]

alr got it ty

.close

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[undone aurora]

Whats the error here?

[mild flower]

very last step i think

you multiply by y - 3 on both sides

but how do you know whether you need to flip the sign?

[undone aurora]

oh I flipped because -y

i multiplied both sides by neg

oh ur saying we dont know whether y - 3 = - or + ?

[mild flower]

yes, you multiplied both sides by a negative number earlier and flipped the sign, that was good

[mild flower]

yes exactly

[undone aurora]

so I need to split that into two cases

[mild flower]

you could...

but what i would do

is consider numerator and denominator separately

it'll still be two cases but it might be easier to work with

[undone aurora]

wym so like

-y >= 0

and y - 3 >= 0?

[mild flower]

i guess ultimately it's the same

for that quotient to be <= 0, the numerator and denominator need to be different signs

[undone aurora]

so y <= 0

and y >= 3 but that wont make sense

[mild flower]

yep so that case doesn't work, try the other way around

[undone aurora]

so -y <= 0

and y - 3 <= 0?

y >= 0

y <= 3

so [0, 3] but y \neq 3 so [0, 3)?

[haughty venture]

hey

[tidal umbra]

yes

right

[haughty venture]

,help me please

[twin meteor]

Command me please not found!
Use the ,list command without arguments to see a list of commands.

Aug 15, 2023, 6:07 PM

[mild flower]

why is y ≠ 3? there's probably a reason i just don't see it

oh right bc we're dividing lol

[undone aurora]

y/y - 3

ye ty I got it now ill split it into cases from now on

.close

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[haughty venture]

heyyyy

can someone help me

[tidal umbra]

sure, but you need to open another channel for this

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[undone aurora]

Is this good enough

[plush temple]

What is math ?

[round plover]

looks good

[plush temple]

I have had difficulties studying analysis 1 and I started to question myself what exactly is my object of study: mathematics

[paper depot]

@plush temple

this question belongs better in #discussion

(sorry, hit enter too soon)

[plush temple]

Oh right

[paper depot]

you are probably expected to prove that h really is a bijection as you claim

[plush temple]

Sorry

[paper depot]

i personally would hesitate to accept such an unproven answer

[undone aurora]

oh whatttt

alr one sec

[paper depot]

i doubt it'll take you "one sec"

it is a moderately involved proof

[loud heath]

One second is over.

Multiple in fact.

[paper depot]

a few hundred i'd say

[mild flower]

in anything but an intro to proofs class i'd say this is a totally reasonable answer

[paper depot]

op is in something like an intro to proofs class, iirc.

[undone aurora]

Injective:

h(n) = h(m) implies that f(n/2) = f(m/2) or g(n+1/2) = g(m+1/2) right

so if f(n/2) = f(m/2) then n = m and if g(n+1/2) = g(m+1/2) then n = m

right?

[undone aurora]

ye

[paper depot]

im gonna ask you to not drop the parentheses around fractions

h(n) = h(m) implies that f(n/2) = f(m/2) or g(n+1/2) = g(m+1/2)
also this i think might warrant more elaboration

[undone aurora]

wait what should I explain abt that

wym

[paper depot]

what if instead it happened that h(n) = f(n/2) but h(m) = g( (m+1)/2)?

can you convince me that can't happen?

[undone aurora]

oh idk i didnt think abt that case

[paper depot]

well for shame.

it could well be that two values of h arising from two different cases within its defn are equal.

maybe it is actually impossible, but that's something you must prove!

[undone aurora]

if n is even and m is odd then h(n) neq h(m) doe

right?

[paper depot]

how do you know that

[undone aurora]

because if n is even then h(n) = f(n/2) and if n is odd then h(m) = g((n+1)/2) and f(n/2) in A and g((n+1)/2) in B and they said that A cap B is emptyset

[paper depot]

right

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[cobalt fiber]

why does the key and mathway have different answers:

[paper depot]

,calc -sin(2) - 2/cos(2)^2

[twin meteor]

Result:

-12.45809583491

[paper depot]

,calc -sin(2 deg) - 2/cos(2 deg)^2

[twin meteor]

Result:

-2.0373384169863

[paper depot]

mathway thinks you are doing degrees.

[cobalt fiber]

ah ok

.close

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[tranquil lodge]

what we thinking, right or nah

[mild flower]

try some points

[vocal sleet]

@tranquil lodge Has your question been resolved?

[tranquil lodge]

uhh how would i do that

[mild flower]

well like

you know f(0) = 0

[tranquil lodge]

yes

[mild flower]

so what value of x would you need to make -f(x + 4) = 0?

[tranquil lodge]

-4?

[mild flower]

yes

and is it true that your red function is 0 at x=-4?

[tranquil lodge]

oh

its not

[tranquil lodge]

so this?

[mild flower]

see that looks better

[frosty gale]

i need more complx quesion's pls

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@tranquil lodge Has your question been resolved?

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[empty linden]

ok so

[empty linden]

theres a bit of a problem for #4

when i try to solve it the usual way

i end up getting a negative in the brackets of the log

x - 3 = 0
x - 3 = 1
x - 3 = 3

since i set x equal to 0, 1 and the base

for these answers, i got x = 3

meaning 3 is the vertical asymptote

for x-3 = 1, i got 4

and for x-3 = 3, i got x=6

but when i try so solve for y,

y = -2log3(3-x)
-2log3(3-3)
-2log3(0)

and logs cant have zeros

so how do i do this?

[hidden kelp]

Well, you just said that x=3 is an asymptote, and then you tried to plug it in. What did you expect?

[empty linden]

oh

yeah mb

kinda new to this

ill solve the other values

but

-2log3(3-x)
-2log3(3-4)
-2log3(-1)

so theres still a problem unfortunately

since logs cant be negative in the brackets

[hidden kelp]

Did you find the domain of that function?

[empty linden]

domain: [3, +inf)

[hidden kelp]

that is not correct

[empty linden]

wait no

domain: [3, -inf)

[hidden kelp]

well, the other way around, but yes. So, is 4 in that domain?

[empty linden]

no

[hidden kelp]

then you can't have a y value for an x that isn't in the domain :)

[empty linden]

for x-3 = 1, i got 4
and for x-3 = 3, i got x=6
so are one of these values incorrect then?

[hidden kelp]

well, you are solving for x-3 = something, and not 3-x = something

so I'm not sure

[empty linden]

ill solve it with 3-x real quick

3-x = 0; x = 3
3-x = 1; x = 2

ohhh

[hidden kelp]

I'm not even sure what you are trying to do, so if you could tell me i'd be able to help you out more

[empty linden]

yeah i guess thats the mistake

[empty linden]

im graphing this log

i tried to rearrange the (3-x)

and did (x-3), but that wasnt correct

ill try these new values

[hidden kelp]

if you want to rearange the 3-x, you have to keep the minus sign: $3-x \ -(-3 + x) \ -(x+3)$

is the bot taking a day off?

$3-x \ -(-3 + x) \ -(x-3)$

[twin meteor]

imTypθ

[hidden kelp]

i'm going to go eat, but if you have more questions, be sure to keep asking. Someone else will come and help you :)

[empty linden]

ok

new values:
vertical asymptote: 3
x = 1: 4
x = 3: 0

im still geting negative logs

leading to undefined logs :/

im just gonna skip this question tbh

[vocal sleet]

@empty linden Has your question been resolved?

[hidden kelp]

i'm back. Again, I could help you much better if you showed me exactly what you're trying to do

[vocal sleet]

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[crude elm]

can i get help with this

[flat whale]

there's no question

[flat whale]

that's the same image

still no question

a question is something like "what's the color of the sky"

or "how much does the moon weigh"

[crude elm]

can i get help with this problem

[frozen bobcat]

What are you trying to do?

[flat whale]

well that's indeed a question

but not a math one

try asking a math question

[crude elm]

bro what

[vast shale]

what's the question?

[crude elm]

how do you set up the problem

[full arch]

what do you want to do with f(x)

[frozen bobcat]

what's the problem?

[vast shale]

What problem?

[flat whale]

lol

[frozen bobcat]

omg

[flat whale]

lol

[full arch]

okay

[vast shale]

that's the number 13

[flat whale]

we're getting trolled

[full arch]

the answer is your mom

[crude elm]

how do you solve 13

[frozen bobcat]

This is a number with a function. There is no question. You've asked "Can i get help with this question?" and basically posted a picture of a tree.

[flat whale]

stop pinging me

[crude elm]

its a picture of a function i need help with

[frozen bobcat]

there's no question to solve. Look at the directions and get back to us.

[vast shale]

A function is not a question

[crude elm]

??? is this not a math discord

[frozen bobcat]

ok, the answer is 7

[vast shale]

No it's a Mathematics server

[flat whale]

it's not a mind reading discord

wait lemme try

[crude elm]

can i get help with question 13

[flat whale]

@frozen bobcat are you thinking of 420

[vast shale]

Where is the question

[frozen bobcat]

that could also work, depends on the question.

[flat whale]

.

[vast shale]

That is not a question

[flat whale]

4 times he sent the same image

[crude elm]

are we seeing the same picture

[frozen bobcat]

look above question 13), are there directions to this problem?

[flat whale]

read this

[frozen bobcat]

THANK YOU!

[flat whale]

LOL

[vast shale]

Yikes

Took you that long

[frozen bobcat]

Now i'm gonna go pull some more teeth somewhere else.

[bleak oak]

@crude elm f(3/2) - f(1) now go fk off

[flat whale]

zybdentist

[crude elm]

be more specific next time my fault og

[vast shale]

Too rude

[flat whale]

also wrong lol

[bleak oak]

😼

yh sorry divide by 1/2

[crude elm]

is f(3/2) - f(1) / (3/2)-1 wrong

[vast shale]

most likely

Basically the question is asking you for the slope of the secant line that passes through f(3/2) and f(1)

[bleak oak]

yeha

its not wrong

its $\int^{1.5}_1 f'(x) dx / (1.5-1) = f(1.5)-f(1)/0.5$

ah w.e.

[twin meteor]

nadav

[crude elm]

huh

[bleak oak]

the slope is f'(x)

[vast shale]

oh yeah you're right, just weight syntax

[bleak oak]

and the average is the integral of function over the range divided by length of range

that should be in ur textbook.

[vast shale]

$$\frac{f(\frac{3}{2})-f(1)}{\frac{3}{2} -1}$$

basically $$\frac{\Delta y}{\Delta x}$$

[twin meteor]

please request a new nickname

[crude elm]

thats what i did so is that right

[twin meteor]

please request a new nickname

[vast shale]

can't you check it yourself

[crude elm]

idk is 3/5 the answer

[vast shale]

you should be able to tell us if the answer is right or not

did you learn about secant lines?

[crude elm]

probably

[vast shale]

if you want to check if the answer is right

just use the point slope formula and check at both of the points

if $$s(x)$$ is the function of the secant line and $$f(x)$$ is the orginial function then $$s(x) = f(x)$$ when $$ x = \frac{3}{2}$$ and $$x = 1$$

[twin meteor]

please request a new nickname

[crude elm]

idk how to check if 3/5 is right mane

[vast shale]

I just told you how

bruh

not even 10 seconds before

[crude elm]

idk what none of that means

[vocal sleet]

@crude elm Has your question been resolved?

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[idle harbor]

Hi is there any way to make chance deviate in a direction
so for example I pick a number 1-100
i want to increase the chance it falls on lowers numbers
also the ability to increase chance it falls on higher numbers
can i do a math for this?

[hybrid flicker]

Depends on what exactly is your process of picking the number 1-100

Here's a possible way :
If your way of picking a number 1-100 (or here 0-100) is tossing a coin 100 times and taking the amount of heads, then by making the side with heads heavier you will have a higher chance of landing on heads, thereby increasing your chance of "higher numbers"

Conversely if you make the side with tails heavier, you will have higher chance of getting lower numbers

[frozen bobcat]

Yes. It depends entirely on how you choose the number (can't be totally random). The method and how you choose the method will depend on how you want the probabilities to form and how specific you want to get.

[vocal sleet]

@idle harbor Has your question been resolved?

[idle harbor]

hmmmm ok

well rightnow i use random number

from coding language it gives function to give random number

so everytime i really get a random number 1-100 without ability to make it more lower or higher so i wanted to know

like is there any maths to make it able to go more in one direction /higher/lower

[sand lodge]

well you can try some stuff with the log function which increases rapidly relatively for smaller x but then starts to "flatten out"

,w log(x)

[idle harbor]

one moment i think i dont give enough info about question

[sand lodge]

or you could make your random number generator generate a random number from 1-10000 and then take the sqrt of the answer and then floor it so it only gives integers

it would give a higher weight towards lower numbers

another crude method is to instead for example have a random number generator for numbers
1-4 and

if it is 1, then output the result of a random number generator from 1-75,
if it is 2,3, or 4 then output the result of a random number generator from 76-100

there's probably coding functions that do this much more efficiently and better dispersed

might be better to ask a coding server lol

[idle harbor]

oh the thing is i want all number

to be able to do 1-100 despite how low chances is

i just want to be able to do maths that make it can give better chance of higher or lower numbers when i want

[sand lodge]

how specific though

[idle harbor]

it can give sort of infinite increase of chance

hold on i think i can give a example

[sand lodge]

oh ok I see

sure

show

[idle harbor]

so for example

i have 1,2,3,4,5

when i want higher chance of higher numbers

i can enter in a container

1,2,2,3,3,3,4,4,4,4,5,5,5,5,5

then the container picks a random out of all entries and since there are more entries

for the higher the number is

the higher the chance of a higher number

now i wanted a way to make this into a math formulae instead

[sand lodge]

oh ok

[idle harbor]

i also want to opposite of ability to make lower numebrs with better chance

i can do the reverse

[sand lodge]

I mean do you want it to be exactly 1,2,2,3,3,3...

[idle harbor]

no i want ability to edit how much the chance is

for example

at a deviation of +1 it will do that

+2 will give it 2 times that etc

-1 will give it opposite

-2 very opposite

so this way i can just need 1 number to say how higher the chance is for higher or lower numbers and then i plug it into the formula

[sand lodge]

lol I still think this is better asked for a coding server

[silent folio]

Well a slightly messy solution is to generate that list of numbers and then pick a random

[idle harbor]

i tried this but couldnt get good response i might try other places

but yes i can make in code a list of all this numbers is just innefficient and might cause code to performance issues so i leave this for last resort

is there math i can do to get total amount of variables if i did this
for example
1,2,2,3,3,3,4,4,4,4,5,5,5,5,5
this is 15 numbers
also is there math i can do to get number at a exact location, example:
1,2,2,3,3,3,4,4,4,4,5,5,5,5,5
number 11 is 5

[vocal sleet]

@idle harbor Has your question been resolved?

[icy spindle]

yes

let arr represent your list

then index i such that arr[i] = n is

n(n-1)/2 <= i <= n(n+1)/2

since you are coding it to solve for n i think you do something like

[idle harbor]

hmmm

[icy spindle]

actually you just hardcode quadratic formula lol

and then floor it

[idle harbor]

idk if it will understand these tho =>

also idk what this quadratic thing is lol

[icy spindle]

oh sorry

[idle harbor]

also is there a way to make this formula add amount of number

so for example

1,2,3
normal just adding 1
1,22,333
normal adding 2
1,222,3333

[icy spindle]

ohh

ok

for your code?

what language are you using

[idle harbor]

very simple languge for lua

i aslo what to learn this to use in other language

for discord custom command bot

[icy spindle]

🙊 i dont know lua sorry haha

[idle harbor]

no if u can give me maths

[icy spindle]

ok

[idle harbor]

i can make it in lua

[icy spindle]

let n be the number that you want to add to your list

e.g., if n = 5, you add five 5s to your list

then the length of your list is equal to:

1+2+3+4+...+n-1+n

or through the arithmetic sequence formula just n(1+n)/2

[idle harbor]

hmmmmmm

[icy spindle]

sorry i dont think im explaining it very well

[idle harbor]

is there formla i can use to just get total that i can for example put in a caluclator

[icy spindle]

total as in total length of list?

[idle harbor]

idk if n(1+n)/2 gives total?

[icy spindle]

yes

it gives total length of list

[idle harbor]

yes total digits

[icy spindle]

lets try it

[idle harbor]

hmmm

[icy spindle]

for n = 5

[idle harbor]

wait

[icy spindle]

the formula says 5(5+1)/2

[idle harbor]

no n is out upscale amount

[icy spindle]

like a large amount?

[idle harbor]

it give me 15

but n for me is just our upscale amount

[icy spindle]

i dont understand this upscale amount sorry

[idle harbor]

one sec

ok here
1,22,333

uspcale amount is 1

[icy spindle]

ohh i see

[idle harbor]

because it add 1 new digit every time it counts

[icy spindle]

upscale = 3 becomes

[idle harbor]

sorry not every time it count

[icy spindle]

give me an example when upscale = 4

i think i understand you

just need to verify

[idle harbor]

ok four is
1,22222,333333333

[icy spindle]

ok yea

i understand

so there is one 1, five 2s, nine 3s

interestingly, our formula still works

[idle harbor]

yea

[icy spindle]

just have to change a bit

[idle harbor]

because it gives 4 then 8

then 12

then 16

[icy spindle]

so you said n is your upscale amount

[idle harbor]

yea

[icy spindle]

what number do you use to represent the largest number

or variable

[idle harbor]

100

we can call him max

[icy spindle]

ok sure

[idle harbor]

this represents highest amount of count actions

[icy spindle]

and 1 always starts at 1?

like one 1

[idle harbor]

yea

im not sure yet if i want the upscaling to start from there

but rightnow i will leave him like that

also the count actions example
1,22222,333333333
is 3 count actions

wait lets make upscaling start at 1

[icy spindle]

so like 1,1,2,2,2,3,3,3,3, etc.

[idle harbor]

yea

this assumes upscaling of 1

also i just realize another example

so upscale is 1, then these are the extra digits per action
1,2,3,4,5,6,7....

so now 1 with 1 extra digit is 11,222,3333,44444,555555

upscale is 2 extra digits become:
2,4,6,8,10

111,22222,3333333,etc

[icy spindle]

ok let me see if i am understanding

it may be a little hard to see and its not in lua

wait no im wrong

you said five 2s

upscale = 4

[idle harbor]

ok at upscale of 4

it should start with 5 1s since uspcale = extra that increases every count

[icy spindle]

ok

there are ten 2s

[idle harbor]

lol

yea it should be 9

one sec

i cna code this for u?

[icy spindle]

yea sure

five 1, nine 2, thirteen 3

the general term for any given index i in this list is

[idle harbor]

yes

thats right

[icy spindle]

(1+upscale)(1+max*upscale)/2

wrong

(2+upscale(1+max))(max)/2

[idle harbor]

this formula is to get total digits right?

[icy spindle]

yea

[idle harbor]

hmmm

we can test it with using 1 as upscale

[icy spindle]

yea sure

oh wait do you want to get what number is at position i?

[idle harbor]

i know after i did some counthing it is 5050 total if im correct

yea i also want to do that

[icy spindle]

ok we can do that after

total digits first

[icy spindle]

what is max

100?

[idle harbor]

yea

sorry was doing something

[icy spindle]

ok then i think you want

(2+101*100)/2

[idle harbor]

hmmm is so close

that gives me 5051

[icy spindle]

something is fishy

let me think about this

sorry i am not so quick

[idle harbor]

nah thats okay

[icy spindle]

ok

upscale of 1 is

1,1,2,2,2,3,3,3,3?

[icy spindle]

i am going with this here

[idle harbor]

yea

wait a min

ur right man

u are correct

i got 5050 but i wasnt making uspcale affect 1

so 5051 makes sense

but ill mke a code to simulate more situations then count them online and then use ur formala and see if its correct for everything

[icy spindle]

ok

u is upscale

i is the index

[idle harbor]

damn that looks complex lol

idk how to type this into the code

[icy spindle]

uh

[idle harbor]

this 2 storey brackets

[icy spindle]

it looks more complicated than it is

[idle harbor]

is there way to put this on 1 line

like u did earlier

[icy spindle]

yea

let me try in lua

[idle harbor]

jsut give me maths

[icy spindle]

haha lua is very similar to python

[idle harbor]

i already have code waiting for it heh

oh i think so

https://steamuserimages-a.akamaihd.net/ugc/490146591486499787/E4AFF98BE00F114EBB5FB7FE77F6F3BB320C8ACC/

that what it loooks like

[icy spindle]

oh wow very similar

we use def for function

our for loop looks a tiny bit different

[idle harbor]

ahh

[icy spindle]

u = 2
i = 9
getPos = math.floor((-(u+2)+math.sqrt((u+2)^2+8*u*i))/(2*u))
print(getPos)

[idle harbor]

oh wait this is the correct thing right
2+101*100)/2?

[icy spindle]

u is upscale

[idle harbor]

this is the correct form for getting the total numbers right
(2+101*100)/2?

[icy spindle]

its (2+upscale + upscale * max) * max/2

for max = 100

it would be

wait no

wait yes (?)

[idle harbor]

(2+(upscale + upscale * max)) * max/2

so like this?

[icy spindle]

yea

[idle harbor]

ok ill put it in one sec

[icy spindle]

lets try

[idle harbor]

il will check the position one rn

its looks complex lol

[icy spindle]

it just looks like that haha

[icy spindle]

yeah ok this

checked

[idle harbor]

wait should 2 be upscale amount?

[icy spindle]

what do you mean

you can change the upscale amount

its a variable

[idle harbor]

(2+(upscale + (upscale * 100))) * 100/2)

i mean the 2+ at the start

[icy spindle]

oh for this

[idle harbor]

should i set it at upscale amount

[icy spindle]

no the 2 is just there from the math

you set upscale = upscale amount

[idle harbor]

yea i set that for every upscale

but i wanna know if the 2+ should be changed to that to

hold on

[icy spindle]

no no

[idle harbor]

(2+(upscale + (upscale * 100))) * 100/2

[icy spindle]

yea

[idle harbor]

ok this how i had to type it in lua
is it correct?

[icy spindle]

errr

the 100 should be your max variable

[idle harbor]

yea its my max rn

[icy spindle]

ok

[idle harbor]

ok so i did it but got value off one sec

wait my bad

it was error on my end

ok it seems to be correct for everything i tested

[icy spindle]

nice!

[idle harbor]

yea thanks man

[icy spindle]

wish you luck 🙏

[idle harbor]

now i jsut gotta do the get position

[icy spindle]

oh

[idle harbor]

sorry lol

its just my main question was to get a random number 1-100 but increase chance of higher numbers or lower

but seems i can get response for this so i am trying now to instead

just give higher numbers more instance in a list like what were doing rn, 11,222,3333 etc

[icy spindle]

OH

[idle harbor]

then pick a random out of all digits and since the higher the more then = higher chance

[icy spindle]

...

[idle harbor]

of it picking a higher number

now i just gotta like find what that number is

[icy spindle]

i think there are some weighted probability functions

[idle harbor]

really

[icy spindle]

but if you still want to try picking the number here is the code:

u = 2
i = 9
getPos = math.floor((-(u+2)+math.sqrt((u+2)^2+8*u*i))/(2*u))
print(getPos)

getPos will be the number at your position

[idle harbor]

lol is there by chance simpler for of this

[icy spindle]

errrrrr

maybe

i am not great at coding

[idle harbor]

i mean this nice function but i like osmething i can aremember

easier

[icy spindle]

haha uhhh.....

[idle harbor]

i will need to keep a notepad of all these magic formals i dont understand lol

[icy spindle]

i dont think so sorry i might have to take some time to think about that

there is a weighted random function in python

[idle harbor]

what does u mean for this

[icy spindle]

u is how much you upscale

[idle harbor]

ok and i

[icy spindle]

i is the position of the number you are getting

[idle harbor]

ooo k

wait am in lol

so this gets the position of a number or gets the number at a certain position?

[icy spindle]

gets the number at a certain position in your list

but you do not have to make a list

so for example the 7th position for upscale = 2:

[idle harbor]

ahhh ok this shold be it then

[icy spindle]

1,1,1,2,2,2,[2],2,3,3,3,3,3,3,3

[idle harbor]

ye

[icy spindle]

uh oh

something is fishy

[idle harbor]

ok so im testing it

[icy spindle]

i think i did something wrong

[idle harbor]

ah probably

cause i put position 4 and it gives 1 instead of well, 11,222 2

brb in a min

back

[icy spindle]

ok im working on the get function

[idle harbor]

ok

[icy spindle]

u = 2
i = 100
getPos = math.floor(u-2 + math.sqrt( (2-u)^2 + 8 * u * (i+1))/(2*u))
print(getPos)

ahhhhhh its not working

[idle harbor]

itsjust 1 off for me

[icy spindle]

changing the upscale value makes things fishy

[idle harbor]

hmmm

man i wish i had a 4d brain to be able to visualize math.sqrt( (2-u)^2 + 8 * u * (i+1))/(2*u)

[icy spindle]

its wrong

😢

sorry let me work on this

[idle harbor]

yea man cook

[icy spindle]

haha

[vast shale]

using a weighted probability distribution, where u assign different probabilities to different numbers based on how likely u want them to be chosen.

example, if u want to increase the chance of picking lower numbers, u could assign higher probabilities to the numbers 1-50 and lower probabilities to the numbers 51-100. similarly, if u want to increase the chance of picking higher numbers, u could assign higher probabilities to the numbers 51-100 and lower probabilities to the numbers 1-50.

there are many ways to assign thse probabilities, and the specific method u choose (example is, linear distribution, exponential distribution, custom distribution, and etc.).

u could use a mathematical formulas (if u want to create a uniform distribution where each number has an equal chance of being chosen, u can prolly use the formula P(x) = 1/n) to calculate the probabilities or u could manually assign them based on ur preference

[idle harbor]

well im taking the distribution route its like making list and putting numbers iwant more likely more times in list

its just i want way to do this with formula, so is efficient because i make list with 20K+ entries every 10 seconds it may cause performance issue

rightnow we already have this list done its now about finding value in the list

[vocal sleet]

@idle harbor Has your question been resolved?

[vocal sleet]

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[royal grotto]

((a^-3)^2))/(2a^-7)^-1

[royal grotto]

a^6)/-2a^7

a^-6-7 )/-2

How am I wrong

<@&286206848099549185>

[tidal umbra]

yeah

wait

[rugged orchid]

Can you write it out

[tidal umbra]

you have to wait 15 mins for that

[royal grotto]

My bad

[royal grotto]

I will wait 15min for punishment then

[tidal umbra]

theres no punishment, lol

[royal grotto]

ah ok

[tidal umbra]

this should be-
a^-6/2a^7

[royal grotto]

Ya

But

My answers says 2a^-13

I have no idea how

[tidal umbra]

it should be a^-13/2

[royal grotto]

What

You mean -2? Btw

[tidal umbra]

no

why will it have negative ssign

sign*

[royal grotto]

Because 2^-1 is -2

[tidal umbra]

wait, no what

[civic otter]

$\frac{{\left(a^{-3}\right)}^2}{{\left(2a^{-7}\right)}^{-1}}$

[twin meteor]

Alberto Z.

[tidal umbra]

okay, yes its correct

i didnt multiply power -1 with 2

[civic otter]

This one? @royal grotto

If yes, your answer is correct @royal grotto

[royal grotto]

Yes this one however

My book says it’s 2a^-13

[civic otter]

Yes exactly

[royal grotto]

Or maybe that is wrong

[civic otter]

I thought that was your answer actually

[civic otter]

No it is the right one

[royal grotto]

No my answer were (a^-13 )/ -2

[civic otter]

Ah ok this is wrong

[royal grotto]

Ah ok

[civic otter]

Because $$2^{-1} = \frac{1}{2^1} = \frac{1}{2}$$

[twin meteor]

Alberto Z.

[royal grotto]

I don’t understand

No sorry I do but

It’s not what I would think to do

[frosty gale]

2/a^13

[civic otter]

And what do you think to do?

[royal grotto]

I will send photo

d)

[civic otter]

The problem is that 2^-1 is not -2 !!

Remember that exponentials (like 2^-1) are always positive

[royal grotto]

What really?

[civic otter]

Of course 😅 didn't they teach you this?

[royal grotto]

I forgot I think

[civic otter]

This is very important: $$a^{-n} = \frac{1}{a^n}$$

[twin meteor]

Alberto Z.

[royal grotto]

Oh I know this rule

But isn’t 2^-1 also (-2)

[civic otter]

Of course not

[civic otter]

(also) for this reason...

A number is unique, it can't be both -2 and 1/2 lol

[royal grotto]

I understand

[civic otter]

Note that if you want to change the sign of a number you must multiply (or divide) by something negative, not exponentiating

[royal grotto]

Correct

But now that I know 2^-1 is 1/2

I’d end up with a^-13/1/2

I still don’t see how it’s 2a^-13

[civic otter]

1/(1/2) = 2

The 1 can be seen as (1/2)⁰

[royal grotto]

Can I see you do it on paper

[civic otter]

Yep, give me a moment

[vocal sleet]

@royal grotto Has your question been resolved?

[royal grotto]

I think I figured it

[civic otter]

@royal grotto

[royal grotto]

Hmm

I don’t know if you used a diffrent method but

I figured out you could do this. 1 second

,flip

,flip image

[vocal sleet]

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[royal grotto]

.reopen

[vocal sleet]

✅

[royal grotto]

Alberto. Can you see the photo

[civic otter]

,rotate

[civic otter]

Yes

Let me look at it

Exactly, that's how I was thinking toi explain it, but then you already made it on your own, perfect 💪

[vocal sleet]

@royal grotto Has your question been resolved?

[vocal sleet]

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• Show your work, and if possible, explain where you are stuck.
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[royal grotto]

Thank you for help

[civic otter]

Anytime

[royal grotto]

Close

.close

[vocal sleet]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[vestal basin]

Hello

[vestal basin]

I somehow got 37/15 for the answer and it is just wrong D:

[dull maple]

!Show

[vocal sleet]

Show your work, and if possible, explain where you are stuck.

[dull maple]

Why is your first column in Da different from B?

[vestal basin]

Ok wait I just redid it on my computer

Does this look right

[frosty gale]

(x,y,z)=(3,-1,2)

pls can somone send questions on polynomials

[dull maple]

This is not Dx. This is just D.

Also, you made some calculation error there as well.

[dull maple]

You have calculated D correctly here. You took the Dx matric wrong.

Remember, In Dx, column for x's coefficient will be replaced by column of constants.

[dull maple]

I don't think this is the place. Create your own channel or sth.

[frosty gale]

ok

[karmic imp]

If you're applying Cramer's rule, for Dx, you never swapped the x column with the solution matrix, as mentioned above

[vocal sleet]

@vestal basin Has your question been resolved?

[vestal basin]

I thought I did it right

I'm so confused 😭

[vestal basin]

What is the solution matrix

[karmic imp]

B matrix

Look up Cramer's rule

[vestal basin]

Can u pls give an example of what I should do

[vocal sleet]

@vestal basin Has your question been resolved?

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[hollow thorn]

Can someone teach me part (b) please?

[vocal sleet]

@hollow thorn Has your question been resolved?

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

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[vast shale]

Give two reasons why "(f : \mathbb{R} \rightarrow \mathbb{R} ) where (f(x) = \pm \sqrt{x})" is not a function.

[twin meteor]

scholablade

[vast shale]

(f(x) = \pm\sqrt{x}) is not a function because in the defintion it has (\pm) which means that we can't determine it's actual value which is important for pure functions and since (f)'s domain is (\mathbb{R}), it can be negative therefore we can have (\sqrt{-2}) as an arbitary value so that is not a valid function.

[twin meteor]

scholablade

[vast shale]

Is this correct?

[round plover]

yeah; the first point is functionality or right-uniqueness (each element of the domain is sent to a single element of the codomain), and the second point is just whether the function is defined on its entire domain (also called totality or left-totality)

[vast shale]

Cool terms, thanks for the help!

I'l close this room now

.close

[vocal sleet]

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[meager garden]

I don’t know how to solve any of the 4 parts

[paper depot]

,rccw

[civic otter]

Given $z^2 - 2z + 25 = 0$, take the conjugate on both sides. What do you get?

[twin meteor]

Alberto Z.

[meager garden]

Ummm so I bring that equation and change signs on the other side of the =?

[civic otter]

No, what does taking the conjugate (of a complex number) mean?

[meager garden]

If the complex number Z=bi where a,b e of R is a root of the polynomial f(Z) with real coefficients (ź)=a-bi is also a root

[civic otter]

Yes, they want you to prove this

But you said you don't know how to do that

[meager garden]

So is a and b -2+25?

Nvm I got it

Thx

[civic otter]

Nope

[civic otter]

How? Without using the conjugate?

[meager garden]

1-2i 1+2i

[civic otter]

They tell you without calculating the roots

[meager garden]

But I didint

Zsquared= a= 1

-2z=b= -2

So the conjugate is 1+2i

[civic otter]

No, that's totally nonsense

[meager garden]

Bruh

So do you do it

[civic otter]

@meager garden ... can you answer this?

[meager garden]

The question?

Not anymore

[civic otter]

I don't get you... @meager garden

[meager garden]

I don’t either

[civic otter]

Don't what? You don't know what I was asking there or you don't know other things?

[meager garden]

When u meant by that you meant by the question?

[meager garden]

I mean I’m solely just for help

Here

[civic otter]

Here you said you can't solve it

[meager garden]

Yep

Because I don’t know how to begin or what to do to start it

[civic otter]

Yes, and I told you how to do it twice

[meager garden]

Not much help

[civic otter]

For the third time... can you do this or not? If no, please tell me what is the thing you do not understand @meager garden

[meager garden]

I cant

I want you for you to point out how I should do part i please

[civic otter]

You mean that I explain you how to do part (i)?

[meager garden]

Please

And hopefully from there I’ll be able To do the remainder

[civic otter]

Given $z^2 - 2z + 25 = 0$, take the conjugate on both sides. What do you get?

[twin meteor]

Alberto Z.

[civic otter]

What exactly are the words you don't understand in this question?

[meager garden]

What is the conjugate in this case?

[civic otter]

Start from the RHS, what's the conjugate?

[meager garden]

I don’t know

[civic otter]

Do you know what RHS stands for?

[meager garden]

Right hand side

Yes

[civic otter]

Alright, so what's the RHS??

[meager garden]

……

I don’t know

[civic otter]

Are you trolling??

You always say "I don't know" even at simple questions...

[meager garden]

Bro why would I spend a whole hour just being here to waste my own time

I don’t get what you mean by it in this question!!

[civic otter]

You are actually making me waste my time, not yours

[meager garden]

Bro I’ll be straight just explain and I’ll understand

[civic otter]

I can't explain things that you even don't understand...

[meager garden]

Just explain and I’ll part ways

[civic otter]

What's the conjugate of the number 0?

[meager garden]

0

[civic otter]

Oh there we go

And the conjugate of z² ?

Do you agree that $$\overline{z^2} = {\overline{z}}^2$$ ?

[twin meteor]

Alberto Z.

[meager garden]

Yh

[civic otter]

Nice, now do you agree that the conjugate of the sum is the sum of the conjugates?

[meager garden]

Yes

So what it’s 0?

[civic otter]

I mean, do you agree that $\overline{\alpha + \beta} = \overline{\alpha} + \overline{\beta}$?

[twin meteor]

Alberto Z.

[civic otter]

The RHS

[meager garden]

Yes

[civic otter]

Ok, so can you answer this now?? @meager garden

[meager garden]

But what is a and b in this LHS

@civic otter

[round plover]

you don't need the value of a or b

use the conjugation rules alberto has pointed out

[meager garden]

Nvm .close

[civic otter]

Lol

[meager garden]

.close

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[eager wadi]

How to find the domain of the function $f(x) = \log_{10}\Bigl( (\log_{10}x)^2 -8\log_{10}x + 15 \Bigr)$

I am able to get part of the answer, I get $(10^5,\infty)$ but I want to understand how to get the rest of it.

[twin meteor]

Anshul

[paper depot]

show work

[eager wadi]

Okay.

@paper depot

[paper depot]

okay first off

missing underscores on all those tens that were meant to be the base

also,

For f to be valid, m - 5 > 0
no

(m-5)(m-3) > 0 is not true ONLY when m-5 > 0.

that's where you lost the other half of your domain.

solve the inequality (m-5)(m-3) > 0 properly.

[eager wadi]

damn, you're right
I need to revisit the inequalities
thank you so much Ann

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[lilac plaza]

What does it mean by "equation of the central horizontal axis" in a sine function?

[round plover]

the line midway between the lines at its maximum and minimum values

so for sin(x), the axis would just be y=0

because it oscillates +/-1 about that line

[lilac plaza]

What about functions like 3sin(x)?

[round plover]

that'd still be y=0

the max and min lines would be +/-3

scaling it vertically or horizontally won't change the axis

only translating it up like sin(x)+3

[lilac plaza]

Ohh you're talking about the translation?

Vertical

[round plover]

yes

[lilac plaza]

Then 3sin(x) + 3 would be y = 3?

[round plover]

yes

[lilac plaza]

Oooo I get it

Thank you!

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[twin meteor]

kebesque integral

[obtuse sierra]

how did you get to this?

how is it 13g

oh nvm

yeah figured that

yeah got T2 ig

um but not in terms of t tf

i balanced the forces on the respective blocks

not on the system

yeah, why not?

the blocks are also at rest

the question says that all objects are at rest, so each block is in equilibrium, so the forces must be balanced on each block

the normal exerted by wedge on block B will be exerted in return to the wedge too, right?

i think

action reaction pair

and the negative x component will result in its motion

which shouldnt occur

the weight of block B is acting on just B

not on wedge

the only interacting force between them is normal

no

its from earth's centre

no

the weight of B will be balanced by the forces acting on B only

no relation with wedge

yes

and B acts on wedge by normal reaction too

normal force is a action reaction pair

they dont both act on one block

so cant cancel

yeah, if the wedge isnt fixed here, it will move too

because of normal reaction by block

oh wait, the question says wedge is fixed, so we need not worry about wedge then

do you know if you place two blocks in contact with each other horizontally, they both exert normal forces on each other?

[lilac plaza]

Because it cancels out with the Normal force maybe?

[lilac plaza]

I mean the calculation should be mg*cos(theta)

[obtuse sierra]

the block is just there on wedge, enjoying his life, sliding

without any external force

see what happens

the red one is acting on wedge and black one on block

cant cancel them

[lilac plaza]

The force affected by gravity affected by angle

[obtuse sierra]

the red one vertical component will push the wedge down

but wont go down because surface below

[lilac plaza]

wdym can't cancel them?

[obtuse sierra]

and the horizontal one will result in its motion

if it isnt fixed

sry, i need to go for dinner, ill be back after half an hour maybe

till then if you want to ask smthng, @scenic ravine he may be able to help